Undergraduate Analysis Tools

Bruce K. Driver

February 21, 2013 File:unanal.tex

Contents

Part I Numbers

1 Natural, integer, and rational Numbers ...... 3 1.1 Limits in Q ...... 4 1.2 The Problem with Q ...... 7 1.3 Peano’s arithmetic (Highly Optional) ...... 9

2 Fields...... 13 2.1 Basic Properties of Fields ...... 13 2.2 OrderedFields...... 15

3 Real Numbers ...... 19 3.1 Extended real numbers ...... 22 3.2 Limsups and Liminfs ...... 24 3.3 Partitioning the Real Numbers ...... 29 3.4 The Decimal Representation of a Real Number ...... 29 3.5 Summary of Key Facts about Real Numbers ...... 30 3.6 (Optional) Proofs of Theorem 3.6 and Theorem 3.3 ...... 31 3.7 Supremum and Inﬁniums of sets ...... 32

4 Complex Numbers ...... 35 4.1 A Matrix Perspective (Optional) ...... 37

5 Set Operations, Functions, and Counting ...... 39 5.1 Set Operations and Functions ...... 39 5.1.1 Exercises...... 40 5.2 Cardinality ...... 41 5.3 FiniteSets...... 41 5.4 Countable and Uncountable Sets ...... 43 4 Contents 5.4.1 Exercises...... 46

Part II Normed and Metric Spaces

6 Metric Spaces ...... 49 6.1 Normed Spaces [Linear Algebra Meets Analysis] ...... 49 6.1.1 Review of Vector Spaces and Subspaces ...... 49 6.1.2 Normed Spaces ...... 50 6.2 Sequences in Metric Spaces ...... 55 6.3 General Limits and Continuity in Metric Spaces ...... 58 6.4 Density and Separability ...... 66 6.5 Test 2: Review Topics ...... 67

7 Series and Sums in Banach Spaces ...... 69

8 More Sums and Sequences ...... 85 8.1 Rearrangements ...... 85 8.2 DoubleSequences ...... 90 8.3 Iterated Limits ...... 93 8.4 Double Sums and Continuity of Sums ...... 94 8.5 Sums of positive functions ...... 95 8.6 Sums of complex functions ...... 97 8.7 Iterated sums and the Fubini and Tonelli Theorems ...... 99 8.8 `p – spaces, Minkowski and H¨olderInequalities ...... 100 8.9 Exercises...... 103 8.9.1 Limit Problems ...... 103 8.9.2 Monotone and Dominated Convergence Theorem Problems ...... 103 8.9.3 `p Exercises ...... 105

9 Topological Considerations ...... 107 9.1 Closed and Open Sets ...... 107 9.2 Continuity Revisited ...... 112 9.3 Metric spaces as topological spaces (Not required Reading!) ...... 113 9.4 Exercises...... 114 9.5 Sequential Compactness ...... 115 9.6 Open Cover Compactness ...... 118 9.7 Equivalence of Sequential and Open Cover Compactness in Metric Spaces ...... 120 9.8 Connectedness ...... 122 9.8.1 Connectedness Problems ...... 123

Part III Calculus

Page: 4 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Contents 5 10 Differential Calculus in One Real Variable ...... 127 10.1 Basics of the Derivative ...... 127 10.2 First Derivative Test and the Mean Value Theorems ...... 129 10.3 Limsup and Liminf revisited ...... 132 10.4 Differentiating past infinite sums ...... 133 10.5 Taylor’s Theorem ...... 139 10.6 Inverse Function Theorem I ...... 141 10.7 L’hopitals Rule ...... 145

11 Riemann (Like) Integration Theory ...... 149 11.1 Banach Spaces of Bounded Functions ...... 149 11.2 Algebras of Sets ...... 151 11.3 Finitely additive measure on A ...... 153 11.4 Simple Functions ...... 155 11.5 Simple Integration ...... 155 11.6 Extending the Integral by Uniform Limits ...... 158

Part IV Appendices

A Appendix: Notation and Logic ...... 163

B Appendix: More Set Theoretic Properties (highly optional) ...... 165 B.1 Appendix: Zorn’s Lemma and the Hausdorﬀ Maximal Principle (optional) ...... 165 B.2 Cardinality ...... 167 B.3 Formalities of Counting ...... 168

C Appendix: Aspects of General Topological Spaces...... 171 C.1 Compactness...... 171 C.2 Exercises...... 176 C.2.1 General Topological Space Problems ...... 176 C.2.2 Metric Spaces as Topological Spaces ...... 176 C.2.3 Compactness Problems ...... 178 C.3 Connectedness in General Topological Spaces ...... 178 C.4 Completions of Metric Spaces ...... 181 C.5 Supplementary Remarks ...... 182 C.5.1 Riemannian Metrics ...... 182

D Appendix: Limsup and liminf of functions ...... 185

Page: 5 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6 Contents E Appendix: Math 140A Topics ...... 195 E.1 Summary of Key Facts about Real Numbers ...... 195 E.2 Test 2 Review Topics: ...... 195 E.3 After Test 2 Review Topics: ...... 195

References ...... 197

Index ...... 199

Page: 6 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Part I

Numbers

1 Natural, integer, and rational Numbers

Notation 1.1 Let N = {1, 2,... } denote the natural numbers, N0 = {0}∪N, Remark 1.4. Let us further observe that the well ordering principle implies the induction axiom. Indeed, suppose that S ⊂ is a subset such that 1 ∈ S and := {0, ±1, ±2,... } = {±n : n ∈ } N Z N0 n + 1 ∈ S whenever n ∈ S. For sake of contradiction suppose that S 6= N so be the integers, and that T := N \ S is not empty. By the well ordering principle there T has a unique minimal element m and in particular T ⊂ {m, m + 1,... } . This implies nm o := : m ∈ and n ∈ that {1, 2, . . . , m − 1} ⊂ S and then by assumption that {1, 2, . . . , m} ⊂ S. But Q n Z N this then implies T ⊂ {m + 1,... } and therefore m∈ / T which violates m being be the rationale numbers. the minimal element of T. We have arrived at the desired contradiction and I am going to assume that the reader is familiar with all the standard arith- therefore conclude that S = N. metic operations (addition, multiplication, inverses, etc.) on , , and . How- N0 Z Q Remark 1.5. Recall that, for q ∈ Q, we deﬁne1 ever let us review the important induction axiom of the natural numbers. q if q ≥ 0 Induction Axiom If S ⊂ N is a subset such that 1 ∈ S and n + 1 ∈ S |q| = −q if q ≤ 0. whenever n ∈ S, then S = N. This axiom takes on two other useful forms which we describe in the next Recall that, for all a, b ∈ Q, Propositions. 1 1 |a + b| ≤ |a| + |b| , |ab| = |a| |b| , and = when a 6= 0. Proposition 1.2 (Strong form of Induction). Suppose S ⊂ N is a subset a |a| such that 1 ∈ S and n + 1 ∈ S whenever {1, 2, . . . , n} ⊂ S, then S = N. It is also often useful to keep in mind that the following statements are equiv- Proof. Let T := {n ∈ N : {1, 2, . . . , n} ⊂ S} . Then 1 ∈ T and if n ∈ T then alent for a, b ∈ Q with b ≥ 0; n + 1 ∈ T by assumption. Therefore by the induction axiom, T = N so that {1, 2, . . . , n} ⊂ S in for all n ∈ N. This suﬃces to show S = N. 1. |a| ≤ b, 2. −b ≤ a ≤ b, and Proposition 1.3 (Well ordering principle). Suppose S ⊂ N is a non-empty subset, then there exists a smallest element m of S. 3. ±a ≤ b, i.e. a ≤ b and −a ≤ b.

Proof. Let S be a subset of N for which there is no smallest element, m ∈ S. Lemma 1.6. If a, b ∈ Q, then Let ||b| − |a|| ≤ |b − a| . (1.1) T = {n ∈ N : n < s for all s ∈ S} . If 1 ∈/ T, then 1 ∈ S and 1 would be a smallest element of S. Hence we must Proof. Since both sides of Eq. (1.1) are symmetric in a and b, we may have 1 ∈ T. Now suppose that n ∈ T so that n < s for all s ∈ S. If n + 1 ∈/ T assume that |b| ≥ |a| so that ||b| − |a|| = |b| − |a| . Since then there exists s ∈ S such that n < s ≤ n+1 which would force s = n+1 ∈ S. But we would then have n + 1 is the minimal element of S which is assumed |b| = |b − a + a| ≤ |b − a| + |a| , not to exist. So we have shown if n ∈ T then n + 1 ∈ T. So by the induction axiom of N it follows that T = N and therefore n∈ / S for all n ∈ N, i.e. S = ∅. it follows that ||b| − |a|| = |b| − |a| ≤ |b − a| . 4 1 Natural, integer, and rational Numbers 1.1 Limits in Q In this course we will often use the abbreviations, i.o. and a.a. which stand for infinitely often and almost always respectively. For example an ≤ bn a.a. n means there exists an N ∈ N such that an ≤ bn for all n ≥ N while an ≤ bn i.o. n means for all N ∈ N there exists a n ≥ N such that an ≤ bn. So for example, 1/n ≤ 1/100 for a.a. n while and (−1)n ≥ 0 i.o. By the way, it should be clear that if something happens for a.a. n then it also happens i.o. n. ∞ Definition 1.9. A sequence {an}n=1 ⊂ Q converges to 0 ∈ Q if for all ε > 0 in Q there exists N ∈ N such that |an| ≤ ε for all n ≥ N. Alternatively put, for all ε > 0 we have |an| ≤ ε for a.a. n. This may also be stated as for all M ∈ N, 1 |an| ≤ M for a.a. n. ∞ Definition 1.10. A sequence {an}n=1 ⊂ Q converges to a ∈ Q if |a − an| → 0 The proof of the previous lemma illustrates one of the key techniques of 1 ∞ as n → ∞, i.e. if for all N ∈ N, |a − an| ≤ N for a.a. n. As usual if {an}n=1 adding 0 to an expression. In this case we added 0 in the form of −a + a converges to a we will write an → a as n → ∞ or a = limn→∞ an. to b. The next remark records a couple of other very important “tricks” in this ∞ subject. Taking to heart the following remarks will greatly aid the student in Proposition 1.11. If {an}n=1 ⊂ Q converges to a ∈ Q, then limn→∞ |an| = real analysis. |a| . Proof. From Lemma 1.6 we have, Remark 1.7( Some basic philosophies of real analysis). Let a, b, ε be numbers (i.e. in Q or later real numbers). We will often prove; ||a| − |an|| ≤ |a − an| .

1. a ≤ b by showing that a ≤ b + ε for all ε > 0. (See the next theorem.) Thus if ε > 0 is given, by definition of an → a there exists N ∈ N such that 2. a = b by proving a ≤ b and b ≤ a or |a − an| < ε for all n ≥ N. From the previously displayed equation, it follows 3. prove a = b by showing |b − a| ≤ ε for all ε > 0. that ||a| − |an|| < ε for all n ≥ N and hence we may conclude that limn→∞ |an| exists and is equal to |a| . Theorem 1.8. The rationale2 numbers have the following properties; ∞ Lemma 1.12 (Convergent sequences are bounded). If {an}n=1 ⊂ Q con- 1. For any p ∈ Q there exists N ∈ N such that p < N. verges to a ∈ , then there exists M ∈ such that |an| ≤ M for all n ∈ . 1 Q Q N 2. For any ε ∈ Q with ε > 0 there exists an N ∈ N such that 0 < < ε. N Proof. Taking ε = 1 in the definition of a = lim a implies there exists 3. If a, b ∈ Q and a ≤ b + ε for all ε > 0, then a ≤ b. n→∞ n N ∈ N such that |an − a| ≤ 1 for all n ≥ N. Therefore, Proof. 1. If p ≤ 0 we may take N = 1. So suppose that p = m with n |a | = |a − a + a| ≤ |a − a| + |a| ≤ 1 + |a| for n ≥ N. m, n ∈ N. In this case let N = m. n n n 2. Write ε = m with m, n ∈ and then take N = 2n. n o n N We may now take M := max |a |N ∪ {1 + |a|} . 3. If a ≤ b is false happens iff a > b which is equivalent to a − b > 0. If we n n=1 a−b ∞ now let ε := 2 > 0, then Theorem 1.13. If {an}n=1 ⊂ Q converges to a ∈ Q \{0} , then a = b + (b − a) > b + ε 1 1 lim = . (1.2) n→∞ an a which would violate the assumption that a ≤ b + ε for all ε > 0. 1 It is possible that an = 0 for small n so that is not defined but for large n 1 Absolute values will be discussed in more generality in Section 2.2 below. an 2 this can not happen and therefore it makes sense to talk about the limit which We will see that the real numbers have these same properties as well. only depends on the tail of the sequences.

Page: 4 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 1.2 The Problem with Q 5 ∞ Proof. Since a 6= 0 we know that |a| > 0. Hence, there exists M := M|a| ∈ N Exercise 1.1. Show that all convergent sequences {an}n=1 ⊂ Q are Cauchy. |a| such that |an − a| < 2 for all n ≥ M. Therefore for n ≥ M ∞ Exercise 1.2. Show all Cauchy sequences {an}n=1 are bounded – i.e. there |a| exists M ∈ N such that |a| = |a − a + a | ≤ |a − a | + |a | < + |a | n n n n 2 n |an| ≤ M for all n ∈ N. 3 |a| ∞ ∞ from which it follows that |an| > for all n ≥ M. If ε > 0 is given arbitrarily, 2 Exercise 1.3. Suppose {an}n=1 and {bn}n=1 are Cauchy sequences in Q. Show ∞ ∞ we may choose N ≥ M such that |a − an| < ε for all n ≥ M. Then for n ≥ N {an + bn}n=1 and {an · bn}n=1 are Cauchy. we have, ∞ ∞ 1 1 a − an |a − an| ε 2ε Exercise 1.4. Assume that {an}n=1 and {bn}n=1 are convergent sequences in − = = < = . ∞ ∞ |a| 2 an a ana |an| |a| Q. Show {an + bn}n=1 and {an · bn}n=1 are convergent in Q and 2 · |a| |a| As ε > 0 is arbitrary it follows that 2ε > 0 is arbitrarily small as well (replace lim (an + bn) = lim an + lim bn and |a|2 n→∞ n→∞ n→∞ 2 ε by ε |a| /2 if you feel it is necessary), and hence we may conclude that Eq. lim (anbn) = lim an · lim bn. (1.2) holds. n→∞ n→∞ n→∞ Variation on the method. In order to make these arguments more rout- ∞ ∞ Exercise 1.5. Assume that {an}n=1 and {bn}n=1 are convergent sequences in ing, it is often a good idea to write an = a + δn where δn := an − a is the error Q such that an ≤ bn for all n. Show A := limn→∞ an ≤ limn→∞ bn =: B. between a and a. By assumption, lim δ = 0 and so for any δ > 0 given n n→∞ n ∞ ∞ there exists N (δ) ∈ such that |δ | ≤ δ. With this notation we have, Exercise 1.6 (Sandwich Theorem). Assume that {an}n=1 and {bn}n=1 are N n ∞ convergent sequences in Q such that limn→∞ an = limn→∞ bn. If {xn} is n=1 1 1 1 1 −δn 1 another sequence in Q which satisfies an ≤ xn ≤ bn for all n, then − = − = · an a a + δn a a + δn a lim xn = a := lim an = lim bn. δ n→∞ n→∞ n→∞ ≤ . |a| (||a| − δ|) Please note that that main part of the problem is to show that limn→∞ xn So if we assume that δ ≤ |a| /2 we find that exists in Q. Hint: start by showing; if a ≤ x ≤ b then |x| ≤ max (|a| , |b|) . Definition 1.15 (Subsequence). We say a sequence, {y }∞ is a subse- 1 1 2 k k=1 quence of another sequence, {x }∞ , provided there exists a strictly increas- − ≤ 2 δ for all n ≥ N (δ) . (1.3) n n=1 an a |a| ing function, N 3 k → nk ∈ N such that yk = xnk for all k ∈ N. [Example, 2 ∞ ∞ nk = k + 3, and {yk := xk2+3}k=1 would be a subsequence of {xn}n=1 .] Taking δ = δ (ε) = min |a| /2, |a|2 ε/2 in Eq. (1.3) shows for n ≥ N (δ (ε)) ∞ Exercise 1.7. Suppose that {xn}n=1 is a Cauchy sequence in Q (or R) which 1 1 1 ∞ that − ≤ ε. Since ε > 0 is arbitrary we may conclude that limn→∞ = an a an has a convergent subsequence, {yk = xnk }k=1 in Q (or R). Show that limn→∞ xn 1 exists and is equal to limk→∞ yk. a . • End of Lecture 1, 9/28/2012 ∞ 1.2 The Problem with Q Definition 1.14. A sequence {an}n=1 ⊂ Q is Cauchy if |an − am| → 0 as m, n → ∞. More precisely we require for each ε > 0 in that |a − a | ≤ ε Q m n The problem with is that it is full of “holes.” To be more precise, is not for a.a. pairs (m, n) , i.e. there should exists N ∈ such that |a − a | ≤ ε for Q Q N m n “complete,” i.e. not all Cauchy sequences are convergent. In fact, according all m, n ≥ N. to Corollary 5.31 below, “most” Cauchy sequences of rational numbers do not 3 The idea is very simple here. If an is near a and a 6= 0 then an must stay away converge to a rational number. Let us demonstrate some examples pointing out from zero. You should draw the picture to go along with the proof. this flaw. We first pause to recall how to sum geometric series.

Page: 5 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6 1 Natural, integer, and rational Numbers Lemma 1.16 (Geometric Series). Let α ∈ , m, n ∈ with n ≤ m, and 1 1 Q Z 0 < ≤ e − Sm ≤ . Pm k (m + 1)! m · m! S := k=n α . Then m − n + 1 if α = 1 Multiplying this inequality by m! then implies, S = m+1 n . α −α if α 6= 1 α−1 1 0 < m!e − m!S ≤ . Moreover if 0 ≤ α < 1, then m m m X 1 − αm−n+1 αn However for m suﬃciently large m!e ∈ (as e is assumed to be rational) and αk = αn ≤ . (1.4) N 1 − α 1 − α m!Sm is always in N and therefore k := m!e − m!Sm ∈ N. But there is no k=n 1 element k ∈ N such that 0 < k < m and hence we must conclude limn→∞ Sn Proof. When α = 1, P∞ 1 1 n can not exist in Q. Moral: the number e = n=0 n! = limn→∞ 1 + n that m you learned about in calculus is not in Q! X S = 1k = m − n + 1. k=n If α 6= 1, then αS − S = αm+1 − αn. Solving for S gives

m X αm+1 − αn S = αk = if α 6= 1. (1.5) α − 1 k=n

Pn 1 Example 1.17. Let Sn := k=0 k! ∈ Q for all n ∈ N. For n > m in N we have,

n n−m X 1 X 1 0 ≤ S − S = = n m k! (m + j)! k=m+1 j=1 1 1 = + ··· + (m + 1)! n! Pn 1 1 n Plotting the partial sums k=0 k! (black curve) and 1 + n (red curve) " # 1 1 1 2 1 n−m which are both converging to “e.” ≤ + + ··· + m! m + 1 m + 1 m + 1 √ Example 1.18 (Square roots√ need not exist). The square root, 2, of 2 does not 1 1 1 1 exist in . Indeed, if 2 = m where m and n have no common factors (in ≤ = , (1.6) Q n 1 particular no common factors of 2 so that either m or n is odd), then m! m + 1 1 − m+1 m · m! 2 wherein we have used Eq. (1.4) for the last inequality. From this inequality it m 2 2 ∞ 2 = 2 =⇒ m = 2n . follows that {Sn}n=0 is a Cauchy sequence and we also have, n 1 1 This shows that m2 is even which would then imply that m = 2k is even (since ≤ S − S ≤ for all n > m. (1.7) 2 2 (m + 1)! n m m · m! odd·odd=odd). However this implies 4k = 2n from which it follows that n2 = 2k2 is even and hence n is even. But this contradicts the assumption that Suppose that e := limn→∞ Sn were to exist in Q. Then letting n → ∞ in Eq. m and n had no common factors (of 2). (1.7) would show,

Page: 6 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 1.3 Peano’s arithmetic (Highly Optional) 7

Exercise 1.8. Use the following outline to construct another Cauchy sequence Induction hypothesis If S ⊂ N0 is a set such that 0 ∈ S and s (n) ∈ S whenever ∞ {qn}n=1 ⊂ Q which is not convergent in Q. n ∈ S, then S = N0. 2 1. Recall that there is no element q ∈ Q such that q = 2. To each n ∈ N let Assuming these axioms one may develop all of the properties or N0 that mn ∈ N be chosen so that you are accustomed to seeing. I will develop the basic properties of addition, multiplication, and the ordering on N0 in this section. For more on this point m2 (m + 1)2 n < 2 < n (1.8) and then the further construction of Z and Q from N0, the reader is referred n2 n2 to the notes; “Numbers” by M. T aylor. You may also consult E. Landau’s

mn book [1] for a very detailed (but perhaps too long winded) exposition of these and let qn := n . 2 ∞ topics. 2. Verify that qn → 2 as n → ∞ and that {qn}n=1 is a Cauchy sequence in Q. 3. Show {q }∞ does not have a limit in . n n=1 Q Lemma 1.20. The map s : N0 → N is a bijection. Example 1.19. It is also a fact that π∈ / where Q Proof. Let S := s (N0) ∪ {0} ⊂ N0. Then 0 ∈ S and s (0) ∈ s (N0) ⊂ S. Moreover if x ∈ ∩ S then s (x) ∈ s ( ) ⊂ S so that x ∈ S =⇒ s (x) ∈ S Z ∞ 1 Z N 1 N N0 and hence S = 0 and therefore s ( 0) = . π = 2 2 dx = 2 lim 2 dx N N N 0 1 + x N→∞ 0 1 + x N 2 Theorem 1.21 (Addition). There exists a function p : N0 × N0 → N0 such X 1 1 = 2 lim · that p (x, 0) = x for all x ∈ N0 and p (x, s (y)) = s (p (x, y)) for all x, y ∈ N0. N→∞ k 2 N k=1 1 + N Moreover, we may construct p so that p (s (x) , y) = p (x, s (y)) for all x, y ∈ N0. N 2 This function p satisﬁes the following properties; X 2N = lim . N→∞ N 2 + k2 1. p (x, 0) = x = p (0, x) for all x ∈ N0, k=1 2. p (x, 1) = p (1, x) = s (x) for all x ∈ N0, The point is that the basic operations from calculus tend to produce “real 3. p (x, y) = p (y, x) for all x, y ∈ N0, numbers” which are not rational even though we start with only rational num- 4. p (x, p (y, z)) = p (p (x, y) , z) for all x, y, z ∈ N0. bers. Proof. We will construct p inductively. Let • End of Lecture 2, 10/1/2012 S := {x ∈ N : ∃px : N0 → N0 3 px (0) = x and px (s (y)) = s (px (y)) ∀y ∈ N0} .

1.3 Peano’s arithmetic (Highly Optional) Taking p0 (y) = y shows 0 ∈ S. Moreover if x ∈ S we deﬁne

ps(x) (y) := s (px (y)) for all y ∈ N0. This section is for those who want to understand N at a more fundamental level. Here we start with Peano’s rather minimalist axioms for N and show how We then have ps(x) (0) = s (px (0)) = s (x) and they lead to all the standard properties you are used to using for N. Here are the axioms; ps(x) (s (y)) := s (px (s (y))) = s ◦ s (px (y)) = s ps(x) (y) non-empty is a non-empty set which contains a distinguished element, 0. N0 which shows s (x) ∈ S. Thus we may conclude S = N0 and we may now deﬁne We let := \{0} and call these the natural numbers. N N0 p (x, y) := px (y) for all x, y ∈ 0. By construction this function satisﬁes, 4 N Successor Function There is an injective function, s : N0 → N and we let 1 := s (0) ∈ N. p (s (x) , y) = s (p (x, y)) = p (x, s (y)) .

4 Injective is the same as one to one. Thus we are assuming that if s (n) = s (m) then We now verify the properties in items 1. – 4. n = m.

Page: 7 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8 1 Natural, integer, and rational Numbers

1. By construction p (x, 0) = x for all x ∈ N0. Let S = {x ∈ N : p (0, x) = x} , Definition 1.24. Given x, y ∈ N0, we say x < y iff y = x + n for some n ∈ N then 0 ∈ S and if x ∈ S we have p (0, s (x)) = s (p (0, x)) = s (x) so that and x ≤ y iff y = x + n for some n ∈ N0. We further let s (x) ∈ S. Therefore S = N0 and the first item holds. 2. p (x, 1) = p (x, s (0)) = s (p (x, 0)) = s (x) and p (1, x) = p (s (0) , x) = Rx := {x + n : n ∈ N0} s (p (0, x)) = s (x) so that item 2. is proved. so that y ≥ x iff y ∈ Rx. 3. Let S = {x ∈ N0 : p (x, ·) = p (·, x)} . Then by items 1 and 2. it follows that 0, 1 ∈ S. Moreover if x ∈ S, then for all y ∈ N0 we find, Proposition 1.25. If x, y ∈ N0 and x ≤ y and y ≤ x then x = y. Moreover if p (s (x) , y) = s (p (x, y)) = s (p (y, x)) = p (y, s (x)) x ≤ y then either x < y or x = y. Proof. By assumption there exists m, n ∈ such that y = x + m and which shows s (x) ∈ S. Therefore S = and item 3. is proved. N0 N0 x = y + n and therefore y = y + (m + n) . Hence by cancellation it follows 4. Let that m + n = 0. If n 6= 0 then n = s (x) for some x ∈ N0 and we have S := {x ∈ N0 : p (x, p (y, z)) = p (p (x, y) , z) ∀y, z ∈ N0} . m + n = m + s (x) = s (m + x) ∈ N which would imply m + n 6= 0. Thus we Then 0 ∈ S and if x ∈ S we find, conclude that m = 0 = n and therefore x = y. If x ≤ y and x 6= y then y = x + n for some n ∈ N0 with n 6= 0, i.e. x < y. p (s (x) , p (y, z)) = s (p (x, p (y, z))) = s (p (p (x, y) , z)) = p (s (p (x, y)) , z) = p (p (s (x) , y) , z) Proposition 1.26. If x, y ∈ N0 then precisely one of the following three choices must hold, 1) x < y, 2) x = y, 3 y < x. which shows that s (x) ∈ S and therefore S = N0 and item 4. is proved. Proof. Suppose that x ≤ y does not hold, i.e. y∈ / Rx. We wish to show that y < x, i.e. that x = y + n for some n ∈ N. We do this by induction on y. That is let S be the the set of y ∈ N0 such that the statement y∈ / Rx implies y < x Notation 1.22 We now write x + y for p (x, y) and refer to the symmetric holds. If y = 0 ∈/ Rx implies n := x 6= 0 so that x = y + n, i.e. y = 0 < x. This binary operator, +, as addition. shows 0 ∈ S. Now suppose that y ∈ S and that y+1 ∈/ Rx = {x + m : m ∈ N0} . It follows that y + 1 6= x + m + 1 for all m ∈ N0 and hence that y 6= x + m To summarize we have now shown addition satisfies for all x, y, z ∈ N0; for all m ∈ N0, i.e. y∈ / Rx. So by induction y < x and therefore x = y + k for 0 0 1. x + 0 = 0 + x = x, some k ∈ N. Since k ∈ N we know there exists k ∈ N0 such that k = k and it 0 2. s (x) = x + 1 = 1 + x, follows that x = y + 1 + k , i.e. y + 1 ≤ x. Since y + 1 ∈/ Rx we may conclude 3. x + y = y + x, that in fact y + 1 < x and therefore y + 1 ∈ S. So by induction S = N0 and 4.( x + y) + z = x + (y + z) . we have shown if x < y does not hold iff y ≤ x. Combining this statement with 5. The induction hypothesis may now be written as; if S ⊂ N0 is a subset such the Proposition 1.25 completes the proof. that 0 ∈ S and n + 1 ∈ S whenever n ∈ S, then S = N0. We have now set up a satisfactory addition operations and ordering on N0. Our next goal is to define multiplication on N0. Proposition 1.23 (Additive Cancellation). If x, y, z ∈ N0 and x+z = y+z, then x = y. Theorem 1.27. There exists a function M : N0×N0 → N0 such that M (x, 0) = 0 for all x ∈ N0 and M (x, y + 1) = M (x, y)+x for all x, y ∈ N0. This function Proof. Let S be those z ∈ N0 for which the statement x + z = y + z implies M satisfies the following properties; x = y holds. It is clear that 0 ∈ S. Moreover if z ∈ S and x + (z + 1) = y + (z + 1) then (x + 1) + z = (y + 1) + z and so by the inductive hypothesis 1. M (x, 0) = 0 = M (0, x) for all x ∈ N0, s (x) = x + 1 = y + 1 = s (y) . Recall that s is one to one by assumption 2. M (x, 1) = M (1, x) = x for all x ∈ N0, and therefore we may conclude x = y and we have shown s (z) ∈ S. Therefore 3. M (x, y) = M (y, x) for all x, y ∈ N0, 4. M (x, y + z) = M (x, y) + M (x, z) for all x, y, z ∈ . S = N0 and the proposition is proved. N0 5. M (x, M (y, z)) = M (M (x, y) , z) for all x, y, z ∈ N0.

Page: 8 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 1.3 Peano’s arithmetic (Highly Optional) 9

Proof. Let S denote those x ∈ N0 such that there exists a function Mx : 5. Let N0 → N0 satisfying Mx (0) = 0 and Mx (y + 1) = Mx (y) + x for all y ∈ N0. S := {x ∈ : M (x, M (y, z)) = M (M (x, y) , z) ∀y, z ∈ } . Taking M0 (y) := 0 shows 0 ∈ S. Moreover if x ∈ S we define Mx+1 (y) := N0 N0 M (y) + y. Then M (0) = 0 and x x+1 Then 0 ∈ S and if x ∈ S we find, M (y + 1) = M (y + 1) + y + 1 = M (y) + x + y + 1 x+1 x x M (x + 1,M (y, z)) = M (x, M (y, z)) + M (y, z) while while Mx+1 (y) + (x + 1) = Mx (y) + y + x + 1 = Mx+1 (y + 1) . M (M (x + 1, y) , z) = M (M (x, y) + y, z) = M (M (x, y) , z) + M (y, z) . This shows that x + 1 ∈ S and so by induction S = N0 and we may now define M (x, y) := Mx (y) for all x, y ∈ N0. We now prove the properties of M stated The last two equations along with the induction hypothesis shows x+1 ∈ S above. and therefore S = N0 and item 5. is proved. 1. By construction M (x, 0) = 0 for all x. Let S := {x ∈ N0 : M (0, x) = 0} . Then 0 ∈ S and if x ∈ S we have Notation 1.28 We now write x · y for M (x, y) and refer to the symmetric M (0, x + 1) = M (0, x) + 0 = 0 + 0 = 0 binary operator, ·, as multiplication..

which shows x + 1 ∈ S. Therefore by induction S = N0 and M (0, x) = 0 To summarize Theorem 1.27 we have shown multiplication satisﬁes for all for all x ∈ N0. x, y, z ∈ N0; 2. M (x, 1) = M (x, 0 + 1) = M (x, 0) + x = 0 + x = x for all x ∈ N0. Let 1. x · 0 = 0 = 0 · x, S := {x ∈ N0 : M (1, x) = x} . Then 0 ∈ S and if x ∈ S we have 2. x · 1 = x = 1 · x, M (1, x + 1) = M (1, x) + 1 = x + 1 3. x · y = y · x, 4. x · (y + z) = x · y + x · z, which shows x + 1 ∈ S. Therefore S = N0 and M (1, x) = x for all x ∈ N0. 5.( x · y) · z = x · (y · z) . 3. Let S := {x ∈ N0 : M (x, ·) = M (·, x)} . Then by items 1. and 2. we know that 0, 1 ∈ S. Now suppose that x ∈ S, then by construction, Proposition 1.29 (Multiplicative Cancellation). If x, y ∈ N0 and z ∈ N such that x · z = y · z, then x = y. M (x + 1, y) = Mx+1 (y) = M (x, y) + y Proof. If x 6= y, say x < y, then y = x + n for some n ∈ N and therefore while M (y, x + 1) = M (y, x) + y. y · z = (x + n) · z = x · z + n · z. The last two displayed equations along with the induction hypothesis shows Hence if x · z = y · z, then by additive cancellation we must have n · z = 0. As 0 0 0 x + 1 ∈ S and therefore S = N0 and item 3. is proved. n, x ∈ N we may write n = n0 + 1 and z = z + 1 with n , z ∈ N0 and therefore, 4. Let S denotes those x ∈ S such that M (x, y + z) = M (x, y) + M (x, z) for 0 = n · z = (n0 + 1) · (z0 + 1) = n0 · z0 + n0 + z0 + 1 6= 0 all y, z ∈ N0. Then 0, 1 ∈ S and if x ∈ S we have, M (x + 1, y + z) = M (x, y + z) + y + z which is a contradiction. = M (x, y) + M (x, z) + y + z Remark 1.30 (Base 10 counting). The typical method of counting is to use base = M (x, y) + y + M (x, z) + z 10 enumeration of N0. The rules are; = M (x + 1, y) + M (x + 1, z) 0 = 0, 1 = 1, 2 := 1 + 1, 3 := 2 + 1, 4 := 3 + 1, 5 := 4 + 1 6 := 5 + 1, 7 := 6 + 1, 8 := 7 + 1, 9 := 8 + 1, and 10 := 9 + 1. which shows x + 1 ∈ S. Therefore S = N0 and we have proved item 4.

Page: 9 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Once these element of N0 have been deﬁned, then given a0, . . . , an ∈ {0, 1,..., 9} with an 6= 0, we let n X k anan−1 . . . a0 := ak10 . k=0 For example, 35 = 3 · 10 + 5 = 34 + 1, etc.

As mentioned above one can formalize Z and Q using N0 constructed above. I will omit the details here and refer the reader to the references already mentioned. 2 Fields

The basic question we want to eventually address is; What are the real 2.1 Basic Properties of Fields numbers? Our answer is going to be; the real numbers is the essentially unique complete ordered field, see Theorem 3.3 below. In order to make sense of this Here are some sample properties about fields. For more information about Fields answer we need to explain the terms, “complete,” “ordered,” and “field.” We see 5-8 of Rudin. will start with the notion of a field which loosely stated means something that can reasonably be interpreted a “numbers.” Lemma 2.2. Let F be a field, then; 1. There is only one additive and multiplicative inverses. Definition 2.1 (Fields, i.e.“ numbers”). A field is a non-empty set F 2. If x, y, z ∈ with x 6= 0 and xy = xz then y = z. equipped with two operations called addition and multiplication, and denoted F 3. 0 · x = 0 for all x ∈ . by + and ·, respectively, such that the following axioms hold;( subtraction and F 4. If x, y ∈ such that xy = 0 then x = 0 or y = 0. division are defined implicitly in terms of the inverse operations of addition and F 5. (−x) y = − (xy) . multiplication:) 6. − (−x) = x for all x ∈ F. 1. Closure of F under addition and multiplication. For all a, b in F, both a+b 7. (−x)(−y) = xy or all x, y ∈ F. and a · b are in F (or more formally, + and · are binary operations on F). Proof. We take each item in turn. 2. Associativity of addition and multiplication. For all a, b, and c in F, the following equalities hold: a+(b+c) = (a+b)+c and a·(b·c) = (a·b)·c. 1. Suppose that x + y = 0 = x + y0, then adding −x to both sides of this 3. Commutativity of addition and multiplication. For all a and b in F, equation shows y = y0. Taking y = −x then shows y = −x = y0, i.e. the following equalities hold: a + b = b + a and a · b = b · a. additive inverses are unique. Similarly if x 6= 0 and xy = 1 then multiplying 4. Additive and multiplicative identity. There exists an element of F, this equation by x−1 shows y = x−1 and so there is only one multiplicative called the additive identity element and denoted by 0 = 0F, such that for all inverse. a in F, a + 0 = a. Likewise, there is an element, called the multiplicative 2. If xy = xz then multiplying this equation by x−1 shows y = z. identity element and denoted by 1 = 1F, such that for all a in F, a · 1 = a. 3. It is assumed that 0F 6= 1F. 0 · x + x = 0 · x + 1 · x = (0 + 1) · x = 1 · x = x. 5. Additive and multiplicative inverses. For every a in F, there exists an Adding −x to both side of this equation using associativity and commuta- element −a in F, such that a+(−a) = 0. Similarly, for any a ∈ F other than −1 −1 tivity of addition then implies 0 · x = 0. 0, there exists an element a in F, such that a · a = 1. (The elements a + (−b) and a · b−1 are also denoted a − b and a/b, respectively.) In other 4. If x ∈ F\{0} and y ∈ F such that xy = 0, then words, subtraction and division operations exist. 0 = x−1 · 0 = x−1 (xy) = x−1x y = 1y = y. 6. Distributivity of multiplication over addition. For all a, b and c in F, the following equality holds: a · (b + c) = (a · b) + (a · c). 5.( −x) y + xy = (−x + x) · y = 0 · y = 0 =⇒ (−x) y = − (xy) . 6. Since (−x) + x = 0 we have − (−x) = x. (Note that all but the last axiom are exactly the axioms for a commuta- 7.( −x)(−y) = − (x · (−y)) = − (− (xy)) = xy by 6. tive group, while the last axiom is a compatibility condition between the two operations.)

Example 2.3. Here are a few examples of Fields; 12 2 Fields

1. F2 = {0, 1} with 0 + 0 = 0 = 1 + 1, and 0 + 1 = 1 + 0 = 0 and 0 · 1 = 1 · 0 = ϕ ((m + 1) + n) = ϕ (m + n + 1) = ϕ (m) + ϕ (n + 1) 0 · 0 = 0 and 1 · 1 = 1. In this case −1 = 1, 1−1 = 1 and −0 = 0. = ϕ (m) + ϕ (n) + 1F 2. Q – the rational numbers with the usual addition and multiplication of m −1 n m −m = ϕ (m) + 1F + ϕ (n) = ϕ (m + 1) + ϕ (n) fractions. n = m if m 6= 0 and − n = n . 3. = (t) where F Q which shows m + 1 ∈ S. Therefore by induction, S = N0 and ϕ (m + n) = ϕ (m) + ϕ (n) for all m, n ∈ . p (t) N0 (t) = : p (t) and q (t) are polynomials over 3 q (t) 6= 0 . If m ∈ N0 we have ϕ (−m) = −ϕ (m) by construction. If n > m ∈ N0, then Q q (t) Q ϕ (n + (−m)) + ϕ (m) = ϕ (n − m) + ϕ (m) = ϕ (n) Again the multiplication and addition are as usual. so that Example 2.4. Z is not a field. For example, 2 has no multiplicative inverse in Z. −1 1 ϕ (n + (−m)) = ϕ (n) + (−ϕ (m)) = ϕ (n) + ϕ (−m) . The inverse to 2, 2 , should be 2 but this is not in Z. If n < m ∈ N0, then Definition 2.5. We say a map ϕ : Z → F is a (ring) homomorphism iff ϕ (1) = 1F, ϕ (0) = 0F, and for all x, y ∈ Z; ϕ (n + (−m)) = −ϕ (m − n) = − [ϕ (m) − ϕ (n)] = ϕ (n) + ϕ (−m) ϕ (x + y) = ϕ (x) + ϕ (y) and ϕ (xy) = ϕ (x) ϕ (y) . and if m, n ∈ N0, then

[The assumption that ϕ (0) = 0F is redundant since ϕ (0) = ϕ (0 + 0) = ϕ (0) + ϕ (−n + (−m)) = ϕ (− (n + m)) = −ϕ (n + m) ϕ (0) and therefore ϕ (0) = 0 .] F = − [ϕ (n) + ϕ (m)] = −ϕ (n) − ϕ (m) Lemma 2.6. For every field F there a unique (ring) homomorphism, ϕ : Z → F. = ϕ (−n) + ϕ (−m) . In fact, ϕ (n) = n1F for all n ∈ Z where 0 · 1F = 0F, Putting all of this together shows ϕ is an additive homomorphism. n times Multiplicative homomorphism: First suppose that m, n ∈ and let z }| { N0 n1F := 1F + ··· + 1F if n ∈ N and S := {m ∈ N0 : ϕ (mn) = ϕ (m) ϕ (n) for all n ∈ N0} . (−n) 1F := − (n1F) if n ∈ N. It is easily seen that 0, 1 ∈ S. Moreover if m ∈ S and n ∈ N0, then [The map ϕ need not be injective as is seen by taking F = F2.] ϕ ((m + 1) n) = ϕ (mn + n) = ϕ (mn) + ϕ (n) Proof. Let us first work on N0 ⊂ Z. We must define ϕ (0) = 0 and ϕ (1) = 1 = ϕ (m) ϕ (n) + ϕ (n) = (ϕ (m) + 1 ) ϕ (n) and then ϕ inductively by ϕ (n + 1) = ϕ (n) + ϕ (1) = ϕ (n) + 1F so that F = ϕ (m + 1) ϕ (n) , n times z }| { ϕ (n) = 1F + ··· + 1F. which shows m + 1 ∈ S. Therefore by induction, S = N0 and ϕ (mn) = ϕ (m) ϕ (n) for all m, n ∈ N0. We now write n1 for ϕ (n) with the convention that 01 = 0 . For n ∈ N we F F F If m, n ∈ N0, then must set ϕ (−n) = −ϕ (n) = − (n1F) . Thus we have ϕ (n) = n1F for all n ∈ Z. We now must show ϕ is a homomorphism. ϕ ((−m) n) = ϕ (−mn) = −ϕ (mn) = − [ϕ (m) ϕ (n)] = [−ϕ (m)] ϕ (n) = ϕ (−m) ϕ (n) Additive homomorphism: First suppose that m, n ∈ N0 and let and S := {m ∈ N0 : ϕ (m + n) = ϕ (m) + ϕ (n) for all n ∈ N0} . ϕ ((−m)(−n)) = ϕ (mn) = ϕ (mn) = (−ϕ (m)) (−ϕ (n)) = −ϕ (m) ϕ (−n) One easily sees that 0 ∈ S and that 1 ∈ S by construction. Moreover if m ∈ S, then which completes the verification that ϕ is a multiplicative homomorphism.

Page: 12 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 2.2 Ordered Fields 13 n 2.2 Ordered Fields More generally if {ai}i=1 ⊂ F we let

Definition 2.7 (Ordered Field). We say F is an ordered field if there ex- a1 ∧ · · · ∧ an := min (a1, . . . , an) and ists, P ⊂ F, called the positive elements, such that a1 ∨ · · · ∨ an := max (a1, . . . , an) Ord 1. is the disjoint union of P, {0} , and −P, i.e. if x ∈ then precisely F F be the smallest and largest element in the finite list (a1, . . . , an) . one of following happens; x ∈ P, x = 0, or −x ∈ P. Ord 2. P + P ⊂ P and P · P ⊂ P. Definition 2.13. Suppose that F and G are fields. A map, ϕ : F → G is a (field) homomorphism iff ϕ (1F) = 1G, ϕ (0F) = 0G, and for all x, y ∈ F; Lemma 2.8. Let (F,P ) be an ordered field, then; ϕ (x + y) = ϕ (x) + ϕ (y) and ϕ (xy) = ϕ (x) ϕ (y) . 1. For all x ∈ F \{0} , x2 ∈ P. In particular 1 = 12 ∈ P. 2. If x ∈ P and y ∈ −P then xy ∈ −P. Lemma 2.14( Q embeds into an ordered field). For every ordered field 3. If x ∈ P then x−1 ∈ P. (F,P ) , there a unique field homomorphism, ϕ : Q → F. In fact, Proof. If x ∈ P then x2 ∈ P · P ⊂ P while if x ∈ −P then −x ∈ P and m m −1 2 2 −1 ϕ = · 1 := m1 · (n1 ) (2.1) x = (−x) ∈ P. For item 3. we have x · x = 1 n n F F F Example 2.9. The field F = {0, 1} is not ordered. The only possible choice for P n times is P = {1} which does not work since 1 + 1 = 0 ∈/ P. z }| { where n1F := 1F + ··· + 1F and (−n) 1F := − (n1F) for n ∈ N and 0 · 1F = 0F. m Moreover; Example 2.10. Take F = Q and P = n : m, n > 0 . This is in fact the unique choice we can make for P in this case. Indeed suppose that P is any order on 1. ϕ (x) ∈ P whenever x > 0, . By Lemma 2.8, we know 1 ∈ P and then by induction it follows that ⊂ P. Q N 2. and ϕ is injective. Thus we may identify Q with ϕ (Q) and consider Q as a Then again by Lemma 2.8 we must have m · n−1 ∈ P for all m, n ∈ . Q sub-field of F. Example 2.11. Take F = Q (t) and [In particular, ordered fields must be fields with an infinite number of elements in it.] p (t) p (t) P = ∈ F : > 0 for t > 0 large , q (t) q (t) Proof. From Lemma 2.6 we know there is a unique ring homomorphism, ϕ : → , given by ϕ (m) = m · 1 . So for m ∈ and n ∈ we must have p(t) Z F F Z N i.e. q(t) ∈ P iff the highest order coefficients of p (t) and q (t) have the same 2 2 m m m t −25t+7 −t +25t+7 ϕ · n1 = ϕ · ϕ (n) = ϕ · n = ϕ (m) = m1 sign. For example t4−1026 ∈ P while t4−1026 ∈ −P. F F 1 1 n n n Notice that t > n for all n ∈ N and t < n for all n ∈ N. This is kind of strange and explains why you have to prove the “obvious” in this course!!! which forces us to define ϕ as in Eq. (2.1). Notice that is easy to verify by Moral: obvious statements are often false. induction that n1F = ϕ (n) ∈ P for all n ∈ N and in particular n1F 6= 0 for m −1 n ∈ N. In particular if x = m/n > 0 then ϕ n = m1F · (n1F) ∈ P by Notation 2.12 (Max and Min) We will often use the following notation in Lemma 2.8. We must still check that ϕ is well defined homomorphism. the sequel. If a, b are elements of an ordered field, let Well defined. Suppose that k ∈ N, we must show a if a ≤ b −1 −1 a ∧ b := min (a, b) = (km) 1 · ((kn) 1 ) = m1 · (n1 ) . b if b ≤ a F F F F By cross multiplying, this will happen iff and b if a ≤ b a ∨ b := max (a, b) = . (km) 1 · (n1 ) = ((kn) 1 ) · m1 a if b ≤ a F F F F

Page: 13 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 14 2 Fields which is the case as ϕ : Z → F is a ring homomorphism. Exercise 2.1. Let (F,P ) be an ordered field and x, y ∈ F with y > x. Show; Homomorphism property. We have 1. y + a > x + a for all a ∈ F, m p m + p −1 2. −x > −y, ϕ + = ϕ = ϕ (m + p) · ϕ (n) 1 1 n n n 3. if we further suppose x > 0, show x > y . −1 = [ϕ (m) + ϕ (p)] · ϕ (n) Definition 2.17. Given x ∈ F, we say that y ∈ F is a square root of x if y2 = x. −1 −1 [From Lemma 2.8, it follows that if x ∈ has a square root then x ≥ 0.] = ϕ (m) · ϕ (n) + ϕ (p) · ϕ (n) F m p Lemma 2.18. Suppose x, y ∈ with x2 = y2, then either x = y or x = −y. In = ϕ + ϕ F n n particular, there are at most 2 square roots of any number x ≥ 0 in F. and Proof. Observe that m q (x − y)(x + y) = (x − y) x + (x − y) y ϕ ϕ = ϕ (m) ϕ (n)−1 ϕ (q) ϕ (p)−1 n p = x2 − yx + xy − y2 = x2 − y2 = 0. = ϕ (m) ϕ (q)[ϕ (n) ϕ (p)]−1 Thus it follows that either x − y = 0 or x + y = 0, i.e. x = y or x = −y. −1 mq √ = ϕ (mq)[ϕ (np)] = ϕ . Definition 2.19. If x > 0 admits a square root we let x be the unique positive np √ root. We also define 0 = 0. m 2 2 Injectivity. If 0 = ϕ n then Lemma 2.20. Suppose that 0 < x < y, i.e. x, y − x ∈ P, then x < y .

0 = ϕ (m) · ϕ (n)−1 Proof. By Lemma 2.16, we know x · x < x · y and x · y < y · y and therefore x2 < y2. which implies ϕ (m) = 0 which happens iff m = 0, i.e. m/n = 0. √ √ √ √ Corollary 2.21. If 0 ≤ x < y and x and y exists, then 0 ≤ x < y. √ √ √ √ 2 Proof. If x = y then x = ( x)2 = y = y which is impossible. • End of Lecture 3, 10/3/2012 √ √ Similarly if x > y then Notation 2.15 If ( ,P ) is an ordered field we write x > y iff x − y ∈ P. We F √ 2 √ 2 also write x ≥ y iff x > y or x = y. x = x > ( y) = y which is again false. Notice that if x, y ∈ F then either x − y = 0 (i.e. x = y), or x − y ∈ P (i.e. Alternatively: starting with y2 − x2 = (y − x)(y + x) and then replacing x > y), or x − y ∈ −P (i.e. y − x ∈ P and y > x). Also in this notation we have √ √ y and x by y and x respectively (assuming they exist) shows, P = {x ∈ F : x > 0} , √ √ √ √ √ √ √ √ −1 y − x = y − x y + x =⇒ y − x = (y − x) y + x −P = {x ∈ F : x < 0 (i.e. 0 > x)} . √ √ from which it follows that y − x ∈ P if (y − x) ∈ P. More importantly this Lemma 2.16. Suppose that x < y and y < z and a > 0. Then x < z and √ shows y depends “continuously” in on y. ax < ay. Definition 2.22. The absolute value, |x| , of x in ordered field F is defined by Proof. By assumption y − x ∈ P and z − y ∈ P, therefore z − x = (y − x) + x if x ≥ 0 (z − y) ∈ P, i.e. z > x. Moreover, a ∈ P and (y − x) ∈ P implies |x| = −x if x < 0 P 3 a (y − x) = ay − ax. Alternatively we may define √ That is ay > ax. |x| = x2.

Page: 14 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 2.2 Ordered Fields 15

Proposition 2.23. For all x, y ∈ F, then 2. S = N is not bounded from above while inf (S) = 1. 3. S = 1 − 1 : n ∈ is bounded from above and 1 = sup (S) while inf (S) = 1. |x| ≥ 0 n N 1 . 2. |xy| = |x| |y| 2 4. Let 3. |x + y| ≤ |x| + |y| . Proof. 1. holds by deﬁnition since −x > 0 if x < 0. S = {1, 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414213, 1.4142135,... } 2 2 2 2 2 2 √ 2. As |x| |y| ≥ 0 and (|x| |y|) = |x| |y| = x y = (xy) , we have where I am getting these digits from the decimal expansion of 2; q √ |x| |y| = (xy)2 = |xy| . 2 =∼ 1.41421356237309504880168872420969807856967187537694807317667973799...

3. It suffices to show |x + y|2 ≤ (|x| + |y|)2 . However, √In this case S is bounded above by 2, or 1.42, or 1.415, etc. Nevertheless 2 = sup (S) does not exists in Q. |x + y|2 = (x + y)2 = x2 + y2 + 2xy Example 2.26. Now let F = Q (t) be the field of rational functions described in ≤ x2 + y2 + 2 |xy| (xy ≤ |xy|) Example 2.11, then; S = N is bounded from above. For example t is an upper 2 2 1 = |x| + |y| + 2 |x| |y| bound but there is not least upper bound. For example m t is also an upper bound for S. = (|x| + |y|)2 . • End of Lecture 4, 10/5/2012 Definition 2.27 (Dedekind Cuts). A subset α ⊂ is called a cut (see [2, p. Definition 2.24. Let ( ,P ) be an ordered field and S be a subset of . Q F F 17]) if; 1. We say that S ⊂ is bounded from above (below) if there exists x ∈ F F 1. α is a proper subset of , i.e. α 6= ∅ and α 6= , such that x ≥ s (x ≤ s) for all s ∈ S. Any such x is called an upper (lower) Q Q 2. if p ∈ α and q ∈ and q < p, then q ∈ α, bound of S. Q 3. if p ∈ α, then there exists r ∈ α with r > p. 2. If S is bounded from above (below), we say that y ∈ F is a least upper bound (greatest lower bound) for S if y is an upper (lower) bound for Example 2.28. To each a ∈ Q, let αa := {q ∈ Q : q < a} . Then αa is a cut and S and y ≤ x (y ≥ x) for any other upper (lower) bound, x, of S. a is the least upper bound of αa in Q. Notice that least upper bounds and greatest lower bounds are unique if they ∞ Example 2.29. Let {Sn}n=0 ⊂ Q be any bounded sequence such that Sn ≤ Sn+1 exists. We will write and for all n. Then

y = l.u.b. (S) = sup (S) ∞ α := ∪n=0αSn = {q ∈ Q : q < Sn a.a. n} if y is the least upper bound for S and is a cut as the reader should verify. Let us further suppose that limn→∞ Sn does Pn 1 y = g.l.b. (S) = inf (S) not exist in Q. [For example from Example 1.17 we may take Sn := k=0 k! ∈ Q.] If m ∈ Q is an upper bound for α, then m ≥ Sn for all n since if m < Sn if y is the greatest lower bound for S. 1 for some n then q := 2 (m + Sn) ∈ α with q > m. Since limn→∞ Sn 6= m as m ∈ Q there must exists ε > 0 such that Example 2.25. Let F = Q, then; 1. max (a, b) and min (a, b) are least upper respectively greatest lower bounds m − Sn = |m − Sn| ≥ ε i.o. n. respectively for S = {a, b} . More generally, if S = {a1, . . . , an} , then As m − Sn is decreasing we may conclude that m − Sn ≥ ε for all n, i.e. S ≤ m − ε for all n. From this it now follows that m − ε is an upper bound sup (S) = a1 ∨ · · · ∨ an := max (a1, . . . , an) and n for α which is strictly smaller that m. So there can be no least upper bound. inf (S) = a1 ∧ · · · ∧ an := min (a1, . . . , an) .

Page: 15 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14

3 Real Numbers

As we saw in Section 1.2, Q is full of holes and calculus tends to produce 1. With these definitions, R, satisfies the axioms of a field. answers which live in these holes. So it is imperative that we fill the holes. Doing 2. Moreover, we can make this into an ordered field by setting P := so will lead to the real numbers provided we fill in the holes without adding too {α ∈ R : α > 0} where we say α > 0 iff there exists an N ∈ N such that 1 1 much extra filler along the way. One good answer to the question, What are an > N for a.a. n. the real numbers?, is contained in the statement of Theorem 3.3. 3. The ordered field (R,P ) is complete, i.e. has the least upper bound property. Definition 3.1. An order preserving field isomorphism between two or- The proof of Theorem 3.6 and Theorem 3.3 will be relegated to Section dered fields, ( ,P ) and ( ,P ) , is a bijection, f : → such that F1 1 F2 2 F1 F2 3.6 at the end of this chapter. For an alternative existence proof of R using f (0) = 0, f (1) = 1, f (P ) = P , and 1 2 Dedekind cuts as the elements of R is covered in Rudin [2, pages 17-21.]. One may also construct the Real numbers using decimal expansions, see T. Gower’s f (x + y) = f (x) + f (y) and f (xy) = f (x) f (y) for all x, y ∈ F1. notes on real numbers as decimals. We will prove the uniqueness assertion of Definition 3.2. An ordered field (F,P ) is has the least upper bound prop- Theorem 3.3 in Section at the end of this section. From now on we are going erty (or is complete) if every non-empty subset, S ⊂ F, which is bounded from to take Theorem 3.3 for granted and derive from this the “familiar”properties above possesses a least upper bound in F. [As we have seen in examples above, of the real numbers. Q does not have the least upper bound property.] Observe that Q, Q (t) , R (t) are not complete and hence are not the real Theorem 3.3 (The real numbers). Up to order preserving field isomorphism numbers, R. For example N ⊂ Q (t) (or N ⊂ R (t) ) is bounded by t say but (see Definition 3.1), there is exactly one complete ordered field. It is this field has no least upper bound. However, we do know that Q ⊂ R by Lemma 2.14. We will soon see that is “dense” in . We now pause to discuss some of the that we refer to as the real numbers and denote by R. Q R ∞ ∞ basic properties of R. Definition 3.4. We say two Cauchy sequences {an}n=1 and {bn}n=1 of ratio- ∞ ∞ nal numbers are equivalent and write {an}n=1 ∼ {bn}n=1 iff Theorem 3.7. Suppose that R is a complete ordered field which we assume we have already embedded Q into R as in Lemma 2.14. Then; lim |an − bn| = 0. n→∞ ∞ 1. For all x ≥ 0 there exists n ∈ N such that n ≥ x. We then define α := [{a } ] to be the equivalence class of the Cauchy sequence 1 n n=1 2. For all ε > 0 in R there exists n ∈ N such that 0 < ≤ ε. {a }∞ and refer to the collection of these equivalence classes as the real n n n=1 3. If ε ≥ 0 satisfies ε ≤ 1/n for all n ∈ N then ε = 0. numbers. The set of real numbers will be denoted by R. 1 4. If a, b ∈ R and a ≤ b + n for all n ∈ N or a ≤ b + ε for all ε > 0, then Notation 3.5 Let i : Q → R be defined by i (a) := [(a, a, a, . . . )] , i.e. i (a) is a ≤ b. the equivalence class of the constant sequence a. Proof. We take each item in turn. Notice that if i (a) = i (b) iff a = limn→∞ a = limn→∞ b = b. Thus the map, 1. If n < x for all n ∈ , then is bounded from above and so a := sup ( ) i : Q → R is injective and we will often simply identify a with i (a) and in this N N N exists in by the completeness axiom. As a is the least upper bound for way consider Q as a subset of R. R N ∞ there must be an n ∈ N such that n > a − 1. However this implies n + 1 > a Theorem 3.6. Let R be as in Theorem 3.3. For α := [{an}n=1] and β := ∞ which violates a be an upper bound for N. [{βn}n=1] in R we define 1 ∞ ∞ Roughly speaking here, you should think of α = limn→∞ an and so α > 0 should α + β = [{an + bn} ] and α · β = [{an · bn} ] . 1 1 n=1 n=1 happen iff α > N for some N ∈ N which then implies an ≥ N for a.a. n. 18 3 Real Numbers 1 1 2. If ε > 0 in R and n > ε for all n ∈ N, then n < ε for all n ∈ N which is impossible by item 1. 1 3. If there exists ε > 0 such that ε ≤ n for all n then n ≤ 1/ε for all n which is again impossible by item 1. 4. It suffices to prove the first assertion. We may assume a ≥ b for otherwise 1 1 we are done. If a ≤ b + n for all n, then 0 ≤ a − b ≤ n for all n ∈ N and hence a = b and in particular a ≤ b.

Proposition 3.8. If R is a complete ordered field, then every subset S ⊂ R which is bounded from below has a greatest lower bound, glb (S) = inf (S) . In fact, inf (S) = − sup (−S) . Theorem 3.13 (Basic Limit Results). Suppose that {an} and {bn} are sequences of real numbers such that A := limn→∞ an and B := limn→∞ bn exists Proof. We let m := − sup (−S) . Then we have −s ≤ −m for all s ∈ S, i.e. in R. Then; s ≥ m for all s ∈ S so that m is a lower bound for S. Moreover if ε > 0 is given there exists sε ∈ S such that −sε ≥ −m − ε, i.e. sε ≤ m + ε. This shows that 1. limn→∞ |an| = |A| . 1 1 2. If A 6= 0 then limn→∞ = . any lower bound, k of S must satisfy, k ≤ m + ε for all ε > 0, i.e. k ≤ m. This an A shows that m is the greatest lower bound for S. 3. limn→∞ (an + bn) = A + B. Let me sketch one way to construct R based on Cauchy sequences of rational 4. limn→∞ (anbn) = A · B. numbers. 5. If an ≤ bn for all n, then A ≤ B. 6. If {xn} ⊂ is another sequence such that an ≤ xn ≤ bn and A = B, then ∞ R Definition 3.9. A sequence {qn}n=1 ⊂ R converges to 0 ∈ R if for all ε > 0 limn→∞ xn = A = B. in R there exists N ∈ N such that |qn| ≤ ε for all n ≥ N. Alternatively put, for all M ∈ we have |q | ≤ 1 for a.a. n. Theorem 3.14. If S ⊂ R is a non-empty set which is bounded from above, then N n M ∞ there exists {xn} ⊂ S such that xn ↑ sup S as n → ∞, i.e. xn ≤ xn+1 for ∞ n=1 Definition 3.10. A sequence {qn}n=1 ⊂ R converges to q ∈ R if |q − qn| → 0 all n and limn→∞ xn = sup S. 1 ∞ as n → ∞, i.e. if for all N ∈ N, |q − qn| ≤ N for a.a. n. As usual if {qn}n=1 converges to q we will write qn → q as n → ∞ or q = limn→∞ qn. Proof. Let M := sup S. For each n ∈ N, there exists yn ∈ S such that 1 M ≥ yn ≥ M − n . We now let xn := max ({y1, . . . , yn}) in which case M ≥ Definition 3.11. A sequence {q }∞ ⊂ is Cauchy if |q − q | → 0 as 1 n n=1 R n m xn ≥ M − n and xn is increasing. By the Sandwich theorem it follows that m, n → ∞. More precisely we require for each ε > 0 in R that |qm − qn| ≤ ε for limn→∞ xn = M. a.a. pairs (m, n) , i.e. there should exists N ∈ N such that |qm − qn| ≤ ε for all m, n ≥ N. • End of Lecture 5, 10/8/2012

The next few results are analogous to what you have already shown in the ∞ Theorem 3.15. If {xn}n=1 ⊂ R is bounded from above and xn is non- case R is replaced by Q. As the proofs are essentially identical to those of ∞ Theorem 1.13 and Exercise 1.1 – 1.6. decreasing, then limn→∞ xn = supn xn. Similarly if {xn}n=1 ⊂ R is bounded from below and xn is non-increasing, then limn→∞ xn = infn xn. ∞ 2 Proposition 3.12. If {an}n=1 ⊂ R is a convergent sequence then it is Cauchy. ∞ Proof. Let M := supn xn, then for all ε > 0, there exists Nε ∈ such that If {an}n=1 is Cauchy sequence then it is bounded. N M ≥ xNε ≥ M − ε. As xn is non-decreasing it follows that M ≥ xn ≥ M − ε 2 We are going to show shortly that the converse is true as well! for all n ≥ Nε, i.e. |M − xn| ≤ ε for all n ≥ Nε.

Page: 18 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 3 Real Numbers 19

2. By item 1. and Theorem 3.14 we can choose q ∈ αb to be as close to b as we choose and in particular q can be chosen to be in αb with q > a. 3. You are asked to prove this in Exercise 3.1 below.

Exercise 3.1. Suppose that α ⊂ Q is a cut as in Deﬁnition 2.27. Show α is bounded from above. Then let m := sup α and show that α = αm, where αm

αm := {y ∈ Q : y < m} .

Also verify that αm is a cut for all m ∈ R. [In this way we see that we may As ε > 0 was arbitrary, we may conclude that limn→∞ xn = M. If xn is decreas- identify R with the cuts of Q. This should motivate Dedekind’s construction of ing instead, then −xn ↑ and we have − limn→∞ xn = supn (−xn) = − infn xn. the real numbers as described in Rudin.]

Proposition 3.17 (Rationals are dense in the reals). For all b ∈ R, there Theorem 3.16. Suppose that R is a complete ordered field which we assume exists qn ∈ Q such that qn ↑ b. Similarly there exists pn ∈ Q such that pn ↓ b. we have already embedded Q into R as in Lemma 2.14. Then; Proof. Given b ∈ Q we know that b = sup αb by Theorem 3.16. Then by 1. For all m ∈ R, if Theorem 3.14 there exists qn ∈ αb such that qn ↑ b as n → ∞. The second αm := {y ∈ Q : y < m} , assertion can be proved in much the same way as the first. Alternatively, let qn ∈ Q such that qn ↑ −b and set pn := −qn ∈ Q. Then pn ↓ b. then sup αm = m. 2. If a, b ∈ R with a < b, then there exists q ∈ Q such that a < q < b. 3. If α ⊂ Q is a cut and m := sup α, then α = αm. Definition 3.18. The real numbers which are not rational are called irrational,3 so the irrational numbers are \ . Proof. We take each item in turn. R Q Pn 1 Example 3.19 (Euler’s number). Let Sn := for all n ∈ 0. We define 1. Proof 1. Let αm := {y ∈ Q : y < m} and M := sup αm ∈ R. Then M ≤ m. k=0 k! N If M 6= m then M < m. To see this last case is not possible ε := m−M > 0 Euler’s number to be, 1 and choose n ∈ N such that 0 < n ≤ ε. Then choose y ∈ Q such that e := lim Sn = sup {Sn : n ∈ 0} ∈ . n→∞ N R 1 M − < y < M. 2n From Example 1.17, we have seen that e ∈ R \ Q. From this it follows that th Theorem 3.20( n – roots). Let n ∈ N and x > 0 in R, then there exists√ n 1/n n 1 1 a unique y ∈ R+ such that y = x. We of course denote y by x for x. M < y + < M + < m The function x → x1/n is increasing.[See Rudin for more properties of x1/n and 2n 2n xm/n where m ∈ Z and n ∈ N.] 1 which shows y + 2n ∈ αm is greater than M violating the assumption that n n M is an upper bound for α . Proof. Uniqueness. First of t > s ≥ 0 then t > s ≥ 0 as can be proved m by induction.4 Thus if x, y ≥ 0 and xn = yn then x = y for otherwise x > y or Proof 2. [Here is a slight rewriting of the above argument.] Choose yn ∈ αm 1 3 such that ym ↑ M as m → ∞. Choose n ∈ N so that m − M > n . Then For what it is worth, as dictionary definition of irrational is “not consistent with 1 1 1 ym + n ↑ M + n < m as m → ∞. So for large m, ym + n < m while or using reason.” Lets try to use irrational numbers in a rational way! 1 1 1 4 n n n+1 n ym + n > M, i.e. ym + n ∈ αm yet ym + n > M. This violates the assumption The statement holds for n = 1 by assumption and if t > s , then t > ts > n+1 n n that M is an upper bound for αm. s . For the last equality we used t > s implies ts > s · s .

Page: 19 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 20 3 Real Numbers y > x in which case xn > yn or yn > xn respectively. This shows that there is Theorem 3.22 (Sup Sup Theorem). Suppose that Λ is a subset of R such th 1/n 1/n at most one n – root if it exists. I also claim that x < y if x < y. If not that Λ = ∪α∈I Λα where Λα ⊂ Λ and I is some index set. Then n n then x1/n ≥ y1/n and this would then imply x = x1/n ≥ y1/n = y which contradicts x < y. sup Λ = sup sup Λα. n x n α∈I Existence. Let Λ := {t ∈ R+ : t ≤ x} . If t = 1+x ∈ (0, 1) , then t ≤ t ≤ x so that t ∈ Λ and Λ 6= ∅. If t = 1 + x, then tn = (1 + x)n ≥ 1 + nx > x and The convention here is that the supremum of a set which is not bounded from therefore Λ is bounded from above. Hence we may define y := sup Λ. We will above is ∞ and the sup ∅ = −∞. now show that yn = x. Proof. Let M := sup Λ and Mα := sup Λα for all α ∈ I. As Λα ⊂ Λ we By Theorem 3.14, there exists tk ∈ Λ such that tk ↑ y as k → ∞. By n have Mα ≤ M for all α ∈ I and therefore supα∈I Mα ≤ M. Conversely, if definition of Λ, tk ≤ x for all k. Passing to the limit as k → ∞ in this inequality n n λ ∈ Λ, then λ ∈ Mα for some α ∈ I and therefore λ ≤ Mα. From this it implies y = limk→∞ tk ≤ x. If yn < x then (using the Binomial theorem) and properties of limits, follows that λ ≤ supα∈I Mα and as λ ∈ Λ is arbitrary we may conclude that M = sup Λ ≤ supα∈I Mα. n n k The next corollary records a typical way the Sup Sup theorem is used. 1 X n 1 y + = yn + yn−k → yn < x as m → ∞. m k m k=1 Corollary 3.23. Suppose that X and Y are sets and S : X × Y → R is a function. Then 1 n Hence for sufficiently large m we will have y + m < x. But this shows that y + 1 ∈ Λ which violates y being an upper bound for Λ. Therefore we conclude sup sup S (x, y) = sup S (x, y) = sup sup S (x, y) . m x∈X y∈Y y∈Y x∈X that yn = x. (x,y)∈X×Y In particular, if S ∈ for all m, n ∈ , then • End of Lecture 6, 10/10/2012. m,n R N

sup sup Sm,n = sup Sm,n = sup sup Sm,n. m n (m,n) n m 3.1 Extended real numbers Proof. Let Λ := {S (x, y):(x, y) ∈ X × Y } , and for x ∈ X let Λx := {S (x, y): y ∈ Y } . Then Λ = ∪ Λ and therefore, Notation 3.21 The extended real numbers is the set R¯ := R∪ {±∞} , i.e. it x∈X x is R with two new points called ∞ and −∞. We use the following conventions, ¯ ¯ sup S (x, y) = sup Λ = sup sup Λx = sup sup S (x, y) . ±∞ · a = ±∞ if a ∈ R with a > 0, ±∞ · a = ∓∞ if a ∈ R with a < 0, (x,y)∈X×Y x∈X x∈X y∈Y ±∞ + a = ±∞ for any a ∈ R, ∞ + ∞ = ∞ and −∞ − ∞ = −∞ while the following expressions are not deﬁned; The same reasoning also shows,

∞ − ∞, − ∞ + ∞, ∞/∞, 0 · ∞, and ∞ · 0. sup S (x, y) = sup sup S (x, y) . (x,y)∈X×Y y∈Y x∈X ¯ A sequence an ∈ R is said to converge to ∞ (−∞) if for all M ∈ R there exists m ∈ N such that an ≥ M (an ≤ M) for all n ≥ m. In these case we write The next Lemma records some basic limit theorems involving the extended limn→∞ an = ±∞ or an → ±∞ as n → ∞. real numbers. For any subset Λ ⊂ ¯, let sup Λ and inf Λ denote the least upper bound and ∞ ∞ 5 ¯ R Lemma 3.24. Suppose {an}n=1 and {bn}n=1 are convergent sequences in R, greatest lower bound of Λ respectively. The convention being that sup Λ = ∞ if then: ∞ ∈ Λ or Λ is not bounded from above and inf Λ = −∞ if −∞ ∈ Λ or Λ is not bounded from below. We will also use the conventions that sup ∅ = −∞ and 1. If an ≤ bn for a.a. n then limn→∞ an ≤ limn→∞ bn. inf ∅ = +∞. The next theorem is a fairly simple but often useful result about 2. If c ∈ R, limn→∞ (can) = c limn→∞ an. computing least upper bounds. 5 The only sequences that do not converge in R¯ are those which oscillate too much.

