Functional Analysis Lecture Notes Chapter 3. Banach

Home , Banach space, Lp space, Normed vector space

CHAPTER 3. BANACH SPACES

CHRISTOPHER HEIL

1. Elementary Properties and Examples Notation 1.1. Throughout, F will denote either the real line R or the complex plane C. All vector spaces are assumed to be over the ﬁeld F.

Definition 1.2. Let X be a vector space over the field F. Then a semi-norm on X is a function k · k: X → R such that (a) kxk ≥ 0 for all x ∈ X, (b) kαxk = |α| kxk for all x ∈ X and α ∈ F, (c) Triangle Inequality: kx + yk ≤ kxk + kyk for all x, y ∈ X. A norm on X is a semi-norm which also satisfies: (d) kxk = 0 =⇒ x = 0. A vector space X together with a norm k · k is called a normed linear space, a normed vector space, or simply a normed space.

Definition 1.3. Let I be a finite or countable index set (for example, I = {1, . . . , N} if finite, or I = N or Z if infinite). Let w : I → [0, ∞). Given a sequence of scalars x = (xi)i∈I , set 1/p |x |p w(i)p , 0 < p < ∞,  i kxkp,w =  Xi∈I  sup |xi| w(i), p = ∞, i∈I  where these quantities could be infinite. Then we set p `w(I) = x = (xi)i∈I : kxkp < ∞ . n o p p p We call `w(I) a weighted ` space, and often denote it just by `w (especially if I = N). If p p w(i) = 1 for all i, then we simply call this space ` (I) or ` and write k · kp instead of k · kp,w.

Date: April 11, 2006. These notes closely follow and expand on the text by John B. Conway, “A Course in Functional Analysis,” Second Edition, Springer, 1990. 1 2 CHRISTOPHER HEIL

p Exercise 1.4. Show that if 1 ≤ p ≤ ∞ then k · kp,w deﬁnes a semi-norm on `w, and it is a norm if w(i) > 0 for all i. p n In particular, if I = {1, . . . , n} then `w = F , and each choice of p and w gives a semi-norm or norm on Fn. Hints: The Triangle Inequality on `p is often called Minkowski’s Inequality. It is easy to prove if p = 1 or p = ∞. There are several ways to prove it for other p. One ways is to begin with p p p−1 kx + ykp = |xi + yi| = |xi + yi| |xi + yi| Xi∈I Xi∈I

p−1 p−1 ≤ |xi + yi| |xi| + |xi + yi| |yi|. Xi∈I Xi∈I Then apply H¨older’s Inequality to each sum using the exponent p0 on the ﬁrst factor and p 0 p−1 for the second (recall that p = p/(p − 1)). Then divide both sides by kx + ykp .

Definition 1.5. Let (X, Ω, µ) be a measure space. Given a measurable f : X → [−∞, ∞] (if F = R) or f : X → C (if F = C), set 1/p p  |f(x)| dµ(x) , 0 < p < ∞, kfk = ZX p  ess sup |f(x)|, p = ∞, x∈X  where these quantities could beinfinite. Define

p L (X) = f : X → [−∞, ∞] or C : kfkp < ∞ . n o Other notations for Lp(X) are Lp(µ), Lp(X, µ), Lp(dµ), Lp(X, dµ), etc. When we write Lp(Rn), it will be assumed that µ is Lebesgue measure on Rn, unless speciﬁcally stated otherwise. In this case we will write dx instead of dµ(x). p p The space `w(I) is a special case of L (X), where X = I and µ is a weighted counting measure on I.

p Exercise 1.6. Show that if 1 ≤ p ≤ ∞ then k · kp is a semi-norm on L (X), and it is a norm if we identify functions that are equal almost everywhere. The Triangle Inequality on Lp is often called Minkowski’s Inequality, and its proof is similar to the proof of Minkowski’s Inequality for `p.

Exercise 1.7. Show that every subspace of a normed space is itself a normed space (using the same norm).

Deﬁnition 1.8 (Distance). Let k · k be a norm on X. Then the distance from x to y in X is d(x, y) = kx − yk. CHAPTER 3. BANACH SPACES 3

Exercise 1.9. Show that d(·, ·) deﬁnes a metric on X (see Appendix A).

Since X has a metric and hence has an associated topology, all the standard topological notions (open/closed sets, convergence, etc.) apply to X. For convenience, we give some explicit deﬁnitions and facts relating to these topics for the setting of normed spaces.

Deﬁnition 1.10 (Convergence). Let X be a normed linear space (such as an inner product space), and let {fn}n∈N be a sequence of elements of X.

(a) We say that {fn}n∈N converges to f ∈ X, and write fn → f, if

lim kf − fnk = 0, n→∞ i.e., ∀ ε > 0, ∃ N > 0 such that n > N =⇒ kf − fnk < ε.

(b) We say that {fn}n∈N is Cauchy if

∀ ε > 0, ∃ N > 0 such that m, n > N =⇒ kfm − fnk < ε.

Exercise 1.11. Let X be a normed linear space. Prove the following. (a) Reverse Triangle Inequality: kfk − kgk ≤ kf − gk.

(b) Continuity of the norm: fn → f =⇒ k fnk → kfk.

(d) Continuity of scalar multiplication: fn → f and αn → α =⇒ αnfn → αf. (e) All convergent sequences are bounded, and the limit of a convergent sequence is unique. (f) Every Cauchy sequence is bounded. (g) Every convergent sequence is Cauchy.

Exercise 1.12. Let {fn}n∈N be a sequence of vectors in a normed space X. Show that if −n kfn − fn+1k < 2 for every n, then {fn}n∈N is Cauchy.

p p Exercise 1.13. Let `w(I) be the weighted ` space deﬁned in Exercise 1.3, where we assume p w(i) > 0 for all i ∈ I. Let {xn}n∈N be a sequence of vectors in `w(I), and let x be a vector in p `w(I). Write the components of xn and x as xn = (xn(1), xn(2), . . . ) and x = (x(1), x(2), . . . ).

(a) Prove that if xn → x (i.e., kx − xnkp,w → 0), then xn converges componentwise to x, i.e., for each fixed k we have limn→∞ xn(k) = x(k). (b) Prove that if I is finite then the converse is also true, i.e., componentwise convergence implies convergence with respect to the norm k · kp,w. (c) Prove that if I is infinite then componentwise convergence does not imply convergence w in the norm of `p (I). 4 CHRISTOPHER HEIL

It is not true in an arbitrary normed space that every Cauchy sequence must converge. Normed spaces which do have the property that all Cauchy sequences converge are given a special name.

Deﬁnition 1.14 (Banach Space). A normed space X is called a Banach space if it is complete, i.e., if every Cauchy sequence is convergent. That is,

{fn}n∈N is Cauchy in X =⇒ ∃ f ∈ X such that fn → f.

p p Exercise 1.15. Show that the weighted ` space `w(I) deﬁned in Exercise 1.3 is a Banach space if w(i) > 0 for all i ∈ I. p Hints: Consider the case I = N. Suppose that {xn}n∈N is a Cauchy sequence in `w. Each xn is a sequence of scalars. Write out the components of xn as

xn = (xn(1), xn(2), . . . ). Prove that for a ﬁxed component k we have

|xm(k) − xn(k)| ≤ C kxm − xnkp,w, where C is a fixed constant (determined by the weight and by k but independent of m and n). Conclude that with k fixed, {xn(k)}n∈N is a Cauchy sequence of scalars, and hence converges. Define x(k) = limn→∞ xn(k) and set x = (x(1), x(2), . . . ). Then we have constructed a candidate limit for the sequence {xn}n∈N. However, so far we only have that each indi- vidual component of xn converges to the corresponding component of x, i.e., xn converges componentwise to x. This is not enough: to complete the proof you must show that xn → x p in the norm of `w. Use the fact that {xn}n∈N is Cauchy together with the componentwise p convergence to show that kx − xnk`w → 0 as n → ∞. Compare this proof to the proof of Theorem 2.18 given below.

Exercise 1.16. Show that the space Lp(X) deﬁned in Example 1.5 is a Banach space if we identify functions that are equal almost everywhere. This is called the Riesz–Fisher Theorem. Hint: The argument is similar in spirit but more subtle than the one used to prove that p `w(I) is a Banach space. First ﬁnd a candidate limit and then show that the sequence converges in norm to this limit.

The next two exercises will be useful to us later.

Exercise 1.17. Let X be a normed linear space. If {fn}n∈N is a Cauchy sequence in X and there exists a subsequence {fnk }k∈N that converges to f ∈ X, then fn → f.

Solution ε N Choose any ε > 0. Since {fn}n∈ is Cauchy, there is an N such that kfm − fnk < 2 for m, ε n > N. Also, there is a k such that nk > N and kf − fnk k < 2 . Hence for n > N we have ε ε kf − f k ≤ kf − f k + kf − f k < + = ε. n nk nk n 2 2 CHAPTER 3. BANACH SPACES 5

Thus fn → f.

Exercise 1.18. Let {fn}n∈N be a Cauchy sequence in a normed space X. Show that there

exists a subsequence {fnk }k∈N such that −k ∀ k ∈ N, kfnk+1 − fnk k < 2 .

Solution Since {fn}n∈N is Cauchy, we can ﬁnd an n1 such that −1 m, n ≥ n1 =⇒ kfm − fnk < 2 .

Then we can ﬁnd an n2 > n1 such that −2 m, n ≥ n2 =⇒ kfm − fnk < 2 .

