CONTENTS. 4 Hausdorff Measure Introduction The Cantor Set Rectifiable Curves Cantor Set-Like Objects...

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1 Contents 1 Functional Analysis Hilbert Spaces Spectral Theorem Normed Vector Spaces Banach Spaces Dual Spaces L p spaces The Hahn-Banach Theorem The Baire Category Theorem Meager Sets and Mapping Theorems Weak Topologies Distribution Theory Introduction Distributions Differentiation Operations on Distributions Approximation of Distributions Sobolev Spaces Notions of Convergence Applications Tempered Distributions i

2 ii CONTENTS Continuous Operations Convolutions On Tempered Distributions Fourier Series Introduction Fourier Series Convolution Summability Kernels Homogeneous Banach Spaces Convergence in Norm Fourier Transform Properties and Inversion Summability Kernels Fourier-Stieltjes Transform Fourier Transform of Borel Measures Fourier Transform on L 2 (R) Plancherel s Theorem Applications Extending to L p (R), p > Hausdorff Measure Introduction The Cantor Set Rectifiable Curves Cantor Set-Like Objects Sobolev Applications Preliminaries Solvability of Lu = f

3 Chapter 1 Functional Analysis 1.1 Hilbert Spaces Our ultimate goal will be to prove the spectral theorem for compact operators. We will denote Hilbert spaces by (H,, ). Definition Let H be a Hilbert space. A subset X H is said to be compact if for every sequence {f n } n=1 with f n X then there exists a convergent subsequence {f nk } k=1. Since all Hilbert spaces are metric spaces, the above definition of compactness agrees with the same definition involving open covers. However, for the purposes of functional analysis, analyzing covers by sets is not as useful as dealing with sequences. The closed unit ball is compact in finite dimensions, but is never compact in infinite dimensions. Indeed, consider an orthonormal basis as the sequence. If this sequence contained a convergent subsequence, it would have to be Cauchy, which would not be possible since the elements are orthogonal. Definition Let T : H H be a linear operator. We say that T is bounded if there exists M > such that x H, T x M x. We define the operator norm as the infimum of all bounds for the linear operator. 1

4 1.1. HILBERT SPACES CHAPTER 1. FUNCTIONAL ANALYSIS Proposition If T : H H is a bounded linear operator, then any of the following define the operator norm: T = inf {M : T x M x, x H} = sup { T x : x H, x 1} = sup { T x : x H, x = 1} { } T x = sup x : x H, x. Definition We say that a linear operator T : H H is compact if for every sequence {f n } n=1 with f n M then T f n has a convergent subsequence; that is, there exists a convergent subsequence such that T f nk g for some g H. Since we would like the image of every uniformly bounded sequence to have a convergent subsequence, we note that the homogeneity of the norm allows us to consider elements which lie entirely within the unit ball. Proposition An operator is compact if and only if T B 1 is compact. Proof. Let T : H H be a linear operator. We begin by assuming that T is compact for which we want to show that the closure of the image of the unit ball is compact. Since every Hilbert space defines a metric space, to show that T B 1 is compact it is sufficient to show that every sequence in T B 1 has a convergent subsequence. Let {g n } n=1 be any sequence in T B 1 and choose {f n } n=1 B 1 such that T f n = g n. Clearly the f n are bounded as they lie in B 1, so by compactness of T it then follows that {T f n } has a convergent subsequence, so T B 1 is compact as required. On the other hand, assume that T B 1 is compact in H. Let {f n } be a bounded sequence in H, with upper bound say M. Define ˆf n = M 1 f n so that {T ˆf } n is now a sequence in T B 1. By sequential compactness, {T ˆf } n has a convergent subsequence and so we conclude that {T f n } does as well. Example Let X H be a finite dimensional linear subspace, and define T : H H as T f = proj X f. Since T is a projection operator, it can only reduce the norm in the image; 2