Page: 20 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 3.2 Limsups and Liminfs 21 ∞ 3. If {an + bn}n=1 is convergent and for all n ≥ N. Since n is arbitrary, it follows that limn→∞ (an + bn) = a + b. Proof of Eq. (3.2). It will be left to the reader to prove the case where lim an lim (an + bn) = lim an + lim bn (3.1) n→∞ n→∞ n→∞ and lim bn exist in R. I will only consider the case where a = limn→∞ an 6= 0 and limn→∞ bn = ∞ here. Let us also suppose that a > 0 (the case a < 0 is provided the right side is not of the form ∞ − ∞. handled similarly) and let α := min a , 1 . Given any M < ∞, there exists ∞ 2 4. {anbn}n=1 is convergent and N ∈ N such that an ≥ α and bn ≥ M for all n ≥ N and for this choice of N, anbn ≥ Mα for all n ≥ N. Since α > 0 is ﬁxed and M is arbitrary it follows lim (anbn) = lim an · lim bn (3.2) n→∞ n→∞ n→∞ that limn→∞ (anbn) = ∞ as desired.

n 1 provided the right hand side is not of the for ±∞ · 0 of 0 · (±∞) . Exercise 3.2. Show limn→∞ α = ∞ and limn→∞ αn = 0 whenever α > 1.

Before going to the proof consider the simple example where an = n and Exercise 3.3. Suppose α > 1 and k ∈ N, show there is a constant c = c (α, k) > 6 n k n bn = −αn + c with α > 0 and c ∈ R. Then 0 such that α ≥ cn for all n ∈ N. [In words, α grows in n faster than any polynomial in n.]  ∞ if α < 1  ∞ ¯ lim (an + bn) = c if α = 1 Lemma 3.25. Suppose that {an}n=1 ⊂ R and limn→∞ an = A ∈ R. Then ∞  −∞ if α > 1 every subsequence, {bk := ank }k=1 , also converges to A. while Exercise 3.4. Prove Lemma 3.25. lim an + lim bn“ = ”∞ − ∞. n→∞ n→∞ This shows that the requirement that the right side of Eq. (3.1) is not of form 3.2 Limsups and Liminfs ∞ − ∞ is necessary in Lemma 3.24. Similarly by considering the examples −α ∞ an = n and bn = n with α > 0 shows the necessity for assuming right hand ¯ Notation 3.26 Suppose that {xn}n=1 ⊂ R is a sequence of numbers. Then side of Eq. (3.2) is not of the form ∞ · 0. Proof. The proofs of items 1. and 2. are left to the reader. lim inf xn = lim inf{xk : k ≥ n} and (3.3) n→∞ n→∞ Proof of Eq. (3.1). Let a := limn→∞ an and b = limn→∞ bn. lim sup xn = lim sup{xk : k ≥ n}. (3.4) Case 1. Suppose b = ∞ in which case we must assume a > −∞. In this n→∞ n→∞ case, for every M > 0, there exists N such that bn ≥ M and an ≥ a − 1 for all n ≥ N and this implies We will also write lim for lim inf and lim for lim sup .

an + bn ≥ M + a − 1 for all n ≥ N. Remark 3.27. Notice that if ak := inf{xk : k ≥ n} and bk := sup{xk : k ≥ n}, then {ak} is an increasing sequence while {bk} is a decreasing sequence. Since M is arbitrary it follows that an + bn → ∞ as n → ∞. The cases where Therefore the limits in Eq. (3.3) and Eq. (3.4) always exist in R¯ (see Theorem b = −∞ or a = ±∞ are handled similarly. 3.15) and Case 2. If a, b ∈ R, then for every ε > 0 there exists N ∈ N such that lim inf xn = sup inf{xk : k ≥ n} and n→∞ n |a − an| ≤ ε and |b − bn| ≤ ε for all n ≥ N. lim sup xn = inf sup{xk : k ≥ n}. Therefore, n→∞ n Owing to the following exercise, one may reduce properties of the lim inf to |a + b − (a + b )| = |a − a + b − b | ≤ |a − a | + |b − b | ≤ 2ε n n n n n n those of the lim sup . 6 This example shows that if you formally arrive at an expression like ∞ − ∞, then you should work harder to decide what it really means! Exercise 3.5. Show lim infn→∞(−an) = − lim supn→∞ an.

Page: 21 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 22 3 Real Numbers ∞ Exercise 3.6. Let {an}n=1 be the sequence given by, M ∈ N we have supk≥n ak ≤ −M for a.a. n. Therefore ak ≤ −M for a.a. k and we have shown limk→∞ ak = −∞. (−1, 2, 3, −1, 2, 3, −1, 2, 3,... ) . ¯ ( =⇒ ) Conversely, suppose that limn→∞ an = A ∈ R exists. If A ∈ R, then for every ε > 0 there exists N(ε) ∈ such that |A − a | ≤ ε for all n ≥ N(ε), Find lim sup a and lim inf a . N n n→∞ n n→∞ n i.e. ∞ ∞ A − ε ≤ an ≤ A + ε for all n ≥ N(ε). Exercise 3.7. If {an}n=1 and {bn}n=1 are two sequences such that an ≤ bn for a.a. n, then From this we learn that

lim sup an ≤ lim sup bn and lim inf an ≤ lim inf bn. (3.5) A − ε ≤ inf a ≤ sup a ≤ A + ε for a.a. n n→∞ n→∞ k k n→∞ n→∞ k≥n k≥n

The following proposition contains some basic properties of liminfs and lim- and so passing to the limit as n → ∞ implies sups. A − ε ≤ lim inf an ≤ lim sup an ≤ A + ε. ∞ ∞ n→∞ Proposition 3.28. Let {an}n=1 and {bn}n=1 be two sequences of real numbers. n→∞ Then Since ε > 0 is arbitrary, it follows that

1. lim infn→∞ an ≤ lim supn→∞ an. ¯ A ≤ lim inf an ≤ lim sup an ≤ A, 2. limn→∞ an exists in R iﬀ n→∞ n→∞ ¯ lim inf an = lim sup an ∈ R. i.e. that A = lim infn→∞ an = lim supn→∞ an. n→∞ n→∞ If A = ∞, then for all M > 0 there exists N = N(M) such that an ≥ M 3. for all n ≥ N. This show that lim infn→∞ an ≥ M and since M is arbitrary it lim sup (an + bn) ≤ lim sup an + lim sup bn (3.6) follows that n→∞ n→∞ n→∞ ∞ ≤ lim inf an ≤ lim sup an. whenever the right side of this equation is not of the form ∞ − ∞. n→∞ n→∞ 4. If an ≥ 0 and bn ≥ 0 for all n ∈ N, then The proof for the case A = −∞ is analogous to the A = ∞ case. Exercise 3.8. Show that lim sup (anbn) ≤ lim sup an · lim sup bn, (3.7) n→∞ n→∞ n→∞ lim sup(an + bn) ≤ lim sup an + lim sup bn (3.8) provided the right hand side of (3.7) is not of the form 0 · ∞ or ∞ · 0. n→∞ n→∞ n→∞

Proof. Items 1. and 2. will be proved here leaving the remaining items as provided that the right side of Eq. (3.8) is well deﬁned, i.e. no ∞−∞ or −∞+∞ an exercise to the reader. For item 1. we have type expressions. (It is OK to have ∞ + ∞ = ∞ or −∞ − ∞ = −∞, etc.) Exercise 3.9. Suppose that a ≥ 0 and b ≥ 0 for all n ∈ . Show inf{ak : k ≥ n} ≤ sup{ak : k ≥ n} ∀ n, n n N

lim sup(anbn) ≤ lim sup an · lim sup bn, (3.9) and therefore by the Sandwich theorem, lim infn→∞ an ≤ lim supn→∞ an. ¯ n→∞ n→∞ n→∞ 2. (⇐=) Let A := lim infn→∞ an = lim supn→∞ an ∈ R. Since provided the right hand side of (3.9) is not of the form 0 · ∞ or ∞ · 0. inf a ≤ a ≤ sup a , k n k ∞ ∞ k≥n k≥n Exercise 3.10. Suppose that {an}n=1 and {bn}n=1 are two non-negative sequences and assume A = limn→∞ an exists in (0, ∞) . Show if A ∈ R then it follows by the sandwich theorem that limn→∞ an = A. If A = ∞, then for all M ∈ we have M ≤ inf a for a.a. n. Therefore N k≥n k lim sup anbn = A · lim sup bn. ak ≥ M for a.a. k and we have shown limk→∞ ak = ∞. If A = −∞ then for all n→∞ n→∞

Page: 22 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 3.2 Limsups and Liminfs 23 ∞ Definition 3.29. A sequence, {an}n=1 , of positive numbers is said to have • End of Lecture 7, 10/12/2012. sub-geometric growth iff for all α > 1 there exists c = c (α) < ∞ such that Exercise 3.12. If an ≥ 0, then limn→∞ an = 0 iff lim supn→∞ an = 0. n an ≤ cα for a.a. n ∈ . (3.10) ∞ N Proposition 3.32. Suppose that {an}n=1 is a sequence of real numbers and let ∞ Lemma 3.30. Suppose that {an}n=1 is a sequence of non-negative numbers. B := {y ∈ R : an ≥ y for i.o. n} . ∞ 1/n Then sup B = lim sup a with the convention that sup B = −∞ if B = ∅. 1. If {an}n=1 has sub-geometric growth, then lim supn→∞ an ≤ 1. n→∞ n ∞ 2. If a > 0 for a.a. n and a−1 has sub-geometric growth (i.e. for all ∞ n n n=1 Proof. If {an} is not bounded from above, then B is not bounded from α > 1 there exists ε = ε (α) < ∞ such that a ≥ εα−n for a.a. n), then n=1 n above and sup B = ∞ = lim sup an. If B = ∅ so that sup B = −∞, then for 1/n n→∞ lim infn→∞ an ≥ 1. all y ∈ we must have a < y for a.a. n. This then implies lim sup a ≤ y ∞ R n n→∞ n 3. In particular if a > 0 for a.a. n and both {a }∞ and a−1 have n n n=1 n n=1 for all y ∈ R from which we conclude that lim supn→∞ an = −∞. So let us now 1/n ∞ sub-geometric growth, then limn→∞ an = 1. assume that B 6= ∅ and {an}n=1 is bounded in which case B is bounded from ∗ above. Let us set β := sup B ∈ R and a := lim supn→∞ an. Proof. 1. For any β > 1 choose α ∈ (1, β) and let c = c (α) . Since If y > β, then a < y for a.a. n from which it follows that a∗ := n n n α n α n n ∗ 7 limn→∞ c = 0 < 1, it follows that an ≤ cα = c β ≤ β for lim supn→∞ an ≤ y. We may now let y ↓ β in order to see that a ≤ β. Now β β ∗ a.a. n and therefore suppose that y < β, then an ≥ y for a.a. n and hence a = lim supn→∞ an ≥ y. Letting y ↑ β then shows a∗ ≥ β. Thus we have shown a∗ = β. 1/n 1/n an ≤ β for a.a. n =⇒ lim sup an ≤ β. ∞ ∞ n→∞ Theorem 3.33. There is a subsequence {ank }k=1 of {an}n=1 such that ∞ limk→∞ ank = lim supn→∞ an. Similarly, there is a subsequence {ank }k=1 of 1/n ∞ As β > 1 is arbitrary we may conclude that lim supn→∞ an ≤ 1. {an}n=1 such that limk→∞ ank = lim infn→∞ an. Moreover, every convergent ∞ ∞ 2. This may be proved similarly to item 1. or it can be reduced to item 1 as subsequence, {bk := ank }k=1 of {an}n=1 satisfies, −1 we show here. By item 1. applies to an n , lim inf an ≤ lim bk ≤ lim sup an. n→∞ k→∞ n→∞ 1 −1/n lim sup = lim sup an ≤ 1. 1/n Proof. Let me prove the last assertion first. Suppose that bk := ank is some n→∞ an n→∞ ∞ convergent subsequence of {an}n=1 . We then have, 1 1 The proof is completed because lim supn→∞ 1/n = 1/n as the reader an lim infn→∞ an inf an ≤ bk ≤ sup an for all k ∈ N. n≥n should prove. k n≥nk 3. From items 1. and 2. we learn that Passing to the limit in this equation then implies, 1/n 1/n 1 ≤ lim inf an ≤ lim sup an ≤ 1 lim inf a = lim inf an ≤ lim bk ≤ lim sup an = lim sup an. n→∞ n→∞ n→∞ k→∞ n≥n k→∞ k→∞ k n≥nk n→∞ ∞ from which the result follows. We have used, {inf a }∞ and sup a are subsequence of the n≥nk n k=1 n≥nk n k=1 ∞ −1 −p p convergent sequences of {inf a }∞ and sup a respectively and Example 3.31. If c n ≤ an ≤ cn for some p ∈ N and c < ∞, then n≥k n k=1 n≥k n k=1 1/n therefore converge to the same limits respectively, see Lemma 3.25. limn→∞ an = 1. Indeed using Exercise 3.3 we may easily verify the hypothesis of Lemma 3.30. Now let us prove the first assertions. I will cover the lim sup case here as the lim inf case is similar or can be deduced from the lim sup case with the aid Exercise 3.11. Suppose that p (t) is a non-zero polynomial of t ∈ R with (pos- of Exercise 3.5. Let A := lim supn→∞ an. We will need to consider three case, sibly) complex coefficients. Show A ∈ R,A = ∞, and A = −∞. 1/n 7 ∞ This can be done more formally by choosing a sequence {yk}k=1 such that yk ↓ α lim |p (n)| = 1. ∗ ∗ n→∞ so that a ≤ yk. Now let k → ∞ to conclude a ≤ limk→∞ yk = α.

Page: 23 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 24 3 Real Numbers 1 ∞ i) A ∈ R, then by Proposition 3.32, for all k ∈ N we have A − k ≤ an First proof. By Corollary 3.34, there is a subsequence, {ank }k=1, such that

for infinitely many n. In particular we can choose n1 < n2 < n3 < . . . limk→∞ ank = L ∈ R. As in the proof of Exercise 1.7, it follows that limn→∞ an 1 inductively so that A − k ≤ ank for all k. Since exists and is equal to L. Second proof. Let a := lim infn→∞ an and b := lim sup an. It suffices 1 n→∞ A − ≤ a ≤ sup a to show a = b. As we always know that a ≤ b it will suffice to show b ≤ a. k nk m m≥nk Given ε > 0, there exists N ∈ N such that and the limit as k → ∞ of both extremes of this inequality are A, it follow |am − an| ≤ ε for all m, n ≥ N. from the sandwich inequality that limk→∞ ank = A. ii) If A = lim supn→∞ an = ∞, then supk≥n ak = ∞ for all n ∈ N which In particular, for m, n ≥ k ≥ N we have am ≤ an + ε and hence implies for all M < ∞ that ak ≥ M i.o. k. Working similarly to case i) we can choose n1 < n2 < n3 < . . . so that an ≥ k for all k and therefore b ≤ sup am ≤ an + ε for all n ≥ k. k m≥k limk→∞ ank = ∞. iii) Finally suppose that A = lim sup an = −∞ so that for all M ∈ N there From this inequality we may further conclude, exists N ∈ N such that supk≥n ak ≤ −M for all n ≥ N, i.e. an ≤ −M for all n ≥ N. In this case it follows that in fact limn→∞ an = −∞ and we do b ≤ inf an + ε ≤ a + ε. not have to even choose as subsequence. n≥k As ε > 0 is arbitrary, we have indeed shown b ≤ a.

Corollary 3.34 (Bolzano–Weierstrass Property / Compactness). Every ∞ • End of Lecture 8, 10/15/2012. bounded sequence of real numbers, {an}n=1 , has a convergent in R subsequence, ∞ ∞ {an } . If we drop the bounded assumption then we may only assert that there k k=1 Exercise 3.13. Suppose that {an}n=1 is a sequence of real numbers and let is a subsequence which is convergent in R¯. A := {y ∈ R : an ≥ y for a.a. n} . Proof. Let M < ∞ such that |an| ≤ M for all n ∈ N, i.e. −M ≤ an ≤ M for all n. We may then conclude from Exercise 3.7 that, Then sup A = lim infn→∞ an with the convention that sup A = −∞ if A = ∅.

∞ −M ≤ lim sup an ≤ M. Exercise 3.14. Suppose that {an}n=1 is a sequence of real numbers. Show n→∞ ∗ lim supn→∞ an = a ∈ R iﬀ for all ε > 0, ∞ It now follows from Theorem 3.33 that there exists a subsequence, {ank }k=1 , ∗ ∞ an ≤ a + ε for a.a. n. and of {an} such that n=1 ∗ a − ε ≤ an i.o. n.

lim ank = lim sup an ∈ [−M,M] ⊂ R. k→∞ n→∞ Similarly, show lim infn→∞ an = a∗ ∈ R iﬀ for all ε > 0,

an ≤ a∗ + ε i.o.. n. and Theorem 3.35( is Cauchy complete). If {a }∞ ⊂ is a Cauchy se- R n n=1 R a∗ − ε ≤ an for a.a. n. quence, then limn→∞ an exists in R and in fact, Notice that this exercise gives another proof of item 2. of Proposition 3.28 in lim an = lim sup an = lim inf an. the case all limits are real valued. n→∞ n→∞ n→∞ Proof. We will give two proofs of this important theorem. Each proof uses Exercise 3.15 (Cauchy Complete =⇒ L.U.B. Property). Suppose that ∞ ∞ denotes any ordered ﬁeld which is Cauchy complete. Show has the least the fact that {an}n=1 ⊂ R is Cauchy implies {an}n=1 is bounded. This is proved R R exactly in the same way as the solution to Exercise 1.2. upper bound property and therefore is the ﬁeld of real numbers.

Page: 24 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 3.4 The Decimal Representation of a Real Number 25 N 3.3 Partitioning the Real Numbers Proposition 3.40. Suppose that −∞ < a < b < ∞ and {Sn}n=0 ⊂ [a, b] such that a = S0 < S1 < ··· < SN−1 < SN = b, then Notation 3.36 (Intervals) For a, b ∈ R with a < b we deﬁne, N X (a, b) := {x ∈ R : a < x < b} , [a, b) = [Sn−1,Sn). [a, b] := {x ∈ R : a ≤ x ≤ b} , n=1

(a, b] := {x ∈ R : a < x ≤ b} , and This result also holds if N = ∞ provided we now assume Sn < Sn+1 for all n, [a, b) := {x ∈ R : a ≤ x < b} . a = S0, and Sn ↑ b as n → ∞. We also also a = −∞ in the intervals, (a, b) and (a, b] and allows b = +∞ in Proof. This proof is very similar to the proof of Proposition 3.39 and so the intervals (a, b) and [a, b). will be omitted.

Notation 3.37 (Pairwise disjoint unions) If X is a set and Aα ⊂ X for P α ∈ I, we write X = α∈I Aα to mean; X = ∪α∈I Aα and Aα ∩ Aβ for all α 6= β. 3.4 The Decimal Representation of a Real Number

Exercise 3.16. Suppose that a, b, c, d ∈ R such that a < b ≤ c < d. Show Lemma 3.41 (Geometric Series). Let α ∈ R or α ∈ Q, m, n ∈ Z and Pm k (a, b] ∩ (c, d] = ∅ and [a, b) ∩ [c, d) = ∅. S := k=n α . Then m − n + 1 if α = 1 Lemma 3.38 (Well Ordering II). Suppose that S is a non-empty subset of Z S = αm+1−αn . which is bounded from below, then inf (S) ∈ S, i.e. S has a (unique) minimizer. α−1 if α 6= 1

Proof. As S is bounded from below, there exists k ∈ Z such that k ≤ s Proof. When α = 1, ˜ for all s ∈ S. Therefore S := {s − k + 1 : s ∈ S} ⊂ N and hence by the Well m X ordering principle, min S˜ := m ∈ N exists. That is m ≤ s−k +1 for all s ∈ S S = 1k = m − n + 1. and there exists s0 ∈ S such that m = s0 − k + 1. These last statements are k=n equivalent to saying, If α 6= 1, then m+1 n s0 = m + k − 1 ≤ s for all s ∈ S, αS − S = α − α . Solving for S gives which is to say s0 = min (S) . ∞ m m+1 n Proposition 3.39. Suppose that {Sn} ⊂ such that Sn < Sn+1 for all X α − α n=−∞ R S = αk = if α 6= 1. (3.12) n ∈ Z, limn→∞ Sn = ∞ and limn→−∞ Sn = −∞. Then α − 1 k=n X X (Sn−1,Sn] = R = [Sn,Sn+1). (3.11) −1 n∈Z n∈Z Taking α = 10 in Eq. (3.13) implies

Proof. The fact that (Sn,Sn+1] ∩ (Sm,Sm+1] = ∅ follows from Exercise m X 10−(m+1) − 10−n 1 1 − 10−(m−n+1) 3.16. For x ∈ R, let 10−k = = 10−1 − 1 10n−1 9 n0 := min ({n ∈ Z : x ≤ Sn}) k=n which exists since {n ∈ Z : x ≤ Sn} is non-empty as Sn → ∞ as n → ∞ and is and in particular, for all M ≥ n, bounded from below since Sn → −∞ as n → −∞. It then follows that x ≤ Sn0 while x S , i.e. S < x ≤ S and we have shown x ∈ (S ,S ] m M n0−1 n0−1 n0 n0−1 n0 X 1 X which completes the proof of the ﬁrst equality in Eq. (3.12). The proof of the lim 10−k = ≥ 10−k. m→∞ 9 · 10n−1 second equality is similar and so will be omitted. k=n k=n

Page: 25 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 26 3 Real Numbers 9 Definition 3.42 (Decimal Numbers). Let D denote those sequences α ∈ α1 α1 + 1 X α1 l α1 l + 1 [m + , m + ) [m + + , m + + ) {0, 1, 2,..., 9}Z with the following properties: 10 10 10 100 10 100 l=0 1. there exists N ∈ such that α = 0 for all n ≥ N and N −n α1 l α1 l+1 and set α2 = l for x ∈ [m + 10 + 100 , m + 10 + 100 ). Continuing this way 2. αn 6= 0 for some n ∈ Z. ∞ inductively we construct {αk}k=1 such that A decimal number is then an expression of the form k k−1 X αj X αj αk + 1 α α . . . α .α α α .... x ∈ [m + , m + + ). −N −N+1 0 1 2 3 10j 10j 10k j=1 j=1 For example √ It is now easy to see that x = a (α) . 52 + 2 =∼ 53.41421356237309504880168872420969807856967187537694807 ... Remark 3.44. The representation of x ≥ 0 as a decimal number may not be To every decimal number α ∈ D is the sequence an = an (α) defined for n ∈ N unique. For example, by ∞ n X 9 9 1 X −k 0.999 = = · = 1.000. an := αk10 . (a finite sum). 10k 10 1 − 1 k=1 10 k=−∞ Since for m > n, [Or note that

m m n-times n–1 times X −k X −k 1 1 |a − a | = α 10 ≤ 9 10 ≤ 9 = , z }| { z }| { −n m n k 9 · 10n 10n 1 − 0. 9 ··· 9 = 0 . 0 ··· 0 1 = 10 → 0 as n → ∞.] k=n+1 k=n+1 On the other hand if we agree to not allow a tail of repeated 9’s as an element it follows that of D, then the representation would be unique. 1 |a − a | ≤ → 0 as m, n → ∞ m n 10min(m,n) ∞ 3.5 Summary of Key Facts about Real Numbers which shows {an}n=1 is a Cauchy sequence. Thus to every decimal number we may associate the real number 1. The real numbers, R, is the unique (up to order preserving field isomor- a (α) := lim an. phism) ordered field with the least upper bound property or equivalently n→∞ which is Cauchy complete. Theorem 3.43. If x ≥ 0 is a real number, there exists α ∈ D such that x = 2. Informally the real numbers are the rational numbers with the (irrational) a (α) , i.e. all real numbers can be represented in decimal form. hole filled in. 3. Monotone bounded sequence always converge in R. Proof. If x = 0, we can take αn = 0 for all n so that 0 = a (α) . So 4. A sequence converges in R iff it is Cauchy. suppose that x > 0 and let p := min ({n ∈ N : x < n}) . Set m = p − 1, then 5. Cauchy sequences are bounded. m ≤ x < m + 1. We then define αk for k ≤ 0 so that m = α−N . . . α0. We now 6. N is unbounded from above in R. construct α for k ≥ 1. For k = 1 we write 1 k 7. For all ε > 0 in R there exists n ∈ N such that n < ε. 9 8. Q and R \ Q is dense in R. In particular, between any two real numbers X l l + 1 [m, m + 1) = [m + , m + ) a < b, there are infinitely many rational and irrational numbers. 10 10 9. Decimal numbers map (almost 1-1) into the real numbers by taking the l=0 limit of the truncated decimal number. l l+1 and then choose α1 = l if x ∈ [m + 10 , m + 10 ). We then construct α2 using, 10. If a, b, ε ∈ R, then

Page: 26 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 3.6 (Optional) Proofs of Theorem 3.6 and Theorem 3.3 27 ∞ ∞ a) a ≤ b by showing that a ≤ b + ε for all ε > 0. Proposition 3.45. Suppose that α := [{an}n=1] and β := [{bn}n=1] are real b) a = b by proving a ≤ b and b ≤ a or numbers. Then precisely one of the following three cases can happen; c) a = b by showing |b − a| ≤ ε for all ε > 0. 1. lim (a − b ) = 0, i.e. α = β, 11. A number of standard limit theorems hold, see Theorem 3.13. n→∞ n n 2. there exists ε = 1 > 0 such that a ≥ b +ε for a.a. n in which case α > β, 12. Unlike limits, lim sup and lim inf always exist. Moreover we have; N n n or lim inf a ≤ lim sup a with equality iﬀ lim a exists in n→∞ n n→∞ n n→∞ n 3. there exists ε = 1 > 0 such that b ≥ a +ε for a.a. n in which case β > α. which case N n n lim inf an = lim an = lim sup an. Proof. If case 1. does not hold then there exists δ > 0 such that |an − bn| ≥ δ n→∞ n→∞ n→∞ for inﬁnitely many n. There are now two possibilities (which will turn out to We may allow the values of ±∞ in these statements. me mutually exclusive; ∞ 13. If bk := {ank }k=1 is a convergent subsequence of {an} , then i) an − bn ≥ δ i.o. n, ii) bn − an ≥ δ i.o. n. lim inf an ≤ lim bk ≤ lim sup an n→∞ k→∞ n→∞ Since {an} and {bn} are Cauchy sequences, there exists N ∈ N such that

and we may choose {bk} so that limk→∞ bk = lim supn→∞ an or |an − am| ≥ δ/3 and |bn − bm| ≥ δ/3 for all m, n ≥ N. limk→∞ bk = lim infn→∞ an. 14. Bounded sequences of real numbers always have convergence subsequences. If case i) holds, we may choose an m ≥ N such that am − bm ≥ δ and so for ∞ n ≥ N we ﬁnd, 15. If S ⊂ R and A := sup (S) , then there exists {an}n=1 ⊂ S such that an ≤ an+1 for all n and limn→∞ an = sup (S) . ∞ δ ≤ am − bm = am − an + an − bn + bn − bm 16. If S ⊂ R and A := inf (S) , then there exists {an}n=1 ⊂ S such that ≤ |am − an| + an − bn + |bn − bm| an+1 ≤ an for all n and limn→∞ an = inf (S) . = δ/3 + an − bn + δ/3

from which it follows that an − bn ≥ ε := δ/3 for all n ≥ N and we are in case 3.6 (Optional) Proofs of Theorem 3.6 and Theorem 3.3 2. Similarly if case ii) holds then we are in fact in case 3. of the proposition. ∞ ∞ In this section, we assume that R is as describe in Theorem 3.6. The next Corollary 3.46. Suppose that α := [{an}n=1] and β := [{bn}n=1] are real exercise is relatively straight forward. numbers, then α ≥ β iff for all N ∈ N, 1 Exercise 3.17. Prove the following properties of R. a − b ≥ − for a.a. n. (3.13) n n N 1. Show addition and multiplication in Theorem 3.6 are well defined. Alternatively put, α ≥ β iff for all N ∈ , 2. Show (R, +, ·) satisfies the axioms of a field. Hint: for constructing mul- N tiplicative inverses, make use of Proposition 3.45 below to conclude if 1 ∞ bn ≤ an + for a.a. n. α := [{an}n=1] ∈ R and a 6= 0 = i (0) , then there exists N ∈ N such N 1 that |an| ≥ N for a.a. n. By redefining the first few terms of an if necessary, 1 Proof. If α = β, then limn→∞ (an − bn) = 0 and therefore Eq. (3.14) holds. you may assume that |an| ≥ N for all n and then take If α > β, then in fact an − bn ≥ ε > 0 > −1/N for a.a. n. −1 h −1 ∞ i Conversely, if α < β, then there exists ε > 0 such that bn ≥ an + ε for a.a. α = an . n=1 n. Thus if Eq. (3.14) were to also hold we could conclude for each N ∈ N that 3. Show i : Q → R is injective homomorphism of fields. 1 1 a ≥ b − ≥ a + ε − for a.a. n. n n N n N To finish the proof of Theorem 3.6 we must show that P is an ordering on R with the least upper bound property. This will be carried out in the remainder This leads to a contradiction as soon as we choose N so large as to make of this section. 1/N < ε. Thus if Eq. (3.14) holds we must have α ≥ β.

Page: 27 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 28 3 Real Numbers ∞ Proposition 3.47. Suppose that λ ∈ R, {an}n=1 be a Cauchy sequence in Q, (Sketch). Suppose that ϕ : F → G is an order preserving homomorphism. ∞ and α := [{an}n=1] . If λ ≤ i (ak) for all k then λ ≤ α. Similarly if i (ak) ≤ λ The usual arguments show that any homomorphism, ϕ : F → G must satisfy for all k then α ≤ λ. ϕ (q1F) = q1G. We know that {q · 1F : q ∈ Q} and {q · 1G : q ∈ Q} are dense ∞ copies of Q inside of F and G respectively. Now for general a ∈ F choose Proof. Let λ = [{λn}n=1] and suppose that λ ≤ i (an) for all n. For sake of qn, pn ∈ Q that qn1F ↑ a and pn1F ↓ a. Since ϕ is order preserving we must have contradiction, suppose that λ > α, i.e. there exists an N ∈ N such that λn ≥ q 1 = ϕ (q 1 ) is increasing and p 1 = ϕ (p 1 ) is decreasing. Moreover, 1 1 n G n F n G n F an + N for a.a. n. The assumption that λ ≤ i (ak) implies that λn ≤ ak + 3N since pn − qn → 0 we must have limn→∞ ϕ (qn1F) = limn→∞ ϕ (pn1F) . Since for a.a. n. Because {ak} is Cauchy, we may conclude there exists M ∈ such N ϕ (qn1 ) ≤ ϕ (a) ≤ ϕ (pn1 ) for all n it then follows that ϕ (a) = limn→∞ qn1 = that F F G 1 limn→∞ pn1G and we have shown ϕ is uniquely determined. λn ≤ ak + for all n, k ≥ M. For the converse, if q ∈ we know that 2N n Q 1 By making M even larger if necessary, we may assume that λn ≥ an + for N |qn1F − qm1F| = |qn − qm| 1F and all n ≥ M as well. From these two inequalities with k = n ≥ M we learn |qn1G − qm1G| = |qn − qm| 1G. 1 1 1 1 ∞ ∞ an + ≤ λn ≤ an + =⇒ ≥ Thus if {q 1 } is convergent in iff {q 1 } is convergent in . Thus N 2N 2N N n F n=1 F n G n=1 G for any a ∈ F we choose qn ∈ Q such that qn1F → a and then define ϕ (a) := and we have reached the desired contradiction. The fact that i (ak) ≤ λ for all k limn→∞ qn1G. One now checks that this formula is well defined (independent implies α ≤ λ is proved similarly. Alternatively if i (ak) ≤ λ then −λ ≤ i (−ak) of the choice of {qn} ⊂ Q such that qn1F → a) and defines an order preserving which implies −λ ≤ −α, i.e. α ≤ λ. isomorphism. For example, if a ≤ b we may choose {qn} ⊂ Q and {pn} ⊂ Q With these results in hand, let us now show that as defined in Theorem R such that qn1F ↑ a and pn1F ↓ b. Then qn1G ≤ pn1G for all n and letting n → ∞ 3.6 has the least upper bound property. shows, Proof of the least upper bound property. So suppose that Λ ⊂ R is a ϕ (a) = lim qn1 ≤ lim pn1 = ϕ (b) . n→∞ G n→∞ G non empty set which is bounded from above. For each m ∈ N, let km ∈ Z be km The other properties of ϕ are proved similarly. the smallest integer such that i (am) := i 2m is an upper bound for Λ. Since, −m for all n ≥ m, am − 2 ≤ an ≤ am, we may conclude that

− min(n,m) |an − am| ≤ 2 → 0 as n, m → ∞. 3.7 Supremum and Infiniums of sets ∞ This shows {an}n=1 is Cauchy and hence we defined an element α := ∞ Definition 3.49. Given a set A ⊂ X, let [{an}n=1] ∈ R. We now will show α = sup Λ. If λ ∈ Λ, then λ ≤ i (an) for all n and so by Proposition 3.47 we conclude 1 if x ∈ A 1 (x) = that λ ≤ α, i.e. α is an upper bound for Λ. Now suppose that β is another A 0 if x∈ / A −n upper bound for Λ. As i (an − 2 ) is not an upper bound for Λ there exists λ ∈ Λ such that be the indicator function of A. −n i an − 2 < λ ≤ β. Lemma 3.50 (Properties of inf and sup). We have: So by another application of Proposition 3.47 we learn that c c ∞ −n ∞ 1.( ∪nAn) = ∩nAn, α = [{an}n=1] = an − 2 ≤ β. c c n=1 2. {An i.o.} = {An a.a.} , P∞ 3. lim sup An = {x ∈ X : n=1 1An (x) = ∞} , This shows that α is in fact the least upper bound for Λ. n→∞ P∞ 4. lim infn→∞ An = x ∈ X : 1Ac (x) < ∞ , Theorem 3.48 (Real numbers are unique). Suppose that and are two n=1 n F G 5. sup 1 (x) = 1 = 1 , complete ordered fields. Then there is a unique order preserving isomorphism, k≥n Ak ∪k≥nAk supk≥n Ak 6. infk≥n 1A (x) = 1∩ A = 1inf A , ϕ : F → G. k k≥n k k≥n k

Page: 28 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 7.1 lim sup An = lim sup 1An , and n→∞ n→∞

8.1 lim infn→∞ An = lim infn→∞ 1An . Proof. These results follow fairly directly from the deﬁnitions and so the proof is left to the reader. (The reader should deﬁnitely provide a proof for herself.)

4 Complex Numbers

Definition 4.1 (Complex Numbers). Let C = R2 equipped with multiplica- gives implies the result in Eq. (4.3). tion rule Probably the most painful thing to check directly is the associative law, (a, b)(c, d) ≡ (ac − bd, bc + ad) (4.1) namely that [z1z2] z3 = z1 [z2z3] for all z1, z2, z3 ∈ C. This is equivalent to showing for all a, b, u, v, x, y ∈ that and the usual rule for vector addition. As is standard we will write 0 = (0, 0) , R 1 = (1, 0) and i = (0, 1) so that every element z of C may be written as z = [(a + ib)(u + iv)] (x + iy) = (a + ib) [(u + iv)(x + iy)] . x1 + yi which in the future will be written simply as z = x + iy. If z = x + iy, let Re z = x and Im z = y. We do this by working out both sides as follows; Writing z = a + ib and w = c + id, the multiplication rule in Eq. (4.1) becomes LHS = [(au − bv) + i (av + bu)] (x + iy) (a + ib)(c + id) ≡ (ac − bd) + i(bc + ad) (4.2) = (au − bv) x − (av + bu) y + i [(av + bu) x + (au − bv) y]; 2 2 and in particular 1 = 1 and i = −1. RHS = (a + ib) [(ux − vy) + i (uy + vx)] Proposition 4.2. The complex numbers C with the above multiplication rule = a (ux − vy) − b (uy + vx) + i [b (ux − vy) + a (uy + vx)] . satisfies the usual definitions of a field – see Definition 2.1. For example z1z2 = The reader should now easily see that both of these expressions are in fact z2z1, and z (w1 + w2) = zw1 + zw2, etc. Moreover if z = a + ib 6= 0, then z has a multiplicative inverse given by equal. The remaining axioms of a field are checked similarly. a b • End of Lecture 9, 10/17/2012. z−1 = − i . (4.3) • Test 1 took place of lecture 10, 10/22/2012. a2 + b2 a2 + b2 Notation 4.3 We will write 1/z for z−1 and w/z to mean z−1 · w. Moreover C contains R as sub-field under the identification Notation 4.4 (Conjugation and Modulous) If z = a + ib with a, b ∈ let 3 a → a1 + 0i = (a, 0) ∈ . R R C z¯ = a − ib and 2 2 2 Proof. Suppose z = a + ib 6= 0, we wish to find w = c + id such that zw = 1 |z| ≡ zz¯ = a + b . and this happens by Eq. (4.2) iff Notice that ac − bd = 1 and (4.4) 1 1 Re z = (z +z ¯) and Im z = (z − z¯) . (4.6) bc + ad = 0. (4.5) 2 2i

Solving these equations as follows a Proposition 4.5. Complex conjugation and the modulus operators satisfy; a(4.4) + b (4.5) =⇒ a2 + b2 c = a =⇒ Re w = c = a2 + b2 1. z¯ = z, b −b(4.4) + a (4.5) =⇒ a2 + b2 d = −b =⇒ Im w = d = − , 2. zw =z ¯w¯ and z¯ +w ¯ = z + w. a2 + b2 3. |z¯| = |z| 32 4 Complex Numbers 2. Say z = a + ib and w = c + id, thenz ¯w¯ is the same as zw with b replaced by −b and d replaced by −d, and looking at Eq. (4.2) we see that

z¯w¯ = (ac − bd) − i(bc + ad) = zw.

4. |zw|2 = zwz¯w¯ = zzw¯ w¯ = |z|2 |w|2 as real numbers and hence |zw| = |z| |w| . 5. Geometrically obvious or also follows from q |z| = |Re z|2 + |Im z|2.

6. This is the triangle inequality which may be understood geometrically or by the computation

|z + w|2 = (z + w)(z + w) = |z|2 + |w|2 + wz¯ +wz ¯ = |z|2 + |w|2 + wz¯ + wz¯ = |z|2 + |w|2 + 2 Re (wz¯) ≤ |z|2 + |w|2 + 2 |z| |w| = (|z| + |w|)2 .

7. Obvious. 8. Follows from Eq. (4.3). Alternatively if ρ = ρ + i0 > 0 is a real number then ρ−1 = ρ−1 + i0 as is easily verified since R is a sub-field of C. Thus since zz¯ = |z|2 we find 1 1 1 Re z Im z zz¯ = |z|2 = 1 =⇒ z−1 = z¯ = − i . |z|2 |z|2 |z|2 |z|2 |z|2

9. z−1 = z¯ = 1 |z¯| = 1 . |z|2 |z|2 |z| Corollary 4.6. If w, z ∈ C, then

n n ||z| − |w|| ≤ |z − w| . 4. |zw| = |z| |w| and in particular |z | = |z| for all n ∈ N. 5. |Re z| ≤ |z| and |Im z| ≤ |z| Proof. Just copy the proof of Lemma 1.6. 6. |z + w| ≤ |z| + |w| . 7. z = 0 iff |z| = 0. Lemma 4.7. For complex numbers u, v, w, z ∈ C with v 6= 0 6= z, we have 8. If z 6= 0 then z¯ 1 1 1 −1 z−1 := = , i.e. u−1v−1 = (uv) |z|2 u v uv u w uw 1 = and (also written as z ) is the inverse of z. v z vz −1 −1 n n 9. z = |z| and more generally |z | = |z| for all n ∈ Z. u w uz + vw + = . v z vz Proof. 1. and 3. are geometrically obvious as well as easily verified. [These statements hold in any field.]

Page: 32 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 4.1 A Matrix Perspective (Optional) 33

Proof. For the first item, it suffices to check that Proof. By assumption there exists M < such that |zn| ≤ M for all n ∈ N. Writing zn = an + ibn with an, bn ∈ R we may conclude that |an| , |bn| ≤ M. (uv) u−1v−1 = u−1uvv−1 = 1 · 1 = 1. According to Corollary 3.34 there exists an increasing function N 3 k → nk ∈ N such that lim a =: A exists. Similarly, we can apply Corollary 3.34 again The rest follow using k→∞ nk to find an an increasing function N 3 l → kl ∈ N such that liml→∞ bnk =: B u w uw ∞ l −1 −1 −1 −1 −1 exists. We now let wl := zn for l ∈ . Then {wl} is a subsequence of = uv wz = uwv z = uw (vz) = . kl N l=1 v z vz ∞ {zn}n=1 which is convergent to A + iB ∈ C. Indeed, u w z u v w zu vw |wl − (A + iB)| = an − A + i bn − B + = + = + kl kl v z z v v z zv vz −1 uz + vw ≤ an − A + bn − B → 0 as l → ∞. = (vz) (zu + vw) = . kl kl vz

n ∞ Notation 4.12 (Euclidean Spaces) Let C := {a := (a1, . . . , an): ai ∈ C} Definition 4.8. A sequence {zn}n=1 ⊂ C is Cauchy if |zn − zm| → 0 as n n n and R := {a := (a1, . . . , an): ai ∈ R} ⊂ C . For a, b ∈ C we let a¯ = m, n → ∞ and is convergent to z ∈ C if |z − zn| → 0 as n → ∞. As usual if ∞ (¯a1,..., a¯n) , {zn}n=1 converges to z we will write zn → z as n → ∞ or z = limn→∞ zn. n Theorem 4.9. The complex numbers are complete, i.e. all Cauchy sequences X a · b := a1b1 + ··· + anbn = aibi, and are convergent. i=1 v Proof. This follows from the completeness of real numbers and the easily u n √ uX 2 kak = kak = a · a¯ = t |ai| . (4.7) proved observations that if zn = an + ibn ∈ C, then 2 i=1 ∞ ∞ ∞ 1. {zn}n=1 ⊂ C is Cauchy iff {an}n=1 ⊂ R and {bn}n=1 ⊂ R are Cauchy and Exercise 4.1. If a, b ∈ n and λ ∈ , then a · b = b · a, 2. zn → z = a + ib as n → ∞ iff an → a and bn → b as n → ∞. C C (λa) · b = a · (λb) = λ (a · b) and kλak = |λ| kak = |λ| ka¯k .

The complex numbers satisfy all the same limit theorems as the real num- n bers. If we further assume that c ∈ C , then (a + b) · c = a · c + b · c.

∞ ∞ Theorem 4.10. If {wn}n=1 and {zn}n=1 are convergent sequences of complex numbers, then 4.1 A Matrix Perspective (Optional) 1. {w }∞ and {z }∞ are Cauchy sequences. n n=1 n n=1 Here is a way to understand some of the basic properties of C using your 2. limn→∞ (wn + zn) = limn→∞ wn + limn→∞ zn. knowledge of linear algebra. Let Mz : C → C denote multiplication by z = a+ib. 3. lim (w · z ) = lim w · lim z . 2 n→∞ n n n→∞ n n→∞ n We now identify C with R by 4. if we further assume that limn→∞ zn 6= 0, then c 2 w lim w C 3 c + id =∼ ∈ R . lim n = n→∞ n . d n→∞ zn limn→∞ zn Using this identiﬁcation, the product formula • End of Lecture 11, 10/24/2012. zw = (ac − bd) + i (bc + ad) , Lemma 4.11 (Bolzano–Weierstrass property). Every bounded sequence, ∞ {zn}n=1 ⊂ C, has a convergent subsequence. becomes

Page: 33 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 ac − bd a −b c M w = = z bc + ad b a d so that a −b M = = aI + bJ z b a where 0 −1 J := . 1 0 We now have the following simple observations; 1. J 2 = −I and J tr = −J, 2. MzMw = MwMz because J and I commute, 3. we have

MzMw = (aI + bJ)(cI + dJ) = (ac − bd) I + (ad + bc) J = Mzw,

4. the associativity of complex multiplication follows from the associativity properties of matrix multiplication, tr 5. Mz = aI − bJ = Mz¯ and in particular tr tr tr 6. Mzw = (MzMw) = Mw Mz = Mw¯Mz¯ = Mw¯z¯, tr 7. Mz Mz = Mzz¯ = M|z|2 = det (Mz) , 8. |wz| = det (Mwz) = det (MwMz) = det (Mw) det (Mz) = |w| |z| , 2 9. Mz is invertible iﬀ det (Mz) 6= 0 which happens iﬀ |z| 6= 0 and in this case we know from basic linear algebra that −1 1 a b 1 tr 1 Mz = 2 2 = 2 Mz = M z¯, a + b −b a |z| |z|2

10. With this notation we have MzMw = Mzw and since I and J commute it follows that zw = wz. Moreover, since matrix multiplication is associative so is complex multiplication. Also notice that Mz is invertible iﬀ det Mz = a2 + b2 = |z|2 6= 0 in which case −1 1 a b Mz = = M 2 |z|2 −b a z/¯ |z|

as we have already seen above. 5 Set Operations, Functions, and Counting

c Let N denote the positive integers, N0 := N∪ {0} be the non-negative inte- B \ A := {x ∈ B : x∈ / A} = A ∩ B . gers and Z = N0 ∪ (−N) – the positive and negative integers including 0, Q the rational numbers, R the real numbers, and C the complex numbers. We will We also define the symmetric difference of A and B by also use F to stand for either of the fields R or C. A4B := (B \ A) ∪ (A \ B) .

As usual if {Aα}α∈I is an indexed collection of subsets of X we define the union 5.1 Set Operations and Functions and the intersection of this collection by Notation 5.1 Given two sets X and Y, let Y X denote the collection of all ∪α∈I Aα := {x ∈ X : ∃ α ∈ I 3 x ∈ Aα} and functions f : X → Y. If X = N, we will say that f ∈ Y N is a sequence ∞ ∩α∈I Aα := {x ∈ X : x ∈ Aα ∀ α ∈ I }. with values in Y and often write fn for f (n) and express f as {fn}n=1 . If N {1,2,...,N} N X = {1, 2,...,N}, we will write Y in place of Y and denote f ∈ Y Example 5.4. Let A, B, and C be subsets of X. Then by f = (f1, f2, . . . , fN ) where fn = f(n). A ∩ (B ∪ C) = [A ∩ B] ∪ [A ∩ C] . Notation 5.2 More generally if {Xα : α ∈ A} is a collection of non-empty sets, Q let XA = Xα and πα : XA → Xα be the canonical projection map defined Indeed, x ∈ A ∩ (B ∪ C) ⇐⇒ x ∈ A and x ∈ B ∪ C ⇐⇒ x ∈ A and x ∈ B or α∈A Q by πα (x) = xα. If If Xα = X for some fixed space X, then we will write Xα x ∈ A and x ∈ C ⇐⇒ x ∈ A ∩ B or x ∈ A ∩ C ⇐⇒ x ∈ [A ∩ B] ∪ [A ∩ C] . α∈A A ` as X rather than XA. Notation 5.5 We will also write α∈I Aα for ∪α∈I Aα in the case that {Aα}α∈I are pairwise disjoint, i.e. Aα ∩ Aβ = ∅ if α 6= β. Recall that an element x ∈ XA is a “choice function,” i.e. an assignment xα := x(α) ∈ Xα for each α ∈ A. The axiom of choice states that XA 6= ∅ Notice that ∪ is closely related to ∃ and ∩ is closely related to ∀. For example ∞ provided that Xα 6= ∅ for each α ∈ A. let {An}n=1 be a sequence of subsets from X and define X Notation 5.3 Given a set X, let 2 denote the power set of X – the collection {An i.o.} := {x ∈ X :# {n : x ∈ An} = ∞} and of all subsets of X including the empty set. {An a.a.} := {x ∈ X : x ∈ An for all n sufficiently large}. The reason for writing the power set of X as 2X is that if we think of 2 meaning {0, 1} , then an element of a ∈ 2X = {0, 1}X is completely determined (One should read {An i.o.} as An infinitely often and {An a.a.} as An almost by the set always.) Then x ∈ {An i.o.} iff A := {x ∈ X : a (x) = 1} ⊂ X. ∀N ∈ N ∃ n ≥ N 3 x ∈ An In this way elements in {0, 1}X are in one to one correspondence with subsets of X. and this may be expressed as For A ∈ 2X let {A i.o.} = ∩∞ ∪ A . Ac := X \ A = {x ∈ X : x∈ / A} n N=1 n≥N n and more generally if A, B ⊂ X let Similarly, x ∈ {An a.a.} iff 36 5 Set Operations, Functions, and Counting

∃ N ∈ N 3 ∀ n ≥ N, x ∈ An 5.1.1 Exercises which may be written as Let f : X → Y be a function and {Ai}i∈I be an indexed family of subsets of Y, verify the following assertions. {A a.a.} = ∪∞ ∩ A . n N=1 n≥N n c c Exercise 5.1. (∩i∈I Ai) = ∪i∈I Ai . We end this section with some notation which will be used frequently in the Exercise 5.2. Suppose that B ⊂ Y, show that B \ (∪ A ) = ∩ (B \ A ). sequel. i∈I i i∈I i −1 −1 Exercise 5.3. f (∪i∈I Ai) = ∪i∈I f (Ai). Deﬁnition 5.6. If f : X → Y is a function and B ⊂ Y, then −1 −1 Exercise 5.4. f (∩i∈I Ai) = ∩i∈I f (Ai). f −1 (B) := {x ∈ X : f (x) ∈ B} . Exercise 5.5. Find a counterexample which shows that f(C ∩ D) = f(C) ∩ If A ⊂ X we also write, f(D) need not hold.

f (A) := {f (x): x ∈ A} ⊂ Y.

c 5.2 Cardinality Example 5.7. If f : X → Y is a function and B ⊂ Y, then f −1 (Bc) = f −1 (B) or to be more precise, In this section, X and Y be sets.

f −1 (Y \ B) = X \ f −1 (B) . Deﬁnition 5.10 (Cardinality). We say card (X) ≤ card (Y ) if there exists an injective map, f : X → Y and card (Y ) ≥ card (X) if there exists a surjective To prove this notice that map g : Y → X. We say card (X) = card (Y ) (also denoted as X ∼ Y ) if there

c exists a bijections, f : X → Y. x ∈ f −1 (Bc) ⇐⇒ f (x) ∈ Bc ⇐⇒ f (x) ∈/ B ⇐⇒ x∈ / f −1 (B) ⇐⇒ x ∈ f −1 (B) .

On the other hand, if A ⊂ X it is not necessarily true that f (Ac) = [f (A)]c . For example, suppose that f : {1, 2} → {1, 2} is the deﬁned by f (1) = f (2) = 1 and A = {1} . Then f (A) = f (Ac) = {1} where [f (A)]c = {1}c = {2} .

Notation 5.8 If f : X → Y is a function and E ⊂ 2Y let

f −1E := f −1 (E) := {f −1(E)|E ∈ E}.

If G ⊂ 2X , let Y −1 f∗G := {A ∈ 2 |f (A) ∈ G}.

X Deﬁnition 5.9. Let E ⊂ 2 be a collection of sets, A ⊂ X, iA : A → X be the inclusion map (iA (x) = x for all x ∈ A) and

−1 EA = iA (E) = {A ∩ E : E ∈ E} .

Page: 36 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 5.3 Finite Sets 37 Proposition 5.11. If X and Y are sets, then card (X) ≤ card (Y ) iﬀ Then f is the desired bijection.2 card (Y ) ≥ card (X) . Alternatively. If n = 1, then J2 = {1, 2} and either J2 \{k} = J1 or J2 \{k} = {2} , either way card (J2 \{k}) = card (J1) . Now suppose that result Proof. If f : X → Y is an injective map, deﬁne g : Y → X by g| = f −1 f(X) holds for a given n ∈ N and k ∈ Jn+2. If k = {n + 2} we have Jn+2 \{k} = Jn+1 and g| = x ∈ X chosen arbitrarily. Then g : Y → X is surjective. Y \f(X) 0 so card (Jn+2 \{k}) = card (Jn+1) while if k ∈ Jn+1 ⊂ Jn+2, then Jn+2 \{k} = −1 If g : Y → X is a surjective map, then Yx := g ({x}) 6= ∅ for all x ∈ X (J \{k}) ∪ {n + 2} ∼ J ∪ {n + 2} ∼ J ∪ {n + 1} = J . Q n+1 n n n+1 and so by the axiom of choice there exists f ∈ x∈X Yx. Thus f : X → Y such that f (x) ∈ Yx for all x. As the {Yx}x∈X are pairwise disjoint, it follows that Lemma 5.16. If m, n ∈ N with n > m, then every map, f : Jn → Jm, is not f is injective. injective.

Theorem 5.12 (Schröder-BernsteinTheorem). If X and Y are sets then Proof. If f : Jn → Jm were injective, then f|Jm+1 : Jm+1 → Jm would either card (X) ≤ card (Y ) or card (Y ) ≤ card (X) . Moreover, if card (X) ≤ be injective as well. Therefore it suffices to show there is no injective map, card (Y ) and card (Y ) ≤ card (X) , then card (X) = card (Y ) . [Stated more f : Jm+1 → Jm for all m ∈ N. We prove this last assertion by induction on m. explicitly; if there exists injective maps f : X → Y and g : Y → X, then there The case m = 1 is trivial as J1 = {1} so the only function, f : J2 → J1 is the exists a bijective map, h : X → Y.] function, f (1) = 1 = f (2) which is not injective. Now suppose m ≥ 1 and there were an injective map, f : Jm+2 → Jm+1. Proof. These results are proved in the appendices. For the first assertion Letting k := f (m + 2) we would have, f|J : Jm+1 → Jm+1\{k} ∼ Jm, which see B.8 and for the second see Theorem B.11. m+1 would produce and injective map from Jm+1 to Jm. However this contradicts the induction hypothesis and thus completes the proof. Exercise 5.6. If X = X1 ∪X2 with X1 ∩X2 = ∅,Y = Y1 ∪Y2 with Y1 ∩Y2 = ∅, and X ∼ Y for i = 1, 2, then X ∼ Y. This exercise generalizes to an arbitrary i i Theorem 5.17. If m, n ∈ N, then card (Jm) ≤ card (Jn) iff m ≤ n. Moreover, number of factors. card (Jn) = card (Jm) iff m = n and hence card (Jm) < card (Jn) iff m < n.

Proof. As Jm ⊂ Jn if m ≤ n and Jm = Jn if m = n, it is only the 5.3 Finite Sets forward implications that have any real content. If card (Jm) ≤ card (Jn) , there exists an injective map, g : Jm → Jn. According to Lemma 5.16 this can only Notation 5.13 (Integer Intervals) For n ∈ N we let happen if m ≤ n. If card (Jn) = card (Jm) , then card (Jn) ≤ card (Jm) and card (Jm) ≤ card (Jn) and hence m ≤ n and n ≤ m, i.e. m = n. Jn := {1, 2, . . . , n} := {k ∈ N : k ≤ n} . Proposition 5.18. If X is a finite set with #(X) = n and S is a non-empty Definition 5.14. We say a non-empty set, X, is finite if card (X) = card (Jn) subset of X, then S is a finite set and #(S) = k ≤ n. Moreover if #(S) = n, for some n ∈ N. We will also write #(X) = n1 to indicate that card (X) = then S = X. card (Jn) . [It is shown in Theorem 5.17 below that #(X) is well defined, i.e. Proof. It suffices to assume that X = Jn and S ⊂ Jn. We now give two it is not possible for card (X) = card (Jn) and card (X) = card (Jm) unless m = n.] proofs of the result. Proof 1. Let S1 = S and f (1) := min S ≥ 1. If S2 := S1 \{f (1)} is not Lemma 5.15. Suppose n ∈ N and k ∈ Jn+1, then card (Jn+1 \{k}) = empty, let f (2) := min S2 ≥ 2. We then continue this construction inductively. card (Jn) . So if f (k) = min Sk ≥ k has been constructed, then we define Sk+1 := Sk \ {f (k)} . If Sk+1 6= ∅ we define f (k + 1) := min Sk+1 ≥ k + 1. As f (k) ≥ k Proof. Let f : Jn → Jn+1 \{k} be defined by 2 The inverse map, g : Jn+1 \{k} → Jn, to f is given by x if x < k f (x) = y if y < k x + 1 if x ≥ k g (y) = y − 1 if y > k. 1 You should read # (X) = n, as X is a set with n elements.

Page: 37 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 38 5 Set Operations, Functions, and Counting for all k that f is defined, this process has to stop after at most n steps. Say equality iff S \{k} = Jn. If sup (S \{k}) > n then sup (S) ≥ sup (S \{k}) ≥ it stops at k so that Sk+1 = ∅. Then f : Jk → S is a bijection and therefore S n + 1 as desired. If sup (S \{k}) = n then S \{k} = Jn therefore S 3 k > n. is finite and # (S) = k ≤ n. Moreover, the only way that k = n is if f (k) = k Hence it follows that sup (S) = k ≥ n + 1. at each step of the construction so that f : Jn → S is the identity map in this Corollary 5.21. Suppose S is a non-empty subset of . Then S is an un- case, i.e. S = Jn. N Proof 2. We prove this by induction on n. When n = 1 the only no-empty bounded subset of N iff card (Jn) ≤ card (S) for all n ∈ N. subset of S of J is J itself. Thus # (S) = 1 and S = J . Now suppose that the 1 1 1 Proof. If S is bounded we know card (S) = card (J ) for some k ∈ which result hold for some n ∈ and let S ⊂ J . If n + 1 ∈/ S, then S ⊂ J and by k N N n+1 n would violate the hypothesis that card (J ) ≤ card (S) for all n ∈ . Conversely the induction hypothesis we know # (S) = k ≤ n < n + 1. So now suppose that n N 0 if card (S) ≤ card (Jn) for some n ∈ N, then there exists and injective map, n + 1 ∈ S and let S := S \{n + 1} ⊂ Jn. Then by the induction hypothesis, 0 0 0 0 f : S → Jn. Therefore card (S) = card (f (S)) = card (Jk) for some k ≤ n. So S is a finite set and # (S ) = k ≤ n, i.e. there exists a bijection, f : Jk → S 0 0 S is finite and hence bounded in N by Theorem 5.20. and S = Jn is k = n. Therefore f : Jk+1 → S given by f = f on Jk and f (k + 1) = n+1 is a bijections from J to S. Therefore # (S) = k +1 ≤ n+1 k+1 Exercise 5.7. Suppose that m, n ∈ N, show Jm+n = Jm ∪ (m + Jn) and with equality iff S0 = J which happens iff S = J . n n+1 (m + Jn) ∩ Jm = ∅. Use this to conclude if X is a disjoint union of two non- empty finite sets, X1 and X2, then # (X) = # (X1) + # (X2) . Proposition 5.19. If f : Jn → Jn is a map, then the following are equivalent, 1. f is injective, Exercise 5.8. Suppose that m, n ∈ N, show Jm × Jn ∼ Jmn. Use this to conclude if X and Y are two non-empty sets, then # (X × Y ) = # (X) · #(Y ) . 2. f is surjective, 3. f is bijective.

Proof. If n = 1, the only map f : J1 → J1 is f (1) = 1. So in this case 5.4 Countable and Uncountable Sets there is nothing to prove. So now suppose the proposition holds for level n and f : Jn+1 → Jn+1 is a given map. Definition 5.22 (Countability). A set X is said to be countable if X = ∅ If f : Jn+1 → Jn+1 is an injective map and f (Jn+1) is a proper subset of or if there exists a surjective map, f : N → X. Otherwise X is said to be Jn+1, then card (Jn+1) < card (f (Jn+1)) = card (Jn+1) which is absurd. Thus uncountable. f is injective implies f is surjective. Remark 5.23. From Proposition 5.11, it follows that X is countable iff there Conversely suppose that f : Jn+1 → Jn+1 is surjective. Let g : Jn+1 → Jn+1 be a right inverse, i.e. f ◦ g = id, which is necessarily injective, see the proof exists an injective map, g : X → N. This may be succinctly stated as; X is of Proposition 5.11. By the pervious paragraph we know that g is necessarily countable iff card (X) ≤ card (N) . From a practical point of view as set X is surjective and therefore f = g−1 is a bijection. countable iff the elements of X may be arranged into a linear list,

Theorem 5.20. A subset S ⊂ N is finite iff S is bounded. Moreover if #(S) = X = {x1, x2, x3,... } . n ∈ N then the sup (S) ≥ n with equality iff S = Jn. Example 5.24. The integers, Z, are countable. In fact N ∼ Z, for example define Proof. If S is bounded then S ⊂ Jn for some n ∈ N and hence S is a finite f : N → Z by set by Proposition 5.18. Also observe that if # (S) = n = sup (S) , then S ⊂ Jn (f (1) , f (2) , f (3) , f (4) , f (5) , f (6) , f (7) ,... ) := (0, 1, −1, 2, −2, 3, −3,... ) . and # (S) = n = # (Jn) . Thus it follows from Proposition 5.18 that S = Jn. Conversely suppose that S ⊂ N is a finite set and let n = # (S) . We will Remark 5.25 (Countability in a nutshell). If f : N → X is surjective, then now complete the proof by induction. If n = 1 we have S ∼ J1 and therefore −1 S = {k} for some k ∈ . In particular sup S = k ≥ 1 with equality iff S = J . g (x) := min f ({x}) defines an injective map, g : X → N. If g : X → N is N 1 injective, then f (n) := g−1 (n) for n ∈ g (X) =: S and f (n) = x ∈ X for Suppose the truth of the statement for some n ∈ N and let S ⊂ N be a set 0 with # (S) = n + 1. If we choose a point, k ∈ S, we have by Lemma 5.15 that n∈ / S defines a surjective map, f : N → X. Moreover, if S is a subset of N we #(S \{k}) = n. Hence by the induction hypothesis, sup (S \{k}) ≥ n with may list its elements in increasing order so that either

Page: 38 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 5.4 Countable and Uncountable Sets 39

S = {n1 < n2 < ··· < nk} for some k ∈ N or Theorem 5.27. The following properties hold; S = {n < n < ··· < n < . . . } . 1 2 k 1. N × N is countable and in fact N × N ∼ N, i.e. there exists a bijective map, In the first case card (X) = card (J ) while in the second card (S) = card ( ) . h, from N to N × N. k N 2. If X and Y are countable, then X × Y is countable. [Define f (j) := nj to set up the bijections between Jk or N and S.] ∞ 3. If {Xn} are countable sets then X := ∪ Xn is a countable set. The above arguments demonstrate that the following statements are equiv- n∈N n=1 alent; 4. If X is countable, then either there exists n ∈ N such that X ∼ Jn or X ∼ N. 1. X is countable, i.e. there exists a surjective map f : N → X. 5. If S ⊂ N and S ∼ Jn for some n ∈ N then S is bounded. 2. card (X) ≤ card (N) , i.e. there exists an injective map, g : X → N. 6. If X is a set and card Jn ≤ card X for all n ∈ N then card N ≤ card X. 3. There exists S ⊂ N such that card (X) = card (S) . Furthermore card (S) = 7. If A ⊂ X is a subset of a countable set X then A is countable. card (Jk) for some k ∈ N iff S is bounded and card (S) = card (N) iff S is unbounded. Proof. We take each item in turn. 4. Either card (X) = card (N) for card (X) = card (Jk) for some k ∈ N. 1. Put the elements of N × N into an array of the form Formal proofs of these observations are given above and below.  (1, 1) (1, 2) (1, 3) ...   (2, 1) (2, 2) (2, 3) ...  Lemma 5.26. If S ⊂ N is an unbounded set, then card (S) = card (N) .    (3, 1) (3, 2) (3, 3) ...  Proof. The main idea is that any subset, S ⊂ , may be given as an finite  . . . .  N . . . .. or infinite list written in increasing order, i.e. and then “count” these elements by counting the sets {(i, j): i + j = k} S = {n1, n2, n3,... } with n1 < n2 < n3 < . . . . one at a time. For example let h (1) = (1, 1) , h(2) = (2, 1), h (3) = (1, 2), h(4) = (3, 1), h(5) = (2, 2), h(6) = (1, 3) and so on. In other words we put If the list is finite, say S = {n1, . . . , nk} , then nk is an upper bound for S. So S × into the following list form, will be unbounded iff only if the list is infinite in which case f : N → S defined N N by f (k) = n defines a bijection. k N × N = {(1, 1) , (2, 1) , (1, 2) , (3, 1) , (2, 2) , (1, 3) , (4, 1) ,..., (1, 4) ,... } . Formal proof. Define f : N → S via, let 2. If f : N → X and g : N → Y are surjective functions, then the function S := S and f (1) := min S , 1 1 (f × g) ◦ h : N → X × Y is surjective where (f × g)(m, n) := (f (m), g(n)) S2 := S1 \{f (1)} and f (2) := min S2, for all (m, n) ∈ N × N. S3 := S2 \{f (2)} and f (3) := min S3 3. By assumption there exists surjective maps, fn : N → Xn, for each n ∈ N. Let h (n) := (a (n) , b (n)) be the bijection constructed for item 1. Then . . f : N → X defined by f (n) := fa(n) (b (n)) is a surjective map. 4. To see this let f : N → X be a surjective map and let g (x) := min f −1 ({x}) In more detail, let T denote those n ∈ N such that there exists f : Jn → S for all x ∈ X. Then g : X → N is an injective map. Let S := g (X) , then and {S ⊂ S}n satisfying, S = S, f (k) = min S and S = S \{f (k)} k k=1 1 k k+1 k g : X → S ⊂ N is a bijection. So it remains to show S ∼ N or S ∼ Jn for 1 ≤ k < n. If n ∈ T, we may define Sn+1 := Sn \{f (n)} and f (n + 1) := for some n ∈ N. If S is unbounded, then S ∼ N as we have already seen. min Sn+1 in order to show n + 1 ∈ T. Thus T = N and we have constructed an So it suffices to consider the case where S is bounded. If S is bounded by injective map, f : N → S. Moreover ∩k∈NSk ⊂ N \ Jn for all n and therefore 1 then S = {1} = J1 and we are done. Now assume the result is true if ∩ S = ∅. Thus it follows that f is a bijection. k∈N k S is bounded by n ∈ N and now suppose that S is bounded by n + 1. If • End of Lecture 14, 10/31/2012. n + 1 ∈/ S, then S is bounded by n and so by induction, S ∼ Jk for some k ≤ n < n + 1. If n + 1 ∈ S, then from above, S \{n + 1} ∼ Jk for some The following theorem summarizes most of what we need to know about k ≤ n, i.e. there exists a bijection, f : Jk → S \{n + 1} . We then extend counting and countability. f to Jk+1 by setting f (k + 1) := n + 1 which shows Jk+1 ∼ S.