Continuing in this way, we inductively construct n1 < n2 < · · · such that for each k, −k m, n ≥ nk =⇒ kfm − fnk < 2 . −k In particular, since nk+1 > nk, we have kfnk+1 − fnk k < 2 .

Deﬁnition 1.19 (Convergent Series). Let {fn}n∈N be a sequence of elements of a normed ∞ linear space X. Then the series n=1 fn converges and equals f ∈ X if the partial sums N sN = n=1 fn converge to f, i.e., Pif P N

kf − sN k = f − fn → 0 as N → ∞.

Xn=1

Deﬁnition 1.20 (Absolutely Convergent Series). Let X be a normed space and let {fn}n∈N be a sequence of elements of X. If ∞

kfnk < ∞, Xn=1

then we say that the series n fn is absolutely convergent in X. P Note that the deﬁnition of absolute convergence does not by itself tell us that the series

n fn actually converges. That is always true if X is a Banach space, but need not be true Pif X is not complete.

Exercise 1.21. Let X be a Banach space and let {fn}n∈N be a sequence of elements of X. Prove that if n kfnk < ∞ then the series n fn does converge in X. Hint: You mPust show that the sequence ofPpartial sums {sN }N∈N converges. Since X is a Banach space, you just have to show that this sequence is Cauchy.

The converse of this exercise is also true, and is often a useful method for proving that a given normed space is a Banach space. 6 CHRISTOPHER HEIL

Proposition 1.22. Let X be a normed space. Prove that X is a Banach space if and only if every absolutely convergent series in X converges in X. Proof. ⇒. This is Exercise 1.21.

⇐. Suppose that every absolutely convergent series is convergent. Let {fn}n∈N be a

Cauchy sequence in X. By Exercise 1.18, there exists a subsequence {fnk }k∈N such that −k kfnk+1 − fnk k < 2 for every k. The series k(fnk+1 − fnk ) is absolutely convergent, because ∞ ∞P −k kfnk+1 − fnk k ≤ 2 = 1 < ∞. Xk=1 Xk=1

By hypothesis, we conclude that k(fnk+1 − fnk ) converges in X, say to f. In terms of the partial sums, this says that P k

fnk+1 − fn1 = (fnj+1 − fnj ) → f as k → ∞, Xj=1 or fnk → g = f +fn1 as k → ∞. Thus {fn}n∈N is a Cauchy sequence which has a subsequence that converges to g. It therefore follows from Exercise 1.17 that fn → g. Therefore X is complete.

Example 1.23 (The Harmonic Series). To illustrate the difference between convergence and absolute convergence, consider the one-dimensional case, i.e., X = F, the field of scalars. Let 1 1 N 1 xn = n . Then n xn = n n is the harmonic series, and the partial sums sN = n=1 n of this series are unPbounded.P Thus the harmonic series does not converge. Since sNP→ ∞, we 1 1 usually say that n n diverges to infinity, and write n n = ∞. n 1 On the other hand,P consider the alternating series Pn(−1) n . Since the terms alternate 1 signs and since n → 0, it follows that this series doesPconverge to a finite scalar (in fact, it converges to − ln 2). However, it does not converge absolutely.

Deﬁnition 1.24 (Unconditionally Convergent Series). Let X be a Banach space and let ∞ {fn}n∈N be a sequence of elements of X. The series f = n=1 fn is said to converge uncon- ∞ ditionally if every rearrangement of the series converges.PThat is, f = n=1 fn converges unconditionally if for each bijection σ : N → N the series P ∞

fσ(n) Xn=1 converges.

∞ Remark 1.25. It is not obvious, but it can be shown that if n=1 fσ(n) is unconditionally convergent, then every rearrangement of the series must convergePto the same sum, i.e., there ∞ is a single f such that f = n=1 fσ(n) for every permutation σ. P CHAPTER 3. BANACH SPACES 7

∞ Exercise 1.26. Let X be a Banach space. Prove that if a series f = n=1 fn converges absolutely, then it converges unconditionally. P

Remark 1.27. In finite dimensions the converse to Exercise 1.26 is true, i.e., if X is finite- ∞ dimensional and a series f = n=1 fn converges unconditionally, then it converges absolutely. However, this fails in infinitePdimensions.

Example 1.28 (The Harmonic Series Revisited). To illustrate the importance of uncondi- n 1 tional convergence, again consider X = F and the alternating series n(−1) n . We know that this series converges, but does not converge absolutely. P 1 Now consider what happens if we change the order of summation. Let pn = 2n and 1 qn = 2n+1 , i.e., the pn are the positive terms from the alternative series and the qn are the absolute values of the negative terms. Each series n pn and n qn diverges. Hence there must exist an m1 > 0 such that P P

p1 + · · · + pm1 > 1.

Then, there must exist an m2 > m1 such that

p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 > 2. Continuing in this way, we see that

p1 + · · · + pm1 − q1 + pm1+1 + · · · + pm2 − q2 + · · · n 1 is a rearrangement of n(−1) n which diverges to +∞. In the same way, wePcan construct a rearrangement which diverges to −∞, which converges to any given real number r, or which simply oscillates without ever converging. Moreover, the same can be done for any series of real scalars which converges conditionally.

The following is an equivalent formulation of unconditional convergence.

Proposition 1.29. Let X be a Banach space and let {fn}n∈N be a sequence of elements ∞ of X. Then the series f = n=1 fn converges unconditionally if and only if it converges with respect to the net of ﬁnitePsubsets of N, i.e., if

∀ ε > 0, ∃ ﬁnite F0 ⊆ N such that ∀ ﬁnite F ⊇ F0, f − fn < ε.

nX∈F

Deﬁnition 1.30 (Topology). Let X be a normed linear space. (a) The open ball in X centered at x ∈ X with radius r > 0 is

Br(x) = B(x, r) = {y ∈ X : kx − yk < r}. (b) A subset U ⊆ X is open if

∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U. (c) A subset F ⊆ X is closed if X \ F is open. 8 CHRISTOPHER HEIL

Deﬁnition 1.31 (Limit Points, Closure, Density). Let X be a normed linear space and let A ⊆ X.

(a) A point f ∈ A is called a limit point of A if there exist fn ∈ A with fn =6 f such that fn → f. (b) The closure of A is the smallest closed set A¯ such that A ⊆ A¯. Speciﬁcally, A¯ = {F ⊆ X : F is closed and F ⊇ A}. T (c) We say that A is dense in X if A¯ = X.

Exercise 1.32. (a) The closure of A equals the union of A and all limit points of A:

A¯ = A ∪ {x ∈ X : x is a limit point of A} = {z ∈ X : ∃ yn ∈ A such that yn → z}.

(b) If X is a normed linear space, then the closure of an open ball Br(x) is the closed ball Br(x) = {y ∈ X : kx − yk ≤ r}. (c) Prove that A is dense if and only if ∀ x ∈ X, ∀ ε > 0, ∃ y ∈ A such that kx − yk < ε.

Example 1.33. The set of rationals Q is dense in the real line R.

Exercise 1.34. Let X be a normed linear space and let F ⊆ X. Then F is closed ⇐⇒ F contains all its limit points.

Solution ⇒. Suppose that F is closed but that there exists a limit point f that does not belong to F . By deﬁnition, there must exist fn ∈ F such that fn → f. However, f ∈ X \ F , which is open, so there exists some r > 0 such that B(f, r) ⊆ X \ F . Yet there must exist some fn with kf − fnk < r, so this fn will belong to X \ F , which is a contradiction. ⇐. Exercise.

Remark 1.35. Some authors make the restriction that, when dealing with normed spaces, the terminology “subspace” is used only for closed subspaces. Other authors use the terminology “linear manifold” to denote a subspace that need not be closed. To avoid ambiguity, we will use the following terminology.

Deﬁnition 1.36. (a) A subset Y of a vector space X is a subspace of X if it is closed under both vector addition and scalar multiplication, i.e., if for all u, v ∈ Y and α, β ∈ F we have αu + βv ∈ Y . (b) A subset Y of a normed linear space X is a closed subspace of X if it is a subspace and it is closed with respect to the norm of X. CHAPTER 3. BANACH SPACES 9

Exercise 1.37. In ﬁnite dimensions, all subspaces are closed sets (this will be proved in Proposition 3.5). This exercise demonstrates that, in inﬁnite dimensions, subspaces need not be closed sets. (a) Fix 1 ≤ p ≤ ∞. Prove that

c00 = {x = (x1, . . . , xN , 0, 0, . . . ) : N > 0, x1, . . . , xN ∈ F} p p is a subspace of ` (N) that is not closed (with respect to the ` -norm). Prove that c00 is dense in `p(N) if p < ∞, but that it is not dense in `∞(N). (b) Deﬁne ∞ c0 = {x = (xk)k=1 : lim xk = 0}. k→∞ ∞ ∞ Prove c0 is a closed subspace of ` (N) (in ` -norm). Prove that c0 is the closure of c00 (under the `∞-norm).

n (c) Fix 1 ≤ p ≤ ∞. Prove that Cc(R ), the space of continuous, compactly supported n p n n functions on R , is a subspace of L (R ) that is not closed. Prove that Cc(R ) is dense in Lp(Rn) if p < ∞, but not if p = ∞. Hints: For a continuous function we have kfk∞ = sup |f(x)|. Hence, if {fn}n∈N is a sequence of continuous functions in L∞(Rn) that converges in L∞-norm, then it converges uniformly. From undergraduate real analysis, we know that the limit of a uniformly convergent sequence of continuous functions is continuous.

n n (d) Let C0(R ) be the set of all continuous functions f : R → F such that lim f(x) = 0. (1.1) |x|→∞ More precisely, (1.1) means that for every ε > 0 there exists a compact set K such that n ∞ n |f(x)| < ε for all x ∈/ K. Prove that C0(R ) is a closed subspace of L (R ) (closed in ∞ n n ∞ L -norm). Prove that C0(R ) is the closure of Cc(R ) (under the L -norm). n n (e) Let Cb(R ) be the set of all bounded, continuous functions f : R → F. Prove that n ∞ n Cb(R ) is a closed subspace of L (R ). (f) Fix 1 ≤ p ≤ ∞, and let E be a (Lebesgue) measurable subset of Rn. Let M = {f ∈ Lp(Rn) : supp(f) ⊆ E}. Prove that M is a closed subspace of Lp(Rn).