5 CHAPTER 1. FUNCTIONAL ANALYSIS 1.1. HILBERT SPACES that is, T f f. T is then compact because the closure of the image of B 1 in X will be contained within B 1, which in finite dimensions is compact. Proposition Every bounded linear operator with finite dimensional range is compact. Proof. Insert Proof Here. Proposition Let T : H H be a bounded linear operator. 1. Suppose that S : H H is compact. Then T S and ST are compact. 2. Suppose there exists a sequence T n : H H such that each T n is compact and that T n T n. In this case, T is compact. 3. Suppose that T is compact, then there exists a sequence T n : H H with finite dimensional range (hence they are compact) such that T T n n. 4. If T is compact then its adjoint T is also compact. Proof. 1. Let f n be a sequence then T f n is bounded and since S is compact then ST f n has a convergent subsequence, so ST is compact. The other direction follows similarly. 2. Consider any sequence {f n } and assume without loss of generality that f n 1. Our goal is to show that T f n has a convergent subsequence. Since H is complete, it is sufficient to find a Cauchy subsequence; that is, for every ɛ > we have that T f nk T f nl < ɛ. Since T 1 is compact, we know that there is a subsequence f n such that T f 1,k is convergent. Similarly, since T 2 is compact there is a subsequence f 2,k such that T f 2,k converges. We continue inductively, creating a sequence f n,k for each n N. Let g k = f k,k : we claim that T g k is Cauchy. Let ɛ > be given so that for any m N we can write T g k T g l T g k T m g k + T m g k T m g l + T m g l T g l. (1.1) Now since the T n converge to T, we have that T g k T m g k T T m g k, T m g l T g l T T m g l (1.2) so we can make the first and last terms as small as necessary. On the other hand by construction, we have the T m g i are Cauchy and hence T m g k T m g l. By an ɛ 3 argument, we have shown that T g k are Cauchy. 3

6 1.1. HILBERT SPACES CHAPTER 1. FUNCTIONAL ANALYSIS 3. Choose an orthonormal basis {e n } of H and choose P k to be the projection on to span {e 1,..., e k }. Further define T k = P k T which is the required sequence of compact operators. 4. Since T is compact, we can construct a convergent sequence of compact operators from (3) such that P k T T. By taking the adjoint, we have (P k T ) T = T T P k. (1.3) But each T P k is compact since it is the composition of a compact operator with a bounded operator, and hence T is the limit of a sequence of compact operators Spectral Theorem Theorem (Spectral Theorem). Let T be a symmetric, compact operator on a Hilbert space H. Then there exists an orthonormal basis {e k } k=1 of H consisting of eigenvectors of T. Moreover, if T e k = λ k e k so that the λ k are eigenvalues of T, then k. λ k In order to show this result, we will give three lemmas which, when combined, will give the spectral theorem. Lemma Let T : H H be a symmetric and bounded linear operator. Then 1. All eigenvalues of T are real. 2. Eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal. Proof. 1. Given a fixed eigenvalue λ and corresponding eigenvector e we have that λ e, e = λe, e = T e, e = e, T e = e, λe = λ e, e (1.4) and so λ = λ which implies that λ R. 2. Let e i, e j be eigenvectors corresponding to λ i, λ j respectively, where λ i λ j. Since T is symmetric, we know that T e i, e j = e i, T e j. Computing we find that = T e i, e j e i, T e j = λ i e i, e j λ j e i, e j since λ j R = (λ i λ j ) e i, e j. 4

7 CHAPTER 1. FUNCTIONAL ANALYSIS 1.1. HILBERT SPACES Since λ i λ j, we may then conclude that e i, e j =. Lemma Let T be compact and symmetric and suppose that τ >. Then the following hold: 1. The number of eigenvalues λ k of T satisfying λ k τ is finite. 2. For a fixed non-zero eigenvalue λ with eigenspace E λ, the dimension of E λ is finite. Proof. 1. For the sake of contradiction, assume that this is not the case and choose {λ k } k=1 to be a set of distinct eigenvalues such that λ k τ for all k. Since all eigenvalues are distinct, we may use Lemma to choose an orthonormal set of eigenvectors {e k } k=1 such that {λ k} k=1 satisfies T e k = λ k e k, which will then have a convergent subsequence since T is compact. However λ k e k λ j e j 2 = λ 2 k + λ 2 j 2τ 2 > (1.5) and so the distance between any two eigenvectors λ k is bounded below by some positive number, meaning that it is impossible for them to Cauchy. This is a contradiction and so the set of eigenvalues whose norm exceeds τ must be finite. 2. For the sake of contradiction, assume that dim E λ = so that there exists an infinite set of orthonormal eigenvectors {e k } k=1 of λ. By compactness, we know that λe k = T e k must have a convergent subsequence. However, by precisely the same reasoning as above, this sequence cannot possibly be Cauchy and so cannot converge.this yields a contradiction, showing that the dimension of the eigenspaces are necessarily finite. In particular, this Lemma implies that it does not matter whether we count the eigenvectors with or without respect to multiplicity, since there are ultimately only countably many k of them. Furthermore, since we can choose τ > arbitrarily, λ k. Lemma Let T be a compact and symmetric, non-zero, linear operator on a Hilbert space H. Then T has an eigenvalue satisfying λ = T. Proof. Recall that if T is symmetric then in fact T = sup { T f, f : f = 1} (1.6) 5

