Math 104: Introduction to Analysis

Contents

1 Lecture 1 3 1.1 The natural numbers ...... 3 1.2 Equivalence relations ...... 4 1.3 The integers ...... 5 1.4 The rational numbers ...... 6

2 Lecture 2 6 2.1 The real numbers by axioms ...... 6 2.2 The real numbers by Dedekind cuts ...... 7 2.3 Properties of R ...... 8

3 Lecture 3 9 3.1 Metric spaces ...... 9 3.2 Topological definitions ...... 10 3.3 Some topological fundamentals ...... 10

4 Lecture 4 11 4.1 Sequences and convergence ...... 11 4.2 Sequences in R ...... 12 4.3 Extended real numbers ...... 12

5 Lecture 5 13 5.1 Compactness ...... 13

6 Lecture 6 14 6.1 Compactness in Rk ...... 14

7 Lecture 7 15 7.1 Subsequences ...... 15 7.2 Cauchy sequences, complete metric spaces ...... 16 7.3 Aside: Construction of the real numbers by completion ...... 17

8 Lecture 8 18 8.1 Taking powers in the real numbers ...... 18 8.2 “Toolbox” sequences ...... 18

9 Lecture 9 19 9.1 Series ...... 19 9.2 Adding, regrouping series ...... 20 9.3 “Toolbox” series ...... 21

1 10 Lecture 10 22 10.1 Root and ratio tests ...... 22 10.2 Summation by parts, alternating series ...... 23 10.3 Absolute convergence, multiplying and rearranging series ...... 24

11 Lecture 11 26 11.1 Limits of functions ...... 26 11.2 Continuity ...... 26

12 Lecture 12 27 12.1 Properties of continuity ...... 27

13 Lecture 13 28 13.1 Uniform continuity ...... 28 13.2 The derivative ...... 29

14 Lecture 14 31 14.1 Mean value theorem ...... 31

15 Lecture 15 32 15.1 L’Hospital’s Rule ...... 32 15.2 Power series ...... 33 15.3 Taylor series ...... 34

16 Lecture 16 35 16.1 The Riemann-Stieltjes integral ...... 35 16.2 Some Riemann-integrable functions ...... 37

17 Lecture 17 38 17.1 Properties of the integral ...... 38

18 Lecture 18 41 18.1 The fundamental theorem of calculus ...... 41

19 Lecture 19 42 19.1 Things that aren’t true ...... 42 19.2 Uniform convergence ...... 44

20 Lecture 20 46 20.1 Basic criteria for uniform convergence ...... 46 20.2 Uniform convergence and continuity ...... 46

21 Lecture 21 48 21.1 Uniform convergence and differentiation ...... 48 21.2 An everywhere continuous but nowhere differentiable function ...... 48 21.3 Differentiation and integration of power series ...... 49

22 Lecture 22 50 22.1 The Stone-Weierstrass theorem ...... 50

2 Though it may seem (or not!) as though I put care into these notes, they are actually very sloppily written. So I guarantee you that there will be errors, typos, flat out lies, and other varieties of mistakes. You may alert me if you find one, but the best thing to do is probably to read it with a skeptical and also not-so-literal eye. You should be especially skeptical about anything I say that seems like it has to do with mathematical logic, since I literally know nothing about that and am just making stuff up. Much of the content of these notes is taken from Walter Rudin’s Principles of Mathematical Analysis and to a lesser extent Kenneth Ross’ Elementary Analysis, though of course the errors are all mine.

1 Lecture 1

Remark 1.1 (Proof by contradiction). This is something you should become acquainted with, if you have not already.

1.1 The natural numbers We will take the following as our axiomatic definition of the natural numbers. There are constructions of the natural numbers from more basic principles but this is our starting point. As Bertrand Russell (maybe) said: it’s turtles all the way down.

Definition 1.2 (The Peano axioms for the natural numbers). The set of natural numbers, denoted N, is defined axiomatically by:

1. There is a distinguished element which we denote 1 P N. It is the “first element” of the natural numbers.

2. There is a function S : N Ñ N called the successor function. This should be thought of as the function Spxq “ x ` 1. In other words, every element has a (unique) well-defined successor.

3. The first element 1 is not a successor of any element, i.e. 1 is not in the image of S.

4. S is injective, i.e. an element can succeed at most one element.

5. Let P be a property. If 1 has P , and also if x has P implies Spxq has P , then every x P N has P . This is the induction principle.

Remark 1.3. The inductive axiom is important. Otherwise, we may be allowed things like two copies of the natural numbers: N Y N. In some ways it specifies that there is a unique first element. It does more than that though: it tells us that “arguments by induction are allowed.” If your next question is: what other kinds of arguments are allowed or not allowed, my answer is I don’t know, and that you should consult a logician. I do not worry much about issues like this.

This is a good starting point, but we are not only interested in N as a set but also in the addition and multiplication operations on it, which give it some additional structure. We should define these in terms of the axioms above.

Definition 1.4 (Addition, multiplication). One can define addition on N recursively as follows. a ` 1 :“ Spaq a ` Spbq :“ Spa ` bq

One can also define multiplication on N rescursively as follows. a ¨ 1 :“ a a ¨ Spbq :“ a ` ab

One can then prove that addition and multiplication as defined above are commutative, associative, and distributive, i.e. that the following are true

3 a ` b “ b ` a pa ` bq ` c “ a ` pb ` cq ab “ ba pabqc “ apbcq apb ` cq “ ab ` ac

Since I am taking the natural numbers as we know them “intuitively” as a starting point, I won’t actually prove these things. If you are interested you can try it.

Remark 1.5. Sometimes people define the natural numbers to include zero. I am making an arbitrary decision here not to invite zero to the party. (It wasn’t a very positive number to be around... ha ha ha...)

1.2 Equivalence relations I will introduce a notion which is very useful in making constructions.

Definition 1.6. Given a set S, an equivalence relation on the set S is a subset E Ă S ˆ S. If pa, bq P E, we often write a „ b. This subset E satisfies the following properties: (reflexivity) for all s P S, we have ps, sq P E (i.e. s „ s), (symmetry) for all ps, tq P E, we have pt, sq P E (i.e. if s „ t then t „ s), (transitivity) for all ps, tq P E and pt, uq P E, we have ps, uq P E (i.e. if s „ t and t „ u then s „ u).

This notion is supposed to generalize equality, as equality is very rigid and literal, but there are situations in which we might want to consider two non-equal things to be “the same.”

Example 1.7. Examples of equivalence relations: (1) Equality is an equivalence relation. e(2) Similarity and congruence in Euclidean geometry. (3) Similarity of matrices in linear algebra. (4) Congruence modulo n.

Definition 1.8 (Equivalence classes). We will introduce the following notation first. For s P S, denote rss to be the subset of S containing all elements equivalent to s, i.e.

rss “ tt P S | s „ tu.

An equivalence class is a subset of the form rss for some s P S.

Remark 1.9 (More intrinsic characterization of equivalence classes). Another characterization of equivalence classes is the following: a subset T Ă S is an equivalence class such that (1) every t, t1 P T has t „ t1; and (2) if s R T then for all t P T we have t  s.

Remark 1.10 (Equivalence classes partition S). Let S be a set with an equivalence relation „. Then the equivalence classes partition S. Check this as an exercise.

Definition 1.11 (Quotients). Let „ be an equivalence relation on S. Then the quotient S{„ is the set of equivalence classes.

Example 1.12. Examples of quotients (1) Under equality, for any set S, S{ „“ S. (2) Left as exercise.

(3) Let Mn be the set of all n ˆ n matrices. Then Mn{„ is the set of equivalence classes with representatives all possible Jordan normal forms, up to reordering. (4) Z{ „“ tr0s, r1s,..., rn ´ 1su

4 1.3 The integers We will define the integers from the natural numbers as follows. The idea is to let a pair pa, bq represent the quantity a ´ b. However, sometimes different pairs will be “equal” under this assignment so we want to capture this using an equivalence relation.

Definition 1.13. The set of integers Z is defined as the quotient pN ˆ Nq{ „ under the equivalence relation pa, bq „ pc, dq if a ` d “ c ` b. The set of integers has an operation, called addition, which is defined by rpa, bqs ` rpc, dqs “ rpa ` c, b ` dqs. This operation is well-defined in the quotient (check as exercise). The set of integers also has a additive identity (i.e. zero), which is rp1, 1qs. The set of integers also has an operation called multiplication defined by pa, bq ¨ pc, dq “ pac ` bd, ad ` bcq. It also has a multiplicative identity given by rp2, 1qs. Finally, there is an operation called additive inversion, where ´pa, bq “ pb, aq.

Remark 1.14 (Motivating the equivalence relation). We want to think of the pair pa, bq of natural numbers as “representing” the integer a ´ b. However, doing this means that every integer, e.g. 0, has many representatives, e.g. p1, 1q, p2, 2q, p3, 3q,.... Thus we must make all these representatives “equal” using an equivalence relation. The equivalence relation that captures when two pairs pa, bq „ pc, dq is that a ´ b “ c ´ d. However, since subtraction is not well-defined in the natural numbers, we must rewrite this as a ` d “ c ` b.

Remark 1.15 (Checking operations are well defined). We defined the addition operation by

rpa, bqs ` rpc, dqs “ rpa ` c, b ` dqs.

What we’ve done is, for each equivalence class, we chose particular representatives and our operation returns back an element of NˆN. Then we take the equivalence class of that element. To show this is well-defined, one must show that if we chose different representatives, we what we get by applying the operation “rule” gives us an equivalent element.

The set of integers forms what is called a commutative ring.

Definition 1.16. A commutative ring is a set R with addition and multiplication operations, denoted ` and ¨; additive and multiplicative identities denoted 0 and 1; and additive inverses, denoted ´. The following properties are satisfied:

1. (associativity of addition) pa ` bq ` c “ a ` pb ` cq

2. (commutativity of addition) a ` b “ b ` a

3. (additive identity) 0 ` a “ a

4. (additive inverses) a ` p´aq “ 0

5. (associativity of multiplication) pabqc “ apbcq

6. (commutativity of multiplication) ab “ ba

7. (multiplicative identity) 1 ¨ a “ a

8. (distributivity) apb ` cq “ ab ` ac

Remark 1.17. This is not a very important definition for this course; it appears frequently in algebra. It will not appear beyond this lecture. You can think of this definition to just be a word which is shorthand for saying: a set which has addition and multiplication and their respective identity elements which satisfies commutativity, associativity and distributativity properties. Or, a set that has all the operations that the integers do.

Proposition 1.18. The integers form a ring.

5 Proof. We will prove one of the properties above: additive inverses. The rest are left as exercises. What we want to show is that:

pa`, a´q ` pa´, a`q „ p1, 1q Using the above formula for addition, we have:

pa` ` a´, a` ` a´q „ p1, 1q

This is true, since using our definition of „, we check that indeed:

a` ` a´ ` 1 “ a` ` a´ “ 1.

Example 1.19 (Examples of other commutative rings). The (multivariate) polynomials with integer or rational or real or complex coefficients, under the usual addition and multiplication of polynomials, form a ring, denoted krxs for one variable, krx, ys for two variables x and y, et cetera, where k “ Z, Q, R, C.

1.4 The rational numbers The construction of the rational numbers is left as a homework exercise. The rational numbers form a field, which is to say all numbers except 0 have a multiplcative inverse.

Definition 1.20. A field k is a commutative ring with multiplicative inverses for all nonzero elements, denoted a´1, i.e. such that aa´1 “ 1.

We will make the following definition as well. We could have made similar definitions before, but I am lazy.

Definition 1.21. A (total) order on a set S is a subset of L Ă S ˆS satisfying the following properties. If pa, bq P L then we will write for shorthand a ď b. (antisymmetry) if a ď b and b ď a then a “ b (transitivity) if a ď b adb b ď c then a ď c (totality) at least one of the following must be true: a ď b or b ď a.

An ordered field k is a field with a total order such that: (a) if a ď b then a ` c ď b ` c for any c P k (b) if a ě 0 and b ě 0 then ab ě 0

Remark 1.22. There are other notions of orders: for example, there are orders on sets where two elements might be incomparable. You can check Wikipedia if you’re interested. We are interested in total orders, but since there are no other orders we will consider, I might be lazy about writing “total.”

Example 1.23. The rationals and reals, Q, R, are both ordered fields. The complex numbers, C is a field but it is not an ordered field.

2 Lecture 2

2.1 The real numbers by axioms There are three approaches we will take to the real numbers in this class. The first is axiomatic.

Definition 2.1 (Axiomatic definition of the real numbers). The real numbers are an ordered field with the least upper bound (LUB) property. The LUB property is as follows: every subset which has an upper bound has a supremum (least upper bound). (We will define these terms now).

6 Definition 2.2. Let X be an ordered set, and S Ă X. The maximum of S is an element m P S such that for all s P S, s ď m (exercise: prove maximums must be unique). An upper bound on S is an element x P X such that for all s P S, s ď x.

Remark 2.3. Notice that maximums must be in the set S, whereas upper bounds do not have to be. Further, notice that upper bounds are usually not unique.

Example 2.4. Let X “ R and S “ r0, 1s. Then, 1 is a maximum of S and any number greater than 1 is an upper bound.

Remark 2.5. All finite sets have maximums. However, not all sets in general have a maximum. For a trivial example, consider p0, 8q Ă R. This fails to have a maximum, and fails to have an upper bound. I can live with this. What’s kind of bad is that not all bounded sets have maximums. For example, the interval p´1, 0q is bounded 1 above by 0 but does not have a maximum. Let m P p´1, 0q be a possible maximum. But, 2 m is also in the set and is even bigger (since m is negative). So there can’t be a maximum. In this way asking for maximums is somehow the wrong thing to do for sets which might be infinite. This will motivate the next definition.

Definition 2.6. A least upper bound on a subset S of an ordered set X is an element x P X which is an upper bound on S, and such that there is no smaller upper bound. More mathematically, it is an upper bound such that if x1 is another upper bound, the necessarily, x ď x1. The least upper bound is also called the supremum and is denoted suppSq. Likewise, the greatest lower bound is called the infimum and is denoted infpSq.

Remark 2.7. In our previous example, the supremum is 0. However, it is still not true that every subset of a well- ? ordered set has a supremum. For example, consider p0, 2q as a subset of Q (not R!) does not have a supremum. Showing this is actually somewhat involved and involves some elementary number theoretic arguments. I will just appeal to intuition here. In a way, though, this problem is less of a defect of our definition of least upper bound ? than it is a defect of the rational numbers: the subset p0, 2q of the rational numbers doesn’t have anything that even resembles a “generalized maximum” if we stay in Q. This motivates the next definition. Definition 2.8. A well ordered set has the least upper bound property if for every bounded above subset S, S has a supremum (least upper bound).

Remark 2.9. The least upper bound property is equivalent to the greatest lower bound property for an ordered field, since for any set S we can always take its negation ´S, and the supremum of S is the infimum of ´S and vice versa.

Remark 2.10. Thus the real numbers are defined as an ordered field which has the LUB property. Vaguely, this means “all its holes are filled in.” However, it’s still not clear such an object exists, since we haven’t constructed it. In the next section we will do that. In a somewhat later section, we will show that there is a general construction one can make to “fill in the holes” on certain kinds of spaces called metric spaces, and that the real numbers are what we get when we do this to Q.

2.2 The real numbers by Dedekind cuts In this section we will give a construction of the real numbers by Dedekind cuts. The way to “think about” Dedekind cuts is that to a real number α, we associate the two subsets p´8, αq and pα, 8q, as subsets of Q (i.e. intersect with Q). Though this is how we “think” of these cuts, in making arguments we cannot actually refer to things like p´8, αq for α irrational, since this would make our construction circular!

Definition 2.11. A Dedekind cut of the rational numbers Q is a partition of Q into subsets A and B that satisfies the following properties: (1) (definition of partition) A Y B “ Q and A X B “H (2) (A is closed downward) if a P A, then p´8, aq Ă A (3) (B is closed upward) if b P B¡ then pb.8q Ă B (4) (infinity is not a number) neither A nor B are empty nor all of Q (5) (a number is cut out) A does not have a maximum, and B does not have a minimum

7 We can give an ordering on the Dedekind cuts as follows. We say pA, Bq ď pA1,B1q if A Ă A1 and B1 Ă B. There is also a way to make Dedekind cuts into a field, but this is left as a homework exercise.

Remark 2.12. Note that the the data of B is extraneous. Given A that satisfies the above properties, it is automatic that B “ Q ´ A does. If this B has a minimum, then remove it from B. (Exercise: show that this new B cannot have a minimum. Hint: show that between any two rational numbers, there is a rational number.) I made the above definition to illustrate why this is a “cut,” but from now on we will only be using the data of A.

Remark 2.13. Property (5) is made because otherwise, the cuts p´8, qq and p´8, qs for q P Q would look like they should be the same. Try to prove the field axioms without this property and see why it’s necessary. For example, we can’t have two candidates for zero in a field!