Remark 1.38. (a) We took the domain in the preceding exercise to be Rn just for convenience; the deﬁnitions of the spaces Cc, C0, Cb can be extended to domains that are more general topological spaces.

(b) Beware that some authors use the notation C0 for the space that we are calling Cc!

Exercise 1.39. Let X be a Banach space and let M be a subspace of X. Then M is itself a Banach space (using the norm from X) if and only if M is closed. 10 CHRISTOPHER HEIL

Solution ⇒. Suppose that M is a Banach space. Let f be any limit point of M, i.e., suppose fn ∈ M and fn → f. Then {fn} is a convergent sequence in X, and hence is Cauchy in X. Since each fn belongs to M, it is also Cauchy in M. Since M is a Banach space, {fn} must therefore converge in M, i.e., fn → g for some g ∈ M. However, limits are unique, so f = g ∈ M. Therefore M contains all its limit points and hence is closed. ⇐. Exercise.

Notation 1.40 (Notation for Closed Subspaces). Since we will often deal with closed subspaces of a Banach space, we declare that the notation M ≤ X means that M is a closed subspace of the Banach space X.

Exercise 1.41. Find examples of normed spaces that are not Banach spaces. Hint: Look for examples of normed spaces Y which are subspaces of a larger Banach space X.

Remark 1.42. Is every normed space Y a subspace of a larger Banach space? The answer is yes, given a normed space Y it is always possible to construct a Banach space X ⊇ Y such that the norm on X, when restricted to Y , is the same as the norm on Y , and Y is dense in X with respect to that norm. This space is called the completion of Y .

Deﬁnition 1.43. Suppose that X is a normed linear space with respect to a norm k · ka and also with respect to another norm k · kb. Then we say that these norms are equivalent if there exist constants C1, C2 > 0 such that

∀ f ∈ X, C1 kfka ≤ kfkb ≤ C2 kfka. (1.2)

Observe that equation (1.2) can be rearranged to read 1 kfk ≤ kfk ≤ 1 kfk . C2 b a C1 b

Exercise 1.44. Let X be a vector space. If k · ka and k · kb are two norms on X, deﬁne k · ka ∼ k · kb if k · ka and k · kb are equivalent. Prove that ∼ is an equivalence relation on the class of norms on X.

The next result will show that equivalent norms deﬁne the same topology and the same convergence criterion.

Proposition 1.45. Let k·ka and k·kb be two norms on a vector space X. Then the following statements are equivalent.

(a) k · ka and k · kb are equivalent norms.

(b) k · ka and k · kb deﬁne the same topologies on X. That is, if U ⊆ X, then U is open with respect to k · ka if and only if it is open with respect to k · kb. CHAPTER 3. BANACH SPACES 11

lim kf − fnka = 0 ⇐⇒ lim kf − fnkb = 0. n→∞ n→∞ a b Proof. (b) ⇒ (a). Assume that statement (b) holds. Let Br (f) and Br(f) denote the open a balls of radius r centered at f ∈ X with respect to k · ka and k · kb. Since B1 (0) is open a with respect to k · ka, the hypothesis that statement (b) holds implies that B1 (0) is open a with respect to k · kb. Therefore, since 0 ∈ B1 (0), there must exist some r > 0 such that b a Br(0) ⊂ B1 (0). Now choose any f ∈ X and any ε > 0. Then

(r − ε) f b a ∈ Br(0) ⊆ B1 (0), kfkb so (r − ε) f < 1. kfkb a

Rearranging, this implies (r − ε) kfk a < kfkb. Since this is true for every ε, we conclude that r kfka ≤ kfkb. A symmetric argument, interchanging the roles of the two norms, shows that there exists an s > such that kfkb ≤ s kfka for every f ∈ X. Hence the two norms are equivalent The remaining implications are exercises. Hints on (c) ⇒ (a): Suppose that statement (c) holds. Show that this implies that the function ν : X → R given by ν(f) = kfkb is continuous with respect to the norm k·ka. Since (−1, 1) is an open subset of R, it follows that ν−1(−1, 1) is an open subset of X (with respect −1 b −1 to k · ka). Since 0 ∈ ν (−1, 1), there must exist an r > 0 such that Br(0) ⊂ ν (−1, 1). −1 a But ν (−1, 1) = B1 (0), so the remainder of the proof proceeds exactly like the proof of (b) ⇒ (a).

2. Linear Operators on Normed Spaces

Definition 2.1 (Notation for Operators). Let X, Y be vector spaces. Let T : X → Y be a function (= operator = transformation) mapping X into Y . We write either T (f) or T f to denote the image of an element f ∈ X. (a) T is linear if T (αf + βg) = αT (f) + βT (g) for every f, g ∈ X and α, β ∈ F. (b) T is injective if T (f) = T (g) implies f = g. (c) The kernel or nullspace of T is ker(T ) = {f ∈ X : T (f) = 0}. (c) The range of T is range(T ) = {T (f) : f ∈ X}. (d) The rank of T is the dimension of its range, i.e., rank(T ) = dim(range(T )). In particular, T is finite-rank if range(T ) is finite-dimensional. (d) T is surjective if range(T ) = Y . 12 CHRISTOPHER HEIL

(e) T is a bijection if it is both injective and surjective.

(f) We use the notation I or IX to denote the identity map of a space X onto itself.

Deﬁnition 2.2 (Continuous and Bounded Operators). Let X, Y be normed linear spaces, and let L: X → Y be a linear operator.

(a) L is continuous at a point f ∈ X if fn → f in X implies Lfn → Lf in Y .

(b) L is continuous if it is continuous at every point, i.e., if fn → f in X implies Lfn → Lf in Y for every f. (c) L is bounded if there exists a ﬁnite K ≥ 0 such that ∀ f ∈ X, kLfk ≤ K kfk. Note that kLfk is the norm of Lf in Y , while kfk is the norm of f in X. (d) The operator norm of L is kLk = sup kLfk. kfk=1

(e) We let B(X, Y ) denote the set of all bounded linear operators mapping X into Y , i.e., B(X, Y ) = {L: X → Y : L is bounded and linear}. If X = Y then we write B(X) = B(X, X). (f) If Y = F then we say that L is a functional. The set of all bounded linear functionals on X is the dual space of X, and is denoted X0 = B(X, F) = {L: X → F : L is bounded and linear}. Another common notation for the dual space is X ∗.

Note that since the norm on F is just absolutely value, the operator norm of a linear functional L ∈ X0 = B(X, F) is kLk = sup |Lf|. kfk=1

Exercise 2.3. Show that if T : X → Y is linear and continuous, then ker(T ) is a closed subspace of X and that range(T ) is a subspace of Y . Must range(T ) be a closed subspace?

Exercise 2.4. Let X, Y be normed linear spaces. Let L: X → Y be a linear operator. (a) L is injective if and only if ker L = {0}. (b) If L is a bijection then L−1 : Y → X is also a linear bijection. (c) L is bounded if and only if kLk < ∞. (d) If L is bounded then kLfk ≤ kLk kfk for every f ∈ X (note that three diﬀerent meanings of the symbol k · k appear in this statement!). CHAPTER 3. BANACH SPACES 13

(e) If L is bounded then kLk is the smallest value of K such that kLfk ≤ Kkfk holds for all f ∈ X. kLfk (f) kLk = sup kLfk = sup . kfk≤1 f6=0 kfk

Exercise 2.5. Show that B(X, Y ) is a subspace of the vector space V consisting of ALL functions A: X → Y . Moreover, show that the operator norm is a norm on the space B(X, Y ), i.e., (a) 0 ≤ kLk < ∞ for all L ∈ B(X, Y ), (b) kLk = 0 if and only if L = 0 (the zero operator that sends every element of X to the zero vector in Y ), (c) kαLk = |α| kLk for every L ∈ B(X, Y ) and every α ∈ F, (d) kL + Kk ≤ kLk + kKk for every L, K ∈ B(X, Y ).

The preceding exercise shows that B(X, Y ) is a normed linear space, and we will show in Theorem 2.18 that it is a Banach space if Y is a Banach space. However, in addition to vector addition and scalar multiplication operations, there is a third operation that we can perform with functions: composition.

Exercise 2.6. Prove that the operator norm is submultiplicative, i.e., prove if A ∈ B(X, Y ) and B ∈ B(Y, Z), then BA ∈ B(X, Z) and kBAk ≤ kBk kAk. (2.1)

In particular, when X = Y = Z, we see that B(X) is closed under compositions. The space B(X) is an example of an algebra.