8 1.1. HILBERT SPACES CHAPTER 1. FUNCTIONAL ANALYSIS hence either T = sup { T f, f : f = 1} or T = inf { T f, f : f = 1}. (1.7) Assume this is true and choose a sequence f n with f n = 1 such that T = lim n T f n, f n. (1.8) By dropping to a subsequence if necessary, we may assume without loss of generality that there is some g such that T f n g via compactness of T. Calculating we find T f n T f n 2 = T f n 2 + T 2 1 f n 2 T f n, T f n T f n, T f n = T f n 2 + T 2 2 T T f n, f n 2 T 2 f 1 n 2 2 T T f n, f n therefore T f n g. Now and clearly g since T g = lim n T (T f n ) = T g (1.9) g = lim n T f n = T. (1.1) Now we can prove the Spectral Theorem Proof of the Spectral Theorem. Let S be the span of the eigenvectors of T and suppose that S is a proper subspace of H so that, by the projection theorem, we may write H = S S and S {}. Consider T which is symmetric and compact. Notice that if v S and e is an eigenvector of T then S T v, e = v, T e = λ v, e = (1.11) and so T v S. Hence T : S S and this is a symmetric, compact operator. By S Lemma we have that there exists v S \ {} such that T (v) = ± T v which is S a contradiction to the definition of S. 6

9 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES 1.2 Normed Vector Spaces Banach Spaces Definition Let X be a normed vector space. If X is complete in the induced metric, we say that X is a Banach space. Theorem (Folland). A normed vector space X is complete if and only if every absolutely convergent series in X converges. Proof. Assume that X is a completed normed space and let {x n } n=1 X be a sequence whose series converges absolutely; that is, x n <. Let S n denote the partial sums S n = n x n (1.12) and notice that absolute convergence implies that the S n are Cauchy. By completeness, the S n then converge as required. k=1 On the other hand, now assume that every absolutely convergent sequence converges in X. Let {x n } n=1 X be a Cauchy sequence and if necessary, drop to a subsequence so that x n x m < 2 j whenever m, n > n j. Imitating the proof to show completeness in L p spaces, we construct a telescoping series by defining y 1 = x n1 and y j = x nj x nj 1 for all j > 1. This series is absolutely convergent since y j y 1 + j=1 2 j = y <. (1.13) j=1 By hypothesis, since the {y n } n=1 are absolutely convergent, there exists y X such that y n y. It is easily confirmed that x n y by exploiting the telescoping nature of the series. Exercise Let (X, M) be a measurable space, and let M(X) be the space of complex measures on (X, M). Show that µ = µ (X) is a norm on M(X) that makes M(X) into a Banach space. 7

10 1.2. NORMED VECTOR SPACES CHAPTER 1. FUNCTIONAL ANALYSIS Fix a σ-finite, positive measure ν on X so that for each µ M(X) we can write dµ = f µ dν for some f µ L 1 (X, ν). To see that µ = µ (X) is a norm, we need to show that it satisfies non-degeneracy, homogeneity, and the triangle inequality. 1. Non-Degeneracy: Consider the zero-measure M(X) which assigns to each measurable set on X the value zero. In this case, we have that = dν and so = (X) = dν =. (1.14) On the other hand, assume that µ M(X) satisfies µ = and write µ = fdν. Evaluating the integral reveals to us that = µ = µ (X) = f dν. (1.15) Since the integrand is everywhere positive, the only way this is zero is if f = which implies that µ =, so non-degeneracy holds. 2. Homogeneity: Let λ C and fix µ = fdν M(X) so that λµ = λµ (X) = λf dν X = λ f dν X = λ µ showing that the norm is indeed homogeneous. 3. Triangle Inequality: Let µ 1, µ 2 M(X) and write µ 1 = f 1 dν and µ 2 = f 2 dν 2. Then µ 1 + µ 2 = µ 1 + µ 2 (X) = f 1 + f 2 dν X f 1 + f 2 dν X X = µ 1 + µ 2. X So we have shown that µ = µ (X) is indeed a norm. To show that M(X) is a Banach space, we hope to take advantage Theorem Let {µ n } n=1 be a collection of complex measures whose series is absolutely convergent; that is µ k <. (1.16) k=1 8

11 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES Write µ k = f k dν for each k N, so that absolute convergence implies µ k = k=1 µ k (X) = k=1 = X k=1 k=1 f k dν < X f k dν where in order to interchange the integrand and the summation, we have used the monotone convergence theorem applied to non-negative functions (See Corollary 1.1 of Stein & Shakarchi). This implies that k f k is integrable over X and hence is finite almost everywhere, implying that k f k converges ν-almost everywhere in X. This is an absolutely convergent series in L 1 (X, ν) which, by the Riesz-Fischer theorem, is complete. By Theorem we then know that lim n n f k = f (1.17) k=1 for some f L 1 (X, ν). Set µ = fdν and notice that so µ M(X). Furthermore we have that n µ = µ (X) = f dν X = f k dν X k=1 f k dν < X k=1 n lim µ n µ k = lim n µ µ k (X) k=1 k=1 n = lim n X f f k dν k=1 = lim f k dν n X =, k=n+1 thus k µ k µ in norm, so M(X) is complete and hence a Banach space. 9