Remark 2.14. The Dedekind cuts need to be made into an ordered field. Unfortunately this can get kind of hairy for multiplication, and I won’t do it here. Interested readers should refer to the constructions in Rudin.

Proposition 2.15. The ordered set of Dedekind cuts has the LUB property. In particular, let tAiuiPI be a collection of Dedekind cuts which is bounded above. Then, its supremum is iPI Ai.

Proof. We need to check that (1) the claimed supremum is alsoŤ a Dedekind cut (in particular, it is not all of Q) and that (2) it really is the supremum.

For (1), it is easy to see that i Ai is still closed downwards. The fact that the Dedekind cuts are bounded above means there is some cut A such that A Ă A for all i P I. This means that A Ă A so A ‰ . Finally, Ť i i i Q suppose that A had a maximum, say m. Then m P A for some fixed p P I, but since it is a maximum of the i i p Ť union it would have to be a maximum for A as well, contradicting that A is a Dedekind cut. Ť p p

For (2), Suppose there is some Y such that Ai Ă Y for i P I (Ai ă Y for all i, so Y is an upper bound) but

Y Ă iPI Ai (Y is lower than the claimed supremum). Let x P iPI Ai; then x P Ap for some p P I. But this contradicts the first statement, so this cannot be. One can sum this argument up in a sentence like, “the union of Ť Ť a collection of subsets is the minimal subset containing all the subsets in the collection.”

Remark 2.16 (Why we like axioms). This construction is nice in reassuring us that the real numbers exist (yay!). I think the logic-y thing to say would be that Dedekind cuts are a model for the real numbers. The nice thing about axioms is that we can build theories without thinking too hard about specific constructions, as long as our axioms are well-designed. The specific model doesn’t really matter as long as we stay within the bounds of the axioms. One might complain that if we want to call something the real numbers, there should be a question about uniqueness, which I know nothing about. So this question is also going under the rug. In practice this is not important.

2.3 Properties of R Proposition 2.17 (Archimedian property). Let a, b P R. If a ą 0 and b ą 0, then there is a positive integer n such that na ą b.

Proof. Suppose otherwise, i.e. that there is some a, b P R such that na ď b for all n P N. In particular, this means that b is an upper bound for S :“ tna | n P Zu. Thus, sup S is a number. In a homework exercise, you will show that suppSq ` suppT q “ suppS ` T q. Let S be as above and T “ tau. Then notice that S ` T “ S, and suppT q “ a, so we should have suppSq ` a “ suppSq which contradicts that a ą 0. Thus, the set must be unbounded and there is some n such that na ą b.

Proposition 2.18 (Q is dense in R). For any a, b P R, there is a x P Q such that x P pa, bq. p Proof. Write x “ q , for p, q P Z. Then we want to find p, q such that aq ă p ă bq. We can use the Archimedian property to see that there is an integer q such that qpb ´ aq “ bq ´ aq ą 1. We claim that if β ´ α ą 1, then there is an integer in pα, βq. This would complete the proof, letting β “ bq and α “ aq, and p the integer.

8 Lemma 2.19. Let α, β P R. If β ´ α ą 1, then there is an integer p P pα, βq. Proof. If α ă 0 and β ą 0, we can prove this result for the smaller interval p0, βq. If α, β ă 0, we can prove this result for the interval p´β, ´αq. So, without loss of generality, we can assume that α, β ě 0: The strategy is as follows. We want to take the set

tn P Z | α ă nu and find its minimum value, say p. By construction we have α ă p and since it is the minimum, we kno wthat p ´ 1 ă α and so p ă α ` 1 ă β as well, completing the proposition. The issue is that this set may not have a minimum and it may also be empty. First, let’s address the existence of a minimum. We can force a set to have a minimum by making it finite, i.e. by adding some conditions bounding the set:

tn P Z | α ă n and n P r0,Nsu for some integer N. This is true so long as α ă N, and we can use the Archimedian property to find an integer N such that N ¨ 1 ą maxp|α|q.

3 Lecture 3

3.1 Metric spaces Definition 3.1. Let k be a totally ordered field. A k-metric space is a set X along with a distance function d : X ˆ X Ñ k satisfying the following properties: (non-negativity) dpx, yq ě 0 for all x, y P X, (identity of indiscernables) dpx, yq “ 0 if and only if x “ y, (symmetry) dpx, yq “ dpy, xq for all x, y P X, (triangle inequality) dpx, yq ` dpy, zq ě dpx, zq for all x, y, z P X.

Remark 3.2. Most texts fix k “ R. In general, we will assume k “ R, but later when we give a consruction of R we will require k “ Q to avoid a circular argument. Remark 3.3. I should be more careful about this definition than I am. I am not sure whether any completeness properties of R are used in results in the literature on metric spaces. I am also not sure how much in the literature applies to loosening the restriction on k, i.e. if we consider ordered abelian groups or ordered rings instead. I hope that these notes at least are internally consistent.

n Example 3.4 (The Euclidean metric). R is a metric space under the Euclidean metric. Let x “ px1, . . . , xnq and y “ py1, . . . , ynq, and define: 2 2 dpx, yq “ px1 ´ y1q ` ¨ ¨ ¨ ` pxn ´ ynq One can think of this is a “generalized Pythagoreana theorem” way of finding distance. For n “ 1 it is not hard to see this is a metric. For larger n, the proofs I know use Cauchy-Schwartz. Note that this metric doesn’t quite make sense when k “ Q for Qn when n ‰ 1, since square roots might not exist in Q. Example 3.5 (The discrete metric). Let X be any set. Then the discrete metric is defined by:

1 if x ‰ y dpx, yq “ #0 if x “ y

One can easily verify this is a metric.

9 3.2 Topological definitions Definition 3.6. Let pX, dq be a metric space. The open ball of radius  at x is defined by:

Bpxq :“ tp P X | dpx, pq ă u and the closed ball is defined by:

Bpxq :“ tp P X | dpx, pq ď u. We will sometimes use the word neighborhood of x to refer to any open ball of any radius at x.

Example 3.7. Let X “ R with the Euclidean metric. Then Bpxq “ px ´ , x ` q.

Example 3.8. Let X have the discrete metric. Then Bpxq “ txu if  ă 1 and Bpxq “ X otherwise.

Definition 3.9. Let pX, dq be a metric space, and E Ă X. A point p P E is called an interior point or E if there is a neighborhood U of x such that U Ă E. A subset E is called open if every point is interior.

Example 3.10 (Open balls are open). Every open ball is open. To show that every p P Bpxq is an interior point, take a “  ´ dpp, xq ą 0 note that by the triangle inequality, Bappq Ă Bpxq.

Definition 3.11. Let pX, dq be a metric space and E Ă X. A point p P X is a limit point of E if every neighborhood U of p has a point q ‰ p such that q P E. A subset E which contains all its limit points is called closed. The closure of E, denote E, is the union of E with all of its limit points.

Example 3.12 (Closed balls are closed). Every closed ball is closed. Consider Bpxq and let p be a possible limit point. If dpp, xq ď r then we are happy. Otherwise, let a ă dpp, xq ´ r. Then Bappq and Bpxq are disjoint by the triangle inequality, and thus p cannot be a limit point.

Proposition 3.13 (The closure of any subset is closed). Let pX, dq be a metric space and E Ă X. The closure E is closed.

Proof. This might seem obvious, but there is something to prove. One has to show that one does not introduce more limit points by including the limit points of E. Suppose p is a limit point of E; then every neighborhood N of p intersects E. If N contains a point of E we are done. If N contains a limit point of E, say q, then since N is open, q is an interior point, so there is a neighborhood of q contained in N which contains a point of E, concluding the proof.

Remark 3.14. The closed ball isn’t really used that much as a topological notion, as far as I know. I include it here for fun.

Definition 3.15. Let pX, dq be a metric space. A subset E Ă X is dense if E “ X.

Example 3.16. As we’ve shown before, Q is dense in R. To see why, we claim that every real number is a limit point of rational numbers: let a P R; every ball Bpaq “ pa ´ , a ` q contains a rational number.

3.3 Some topological fundamentals Proposition 3.17 (Complements). A set is open if and only if its complement is closed.

Proof. This proof is mostly just juggling the logic in the definitions. Suppose that E is open. Let p be a limit point of Ec. Then every neighborhood of p intersects Ec nontrivially, i.e. every neighorhood is not contained in E, so p R E since it cannot be an interior point, so p P Ec, so Ec is closed. Suppose that E is closed. Let p P Ec, so p is not a limit point of E, so their is a neighborhood of p disjoint from E, so p is an interior point, so Ec is open.

Proposition 3.18 (Unions and intersections). Any (possibly infinite) union of open sets is open. Any (possible infinite) intersection of closed sets is closed. Any finite union or intersection of open or closed sets is open or closed, respectively.

10 Proof. By the previous proposition, we only need to prove the statements for open sets. I will leave the statement that an infinite union of open sets is open as an exercise. For a finite intersection of open sets, take open sets n G1,...,Gn, and x P i“1 Gi. Since x P Gi for all i is interior, we have Bi pxq Ă Gi for all i and for some i. Then, B pxq Ă n G and the result follows. minpiq i“1 Şi PropositionŞ 3.19 (Closure). The closure E is the intersection of all closed sets that contain E. Equivalently, it is the smallest closed set containing E.

Proof. Let F “ EĂV closed V be the intersection of all closed sets containing E. Clearly, F Ă E since E is closed and contains E. We want to show that every closed set containing E also contains E. Let V be closed and E Ă V . Ş Taking the closure of both sides, we find that E Ă V .

4 Lecture 4

4.1 Sequences and convergence Remark 4.1. Warning: the sequence notion of limit is not quite the same as than the notion of limit you learned in calculus – the context is different.

Definition 4.2. Let pX, dq be a metric space. A sequence tpnu in X converges to p P X if for every  ą 0, there is some N such that dppn, pq ă  when n ą N. Equivalently, tpnuněN Ă Bpxq X tpnu. In this case, we write:

lim pn “ p nÑ8 If a sequence does not converge to any point, it diverges.

Proposition 4.3. Limits of sequences are unique.

Proof. Let p, q be two limits of a sequence tpnu. We will show that dpp, qq ă  for any  ą 0. Choose any  ą 0;   then there is an N such that dppn, pq ă 2 for n ą N and an M such that dpqn, pq ă 2 for n ą M. Then, for n ą maxpN,Mq, we have that dpp, qq ď dpp, pnq ` dppn, qq ă  as desired.

Proposition 4.4 (Connections to topology). If E Ă X and p is a limit point of E, then there is a sequence in E converging to p.

Proof. For every n, choose some sn P B1{nppq X E, which we know is nonempty since p is a limit point of E. We 1 claim this sequence converges to p. For  ą 0, choose N such that N ă . Then, one sees that for n ą N, we have 1 dpsn, pq ă N ă . Definition 4.5. Let pX, dq be a metric space. A subset E Ă X is bounded if there is a number L such that tdpp, qq | p, q P Eu is bounded above (in the order-theoretic sense) by L.

Proposition 4.6 (Convergent sequences are bounded). Convergent sequences are bounded (considered as sets).

Proof. Let tpnu Ñ p be the convergent sequence. We will break up the sequence into two “groups” for which we can find a bound on dppn, pmq: one for m, n ď N and one for m, n ě N. Fix any  (for example,  “ 1) and take an N such that n ě N implies dppn, pq ă . By the triangle inequality, dppn, pmq ă 2 for n, m ě N. This is the first “group.” For the second, we know that for n, m ď N, dppn, pmq ě m :“ maxptdppn, pmu | n, m ď Nu. Then, using the triangle inequality, we know that

dppn, pmq ď dppn, pN q ` dppN , pmq ď 2 maxpm, 2q for any m, n, establishing a bound.

11 4.2 Sequences in R

Proposition 4.7. Let tsnu and ttnu be sequences in R, with limits s and t respectively. Let c P R be a constant. Then:

(a) limnÑ8pcsnq “ cs and limnÑ8pc ` snq “ c ` s. (b) limnÑ8psn ` tnq “ s ` t (c) limnÑ8psntnq “ st (d) limnÑ8psn{tnq “ s{t if for all n, sn ‰ 0 and s ‰ 0 Proof. The proof of (a) is easy and left as an exercise. For (b), fix an  ą 0. We want to find an N such that if n ą N then |ps ` tq ´ psn ` tnq| ă . By the triangle inequality, we have that

|ps ` tq ´ psn ` tnq| ď |s ´ sn| ` |t ´ tn|.

1 By the limits we already know, we can find an Ns such that n ą Ns implies |s ´ sn| ă 2  and likewise an Nt. Then, let N “ maxpNs,Ntq and we have the result. For (c), rewrite the expression:

st ´ sntn “ pt ´ tnqps ` snq ` tns ´ snt “ pt ´ tnqps ` snq ` ptn ´ tqs ´ psn ´ sqt

The strategy is as follows: in each of the three additive terms, we have one factor that should “go to zero.” We also have bounds on the other factors, which guarantees that their products “go to zero.” More precisely let L be a bound on the sequence ts ` snu. Fix  ą 0; we want to find an N such that if n ą N then |st ´ sntn| ă . Applying the triangle inequality to our rewriting, we find that

|st ´ sntn| ď |L||t ´ tn| ` |s||t ´ tn| ` |t||s ´ sn|

  There is an N such that for n ą N, we have |s ´ sn| ă |L|`|s|`|t| and |t ´ tn| ă | L| ` |s| ` |t|. Then, one finds that |st ´ sntn| ă  as desired. For (d), we only need to show that 1 1 lim “ . nÑ8 sn s Fix  ą 0. We want to find N such that for n ą N, we have:

1 1 s ´ s | ´ | “ | n | ă  sn s sns

2 The idea is as follows: the denominator sns tends toward s which is a constant we can account for. We want to 2 “overestimate” this quantity with respect to s since it is in the denominator. So, we can find N1 such that n ą N1 1 2 implies that 2|sn| ą |s| (check this), and so |sns| ą 2 s . Thus our expression becomes:

sn ´ s 1 | | ă 2 2 |sn ´ s| sns s

s2 Now, we can find N2 such that |sn ´ s| ă 2 and the result follows.

4.3 Extended real numbers Definition 4.8. We write

lim sn “ 8 nÑ8 to mean that for each M there is some N such that n ą N implies that sn ą M. Likewise, we write

lim sn “ ´8 nÑ8 to mean that for each M there is some N such that n ą N implies that sn ă N.

12 Remark 4.9. The “extended reals” are not a metric space, nor a field, though it does have an total order. However, the following arithmetical sentences still “make sense” for any x P R: x x x ` 8 “ 8, x ´ 8 “ ´8, 8 “ ´8 “ 0 if x ą 0 then x ¨ 8 “ 8 and x ¨ p´8q “ ´8 if x ă 0 then x ¨ p´8q “ 8 and x ¨ 8 “ ´8 The statements of the above proposition are still true in these cases.

5 Lecture 5

5.1 Compactness

Remark 5.1 (Induced metric). Let pX, dq be a metric space, and let Y Ă X. Then, pY, d|Y q is a metric space, where d|Y is the distance function when restricted to Y . Proposition 5.2. Let pX, dq be a metric space, and E Ă X. Let U be open in X. Then U X E is open in E. Proof. This is left as an exercise. Just check the definition, and note that open balls in Y are open balls in X intersected with Y .

Remark 5.3 (Openness and closedness as properties of subsets). Asking whether a set is opened or closed only makes sense once we identify it as an subset of a metric space, and the answer depends on its embedding as a subset. For example, r0, 1s is an open subset of the metric space r0, 1s, but it is not an open subset of the metric space R. The notion we will now introduce, compactness, is intrinsic: it does not depend on any embedding. Definition 5.4. Let pX, dq be a metric space. Then, X is compact if every open cover has a finite subcover. An open cover of a space X is a (possibly infinite) collection of open subsets of X, say tUiu such that they cover X, i.e. Ui “ X.A finite subcover is a finite subcollection which is still a cover. ExampleŤ 5.5 (Open covers of R). Here are many many examples of open covers of R. Here are some. (1) Un “ p´n, nq for n P Z, (2) Un “ p´8, nq for n P N, (3) Un “ p´8, 8q for n P Z, (4) Un “ pa ` 1, a ´ 1q for a P R, (5) Un “ pa ` , a ´ q for a,  P R. I claim that none of these have finite subcovers. I’ll prove this for (1) and leave the rest as an exercise. Proof. Suppose there was a finite subcover; then it would be given by a finite subset of the index set of integers, call it I Ă Z. Check that iPI p´n, nq “ p´| maxpIq|, | maxpIq|q, which is not R. 1 Example 5.6 (Not openŤ covers). Let X “ p0, 1s Ă R. Then p0, n q is not an open cover, since 1 P X is not in any of those open sets.

Corollary 5.7. R is not compact. Proof. To be compact, every open cover must have a finite subcover. Since we’ve given an open cover which does not have a finite subcover, we are done.