Exercise 2.7. Let Fn be n-dimensional Euclidean space over F, under the Euclidean norm, and let Y be any normed linear space. Prove that if L: Fn → Y is linear, then L is bounded. n Hint: If x = (x1, . . . , xn) ∈ F then x = x1e1 +· · ·+xnen where {e1, . . . , en} is the standard basis for Fn. Use the Triangle and the Cauchy-Schwarz Inequalities.

Solution Given x ∈ Fn, we have n n n 1/2 n 1/2 kLxk = L(x e ) ≤ |x | kLe k ≤ |x |2 kLe k2 = C kxk, k k k k k k Xk=1 Xk=1 Xk=1 Xk=1

n 2 1/2 where C = k=1 kLekk . Hence L is bounded. P Remark 2.8. We will prove later that if X is any ﬁnite-dimensional vector space and k·k is any norm on X, then any linear function L: X → Y is bounded. To do this we will use the fact (that we will prove later) that all norms on a ﬁnite-dimensional space are equivalent. 14 CHRISTOPHER HEIL

The next lemma is a standard fact about continuous functions. L−1(U) denotes the inverse image of U ⊆ Y , i.e., L−1(U) = {f ∈ X : Lf ∈ U}.

Exercise 2.9. Let X, Y be normed linear spaces. Let L: X → Y be linear. Then L is continuous ⇐⇒ U open in Y =⇒ L−1(U) open in X . Solution ⇒. Suppose that L is continuous and that U is an open subset of Y . We will show that X \ L−1(U) is closed by showing that it contains all its limit points. −1 −1 Suppose that f is a limit point of X \L (U). Then there exist fn ∈ X \L (U) such that −1 fn → f. Since L is continuous, this implies Lfn → Lf. However, fn ∈/ L (U), so Lfn ∈/ U, −1 i.e., Lfn ∈ Y \ U, which is a closed set. Therefore Lf ∈ Y \ U, and hence f ∈ X \ L (U). Thus X \ L−1(U) is closed, so L−1(U) is open. ⇐. Exercise.

Theorem 2.10 (Equivalence of Bounded and Continuous Linear Operators). Let X, Y be normed linear spaces, and let L: X → Y be a linear mapping. Then the following statements are equivalent. (a) L is continuous at some f ∈ X. (b) L is continuous at f = 0. (c) L is continuous. (d) L is bounded. Proof. (c) ⇒ (d). Suppose that L is continuous but unbounded. Then kLk = ∞, so there must exist fn ∈ X with kfnk = 1 such that kLfnk ≥ n. Set gn = fn/n. Then kgn − 0k = kgnk = kfnk/n → 0, so gn → 0. Since L is continuous and linear, this implies Lgn → L0 = 0, and therefore kLgnk → k0k = 0. But 1 1 kLg k = kLf k ≥ · n = 1 n n n n for all n, which is a contradiction. Hence L must be bounded. (d) ⇒ (c). Suppose that L is bounded, so kLk < ∞. Suppose that f ∈ X and that fn → f. Then kfn − fk → 0, so

kLfn − Lfk = kL(fn − f)k ≤ kLk kfn − fk → 0,

i.e., Lfn → Lf. Thus L is continuous. The remaining implications are exercises.

Deﬁnition 2.11 (Isometries and Isometric Isomorphisms). Let X, Y be normed linear spaces and let L: X → Y be linear. CHAPTER 3. BANACH SPACES 15

(a) If kLfk = kfk for all f ∈ X then L is called an isometry or is said to be norm- preserving. (b) An isometry L: X → Y that is a bijection is called an isometric isomorphism. In this case we say that X and Y are isometrically isomorphic. Exercise 2.12. (a) Suppose that L: X → Y is an isometry. Prove that L is injective and ﬁnd kLk. (b) Find an example of an isometry that is not surjective. Contrast this with the fact that if A: Cn → Cn is linear, then A is injective if and only if it is surjective. (c) Prove that if L: X → Y is an isometric isomorphism, then L−1 : Y → X is also an isometric isomorphism.

Exercise 2.13 (Unilateral Shift Operators). Fix 1 ≤ p ≤ ∞. p p p (a) Define L: ` (N) → ` (N) by L(x) = (x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ ` (N). Prove that this left-shift operator is bounded, linear, surjective, not injective, and is not an isometry. Find kLk. p p p (b) Define R: ` (N) → ` (N) by R(x) = (0, x1, x2, x3, . . . ) for x = (x1, x2, . . . ) ∈ ` (N). Prove that this right-shift operator is bounded, linear, injective, not surjective, and is an isometry. Find kRk. (c) Compute LR and RL. Contrast this computation with the fact that in finite dimensions, if A, B : Cn → Cn are linear maps (hence correspond to multiplication by n × n matrices), then AB = I implies BA = I and conversely.

Deﬁnition 2.14 (Topological Isomorphisms). Let X, Y be normed linear spaces. If L: X → Y is a linear bijection such that both L and L−1 are bounded, then L is called a topological isomorphism. In this case we say that X and Y are topologically isomorphic.

Remark 2.15. We will see later that if X and Y are Banach spaces and L: X → Y is a bounded bijection, then L−1 is automatically bounded and hence L is a topological isomorphism. Thus, when X and Y are Banach spaces, every continuous invertible map is a topological isomorphism. Sometimes the abbreviation isomorphism or invertible map is used to mean a topological isomorphism, but it should be noted that these terms are ambiguous.

Exercise 2.16. Prove that if L: X → Y is a topological isomorphism, then 1 ∀ f ∈ X, kfk ≤ kLfk ≤ kLk kfk. kL−1k

Exercise 2.17. Let X be a Banach space and Y a normed linear space. Suppose that L: X → Y is bounded and linear. Prove that if there exists c > 0 such that kLfk ≥ ckfk for all f ∈ X, then L is injective and range(L) is closed. 16 CHRISTOPHER HEIL

The next theorem shows that B(X, Y ) is a Banach space whenever Y is a Banach space.

Theorem 2.18. If X is a normed space and Y is a Banach space, then B(X, Y ) is a Banach space.

Proof. Assume that {An}n∈N is a sequence of operators An ∈ B(X, Y ) that is Cauchy in operator norm. For any given f ∈ X, we have

kAmf − Anfk ≤ kAm − Ank kfk,

so we conclude that {Anf}n∈N is a Cauchy sequence of vectors in Y . Since Y is complete, this sequence must converge, say Afn → g ∈ Y . Define Af = g. This gives us a candidate limit operator A. Exercise: Show that A defined in this way is a linear operator. To show that A is bounded, first recall that all Cauchy sequences are bounded. Hence we must have C = sup kAnk < ∞. If f ∈ X, then since Anf → Af we have

kAfk = lim kAnfk ≤ sup kAnfk ≤ sup kAnk kfk = C kfk. n→∞ n∈N n∈N Hence A is bounded, and kAk ≤ C. Finally, we must show that An → A in operator norm. Fix any ε > 0. Then there exists an N such that ε m, n > N =⇒ kA − A k < . m n 2 Choose any f ∈ X with kfk = 1. Then since Amf → Af, there exists an m > N such that ε kAf − A fk < . m 2 Hence for any n > N we have ε ε kAf −A fk ≤ kAf −A fk+kA f −A fk ≤ kAf −A fk+kA −A k kfk < + = ε. n m m n m m n 2 2 Taking the supremum over all unit vectors, we conclude that kA − Ank ≤ ε for all n > N. Thus An → A.

Corollary 2.19. If X is a normed space, then its dual space X 0 = B(X, F) is a Banach space.

The next exercise deals with the problem of extending an operator deﬁned only a dense subspace to the entire space.

Exercise 2.20 (Extension of Bounded Operators). Let Y be a dense subspace of a normed space X, and let Z be a Banach space. (a) Suppose that L: Y → Z is a bounded linear operator. Show that there exists a unique bounded linear operator L˜ : X → Z whose restriction to Y is L. Prove that kL˜k = kLk. (b) Prove that if L: Y → range(L) is a topological isomorphism, then L˜ : X → range(L) is also. CHAPTER 3. BANACH SPACES 17

Solution (a) Fix any f ∈ X. Since Y is dense in X, there exist gn ∈ Y such that gn → f. Since L is bounded, we have kLgm − Lgnk ≤ kLk kgm − gnk. But {gn}n∈N is Cauchy in X, so this implies that {Lgn}n∈N is Cauchy in Z. Since Z is a Banach space, we conclude that there exists an h ∈ Z such that Lgn → h. Define L˜f = h. ˜ 0 0 To see that L is well-defined, suppose that we also had gn → f for some gn ∈ Y . Then 0 0 0 0 kLgn −Lgnk ≤ kLk kgn −gnk → 0. Since Lgn → h, it follows that Lgn = Lgn +(Lgn −Lgn) → h + 0 = h. Thus L˜ is well-defined. To see that L˜ is linear, suppose that f, g ∈ X are given and c ∈ F. Then there exist fn, gn ∈ Y such that fn → f and gn → g. Since cfn + gn → cf + g and

L(cfn + gn) = cLfn + Lgn → cL˜f + Lg˜ , by definition we have that L˜(cf + g) = cLf˜ + Lg˜ . To see that L˜ is an extension of L, suppose that g ∈ Y is fixed. If we set gn = g, then gn → g and Lgn → Lg, so by definition we have L˜g = Lg. Hence the restriction of L˜ to Y is L. Consequently, kL˜k = sup kLf˜ k ≥ sup kLf˜ k = sup kLfk = kLk. f∈X, kfk=1 f∈Y, kfk=1 f∈Y, kfk=1