12 1.2. NORMED VECTOR SPACES CHAPTER 1. FUNCTIONAL ANALYSIS Exercise If < α < 1, let Λ α ([, 1]) be the space of Hölder continuous functions of exponent α on [, 1]. That is, f Λ α ([, 1]) if and only if f Λα < where Show that f Λα = f() + f(x) f(y) sup. (1.18) x,y [,1] x y α x y (a) Λα is a norm which makes Λ α ([, 1]) into a Banach space, (b) If λ α ([, 1]) is the set of all f Λ α ([, 1]) such that f(x) f(y) x y α as x y, y [, 1] (1.19) then when α < 1, λ α ([, 1]) is an infinite dimensional closed subspace of Λ α ([, 1]) and if α = 1 then λ α ([, 1]) contains only constant functions. a. Since there is no ambiguity, we will denote Λα as. To see that this is a norm, we simply check the three conditions defining a norm. (a) Non-Degeneracy: Let f be the zero function. In this case we have that f() = and f(x) f(y) = for all x, y [, 1] and so f =. On the other hand, assume that f. If f() then we are done, since then f > ; otherwise, there exists x [, 1] such that f(x) in which case < f(x) f() f(x) f(y) < sup (1.2) x α x,y [,1] x y α x y so again f >. We conclude that is non-degenerate. (b) Homogeneity: Let γ R, in which case homogeneity follows from homogeneity of the absolute value. More explicitly, we have γf = γf() + = γ f() + = γ f. γf(x) γf(y) sup x,y [,1] x y α x y sup x,y [,1] x y f(x) f(y) x y α 1

13 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES (c) Triangle Inequality: This again follows from the triangle inequality on. Indeed, if f, g Λ α ([, 1]) then f + g = f() + g() + f() + g() + = f + g f(x) + g(x) f(y) g(y) sup x,y [,1] x y α x y sup x,y [,1] x y f(x) f(y) g(x) g(y) + sup x y α x,y [,1] x y α x y To see that this is a Banach space, we employ Theorem Let {f k } k=1 Λ α([, 1]) be a sequence with an absolutely convergent series. Naively, we define f = k f k which we do not a priori know is finite on [, 1]. However, we see that f f k < (1.21) k= by assumption of absolute convergence. Hence not only is f bounded but the function f must be everywhere finite, since otherwise f(x) f(y) sup = (1.22) x,y [,1] x y α x y which would in turn make the norm infinite, which we know cannot happen. b. Let us begin by consider the case when α = 1. The condition that f(x) f(y) lim x y x y =, y [, 1] (1.23) is the definition for the function f to be differentiable with derivative at each point y [, 1]. Since differentiability implies continuity, the only such functions are constant functions which is what we wanted to show. Now consider the case when < α < 1. We recall that for all natural numbers n N we have x n y n n 1 x y = x k y n 1 k (1.24) so that if p n (x) = x n then k= p n (x) p n (y) = xn y n x y α x y α x n y n = x y x y α 1 n 1 = x k y n 1 k x y 1 α x y k= 11

14 1.2. NORMED VECTOR SPACES CHAPTER 1. FUNCTIONAL ANALYSIS where we have used the fact that < α < 1 to ensure that 1 α >. Since this holds for all n N and monomials of different degrees are linearly independent, λ α ([, 1]) contains an infinite, linearly independent set and so must be infinite dimensional. To see that λ α ([, 1]) is closed, let {f n } n=1 λ α([, 1]) such that f f n n. We want to show that f λ α ([, 1]). Notice that f f n implies that f() f n () n f(x) f n (x) f(y) + f n (y) n, and sup. x,y [,1] x y α x y (1.25) Let ɛ > be given. Choose N N such that f(x) f N (x) f(y) + f N (y) sup < ɛ x,y [,1] x y α 2 x y (1.26) and choose δ > such that if x y < δ then f N (x) f N (y) x y α < ɛ 2. (1.27) Then for any x y < δ we have ( ) f(x) f(y) f(x) f N (x) f(y) + f N (y) + = x y α x y α + x y α f(x) f N (x) f(y) + f N (y) < ɛ 2 + ɛ 2 = ɛ. ( f N (x) f N (y)) f N (x) f N (y) x y α Since ɛ was arbitrary, we conclude that f λ α ([, 1]), so λ α ([, 1]) is closed. Theorem If X and Y are normed vector spaces and T : X Y is a linear map, then the following are equivalent: 1. T is bounded. 2. T is uniformly continuous. 3. T is continuous. 4. T is continuous at 12