Page: 39 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 40 5 Set Operations, Functions, and Counting

N 5. We again prove this by induction on n. If n = 1, then S = {m} for some (f1 (n) , f2 (n) , f3 (n) ,... ) . Now define a ∈ {0, 1} by an := 1 − fn(n). By m ∈ N which is bounded. So suppose for some n ∈ N, every subset S ⊂ N construction fn (n) 6= an for all n and so a∈ / f (N) . This contradicts the with S ∼ Jn is bounded. Now suppose that S ⊂ N with S ∼ Jn+1. Then assumption that f is surjective and shows 2N is uncountable. For the general X X X f (Jn) ∼ Jn and hence f (Jn) is bounded in N. Then max f (Jn)∨{f (n + 1)} case, since Y0 ⊂ Y for any subset Y0 ⊂ Y, if Y0 is uncountable then so X is an upper bound for S. This completes the inductive argument. is Y . In this way we may assume Y0 is a two point set which may as well 6. For each n ∈ N there exists and injection, fn : Jn → X. By replacing X be Y0 = {0, 1} . Moreover, since X is an infinite set we may find an injective N X by X0 := ∪n∈Nfn (Jn) we may assume that X = ∪n∈Nfn (Jn) . Thus there map x : N → X and use this to set up an injection, i : 2 → 2 by setting X exists a surjective map, f : N → X by item 3. Let g : X → N be defined i (A) := {xn : n ∈ N} ⊂ X for all A ⊂ N. If 2 were countable we could by g (x) := min f −1 ({x}) for all x ∈ X and let S := g (X) . To finish the find a surjective map f : 2X → N in which case f ◦ i : 2N → N would be proof we need only show that S is unbounded. If S were bounded, then surjective as well. However this is impossible since we have already seen that we would find k ∈ N such that Jk ∼ S ∼ X. However this is impossible 2N is uncountable. since card Jn ≤ card X = card Jk would imply n ≤ k even though n can be Corollary 5.31. The set (0, 1) := {a ∈ : 0 < a < 1} is uncountable while chosen arbitrarily in N. R ∩ (0, 1) is countable. More generally, for any a card (N) . Proof. From Section 3.4, the set {0, 1, 2 ..., 8}N can be mapped injec- tively into (0, 1) and therefore it follows from Theorem 5.30 that (0, 1) is Lemma 5.28. If X is a countable set which contains Y ⊂ X with Y ∼ , then n N uncountable. For each m ∈ N, let Am := m : n ∈ N with n < m . Since X ∼ . ∞ N Q ∩ (0, 1) = ∪m=1Xm and # (Xm) < ∞ for all m, another application of The- orem 5.27 shows ∩ (0, 1) is countable. Proof. By assumption there is an injective map, g : X → and a bijective Q N The fact that these results hold for any other finite interval follows from the map, f : → Y. It then follows that g ◦ f : → is injective from which it N N N fact that f : (0, 1) → (a, b) defined by f (t) := a + t (b − a) is a bijection. follows that g (X) is unbounded. [Indeed, (g ◦ f)(Jn) ⊂ g (X) for all n implies card (Jn) ≤ card (g (X)) for all n which implies g (X) is unbounded by Corollary Definition 5.32. We say a non-empty set X is infinite if X is not a finite 5.21.] Therefore X ∼ g (X) ∼ N by Lemma 5.26. set.

Corollary 5.29. We have card (Q) = card (N) and in fact, for any a < b in R, Example 5.33. Any unbounded subset, S ⊂ N, is an inﬁnite set according to card (Q∩ (a, b)) = card (N) . Theorem 5.20.

Proof. First off Q is a countable since Q may be expressed as a countable Theorem 5.34. Let X be a non-empty set. The following are equivalent; union of countable sets; 1. X is an infinite set, [ n n o = : n ∈ . 2. card (Jn) ≤ card (X) for all n ∈ N, Q m∈N m Z 3. card (N) ≤ card (X) , 4. card (X \{x}) = card (X) for some (or all) x ∈ X. From this it follows that Q∩ (a, b) is countable for all a < b in R. As these sets are not finite, they must have the cardinality of . N Proof. 1. =⇒ 2. Suppose that X is an infinite set. We show by induction Theorem 5.30 (Uncountabilty results). If X is an infinite set and Y is that card (Jn) ≤ card (X) for all n ∈ N. Since X is not empty, there exists a set with at least two elements, then Y X is uncountable. In particular 2X is x ∈ X and we may define f : J1 → X by f (1) = x in order to learn card (J1) ≤ uncountable for any infinite set X. card (X) . Suppose we have shown card (Jn) ≤ card (X) for some n ∈ N, i.e. there exists and injective map, f : Jn → X. If f (Jn) = X it would follows N Proof. Let us begin by showing 2N = {0, 1} is uncountable. For sake that card (X) = card (Jn) and would violated the assumption that X is not a 0 of contradiction suppose f : N → {0, 1}N is a surjection and write f (n) as finite set. Thus there exists x ∈ X \ f (Jn) and we may define f : Jn+1 → X

Page: 40 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 5.4 Countable and Uncountable Sets 41

0 0 0 by f |Jn = f and f (n + 1) = x. Then f : Jn+1 → X is injective and hence Exercise 5.10. Let Q [t] be the set of polynomial functions, p, such that p has card (Jn+1) ≤ card (X) . rationale coefficients. That is p ∈ Q [t] iff there exists n ∈ N0 and ak ∈ Q for 2. ⇐⇒ 3. This is the content of Theorem B.19. 0 ≤ k ≤ n such that n 3. =⇒ 4. Let x1 ∈ X and f : N → X be an injective map such that X k p (t) = akt for all t ∈ R. (5.1) f (1) = x1. We now define a bijections, ψ : X → X \{x1} by k=0 x if x∈ / f ( ) Show [t] is a countable set. ψ (x) = N Q f (i + 1) if x = f (i) ∈ f (N) . Definition 5.37 (Algebraic Numbers). A real number, x ∈ R, is called al- 4. =⇒ 1. We will prove the contrapositive. If X is a finite and x ∈ X, we gebraic number, if there is a non-zero polynomial p ∈ Q [t] such that p (x) = 0. have seen that card (X \{x}) < card (X) , namely # (X \{x}) = # (X) − 1. [That is to say, x ∈ R is algebraic if it is the root of a non-zero polynomial with The next two theorems summarizes the properties of cardinalities that have coefficients from Q.] been proven above. Note that for all q ∈ Q, p (t) := t − q satisfies p (q) = 0. Hence all rational Theorem 5.35 (Cardinality/Counting Summary I). Given a non-empty numbers are algebraic. But there are many more algebraic numbers, for example set X, then one and only one of the following statements holds; y1/n is algebraic for all y ≥ 0 and n ∈ N.

1. There exists a unique n ∈ N such that card X = card Jn. Exercise 5.11. Show that the set of algebraic numbers is countable. [Hint: 2. card X = card N, any polynomial of degree n has at most n – real roots.] In particular, “most” 3. card X > card N. irrational numbers are not algebraic numbers, i.e. there is still an uncountable Cases 2. or 3. hold iff card Jn ≤ card X for all n ∈ N which happens iff number of non-algebraic numbers.. card N ≤ card X. If X satisfies case 1. we say X is a finite set. If X satisfies case 2 we say X is a countably infinite set and if X satisfies case 3. we say X is an uncountably infinite set.

Theorem 5.36 (Cardinality/Counting Summary II). Let X and Y be sets and S be a subset of N. 1. If S ⊂ N is an unbounded set, then card (S) = card (N) . 2. If S ⊂ N is a bounded set then card (S) = card (Jn) for some n ∈ N. ∞ 3. If {Xk}k=1 are subsets of X such that card (Xk) ≤ card (N) , then ∞ card (∪k=1Xk) ≤ card (N) . 4. If X and Y are sets such that card (X) ≤ card (N) and card (Y ) ≤ card (N) , then card (X × Y ) ≤ card (N) . 5. card (Q) = card (N) = card (Z) . 6. For any a card (N) .

5.4.1 Exercises

Exercise 5.9. Show that Qn is countable for all n ∈ N.

Page: 41 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14

Part II

Normed and Metric Spaces

6 Metric Spaces

Definition 6.1. A function d : X × X → [0, ∞) is called a metric if 4. Inverse elements of addition: −v + v = 0 for all v ∈ Z. (In fact −v = (−1) · v.) 1. (Symmetry) d(x, y) = d(y, x) for all x, y ∈ X, 5. Distributivity of scalar multiplication with vector addition: a· 2. (Non-degenerate) d(x, y) = 0 if and only if x = y ∈ X, and (v + w) = a · v + a · w. 3. (Triangle inequality) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X. 6. Distributivity of scalar multiplication with respect to field addi- Example 6.2. Here are a few immediate examples of metric spaces; tion (a + b) · v = a · v + b · v. 7. Compatibility of scalar multiplication with the multiplication on 1. Let X be any set and then define, R a · (b · v) = (ab) · v. 8. Identity element of scalar multiplication 1 · v = v for all v ∈ Z. 0 if x = y d (x, y) = . 1 if x 6= y Example 6.4. Here are two fundamental examples of vector spaces. 1. with usual vector addition and scalar multiplication is vector space over 2. Let X = R with d (x, y) := |y − x| . Notice that R R. d (x, z) = |x − y + y − z| 2. Cn with usual vector addition and scalar multiplication is vector space over ≤ |x − y| + |y − z| = d(x, y) + d(y, z). C. Notation 6.5 If T and X are sets, let XT denote the collection of functions, 3. Let X be any subset of and define d (w, z) := |z − w| . C f : T → X. In general our typical example of a metric space will often be a generalization Example 6.6 (The Main Umbrella Example). Let T be a non-empty set and let of the last example above, see Example 6.12. Z := RT . For f, g ∈ Z and λ ∈ R we define f + g and λ · f by

(f + g)(t) = f (t) + g (t) (addition in R) for all t ∈ T 6.1 Normed Spaces [Linear Algebra Meets Analysis] (λ · f)(t) = λf (t) (multiplication in R) for all t ∈ T. 6.1.1 Review of Vector Spaces and Subspaces It can now be checked that Z is a vector space so that functions have now become vectors! Essentially all other examples of vector spaces we give will be Deﬁnition 6.3 (Vector Space). A vector space is a non-empty set Z of related to an example of this form. The same observations show T is a complex objects, called vectors, equipped with an addition operation “+” and scalar (= C R vector space. or maybe C) multiplication “·” satisfying all of the properties above: i.e. For all u, v, w ∈ Z and a, b ∈ R; Example 6.7. For example; R3 = R{1,2,3} and more generally Rn = RJn where J := {1, 2, . . . , n} . In this setting we usually specify x ∈ Jn by listing its 1. Associativity of addition: u + (v + w) = (u + v) + w. n R values (x (1) , . . . , x (n)) . To abbreviate notation a bit more we will usually 2. Commutativity of addition: v + w = w + v. write x (i) as x so that (x (1) , . . . , x (n)) becomes (x , . . . , x ) . 3. Identity element of addition: There is a element 0 ∈ Z such that i 1 n 0 + v = v for all v. 46 6 Metric Spaces n Example 6.8. The vector space of 2 × 2 matrices; X kz + wk1 = |zi + wi| i=1 M2×2 = {A : A is a 2 × 2 – matrix } = {A : {(1, 1) , (1, 2) , (2, 1) , (2, 2)} → } . R n X This can be generalized. ≤ [|zi| + |wi|] i=1 n n Deﬁnition 6.9 (Subspace). Let Z be a vector space. A non-empty subset, X X H ⊂ Z, is a subspace of Z if H is closed under addition and scalar multipli- = |zi| + |wi| = kzk1 + kwk1 . cation. Note, if H is a subspace and v ∈ H, then 0 = 0 · v ∈ H. i=1 i=1

T T 4. Let X be a set and for any function f : X → C, let The vector space R and C are typically the “largest” vector spaces we will consider in this course. kfku := sup |f (x)| . x∈X Example 6.10. Here are three common subspaces of RR; Then Z := {f : X → C : kfku < ∞} is a vector space and k·ku is a norm 1. H = {f ∈ Z : f is continuous} . on Z. 2. H = {f ∈ Z : f is continuously differentiable.} 3. H = {f ∈ Z : f is differentiable at π} . Exercise 6.1. Verify the last item of Example 6.13. That is let X be a set and for any function f : X → C, let 6.1.2 Normed Spaces kfku := sup |f (x)| . x∈X Definition 6.11. A norm on a vector space Z is a function k·k : Z → [0, ∞) such that Show Z := {f : X → C : kfku < ∞} is a vector space and k·ku is a norm on Z.

n 1. (Homogeneity) kλfk = |λ| kfk for all λ ∈ F and f ∈ Z. Our next goal is to show that k·k2 defined in Eq. (4.7) defines a norm on R 2. (Triangle inequality) kf + gk ≤ kfk + kgk for all f, g ∈ Z. and Cn. We will begin by proving the important Cauchy-Schwarz inequality. 3. (Positive definite) kfk = 0 implies f = 0. Lemma 6.14. If x, y ≥ 0 and ρ > 0, then A pair (Z, k·k) where Z is a vector space and k·k is a norm on Z is called a 1 1 normed vector space or normed space for short. xy ≤ ρx2 + y2 (6.1) 2 ρ Example 6.12. If (Z, k·k) is a normed space, then d (x, y) := kx − yk is a metric on Z and restricts to a metric on any subset of Z. with equality when ρ = y/x in the case x > 0.

Example 6.13 (Normed Spaces). The following are normed spaces; Proof. Since √ y 2 1 1. Z = R with kxk = |x| . 0 ≤ ρx − √ = ρx2 + y2 − 2xy, 2. Z = C with kzk = |z| . ρ ρ n 3. Z = C with √ y with equality iff ρx = √ , i.e. iff ρ = y/x, we see that n ρ X kzk1 := |zi| for z = (z1, . . . , zn) ∈ Z. 1 2 1 2 i=1 xy ≤ ρx + y 2 ρ The triangle inequality is easily verified here since, with equality iff ρ = y/x.

Page: 46 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6.1 Normed Spaces [Linear Algebra Meets Analysis] 47 n n Definition 6.15. The two norm, k·k2 on C is the function defined by Fact 6.18 For 0 < p < ∞, let k·kp : C → [0, ∞) be defined by v u n v uX 2 n u n kzk2 = t |zi| for all z ∈ C . up X p n kzkp = t |zi| for all z ∈ C . i=1 i=1 Theorem 6.16 (Cauchy-Schwarz Inequality). For a, b ∈ n, |a · b| ≤ kak · C 2 n kbk . Then k·kp is a norm on C for all 1 ≤ p < ∞ but is not a norm (the triangle 2 inequality fails) for 0 0 that Definition 6.19 (Balls). Let (X, d) be a metric space. For x ∈ X and r ≥ 0 n n n let X X X |a · b| ≤ aibi ≤ |aibi| = |ai| · |bi| Bx (r) := {y ∈ X : d (x, y) < r} and i=1 i=1 i=1 n Cx (r) := {y ∈ X : d (x, y) ≤ r} . X 1 2 1 2 1 2 1 ≤ ρ |a | + |b | = ρ kak + kbk . 2 i ρ i 2 2 ρ 2 i=1 We refer to Bx (r) and Cx (r) as the open and closed ball respectively about x with radius r. Taking ρ = kbk2 / kak2 then completes the proof. n Example 6.20. Let us consider k·k to 2. Figure 6.1 shows the boundary of the Theorem 6.17 (Triangle Inequality). The function, k·k2 , on C is a norm p R and in particular for a, b ∈ Cn, balls of radius 1 centered at 0 for some of the different p – norms.

ka + bk2 ≤ kak2 + kbk2 . Proof. The main point is to prove the triangle inequality. The proof is as follows; 2 ¯ ka + bk2 = (a + b) · a + b = (a + b) · a¯ + b ¯ 2 ¯ 2 = (a + b) · a¯ + (a + b) · b = kak2 + b · a¯ + a · b + kbk2 2 2 ¯ 2 2 ¯ = kak2 + kbk2 + 2 Re a · b ≤ kak2 + kbk2 + 2 Re a · b 2 2 ¯ ≤ kak2 + kbk2 + 2 a · b 2 2 ¯ ≤ kak2 + kbk2 + 2 kak2 · b 2 (Theorem 6.16) 2 2 2 = kak2 + kbk2 + 2 kak2 · kbk2 = (kak2 + kbk2) . The remaining properties of a norm are easily checked. For example, if λ ∈ C and a ∈ Cn, then v v u n u n Fig. 6.1. Balls in 2 corresponding to p ∈ 1 , 1, 2, 5, 20 . The case p = 1 is not uX 2 uX 2 2 R 2 2 kλak2 = t |λai| = t |λ| |ai| convex like the other balls which is an indication that the triangle inequality fails. i=1 i=1 v v u n q u n u 2 X 2 2uX 2 = t|λ| |ai| = |λ| t |ai| = |λ| kak2 . i=1 i=1 Example 6.21. The next ﬁgures explain how to understand balls in C ([0, 1] , R) equipped with the uniform norm.

Page: 47 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 48 6 Metric Spaces

2 Fig. 6.3. Here are some unit balls in R relative to some weighted 2 – norms. Each 2 Fig. 6.2. The ball in C ([0, 1] , R) of radius 1/4 centered at f (x) = sin x are all of these “balls” is the interior of an ellipse. It would be possible to construct norms the continuous functions whose graphs lie between the green envelope. where these ellipses are rotated as well.

For the next two exercise you will be using some concepts from calculus which we will developed in detail next quarter. For now, I assume you know what the Riemann integral is for continuous functions on [0, 1] with values in R. Let Z denote the continuous functions1 on [0, 1] with values in R. The only properties that you need to know about the Riemann integral are;

1. The integral is linear, namely for all f, g ∈ Z and λ ∈ R, Z 1 Z 1 Z 1 (f(t) + λg (t))dt = f (t) dt + λ g (t) dt. 0 0 0 2. If f, g ∈ Z and f (t) ≤ g (t) for all t ∈ [0, 1] , then Z 1 Z 1 f (t) dt ≤ g (t) dt. 0 0 Exercise 6.2 (Weighted 2 – norms). Suppose that ρ ∈ (0, ∞) for 1 ≤ i ≤ n 3. For all f ∈ Z, i Z 1 Z 1 n and for a, b ∈ C let f (t) dt ≤ |f (t)| dt.

v 0 0 n u n In fact this item follows from items 1. and 2. Indeed, since ±f (t) ≤ |f (t)| X √ uX 2 a ∗ b := aibiρi and kak := a ∗ a¯ = t |ai| ρi. for all t, we ﬁnd i=1 i=1 Z 1 Z 1 Z 1 Z 1 Z 1

n n ± f (t) dt = ±f (t) dt ≤ |f (t)| dt ⇐⇒ f (t) dt ≤ |f (t)| dt. Show, for all a, b ∈ C that |a ∗ b| ≤ kak · kbk and that k·k is a norm on C . 0 0 0 0 0 [Hint: reduce to the case where ρi = 1 for all i.] 1 The notion of continuity will be formally developed shortly.

Page: 48 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6.2 Sequences in Metric Spaces 49

Exercise 6.3. Let Z denote the continuous function on [0, 1] with values in R collection of metric spaces and for x = (x1, x2, . . . , xn) and y = (y1, . . . , yn) in Qn and for f ∈ Z let X := i=1 Xi, let Z 1 kfk1 := |f (t)| dt. d(x, y) = k(d1 (x1, y1) , d2 (x2, y2) , . . . , dn (xn, yn))k . 0

Show k·k1 satisﬁes, Show (X, d) is a metric space. 1. (Homogeneity) kλfk = |λ| kfk for all λ ∈ R and f ∈ Z. • End of Lecture 13, 10/28/2012. [We started Section 5.1 above as well this 2. (Triangle inequality) kf + gk ≤ kfk + kgk for all f, g ∈ Z. day.]

Remark 6.22 (An interpretation of kfk1 .). If we interpret f (t) as the speed of a particle on the real line at time t, then kfk represents the total distance 1 6.2 Sequences in Metric Spaces (including retracing of its path) the particle travels over the time interval [0, 1] . ∞ Exercise 6.4. Let Z denote the continuous function on [0, 1] with values in R Definition 6.24. A sequence {xn}n=1 in a metric space (X, d) is said to be and for f ∈ Z let convergent if there exists a point x ∈ X such that limn→∞ d(x, xn) = 0. In s Z 1 this case we write limn→∞ xn = x or xn → x as n → ∞. 2 kfk2 := |f (t)| dt. 0 Exercise 6.6. Show that x in Definition 6.24 is necessarily unique. Show; ∞ Definition 6.25 (Cauchy sequences). A sequence {xn}n=1 in a metric space 1. for f, g ∈ Z that (X, d) is Cauchy provided that limm,n→∞ d(xn, xm) = 0, i.e. for all ε > 0 there Z 1 exists N (ε) ∈ such that N f (t) g (t) dt ≤ kfk2 · kgk2 , 0 d(xn, xm) ≤ ε when m, n ≥ N (ε) . 2. Homogeneity) kλfk = |λ| kfk for all λ ∈ and f ∈ Z, and 2 2 R • End of Lecture 15, 11/2/2012. 3. (Triangle inequality) kf + gk2 ≤ kfk2 + kgk2 for all f, g ∈ Z. Exercise 6.7. Show that convergent sequences in metric spaces are always Remark 6.23 (An interpretation of kfk2 .). Let us suppose that f (t) is the voltage across a 1 Ohm resistor. By Ohms law the current through this re- Cauchy sequences. The converse is not always true. For example, let X = Q be the set of rational numbers and d(x, y) = |x − y|. Choose a sequence sistor is f (t) /1 = f (t) and the power dissipated by the resistor at time t is ∞ √ ∞ 2 {x } ⊂ which converges to 2 ∈ , then {x } is ( , d) – Cauchy (Voltage·Current) is f (t) . The work done over the time interval, [0, 1] is then n n=1 Q R n n=1 Q but not (Q, d) – convergent. Of course the sequence is convergent in R. Z 1 Z 1 2 2 ∞ Power (t) dt = f (t) dt = kfk2 . Exercise 6.8. If {xn}n=1 is a Cauchy sequence in a metric space (X, d), 0 0 ∞ limn→∞ d (xn, y) exists in R for all y ∈ X. In particular, {d (xn, y)}n=1 is a bounded sequence in for all y ∈ X. On the other hand if we had a constant voltage of kfk2 across the resistor R over the time interval [0, 1] , the work done over this period would again be 2 Definition 6.26. A metric space (X, d) (or normed space (X, k·k)) is com- kfk2 . Thus kfk2 is often referred to as the RMS voltage (root mean squared plete if all Cauchy sequences are convergent sequences. A complete normed voltage) and represents the equivalent DC (Direct Current, i.e. constant) voltage space is called a Banach space. necessary to produce the same amount of work over the time interval [0, 1] . Lemma 6.27. Let X be a non-empty set and Exercise 6.5. Let k·k be a norm on Rn such that kak ≤ kbk whenever 0 ≤ 2 X ai ≤ bi for 1 ≤ i ≤ n. Further suppose that (Xi, di) for i = 1, . . . , n is a finite kfku := sup |f (x)| for all f ∈ C . x∈X 2 n For example we could take k·k to be k·ku , k·k1 , or k·k2 on R . Not all norms 2 X satisfy this assumption though. For example, take k(x, y)k = |x − y| + |y| on R , Then the subspace, Z := f ∈ f ∈ C : kfku < ∞ is a Banach space, i.e. then k(x, y)k is decreasing in x when 0 ≤ x < y. (Z, k·ku) is a complete normed space.

Page: 49 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 50 6 Metric Spaces

∞ n Proof. Let {fn}n=1 ⊂ Z be a Cauchy sequence. Since for any x ∈ X, we Exercise 6.10 (Equivalence of 3 norms on C ). Let k·k1 , k·ku , and k·k2 have be the three norms on Cn given above. Show for all z ∈ Cn that |fn (x) − fm (x)| ≤ kfn − fmku (6.2) kzku ≤ kzk1 ≤ n kzku , which shows that {f (x)}∞ ⊂ is a Cauchy sequence of complex numbers. √ n n=1 C kzk ≤ n kzk , (Hint: use Cauchy Schwarz.) Because is complete, f (x) := lim f (x) exists for all x ∈ X. Passing to 1 2 C n→∞ n q the limit n → ∞ in Eq. (6.3) implies kzk2 ≤ kzku · kzk1 ≤ kzk1 and √ |f (x) − fm (x)| ≤ lim inf kfn − fmk kzk2 ≤ n kzku . n→∞ u It follows from these inequalities that k·k , k·k , and k·k are equivalent norms and taking the supremum over x ∈ X of this inequality implies 1 u 2 on Cn.

kf − fmk ≤ lim inf kfn − fmk → 0 as m → ∞ n u n→∞ u Theorem 6.30 (Completeness of C ). Let n ∈ N and k·k denote any one of n n the norms, k·k1 , k·k2 , or k·ku on C . Then (C , k·k) is complete. showing fm → f in Z. Proof. By Exercise 6.10 all of these norms are equivalent to k·ku and hence it suﬃces to show that k·k is a complete norm on n. This is a special case of Deﬁnition 6.28. We say that two norms, k·ka and k·kb , on a vector space X u C are equivalent if there are constants C1,C2 ∈ (0, ∞) such that Lemma 6.27 with X = {1, 2, . . . , n} .

kxk ≤ C1 kxk and kxk ≤ C2 kxk for all x ∈ X. a b b a Exercise 6.11. Let X be a set and (Y, ρ) be a complete metric space. Suppose

Similarly two metrics, da and db on a set X are said to be equivalent if there that fn : X → Y are functions such that are constants C1,C2 ∈ (0, ∞) such that δm,n := sup d (fn (x) , fm (x)) → 0 as m, n → ∞. x∈X da (x, y) ≤ C1db (x, y) and db (x, y) ≤ C2da (x, y) for all x, y ∈ X. Show there exists a (unique) functions, f : X → Y such that

Exercise 6.9. Show that two norms, k·ka and k·kb on a vector space X are lim sup d (fn (x) , f (x)) = 0. equivalent iff the corresponding metrics, da (x, y) := ky − xka and db (x, y) := n→∞ x∈X ky − xkb , on X are equivalent metrics. Hint: mimic the proof of Lemma 6.27. Corollary 6.29. If d and d are two equivalent metrics on a set X then (X, d ) a b a Exercise 6.12. Let Z denote the continuous functions on [0, 1] with values in is a complete metric space iff (X, db) is a complete metric space. R and as above let Proof. Suppose that (X, d ) is complete. If {x }∞ is d – Cauchy implies s b n n=1 a Z 1 Z 1 2 kfk1 := |f (t)| dt, kfk2 := |f (t)| dt, and kfku = sup |f (t)| . db (xn, xm) ≤ C2da (xn, xm) → 0 as m, n → ∞ 0 0 0≤t≤1 ∞ Show for all f ∈ Z that; which shows that {xn}n=1 is db – Cauchy. As (X, db) is complete, there exists x ∈ X such that db (x, xn) → 0 as n → ∞. Since kfk1 ≤ kfk2 and kfk2 ≤ kfku . da (x, xn) ≤ C1db (x, xn) → 0 as n → ∞ [Hint: for the first inequality use Cauchy Schwarz.] Also show there is no constant C < ∞ such that we see that xn → x in the da – metric as well. This shows (X, da) is complete. The reverse implication is proved the same way. kfku ≤ C kfk2 for all f ∈ Z. (6.3) n [Hint: consider the sequence, fn (t) = t .]

Page: 50 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6.3 General Limits and Continuity in Metric Spaces 51

Example 6.31. Let Z = C ([0, 1] , R) be the vector space of continuous functions lim f (t) = lim h (t) = 1 and t↓ 1 t↓ 1 on [0, 1] with values in R and for f ∈ Z let 2 2 lim f (t) = lim h (t) = 0 1 1 1 Z t↑ 2 t↑ 2 kfk1 := |f (t)| dt. 0 1 there is not choice for f (1/2) for which f would be continuous at 2 . Hence we have shown that fn can not converge to any element, f ∈ Z. Let us show that (Z, k·k1) is not complete. To this end let In fact, (Z, k·k1) is full of uncountably many “holes” and h is the location 2t if t ≤ 1/2 of just one of these holes. In the third quarter of this course we will ﬁll these g (t) := 1 if 1/2 ≤ t ≤ 1 holes.

n and then set fn (t) = g (t) or all t ∈ [0, 1] . Then 6.3 General Limits and Continuity in Metric Spaces

Suppose now that (X, ρ) and (Y, d) are two metric spaces and f : X → Y is a function..

Deﬁnition 6.32 (Limits of functions). If x0 ∈ X and f : X \{x0} → Y

is a function, then we say limx→x0 f (x) = y0 ∈ Y iﬀ for all ε > 0 there is a δ = δ (ε, x0) > 0 such that

d(f (x) , y0) ≤ ε provided that 0 < ρ(x, x0) ≤ δ (ε, x0) . (6.4)

[In generally when we write limx→x0 f (x) we do not need to assume that f (x0) is deﬁned.]

Fig. 6.4. Here is a plot of f1, f2, f4, f8, f32, and the pointwise limit function h.

1 0 if t < 2 lim fn (t) = h (t) = 1 n→∞ 1 if t ≥ 2

1 which is discontinuous at 1/2. Let us now observe that khk1 = 2 and

1 Z 2 n 1 1 kfn − hk1 = (2t) dt = 0 2 (n + 1) so that fn → h in k·k1 . If there were some f ∈ Z so that kf − fnk1 → 0 we 1 would have to have kf − hk1 = 0. If ε := |f (t0) − h (t0)| for some t0 6= 2 , by Theorem 6.33 (Computing Limits Using Sequences). If x0 ∈ X and continuity we would have |f (t) − h (t)| ≥ ε/2 for t near t0 from which it would f : X \{x0} → Y is a function as above, then limx→x0 f (x) = y0 ∈ Y ∞ follow that kf − hk1 > 0. Therefore we must have f (t) = h (t) for all t 6= t0. iﬀ limn→∞ f (xn) = y0 for all sequences {xn}n=1 ⊂ X \{x0} such that Since limn→∞ xn = x0.

Page: 51 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 52 6 Metric Spaces ∞ Proof. Suppose that limx→x0 f (x) = y0 ∈ Y and {xn}n=1 ⊂ X \{x0} with Example 6.37. The functions f : C \{0} → C deﬁned by f (z) = 1/z is contin- ∞ limn→∞ xn = x0. Then all ε > 0 there is a δ > 0 such that Eq. (6.5) holds. uous. Indeed, if {zn}n=1 ⊂ C \{0} and limn→∞ zn = z ∈ C \{0} , then Since xn → x0 as n → ∞ it follows that 0 < ρ (xn, x0) ≤ δ for a.a. n and therefore d (f (x ) , y ) ≤ ε for a.a. n. As ε > 0 was arbitrary it follows that 1 1 n 0 lim f (zn) = lim = = f (z) . n→∞ n→∞ limn→∞ d (f (xn) , y0) = 0, i.e. that limn→∞ f (xn) = y0. zn z

Conversely if limx→x0 f (x) 6= y0, then there exists ε > 0 such that for 1 Example 6.38. Let f : R → R be the function deﬁned by any δ = n > 0 there exists xn ∈ X \{x0} such that d(f(xn), y0) ≥ ε while 1 ρ (x, xn) < . We then have limn→∞ xn = x while limn→∞ f (xn) 6= y0. 1 if x ∈ n f (x) = Q 0 if x∈ / . Deﬁnition 6.34 (Continuity). A function f : X → Y is continuous at Q x ∈ X if for all ε > 0 there is a δ > 0 such that The function f is discontinuous at all points in R. For example, if x0 ∈ Q we may choose x ∈ \ such that lim x = x while d(f (x) , f(x0)) < ε provided that ρ(x, x0) < δ. (6.5) n R Q n→∞ n 0

lim f (xn) = lim 0 = 0 6= 1 = f (x0) . The function f is said to be continuous if f is continuous at all points x ∈ X. n→∞ n→∞ We will write C (X,Y ) for the collection of continuous functions from X to Y. Similarly if if x0 ∈ R \ Q we may choose xn ∈ Q such that limn→∞ xn = x0 Deﬁnition 6.35 (Seqeuential Continuity). A function f : X → Y is con- while lim f (xn) = lim 1 = 1 6= 0 = f (x0) . tinuous at x ∈ X if n→∞ n→∞ ∞ lim f (xn) = f (x) for all {xn} ⊂ X with lim xn = x. Exercise 6.13. Consider as a metric space with d (m, n) := |m − n| and n→∞ n=1 n→∞ N suppose that (Y, d) is a metric space. Show that every function, f : N → Y is We say f is sequentially continuous on X if it is continuous at all points in X. continuous.

Exercise 6.14. Suppose that (X, d) is a metric space and f, g : X → C are two continuous functions on X. Show; 1. f + g is continuous, 2. f · g is continuous, 3. f/g is continuous provided g (x) 6= 0 for all x ∈ X.

Exercise 6.15. Show the following functions from C to C are continuous. 1. f (z) = c for all z ∈ C where c ∈ C is a constant. 2. f (z) = |z| . 3. f (z) = z and f (z) =z. ¯ 4. f (z) = Re z and f (z) = Im z. PN m n 5. f (z) = m,n=0 am,nz z¯ where am,n ∈ C. Exercise 6.16. Suppose now that (X, ρ), (Y, d), and (Z, δ) are three metric spaces and f : X → Y and g : Y → Z. Let x ∈ X and y = f (x) ∈ Y, show g ◦ f : X → Z is continuous at x if f is continuous at x and g is continuous Corollary 6.36. Continuity and sequential continuity are the same notions. at y. Recall that (g ◦ f)(x) := g (f (x)) for all x ∈ X. In particular this implies that if f is continuous on X and g is continuous on Y then f ◦ g is continuous Proof. This follows rather directly from Theorem 6.33. on X.

Page: 52 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6.3 General Limits and Continuity in Metric Spaces 53

Example 6.39. If f : X → C is a continuous function then |f| is continuous and limn→∞ f (xn) 6= y0. The equivalent of statements a.–c. are proved similarly and so the proofs will be omitted. N For the last statement it is clear that lim f (x) = y implies X m ¯n x→x0 0 F := amnf · f limx↓x0 f (x) = y0 = limx↑x0 f (x) . For the converse assertion, suppose that m,n=0 ε > 0 is given, then choose δ± (ε) > 0 such that is continuous. d (f (x) , y0) ≤ ε when 0 < x − x0 ≤ δ+ (ε) or 0 < x0 − x ≤ δ− (ε) . • End of Lecture 16, 11/5/2012. If we take δ (ε) := min (δ+ (ε) , δ− (ε)) , then we will have Deﬁnition 6.40 (One sided limits). Suppose (Y, d) is a metric space, −∞ < d (f (x) , y0) ≤ ε when 0 < |x − x0| ≤ δ (ε) a < b < ∞, and f :(a, b) → Y is a function. For x0 ∈ (a, b) we say

and since ε > 0 was arbitrary, it follows that limx→x0 f (x) = y0. lim f (x) = y0 ⇐⇒ ∀ε > 0 ∃δ = δ (ε) > 0 3 d (f (x) , y0) ≤ ε if 0 < x−x0 ≤ δ x↓x0 Corollary 6.42 (A monotone continuity criteria). Suppose that (Y, d) is and a metric space, X = (a, b) ⊂ R, and f : X → Y is a function. Then f is con- ∞ tinuous at x ∈ X iff limn→∞ f (xn) = f (x) whenever {xn}n=1 ⊂ X converges lim f (x) = y0 ⇐⇒ ∀ε > 0 ∃δ = δ (ε) > 0 3 d (f (x) , y0) ≤ ε if 0 < x0−x ≤ δ. monotonically to x as n → ∞. In other words, it is sufficient to check sequential x↑x0 continuity along sequences which are either increasing or decreasing. Theorem 6.41 (One sided limit criteria). Suppose that (Y, d) is a metric Proof. This is a direct consequence of Theorem 6.41. space, (a, b) ⊂ R, f :(a, b) → Y is a function, and x0 ∈ (a, b) . Then the following are equivalent; Exercise 6.17 (Continuity of x1/m). Show for each m ∈ N that the function 1. limx↓x0 f (x) = y0 1/m ∞ f (x) := x is continuous on [0, ∞). 2. limn→∞ f (xn) = y0 for all {xn}n=1 ⊂ (x0, b) such that limn→∞ xn = x0. ∞ 1/m 3. limn→∞ f (xn) = y0 for all {xn}n=1 ⊂ (x0, b) such that xn ↓ x0, i.e. xn+1 ≤ Exercise 6.18 (Differentiability of x ). Show for each m ∈ N that the 1/m xn and x0 < xn for all n and limn→∞ xn = x0. function f (x) := x is differentiable on (0, ∞) and that

We also have the following equivalent statements; d y1/m − x1/m 1 x1/m := lim = x1/m−1. dx y→x y − x m a. limx↑x0 f (x) = y0 ∞ b. limn→∞ f (xn) = y0 for all {xn}n=1 ⊂ (a, x0) such that limn→∞ xn = x0. Exercise 6.19 (Intermediate value theorem). Suppose that −∞ < a < ∞ c. limn→∞ f (xn) = y0 for all {xn}n=1 ⊂ (a, x0) such that xn ↑ x0, i.e. xn+1 ≥ b < ∞ and f :[a, b] → R is a continuous function such that f (a) ≤ f (b) . Show xn and xn < x0 for all n and limn→∞ xn = x0. for any y ∈ [f (a) , f (b)] , there exists a c ∈ [a, b] such that f (c) = y.3 Hint: Let S := {t ∈ [a, b]: f (t) ≤ y} and let c := sup (S) . Moreover, limx→x0 f (x) = y0 iﬀ limx↓x0 f (x) = y0 = limx↑x0 f (x) . Exercise 6.20 (Inverse Function Theorem I). Let f :[a, b] → [c, d] be a Proof. (1. =⇒ 2.) If lim f (x) = y then for all ε > 0 there exits δ = x↓x0 0 strictly increasing (i.e. f (x ) < f (x ) whenever x < x ) continuous function δ (ε) > 0 such that d (f (x) , y ) ≤ ε if 0 < x−x ≤ δ. Hence if {x }∞ ⊂ (x , b) 1 2 1 2 0 0 n n=1 0 such that f (a) = c and f (b) = d. Then f is bijective and the inverse function, such that x → x then 0 < x − x ≤ δ for a.a. n and hence d (f (x ) , y ) ≤ ε n 0 n 0 n 0 g := f −1 :[c, d] → [a, b] , is strictly increasing and is continuous. for a.a. n. Since ε > 0 was arbitrary, it follows that limn→∞ d (f (xn) , y0) = 0, i.e. limn→∞ f (xn) = y0. It is trivial that (2. =⇒ 3.) Notations 6.43 Let (X, ρ) and (Y, d) be metric spaces and f : X → Y be a 1 (3. =⇒ 1.) If limx↓x0 f (x) 6= y0 there exists ε > 0 such that for all δ = n > function. 0 0 1 0 0 there exists xn ∈ (x0, b) such that 0 ≤ xn − x0 ≤ n while d (f (xn) , y0) ≥ ε. 3 Then take x = min (x0 , . . . , x0 ) . Then x ↓ x while d (f (x ) , y ) ≥ ε, i.e., The same result holds for y ∈ [f (b) , f (a)] if f (b) ≤ f (a) – just replace f by −f n 1 n n n 0 in this case.

Page: 53 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 54 6 Metric Spaces 1. We say f is uniformly continuous, iff for all ε > 0 there exists δ = Proof. Let a ∈ A and x, y ∈ X, then δ (ε) > 0 such that dA (x) ≤ d(x, a) ≤ d(x, y) + d(y, a). ∀ x, x0 ∈ X with ρ (x, x0) ≤ δ =⇒ d (f (x) , f (x0)) ≤ ε. Take the infimum over a in the above equation shows that 2. A function, f : X → Y, is said to be Lipschitz if there is a constant C < ∞ such that dA (x) ≤ d(x, y) + dA(y) ∀x, y ∈ X. d (f (x) , f (x0)) ≤ Cρ (x, x0) for all x, x0 ∈ X. Therefore, dA (x) − dA(y) ≤ d(x, y) and by interchanging x and y we also have Recall that a function f : X → Y is continuous at x0 ∈ X if for all ε > 0 that dA(y) − dA (x) ≤ d(x, y) which implies Eq. (6.8). there exists δ = δ (ε, x ) > 0 such that 0 Corollary 6.46. The function d satisfies, ∀ x ∈ X with ρ (x, x ) ≤ δ =⇒ d (f (x) , f (x )) ≤ ε. 0 0 |d(x, y) − d(x0, y0)| ≤ d(y, y0) + d(x, x0). Thus we see that a function is uniformly provided we can take δ (ε, x ) > 0 to 0 Therefore d : X × X → [0, ∞) is continuous in the sense that d(x, y) is close be independent of x . If f is Lipschitz and ε > 0, we may take δ := ε/C in 0 to d(x0, y0) if x is close to x0 and y is close to y0. In particular, if x → x and order to see that if n yn → y then ρ (x, x0) ≤ δ =⇒ d (f (x) , f (x0)) ≤ Cρ (x, x0) ≤ Cδ = ε lim d (xn, yn) = d (x, y) = d lim xn, lim yn . n→∞ n→∞ n→∞ which shows f is uniformly continuous. Proof. First Proof. By Lemma 6.45 for single point sets and the triangle Example 6.44. Any function, f : R → R which is everywhere differentiable is inequality for the absolute value of real numbers, Lipschitz iff K := sup |f 0 (t)| < ∞. Indeed if t∈R |d(x, y) − d(x0, y0)| ≤ |d(x, y) − d(x, y0)| + |d(x, y0) − d(x0, y0)| |f (y) − f (x)| ≤ K |y − x| for all x, y ∈ R ≤ d(y, y0) + d(x, x0). then Second Proof. By the triangle inequality, 0 |f (y) − f (x)| |f (x)| = lim ≤ K for all x ∈ R. y→x |y − x| d (x, y) ≤ d (x, x0) + d (x0, y) ≤ d (x, x0) + d (x0, y0) + d (y0, y) Conversely, if K := sup |f 0 (t)| < ∞, then by the mean value Theorem 10.16, t∈R for all y > x there exists c ∈ (x, y) such that from which it follows that d (x, y) − d (x0, y0) ≤ d (x, x0) + d (y0, y) . f (y) − f (x) 0 = |f (c)| ≤ K. y − x Interchanging x with x0 and y with y0 in this inequality shows, It turns out that every metric spaces with an infinite number of elements d (x0, y0) − d (x, y) ≤ d (x, x0) + d (y0, y) comes equipped with a large collection of Lipschitz functions.

Lemma 6.45 (Distance to a Set). For any non empty subset A ⊂ X, let and the result follows from the last two inequalities. Exercise 6.21 (Continuity of integration). Let Z = C ([0, 1] , R) be the dA (x) := inf{d(x, a)|a ∈ A}, continuous functions from [0, 1] to R and k·ku be the uniform norm, kfku := then sup0≤t≤1 |f (t)| . Deﬁne K : Z → Z by |dA (x) − dA(y)| ≤ d(x, y) ∀ x, y ∈ X. (6.6) Z x K (f)(x) := f (t) dt for all x ∈ [0, 1] . In particular, dA : X → [0, ∞) is continuous. 0

Page: 54 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6.3 General Limits and Continuity in Metric Spaces 55 Show that K is a Lipschitz function. In more detail, show

kK (f) − K (g)ku ≤ kf − gku for all f, g ∈ Z. In this problem please take for granted the standard properties of the integral including 1. The function x → K (f)(x) is indeed continuous (in fact differentiable by the fundamental theorem of calculus). 2. K : Z → Z is a linear transformation. R x R x 3. If f (t) ≤ g (t) for all t ∈ [0, 1] , then 0 f (t) dt ≤ 0 g (t) dt for all x ∈ X. R x R x 4. From 3. it follows that 0 f (t) dt ≤ 0 |f (t)| dt. Exercise 6.22 (Discontinuity of differentiation). Let Z be the polynomial Pn k functions in C ([0, 1] , R) , i.e. functions of the form p (t) = k=0 akt with ak ∈ R. As above we let kpku := sup0≤t≤1 |p (t)| . Define D : Z → Z by 0 Pn k Fig. 6.5. Here is a sequence of continuous functions converging pointwise to a dis- D (p) = p , i.e. if p (t) = akt then k=0 continuous function. This convergence is not uniform. n X k−1 D (p)(t) = kakt . k=1 Hopefully it is clear to the reader the uniform convergence implies pointwise 1. Show D is discontinuous at 0 – where 0 represents the zero polynomial. convergence. The next theorem is a basic fact about uniform convergence which 2. Show D is discontinuous at all points p ∈ Z. does not hold in general for pointwise convergence. The proof of this theorem is a bit subtle but well worth mastering as the method will arise over and over Exercise 6.23 (Continuity of integration II). Let Z = C ([0, 1] , R) and K be as in Exercise 6.21. Further let again. s Z 1 Theorem 6.49 (Uniform Convergence Preserves Continuity). Suppose 2 ∞ kfk2 := |f (t)| dt. that {fn}n=1 are continuous functions from X to Y and fn converges uniformly 0 to f : X → Y. If fn is continuous at x ∈ X for all n then f is continuous at Show x as well. In particular if fn is continuous on X for all n then f is continuous 1 kK (f) − K (g)k ≤ √ kf − gk for all f, g ∈ Z. on X as well. 2 2 2 Proof. We will give three proofs of this important theorem. In these proofs Definition 6.47 (Pointwise Convergence). Let (X, d) and (Y, ρ) be metric we will let spaces and fn : X → Y be functions for each n ∈ . We say that fn converges N δn := sup d (f (x) , fn (x)) . pointwise to f : X → Y provided limn→∞ fn (x) = f (x) for all x ∈ X, i.e. x∈X provided First Proof. Suppose that f were discontinuous at some point x0 ∈ X. lim d (f (x) , fn (x)) = 0 for each x ∈ X. n→∞ Then there would exists ε > 0 and xk ∈ X \{x0} such that limk→∞ xk = x0 while ρ (f (x ) , f (x )) ≥ ε for all ε > 0. Let n ∈ and set g := f , then Definition 6.48 (Uniform Convergence). Let (X, d) and (Y, ρ) be metric k 0 N n spaces and f : X → Y be functions for each n ∈ . We say that f converges n N n ε ≤ ρ (f (xk) , f (x0)) ≤ ρ (f (xk) , g (xk)) + ρ (g (xk) , g (x0)) + ρ (g (x0) , f (x0)) uniformly to f : X → Y provided ≤ δn + ρ (g (xk) , g (x0)) + δn = 2δn + ρ (g (xk) , g (x0)) . δn := sup d (f (x) , fn (x)) → 0 as n → ∞. x∈X Letting k → ∞ in this inequality implies, ε ≤ 2δn and then letting n → ∞ implies ε = 0 and we have reached the desired contradiction, see Figure 6.6.

Page: 55 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 56 6 Metric Spaces Example 6.50 (Non-uniform convergence). For an example of nonuniform con- 2 vergence, suppose that g (x) = max 1 − 4x , 0 and fn (x) := g (x − 3n) for all n, see Figure 6.7 Notice that for each x ∈ R+, fn (x) = 0 for a.a. n and therefore

Fig. 6.6. Showing that f can not have any discontinuities.

∞ Second Proof. We must show limk→∞ f (xk) = f (x) whenever {xk}k=1 ⊂ X is a convergent sequence such that x := limk xk ∈ X. So assume we are given ∞ such a sequence {xk}k=1 . Then for any n ∈ N we have,

ρ (f (x) , f (xk)) ≤ ρ (f (x) , fn (x)) + ρ (fn (x) , f (xk)) Fig. 6.7. Here is a plot of {f }3 . ≤ ρ (f (x) , fn (x)) + ρ (fn (x) , fn (xk)) + ρ (fn (xk) , f (xk)) n n=0

≤ δn + ρ (fn (x) , fn (xk)) + δn. Therefore, limn→∞ fn (x) = 0. On the other hand, kfnku = 1 for all n so fn converges to lim sup ρ (f (x) , f (xk)) ≤ lim sup ρ (fn (x) , fn (xk)) + 2δn = 2δn, 0 pointwise but not uniformly in x. Nevertheless the limiting function is still k→∞ k→∞ continuous. This is not always the case as you will see in the next exercise. wherein we have used the continuity of f for the last equality. Thus we have n Exercise 6.24. Let f : [0, 1] → be defined by f (x) = xn for x ∈ [0, 1] . shown n R n Show f (x) := limn→∞ fn (x) exists for all x ∈ [0, 1] and find f explicitly. Show lim sup ρ (f (x) , f (xk)) ≤ 2δn k→∞ that fn does not converge to f uniformly. which upon passing to the limit as n → ∞ shows lim supk→∞ ρ (f (x) , f (xk)) = 0. This suffices to show limk→∞ f (xk) = f (x) . Third Proof. Let x ∈ X and ε > 0 be given. Choose n ∈ N so that 6.4 Density and Separability δn ≤ ε and let g := fn. Since g is continuous there exists δ > 0 such that ρ (g (x) , g (x0)) ≤ ε when d (x, x0) ≤ δ. So if d (x, x0) ≤ δ, then Definition 6.51. Let (X, d) be a metric space. We say A ⊂ X is dense in X if for all x ∈ X, there exists {x }∞ ⊂ A such that x = lim x . [In words, 0 0 n n=1 n→∞ n ρ (f (x) , f (x )) ≤ ρ (f (x) , g (x)) + ρ (g (x) , f (x )) all points in X are limit points of sequences in A.] A metric space is said to be ≤ ρ (f (x) , g (x)) + ρ (g (x) , g (x0)) + ρ (g (x0) , f (x0)) separable if it contains a countable dense subset, A. 0 ≤ δn + ρ (g (x) , g (x )) + δn ≤ ε + ε + ε = 3ε. Example 6.52. The spaces Rn and Cn with their Euclidean metrics are sepa- n As ε > 0 and x ∈ X were arbitrary, we have shown f is continuous on X. rable. Indeed we can take D = Qn and D = (Q + iQ) respectively for the

Page: 56 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 6.5 Test 2: Review Topics 57

n countable dense subsets. For example given k ∈ N and x = (x1, . . . , xn) ∈ R , 1. Show that if F : X → Y and G : X → Y are two continuous functions such k k k k 1 we may choose q = q1 , . . . , qn ∈ D such that xi − qi ≤ k for 1 ≤ i ≤ n. that F = G on A then F = G on X. We then have, 2. Now suppose that (Y, d) is complete and f : A → Y is a function which is uniformly continuous (Notation 6.43). Recall this means for every ε > 0 v v u n u n 2 √ there exists a δ > 0 such that k uX k 2 uX 1 1 d x, q = t xi − q ≤ t = n · → 0 as k → ∞. i k k i=1 i=1 d (f (a) , f (b)) ≤ ε for all a, b ∈ A with ρ (a, b) ≤ δ. (6.7)

The Cn case now follows from this case as Cn is really R2n in disguise. Show there is a unique continuous function F : X → Y such that F = f on A. 0 ∞ Example 6.53. Let Y := R \ Q which we equip with the usual metric, d (y, y ) = Hint: Deﬁne F (x) = limn→∞ f (xn) where {xn} ⊂ A is chosen to 0 0 n=1 |y − y | for all y, y ∈ Y. I now claim that Y is separable. We can no longer use Q converge x ∈ X. You must show the limit exists and is independent of the as the countable dense subset of Y since is not contained in Y ! On the other ∞ Q choice of sequence {xn}n=1 ⊂ A which converges for x. hand, for each q ∈ Q we may choose yn (q) ∈ Y such that limn→∞ yn (q) = q. 3. Let X = = Y and A = ⊂ X, ﬁnd a function f : → which is 1 R Q Q R Then if y ∈ Y and ε = k > 0 is given, we may choose q ∈ Q such that continuous on but does not extend to a continuous function on . 1 1 Q R |y − q| ≤ 2k . We then take ak := yn (q) for some large n so that |ak − q| ≤ 2k . 1 It then follows that |y − ak| ≤ k → 0 as k → ∞ and this shows that A := ∪ {y (q): n ∈ } is a dense subset of Y. Moreover A is countable, why? q∈Q n N 6.5 Test 2: Review Topics

Remark 6.54. An equivalent way to say that A ⊂ X is dense is to say dA ≡ 0, 1. Understand the basic properties of complex numbers. i.e. dA (x) = 0 for all x ∈ X. Indeed if x ∈ X, you should show that dA (x) = 0 2. Coutability. Key facts are that countable union of countable sets is count- iff there exists an ∈ A such that d (x, an) → 0 as n → ∞. able and the finite product of countable sets is countable. Exercise 6.25. Suppose that (X, d) is a separable metric space and Y is a non- 3. Definitions of metric and normed spaces and their basic properties which empty subset of X which is also a metric space by restricting d to Y. Show (Y, d) in the end of the day typically follow from the triangle inequality. is separable. [Hint: suppose that A ⊂ X be a countable dense subset of X. For 4. Be aware of different norms, k·ku , k·k1 , and k·k2 . ∞ 1 5. Understand the notion of; limits of sequences, Cauchy sequences, com- each a ∈ A choose {yn (a)}n=1 ⊂ Y so that dY (a) ≤ d (a, yn (a)) ≤ dY (a) + n . Now show AY := ∪a∈A {yn (a): n ∈ N} is a countable dense subset of Y.] pleteness, limits and continuity of functions. 6. Know what is meant by pointwise and uniform convergence. You should Exercise 6.26. Let n ∈ N. Show any non-empty subset Y ⊂ Cn equipped with be able to compute pointwise limits and know how to test if the limit is the metric, uniform or not. A key theorem is the uniform limit of continuous functions d (x, y) = ky − xk for all x, y ∈ Y is still continuous. is separable, where k·k is either k·ku , k·k1 , or k·k2 .

Exercise 6.27. For x, y ∈ R, let 0 if x = y d (x, y) := . 1 if x 6= y

Then (R, d) is a non-separable metric space. (This metric space is also complete.)

Exercise 6.28. Suppose (X, ρ) and (Y, d) are metric spaces and A is a dense subset of X.

Page: 57 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14

7 Series and Sums in Banach Spaces

∞ Definition 7.1. Suppose (X, k·k) is a normed space and {xn}n=1 is a sequence We also have, P∞ PN in X. Then we say n=1 xn converges in X iff s := limN→∞ n=1 xn exists N P∞ PN X in X otherwise we say n=1 xn diverges. We often let SN := n=1 xn and [f (n + 1) − f (n)] = f (N + 1) − f (M) for N ≥ M. (7.2) ∞ refer to {SN }N=1 ⊂ X as the sequence of partial sums. n=M P∞ PN Proof. When N = 3 we have, If X = R and xn ≥ 0, then n=1 xn diverges iff limN→∞ n=1 xn = ∞ and so we will write P∞ x = ∞ in this case to indicate that P∞ x diverges n=1 n n=1 n 3 X to infinity. (f (n + 1) − f (n)) = (f (2) − f (1)) + (f (3) − f (2)) + (f (4) − f (3)) ∞ ∞ n=1 Theorem 7.2 (Comparison Theorem). Suppose that {an}n=1 and {bn}n=1 are sequences in [0, ∞). If an ≤ bn for all n, then = f (4) − f (1) .

∞ ∞ X X In general, Eqs. (7.1) and (7.2) are easily verified by a simple induction argu- an ≤ bn ment. The rest of the theorem is now evident. n=1 n=1 Example 7.4 (Geometric Series). Suppose that f (n) = αn where α ∈ C. Then where we allow for these sums to be infinite. Moreover if an ≤ bn for a.a. n then P∞ P∞ P∞ P∞ n+1 n n=1 bn < ∞ implies n=1 an < ∞ and if n=1 an = ∞ then n=1 bn = ∞. f (n + 1) − f (n) = α − α n Pk Pk = α (α − 1) Proof. Let Ak := n=1 an and Bk := n=1 bn. Then a simple induction argument shows that Ak ≤ Bk for all k and therefore and we find, N ∞ ∞ X X X (α − 1) αn = αN+1 − α. an = lim Ak ≤ lim Bk = bn k→∞ k→∞ n=1 n=1 n=1 If α 6= 1 it follows that by the sandwich lemma. N X αN+1 − α αn = Theorem 7.3 (Telescoping Series / Fundamental Theorem of Summa- α − 1 ∞ n=1 tion). Let {f (n)}n=1 ⊂ X be a sequence, then N+1 and if |α| < 1, it follows that αN+1 = |α| → 0 as N → ∞ and therefore N X [f (n + 1) − f (n)] = f (N + 1) − f (1) for all N ∈ N (7.1) ∞ X n α n=1 α = . 1 − α P∞ n=1 and n=1 [f (n + 1) − f (n)] is convergent in X iff limN→∞ f (N) exists in X in which case, Theorem 7.5 (Integral test). Let f : (0, ∞) → R be a C1 function such 0 0 00 ∞ that f (x) ≥ 0 and f (x) is decreasing in x (i.e. f (x) ≤ 0 if it exists) and X (f (n + 1) − f (n)) = lim f (N) − f (1) . f (∞) := limx→∞ f (x) , see Figure 7.1. (As f is an increasing function, f (∞) N→∞ n=1 exists in (−∞, ∞].) Then 60 7 Series and Sums in Banach Spaces N X Summing these inequalities on n, using [f (N + 1) − f (M)] ≤ f 0 (n) ≤ [f (N + 1) − f (M)] + f 0 (M) − f 0 (N + 1) N+1 n=M X (7.3) [f (n) − f (n − 1)] = f (N + 1) − f (M) , ≤ [f (N + 1) − f (M)] + f 0 (M) n=M+1 for all M ≤ N. Letting N → ∞ in these inequalities also gives, shows,

∞ N+1 N+1 N X X 0 X 0 X 0 f (∞) − f (M) ≤ f 0 (n) ≤ f (∞) − f (M) + f 0 (M) (7.4) f (n) ≤ f (N + 1) − f (M) ≤ f (n − 1) = f (n) . n=M n=M+1 n=M+1 n=M

P∞ 0 and in particular, n=1 f (n) < ∞ iﬀ f (∞) < ∞. This inequality is equivalent to Eq. (7.3) because it gives,

N N+1 X X [f (N + 1) − f (M)] ≤ f 0 (n) = f 0 (n) + f 0 (M) − f 0 (N + 1) n=M n=M+1 ≤ [f (N + 1) − f (M)] + f 0 (M) − f 0 (N + 1) .

Corollary 7.6( p – series). Let p ∈ R, then

∞ X 1 ∞ if p ≤ 1 = . np < ∞ if p > 1 n=1

1 Proof. If p ≤ 0, then limn→∞ np 6= 0 and hence the series diverges and we may assume that p > 0. For p > 1, let f (x) = −x−(p−1) so that f 0 (x) = (p − 1) x−p which is decreasing in x and therefore by Theorem 7.5, √ Fig. 7.1. The items plotted here are ln x and −1/ x both of which satisfy the ∞ X p − 1 1 p − 1 hypothesis of the propostion. f (∞) − f (M) ≤ ≤ f (∞) − f (M) + f 0 (M) = + < ∞ np M p−1 M p n=M and so ∞ 1 X 1 1 1 Proof. By the mean value Theorem 10.16, there exists cn ∈ (n − 1, n) such ≤ ≤ + . 0 0 (p − 1) M p−1 np (p − 1) M p−1 M p that f (n) − f (n − 1) = f (cn) . Since f is decreasing, it follows that n=M 0 0 0 In particular for M = 1, f (n) ≤ f (cn) ≤ f (n − 1) for all n, ∞ or equivalently1 that 1 X 1 1 ≤ ≤ + 1. p − 1 np p − 1 f 0 (n) ≤ f (n) − f (n − 1) ≤ f 0 (n − 1) for all n. n=1

1 Since This is a standard inequality involving convex functions. The reader should draw ∞ ∞ 1 X 1 X 1 the picture and note that f (n) − f (n − 1) is the slope of the cord joining ≤ ≤ (n − 1, f (n − 1)) to (n, f (n)) . p − 1 np n n=1 n=1

Page: 60 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 7 Series and Sums in Banach Spaces 61 P∞ 1 Pn we may let p ↓ 1 in order to show, ∞ ≤ n=1 n . This complete the proof as Sn −Sm−1 = k=m xk. For the second item apply the ﬁrst with n = m+1. For ∞ for any p < 1 we have , P ∞ ∞ the third item let S := xk, then limN→∞ SN = S and so by very deﬁnition, X 1 X 1 k=1 ∞ ≤ ≤ . n np n=1 n=1 ∞ X xk = S − SN+1 → 0 as N → ∞. k=N Exercise 7.1. Take f (x) = ln x in Theorem 7.5 in order to directly conclude P∞ 1 that n=1 n = ∞. Exercise 7.3. Let (X, d) be a metric space. Suppose that {x }∞ ⊂ X is a Exercise 7.2. Let 0 ≤ p < ∞. Prove, n n=1 sequence and set εn := d(xn, xn+1). Show that for m > n that ∞ X 1 ∞ if 0 ≤ p ≤ 1 = . m−1 ∞ n (ln n)p < ∞ if p > 1 X X n=2 d(xn, xm) ≤ εk ≤ εk. k=n k=n by applying Theorem 7.5 with the following functions; Conclude from this that if 1−p 1. for 0 ≤ p < 1 take f (x) = (ln x) , ∞ ∞ X X 2. for p = 1 take f (x) = ln ln x, and ε = d(x , x ) < ∞ 1−p k n n+1 3. for p > 1 take f (x) = − (ln x) . k=1 n=1

∞ ∞ ∞ Theorem 7.7. Let (X, k·k) be a Banach space and {xk}k=1 ⊂ X be a sequence. then {xn}n=1 is Cauchy. Moreover, show that if {xn}n=1 is a convergent se- Then; quence and x = limn→∞ xn then ∞ P ∞ 1. xk converges iff X k=1 d(x, xn) ≤ εk. k=n n X xk → 0 as n, m → ∞ with n ≥ m. Theorem 7.8 (Absolute Convergence Implies Convergence). Let (X, k · ∞ k=m ∞ P k) be a Banach space {xk}k=1 ⊂ X be a sequence. Then kxkk < ∞ implies k=1 ∞ ∞ ∞ ∞ P P P P 2. If xk converges then limk→∞ xk = 0 or alternatively if limk→∞ xk 6= 0 xk is convergent. [We say xk is absolutely convergent if kxkk < k=1 k=1 k=1 k=1 ∞ P ∞.] then xk diverges. k=1 ∞ ∞ ∞ Exercise 7.4. Prove Theorem 7.8. Namely if (X, k · k) is a Banach space and 3. If P x converges then lim P x = 0, i.e. the N - tail, P x , of a ∞ ∞ k N→∞ k k ∞ P P k=1 k=N k=N {xk} ⊂ X is a sequence, then kxkk < ∞ implies xk is convergent. ∞ k=1 P k=1 k=1 convergent series, xk, go to zero as N → ∞. k=1 ∞ Proposition 7.9 (Alternating Series Test). If {ak}k=1 ⊂ [0, ∞) is a non- Remark: the only place we use X is complete is in the implication (⇐=) increasing sequence (i.e. ak ≥ ak+1 for all k) such that limk→∞ ak = 0, then P∞ k+1 in item 1. All of the remaining statements hold for any normed space X. s := k=1 (−1) ak is convergent. Moreover, for all n ∈ N ∞ n n ∞ Proof. Let S := P x so that P x converges iff lim S exists X k+1 X k+1 n k=1 k k n→∞ n s − (−1) a = (−1) a ≤ a . (7.5) k=1 k k n+1 ∞ k=1 k=n+1 iff {Sn}n=1 is Cauchy since X is a Banach space which gives item 1. since

Page: 61 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 62 7 Series and Sums in Banach Spaces a 1 which imply Eq. (7.6). a2 Alternatively, here is another way to think about the error estimates; a3 a 4 0 ≤ a1 − a2 = S2 ≤ S ≤ S1 ≤ a1 a5 ∞ k+1 a6 P . so that 0 ≤ S ≤ a1. Applying these results to k=n+1 (−1) ak shows, . n ∞ X k+1 X k+1 s − (−1) a = (−1) a ≤ a . 0 S2 S4 S6 S S5 S3 S1 k k n+1 k=1 k=n+1 Fig. 7.2. The picture proof of the alternating series test.

Proof. The proof is essentially contained in Figure 7.2. Let Sn := Example 7.10. By the alternating series test, Pn (−1)k+1 a . For n ∈ , k=1 k N ∞ X n 1 (−1) S2n+1 = S2n−1 − (a2n − a2n+1) ≤ S2n−1 and np n=1 S2(n+1) = S2n + (a2n+1 − a2n+2) ≥ S2n, is convergent for all p > 0. On the other hand, by Corollary 7.6, the series is wherein we have used the fact that {ak} is decreasing to conclude the terms absolutely convergent iﬀ p > 1. in parenthesis above are non-negative. From these inequalities we learn that ∞ ∞ {S2n}n=1 is an increasing sequence while {S2n+1}n=1 is a decreasing sequence Theorem 7.11 (Uniform convergence and the Weierstrass M – test). and hence Suppose that (X, k·k) is a Banach space, (Y, d) is a metric space, for each n ∈ N, ∞ fn : Y → X is a function, and there exists {Mn}n=1 ⊂ [0, ∞) satisfying; sev := lim S2n exists in (−∞, ∞] and n→∞ ∞ X sodd := lim S2n−1 exists in (−∞, ∞]. sup kf (y)k ≤ M ∀ n ∈ and M < ∞. n→∞ n X n N n y∈Y n=1 Since S2n+1 − S2n = a2n+1 → 0 as n → ∞ it follows that sev = sodd ∈ . It is R PN now easy to conclude (You prove!) that Then SN (y) := n=1 fn (y) converges absolutely and uniformly to S (y) = P∞ ∞ fn (y) . Moreover, if we further assume that {fn} ⊂ C (Y,X) (i.e. fn ∞ n=1 n=1 X k+1 is continuous for all n), then the function S : Y → X is also continuous. S = (−1) ak = lim Sn = sev = sodd ∈ . n→∞ R k=1 Proof. For any y ∈ Y,

In summary, S2n ↑ S while S2n+1 ↓ S as n → ∞ and in particular, for all n ∈ N, ∞ ∞ X X kfn (y)kX ≤ Mn < ∞ S2n+1 − a2n = S2n ≤ S ≤ S2n+1 = S2n + a2n+1. n=1 n=1 These inequalities then implies P∞ and therefore S (y) = n=1 fn (y) converges absolutely. Moreover we have, ∞ X k+1 M 0 ≤ S − S2n = (−1) ak ≤ a2n+1 X kS (y) − SN (y)k = lim kSM (y) − SN (y)k = lim fn (y) k=2n+1 M→∞ M→∞ n=N+1 and M ∞ ∞ ∞ X X X X k+1 ≤ lim inf kfn (y)k = kfn (y)k ≤ Mn. 0 ≤ S2n+1 − S = (−1) ak ≤ a2n. M→∞ n=N+1 n=N+1 n=N+1 k=2n

Page: 62 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 7 Series and Sums in Banach Spaces 63 ∞ As the last member of this inequality does not depend on y we have, Theorem 7.13 (Ratio test). Suppose that {an}n=1 is a sequence in C such that a 6= 0 for a.a. n. Then ∞ n X sup kS (y) − SN (y)k ≤ Mn → 0 as N → ∞ an+1 P∞ P∞ y∈Y 1. If α := lim supn→∞ < 1 then n=1 |an| < ∞ and n=1 an is abso- n=N+1 an lutely convergent. because tails of convergent series vanish, Theorem 7.7. The continuity of S now an+1 P∞ 2. If ≥ 1 for a.a. n then limn→∞ |an| > 0 and an diverges. an n=1 follows form the continuity of SN , the uniform convergence just proved, and an+1 P∞ 3. If lim infn→∞ > 1 then limn→∞ |an| = ∞ and an diverges. Theorem 6.48. an n=1 a ∞ n+1 4. If lim supn→∞ a = 1, the test fails, i.e. you must work harder!] Theorem 7.12 (Root test). Suppose that {an}n=1 is a sequence in C and let n α := lim sup |a |1/n . Then n→∞ n Proof. We take each item in turn. P∞ P∞ 1. If α < 1 then n=1 |an| < ∞ and n=1 an is absolutely convergent. an+1 ∞ 1. If α < 1, let β ∈ (α, 1) , then ≤ β for a.a. n, i.e. there exists N ∈ P an N 2. If α > 1, then lim supn→∞ |an| = ∞ and n=1 an diverges. 3. If α = 1, the test fails, i.e. you must work harder!] such that |an+1| ≤ β |an| for all n ≥ N. A simple induction argument shows, n−N −N n Proof. We take each item in turn. |an| ≤ |aN | β = β |aN | β for n ≥ N.