Finally, suppose that f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → Lf˜ , so

kLf˜ k = lim kLgnk ≤ lim kLk kgnk = kLk kfk. n→∞ n→∞ Hence kL˜k ≤ kLk. Combining this with the opposite inequality derived above, we conclude that kL˜k = kLk. (b) Suppose that L: Y → range(Y ) is a topological isomorphism. We already know that L˜ : X → range(L˜) is bounded. We need to show that L˜ is injective, that L˜−1 : range(L˜) → X is bounded, and that range(L˜) = range(L). Fix any f ∈ X. Then there exist gn ∈ Y such that gn → f and Lgn → L˜f. Since L is a −1 topological isomorphism, we have by Exercise 2.16 that kgnk ≤ kL k kLgnk. Hence

kgnk kfk kLf˜ k = lim kLgnk ≥ lim = . n→∞ n→∞ kL−1k kL−1k Consequently, L˜ is injective and for any h ∈ range(L˜) we have kL˜−1hk ≤ kL−1k kL˜(L˜−1h)k = kL−1k khk. Therefore L˜−1 : range(L˜) → X is bounded. It remains only to show that the range of L˜ is the closure of the range of L. If f ∈ X, then by deﬁnition there exist gn ∈ Y such that gn → f and Lgn → Lf˜ . Hence L˜f ∈ range(L), so range(L˜) ⊂ range(L). 18 CHRISTOPHER HEIL

On the other hand, suppose that h ∈ range(L) Then there exist gn ∈ Y such that Lgn → h. −1 −1 −1 Since L˜ is bounded and L˜ extends L, we conclude that gn = L˜ (Lgn) → L˜ (h). Hence f = L˜−1(h), so f ∈ range(L˜).

3. Finite-Dimensional Normed Spaces In this section we will prove some basic facts about finite-dimensional spaces. First, recall that a finite-dimensional vector space has a finite basis, which gives us a natural notion of coordinates of a vector, which in turn yields a linear bijection of X onto Fn for some n. Example 3.1 (Coordinates). Let X be a finite-dimensional vector space over F. Then X has a finite basis, say B = {e1, . . . , en}. Every element of X can be written uniquely in this basis, say,

x = c1(x) e1 + · · · + cn(x) en, x ∈ X. Deﬁne the coordinates of x with respect to the basis B to be

c1(x) . [x]B =  .  . c (x)  n  n Then the mapping T : X → F given by x 7→ [x]B is, by deﬁnition of basis, a linear bijection of X onto Fn.

Since we already know how to construct many norms on Fn, by transferring these to X we obtain a multitude of norms for X.

p Exercise 3.2 (`w Norms on X). Let X be a ﬁnite-dimensional vector space over F and let B = {e1, . . . , en} be any basis. Fix any 1 ≤ p ≤ ∞ and any weight w : {1, . . . , n} → (0, ∞). Using the notation of Example 3.1, given x ∈ X deﬁne

n 1/p p p  |ck(x)| w(k) , 1 ≤ p < ∞, kxk = [x] = p,w B p,w  Xk=1  max |ck(x)| w(k), p = ∞. k   n Note that while we use the same symbol k · kp,w to denote a function on X and on F , by context it has diﬀerent meanings depending on whether it is being applied to an element of X or to an element of Fn. Prove the following.

(a) k · kp,w is a norm on X. n (b) x 7→ [x]B is a isometric isomorphism of X onto F (using the norm k · kp,w on X and n the norm k · kp,w on F ). CHAPTER 3. BANACH SPACES 19

(c) Let {xn}n∈N be a sequence of vectors in X and let x ∈ X. Prove that xn → x with respect to the norm k · kp,w on X if and only if the coordinate vectors [xn]B converge componentwise to the coordinate vector [x]B.

(d) X is complete in the norm k · kp,w. Now we can show that all norms on a ﬁnite-dimensional space are equivalent. Theorem 3.3. If X is a ﬁnite-dimensional vector space over F, then any two norms on X are equivalent.

Proof. Let B = {e1, . . . , en} be any basis for X, and let k · k∞ be the norm on X deﬁned in Exercise 3.2. Since equivalence of norms is an equivalence relation, it suﬃces to show that an arbitrary norm k · k on X is equivalent to k · k∞. Using the notation of Exercise 3.1, given x ∈ X we can write x uniquely as x = c1(x) e1 + · · · + cn(x) en. Therefore, n n

kxk ≤ |ck(x)| kekk ≤ kekk max |ck(x)| = C2 kxk∞, k Xk=1 Xk=1 n where C2 = k=1 kekk is a nonzero constant independent of x. It remainsPto show that there is a constant C1 > 0 such that C1 kxk∞ ≤ kxk for every x. First, let D = {x ∈ X : kxk∞ = 1} be the `∞-unit circle in X. Exercise: Show that D is compact (with respect to the norm k · k∞). Hints: Suppose that {xn}n∈N is a sequence of vectors in D. Then for each n, we have |ck(xn)| = 1 for some k ∈ {1, . . . , n}. Hence there must be some k such that |ck(xnj )| = 1 for inﬁnitely many j.

Since {c1(xnj )}j∈N is an inﬁnite sequence of scalars in the compact set {c ∈ F : |c| ≤ 1}, we can select a subsequence whose ﬁrst coordinates converge. Repeat for each coordinate, and remember that the kth coordinate is always 1. Hence we can construct a subsequence that converges to x ∈ D with respect to the `∞-norm. Our next goal is to show that D is also compact with respect to the norm k · k. Let {xn}n∈N be any sequence of vectors in D. Since D is compact with respect to k · k∞,

there exists a subsequence {xnk }k∈N and an x ∈ D such that kx − xnk k∞ → 0. But then

kx − xnk k ≤ C2 kx − xnk k∞ → 0, so {xnk }k∈N is a subsequence that converges to x ∈ X with respect to k · k. Hence D is compact with respect to k · k. Now, k · k is a continuous function with respect to the convergence criteria deﬁned by k · k (this is part (b) of Exercise 1.11). The set D is compact with respect to the topology deﬁned by k · k. A real-valued continuous function on a compact set must achieve a maximum and minimum on that set. Hence, there must exist constants m and M such that m ≤ kxk ≤ M for all x ∈ D. Since x ∈ D if and only if kxk∞ = 1, this implies that

∀ x ∈ X, m kxk∞ ≤ kxk ≤ M kxk∞. If we had m = 0, then this would imply that there is an x ∈ D such that kxk = 0. But then x = 0, which implies kxk∞ = 0, contradicting the fact that x ∈ D. Hence m > 0, so we can take C1 = m. 20 CHRISTOPHER HEIL

Consequently, from now on we need not specify the norm on a ﬁnite-dimensional vector space X—we can take any norm that we like whenever we need it.

Exercise 3.4. Let Fm×n be the space of all m × n matrices with entries in F. Fm×n is naturally isomorphic to Fmn. m×n n×k Prove that if k · ka is any norm on F , k · kb is any norm on F , and k · kc is any norm on Fm×k, then there exists a constant C > 0 such that m×n n×k ∀ A ∈ F , ∀ B ∈ F , kABkc ≤ C kAka kBkb.

Proposition 3.5. If M is a subspace of a ﬁnite-dimensional vector space X, then M is closed.

Proof. Let k · k be any norm on X. Suppose that xn ∈ M and that xn → y ∈ X. Set

M1 = span{M, y} = {m + cy : m ∈ M, c ∈ F}.

Then M1 is ﬁnite-dimensional subspace of X. Moreover, every element z ∈ M1 can be written uniquely as z = m(z) + c(z)y where m(z) ∈ M and c(z) is a scalar. For z ∈ M1 deﬁne

kzkM1 = km(z)k + |c(z)|.

Exercise: Show that k · kM1 is a norm on M1. Since k · k is also a norm on M1 and all norms on a ﬁnite-dimensional space are equivalent, we conclude that there is a constant C > 0 such that kzkM1 ≤ C kzk for all z ∈ M1. Since c(xn) = 0 for every n, we therefore have

|c(y)| = |c(y) − c(xn)| = |c(y − xn)|

≤ km(y − xn)k + |c(y − xn)|

= ky − xnkM1

≤ C ky − xnk → 0. Therefore c(y) = 0, so y ∈ M.

If X is any normed vector space and M is a ﬁnite-dimensional subspace of X, then a proof identical to the one used in the preceding proposition, except using the given norm k · k on X, shows that M is closed.

Exercise 3.6. Let X be a normed linear space. If M is a ﬁnite-dimensional subspace of X, then M is closed.

Exercise 3.7. Let X be a ﬁnite-dimensional normed space, and let Y be a normed linear space. Prove that if L: X → Y is linear, then L is bounded.