15 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES Proof. The implications (2) (3) (4) are trivial, so we endeavor to show that (1) (2) and (4) (1). Assume then that T is bounded and recall that T x T y T x y (1.28) and hence for any ɛ > and any points x, y X, choosing δ = ɛ T 1 will be sufficient to guarantee uniform continuity. To show that (4) (1) we will proceed via contrapositive. In particular, we will show that if T is not bounded then T is not continuous at. If T is unbounded, then we can find x n X such that x n = 1 and T x n > n. By defining y n = x n n 1 2 we then have n y n but T y n > T x n n 1 2 > n (1.29) so T does not map convergent sequences to convergent sequences, and so cannot be continuous. Definition If X and Y are normed vector spaces, define L(X, Y ) = {T : X Y : T linear and bounded }. (1.3) Theorem If X and Y are normed vector spaces and Y is complete then L(X, Y ) is also complete. Proof. Let {T n } n=1 L(X, Y ) be a Cauchy sequence and note that for any x X, {T nx} n=1 is a Cauchy sequence since Define T : X Y as T n x T m x T n T m x m,n. (1.31) T (x) = lim n T n (x) (1.32) which is well defined by our previous discussion. Now T n T since T T n = sup (T T n )x sup T T n x (1.33) x =1 x =1 so all that remains to be shown is that T L(X, Y ). Note that T is linear as the limits preserve linearity, so we want to show that T is bounded. Thus L(X, Y ) is complete as required. 13

16 1.2. NORMED VECTOR SPACES CHAPTER 1. FUNCTIONAL ANALYSIS For an element T L(X, Y ), it may be useful to consider when we can construct an inverse element T 1 L(Y, X) such that T T 1 = I Y and vice-versa. Interestingly, it turns out that we can invert elements so long as they are not too far from the identity element I. Lemma If X is a Banach space and T L(X, X ) then T 1 is bounded if and only if C > such that T x C x for all x X. Proof. Assume that T 1 is bounded and let M > be such that for all x X we have that T 1 y M y ; that is, T 1 M. Let x be such that T x = y and notice that x = T 1 T x T 1 T x. (1.34) By rearranging we get that T x 1 x (1.35) T 1 so by choosing C = T 1 1 we get the desired result. On the other hand, suppose that there exists C > such that for all x X we have T x C x. Then x = T T 1 x C T 1 x (1.36) which by rearranging yields T 1 x 1 x. (1.37) C Since x was arbitrary, this holds for all x X so T 1 is bounded. Theorem Let X be a Banach space and consider T L(X, X ). 1. If T L(X, X ) and I T < 1 where I is the identity operator, then T is invertible; in fact, the series (I T ) n (1.38) converges in L(X, X ) to T 1. k= 2. If T L(X, X ) is invertible and S T < T 1 1, then S is invertible. Thus the set of invertible linear operators is open in L(X, X ). 14

17 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES Proof. 1. Since the operator norm is submultiplicative, we note that for any T L(X, X ) we have T n T n. The assumption that I T < 1 implies that (I T ) n I T n Define S = I T for which we claim that n. (1.39) T 1 = (I (I T )) 1 = (I S) 1 = S n. (1.4) k= The first two equivalences are clear, so we endeavor to show the final equivalence. The right hand side converges since S n S n S n k= = k= 1 1 S k= since S = I T < 1. Let the limit of the partial sums be given by U and notice that (I S)U = lim n (I S) = lim n n k= n S n k= S n k= = lim n I S n+1 = I S n+1 so that U = (I S) 1. We conclude that T is invertible and is given by T 1 = (I T ) n. (1.41) k= 2. Note that I T 1 S = T 1 (T S) and so I T 1 S = T 1 (T S) T 1 S T < T 1 T 1 < 1. By part (a) we may conclude that T 1 S is invertible. By Lemma we know C 1, C 2 > such that T x C 1 x and T 1 Sx C 2 x so Sx = T T 1 Sx C1 T 1 Sx C1 C 2 x. (1.42) 15