1/n 1. If α < 1, let β ∈ (α, 1) , then |an| ≤ β for a.a. n which implies that The result follows by the comparison Theorem 7.2 and the fact that |a | ≤ βn for a.a. n and so the result follows by the comparison Theorem n ∞ 7.2 as X −N n |aN | ∞ β |aN | β = < ∞. X n β 1 − β β = < ∞. n=N 1 − β n=1 an+1 1/n n 2. Suppose there exists N ∈ N such that a ≥ 1 for all n ≥ N. 2. If α > 1 and β ∈ (1, α) , then |an| ≥ β i.o. n and hence |an| ≥ β i.o. n. n n This inequality says that |an| is non-decreasing for large n and therefore As β → ∞ as n → ∞ it follows that lim supn→∞ |an| = ∞. P∞ 1 P∞ 1 limn→∞ |an| ≥ |am| for any m ≥ N. We may now choose m ≥ N such that 3. From Corollary 7.6 we know that n=1 n = ∞ while n=1 n2 < ∞. |am|= 6 0. However from Lemma 3.30, an+1 an+1 3. If lim infn→∞ > β > 1, then ≥ β for a.a. n. Working as above 1/n 1/n an an 1 1 n−N lim = 1 = lim there exists N ∈ N such that |aN |= 6 0 and |an| ≥ |aN | β for all n ≥ N. n→∞ n→∞ 2 n n From this it follows that limn→∞ |an| = ∞. P∞ 1 4. From Corollary 7.6 we know that p < ∞ iﬀ p > 1. However which shows the test has failed. In fact, if 0 < p < ∞ and k ∈ N such that n=1 n 1 1 p k ≥ p, then k ≤ np ≤ 1 which implies   n 1 p p n+1 n n p 1/n 1/n lim  1 p  = lim = lim = 1 = 1 1 1 1/n n→∞ n→∞ n + 1 n→∞ n + 1 ≤ ≤ (1) = 1. n nk np which shows the test has failed and does not resolve when P∞ 1 < ∞. 1 1/n n=1 np By Lemma 3.30 we know limn→∞ nk = 1 and therefore by the sand- 1 1/n wich lemma, limn→∞ p = 1. n In what follows let (Z, k·k) be a complex Banach space. For example, Z = C and kzk = |z| is an important special case.

P∞ n/p n Exercise 7.5. For every p ∈ N, show n=0 (n) z is convergent iﬀ z = 0.

Page: 63 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 64 7 Series and Sums in Banach Spaces

kan+1k Remark 7.14. The root and ratio tests extend to series in Banach spaces Corollary 7.17. If µ = limn→∞ exists, the radius (R) of convergence ∞ kank ({an} ⊂ Z) with the only changes being that we replace |an| by kank and 1 kank n=1 of a power series in Eq. (7.7) is R = = limn→∞ . µ kan+1k an+1 by kan+1k wherever these occur. The point is that we can apply the root an kank ∞ kan+1k Proof. If µ = limn→∞ , then and ratio test to the sequence of real numbers {kank}n=1 in order to make kank P∞ conclusion about the absolute convergence or divergence of n=1 an. n+1 an+1 (z − z0) Definition 7.15. Given z ∈ and {a }∞ ⊂ Z, the series of the form kan+1k 0 C n n=0 ρ (z) := lim n = |z − z0| lim = µ · |z − z0| . n→∞ kan (z − z0) k n→∞ kank ∞ X n a (z − z ) (7.6) P∞ n n 0 The the Ratio test in Theorem 7.13 now implies n=0 an (z − z0) converges n=0 1 1 1 if |z − z0| < µ (ρ (z) < 1) and diverges if |z − z0| > µ (ρ (z) > 1) , i.e. R = µ is called a power series. If z0 = 0 we call it a Maclaurin series, i.e. a series from Eq. (7.9). of the form ∞ Theorem 7.18. If the radius of convergence (R) of a power series X n P∞ n anz . (7.7) n=0 an (z − z0) is positive (R > 0) , then the functions n=0 S : D (z ,R) := {z ∈ : |z − z < R|} → Z 1 0 C 0 The radius of convergence of either of these series is defined to be R := α ∈ P∞ n [0, ∞] where defined by S (z) := n=0 an (z − z0) is continuous and the series is uniformly 1/n α := lim sup kank ∈ [0, ∞] . convergent on D (z0, ρ) for all ρ < R. n→∞ n P∞ 1 Proof. For any ρ < R let Mn := kank ρ . Then n=0 Mn < ∞ by the root By definition, 0 := ∞ in this formula. test or the definition of R. So if z ∈ D (z0, ρ) we find, The next theorem shows that R is the critical radius governing the conver- n n gence of Eqs. (7.7) and (7.8). kan (z − z0) k ≤ kank ρ = Mn. Proposition 7.16. If R is the radius of convergence of a power series in Eq. It follows from the Weierstrass M – test, Theorem 7.11, that the series is (7.7) then: uniformly convergent and S is continuous on D (z0, ρ) for all ρ < R. Since D (z0,R) = ∪0<ρ R, the series diverges. 3. If |z − z0| = R, the series may or may not converge. Exercise 7.6. Show that each of the following power series have an infinite radius of convergence and hence define continuous functions from C to C. We may also characterize R as P∞ zn 1. exp (z) := n=0 n! , ( ∞ ) 2n+1 X n 2. sin (z) := P∞ (−1)2n+1 z , and R := sup |z − z0| : z ∈ C and an (z − z0) converges . (7.8) n=0 (2n+1)! 2n n=0 P∞ 2n z 3. cos (z) := n=0 (−1) (2n)! . Proof. Let ∞ Proposition 7.19. Suppose that (X, k·k) is a normed space and {an}n=1 , n 1/n ∞ P∞ P∞ ρ (z) := lim sup kan (z − z0) k {bn}n=1 ⊂ X such that A := n=1 an and B := n=1 bn exist in X. Then n→∞ for all λ ∈ F (F = R or C if X is a real or complex vector space respectively), P∞ 1/n |z − z0| we have the series, (an + λbn) , is convergent and = |z − z0| · lim sup kak = . n=1 n→∞ R ∞ X By the root test, Theorem 7.12, we know that power series in Eq. (7.7) converges (an + λbn) = A + λB. if ρ (z) < 1 and diverges if ρ (z) > 1 and the test fails if ρ (z) = 1. n=1

Page: 64 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 7 Series and Sums in Banach Spaces 65 The easy proof of this proposition is left to the reader. Our next goal is to consider the multiplication of series. In this case we will need a multiplication ∞ on X itself. We start with the case where X = C or X = R. Let {an}n=0 , ∞ {bn}n=0 ⊂ C and let us consider the power series,

∞ ∞ X n X n A (z) := anz and B (z) := bnz . n=0 n=0 Working formally,

∞ ! ∞ ! ∞ X n X m X n m A (z) B (z) = anz bmz = anbmz z n=0 m=0 m,n=0 ∞ ∞ ! X X n m X X k = anbmz z = anbm z . k=0 m+n=k k=0 m+n=k

Thus we expect A (z) B (z) to be represented by a power series

∞ k X k X X C (z) = ckz where ck := anbm = ak−mbm. k=0 m+n=k m=0 PN Fig. 7.3. The red region represents the indices contributing to k=0 ck while the ∞ PN Theorem 7.20 (Multiplication of Series). Suppose that {an}n=0 , N × N square represents the indices contributing to m,n=0 ambn. The blue and ∞ P∞ P∞ green indices include all of the indices in the square minus those contributing to {bn}n=0 ⊂ C and n=0 an and n=0 bn are absolutely convergent series. Let ∞ PN ck. {ck}k=0 be deﬁned by k=0 k X ck := ak−mbm. (7.9) m=0 N N X X X X P∞ RN ≤ |anbm| + |anbm| Then k=0 ck converges absolutely and N n=0 N m=0 2 N , ≤ |bm| · A + |an| · B → 0 as N → ∞ and m> N n> N X 2 2 RN = |anbm| .

(m,n)∈ΓN because the tails of convergent series tend to zero. P∞ We now have the identity, I now claim that limN→∞ RN = 0. To see this let A := n=0 |an| ,B := P∞ m=0 |bm| and observe m + n > N implies either m > N/2 or n > N/2, see Figure 7.3. Hence we ﬁnd,

Page: 65 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 66 7 Series and Sums in Banach Spaces N N P∞ k X X X then k=0 ckz is convergent for |z| < R and an · bm = anbm ∞ ∞ ! ∞ ! n=0 m=0 (m,n)∈JN ×JN X X X c zk = a zn b zm . N k n m X X k=0 n=0 m=0 = anbm + rN = ck + rN , m+n≤N k=0 Proof. For |z| < R, the root test shows P ∞ ∞ where rN := (m,n)∈Γ anbm.Since |rN | ≤ RN → 0, it follows that X n X m N |anz | < ∞ and |bmz | < ∞. ∞ N " N N # n=0 m=0 X X X X n ck = lim ck = lim an · bm − rN The result now follows by applying Theorem 7.20 with an → |anz | and bn → N→∞ N→∞ n k=0 k=0 n=0 m=0 |bnz | . N N ∞ ! ∞ ! X X X X Example 7.22. For |z| < 1 we have = lim an · lim bm = an · bn . N→∞ N→∞ ∞ n=0 m=0 n=0 n=0 X (1 − z) · zn = 1. Lastly observe that n=0 k X P∞ k |ck| ≤ |ak−m| |bm| =:c ¯k This shows that power series, k=0 ckz , in Corollary 7.21 may in fact have m=0 radius of convergence larger than R in Eq. (7.12). and so applying what we have just proved to {|a |}∞ and {|b |}∞ shows n n=0 n n=0 Theorem 7.23. The function exp : C → C satisﬁes the following properties; ∞ ∞ ∞ ! ∞ ! X X X X 1. exp (0) = 1, |ck| ≤ c¯k = |an| · |bn| < ∞. 2. exp (w) exp (z) = exp (w + z) for all w, z ∈ C, n=0 n=0 −1 k=0 k=0 3. exp (w) is never 0 and [exp (w)] = exp (−w) for all w ∈ C. Here is the short summary of the argument; Proof. The ﬁrst assertion is clear and item 3. easily follows from item 2. with z = −w. We now prove item 2. by applying Theorem 7.20 with a := 1 zn n n! N N 1 n X X X and bn := n! w . In this case, an · bm − ck = an · bm

m,n=0 k=0 (m,n)∈ΓN k X 1 m 1 k−m X ck = z · w ≤ |an · bm| m! (k − m)! m=0 (m,n)∈Γ N k X X 1 X k 1 k ≤ |a · b | + |a · b | = zmwk−m = (z + w) n m n m k! m k! N N m=0 2 Binomial theorem for the last equality, see Exercise ≤ |bm| · A + |an| · B → 0 as N → ∞. N N 7.10. It now follows from Theorem 7.20 that m> 2 n> 2 ∞ ∞ X zn X wn exp (w) exp (z) = · n! n! ∞ ∞ P n=0 n=0 Corollary 7.21. Let {an}n=0 , {bn}n=0 ⊂ C and ck := m+n=k ambn. If ∞ X 1 k = (z + w) = exp (w + z) . ! k! 1 1 k=0 R := min , > 0, (7.11) 1/n 1/n lim supn→∞ |an| lim supn→∞ |bn|

Page: 66 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 7 Series and Sums in Banach Spaces 67 Deﬁnition 7.24 (Euler’s Number). Euler’s number is deﬁned by and

∞ ∞ sin (θ + α) = Im exp (i (θ + α)) = Im [exp (iθ) exp (iα)] X 1m X 1 e := exp (1) = = . m! m! = Im [(cos θ + i sin θ) · (cos α + i sin α)] m=0 m=0 = cos θ sin α + sin θ cos α. (7.14) If m ∈ , then N We also define m ∞ 1 1 ez − e−z X z2n+1 e = exp (1) = exp m = exp . sinh (z) = = and m m 2 (2n + 1)! n=0 ∞ As exp 1 > 0 we may conclude that exp 1 = e1/m. Now if n ∈ we find, ez + e−z X z2n m m N0 cosh (z) = = . 2 (2n)!  n times  n=0 n n z }| { The following identities are now easily verified; n h 1/mi 1  1 1  n e m = e = exp = exp  + ··· +  = exp . m m m m cos (−z) = cos (z) and cosh (−z) = cosh (z) sin (−z) = − sin (z) and sinh (−z) = − sinh (z) , As ez = cosh (z) + sinh (z) , n 1 1 n − m iz −iz e = n = n = exp − e − e 1 e m exp m sin (z) = = sinh (iz) , and m 2i i we have in fact shown that eq = exp (q) for all q ∈ Q. Owing to this identity, eiz + e−iz z cos (z) = = cosh (iz) . we will often write e for exp (z) . 2 With the definitions in Exercise 7.6, we have Exercise 7.7 (Addition Formulas). Prove the addition formulas in Eqs. ∞ n ∞ 2n ∞ 2n+1 (7.14) and (7.15) extend to α, θ ∈ , i.e. show for all w, z ∈ that X (iz) X (iz) X (iz) C C exp (iz) = = + n! (2n)! (2n + 1)! cos (w + z) = cos (w) cos (z) − sin (w) sin (z) and n=0 n=0 n=0 ∞ 2n ∞ 2n+1 sin (w + z) = cos (w) sin (z) + cos (z) sin (w) . X n z X n (z) = (−1) + i (−1) (2n)! (2n + 1)! Exercise 7.8. Suppose that p (t) and q (t) are non-zero polynomials of t ∈ n=0 n=0 R with (possibly) complex coefficients. Further assume that q (n) 6= 0 for all n ∈ = cos (z) + i sin (z) , ∞ N0. Show for any sequence {an}n=0 ⊂ C that the power series; ∞ ∞ i.e. X p (n) X a zn and a zn exp (iz) = cos (z) + i sin (z) for all z ∈ C. (7.12) q (n) n n n=0 n=0 This is called Euler’s formula. If z = θ ∈ , it states that R have the same radius of convergence. Re exp (iθ) = cos θ and Im exp (iθ) = sin θ. Theorem 7.25 (Hilbert Schmidt norm). Let Z = CJN ×JN denote the space N of N × N complex matrices, A = (Aij) with Aij ∈ C. We let Thus if α, θ ∈ R, we have the following addition formulas; i,j=1 v u N cos (θ + α) = Re exp (i (θ + α)) = Re [exp (iθ) exp (iα)] u X 2 kAk2 := t |Aij| = Re [(cos θ + i sin θ) · (cos α + i sin α)] i,j=1 = cos θ cos α − sin θ sin α (7.13) which is called the Hilbert Schmidt norm on Z. This norm satisfies,

Page: 67 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 68 7 Series and Sums in Banach Spaces √ 1. kIk2 = N where I is the N × N – identity matrix. and 2. kABk2 ≤ kAk2 · kBk2 for all A, B ∈ Z. n n lim (f + g)(xn) = lim [f (xn) + g (xn)] = f (x) + g (x) = (f + g)(x) . 3. kA k2 ≤ kAk2 for all n ∈ N. n→∞ n→∞ 4. If An → A and Bn → B then AnBn → AB and An + Bn → A + B as n → ∞. Thus matrix multiplication and addition are continuous operations This shows f · g and f + g are continuous.

on (Z, k·k2) . Item 6. follows from item 5 with X = Z along with an induction argument. N 2 5. If (X, d) is a metric space and f : X → Z and g : X → Z are continuous Item 7. follows from the fact that (Z, k·k2) is C with the 2 – norm in a functions then so is f · g (order matters) and f + g. slight disguise. 6. The functions f (A) = An is continuous on Z for all n ∈ , where by N0 P∞ n convention A0 = I. Definition 7.26 (Matrix Functions). If f (z) = n=0 cnz for some cn ∈ C and A ∈ CJN ×JN , then we define 7. (Z, k·k2) is a Banach space. ∞ Proof. Item 1. is clear. From the definition of matrix multiplication and the X n f (A) := cnA provided the sum is convergent. Cauchy Schwarz inequality we find n=0 2 2 N N N For example, X X 2 X 2 ∞ 2n+1 (AB) = AikBkj ≤ |Aik| · |Bkj| . ij X (−1) 2n+1 sin (A) := A . k=1 k=1 k=1 (2n + 1)! n=0

Therefore, ∞ Theorem 7.27. Suppose that {cn}n=1 is a sequence in C and n n N N ! −1 2 1/n JN ×JN 2 X X X 2 X 2 R := lim supn→∞ |cn| . If A ∈ C with kAk < R, then kABk2 = (AB)ij ≤ |Aik| · |Bkj| i,j=1 i,j=1 k=1 k=1 ∞ N N X n X 2 X 2 2 2 f (A) := cnA = |A | · |B | = kAk · kBk ik kj 2 2 n=0 i,k=1 j,k=1 JN ×JN is convergent and f : B0 (R) := {A : kAk < R} → C is continuous. which proves item 2. Item 3. follows by an easy induction argument. Moreover, for any ρ < R, the sum converges absolutely and uniformly on B0 (ρ) . Item 4. Let δAn := An − A and δBn := Bn − B so that An = A + δAn and Bn = B + δBn where kδAnk → 0 and kδBnk → 0 as n → ∞. Then n 1/n 2 2 Proof. Let ρ ∈ (kAk ,R) and let Mn := |cn| ρ . Then lim supn→∞ Mn = P∞ ρ/R < 1 and therefore by the Root test, n=0 Mn < ∞. Since, for n ≥ 1, kAnBn − ABk2 = k(A + δAn)(B + δBn) − ABk2 n n n = kδAnB + AδBn + δAnδBnk2 kcnA k = |cn| kA k ≤ |cn| kAk ≤ Mn, ≤ kδAnBk + kAδBnk + kδAnδBnk 2 2 2 the results are now again a consequence of the Weierstrass M – test in Theorem ≤ kδAnk2 kBk2 + kAk2 kδBnk2 + kδAnk2 kδBnk2 → 0 as n → ∞. 7.11.

Similarly, Theorem 7.28 (Matrix Exponentials and etc.). The series for eA, sin (A) , cos (A) , sinh (A) , and cosh (A) are all absolutely convergent deﬁne continuous kAn + Bn − (A + B)k = kδAn + δBnk → 0 as n → ∞. 2 2 functions from CJN ×JN to CJN ×JN in the Hilbert Schmidt norm. ∞ Item 5. Suppose that {xn}n=1 ⊂ X and xn → x as n → ∞, then using item 4., Proof. Since the corresponding numerical series all of inﬁnite radius of convergence the result follows from Theorem 7.27. lim (f · g)(xn) = lim [f (xn) g (xn)] = f (x) g (x) = (f · g)(x) n→∞ n→∞

Page: 68 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 7 Series and Sums in Banach Spaces 69 1 0 Exercise 7.9 (Inverting perturbations of the identity). For kAk2 < 1 in J 2 = −I where I = , CJN ×JN ,I − A is invertible and 0 1 n n n ∞ it follows that J 2n = (−I) = (−1) I and J 2n+1 = J 2nJ = (−1) J for all −1 X n (I − A) = A n ∈ N0. Therefore, n=0 ∞ X θn −1 eA+B = eθJ = J n where the sum is absolutely convergent. Moreover the function A → (I − A) n! is continuous on the ball, B := {A : kAk < 1} and n=0 2 ∞ ∞ X 1 2n 2n X 1 2n+1 2n+1 ∞ = θ J + θ J −1 X kAk √ (2n)! (2n + 1)! n 2 n=0 n=0 (I − A) ≤ kA k2 ≤ + N. 2 1 − kAk n=0 2 ∞ ! ∞ ! X 1 n X 1 n = θ2n (−1) · I + θ2n+1 (−1) J J ×J (2n)! (2n + 1)! Exercise 7.10 (Binomial Theorem). Suppose that A, B ∈ C N N are n=0 n=0 commuting matrices, i.e. AB = BA. For any n ∈ N, show by induction that cos θ sin θ = cos θ · I + sin θ · J = . − sin θ cos θ n X n n AkBn−k = (A + B) . (7.15) k From these formula it is evident (using facts that you know about sin θ and k=0 cos θ) that eAeB 6= eA+B unless θ = 0. There is no contradiction to Exercise Exercise 7.11 (Matrix Addition Formula). If A, B ∈ CJN ×JN are com- 7.11 since muting matrices, show exp (A + B) = exp (A) · exp (B) . Hint: mimic the proof 0 θ 0 0 0 0 0 θ of Theorem 7.23. [A, B] = − 0 0 −θ 0 −θ 0 0 0 Example 7.29( exp (A + B) 6= exp (A) · exp (B)). Let θ ∈ and −θ2 0 −1 0 R = = −θ2 0 θ2 0 1 0 θ 0 0 A = and B = . 0 0 −θ 0 which is zero iﬀ θ = 0.

0 0 Exercise 7.12. Let (X, k·k) be a normed space and T : X → X be a linear Since A2 = = B2 it follows that 0 0 transformation and suppose T is continuous at 0 ∈ X. Show; 1. There exists C < ∞ such that kT (x)k ≤ C kxk for all x ∈ X. Hint: for A 1 θ B 1 0 e = I + A = and e = I + B = sake of contradiction suppose that no such C < ∞ exists. Then construct 0 1 −θ 1 ∞ a sequence {yn}n=1 ⊂ X such that limn→∞ yn = 0 while kT (yn)k = 1 for and in particular, all n. 2. Show T is continuous on all of X. 1 θ 1 0 1 − θ2 θ eAeB = = . 0 1 −θ 1 −θ 1

Let us now compute eA+B. First observe that

0 1 A + B = θJ where J := . −1 0

Since

Page: 69 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14

8 More Sums and Sequences

Warning: this chapter is a bit rough and will be edited later when it be- Corollary 8.3. Again suppose that Λ is a countable set and a : Λ → [0, ∞] is comes needed. a function. If Λn ⊂ Λ and Λn ↑ Λ, then X X a (x) = lim a (x) . n→∞ 8.1 Rearrangements Λ Λn

Proof. Apply Theorem 8.2 with an (x) := a (x) 1Λn (x) . [The stuﬀ about sums of positive numbers could go much earlier in fact.] Corollary 8.4. If Λ = A ∪ B with A ∩ B = ∅, then Deﬁnition 8.1. If Λ is a countable set and a : Λ → [0, ∞] is a function, let X X X a (x) = a (x) + a (x) . X X x∈Λ x∈A x∈B a (x) := sup a (x) . Λ0⊂f Λ x∈Λ x∈Λ0 Proof. Choose Λn ⊂f Λ such that Λn ↑ Λ. Then Λn ∩A ↑ A and Λn ∩B ↑ B and hence, Notice that if 0 ≤ a (x) ≤ b (x) , then P a (x) ≤ P b (x) . Moreover x∈Λ x∈Λ X X if B ⊂ Λ, then a (x) = lim a (x) X X n→∞ a (x) = a (x) 1B (x) . x∈Λ x∈Λn " # x∈B x∈Λ X X = lim a (x) + a (x) Theorem 8.2 (Monotone Convergence Theorem for Sums). If Λ is a n→∞ x∈Λn∩A x∈Λn∩B countable set and an : Λ → [0, ∞] is an increasing sequence of functions, then X X = lim a (x) + lim a (x) n→∞ n→∞ X X x∈Λn∩A x∈Λn∩B lim an (x) := lim an (x) . n→∞ n→∞ X X x∈Λ x∈Λ = a (x) + a (x) . x∈A x∈B Proof. This is an easy consequence of the sup – sup theorem. Indeed, P x∈Λ an (x) is an increasing sequence of numbers, therefore, Corollary 8.5. If Λ is a countable set, a : Λ → [0, ∞] is a function, and X X X P∞ lim an (x) = sup an (x) = sup sup an (x) Λ = Λn, then n→∞ n=1 n n Λ0⊂f Λ ∞ x∈Λ x∈Λ x∈Λ0 X X X X X a (x) = a (x) = sup sup an (x) = sup lim an (x) n→∞ Λ n=1 Λn Λ0⊂f Λ n Λ0⊂f Λ x∈Λ0 x∈Λ0 X X Proof. We have = sup lim an (x) = lim an (x) . n→∞ n→∞ Λ0⊂f Λ ∞ N x∈Λ0 x∈Λ X X X X X X a (x) = lim a (x) = lim a (x) = a (x) . N→∞ N→∞ n=1 Λ n=1 Λ N x∈Λ n n ∪n=1Λn 72 8 More Sums and Sequences ∞ Corollary 8.6. If {xn}n=1 is any enumeration of Λ then Theorem 8.9 (Dominated Convergence Theorem for Sums I). If Λ is a countable set and an : Λ → [0, ∞) is a sequence of functions such that ∞ P X X sup an (x) < ∞ and limn→∞ an (x) = 0 for all x ∈ Λ, then a (x) = a (xn) . x∈Λ n x∈Λ n=1 X lim an (x) := 0. P∞ n→∞ Proof. Λ = n=1 {xn} and therefore x∈Λ ∞ ∞ P X X X X Proof. For all ε > 0 there Λε ⊂f Λ such that sup an (x) ≤ ε. a (x) = a (x) = a (x ) . x/∈Λε n n Therefore, x∈Λ n=1 x∈{xn} n=1 X X X an (x) = an (x) + an (x) x∈Λ x∈Λ x/∈Λ Corollary 8.7 (Tonelli Theorem for Sums). Suppose that Λ is a countable ε ε X set and an : Λ → [0, ∞] is a sequence of functions, then ≤ an (x) + ε x∈Λ ∞ ∞ ε X X X X an (x) = an (x) . and therefore, n=1 x∈Λ x∈Λ n=1 X X X Proof. This is a simple consequence of MCT, lim sup an (x) ≤ lim sup an (x) + ε = lim an (x) + ε = ε. n→∞ n→∞ n→∞ x∈Λ x∈Λε x∈Λε ∞ N N X X X X X X an (x) = lim an (x) = lim an (x) Since ε > 0 is arbitrary, it follows that N→∞ N→∞ n=1 x∈Λ n=1 x∈Λ x∈Λ n=1 X N ∞ lim sup an (x) = 0. n→∞ X X X X x∈Λ = lim an (x) = an (x) . N→∞ x∈Λ n=1 x∈Λ n=1 Alternative Proof. Let α (x) := supn an (x) , then α (x) − an (x) ≥ 0 for all x ∈ Λ. Therefore by Fatou’s Lemma 8.8,

Lemma 8.8 (Fatou’s Lemma). If Λ is a countable set and an : Λ → [0, ∞] X X lim inf [α (x) − an (x)] ≤ lim inf [α (x) − an (x)] is a sequence of functions, then n→∞ n→∞ x∈Λ x∈Λ X X " # lim inf an (x) ≤ lim inf an (x) . n→∞ n→∞ X X = α (x) + lim inf − an (x) x∈Λ x∈Λ n→∞ x∈Λ x∈Λ Proof. By deﬁnition and MCT, " # X X X X = α (x) − lim sup an (x) lim inf an (x) = lim inf ak (x) n→∞ n→∞ n→∞ k≥n x∈Λ x∈Λ x∈Λ x∈Λ X while similarly, = lim inf ak (x) n→∞ k≥n x∈Λ X X h i X lim inf [α (x) − an (x)] = α (x) + lim inf [−an (x)] = lim inf inf ak (x) n→∞ n→∞ n→∞ k≥n x∈Λ x∈Λ x∈Λ X X X ≤ lim inf a (x) . = α (x) − lim sup an (x) = α (x) . n n→∞ n→∞ x∈Λ x∈Λ x∈Λ This then implies that

Page: 72 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.1 Rearrangements 73 " # X X X 2. Choose Λn ⊂f Λ such that Λn ↑ Λ and set an (x) = a (x) 1Λn (x) for all α (x) ≤ α (x) − lim sup an (x) n→∞ n ∈ N and x ∈ Λ. Then x∈Λ x∈Λ x∈Λ and hence that lim ka − ank = 0 by DCT. " # n→∞ 1 X lim sup an (x) ≤ 0 n→∞ 3. This is a consequence of the BLT theorem. x∈Λ as before. Now suppose that (Z, k·k) is a Banach space and a : Λ → Z is a function P ∞ 1 such that x∈Λ ka (x)k < ∞. Theorem 8.11 (DCT II). Suppose that {an}n=1 ⊂ ` (Λ, Z) and A (x) := Theorem 8.10. Let `1 (Λ, Z) be the space of functions a : Λ → Z such that supn kan (x)k satisfy; X kak := ka (x)k < ∞. 1. a (x) := limn→∞ an (x) existing in Z for all x ∈ Λ, [Conflict of notation]and 1 P x∈Λ 2. x∈Λ A (x) < ∞, Then; then X X 1 lim ka − ank = 0 and lim an (x) = a (x) . 1. ` (Λ, Z) , k·k1 is a Banach space. n→∞ 1 n→∞ 2. The set D = {a : Λ → Z :#({x ∈ Λ : a (x) 6= 0} < ∞)} is dense subspace x∈Λ x∈Λ 1 of ` (Λ, Z) . Proof. It suffices to prove the first assertion since the sum operation is 3. For a ∈ D we may define 1 continuous relative to the k·k1 – norm on ` (Λ, Z) . Since X X a (x) = a (x) ∈ Z. ka (x)k = lim kan (x)k ≤ A (x) x∈Λ x:a(x)6=0 n→∞ Since it follows that X X a (x) ≤ ka (x)k = kak , 1 ka (x) − a (x)k ≤ ka (x)k + ka (x)k ≤ 2A (x) . x∈Λ x∈Λ n n this linear operator has a unique continuous extension to a linear operator Since Σ : `1 (Λ, Z) → Z.

Proof. We take each item in turn. lim ka (x) − an (x)k = a (x) − lim an (x) = ka (x) − a (x)k = 0 for all x ∈ Λ, n→∞ n→∞ 1. Let {a }∞ be a Cauchy sequence in `1 (Λ, Z) . Since n n=1 we may use the dominated convergence theorem for sums (Theorem 8.9) to kan (x) − am (x)k ≤ kan − amk1 → 0 as m, n → ∞, conclude, it follows that limn→∞ an (x) =: a (x) exists for all x ∈ Λ. Moreover, using X X lim ka − ank = lim ka (x) − an (x)k = lim ka (x) − an (x)k = 0. Fatou’s Lemma 8.8, n→∞ 1 n→∞ n→∞ X x∈Λ x∈Λ ka − ank1 = ka (x) − an (x)k x∈Λ X 1 = lim inf kam (x) − an (x)k Corollary 8.12. Suppose (Z, k·k) and a ∈ ` (Λ, Z) . If Λn ↑ Λ, then m→∞ x∈Λ X X X lim a (x) = a (x) . ≤ lim inf kam (x) − an (x)k n→∞ m→∞ x∈Λ x∈Λ x∈Λ n

= lim inf kam − ank → 0 as n → ∞. m→∞ 1

Page: 73 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 74 8 More Sums and Sequences

Proof. Let an (x) := 1Λn (x) a (x) for all n. Then kan (x)k ≤ ka (x)k and Proof. First observe that P ka (x)k < ∞. Since lim a (x) = a (x) for all x ∈ Λ, it follows by x∈Λ n→∞ n N N ∞ DCT that X X X 1 an (x) ≤ kan (x)k ≤ kan (x)k ∈ ` (Λ, Z) X X X X n=1 n=1 n=1 lim a (x) = lim an (x) = lim an (x) = a (x) . n→∞ n→∞ n→∞ PN P∞ x∈Λn x∈Λ x∈Λ x∈Λ and limN→∞ n=1 an (x) = n=1 an (x) . Therefore by DCT, ∞ N N X X X X X X an (x) = lim an (x) = lim an (x) N→∞ N→∞ Corollary 8.13. Suppose (Z, k·k) is a Banach space, Λ is a countable set, and n=1 x∈Λ n=1 x∈Λ x∈Λ n=1 1 P∞ a ∈ ` (Λ, Z) . If Λ = Λn, then N ∞ n=1 X X X X ∞ = lim an (x) = an (x) . X X X N→∞ a (x) = a (x) . x∈Λ n=1 x∈Λ n=1

x∈Λ n=1 x∈Λn Better proof. By assumption, ∞ ∞ [We permit Λn = ∅ for some n in this result with the convention that X X 1 P kank1 < ∞ =⇒ an converges in ` (Λ, Z) . x∈∅ a (x) = 0.] n=1 n=1 Proof. We have Therefore, ∞ ! ∞ ∞ N X X X X X X X X X X a = a . a (x) = lim a (x) = lim a (x) = a (x) . n n N→∞ N→∞ Λ n=1 n=1 Λ n=1 Λ n=1 Λ N x∈Λ n n ∪n=1Λn

N because ∪n=1Λn ↑ Λ as N → ∞. [Bruce: we need to prove the finite additivity Exercise 8.1 (Differentiating past an infinite sum). Suppose that here, namely that if Λ = A ∪ B, then ∞ ∞ X 0 X X X X sup |fn (x)| < ∞ and fn (0) exists. x a (x) = a (x) + a (x) . n=0 n=0 x∈Λ x∈A x∈B P∞ Then S (x) := n=0 fn (x) exists and But this is simple, since a = 1 a + 1 a and A B ∞ 0 X 0 X X S (x) = fn (x) . (8.1) 1Aa = a n=0 x∈Λ x∈A Exercise 8.2 (Differentiating past a limit). Suppose limn→∞ fn (x) = where this last equality follow by choosing Λn ⊂f Λ such that Λn ↑ Λ so that 0 0 f (x) and fn → g uniformly. Show f (x) = g (x) , i.e. we have in this case X X X X that 1Aa = lim 1Aa = lim a = a. d d n→∞ n→∞ lim fn (x) = lim fn (x) . x∈Λ x∈Λn x∈Λn∩A x∈A dx n→∞ n→∞ dx Exercise 8.3. Suppose that ∞ ∞ Corollary 8.14. If P P ka (x)k < ∞, then X n n=1 x∈Λ n f (z) := anz ∞ ∞ n=0 X X X X a (x) = a (x) ∈ Z. 0 P∞ n n n has radius of convergence R. Show f (z) = n=0 nanz for all |z| < R and the n=1 x∈Λ x∈Λ n=1 radius of convergence of f 0 is still R.

Page: 74 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.2 Double Sequences 75 ∞ Theorem 8.15 (Changing the center of a power series). Suppose that Definition 8.16. Suppose that {Sm,n}m,n=1 is a sequence of complex numbers P∞ an n A (z) = n=0 n! z has radius of convergence R > 0. For |z| < R and |h| < (or more generally elements of a metric space). We say limm∧n→∞ Sm,n = L R − |z| we have iff for all ε > 0 there exists N ∈ such that ∞ N X 1 A (z + h) = a (z) hm m! m |L − Sm,n| ≤ ε for all m ∧ n ≥ N. m=0 where We say limm∨n→∞ Sm,n = L exists iff for all ε > 0 there exists N ∈ such ∞ N X an n−m (m) that am (z) := z = A (z) . (n − m)! |L − Sm,n| ≤ ε for all m ∨ n ≥ N. n=m

Proof. Working formally for the moment, Clearly limm∨n→∞ Sm,n = L implies limm∧n→∞ Sm,n = L but the converse ∞ ∞ n is not true. We are mostly interested in ﬁnding suﬃcient conditions in order for X an n X an X n A (z + h) = (z + h) = zn−mhm iterated limits to be equal, i.e. for n! n! m n=0 n=0 m=0 X an lim lim Sm,n = lim lim Sm,n. = zn−mhm m→∞ n→∞ n→∞ m→∞ m! · (n − m)! 0≤m≤n<∞ 1 Example 8.17 (Switching Limits is Dangerous I). If Sm,n = m , then ∞ " ∞ # 1+ n X 1 X an = zn−m hm limm→∞ Sm,n = 0 (but not uniformly in n) so that limn→∞ limm→∞ Sm,n = 0 m! (n − m)! m=0 n=m while limn→∞ Sm,n = 1 and limm→∞ limn→∞ Sm,n = 1. In order to visualize 1 better what is going on here let us make the change of variables, x = m and which gives the result provided the interchange of order of sums was permissible. 1 y = n , i.e. let To check this we consider, 1 x S (x, y) = S = = m,n 1 + y x + y X an n−m m X |an| n−m m x z h = |z| |h| m! · (n − m)! m! · (n − m)! 0≤m≤n<∞ 0≤m≤n<∞ whose plot appears in Figure 8.17. ∞ X |an| n = (|z| + |h|) < ∞ n! n=0

P∞ |an| n as we know for all 0 ≤ ρ < R that n=0 n! ρ < ∞. Hence we may apply Fubini’s theorem for sums in order to conclude the result.

8.2 Double Sequences

∞ In this chapter we will consider doubly indexed sequences, {Sm,n}m,n=1 , of 1 ∞ complex numbers. To be more precise {Sm,n}m,n=1 is simply a function from N2 to C. In this chapter we are interested in the following limits;

lim lim Sm,n, lim lim Sm,n, lim Sm,n, and lim Sm,n, m→∞ n→∞ n→∞ m→∞ m∨n→∞ m∧n→∞ where the last two limits are deﬁned as follows. 1 Much of what we will say holds for sequences taking values in “complete metric spaces” to be covered later.

Page: 75 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 76 8 More Sums and Sequences

cos(πm) 1 1 A plot of Sm,n = − m in tems of the variables and . 1+ n m n Example 8.19 (Switching Limits is Dangerous III). If 1 1 1 A plot of Sm,n = 1+ m in tems of the variables m and n . n (−1)n (−1)m cos (nπ) cos (mπ) With this change of variables, m → ∞ ⇐⇒ x → 0 and n → ∞ ⇐⇒ y → 0 Sm,n := + = − − m n m n and m = ∞ and n = ∞ correspond to x = 0 and y = 0 respectively. It is now quite clearly that lim A (x, y) = 0 for all y > 0 and lim A (x, y) = 1 for 1 1 x→0 y→0 and we let x = m and y = n . Then all x > 1. S (x, y) = S = −x cos (π/y) − y cos (π/x) Example 8.18 (Switching Limits is Dangerous II). Let m,n m+1 whose plot appears in Figure 8.19. (−1) cos (πm) Sm,n := m = − m . 1 + n 1 + n

Then limm→∞ Sm,n = 0 (but not uniformly in n) so that m limn→∞ limm→∞ Sm,n = 0. We also have limn→∞ Sm,n = (−1) and m limm→∞ limn→∞ Sm,n = limm→∞ (−1) does not exists. In order to visualize 1 better what is going on here let us again make the change of variables, x = m 1 and y = n , i.e. let x π S (x, y) = S = − cos m,n x + y x whose plot appears in Figure 8.18.

Page: 76 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.2 Double Sequences 77

cos(nπ) cos(mπ) 1 1 1 1 1 A plot of − − in tems of the variables and . A plot of 1+ 1 + 1 in tems of the variables m and n . m n m n n m2 In this case, limm∧n→∞ Sm,n = 0 exists but limm→∞ Sm,n and limn→∞ Sm,n This example is covered by Theorem 8.22 below. do not exist! Theorem 8.22 (Equality of monotone iterated limits). Suppose that Remark 8.20. However, if limm∧n→∞ Sm,n exists then we do always have, Sm,n ≥ 0 for all m, n ∈ N and Sm+1,n ≥ Sm,n and Sm,n+1 ≥ Sm,n for all lim Sm,n = lim lim sup Sm,n = lim lim inf Sm,n. m, n ∈ N. Then m∧n→∞ m→∞ n→∞ m→∞ n→∞

On the other hand if limm∨n→∞ Sm,n exists then we will have L := lim lim Sm,n = lim lim Sm,n = sup Sm,n (8.2) n→∞ m→∞ m→∞ n→∞ (m,n) lim Sm,n = lim lim Sm,n = lim lim Sm,n. m∨n→∞ m→∞ n→∞ n→∞ m→∞ where all limits exist but may take on the value inﬁnity. Moreover One way to avoid these types of behaviors is to assume Sm,n ≥ 0 and is limm∧n→∞ Sm,n = L. increasing in each index. Proof. Because Sm,n is increasing in each of its variables, we ﬁnd, Example 8.21 (Monotinicity is good). If lim lim Sm,n = sup sup Sm,n and lim lim Sm,n = sup sup Sm,n 1 n→∞ m→∞ n m m→∞ n→∞ m n Sm,n := 1 1 1 + + 2 n m and therefore Eq. (8.2) follows from Corollary 8.2 of the sup-sup Theorem. So 1 1 and we let x = m and y = n . Then it only remains to show limm∧n→∞ Sm,n = L. 1 Cases 1. If L = ∞, then for any N ∈ N, there exists (mN , nN ) such that S (x, y) = Sm,n = S ≥ N and since S is increasing in each of its variables it follows that 1 + y + x2 mN ,nN m,n whose plot appears in Figure 8.21. Sm,n ≥ N for all m ≥ mN and n ≥ nN

Page: 77 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 78 8 More Sums and Sequences and it follows that limm∧n→∞ Sm,n = ∞. Proof. Let ε > 0 be given. Choose N ∈ N such that Case 2. L < ∞. In this case if ε > 0 is given, there exists (mε, nε) such that Sm ,n ≥ L − ε. Since Sm,n is increasing in each of its variables it follows that sup |Sm,∞ − Sm,n| ≤ ε for all n ≥ N. ε ε m

L ≥ Sm,n ≥ L − ε for all m ≥ mε and n ≥ nε Now choose M ∈ N such that and so |L − Sm,n| ≤ ε for all m, n ≥ mε∨nε. So by deﬁnition, limm∧n→∞ Sm,n = L. |S∞,N − Sm,N | ≤ ε for all m ≥ M. ∞ Exercise 8.4. If {fn (x)}n=1 is a sequence of increasing continuous functions Then for n ≥ N and m ≥ M we have, on R such that fn (x) ↑ f (x) < ∞ as n → ∞, then f (x) is continuous and increasing as well. |Sm,n − S∞,N | ≤ |Sm,n − Sm,∞| + |Sm,∞ − S∞,N | ≤ |Sm,n − Sm,∞| + |Sm,∞ − Sm,N | + |Sm,N − S∞,N | Exercise 8.5. Show limm∧n→∞ Sm,n = L iﬀ for all ε > 0 there exists M,N ∈ N such ≤ ε + ε + ε = 3ε. |L − Sm,n| ≤ ε for all m ≥ M and n ≥ N. Therefore it follows that ∞ Lemma 8.23. Suppose that {Sm,n}m,n=1 is Cauchy in the sense that for all 0 0 ε > 0 there exists M,N ∈ N such that |Sm,n − Sm0,n0 | ≤ 6ε for all m, m ≥ M and n, n ≥ N. 0 0 ∞ |Sm,n − Sm0,n0 | ≤ ε for all m, m ≥ M and n, n ≥ N. (8.3) Hence {Sm,n}m,n=1 is Cauchy and therefore by Lemma 8.23 we know that L := limm∧n→∞ Sm,n exists, i.e. for all ε > 0 there exists M,N ∈ N such that Then limm∧n→∞ Sm,n = L exists. ∞ |S − L| ≤ ε for all m ≥ M and n ≥ N. Proof. Let sn := Sn,n. Then the assumption shows that {sn}n=1 is a Cauchy m,n 0 0 sequence and hence convergent. Let L := limn→∞ sn. Now take m = n and then let n0 → ∞ in Eq. (8.3) in order to learn, Letting m → ∞ above then shows,

|Sm,n − L| ≤ ε for all m ≥ M and n ≥ N. |S∞,n − L| ≤ ε for all and n ≥ N

From this we conclude that limm∧n→∞ Sm,n = L. and therefore limn→∞ S∞,n = L. Similarly one shows limm→∞ Sm,∞ = L as Another way to ensure that the iterated limits are equal is to assume some well. uniformity in one of the limits as in the next key theorem. [This theorem will Metric space proof. Let ε > 0 be given. Choose N ∈ N such that be used in one guise or another repeatedly throughout these notes.] ∞ sup ρ (Sm,∞,Sm,n) ≤ ε for all n ≥ N. Theorem 8.24. Suppose that {Sm,n}m,n=1 is a sequence of complex numbers m (or more generally elements of a complete metric space (X, ρ)). Assume that Now choose M ∈ N such that Sm,∞ := lim Sm,n exists uniformly in m and n→∞ ρ (S∞,N ,Sm,N ) ≤ ε for all m ≥ M. S∞,n := lim Sm,n exists pointwise in n. m→∞ Then for n ≥ N and m ≥ M we have, Then limm→∞ limn→∞ Sm,n, limn→∞ limm→∞ Sm,n, and limm∧n→∞ Sm,n all exist and are equal, i.e. ρ (Sm,n,S∞,N ) ≤ ρ (Sm,n,Sm,∞) + ρ (Sm,∞,S∞,N )

≤ ρ (Sm,n,Sm,∞) + ρ (Sm,∞,Sm,N ) + ρ (Sm,N ,S∞,N ) L := lim lim Sm,n = lim lim Sm,n = lim Sm,n. m→∞ n→∞ n→∞ m→∞ m∧n→∞ ≤ ε + ε + ε = 3ε. [In words, the existence of limits in both variables along with uniformity in one of the variables implies the iterated limits exists and agree.] Therefore it follows that

Page: 78 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.3 Iterated Limits 79

then limm∧n→∞ am,n = L and in particular,

lim lim am,n = lim lim am,n. m→∞ n→∞ n→∞ m→∞ ∼ Idea.. The ﬁrst assumption guarantees the rows of am,· = A· for large ∼ m. The second assertion say that An = L for large n. Thus we must have ∼ ∼ am,n = An = L for large m and n. ∞ Theorem 8.27. Let {am,n}m,n=1 be a sequence of complex numbers and assume that

lim am,n = an exists uniformly in n m→∞

lim am,n = bm exists n→∞

lim bm = B exists . m→∞

Then, limn→∞ an = B. In other words under the above conditions,

lim lim am,n = lim lim am,n. m→∞ n→∞ n→∞ m→∞

Moreover limm∧n→∞ am,n = B as well. ∼ Idea. As above we know that am,· = a· for m large. We are also given that 0 0 ∼ ∼ ∼ ρ (Sm,n,Sm0,n0 ) ≤ 6ε for all m, m ≥ M and n, n ≥ N. limn→∞ am,n = B for m large. Thus B = limn→∞ am,n = limn→∞ an for m large. Hence we may now apply the previous theorem. ∞ Hence {Sm,n}m,n=1 is Cauchy and therefore by Lemma 8.23 we know that [Details] We have L := limm∧n→∞ Sm,n exists, i.e. for all ε > 0 there exists M,N ∈ N such that |B − an| 6 |B − bm| + |bm − am,n| + |am,n − an| ρ (S ,L) ≤ ε for all m ≥ M and n ≥ N. m,n 6 |B − bm| + |bm − am,n| + sup |am,k − ak| k Letting m → ∞ above then shows, and then taking lim supn→∞ of this inequality shows, ρ (S∞,n,L) ≤ ε for all and n ≥ N lim sup |B − an| 6 |B − bm| + sup |am,k − ak| → 0 as m → ∞. n→∞ k and therefore limn→∞ S∞,n = L. Similarly one shows limm→∞ Sm,∞ = L as well. We now prove the second assertion. For this we have, |B − a | |B − a | + |a − a | Remark 8.25. The previous theorem is rather easy to understand intuitively as m,n 6 n n m,n the following picture indicates the strategy of the proof. ≤ |B − an| + sup |ak − am,k| . k Thus given ε > 0 there exists M ∈ N and N ∈ N such that 8.3 Iterated Limits |B − am,n| 6 |B − an| + sup |ak − am,k| ≤ ε + ε k ∞ Theorem 8.26. Let {am,n}m,n=1 be a sequence of complex numbers and as- for n ≥ N and m ≥ M. This proves the stronger statement that sume that limm∧n→∞ am,n = B, i.e. am,n is near B as long as both m, n are suﬃciently large. lim am,n = An exists uniformly in n m→∞ Exercise 8.6. Suppose that fn → f uniformly and fn is continuous for all n, lim An = L exists n→∞ then f is continuous. [Use sequential notion of continuity here.]

Page: 79 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 80 8 More Sums and Sequences

8.4 Double Sums and Continuity of Sums ∞ ∞ X X sup Sm,n − ak,n ≤ Mk → 0 as m → ∞. n Here are a couple of very useful consequences of these theorems. k=1 k=m+1 Therefore S → P∞ a uniformly in n as m → ∞. Moreover, Theorem 8.28 (Monotone convergence theorem for sums). Let m,n k=1 k,n ∞ {a } be a sequence of non-negative numbers, assume that a ≥ a m m ∞ k,n k,n=1 k,n+1 k,n X X X for all k, n. Then lim Sm,n = lim ak,n = ak → ak as m → ∞. n→∞ n→∞ k=1 k=1 k=1 ∞ ∞ m X X X lim ak,n = lim ak,n = lim ak,n. Thus the hypothesis of Theorem 8.27 are satisﬁed and so we may conclude, n→∞ n→∞ m∧n→∞ k=1 k=1 k=1 ∞ m ∞ X Proof. Let S := P a , then {S } satisﬁes the hypothesis of lim lim Sm,n = lim lim Sm,n = lim Sm,n = ak m,n k=1 k,n m,n m,n=1 m→∞ n→∞ n→∞ m→∞ m∧n→∞ Theorem 8.22 and the conclusions now follows from that Theorem upon noting k=1 that which is exactly Eq. (8.4). ∞ ∞ X X ∞ lim ak,n = lim lim Sm,n and lim ak,n = lim lim Sm,n. Theorem 8.31 (Fubini’s Theorem for sums). Let {ak,l}k,l=1 be any se- n→∞ n→∞ m→∞ n→∞ m→∞ n→∞ P∞ k=1 k=1 quence of complex numbers. If k,l=1 |ak,l| < ∞, then

∞ ∞ ∞ ∞ ∞ X X X X X a = a = a . Theorem 8.29 (Tonelli’s Theorem for sums). Let {a }∞ be any se- k,l k,l k,l k,l k,l=1 k=1 l=1 k,l=1 l=1 k=1 quence of non-negative numbers, then

∞ ∞ ∞ ∞ X X X X ak,l = ak,l. 8.5 Sums of positive functions k=1 l=1 l=1 k=1 In this and the next few sections, let X and Y be two sets. We will write α ⊂⊂ X Proof. Apply Theorem 8.22 with S := Pm Pn a . X m,n k=1 l=1 k,l to denote that α is a finite subset of X and write 2f for those α ⊂⊂ X. Theorem 8.30 (Domintated convergence theorem for sums). Let Definition 8.32. Suppose that a : X → [0, ∞] is a function and F ⊂ X is a ∞ {ak,n}k,n=1 be a sequence of complex numbers such that limn→∞ ak,n = ak subset, then exists for all n and there exists a summable dominating sequence, {M } , such k ( ) that |ak,n| ≤ Mk for all k, n ∈ . Then X X X N a = a (x) := sup a (x): α ⊂⊂ F . ∞ ∞ m x∈α X X X F x∈F lim akn = lim akn = lim ak,n. (8.4) n→∞ n→∞ m∧n→∞ k=1 k=1 k=1 Remark 8.33. Suppose that X = N = {1, 2, 3,... } and a : X → [0, ∞], then Pm ∞ ∞ N Proof. Let Sm,n := ak,n, then {Sm,n} satisfies the hypothesis X X X k=1 m,n=1 a = a(n) := lim a(n). of Theorem 8.27. Indeed, N→∞ N n=1 n=1 ∞ ∞ ∞ ∞ X X X X PN P S − a = a ≤ |a | ≤ M Indeed for all N, a(n) ≤ a, and thus passing to the limit we learn m,n k,n k,n k,n k n=1 N that k=1 k=m+1 k=m+1 k=m+1 ∞ X X and hence, a(n) ≤ a. n=1 N

Page: 80 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.5 Sums of positive functions 81

Conversely, if α ⊂⊂ N, then for all N large enough so that α ⊂ {1, 2,...,N}, Lemma 8.36. Let X and Y be sets, R ⊂ X × Y and suppose that a : R → R¯ P PN is a function. Let R := {y ∈ Y :(x, y) ∈ R} and R := {x ∈ X :(x, y) ∈ R} . we have α a ≤ n=1 a(n) which upon passing to the limit implies that x y Then ∞ X X a ≤ a(n). sup a(x, y) = sup sup a(x, y) = sup sup a(x, y) and α n=1 (x,y)∈R x∈X y∈xR y∈Y x∈Ry Taking the supremum over α in the previous equation shows inf a(x, y) = inf inf a(x, y) = inf inf a(x, y). (x,y)∈R x∈X y∈xR y∈Y x∈Ry ∞ X X a ≤ a(n). (Recall the conventions: sup ∅ = −∞ and inf ∅ = +∞.) N n=1 P Proof. Let M = sup a(x, y),N := sup a(x, y). Then a(x, y) ≤ Remark 8.34. Suppose a : X → [0, ∞] and a < ∞, then {x ∈ X : a (x) > 0} (x,y)∈R x y∈xR X M for all (x, y) ∈ R implies N = sup a(x, y) ≤ M and therefore that is at most countable. To see this first notice that for any ε > 0, the set {x : x y∈xR P a (x) ≥ ε} must be finite for otherwise X a = ∞. Thus sup sup a(x, y) = sup Nx ≤ M. (8.5) [ ∞ x∈X y∈xR x∈X {x ∈ X : a (x) > 0} = k=1{x : a (x) ≥ 1/k} Similarly for any (x, y) ∈ R, which shows that {x ∈ X : a (x) > 0} is a countable union of finite sets and thus countable by Lemma ??. a(x, y) ≤ Nx ≤ sup Nx = sup sup a(x, y) Lemma 8.35. Suppose that a, b : X → [0, ∞] are two functions, then x∈X x∈X y∈xR X X X and therefore (a + b) = a + b and M = sup a(x, y) ≤ sup sup a(x, y) (8.6) X X X (x,y)∈R x∈X y∈xR X X λa = λ a Equations (8.5) and (8.6) show that X X for all λ ≥ 0. sup a(x, y) = sup sup a(x, y). (x,y)∈R x∈X y∈xR I will only prove the first assertion, the second being easy. Let α ⊂⊂ X be a finite set, then The assertions involving infimums are proved analogously or follow from what X X X X X we have just proved applied to the function −a. (a + b) = a + b ≤ a + b α α α X X Theorem 8.37 (Monotone Convergence Theorem for Sums). Suppose which after taking sups over α shows that that fn : X → [0, ∞] is an increasing sequence of functions and X X X (a + b) ≤ a + b. f (x) := lim fn (x) = sup fn (x) . n→∞ n X X X Then Similarly, if α, β ⊂⊂ X, then X X lim fn = f X X X X X X n→∞ a + b ≤ a + b = (a + b) ≤ (a + b). X X α β α∪β α∪β α∪β X Proof. We will give two proofs. Taking sups over α and β then shows that First proof. Let X X X X 2f := {A ⊂ X : A ⊂⊂ X}. a + b ≤ (a + b). X X X Then

Page: 81 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 82 8 More Sums and Sequences

Lemma 8.38 (Fatou’s Lemma for Sums). Suppose that fn : X → [0, ∞] is a sequence of functions, then X X lim inf fn ≤ lim inf fn. n→∞ n→∞ X X

Proof. Define gk := inf fn so that gk ↑ lim infn→∞ fn as k → ∞. Since n≥k gk ≤ fn for all n ≥ k, X X gk ≤ fn for all n ≥ k X X and therefore X X gk ≤ lim inf fn for all k. Fig. 8.1. The x and y – slices of a set R ⊂ X × Y. n→∞ X X We may now use the monotone convergence theorem to let k → ∞ to find X X X X lim fn = sup fn = sup sup fn = sup sup fn X X MCT X X n→∞ n n X X n lim inf fn = lim gk = lim gk ≤ lim inf fn. X X α∈2f α α∈2f α n→∞ k→∞ k→∞ n→∞ X X X X X X = sup lim fn = sup lim fn X n→∞ X n→∞ α∈2 α α∈2 α f f P X X Remark 8.39. If A = a < ∞, then for all ε > 0 there exists αε ⊂⊂ X such = sup f = f. X that α∈2X f α X X A ≥ a ≥ A − ε P P α Second Proof. Let Sn = X fn and S = X f. Since fn ≤ fm ≤ f for all n ≤ m, it follows that for all α ⊂⊂ X containing αε or equivalently, Sn ≤ Sm ≤ S which shows that lim S exists and is less that S, i.e. X n→∞ n A − a ≤ ε (8.9) X X α A := lim fn ≤ f. (8.7) n→∞ for all α ⊂⊂ X containing α . Indeed, choose α so that P a ≥ A − ε. X X ε ε αε P P Noting that α fn ≤ X fn = Sn ≤ A for all α ⊂⊂ X and in particular, X 8.6 Sums of complex functions fn ≤ A for all n and α ⊂⊂ X. α Definition 8.40. Suppose that a : X → C is a function, we say that Letting n tend to infinity in this equation shows that X X X a = a (x) f ≤ A for all α ⊂⊂ X X x∈X α and then taking the sup over all α ⊂⊂ X gives exists and is equal to A ∈ C, if for all ε > 0 there is a finite subset αε ⊂ X such that for all α ⊂⊂ X containing α we have X X ε f ≤ A = lim fn (8.8) n→∞ X X X A − a ≤ ε. which combined with Eq. (8.7) proves the theorem. α

Page: 82 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.6 Sums of complex functions 83 The following lemma is left as an exercise to the reader. lim P a = ∞, and therefore P a can not exist as a number in . n→∞ αn∪α X R P iθ P Finally if a is summable, write X a = ρe with ρ ≥ 0 and θ ∈ R, then Lemma 8.41. Suppose that a, b : X → C are two functions such that X a and P b exist, then P (a + λb) exists for all λ ∈ and X X C X X X a = ρ = e−iθ a = e−iθa X X X (a + λb) = a + λ b. X X X X X + X X X = Re e−iθa ≤ Re e−iθa Definition 8.42 (Summable). We call a function a : X → C summable if X X X X X ≤ Re e−iθa ≤ e−iθa ≤ |a| . X |a| < ∞. X X X X Alternatively, this may be proved by approximating P a by a finite sum and P P X Proposition 8.43. Let a : X → C be a function, then X a exists iff X |a| < then using the triangle inequality of |·| . ∞, i.e. iff a is summable. Moreover if a is summable, then Remark 8.44. Suppose that X = and a : → is a sequence, then it is not N N C X X necessarily true that a ≤ |a| . ∞ X X X X a(n) = a(n). (8.11) n=1 n∈N Proof. If P |a| < ∞, then P (Re a)± < ∞ and P (Im a)± < ∞ X X X This is because and hence by Remark 8.39 these sums exists in the sense of Definition 8.40. ∞ N P X X Therefore by Lemma 8.41, X a exists and a(n) = lim a(n) N→∞ ! n=1 n=1 X X + X − X + X − P a = (Re a) − (Re a) + i (Im a) − (Im a) . depends on the ordering of the sequence a where as n∈ a(n) does not. For n P PN X X X X X example, take a(n) = (−1) /n then |a(n)| = ∞ i.e. a(n) does not n∈N n∈N exist while P∞ a(n) does exist. On the other hand, if P n=1 Conversely, if X |a| = ∞ then, because |a| ≤ |Re a| + |Im a| , we must have ∞ X X X X |Re a| = ∞ or |Im a| = ∞. |a(n)| = |a(n)| < ∞ X X n∈N n=1

Thus it suﬃces to consider the case where a : X → R is a real function. Write then Eq. (8.11) is valid. a = a+ − a− where Theorem 8.45 (Dominated Convergence Theorem for Sums). Sup- + − a (x) = max(a (x) , 0) and a (x) = max(−a (x) , 0). (8.10) pose that fn : X → C is a sequence of functions on X such that f (x) = limn→∞ fn (x) ∈ C exists for all x ∈ X. Further assume there is a dominating Then |a| = a+ + a− and function g : X → [0, ∞) such that

X X + X − ∞ = |a| = a + a |fn (x) | ≤ g (x) for all x ∈ X and n ∈ N (8.12) X X X and that g is summable. Then P + P − which shows that either X a = ∞ or X a = ∞. Suppose, with out loss P + 0 X X of generality, that X a = ∞. Let X := {x ∈ X : a (x) ≥ 0}, then we know lim fn (x) = f (x) . (8.13) P 0 n→∞ that X0 a = ∞ which means there are ﬁnite subsets αn ⊂ X ⊂ X such x∈X x∈X that P a ≥ n for all n. Thus if α ⊂⊂ X is any ﬁnite set, it follows that αn

Page: 83 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 84 8 More Sums and Sequences

Proof. Notice that |f| = lim |f | ≤ g so that f is summable. By considering n X X X the real and imaginary parts of f separately, it suﬃces to prove the theorem in f − fn ≤ |f − fn| + 2ε the case where f is real. By Fatou’s Lemma, X X α X X X which then implies that (g ± f) = lim inf (g ± fn) ≤ lim inf (g ± fn) n→∞ n→∞ X X X X X X ! lim sup f − fn ≤ lim sup |f − fn| + 2ε n→∞ n→∞ X X X X α = g + lim inf ± fn . n→∞ X X = 2ε.

Since lim infn→∞(−an) = − lim supn→∞ an, we have shown, Because ε > 0 is arbitrary we conclude that P X X X lim infn→∞ X fn X X g ± f ≤ g + P lim sup f − fn = 0. − lim supn→∞ X fn n→∞ X X X X X and therefore which is the same as Eq. (8.13). X X X lim sup fn ≤ f ≤ lim inf fn. n→∞ n→∞ X X X Remark 8.46. Theorem 8.45 may easily be generalized as follows. Suppose P P fn, gn, g are summable functions on X such that fn → f and gn → g pointwise, This shows that lim X fnexists and is equal to X f. P P n→∞ |fn| ≤ gn and X gn → X g as n → ∞. Then f is summable and Eq. (8.13) Proof. (Second Proof.) Passing to the limit in Eq. (8.12) shows that |f| ≤ g still holds. For the proof we use Fatou’s Lemma to again conclude and in particular that f is summable. Given ε > 0, let α ⊂⊂ X such that X X X (g ± f) = lim inf (gn ± fn) ≤ lim inf (gn ± fn) X n→∞ n→∞ g ≤ ε. X X X ! X\α X X = g + lim inf ± fn n→∞ Then for β ⊂⊂ X such that α ⊂ β, X X

and then proceed exactly as in the ﬁrst proof of Theorem 8.45. X X X f − fn = (f − fn)

β β β X X X 8.7 Iterated sums and the Fubini and Tonelli Theorems ≤ |f − fn| = |f − fn| + |f − fn| β α β\α Let X and Y be two sets. The proof of the following lemma is left to the reader. X X ≤ |f − fn| + 2 g Lemma 8.47. Suppose that a : X → is function and F ⊂ X is a subset such α β\α C that a (x) = 0 for all x∈ / F. Then P a exists iﬀ P a exists and when the X F X ≤ |f − fn| + 2ε. sums exists, X X α a = a. and hence that X F

X X X Theorem 8.48 (Tonelli’s Theorem for Sums). Suppose that a : X × Y → f − fn ≤ |f − fn| + 2ε. [0, ∞], then β β α X X X X X a = a = a. Since this last equation is true for all such β ⊂⊂ X, we learn that X×Y X Y Y X

Page: 84 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.8 `p – spaces, Minkowski and H¨olderInequalities 85 P P Proof. It suﬃces to show, by symmetry, that 3. Y X |a| < ∞. Then X X X X X X X X a = a a = a = a. X×Y X Y X×Y X Y Y X

Let Λ ⊂⊂ X × Y. Then for any α ⊂⊂ X and β ⊂⊂ Y such that Λ ⊂ α × β, we Proof. If a : X → R is real valued the theorem follows by applying Theorem have 8.48 to a± – the positive and negative parts of a. The general result holds for X X X X X X X X a ≤ a = a ≤ a ≤ a, complex valued functions a by applying the real version just proved to the real Λ α×β α β α Y X Y and imaginary parts of a. P P P i.e. Λ a ≤ X Y a. Taking the sup over Λ in this last equation shows X X X p a ≤ a. 8.8 ` – spaces, Minkowski and HölderInequalities X×Y X Y In this chapter, let µ : X → (0, ∞) be a given function. Let F denote either R x x For the reverse inequality, for each x ∈ X choose βn ⊂⊂ Y such that βn ↑ Y as or C. For p ∈ (0, ∞) and f : X → F, let n ↑ ∞ and X X !1/p a(x, y) = lim a(x, y). X p n→∞ kfk := |f (x)| µ (x) x p y∈Y y∈βn x∈X If α ⊂⊂ X is a given finite subset of X, then and for p = ∞ let X X a(x, y) = lim a(x, y) for all x ∈ α kfk∞ = sup {|f (x)| : x ∈ X} . n→∞ y∈Y y∈βn Also, for p > 0, let where β := ∪ βx ⊂⊂ Y. Hence p n x∈α n ` (µ) = {f : X → F : kfkp < ∞}. X X X X X X a(x, y) = lim a(x, y) = lim a(x, y) In the case where µ (x) = 1 for all x ∈ X we will simply write `p(X) for `p(µ). n→∞ n→∞ x∈α y∈Y x∈α y∈βn x∈α y∈βn X X Definition 8.50. A norm on a vector space Z is a function k·k : Z → [0, ∞) = lim a(x, y) ≤ a. n→∞ such that (x,y)∈α×βn X×Y 1. (Homogeneity) kλfk = |λ| kfk for all λ ∈ F and f ∈ Z. Since α is arbitrary, it follows that 2. (Triangle inequality) kf + gk ≤ kfk + kgk for all f, g ∈ Z. X X X X X 3. (Positive definite) kfk = 0 implies f = 0. a(x, y) = sup a(x, y) ≤ a α⊂⊂X x∈X y∈Y x∈α y∈Y X×Y A function p : Z → [0, ∞) satisfying properties 1. and 2. but not necessarily 3. above will be called a semi-norm on Z. which completes the proof. A pair (Z, k·k) where Z is a vector space and k·k is a norm on Z is called a normed vector space. Theorem 8.49 (Fubini’s Theorem for Sums). Now suppose that a : X × Y → C is a summable function, i.e. by Theorem 8.48 any one of the following The rest of this section is devoted to the proof of the following theorem. equivalent conditions hold: Theorem 8.51. For p ∈ [1, ∞], (`p(µ), k · k ) is a normed vector space. P p 1. X×Y |a| < ∞, P P Proof. The only difficulty is the proof of the triangle inequality which is 2. X Y |a| < ∞ or the content of Minkowski’s Inequality proved in Theorem 8.58 below.

Page: 85 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 86 8 More Sums and Sequences Proposition 8.52. Let f : [0, ∞) → [0, ∞) be a continuous strictly increasing Z g(t) st − F (s) = h(s) ≤ (t − f(σ))dσ = h(g(t)) function such that f(0) = 0 (for simplicity) and lim f(s) = ∞. Let g = f −1 s→∞ 0 and for s, t ≥ 0 let R g(t) with equality when s = g(t). To finish the proof we must show 0 (t − Z s Z t f(σ))dσ = G(t). This is verified by making the change of variables σ = g(τ) F (s) = f(s0)ds0 and G(t) = g(t0)dt0. 0 0 and then integrating by parts as follows: Z g(t) Z t Z t Then for all s, t ≥ 0, 0 0 st ≤ F (s) + G(t) (t − f(σ))dσ = (t − f(g(τ)))g (τ)dτ = (t − τ)g (τ)dτ 0 0 0 and equality holds iff t = f(s). Z t = g(τ)dτ = G(t). Proof. Let 0

As := {(σ, τ) : 0 ≤ τ ≤ f(σ) for 0 ≤ σ ≤ s} and

Bt := {(σ, τ) : 0 ≤ σ ≤ g(τ) for 0 ≤ τ ≤ t} then as one sees from Figure 8.2, [0, s] × [0, t] ⊂ As ∪ Bt. (In the ﬁgure: s = 3, t = 1,A3 is the region under t = f(s) for 0 ≤ s ≤ 3 and B1 is the region to the left of the curve s = g(t) for 0 ≤ t ≤ 1.) Hence if m denotes the area of a region in the plane, then

st = m ([0, s] × [0, t]) ≤ m(As) + m(Bt) = F (s) + G(t).

As it stands, this proof is a bit on the intuitive side. However, it will become rigorous if one takes m to be “Lebesgue measure” on the plane which will be introduced later. Fig. 8.2. A picture proof of Proposition 8.52. We can also give a calculus proof of this theorem under the additional assumption that f is C1. (This restricted version of the theorem is all we need in this section.) To do this ﬁx t ≥ 0 and let Z s h(s) = st − F (s) = (t − f(σ))dσ. Deﬁnition 8.53. The conjugate exponent q ∈ [1, ∞] to p ∈ [1, ∞] is q := p 0 p−1 with the conventions that q = ∞ if p = 1 and q = 1 if p = ∞. Notice that q is If σ > g(t) = f −1(t), then t − f(σ) < 0 and hence if s > g(t), we have characterized by any of the following identities:

Z s Z g(t) Z s 1 1 q p + = 1, 1 + = q, p − = 1 and q(p − 1) = p. (8.14) h(s) = (t − f(σ))dσ = (t − f(σ))dσ + (t − f(σ))dσ p q p q 0 0 g(t) Z g(t) p Lemma 8.54. Let p ∈ (1, ∞) and q := p−1 ∈ (1, ∞) be the conjugate exponent. ≤ (t − f(σ))dσ = h(g(t)). Then 0 sp tq st ≤ + for all s, t ≥ 0 (8.15) Combining this with h(0) = 0 we see that h(s) takes its maximum at some p q point s ∈ (0, g(t)] and hence at a point where 0 = h0(s) = t − f(s). The only with equality if and only if tq = sp. (See Example ?? below for a generalization solution to this equation is s = g(t) and we have thus shown of the inequality in Eq. (8.15).)