The following lemma will be needed for Exercise 3.9. CHAPTER 3. BANACH SPACES 21

Lemma 3.8 (F. Riesz’s Lemma). Let M be a proper, closed subspace of a normed space X. Then for each ε > 0, there exists g ∈ X with kgk = 1 such that

dist(g, M) = inf kg − fk > 1 − ε. f∈M

Proof. Choose any u ∈ X \ M. Since M is closed, we have

a = dist(u, M) = inf ku − fk > 0. f∈M

a Fix δ > 0 small enough that a+δ > 1 − ε. By deﬁnition of inﬁmum, there exists v ∈ M such that a ≤ ku − vk < a + δ. Set u − v g = , ku − vk and note that kgk = 1. Given f ∈ M we have h = v + ku − vk f ∈ M, so

u − v − ku − vk f ku − hk a kg − fk = = > > 1 − ε. ku − vk ku − vk a + δ

Exercise 3.9. Let X be a normed linear space. Let B = {x ∈ X : kxk ≤ 1} be the closed unit ball in X. Prove that if B is compact, then X is finite-dimensional. Hints: Suppose that X is infinite-dimensional. Given any nonzero e1 with ke1k ≤ 1, by 1 Lemma 3.8 there exists e2 ∈ X \span{e1} with ke2k ≤ 1 such that ke2 −e1k > 2 . Continue in this way to construct vectors ek such that {e1, . . . , en} are independent for any n. Conclude that X is infinite-dimensional.

Deﬁnition 3.10. We say that a normed linear space X is locally compact if for each f ∈ X there exists a compact K ⊂ X with nonempty interior K ◦ such that f ∈ K◦.

In other words, X is locally compact if for every f ∈ X there is a neighborhood of f that is contained in a compact subset of X. For example, Fn is locally compact. With this terminology, we can reword Exercise 3.9 as follows.

Exercise 3.11. Let X be a normed linear space.

(a) Prove that if X is locally compact, then X is ﬁnite-dimensional.

(b) Prove that if X is inﬁnite-dimensional, then no nonempty open subset of X has compact closure. 22 CHRISTOPHER HEIL

4. Quotients and Products of Normed Spaces Any vector space is an abelian group under the operation of vector addition. So, if you are familiar with the basic notions of abstract algebra, the concept of a coset will be familiar to you. However, even if you have not studied abstract algebra, the idea of a coset in a vector space is very natural.

Example 4.1 (Cosets in R2). Consider the vector space X = R2. Let M be any one- dimensional subspace of R2, i.e., M is a line in R2 through the origin. A coset of M is simply a rigid translate of M by a vector in R2. For concreteness, let us speciﬁcally consider 2 the case where M is the x1-axis in R , i.e., M = {(x1, 0) : x1 ∈ R}. Then given a vector 2 y = (y1, y2) ∈ R , the coset y + M is the set

y + M = {y + m : m ∈ M} = {(y1 + x1, y2 + 0) : x1 ∈ R} = {(x1, y2) : x1 ∈ R}, 2 which is the horizontal line at height y2. This is not a subspace of R , but it is a rigid translate of the x1-axis. Note that there are inﬁnitely many diﬀerent choices of y that give the same coset. Furthermore, we have the following facts for this particular setting. (a) Two cosets are either identical or entirely disjoint. (b) The union of all the cosets is all of R2. (c) The set of distinct cosets is a partition of R2.

The preceding example is entirely typical.

Deﬁnition 4.2 (Cosets). Let M be a subspace of a vector space X. Then the cosets of M are the sets f + M = {f + m : m ∈ M}, f ∈ M.

Exercise 4.3. Let X be a vector space, and let M be a subspace of X. Given f, g ∈ M, deﬁne f ∼ g if f − g ∈ M. Prove the following. (a) ∼ is an equivalence relation on X. (b) The equivalence class of f under the relation ∼ is [f] = f + M. (c) If f, g ∈ M then either f + M = g + M or (f + M) ∩ (g + M) = ∅. (d) f + M = g + M if and only if f − g ∈ M. (e) f + M = M if and only if f ∈ M. (f) If f ∈ X and m ∈ M then f + M = f + m + M. (g) The set of distinct cosets of M is a partition of X.

Deﬁnition 4.4 (Quotient Space). If M is a subspace of a vector space X, then the quotient space X/M is X/M = {f + M : f ∈ X}. CHAPTER 3. BANACH SPACES 23

Since two cosets of M are either identical or disjoint, the quotient space X/M is simply the set of all the distinct cosets of M.

2 Example 4.5. Again let M = {(x1, 0) : x1 ∈ R} be the x1-axis in R . Then, by Example 4.1, we have that 2 2 R /M = {y + M : y ∈ R } = {(x1, 0) + M : x1 ∈ R}, i.e., R2/M is the set of all horizontal lines in R2. Note that R2/M is in 1-1 correspondence with the set of distinct heights, i.e., there is a natural bijection of R2/M onto R. This is a special case of a more general fact that we will explore.

Next we deﬁne two natural operations on the set of cosets: addition of cosets and multiplication of a coset by a scalar. These are deﬁned formally as follows.

Definition 4.6. Let M be a subspace of a vector space X. Given f, g ∈ X, define addition of cosets by (f + M) + (g + M) = (f + g) + M. Given f ∈ X and c ∈ F, define scalar multiplication by c(f + M) = cf + M.

Remark 4.7. Before proceeding, we must show that these operations are actually well- defined. After all, there need not be just one f that determines the coset f + M—how do we know that if we choose different vectors that determine the same cosets, we will get the same result when we compute (f + g) + M? We must show that f1 + M = f2 + M and g1 + M = g2 + M then (f1 + g1) + M = (f2 + g2) + M in order to know that Definition 4.6 makes sense.

Proposition 4.8. If M is a subspace of a vector space X, then the addition of cosets of M given in Deﬁnition 4.6 is well-deﬁned.

Proof. Suppose that f1 + M = f2 + M and g1 + M = g2 + M. Then by Exercise 4.3(d) we know that f1 − f2 = k ∈ M and g1 − g2 = l ∈ M. If h ∈ (f1 + g1) + M then we have h = f1 + g1 + m for some m ∈ M. Hence

h = (f2 + k) + (g2 + l) + m = (f2 + g2) + (k + l + m) ∈ (f2 + g2) + M.

Thus (f1 + g1) + M ⊂ (f2 + g2) + M, and the converse inclusion is symmetric.

Exercise 4.9. Show that scalar multiplication is likewise well-deﬁned.

Now we can show that the quotient space is actually a vector space under the operations just deﬁned.

Proposition 4.10. If M is a subspace of a vector space X, then X/M is a vector space with respect to the operations given in Deﬁnition 4.6. 24 CHRISTOPHER HEIL

Proof. Addition of cosets is commutative because (f + M) + (g + M) = (f + g) + M = (g + f) + M = (g + M) + (f + M). The zero vector in X/M is the coset 0+M = M, because (f +M)+(0+M) = (f +0)+M = f + M. Exercise: Show that the remaining axioms of a vector space are satisﬁed.

Deﬁnition 4.11 (Codimension). If M is a subspace of a vector space X, then the codimension of M is the dimension of X/M, i.e., codim(M) = dim(X/M).

Example 4.12. Let C(R) be space of continuous functions on R, and let P be the subspace containing the polynomials. Given f ∈ C(R), the coset f + P is f + P = {f + p : p is a polynomial}. Further, f + P = g + P if and only if f − g is a polynomial. Thus, f + P can be thought of as “f modulo the polynomials,” i.e., it is the equivalence class obtained by identifying functions which diﬀer by a polynomial.

In the same way, a coset f + M can be thought of as the equivalence class obtained by identifying vectors which diﬀer by an element of M. We can imagine the mapping that takes f to f + M as “collapsing information modulo M.”1

Deﬁnition 4.13. If M is a subspace of a vector space X, then the canonical projection or the canonical mapping of X onto X/M is π : X → X/M deﬁned by π(f) = f + M, f ∈ X.

Exercise 4.14. Let M be a subspace of a vector space X. (a) Prove that the canonical projection π is linear. (b) Prove that π is surjective and ker(π) = M. (c) Prove that if E ⊂ X, then the inverse image of π(E) is π−1 π(E) = E + M = {u + m : u ∈ E, m ∈ M}. Solution (c) Suppose that u ∈ E and m ∈ M are given. Then π(u + m) = u + m + M = u + M = π(u) ∈ π(E). Hence u + m ∈ π−1 π(E) . Now suppose that v ∈ π−1 π(E) . Then, by deﬁnition, π(v) ∈ π(E) = {u + M : u ∈ E}. Hence v + M = π(v) = u + M forsome u ∈ E. But then m = v − u ∈ M, so v = u + m with u ∈ E and m ∈ M.

1Conway calls this map Q, but I prefer to call it π for “projection.” CHAPTER 3. BANACH SPACES 25

We will mostly be interested in the case where X is a normed space. The following result shows that X/M is a semi-normed space in general, and is a normed space if M is closed.

Proposition 4.15. Let M be a subspace of a normed linear space X. Given f ∈ X, deﬁne kf + Mk = dist(f, M) = inf kf − mk. m∈M Then the following statements hold. (a) k · k is well-deﬁned. (b) k · k is a semi-norm on X/M. (c) If M is closed, then k · k is a norm on X/M. Proof. (a) Exercise. Hint: Show that if f1 + M = f2 + M then {f1 − m : m ∈ M} = {f2 − m : m ∈ M}. (b) Exercise.

(c) Suppose that M is closed, and that kf + Mk = 0. Then inf m∈M kf − mk = 0. Hence there exist vectors gn ∈ M such that kf − gnk → 0 as n → ∞. But M is closed, so this implies f ∈ M. By Exercise 4.3(e), we therefore have f + M = M = 0 + M, which is the zero vector in X/M.

Now we derive some basic properties of the canonical projection π of X onto X/M.