18 1.2. NORMED VECTOR SPACES CHAPTER 1. FUNCTIONAL ANALYSIS Applying Lemma again we conclude that S is invertible. In general, we conclude that B T 1 1(T ) = {S L(X, X ) : S T < T 1 1} (1.43) is an open ball around T contained in the set of invertible, bounded, linear operators, so the set of invertible operators is open Dual Spaces Definition If X is a normed vector space, we define its dual space X to be the collection of all bounded linear maps l : X R. Proposition If X is a Banach space then X is also a Banach space in the operator norm. Proof. This is immediately obvious by Theorem if we realize that X = L(X, R) and that R is certainly complete. However, since this special case is important, we shall re-iterate the important points. Suppose {l n } n=1 is a Cauchy sequence in X. Fix x X and consider the sequence {l n (x)}. Since the l n are Cauchy, it must follow that {l n (x)} are Cauchy in R, but since X is complete there exists l : X R such that l n (x) l(x). Now l is clearly linear and bounded, so in particular l X. Let ɛ > be given and choose N N such that l n l m < ɛ for all n, m > N. Fix x X, so that for any n > N and any m > n we have 2 l(x) l n (x) l(x) l m (x) + l m (x) l n (x) l(x) l m (x) + ɛ 2 x therefore l(x) l n (x) lim sup l(x) l m (x) + ɛ n 2 x = ɛ x (1.44) 2 so we conclude that l l n ɛ 2. 16

19 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES L p spaces Theorem (Riesz Representation Theorem for L p ). Suppose that 1 p < and q is the Hölder dual of q so that = 1. Then p q Lp (R n ) = L q (R n ). If l L p (R n ) then there exists a unique g L q (R n ) such that l(f) = g(x)f(x)dx, and l = g L q (R n ) (1.45) R n Notice that equivalently, this theorem means that for any linear operator l there exists an element g such that l g (f) = fg dx. This implies that the map g l g is an isometric isomorphism of Banach spaces between L p (R d ) and L q (R d ). To prove this, we will need the following Lemma: Lemma If 1 q with p its Hölder conjugate, suppose that g L q (R d ) and let { } M = sup fg : f Lp (R d ), f p = 1, (1.46) then g q = M. More generally, if we do not know a priori that g is in L q (R d ) but we do know that g is locally integrable and M < then g L q (R d ) with g q = M. This theorem tells us that any linear functional defined by the weighted integration against a q-integrable function has operator norm equal to the weighting function. Proof. By Hölder s inequality we know that fg g q f p = g q which implies that M g q. Now we want to find functions with unit norm which saturate this bound for which we will deal with the extremal cases first. Assume that q = 1 and define f = sgn(g(x)). It is easily seen that f L (R d ) with f = 1 and fg = g(x) dx = g 1. (1.47) Hence this upper bound is saturated and we conclude that M = g 1. If q = let ɛ > be given and find E R d of positive measure such that g(x) g ɛ on E. Define f(x) = χ E(x) sgn(g(x)) (1.48) m(e) 17

20 1.2. NORMED VECTOR SPACES CHAPTER 1. FUNCTIONAL ANALYSIS which again is seen to have unit norm, so that we get g(x) fg = m(e) dx g ɛ. (1.49) By taking ɛ we get the desired saturation so M = g. More generally, assume that 1 < q < and define It is easy to check that check that f p = 1 and E f(x) = g(x) q 1 sgn(g(x)) g q 1. (1.5) q fg = g q q g q 1 q = g q (1.51) allowing us to conclude that M = g q, and as all cases have been covered our result holds in general. In the case that g is locally integrable, a similar proof holds. One takes the approximation to g by simple functions and follows a similar procedure (exercise). Theorem (Radon-Nikodym). Let (X, M) be a measure space and let µ be a finite positive measure on (X, M). If ν is a finite signed measure on (X, M) such that ν(e) = whenever µ(e) = then there exists a unique g L 1 (X, µ) such that ν(e) = χ E (x)g(x) dµ(x). (1.52) X Proof of Riesz Representation (Theorem ). Consider X = B R () = B r R d the ball of radius R centred at the origin, and let m be the Lebesgue measure restricted to X. Define ν(e) = l(χ E ) for all sets E X. The measure ν is finitely additive since ν(a B) = l(χ A + χ B ) = ν(a) + ν(b) (1.53) and so we need only now show that ν is countable additive. Suppose that E = n E n for E X. Define E >N = E j (1.54) in which case j=n+1 ( ) ν(e) = l χ N + χ i=1 En E >N = 18 n ν(e n ) + V (E >N ) (1.55) i=1