Page: 86 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.8 `p – spaces, Minkowski and H¨olderInequalities 87

p 1 s p−1 p−1 Proof. Let F (s) = p for p > 1. Then f(s) = s = t and g(t) = t = Proof. Since f (λ) → ∞ if λ ↓ 0 or λ ↑ ∞, it follows that f has a minimum tq−1, wherein we have used q − 1 = p/ (p − 1) − 1 = 1/ (p − 1) . Therefore in (0, ∞) . We can find this minimum using the first derivative test. So G(t) = tq/q and hence by Proposition 8.52, b λp−1 0 set= f 0 (λ) = aλp−1 − bλ−q−1 =⇒ = = λp+q. sp tq a λ−q−1 st ≤ + p q Evaluating f at this λ gives the desired minimum; with equality iff t = sp−1, i.e. tq = sq(p−1) = sp. a b a b b f (λ) = λ−q λp+q + = λ−q + = λ−qb ** For those who do not want to use Proposition 8.52, here is a direct p q p a q q q calculus proof. Fix t > 0 and let − p+q − p+q p b b p+q q p+q 1 1 = b = b p+q = a p+q b = a p b q . sp a a h (s) := st − . p where we have used 0 p−1 Then h (0) = 0, lims→∞ h (s) = −∞ and h (s) = t − s which equals zero iff q q/ (qp) 1 p 1 1 = = and = . s = t p−1 . Since 1 1 q + p p + q p q + p q p p p−1 p−1 q 1 1 t p t q 1 t h t p−1 = t p−1 t − = t p−1 − = t 1 − = , p p p q Theorem 8.56 (Hölder’sinequality). Let p, q ∈ [1, ∞] be conjugate expo- it follows from the first derivative test that nents. For all f, g : X → F, q q n 1 o t t kfgk1 ≤ kfkp · kgkq. (8.16) max h = max h (0) , h t p−1 = max 0, = . q q If p ∈ (1, ∞) and f and g are not identically zero, then equality holds in Eq. So we have shown (8.16) iff |f| p |g| q sp tq = . (8.17) st − ≤ with equality iff t = sp−1. kfkp kgkq p q Proof. The proof of Eq. (8.16) for p ∈ {1, ∞} is easy and will be left to the reader. The cases where kfkp = 0 or ∞ or kgkq = 0 or ∞ are easily dealt We also have the following closely related calculus result. with and are also left to the reader. So we will assume that p ∈ (1, ∞) and 0 < kfk , kgk < ∞. Letting s = |f (x)| /kfk and t = |g|/kgk in Lemma 8.54 Lemma 8.55. Let a, b > 0 and p, q ∈ [1, ∞) such that 1 + 1 = 1 and define p q p q p q implies |f (x) g (x)| 1 |f (x)|p 1 |g (x)|q a b ≤ + f (λ) := λp + λ−q for λ > 0. kfk kgk p kfkp q kgkq p q p q p q with equality iff Then |f (x)|p |g (x)|q 1 1 = sp = tq = . (8.18) min f (λ) = a p b q . kfkp kgkq λ>0 p q which in particular implies [let λ = 1 and repalce a → ap and b → bp] that Multiplying this equation by µ (x) and then summing on x gives kfgk 1 1 ap bq 1 ≤ + = 1 ab ≤ + for all a, b ≥ 0. kfk kgk p q p q p q with equality iff Eq. (8.18) holds for all x ∈ X, i.e. iff Eq. (8.17) holds.

Page: 87 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 88 8 More Sums and Sequences X X X Definition 8.57. For a complex number λ ∈ C, let |f + g|pµ ≤ |f| |f + g|p−1µ + |g| |f + g|p−1µ X X X λ |λ| if λ 6= 0 p−1 sgn(λ) = ≤ (kfkp + kgkp) |f + g| . (8.21) 0 if λ = 0. q Since q(p − 1) = p, as in Eq. (8.14), For λ, µ ∈ C we will write sgn(λ) $ sgn(µ) if sgn(λ) = sgn(µ) or λµ = 0. p−1 q X p−1 q X p p Theorem 8.58 (Minkowski’s Inequality). If 1 ≤ p ≤ ∞ and f, g ∈ `p(µ) k|f + g| kq = (|f + g| ) µ = |f + g| µ = kf + gkp. (8.22) then X X kf + gkp ≤ kfkp + kgkp. (8.19) Combining Eqs. (8.21) and (8.22) shows Moreover, assuming f and g are not identically zero, equality holds in Eq. (8.19) kf + gkp ≤ (kfk + kgk ) kf + gkp/q (8.23) iff p p p p and solving this equation for kf + gkp (making use of Eq. (8.14)) implies Eq. sgn(f) $ sgn(g) when p = 1 and (8.19). Now suppose that f and g are not identically zero and p ∈ (1, ∞) . f = cg for some c > 0 when p ∈ (1, ∞). Equality holds in Eq. (8.19) iff equality holds in Eq. (8.23) iff equality holds in Eq. (8.21) and Eq. (8.20). The latter happens iff Proof. For p = 1, sgn(f) $ sgn(g) and X X X X p p kf + gk1 = |f + g|µ ≤ (|f| µ + |g|µ) = |f| µ + |g|µ |f| |f + g|p |g| X X X X = p = . (8.24) kfkp kf + gkp kgkp with equality iff wherein we have used p−1 q p |f| + |g| = |f + g| ⇐⇒ sgn(f) $ sgn(g). |f + g| |f + g| p−1 = p . k|f + g| kq kf + gkp For p = ∞, Finally Eq. (8.24) is equivalent to |f| = c |g| with c = (kfkp/kgkp) > 0 and this kf + gk∞ = sup |f + g| ≤ sup (|f| + |g|) equality along with sgn(f) $ sgn(g) implies f = cg. X X

≤ sup |f| + sup |g| = kfk∞ + kgk∞. X X 8.9 Exercises Now assume that p ∈ (1, ∞). Since Exercise 8.7. Now suppose for each n ∈ N := {1, 2,...} that fn : X → R is a |f + g|p ≤ (2 max (|f| , |g|))p = 2p max (|f|p , |g|p) ≤ 2p (|f|p + |g|p) function. Let D := {x ∈ X : lim fn (x) = +∞} it follows that n→∞ kf + gkp ≤ 2p kfkp + kgkp < ∞. show that p p p ∞ ∞ D = ∩M=1 ∪N=1 ∩n≥N {x ∈ X : fn (x) ≥ M}. (8.25) Eq. (8.19) is easily veriﬁed if kf + gkp = 0, so we may assume kf + gkp > 0. Multiplying the inequality, Exercise 8.8. Let fn : X → R be as in the last problem. Let

p p−1 p−1 p−1 C := {x ∈ X : lim fn (x) exists in }. |f + g| = |f + g||f + g| ≤ |f| |f + g| + |g||f + g| (8.20) n→∞ R by µ, then summing on x and applying Holder’s inequality on each term gives Find an expression for C similar to the expression for D in (8.25). (Hint: use the Cauchy criteria for convergence.)

Page: 88 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 8.9 Exercises 89 Pm n P∞ 8.9.1 Limit Problems 2. Letq ˜m (x) := 1− n=1 cnx . Use (8.30) to show n=1 cn = 1 and conclude from this that √ Exercise 8.9. Show lim infn→∞(−an) = − lim supn→∞ an. lim sup | 1 − x − q˜m (x) | = 0. (8.31) m→∞ |x|≤1 ¯ Exercise 8.10. Suppose that lim sup an = M ∈ , show that there is a √ n→∞ R 2 ∞ ∞ 3. Conclude that qn (t) := Mqñ (1 − t/M) and pn (t) := qn t for n ∈ N subsequence {ank }k=1 of {an}n=1 such that limk→∞ ank = M. are polynomials verifying Eqs. (8.28) and (8.29) respectively. Exercise 8.11. Show that n n Notation 8.59 For u0 ∈ R and δ > 0, let Bu0 (δ) := {x ∈ R : |x − u0| < δ} n lim sup(an + bn) ≤ lim sup an + lim sup bn (8.26) be the ball in R centered at u0 with radius δ. n→∞ n→∞ n→∞ n Exercise 8.16. Suppose U ⊂ R is a set and u0 ∈ U is a point such that provided that the right side of Eq. (8.26) is well defined, i.e. no ∞ − ∞ or U ∩ (Bu0 (δ) \{u0}) 6= ∅ for all δ > 0. Let G : U \{u0} → C be a function −∞ + ∞ type expressions. (It is OK to have ∞ + ∞ = ∞ or −∞ − ∞ = −∞, 3 on U \{u0}. Show that limu→u0 G(u) exists and is equal to λ ∈ C, iff for all etc.) ∞ sequences {un}n=1 ⊂ U \{u0} which converge to u0 (i.e. limn→∞ un = u0) we have limn→∞ G(un) = λ. Exercise 8.12. Suppose that an ≥ 0 and bn ≥ 0 for all n ∈ N. Show Exercise 8.17. Suppose that Y is a set, U ⊂ Rn is a set, and f : U × Y → C lim sup(anbn) ≤ lim sup an · lim sup bn, (8.27) n→∞ n→∞ n→∞ is a function satisfying: provided the right hand side of (8.27) is not of the form 0 · ∞ or ∞ · 0. 1. For each y ∈ Y, the function u ∈ U → f(u, y) is continuous on U.4 2. There is a summable function g : Y → [0, ∞) such that Exercise 8.13. Prove Lemma 8.41. |f(u, y)| ≤ g(y) for all y ∈ Y and u ∈ U. Exercise 8.14. Prove Lemma 8.47. Show that X 8.9.2 Monotone and Dominated Convergence Theorem Problems F (u) := f(u, y) (8.32) y∈Y

Exercise 8.15. Let M < ∞, show there are polynomials pn(t) and qn (t) for is a continuous function for u ∈ U. n ∈ N such that √ lim sup t − qn(t) = 0 (8.28) Exercise 8.18. Suppose that Y is a set, J = (a, b) ⊂ R is an interval, and n→∞ 0≤t≤M f : J × Y → C is a function satisfying: and 1. For each y ∈ Y, the function u → f(u, y) is diﬀerentiable on J, lim sup ||t| − pn(t)| = 0 (8.29) n→∞ |t|≤M 2. There is a summable function g : Y → [0, ∞) such that using the following outline. ∂ √ f(u, y) ≤ g(y) for all y ∈ Y and u ∈ J. 1. Let f (x) = 1 − x for |x| ≤ 1 and use Taylor’s theorem with integral ∂u remainder (see Eq. ?? of Appendix ??), or analytic function theory if you 3 More explicitly, limu→u G(u) = λ means for every every > 0 there exists a δ > 0 2 0 know it, to show there are constants cn > 0 for n ∈ N such that such that |G(u) − λ| < whenever u ∈ U ∩ (Bu (δ) \{u0}) . √ ∞ 0 X n 1 − x = 1 − cnx for all |x| < 1. (8.30) 4 n=1 To say g := f(·, y) is continuous on U means that g : U → C is continuous relative n to the metric on R restricted to U. 2 (2n−3)!! In fact cn := 2nn! , but this is not needed.

Page: 89 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 90 8 More Sums and Sequences P 3. There is a u0 ∈ J such that y∈Y |f(u0, y)| < ∞. Show: a) for all u ∈ J that P |f(u, y)| < ∞. P y∈Y b) Let F (u) := y∈Y f(u, y), show F is diﬀerentiable on J and that X ∂ F˙ (u) = f(u, y). ∂u y∈Y

(Hint: Use the mean value theorem.)

8.9.3 `p Exercises

Exercise 8.19. Generalize Proposition 8.52 as follows. Let a ∈ [−∞, 0] and Fig. 8.3. Comparing areas when t ≥ b goes the same way as in the text. f : R ∩ [a, ∞) → [0, ∞) be a continuous strictly increasing function such that lim f(s) = ∞, f(a) = 0 if a > −∞ or lims→−∞ f(s) = 0 if a = −∞. Also let s→∞ g = f −1, b = f(0) ≥ 0, Z s Z t F (s) = f(s0)ds0 and G(t) = g(t0)dt0. 0 0 Then for all s, t ≥ 0,

st ≤ F (s) + G(t ∨ b) ≤ F (s) + G(t) and equality holds iﬀ t = f(s). In particular, taking f(s) = es, prove Young’s inequality stating

st ≤ es + (t ∨ 1) ln (t ∨ 1) − (t ∨ 1) ≤ es + t ln t − t, where s ∨ t := min (s, t) . Hint: Refer to Figures 8.3 and 8.4.. Fig. 8.4. When t ≤ b, notice that g(t) ≤ 0 but G(t) ≥ 0. Also notice that G(t) is no longer needed to estimate st. Exercise 8.20. Using diﬀerential calculus, prove the following inequalities

1. For y > 0, let g (x) := xy − ex for x ∈ R. Use calculus to compute the maximum of g (x) and use this prove Young’s inequality; 3. Suppose now that u : [0, ∞) → [0, ∞) is a C1 - function such that: u (0) = 0, u(x) 0 x limx→∞ = ∞, and u (x) > 0 for all x > 0. Show xy ≤ e + y ln y − y for x ∈ R and y > 0. x 2. For p > 1 and y ≥ 0, let g (x) := xy − xp/p for x ≥ 0. Again use calculus to xy ≤ u (x) + v (y) for all x, y ≥ 0, compute the maximum of g (x) and show that your result gives the following 0 −1 0 −1 inequality; where v (y) = y (u ) (y) − u (u ) (y) . Hint: consider the function, xp yq xy ≤ + for all x, y ≥ 0. g (x) := xy − u (x) . p q p 1 1 where q = p−1 , i.e. q = 1 − p .

Page: 90 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9 Topological Considerations

9.1 Closed and Open Sets

Let (X, d) be a metric space.

Deﬁnition 9.1. Let (X, d) be a metric space. The open ball B(x, δ) ⊂ X centered at x ∈ X with radius δ > 0 is the set

B(x, δ) := {y ∈ X : d(x, y) < δ}.

We will often also write B(x, δ) as Bx(δ). We also deﬁne the closed ball centered at x ∈ X with radius δ > 0 as the set Cx(δ) := {y ∈ X : d(x, y) ≤ δ}.

2 Fig. 9.2. The ball in C ([0, 1] , R) of radius 1/4 centered at f (x) = sin x are all the continuous functions whose graphs lie between the green envelope.

2 1 Fig. 9.1. Balls in R corresponding to the 1 – norm, 2− norm, 5 – norm, and 2 – “norm.” 92 9 Topological Considerations

Definition 9.2. A set E ⊂ X is bounded if E ⊂ B (x, R) for some x ∈ X and 2. If {Cα}α∈I is a collection of closed subsets of X, then ∩α∈I Cα is closed in ∞ R < ∞. A set F ⊂ X is closed iff every convergent sequence {xn}n=1 which is X. contained in F has its limit back in F. A set V ⊂ X is open iff V c is closed. 3. If A and B are closed sets then A ∪ B is closed. We will write F @ X to indicate F is a closed subset of X and V ⊂o X to Proof. 3. Let {z }∞ ⊂ A ∪ B such that lim z =: z exists. Then indicate the V is an open subset of X. We also let τd denote the collection of n n=1 n→∞ n open subsets of X relative to the metric d. zn ∈ A i.o. or zn ∈ B i.o. For sake of definiteness say zn ∈ A i.o. in which case

we may choose a subsequence, wk := znk ∈ A for all k. Since limk→∞ wk = z Lemma 9.3. If f : X→ R is a continuous function and k ∈ R, then the follow- and A is closed it follows that z ∈ A and hence z ∈ A ∪ B. Thus we have shown ing sets are closed, A ∪ B is closed.

A := {x ∈ X : f (x) ≤ k} ,B := {x ∈ X : f (x) = k} , and Exercise 9.1. Prove item 2. of Theorem 9.7. If {Cα}α∈I is a collection of closed C := {x ∈ X : f (x) ≥ k} . subsets of X, then ∩α∈I Cα is closed in X. ∞ Exercise 9.2. Give an example of a collection of closed subsets, {An}n=1 , of Proof. The proof that A, B, and C are closed all go the same way so let ∞ ∞ C such that ∪n=1An is not closed. me just check that A is closed. To this end, suppose that {xn}n=1 is a sequence in A such that x := limn→∞ xn exists in X. Since xn ∈ A, f (xn) ≤ k and Corollary 9.8. Let (X, d) be a metric space. Then the collection of open sub- therefore, sets, τd, of X satisfy; k ≥ lim f (xn) = f (x) n→∞ 1. X and ∅ are in τd. wherein the last equality we have used the definition of f being continuous. By 2. τd is closed under taking arbitrary unions. i.e. if {Uα} is a collection of definition of A it then follows that x ∈ A and so we have checked that A is α∈I open sets then ∪α∈I Uα is open.. closed. 3. τd is closed under taking finite intersections, i.e. if U and V are open sets then then U ∩ V is open as well. Example 9.4 (Closed Balls). Closed balls are closed. Indeed, we have seen f (y) := d (x, y) is continuous and therefore Exercise 9.3. Prove Corollary 9.8.

Cx(δ) := {y ∈ X : d (x, y) ≤ δ} = {y ∈ X : f (y) ≤ δ} Exercise 9.4. Let U be a subset of a metric space (X, d) . Show the following are equivalent; is a closed set. Notice that {x} = Cx (0) is a closed set for all x ∈ X. 1. U is open, Example 9.5. The following subsets of are closed; C 2. for all z ∈ U there exists ρ > 0 such that Bz (ρ) ⊂ U. 3. U can be written as a union of open balls. 1. {z ∈ C : a ≤ Im z ≤ b} for all a ≤ b in R. 2. {z ∈ : a ≤ Re z ≤ b} for all a ≤ b in . C R Exercise 9.5 (Redundant now). Show that V ⊂ X is open iff for every 3. {z ∈ C : Im z = 0 and a ≤ Re z ≤ b} for all a ≤ b in R. x ∈ V there is a δ > 0 such that Bx(δ) ⊂ V. In particular show Bx(δ) is open c Example 9.6 (Open Balls). Open balls in metric spaces are open sets. Indeed for all x ∈ X and δ > 0. Hint: by definition V is not open iff V is not closed. let f (y) := d (x, y) , then Exercise 9.6. Let (X, d) be a metric space and {x1, . . . , xn} be a finite subset c of X. Show X \{x1, . . . , xn} is an open subset. Bx(δ) := X \ Bx(δ) = {y ∈ X : d (x, y) ≥ δ} = {y ∈ X : f (y) ≥ δ} Exercise 9.7. Let (X, d) be a complete metric space. Let A ⊂ X be a subset which is closed since f is continuous. Thus Bx (δ) is open. of X viewed as a metric space using d|A×A. Show that (A, d|A×A) is complete Theorem 9.7. The closed subsets of (X, d) have the following properties; iff A is a closed subset of X. 1. X and ∅ are closed.

Page: 92 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9.1 Closed and Open Sets 93

Lemma 9.9. For any non empty subset A ⊂ X, let dA (x) := inf{d(x, a)|a ∈ Definition 9.13. Given a set A contained in a metric space X, let A¯ ⊂ X be A}, then the closure of A defined by |dA (x) − dA(y)| ≤ d(x, y) ∀ x, y ∈ X (9.1) A¯ := {x ∈ X : ∃ {xn} ⊂ A 3 x = lim xn}. n→∞ and in particular if xn → x in X then dA (xn) → dA (x) as n → ∞. Moreover the set Fε := {x ∈ X|dA (x) ≥ ε} is closed in X. That is to say A¯ contains all limit points of A. We say A is dense in X if A¯ = X, i.e. every element x ∈ X is a limit of a sequence of elements from A. Proof. Let a ∈ A and x, y ∈ X, then A metric space is said to be separable if it contains a countable dense subset, D. dA (x) ≤ d(x, a) ≤ d(x, y) + d(y, a). Lemma 9.14. For any A ⊂ X, then Take the infimum over a in the above equation shows that 1. A = A¯ if A is closed. dA (x) ≤ d(x, y) + dA(y) ∀x, y ∈ X. 2. A¯ = {x : dA (x) = 0} and A¯ is closed. 3. A¯ = {x ∈ X : A ∩ Bx (ρ) 6= ∅ ∀ ρ > 0} . c Therefore, dA (x) − dA(y) ≤ d(x, y) and by interchanging x and y we also have 4. dA (x) > 0 for all x ∈ A if A is closed. that d (y) − d (x) ≤ d(x, y) which implies Eq. (9.1). If x → x ∈ X, then by A A n ¯ Eq. (9.1), Proof. 1. We always have A ⊂ A. If A is closed we can not leave A by taking limits and hence A¯ ⊂ A, i.e. A = A¯ if A is closed. |dA (x) − dA(xn)| ≤ d(x, xn) → 0 as n → ∞ 2. Let F := {x : dA (x) = 0} which is a closed set since dA is continuous. If so that lim d (x ) = d (x) . Now suppose that {x }∞ ⊂ F and x → x n→∞ A n A n n=1 ε n x ∈ F (i.e. dA (x) = 0), there exists xn ∈ A such that d (x, xn) ≤ 1/n for all in X, then ¯ ¯ n ∈ N. Thus limn→∞ xn = x and so x ∈ A. This shows F ⊂ A. Conversely if dA (x) = lim dA (xn) ≥ ε ¯ n→∞ x ∈ A, there exists {xn} ⊂ A such that limn→∞ xn = x and so since dA (xn) ≥ ε for all n. This shows that x ∈ Fε and hence Fε is closed. dA (x) = dA lim xn = lim dA (xn) = lim 0 = 0 n→∞ n→∞ n→∞ Definition 9.10. A subset A ⊂ X is a neighborhood of x if there exists an which shows x ∈ F. open set V ⊂ X such that x ∈ V ⊂ A. We will say that A ⊂ X is an open o 2. ⇐⇒ 3. Since A ∩ B (ρ) 6= ∅ happens iff d (x) < ρ we see that A ∩ neighborhood of x if A is open and x ∈ A. x A Bx (ρ) 6= ∅ ∀ ρ > 0 iff dA (x) < ρ for all ρ > 0, i.e. iff dA (x) = 0. c 4. If A is closed then A = A¯ = {x ∈ X : d (x) = 0} and therefore Ac = Example 9.11. Let x ∈ X and δ > 0, then Cx (δ) and Bx (δ) are closed subsets A ∞ {x ∈ X : dA (x) > 0} . of X. For example if {yn}n=1 ⊂ Cx (δ) and yn → y ∈ X, then d (yn, x) ≤ δ for all n and using Corollary 6.46 it follows d (y, x) ≤ δ, i.e. y ∈ Cx (δ) . A similar c Exercise 9.8. If A is a non-empty subset of X, then dA = dA¯. proof shows Bx (δ) is closed, see Exercise 9.5. Exercise 9.9. Given A ⊂ X, show A¯ is a closed set and in fact Lemma 9.12 (Approximating open sets from the inside by closed A¯ = ∩{F : A ⊂ F ⊂ X with F closed}. (9.2) sets). Let U ⊂ X be an open set and Fε := {x ∈ X|dU c (x) ≥ ε} @ X be as in Lemma 9.9. Then Fε ↑ U as ε ↓ 0. That is to say A¯ is the smallest closed set containing A. c Proof. It is clear that dU c (x) = 0 for x ∈ U so that Fε ⊂ U for each ε > 0 Definition 9.15. Let (X, d) be a metric space and A be a subset of X. and hence ∪ε>0Fε ⊂ U. Now suppose that x ∈ U ⊂o X. By Exercise 9.5 there c 1. The closure of A is the smallest closed set A¯ containing A, i.e. exists an ε > 0 such that Bx(ε) ⊂ U, i.e. d(x, y) ≥ ε for all y ∈ U . Hence x ∈ Fε and we have shown that U ⊂ ∪ε>0Fε. Finally it is clear that Fε ⊂ Fε0 A¯ := ∩ {F : A ⊂ F X} . whenever ε0 ≤ ε. @ (Because of Exercise 9.9 this is consistent with Definition 9.13 for the closure of a set in a metric space.)

Page: 93 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 94 9 Topological Considerations 2. The interior of A is the largest open set Ao contained in A, i.e. Exercise 9.11. Suppose that A and B are subsets of a metric space, show A ∪ B = A¯ ∪ B.¯ Ao = ∪ {V ∈ τ : V ⊂ A} . ∞ 3. A ⊂ X is a neighborhood of a point x ∈ X if x ∈ Ao. Exercise 9.12. Given an example showing that ∪n=1An need not be equal to ∞ 4. The accumulation points of A is the set ∪n=1An.

acc(A) = {x ∈ X : V ∩ [A \{x}] 6= ∅ for all V ∈ τx}. Remark 9.18. The relationships between the interior and the closure of a set are: 5. The boundary of A is the set bd(A) := A¯ \ Ao. \ \ 6. A is dense in X if A¯ = X and X is said to be separable if there exists a (Ao)c = {V c : V ∈ τ and V ⊂ A} = {C : C is closed C ⊃ Ac} = Ac countable dense subset of X. ¯ c c o Lemma 9.16. Let (X, d) be a metric space and A be a subset of X, then and similarly, (A) = (A ) . o A = {x ∈ X : Bx (ρx) ⊂ A for some ρx > 0} . Proposition 9.19. If A is a subset of a metric space, (X, d) , then bd (A) may be computed using either; Proof. Let V := {x ∈ X : Bx (ρ) ⊂ A for some ρ = ρx > 0} . If Bx (ρ) ⊂ A ¯ c and y ∈ Bx (ρ) and δ = ρ − d (x, y) , then By (δ) ⊂ Bx (ρ) ⊂ A which shows 1. bd (A) = A ∩ A , that y ∈ V, i.e. Bx (ρ) ⊂ V. Thus we may write 2. bd (A) = {x ∈ X : dA (x) = 0 = dAc (x)} , c 3. bd (A) = {x ∈ X : Bx (ρ) ∩ A 6= ∅= 6 Bx (ρ) ∩ A for all ρ > 0} , or V = ∪ {Bx (ρ): Bx (ρ) ⊂ A} . c 4. bd (A) = {x ∈ X : ∃ {xn} ⊂ A and {yn} ⊂ A 3 limn→∞ xn = x = limn→∞ yn} . This shows that V is an open subset of A. Moreover if W is another open subset ¯ c c o of A, then Proof. From the deﬁnition of bd (A) and the relation, (A) = (A ) , we ﬁnd o o c c W = ∪ {Bx (ρ): Bx (ρ) ⊂ W } ⊂ ∪ {Bx (ρ): Bx (ρ) ⊂ A} = V bd (A) = A¯ \ A = A¯ ∩ (A ) = A¯ ∩ A . (9.3) so that V is the largest open subset contained in A. This completes the proof. Item 2. now follow from item 1. and the fact that A¯ = {x ∈ X : dA (x) = 0} . I leave it to the reader to check that 2. =⇒ 3. =⇒ 4. =⇒ 1.

Exercise 9.10. Let (X, k·k) be a normed space and d (x, y) := ky − xk . Show; Exercise 9.13. If D is a dense subset of a metric space (X, d) and E ⊂ X is o a subset such that to every point x ∈ D there exists {x }∞ ⊂ E with x = 1. Cx (ρ) = Bx (ρ) , n n=1 lim x , then E is also a dense subset of X. If points in E well approximate 2. Bx (ρ) = Cx (ρ) , n→∞ n every point in D and the points in D well approximate the points in X, then 3.bd( Cx (ρ)) = bd (Cx (ρ)) = {y ∈ X : ky − xk = ρ} . the points in E also well approximate all points in X. Example 9.17 (Words of Caution). Let (X, d) be a metric space. It is always true that Bx(ε) ⊂ Cx(ε) since Cx(ε) is a closed set containing Bx(ε). However, Exercise 9.14. Suppose (X, d) is a metric space which contains an uncountable it is not always true that Bx(ε) = Cx(ε). For example let X = {1, 2} and subset Λ ⊂ X with the property that there exists ε > 0 such that d (a, b) ≥ ε d(1, 2) = 1, then B1(1) = {1} , B1(1) = {1} while C1(1) = X. For another for all a, b ∈ Λ with a 6= b. Show that (X, d) is not separable. counterexample, take Exercise 9.15. Let Y = BC (R, C) be the Banach space of continuous bounded X = (x, y) ∈ 2 : x = 0 or x = 1 R complex valued functions on R equipped with the uniform norm, kfku := sup |f (x)| . Further let C ( , ) denote those f ∈ C ( , ) such that vanish with the usually Euclidean metric coming from the plane. Then x∈R 0 R C R C at inﬁnity, i.e. limx→±∞ f (x) = 0. Also let Cc (R, C) denote those f ∈ C (R, C) 2 B(0,0)(1) = (0, y) ∈ R : |y| < 1 , with compact support, i.e. there exists N < ∞ such that f (x) = 0 if |x| ≥ N. 2 Show C0 ( , ) is a closed subspace of Y and that B(0,0)(1) = (0, y) ∈ R : |y| ≤ 1 , while R C

C(0,0)(1) = B(0,0)(1) ∪ {(1, 0)} . Cc (R, C) = C0 (R, C) .

Page: 94 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9.2 Continuity Revisited 95 9.2 Continuity Revisited Lemma 9.22 (Global Continuity Lemma). Suppose that (X, ρ) and (Y, d) are two metric spaces and f : X → Y is a function defined on all of X. Then Suppose now that (X, ρ) and (Y, d) are two metric spaces and f : X → Y is a the following are equivalent: function. 1. f is continuous. −1 −1 Definition 9.20. A function f : X → Y is continuous at x ∈ X if for all 2. f (V ) ∈ τρ for all V ∈ τd, i.e. f (V ) is open in X if V is open in Y. ε > 0 there is a δ > 0 such that 3. f −1(C) is closed in X if C is closed in Y. 4. For all convergent sequences {xn} ⊂ X, {f(xn)} is convergent in Y and d(f (x) , f(x0)) < ε provided that ρ(x, x0) < δ. (9.4) lim f(xn) = f lim xn . The function f is said to be continuous if f is continuous at all points x ∈ X. n→∞ n→∞ −1 c −1 c The following lemma gives two other characterizations of continuity of a Proof. Since f (A ) = f (A) , it is easily seen that 2. and 3. are function at a point. equivalent. So because of Lemma 9.21 it only remains to show 1. and 2. are equivalent. If f is continuous and V ⊂ Y is open, then for every x ∈ f −1 (V ) ,V Lemma 9.21 (Local Continuity Lemma). Suppose that (X, ρ) and (Y, d) is a neighborhood of f (x) and so f −1 (V ) is a neighborhood of x. Hence f −1 (V ) are two metric spaces and f : X → Y is a function defined in a neighborhood is a neighborhood of all of its points and from this and Exercise 9.5 it follows that of a point x ∈ X. Then the following are equivalent: f −1 (V ) is open. Conversely, if x ∈ X and A ⊂ Y is a neighborhood of f (x) then −1 −1 there exists V ⊂o X such that f (x) ∈ V ⊂ A. Hence x ∈ f (V ) ⊂ f (A) 1. f is continuous at x ∈ X. and by assumption f −1 (V ) is open showing f −1 (A) is a neighborhood of x. −1 2. For all neighborhoods A ⊂ Y of f (x) , f (A) is a neighborhood of x ∈ X. Therefore f is continuous at x and since x ∈ X was arbitrary, f is continuous. ∞ 3. For all sequences {xn}n=1 ⊂ X such that x = limn→∞ xn, {f(xn)} is convergent in Y and Exercise 9.16. Use Example 9.28 and Lemma 9.22 to recover the results of lim f(xn) = f lim xn . n→∞ n→∞ Example 9.11. Proof. 1 =⇒ 2. If A ⊂ Y is a neighborhood of f (x) , there exists ε > 0 The next result shows that there are lots of continuous functions on a metric such that Bf(x) (ε) ⊂ A and because f is continuous there exists a δ > 0 such space (X, d) . that Eq. (9.4) holds. Therefore Lemma 9.23 (Urysohn’s Lemma for Metric Spaces). Let (X, d) be a met- −1 −1 Bx (δ) ⊂ f Bf(x) (ε) ⊂ f (A) ric space and suppose that A and B are two disjoint closed subsets of X. Then

−1 showing f (A) is a neighborhood of x. dB (x) ∞ f (x) = for x ∈ X (9.5) 2 =⇒ 3. Suppose that {xn}n=1 ⊂ X and x = limn→∞ xn. Then for any ε > dA (x) + dB (x) −1 0,Bf(x) (ε) is a neighborhood of f (x) and so f Bf(x) (ε) is a neighborhood deﬁnes a continuous function, f : X → [0, 1], such that f (x) = 1 for x ∈ A and of x which must contain Bx (δ) for some δ > 0. Because xn → x, it follows that −1 f (x) = 0 if x ∈ B. xn ∈ Bx (δ) ⊂ f Bf(x) (ε) for a.a. n and this implies f (xn) ∈ Bf(x) (ε) for a.a. n, i.e. d(f (x) , f (xn)) < ε for a.a. n. Since ε > 0 is arbitrary it follows that Proof. By Lemma 9.9, dA and dB are continuous functions on X. Since limn→∞ f (xn) = f (x) . A and B are closed, dA (x) > 0 if x∈ / A and dB (x) > 0 if x∈ / B. Since 3. =⇒ 1. We will show not 1. =⇒ not 3. If f is not continuous at x, −1 A ∩ B = ∅, d (x) + d (x) > 0 for all x and (d + d ) is continuous as well. there exists an ε > 0 such that for all n ∈ there exists a point x ∈ X with A B A B N n The remaining assertions about f are all easy to verify. ρ (x , x) < 1 yet d (f (x ) , f (x)) ≥ ε. Hence x → x as n → ∞ yet f (x ) does n n n n n Sometimes Urysohn’s lemma will be use in the following form. Suppose not converge to f (x) . F ⊂ V ⊂ X with F being closed and V being open, then there exists f ∈ Here is a global version of the previous lemma. C (X, [0, 1])) such that f = 1 on F while f = 0 on V c. This of course follows from Lemma 9.23 by taking A = F and B = V c.

Page: 95 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 96 9 Topological Considerations

9.3 Metric spaces as topological spaces (Not required Definition 9.26 (Global continuity in topological terms). Let (X, τX ) Reading!) and (Y, τY ) be topological spaces. A function f : X → Y is continuous if −1 −1 f (τY ) := f (V ): V ∈ τY ⊂ τX . Let (X, d) be a metric space and let τ = τd denote the collection of open subsets of X. (Recall V ⊂ X is open iff V c is closed iff for all x ∈ V there We will also say that f is τX /τY –continuous or (τX , τY ) – continuous. Let exists an ε = εx > 0 such that B (x, εx) ⊂ V iff V can be written as a (possibly C(X,Y ) denote the set of continuous functions from X to Y. uncountable) union of open balls.) Although we will stick with metric spaces in this chapter, it will be useful to introduce the definitions needed here in the Exercise 9.17. Show f : X → Y is continuous (Definition ??) iff f is continu- more general context of a general “topological space,” i.e. a space equipped ous at all points x ∈ X. with a collection of “open sets.” Exercise 9.18. Show f : X → Y is continuous iff f −1(C) is closed in X for all Definition 9.24 (Topological Space). Let X be a set. A topology on X is closed subsets C of Y. a collection of subsets (τ) of X with the following properties; Definition 9.27. A map f : X → Y between topological spaces is called a 1. τ contains both the empty set (∅) and X. homeomorphism provided that f is bijective, f is continuous and f −1 : Y → 2. τ is closed under arbitrary unions. X is continuous. If there exists f : X → Y which is a homeomorphism, we say 3. τ is closed under finite intersections. that X and Y are homeomorphic. (As topological spaces X and Y are essentially the same.) The elements V ∈ τ are called open subsets of X. A subset F ⊂ X is said c to be closed if F is open. I will write V ⊂o X to indicate that V ⊂ X and Example 9.28. The function dA defined in Lemma 9.9 is continuous for each V ∈ τ and similarly F @ X will denote F ⊂ X and F is closed. Given x ∈ X A ⊂ X. In particular, if A = {x} , it follows that y ∈ X → d(y, x) is continuous we say that V ⊂ X is an open neighborhood of x if V ∈ τ and x ∈ V. Let for each x ∈ X. τx = {V ∈ τ : x ∈ V } denote the collection of open neighborhoods of x. Definition 9.25 (Continuity at a point in topological terms). Let 9.4 Exercises (X, τX ) and (Y, τY ) be topological spaces. A function f : X → Y is continuous at a point x ∈ X if for every open neighborhood V of f (x) there is an Exercise 9.19. Show that (X, d) is a complete metric space iff every sequence open neighborhood U of x such that U ⊂ f −1(V ). See Figure 9.3. ∞ P∞ {xn}n=1 ⊂ X such that n=1 d(xn, xn+1) < ∞ is a convergent sequence in X. You may find it useful to prove the following statements in the course of the proof.

X 1. If {xn} is Cauchy sequence, then there is a subsequence yj := xnj such that ∞ Y P d(y , y ) < ∞. V j=1 j+1 j U 2. If {x }∞ is Cauchy and there exists a subsequence y := x of {x } such f n n=1 j nj n that x = lim y exists, then lim x also exists and is equal to x. x f(x) j→∞ j n→∞ n Exercise 9.20. Suppose that f : [0, ∞) → [0, ∞) is a C2 – function such that f(0) = 0, f 0 > 0 and f 00 ≤ 0 and (X, ρ) is a metric space. Show that d(x, y) = f(ρ(x, y)) is a metric on X. In particular show that f −1(V ) ρ(x, y) d(x, y) := 1 + ρ(x, y) Fig. 9.3. Checking that a function is continuous at x ∈ X. is a metric on X. (Hint: use calculus to verify that f(a + b) ≤ f(a) + f(b) for all a, b ∈ [0, ∞).)

Page: 96 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9.5 Sequential Compactness 97 ∞ Exercise 9.21. Let {(Xn, dn)}n=1 be a sequence of metric spaces, X := Theorem 9.33 (Bolzano–Weierstrass / Heine–Borel theorem). As sub- Q∞ ∞ ∞ D n=1 Xn, and for x = (x(n))n=1 and y = (y(n))n=1 in X let set K ⊂ C is sequentially compact iff it is closed and bounded. ∞ Proof. In light of Examples 9.30 and 9.31 we are left to show that closed X dn(x(n), y(n)) d(x, y) = 2−n . (9.6) D 1 + d (x(n), y(n)) and bounded sets are sequentially compact. So let K ⊂ C be a closed and n=1 n ∞ bounded set and {zn}n=1 be any sequence in K. According toe Lemma 9.32, ∞ ∞ Show: {zn}n=1 has a convergent subsequence, {wk := znk }k=1 . Since wk ∈ K for all k and K is closed it necessarily follows that limk→∞ wk ∈ K which shows K is 1.( X, d) is a metric space, sequentially compact. ∞ 2. a sequence {xk}k=1 ⊂ X converges to x ∈ X iff xk(n) → x(n) ∈ Xn as k → ∞ for each n ∈ N and Example 9.34 (Warning!). It is not true that a closed and bounded subset of an arbitrary metric space (X, d) is necessarily sequentially compact. For example 3. X is complete if Xn is complete for all n. let Z denote the vector space of continuous functions on [0, 1] with values in R ∞ and for f ∈ Z let kfk = supt∈[0,1] |f (t)| . Then the set C := {fn}n=0 where 9.5 Sequential Compactness  0 if t ∈ [0, 2−(n+1)] ∪ [2−n, 1] (n+2)  −(n+1) −(n+1) −(n+2) Suppose that (X, d) and (Y, ρ) are metric spaces. fn (t) = 2 t − 2 if 2 ≤ t ≤ 3 · 2  2−n − t if 3 · 2−(n+2) ≤ t ≤ 2−n. Definition 9.29. As subset K ⊂ X is (sequentially) compact if every se- ∞ ∞ [So f (t) is a shark tooth over the interval 2−(n+1), 2−n .] Notice that kf k = 1 quence {zn}n=1 ⊂ K has a convergent subsequence, {wk := znk }k=1 such that n n limk→∞ wk ∈ K. for all n so that C is bounded. Moreover kfn − fmk = 1 for all m 6= n, therefore there are no convergent subsequence of C. The reader should use this fact to Example 9.30. Suppose that F ⊂ X is an unbounded set, i.e. for all n ∈ N ∞ see that C is closed and bounded but not sequentially compact! there exists zn ∈ F such that d (x, zn) ≥ n. The sequence {zn}n=1 and all of its subsequences are unbounded and therefore not Cauchy in X and hence not convergent in X. This shows that sequentially compact sets must be bounded.

∞ Example 9.31. Suppose that F ⊂ X is not closed. Then there exists {zn}n=1 ⊂ F such that z := limn→∞ zn ∈/ F. Moreover, although every subsequence of ∞ {zn}n=1 is convergent, they all still converge to z∈ / F. This shows that a sequentially compact set must be closed.

Lemma 9.32 (Bolzano–Weierstrass property for CD). Let D ∈ N. Every ∞ D bounded sequence, {z (n)}n=1 ⊂ C , has a convergent subsequence. Proof. By assumption there exists M < ∞ such that kz (n)k = D d (z (n) , 0) ≤ M for all n ∈ N. Writing z (n) = (z1 (n) , . . . , zD (n)) ∈ C . Since ∞ |zi (n)| ≤ kz (n)k it follows that {zi (n)}n=1 is a bounded sequence in C. Hence by the Bolzano–Weierstrass property for C we replace z (n) by a subsequence z (nk) such that limk→∞ z (nk) = z1 exists. We may now replace the original z Fig. 9.4. Here are the plots of f0 and f1. by this new subsequence and then ﬁnd a further subsequence z (nk) such that limk→∞ zi (nk) = zi exists for i = 1, 2. We may continue this way inductively to ﬁnd a subsequence such that limk→∞ zi (nk) = zi exists for all 1 ≤ i ≤ D. It then follows that limk→∞ kz − z (nk)k = 0 as desires where z := (z1, . . . , zD) . Exercise 9.22 (Extreme value theorem). Let K be sequentially compact subset of X and f : K → R be a continuous function. Show −∞ <

Page: 97 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 98 9 Topological Considerations iθ l l+1 n infx∈K f (x) ≤ supx∈K f (x) < ∞ and there exists a, b ∈ K such that |α0| ≤ q re = |α0| 1 − εr + cl+1r + ··· + cnr f(a) = infx∈K f (x) and f(b) = supx∈K f (x) . Hint: first argue that there l l+1 n ∞ ≤ |α0| 1 − εr + cl+1r + ··· + cnr exists {zn} ⊂ K such that f (zn) ↑ sup f (x) as n → ∞. n=1 x∈K l l+1 l ≤ |α0| 1 − εr + Cr = |α0| 1 − r [ε − Cr] Exercise 9.23 (Extreme value theorem II). Suppose that f : X → R is a continuous function on a metric space (X, d) . Further assume there exists a where {ck} and C are certain appropriate constants. Choosing r ∈ (0, ε/C) sequentially compact subset K ⊂ X and x0 ∈ K such that f (x0) ≤ f (x) for then gives the |α0| < |α0| which is absurd and we have reached the desired all x ∈ X \ K. Show there exists k ∈ K such that infx∈X f (x) = f (k) . contradiction. Theorem 9.35 (Fundamental Theorem of Algebra). Suppose that p (z) = Remark 9.37. The fundamental theorem of algebra does not hold for polynomi- Pn k als in z andz ¯ or in x and y. For example consider the polynomial k=0 akz is a polynomial on C with an 6= 0 and n > 0. Then there exists z0 ∈ C such that p (z0) = 0. p (z, z¯) = 1 + z z¯ = 1 + x2 + y2. Proof. Since The point is that Lemma 9.36 does not hold for these more general classes of n−1 |p (z)| X ak polynomials. lim n = lim + an = |an| , |z|→∞ |z| |z|→∞ zn−k k=0 Exercise 9.24 (Uniform Continuity). Let K be sequentially compact subset of X and f : K → be a continuous function. Show that f is uniformly if R > 0 is sufficiently large, then C continuous, i.e. if ε > 0 there exists δ > 0 such that |f(z) − f(w)| < ε if n n w, z ∈ K with d (w, z) < δ. Hint: prove the contrapositive. |p (z)| ≥ |an| |z| /2 ≥ |an| R /2 ≥ |a0| = |p (0)| for |z| > R. Exercise 9.25. If (X, d) is a metric space and K ⊂ X is sequentially compact. Applying Exercise 9.23 with K = {z ∈ : |z| ≤ R} and x = 0 there exists C 0 Show subset, C ⊂ K, which is closed is sequentially compact as well. z0 ∈ K ⊂ C such that |p (z0)| ≤ |p (z)| for all z ∈ C. It now follows from Lemma 9.36 below that p (z ) = 0. 0 Exercise 9.26. If K ⊂ R is sequentially compact then sup (K) ∈ K, i.e. sup (K) = max (K) . Pn k Lemma 9.36. Suppose that p (z) = k=0 akz is a polynomial on C with an 6= Exercise 9.27. Let (X, d) and (Y, ρ) be metric spaces, K ⊂ X be a sequen- 0 and n > 0. If |p (z)| has a minimum at z0 ∈ C, then |p (z0)| = 0. tially compact set, and f : K → Y be a continuous function. Show f (K) is sequentially compact in Y. In particular, for C ⊂ K closed, we have f (C) is Proof. For sake of contradiction, let us suppose that p (z ) 6= 0 and set 0 closed and in fact sequentially compact in Y. n n X k X k Exercise 9.28. Let f :[a, b] → [c, d] be a strictly increasing continuous func- q (z) := p (z0 + z) = ak (z + z0) = αkz . −1 k=0 k=0 tion such that f (a) = c and f (b) = d and g := f :[c, d] → [a, b] as in Exercise 6.20. Give one or better yet two alternative proofs that g is continuous based Then 0 < |q (0)| = |α0| ≤ |q (z)| for all z ∈ C and αn = an 6= 0. Let l ≥ 1 be on sequentially compactness arguments. the first index such that αl 6= 0 so that Definition 9.38. Let Z be a vector space. We say that two norms, |·| and k·k , l n q (z) = α0 + αlz + ··· + αnz on Z are equivalent if there exists constants α, β ∈ (0, ∞) such that αl l αn n = α0 1 + z + ··· + z kfk ≤ α |f| and |f| ≤ β kfk for all f ∈ Z. α0 α0 Theorem 9.39. Let Z be a finite dimensional vector space. Then any two Evaluating this at z = reiθ with θ chosen so that αl eilθ = − αl = −ε implies α0 α0 norms |·| and k·k on Z are equivalent. (This is typically not true for norms for r > 0 small that on infinite dimensional spaces.)

Page: 98 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9.6 Open Cover Compactness 99 n Proof. Let {fi}i=1 be a basis for Z and deﬁne a new norm on Z by 9.6 Open Cover Compactness v Deﬁnition 9.42. Let A be a subset of a metric space, (X, d) . An open cover n u n X uX 2 of A is a collection of, U, of open subsets of X such that A ⊂ ∪ V. aifi := t |ai| for ai ∈ F. V ∈U

i=1 2 i=1 Definition 9.43. The subset A ⊂ X is open cover compact1 if every open By the triangle inequality for the norm |·| , we find cover (Definition 9.42) of A has finite a sub-cover, i.e. if U is an open cover of A there exists finite subcollection, U0 ⊂f U such that U0 is still a cover of v v n n n n n A. (We will write A @@ X to denote that A ⊂ X and A is compact.) A subset u u X X uX 2uX 2 X A ⊂ X is precompact if A¯ is compact. aifi ≤ |ai| |fi| ≤ t |fi| t |ai| ≤ M aifi i=1 i=1 i=1 i=1 i=1 2 Remark 9.44. If (X, d) is a metric space, we will see below in Theorem 9.53 that q X is sequentially compact iff it is open cover compact. We will not use this fact Pn 2 where M = i=1 |fi| . Thus we have |f| ≤ M kfk2 for all f ∈ Z and this in this section however. In the future though we will just refer to compact sets with out the any extra adjectives. inequality shows that |·| is continuous relative to k·k2 . Since the normed space (Z, k·k ) is homeomorphic and isomorphic to n with the standard euclidean 2 F Example 9.45. Suppose that A is an unbounded subset of X. Pick a1 ∈ A and norm, the closed bounded set, S := {f ∈ Z : kfk = 1} ⊂ Z, is a sequentially ∞ 2 then choose {a } inductively so that d (a ) ≥ 1 for all n. This compact subset of Z relative to k·k . Therefore by Exercise 9.22 there exists n n=2 {a1,...,an} n+1 2 sequence then has the property that d (ak, al) ≥ 1 for all k 6= l and from this it f0 ∈ S such that follows that F := {a1, a2,... } is a closed set. We then define an open cover of m = inf {|f| : f ∈ S} = |f0| > 0. A by taking, f Hence given 0 6= f ∈ Z, then ∈ S so that c kfk2 U = {F ,Ba1 (1/3) ,Ba2 (1/3) ,Ba3 (1/3) ,... } .

f 1 This cover has no ﬁnite subcover. Therefore A can not be open cover compact. m ≤ = |f| kfk2 kfk2 Alternatively. Given x ∈ X, the collection U := {Bx (n): n ∈ N} is an open cover of X. So if K is an open cover compact subset of X, there must or equivalently exist n < n < ··· < n so that 1 1 2 l kfk2 ≤ |f| . m K ⊂ Bx (n1) ∪ Bx (n2) ∪ · · · ∪ Bx (nl) = Bx (nl) .

This shows that |·| and k·k2 are equivalent norms. Similarly one shows that k·k This shows K is bounded. and k·k2 are equivalent and hence so are |·| and k·k . Lemma 9.46. Suppose that K ⊂ X is an open cover compact set, then K is Corollary 9.40. If (Z, k·k) is a finite dimensional normed space, then A ⊂ Z closed. is sequentially compact iff A is closed and bounded relative to the given norm, Proof. We will shows that Kc is open. To this end suppose x ∈ Kc. Then k·k . 1 let εk := 3 d (x, k) > 0 for all k ∈ K. It then follows that Bx (εk) ∩ Bk (εk) = ∅ Corollary 9.41. Every finite dimensional normed vector space (Z, k·k) is com- for all k ∈ K. As {Bk (εk)}k∈K is an open cover K, there exists Λ ⊂f K such plete. In particular any finite dimensional subspace of a normed vector space is that K ⊂ ∪k∈ΛBk (εk) . If we now let δ := mink∈Λ εk > 0, then automatically closed. Bx (δ) ∩ Bk (εk) ⊂ Bx (εk) ∩ Bk (εk) = ∅ for all k ∈ Λ Proof. If {f }∞ ⊂ Z is a Cauchy sequence, then {f }∞ is bounded and n n=1 n n=1 and therefore hence has a convergent subsequence, gk = fn , by Corollary 9.40. It is now k Bx (δ) ∩ K ⊂ Bx (δ) ∩ [∪k∈K Bk (εk)] = ∅. routine to show limn→∞ fn = f := limk→∞ gk.

1 We will usually simply say A is compact in this case.

Page: 99 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 100 9 Topological Considerations

Theorem 9.47. The open cover compact subsets of Rn are the closed and bounded sets.

Proof. Let us ﬁrst suppose that K = [−M,M]n for some positive integer M and for sake of contradiction let us suppose that U is an open cover of K −k with not ﬁnite subcover. For any k ∈ N let Λk := [−M,M) ∩ `2 : ` ∈ Z n and for x ∈Λk , let

k −k −k Cx := x1, x1 + 2 × · · · × xn, xn + 2 .

k k n We now let C = Cx : x ∈ Λk . The cubes have the following properties; k 1. K = ∪C∈Ck C√for all k and if C ∈ C , then C = ∪F ∈Ck+1 F. 2. diam (C) = n · 2−k for all C ∈ Ck. Fig. 9.5. Nested Sequence of cubes.

1 By item 1. there must be a C1 ∈ C such that U has no finite subcover of 2 C1. Similarly there exists C2 ∈ C such that C2 ⊂ C1 such that U has no finite k Exercise 9.29. Suppose f : X → Y is continuous and K ⊂ X is open cover subcover of C2. Continuing this way inductively we may construct Ck ∈ C such compact, then f(K) is an open cover compact subset of Y. Give an example of that C1 ⊃ C2 ⊃ C3 ⊃ ... and U has no finite subcover of Ck for any k. Choose √ −k continuous map, f : X → Y, and an open cover compact subset K of Y such a point zk ∈ Ck for all k ∈ N. Because diam (Ck) = n · 2 → 0 one learns ∞ that f −1(K) is not open cover compact. that {zk}k=1 is a Cauchy sequence and hence convergent. Let z = limk→∞ zk which is in C for all k as each of these cubes are closed sets. Since U is an open k Exercise 9.30 (Extreme value theorem III). Let (X, d) be an open cover cover of K, there exists V ∈ U such that z ∈ V. As V is open, there exists ε > 0 compact metric space and f : X → be a continuous function. Show −∞ < such that B (ε) ⊂ V. On the other hand because diam (C ) → 0 as k → ∞, R z k inf f ≤ sup f < ∞ and there exists a, b ∈ X such that f(a) = inf f and it follows that C ⊂ B (ε) ⊂ V for all k sufficiently large which violates the k z f(b) = sup f. Hint: use Exercise 9.29 and Theorem 9.47. condition that U has no finite subcover of Ck for any k. Now suppose that K is a general closed and bounded subset of Rn and U Exercise 9.31 (Uniform Continuity). Let (X, d) be an open cover compact is an open cover of K. Since K is bounded, K ⊂ [−M,M]n for some M ∈ N. metric space, (Y, ρ) be a metric space and f : X → Y be a continuous function. Since U 0 := U∪ {Kc} is an open cover of [−M,M]n it has a finite subcover, say Show that f is uniformly continuous, i.e. if ε > 0 there exists δ > 0 such that c c {U1,...,Ul} ∪ {K } . As K ∩ K = ∅ we must have that {U1,...,Ul} covers K ρ(f(y), f (x)) < ε if x, y ∈ X with d(x, y) < δ. – i.e. {U1,...,Ul} ⊂ U is a finite subcover of K. Exercise 9.32. Suppose f : X → Y is continuous and K ⊂ X is open cover compact, then f(K) is an open cover compact subset of Y. Give an example of Proposition 9.48. Suppose that K ⊂ X is an open cover compact set and continuous map, f : X → Y, and an open cover compact subset K of Y such n −1 F ⊂ K is a closed subset. Then F is open cover compact. If {Ki}i=1 is a finite that f (K) is not open cover compact. n collections of open cover compact subsets of X then K = ∪i=1Ki is also an open cover compact subset of X. Exercise 9.33 (Dini’s Theorem). Let X be an open cover compact metric space and fn : X → [0, ∞) be a sequence of continuous functions such that c Proof. Let U ⊂ τ be an open cover of F, then U∪ {F } is an open cover of fn (x) ↓ 0 as n → ∞ for each x ∈ X. Show that in fact fn ↓ 0 uniformly in x, K. The cover U∪ {F c} of K has a finite subcover which we denote by U ∪ {F c} 0 i.e. supx∈X fn (x) ↓ 0 as n → ∞. Hint: Given ε > 0, consider the open sets c where U0 ⊂⊂ U. Since F ∩ F = ∅, it follows that U0 is the desired subcover Vn := {x ∈ X : fn (x) < ε}. of F. For the second assertion suppose U ⊂ τ is an open cover of K. Then U covers each compact set Ki and therefore there exists a finite subset Ui ⊂⊂ U Definition 9.49. A collection F of closed subsets of a metric space (X, d) has n for each i such that Ki ⊂ ∪Ui. Then U0 := ∪i=1Ui is a finite cover of K. the finite intersection property if ∩F0 6= ∅ for all F0 ⊂⊂ F.

Page: 100 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9.7 Equivalence of Sequential and Open Cover Compactness in Metric Spaces 101 The notion of open cover compactness may be expressed in terms of closed Proof. The proof will consist of showing that a ⇒ b ⇒ c ⇒ a. sets as follows. (a ⇒ b) We will show that not b ⇒ not a. Suppose there exists ∞ {xn}n=1 ⊂ X which has no convergent subsequence. In this case the set Proposition 9.50. A metric space X is open cover compact iff every family of S := {xn ∈ X : n ∈ } must be an inifnite set as we have already seen finite X T N closed sets F ⊂ 2 having the finite intersection property satisfies F 6= ∅. sets are sequentially compact. For every x ∈ X we must have Proof. The basic point here is that complementation interchanges open ε (x) := lim inf d (xn, x) > 0 and closed sets and open covers go over to collections of closed sets with empty n→∞ intersection. Here are the details. ∞ X since otherwise there would be a subsequence {xnk }k=1 such that (⇒) Suppose that X is open cover compact and F ⊂ 2 is a collection of lim d (x , x) = 0, i.e. lim x = x. We now let V := B 1 ε (x) closed sets such that T F = ∅. Let k→∞ nk k→∞ nk x x 2 and observe that xn can be in Vx for only finitely many n – otherwise we would conclude that lim inf d (x , x) ≤ ε (x) /2. From these observations, U = F c := {Cc : C ∈ F} ⊂ τ, n→∞ n U := {Vx : x ∈ X} is an open cover of X with no finite subcover. Indeed, if c Λ ⊂f X, then we must still have xn ∈ ∪x∈ΛVx for only finitely many n and in then U is a cover of X and hence has a finite subcover, U0. Let F0 = U0 ⊂⊂ F, particular ∪x∈ΛVx can not cover S. then ∩F0 = ∅ so that F does not have the finite intersection property. ∞ (b ⇒ c) Suppose {xn}n=1 ⊂ X is a Cauchy sequence. By assumption there (⇐) If X is not open cover compact, there exists an open cover U of X with ∞ exists a subsequence {xnk }k=1 which is convergent to some point x ∈ X. Since no finite subcover. Let ∞ {xn}n=1 is Cauchy it follows that xn → x as n → ∞ showing X is complete. F = U c := {U c : U ∈ U} , Now for sake of contradiction suppose that X is not totally bounded. There there exists ε > 0 for which X is not ε – bounded. In particular, then F is a collection of closed sets with the finite intersection property while U := {Bx (ε): x ∈ X} is an open cover of X with no finite subcover. We now T ∞ F = ∅. use this to construct a sequence {xn}n=1 ⊂ X. Choose x1 ∈ X at random, then

choose x2 ∈ X \ Bx1 (ε) , then x3 ∈ X \ [Bx1 (ε) ∪ Bx2 (ε)] ... x ∈ X \ B (ε) ∪ B (ε) ∪ · · · ∪ B (ε) ,.... 9.7 Equivalence of Sequential and Open Cover n x1 x2 xn−1 Compactness in Metric Spaces The process may be continued indefinitely as U has no finite subcover. By ∞ construction we have chosen {xn} such that d{x ,...,x }(xn) ≥ ε for all n Definition 9.51. A metric space (X, d) is ε – bounded (ε > 0) if there exists a n=1 1 n−1 and therefore d (xk, xl) ≥ ε for all k 6= l. Every subsequence will share this finite cover of X by balls of radius ε. We further say (X, d) is totally bounded property, i.e. not be Cauchy, and hence can not be convergent. if it is ε – bounded for all ε > 0. (c ⇒ a) For sake of contradiction, assume there exists an open cover V = {V } of X with no finite subcover. Since X is totally bounded for each Remark 9.52 (Totally bounded means almost finite). An equivalent way to state α α∈A n ∈ N there exists Λn ⊂⊂ X such that that (X, d) is ε – bounded is to say there exists a finite subset Λ = Λε ⊂f X such that dΛ (x) < ε for all x ∈ X. In other words, (X, d) is ε – bounded iff X [ [ X = Bx(1/n) ⊂ Cx(1/n). is a finite subset within an ε – error. x∈Λn x∈Λn Theorem 9.53. Let (X, d) be a metric space. The following are equivalent. Choose x1 ∈ Λ1 such that no finite subset of V covers K1 := Cx1 (1). Since K = ∪ K ∩ C (1/2), there exists x ∈ Λ such that K := K ∩ C (1/2) (a) X is open cover compact. 1 x∈Λ2 1 x 2 2 2 1 x2 (b) X is sequentially compact. can not be covered by a finite subset of V, see Figure 9.6. Continuing this way inductively, we construct sets K = K ∩C (1/n) with x ∈ Λ such that no (c) X is totally bounded and complete. n n−1 xn n n Kn can be covered by a finite subset of V. Now choose yn ∈ Kn for each n. Since ∞ {Kn}n=1 is a decreasing sequence of closed sets such that diam(Kn) ≤ 2/n, it follows that {yn} is a Cauchy and hence convergent with

Page: 101 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 102 9 Topological Considerations ∞ y = lim yn ∈ ∩m=1Km. 9.8 Connectedness n→∞

Since V is a cover of X, there exists V ∈ V such that y ∈ V. Since Kn ↓ {y} Deﬁnition 9.55. Let (X, d) be a metric space. Two subset A and B of X are 2 and diam(Kn) → 0, it now follows that Kn ⊂ V for some n large. But this separated if A¯ ∩ B = ∅ = A ∩ B.¯ A set E ⊂ X is disconnected if E = A ∪ B violates the assertion that Kn can not be covered by a ﬁnite subset of V. where A and B are two non-empty separated sets, otherwise E is said to be connected.

Theorem 9.56 (The Connected Subsets of R). The connected subsets of R are intervals. Proof. We will break the proof into two parts. First we show if E is connected then E is an interval. Then we show if E is disconnected then E is not an interval. 1) Suppose that E ⊂ R is a connected subset and that a, b ∈ E with a < b. If there exists c ∈ (a, b) such that c∈ / E, then A := (−∞, c) ∩ E and B := (c, ∞) ∩ E would be two non-empty separated subsets such that E = A ∪ B. Hence (a, b) ⊂ E. Let α := inf(E) and β := sup(E) and choose αn, βn ∈ E such that αn < βn and αn ↓ α and βn ↑ β as n → ∞. By what we ∞ have just shown, (αn, βn) ⊂ E for all n and hence (α, β) = ∪n=1(αn, βn) ⊂ E. Fig. 9.6. Nested Sequence of cubes. From this it follows that E = (α, β), [α, β), (α, β] or [α, β], i.e. E is an interval. 2) Now suppose that E is a disconnected subset of R. Then E = A∪B where A and B are non-empty separated sets and let a ∈ A and b ∈ B. After relabelling A and B if necessary we may assume that a < b. Let p = sup ([a, b] ∩ A) . Since p ∈ A¯ ∩ [a, b] it follows that p∈ / B and a ≤ p ≤ b. Since b ∈ B, p 6= b and hence Corollary 9.54. The compact subsets of Rn are the closed and bounded sets. a ≤ p < b. i) if p∈ / A then p∈ / E and a < p < b which shows E is not an interval. 3 Proof. If K is closed and bounded then K is complete (being the closed ii) if p ∈ A, then p∈ / B¯ and there exists p1 such that a ≤ p < p1 0, let again E is not an interval. n n n Λδ = δZ ∩ [−M,M] := {δx : x ∈ Z and δ|xi| ≤ M for i = 1, 2, . . . , n}. Lemma 9.57. Suppose that f : X → Y is a continuous function between two We will show, by choosing δ > 0 suﬃciently small, that metric spaces (X and Y ) and A and B are separated subsets of Y. Then α := n f −1 (A) and β := f −1 (B) are separated subsets of X. In particular, if E ⊂ X K ⊂ [−M,M] ⊂ ∪x∈Λδ B(x, ε) (9.7) is connected, then f (E) is connected in Y. which shows that K is totally bounded. Hence by Theorem 9.53, K is compact. n −1 −1 ¯ −1 ¯ Suppose that y ∈ [−M,M] , then there exists x ∈ Λδ such that |yi − xi| ≤ δ Proof. Since α := f (A) ⊂ f A and f A is the closed being the for i = 1, 2, . . . , n. Hence inverse image under a continuous function of the closed set A,¯ it follows that α¯ ⊂ f −1 A¯ . Thus if x ∈ α¯ ∩ β, then f (x) ∈ A¯ ∩ B = ∅ and henceα ¯ ∩ β = ∅. n 2 X 2 2 Similarly one shows α ∩ β¯ = ∅ as well. d (x, y) = (yi − xi) ≤ nδ i=1 2 Notice that separated sets are disjoint. The sets A = (0, 1) and B = [1, ∞) are √ √ disjoint but not separated. which shows that d(x, y) ≤ nδ. Hence if choose δ < ε/ n we have shows that 3 This is because B¯c is an open subset of . d(x, y) < ε, i.e. Eq. (9.7) holds. R

Page: 102 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 9.8 Connectedness 103 If f (E) disconnected, there exists non-empty separated subsets, A and B Definition 9.63. A subset, C, of a vector space, X, is convex if for all a, b ∈ C of f (E) , so that f (X) = A ∪ B. The sets α := f −1 (A) and β := f −1 (B) are the path, now non-empty separated subsets of X. It now follows that α0 := α ∩ E and σ (t) = a + t (b − a) = (1 − t) a + tb for 0 ≤ t ≤ 1 β0 := β ∩ E are non-empty sets such that is contained in C. ¯ ¯ α¯0 ∩ β0 ⊂ α¯ ∩ β = ∅ and α0 ∩ β0 ⊂ α ∩ β = ∅ Example 9.64. Every convex subset, C, of a normed vector space (X, k·k) is path connected and hence connected. which would imply E is disconnected. Thus if E is connected we must have that f (E) is connected. Exercise 9.35. Suppose that (X, k·k) is a normed space, x ∈ X, and R > 0. Show the open and closed balls in X,Bx (R) and Cx (R) , are both convex sets Theorem 9.58 (Intermediate Value Theorem). Suppose that (X, d) is a and hence path connected. connected metric space and f : X → R is a continuous map. Then f satisfies the intermediate value property. Namely, for every pair x, y ∈ X such that Definition 9.65. A metric space X is locally path connected if for each f (x) < f(y) and c ∈ (f (x) , f(y)), there exists z ∈ X such that f(z) = c. x ∈ X, there is an open neighborhood V ⊂ X of x which is path connected. Proposition 9.66. Let X be a metric space. Proof. By Lemma 9.57, f (X) is a connected subset of R. So by Theorem 9.56, f (X) is a subinterval of R and this completes the proof. 1. If X is connected and locally path connected, then X is path connected. n Lemma 9.59. If E ⊂ X is a connected set and E ⊂ A ∪ B where A and B are 2. If X is any connected open subset of R , then X is path connected. two separated sets, then E ⊂ A or E ⊂ B. Exercise 9.36. Prove item 1. of Proposition 9.66, i.e. if X is connected and locally path connected, then X is path connected. Hint: fix x ∈ X and let Proof. Let α := E ∩ A and β := E ∩ B, thenα ¯ ∩ β ⊂ A¯ ∩ B = ∅ and 0 W denote the set of x ∈ X such that there exists σ ∈ C([0, 1],X) satisfying similarly α ∩ β¯ = ∅. Thus E = α ∪ β with α and β being separated sets. Since σ(0) = x and σ(1) = x. Then show W is both open and closed. E is connected we must have α = ∅ or β = ∅, i.e. E ⊂ B or E ⊂ A. 0 Exercise 9.37. Prove item 2. of Proposition 9.66, i.e. if X is any connected Proposition 9.60. Suppose that F and G are connected subsets of X such that open subset of n, then X is path connected. F¯ ∩ G 6= ∅, then E = F ∪ G is connected in X as well. R Proof. Suppose that E = A ∪ B where A and B are separated sets. Then 9.8.1 Connectedness Problems from Lemma 9.59 we know that F ⊂ A or F ⊂ B and similarly G ⊂ A or G ⊂ B. If both F and G are in the same set (say A), then E ⊂ A ⊂ E and B Exercise 9.38. In this exercise we will work inside the metric space Q with must be empty. On the other hand if F ⊂ A and G ⊂ B, then ∅= 6 F¯∩G ⊂ A¯∩B d (x, y) := |x − y| for all x, y ∈ Q. Let a, b ∈ Q with a < b and let which would violate A and B being separated. Thus we have shown there does J := [a, b] ∩ Q = {x ∈ Q : a ≤ x ≤ b} . not exists two non-empty separated sets A and B such that E = A ∪ B, i.e. E is connected. Show J is disconnected in Q. Definition 9.61. A subset E of a metric space X is path connected if to Exercise 9.39. Suppose a < b and f :(a, b) → R is a non-decreasing function. Show if f satisfies the intermediate value property (see Theorem 9.58), then f every pair of points {x0, x1} ⊂ E there exists σ ∈ C([0, 1],E), such that σ(0) = is continuous. x0 and σ(1) = x1. We refer to σ as a path joining x0 to x1. Proposition 9.62. Every path connected subset, E, of a metric space X is Exercise 9.40. Suppose −∞ < a < b ≤ ∞ and f :[a, b) → R is a strictly connected. increasing continuous function. Using the intermediate value theorem, one sees that f ([a, b)) is an interval and since f is strictly increasing it must of the Exercise 9.34. Prove Proposition 9.62, i.e. if E ⊂ X is path connected then form [c, d) for some c ∈ R and d ∈ R¯ with c < d. Show the inverse function E is connected.. Hint: for sake of contradiction suppose that A and B are two f −1 :[c, d) → [a, b) is continuous and is strictly increasing. In particular if non-empty separated subsets of X such that E = A ∪ B and choose a path n ∈ N, apply this result to f (x) = xn for x ∈ [0, ∞) to construct the positive connecting a point in A to a point in B. nth – root of a real number. Compare with Exercise ??.