Proposition 4.16. Let M be a closed subspace of a normed linear space X. Then the following statements hold. (a) kπ(f)k = kf + Mk ≤ kfk for each f ∈ X. X X/M (b) Let Br (f) denote the open ball of radius r in X centered at f, and let Br (f + M) denote the open ball of radius r in X/M centered at f + M. Then for any f ∈ X and r > 0 we have X X/M π Br (f) = Br (f + M). (c) W ⊂ X/M is open in X/M if and only if π−1(W ) = {f ∈ X : f + M ∈ W } is open in X. (d) π is an open mapping, i.e., if U is open in X then π(U) is open in X/M. Proof. (a) Choose any f ∈ X. Since 0 is one of the elements of M, we have kπ(f)k = kf + Mk = inf kf − mk ≤ kf − 0k = kfk. m∈M

X (b) First consider the case f = 0 and r > 0. Suppose that g + M ∈ π Br (0) . Then X g + M = h + M for some h ∈ Br (0), i.e., khk < r. Hence kg + Mk = kh +Mk ≤ khk < r, X/M so g + M ∈ Br (0 + M). X/M Now suppose that g + M ∈ Br (0 + M). Then infm∈M kg − mk = kg + Mk < r. Hence X there exists m ∈ M such that kg − mk < r. Thus g − m ∈ Br (0), so X g + M = g − m + M = π(g − m) ∈ π Br (0) . 26 CHRISTOPHER HEIL

Exercise: Show that statement (b) holds for an arbitrary f ∈ X. (c) ⇒. Part (a) implies that π is continuous. Hence π−1(W ) must be open in X if W is open in X/M. ⇐. Suppose that W is a subset of X/M such that π−1(W ) is open in X. We must show that W is open in X/M. Choose any point f + M ∈ W . Then f ∈ π−1(W ), which is open X −1 in X. Hence, there exists an r > 0 such that Br (f) ⊂ π (W ). By part (b) we therefore have X/M X −1 Br (f + M) = π Br (f) ⊆ π π (W ) = W. Therefore W is open. (d) Suppose that U is an open subset of X. Then by Exercise 4.14(c), we have π−1 π(U) = U + M = {u + m : u ∈ U, m ∈ M} = (U + m). mS∈M But each set U + m, being the translate of the open set U, is itself open. Hence π−1 π(U) is open, since it is a union of open sets. Part (c) therefore implies that π(U) is op en in X/M.

Exercise 4.17. Let M be a closed subspace of a normed space X, and let π be the canonical projection π of X onto X/M. Prove that kπk = 1. Hint: Lemma 3.8.

Now we can prove that if X is a Banach space, then X/M inherits a Banach space structure from X.

Theorem 4.18. If M is a closed subspace of a Banach space X, then X/M is a Banach space. Proof. We have already shown that X/M is a normed space, so we must show that it is complete in that norm. Suppose that {fn + M}n∈N is a Cauchy sequence in X/M. It would be convenient if this implies that {fn}n∈N is a Cauchy sequence in X, but this need not be the case. For, the vectors fn are not unique in general: if we replace fn by any vector fn + m with m ∈ M, then we obtain the same coset. We will show that by choosing an appropriate subsequence

{fnk }k∈N and replacing the fnk by appropriate vectors that determine the same cosets fnk +M, we can create a sequence in X that is Cauchy and hence converges, and then use this to show that the original sequence of cosets {fn + M}n∈N converges in X/M.

We begin by applying Exercise 1.18: there exists a subsequence {fnk + M}k∈N such that −k ∀ k ∈ N, k(fnk+1 − fnk ) + Mk = k(fnk+1 + M) − (fnk + M)k < 2 .

Now we seek to create vectors gk ∈ M so that {fnk − gk}k∈N will converge in X. Note that

the cosets determined by fnk and by fnk − gk are identical. Set g1 = 0. Then 1 inf k(fn1 − g1) − (fn2 − g)k = inf k(fn1 − fn2 ) + gk = k(fn1 − fn2 ) + Mk < . g∈M g∈M 2 CHAPTER 3. BANACH SPACES 27

Therefore, there exists a g2 ∈ M such that 1 k(f − g ) − (f − g )k < 2 · . n1 1 n2 2 2

Then, since g2 ∈ M, 1 inf k(fn2 − g2) − (fn3 − g)k = inf k(fn2 − fn3 ) + gk = k(fn2 − fn3 ) + Mk < . g∈M g∈M 22

Therefore, there exists a g3 ∈ M such that 1 k(f − g ) − (f − g )k < 2 · . n2 2 n3 3 22

Continuing in this way, by induction we construct hk = fnk − gk such that 1 kh − h k < . k k+1 2k−1

Exercise 1.12 therefore implies that {hk}k∈N is a Cauchy sequence in X. Since X is complete, this sequence converges, say hk → h. Since

k(fnk + M) − (h + M)k = kfnk − gk − h + Mk (since gk ∈ M)

= khk − h + Mk

≤ khk − hk → 0, we see that {fnk + M}k∈N is a convergent subsequence of {fn + M}n∈N. Since we know that

{fn + M}n∈N is Cauchy, it follows from Exercise 1.17 that fnk + M → h + M. Thus X/M is complete.

The following exercise shows that the converse of the preceding theorem is true as well.

Exercise 4.19. Let M be a closed subspace of a normed space X. Prove that if M and X/M are both complete, then X must be complete. Hints: Suppose that {fn}n∈N is a Cauchy sequence in X. Show that {fn + M}n∈N is a Cauchy sequence in X/M, hence converges to some coset f + M. Thus kf − fn + Mk → 0. Does this imply that there exists a g ∈ M such that kf − fn + gk → 0? Or perhaps vectors gn ∈ M such that kf − fn + gnk → 0?

The quotient space and canonical map will be useful tools for proving many later results. The following proof illustrates their utility.

Proposition 4.20. Let X be a normed linear space. If M is a closed subspace of X and N is finite-dimensional, then M + N is a closed subspace of X. Proof. Let π be the canonical projection of X onto X/M. Since N is finite-dimensional, it has a finite basis, say {e1, . . . , en}. Then since π is linear,

π(N) = π span{e1, . . . , en} = span{π(e1), . . . , π(en)} = span{e1 + M, . . . , en + M}. 28 CHRISTOPHER HEIL

Thus π(N) is a ﬁnite-dimensional subspace of X/M, and therefore is closed by Proposi- tion 3.6. Since π is continuous, it follows that π−1(π(N)) is closed in X. However, by Exercise 4.14(c), we have π−1(π(N)) = M + N.

Exercise 4.21. Let M be a closed subspace of a normed linear space X. (a) Prove that if X is separable, then X/M is separable. (b) Prove that if X/M and M are both separable, then X is separable. (c) Give an example of X, M such that X/M is separable, but X is not separable.

Solution (b) Let {fn + M}n∈N be a countable dense subset of X/M, and let {gn}n∈N be a countable dense subset of M. Then S = {fm + gn}m,n∈N is a countable subset of X, and we claim that it is dense. To see this, ﬁx any f ∈ X. Then there exists an m such that ε inf kf − fm + hk = kf − fm + Mk = k(f + M) − (fm + M)k < . h∈M 2 ε Hence, there exists some h ∈ M such that kf − fm + hk < 2 . Since h ∈ M, there exists an ε n such that kh − gnk < 2 . Therefore ε ε kf − (f + g )k ≤ kf − f − hk + kh − g k < + = ε. m n m n 2 2

∞ Exercise 4.22. Recall from Exercise 1.37 that c0 is a closed subspace of the Banach space ` . (a) Prove that `∞ is not separable. Hint: Consider

S = {(x1, x2, . . . ) : xk = 0 or 1 for every k}. What is the distance between two distinct elements of S?

(b) Let x, y be vectors in S Prove that if xk =6 yk for at most ﬁnitely many k, then x + c0 = y + c0. Prove that if xk =6 yk for inﬁnitely many k, then kx − y + c0k = 1. ∞ (c) Use part (b) to prove directly that ` /c0 is not separable.

(e) Prove that the standard basis {en}n∈N is a Schauder basis for c0 (with respect to the `∞-norm).

(f) Use part (e) to show that c0 is separable. Hint: Use one of the exercises about Schauder bases from the Chapter 1 lecture notes. ∞ (g) Use part (e) and Exercise 5.4 to show that ` /c0 is not separable.

In the Hilbert space case, there is a very close relationship between π and the orthogonal projection of H onto M ⊥. CHAPTER 3. BANACH SPACES 29

Exercise 4.23. Let M be a closed subspace of a Hilbert space H, and let π be the canonical projection of H onto H/M. Prove that the restriction of π to M ⊥ is an isometric isomorphism of M ⊥ onto H/M.

Remark 4.24. There is no analog of the preceding result for arbitrary Banach spaces. If X is a Banach space and M is a closed subspace then we say that M is complemented in X if there exists another closed subspace N such that M ∩ N = {0} and M + N = X. It is not true that every closed subspace of every Banach space is complemented. In ∞ p particular, c0 is not complemented in ` . Also, if 1 < p ≤ ∞ and p =6 2, then ` has uncomplemented subspaces.

Next we will define the product or direct sum of normed spaces. An analogous definition holds for the case of a finite collection of spaces.

Deﬁnition 4.25. Let {Xi}i∈N be a countable family of Banach spaces, and let k · ki denote the norm on Xi. Deﬁne ∞

Xi = f = (f1, f2, . . . ) : fi ∈ Xi . Yi=1 For 1 ≤ p < ∞, deﬁne ∞ ∞ 1/p X = f ∈ X : kfk = kf kp < ∞ . p i k p i i L Yi=1 Xi=1 For p = ∞, deﬁne ∞

∞Xi = f ∈ Xk : kfk∞ = sup kfiki < ∞ . i L Yi=1

Exercise 4.26. Let {Xi}i∈N be a countable family of Banach spaces and ﬁx 1 ≤ p ≤ ∞. Let X = pXi. Prove the following. L (a) X is a normed space.