21 CHAPTER 1. FUNCTIONAL ANALYSIS 1.2. NORMED VECTOR SPACES by finite additivity. But we know that and so and so we conclude that ν(e) l χ E 1 = l m(e) 1 p (1.56) N ( ν(e) ν(e n ) ν(e >N) l n=1 N ν(e) = so that ν is countably additive as required. n=n+1 m(e n ) ) 1 p ν(e n ) (1.57) n=1 If m(e) = then ν(e) = as well by (1.56). Employing Radon-Nikodym we know that there exists a unique g L 1 (X, m) such that l(χ E ) = χ E gdm(x). (1.58) X By linearity, l(f) = fgdm(x) for all simple functions f with supp(f) X. By density of X simple function and continuity of both sides of this equation, we get that l(f) = fgdm(x) X for all f L p (R d ) such that supp(f) X. Notice by Lemma that g q l and supp(g) X. By this reasoning, we know that for each n 1 there exists a unique g n L q (R d ) with supp(g n ) B n () such that l(f) = B n fg n dm(x) for all f L p (R d ) with supp(f) B n, and g n q l. Hence if m n we have that fg n = fg m, f L p (B n ) (1.59) B n B n and so g n = g m in L q (B n ). In particular, g n (x) = g m (x) almost everywhere on B n for n m. Therefore, we can find a set Z with m(z) = such that g n (x) = g m (x) for all n, m x and x / Z. Hence there exists a function g(x) such that lim g n(x) = g(x), x / Z. (1.6) n By Fatou s Lemma, g q l. Note that g(x) = g n (x) for x B n \ Z and so fgdm(x) = fg n dm(x) (1.61) for all f supported in B n. We get from this that l(f) = fgdm(x) for all f with bounded support. By density and continuity we have that l(f) = fgdm(x) (1.62) for all f L p (R d ). 19

22 1.3. THE HAHN-BANACH THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS 1.3 The Hahn-Banach Theorem Definition If X is a vector space, a Minkowski or sublinear functional is a map p : X R + such that for all x, y X and λ > p(x + y) p(x) + p(y), p(λx) = λp(x). (1.63) The most important example of a sublinear functional the norm function p(x) = x. Theorem (Hahn-Banach Theorem). Let X be a vector space with a sublinear functional p : X R +. If M X is a closed linear subspace and f : M R is a linear functional such that f(x) p(x) then there exists an extension F of f such that F : X R, F = f, and F (x) p(x) for all x X. M To do this, we will need Zorn s Lemma, which we recall as follows: Lemma (Zorn s Lemma). Consider any set S and assume that S is equipped with a partial ordering. If every chain has an upper bound then there exists an element S which is maximal. Proof of Hahn-Banach. Our program will follow a two-step process. In our first step, we shall consider one dimensional extensions of M will allow our function f to be extended. It should be clear at this point that an inductive process will allow us to continue adding one-dimensional spaces until our vector space has been saturated. The only problem arises if our vector space is in fact infinite dimensional, which is the part requiring Zorn s lemma and is addressed in the second step. Step 1: Pick any x X \ M and consider the one dimensional extension of M to M + Rx = {y + λx : y X, λ R}. (1.64) We claim that there exists an extension g of f to M + Rx satisfying g(y + λx) p(x + λx). Note that by demanding that g be a linear functional, the action of g is entirely determined by the image of x. In particular, if g(x) = α then g(y + λx) = f(y) + λα is the only possible choice for g. However, our choice for α is constrained by the condition 2

23 CHAPTER 1. FUNCTIONAL ANALYSIS 1.3. THE HAHN-BANACH THEOREM that g(y + λx) p(y + λx) for all y M and λ R. Necessarily, it then follows that f(y) + λα p(y + λx). (1.65) If λ > then we may divide by λ to conclude that for all y M we have ( y ( y ) f + α p λ) λ + x. (1.66) However, since this must hold for all possible y, we may define an auxiliary variable y = y and bring the f(y) to the other side of the inequality to yield λ α p(y + x) f(y ). (1.67) Proceeding in precisely the same manner but with λ <, we divide by λ to get ( ( y f + α λ) 1 ) p(x + λx) λ f(y ) + α p(y x) α f(y ) p(y x) for all y M. With this in mind we observe that f(y 1 ) + f(y 2 ) = f(y 1 + y 2 ) p(y 1 + y 2 ) = p and so by simply re-arranging, we find that ( ) (y 1 x) + (x + y 2 ) p(y 1 x) + p(x + y 2 ) f(y 2 ) p(x + y 2 ) p(y 1 x) f(y 1 ). (1.68) Let us examine the infimum over y 1 M and the supremum of y 2 M. Choose an α such that sup f(y 2 ) p(x + y 2 ) α inf p(y 1 x) f(y 1 ) (1.69) y 2 M y 1 M for which we set g(x) = α so that g(y + λx) = f(y) + λα, and this choice of g(x) satisfies all the desired properties. Indeed, for λ > we have that f(y) + λα p(y + λx) is true if and only if α p ( y + x) f ( y λ λ) which is true by (1.69). Precisely the same argument holds for λ <. Step 2: We claim that having shown the result in Step 1, the Hahn-Banach theorem is true. In order to show this, we will need to use Zorn s Lemma. Consider the set S of all extensions F of f which satisfy F (x) p(x). There is a natural partial ordering on this collection as follows: If F 1 : M 1 R and F 2 : M 2 R then F 1 < F 2 if M 1 M 2 and F 2 M1 = F 1. This set S satisfies the axioms of Zorn s Lemma and so there exists 21