Remark 5.8 (R has an open cover with a finite subcover). Note that example (3) is an open cover which has a finite subcover. This does not contradict that R is not compact, since only one open cover can break compactness. Example 5.9 (Compact intervals in R). Some more examples which we can do explicitly. (1) p´1, 1q is not compact. (2) r0, 1s is compact. (3) p0, 8q is not compact. (4) r0, 8q is not compact. (5) Z is not compact. (6) Q X r0, 1s is not compact.

13 1 1 Proof. (1) Use the open subcover given by Un “ p´1 ` n , 1 ´ n q for n P N. (2) We will prove a general theorem later. (3) Use the open subcover p0, nq for n P N. (4) Use the open subcover r0, nq for n P N. Note that while these are not open in R, they are open in r0, 8q. (5) Use the open subcover tnu for n P Z. 1 (6) This is a little trickier. Choose an irrational number α P r0, 1s. Let sn be a rational number in pα ´ n , αq. Then, take the open subcover Q X pr0, snq Y pα, 1sq for n P N. Proposition 5.10. Let pX, dq be a metric space and E Ă X. If E is compact then E is a closed and bounded subset of X.

Proof. We can do the following for any metric space. For every x P E (and x ‰ p), take a neighborhood Ux which is disjoint from some neighborhood of p. For example, one could take Ux “ B dpx,pq pxq X E. If p ‰ E, this is an 2 open cover, since it contains every point in E. By compactness, take a finite subcover, say indexed by the finite n points x1, . . . , xn P E. Then, notice that i“1 Uxn is disjoint from B dpxi,pq ppq, so p is not a limit point. Thus minp 2 q E contains all of its limit points. Ť For boundedness, let x P E be any point, and let consider the open cover Un “ Bnpxq X E. By compactness this has an open subcover, and let N be the maximum n in this subcover. Then E Ă BN pxq and is bounded.

Remark 5.11. It is a little remarkable that an intrinsic property of a space tells us a property about every embedding of that space.

Proposition 5.12. Let X be a compact metric space. A closed subset of X is compact.

Proof. Let E Ă X be closed. Take a finite subcover Ui of E. First, we claim (and will prove later) that an open set of E can be written as the intersection of E with an open set of X. Given this claim, for each Ui take Vi Ă X open such that Ui “ Vi X E. Now take the open cover tViu Y tX ´ Eu of X. Since X is compact, this has a finite subcover – if X ´ E is in the finite subcover, remove it to obtain a finite subcover of E.

Lemma 5.13. Let pX, dq be a metric space and let E Ă X be a subset considered as a metric space. Let U be an open set of E. Then there is an open set V of X such that U “ E X V .

Proof. Recall that every open set can be written as the union of open balls. Take the same open balls but in X instead.

Proposition 5.14. Any finite union of compact subsets is compact.

Proof. Left as an exercise.

6 Lecture 6

6.1 Compactness in Rk We will make the following definition, which will only be used in this section.

Definition 6.1. A k-cell is a subset of Rk of the form

ra1, b1s ˆ ra2, b2s ˆ ¨ ¨ ¨ ˆ rak, bks i.e. a “box” with given vertices.

k Lemma 6.2 (Cantor intersection lemma for R ). Let E0 Ą E1 Ą E2 Ą ¨ ¨ ¨ be a descending chain of nested k nonempty k-cells of R . Then, i Ei is nonempty. Ş

14 1 Proof. First, we can prove this for R . Consider a sequence where Ei “ rai, bis. To be nested means a1 ď a2 ď ¨ ¨ ¨ and b1 ě b2 ě ¨ ¨ ¨ . Let a “ suppa1, a2,...q and b “ infpb1, b2, ¨ ¨ ¨ q. We claim that a ď b. Then any x P ra, bs would suffice. To prove the claim, suppose that a ą b; let x P pb, aq. Then there is some n for with x ă an and also some k m for with x ą bm. Then, bmaxpm,nq ă x ă amaxpm,nq, which cannot happen. To prove this for R , one makes a similar argument for each component of the k-tuple, and is left as an exercise.

Theorem 6.3 (Heine-Borel theorem). A subset of Rn is compact if and only if it is closed and bounded.

Proof. We’ve already proven one direction. Now let X Ă Rn be closed and bounded. Since X is bounded, it is n contained in a closed box, T0 :“ r´N,Ns for large N. Since a closed subset of a compact set is compact, it suffices to show that this box is compact. n Fix an open cover of the box T0. If one cuts this box in half in all possible directions, one obtains 2 smaller n boxes with half the side length of the original. If every one of the 2 half-boxes had a finite subcover, then T0 would have a finite subcover (by taking the finite union of the finite subcovers). So, suppose otherwise, and choose a box

T1 which does not have a finite subcover. We can repeat this process to obtain a descending chain of nested boxes:

T0 Ą T1 Ą T2 Ą ¨ ¨ ¨ .

By the previous lemma, this is nonempty and contains a point p. This point p must be in one of the open sets of the cover, say U, and thus there is some ball Bppq Ă U. But, this ball contains all Ti for i ą M for some large M, contradicting that each of the Ti does not have a finite subcover. This proves the theorem.

Example 6.4 (Not true for all metric spaces). The Heine-Borel theorem is not true for all metric spaces. For example, if X is an infinite metric space with the discrete topology, then a subset is compact if and only if it is finite. However, every subset is closed and bounded.

Remark 6.5. This theorem might convince you that compactness is a redundant notion. In the following sections we will see that this is very untrue, if we want to work in the general context of metric spaces or topological spaces.

7 Lecture 7

7.1 Subsequences I will not give a definition of subsequences, and convergence of a subsequence. It is exactly what you think it is. However, we will make a few definitions anyway.

Definition 7.1. A sequence s in R is monotonically increasing if si ď si`1 for all i. It is monotonically decreasing Proposition 7.2. All bounded monotone sequences converge.

Proof. Suppose sn is monotonically increasing. We claim sn converges to s :“ supptsnuq. For  ą 0, there is some sN P ps ´ , sq by definition of supremum. Since the sequence is mootonic, this is true for all sn and n ą N.

Proposition 7.3. Every sequence has a monotone subsequence.

Proof. Let sn be a sequence. We say the kth term, sk, is dominant if sk ą si for all i ą k, i.e. bigger than all its subsequent terms. There are two cases to consider. In the first case, there are infinitely many dominant terms; then, the subsequence consisting of dominant terms is a monotonically decreasing sequence. In the second case, there are only finitely many. Then, in choosing a subsequence, we follow the rule: never choose a dominant term. In doing so, we can always choose some term in the sequence following any choice which is bigger, obtaining a monotonically increasing sequence.

Theorem 7.4 (Bolzano-Weierstrass for R). Every bounded sequence has a convergent subsequence. Proposition 7.5 (Bolzano-Weierstrass, generalization). Let X be a compact metric space. Every sequence has a convergent subsequence.

15 Proof. Let txnu be the sequence, and let E “ tx1, x2,...u i.e. the set of points in the sequence. If E is finite then there must be a point that reoccurs infinitely many times, and that is our convergent subsequence. If E is infinite, then we claim it has a limit point, say p. Given this claim, for each ball B 1 ppq we can choose successive elements n in smaller balls converging to p. More precisely, there is some Nn such that si is in the ball B 1 ppq for i ą Nn. n Choose successively in such that in ą in´1 and in ą Nn to obtain the subsequence. Lemma 7.6. If E is an infinite subset of a compact metric space X, then E has a limit point in X.

Proof. Suppose to the contrary; if no point in X is a limit point, every point in x P X has a neighborhood which contains at most one point of E (namely, the point x itself if x P E). This gives an open cover of X, but it cannot have a finite subcover since each open contains at most one point of E.

Remark 7.7. Note that this proposition is not true for just closed and bounded subsets. For example, take X “ Z under the discrete topology, so that it is closed and bounded but not compact. The sequence sn “ n has no convergent subsequence. So this is another piece of evidence in favor of compactness as a “good notion.”

Definition 7.8. The limit supremum of a sequence sn is defined as

lim sup “ lim suppsn, sn`1, sn`2,...q nÑ8 nÑ8

The limit infimum is defined similarly

lim inf “ lim infpsn, sn`1, sn`2,...q nÑ8 nÑ8

Proposition 7.9 (Characterization of limit supremum). Let sn be a bounded sequence. Let E be the set of subse- quential limits. Then lim sup sn exists and is equal to suppEq, and lim inf sn exists and is equal to infpEq.

Proof. I will prove the statement for lim sup. Since sn is bounded above, suppEq exists. ehhhhhhhhh

Proposition 7.10. For any sequence sn we have that lim inf sn ď lim sup sn. If they are equal, then the limit exists and limnÑ8 sn “ lim inf sn “ lim sup sn. Conversely, if the limit exists, then limnÑ8 sn “ lim inf sn “ lim sup sn.

Proof. The first statement is left as an exercise. For the second, let s “ lim sup sn “ lim inf sn. Fix  ą 0. There is an N such that for n ą N, we have that suppsn, sn`1,...q P ps ´ , s ` q. Thus, sn ă s `  for all n ą N.A similar argument with the infimum shows that s ´  ă sn for all n ą N. Thus, |sn ´ s| ă  for n ą N. For the third statement, we have that for every  ą 0, there is some N such that for n ą N we have sn P ps ´ , s ` q. Thus, lim suppsnq ă s `  and lim inf ą s ´  for every , and the result follows.

Corollary 7.11 (Application to sequences: “squeezing”). Let sn, tn be sequences in R that converge to s, t, such that sn ď tn for n ą N for some N. Then, lim sn ď lim tn.

Proof. One can easily check that lim sup sn ď lim inf tn. The rest follows by the previous proposition.

7.2 Cauchy sequences, complete metric spaces

1 Example 7.12. Let X “ R ´ t0u be a metric space. The sequence t n u does not converge, because 0 R X. But it “looks like” a convergent sequence. We want to make a definition that captures this phenomenon.

Definition 7.13. A sequence is Cauchy if for every  ą 0 there is an integer N such that dppn, pmq ă  for m, n ą N. A metric space for which every Cauchy sequence is convergent is called complete.

Remark 7.14. As we saw above, not every Cauchy sequence converges, and in our example was because there was a “hole.”

Proposition 7.15. Cauchy sequences are bounded

Proof. Left as an exercise.

16 Proposition 7.16. Every convergent sequence is Cauchy.

 Proof. Suppose sn converges to s. Then, for every  ą 0 there is an N such that dpsn, sq ă 2 for n ą N. Then, for m, n ą N, dpsn, smq ď dpsn, sq ` dpsm, sq ă .

Proposition 7.17 (Cantor intersection lemma). Let E0 Ą E1 Ą E2 Ą ¨ ¨ ¨ be a descending chain of nonempty compact subsets of a metric space X. Then, i Ei is nonempty. Further, define the diameter of a set E by

diampŞEq :“ supptdpx, yq | x, y P Eu.

If limnÑ8 diampEiq “ 0, then Ei consists of exactly one point. Ş Proof. For the first statement, if the intersection is empty, then tE0 ´ EiuiPN is an open cover of E0 (open because Ei is compact in compact E0 and thus closed). It has a finite subcover, since E0 is compact. This means that eventually the sequence stabilizes, i.e. Ei “ Ei`1. This contradicts that the Ei are nonempty, so the intersection must be nonempty. For the second statement, this must be true since if the intersection contained two points, say x, y, then diampEiq ě dpx, yq for all i.

Proposition 7.18. Every compact space is complete.

Proof. Let si be a Cauchy sequence. Define Ei “ tsi, si`1,...u, in the notation of the previous proposition. Since the sequence is Cauchy, limnÑ8 diampEnq “ 0. Thus there is a single point in the intersection, say s. I claim this is the limit (the “hard part” is over, since the failure for a Cauchy sequence to converge is somehow due to the point it should converge to “not being there”). To see this, fix  ą 0; if diampEN q ă  then this implies that dpsn, sq ă  for all n ą N.

Corollary 7.19. Rk is complete. Proof. Consider a Cauchy sequence; it is bounded (check as exercise) so we may enclose it in a bounded and closed box. This is compact by Heine-Borel, and thus every bounded Cauchy sequence has a limit.

Definition 7.20. Let pX, dq be a metric space. We will define the completion of X, denoted X. As a set, the elements of X are Cauchy sequences (indexed by N) in X, modulo the an equivalence relation which we will define later. For now, define the distance function between two Cauchy sequences by:

dpx, yq “ lim dpxn, ynq n

Check as an exercise that this is well-defined since the sequences are Cauchy. Then, two sequences are equivalent x „ y if dpx, yq “ 0. This distance function is also well-defined on equivalence classes and satisfies the axioms of a distance function.

7.3 Aside: Construction of the real numbers by completion

In this section we will backtrack and pretend we don’t know what R is. We will treat Q as a Q-metric space. This will be the only time in these notes that we think of distances as taking values in any field other than R. We will use Cauchy sequences in Q to construct R.

Definition 7.21. The set of real numbers is defined to be the set of Cauchy sequences in Q (using the absolute value metric) modulo the equivalence relation where x „ y if limnÑ8 xn ´ yn “ 0. Addition and multiplication are defined on Cauchy sequences by px ` yqn “ xn ` yn and likewise. The additive identity is the constant zero sequence, and the multiplicative identity is the constant one sequence. The order is defined by x ď y if x „ y of if there is some N such that xn ă yn for all n ą N. Finally, the rational numbers are considered as a subset of R by taking constant sequences.

17 Proof. Check that all operations and orders are well-defined and the field axioms as an exercise. One is assigned as homework.

The upper bound property is a little more work. Let S be a subset of real numbers, and let u0 be an upper bound of S. Choose `0 such that there is some s P p`0, u0q X S. One should quickly check that this can be done: take some s P S and find the N such that m, n ą N implies |sn ´ sm| ă 1, and we can take `0 “ sn ´ 2. Now, `n´1`u0 repeat the following process to construct a sequence: consider mn “ 2 . If this is an upper bound for S, then set un “ mn and `n “ `n´1. Otherwise, set un “ un´1 and `n “ mn. This defines two sequences, un and `n. These are both Cauchy sequences; I will leave this as an exercise.

First, notice that every un is an upper bound on S by construction. Thus the sequence u thought of as an element of R is an upper bound on S by definition. I claim that u is the supremum. To see this, suppose there is 1 1 a smaller upper bound, u . This means that limnÑ8 un ´ un “ a for some positive number a. Notice that the the quantity un ´ `n is cut in half each time we increase n by one. Also notice that by construction, there is always a 1 element in S greater than `n. Thus, when un ´ `n ă a, we find that this contradicts that un is an upper bound.

8 Lecture 8

8.1 Taking powers in the real numbers We haven’t actually define what it means to take powers in the positive real numbers to numbers other than integers. We will first define fractional powers, which we could have done earlier as soon as we defined Dedekind cuts. Definition 8.1. Let α P R be have α ą 1 and p P N. We define α1{p to be the unique positive real number x such that xp “ α. Constructively, this is the Dedekind cut:

p tx P Q | x ď 1 or x ă αu.

Now, let p P R be positive, and α ą 1. We define αp to be the Dedekind cut:

q tx P Q | x ď 1 or x ă α for 0 ă q ă pu.

For the case 0 ă α ă 1, one uses that α´1 ą 1. One does have to prove that this cut does have the given property, and that it is unique in this sense. The first is left as an exercise, and the second follows from the order axioms. We won’t prove the following. Proposition 8.2. For a, b ą 0 and p real numbers, using the above definitions, we have that pabqp “ apbp.

8.2 “Toolbox” sequences Recall: Theorem 8.3 (Binomial theorem). Let n P N. Then, n p1 ` xqn “ 1 ` nx ` x2 ` ¨ ¨ ¨ ` nxn´1 ` xn 2 ˆ ˙ Proposition 8.4. The following are useful sequences to know. 1 (a) If p ą 0, then limn p “ 0. Ñ8 ?n (b) If p ą 0, then limnÑ8 n p “ 1. ?n (c) limnÑ8 n “ 1. nα (d) If p ą 0, then limnÑ8 p1`pqn “ 0. n (e) If |x| ă 1, then limnÑ8 x “ 0. 1 1 1{p Proof. For (a), we want to verify that we can make np ă  for large n. Doing some rearranging, we want n ą p  q . 1 1 α Here we use two facts: that 0 ă x ă y implies 0 ă y ă x and that the function fpxq “ x is monotonically increas- ing for α ą 0 (i.e. x ď y implies xα ď yα).

18 ? ? For (b), we look at three cases: p ą 1, p “ 1, and 0 ă p ă 1. If p ą 1, then n p ą 1 and take xn “ n p ´ 1 ą 0. By the binomial theorem: n 1 ` nx ď 1 ` nx ` x2 ` ¨ ¨ ¨ “ p1 ` x qn “ p n n 2 n n ˆ ˙ So: p ´ 1 0 ă x ď n n ? Thus, xn Ñ 0 so n p Ñ 1. If p “ 1 the statement is trivial. If 0 ă p ă 1 then we take reciprocals and proceed.