(b) For each i, the projection Pi : X → Xi given by Pi(f1, f2, . . . ) = fi is continuous, and kPik = 1.

Exercise 4.27. Let X1, . . . , Xn be ﬁnitely many normed spaces. Prove that the spaces ⊕pXi are equal for 1 ≤ p ≤ ∞, and that all the norms k · kp are equivalent. For this reason, we often denote this space by X1 × · · · × Xn. 30 CHRISTOPHER HEIL

Exercise 4.28. Let X and Y be normed vector spaces. Deﬁne T : B(X, Y ) × X → Y by T (A, f) = Af.

(a) Prove that T is continuous if and only if An → A and fn → f implies Anfn → Af. (b) Prove that T is continuous. Conclude that T ∈ B(B(X, Y ) × X, Y ).

5. Linear Functionals Deﬁnition 5.1 (Hyperplane). Let M be a subspace of a vector space X. Then we say that M is a hyperplane if it has codimension 1, i.e., if codim(M) = dim(X/M) = 1.

Example 5.2. Let H be a Hilbert space, and let L be a bounded linear functional on H. That is, L: H → F is a bounded linear operator, so L∗ : F → H is also a bounded linear operator. Now, F is one-dimensional, and a linear operator is entirely determined by what it does to the basis elements. In particular, {1} is a basis for F, so L∗(c) = L∗(c1) = c L∗(1). Hence range(L∗) = span{L∗(1)}. If L∗(1) = 0 then L∗ is the zero operator, and hence L is the zero operator as well (why?). Otherwise, range(L∗) must be one-dimensional, and hence is a closed subspace of H. Therefore, H = range(L∗) ⊕ range(L∗)⊥ = range(L∗) ⊕ ker(L). Hence, every vector f ∈ H can be written uniquely as f = c L∗(1) + k for some scalar c ∈ range(L∗) and k ∈ ker(L). Hence, if π is the canonical projection of H onto H/ ker(L), then since k ∈ ker(L) we have f + ker(L) = π(f) = cL∗(1) + k + ker(L) = cL∗(1) + ker(L). Therefore H/ ker(L) = {f + ker(L) : f ∈ H} = {cL∗(1) + k + ker(L) : c ∈ F, k ∈ ker(L)} = {cL∗(1) + ker(L) : c ∈ F} = span{L∗(1) + ker(L)}, which is one-dimensional. Thus, if L is a bounded linear functional on a Hilbert space H, then we conclude that ker(L) is a hyperplane in H.

We will show that a similar result holds for arbitrary normed spaces. We will need the following tool. In abstract algebra, the group version of the next result is called the First Isomorphism Theorem or the Homomorphism Theorem. In the group setting, an isomorphism is a bijective homomorphism. In the vector space setting, an isomorphism is a linear bijection. CHAPTER 3. BANACH SPACES 31

Exercise 5.3 (Isomorphism Theorem). Let X and Y be vector spaces, and let ϕ: X → Y be a linear surjection. Let M = ker(ϕ). Prove that ψ : X/M → Y given by ψ(f +M) = ϕ(f) is a well-deﬁned linear bijection, and that ϕ = ψ ◦ π.

Exercise 5.4. Fix 1 ≤ p ≤ ∞, and set M = {x ∈ `p : x(2k) = 0 for every k}. Prove that M is a closed subspace of `p, and that `p/M is isometrically isomorphic to `p.

Now we can show that every hyperplane is the kernel of some (not necessarily continuous) linear functional.

Proposition 5.5. Let M be a subspace of a normed space X. Then the following statements are equivalent. (a) M is a hyperplane. (b) M = ker(µ) for some nonzero linear functional µ: X → F. Proof. (b) ⇒ (a). Suppose that M = ker(µ) where µ is a nonzero linear functional. Then by the Isomorphism Theorem, there is a linear bijection ψ : X/ ker(µ) → F. Since F is one-dimensional, we conclude that X/ ker(µ) is as well. (a) ⇒ (b). Assume that M is a hyperplane. Then X/M is one-dimensional, so there exists a linear bijection ψ : X/M → F. Set ϕ = ψ ◦ π, then ψ : X → F. If f ∈ M, then π(f) = f + M = 0 + M, so ψ(f) = ψ(π(f)) = ψ(0 + M) = 0, so f ∈ ker(ϕ). Conversely, if f ∈ ker(ϕ) then we have ψ(π(f)) = ϕ(0 + M) = 0. But ψ is a bijection, so this implies f + M = π(f) = 0 + M. Hence f ∈ M, so ker(ϕ) ⊂ M.

Proposition 5.6. Let X be a normed linear space. If µ, ν : X → F are nonzero linear functionals, then ker(µ) = ker(ν) ⇐⇒ µ = cν for some nonzero scalarc. Proof. ⇐. Trivial. ⇒. Suppose that ker(µ) = ker(ν). Since µ =6 0, there exists some f ∈ X such that µ(f) =6 0, and by rescaling, we can assume that µ(f) = 1. Since f ∈/ ker(µ) = ker(ν), we have ν(f) =6 0. Given any g ∈ X, we have µ g − µ(g)f = µ(g) − µ(g) · µ(f) = 0, so g − µ(g)f ∈ ker(µ) = ker(ν). Therefore ν(g) − µ(g) · ν(f) = ν g − µ(g)f = 0, so after rearranging we see that ν = ν(f)µ.

Proposition 5.7. If M is a hyperplane in a normed linear space X, then M is either closed or is dense in X. 32 CHRISTOPHER HEIL

Proof. We are given that X/M is one-dimensional. Let π be the canonical projection of X onto X/M. The closure M of M is a subspace of X, so since π is linear we know that π(M) is a subspace of X/M. But X/M is one-dimensional, so there only two possibilities. First, we could have π(M) = {0+M}. In this case, M ⊂ ker(π) = M, so we have M = M and M is closed. Second, we could have π(M) = X/M. In this case, we have by Exercise 4.14(c) that

X = π−1(X/M) = π−1(π(M)) = M + M = M, and thus M is dense.

Proposition 5.8. Let µ: X → F be a linear functional on a normed space X. Then:

µ is continuous ⇐⇒ ker(µ) is closed.

Proof. ⇒. Exercise.

⇐. Suppose that ker(µ) is closed. We know by Proposition 5.5 that ker(µ) is a hyperplane, so X/ ker(µ) is one-dimensional. Let π be the canonical projection of X onto X/ ker(µ). Because M is closed, we know that π is continuous. By the Isomorphism Theorem, there exists a linear bijection ψ : X/ ker(µ) → F. Since X/ ker(µ) is a one-dimensional normed linear space, and linear map from X/ ker(µ) into another normed space is continuous by Exercise 3.7. Therefore ψ is continuous, and hence µ = ψ ◦ π is continuous.

Recall now the deﬁnition of the dual space of a normed space X:

X∗ = X0 = B(X, F) = {L: X → F : L is bounded and linear}.

Since the norm on F is just absolute value, the operator norm of a linear functional L ∈ X0 = B(X, F) is kLk = sup |Lf|. kfk=1 Since F is a Banach space, the dual space X 0 is a Banach space even if X is not. In order to give some examples of dual spaces, we recall H¨older’s Inequality.

Theorem 5.9 (H¨older’s Inequality). Let (X, Ω, µ) be a measure space. If 1 ≤ p ≤ ∞ and 1 1 p + p0 = 1, then

p p0 ∀ f ∈ L (X), ∀ g ∈ L (X), kfgk1 ≤ kfkp kgkp0 .

0 In particular, if f ∈ Lp(X) and g ∈ Lp (X), then fg ∈ L1(X). CHAPTER 3. BANACH SPACES 33

Appendix A. Appendix: Topological and Metric Spaces Deﬁnition A.1 (Topological Space). A topological space (X, T ) is a nonempty set X together with a family T of subsets of X such that the following statements hold. (a) ∅, X ∈ T . (b) Closure under ﬁnite intersections: If U, V ∈ T , then U ∩ V ∈ T .

(c) Closure under arbitrary unions: If I is any index set and Ui ∈ T for i ∈ I, then ∪i Ui ∈ T . We call T a topology on X. The elements of T are the open subsets of X.

Deﬁnition A.2 (Metric Space). Let X be a nonempty set. A metric on X is a function d(·, ·) → R such that (a) d(f, g) ≥ 0 for all f, g ∈ X, (b) d(f, g) = 0 if and only if f = g, (c) Triangle Inequality: d(f, h) ≤ d(f, g) + d(g, h) for all f, g, h ∈ X. A space X together with a metric d(·, ·) is called a metric space.

Deﬁnition A.3 (Topology on a Metric Space). Let X be a metric space. (a) The open ball in X centered at x ∈ X with radius r > 0 is

Br(x) = B(x, r) = {y ∈ X : d(x, y) < r}. (b) A subset U ⊆ X is open if

∀ x ∈ U, ∃ r > 0 such that Br(x) ⊆ U. (c) The topology on X is T = {U ⊆ X : U is open}.

Exercise A.4. Prove that if X is a metric space, then (X, T ) is a topological space using the preceding deﬁnition.