Page: 103 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Exercise 9.41. Let

2 −1 X := (x, y) ∈ R : y = sin(x ) with x 6= 0 ∪ {(0, 0)} equipped with the relative topology induced from the standard topology on R2. Show X is connected but not path connected.

Remark 9.67 (Structure of open sets in R). Let V ⊂ R be an open set. For x ∈ V, let ax := inf {a :(a, x] ⊂ V } and bx := sup {b :[x, b) ⊂ V } . Since V is open, ax < x < bx and it is easily seen that Jx := (ax, bx) ⊂ V. Moreover if y ∈ V and Jx ∩ Jy 6= ∅, then Jx = Jy. The collection, {Jx : x ∈ V } , is at most countable since we may label each J ∈ {Jx : x ∈ V } by choosing a rational number r ∈ J. Letting {Jn : n < N}, with N = ∞ allowed, be an enumeration ` of {Jx : x ∈ V } , we have V = n

Calculus

10 Diﬀerential Calculus in One Real Variable

10.1 Basics of the Derivative Notation 10.5 (ε (·) ,O (·) , o (·) Notation) To simplify proofs below, we will often let ε (h) ,O (h) , and o (h) denote generic X or R – valued functions defined Definition 10.1. If f : J → X is a function we say lim f (t) = x ∈ X iff t→t0 in some deleted neighborhood of 0 ∈ R (i.e. for h ∈ (−δ, δ) \{0}) such that for all ε > 0 there exists δ > 0 such that 1. limh→0 ε (h) = 0, kf (t) − xk ≤ ε whenever 0 < |t − t0| ≤ δ. 2. there exists C < ∞ such that kO (h)k ≤ C |h| for h near 0 or equivalently, kO(h)k f(t)−f(t0) lim suph→0 |h| < ∞, A function f : J → X is said to be differentiable at t0 ∈ J iff limt→t 0 t−t0 o(h) ˙ 0 df 3. limh→0 = 0, i.e. for all c > 0 there exists δ = δ (c) > 0 such that exists in X. We denote the limit by f (t0) or f (t0) or by (t0) , i.e. h dt ko (h)k ≤ c |h| for 0 < |h| ≤ δ. ˙ f (t) − f (t0) f (t0) = lim We will often assume that each of these functions has been extended to h = 0 t→t0 t − t 0 by setting there values at 0 to be 0. Example 10.2. If f (t) = at + b, then f˙ (t) = a for all t ∈ . Indeed, R Remark 10.6. Using the above notation we have, O (h) ε (h) = o (h) ,O (h) + f (t) − f (t0) a (t − t0) O (h) = O (h) = O (h) + o (h) ,O (h) + ε (h) = ε (h) = ε (h) + ε (h) , etc. etc. We = = a for all t 6= t0. t − t0 t − t0 may also write, limt→t0 f (t) = x as f (t0 + h) = x + ε (h) and f is differentiable at t with derivative f˙ (t ) iff Remark 10.3. One may also define the derivative of f by 0 0 ˙ f (t0 + h) = f (t0) + f (t0) h + o (h) . ˙ f (t0 + h) − f (t0) f (t0) = lim h→0 h Lemma 10.7. If f is differentiable at t0 then it is continuous at t0. f(t)−f(t ) Proof. By definition, provided the limit exists. To see this is the case let ∆ (t) := 0 − L . t−t0 0 Then limh→0 ∆ (t0 + h) = 0 iff for all ε > 0 there exists δ > 0 such that f (t0 + h) = f (t0) + f (t0) h + o (h) = f (t0) + ε (h) ∆ (t0 + h) ≤ ε if 0 < |h| ≤ δ. Letting t = t0 + h, we see this is equivalent to and so limh→0 f (t0 + h) = f (t0) . ∆ (t) ≤ ε if 0 < |h| = |t − t0| ≤ δ, i.e. iff limt→t0 ∆ (t) = 0. −1 d Proposition 10.4. The function f (t) := 1/t = t defined for t 6= 0 is differ- Theorem 10.8 (Linearity of dt ). Suppose that f :(a, b) → X and g :(a, b) → ˙ 2 entiable and f (t) = −1/t for all t 6= 0. X are differentiable at t0 ∈ (a, b) and λ ∈ R, then

Proof. We have d 0 0 (f + λg)(t0) = f (t0) + λg (t0) . f (t + h) − f (t) 1 1 1 1 t − (t + h) dt = − = h h t + h t h (t + h) t Proof. We are given, 1 1 = − → − as h → 0. ˙ (t + h) t t2 f (t0 + h) = f (t0) + f (t0) h + o (h) and g (t0 + h) = g (t0) +g ˙ (t0) h + o (h) 1 Here we have used the fact that t is continuous for the last limit. and therefore, 108 10 Differential Calculus in One Real Variable ˙ f (t0 + h) + λg (t0 + h) = f (t0) + f (t0) h + o (h) + λg (t0) + λg˙ (t0) h + λo (h) Lemma 10.10. We have the following estimates, o (O (h)) = o (h) and h ˙ i O (o (h)) = o (h) . = f (t0) + λg (t0) + f (t0) + λg˙ (t0) h + o (h) Proof. Suppose that f (h) and g (h) are functions such that f (h) = o (h) from which the result follows. and g (h) = O (h) . These statements are equivalent to the existence of a constant C < ∞ and a function ε (h) such that lim ε (h) = 0 and Theorem 10.9 (Product rule I). Suppose that f :(a, b) → R and g :(a, b) → h→0 X are differentiable at t0 ∈ (a, b) , then |f (h)| ≤ |ε (h)| |h| and |g (h)| ≤ C |h| for h near 0. d (f · g)(t ) = f 0 (t ) g (t ) + f (t ) g0 (t ) . (10.1) With these estimates in hand we have dt 0 0 0 0 0 |f (g (h))| ≤ |ε (g (h))| |g (h)| ≤ C |ε (g (h))| |h| Proof. We have, and (f · g)(t + h) = f (t + h) · g (t + h) 0 0 0 |g (f (h))| ≤ C |f (h)| ≤ C |ε (h)| |h| . = f (t ) + f˙ (t ) h + o (h) (g (t ) +g ˙ (t ) h + o (h)) 0 0 0 0 Since limh→0 C |ε (g (h))| = 0 we see that f (g (h)) = o (h) and since h ˙ i limh→0 C |ε (h)| = 0 it follows that g (f (h)) = o (h) . = f (t0) g (t0) + f (t0) g (t0) + f (t0)g ˙ (t0) h + o (h) (10.2) Theorem 10.11 (Chain Rule). Suppose that g :(c, d) → X and f :(a, b) → 2 ˙ 0 wherein we have used h = o (h) , ho (h) = o (h) , and o (h) o (h) = o (h) . From (c, d) are given functions such that f (t0) at t0 ∈ (a, b) and g (w0) exists at Eq. (10.2) we deduce Eq. (10.1). z0 := f (t0) . Then n n Exercise 10.1. Suppose that f :(a, b) → R and g :(a, b) → R are differen- d 0 0 0 0 g (f (t)) |t=t0 = g (z0) f (t0) = g (f (t0)) f (t0) . tiable at t0 ∈ (a, b) , show dt

d 0 0 Proof. Let h (t) := g (f (t)) . Then by assumption, (f · g)(t0) = f (t0) · g (t0) + f (t0) · g (t0) dt 0 f (t0 + ∆t) = f (t0) + f (t0) ∆t + o (∆t) and n where now “·” denotes the usual dot product in , i.e. 0 R g (z0 + ∆z) = g (z0) + g (z0) ∆z + o (∆z) . n X x · y := x y for all x, y ∈ n. (10.3) Letting i i R 0 i=1 ∆z := f (t0 + ∆t) − f (t0) = f (t0) ∆t + o (∆t) n which is small for ∆t small, we find, Exercise 10.2. Let f :(a, b) → R and write f (t) = (f1 (t) , . . . , fn (t)) where ˙ fk :(a, b) → R are functions for 1 ≤ k ≤ n. Show; if f (t) exists at t ∈ (a, b) , h (t + ∆t) = g (f (t + ∆t)) = g (z + ∆z) ˙ 0 0 0 then fk (t) exists for all k and 0 = g (z0) + g (z0) ∆z + o (∆z) ˙ ˙ ˙ 0 0 0 f (t) = f1 (t) ,..., fn (t) . = h (t0) + g (z0)[f (t0) ∆t + o (∆t)] + o (f (t0) ∆t + o (∆t)) 0 0 = h (t0) + g (z0) f (t0) ∆t + o (∆t) . (10.4) th n Hint: fk (t) = ek · f (t) where ek is the k standard basis vector for R wherein we have used Lemma 10.10 to conclude, o (f 0 (t ) ∆t + o (∆t)) = n 0 Exercise 10.3. Suppose that f :(a, b) → R and write f (t) = o (∆t) . The result follows Eq. (10.4) and the definition of the derivative. ˙ (f1 (t) , . . . , fn (t)) where fk :(a, b) → R. Show; if fk (t) exists at t ∈ (a, b) for all k, then f˙ (t) exists and Corollary 10.12 (Quotient Rule). Suppose that f :(a, b) → R and g : (a, b) → X are differentiable at t0 ∈ (a, b) and f (t0) 6= 0, then f˙ (t) = f˙ (t) ,..., f˙ (t) . 1 n d g f (t ) g0 (t ) − f 0 (t ) g (t ) (t ) = 0 0 0 0 . n 0 2 P dt f f (t0) Hint: f (t) = k=1 fk (t) ek.

Page: 108 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.2 First Derivative Test and the Mean Value Theorems 109

1 1 n Proof. First of let ψ (z) := z , then f = ψ (f) and so by the chain rule, 2. Show q (x) := Ax · x is a continuous function on R . Let S := {x ∈ Rn : kxk = 1} be the closed and bounded and hence compact n d 1 0 0 1 0 unit sphere in R . By the extreme value theorem, λ := maxx∈S q (x) exists. (t0) = ψ (f (t0)) f (t0) = − 2 f (t0) . dt f f (t0) Let x0 ∈ S be chosen so that

Hence by the product rule, λ = q (x0) = Ax0 · x0.

Ax·x x n d g d 1 d 1 1 0 3. Explain why R (x) = = q is a continuous function on \{0} (t ) = · g (t ) = (t ) · g (t ) + g (t ) kxk2 kxk R dt f 0 dt f 0 dt f 0 0 f (t ) 0 0 and show x0 is also a maximizer of R. 1 1 n = − f 0 (t ) · g (t ) + g0 (t ) 4. For any v ∈ R let fv (t) = R (x0 + tv) which is deﬁned for t such that 2 0 0 0 ˙ f (t0) f (t0) kx0 + tvk= 6 0 and in particular for t near zero. Compute fv (t) for t near 0 0 0 f (t0) g (t0) − f (t0) g (t0) and in particular show = 2 . f (t0) ˙ fv (0) = 2 (Ax0 − λx0) · v

5. Noting that fv (t) has a maximum at t = 0, the first derivative implies ˙ fv (0) = 0. By making a judicious choice for v, show Ax0 = λx0, i.e. x0 is an eigenvector of A with eigenvalue λ. 10.2 First Derivative Test and the Mean Value Theorems Exercise 10.5 (Multiplication Operators). Let V := C ([0, 1] , R) and A : Definition 10.13. A function f :(a, b) → R has as local minimum (maximum) V → V be the operation of multiplication by t, i.e. for f ∈ V let Af ∈ V be at t0 ∈ (a, b) if there exists δ > 0 such that f (t) ≥ f (t0)(f (t) ≤ f (t0)) when define by |t − t0| ≤ δ. (Af)(t) = tf (t) for all t ∈ [0, 1] . Show; if there is a λ ∈ and f ∈ V such that Af = λf, then f ≡ 0, i.e. A has Theorem 10.14 (First Derivative Test). If f :(a, b) → R has a local ex- R no eigenvectors (or eigenfunctions if you prefer). tremum (i.e. local minimum or maximum) at t0 ∈ (a, b) and f is differentiable 0 [Moral: The reader should compare this exercise with Exercise 10.4. Notice at t0, then f (t0) = 0. that if we define an inner product on V by Proof. For sake of definiteness, suppose that t is a local minimum for f. 0 Z 1 Then for h > 0 small, (f, g) = f (t) g (t) dt 0 f (t0 + h) − f (t0) f (t0 − h) − f (t0) ≥ 0 and ≤ 0. then (Af, g) = (f, Ag) , i.e. A is symmetric. Nevertheless, A has no eigenvectors. h −h This may be considered as another proof the that unit sphere in V is not 0 0 Therefore letting h ↓ 0 in these two inequalities leads to f (t0) ≥ 0 and f (t0) ≤ compact!] 0 0 which implies f (t0) = 0. Theorem 10.15 (Rolle’s Theorem). Suppose that f :[a, b] → R is a con- Exercise 10.4 (Existence of Eigenvectors for Symmetric Matrices). tinuous function such that f (a) = L = f (b) and f is differentiable on (a, b) . 0 Let A be an n × n symmetric real matrix. Recall this implies (and is equiv- Then there exists t0 ∈ (a, b) such that f (t0) = 0. n n alent to) Ax · y = x · Ay for all x, y ∈ R where “·” is the dot product on R as Proof. If f is not constant then there exists t ∈ (a, b) such that either in Eq. (10.3). In this problem you are going to show that A has an eigenvector f (t) > L or f (t) < L. For definiteness let assume the first case. Then by the by proving the following statements. extreme value theorem, there exists t0 ∈ [a, b] such that f (t0) ≥ f (t) for all qP 2 t ∈ [a, b] . Moreover this maximizer can not be at a or b as max f ([a, b]) > L = 1. Show |Ax · y| ≤ C kxk kyk where C := i,j Aij – the Hilbert-Schmidt f (a) = f (b) . Therefore f has a maximum at some point t0 ∈ (a, b) . It then norm of A. 0 follows by the first derivative test that f (t0) = 0.

Page: 109 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 110 10 Diﬀerential Calculus in One Real Variable

Theorem 10.16 (Mean Value Theorem). Suppose that f :[a, b] → R is In particular, if f :[a, b] → C =∼ R2 then there exists c ∈ (a, b) such that a continuous function such that f is diﬀerentiable on (a, b) . Then there exists ˙ t0 ∈ (a, b) such that |f (b) − f (a)| ≤ f (c) (b − a) . f (b) − f (a) f˙ (t ) = . 0 b − a [Although we do not prove it now, this result remains true when X is any normed vector space however the proof would require the “Hahn-Banach” theorem. We Proof. Let will give a proof of a closely related result later based on the fundamental theorem t − a of calculus.] h (t) := f (t) − f (a) + (f (b) − f (a)) . b − a n Proof. For v ∈ R , let fv (t) := v · f (t) . Applying the mean value theorem to fv shows there exists c (v) ∈ (a, b) such that Then h (a) = 0 = h (b) and so by Rolle’s theorem, there exist t0 ∈ (a, b) such that d v · [f (b) − f (a)] = v · f (b) − v · f (a) = v · f (t) |t=c(v) (b − a) 1 dt 0 = h˙ (t ) = f˙ (t ) − (f (b) − f (a)) h i 0 0 b − a = v · f˙ (c (v)) (b − a) (10.5) f (b) − f (a) = f˙ (t ) − . 0 b − a wherein we have made use of Exercise 10.1. If f (b) = f (a) there is nothing to prove so we will assume that f (b) 6= f (a) . In the latter case we take, Alternatively, apply the generalized mean value Theorem 10.46 below with f (b) − f (a) g (t) = t. v := kf (b) − f (a)k Corollary 10.17. Suppose that f :(a, b) → R is diﬀerentiable on (a, b) . in Eq. (10.5) in order to learn, 1. If f 0 ≥ 0 on (a, b) then f is non-decreasing on (a, b) . h i 2. If f 0 ≤ 0 on (a, b) then f is non-increasing on (a, b) . kf (b) − f (a)k = v · [f (b) − f (a)] = v · f˙ (c (v)) (b − a) 0 3. If f = 0 on (a, b) , then f is constant. ˙ ˙ ≤ kvk f (c (v)) (b − a) = f (c (v)) (b − a) . Proof. For a < x1 < x2 < b there exists c = c (x1, x2) ∈ (x1, x2) such that

0 f (x2) − f (x1) = f (c (x1, x2)) (x2 − x1) . Corollary 10.20. Suppose that f :[a, b] → Rn is a continuous function such The result easily follow from this identity. that f 0 (t) = 0 for all t ∈ (a, b) , then f (t) = f (a) for all t ∈ [a, b] , i.e. f is constant on [a, b] . Remark 10.18. The mean value Theorem 10.16 does not hold for vector valued Proof. By the mean value inequality for all β ∈ [a, b] we have functions and in particular not for C – valued functions. For example let f (t) = (u (t) , v (t)) ∈ R2 be differentiable function such that f (0) = f (1) = 0 ∈ R2. 0 1 kf (β) − f (a)k ≤ kf (c)k (β − a) = 0 We can arrange foru ˙ (t) to only be zero at 2 andv ˙ (t) to be zero only at 3/4. In this case both u and v satisfy the mean value theorem with two different for some c ∈ (a, β) . This shows that f (β) = f (a) for all β ∈ [a, b] . times and there is no common time t0 such thatu ˙ (t0) = 0 =v ˙ (t0) . However, the following mean value inequality does still hold. 10.3 Limsup and Liminf revisited Theorem 10.19 (Mean Value Inequality). Suppose that X = Rn and f : [a, b] → X = Rn is a continuous function which is differentiable on (a, b) . Then For proofs to follow it will be convenient to introduce the notion of lim sup and there exists c ∈ (a, b) such that lim inf for function g : Y \{y0} → R, where Y is a metric space and y0 ∈ Y. [The reader will find many more details regarding the material in this section ˙ kf (b) − f (a)k ≤ f (c) (b − a) . in the Appendix D which has been generously supplied by Ali Behzadan.]

Page: 110 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.3 Limsup and Liminf revisited 111 Notation 10.21 The deleted open ball at y with radius δ > 0 is deﬁned by L − ε ≤ g (y) ≤ L + ε for all y ∈ B0 (δ) 0 y0

0 which, by definition, implies limy→y g (y) = L. B (δ) := By (δ) \{y0} = {y ∈ Y : 0 < d (y, y0) < δ} . 0 y0 0 Next suppose that Definition 10.22. Given g : Y \{y0} → R we let ∞ = L = lim inf g (y) = lim inf g (y) . y→y0 δ↓0 y∈B0 (δ) lim sup g (y) := lim sup g (y) y0 y→y0 δ↓0 y∈B0 (δ) y0 Then for all 0 < M < ∞ there exists δ > 0 such that infy∈B0 (δ) g (y) ≥ M, i.e. y0 and g (y) ≥ M if 0 < d (y, y0) ≤ δ. By definition, this implies limy→y0 g (y) = ∞. lim inf g (y) := lim inf g (y) . The case L = −∞ is similar and will be omitted. y→y0 δ↓0 y∈B0 (δ) y0 ¯ ( =⇒ ) Suppose that L := limy→y0 g (y) exists in R. If L ∈ R then for all ε > 0 there exists δ > 0 such that These limits always exist since supy∈B0 (δ) g (y) decreases as δ decreases y0 while infy∈B0 (δ) g (y) increases as δ decreases. Moreover y0 |g (y) − L| ≤ ε if 0 < d (y, y0) < δ, inf g (y) ≤ sup g (y) for all δ > 0, i.e. 0 y∈By (δ) y∈B0 (δ) 0 0 y0 L − ε ≤ g (y) ≤ L + ε if y ∈ B (δ) . y0 and hence by passing to the limit as δ ↓ 0, From this it follows that

lim inf g (y) ≤ lim sup g (y) . L − ε ≤ inf g (y) ≤ sup g (y) ≤ L + ε y→y 0 0 0 y→y0 y∈By (δ) y∈B (δ) 0 y0 The following proposition is analogous to Proposition 3.28 and the reader may and therefore letting δ ↓ 0 shows find a more general version in Proposition D.9. L − ε ≤ lim inf g (y) ≤ lim sup g (y) ≤ L + ε. y→y0 Proposition 10.23. Keeping the notation introduced above, limy→y0 g (y) ex- y→y0 ists in R¯ iff lim inf g (y) = lim sup g (y) We may now let ε ↓ 0 to conclude that y→y 0 y→y0 L = lim inf g (y) ≤ lim sup g (y) = L y→y in which case 0 y→y0 lim g (y) = lim inf g (y) = lim sup g (y) . (10.6) y→y y→y 0 0 y→y0 which suffices to prove omitted (10.6) in this case. Next suppose that L = ∞. Then for every M ∈ (0, ∞) there exists δ = δ (M) Proof. (⇐=) Suppose that L := lim infy→y g (y) = lim sup g (y) . If 0 y→y0 such that g (y) ≥ M if y ∈ B0 (δ) and therefore, L ∈ R, then given ε > 0 there exists δ > 0 such that y0 lim inf g (y) = lim inf g (y) ≥ M. y→y0 δ↓0 y∈B0 (δ) inf g (y) − L ≤ ε and sup g (y) − L ≤ ε. y0 y∈B0 (δ) 0 y0 y∈By (δ) 0 We may now let M ↑ ∞ to learn In particular we will have lim g (y) = ∞ ≤ lim inf g (y) ≤ lim sup g (y) y→y0 y→y0 y→y L − ε ≤ inf g (y) ≤ sup g (y) ≤ L + ε 0 0 y∈By (δ) y∈B0 (δ) 0 y0 which again proves Eq. (10.6). The case where L = −∞ is proved similarly so and so will be omitted.

Page: 111 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 112 10 Diﬀerential Calculus in One Real Variable

Corollary 10.24. If g : Y \{y0} → [0, ∞) and lim supy→y0 g (y) = 0, then 10.4 Diﬀerentiating past inﬁnite sums limy→y0 g (y) = 0. Theorem 10.28 (Weierstrass M – test II). Suppose that (X, k·k) is a Ba- Proof. This is a direct consequence of Proposition 10.23 and the observation nach space, (Y, d) is a metric space, for each n ∈ N, fn : Y \{y0} → X is a that ∞ P∞ function, and there exists {Mn}n=1 ⊂ [0, ∞) with n=1 Mn < ∞ such that 0 ≤ lim inf g (y) ≤ lim sup g (y) ≤ 0. y→y 0 y→y0 sup kfn (y)kX ≤ Mn ∀ n ∈ N. y∈Y

Under the above assumptions, if limy→y0 fn (y) exists in X for all n ∈ N, then Lemma 10.25. If n ∈ N and gk : Y \{y0} → [0, ∞] are given functions for 1 ≤ k ≤ n, then ∞ ∞ n n X X X X lim fn (y) = lim fn (y) . lim sup gk (y) ≤ lim sup gk (y) . (10.7) y→y0 y→y0 y→y y→y n=1 n=1 0 k=1 k=1 0

Proof. Let xn := limy→y0 fn (y) . Ass the norm on X is continuous it follows Proof. Since P∞ that kxnk ≤ Mn for all n and therefore n=1 xn is absolutely convergent. To n n ﬁnish the proof we must show lim P∞ f (y) = x. We begin with the X X y→y0 n=1 n g (z) ≤ sup g (y) for all z ∈ B0 (δ) , k k y0 estimate; y∈B0 (δ) k=1 k=1 y0 ∞ ∞ X X it follows that x − f (y) = [x − f (y)] n n n n n X X n=1 n=1 sup gk (z) ≤ sup gk (y) . 0 0 N ∞ z∈By (δ) y∈By (δ) 0 k=1 k=1 0 X X = [xn − fn (y)] + [xn − fn (y)] Passing the the limit as δ ↓ 0 in this inequality gives Eq. (10.7). n=1 n=N+1 N ∞ Notation 10.26 (One sided limits) Given a function, f :(a, b) → R we let X X ≤ k[xn − fn (y)]k + [kxnk + kfn (y)k] lim sup f (x) := lim sup f (y) and lim inf f (x) := lim inf f (y) . n=1 n=N+1 x↑b x↑b x↑b x

lim sup f (x) := lim sup f (y) and lim inf f (x) := lim inf f (y) . Taking the lim supy→y0 of this inequality and letting N → ∞ shows x↓a x↓a a

supremum over y such that sup0

Page: 112 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.4 Diﬀerentiating past inﬁnite sums 113

∞ N ∞ Proof. For this proof we will need the mean value inequality (Theorem X X X sup x − fn (y) ≤ sup k[xn − fn (y)]k + 2 Mn. 10.19 or Proposition ?? below) for X – valued functions. As of yet we have only 0 0 be given and then choose N = N (ε) large (but now ﬁxed) so that P∞ ˙ is convergent as well. Furthermore the assumptions imply G (t) := n=1 fn (t) ∞ is absolutely convergent and that G is a continuous function by the Weierstrass X ε 2 M ≤ . M - test, Theorem 7.11. Since for all n ∈ N and h 6= 0 but small, n 2 n=N+1 fn (t + h) − fn (t) ≤ Mn, Then choose δ = δ (ε, N (ε)) > 0 (so δ only depends on ε) so small that h ε k[x − f (y)]k ≤ when 0 < d (y, y ) ≤ δ. it follows from Theorem 10.28 that n n 2N (ε) 0 ∞ F (t + h) − F (t) X fn (t + h) − fn (t) Using these estimates in Eq. Eq. (10.8) shows, gives lim = lim h→0 h h→0 h n=1

∞ N ε ε ∞ X X X fn (t + h) − fn (t) x − fn (y) ≤ + 2 · = ε when 0 < d (y, y0) ≤ δ. = lim 2N 2 h→0 h n=1 n=1 n=1 P∞ ∞ Thus by the definition of the limit we have shown limy→y0 n=1 fn (y) = x. X ˙ = fn (t) = G (t) Theorem 10.29 (Differentiating past a sum). Suppose that fn : J := n=1 (a, b) → X is a sequence of functions satisfying; which completes the proof of Eq. (10.9). ∞ P∞ P Alternative computation of F.˙ Let F (t) = fn (t) as above. For 1. there exists c ∈ J such that F (c) := n=1 fn (c) is convergent in X. n=1 2. For all n ∈ , f is differentiable on J and f˙ is continuous on J. t ∈ J and ε > 0 small so that (t − ε, t + ε) ⊂ J, let N n n ˙ P∞ 3. If Mn := supt∈J fn (t) , then n=1 Mn < ∞. fn(t+h)−fn(t) X h if h 6= 0 gn (h) = ˙ . P∞ fn (t) if h = 0 Then F (t) := n=1 fn (t) is convergent for t ∈ J, F is differentiable on J, F˙ is continuous on J, and ˙ By the definition of the derivative, limh→0 gn (h) = fn (t) = gn (0) which shows ∞ gn is continuous at 0 and hence gn is continuous for h ∈ (−ε, ε) . By the mean X ˙ F˙ (t) = fn (t) for all t ∈ J. (10.9) value inequality (Theorem 10.19 or Proposition ?? below), there exists c be- n=1 tween t and t + h such that

Page: 113 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 114 10 Diﬀerential Calculus in One Real Variable

˙ order to conclude Eq. (10.10) holds for |x − x | < r. As r < R was arbitrary, kgn (h)k ≤ fn (c) ≤ Mn for h 6= 0. 0 we may conclude that Eq. (10.10) holds for |x − x0| < R. The computation 0 This inequality holds at h = 0 as well. We may now apply the Weierstrass M - above shows the series for f (x) converges of |x − x0| < R. Thus these same test of Theorem 7.11 to conclude that results apply to f 0 (x) and then f 00 (x) , etc. etc. So the result now follows by a ∞ straightforward induction argument. X G (h) := gn (h) is continuous in h ∈ (−ε, ε) . Recall for a ∈ N that the binomial theorem states, n=1 a a X a k Since G (h) = [F (t + h) − F (t)] /h for 0 < |h| < ε it follows that (1 + x) = x (10.11) k k=0 ∞ ∞ F (t + h) − F (t) X X ˙ lim = lim G (h) = G (0) = gn (0) = fn (t) . where a := 1 and h→0 h h→0 0 n=1 n=1 a a! a (a − 1) ... (a − k + 1) This shows that F is differentiable at t and F˙ (t) = P∞ f˙ (t) . := = . n=1 n k k!(a − k)! k! Definition 10.30. We say f :(a, b) → X is n – times differentiable on (a, b) if the iterated derivatives of f up to order n exist. That is we assume f 0 exists We are going to generalize this theorem to all a ∈ R, see Corollary 10.33 and 0 and then f (2) := (f 0)0 exists on (a, b) , and f (3) = f (2) exists on (a, b) ,..., Exercise 10.15 below for the general case. (n) (n−1)0 f = f exists on (a, b) . Definition 10.32 (Generalized Binomial Coefficients). For a ∈ R \{0} let Theorem 10.31 (Differentiation of Power Series). Suppose a(a−1)...(a−k+1) a if k ∈ x ∈ and {a }∞ is a sequence of complex numbers such that := k! N , 0 R n n=0 k 1 if k = 0 h i−1 R := lim sup |a |1/n > 0. Then f (x) := P∞ a (x − x )n de- n→∞ n n=0 n 0 i.e. fines an infinitely differentiable function f :(x0 − R, x0 + R) → C. Moreover, a a a a a a − 1 a a a − 1 a − 2 ∞ ∞ = 1, = , = , = ,.... 0 X n−1 X n−1 0 1 1 2 1 2 3 1 2 3 f (x) = nan (x − x0) = nan (x − x0) (10.10) k n=0 n=1 a Y a − l + 1 a a − 1 a − 2 a − k + 1 = = ... . (10.12) k l 1 2 3 k and this series has the same radius of convergence as the series for f. [This l=1 ∞ theorem holds more generally when {an}n=0 is a sequence in a Banach space X h i−1 Notice that if a ∈ , we will have a = 0 if k > a and for 0 ≤ k ≤ a, so such that R := lim sup ka k1/n > 0.] N k n→∞ n X that Eq. (10.11) may be written as Proof. We are going to apply Theorem 10.29. To this end suppose that ∞ n a X a k r ∈ (0,R) and that x satisfies |x − x0| < r. Letting fn (x) := an (x − x0) we (1 + x) = x for x ∈ and a ∈ . k R N have k=0 n |f 0 (x)| = na (x − x )n−1 ≤ n |a | rn−1 =: M = |a | rn−1 It is also often useful to observe that n n 0 n n r n where a a − k a 1/n = , (10.13) 1/n n 1/n r k + 1 k + 1 k lim sup Mn = r · lim sup |an| = < 1 n→∞ n→∞ r R and n 1/n wherein we have used Lemma 3.30 to see that limn→∞ r = 1. So by the P∞ root test it follows that n=1 Mn < ∞ and so we may apply Theorem 10.29 in

Page: 114 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.4 Differentiating past infinite sums 115 a + 1 a + 1 a a − 1 a − k + 2 k = ... To complete the proof we observe that the coefficient of x may be expressed k 1 2 3 k as, a + 1 a a − 1 a − k + 2 = ... (n + k − 1)! (n + k − 1) · (n + k − 2) . . . n k 1 2 k − 1 (−1)k = (−1)k k! · (n − 1)! k! a + 1 a = . (10.14) (a − k + 1) · (a − k + 2) . . . a k k − 1 = k! Letting a → a − 1 and k → k + 1 in this last identity gives the identity, k Y a − l + 1 a = = . a a a − 1 l k = . (10.15) l=1 k + 1 k + 1 k Alternatively, we can prove Eq. (10.16) by induction; Corollary 10.33. For a ∈ −N, ∞ ∞ a−1 d a d X a k a X a a (1 + x) = (1 + x) = x (1 + x) = xk for |x| < 1. (10.16) dx dx k k k=0 k=0 ∞ ∞ X a X a = kxk−1 = (k + 1) xk Proof. Let n = −a ∈ N. We are going to arrive at this result by differenti- k k + 1 ating the geometric series, k=1 k=0 ∞ X a a − 1 ∞ = (k + 1) xk 1 X k k + 1 k = x for |x| < 1 k=0 1 − x ∞ k=0 X a − 1 = a xk, repeatedly. For example, k k=0 1 2 d 1 1 · = where we have made us of Eq. (10.15). This completes the induction step. 1 − x dx 1 − x ∞ ∞ X X = kxk−1 = (k + 1) xk Exercise 10.6. For z ∈ C recall we have defined k=1 k=0 ∞ X zn ez := . and then similarly n! n=0 3 ∞ ∞ 1 X X 2! · = (k + 1) kxk−1 = (k + 2) (k + 1) xk, The goal here is to give another proof of Proposition 7.23 using the following 1 − x k=1 k=0 outline. Show; . d tz tz 0z . 1. dt e = e with e = 1. n ∞ 2. Use this results and the obvious product rule to show 1 X (n − 1)! · = (k + n − 1) ... (k + 2) (k + 1) xk. 1 − x d h tz tw −t(z+w)i k=0 e e e = 0 for all z, w ∈ and t ∈ . (10.17) dt C R So replacing x by −x in the above expression implies 3. Use Eq. (10.17) to show ezewe−(z+w) = 1 for all z, w ∈ C. ∞ −z z −1 z w z+w a X (k + n − 1) ... (k + 2) (k + 1) k 4. Now, using item 3., show e = (e ) and then that e e = e for all (1 + x) = (−1) xk for |x| < 1. (n − 1)! z, w ∈ . k=0 C

Page: 115 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 116 10 Diﬀerential Calculus in One Real Variable

Corollary 10.34. For x ∈ R, the functions 2. The functions eiθ, cos (θ) , and sin (θ) are 2π – periodic. Here we say a function, f : → is 2π – periodic iff f (θ + 2π) = f (θ) for all θ ∈ . ∞ R C R X xn 3. eiθ 6= 1 for 0 < θ < 2π and hence [0, 2π) 3 θ → eiθ ∈ is one to one. ex := , (10.18) C n! 4. For all z ∈ with |z| = 1, there exists a unique θ ∈ [0, 2π) such that n=0 C eiθ = z. ∞ 2n+1 X n x iθ sin x := (−1) and (10.19) 5. The arc length of the curve [0, α] 3 θ → e is α for all α ≥ 0. [We have (2n + 1)! n=0 to wait until we do integration theory in order to define arc-length.] In ∞ 2n particular, 2π is the arc-length of the unit circle in C. X n x cos x = (−1) (10.20) ∞ (2n)! Exercise 10.8 (Theory behind separation of variables). Let {an}n=−∞ n=0 P∞ be a summable sequence of complex numbers, i.e. n=−∞ |an| < ∞. For t ≥ 0 are infinitely differentiable and they satisfy and x ∈ R, define ∞ ∞ d x x 0 X 2 X 2 e = e with e = 1 F (t, x) = a e−tn einx = a e−tn +inx, dx n n d n=−∞ n=−∞ sin x = cos x with sin (0) = 0 dx where as usual eix = cos (x) + i sin (x) , this is motivated by replacing x in Eq. d (10.18) by ix and comparing the result to Eqs. (10.19) and (10.20). cos x = − sin x with cos (0) = 1. dx 1. F (t, x) is continuous for (t, x) ∈ [0, ∞) × R. Hint: Weierstrass M - test, Proof. Uses Theorem 10.31 to justify differentiating the power series ex- Theorem 7.11. pansions term by term to arrive at the desired results. Alternatively, take 2. ∂F (t, x)/∂t, ∂F (t, x)/∂x and ∂2F (t, x)/∂x2 exist for t > 0 and x ∈ R. d x x z = 1 in Exercise 10.6 in order to show dx e = e . Then take z = i in Exercise Hint: Use Theorem 10.29 to compute these derivatives. In computing the 10.6 to learn that t derivative, you should let ε > 0 and apply Theorem 10.29 with t > ε and d eix = ieix. (10.21) then afterwards let ε ↓ 0. dx 3. F satisfies the heat equation, namely ix Recall from Euler’s formula, Eq. (7.13), that e = cos x + i sin x. Then by 2 2 Exercise 10.2, Eq. (10.21) is equivalent to ∂F (t, x)/∂t = ∂ F (t, x)/∂x for t > 0 and x ∈ R.

d d cos x + i sin x = i (cos x + i sin x) = − sin x + i cos x. dx dx 10.5 Taylor’s Theorem Comparing the real and imaginary parts of this equation proves the derivative P∞ n Remark 10.36. If f (x) := n=0 anx converges of x ∈ (−R,R) , then by in- formulas for sin x and cos x. duction along with Theorem 10.31, one shows for |x| < R that, Exercise 10.7. Show; ∞ (k) X n−k f (x) = an · n (n − 1) (n − 2) ... (n − k + 1) x 2 2 d 2 2 1. cos y + sin y = 1 for all y ∈ R. Hint: compute dy cos y + sin y = 0. n=k iy 2. Conclude that e = 1 for all y ∈ R. and in particular that 3. Show |ez| = eRe z for all z ∈ . (k) C f (0) = k! · ak. Remark 10.35( cos and sin are the functions you think they are). Please see Therefore we may express f as, page 182-184 of Rudin where he shows; ∞ X f (k) (0) f (x) = xk. 1. There is a number called π/2 which is the ﬁrst positive zero of cos (θ) . k! k=0

Page: 116 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.5 Taylor’s Theorem 117 3 Exercise 10.9. Suppose f : R → R is a continuous function such that f is f (x) = p2 (x) + M (x − x0) . (n) differentiable to all orders on R\{0} . Further suppose that limt→0 f (t) = Ln Our goal is to show, for some c between x and x, that exists in R for each n ∈ N. Show f is differentiable at t = 0 to all orders and 0 that f (n) (0) = L . n f (3) (c) d x x M = . Exercise 10.10. [In this exercise we will take for granted that dx e = e . This 3! will be proved a little later in this chapter.] Let f : R → R be the function defined by To prove this we let 0 if t ≤ 0 3 f (t) := . h (t) := f (t) − p2 (t) − M (t − x0) e−1/t if t > 0 (k) (n) and observe that h (x) = 0 by construction of M and that h (x ) = 0 for Show f is differentiable to all orders on R and that f (0) = 0 for all n ∈ N. 0 0 ≤ k ≤ 2 and Hints: first show there exists a polynomial functions, pn (x) , such that 3 3 (n) −1 x PN xn h (t) = f (t) − M · 3!. f (t) = pn t f (t) for t > 0. You may also use e ≥ n=0 n! when x ≥ 0 xn So to finish the proof it suffices to find a c between x and x such that h3 (c) = 0 and therefore limx→∞ ex = 0 for all n ∈ N. 0 as this will imply, Remark 10.37. Exercise 10.10 shows that not all infinitely differentiable functions may be represented as a power series! Nevertheless we can try to approx- f (3) (c) 0 = f 3 (c) − M · 3! =⇒ M = imate a function f :(a, b) → X near x0 ∈ (a, b) by its “Taylor Polynomial of 3! degree n” defined by To prove this last assertion we are repeatedly going to use the Mean value n (k) X f (x0) k theorem (or use Rolle’s theorem directly); 1) as h (x) − h (x0) = 0 − 0 = 0, pn (x) := (x − x0) . (10.22) 0 0 k! there exists c1 between x0 and x so that h (c1) = 0. 2) Similarly since h (c1) − k=0 0 00 h (x0) = 0 − 0 = 0, there exists c2 between x0 and c1 such that h (c2) = 0. 3) It is easy to check that pn is the unique degree n – polynomial such that Since h00 (c ) − h00 (x ) = 0 − 0 = 0, there exists c between x and c such that (k) 2 0 3 0 2 p (x ) = f (k) (x ) for 0 ≤ k ≤ n with the convention that f (0) = f. Indeed if 000 n 0 0 h (c3) = 0. Thus we may take c = c3 which is between x0 and x. n For the general case, let x ∈ (a, b) with x 6= x0 and define M = Mx so that X k p (x) = pk (x − x0) where pk ∈ X, n+1 k=0 f (x) − pn (x) = M (x − x0) (10.24) (k) (k) then a simple computation shows p (x0) = pk · k!. So if we want p (x0) = and set (k) (k) n+1 f (x0) for 0 ≤ k ≤ n we must require pk := f (x0) /k!. The question now h (t) := f (t) − pn (t) − M (t − x0) . becomes what is the error between f and p . n (k) By construction; h (x) = 0, h (x0) = 0 for 0 ≤ k ≤ n, and Theorem 10.38 (Taylor’s Theorem). Suppose that f : R → R is n+1 times (n+1) (n+1) differentiable on (a, b) and x0 ∈ (a, b) . Let pn (x) be as in Eq. (10.23). Then h (t) = f (t) − M · (n + 1)!. for all x ∈ (a, b) there exists a real number c between x0 and x such that For sake of definiteness, suppose that x > x0. Since h (x) − h (x0) = 0 − 0 = 0, (n+1) f (c) n+1 then by the Mean value theorem (or use Rolle’s theorem directly) there exists c1 f (x) = pn (x) + (x − x0) (10.23) 0 0 0 (n + 1)! between x0 and x so that h (c1) = 0. Similarly since h (c1)−h (x0) = 0−0 = 0, 00 Proof. Let me first give the proof in the special case that n = 2. Given there exists c2 between x0 and c1 such that h (c2) = 0. By repeating this 1 procedure we find x ∈ (a, b) with x 6= x0 and let M = Mx so that x0 < cn+1 < cn < ··· < c1 < x 1 That is, (k) f (x) − p (x) so that h (ck) = 0 for 1 ≤ k ≤ n + 1. Setting c = cn+1, we then have M := 2 . (x − x )3 0 0 = h(n+1) (c) = f (n+1) (c) − M · (n + 1)!

Page: 117 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 118 10 Differential Calculus in One Real Variable which implies, 10.6 Inverse Function Theorem I f (n+1) (c) M = . (n + 1)! Theorem 10.40 (Converse to the Chain Rule). Let U and V be open subset of and suppose f : U → V and g : V → are continuous functions Using this back in Eq. (10.25) gives Eq. (10.24). R R such that g is differentiable on V and h := g ◦ f is differentiable on U. Then f is differentiable at all point x ∈ U where g0 (f (x)) 6= 0 and at such a point,

Example 10.39 (Exponential Function revisited). Suppose f : R → R, such that −1 f 0 (x) = g0 (f (x)) h0 (x) . (10.25) f 0 (x) = f (x) for all x ∈ R and f (0) = 0. Then by Taylor’s theorem with α = 0, for all x ∈ and n ∈ there exists a point c = c (x) between 0 and x such R N0 n Proof. Suppose that x ∈ U and g0(f (x)) 6= 0. Let ∆f = f (x + ∆x) − f (x) that n n+1 and notice that ∆f = ε (∆x) because f is continuous at x. Then on one hand, X 1 k x f (x) = x + f (cn (x)) k! (n + 1)! 0 k=0 ∆h := g (f (x) + ∆f) − g (f (x)) = g (f (x)) ∆f + o (∆f) (n) wherein we have used f = f for all n. Letting Mx := max {|f (s)| : |s| ≤ |x|} while on the other hand, we may conclude that ∆h = h (x + ∆x) − h (x) = h0 (x) ∆x + o (∆x) . n X 1 xn+1 xn+1 f (x) − xk = |f (c (x))| ≤ M → 0 as n → ∞. k! (n + 1)! n (n + 1)! x Comparing these equations implies, k=0 0 0 Thus there is at most one such function and if it exists it must be given by h (x) ∆x + o (∆x) = g (f (x)) ∆f + o (∆f)

∞ or equivalently, X 1 f (x) = xk, k! 0 −1 0 k=0 ∆f = g (f (x)) [h (x) ∆x + o (∆x) − o (∆f)] 0 −1 0 i.e. the function we have defined previously to be exp (x) = ex. We have seen in = g (f (x)) h (x) ∆x + o (∆x) + o (∆f) . (10.26) Corollary 10.34 that exp (x) is differentiable and the d exp (x) = exp (x) with dx Since f is continuous, ∆f → 0 as ∆x → 0. Therefore there exists δ > 0 such exp (0) = 1 as required. 1 that |o (∆f)| ≤ 2 |∆f| if |∆x| ≤ δ. Using this in Eq. (10.27) implies, for some d C < ∞ and |∆x| ≤ δ that Exercise 10.11 (Taylor’s Theorem II). Suppose that f : R → R is n + 1 times differentiable on (a, b) and x0 ∈ (a, b) . Let 0 −1 0 |∆f| = g (f (x)) h (x) ∆x + o (∆x) + o (∆f) n (k) X f (x0) k p (x) := (x − x ) 0 −1 0 1 n k! 0 ≤ g (f (x)) h (x) ∆x + o (∆x) + |∆f| k=0 2 1 ≤ C |∆x| + |∆f| , as in Eq. (10.23). Then for all x ∈ (a, b) there exists a c between x0 and x such 2 that |x − x |n+1 i.e. |∆f| ≤ 2C |∆x| for |∆x| ≤ δ. From this we may now conclude that o (∆f) = kf (x) − p (x)k ≤ f (n+1) (c) 0 . n (n + 1)! o (∆x) and therefore Eq. (10.27) asserts,

Hint: use the idea in the proof of Theorem 10.19 to reduce this to the case f (x + ∆x) − f (x) = g0 (f (x))−1 h0 (x) ∆x + o (∆x) . where d = 1. i.e. f 0 (x) exists and is give by Eq. (10.26).

Page: 118 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.6 Inverse Function Theorem I 119 d 1 Theorem 10.41 (Inverse Function Theorem). Suppose that g :(a, b) → R ln y = for all y > 0. is a differentiable function such that g0 (x) > 0 (or g0 (x) < 0) for all a < dy y x < b. Let α = inf {g (x): a < x < b} and β := sup {g (x): a < x < b} , then 2. For a, b ∈ (0, ∞) , ln (ab) = ln a + ln b. g :(a, b) → (α, β) is a bijection and there exists a unique inverse function, 3. For a > 0, ln a−1 = − ln a. f :(α, β) → (a, b) such that g (f (y)) = y for all y ∈ (α, β) and f (g (x)) = x 4. For any k ∈ and a > 0, ln ak = k ln a or equivalently ak = ek ln a. for all x ∈ (a, b) . Moreover the function f is differentiable on (α, β) and Z 5. For any q ∈ Q and a > 0, ln aq = q ln a or equivalently aq = eq ln a. 1 0 Proof. 1. The first assertion follows directly from Theorem 10.41 noting f (y) = 0 for all y ∈ (α, β) . g (f (y)) that d d d 0 1 = y = eln y = eln y ln y = y ln y Proof. I will assume that g > 0, if not apply the results to −g instead. dy dy dy It follows by the mean value theorem that g (x) < g (x0) if x < x0. i.e. g is 0 strictly increasing. Since g is continuous and (a, b) is connected we know that and e = 1 so that ln 1 = 0. x y x y x+y J := g (a, b) is connected and hence equal to an interval. If β = sup J ∈ J, 2. If x = ln a and y = ln b, then a = e and b = e so that ab = e e = e then there would exists x ∈ (a, b) such that g (x) = β and then for x0 ∈ (x, b) and hence ln (ab) = x + y = ln a + ln b. −1 −1 we would have g (x0) ∈ J while g (x0) > β which violates the fact that β is 3. Take b = a in item 2. to conclude that 0 = ln 1 = ln a · a = −1 −1 an upper bound for J. Similarly we see that α∈ / J and therefore J = (α, β) . ln a + ln a , i.e. ln a = − ln a. 0 0 Hence we have shown g :(a, b) → (α, β) is a bijection and hence there exists a 4. If k = 0 and a > 0 then a = 1 so that ln a = ln 1 = 0 = 0 · ln a. For k unique inverse map, f :(α, β) → (a, b) and this map is also strictly increasing. k ∈ N we show using item 1. and induction that ln a = k ln a. If k ∈ −N then The continuity of f has already been addressed in Exercise 6.20 and Exercise −1 9.40. Here is yet another variant of the continuity proof. Suppose y ∈ (α, β) , ln ak = ln a|k| = − ln a|k| = − |k| ln a = k ln a. α < α0 < y < β0 < β and yn ∈ [α0, β0] is a sequence such that limn→∞ yn = 1 1 ln a y. Then x := f (y ) ∈ [f (α ) , f (β )] and we wish to show lim x = n n n 0 0 n→∞ n 5. If b = n , then e > 0 and f (y) . For sake of contradiction suppose not, then there exists ε > 0 and as n 1 ln a n· 1 ln a ln a subsequence,x ˜k := xnk such that |x˜k − f (y)| ≥ ε. By passing to a subsequence e n = e n = e = a. if necessary, using the sequential compactness of [f (α0) , f (β0)] , we may assume that limk→∞ x˜k = x0 exists. Since g is continuous it follows that As we know there is a unique positive nth – root of a (we have already seen 1 ln a 1/n this but it also follows from Theorem 10.41) and therefore, e n = a . So if g (x0) = lim g (˜xk) = lim g (f (yn )) = lim yn = y k→∞ k→∞ k k→∞ k q = m/n with m ∈ Z and n ∈ N we find, m m which shows x = f (y) . But this then leads to the contradiction; q 1/n 1 ln a m ln a q ln a 0 a = a = e n = e n = e .

0 < ε ≤ lim |x˜k − f (y)| = |x0 − f (y)| = 0. k→∞ The differentiability assertions now follows from Theorem 10.40. Definition 10.43. For a > 0 and b ∈ R we define ab := eb ln a. [This agrees x 0 x b Since the function g (x) = e satisfies g (x) > 0 for all x ∈ R, limx→−∞ e = with our previous definition of a when b ∈ Q by item 5. of Theorem 10.42.] x 0, and limx→∞ e = ∞ there is a unique function, ln : (0, ∞) → R such that ln y x Corollary 10.44. For a > 0 and b ∈ R, e = y for all y > 0 and ln (e ) = x and x ∈ R. We refer to this function as the natural logarithm. d ax = ln a · ax for all x ∈ dx R Theorem 10.42 (Natural logarithm). Let ln : (0, ∞) → R be the inverse function to exp (x) = ex. Then; and d xb = bxb−1 for all x > 0. 1. ln is an increasing differentiable function such than ln 1 = 0 and dx

Page: 119 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 120 10 Diﬀerential Calculus in One Real Variable

Exercise 10.12. Prove Corollary 10.44, i.e. for a > 0 and b ∈ R, Thus the inverse function theorem asserts there exists a differentiable increasing function, f :(−1, 1) → (−π/2, π/2) such that sin (f (x)) = x. [Of d ax = ln a · ax for all x ∈ course f (x) = arcsin (x) .] Differentiating this equation then implies dx R 1 and 1 = cos (f (x)) f 0 (x) , i.e. f 0 (x) = . d cos (f (x)) xb = bxb−1 for all x > 0. dx √ Letting y = f (x) , we have x = sin y and so cos y = ± 1 − x2. We know Example 10.45. In a similar fashion one may use Theorem 10.41 to prove the we must take the + sign here since cos y > 0 for y ∈ (−π/2, π/2) . Thus we usual differentiation formulas for the inverse trigonometric functions; have shown, d 1 d 0 1 1 arctan (x) = 2 , arcsin (x) = f (x) = = √ . dx x + 1 dx cos (f (x)) 1 − x2 d 1 arcsin (x) = √ , dx 1 − x2 Exercise 10.13. Use Taylor’s Theorem 10.38 to show d 1 ∞ n+1 arccos (x) = −√ , and X (−1) 1 dx 1 − x2 ln (1 + x) = xn for − < x < 1. n 2 d 1 n=1 arcsec (x) = p dx |x| (x2 − 1) Exercise 10.14. Show the series

∞ n+1 (where they are deﬁned). X (−1) f (x) := xn n 1. For example, n=1

d d sin y cos2 y + sin2 y 1 converges for |x| < 1 and that f (x) = ln (1 + x) for |x| < 1. Hint: this problem tan (y) = = = dy dy cos y cos2 y cos2 y can be done independently from Exercise 10.13. Your goal is to show g (x) := f (x) − ln (1 + x) = 0 for |x| < 1.

which is positive on (−π/2, π/2) . Moreover we have limy↑π/2 tan y = ∞ a and lim tan y = −∞. Thus by the inverse function theorem asserts Exercise 10.15 (Binomial Expansions). Let a ∈ R and f (x) = (1 + x) . y↓π/2 0 there exist a increasing diﬀerentiable function, f : → (−π/2, π ) such By direct computation using induction one shows, f (0) = 1, f (x) = R 2 a−1 that tan (f (x)) = x for all x ∈ R. Diﬀerentiating this equation then implies a (1 + x) , 00 1 f (x) a (a − 1) a−2 1 = f 0 (x) , i.e. f 0 (x) = cos2 (f (x)) . = (1 + x) , cos2 (f (x)) 2! 2 f (3) (x) a (a − 1) (a − 2) = (1 + x)a−2 Letting y = f (x) , we have 3! 3! 2 . 2 2 sin y 2 2 2 2 1 . x = tan y = 2 =⇒ 1 − cos y = x cos y =⇒ cos y = 2 . cos y 1 + x f (n) (x) a (a − 1) ... (a − n + 1) = (1 + x)a−n Therefore it follows that f 0 (x) = 1 . The functions f (x) is more com- n! n! 1+x2 a monly known as arctan (x) . = (1 + x)a−n . d n 2. As another example, let us observe that dx sin (x) = cos (x) > 0 on (−π/2, π/2) and in this case lim sin y = 1 and lim sin y = −1. y↑π/2 y↓π/2 So we expect, based on Taylor’s Theorem 10.38, that

Page: 120 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 10.7 L’hopitals Rule 121 ∞ ¯ a X a Theorem 10.47 (L’hopitals Rule). Let c and L be in . The real valued (1 + x) = xn. (10.27) R n functions f and g are assumed to be differentiable on an open interval J with n=0 endpoint c, and additionally g0 (x) 6= 0 and g (x) 6= 0 for x ∈ J. It is also for |x| < R where R is the radius of convergence of the series above. Show; assumed that f 0 (x) 1. R = 1. Hint: you might use the ratio test here. lim = L, x→c g0 (x) 2. Show (1 + x)a solves the differential equation, where limx→c means limx↑c if c is the right end point of J and limx↓c if c is the d (1 + x) (1 + x)a = a (1 + x)a for x > −1. (10.28) left endpoint of J. If either; dx 1) lim f (x) = lim g (x) = 0 or 3. Show Eq. (10.29) is valid for |x| < 1 as follows. x→c x→c a) Show 2) lim |f (x) | = lim |g (x) | = ∞, ∞ x→c x→c X a g (x) := xn (10.29) n then n=0 f (x) 0 lim = L. satisfies (1 + x) g (x) = ag (x) for |x| < 1, the same differential equation x→c g (x) as (1 + x)a . b) Next compute the derivative of h (x) := (1 + x)−a g (x) and use to show The following proof is taken fairly directly from Wikipedia.. For h (x) = 1 for |x| < 1. definiteness we assume that J = (a, c) for some −∞ < a < c, we allow for c = ∞ here. The case where c is the left endpoint of J is proved similarly. For Exercise 10.16. Rudin problem 5.15 on page 115. x ∈ J let 0 0 f (ξ) ¯ f (ξ) ¯ m (x) = inf 0 ∈ R and M (x) = sup 0 ∈ R. 10.7 L’hopitals Rule ξ∈(x,c) g (ξ) ξ∈(x,c) g (ξ)

Theorem 10.46 (Cauchy’s Generalized Mean Value Theorem). Suppose By Cauchy’s mean value Theorem 10.46, if a < x < y < c there exist ξ ∈ (x, y) that f, g :[a, b] → R are continuous functions which are diﬀerentiable on (a, b) . such that f (x) − f (y) f 0(ξ) Then there exists t0 ∈ (a, b) such that = 0 . 0 0 g (x) − g (y) g (ξ) [f (b) − f (a)] g (t0) = [g (b) − g (a)] f (t0) . [Notice g (x) − g (y) 6= 0 by the mean value theorem and the assumption that Proof. Let g0 (ξ) 6= 0 for ξ ∈ J.] In particular we have h (t) := [f (b) − f (a)] g (t) − [g (b) − g (a)] f (t) . f (x) − f (y) m (x) ≤ ≤ M (x) for all a < x < y < c. (10.30) Then g (x) − g (y)

h (b) = [f (b) − f (a)] g (b) − [g (b) − g (a)] f (b) = g (a) f (b) − f (a) g (b) Case 1. Suppose that limx→c f (x) = limx→c g (x) = 0. In this case we may let y ↑ c in Eq. (10.33) in order to learn and f (x) h (a) = [f (b) − f (a)] g (a) − [g (b) − g (a)] f (a) = g (a) f (b) − f (a) g (b) m (x) ≤ ≤ M (x) . (10.31) g (x) so that h (a) = h (b) . Applying Rolle’s theorem shows there exists t0 ∈ (a, b) such that Case 2. If limx→c |g (x) | = ∞, then

0 0 0 f(y) f(x) 0 = h (t0) = [f (b) − f (a)] g (t0) − [g (b) − g (a)] f (t0) . f (y) − f (x) − m (x) ≤ = g(y) g(y) ≤ M (x) . (10.32) g (y) − g (x) g(x) 1 − g(y)

Page: 121 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Since f (x) g (x) lim = 0 = lim y↑c g (y) y↑c g (y) we may pass to the limit in Eq. (10.35) in order to ﬁnd,

f (y) f (y) m (x) ≤ lim inf ≤ lim sup ≤ M (x) . (10.33) y↑c g (y) y↑c g (y)

Since f 0 (x) f 0 (x) lim m (x) = lim inf 0 = L = lim sup 0 = lim M (x) x↑c x↑c g (x) x↑c g (x) x↑c we may pass to the limit in Eqs. (10.34) and (10.36) using the squeeze theorem f(x) in order to learn L = limx↑c g(x) in case 1) and

f (y) f (y) L ≤ lim inf ≤ lim sup ≤ L y↑c g (y) y↑c g (y)

f(y) in Case 2. The last equation implies limy↑c g(y) exists and is equal to L. Example 10.48. By L’hopitals Rule,

sin (x) d sin (x) cos (x) lim = lim dx = lim = 1 x→0 x x→0 d x→0 1 dx x and by two applications of L’hopitals Rule,

1 − cos (x) sin (x) cos (x) 1 lim = lim = lim = . x→0 x2 x→0 2x x→0 2 2 Technically one needs to work backwards in these string of equalities to justify the use of L’hopitals. Let me point out you would be wrong to use L’hopitals cos(x) rule for the following limit; limx↓0 x = ∞. If we were to mistakenly use L’hopitals rule in this ∞/0 situation we would conclude mistakenly that

cos (x) d cos (x) − sin (x) ∞ = lim = lim dx = lim = 0. x↓0 x x↓0 d x↓0 1 dx x Exercise 10.17. Rudin problem 5.11 on page 115.

Exercise 10.18. Rudin problem 5.13 on page 115. In this problem assume that x ∈ [0, 1] rather than [−1, 1] and assume a 6= 0 and c > 0. 11 Riemann (Like) Integration Theory

In this chapter we will let F be either R or C and X and Y will be Banach Notation 11.5 Given a function, f : Ω → Y and y ∈ Y, let spaces over F. {f = y} := f −1 ({y}) := {ω ∈ Ω : f (ω) = y} .

Example 11.6 (Simple functions). Given subsets Ak ⊂ Ω and yk ∈ Y for 1 ≤ 11.1 Banach Spaces of Bounded Functions k ≤ n, the function, n X Notation 11.1 Let Ω be a set and Y be a Banach space and for any function f = yk1Ak ∈ B (Ω,Y ) . f : Ω → Y let k=1 kfku := sup kf (ω)kY . Notice that if S = z = P y : Λ ⊂ {{1, 2, . . . , n}} where by convention ω∈Ω k∈Λ k z = 0 if Λ = ∅, then f (Ω) ⊂ S and in particular f (Ω) is a finite set with no Also let B (Ω,Y ) denote those functions, f : Ω → Y such that kfku < ∞. We more that 2n – elements. Conversely if f : Ω → Y is a function such that f (Ω) refer to B (Ω,Y ) as the space of Y – valued bounded functions on Ω. is a finite subset of Y, then Fact 11.2 The collection of all functions f : Ω → Y, denoted by Y Ω, is an F - X f = y · 1{f=y} vector space where for g : Ω → Y and λ ∈ F we define y∈f(Ω) (f + λg)(ω) := f (ω) + λg (ω) for all ω ∈ Ω. where −1 We are going to take this fact for granted. To prove one needs to open a book {f = y} := {ω ∈ Ω : f (ω) = y} = f ({y}) . on linear algebra and verify that Y Ω satisfies all the axioms of a vector space. A function f : Ω → Y such that f (Ω) is a finite set is called a simple function.

Ω Proposition 11.3. The space B (Ω,Y ) is a subspace of Y , k·ku is a norm on We are now going to take Ω = [a, b] where −∞ < a < b < ∞. B (Ω,Y ) , and (B (Ω,Y ) , k·ku) is complete, i.e. (B (Ω,Y ) , k·ku) is a Banach space. Definition 11.7 (Patititions). A partition of [a, b] is a finite subset, Π, of [a, b] such that a, b ∈ Π. We will typically write Π as, Exercise 11.1. Prove the first part of Proposition 11.3. That is show B (Ω,Y ) Ω Π = {a = t0 < t1 < ··· < tn = b} . (11.1) is a subspace of Y , k·ku is a norm on B (Ω,Y ) . Hint: model your proof on Exercise 6.1. The mesh size of Π is defined to be, Exercise 11.2. Prove the second part of Proposition 11.3. That is show |Π| := max |ti − ti−1| . (B (Ω,Y ) , k·ku) is complete, i.e. (B (Ω,Y ) , k·ku) is a Banach space. Hint: 1≤i≤n model your proof on Lemma 6.27. Notation 11.8 Given a partition, Π, of [a, b] as in Eq. (11.2) and c = Notation 11.4 (Indicator Functions) The indicator function, 1A, of a sub- (c1, . . . , cn) ∈ [a, b] such that ti−1 ≤ ci ≤ ti for 1 ≤ i ≤ n [we will say that set A ⊂ Ω is the function defined by, c interlaces Π] let n 1 if ω ∈ A X 1A (ω) = . f := f (c ) 1 + f (c ) 1 . 0 if ω∈ / A Π,c 1 {a} i (ti−1,ti] i=1 124 11 Riemann (Like) Integration Theory

In other words, fΠ,c (t) = c1 if a = t0 ≤ t ≤ t1 and fΠ,c (t) = ci if ti−1 < t ≤ ti 11.2 Algebras of Sets with 2 ≤ i ≤ n. Notice that fΠ,c :[a, b] → Y is a simple function. See Figure 11.1. Notation 11.10 (Pairwise Disjoint Unions) If {Aα} and A are subsets P α∈I of Ω, we write A = Aα to mean that A = ∪α∈I Aα with the added restric- α∈I ` tion that Aα ∩ Aβ = ∅ if α 6= β. We will also write A B for A ∪ B with the added condition that A ∩ B = ∅. Similarly A ` B ` C = A ∪ B ∪ C with the added requirement that A ∩ B = A ∩ C = B ∩ C = ∅. Definition 11.11. A set S ⊂ 2Ω is said to be an semialgebra or elementary class provided that • ∅∈S •S is closed under finite intersections • if E ∈ S, then Ec is a finite disjoint union of sets from S. (In particular Ω = ∅c is a finite disjoint union of elements from S.)

Notation 11.12 A subset J ⊂ R is said to be a half open interval it J = (a, b] ∩ R for some −∞ ≤ a ≤ b ≤ ∞. In other words a half open interval is a set of the form (a, b] with −∞ ≤ a ≤ b < ∞ or of the form (a, ∞) with Fig. 11.1. The usual picture associated to the Riemann integral. −∞ ≤ a < ∞. If J and K are two half open intervals we will write J < K iﬀ s < t for all s ∈ J and t ∈ K. More concretely, (a, b] ≤ (c, d] ∩ R iﬀ b ≤ c. Example 11.13. Let Ω = R, then ¯ Exercise 11.3 (Uniform Continuity). Show that every continuous function, S := (s, t] ∩ R : s, t ∈ R f :[a, b] → Y is uniformly continuous, i.e. if ε > 0 there exists δ > 0 such that = {(s, t]: s ∈ [−∞, ∞) and s < t < ∞} ∪ {∅, R} kf(t) − f(s)k ≤ ε if s, t ∈ [a, b] with |t − s| ≤ δ. Hint: see Exercise 9.24. Y is an elementary class. Proposition 11.9. If f :[a, b] → Y is a continuous function then f ∈ ∞ Example 11.14. Let −∞ < a ≤ b < ∞,Ω = (a, b], and B ([a, b] ,Y ) . Let us further suppose that {Πm}m=1 is any collection of partitions of [a, b] such that limm→∞ |Πm| = 0 and that fm := fΠm , cm where cm S := {(s, t]: a ≤ s ≤ t ≤ b} interlaces Πm, then limm→∞ kf − fmku = 0. In particular, every continuous function is the uniform limit of simple functions. is an elementary class.

Exercise 11.4. Prove Proposition 11.9. Suppose that f :[a, b] → Y is a con- Example 11.15. Suppose that Ω is a ﬁnite set, then tinuous function. S := {{ω} : ω ∈ Ω} ∪ {∅} 1. Show f ∈ B ([a, b] ,Y ) . is an elementary class. 2. Show for every ε > 0 there exists a δ > 0 such that for every partition Π 2 of [a, b] with |Π| ≤ δ we have kf − fΠ , cku ≤ ε for any collections of times Example 11.16. Suppose that Ω = (a1, b1] × (a2, b2] ⊂ . Then ∞ R c which interlace Π. In particular if {Πm}m=1 is a collection of partitions S := {(s1, t1] × (s2, t2]: ai ≤ si ≤ ti ≤ bi for i = 1, 2} of [a, b] such that limm→∞ |Πm| = 0 and that fm := fΠm , cm where cm interlaces Π , then lim kf − f k = 0. Hint: make use of Exercise m m→∞ m u is an elementary class as follows from from Example 11.14 and Exercise 11.5 11.3. below or may be veriﬁed directly. This example has an obvious generalization to any dimension.

Page: 124 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 11.2 Algebras of Sets 125

Exercise 11.5. Let Ω and Y be sets and A ⊂ 2Ω and B ⊂ 2Y be elementary Example 11.18. Let Ω = R, then A consisting of finite disjoint unions of ele- classes. Show the collection ments from S := {A × B : A ∈ A and B ∈ B} S := (a, b] ∩ R : a, b ∈ R¯ = {(a, b]: a ∈ [−∞, ∞) and a < b < ∞} ∪ {∅, R} is also a elementary class. is an algebra For example, Proposition 11.17( A is an algebra of sets). Let S be a semi-algebra on Ω and let A = A(S)1 denote the subsets of Ω which may be written as a finite A = (0, π] ∪ (2π, 7] ∪ (11, ∞) ∈ A (S) . disjoint union of elements from S. So A ∈ A iff there Ei ∈ S such that A = Pn Example 11.19 (Algebra of half open intervals). Let Ω = (a, b]. The algebra of i=1 Ei. Then A is an algebra of subsets of Ω, i.e.; half open intervals, A, on Ω is the collection of subsets of (a, b] which may 1. ∅ and Ω belong to A, be written as a union of pairwise disjoint half open subintervals of (a, b]. [The 2. if A, B ∈ A, then A ∩ B ∈ A, Pn general element of A may be written as A = i=1 Ji with J1 < J2 < ··· < Jn c 3. if A ∈ A then A = Ω \ A ∈ A, and and Ji ⊂ (a, b] for all i.] 4. if A, B ∈ A then A ∪ B ∈ A. Lemma 11.20. Suppose that F is a finite collection of half open intervals which Proof. We chech each assertion in turn. are pairwise disjoint, i.e. if J, K ∈ F with J 6= K then J ∩K = ∅. If n = # (F) , there exists a unique listing of F as {J1,...,Jn} with J1 < J2 < ··· < Jn. 1. By the properties of S, we know that ∅,Ω ∈ A. Pn 2.( A is closed under intersections.) Suppose that A = i=1 Ei ∈ A and Proof. The proof is by induction on n. If n = 1, F = {J1} and we are Pm B = j=1 Fj ∈ A where Ei,Fj ∈ S. Then done. Now suppose the statement is true when # (F) = n − 1 for some n ≥ 2 and now suppose that # (F) = n. Choose J1 = (a1, b1] ∈ F so that it has   n ! m the smallest possible left end point. By in the induction hypothesis we can list X X X A ∩ B = Ei ∩  Fj = Ei ∩ Fj F\{J1} as {J2,...,Jn} with J2 < J3 < . . . < Jn. It now only remains to show i=1 j=1 i,j J1 = (a1, b1] < J2 = (a2, b2]. By construction we know that a2 ≥ a1 and by assumption that J ∩ J = ∅. These two conditions can only hold if a ≥ b , i.e. P 1 2 2 1 where we are justified in writing i,j since, J1 < J2.