?n For (c), proceed similarly as (b) but let xn “ n ´ 1. We take a different term in the binomial theorem:

n x2 ď p1 ` x qn “ n 2 n n ˆ ˙ and find that 2 0 ă xn ă n ´ 1 c so ? 1 0 ă x ă 2p q1{2 n`1 n and use part (a).

For (d), let k be a positive integer such that k ą α (we will need to take binomial terms up to k to dominate α). For n ą 2k, nkpk npn ´ 1q ¨ ¨ ¨ pn ´ k ` 1q n ă pk “ pk ď p1 ` pqn 2kk! k! k ˆ ˙ and so nα 2kk! 1 0 ă ă p qk´α p1 ` pqn pk n and use part (a).

For (e), just take α “ 0 and 1 ` p “ x from (d).

9 Lecture 9

9.1 Series

Definition 9.1. Let tsnu be a sequence in R. The nth partial sum is

n Sn :“ si i“1 ÿ and we write 8 sn :“ lim Sn nÑ8 i“1 ÿ to be the series of sn.

Remark 9.2. Much of our study of series will be to find necessary and sufficient conditions for convergence in different situations, based on our knowledge of sequences. This is analogous to studying integrals in terms of functions.

The following is an easy but useful criterion for the convergence of series. It is essentially the Cauchy criterion.

19 Proposition 9.3. The series for sn converges if and only if for every  ą 0 there is an integer N such that m, n ą N implies that m | si| ă  i“n ÿ Remark 9.4. Note that while this seems like an easy result, we had to do a lot of work to prove that Cauchy sequences converge. In general it’s harder for us to identify the exact number that a series converges to versus a sequence (maybe in the same way it’s hard to take integrals), so the Cauchy criterion is really useful for series.

Corollary 9.5. If sn converges, then sn Ñ 0.

Proof. Take n “ mřabove.

1 Example 9.6 (Counterexample to converse). Consider sn “ n , which has limnÑ8 sn “ 0. The corresponding series does not converge. To see why, consider “grouping“ the sums as follows:

1 1 1 1 1 1 1 1 1 1 1 ` ` p ` q ` p ` ` ` q ` ¨ ¨ ¨ ą 1 ` ` ` ` ¨ ¨ ¨ 2 3 4 5 6 7 8 2 2 2 Note that in this argument I am not rearranging the series or changing it any way. Rearranging the terms in series must be done delicately, as we will show later.

Proposition 9.7 (Comparison test). If |an| ď cn for n ą N0 (some N0), and cn converges, then an converges. If a ě d ě 0 for n ą N , and d diverges, then a diverges. n n 0 n n ř ř Proof. Note that this is a little deeperř than a simpleř application of the comparison of sequences result we proved earlier, since there is no mention of a lower bound converging to the same number. Let N ě N0 be such that m m, n ą N implies | i“n ci| ă . Then we have ř m m m | ai| ď |ai| ď ci ă . i“n i“n i“n ÿ ÿ ÿ The second statement follows from comparison of sequences. Note that since all the dn terms are positive, the partial sums form a monotonically increasing sequence, and since it diverges it must in fact diverge to infinity.

Since an ě dn, then the series for an must also diverge to infinity.

Example 9.8. Note that it is important for the above cn to have only positive terms, since must serve as an n 1 “overestimate” of a convergent series. For example, if we take cn “ p´1q n , we find that cn converges, whereas 1 does not. Likewise, it is important for d and a to be entirely positive or entirely negative. n n n ř 9.2 Adding, regrouping series

Proposition 9.9. Let an “ A and bn “ B. Then, pan ` bnq “ A ` B and can “ cA for any c.

Proof. Straightforward,ř left as exercise.ř ř ř

Definition 9.10. A regrouping of a series an is defined as a subsequence of the sequence of partial sums Sn “ n s . This is best exhibited by example. Recall our regrouping of a “ 1 : i“1 i ř n n ř 1 1 1 1 1 1 1 1 1 1 1 ` ` p ` q ` p ` ` ` q ` ¨ ¨ ¨ ą 1 ` ` ` ` ¨ ¨ ¨ 2 3 4 5 6 7 8 2 2 2 This corresponds to taking the partial sums:

S1,S2,S4,S8,S16,...

The way to think of this is that we want to treat the parentheses as one “group” and ignore any intermediate terms one might obtain by only summing some part. Note that this only allows for finitely many terms in each group.

20 Proposition 9.11. Let ai be a series of non-negative terms. Then any regrouping of the series has the same behavior as the series itself. ř Proof. Since the terms are nonnegative, the sequence is monotonically increasing, and thus the set of subsequential limit has at most one point.

n Example 9.12. Let an “ p´1q . Note that the following regrouping:

p1 ` ´1q ` p1 ` ´1q ` p1 ` ´1q ` ¨ ¨ ¨ would change the behavior of this series, since it makes this divergent series converge to zero.

Definition 9.13. Let σ : N Ñ N be a injective and surjective function. Then the rearrangement of the sequence tanu corresponding to σ is given by: σ pa qk “ aσpkq Note that I do not think this notation is standard.

Proposition 9.14. Suppose σ fixes all but finitely many numbers, i.e. σpnq “ n for all but finitely many n. Then, the convergence behavior and also limits of two sequences aσ and a are the same. Proof. This is left as an exercise.

9.3 “Toolbox” series Of course, we need some series to compare to.

Proposition 9.15 (Geometric series). If x P r0, 1q then

8 1 xn “ 1 ´ x n“0 ÿ and if x ě 1 then the series diverges. Proof. We can compute by induction the partial sums:

k 1 ´ xk`1 xn “ 1 ´ x n“0 ÿ and use previous results on sequences. For x “ 1 we see immediately it diverges, and we can use the comparison test for other cases.

1 Proposition 9.16 (Negative power series). n np converges if p ą 1 and diverges if p ď 1. Proof. We have shown divergence for p “ 1 andř the comparison test takes care of p ď 1. For p ą 1 we will use a similar technique as for p “ 1, i.e. we will group: 1 1 1 1 1 1 1 1 2 4 8 1 ` p ` q ` p q ` ` ` q ` p ` ¨ ¨ ¨ q ` ¨ ¨ ¨ ď 1 ` ` ` p ` ` ¨ ¨ ¨ q 2p 3p 4p 5p 6p 7p 8p 2p 2p 4p 8p

1 The term in the parentheses is a geometric series with ratio 2p´1 which is less than 1 if p ´ 1 ą 0, and the result follows.

Proposition 9.17 (n logpnq). If p ą 1 then 8 1 nplogpnqqp n“2 ÿ converges. If p ď 1 the series diverges. Proof. This is left as an exercise; use the grouping technique.

21 10 Lecture 10

10.1 Root and ratio tests

n Theorem 10.1 (Root test). Consider a sequence an and let α “ lim sup |an|. Then, (a) if α ă 1 then a converges, n a (b) if α ą 1 then a diverges. ř n 1 1 Remark 10.2. Noteř that if α “ 1 we cannot conclude anything. For example, n diverges, but n2 converges, but they both have α “ 1. ř Proof. If α ă 1, then there is some  such that α `  ă 1. By definition of lim sup, there is some N such that N N`1 n 8 supt |aN |, |aN`1|,...u ă α `  ă 1. In other words, |an| ă pα ` q for every n ą N. Thus, n“N |an| converges by the comparison test. a a ř If α ą 1, then take  “ α´1 and let γ “ α ´ . There are is an N such that for n ą N, we have ? 2 n ? supt an, n`1 an`1,...u ą γ ą 1. Thus we have a subsequence of an whose terms are all greater than 1, so the sequence an cannot converge to zero. Note that this argument fails when α “ 1; in this case we have  “ 0. Also note that we cannot use the comparison test and geometric series in analogy to the α ă 1 case, due to subtle differences in the comparison test 1 for divergence requiring all terms to be positive or all terms to be negative. If α “ 1, note that n diverges but 1 converges. Thus the test cannot be conclusive in this case. n2 ř

Theoremř 10.3 (Ratio test). The series an: (a) converges if lim sup | an`1 | ă 1 an an`1 ř (b) diverges if | | ě 1 for all n ą N0 for some fixed N0 an Proof. For (a), fix some β P plim sup | an`1 |, 1q; there is some N such that if n ě N we have that | an`1 | ă β. Thus, an an we find that:

|aN`1| ă β|aN | and inductively we find k aN`k| ă β |aN | i.e. n´N |an| ă β |aN |

n for n ą N. Since β converges, by the comparison test, an converges. For (b), check that an cannot limit to zero. ř ř Remark 10.4 (Similarities and differences). The techniques for both tests are relatively similar: for convergence we use comparison with a geometric series, and for divergence we show that the sequence does not limit to zero. However, there is a difference in divergence for the ratio test, which insists that the ratio must be greater than or equal to one for all terms past a point, whereas the ratio test asks for a limit supremum to be positive. If we try 1 to apply the ratio test to the series n , we find that it falls under neither (a) nor (b), since the limit supremum of | n | is 1 but is never greater than 1. n`1 ř Example 10.5. This example will show that the divergence part of the ratio test is extremely blunt. Take the series: 1 1 1 1 1 1 ` ` ` ` ` ` ¨ ¨ ¨ 2 3 22 32 23 33 1 and note that (index a2 “ 2 ): ? 1 1 lim sup n a “ lim 2n “ ? n 2n c 2 whereas a lim sup | n`1 | “ 8 an 1 1 since we always have consecutive terms of the form 2k , 3k`1 .

22 Example 10.6. This example is a “rearrangement” of a convergent series

1 1 1 1 1 ` 1 ` ` ` ` ` ¨ ¨ ¨ 2 8 4 32 16 obtained by swapping pairs of terms. Notice that:

? 1 lim sup n a “ n 2 but a lim sup | n | “ 2 an`1 Notice that: a 1 lim inf | n | “ an`1 8 1 1 and that the geometric mean of 2 and 8 is 2 . In some really vague sense, the ratio test is too sensitive to the ordering of the sequence.

Proposition 10.7 (The root test is stronger). For any sequnece an of positive numbers, ? a n n`1 lim sup an ď lim sup an

Proof. Choose some β ą lim sup cn`1 . There is an N such that n ě N implies that an`1 ă β. Thus one finds that cn an n´N ? n N N an ă β aN . Taking nth roots, one obtains that n an ă β β aN . Note that on the right hand side β aN is ? an`1 a constant, and taking lim sup we find that lim sup n an ă β. Since this is true for every β ą lim sup , one a an obtains the result.

10.2 Summation by parts, alternating series

n Theorem 10.8 (Summation by parts). Let an and bn be two sequences, and write An “ sumi“0ai for n ě 0 and A´1 “ 0. Then for 0 ď m ď n one has:

n n´1 anbn “ Aipbi ´ bi`1q ` Anbn ´ Am´1bm i“m i“m ÿ ÿ

Proof. Write an “ An`1 ´ An. Proposition 10.9. If,

(a) the partial sums An of an are bounded, (b) bn is monotonically decreasing, (c) bn converges to zero, then anbn converges.

Proof.řSuppose |An| ă M, i.e. M is a bound. We want to use the Cauchy criterion, i.e. we want to make the following expression ă :

n n´1 n´1 | anbn| “ | Aipbi ´ bi`1q ` Anbn ´ Am´1bm| ă Mp |bi ´ bi`1| ` |bn| ` |bm|q i“m i“m i“m ÿ ÿ ÿ

Since bi ´ bi`1 ě 0 and bi ě 0, this is:

n´1 Mp bi ´ bi`1 ` bn ` bmq “ 2Mbm i“m ÿ

23  Now, by convergence of bn, for any  ą 0, we can find an N such that m ą N implies that bm ă 2M , completing the proof.

Proposition 10.10 (Alternating series test). A sequence is alternating if its signs alternate, i.e. if ai ě 0 then ai`1 ď 0 and ai`2 ě 0, et cetera. Let an be an alternating series such that |an| is monotonically decreasing and an converges to 0. Then, an converges.

n Proof. Apply the previousř proposition with an “ p´1q and bn “ |cn|.

Example 10.11 (Why monotonically decreasing). Otherwise, one could make the negative terms converge much 1 1 faster than the positive terms. For example, take a2n “ n and a2n`1 “ ´ 2n . One can check by regrouping: 1 1 1 1 1 1 1 1 ´ ` ´ ` p ´ ` ´ q ` ¨ ¨ ¨ 2 2 4 3 8 4 16

1 that the groups are all ě 4 .

10.3 Absolute convergence, multiplying and rearranging series

Definition 10.12. A series an converges absolutely if |an| converges. It is easy to see an absolutely convergent series is convergent. A series that converges but not absolutely is called conditionally convergent. ř ř

Remark 10.13. The root test and ratio tests all use |an| rather than an, and thus are really testing for absolute convergence. Summation by parts does handle some non-absolutely convergent series.

n 1 1 Example 10.14 (A conditionally convergent series). By the alternating series test, p´1q n converges, but n does not. ř ř

Definition 10.15. For series an and bn, define their product, which we will write by informal convention as cn, by ř ř n ř cn “ aibj “ akbn´k i`j“n k“0 ÿ ÿ This definition is motivated by imagining products of power series in the formal variasble x:

8 n n n p anx qp bnx q “ aibjx n“1 i`j“n ÿ ÿ ÿ ÿ Remark 10.16. Notice that we can’t apply sequence techniques to conclude that AB “ C (the infinite sums), since AnBn ‰ Cn (the partial sums).

p´1qn ? Example 10.17. Take an “ bn “ n . These both converge by the alternating series test. However, the product has: 1 c “ p´1qn ? . n ij i`j“n ÿ n Now, notice that ij is maximized when i “ j “ 2 . Thus we have that n ´ 2 4 cn ě “ 2 ´ n2{4 n a and thus cn does not converge to 0, so the series cannot converge.

Proposition 10.18. Using the previous notation, if one of an, bn converges absolutely, then AB “ C.

Proof. For shorthand, write βn “ Bn ´ B. We can write ř ř

Cn “ a0Bn ` a1Bn´1 ` ¨ ¨ ¨ ` anB0 “ a0pB ` βnq ` a1pB ` βn´1q ` ¨ ¨ ¨ ` anpB ` β0q

24 “ AnB ` a0βn ` a1βn´1 ` ¨ ¨ ¨ ` anβ0 To prove the result, we need to show that the “tail:”

γn “ aiβj “ a0βn ` a1βn´1 ` ¨ ¨ ¨ ` anβ0 i`j“n ÿ converges to zero.

Since an converges absolutely, let α “ |an|. Since β converges to 0, let N be such that n ě N implies that β ă . Then, we can break up the sum into two groups using the triangle inequality: n ř ř

|γn| ď |β0αn ` ¨ ¨ ¨ ` βN an´N | ` |βN`1an´N´1 ` ¨ ¨ ¨ ` βna0|

ď |β0αn ` ¨ ¨ ¨ ` βN an´N | ` α

Taking lim sup of both sides, we find that lim sup |γn| ď α (we can’t take the limit because we don’t know it exists). Since  was chosen arbitrarily, it must follow that lim sup γn “ 0, and the result follows.

Definition 10.19. Let σ : N Ñ N be a injective and surjective function. Then the rearrangement of the sequence tanu corresponding to σ is given by: σ pa qk “ aσpkq Note that I do not think this notation is standard.

p´1qn`1 1 1 Example 10.20. Take an “ n “ 1 ´ 2 ` 3 ´ ¨ ¨ ¨ . The series converges by the alternating series test, to a 1 1 5 number A ă 1 ´ 2 ` 3 “ 6 . Now, we can rearrange this series so that it has two positive numbers followed by one negative:

1 1 1 1 1 1 1 1 r1 ` ´ s ` r ` ´ s ` r ` ´ s ¨ ¨ ¨ 3 2 5 7 4 9 11 6 (the brackets are only for visual ease). Each three term group is:

1 1 1 ` ´ 4k ´ 3 4k ´ 1 2k

3 and one can verify that this is positive when k ě 8 . This the partial sums are monotonically increasing after a 5 point, and one can check that eventually the sum becomes greater than 6 . Thus the two cannot converge to the same number.

Proposition 10.21. Suppose σ fixes all but finitely many numbers, i.e. σpnq “ n for all but finitely many n. Then, the convergence behavior and also limits of two sequences aσ and a are the same.

Proof. This is left as an exercise.

Proposition 10.22. If an converges absoutely, then every rearrangement converges to the same number. 1 ř n Proof. Let an be the rearrangement. For  ą 0, choose N such that for m, n ą N, one has that i“m |ai| ă  n 1 and i“m |ai| ă . So, we need to “take care” of terms a1, . . . , aN . ř 1 1 ř 1 Now, choose M such that a1, . . . , aM contains all the a1, . . . , aN terms. Then, in the quantity |An ´ An|, the ř 1 first N of the original series will cancel in both An and An. Using the triangle inequality, this quantity is ă 2.

Proposition 10.23. Suppose an converges conditionally. Let α ď β. Then there is a rearrangement of an which has lim inf “ α and lim sup “ β. One can take α, β to be infinite. ř Proof. I will skip this. Refer to Rudin.

25 11 Lecture 11

11.1 Limits of functions Definition 11.1. Let X,Y be metric spaces, and E Ă X, and p a limit point of E. We write the limit

lim fpxq “ q xÑp if for every  ą 0 there exists a δ ą 0 such that x P Bδppq ´ tpu implies that fpxq P Bpqq. Note that it is not relevant whether or not p P E. Also, I will use the same symbols for this set-up throughout this section.

1 Example 11.2 (Sequences). We can realize N “ N Y t8u as the subset t n | n P Nu Y t0u Ă R. Here, 8 is the only limit point, and the definitions agree.

Example 11.3 (One-sided limits in R). In calculus sometimes we take one-sided limits, i.e. limxÑ0` txu “ 0 but limxÑ0´ txu “ ´1. Here, the subset E “ p0, 8q and p´8, 0q respectively.

Proposition 11.4. One has limxÑp fpxq “ q if and only if for every sequence tpnu such that pn ‰ p (for all n) and pn Ñ p, one has limnÑ8 fppnq “ q.

Proof. Suppose that limxÑp fpxq “ q. Then, for every  ą 0 one can find a δ ą 0 such that x P Bδppq ´ tpu implies that fpxq P Bpqq. Any sequence pn converging to p and such that pn ‰ p has an N such that pn P Bδppq ´ tpu for n ą N, and so fppnq P Bpqq. Thus we have shown that for every  there is an N such that fppnq P Bpqq. If one did not have that limxÑp fpxq “ q, then there is some  ą 0 for which one could find a pn arbitrarily close to p (i.e. for every δ) such that fppnq R Bpqq. Fix this ; choose pn P B 1 ppq ´ tpu such that fppnq R Bpqq. This is n a sequence such that fppnq does not converge to q. Corollary 11.5. Limits of functions are unique, if they exist.

Proposition 11.6. Let limxÑp fpxq “ A and limxÑp gpxq “ B. Then, (a) limpf ` gqpxq “ A ` B, (b) limpfgqpxq “ AB, f (c) lim g pxq “ A{B if B ‰ 0 Proof. Follows from the result in sequences and the previous proposition.

Example 11.7. We can check explicitly: limxÑa x “ a. Fix an  ą 0; we are concerned with the interval pa´, a`q. We want to show there is δ such that x P pa ´ δ, a ` δq implies that fpxq P pa ´ , a ` q. But we can just take δ :“ !

Example 11.8. Define fpxq to be 0 if x P Q and 1 if x R Q. The limit limxÑ0 fpxq does not exist, since in every interval p´, q there are points such that |fpxq ´ fp0q| “ 1 and |fpxq ´ fp0q| “ 0.

11.2 Continuity Definition 11.9. Let f : X Ñ Y be a function, and U Ă Y a subset. Then define the inverse image:

f ´1pUq :“ tx P X | fpxq P Uu

Let V Ă X be a subset, and define the image

fpV q :“ tfpxq P Y | x P Xu

Definition 11.10. Let X and Y be metric spaces. A function f : X Ñ Y is continuous at p P X if the following equivalent are true:

(a) limxÑp fpxq “ fppq, (b) for every  ą 0, there is a δ ą 0 such that fpxq P Bpfppqq for all x P Bδppq, (c) for every  ą 0, there is a δ ą 0 such that fpBδppqq Ă Bpfppqq.

26 Definition 11.11. Let X and Y be metric spaces. A function f : X Ñ Y is continuous if the following equivalent are true:

(a) for every p P X, limxÑp fpxq “ fppq, (b) for every p P X and every  ą 0, there is a δ ą 0 such that fpxq P Bpfppqq for all x P Bδppq, (c) for every open subset U Ă Y , f ´1pUq is open, (d) for every closed subset Z Ă Y , f ´1pZq is closed. Proof. The equivalence of (c) and (d) is a set-theoretic exercise, knowing that U is open if and only if U c is closed. The equivalence of (a) and (b) is simply untangling the definition. I will show that (b) and (c) are equivalent. Assume (a). Let U Ă Y be open, and let p P f ´1pUq. We want to show that p is an interior point. Since fppq P U it is an interior point, so choose  such that Bpfppqq Ă U. Then there is a δ such that fpBδppqq Ă Bpfppqq Ă U ´1 ´1 and thus Bδppq Ă f pUq. Now assume (c). Then, f pBpfppqq is open, so p is an interior point, so there is some ´1 Bδppq Ă f pBpfppqq.

Example 11.12. In the discrete metric, all functions are continuous.

Remark 11.13. Note that it is not true that if f is continuous at x, then it is continuous in a neighborhood of x. As a counterexample, consider the function f : R Ñ R such that fpxq “ 0 if x R Q and fpxq “ x if x P Q. Check that f is continuous only at 0.

Proposition 11.14 (Composition of continuous functions). Let f : X Ñ Y and g : Y Ñ Z. If f is continuous at p P X and g is continuous at fppq P Y , then g ˝ f is continuous at p.

Proof. This proof is just unwinding a lot of definitions. For any  ą 0 we want to show that there is a δ ą 0 such 1 1 that gpfpBδppqq Ă Bpgpfppqq. We know that we can find a δ ą 0 such that gpBδ1 pfppqq Ă Bpgpfppqq. Using δ as the “” for f, we then find a δ such that fpBδppqq Ă Bδ1 pfppqq.

Corollary 11.15. Let f : X Ñ Y and E Ă X. Suppose L “ limxÑa fpxq exists. If g : fpXq Y tLu Ñ Z is continuous at L, then lim gpfpxqq “ gpLq. xÑa Proposition 11.16 (Algebraic operations on continuous functions in R). If f and g are continuous then f ` g, fg, f{g are continuous (assume gpxq ‰ 0 for the last one).

12 Lecture 12

12.1 Properties of continuity Proposition 12.1. Let X be compact, and f : X Ñ Y a continuous function. Then the image, fpXq, is compact.

Proof. Take an open cover of fpXq and pull it back using f ´1. Since f is continuous, this is an open cover of X. It has a finite subcover since X is compact. The corresponding subcollection in the original cover of fpXq is thus a subcover.

Remark 12.2. Note that it is not true that fpUq is open for U open. For example, let f : R Ñ R2 send x ÞÑ px, 0q. Let U be any open interval; the image is not open in R2. It is also not true that fpZq is closed for Z closed. For example, let f : R Ñ R by fpxq “ ex, and consider the image fpp´8, asq “ p0, eas for any a. It is also not true 1 that fpEq is bounded for E bounded. Let f : R ´ t0u Ñ R be defined by fpxq “ x , and consider fpp0, 1qq “ p1, 8q. Nor is it true that f ´1pEq is bounded for E bounded. Let f : RR be a constant function, say fpxq “ 0. Then f ´1pt0uq “ R is not bounded.

Theorem 12.3 (Maximum value theorem). Let ra, bs be any nonempty closed interval. Suppose that f : ra, bs Ñ R is a continuous function of real numbers. Then, f attains maximum and minimum values on ra, bs.

Proof. This follows immediately; fpra, bsq is closed and bounded.

In the homework, the student explored connectedness. Recall the definition.

27 Definition 12.4. Two subsets of X are separated if A X B “ A X B “H. A subset E Ă X is connected if it is not the union of two nonempty separated sets.

Proposition 12.5. A subset E of R is connected if and only if for all pairs, x, y P E, one has px, yq Ă E. Proof. If the condition is false then one can take z P px, yq and one has p´8, zq and pz, 8q separating E. Conversely, if E is not connected, take nonempty separated sets A and B covering E. Choose x P A and y P B and take z “ sup A X rx, ys. Since z P A, z ‰ B. If z R A then z P px, yq breaks the condition. If z P A then z R B, so one can find z1 P pz, yq such that z1 R B, but also z1 R A, so z1 R E, breaking the condition.

Proposition 12.6. Let E Ă X be connected, and f : X Ñ Y continuous. Then fpEq is connected.

Proof. Suppose fpEq is not connected; then it has separating sets A and B. We claim that f ´1pAq and f ´1pBq are separating sets for E. They are clearly nonempty and cover E. Using two facts: (1) f ´1pAq Ă f ´1pAq and (2) f ´1pA X Bq “ f ´1pAq X f ´1pBq, one has that f ´1pAq X f ´1pBq “ f ´1pAq X f ´1pBq “ f ´1pA X Bq “ H. Check the two facts as an exercise.

Theorem 12.7 (Intermediate value theorem). Let f : ra, bs Ñ R be continuous. If fpaq ă fpbq, and c P pfpaq, fpbqq, then there is some x P pa, bq such that fpxq “ c.

Proof. Apply the previous two propositions.

Corollary 12.8 (Fixed point theorem for R). Let f : r0, 1s Ñ r0, 1s be a continuous map. Then f has a fixed point, i.e. some p such that fppq “ p.

Proof. Apply the IVT to gpxq “ fpxq ´ x. Note that gp0q “ fp0q ě 0 and gp1q “ fp1q ´ 1 ď 0, one has some p such that gppq “ 0.

Proposition 12.9. Let f be a one-to-one continuous function on an interval I, i.e. f : I Ñ R. Then f is either strictly increasing or strictly decreasing. Recall that f : I Ñ R is strictly increasing if for all a ă b P I, one has that fpaq ă fpbq.

Proof. First, we will show the result for a closed interval I “ ra, bs. Suppose that fpaq ă fpbq (for fpbq ă fpaq the proof is the same). We want to show that every c P pa, bq has fpcq P pfpaq, fpbqq. If not, then if fpcq ă fpaq, then fpaq P pfpcq, fpbqq, so there is some p P pc, bq such that fppq “ fpaq by the IVT, contradicting injectivity. If fpcq ą fpbq a similar argument applies. Thus, fpcq P pfpaq, fpbqq. Now, if the interval were open, say I “ pa, bq, then one could take a closed subinterval ra1, b1s Ă pa, bq and extend the argument as follows. Let c P pa, a1s. We wish to show that fpcq ă fpa1q. If not, i.e. fpcq ą fpa1q, then choose p such that fppq P pfpa1q, fpcqq. Then apply the IMT to contradict injectivity, and this completes the proof.

13 Lecture 13

13.1 Uniform continuity Definition 13.1. Let f : X Ñ Y be a function on metric spaces. We say that f is uniformly continuous on X if for every  ą 0 there exists δ ą 0 such that dpfpxq, fpx1qq ă  for all x, x1 such that dpx, x1q ă δ.

Remark 13.2. Unlike the notion of continuity, which only had a topological characterization (i.e. in terms of open sets only), uniform continuity requires the use of a metric. There are ways around this (e.g. uniform spaces) but we won’t get into that here.

28 1 Example 13.3. Consider fpxq “ x . We claim that fpxq is uniformly continuous on ra, 8q for any a ą 0. Fix  ą 0; one can show because fpxq is monotonically decreasing, that for fixed x, dpfpxq, fpx1qq is maximized when x1 “ x ` δ or x1 “ x ´ δ. Then,

1 1 δ |fpxq ´ fpx ` δq| “ | ´ | “ . x x ` δ xpx ` δq

δ This is maximized when x is minimal, i.e. at x “ a, and takes value a2`δa . Thus, for  ą 0 we want to find a δ such that δ ă  a2 ` δa Check that a2 δ ă 1 ´ a 1 are all good choices. One has to check that the right hand side is positive, which is true as long as  ă a . However for larger  check that in fact any δ will do. 1 However, fpxq “ x is not uniformly continuous on the interval p0, 8q. Fix  ą 0. I claim that we can find 1 1 1 1 1 two x, x arbitrarily close to each other such that dpfpxq, fpx qq ą . Note that for x “ n and x “ n`1 one has that |fpxq ´ fpx1q| “ n. For n large enough, one can make |x ´ x1| arbitrarily small, and at the same time make n arbitrarily large.

Proposition 13.4. Let X be compact and f : X Ñ Y be continuous. Then f is uniformly continuous.

Proof. Fix  0. By continuity of f, for each x X choose δ 0 such that f B x B  f x . Consider the ą P x ą p δx p qq Ă 2 p p qq open cover:

tJx :“ B δx pxquxPX 2 1 This has a finite subcover by compactness. Let that finite subcover by indexed by I; then take δ “ 2 minpδx | x P Iq. Now, we claim that for p, q P X, that dpp, qq ă δ implies that dpfppq, fpqqq ă . Suppose p be in the open given by i P I and q in the open given by j P I. Since dpp, qq ă δ, there must be a point in common between the balls Ji and Jj, call it x. Then, dpfppq, fpqqq ď dpfppq, fpxqq ` dpfpxq, fpqqq ă  as desired.

Proposition 13.5. Let f : X Ñ Y be uniformly continuous. Let sn be a Cauchy sequence in X. Then fpsnq is a Cauchy sequence in Y .

Remark 13.6. Note that this is untrue for continuous functions. For continuous functions, we insist that sn is convergent.

Proof. Fix  ą 0. We want to find an N such that m, n ą N implies that |fpsnq´fpsmq| ă . By uniform continuity there is a δ such that |sn ´ sm| ă δ implies this, and by Cauchy-ness of sn there is an N that guarantees this.

13.2 The derivative

Definition 13.7. Let f : ra, bs Ñ R be a real valued function. Define the derivative:

fptq ´ fpxq f 1pxq “ lim tÑx t ´ x

Note that this limit may not always be well-defined at every point. If f 1pxq exists at x we say f is differentiable at x.

Proposition 13.8. If f is differentiable at x, then f is continuous at x.

29 Proof. Using limit laws, fpxq ´ fptq lim fpxq ´ fptq “ lim px ´ tq “ f 1pxq ¨ 0 “ 0 tÑ0 tÑ0 x ´ t

Remark 13.9. Note that it is not true that if f is differentiable at x, then f is continuous in a neighborhood of x. For example, take x2 x P Q fpxq “ #´x2 x R Q This function is continuous only at zero, and differentiable only at zero. Example 13.10. Take x sinp 1 q x ‰ 0 fpxq “ x #0 x “ 0 We claim that the derivative does not exist at 0. To see this, note that

fpxq ´ fp0q 1 “ sinp q x x But this has no limit as x Ñ 0. Example 13.11. Now take x2 sinp 1 q x ‰ 0 fpxq “ x #0 x “ 0 We claim that f 1p0q “ 0. Proposition 13.12 (Arithmetic operations and the derivative). Let f, g : ra, bs Ñ R and suppose f, g are differen- tiable at x P ra, bs. Then the following are differentiable at x and are given by: (a) pf ` gq1pxq “ f 1pxq ` g1pxq (b) pfgq1pxq “ f 1pxqgpxq ` fpxqg1pxq 1 1 1 f pxqgpxq´fpxqg pxq (c) pf{gq pxq “ gpxq2

Proof. It is easy to prove (a) from the definition. We will prove (b).

fptqgptq ´ fpxqgpxq fptqrgptq ´ gpxqs ` gpxqrfptq ´ fpxqs gptq ´ gpxq fptq ´ fpxq “ “ fptq ` gpxq t ´ x t ´ x t ´ x t ´ x

Taking the limit as t Ñ x, one has the result. See the textbook for the proof of (c).

Example 13.13. This allows us to differentiate all polynomials. It is easy to show directly that if fpxq “ 1 then f 1pxq “ 0. It is also easy to show that if gpxq “ x then g1pxq “ 1. Using the product rule, one has that pxnq1 “ xn´1pxq1 ` pxn´1q1x “ xn´1 ` pxn´1q1x, and one can use an argument by induction to show that pxnq1 “ nxn´1. Then one uses scaling and addition to get all polynomials. Proposition 13.14 (Chain rule). Let f : ra, bs Ñ R be differentiable at x P ra, bs, and g : I Ñ R where fpra, bsq Ă I, and g is differentiable at fpxq. Then pg ˝ fq1pxq “ g1pfpxqqf 1pxq Remark 13.15. What one wants to do is write

gpfptqq ´ gpfpxqq gpfptqq ´ gpfpxqq fptq ´ fpxq “ t ´ x fptq ´ fpxq t ´ x and take the limit. The problem here is we cannot guarantee that fptq ´ fpxq ‰ 0 near t “ x. For example, 2 1 considering fpxq “ x sinp x q for x ‰ 0 and fp0q “ 0, one that for x P Bδp0q for any δ, f has a zero. So there are functions for which one cannot avoid this problem by taking a sufficiently small δ.

30 Proof. Put y “ fpxq. We can find functions u and v such that:

fptq ´ fpxq “ pt ´ xqpf 1pxq ` uptqq

gpsq ´ gpyq “ ps ´ yqpg1pyq ` vpsqq and upxq “ 0 and vpyq “ 0. Note that u and v are continuous at x and y respectively, as

fptq ´ fpxq 1 limt 0 ´ f pxq “ 0 Ñ t ´ x and likewise for v. Then,

gpfptqq ´ gpfpxqq “ pfptq ´ fpxqqpg1pfpxqq ` vpfptqq “ pt ´ xqpf 1pxq ` uptqqpg1pfpxqq ` vpfptqqq and for t ‰ x, one has gpfptqq ´ gpfpxqq “ pf 1pxq ` uptqqpg1pfpxqq ` vpfptqqq t ´ x Taking the limit t Ñ x, one finds has the result.