[Ei ∩ Fj] ∩ [Ei0 ∩ Fj0 ] = [Ei ∩ Ei0 ] ∩ [Fj ∩ Fj0 ] = ∅ Definition 11.21. Let Ω be a set. A collection, Π, of non empty subsets of Ω P is a partition of Ω if Ω = A∈Π A. In other words, every point of x ∈ Ω, unless i = i0 and j = j0. Therefore A is closed under intersections. belongs precisely to one and only one element, A, in Π. We say that Π is a Pn 3.( A is closed under complementation.) If A = i=1 Ei is in A with Ei ∈ S, finite partition if we further assume that #(Π) < ∞. c n c then A = ∩i=1Ei ∈ A from item 2. and the property of S that implies Ec ∈ A whenever E ∈ S. Corollary 11.22. If Π is a finite partition of (a, b] consisting of half open i i n 4.( A is closed under unions.) If A, B ∈ A, then sub-intervals of (a, b]. Then there exists {tk}k=0 ⊂ (a, b] such that Π = n {(ti−1, tk]}k=1 where n = # (Π) and a = t0 < t1 < t2 < ··· < tn = b. c c c c c A ∪ B = [[A ∪ B] ] = [A ∩ B ] ∈ A n Proof. By Lemma 11.20 we know that Π = {Jk = (ak, bk]}k=1 with J1 < c as Ac,Bc ∈ A and so Ac ∩ Bc ∈ A and therfore [Ac ∩ Bc] ∈ A. J2 < ··· < Jn. With this ordering we must have a = a1 and b = bn. If bk < ak+1 for some k < n then ∅= 6 (bk, ak+1) ⊂ (a, b] ∩ R while on the other hand, (b , a ) ∩ (a, b] ∩ = (b , a ) ∩ [∪ J] = ∅. 1 We will refer to A(S) as the algebra generated by S. k k+1 R k k+1 J∈Π

Thus we conclude that bk = ak+1 for k = 1, 2, . . . , n − 1 and therefore the theorem holds with t0 = a and tk = bk for 1 ≤ k ≤ n.

Page: 125 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 126 11 Riemann (Like) Integration Theory n Pn Exercise 11.6 (Structure of finite algebras). Suppose that Ω is a set and 4.( µ is finitely subadditive) µ(∪j=1Aj) ≤ j=1 µ(Aj). Hint: Define B1 := Ω A is a sub-algebra of 2 with only finitely many elements, i.e. # (A) < ∞. For A1 and then Bj := Aj \ [A1 ∪ ...Aj−1] for 2 ≤ j ≤ n. Then observe that n Pn each x ∈ Ω let Bj ⊂ Aj for all j and that ∪j=1Aj = j=1 Bj. Ax = ∩ {A ∈ A : x ∈ A} ∈ A, Definition 11.24. If S ⊂ 2Ω is a semi-algebra, we say a function µ : S → Z wherein we have used A is finite to insure Ax ∈ A. Hence Ax is the smallest set Pn is additive on S if whenever n ∈ N and E = i=1 Ei with Ei,E ∈ S, then in A which contains x, i.e. if A ∈ A and x ∈ A then Ax ⊂ A. Show; n X 1. If x, y ∈ Ω with Ax∩Ay 6= ∅ then Ax = Ay so either Ax = Ay or Ax∩Ay = ∅. µ (E) = µ (Ei) . Hint: if x∈ / Ay, then Ax \ Ay would be a set in A which is strictly smaller i=1 than Ax – this would violate the definition of Ax. 2 Lemma 11.25. Let Ω = (a, b], α :[a, b] → Z be a function, and S be the 2. Let Π := {A1,...,An} ⊂ A be the list of distinct elements in P collection of half open intervals in Ω. Then µ : S → Z defined by {Ax : x ∈ Ω} . Show that every A ∈ A may be written as A = j∈Λ Aj for some finite subset Λ ⊂ {1, 2, . . . , n} . By convention, if Λ = ∅ we define P Ω Pn µ ((s, t]) = α (t) − α (s) for all a ≤ s ≤ t ≤ b j∈Λ Aj = ∅ ∈ 2 . In particular this shows Ω = i=1 Ai and hence Π is a partition of Ω. [Conclusion, finite subalgebras of 2Ω are in one to one is additive on S. correspondence with finite partitions of Ω.] Pn 3. Explain why there is no algebra A ⊂ 2Ω such that # (A) = 6. Proof. Suppose that J = i=1 Ji with J = (u, v] and Ji := (ui, vi] in S. We must show n X µ (J) = µ (Ji) . (11.3) 11.3 Finitely additive measure on A i=1 As the sum in Eq. (11.4) is independent of the ordering we may reorder the Let Z be a vector space (typically we are going to take Z = ). R terms so that J1 < J2 < ··· < Jn and then use Corollary 11.22 to deduce that J = (t , t ] where u = t < t < ··· < t = v. With this notation, we find by Definition 11.23. An Z – valued finitely additive measure on A is a func- i i−1 i 0 1 n a telescoping series arguments (see Theorem 7.3) that tion, µ : A → Z such that; n n 1. µ (∅) = 0 ∈ Z, and X X µ (Ji) = [α (ti) − α (ti−1)] 2. µ (A ∪ B) = µ (A)+µ (B) whenever A, B ∈ A with A∩B = ∅. [By induction i=1 i=1 Pn Pn it then follows that µ j=1 Aj = j=1 µ (Aj) where Aj ∈ A.] = α (tn) − α (t0) = α (v) − α (u) = µ (J) .

Exercise 11.7. Let Ω be a set, A be a sub-algebra of 2Ω, and µ : A → [0, ∞) be a finitely additive measure. Further suppose that A, B ∈ A and {A }n ⊂ A. j j=1 Theorem 11.26 (Construction of Finitely Additive Measures). Suppose Show; S ⊂ 2Ω is a semi-algebra (see Definition 11.11) and A = A(S) is the algebra 1.( µ is monotone) If A ⊂ B, then µ (A) ≤ µ(B) and more precisely, µ (B) − generated by S, see Proposition 11.17. Then every additive function µ : S → Z µ (A) = µ (B \ A) . such that µ (∅) = 0 extends uniquely to an additive measure on A. This unique Pn Pn extension will still be denoted by µ. 2. Show µ j=1 Aj = j=1 µ (Aj) if Aj ∩ Ai = ∅ when i 6= j. P 3. For any A, B ∈ A, Proof. Since every element A ∈ A is of the form A = i Ei for a finite collection of Ei ∈ S, it is clear that if µ extends to a measure on A then this µ (A ∪ B) = µ (A) + µ (B) − µ (A ∩ B) . (11.2) extension is unique and we must define µ (A) by 2 We know this list is finite since A is a finite collection of subsets of Ω. X µ(A) = µ(Ei). (11.4) i

Page: 126 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 11.4 Simple Functions 127 n n To prove existence, the main point is to show that µ(A) in Eq. (11.5) is well X X P µ (A) = µ ((ai, bi]) = [α (bi) − α (ai)] (11.6) defined; i.e. if we also have A = Fj with Fj ∈ S, then we must show j i=1 i=1 X X µ(Ei) = µ(Fj). (11.5) which shows µ is uniquely determined by its cumulative distribution func- i j tion, α. P P But Ei = j (Ei ∩ Fj) and the additivity of µ on S implies µ(Ei) = j µ(Ei ∩ Example 11.29. Suppose that Ω is a finite set and let S := {{ω} : ω ∈ Ω} ∪ ∅. Ω Fj) and hence Then S is a semi-algebra and the associated algebra is A = 2 . Given any function, α : Ω → Z we may define µ : S → Z by µ (∅) = 0 and µ ({ω}) = α (ω) . X X X X µ(Ei) = µ(Ei ∩ Fj) = µ(Ei ∩ Fj). Then an application of Theorem 11.26 states that µ extends to a finitely additive i i j i,j measure on A given by ! Similarly, X X X X X µ (A) = µ {ω} = µ ({ω}) = α (ω) . µ(Fj) = µ(Ei ∩ Fj) ω∈A ω∈A ω∈A j i,j which combined with the previous equation shows that Eq. (11.6) holds. P Pm Finally if A = i Ei and B = j=1 Fj with A ∩ B = ∅, then A ∪ B = 11.4 Simple Functions P ` Pm ( i Ei) j=1 Fj so that In this section, let F be either R or C and let Y be a vector space over F. Further X X suppose that Ω is a set and A ⊂ 2Ω is an algebra of sets. µ (A ∪ B) = µ(Ei) + µ(Fj) = µ (A) + µ (B) . i j Definition 11.30 (Simple functions). A function, f : Ω → Y is said to be simple if f (Ω) ⊂ Y is a finite set. [See Example 11.6 for the structure of simple This shows that µ is a finitely additive measure on A. functions.] We further say that f is an A – simple function if {f = y} ∈ A Corollary 11.27. Let α :[a, b] → Z be any function, Ω = (a, b], and A be the for all y ∈ f (Ω) ⊂ Y. Let S (A,Y ) denote the collection of Y – valued simple algebra of half open intervals in Ω. Then there exists a unique finitely additive functions on Ω. measure µ = µ : A → Z such that α Lemma 11.31. The set of simple functions, S (A,Y ) , is a vector subspace of µ ((s, t]) = α (t) − α (s) for all a ≤ s ≤ t ≤ b. all functions from Ω to Y. Moreover, f ∈ S (A, F) and g ∈ S (A,Y ) then f · g ∈ S (A,Y ) . [In algebraic jargon, S (A, F) is an algebra of functions and S (A,Y ) Proof. This corollary follows directly from Lemma 11.25 and Theorem is a module over S (A, F) .] 11.26. Proof. Let us observe that 1Ω = 1 and 1∅ = 0 are in S (A,Y ) . If f, g ∈ Remark 11.28. Let Ω = (a, b] and A be the algebra of half open intervals. S (A,Y ) and c ∈ F\{0} , then for any y ∈ Y, Given a finitely additive measure µ : A → Z, let α :[a, b] → Z be defined by [ α (t) := µ ((a, t]) ∈ Z. Then if J = (s, t] ⊂ (a, b], then (a, t] = (a, s] ∪ (s, t] and {f + cg = y} = ({f = a} ∩ {g = b}) ∈ A. (11.7) so a,b∈C:a+cb=y α (t) = µ ((a, t]) = µ ((a, s]) + µ ((s, t]) = α (s) + µ ((s, t]) This shows that S (A,Y ) is a vector subspace. and hence We first observe that the f · g can only take values in µ (J) = µ ((s, t]) = α (t) − α (s) . {ay : a ∈ f (Ω) and y ∈ g (Ω)} which is a finite set. Therefore f · g is a Pn simple function. If y ∈ Y \{0} , then Hence if A = i=1(ai, bi] we find {f · g = y} = ∪a·z=y ({f = a} ∩ {g = z}) ∈ A (11.8)

Page: 127 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 128 11 Riemann (Like) Integration Theory where the union is over those a ∈ f (Ω) \{0} and z ∈ g (Ω) \{0} such that 11.5 Simple Integration az = y. We also have We now suppose that Y is a normed space, Ω is a set, A is a sub-algebra of 2Ω, {f · g = 0} = {f = 0} ∪ {g = 0} ∈ A and µ : A → R+ be the unique finitely additive measure. and thus we have shown f · g ∈ S (A, F) . Definition 11.34 (Simple Integral). For f ∈ S (A,Y ) the integral (written as R fdµ or R f (ω) dµ (ω) is defined by Definition 11.32. When Ω = (a, b], we say that f : Ω → Y is a step function Ω Ω if there is a partition, Π = {a = t0 < t1 < ··· < tn = b} of Ω such that f is Z X fdµ = µ(f = y) · y ∈ Y. (11.11) constant on (ti−1, ti] for each i, i.e. there exists yi ∈ Y such that Ω y∈f(Ω) n X f = yi1(ti−1,ti]. (11.9) where µ (f = y) is shorthand for µ ({f = y}) . It is often convenient to write the i=1 last sum as X µ(f = y) · y Lemma 11.33. Let Ω = (a, b] and f : Ω → Y be a function, then; y∈Y

1. f is an A – simple function iff there exists yi ∈ Y and Ai ∈ A for 1 ≤ i ≤ n with the understanding that this is really a finite sum since µ (f = y) 6= 0 hap- for some n ∈ N such that pens only when y ∈ f (Ω) which is a finite set. n X f = yi1A . (11.10) R R i To simplify notation I will often write f for Ω fdµ in this section. i=1 2. f is an A – simple function iff f is a step function. Example 11.35. Suppose that A ∈ A and y ∈ Y, then 3. Suppose X is another vector space and F : Y → X is any function then Z Z F ◦ f ∈ S (A,X) for all f ∈ S (A,Y ) . In particular, kfkY ∈ S (A, R) when y1A = y1Adµ = µ (A) y. (11.12) f ∈ S (A,Y ) . Ω R Proposition 11.36. The integration operator, : S (A,Y ) → Y, satisfies: Proof. 1. Since S (A,Y ) is a module over S (A, F) (see Lemma 11.31), every f of the form in Eq. (11.11) or Eq. (11.10) is in S (A,Y ) . Conversely if f ∈ 1. If f ∈ (A,Y ) and λ ∈ , then P S F S (A,Y ) it follows by definition that f = y1{f=y} which is of the form y∈f(Ω) Z Z in Eq. (11.11). λf = λ f. (11.13) P Pny y 2. Moreover as Ω = y {f = y} and {f = y} = k=1 Jk we P Pny y have Ω = y∈f(Ω) k=1 Jk and so that collection of half open in- y 2. If f, g ∈ S (A,Y ) , then tervals ∪y∈f(Ω) {Jk : 1 ≤ k ≤ ny} forms a partition of Ω. Therefore there exits Π = {a = t0 < t1 < ··· < tN = b} such that (ti−1, ti] is in Z Z Z y (f + g) = g + f. (11.14) ∪y∈f(Ω) {Jk : 1 ≤ k ≤ ny} for each i and therefore f is constant on (ti−1, ti], i.e. f is a step function. Conversely if f is a step function, then Eq. (11.10) holds for some Π and so f is an A – simple function. Items 1. and 2. say that integration is a linear functional on S (A,Y ) . PN 3. If f ∈ S (A,Y ) and F : Y → X is any function, then 3. If f = j=1 yj1Aj for some yj ∈ Y and some Aj ∈ A, then

  N X X Z X F ◦ f = F ◦  y · 1{f=y} = F (y) · 1{f=y} ∈ S (A,Y ) . f = yjµ (Aj) . (11.15) y∈f(Ω) y∈f(Ω) j=1

Page: 128 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 11.5 Simple Integration 129 4. For all f ∈ (A,Y ) , and similarly, S Z Z Z X wµ ({f = v, g = w}) = g. f ≤ kf (t)k dµ (t) ≤ kfk∞ · µ (Ω) . (11.16) v,w Y Ω where Equation (11.15) is now a consequence of the last three displayed equations. PN 3. If f = yj1A , then using the linearity of the integral and Example kfk∞ := sup kf (ω)kY . j=1 j ω∈Ω 11.35 we ﬁnd 5. If X is another vector space and T : Y → X is a linear operator, then  N  N N Z Z X X Z X Z Z f = y 1 = y 1 = y µ (A ) . T fdµ = T (f) dµ.  j Aj  j Aj j j Ω Ω j=1 j=1 j=1

Proof. 4. By the triangle inequality, 1. If λ ∈ and λ 6= 0, then F Z Z Z X X X X λf = y µ(λf = y) = y µ(f = y/λ) f = yµ (f = y) ≤ kyk µ (f = y) = kfk dµ, Ω y∈Y y∈Y y∈f(Ω) y∈f(Ω) X Z = λz µ(f = z) = λ f. wherein the last equality we have used Eq. (11.16) and the fact that |f| = z∈Y P y∈Y |y| 1f=y. Since for y ∈ f (Ω) we have kyk ≤ kfk∞ , we may conclude The case λ = 0 is trivial. that 2. Writing {f = v, g = w} for f −1({v}) ∩ g−1({w}), then X X kyk µ (f = y) ≤ kfk µ (f = y) Z ∞ X y∈f(Ω) y∈f(Ω) (f + g) = z µ(f + g = z) X z∈Y = kfk∞ µ (f = y) = kfk∞ · µ (Ω) ! X X y∈f(Ω) = z µ {f = v, g = w} z∈Y v+w=z from which we see the second inequality in Eq. (11.17) holds. X X 5. Since T is linear, = z µ ({f = v, g = w}) X z∈Y v+w=z T ◦ f = T (y) 1{f=y} X X y∈f(Ω) = (v + w) µ ({f = v, g = w}) z∈Y v+w=z and therefore, X = (v + w) µ ({f = v, g = w}) . Z X Z X v,w T (f) dµ = T (y) 1{f=y}dµ = T (y) µ (f = y) Ω y∈f(Ω) Ω y∈f(Ω) But   X X X X Z Z vµ ({f = v, g = w}) = v µ ({f = v, g = w}) = T  yµ (f = y) = T fdµ = T fdµ. v,w v w y∈f(Ω) Ω Ω X = vµ (∪w {f = v, g = w}) v X Z Proposition 11.37. If Y = = , then R is positive, i.e. R f ≥ 0 for all = vµ ({f = v}) = f F R 0 ≤ f ∈ (A, ) . More generally, if f, g ∈ (A, ) and f ≤ g, then R f ≤ R g. v S R S R

Page: 129 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 130 11 Riemann (Like) Integration Theory Proof. If f ≥ 0 then Secondly suppose that α ≤ γ ≤ β, then by the ﬁrst case, Z X Z β Z γ Z β f = aµ(f = a) ≥ 0 fdµ = fdµ + fdµ a≥0 α α γ and if f ≤ g, then g − f ≥ 0 so that and so Z γ Z β Z β Z β Z γ Z Z Z fdµ = fdµ − fdµ = fdµ + fdµ. g − f = (g − f) ≥ 0. α α γ α β The remaining possibilities are veriﬁed by similar means. For the last assertion, we may suppose that α ≤ β and the result follows Notation 11.38 If A ∈ A and f ∈ S (A,Y ) , we let from Eq. (11.17) since, Z Z fdµ := 1 fdµ. Z β Z b A A Ω fdµ = 1(α,β]fdµ α a Notation 11.39 When Ω := (a, b], α :[a, b] → R is an increasing function, Z b Z b Z β

A is the algebra of half open intervals on Ω, and µ : A → R+ is the unique ≤ 1(α,β]f dµ = 1(α,β] kfk dµ = kfk dµ. measure such that µ ((s, t]) = α (t) − α (s) for all a ≤ s ≤ t ≤ b, we will denote a a α R R b R b Ω fdµ by a fdµ or a fdα. Moreover, if f ∈ S (A,Y ) and α, β ∈ [a, b] we let Z β ( R b a 1(α,β]fdµ if α ≤ β fdµ = R b . α − a 1(β,α]fdµ if α ≥ β 11.6 Extending the Integral by Uniform Limits Proposition 11.40. Let us continue using the setup in Notation 11.39. If f ∈ We now suppose that Y is a Banach space, Ω is a set, A is a subalgebra of 2Ω, (A,Y ) and α, β, γ ∈ [a, b] , then S and µ : A → R+ is a ﬁnitely additive measure on A. Z γ Z β Z γ Theorem 11.41. Suppose that f : Ω → Y is a function such that there exists fdµ = fdµ + fdµ. R α α β fn ∈ S (A,Y ) with kf − fnk∞ → 0 as n → ∞, then limn→∞ Ω fndµ exists in Y. Moreover if g ∈ (A,Y ) with kf − g k → 0 as n → ∞, then Moreover, n S n ∞

Z β Z β Z Z

fdµ ≤ kfk dµ . lim gndµ = lim fndµ. α α n→∞ Ω n→∞ Ω R Proof. There are a number of cases to consider. I will only check two of We denote this common limit by Ω fdµ and call it the integral of f relative to them. µ. First suppose that α ≤ β ≤ γ. In this case we have Proof. To show the limit exists we need only need to use the completeness 1(α,γ]f = 1(α,β]f + 1(β,γ]f of Y along with the observation that Z Z and therefore, fndµ − fmdµ ≤ kfn − fmk∞ µ (Ω) → 0 as m, n → ∞. Z γ Z Z Z Ω Ω fdµ = 1(α,γ]f = 1(α,β]f + 1(β,γ]f α Similarly, Z β Z γ Z Z = fdµ + fdµ. α β fndµ − gmdµ ≤ kfn − gnk∞ µ (Ω) → 0 as m, n → ∞ Ω Ω

Page: 130 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 11.6 Extending the Integral by Uniform Limits 131 since Notation 11.43 For u, v ∈ [a, b] and ∈ S¯ (A,Y ) we let Z v R kf − g k ≤ kf − fk + kf − g k → 0 + 0 = 0 as n → ∞. 1(u,v]fdµ if u ≤ v n n ∞ n ∞ n ∞ fdµ := ΩR . u − Ω 1(v,u]fdµ if u ≥ v Theorem 11.44. Suppose that Ω = (a, b] and let C ([a, b] ,Y ) be the space of In light of Theorem 11.41 we have extended the integral from S¯ (A,Y ) – the continuous functions from [a, b] → Y. Then C ([a, b] ,Y ) ⊂ ¯ (A,Y ) . Further closure of S (A,Y ) in the collection of bounded functions from Ω → Y relative S suppose that f ∈ C ([a, b] ,Y ) and to each partition Π := {a = t0 < t1 < ··· < to the k·k∞ . Π tn = b} of [a, b] we have chosen ci ∈ [ti−1, ti] for 1 ≤ i ≤ n. Then ¯ ¯ ¯ Exercise 11.8. If f ∈ S (A, F) and g, h ∈ S (A,Y ) , then h + fg ∈ S (A,Y ) and b n ¯ Z X kgk ∈ (A, ) . You should prove this by showing; if fn ∈ (A, ) and gn, hn ∈ Π S R S F fdµ = lim f(ci )(α (ti) − α (ti−1)). |Π|→0 S (A,Y ) are chosen so that fn → f, hn → h, and gn → g uniformly on Ω then a i=1 h + f g → h + fg and kg (·)k → kg (·)k uniformly on Ω. n n n n In particular if α (t) = t, then we write dt for dµ or dα and we have, Theorem 11.42 (Linearity of the Integral). The extended integral, R : Ω Z b n ¯ (A,Y ) → Y is still linear and satisﬁes the estimates, X Π S f (t) dt = lim f(ci )(ti − ti−1). a |Π|→0 Z Z i=1

fdµ ≤ kf (ω)kY dµ (ω) ≤ kfk∞ µ (Ω) . (11.17) Proof. From Exercise 11.4 we know that f may be written as a uniform Ω Y Ω limit of step functions and hence is in S¯ (A,Y ) . Moreover that same exercise ¯ shows; if ε > 0 there exists δ > 0 such that Proof. Let λ ∈ F and g, h ∈ S (A,Y ) and choose gn, hn ∈ S (A,Y ) are chosen so that hn → h, and gn → g uniformly on Ω. Then f − fΠ,cΠ u ≤ ε if |Π| ≤ δ. Z Z (h + λg) dµ = lim (hn + λgn) dµ Thefore if |Π| ≤ δ, then Ω n→∞ Ω Z b Z b Z Z fdµ − f Π dµ ≤ f − f Π · µ (Ω) ≤ ε · µ (Ω) = lim hndµ + λ gndµ Π,c Π,c u n→∞ Ω Ω a a Z Z = hdµ + λ gdµ. and this suﬃces to show Ω Ω Z b Z b lim fΠ,cΠ dµ = fdµ. Similarly using khnk → khk uniformly on Ω we have, |Π|→0 a a Z Z Z Z This completes the proof since, hdµ = lim hndµ ≤ lim khnk dµ = khk dµ n→∞ n→∞ n Ω Ω Ω Ω X f = f (c ) 1 on Ω = (a, b] and Π,c i (ti−1,ti] Z i=1 khnk dµ ≤ khnk∞ µ (Ω) Ω and hence so that Z b n n Z X X Π fΠ,cΠ dµ = f (ci) µ ((ti−1, ti]) = f(ci )(α (ti) − α (ti−1)). lim khnk dµ ≤ lim khnk µ (Ω) = khk µ (Ω) . ∞ ∞ a i=1 i=1 n→∞ Ω n→∞

¯ Let us now specialize to the case where Ω := (a, b], α :[a, b] → is an If fn ∈ S and f ∈ S such that limn→∞ kf − fnk = 0, then for a ≤ u < R ∞ increasing function, A is the algebra of half open intervals on Ω, and µ : A → R+ v ≤ b, then 1(u,v]fn ∈ S and limn→∞ 1(u,v]f − 1(u,v]fn ∞ = 0. This shows ¯ ¯ is the unique measure such that µ ((s, t]) = α (t) − α (s) for all a ≤ s ≤ t ≤ b. 1(u,v]f ∈ S whenever f ∈ S.

Page: 131 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 ¯ ¯ Notation 11.45 For f ∈ S and a ≤ α ≤ β ≤ b we will denote I(1(α,β]f) by R f(t) dt or (α,β] f(t)dt. Also following the usual convention, if a ≤ β ≤ α ≤ b, we will let Z β Z α ¯ f(t) dt = −I(1(β,α]f) = − f(t) dt. α β The next Lemma, whose proof is left to the reader, contains some of the many familiar properties of the integral.

Lemma 11.46. For f ∈ S¯([a, b],X) and α, β, γ ∈ [a, b], the integral satisﬁes:

1. R β f(t) dt ≤ (β − α) sup {kf(t)k : α ≤ t ≤ β} . α X R γ R β R γ 2. α f(t) dt = α f(t) dt + β f(t) dt. R t 3. The function G(t) := a f(τ)dτ is continuous on [a, b]. 4. If Y is another Banach space and T ∈ L(X,Y ), then T f ∈ S¯([a, b],Y ) and

Z β ! Z β T f(t) dt = T f(t) dt. α α

¯ 5. The function t → kf(t)kX is in S([a, b], R) and

Z b Z b

f(t) dt ≤ kf(t)kX dt. a a X

6. If f, g ∈ S¯([a, b], R) and f ≤ g, then Z b Z b f(t) dt ≤ g(t) dt. a a Exercise 11.9. Prove Lemma 11.46. Part IV

Appendices

A Appendix: Notation and Logic

The following abbreviations along with their negations are used throughout these notes.

Symbol Meaning Negation ∀ for all ∃ ∃ there exits ∀ , or “space” then 3 3 such that , or “space” a.a. almost always i.o. i.o. inﬁnitely often a.a. = equals 6= 6= not equals = ≤ less than or equal > > greater than ≤ Here are some examples.

1. an = bn i.o. n ⇐⇒ #({n : an = bn}) = ∞. The negation of #({n : an = bn}) = ∞ is

#({n : an = bn}) < ∞ ⇐⇒ an 6= bn for a.a. n.

2. limn→∞ an = L is by deﬁnition the statement;

∀ε > 0 ∃N ∈ N 3 ∀n ≥ N, |L − an| ≤ ε. This may also be written as

∀ε > 0, |L − an| ≤ ε for a.a. n.

3. The negation of the previous statement is limn→∞ an 6= L which translates to ∃ε > 0 3 ∀N ∈ N, ∃ n ≥ N 3 |L − an| > ε. This last statement is also equivalent to;

∃ε > 0 3 |L − an| > ε i.o. n. It is sometimes useful to reformulate this last statement as; there exists ε > 0 and an increasing function N 3 k → nk ∈ N such that

|L − ank | > ε for all k ∈ N.

B Appendix: More Set Theoretic Properties (highly optional)

B.1 Appendix: Zorn’s Lemma and the Hausdorff Maximal 2. The Hausdorff Maximal Principle: Every partially ordered set has a Principle (optional) maximal (relative to the inclusion order) linearly ordered subset. 3. Zorn’s Lemma: If X is partially ordered set such that every linearly or- Definition B.1. A partial order ≤ on X is a relation with following properties; dered subset of X has an upper bound, then X has a maximal element.1

1. If x ≤ y and y ≤ z then x ≤ z. Proof. (2 ⇒ 3) Let X be a partially ordered subset as in 3 and let F = 2. If x ≤ y and y ≤ x then x = y. {E ⊂ X : E is linearly ordered} which we equip with the inclusion partial 3. x ≤ x for all x ∈ X. ordering. By 2. there exist a maximal element E ∈ F. By assumption, the Example B.2. Let Y be a set and X = 2Y . There are two natural partial orders linearly ordered set E has an upper bound x ∈ X. The element x is maximal, on X. for if y ∈ Y and y ≥ x, then E ∪ {y} is still an linearly ordered set containing E. So by maximality of E,E = E ∪{y} , i.e. y ∈ E and therefore y ≤ x showing 1. Ordered by inclusion, A ≤ B is A ⊂ B and which combined with y ≥ x implies that y = x.2 2. Ordered by reverse inclusion, A ≤ B if B ⊂ A. (3 ⇒ 1) Let {Xα}α∈A be a collection of non-empty sets, we must show Q X is not empty. Let G denote the collection of functions g : D(g) → Definition B.3. Let (X, ≤) be a partially ordered set we say X is linearly or α∈A α ` X such that D(g) is a subset of A, and for all α ∈ D(g), g(α) ∈ X . totally ordered if for all x, y ∈ X either x ≤ y or y ≤ x. The real numbers α∈A α α R Notice that G is not empty, for we may let α ∈ A and x ∈ X and then set with the usual order ≤ is a typical example. 0 0 α D(g) = {α0} and g(α0) = x0 to construct an element of G. We now put a partial Definition B.4. Let (X, ≤) be a partial ordered set. We say x ∈ X is a max- order on G as follows. We say that f ≤ g for f, g ∈ G provided that D(f) ⊂ D(g) imal element if for all y ∈ X such that y ≥ x implies y = x, i.e. there is no and f = g|D(f). If Φ ⊂ G is a linearly ordered set, let D(h) = ∪g∈ΦD(g) and for element larger than x. An upper bound for a subset E of X is an element α ∈ D(g) let h(α) = g(α). Then h ∈ G is an upper bound for Φ. So by Zorn’s x ∈ X such that x ≥ y for all y ∈ E. 1 ∞ If X is a countable set we may prove Zorn’s Lemma by induction. Let {xn}n=1 Example B.5. Let be an enumeration of X, and define En ⊂ X inductively as follows. For n = 1 let E1 = {x1}, and if En have been chosen, let En+1 = En ∪{xn+1} if xn+1 is an upper ∞ X = a = {1} b = {1, 2} c = {3} d = {2, 4} e = {2} bound for En otherwise let En+1 = En. The set E = ∪n=1En is a linearly ordered (you check) subset of X and hence by assumption E has an upper bound, x ∈ X. I ordered by set inclusion. Then b and d are maximal elements despite that fact claim that his element is maximal, for if there exists y = xm ∈ X such that y ≥ x, that b d and d b. We also have; then xm would be an upper bound for Em−1 and therefore y = xm ∈ Em ⊂ E. 1. If E = {a, c, e}, then E has no upper bound. That is to say if y ≥ x, then y ∈ E and hence y ≤ x, so y = x. (Hence we may view 2. If E = {a, e}, then b is an upper bound. Zorn’s lemma as a “ jazzed” up version of induction.) 2 Similarly one may show that 3 ⇒ 2. Let F = {E ⊂ X : E is linearly ordered} 3. If E = {e}, then b and d are upper bounds. and order F by inclusion. If M ⊂ F is linearly ordered, let E = ∪M = S A. If A∈M Theorem B.6. The following are equivalent. x, y ∈ E then x ∈ A and y ∈ B for some A, B ⊂ M. Now M is linearly ordered by set inclusion so A ⊂ B or B ⊂ A i.e. x, y ∈ A or x, y ∈ B. Since A and B 1. The axiom of choice: to each collection, {Xα} , of non-empty sets ` α∈A are linearly order we must have either x ≤ y or y ≤ x, that is to say E is linearly there exists a “choice function,” x : A → Xα such that x(α) ∈ Xα for α∈A ordered. Hence by 3. there exists a maximal element E ∈ F which is the assertion Q in 2. all α ∈ A, i.e. α∈A Xα 6= ∅. 138 B Appendix: More Set Theoretic Properties (highly optional)

Lemma there exists a maximal element h ∈ G. To finish the proof we need only 1. P0 ∈ F1. show that D(h) = A. If this were not the case, then let α0 ∈ A \ D(h) and 2. If Φ ⊂ F1 is a totally ordered (by set inclusion) subset then ∪Φ ∈ F1. ˜ x0 ∈ Xα0 . We may now define D(h) = D(h) ∪ {α0} and 3. If P ∈ F1 then g(P ) ∈ F1. h(α) if α ∈ D(h) Let us call a general subset F 0 ⊂ F satisfying these three conditions a tower h˜(α) = x0 if α = α0. and let 0 0 F0 = ∩ {F : F is a tower} . Then h ≤ h˜ while h 6= h˜ violating the fact that h was a maximal element. (1 ⇒ 2) Let (X, ≤) be a partially ordered set. Let F be the collection of Standard arguments show that F0 is still a tower and clearly is the smallest tower containing P . (Morally speaking F consists of all of the sets we were linearly ordered subsets of X which we order by set inclusion. Given x0 ∈ X, 0 0 trying to constructed in the “idea section” of the proof.) We now claim that {x0} ∈ F is linearly ordered set so that F= 6 ∅. Fix an element P0 ∈ F. If F is a linearly ordered subset of F. To prove this let Γ ⊂ F be the linearly P0 is not maximal there exists P1 ∈ F such that P0 P1. In particular we 0 0 ordered set may choose x∈ / P0 such that P0 ∪ {x} ∈ F. The idea now is to keep repeating this process of adding points x ∈ X until we construct a maximal element P of F. We now have to take care of some details. We may assume with out Γ = {C ∈ F0 : for all A ∈ F0 either A ⊂ C or C ⊂ A} . loss of generality that F˜ = {P ∈ F : P is not maximal} is a non-empty set. Shortly we will show that Γ ⊂ F is a tower and hence that F = Γ. That is For P ∈ F˜, let P ∗ = {x ∈ X : P ∪ {x} ∈ F} . As the above argument shows, 0 0 to say F is linearly ordered. Assuming this for the moment let us finish the P ∗ 6= ∅ for all P ∈ F˜. Using the axiom of choice, there exists f ∈ Q P ∗. 0 P ∈F˜ proof. We now define g : F → F by Let P ≡ ∪F0 which is in F0 by property 2 and is clearly the largest element P if P is maximal in F0. By 3. it now follows that P ⊂ g(P ) ∈ F0 and by maximality of P, we g(P ) = (B.1) P ∪ {f(x)} if P is not maximal. have g(P ) = P, the desired fixed point. So to finish the proof, we must show that Γ is a tower. First off it is clear that P0 ∈ Γ so in particular Γ is not The proof is completed by Lemma B.7 below which shows that g must have a empty. For each C ∈ Γ let fixed point P ∈ F. This fixed point is maximal by construction of g. Φ := {A ∈ F : either A ⊂ C or g(C) ⊂ A} . Lemma B.7. The function g : F → F defined in Eq. (B.1) has a fixed point.3 C 0

Proof. The idea of the proof is as follows. Let P0 ∈ F be chosen arbi- We will begin by showing that ΦC ⊂ F0 is a tower and therefore that ΦC = F0. (n) ∞ 1. P0 ∈ ΦC since P0 ⊂ C for all C ∈ Γ ⊂ F0. 2. If Φ ⊂ ΦC ⊂ F0 is totally trarily. Notice that Φ = g (P0) n=0 ⊂ F is a linearly ordered set and it is ∞ ordered by set inclusion, then A := ∪Φ ∈ F . We must show A ∈ Φ , that S (n) Φ 0 Φ C therefore easily verified that P1 = g (P0) ∈ F. Similarly we may repeat is that A ⊂ C or C ⊂ A . Now if A ⊂ C for all A ∈ Φ, then A ⊂ C and n=0 Φ Φ Φ ∞ ∞ hence A ∈ Φ . On the other hand if there is some A ∈ Φ such that g(C) ⊂ A S (n) S (n) Φ C the process to construct P2 = g (P1) ∈ F and P3 = g (P2) ∈ F, etc. then clearly g(C) ⊂ A and again A ∈ Φ . 3. Given A ∈ Φ we must show n=0 n=0 Φ Φ C C ∞ g(A) ∈ Φ , i.e. that etc. Then take P∞ = ∪n=0Pn and start again with P0 replaced by P∞. Then C keep going this way until eventually the sets stop increasing in size, in which g(A) ⊂ C or g(C) ⊂ g(A). (B.2) case we have found our fixed point. The problem with this strategy is that we There are three cases to consider: either A C,A = C, or g(C) ⊂ A. In the may never win. (This is very reminiscent of constructing measurable sets and case A = C, g(C) = g(A) ⊂ g(A) and if g(C) ⊂ A then g(C) ⊂ A ⊂ g(A) and the way out is to use measure theoretic like arguments.) Eq. (B.2) holds in either of these cases. So assume that A C. Since C ∈ Γ, Let us now start the formal proof. Again let P0 ∈ F and let F1 = {P ∈ either g(A) ⊂ C (in which case we are done) or C ⊂ g(A). Hence we may F : P0 ⊂ P }. Notice that F1 has the following properties: assume that 3 Here is an easy proof if the elements of F happened to all be finite sets and there A C ⊂ g(A). existed a set P ∈ F with a maximal number of elements. In this case the condition Now if C were a proper subset of g(A) it would then follow that g(A) \ A would that P ⊂ g(P ) would imply that P = g(P ), otherwise g(P ) would have more consist of at least two points which contradicts the definition of g. Hence we elements than P.

Page: 138 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 B.2 Cardinality 139 must have g(A) = C ⊂ C and again Eq. (B.2) holds, so ΦC is a tower. It is now which these theorems are stated, while equivalent for finite sets, are inequivalent easy to show Γ is a tower. It is again clear that P0 ∈ Γ and Property 2. may be in the context of infinite sets. checked for Γ in the same way as it was done for ΦC above. For Property 3., if The notion of counting may be extended to them in the sense of establishing C ∈ Γ we may use ΦC = F0 to conclude for all A ∈ F0, either A ⊂ C ⊂ g(C) (the existence of) a bijection with some well understood set. For instance, if a set or g(C) ⊂ A, i.e. g(C) ∈ Γ. Thus Γ is a tower and we are done. can be brought into bijection with the set of all natural numbers, then it is called Here is an example of using Zorn’s lemma. “countably infinite.” This kind of counting differs in a fundamental way from counting of finite sets, in that adding new elements to a set does not necessarily Proposition B.8. Suppose that X and Y are non-empty sets, then either there increase its size, because the possibility of a bijection with the original set is not exists an injective function, f : X → Y, or an injective function g : Y → X. In excluded. For instance, the set of all integers (including negative numbers) can other words, either card (X) ≤ card (Y ) or card (Y ) ≤ card (X) . be brought into bijection with the set of natural numbers, and even seemingly much larger sets like that of all finite sequences of rational numbers are still Proof. Let F be the collection of injective functions, u : D (u) → Y where (only) countably infinite. Nevertheless there are sets, such as the set of real D (u) is a non-empty subset of X. We say that u ≤ v for u, v ∈ F provided numbers, that can be shown to be“too large” to admit a bijection with the D (u) ⊂ D (v) and u = v| . One now checks that (F, ≤) is a partially ordered D(u) natural numbers, and these sets are called ”uncountable.” Sets for which there set such that every linearly ordered subset of F has an upper bound. Therefore, exists a bijection between them are said to have the same cardinality, and in by an application of Zorn’s lemma, F has a maximal element, U. the most general sense counting a set can be taken to mean determining its If D (U) = X, we take f = U and we have constructed an injective map from cardinality. Beyond the cardinalities given by each of the natural numbers, X to Y. If D (U) 6= X, then Ran (U) := U (D (U)) = Y. [Indeed, if not we could there is an infinite hierarchy of infinite cardinalities, although only very few find and x ∈ X \D (U) and y ∈ Y \ Ran (U) and then extend U to D (U) ∪ {x} such cardinalities occur in ordinary mathematics (that is, outside set theory by setting U (x) = y. The extended U is still injective and hence would violate that explicitly studies possible cardinalities). the maximallity of U.] In this case we take g := U −1 : Y → D (U) ⊂ X. Counting, mostly of finite sets, has various applications in mathematics. One important principle is that if two sets X and Y have the same finite number of elements, and a function f : X → Y is known to be injective, then it is also B.2 Cardinality surjective, and vice versa. A related fact is known as the pigeonhole principle, which states that if two sets X and Y have finite numbers of elements n and In mathematics, the essence of counting a set and finding a result n, is that it m with n > m, then any map f : X → Y is not injective (so there exist two establishes a one to one correspondence (or bijection) of the set with the set distinct elements of X that f sends to the same element of Y ); this follows of numbers {1, 2, . . . , n}. A fundamental fact, which can be proved by math- from the former principle, since if f were injective, then so would its restriction ematical induction, is that no bijection can exist between {1, 2, . . . , n} and to a strict subset S of X with m elements, which restriction would then be {1, 2, . . . , m} unless n = m; this fact (together with the fact that two bijec- surjective, contradicting the fact that for x in X outside S, f(x) cannot be in tions can be composed to give another bijection) ensures that counting the the image of the restriction. Similar counting arguments can prove the exis- same set in different ways can never result in different numbers (unless an error tence of certain objects without explicitly providing an example. In the case of is made). This is the fundamental mathematical theorem that gives counting its infinite sets this can even apply in situations where it is impossible to give an purpose; however you count a (finite) set, the answer is the same. In a broader example; for instance there must exists real numbers that are not computable context, the theorem is an example of a theorem in the mathematical field of numbers, because the latter set is only countably infinite, but by definition a (finite) combinatorics—hence (finite) combinatorics is sometimes referred to as non-computable number cannot be precisely specified. ”the mathematics of counting.” The domain of enumerative combinatorics deals with computing the number Many sets that arise in mathematics do not allow a bijection to be estab- of elements of finite sets, without actually counting them; the latter usually lished with {1, 2, . . . , n} for any natural number n; these are called infinite sets, being impossible because infinite families of finite sets are considered at once, while those sets for which such a bijection does exist (for some n) are called such as the set of permutations of {1, 2, . . . , n} for any natural number n. finite sets. Infinite sets cannot be counted in the usual sense; for one thing, the mathematical theorems which underlie this usual sense for finite sets are false for infinite sets. Furthermore, different definitions of the concepts in terms of

Page: 139 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 140 B Appendix: More Set Theoretic Properties (highly optional)

B.3 Formalities of Counting 2. f (XX ) = YX . Indeed if x ∈ XX then again ancestor (f (x)) = (x, ancestor (x)) which ends in X so that f (x) ∈ YX . Moreover if y ∈ YX , Definition B.9. We say card (X) ≤ card (Y ) if there exists an injective map, then ancestor (y) = (y, ancestor (x)) where f (x) = y so that ancestor (x) f : X → Y and card (Y ) ≥ card (X) if there exists a surjective map g : Y → X. ends in X, i.e. y ∈ f (XX ) . Thus we have shown f : XX → YX is a bijection, We say card (X) = card (Y ) if there exists a bijections, f : X → Y. i.e. card (XX ) = card (YX ) . 3. By the same argument as in item 2. it follow that g : YY → XY is a bijection, Proposition B.10. We have card (X) ≤ card (Y ) iff card (Y ) ≥ card (X) . i.e. card (XY ) = card (YY ) . −1 Proof. If f : X → Y is an injective map, define g : Y → X by g|f(X) = f The last three statement implies card (X) = card (Y ) . We may in fact define and g|Y \f(X) = x0 ∈ X chosen arbitrarily. Then g : Y → X is surjective. a bijection, h : X → Y, by −1 If g : Y → X is a surjective map, then Yx := g ({x}) 6= ∅ for all x ∈ X and so by the axiom of choice there exists f ∈ Q Y . Thus f : X → Y such f (x) if x ∈ X∞ ∪ XX x∈X x h (x) = . g|−1 (x) if x ∈ X that f (x) ∈ Yx for all x. As the {Yx}x∈X are pariwise disjoint, it follows that YY Y f is injective.

Theorem B.11 (Schröder-BernsteinTheorem). If card (X) ≤ card (Y ) Definition B.12. We say card (X) < card (Y ) if card (X) ≤ card (Y ) and and card (Y ) ≤ card (X) , then card (X) = card (Y ) . Stated more explicitly; if card (X) 6= card (Y ) , i.e. card (X) < card (Y ) if there exists an injective map, there exists injective maps f : X → Y and g : Y → X, then there exists a f : X → Y, but not bijective map exists. Similarly we say card (Y ) > card (X) bijective map, h : X → Y. if card (Y ) ≥ card (X) and card (Y ) 6= card (X) , i.e. card (Y ) > card (X) if Proof. Starting with an x ∈ X we may form the sequence of “ancestors” of there exists a surjective map g : Y → X but no bijective map exists. −1 −1 x, namely ancestor (x) := (x, y1, x1, y2,... ) where y1 = g (x) , x1 = f (y1) , X −1 Proposition B.13. For any non-empty set X, card (X) < card 2 . y2 = g (x1) , . . . ., i.e. X −1 −1 −1 −1 Proof. Define f : X → 2 by f (x) = {x} . Then f is an injective map and g f g f X X x → y1 → x1 → y2 → ... hence card (X) ≤ card 2 . Now suppose that g : X → 2 is any map. Let X0 = {x ∈ X : x∈ / g (x)} ⊂ X. I claim that X0 ∈/ g (X) . We continue this process of inverse iterates as long as it is possible, i.e. we can Indeed suppose there exists x0 ∈ X such that g (x0) = X0. If x0 ∈ X0, construct yn+1 if xn ∈ g (Y ) and xn+1 if yn+1 ∈ f (X) . There are now three then x0 ∈/ g (x0) = X0 which is impossible. Similarly if x0 ∈/ X0 = g (x0) then possibilities; x0 ∈ X0 and again we have reached a contradiction. Thus we must conclude that X ∈/ g (X) . Thus there are no surjective maps, g : X → 2X so that 1. ancestor (x) has infinite length so the process never gets stuck in which case 0 card (X) 6= card 2X . we say x ∈ X∞, read as start in X and end never get stuck. 2. ancestor (x) is finite and the last term in the sequence is in X, in which case Proposition B.14. If card (X) < card (Y ) and card (Y ) ≤ card (Z) , then we say x ∈ XX (read as start in X and end in (get stuck in) X). card (X) < card (Z) . 3. ancestor (x) is finite and the last term in the sequence is in Y, in which case we say x ∈ XY (read as start in X and end in (get stuck in) Y ). Proof. If there exists an injective map, f : Z → X then composing this with and injective map, g : X → Y gives an injective map, g ◦ f : Z → X and In this way we partition X into three disjoint sets, X∞,XX , and XY . Sim- there for card (Z) ≤ card (X) . But this would imply that card (X) = card (Z) . ilarly we may partition Y into Y∞,YY , and YX . Let us now observe that,

1. f (X∞) = Y∞. Indeed if x ∈ X∞ then ancestor (f (x)) = (x, ancestor (x)) Deﬁnition B.15. Let An := {1, 2, . . . , n} for all n ∈ N and write n for is an inﬁnite sequence, i.e. f (x) ∈ Y∞. Moreover if y ∈ Y∞, then card (An) . ancestor (y) = (y, ancestor (x)) where f (x) = y so that x ∈ X∞ and y ∈ f (X∞) . Thus we have shown f : X∞ → Y∞ is a bijection, i.e. Proposition B.16. We have card (Am) < card (An) for all m < n. Moreover card (X∞) = card (Y∞) , if ∅= 6 X $ An then card (X) = card (Ak) for some k < n.

Page: 140 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 B.3 Formalities of Counting 141

Proof. If f : A1 → A2, then either f (1) = 1 or f (1) = 2. In either case of Proposition B.10. By the pervious paragraph we know that g is necessarily −1 f is injective but not bijective so that card (A2) < card (A1) . Let Sn be the surjective and therefore f = g is a bijection. statement that card (Ak) < card (Al) for all 1 ≤ k < l ≤ n and for any proper It now only remains to prove card (Bijec (An)) = n! which we again do subset X ⊂ An we have card (X) = card (Am) for some m < n. Then we have by induction. For n = 1 the result is clear. So suppose it holds at level n. If just shown that S2 is true. So suppose that Sn is now true. As f : Ak → Al f : An+1 → An+1 is a bijection. Given 1 ≤ k ≤ n + 1 let deﬁned by f (m) = m for all m ∈ Ak is a injection when k < l we always Bij (A ) := {f ∈ Bij (A ): f (n + 1) = k} . have card (Ak) ≤ card (Al) . Now suppose that card (Ak) = card (An+1) for k n+1 n+1 some k ≤ n. Then there exists a bijection, f : An+1 → Ak. In this case f (An) ∼ For f ∈ Bijk (An+1) , we have f : An → An+1 \{k} = An is a bijection. Thus is a proper subset of Ak and therefore card (f (An)) < card (Ak) but on the ∼ Bijk (An+1) = Bij (An) and other hand card (f (An)) = card (An) ≥ card (Ak) which is a contradiction. So no such bijection can exists and we have shown card (A ) < card (A ) n+1 k n+1 X for all k ≤ n. Finally suppose that X ⊂ An+1 is proper subset. If X ⊂ An Bij (An+1) = Bijk (An+1) then card (X) = card (Ak) for some k ≤ n by the induction hypothesis. On k=1 0 the other hand if n + 1 ∈ X, let X := X \{n + 1} $ An. Therefore by the 0 we have induction hypothesis card (X ) = card (Ak) for some k < n. It is then clear that 0 n+1 card (X) = card (Ak+1) where k + 1 < n, indeed we map X := X ∪ {n + 1} → X Ak ∪ {k + 1} = Ak+1. card (Bij (An+1)) = card (Bijk (An+1)) k=1 Example B.17. card (A \{k}) = n − 1 for k ∈ A . Indeed, let f : A → n+1 n+1 n n n−1 X X An \{k} be deﬁned by = card (Bij (An)) = n! k=1 k=1 x if x < k f (x) = . = (n + 1) n! = (n + 1)!. x + 1 if x ≥ k

Then f is the desired bijection. More generally if X ⊂ Y and card (X) = m < n = card (Y ) , then card (Y \ X) = n−m and if X and Y are finite disjoint sets Theorem B.19. Suppose that X is a set. Then card (Jn) ≤ card (X) for all then card (X ∪ Y ) = card (X) + card (Y ) . Similarly, card (X × Y ) = card (X) · n ∈ N iff card (N) ≤ card (X) . card (Y ) . Proof. Since card (Jn) ≤ card (N) for all n ∈ N it suffices to prove Proposition B.18. If f : An → An is a map, then the following are equivalent, card (Jn) ≤ card (X) for all n ∈ N implies card (N) ≤ card (X) . The intuitive idea is as follows. 1. f is injective, Suppose we have constructed fn : Jn → X which is injective. If fn were 2. f is surjective, bijective we would have card (Jn) = card (X) and in particular card (Jm) > 3. f is bijective. card (Jn) = card (X) for all m > n. Thus there exists x ∈ X \ fn (Jn) and we then define fn+1 : Jn+1 → X so that fn+1 (n + 1) = x and fn+1|J = fn. This Moreover card (Bijec (An)) = n!. n process continues indefinitely and so we may construct injective maps fn : Jn →

Proof. If n = 1, the only map f : A1 → A1 is f (1) = 1. So in this case X such that fm = fn|Jm for all m ≤ n. We then deﬁne f (m) := fn (m) where there is nothing to prove. So now suppose the proposition holds for level n and n ∈ N is any integer such that n ≥ m. In this way we construct a function, f : An+1 → An+1 is a given map. f : N → X such that f|Jn = fn for all n. This function is easily seen to be If f : An+1 → An+1 is an injective map and f (An+1) is a proper subset injective. of An+1, then card (An+1) < card (f (An+1)) = card (An+1) which is absurd. Formalities Version 1. Consider the collection of injective maps f : Thus f is injective implies f is surjective. D (f) ⊂ N → X, where D (f) is either Jn for some n ∈ N or is N. We say Conversely suppose that f : An+1 → An+1 is surjective. Let g : An+1 → f ≤ g if D (f) ⊂ D (g) and f = g|D(f). It is easy to see that every linearly or- An+1 be a right inverse, i.e. f◦g = id, which is necessarily injective, see the proof dered collection of such maps has an upper bound and so by Zorn’s lemma (see

Page: 141 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Theorem B.6), there exists a maximal element, f. If D (f) 6= N then D (f) = Jn for some n. By the last paragraph we could extend f to injective map on Jn+1 violating the maximality of f. Thus D (f) = N and we have found an injective map from N to X. Formalities Version 2. (This argument will avoid the use of Zorn’s Lemma.) By assumption, for each n ∈ N there exists an injective map, fn : Jn → X. We now let Y := ∪n∈Nfn (Jn) ⊂ X. We may construct a surjective map (but not necessarily injective map) F : N →Y. From this map we then deﬁne ψ : Y → N by ψ (y) := min F −1 ({y}) so that ψ : Y → N is now injective. Suppose for the sake of contradiction that ψ (Y ) ⊂ JN for some N ∈ N, i.e. ψ (Y ) is a bounded set. Then using our above arguments, we we know that card (ψ (Y )) = card (Jk) for some k ≤ N. On the other hand, fn : Jn → Y being injective implies card (ψ (Y )) ≥ card (Jn) for all n ∈ N. As both of these statements can not be correct at the same time we conclude that ψ (Y ) is unbounded. We may now apply Lemma 5.26 in order to see that card (Y ) = card (ψ (Y )) = card (N) . From this it follows that card (N) ≤ card (X) . Alternate Proof. By assumption, there exists and injective map, fn :

Jn → X for each n ∈ N. By replacing X by X0 := ∪n∈Nfn (Jn) we may assume that X = ∪n∈Nfn (Jn) . As X is the countable union of finite sets it follows that there exists a surjective map, f : N → X by item 2 of Theorem 5.27. . Let g : X → N be defined by g (x) := min f −1 ({x}) for all x ∈ X and let S := g (N) . To finish the proof we need only show that S is unbounded. If S were bounded, then we would find k ∈ N such that Jk ∼ X. However this is impossible since card Jn ≤ card X = card Jk would imply n ≤ k. C Appendix: Aspects of General Topological Spaces

C.1 Compactness Proposition C.4. A topological space X is compact iff every family of closed sets F ⊂ 2X having the finite intersection property satisfies T F 6= ∅. Definition C.1. The subset A of a topological space (X, τ) is said to be compact if every open cover (Definition 9.42) of A has finite a sub-cover, i.e. if U Proof. The basic point here is that complementation interchanges open and closed sets and open covers go over to collections of closed sets with empty is an open cover of A there exists U0 ⊂⊂ U such that U0 is a cover of A. (We will write A X to denote that A ⊂ X and A is compact.) A subset A ⊂ X intersection. Here are the details. @@ X is precompact if A¯ is compact. (⇒) Suppose that X is compact and F ⊂ 2 is a collection of closed sets such that T F = ∅. Let Proposition C.2. Suppose that K ⊂ X is a compact set and F ⊂ K is a closed n U = F c := {Cc : C ∈ F} ⊂ τ, subset. Then F is compact. If {Ki}i=1 is a finite collections of compact subsets n of X then K = ∪i=1Ki is also a compact subset of X. c then U is a cover of X and hence has a finite subcover, U0. Let F0 = U0 ⊂⊂ F, Proof. Let U ⊂ τ be an open cover of F, then U∪ {F c} is an open cover of then ∩F0 = ∅ so that F does not have the finite intersection property. c c K. The cover U∪ {F } of K has a finite subcover which we denote by U0∪ {F } (⇐) If X is not compact, there exists an open cover U of X with no finite c where U0 ⊂⊂ U. Since F ∩ F = ∅, it follows that U0 is the desired subcover subcover. Let c c of F. For the second assertion suppose U ⊂ τ is an open cover of K. Then U F = U := {U : U ∈ U} , covers each compact set Ki and therefore there exists a finite subset Ui ⊂⊂ U then F is a collection of closed sets with the finite intersection property while n for each i such that Ki ⊂ ∪Ui. Then U0 := ∪i=1Ui is a finite cover of K. T F = ∅.

Exercise C.1 (Suggested by Michael Gurvich). Show by example that Exercise C.4. Let (X, τ) be a topological space. Show that A ⊂ X is compact the intersection of two compact sets need not be compact. (This pathology iff (A, τA) is a compact topological space. disappears if one assumes the topology is Hausdorff, see Definition ?? below.) Let (X, d) be a metric space and for x ∈ X and ε > 0 let Exercise C.2. Suppose f : X → Y is continuous and K ⊂ X is compact, then 0 f(K) is a compact subset of Y. Give an example of continuous map, f : X → Y, Bx(ε) := Bx(ε) \{x} and a compact subset K of Y such that f −1(K) is not compact. be the ball centered at x of radius ε > 0 with x deleted. Recall from Definition Exercise C.3 (Dini’s Theorem). Let X be a compact topological space and 9.15 that a point x ∈ X is an accumulation point of a subset E ⊂ X if ∅ 6= fn : X → [0, ∞) be a sequence of continuous functions such that fn(x) ↓ 0 E ∩ V \{x} for all open neighborhoods, V, of x. The proof of the following as n → ∞ for each x ∈ X. Show that in fact fn ↓ 0 uniformly in x, i.e. elementary lemma is left to the reader. supx∈X fn(x) ↓ 0 as n → ∞. Hint: Given ε > 0, consider the open sets Vn := Lemma C.5. Let E ⊂ X be a subset of a metric space (X, d) . Then the fol- {x ∈ X : fn(x) < ε}. lowing are equivalent: Definition C.3. A collection F of closed subsets of a topological space (X, τ) has the finite intersection property if ∩F 6= ∅ for all F ⊂⊂ F. 1. x ∈ X is an accumulation point of E. 0 0 0 2. Bx(ε) ∩ E 6= ∅ for all ε > 0. The notion of compactness may be expressed in terms of closed sets as 3. Bx(ε) ∩ E is an infinite set for all ε > 0. ∞ follows. 4. There exists {xn}n=1 ⊂ E \{x} with limn→∞ xn = x. 144 C Appendix: Aspects of General Topological Spaces

Definition C.6. A metric space (X, d) is ε – bounded (ε > 0) if there exists X which can have no convergent subsequences. Indeed, the xn have been chosen a finite cover of X by balls of radius ε and it is totally bounded if it is ε – so that d (xn, xm) ≥ ε > 0 for every m 6= n and hence no subsequence of ∞ bounded for all ε > 0. {xn}n=1 can be Cauchy. Theorem C.7. Let (X, d) be a metric space. The following are equivalent. (a) X is compact. x5 (b) Every infinite subset of X has an accumulation point. x6 ∞ (c) Every sequence {xn}n=1 ⊂ X has a convergent subsequence. (d) X is totally bounded and complete. x2 x4 x7 Proof. The proof will consist of showing that a ⇒ b ⇒ c ⇒ d ⇒ a.

(a ⇒ b) We will show that not b ⇒ not a. Suppose there exists an infinite x1 x3 x8 subset E ⊂ X which has no accumulation points. Then for all x ∈ X there exists δx > 0 such that Vx := Bx(δx) satisfies (Vx \{x}) ∩ E = ∅. Clearly Fig. C.2. Constructing a set with out an accumulation point. V = {Vx}x∈X is a cover of X, yet V has no finite sub cover. Indeed, for each x ∈ X,Vx ∩ E ⊂ {x} and hence if Λ ⊂⊂ X, ∪x∈ΛVx can only contain a finite (d ⇒ a) For sake of contradiction, assume there exists an open cover V = number of points from E (namely Λ ∩ E). Thus for any Λ ⊂⊂ X,E " ∪x∈ΛVx {Vα}α∈A of X with no finite subcover. Since X is totally bounded for each and in particular X 6= ∪x∈ΛVx. (See Figure C.1.) n ∈ N there exists Λn ⊂⊂ X such that [ [ X = Bx(1/n) ⊂ Cx(1/n). δe x∈Λ x∈Λ e n n δx Choose x1 ∈ Λ1 such that no finite subset of V covers K1 := Cx (1). Since x 1 K1 = ∪x∈Λ2 K1 ∩ Cx(1/2), there exists x2 ∈ Λ2 such that K2 := K1 ∩ Cx2 (1/2) can not be covered by a finite subset of V, see Figure C.3. Continuing this way Fig. C.1. The black dots represents an infinite set, E, with not accumulation points. inductively, we construct sets Kn = Kn−1 ∩Cxn (1/n) with xn ∈ Λn such that no For each x ∈ X \ E we choose δ > 0 so that B (δ ) ∩ E = ∅ and for x ∈ E so that x x x Kn can be covered by a finite subset of V. Now choose yn ∈ Kn for each n. Since Bx (δx) ∩ E = {x} . ∞ {Kn}n=1 is a decreasing sequence of closed sets such that diam(Kn) ≤ 2/n, it follows that {yn} is a Cauchy and hence convergent with ∞ (b ⇒ c) Let {xn}n=1 ⊂ X be a sequence and E := {xn : n ∈ N} . If ∞ ∞ ∞ y = lim yn ∈ ∩m=1Km. n→∞ #(E) < ∞, then {xn}n=1 has a subsequence {xnk }k=1 which is constant and hence convergent. On the other hand if #(E) = ∞ then by assumption E has ∞ Since V is a cover of X, there exists V ∈ V such that y ∈ V. Since Kn ↓ {y} an accumulation point and hence by Lemma C.5, {xn}n=1 has a convergent subsequence. and diam(Kn) → 0, it now follows that Kn ⊂ V for some n large. But this ∞ violates the assertion that Kn can not be covered by a finite subset of V. (c ⇒ d) Suppose {xn}n=1 ⊂ X is a Cauchy sequence. By assumption there ∞ exists a subsequence {xnk }k=1 which is convergent to some point x ∈ X. Since ∞ {xn}n=1 is Cauchy it follows that xn → x as n → ∞ showing X is complete. We Corollary C.8. Any compact metric space (X, d) is second countable and hence now show that X is totally bounded. Let ε > 0 be given and choose an arbitrary also separable by Exercise ??. (See Example ?? below for an example of a com- point x1 ∈ X. If possible choose x2 ∈ X such that d(x2, x1) ≥ ε, then if possible pact topological space which is not separable.) choose x3 ∈ X such that d{x1,x2}(x3) ≥ ε and continue inductively choosing n points {xj}j=1 ⊂ X such that d{x1,...,xn−1}(xn) ≥ ε. (See Figure C.2.) This Proof. To each integer n, there exists Λn ⊂⊂ X such that X = ∞ process must terminate, for otherwise we would produce a sequence {xn}n=1 ⊂ ∪x∈Λn B(x, 1/n). The collection of open balls,

Page: 144 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 C.1 Compactness 145

Example C.10. Let X = `p(N) with p ∈ [1, ∞) and µ ∈ `p(N) such that µ(k) ≥ 0 for all k ∈ N. The set

K := {x ∈ X : |x(k)| ≤ µ(k) for all k ∈ N}

∞ is compact. To prove this, let {xn}n=1 ⊂ K be a sequence. By compactness of ∞ closed bounded sets in C, for each k ∈ N there is a subsequence of {xn(k)}n=1 ⊂ C which is convergent. By Cantor’s diagonalization trick, we may choose a ∞ ∞ subsequence {yn}n=1 of {xn}n=1 such that y(k) := limn→∞ yn(k) exists for all 1 k ∈ N. Since |yn(k)| ≤ µ(k) for all n it follows that |y(k)| ≤ µ(k), i.e. y ∈ K. Finally

∞ ∞ p X p X p Fig. C.3. Nested Sequence of cubes. lim ky − ynk = lim |y(k) − yn(k)| = lim |y(k) − yn(k)| = 0 n→∞ p n→∞ n→∞ k=1 k=1

wherein we have used the Dominated convergence theorem. (Note V := ∪n∈N ∪x∈Λn {B(x, 1/n)} p p p forms a countable basis for the metric topology on X. To check this, suppose |y(k) − yn(k)| ≤ 2 µ (k) that x ∈ X and ε > 0 are given and choose n ∈ such that 1/n < ε/2 0 N and µp is summable.) Therefore y → y and we are done. and x ∈ Λ such that d (x , x) < 1/n. Then B(x, 1/n) ⊂ B (x , ε) because for n n 0 0 Alternatively, we can prove K is compact by showing that K is closed and y ∈ B(x, 1/n), ∞ totally bounded. It is simple to show K is closed, for if {xn}n=1 ⊂ K is a d (y, x0) ≤ d (y, x) + d (x, x0) < 2/n < ε. convergent sequence in X, x := limn→∞ xn, then

n |x(k)| ≤ lim |xn(k)| ≤ µ(k) ∀ k ∈ N. Corollary C.9. The compact subsets of R are the closed and bounded sets. n→∞ Proof. This is a consequence of Theorem ?? and Theorem C.7. Here is This shows that x ∈ K and hence K is closed. To see that K is totally 1/p another proof. If K is closed and bounded then K is complete (being the closed bounded, let ε > 0 and choose N such that P∞ |µ(k)|p < ε. Since subset of a complete space) and K is contained in [−M,M]n for some positive k=N+1 QN C (0) ⊂ N is closed and bounded, it is compact. Therefore there integer M. For δ > 0, let k=1 µ(k) C exists a ﬁnite subset Λ ⊂ QN C (0) such that n n n k=1 µ(k) Λδ = δZ ∩ [−M,M] := {δx : x ∈ Z and δ|xi| ≤ M for i = 1, 2, . . . , n}. N We will show, by choosing δ > 0 suﬃciently small, that Y N Cµ(k)(0) ⊂ ∪z∈ΛBz (ε) n k=1 K ⊂ [−M,M] ⊂ ∪x∈Λδ B(x, ε) (C.1) 1 1 ∞ ∞ which shows that K is totally bounded. Hence by Theorem C.7, K is compact. The argument is as follows. Let {nj }j=1 be a subsequence of N = {n}n=1 such that 2 ∞ 1 ∞ n limj→∞ xn1 (1) exists. Now choose a subsequence {nj }j=1 of {nj }j=1 such that Suppose that y ∈ [−M,M] , then there exists x ∈ Λδ such that |yi − xi| ≤ δ j 3 ∞ 2 ∞ limj→∞ x 2 (2) exists and similarly {nj }j=1 of {nj }j=1 such that limj→∞ x 3 (3) for i = 1, 2, . . . , n. Hence nj nj n exists. Continue on this way inductively to get 2 X 2 2 d (x, y) = (yi − xi) ≤ nδ ∞ 1 ∞ 2 ∞ 3 ∞ {n}n=1 ⊃ {nj }j=1 ⊃ {nj }j=1 ⊃ {nj }j=1 ⊃ ... i=1 √ √ j such that limj→∞ x k (k) exists for all k ∈ . Let mj := n so that eventually which shows that d(x, y) ≤ nδ. Hence if choose δ < ε/ n we have shows that nj N j ∞ k ∞ d(x, y) < ε, i.e. Eq. (C.1) holds. {mj }j=1 is a subsequence of {nj }j=1 for all k. Therefore, we may take yj := xmj .