14 Lecture 14

14.1 Mean value theorem

Definition 14.1. A a point p is a local maximum of f if there is a δ such that fpxq “ maxpfpBδpxqqq. Likewise, for local minimum.

Proposition 14.2. Let f : ra, bs Ñ R. If f has a local maximum or minimum at p, and if fppq exists, then f 1ppq “ 0.

Proof. Suppose f has a local maximum at p. Then we have some δ ą 0 such that fppq ě fpxq for x P pp ´ δ, p ` δq. The derivative is fptq ´ fppq lim tÑp t ´ p Note that this expression is ě 0 when t ď p and ď 0 when t ě p. Thus if this limit exists it must be zero.

Remark 14.3. One consequence of this proposition is that to find the local minima or maxima of a function, one only needs to look at points where f 1pxq “ 0 or does not exist.

Theorem 14.4 (Rolle’s theorem). Let f : ra, bs Ñ R be continuous, and suppose f is differentiable on pa, bq, and that fpaq “ fpbq. Then there is some c P pa, bq such that f 1pcq “ 0.

Proof. By the maximum value theorem, there is a global maximum on this interval, which must be a local maximum as well, and the result follows from the previous proposition.

Theorem 14.5 (Generalized mean value theorem). Let f, g : ra, bs Ñ R be continuous and differentiable in pa, bq. Then there is c P pa, bq such that rfpbq ´ fpaqsg1pcq “ rgpbq ´ gpaqsf 1pxq

Remark 14.6. One way to think of this is to rearrange terms:

fpbq ´ fpaq f 1pxq “ gpbq ´ gpaq g1pxq

Proof. Take hpxq “ rfpbq ´ fpaqsgpxq ´ rgpbq ´ gpaqsfpxq. We want to show that h1pxq “ 0 for some c P pa, bq. Note that hpaq “ hpbq and apply the previous theorem.

31 Theorem 14.7 (Mean value theorem). If f : ra, bs Ñ R is continuous and differentiable on pa, bq, then there is some c P pa, bq such that fpbq ´ fpaq “ pb ´ aqf 1pcq Proof. Take gpxq “ x above.

Here is an application Proposition 14.8. Let f be differentiable on pa, bq. Then, (a) if f is strictly increasing then f 1pxq ą 0 for x P pa, bq, (b) if f is strictly decreasing then f 1pxq ă 0 for x P pa, bq, (c) if f is increasing then f 1pxq ě 0 for x P pa, bq, (d) if f is decreasing then f 1pxq ď 0 for x P pa, bq. Proof. Let a ă x ă y ă b. Then by the mean value theorem,

fpyq ´ fpxq “ f 1pcq ą 0 y ´ x for some c P px, yq. Thus fpyq ą fpxq.

Proposition 14.9 (Intermediate value theorem for derivatives). Let f : ra, bs Ñ R be differentaible, such that f 1paq ă f 1pbq. Let λ P pf 1paq, f 1pbqq. Then there is some x P pa, bq such that f 1pxq “ λ. Remark 14.10. Note that this isn’t an immediate application of the intermediate value theorem, since derivatives of continuous functions need not be continuous. Proof. Take gpxq “ fpxq ´ λx. We want to show that g has a zero in pa, bq. Note that g1paq ă 0 and g1pbq ą 0. Using the definition of derivative, this means there is some point t1 such that gpt1q ă gpaq and a point t2 such that gpt2q ă gpbq. Thus neither a nor b can be global minimums. Since f is continuous in ra, bs, it attains a global in this interval not at the endpoints, and at this point the derivative of g vanishes.

15 Lecture 15

15.1 L’Hospital’s Rule

Theorem 15.1 (L’Hospital’s Rule). Suppose f, g : ra, bs Ñ R are differentiable in pa, bq. Allow ˘8 for the symbols a, b. Suppose: f 1pxq lim “ L xÑa` g1pxq If one of the following are true:

(a) limxÑa` fpxq “ limxÑa` gpxq “ 0 (b) limxÑa` gpxq “ 8 Then one has: fpxq lim “ L xÑa` gpxq The symmetric statements for b´ are also true. Proof. In this proof we will have to choose a lot of symbols. One should think of c, c1, c2 as being increasingly close to a as the number of 1 symbols increase. Case 1: A is finite or A “ ´8. The assumptions of this case allow us to choose some q P pA, 8q. The goal is fpxq to show that there is some interval pa, a ` δq such that if x P pa, a ` δq then gpxq ă q. For technical reasons we will fpxq see later, we want to again choose an r P pA, qq, and we will show that gpxq ď r ă q instead. Since the limit of the expression is A, there is some c such that x P pa, cq implies that

f 1pxq ă r g1pxq

32 Now choose α ă β P pa, cq. Using the generalized mean value theorem, there is some t P pα, βq such that

fpαq ´ fpβq f 1ptq “ ă r gpαq ´ gpβq g1ptq

If (a) holds, we can take the limit α Ñ a to find that for β P pa, cq, one has

fpβq ď r ă q gpβq

This proves the goal in this case. If (b) holds, we can find some c1 P pa, cq such that gppa, c1qq ą maxp0, gpβqq, i.e. such that

gpαq ´ gpβq ą 0 gpαq for α P pa, c1q. Multiplying by this factor, one has:

fpαq ´ fpβq pgpαq ´ gpβq ă r gpαq gpαq so fpαq gpβq fpβq M ă r ´ r ` “ r ´ gpαq gpαq gpαq gpαq

1 2 for some constant M (i.e. we will fix β and r). Let  “ 2 pq ´ rq. Since gpαq Ñ 8 as α Ñ a, there is some c P pa, cq 2 M such that for α P pa, c q one has that | gpαq | ă  so that

fpαq ă r `  ă q gpαq

This proves the goal in this case. Case 2 A is finite or A “ 8. One can repeat the argument for q ă A. Together, this completes the argument.

Remark 15.2. Of course, one could loosen condition (b) to be ´8 by taking ´gpxq.

15.2 Power series

Definition 15.3. A power series is defined given a sequence tanuně0 by:

8 n anz n“0 ÿ It is immaterial whether this sequence converges or diverges for a particular z in this definition, though this is a question we will ask later.

A quick application of the root test gives:

Proposition 15.4. Given a power series 8 n fpzq “ anz n“0 ÿ put n α “ lim sup |an| nÑ8 and a 1 R “ α Then, fpzq converges for |z| ă R and diverges for |z| ą R. This R is called the radius of convergence.

33 Proposition 15.5. Given a power series 8 n fpzq “ anz n“0 ÿ put |a | α “ lim sup n`1 nÑ8 |an| 1 Then the radius of convergence R has R ě α . Example 15.6. We compute the radius of convergence: (a) nnzn has R “ 0 n (b) z has R “ 8 (use ratio test instead) ř n! (c) zn has R “ 1. If |z| “ 1 then the series diverges. ř zn (d) n has R “ 1. It diverges for z “ 1 and converges for z “ ´1. ř zn (e) 2 has R “ 1 and converges for |z| “ 1. ř n ř 15.3 Taylor series Definition 15.7. Let f be a function for which all higher derivatives exist at a. Then, define the Taylor series of f at a to be the formal power series: 8 f pnqpaq T pxq “ px ´ aqn f n! n“0 ÿ Theorem 15.8 (Taylor’s theorem). Let f : ra, bs Ñ R, f pn´1q is continuous on ra, bs (and therefore all lower derivatives are continuous), and f pnq exists on pa, bq. Let α, β P ra, bs be distinct points and define

n´1 f pkqpαq P ptq “ pt ´ αqk k! k“0 ÿ Then there exists a point c between α and β such that

f pnqpcq fpβq “ P pβq ` pβ ´ αqn. n!

Remark 15.9. Let n “ 1. Then P ptq “ fpαq and the statement is that there is c P pα, βq such that

fpβq ´ fpαq “ f 1pcqpβ ´ αq i.e. the mean value theorem. Thus one can think of this as a “higher order mean value theorem.”

Proof. Without loss of generality let α ă β. Choose M such that

fpβq “ P pβq ` Mpβ ´ αqn and let gpxq “ fpxq ´ P pxq ´ Mpt ´ αqn so that gpβq “ 0. We want to show that n!M “ f pnqpxq for some x P pα, βq. One finds that

gpnq “ f pnqpxq ´ n!M so we’ve reduced our claim to showing that gpnqpxq “ 0 for some x P pα, βq. Note that gpαq “ 0. By Rolle’s theorem (here we need continuity of the derivatives at the endpoints) there 1 1 is some c1 P pα, βq such that g pc1q “ 0. Notice that also, g pαq “ 0¡ and in fact, all higher derivatives up to pn´1q 1 2 f pαq “ 0. Thus we can repeat the argument for g and pα, c1q, and then for g and pα, c2q, and at the end we find cn “ c.

34 Remark 15.10. Note that Taylor’s theorem doesn’t say anything about convergence of the Taylor series. Consider the function: ´ 1 e x x ą 0 fpxq “ #0 x ď 0 All higher derivatives of this function exist and are continuous everywhere. However, its Taylor series at 0 is

Tf pxq “ 0.

16 Lecture 16

16.1 The Riemann-Stieltjes integral

We will have to introduce a lot of notation in this section. For the entirety of this section f : ra, bs Ñ R will refer to a bounded function, and α : ra, bs Ñ R a monotonically increasing weight function. We will treat other cases later.

Definition 16.1. Let ra, bs be an interval. Let α : ra, bs Ñ R be a monotonically increasing weight function. If one would prefer to simpliy the exposition on a first reading, one can just fix αpxq “ x. A partition P of ra, bs can be given by a finite set of points a “ x0 ď x1 ď ¨ ¨ ¨ ď xn “ b which we think of as endpoints of subintervals of ra, bs, i.e.

ra, bs “ rx0, x1s Y rx1, x2s Y ¨ ¨ ¨ Y rxn´1, xns

Note that this is not a partition in the sense of equivalence relations since the intervals are not disjoint. Write:

∆αi “ αpxiq ´ αpxi´1q i.e. the length of rxi´1, xis. Unfortunately one does have to remember the convention that ∆xi refers to the interval where xi is the upper endpoint. Let

Mi “ suptfpxq | x P rxi´1, xisu

mi “ inftfpxq | x P rxi´1, xisu Note that both these exist since f is bounded. Define the upper and lower Darboux sums:

n UpP, f, αq “ Miδαi i“1 ÿ n LpP, f, αq “ miδαi i“1 ÿ Define the upper and lower Darboux integrals:

Upf, αq “ inftUpP, f, αq | partitions P u

Lpf, αq “ suptLpP, f, αq | partitions P u where we range over all partitions P . If Upfq “ Lpfq then we say f is Riemann-integrable and define the Riemann-Stiljes integral:

b f dα ża

Definition 16.2. Let P,Q be two partitions, given by points x0, . . . , xn and y0, . . . , ym. We say that P is a

35 refinement of Q if ty0, . . . , ymu Ă tx0, . . . , xnu. Note that we forget about the “multiplicities” here, i.e. P “ p0, 1, 2, 3q is a refinement of p0, 1, 1, 3q. Further, any two partitions P and Q have a common refinement P Y Q. The common refinement is unique only up to repetition of one of the points xi.

Proposition 16.3. If P is a refinement of Q, then

UpP, f, αq ď UpQ, f, αq

LpQ, f, αq ď LpP, f, αq

Proof. We will only prove the first statement. It suffices to prove the case where P has only one more point than

Q. If that point is already a point of Q there is nothing to prove, since δαi “ 0 for that term, so suppose that the extra point of Q is xi P pxi´1, xi`1q. Then,

UpQ, f, αq´UpP, f, αq “ suptfprxi´1, xi`1squpαi`1´αi´1q´suptfprxi´1, xisqupαi´αi´1q´suptfprxi, xi`1squpαi`1´αiq

But notice that L :“ sup fprxi´1, xi`1sq ě sup fpIq for any subinterval I,

UpQ, f, αq ´ UpP, f, αq ě Lpαi`1 ´ αi´1 ´ αi ` αi`1 ´ αi`1 ` αiq “ 2Lpαi`1 ´ αi´1q ě 0

Proposition 16.4. Upf, αq ě Lpf, αq

Proof. It is clear that for any particular P , one has UpP, f, αq ě LpP, f, αq. Further, for any P1 and P2 with common refinement Q, one has:

LpP1, f, αq ď LpQ, f, αq ď UpQ, f, αq ď UpP2, f, αq

Taking the sup over P1 and the inf over P2 gives:

Lpf, αq ď Upf, αq

The following is a sufficient condition for integrability.

Proposition 16.5. f is Riemann integrable (with respect to α) if and only if for every  ą 0 there is a partition P such that UpP, f, αq ´ LpP, f, αq ă 

Proof. Suppose the condition holds. Note that

0 ď Upf, αq ´ Lpf, αq ď UpP, f, αq ´ LpP, f, αq for any P , and the result follows. Conversely, if f is Riemann-integrable, then there exists a P such that  UpP, f, αq ´ Upf, αq ă 2 and a Q such that  Lpf, αq ´ LpP, f, αq ă 2 and thus their common refinement has UpP, f, αq ´ LpP, f, αq ă 

36 Example 16.6 (αpxq ‰ x). We can consider the case where αpxq is a “step” function, i.e. take the floor function:

αpxq “ txu

b Now, what is a f dα, if it exists? Note that the only intervals which contribute to the sums are those which contain an integer, since otherwise one would have ∆α “ 0. Further, we can refine any partition so that each interval ş i contains at most one integer. Then as long as f is continuous at each integer, one has that (say a, b P Z):

b b f dα “ fpiq. a i“a ż ÿ Example 16.7. As for another example, let’s take αpxq “ 2x. Then one can check that:

b b f dα “ 2 f dx ża ża if the integral exists at all. We will see more examples later.

16.2 Some Riemann-integrable functions Proposition 16.8. If f is monotonic and α is continuous, then f is Riemann-integrable.

Proof. Suppose that f is monotonically increasing. Then, in the Darboux sum, for the subinterval rxi´1, xis, one αpbq´αpaq has that Mi “ fpxiq and mi “ fpxi´1q. Take P to be a partition where ∆αi “ n . Then,

αpbq ´ αpaq n pαpbq ´ αpaqqpfpbq ´ fpaq UpP, f, αq ´ LpP, f, αq “ pfpx q ´ fpx qq “ n i i´1 n i“1 ÿ For large n this becomes arbitrarily small.

Proposition 16.9. If f is continuous then f is Riemann-integrable for any α. Proof. Given  ą 0, choose some η ą 0 such that

pαpbq ´ αpaqqη ă 

Since f is continuous on ra, bs and ra, bs is compact, it is uniformly continuous, so there exists some δ ą 0 such that

|fpxq ´ fpyq| ă η for |x ´ y| ă δ. Now, take P to be a partition where xi`1 ´ xi ă δ, so that

Mi ´ mi ď η and so UpP, f, αq ´ LpP, f, αq ď η ∆α “ ηpαpbq ´ αpaqq ă  ÿ

Proposition 16.10. Suppose f has finitely many points of discontinuity, and α is continuous at every discontinuous point of f. Then f is Riemann-integrable.