Page: 145 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 146 C Appendix: Aspects of General Topological Spaces

N N 1 where Bz (ε) is the open ball centered at z ∈ C relative to the Multiplying this equation by 2 then shows `p({1, 2, 3,...,N}) – norm. For each z ∈ Λ, letz ˜ ∈ X be deﬁned by 1 1 1 1 z˜(k) = z(k) if k ≤ N andz ˜(k) = 0 for k ≥ N + 1. I now claim that V ⊂ Y + V = Y + V 2 2 4 4 K ⊂ ∪ B (2ε) (C.2) z∈Λ z˜ and hence 1 1 1 which, when veriﬁed, shows K is totally bounded. To verify Eq. (C.2), let x ∈ K V ⊂ Y + V ⊂ Y + Y + V = Y + V. 2 4 4 and write x = u + v where u(k) = x(k) for k ≤ N and u(k) = 0 for k < N. Continuing this way inductively then shows that Then by construction u ∈ Bz˜(ε) for somez ˜ ∈ Λ and 1 ∞ !1/p V ⊂ Y + n V for all n ∈ N. (C.4) X p 2 kvkp ≤ |µ(k)| < ε. k=N+1 Indeed, if Eq. (C.4) holds, then

So we have 1 1 1 1 V ⊂ Y + V ⊂ Y + Y + V = Y + V. 2 2 2n 2n+1 kx − z˜kp = ku + v − z˜kp ≤ ku − z˜kp + kvkp < 2ε. −n Hence if x ∈ V, there exists yn ∈ Y and zn ∈ B0 (2 ) such that yn + zn → x. Exercise C.5 (Extreme value theorem). Let (X, τ) be a compact topologi- Since limn→∞ zn = 0, it follows that x = limn→∞ yn ∈ Y.¯ Since dim Y ≤ cal space and f : X → R be a continuous function. Show −∞ < inf f ≤ sup f < #(Λ) < ∞, Corollary 9.41 implies Y = Y¯ and so we have shown that V ⊂ Y. ∞ and there exists a, b ∈ X such that f(a) = inf f and f(b) = sup f. Hint: use Since for any x ∈ X, 1 x ∈ V ⊂ Y, we have x ∈ Y for all x ∈ X, i.e. X = Y. Exercise C.2 and Corollary C.9. 2kxk

Exercise C.6 (Uniform Continuity). Let (X, d) be a compact metric space, Lemma C.12. Let H be a normed linear space and H0 a closed proper subspace. (Y, ρ) be a metric space and f : X → Y be a continuous function. Show that f is For any ε > 0, there exists x0 ∈ H such that kx0k = 1 and kx − x0k ≥ 1 − ε uniformly continuous, i.e. if ε > 0 there exists δ > 0 such that ρ(f(y), f(x)) < ε whenever x ∈ H0. if x, y ∈ X with d(x, y) < δ. Hint: you could follow the argument in the proof of Theorem ??. Proof. Can assume ε < 1. Take any z0 ∈/ H0. Let d = infx∈H0 kx − z0k. For εd any δ > 0, there exists z ∈ H0, such that kz − z0k ≤ d + δ. Take δ = . Let Theorem C.11. Suppose that (X, k·k) is a normed vector in which the unit 1−ε x0 = (z − z0)/kz − z0k, where z is determined for this δ. Then kx0k = 1, and if ball, V := B0 (1) , is precompact. Then dim X < ∞. x ∈ H0, An alternate proof is given in Proposition C.13.. Since V¯ is compact, kkz − z0kx − z + z0k d d we may choose Λ ⊂⊂ X such that kx − x0k = ≥ ≥ = 1 − ε. kz − z0k kz − z0k d + δ 1 V¯ ⊂ ∪x∈Λ x + V (C.3) [Here is the proof again at a higher level. Choose h ∈ H0 such that d := 2 ∼ dist (z0,H0) = kh − z0k and then take x0 := (h − z0)/kh − z0k as above. Then where, for any δ > 0, 1 dist (x0,H0) = dist (h − z0,H0) kh − z0k δV := {δx : x ∈ V } = B0 (δ) . 1 ∼ = dist (z0,H0) = 1, Let Y := span(Λ), then Eq. (C.3) implies, kh − z0k 1 where we have used the easily veriﬁed fact that dist (ax + h, H) = |a| dist (x, H) V ⊂ V¯ ⊂ Y + V. 2 for all a ∈ R and h ∈ H.]

Page: 146 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 C.2 Exercises 147 Proposition C.13. A locally compact Banach space is ﬁnite dimensional. Exercise C.11. Suppose that d and ρ are two metrics on X.

Proof. We prove that an infinite dimensional Banach space is not locally 1. Show τd = τρ if d and ρ are equivalent. compact. We construct a sequence x1, x2, . . . , xn,... such that kxnk = 1, kxi − 2. Show by example that it is possible for τd = τρ even though d and ρ are xjk ≥ 1/2, i 6= j. Take x1 to be any unit vector. Suppose vectors x1, . . . , xn inequivalent. are constructed. Let H be the linear span of x , . . . , x . By Corollary 9.41, 0 1 n Exercise C.12. Let (X , d ) for i = 1, . . . , n be a finite collection of metric H is closed. By Lemma C.12, there exists x such that kx − x k ≥ 1/2, i i 0 n+1 i n+1 spaces and for 1 ≤ p ≤ ∞ and x = (x , x , . . . , x ) and y = (y , . . . , y ) in i = 1, . . . , n. Now the sequence just constructed has no Cauchy subsequence. 1 2 n 1 n X := Qn X , let Hence the closed unit ball is not compact. Similarly the closed ball of radius i=1 i r > 0 is also not compact. Pn p 1/p ( [di(xi, yi)] ) if p 6= ∞ ρp(x, y) = i=1 . maxi di(xi, yi) if p = ∞ Exercise C.7. Suppose (Y, k·kY ) is a normed space and (X, k·kX ) is a finite dimensional normed space. Show every linear transformation T : X → Y is 1. Show (X, ρ ) is a metric space for p ∈ [1, ∞]. Hint: Minkowski’s inequality. necessarily bounded. p 2. Show for any p, q ∈ [1, ∞], the metrics ρp and ρq are equivalent. Hint: This can be done with explicit estimates or you could use Theorem 9.39 below.

∞ C.2 Exercises Notation C.15 Let X be a set and p := {pn}n=0 be a family of semi-metrics on X, i.e. pn : X × X → [0, ∞) are functions satisfying the assumptions of C.2.1 General Topological Space Problems metric except for the assertion that pn(x, y) = 0 implies x = y. Further assume that pn(x, y) ≤ pn+1(x, y) for all n and if pn(x, y) = 0 for all n ∈ N then x = y. Exercise C.8. Let V be an open subset of R. Show V may be written as a Given n ∈ N and x ∈ X let disjoint union of open intervals Jn = (an, bn), where an, bn ∈ R∪ {±∞} for n = 1, 2, ··· < N with N = ∞ possible. Bn(x, ε) := {y ∈ X : pn(x, y) < ε} .

0 Exercise C.9. Let (X, τ) and (Y, τ ) be a topological spaces, f : X → Y be a We will write τ(p) form the smallest topology on X such that pn(x, ·): X → n function, U be an open cover of X and {Fj}j=1 be a ﬁnite cover of X by closed [0, ∞) is continuous for all n ∈ N and x ∈ X, i.e. τ(p) := τ(pn(x, ·): n ∈ N sets. and x ∈ X).

0 1. If A ⊂ X is any set and f : X → Y is (τ, τ ) – continuous then f|A : A → Y Exercise C.13. Using Notation C.15, show that collection of balls, 0 is (τA, τ ) – continuous. 0 0 B := {Bn(x, ε): n ∈ , x ∈ X and ε > 0} , 2. Show f : X → Y is (τ, τ ) – continuous iﬀ f|U : U → Y is (τU , τ )– N continuous for all U ∈ U. forms a base for the topology τ(p). Hint: Use Exercise C.10 to show B is a 3. Show f : X → Y is (τ, τ 0) – continuous iﬀ f| : F → Y is (τ , τ 0)– Fj j Fj sub-base for the topology τ(p) and then use Exercise ?? to show B is in fact a continuous for all j = 1, 2, . . . , n. base for the topology τ(p).

Exercise C.10. Suppose that X is a set, {(Yα, τα): α ∈ A} is a family of topo- Exercise C.14 (A minor variant of Exercise 9.21). Let pn be as in Nota- logical spaces and fα : X → Yα is a given function for all α ∈ A. Assum- tion C.15 and ing that S ⊂ τ is a sub-base for the topology τ for each α ∈ A, show ∞ α α α X pn(x, y) −1 d(x, y) := 2−n . S := ∪α∈Afα (Sα) is a sub-base for the topology τ := τ(fα : α ∈ A). 1 + p (x, y) n=0 n ∞ C.2.2 Metric Spaces as Topological Spaces Show d is a metric on X and τd = τ(p). Conclude that a sequence {xk}k=1 ⊂ X converges to x ∈ X iﬀ Deﬁnition C.14. Two metrics d and ρ on a set X are said to be equivalent −1 lim pn(xk, x) = 0 for all n ∈ N. if there exists a constant c ∈ (0, ∞) such that c ρ ≤ d ≤ cρ. k→∞

Page: 147 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 148 C Appendix: Aspects of General Topological Spaces ∞ Exercise C.15. Let {(Xn, dn)}n=1 be a sequence of metric spaces, X := If A ⊂ Y is open, then Y \ A is closed and hence Y \ A = Y ∩ C for some Q∞ ∞ ∞ n=1 Xn, and for x = (x(n))n=1 and y = (y(n))n=1 in X let closed subset, C ⊂ X. Since, ∞ A = Y \ (Y \ A) = Y ∩ [Y ∩ C]c = Y ∩ [Y c ∪ Cc] = Y ∩ Cc X dn(x(n), y(n)) d(x, y) = 2−n . 1 + d (x(n), y(n)) c n=1 n we see that A = Y ∩ V where V := C is open in X. Conversely if A = Y ∩ V, then (See Exercise 9.21.) Moreover, let πn : X → Xn be the projection maps, show Y \ A = Y ∩ [Y ∩ V ]c = Y ∩ [Y c ∪ V c] = Y ∩ V c ∞ τd = ⊗n=1τdn := τ({πn : n ∈ N}). which shows that Y \ A is closed in Y. That is show the d – metric topology is the same as the product topology on Proposition C.17. Let E be a subset of a metric space, (X, d) . Then E is X. Suggestions: 1) show πn is τd continuous for each n and 2) show for each disconnected iﬀ E is the disjoint union of two non-empty relatively open subset ∞ x ∈ X that d (x, ·) is ⊗ τd – continuous. For the second assertion notice of E, i.e. there exists open subset U and V of X such that; n=1 n P∞ −n dn(x(n),·) that d (x, ·) = fn where fn = 2 ◦ πn. n=1 1+dn(x(n),·) 1. U ∩ E 6= ∅= 6 V ∩ E while U ∩ V ∩ E = ∅, 2. and E = (U ∩ E) ∪ [V ∩ E] . C.2.3 Compactness Problems Proof. Suppose that E is a disconnected and write E = A ∪ B where where A and B are two non-empty separated sets. Let U := B¯c and V := A¯c. Then Exercise C.16 (Tychonoﬀ’s Theorem for Compact Metric Spaces). Let A ⊂ U and B ⊂ V. Moreover, E ∩ U = (A ∪ B) ∩ B¯c = A \ B¯ = A 6= ∅ and us continue the Notation used in Exercise 9.21. Further assume that the spaces similarly E ∩ V = B 6= ∅ while U ∩ V ∩ E = A ∩ B = ∅. Therefore E is the X are compact for all n. Show (without using Theorem ?? below) that (X, d) is n disjoint union of two non-empty relatively open sets. compact. Hint: Either use Cantor’s method to show every sequence {x }∞ ⊂ m m=1 Conversely, suppose that E = A ∪ B where A and B are relatively open X has a convergent subsequence or alternatively show (X, d) is complete and non-empty disjoint open subsets of E. Then both A and B are relatively closed totally bounded. (Compare with Example C.10 and see Theorem ?? below for in E and hence, the general version of this theorem.) E n o A = A¯ := x ∈ E : ∃ {an} ⊂ A 3 lim an = x = A¯ ∩ E n→∞ C.3 Connectedness in General Topological Spaces and similarly B = B¯E = B¯ ∩ E. Since

(You may safely skip this section on first reading.) As a warm up suppose that A = A¯ ∩ E = A¯ ∩ [A ∪ B] = A ∪ A¯ ∩ B (X, d) is a metric space and Y be a non-empty subset of X. If A ⊂ Y, the closure of A in the metric space (Y, d) is denote by A¯Y and is defined by and A ∩ B = ∅ we conclude, ¯ ¯ Y n o ∅ = A ∩ B = A ∪ A ∩ B ∩ B = A ∩ B. A¯ := y ∈ Y : y = lim an for some {an} ⊂ A = A¯ ∩ Y. n→∞ Similarly we may show A ∩ B¯ = ∅ and hence E is written as the union of two Lemma C.16. A set A ⊂ Y is closed in Y iff A = A¯Y = A¯ ∩ Y iff there exists non-empty separated sets. C ⊂ X closed such that A = C ∩ Y. Similarly, A is open in Y iff there exists V open in X such that A = V ∩ Y. Definition C.18. (X, τ) is disconnected if there exist non-empty open sets U and V of X such that U ∩ V = ∅ and X = U ∪ V . We say {U, V } is a Proof. The assertion that A ⊂ Y is closed in Y iff A = A¯Y = A¯ ∩ Y follows disconnection of X. The topological space (X, τ) is called connected if it directly from the definition. Hence if A is closed in Y, then A = C ∩Y where we is not disconnected, i.e. if there is no disconnection of X. If A ⊂ X we say may take C = A¯ – a closed subset of X. Conversely, if A = C ∩ Y, then A¯ ⊂ C A is connected iff (A, τA) is connected where τA is the relative topology on and hence A ⊂ A¯Y = A¯ ∩ Y ⊂ C ∩ Y = A which shows that A = A¯Y and so A A. Explicitly, A is disconnected in (X, τ) iff there exists U, V ∈ τ such that is closed in Y. U ∩ A 6= ∅,U ∩ A 6= ∅,A ∩ U ∩ V = ∅ and A ⊂ U ∪ V.

Page: 148 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 C.3 Connectedness in General Topological Spaces 149 The reader should check that the following statement is an equivalent defi- Example C.21. Suppose that f : X → Y is a continuous map between two nition of connectivity. A topological space (X, τ) is connected iff the only sets topological spaces, the space X is connected and the space Y is “T1,” i.e. {y} A ⊂ X which are both open and closed are the sets X and ∅. This version of is a closed set for all y ∈ Y as in Definition ?? below. Further assume f is the definition is often used in practice. locally constant, i.e. for all x ∈ X there exists an open neighborhood V of x in X such that f|V is constant. Then f is constant, i.e. f(X) = {y0} for some Remark C.19. Let A ⊂ Y ⊂ X. Then A is connected in X iff A is connected in −1 y0 ∈ Y. To prove this, let y0 ∈ f(X) and let W := f ({y0}). Since {y0} ⊂ Y Y . is a closed set and since f is continuous W ⊂ X is also closed. Since f is locally Proof. Since constant, W is open as well and since X is connected it follows that W = X, i.e. f(X) = {y0} . τA := {V ∩ A : V ⊂ X} = {V ∩ A ∩ Y : V ⊂ X} = {U ∩ A : U ⊂o Y }, As a concrete application of this result, suppose that X is a connected open subset of Rd and f : X → R is a C1 – function such that ∇f ≡ 0. If x ∈ X and the relative topology on A inherited from X is the same as the relative topology ε > 0 such that B (x, ε) ⊂ X, we have, for any |v| < ε and t ∈ [−1, 1] , that on A inherited from Y . Since connectivity is a statement about the relative topologies on A, A is connected in X iff A is connected in Y. d f (x + tv) = ∇f (x + tv) · v = 0. dt Theorem C.20 (The Connected Subsets of R). The connected subsets of R are intervals. Therefore f (x + v) = f (x) for all |v| < ε and this shows f is locally constant. Hence, by what we have just proved, f is constant on X. Proof. Suppose that A ⊂ R is a connected subset and that a, b ∈ A with a < b. If there exists c ∈ (a, b) such that c∈ / A, then U := (−∞, c) ∩ A Theorem C.22 (Properties of Connected Sets). Let (X, τ) be a topological and V := (c, ∞) ∩ A would form a disconnection of A. Hence (a, b) ⊂ A. Let space. α := inf(A) and β := sup(A) and choose αn, βn ∈ A such that αn < βn and 1. If B ⊂ X is a connected set and X is the disjoint union of two open sets U αn ↓ α and βn ↑ β as n → ∞. By what we have just shown, (αn, βn) ⊂ A for all ∞ and V, then either B ⊂ U or B ⊂ V. n and hence (α, β) = ∪n=1(αn, βn) ⊂ A. From this it follows that A = (α, β), 2. If A ⊂ X is connected, [α, β), (α, β] or [α, β], i.e. A is an interval. a) then A¯ is connected. Conversely suppose that A is a sub-interval of . For the sake of contra- R b) More generally, if A is connected and B ⊂ acc(A) or B ⊂ bd (A) , then diction, suppose that {U, V } is a disconnection of A with a ∈ U, b ∈ V. After A∪B is connected as well. (Recall that acc(A) – the set of accumulation relabelling U and V if necessary we may assume that a < b. Since A is an points of A was defined in Definition 9.15 above. Moreover by Exercise interval [a, b] ⊂ A. Let p = sup ([a, b] ∩ U) , then because U and V are open, ??, we know that acc(A) \ A = bd (A) \ A. What we are really showing a p and p here is that for any B such that A ⊂ B ⊂ A,¯ then B is connected.) can not be in V for otherwise p < sup ([a, b] ∩ U) . From this it follows that p∈ / U ∪ V and hence A 6= U ∪ V contradicting the assumption that {U, V } is a 3. If {Eα}α∈A is a collection of connected sets such that Eα ∩ Eβ 6= ∅ for all 2 S disconnection of A. α, β ∈ A, then Y := α∈A Eα is connected as well. ¯ Alternative proof of the converse. Because of Theorem C.22 below, it 4. Suppose A, B ⊂ X are non-empty connected subsets of X such that A∩B 6= suffices to assume that A is an open interval. For the sake of contradiction, ∅, then A ∪ B is connected in X. suppose that {U, V } is a disconnection of A with a ∈ U, b ∈ V. After relabelling 5. Every point x ∈ X is contained in a unique maximal connected subset Cx of X and this subset is closed. The set Cx is called the connected component U and V if necessary we may assume that a < b. Let Ja = (α, β) be the maximal open interval in U which contains a. (See Exercise C.8 or Remark 9.67 for the of x. structure of open subsets of R.) If β ∈ U we could extend Ja to the right and 2 One may assume much less here. What we really need is for any α, β ∈ A there n still be in U violating the definition of β. Moreover we can not have β ∈ V exists {αi}i=0 in A such that α0 = α, αn = β, and Eαi ∩Eαi+1 6= ∅ for all 0 ≤ i < n. because in this case Ja would not be in U. Therefore β∈ / U ∪ V = A while on Moreover if we make use of item 4. it suffices to assume that the other hand a < β < b and so β ∈ A as A is an interval and we have reached E¯ ∩ E ∪ E ∩ E¯ 6= ∅ for all 0 ≤ i < n. the desired contradiction. αi αi+1 αi αi+1 Here is a typical way these connectedness ideas are used.

Page: 149 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 150 C Appendix: Aspects of General Topological Spaces Proof.

1. Since B is the disjoint union of the relatively open sets B ∩U and B ∩V, we Proposition C.23 (Stability of Connectedness Under Products). Let must have B ∩ U = B or B ∩ V = B for otherwise {B ∩ U, B ∩ V } would (Xα, τα) be connected topological spaces. Then the product space XA = be a disconnection of B. Q α∈A Xα equipped with the product topology is connected. 2. a) Let Y = A¯ be equipped with the relative topology from X. Suppose that U, V ⊂o Y form a disconnection of Y = A.¯ Then by 1. either A ⊂ U Proof. Let us begin with the case of two factors, namely assume that X and or A ⊂ V. Say that A ⊂ U. Since U is both open and closed in Y, it Y are connected topological spaces, then we will show that X × Y is connected follows that Y = A¯ ⊂ U. Therefore V = ∅ and we have a contradiction to as well. Given x ∈ X, let fx : Y → X × Y be the map fx(y) = (x, y) and notice the assumption that {U, V } is a disconnection of Y = A.¯ Hence we must that fx is continuous since πX ◦ fx(y) = x and πY ◦ fx(y) = y are continuous conclude that Y = A¯ is connected as well. maps. From this we conclude that {x} × Y = fx(Y ) is connected by Lemma b) Now let Y = A ∪ B with B ⊂ acc(A), then 9.57. A similar argument shows X × {y} is connected for all y ∈ Y. Let p = (x0, y0) ∈ X × Y and Cp denote the connected component of p. A¯Y = A¯ ∩ Y = (A ∪ acc(A)) ∩ Y = A ∪ B. Since {x0} × Y is connected and p ∈ {x0} × Y it follows that {x0} × Y ⊂ Cp Because A is connected in Y, by (2a) Y = A ∪ B = A¯Y is also connected. and hence Cp is also the connected component (x0, y) for all y ∈ Y. Similarly, S X × {y} ⊂ C(x0,y) = Cp is connected, and therefore X × {y} ⊂ Cp. So we have 3. Let Y := α∈A Eα. By Remark C.19, we know that Eα is connected in Y for each α ∈ A. If {U, V } were a disconnection of Y, by item (1), either shown (x, y) ∈ Cp for all x ∈ X and y ∈ Y, see Figure C.4. By induction the theorem holds whenever A is a finite set, i.e. for products of a finite number of Eα ⊂ U or Eα ⊂ V for all α. Let Λ = {α ∈ A : Eα ⊂ U} then U = ∪α∈ΛEα connected spaces. and V = ∪α∈A\ΛEα. (Notice that neither Λ or A \ Λ can be empty since U and V are not empty.) Since Y X × Y [ ∅ = U ∩ V = α∈Λ,β∈Λc (Eα ∩ Eβ) 6= ∅. we have reached a contradiction and hence no such disconnection exists. 4. (A good example to keep in mind here is X = R,A = (0, 1) and B = [1, 2).) For sake of contradiction suppose that {U, V } were a disconnection of Y = y X × {y0} A∪B. By item (1) either A ⊂ U or A ⊂ V, say A ⊂ U in which case B ⊂ V. 0 p Since Y = A ∪ B we must have A = U and B = V and so we may conclude: A and B are disjoint subsets of Y which are both open and closed. This {x } × Y implies 0 A = A¯Y = A¯ ∩ Y = A¯ ∩ (A ∪ B) = A ∪ A¯ ∩ B X × {y} and therefore y ∅ = A ∩ B = A ∪ A¯ ∩ B ∩ B = A¯ ∩ B 6= ∅ which gives us the desired contradiction. Alternative proof. Let A0 := A ∪ B ∩ A¯ so that A ∪ B = A0 ∪ B. By 0 0 X item 2b. we know A is still connected and since A ∩ B 6= ∅ we may now x0 apply item 3. to finish the proof. Fig. C.4. This picture illustrates why the connected component of p in X × Y must 5. Let C denote the collection of connected subsets C ⊂ X such that x ∈ C. contain all points of X × Y. Then by item 3., the set Cx := ∪C is also a connected subset of X which contains x and clearly this is the unique maximal connected set containing A x. Since C¯x is also connected by item (2) and Cx is maximal, Cx = C¯x, i.e. For the general case, again choose a point p ∈ XA = X and again let Cx is closed. C = Cp be the connected component of p. Recall that Cp is closed and therefore

Page: 150 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 C.5 Supplementary Remarks 151 if Cp is a proper subset of XA, then XA \ Cp is a non-empty open set. By the 2. Define two elements a, b ∈ C to be equivalent (write a ∼ b) when- definition of the product topology, this would imply that XA \ Cp contains a ever dC(a, b) = 0. Show “ ∼ ” is an equivalence relation on C and that 0 0 0 0 non-empty open set of the form dC (a , b ) = dC (a, b) if a ∼ a and b ∼ b . (Hint: see Corollary 6.46.) 3. Given a ∈ C leta ¯ := {b ∈ C : b ∼ a} denote the equivalence class containing −1 V := ∩α∈Λπα (Vα) = VΛ × XA\Λ ⊂ XA \ Cp (C.5) a and let X¯ := {a¯ : a ∈ C} denote the collection of such equivalence classes. ¯ ¯ ¯ Show that d a,¯ b := dC (a, b) is well defined on X¯ × X¯ and verify X,¯ d is where Λ ⊂⊂ A and Vα ∈ τα for all α ∈ Λ. a metric space. On the other hand, let ϕ : XΛ → XA by ϕ(y) = x where 4. For x ∈ X let i (x) =a ¯ where a is the constant sequence, an = x for all n. Verify that i : X → X¯ is an isometric map and that i (X) is dense in X.¯ yα if α ∈ Λ ¯ ¯ ∞ ¯ xα = 5. Verify X, d is complete. Hint: if {a¯(m)}m=1 is a Cauchy sequence in X pα if α∈ / Λ. ¯ ¯ choose bm ∈ X such that d (i (bm) , a¯(m)) ≤ 1/m. Then showa ¯(m) → b ∞ where b = {bm} . If α ∈ Λ, πα ◦ϕ(y) = yα = πα(y) and if α ∈ A\Λ then πα ◦ϕ(y) = pα so that in m=1 every case πα ◦ ϕ : XΛ → Xα is continuous and therefore ϕ is continuous. Since XΛ is a product of a finite number of connected spaces and so is connected C.5 Supplementary Remarks and thus so is the continuous image, ϕ(XΛ) = XΛ × {pα}α∈A\Λ ⊂ XΛ. Since p ∈ ϕ (X ) we must have Λ C.5.1 Riemannian Metrics X × {p } ⊂ C . (C.6) Λ α α∈A\Λ p This subsection is not completely self contained and may safely be skipped.

Hence it follows from Eqs. (C.6) and (C.7) that Lemma C.24. Suppose that X is a Riemannian (or sub-Riemannian) manifold and d is the metric on X deﬁned by VΛ × {pα}α∈A\Λ = XΛ × {pα}α∈A\Λ ∩ V ⊂ Cp ∩ [XA \ Cp] = ∅ d(x, y) = inf {`(σ): σ(0) = x and σ(1) = y} which is a contradiction since V × {p } 6= ∅. Λ α α∈A\Λ where `(σ) is the length of the curve σ. We deﬁne `(σ) = ∞ if σ is not piecewise smooth. Then C.4 Completions of Metric Spaces

Bx(ε) = Cx(ε) and Exercise C.17 (Completions of Metric Spaces). Suppose that (X, d) is a bd(B (ε)) = {y ∈ X : d(x, y) = ε} (not necessarily complete) metric space. Using the following outline show there x ¯ exists a complete metric space X,¯ d and an isometric map i : X → X¯ such where the boundary operation, bd(·) is defined in Definition 9.15 below. that i (X) is dense in X,¯ see Definition 9.13. Proof. Let C := C (ε) ⊂ B (ε) =: B.¯ We will show that C ⊂ B¯ by showing ∞ x x c c c 1. Let C denote the collection of Cauchy sequences a = {an}n=1 ⊂ X. Given B¯ ⊂ C . Suppose that y ∈ B¯ and choose δ > 0 such that By(δ) ∩ B¯ = ∅. In two element a, b ∈ C show dC (a, b) := limn→∞ d (an, bn) exists, dC (a, b) ≥ 0 particular this implies that for all a, b ∈ C and dC satisfies the triangle inequality, By(δ) ∩ Bx(ε) = ∅. dC (a, c) ≤ dC (a, b) + dC (b, c) for all a, b, c ∈ C. We will finish the proof by showing that d(x, y) ≥ ε + δ > ε and hence that c Thus (C, dC) would be a metric space if it were true that dC(a, b) = 0 iff y ∈ C . This will be accomplished by showing: if d(x, y) < ε + δ then By(δ) ∩ a = b. This however is false, for example if an = bn for all n ≥ 100, then Bx(ε) 6= ∅. If d(x, y) < max(ε, δ) then either x ∈ By(δ) or y ∈ Bx(ε). In either dC(a, b) = 0 while a need not equal b. case By(δ) ∩ Bx(ε) 6= ∅. Hence we may assume that max(ε, δ) ≤ d(x, y) < ε + δ. Let α > 0 be a number such that

Page: 151 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Fig. C.5. An almost length minimizing curve joining x to y.

max(ε, δ) ≤ d(x, y) < α < ε + δ and choose a curve σ from x to y such that `(σ) < α. Also choose 0 < δ0 < δ such that 0 < α − δ0 < ε which can be done since α − δ < ε. Let k(t) = d(y, σ(t)) a continuous function on [0, 1] and therefore k([0, 1]) ⊂ R is a connected set which contains 0 and d(x, y). Therefore there exists t0 ∈ [0, 1] such that d(y, σ(t0)) = 0 k(t0) = δ . Let z = σ(t0) ∈ By(δ) then

0 d(x, z) ≤ `(σ|[0,t0]) = `(σ) − `(σ|[t0,1]) < α − d(z, y) = α − δ < ε and therefore z ∈ Bx(ε) ∩ Bx(δ) 6= ∅. Remark C.25. Suppose again that X is a Riemannian (or sub-Riemannian) manifold and

d(x, y) = inf {`(σ): σ(0) = x and σ(1) = y} .

Let σ be a curve from x to y and let ε = `(σ) − d(x, y). Then for all 0 ≤ u < v ≤ 1,

d(x, y) + ε = `(σ) = `(σ|[0,u]) + `(σ|[u,v]) + `(σ|[v,1])

≥ d(x, σ(u)) + `(σ|[u,v]) + d(σ(v), y) and therefore, using the triangle inequality,

`(σ|[u,v]) ≤ d(x, y) + ε − d(x, σ(u)) − d(σ(v), y) ≤ d(σ(u), σ(v)) + ε.

This leads to the following conclusions. If σ is within ε of a length minimizing curve from x to y then σ|[u,v] is within ε of a length minimizing curve from σ(u) to σ(v). In particular if σ is a length minimizing curve from x to y then σ|[u,v] is a length minimizing curve from σ(u) to σ(v).Appendix: Limsup and liminf of functions D Appendix: Limsup and liminf of functions

[Thanks go to Ali Behzadan from providing this appendix.] Our goal in lim inf f1(x) = lim inf f2(x) , lim sup f1(x) = lim sup f2(x). x→p x→p this appendix is to introduce the notion of limit superior and limit inferior x→p x→p for functions. The definitions and properties will be very similar to those that Remark D.5. Notice that if we set, were previously discussed for sequences. Here we will assume the domain is a metric space and the destination set is the set of extended real numbers. One g(δ) := inf{f(x): x ∈ E ∩ Bp(δ) − {p}}, may easiliy generalize the definitions to the case where domain and codomain h(δ) := sup{f(x): x ∈ E ∩ B (δ) − {p}}, are topological spaces and of course the codomain must be an ordered set. By p introducing a proper topology on the set of extended real numbers one can then as δ shrinks (as δ ↓ 0) g(δ) increases and h(δ)decreases ,i.e., if δ1 > δ2 then consider the limit superior and limit inferior of sequences as a special case of g(δ1) 6 g(δ2) and h(δ1) > h(δ2). Therefore the limits in the definition of lim sup what will be discussed here. and lim inf always exist in the set of extended real numbers and we have: First let’s review the definition of limit points in a metric space. lim inf f(x) = sup inf{f(x): x ∈ E ∩ Bp(δ) − {p}}, x→p Definition D.1. Let (X, d) be a metric space and E ⊆ X. A point p ∈ X is a δ>0 limit point of the set E if every open ball about p contains a point q 6= p such lim sup f(x) = inf sup{f(x): x ∈ E ∩ Bp(δ) − {p}}. that q ∈ E. x→p δ>0 Notation D.6 S := E ∩ B (δ) − {p}. Exercise D.1. Prove that p is a limit point of E iff there exists a sequence p,δ p ∞ {xn}n=1 ∈ E − {p} such that limn→∞ xn = p. Remark D.7. In the special case where X = R and E is an open interval in R containing p we have: Example D.2. Consider R(with the Euclidean metric d(x, y) = |x − y|). Let E = (2, 3] ∪ {5}. p = 2 is a limit point of E because 2 = limn→∞(2 + 1/n). lim inf f(x) = lim inf f(x), p = 5 is not a limit point of E because the open ball of radius 1/2 centered at x→p δ↓0 0<|x−p|<δ 5 does not intersect E − {5}. lim sup f(x) = lim sup f(x). x→p δ↓0 0<|x−p|<δ Definition D.3. Let (X, d) be a metric space. Suppose E ⊆ X and p is a limit point of E. Let f be a real (or extended real) valued function on E, i.e., In the following two propositions we list some of the most important properties of limit superior and limit inferior of functions. The reader may check that these f : E→ R (or f : E→R¯). Then properties are completely analogous to those of sequences (Proposition 3.28). lim inf f(x) = lim inf{f(x): x ∈ E ∩ Bp(δ) − {p}}, x→p δ↓0 Proposition D.8. Let (X, d) be a metric space. Suppose E ⊆ X, f : E→R¯, g : E→R¯ and p is a limit point of E. lim sup f(x) = lim sup{f(x): x ∈ E ∩ Bp(δ) − {p}}. x→p δ↓0 1. lim infx→p(−f(x)) = − lim supx→p f(x). ˆ 2. lim infx→p f(x) 6 lim supx→p f(x). Remark D.4. Clearly if ν > 0 is such that Bp(ν) ⊆ E and f := f|Bp(ν) then 3. If f(x) 6 g(x) on the intersection of an open ball about p with E (or on lim inf f(x) = lim inf fˆ(x) , lim sup f(x) = lim sup fˆ(x). Sp,ν for some ν > 0) then x→p x→p x→p x→p lim inf f(x) 6 lim inf g(x) , lim sup f(x) 6 lim sup g(x). x→p x→p x→p x→p So in fact if f1 and f2 agree on an open ball about p in X then 154 D Appendix: Limsup and liminf of functions

Proof. 6. limx→p f(x) exists in R then, 1. From Proposition 3.8 one can easily conclude that if S is a nonempty subset lim sup(f(x) + g(x)) = lim f(x) + lim sup g(x), of the extended real numbers then inf(S) = − sup(−S). Thus x→p x→p x→p lim inf(f(x) + g(x)) = lim f(x) + lim inf g(x). lim inf(−f(x)) = lim inf (−f(x)) = lim(− sup (f(x)) x→p x→p x→p x→p δ↓0 x∈S δ↓0 p,δ x∈Sp,δ 7. If f(x), g(x) > 0 on the intersection of an open ball about p with E (or on = − lim sup (f(x)) = − lim sup f(x). δ↓0 Sp,ν for some ν > 0) then x∈Sp,δ x→p lim sup(f(x).g(x)) 6 (lim sup f(x))(lim sup g(x)), 2. We have x→p x→p x→p ∀δ > 0 inf f(x) sup f(x). x∈S 6 p,δ x∈Sp,δ provided the right hand side of this equation is not of the form 0.∞ or ∞.0. 8. If f(x), g(x) 0 on the intersection of an open ball about p with E (or on Since the limit of both sides as δ ↓ 0 exists (in the extended real numbers) > S for some ν > 0) and A = lim f(x) exists in (0, ∞) then we get p,ν x→p lim inf f(x) lim sup f(x), lim sup(f(x).g(x)) = ( lim f(x))(lim sup g(x)). δ↓0 x∈S 6 δ↓0 p,δ x∈Sp,δ x→p x→p x→p

therefore, lim infx→p f(x) 6 lim supx→p f(x). Proof. 3. By assumption there exists ν > 0 such that f(x) 6 g(x) on Sp,ν . Thus if ¯ 1.( ⇐) Let A := lim infx→p f(x) = lim supx→p f(x) ∈ R. We may consider 3 0 < δ ν then f(x) sup g(y) for all x ∈ Sp,δ. Therefore 6 6 y∈Sp,δ cases: sup f(x) 6 sup g(y). a) A ∈ R. Let {xn} be a sequence in E − {p} such that xn → p as n → x∈S y∈S p,δ p,δ ∞. In what follows we will prove that limn→∞ f(xn) = A and hence by Theorem 6.33 we can conclude that limx→p f(x) = A. Let ε > 0 Similarly, if 0 < δ 6 ν then infx∈Sp,δ f(x) 6 g(y) for all y ∈ Sp,δ and hence be given. By assumption there exists δ1 > 0 be such that A − ε 6 infx∈S f(x) and there exists δ2 > 0 such that sup f(x) A+ε . inf f(x) inf g(y). p,δ1 x∈Sp,δ2 6 x∈S 6 y∈S p,δ p,δ Let δ := min{δ1, δ2}. Since xn → p and {xn} ⊆ E−{p} we can conclude that x ∈ S a.a. n. Therefore, Passing to the limit (as δ ↓ 0) in each of these inequalities gives the desired n p,δ

inequalities. ∀ε > 0∃δ > 0 A−ε inf f(x) f(xn) sup f(x) A+ε a.a.n 6 x∈S 6 6 6 p,δ x∈Sp,δ

This precisely means that limn→∞ f(xn) = A. Proposition D.9. Let (X, d) be a metric space. Suppose E ⊆ X, f : E→ , R b) A = ∞. Let M > 0 be given. Since limδ↓0 infx∈Sp,δ f(x) = ∞ we may g : E→ and p is a limit point of E. R conclude that there exists ν > 0 such that infx∈Sp,δ f(x) > M provided ¯ ¯ 0 < δ 6 ν. Therefore 1. limx→p f(x) exists in R iﬀ lim infx→p f(x) = lim supx→p f(x) ∈ R. In this case limx→p f(x) = lim infx→p f(x) = lim supx→p f(x). ∀M > 0 ∃ν > 0 s.t. ∀x ∈ Sp,ν f(x) > M. 2. lim supx→p(f(x) + g(x)) 6 lim supx→p f(x) + lim supx→p g(x),provided the right side of this equation is not of the form ∞ − ∞ or −∞ + ∞. This precisely means that limx→p f(x) = ∞. c) A = −∞. Let M < 0 be given. Since limδ↓0 sup f(x) = −∞ we 3. lim infx→p f(x)+lim infx→p g(x) 6 lim infx→p(f(x)+g(x)),provided the left x∈Sp,δ may conclude that there exists ν > 0 such that sup f(x) < M side of this equation is not of the form ∞ − ∞ or −∞ + ∞. x∈Sp,δ provided 0 < δ ν. Therefore 4. lim infx→p f(x) + lim supx→p g(x) 6 lim supx→p(f(x) + g(x)), provided the 6 left side of this equation is not of the form ∞ − ∞ or −∞ + ∞. ∀M < 0 ∃ν > 0 s.t. ∀x ∈ Sp,ν f(x) < M. 5. lim infx→p(f(x) + g(x)) 6 lim infx→p f(x) + lim supx→p g(x),provided the right side of this equation is not of the form ∞ − ∞ or −∞ + ∞. This precisely means that limx→p f(x) = −∞.

Page: 154 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 D Appendix: Limsup and liminf of functions 155 ¯ (⇒) Conversely, suppose that A := limx→p f(x) ∈ . We may consider 3 lim inf f(x) + lim inf g(x) = −(lim sup(−f(x)) + lim sup(−g(x))) R x→p x→p cases: x→p x→p a) A ∈ . Let ε > 0 be given. Since lim f(x) = A there exists ν > 0 6 −(lim sup(−(f(x) + g(x)))) R x→p x→p such that if x ∈ S then A − ε < f(x) < A + ε. So for all 0 < δ ν p,ν 6 = lim inf(f(x) + g(x)). we have x→p A − ε inf f(x) sup f(x) A + ε. 6 x∈S 6 6 p,δ x∈Sp,δ 4. We may consider 3 cases: Passing to the limit (as δ ↓ 0) we get a) lim infx→p f(x) ∈ R. We have

A − ε 6 lim inf f(x) 6 lim sup f(x) 6 A + ε. lim sup g(x) = lim sup(g(x) + f(x) − f(x)) x→p x→p x→p x→p 6 lim sup(f(x) + g(x)) + lim sup(−f(x)) Since ε > 0 is arbitrary, it follows that, x→p x→p = lim sup(f(x) + g(x)) − lim inf(f(x)). A 6 lim inf f(x) 6 lim sup f(x) 6 A, x→p x→p x→p x→p Therefore i.e. that A = lim infx→p f(x) = lim supx→p f(x). b) A = ∞. Let M > 0 be given. There exists ν > 0 such that f(x) > M lim inf f(x) + lim sup g(x) 6 lim sup(f(x) + g(x)). x→p x→p x→p provided x ∈ Sp,ν . So for all 0 < δ 6 ν we have infx∈Sp,δ f(x) > M. Passing to the limit (as δ ↓ 0) we get lim infx→p f(x) > M. Since b) lim infx→p f(x) = −∞. By assumption lim infx→p f(x) + M > 0 is arbitrary it follows that lim infx→p f(x) = ∞. Also since lim supx→p g(x) cannot be ∞ − ∞, so either lim supx→p g(x) ∈ R in general lim infx→p f(x) 6 lim supx→p f(x) we can conclude that or lim supx→p g(x) = −∞. In both cases lim infx→p f(x) + lim supx→p f(x) = ∞. c) A = −∞. Let M < 0 be given. There exists ν > 0 such that f(x) < M lim supx→p g(x) = −∞. So obviously the claimed inequality holds true. provided x ∈ S . So for all 0 < δ ν we have sup f(x) M. c) lim infx→p f(x) = ∞. By assumption lim infx→p f(x) + lim supx→p g(x) p,ν 6 x∈Sp,δ 6 cannot be ∞ − ∞, so either lim supx→p g(x) ∈ R or lim supx→p g(x) = Passing to the limit (as δ ↓ 0) we get lim supx→p f(x) 6 M. Since ∞. Also note that in general lim infx→p f(x) 6 lim supx→p f(x), so we M < 0 is arbitrary it follows that lim supx→p f(x) = −∞. Also can conclude that lim supx→p f(x) = ∞ and hence limx→p f(x) = ∞. since in general lim infx→p f(x) 6 lim supx→p f(x) we can conclude that lim inf f(x) = −∞. In what follows we will show that lim supx→p(f(x) + g(x)) = ∞ and x→p hence the desired inequality obviously holds true. First let’s consider 2. Let δ > 0 the case where A := lim supx→p g(x) ∈ R. Let M > 0 be given.

∀x ∈ Sp,δ f(x) + g(x) sup f(y) + sup g(y). 6 lim f(x) = ∞ =⇒ ∃δ1 > 0 3 ∀x ∈ Sp,δ f(x) M + |A|, y∈Sp,δ y∈Sp,δ x→p 1 > So A = lim sup g(x) =⇒ ∃δ2 > 0 3 ∀0 < ν 6 δ2 sup g(x) > A/2. sup (f(x) + g(x)) 6 sup f(y) + sup g(y). x→p x∈Sp,ν x∈Sp,δ y∈Sp,δ y∈Sp,δ Therefore if we let δ := min{δ1, δ2} then Passing to the limit as δ ↓ 0 in this equation then implies ∀0 < ν 6 δ ∃x ∈ Sp,ν 3 f(x) + g(x) > M + |A| + A/2 > M. lim sup(f(x) + g(x)) 6 lim sup f(x) + lim sup g(x). x→p x→p x→p Consequently 3. We have, ∀0 < ν 6 δ sup (f(x) + g(x)) > M. x∈Sp,ν

Page: 155 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 156 D Appendix: Limsup and liminf of functions

Passing to the limit (as ν ↓ 0) we get lim supx→p(f(x) + g(x)) > M. Similarly by item 3. and item 5. we have Since M is arbitrary we can conclude that lim supx→p(f(x)+g(x)) = ∞. Now let’s consider the case where lim sup g(x) = ∞. Let M > 0 be lim inf f(x)+lim inf g(x) 6 lim inf(f(x)+g(x)) 6 lim sup f(x)+lim inf g(x) x→p x→p x→p x→p x→p x→p given. and so

lim f(x) = ∞ =⇒ ∃δ1 > 0 3 ∀x ∈ Sp,δ1 f(x) > M/2, x→p lim f(x) + lim inf g(x) lim inf(f(x) + g(x)) lim f(x) + lim inf g(x) x→p x→p 6 x→p 6 x→p x→p lim sup g(x) = ∞ =⇒ ∃δ2 > 0 3 ∀0 < ν 6 δ2 sup g(x) > M/2. x→p x∈Sp,ν Therefore lim inf(f(x) + g(x)) = lim f(x) + lim inf g(x). Therefore if we let δ := min{δ1, δ2} then x→p x→p x→p

7. By assumption there exists ν > 0 such that f(x), g(x) 0 for all x ∈ Sp,ν . ∀0 < ν 6 δ ∃x ∈ Sp,ν 3 f(x) + g(x) > M/2 + M/2 = M. > For all 0 < δ 6 ν we have Consequently ∀x ∈ Sp,δ f(x).g(x) 6 ( sup f(y))( sup g(y)), y∈Sp,δ y∈Sp,δ ∀0 < ν 6 δ sup (f(x) + g(x)) > M. x∈Sp,ν and therefore, Passing to the limit as ν ↓ 0 we get lim sup (f(x)+g(x)) M. Since x→p > sup (f(x).g(x)) 6 ( sup f(y))( sup g(y)). M is arbitrary we can conclude that lim supx→p(f(x) + g(x)) = ∞. x∈Sp,δ y∈Sp,δ y∈Sp,δ 5. By using item 4. we can write Letting δ ↓ 0 in this inequality implies, lim inf(g(x) + f(x)) = − lim sup(−g(x) − f(x)) x→p x→p lim sup (f(x).g(x)) lim[( sup f(y))( sup g(y))] δ↓0 6 δ↓0 x∈Sp,δ y∈Sp,δ y∈Sp,δ 6 −(lim sup(−f(x)) + lim inf(−g(x))) x→p x→p = lim sup f(y). lim sup g(y) δ↓0 y∈S δ↓0 y∈S = lim inf f(x) + lim sup g(x). p,δ p,δ x→p x→p = (lim sup f(x))(lim sup g(x)). x→p x→p 6. By item 2. and item 4. we have provided (lim supx→p f(x))(lim supx→p g(x)) is not of the form 0·∞ or ∞·0 lim inf f(x)+lim sup g(x) 6 lim sup(f(x)+g(x)) 6 lim sup f(x)+lim sup g(x) 8. Note that by item 1. lim infx→p f(x) = lim supx→p f(x) = limx→p f(x); so x→p x→p x→p x→p x→p considering the previous item we just need to show that But by item 1. we know lim sup(f(x).g(x)) > A. lim sup g(x). x→p x→p lim inf f(x) = lim sup f(x) = lim f(x). x→p x→p x→p Let α ∈ (0,A). By assumption there exists ν1 > 0 such that f(x), g(x) > 0

for all x ∈ Sp,ν1 . Since limx→p f(x) = A there exists ν2 > 0 such that Therefore f(x) α for all x ∈ S and hence f(x)g(x) αg(x) for all x ∈ S > p,ν2 > p,ν where ν = min{v1, ν2}. Therefore by item 3. of Proposition D.8 we can lim f(x) + lim sup g(x) 6 lim sup(f(x) + g(x)) 6 lim f(x) + lim sup g(x) x→p x→p x→p x→p x→p conclude that

Consequently lim sup(f(x).g(x)) > lim sup(αg(x)) = α. lim sup g(x). x→p x→p x→p lim sup(f(x) + g(x)) = lim f(x) + lim sup g(x). x→p x→p x→p Now we may consider two cases:

Page: 156 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 D Appendix: Limsup and liminf of functions 157

a) If lim supx→p g(x) < ∞ we may let α ↑ A in order to see that 3. lim supx→p f(x) = A if and only if for every ε > 0 ∃δ > 0 such that f(x) 6 A+ε provided x ∈ Sp,δ, ∀δ > 0 ∃ xδ,ε ∈ Sp,δ such that f(xδ,ε) > A−ε lim sup(f(x).g(x)) > A. lim sup g(x). . x→p x→p Proof. b) If lim supx→p(g(x)) = ∞ it follows that 1.( ⇒) Let ε > 0 be given. lim sup f(x) A so there exists δ > 0 such ν↓0 x∈Sp,ν 6 lim sup(f(x).g(x)) = A. lim sup g(x) = α. lim sup g(x) = ∞. that for all 0 < ν δ, sup f(x) A+ε. So in particular f(x) A+ε 6 x∈Sp,ν 6 6 x→p x→p x→p provided x ∈ Sp,δ. 2.( ⇐) Conversely, if ε > 0 then by assumption there exists δ > 0 such that f(x) 6 A + ε on Sp,δ. Therefore, by item 3. of Proposition D.8, Remark D.10. By item 1. of the previous proposition if limit exists, then limsup lim supx→p f(x) 6 lim supx→p(A + ε) = limx→p(A + ε) = A + ε. Since and liminf are both equal to the limit. So whenever the limit exists we may ε > 0 is arbitrary we can conclude that lim supx→p f(x) 6 A. replace the limit by limsup and liminf. The important point is that even if the 3.( ⇒)Let ε > 0 be given. Since inf sup f(x) A we can conclude δ>0 x∈Sp,δ > limit does not exist, limsup and liminf always exist. Therefore if we want to that for all δ > 0, sup f(x) A . Therefore for all δ > 0 there exists x∈Sp,δ > take limit but we do not know whether the limit exists we should take limsup x ∈ S such that f(x ) A − ε (otherwise sup f(x) A − ε). δ,ε p,δ δ,ε > x∈Sp,δ 6 or liminf instead. 4.( ⇐)Conversely, if ε > 0 then the assumption implies that for all δ > 0, sup f(x) A − ε. Therefore lim sup f(x) = x∈Sp,δ > x→p Exercise D.2. Prove that if f(x) > 0, then lim supx→p f(x) = 0 if and only inf sup f(x) A − ε. Since ε > 0 is arbitrary we can con- δ>0 x∈Sp,δ > if limx→p f(x) = 0. Conclude that, even if f is not necessarily nonnegative, clude that lim supx→p f(x) > A. limx→p f(x) = c iﬀ lim supx→p |f(x) − c| = 0. 5. This one is a direct consequence of the previous items. Exercise D.3. Prove the followings:

1. If f(x), g(x) > 0 on the intersection of an open ball about p with E (or on Exercise D.4. Let (X, d) be a metric space. Suppose E ⊆ X and p is a limit Sp,ν for some ν > 0) then point of E. Let f be a real (or extended real) valued function on E and suppose A ∈ R. lim inf(f(x).g(x)) (lim inf f(x))(lim inf g(x)), x→p > x→p x→p 1. lim infx→p f(x) > A if and only if for every ε > 0 ∃δ > 0 such that provided the right hand side of this equation is not of the form 0.∞ or ∞.0. f(x) > A − ε provided x ∈ Sp,δ. 2. lim inf f(x) A if and only if for every ε > 0 ∀δ > 0 ∃ x ∈ S such 2. If f(x), g(x) > 0 on the intersection of an open ball about p with E (or on x→p 6 δ,ε p,δ that f(x ) A + ε . Sp,ν for some ν > 0) and A = limx→p f(x) exists in (0, ∞) then δ,ε 6 3. lim infx→p f(x) = A if and only if for every ε > 0 ∃δ > 0 such that lim inf(f(x).g(x)) = ( lim f(x))(lim inf g(x)). f(x) > A−ε provided x ∈ Sp,δ, ∀δ > 0 ∃ xδ,ε ∈ Sp,δ such that f(xδ,ε) 6 A+ε x→p x→p x→p .

The claim of the following theorem is similar to that of Exercise 3.14. Theorem D.12. Let (X, d) be a metric space. Suppose E ⊆ X and p is a limit Theorem D.11. Let (X, d) be a metric space. Suppose E ⊆ X and p is a limit point of E. Let f be a real (or extended real) valued function on E. Suppose ¯ point of E. Let f be a real (or extended real) valued function on E and suppose lim supx→p f(x) = A ∈ R. A ∈ . ∞ R 1. If{xn}n=1 is a sequence in E − {p} such that p = limn→∞ xn, then lim supn→∞ f(xn) 6 A. 1. lim supx→p f(x) 6 A if and only if for every ε > 0 ∃δ > 0 such that ∞ 2. There exists a sequence{xn}n=1in E − {p} such that p = limn→∞ xn and f(x) 6 A + ε provided x ∈ Sp,δ. A = limn→∞ f(xn). 2. lim supx→p f(x) > A if and only if for every ε > 0 ∀δ > 0 ∃ xδ,ε ∈ Sp,δ such that f(xδ,ε) > A − ε . Proof.

Page: 157 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 158 D Appendix: Limsup and liminf of functions

1. We may consider 3 cases: Deﬁnition D.14. Let X = R and E = (a, b)be an open interval in R containing Case 1: A = +∞. Clearly lim supn→∞ f(xn) 6 A = +∞. p (or let E = (a, b) − {p} where p ∈ (a, b)). Let f be a real (or extended real) Case 2: A = −∞. In general lim infx→p f(x) 6 lim supx→p f(x) so we valued function on E. We deﬁne may conclude that lim infx→p f(x) = lim supx→p f(x) = −∞ and hence limx→p f(x) = −∞. Since {xn} ⊆ E − {p} and xn → p, it follows that lim sup f(x) = lim sup f(x), x↓p δ↓0 0 0 be given. By Theorem D.11 there exists δ > 0 lim inf f(x) = lim inf f(x), x↓p δ↓0 0 0 is arbitrary x↑p 0 1 (let δ = ε = 1/n), there exists x↑p δ↓0 0 A − 1/n. Therefore (Note that the set E can be unbounded.)

xn ∈ Sp,1/n =⇒ d(xn, p) < 1/n, {xn} ⊆ E−{p} =⇒ xn → p, {xn} ⊆ E−{p} Remark D.15. Note that clearly if we let f1 := f|(a,p)and f2 := f|(p,b)then Moreover, lim sup f(x) = lim sup f2(x) , lim inf f(x) = lim inf f2(x), x↓p x→p x↓p x→p f(xn) A−1/n =⇒ lim inf f(xn) lim inf(A−1/n) = lim (A−1/n) = A > n→∞ > n→∞ n→∞ lim sup f(x) = lim sup f1(x) , lim inf f(x) = lim inf f1(x). x↑p x→p x↑p x→p But by item i. lim supn→∞ f(xn) 6 A and so So in fact the properties that were previously mentioned for limsup and liminf A 6 lim inf f(xn) 6 lim sup f(xn) 6 A of functions (e.g. Prposition 10) will hold true for one sided limsup and liminf n→∞ n→∞ as well (with obvious modifications). ¯ Consequently lim infn→∞ f(xn) = lim supn→∞ f(xn) = A which implies Also it will be helpful to note that limx↓p f(x) = A ∈ R iff limx→p f2(x) = ¯ ¯ ¯ limn→∞ f(xn) = A. A ∈ R. Similarly limx↑p f(x) = A ∈ R iff limx→p f1(x) = A ∈ R. Corollary D.16. An immediate consequence of the above remark, item 1. of Proposition D.9, and Theorem 6.41 is that if f is a real valued function on the Exercise D.5. Let (X, d) be a metric space. Suppose E ⊆ X and p is a limit open interval E = (a, b) which contains p (or on E = (a, b) − {p} where p ∈ point of E. Let f be a real (or extended real) valued function on E. Suppose ¯ (a, b)) then lim infx→p f(x) = A ∈ R. ¯ ¯ ∞ 1. limx↓p f(x) = A ∈ R iff lim supx↓p f(x) = lim infx↓p f(x) = A ∈ R. 1. If{xn}n=1 is a sequence in E − {p} such that p = limn→∞ xn, then ¯ ¯ lim inf f(x ) A. 2. limx↑p f(x) = A ∈ R iff lim supx↑p f(x) = lim infx↑p f(x) = A ∈ R. n→∞ n > ¯ 2. There exists a sequence{x }∞ in E − {p} such that p = lim x and 3. limx→p f(x) = A ∈ R iff lim supx↓p f(x) = lim infx↓p f(x) = n n=1 n→∞ n ¯ A = limn→∞ f(xn). lim supx↑p f(x) = lim infx↑p f(x) = A ∈ R.

Remark D.13. Considering the above theorem and exercise, one can give the Theorem D.17. Let X = R and E = (a, b)be an open interval in R containing p (or let E = (a, b) − {p} where p ∈ (a, b)). If f is a real valued function on E following interpretation of lim supx→p f(x) and lim infx→p f(x): ∞ then lim supx→p f(x) is the largest limiting value of {f(xn)}n=1 over all sequences ∞ lim sup f(x) = max{lim sup f(x), lim sup f(x)}. {xn}n=1 ⊆ E − {p} that converge to {p}. ∞ x→p x↓p x↑p lim infx→p f(x) is the smallest limiting value of {f(xn)}n=1 over all se- ∞ quences {xn}n=1 ⊆ E − {p} that converge to {p}.

Page: 158 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 D Appendix: Limsup and liminf of functions 159  Proof. We have  1/x if x ∈ Q − {0} f(x) = 1 if x ∈ Qc . ∀δ > 0 sup f(x) 6 sup f(x) =⇒ lim sup f(x) 6 lim sup f(x),  0 if x = 0 0 0 sup f(x) 6 sup f(x) =⇒ lim sup f(x) 6 lim sup f(x). note that for {xn = 1/n} ⊆ (0, 2) we have limn→∞ xn = 0 and 0 c of the sets A1 = {n : xn ∈ (a, p)} or A2 = {n : xn ∈ (p, b)} is in- lim infx↑0 f(x) and therefore lim infx↑0 f(x) = −∞. Also since Q is dense in R, ∞ finite. If A1 is infinite then {xn} has a subsequence {yk}k=1 ⊆ (a, p). So in particular y → p and lim f(y ) = lim sup f(x). Since y → p ∀δ > 0 sup f(x) = 1 =⇒ lim sup f(x) = 1 =⇒ lim sup f(x) = 1, k k→∞ k x→p k 0<0−x<δ δ↓0 0<0−x<δ x↑0 we can conclude that lim supk→∞ f(yk) 6 lim supx↑p f(x) (by Theorem D.12 1 and Remark D.15). But lim supk→∞ f(yk) = limk→∞ f(yk) = lim supx→p f(x) ∀0 < δ < inf f(x) = 1 =⇒ lim inf f(x) = 1 =⇒ lim inf f(x) = 1. 2 0t t>0 x>t lim inf f(x) = min{lim inf f(x), lim inf f(x)}. x→p x↓p x↑p lim sup f(x) := lim sup f(x) = inf sup f(x). t→∞ t>0 x→∞ x>t x>t 0 if x ∈ Q c Example D.18. Let f(x) = c and p ∈ R. Since Q and Q are both Previously mentioned properties for limsup and liminf of a function at a 1 if x ∈ Q finite point hold for limsup and liminf at infinity as well (with obvious modifi- dense in , we have R cations). For completeness here we will state those prpoperties again.

∀δ > 0 sup f(x) = 1 =⇒ lim sup f(x) = 1 =⇒ lim sup f(x) = 1, Proposition D.21. Let f be a real (or extended real) valued function on . 0 0 sup f(x) = 1 =⇒ lim sup f(x) = 1 =⇒ lim sup f(x) = 1, 1. lim infx→∞(−f(x)) = − lim supx→∞ f(x). 0 0 inf f(x) = 0 =⇒ lim inf f(x) = 0 =⇒ lim inf f(x) = 0, ¯ 0 0, then ∀δ > 0 inf f(x) = 0 =⇒ lim inf f(x) = 0 =⇒ lim inf f(x) = 0. 0

Also by Theorem D.17 and Exercise D.6, lim supx→p f(x) = max{1, 1} = 1 and (Proof is completely analogous to the proofs of the corresponding items of lim infx→p f(x) = min{0, 0} = 0. In particular, considering Corollary D.16, we Proposition D.8.) can conclude that limx→p f(x) does not exist. Proposition D.22. Let f and g be a real valued function on R, i.e., f : R→ R Example D.19. Let and g : R→ R .

Page: 159 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 160 D Appendix: Limsup and liminf of functions ¯ ¯ 1. limx→∞ f(x) exists in R iﬀ lim infx→∞ f(x) = lim supx→∞ f(x) ∈ R. In this Theorem D.23. Let f be a real (or extended real) valued function on R and case limx→∞ f(x) = lim infx→∞ f(x) = lim supx→∞ f(x). suppose A ∈ R. 2. lim sup (f(x) + g(x)) lim sup f(x) + lim sup g(x),provided x→∞ 6 x→∞ x→∞ 1. lim sup f(x) A if and only if for every ε > 0 ∃M > 0 such that the right side of this equation is not of the form ∞ − ∞ or −∞ + ∞. x→∞ 6 f(x) A + ε provided x M. 3. lim inf f(x) + lim inf g(x) lim inf (f(x) + g(x)),provided the 6 > x→∞ x→∞ 6 x→∞ 2. lim sup f(x) A if and only if for every ε > 0 ∀M > 0 ∃ x M left side of this equation is not of the form ∞ − ∞ or −∞ + ∞. x→∞ > M,ε > such that f(x ) A − ε . 4. lim inf f(x) + lim sup g(x) lim sup (f(x) + g(x)), provided M,ε > x→∞ x→∞ 6 x→∞ 3. lim sup f(x) = A if and only if for every ε > 0 ∃M > 0 such that the left side of this equation is not of the form ∞ − ∞ or −∞ + ∞. x→∞ f(x) A + ε provided x M, ∀M > 0 ∃ x M such that f(x ) 5. lim inf (f(x)+g(x)) lim inf f(x)+lim sup g(x),provided the 6 > M,ε > M,ε > x→∞ 6 x→∞ x→∞ A − ε . right side of this equation is not of the form ∞ − ∞ or −∞ + ∞. 6. limx→∞ f(x) exists in R then, (Proof is completely analogous to the proof of Theorem D.11.)

lim sup(f(x) + g(x)) = lim f(x) + lim sup g(x), Exercise D.9. Let f be a real (or extended real) valued function on R and x→∞ x→∞ x→∞ suppose A ∈ R. lim inf(f(x) + g(x)) = lim f(x) + lim inf g(x). x→∞ x→∞ x→∞ 1. lim infx→∞ f(x) > A if and only if for every ε > 0 ∃M > 0 such that f(x) > A − ε provided x > M. 7. If f(x), g(x) > 0 on [M, ∞) for some M > 0, then 2. lim infx→∞ f(x) 6 A if and only if for every ε > 0 ∀M > 0 ∃ xM,ε > M such that f(x ) A + ε . lim sup(f(x).g(x)) 6 (lim sup f(x))(lim sup g(x)), M,ε 6 x→∞ x→∞ x→∞ 3. lim infx→∞ f(x) = A if and only if for every ε > 0 ∃M > 0 such that f(x) A − ε provided x M, provided the right hand side of this equation is not of the form 0.∞ or ∞.0. > > ∀M > 0 ∃ xM,ε > M such that f(xM,ε) 6 A + ε . 8. If f(x), g(x) > 0 on [M, ∞) for some M > 0 and A = limx→∞ f(x) exists in (0, ∞) then Theorem D.24. Let f be a real (or extended real) valued function on R. Sup- ¯ pose lim supx→∞ f(x) = A ∈ R. lim sup(f(x).g(x)) = ( lim f(x))(lim sup g(x)). x→∞ x→∞ x→∞ ∞ 1. If{xn}n=1 is a sequence of real numbers such that limn→∞ xn = ∞, then lim supn→∞ f(xn) 6 A. (Proof is completely analogous to the proofs of the corresponding items of ∞ Proposition D.9.) 2. There exists a sequence{xn}n=1of real numbers such that limn→∞ xn = ∞ and A = limn→∞ f(xn). Exercise D.7. Prove that if f(x) 0, then lim sup f(x) = 0 if and only > x→∞ (Proof is completely analogous to the proof of Theorem D.12.) if limx→∞ f(x) = 0. Conclude that, even if f is not necessarily nonnegative, limx→∞ f(x) = c iﬀ lim supx→∞ |f(x) − c| = 0. Exercise D.10. Let f be a real (or extended real) valued function on R. Sup- ¯ pose lim infx→∞ f(x) = A ∈ R. Exercise D.8. Prove the followings: ∞ 1. If{xn}n=1 is a sequence of real numbers such that limn→∞ xn = ∞, then 1. If f(x), g(x) 0 on [M, ∞) for some M > 0 then > lim infn→∞ f(xn) > A. ∞ 2. There exists a sequence{xn}n=1of real numbers such that limn→∞ xn = ∞ lim inf(f(x).g(x)) > (lim inf f(x))(lim inf g(x)), x→∞ x→∞ x→∞ and A = limn→∞ f(xn). provided the right hand side of this equation is not of the form 0.∞ or ∞.0. Example D.25. Let f(x) = sin x. We have 2. If f(x), g(x) > 0 on [M, ∞) for some M > 0 and A = limx→p f(x) exists in ∀t sup f(x) = 1 =⇒ lim sup f(x) = 1 =⇒ lim sup f(x) = 1, (0, ∞) then t→∞ x>t x>t x→∞ lim inf(f(x).g(x)) = ( lim f(x))(lim inf g(x)). ∀t inf f(x) = −1 =⇒ lim inf f(x) = 1 =⇒ lim inf f(x) = −1. x→∞ x→∞ x→∞ x>t t→∞ x>t x→∞

Page: 160 job: unanal macro: svmonob.cls date/time: 21-Feb-2013/17:14 Example D.26. Let f(x) = x sin(x2). n p o∞ For xn = 2nπ + π/2 we have limn→∞ xn = ∞ and n=1 p p lim f(xn) = lim 2nπ + π/2 sin(2nπ + π/2) = lim 2nπ + π/2 = ∞ n→∞ n→∞ n→∞ So by Theorem D.24 we can conclude that

∞ = lim f(xn) = lim sup f(xn) 6 lim sup f(x) n→∞ n→∞ x→∞ and therefore lim supx→∞ f(x) = ∞. n p o∞ Similarly, for xn = 2nπ + 3π/2 we have limn→∞ xn = ∞ and n=1 r 3π 3π p lim f(xn) = lim 2nπ + sin 2nπ + = lim − 2nπ + 3π/2 = −∞ n→∞ n→∞ 2 2 n→∞

So by Exercise D.10 we can conclude that

−∞ = lim f(xn) = lim inf f(xn) lim inf f(x) n→∞ n→∞ > x→∞ and therefore lim infx→∞ f(x) = −∞

E Appendix: Math 140A Topics

∞ E.1 Summary of Key Facts about Real Numbers 16. If S ⊂ R and A := inf (S) , then there exists {an}n=1 ⊂ S such that an+1 ≤ an for all n and limn→∞ an = inf (S) . 1. The real numbers, R, is the unique (up to order preserving field isomorphism) ordered field with the least upper bound property or equivalently which is Cauchy complete. E.2 Test 2 Review Topics: 2. Informally the real numbers are the rational numbers with the (irrational) hole filled in. 1. Understand the basic properties of complex numbers. 3. Monotone bounded sequence always converge in R. 2. Coutability. Key facts are that countable union of countable sets is count- 4. A sequence converges in R iff it is Cauchy. able and the finite product of countable sets is countable. 5. Cauchy sequences are bounded. 3. Definitions of metric and normed spaces and their basic properties which 6. N is unbounded from above in R. in the end of the day typically follow from the triangle inequality. 1 7. For all ε > 0 in R there exists n ∈ N such that n < ε. 4. You should know that metrics and norms are continuous functions that 8. Q and R \ Q is dense in R. In particular, between any two real numbers satisfy, a < b, there are infinitely many rational and irrational numbers. 9. Decimal numbers map (almost 1-1) into the real numbers by taking the |kxk − kyk| ≤ kx − yk and limit of the truncated decimal number. |d (x, y) − d (x0, y0)| ≤ d (x, x0) + d (y, y0) . 10. If a, b, ε ∈ R, then

a) a ≤ b by showing that a ≤ b + ε for all ε > 0. 5. Be aware of different norms, k·ku , k·k1 , and k·k2 . b) a = b by proving a ≤ b and b ≤ a or 6. Understand the notion of; limits of sequences, Cauchy sequences, com- c) a = b by showing |b − a| ≤ ε for all ε > 0. pleteness, limits and continuity of functions. 11. A number of standard limit theorems hold, see Theorem 3.13. 7. Know what is meant by pointwise and uniform convergence. You should 12. Unlike limits, lim sup and lim inf always exist. Moreover we have; be able to compute pointwise limits and know how to test if the limit is lim infn→∞ an ≤ lim supn→∞ an with equality iff limn→∞ an exists in uniform or not. A key theorem is the uniform limit of continuous functions which case is still continuous. lim inf an = lim an = lim sup an. n→∞ n→∞ n→∞ We may allow the values of ±∞ in these statements. ∞ E.3 After Test 2 Review Topics: 13. If bk := {ank }k=1 is a convergent subsequence of {an} , then Let (X, k·k) be a Banach space. lim inf an ≤ lim bk ≤ lim sup an n→∞ k→∞ n→∞ P∞ PN 1. Know n=1 xn = limN→∞ n=1 xn if the limit exists. and we may choose {b } so that lim b = lim sup a or P∞ P∞ k k→∞ k n→∞ n 2. Know n=1 xn converges absolutely iff n=1 kxnk < ∞ and absolute con- limk→∞ bk = lim infn→∞ an. vergence implies convergence in a Banach space. 14. Bounded sequences of real numbers always have convergence subsequences. 3. Telescoping series and geometric series. ∞ 15. If S ⊂ R and A := sup (S) , then there exists {an}n=1 ⊂ S such that 4. Alternating series test. an ≤ an+1 for all n and limn→∞ an = sup (S) . 5. Absolute convergence tests: 1) integral test, 2) root test, 3) ratio test often combined with the 4) comparison test. 6. p – series examples. 7. nth – term test for divergence. 8. Cauchy criteria for convergence and the fact that tails of convergent series tend to zero. i.e. tails of convergent series tend to zero. 9. Pointwise convergence of sums. 10. Uniform convergence of sums and the Weierstrass M – test 11. Power series including radius of convergence notion. 12. The exponential function and its relatives, sin, sinh, cos, cosh . References

1. Edmund Landau, Foundations of Analysis. The Arithmetic of Whole, Rational, Irrational and Complex Numbers, Chelsea Publishing Company, New York, N.Y., 1951, Translated by F. Steinhardt. MR 0038404 (12,397m) 2. Walter Rudin, Principles of mathematical analysis, third ed., McGraw-Hill Book Co., New York, 1976, International Series in Pure and Applied Mathematics. MR 0385023 (52 #5893)

Index

Cauchy, 5, 18 Function Cauchy sequence continuous, 52, 95, 96 in a metric space, 49 continuous at a point, 52, 95 Complete Homeomorphism, 96 Metric space, 49 Continuous function, 96 Summable, 83