Proof. The idea is as follows: where f is continuous we can bound Mi ´ mi. Where α is continuous we can bound αpxiq ´ αpxi´1q. We formalize this as follows. Let p1, . . . , pk be the points of discontinuity for f, and find intervals rci, dis such that pi is the only point of discontinuity in it and such that i αpdiq ´ αpciq ă . Let M be an upper bound on f on ra, bs. Then one has that the Darboux sum on this part is less than ř

pMi ´ miqpαpdiq ´ αpciqq ď 2M αpdiq ´ αpciq “ 2M ÿ ÿ 37 Let K “ ra, bs ´ rci, dis. This is compact, so f is uniformly continuous on this interval, so find a δ such that |fpxq ´ fpyq| ă  if |x ´ y| ă δ. Now, there is a partition P such that Ť (a) ci and di are in P (b) no point in pci, diq is in P (c) every interval not rci, dis has length less than δ

Then, one has that, following the proof of the previous proposition,

UpP, f, αq ´ LpP, f, αq ď pαpbq ´ αpaqq ď 2M

Proposition 16.11. Let f be Riemann-integrable, and fpra, bsq Ă I for some closed interval I. Suppose that g : I Ñ R is continuous. Then h :“ g ˝ f is Riemann-integrable, with respect to any α. Proof. Fix an  ą 0. Now, consider the sum we want to make small for some partition P (i.e. smaller than ):

h h UpP, h, αq ´ LpP, h, αq “ pMi ´ mi qpαpxiq ´ αpxi´1qq ÿ h h f f where the Mi , mi correspond to h “ g ˝ f. Let Mi and mi be the corresponding numbers for f. First, we use uniform continuity of g. There is positive δ ą 0 such that |gpxq ´ gpyq| ă  if |x ´ y| ď δ. Then, if f f h h Mi ´ mi ă δ, i.e. |fpxq ´ fpyq| ă δ for x, y P rxi´1, xis, one has that Mi ´ mi ă . f f f f However, if Mi ´ mi ě δ, since f is Riemann-integrable, the intervals on which Mi ´ mi ě δ have to be “short 2 enough.” We can choose a partition P “ tx0, . . . , xnu such that UpP, f, αq ´ LpP, f, αq ă δ . Then we have

f f 2 δ ∆αi ď pMi ´ mi q∆α ă δ ÿ ÿ ∆αi ă δ

ÿf f where the sum is only taken over intervals where Mi ´ mi ě δ. Note that this says that the intervals on which f f h h Mi ´mi ě δ must be of total weighted length less than δ. Let K be an upper bound on |g|, so that Mi ´mi ď 2K. Then, h h pMi ´ mi q∆αi ă 2Kδ. Putting these two cases together, and takingÿδ such that δ ă , one has that

UpP, h, αq ´ LpP, h, αq ď pαpbq ´ αpaqq ` 2Kδ ă pαpbq ´ αpaq ` 2Kq

17 Lecture 17

17.1 Properties of the integral

Proposition 17.1 (Linearity properties). Let f, g be Riemann integrable and let c P R, and α continuous at c. Then the following are also Riemann-integrable and have values:

b b cf dα “ cf dα ża ża b b b f ` g dα “ f dα ` g dα ża ża ża 1 1 Proof. These follow directly from the definition. Let Mi and mi correspond to f and Mi and mi correspond to g.

38 Then one has that for any P ,

1 UpP, f ` g, αq ď pMi ` Mi q∆αi “ UpP, f, αq ` UpP, g, αq ÿ The inequality is because the maximum of f and g may not occur at the same point. Simialrly,

LpP, f ` g, αq ě LpP, f, αq ` LpP, g, αq and since Upf, αq “ Lpf, αq and Upg, αq “ Lpg, αq, one has that Upf ` g, αq “ Lpf ` g, αq.

Proposition 17.2 (Bounds of integration). If f is Riemann-integrable and c P ra, bs, then the following as also Riemann-integrable and one has: c b b f dα ` f dα “ f dα ża żc ża Proof. Any partition of ra, bs can be refined to include the point c. The rest is left as an exercise.

Proposition 17.3 (Comparison). Let f, g be Riemann integrable such that fpxq ď gpxq on ra, bs. Then

b b f dα ď g dα ża ża Further, if |fpxq| ď M, then b | f dα| ď Mpαpbq ´ αpaqq ża Proof. Since UpP, f, αq ď UpP, g, αq for every partition P , taking inf over all P yields the result. For the second statement, then one has UpP, f, αq ď Mpαpbq ´ αpaq and similarly for LpP, f, αq, for any P .

Proposition 17.4 (Linearity in the weights). If f is Riemann integrable with respect to α1 and also α2, then the following are also Riemann-integrable:

b b b f dpα1 ` α2q “ f dα1 ` f dα2 ża ża ża b b f dpcαq “ c f dα ża ża Proof. This is immediate.

Proposition 17.5 (Closure under products). If f is Riemann integrable, and g is Riemann integrable, then fg is Riemann integrable. Proof. By a previous proposition, if f is Riemann-integrable, then f 2 is. Then note that pf ` gq2 “ f 2 ` 2fg ` g2 1 2 2 2 and so fg “ 2 ppf ` gq ´ f ´ g q, completing the proof. Proposition 17.6 (Triangle inequality). If f is Riemann integrable, then |f| is, and

b b | f dα| ď |f| dα ża ża Proof. For the first statement, note that since gpxq “ |x| is Riemann-integrable, if f is Riemann-integrable, then b b so is |f|. For the second statement, suppose a f dα ě 0. Then since f ď |f| the statement follows. If a f dα ď 0, then since ´f ď |f|, the statement follows. ş ş Proposition 17.7 (Intermediate value theorem for integrals or “average value” theorem). Let f : ra, bs Ñ R be continuous. Then there is an x P ra, bs such that

1 b fpxq “ f dx b ´ a ża

39 Proof. Let M and m be the minimum and maximum values of f on ra, bs. Then,

1 b m ď f dx ď M b ´ a ża Now apply the intermediate value theorem.

Proposition 17.8. Suppose α increases monotonically, is everywhere differentiable, and α1 is Riemann integrable. Then f is Riemann integrable with respect to α if and only if fα1 is Riemann integrable with respect to x. In this case, b b f dα “ fα1 dx ża ża Proof. It suffices to show the equality of the upper and lower integrals of f dα and fα1 dx. We will do the upper; the lower follows in the same way. Since α1 is Riemann-integrable, for every  ą 0 there is a P such that

UpP, α1, xq ´ LpP, α1, xq ă 

By the mean value theorem, we can find points ci P rxi´1, xis such that

1 1 ∆αi “ αpxiq ´ αpxi´1q “ α pciqpxi ´ xi´1q “ α pciq∆xi

For any ti P rxi´1, xis one has:

1 1 α1 α1 |α ptiq ´ α pciq|∆xi ď pMi ´ mi q∆xi ă  ÿ ÿ Let M be a bound for |f|. Then one has:

1 1 1 | fptiq∆αi ´ fptiqα ptiq∆xi| ď |fptiq||∆αi ´ α ptiq∆xi| ď M |∆αi ´ α ptiq∆xi|

ÿ ÿ 1 ÿ1 ÿ ď Mp|∆αi ´ α pciq∆xi| ` |α pciq∆xi ´ αptiq∆xi|q ă Mp ` 0q “ M Lemma 17.9. Let c ą 0. Then |x ` y| ă c if and only if x ă c ` y and y ă c ` x. So, 1 1 fptiq∆αi ď fptiqα ptiq∆xi ` M ď UpP, fα , xq ` M

ÿ 1 ÿ fptiqα ptiq∆xi ď fptiq∆αi ` M ď UpP, f, αq ` M

Since this is true for any choiceÿ of ti, in particularÿt ` i “ Mi, one has:

UpP, fα1, xq ď UpP, f, αq ` M

UpP, f, αq ď UpP, fα1, xq ` M So, |UpP, fα1, xq ´ UpP, f, αq| ă M Since everything above is true for a refinement,

|Upfα1, xq ´ Upf, αq| ă M and so UpF, α1, xq “ Upf, αq

Proposition 17.10. Suppose φ : rA, Bs Ñ ra, bs is strictly increasing and continuous. Suppose α is monotonically

40 increasing on ra, bs and f : ra, bs Ñ R is Riemann integrable with respect to α. Define by change of variables:

βpxq “ αpφpxqq

gpxq “ fpφpxqq Then g is Riemann integrable with respect to beta and

B b g dβ “ f dα żA ża Proof. Since φ is strictly increasing, a partiton of rA, Bs gives a partition of ra, bs and vice versa. Let them be P and Q respectively. Then one finds that

UpP, g, betaq “ UpQ, f, αq and likewise, and the result follows.

Corollary 17.11. Suppose φ1 is Riemann integrable. Then,

b B fpxq dx “ fpφpyqqφ1pyq dy ża żA Proof. Take αpxq “ x and β “ φ above.

18 Lecture 18

18.1 The fundamental theorem of calculus In this section we will take αpxq “ x.

Theorem 18.1 (The fundamental theorem of calculus I). Let f be Riemann integrable on ra, bs. Define a function F on ra, bs by: x F pxq “ fptq dt ża Then F is continuous on ra, bs. Furthermore, f is continuous at a point x0 if and only if F is differentiable at x0 and: 1 F px0q “ fpx0q

Proof. First we will show that F is uniformly continuous, and thus continuous. Let M be a bound on |f|. Then,

y |F pyq ´ F pxq| “ | fptq dt| ď M|x ´ y| żx  Thus, if |x ´ y| ă M , then |F pxq ´ F pyq| ă , finishing the claim. Now, suppose f is continuous at x0. For any  ą 0 there is δ ą 0 such that |fptq ´ fpx0q| ă  for |t ´ x0| ă δ. Then for any t P px0 ´ δ, x0 ` δq with t ‰ x0 and by the intermediate value theorem for integrals, there is a c P pt, sq such that F ptq ´ F px q 1 t 0 “ fpxq dx “ fpcq t ´ x t ´ x 0 0 żs so F ptq ´ F px0q | ´ fpx0q| “ |fpcq ´ fpx0q| ă  t ´ x0 Thus one has that 1 F ptq ´ F px0q F px0q “ lim “ fpx0q tÑx0 t ´ x0

41 as desired. One obtains the converse by reversing these arguments.

Theorem 18.2 (The fundamental theorem of calculus II). If f is Riemann integrable on ra, bs which has an antiderivative, i.e. a function F such that f 1 “ f, then

b fpxq dx “ F pbq ´ F paq ża for any antiderivative F .

Proof. Fix  ą 0. Let P be a partition such that UpP, fq ´ LpP, fq ă . Applying the mean value theorem to each interval, there is some ti P rxi´1, xis such that

1 F pxiq ´ F pxi´1q “ F ptiq∆xi “ fptiq∆xi

And so

fptiq∆xi “ F pbq ´ F paq

b ÿ Since both the above sum and a fpxq is between LpP, fq and UpP, fq, one has that

ş b |F pbq ´ F paq ´ fpxq dx| ă  ża

Corollary 18.3 (Substitution). Let u be a differentiable function on pa, bq such that u1 is continuous, and I be an open interval such that upra, bsq Ă I. If f is continuous on I, then f ˝ u is continuous on J and

b upbq fpupxqqu1pxq dx “ fpuq du ża żupaq Proof. Let F be as in the fundamental theorem. Then if G “ F ˝ u then by the chain rule one has that G1pxq “ F pupxqqu1pxq. Now one has

b b upbq fpupxqqu1pxq dx “ G1pxq dx “ Gpbq ´ GpaqF pupbqq ´ F pupaqq “ fpuq du ża ża żupaq

Corollary 18.4 (Integration by parts). Suppose F and G are differentiable functions on ra, bs and their derivatives f, g are Riemann integrable. Then

b b F pxqgpxq dx “ F pbqGpbq ´ F paqGpaq ´ fpxqGpxq dx ża ża Proof. Put Hpxq “ F pxqGpxq. Then H1pxq “ F pxqgpxq`fpxqGpxq, and apply the second fundamental theorem.

19 Lecture 19

19.1 Things that aren’t true First, I want to introduce the most immediate version of convergence one can ask for of a function.

Definition 19.1. Let fn : I Ñ R be a sequence of functions. Then one says they converge pointwise to a function f : J Ñ R (for J Ă I) if lim fnpxq “ fpxq nÑ8

42 for every fixed x P J. Example 19.2 (Interchanging limits for sequences). Define m sm,n “ m ` n Then,

lim lim sm,n “ lim 0 “ 0 mÑ8 nÑ8 mÑ8 but

lim lim sm,n “ lim 1 “ 1 nÑ8 mÑ8 nÑ8 Example 19.3 (Limit of continuous functions is not continuous). Define

x2 fnpxq “ p1 ` x2qn and the partial sums n Fnpxq “ fnpxq k“1 ÿ Then, 0 x “ 0 lim Fnpxq “ nÑ8 #1 ` x2 x ‰ 0

2 1 since for fixed x, the fn give a geometric sequence with first term x and ratio 1`x2 , and so

2 1 2 x 1 “ 1 ` x 1 ´ 1`x2 This function is not continuous.

Example 19.4 (Limit of derivatives is not derivative of limit). Define

sinpnxq f pxq “ ? n n and define

fpxq “ lim fnpxq “ 0 nÑ8 Then we have that d lim fnpxq “ 0 dx nÑ8 but d ? lim fnpxq “ lim n cospnxq “ 8 nÑ8 dx nÑ8 Example 19.5 (Limit of integrals is not integral of limits). Define

2 2 n fnpxq “ n xp1 ´ x q

Then one has that

1 1 2 2 2 n 2 1 2 n`1 1 n fnpxq dx “ n xp1 ´ x q dx “ n p p1 ´ x q q| “ 2pn ` 1q 0 2n ´ 2 ż0 ż0 So, 1 lim fnpxq “ 8 nÑ8 ż0 43 However, lim n2xp1 ´ x2qn “ 0 nÑ8 for x P r0, 1s, and so 1 lim fnpxq “ 0 nÑ8 ż0 19.2 Uniform convergence

Definition 19.6. Let fn be a sequence of functions which converges pointwise to f. Then we say fn converges uniformly to f on E if for every  ą 0 there is an N such that for n ą N, one has

|fnpxq ´ fpxq| ă  for every x P E.

Example 19.7. Take E “ p´1, 1q and define

n fnpxq “ p1 ´ |x|q

One can check that this converges pointwise to

1 x “ 0 fpxq “ #0 x ‰ 0

However, it does not converge uniformly. When x ‰ 0 one has that

n |fpxq ´ fnpxq| “ p1 ´ |x|q

1 However, for  “ 2 and any n, one can find x ‰ 0 such that

1 x2 ă 1 ´ n 2 c Example 19.8. Define 1 f pxq “ sinpnxq n n One can check that this converges to 0 pointwise. It converges uniformly.

Example 19.9. Take E “ r0, 1s and n fnpxq “ x This converges to 0 x ‰ 1 fpxq “ #1 x “ 1 This does not converge uniformly.

Each of the previous examples do not converge uniformly. I encourage you to do the following as exercises and check against these notes.

Example 19.10. Take E “ and define, N m smpnq “ m ` n This does not converge uniformly in E. For fixed n P E, the limit of the sequence is 1. However, m n | ´ 1| “ | | m ` n m ` n

44 1 For, say,  “ 2 and any m, there is some n such that n | | ą  m ` n for example, m ă n. Further, one can do this the other, i.e. take m snpmq “ m ` n

1 This also does not converge uniformly in E. For any fixed m P E the limit of the sequence is 0. However, for  “ 2 and any n, there is an m such that m 1 ă m ` n 2 for example, choose n ą m. Example 19.11. Define x2 fnpxq “ p1 ` x2qn and the partial sums n Fnpxq “ fnpxq k“1 ÿ so that 0 x “ 0 F pxq “ lim Fnpxq “ nÑ8 #1 ` x2 x ‰ 0 Then one has that 0 x “ 0 |F pxq ´ Fnpxq| “ #p1 ` x2q1´n x ‰ 0

1 The sequence cannot be uniformly continuous, since for say  “ 2 and any n 1 p1 ` x2q1´n ą 2 is equivalent to 2 ą p1 ` x2qn´1 which can be achieved by taking x such that ? x2 ă n´1 2 ´ 1 ? which is possible since n´1 2 ´ 1 ą 0. Example 19.12. Take E “ r0, 1s and define

2 2 n fnpxq “ n xp1 ´ x q so that lim n2xp1 ´ x2qn “ 0 nÑ8 However, one has that n2xp1 ´ x2qn ą 1 1 p1 ´ x2qn ą xp1 ´ x2qn ą n2 1 x2 ă 1 ´ n n2 c

45 20 Lecture 20

20.1 Basic criteria for uniform convergence

Proposition 20.1 (Cauchy criterion). Let E Ă R, and fn : E Ñ R. This sequence converges uniformly if and only if for every  ą 0 there exists an integer N such that for m, n ě N, one has

|fnpxq ´ fmpxq| ă 

Proof. Suppose fn converges uniformly to f. Then for every  there is an N such that for n ą N one has  |fnpxq ´ fpxq| ă 2 . Then, by the triangle inequality,

|fnpxq ´ fmpxq| ď |fnpxq ´ fpxq| ` |fpxq ´ fnpxq| ă  for n, m ą N. For the converse, using the usual Cauchy criterion, we know that fn converges pointwise to some function f. Let  ą 0 be arbitrary, and suppose the condition above holds. Then,

|fnpxq ´ fmpxq| ă  for n, m ą N. Taking m Ñ 8, one has the required condition for uniform convergence.

Proposition 20.2. Suppose fn converges pointwise to f on E. Then fn converges uniformly on E if and only if

lim sup |fnpxq ´ fpxq| “ 0. nÑ8 xPE

Proof. This is a rewording of the definition.

Proposition 20.3 (M-test). Suppose fn is a sequence on E and suppose that we have bounds (for x P E)

|fnpxq| ď Mn for all n. If Mn converges, then fn converges uniformly.

Proof. If Mřn converges, then forř any  ą 0 there is N such that if m, n ą N then

ř m Mi ă  i“n ÿ Then, one has m m | fipxq| ď Mi ă  i“n i“n ÿ ÿ

20.2 Uniform convergence and continuity

Proposition 20.4. Let X be a metric space, and let E Ă X. Let fn, f : E Ñ R, and fn Ñ f uniformly. Let p be a limit point of E, and suppose that

lim fnpxq “ An xÑp

Then lim An converges and lim lim fnpxq “ lim lim fnpxq xÑp nÑ8 nÑ8 xÑp i.e.

lim fpxq “ lim An xÑp nÑ8

46 Proof. FIx  ą 0. By uniform convergence, there is N such that for n, m ą N one has

|fnpxq ´ fmpxq| ă  for all x P E. Taking x Ñ p one has |An ´ Am| ď  and so the sequence An converges. Let its limit be A. Now, we want to show that lim fpxq “ A xÑp Note that

|fpxq ´ A| ď |fpxq ´ fnpxq| ` |fnpxq ´ An| ` |An ´ A|

We can make |fpxq ´ fnpxq| ă  for all x P E by uniorm continuity. We can also make |An ´ A| ă  by convergence. Finally we can make |fnpxq ´ An| ă  in a neighborhood of p and for large n. Together, one has that

|fpxq ´ A| ď 3 for x in a neighborhood of p (note n plays no role now), proving the result.

Corollary 20.5. Let fn Ñ f uniformly, and fn continuous. Then f is continuous.

Proposition 20.6. Let E “ ra, bs. Let α be monotonically increasing. Suppose fn : E Ñ R are integrable and fn Ñ f uniformly. Then f is integrable and

b b fn dα “ lim fn dα nÑ8 ża ża Proof. Let

n “ sup |fnpxq ´ fpxq| xPE

By uniform continuity n Ñ 0. Further one has that

fn ´ n ď f ď fn ` n and taking integrals b b pfn ´ nq dα ď Lpf, αq ď Upf, αq ď pf ` nq dα ża ża and thus b Upf, αq ´ Lpf, αq ď 2 n dα “ 2pαpbq ´ αpaqqn ża Taking the limit n Ñ 8 one has that f is integrable. Further we have that

b b | f dα ´ fn dα| ď npαpbq ´ αpaqq ża ża

Corollary 20.7 (Integration of series). Let fn be integrable and the series fpxq “ sumfnpxq converge uniformly. b n Then, a f dα “ a fn dα. ş ř ş

47 21 Lecture 21

21.1 Uniform convergence and differentiation

Proposition 21.1. Let E “ ra, bs and let fn be differentiable on E. Suppose fn converges pointwise, for some 1 point p P E. If fn converges uniformly on E, then fn converges uniformly on E, and

1 1 f pxq “ lim fnpxq nÑ8

Proof. First, we will show that f converges uniformly on E. Fix  ą 0 and choose N such that m, n ą N implies (by pointwise convergence)  f px q ´ f px q| ă n 0 m 0 2 and also (by uniform convergence)  |f 1 pxq ´ f 1 pxq| ă n m 2pb ´ aq for all x P E. Now, apply the mean value theorem to fn ´ fm at any two points x, y:

|x ´ y|  |fnpxq ´ fmpxq ´ fnpyq ` fmpyq| ď ď 2pb ´ aq 2

Then take y “ x0:

|fnpxq ´ fmpxq| ď |fnpxq ´ fmpxq ´ fnpx0q ` fmpx0q| ` |fnpx0q ´ fmpx0q| ď  and so fn converges uniformly. Let f be its limit, and take p P E. Define

fnpxq ´ fnppq φnpxq “ x ´ p

fpxq ´ fppq φpxq “ x ´ p 1 By definition of derivative, limxÑp φnpxq “ fnpxq. Using the previous inequality, one has that  |φnpxq ´ φmpxq| ď 2pb ´ aq for large n, m, so that φn converges uniformly for x ‰ p, and it must converge pointwise (and thus uniformly) to φ. Thus, by interchanging limits, one has that

1 lim φpxq “ lim fnpxq xÑp nÑ8 and the result follows.

21.2 An everywhere continuous but nowhere differentiable function

Example 21.2. Define φpxq “ |x| for x P r´1, 1s, and define it for the rest of R periodically, i.e. so that φpx ` 2q “ φpxq. Define 8 3 fpxq “ p qnφp4nxq 4 n“0 ÿ 3 n n This might help think of this function: each term p 4 q φp4 q “squeezes” the function φ horizontally by a factor of 1 3 n 4n and also squeezes it vertically by a factor of p 4 q . One can draw φ as a spiky function, and in each summand in fpxq, the spikes (discontinuous points) get shorter and also closer together. They have to get shorter in order for the function to converge, and they have to get closer together in order for the derivative to fail to exist everywhere.

48 More formally, since φ is bounded, the M-test for uniform convergence shows that this series converges uniformly, and so f is continuous. To see that it is not differentiable anywhere, take x P R. Notice that non-differentiable points of φ occur at the integers, so for any k we want to choose a δk such that φ has no integer points in rx, x ` δks. Explicitly, we choose 1 δ “ ˘ 4´k k 2 Define n n φp4 px ` δmqq ´ φp4 xq γn “ δm n when n ą m, δm4 is an even integer, so by periodicity, γn “ 0. Further, one has that for any s, t:

|φpsq ´ φptq| ď |s ´ t|

n m so for n ă m one has |γn| ă 4 , and for n “ m one has that γm “ 4 . Then, take m m´1 fpx ` δmq ´ fpxq 3 n m n 1 m “ p q γn ě 3 ´ 3 “ p3 ` 1q δm ˇ 4 ˇ 2 ˇ ˇ ˇn“0 ˇ n“0 ˇ ˇ ˇ ÿ ˇ ÿ As m Ñ 8 one has δm Ñˇ0. So that: ˇ ˇ ˇ ˇ ˇ ˇ 1 ˇ f 1pxq ě p3m ` 1q 2 for every m, so f is not differentiable at x.

21.3 Differentiation and integration of power series Proposition 21.3. Suppose the series 8 n fpxq “ anx n“0 ÿ has radius of convergence R. Then it converges uniformly on any r´R ` , R ´ s for any  ą 0. Further, f is continuous and differentiable in p´R,Rq, and one can differentiate term-by-term:

8 1 n´1 f pxq “ nanx n“1 ÿ and has radius of convergence R. Further, one can integrate term-by-term:

x 8 a fptq dt “ n xn`1 n ` 1 0 n“0 ż ÿ and has radius of convergence R.

Proof. One has that n n |anx | ď Mn :“ |an|pR ´ q and since Mn converges absolutely by the root test, the M-test tells us that the series converges absolutely. The rest of the facts follow from previous propositions. ř Proposition 21.4 (Abel’s theorem). Take 8 n fpxq “ anx n“0 ÿ Let R be the radius convergence, and suppose f converges at R. Then, f is continious at R, i.e.

8 n lim fpxq “ anR . xÑR n“0 ÿ

49 The same is true for ´R.

Proof. Without loss of generality assume R “ 1 (otherwise, we can replace x with x{R). Define

Sn “ a0 ` ¨ ¨ ¨ ` an

We can rewrite 8 n fpxq “ p1 ´ xq Snx n“0 ÿ  Let S “ lim Sn. For  ą 0, there is N such that for n ą N one has |S ´ Sn| ă 2 . Then,  |fpxq ´ S| “ |p1 ´ xq pS ´ Sqxn| ď p1 ´ xqp |S ´ S||x|n ` q ď  n n 2 ÿ ÿ So as near 1, one has that p1 ´ xq Ñ 0 and since  was arbitrary the result follows.

22 Lecture 22

22.1 The Stone-Weierstrass theorem

Definition 22.1. A family of real-valued functions on a set E is a subset of the set of all functions from E to R. A family A is said to be an algebra if it is closed under addition, multiplication, and scalar multiplication in R.A family A is said to be uniformly closed if for any sequence fn with fn P A, such that fn Ñ f uniformly, one has that f P A. Further, given any family A one can take its uniform closure, i.e. the smallest set family containing A that is uniformly closed.

Example 22.2. For example, the family of polynomial functions on E “ R is an algebra, since the addition and multiplication of polynomials are also polynomials. Also, the family of functions of the form P pxqex form an algebra, where P is any polynomial. Neither of these are uniformly closed, as we will see later.

Proposition 22.3. Let B be the uniform closure of A. Then B consists of all functions which arise as a uniformly convergent limit of functions in A.

Definition 22.4. Let A be a family of functions on a set E. Then A separates points if for any x1, x2 P E, there is a function f P A such that fpx1q ‰ fpx2q; and it vanishes at no point if for every x P E there is some f P A such that fpxq ‰ 0.

Example 22.5. The family of polynomials on R separates points: the function fpxq “ x separates all points. Further, it vanishes at no point, since the function fpxq “ 1 is nowhere zero

Example 22.6. The family of even polynomials on R does not separate points, since fpxq “ ´fpxq for every such f.

Theorem 22.7. Let A be an algebra of real continuous functions on a compact set K. If A separates points and vanishes at no point, then the uniform closure B consists of all real continuous functions on K.

Proof. This is a long proof. We will divide it into five lemmas.

Lemma 22.8 (Stone-Weierstrass theorem). Let A be an algebra of continuous functions ra, bs Ñ R which contains fpxq “ x. Then, B contains every continuous function on ra, bs.

˜ x´a Proof. First, note that we may as well assume that ra, bs “ r0, 1s, since one can take fpxq “ fp b´a q which has domain r0, 1s, find a sequence of polynomials P˜n converging to it, and take Pnpxq “ P˜nppb ´ aq{x ` aq. The idea is as follows: the Dirac delta “function” is a function δapxq which is “defined” by:

8 x “ a δapxq “ #0 x ‰ 1

50 It has the “property” that: 8 δapxqfpxq dx “ fpaq ż´8 I will disgress for a bit to justify this property. Since δa is not really a function, we want to express it as a limit (which does not actually converge) of functions of the form:

n 1 1 2 x P ra ´ n , a ` n s Qnpxq “ #0 else

The constant is chosen so that 8 Qn dx “ 1 ż´8 for all n. Then, one can verify that, if F is the antideriative of f, then

1 8 a` n 1 1 n F pa ` n q ´ F pa ´ n q Qnpxqfpxq dx “ f dx “ 1 1 ´8 2 a´ 1 ´ p´ q ż ż n n n Taking the limit as n Ñ 8, one obtains F 1paq “ fpaq as desired. Further, convolution gives:

8 pf ˚ δaqpxq “ fptqδapx ´ tq dt “ fpx ´ aq ż´8 In particular, if a “ 0: 8 pf ˚ δ0qpxq “ fptqδ0px ´ tq dt “ fpxq ż´8 Our strategy is as follows: the Dirac delta function is not really a function, so we will take a sequence of polynomial functions that “limit” to it. That sequence of functions is:

2 n cnp1 ´ x q x P r´1, 1s Qnpxq “ #0 x R r´1, 1s where cn is chosen such that 8 Qnpxq dx “ 1 ż´8 And we will claim that 8 Pnpxq :“ Qnpx ´ tqfptq dt ż´8 converge uniformly to f (where we extend f to R by zero). First we need to show that Pnpxq is actually a polynomial. To this end, notice that (since f is defined in r0, 1s)

1 Pnpxq “ Qnpx ´ tqfptq dt ż0 Using integration by parts, and letting F be the antiderivative of f, one has that

1 d P pxq “ Qpx ´ tqF ptq|1 ´ Q px ´ tqF ptq dt n 0 dt n ż0 b d The term Qpx ´ tqF ptq|a is a polynomial, and in the integral term the polynomial dt Qnpx ´ tq is of smaller degree than Qnpx ´ tq, so by induction one has that Pnpxq is a polynomial. Note that this relies on the fact that the truncated polynomial Qpx ´ tq is a polynomial for t P r0, 1s and x P r0, 1s Next, we want to show that Pn converges uniformly to f. Let us try to work out the expression first. All

51 integrals below are over all of R:

|Pnpxq ´ fpxq| “ Qnpx ´ tqfptq ´ Qnpx ´ tqfpxq dt ˇż ˇ ˇ ˇ ˇ ˇ since Qnptq dt “ 1. Let us do a change of variables:ˇ ˇ ş Qnpsqpfpx ´ sq ´ fpxqq ds ˇż ˇ ˇ ˇ ˇ ˇ In this expression we are mostly worried aboutˇ the integral near 0, whereˇ Qn becomes very large (recall that the Dirac delta function “takes on” the value 8 at 0), and it won’t be enough to just replace the expression fpx´sq´fpxq with a constant upper bound.

Our first step is to evaluate cn. We compute

1 1 p1 ´ x2qn dx “ 2 p1 ´ x2qn dx ż´1 ż0 At this point one wants to replace p1´x2qn with 1´nx2. One could try to work out the binomial theorem explicitly, but a short cut is to take the function p1 ´ x2qn ´ p1 ´ nxq whose derivative is 2nxp1 ´ p1 ´ x2qn´1q which is positive on p0, 1q, and since they are equal at 0, one has

p1 ´ x2qn ě 1 ´ nx on 0, 1 . Thus, r s ? 1 1 1{ n 4 1 p1 ´ x2qn dx ě 2 1 ´ nx dx ě 1 ´ nx dx “ ? ą ? 3 n n ż´1 ż0 ż0 ? ?1 so, cn ă n. The choice of taking the upper limit of n is a bit arbitrary, but it makes the expression work out nicely.

Next, we will analyze the behavior of Qn away from r´δ, δs. Since the function Qnpxq is decreasing in r0, 1s, one has that for any δ ą 0 and any x P rδ, 1s: ? 2 n Qnpxq ď np1 ´ δ q and thus Qn converges to zero uniformly in rδ, 1s. Further, let M be a bound on f. Now, by uniform continuity f (ra, bs is compact), for every  ą 0 there is a δ ą 0 such that |x ´ y| ă 2δ implies that |fpxq ´ fpyq| ă . Given these, we can evaluate:

´δ 1 δ Qnpsqpfpx ´ sq ´ fpxqq ds ď 2M Qnptq dt ` Qnptq dt `  Qnptq dt ˇż ˇ ˜ż´1 żδ ¸ ż´δ ˇ ˇ ˇ ˇ ? ˇ ďˇ 4M np1 ´ δ2qn `  ă 2 δ for large n, and since ´δ Qnptq dt ď 1, and we are done. ş Lemma 22.9. If f P B, then |f| P B (where B is any uniformly closed algebra).

Proof. fpKq Ă ra, bs Ă R. For any  ą 0, there is a polynomial

n P pxq “ a0 ` a1x ` ¨ ¨ ¨ ` anx

52 such that: |P pxq ´ |x|| ă  for any x P ra, bs. Then, the function n P pfq “ a0 ` a1f ` ¨ ¨ ¨ ` anf is in B and further one has that |P pfpxqq ´ |fpxq|| ă  so that |f| P B by closedness.

Lemma 22.10. If f P B and g P B, then maxpf, gq P B and minpf, gq P B (where B is any uniformly closed algebra).

Proof. This follows from the fact that:

f ` g |f ´ g| maxpf, gq “ ` 2 2

f ` g |f ´ g| minpf, gq “ ´ 2 2

Lemma 22.11. Suppose that A is an algebra of functions on any set E which separates points and vanishes at no point. Let x1 ‰ x2 P E and c1, c2 P R. Then A has a function such that fpx1q “ c1 and fpx2q “ c2.

Proof. This is best done as an exercise, but choose a functions gi which do not vanish at xi and h be a function such that hpx1q ‰ hpx2q. Then take

c g pxqphpxq ´ hpx qq c g pxqphpxq ´ hpx qq fpxq “ 1 1 2 ` 2 2 1 g1px1qphpx1q ´ hpx2qq g2px2qphpx2q ´ hpx1qq the key observation being that g1pxqphpxq ´ hpx2qq is nonzero at x1 and zero at x2 (we need h to separate x1 and x2). Also, notice that constant functions are not automatically in A, so we need to introduce the functions gi to multiply the constants by, i.e. we need gi to not vanish at xi.

The strategy is as follows. Let f be the continuous function we want to approximate. We will find for each point p P K a function that is within  of f on one side, i.e. no less than f ´ . Using compactness we will choose finitely many of these and take the maximum of the functions. We will do the same again on the other side. Lemma 22.12. Let A, B as in the theorem. Let f be continuous on K (where K is compact). Let p P K and  ą 0. There exists a function gp P B such that gpppq “ fppq and gppxq ą fpxq ´ .

Proof. For every q P K, we can find hq P B such that

hqppq “ fppq, hqpqq “ fpqq

Define the open set ´1 Jq :“ pf ´ hqq pp´8, q

The Jq as q P K form a cover, since q P Jq. Since K is compact there is a finite subcover indexed by I, and define the function

gp “ maxphyi qiPI

We will now show that given f : K Ñ R continuous, and  ą 0, there is a function h P B such that for x P K:

|hpxq ´ fpxq| ă 

53 From the previous lemma, for any p P K, we have gp such that

gppxq ą fpxq ´ 

Define ´1 Vp :“ pgp ´ fq pp´8, qq i.e. points such that

gppxq ă fpxq `  By compactness, this open cover has a finite subcover, and take

h “ minpgxi qiPI

This function h has the property that hpxq ą fpxq ´  hpxq ă fpxq `  for x P K.

54