24 1.3. THE HAHN-BANACH THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS a maximal element in S, say F : M R where M contains M. We need only show that M = X. However, if this were not the case then there exists an x M and apply Step 1, which would give us an extension of T. But we assumed that F was maximal, and so no such x can exist. Theorem (Corollary to Hahn-Banach). Let X be a normed vector space. Then the following hold: 1. If M X is a closed subspace and x / M then we can find f X with f = 1, f =, and M f(x) = dist(x, M) = inf { x y : y M} >. (1.7) 2. If x is a point in X there exists f X such that f = 1 and f(x) = x. 3. If x y then there exists f X with f = 1 such that f(x) f(y). 4. Define the function E x : X R by E x (f) = f(x) then the map x E x is a linear isometry of X into X. 5. We can recover the norm in X from the norm in X as follows: x = sup { f(x) : f X, f = 1}. (1.71) Note that the second result stated above is clearly true in a Hilbert space. All we need to do is define l y, x (y) =. In the more general case of Banach spaces, this may be proven x 2 using the Hahn-Banach theorem. Proof. 1. Let our semi-norm be given by the norm function p(x) = x. Consider M+Rx and define f : M + Rx R by f(y + λx) = λδ where δ = dist(x, M). Now on M we have f(y + x) = whilst at x we have f( + 1x) = δ. It is easy to see that f(y + λx) = λ δ λ λ 1 y + x = y + λx = p(y + λx) (1.72) and so we can apply Hahn-Banach to find an extension of f to all of X whose norm is one. 2. This is a special case of property (1) with M = {}. 3. If x y then x y and so by property (2) there exists f X such that f(x y) = x y. But since f is linear this implies that f(x) f(y). 22

25 CHAPTER 1. FUNCTIONAL ANALYSIS 1.3. THE HAHN-BANACH THEOREM 4. Linearity follows for the fact that addition is defined pointwise on X ; that is, E x (f + g) = f(x) + g(x) = E x f + E x g. To see that this map is an isometry, note that E x (f) = f(x) f x (1.73) and so E x (f) f. Conversely, by Hahn-Banach applied now to X we know that for any f we can find E x such that E x (f) = f implying that this bound is saturated, so that x = E x and our map is an isometry. 5. This is immediate by using the identification given in property (4) above. Definition If X and Y are normed vector spaces and T L(X, Y ) we define the adjoint operator as T : Y X by T f = f T. Theorem If X and Y are normed vector spaces and T L(X, Y ) then 1. T = T, 2. Using the canonical injection, identify ˆX = ι X (X) X and Ŷ = ι Y (Y ) Y. If T L(Y, X ) then T X = T. 3. T is injective if and only if T is dense in Y. Proof. 1. Fix T L(X, Y ), for which our goal is to show that T is linear and bounded. To show linearity, let f, g Y be linear functionals on Y and x X so that T (af + bg)(x) = ( ) (af + b) T (x) = af(t (x)) + bg(t (x)) = (at f + bt g)(x). Since this holds for all x X we conclude that T (af + bg) = at f + bt g and so T is linear as required. To show boundedness, we note that since T is bounded it is 23

26 1.3. THE HAHN-BANACH THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS sufficient to show that T = T. Now T = sup T f f =1 = sup sup f =1 x =1 sup sup f =1 x =1 f(t x) f T x = sup T x = T x =1 so T T. On the other hand, T (X) Y is a closed subspace of Y, so for any y T (X) there is a linear functional f y Y such that f = 1 and f(x) = x, implying that f(t x) = T x so that T T f = sup f(t x) x =1 = sup T x = T x =1 and both inequalities imply that T = T. 2. This is a simple matter of computation. Let ι X : X X and ι Y : Y Y so that ι X (x)(f) = f(x) and ι Y (g) = g(y) in which case ˆX = ι X (X) and Ŷ = ι Y (Y ). Fix ˆx ˆX and g Y (T ˆx)(g) = ˆx(T g) = (T g)(x) = g(t x) = (T x)g. With the identification of T x with T x and the fact that g and ˆx where arbitrary, we have that T ˆX = T. 3. Let us begin by assuming that the range of T is dense. Since T is linear, it is sufficient to show that it has trivial kernel. Let f ker T and x X so that = (T f)(x) = (f T )(x) = f(t x). (1.74) However, since the range of T is dense in Y and x was chosen arbitrarily, this means that f is zero on a dense subset of Y. This is only possible if f and so ker T = {} and we conclude that T is injective. On the other hand, suppose that T is injective and let M = T (X) Y be the closure of the range of T. For the sake of contradiction, assume that M Y so that we can 24

27 CHAPTER 1. FUNCTIONAL ANALYSIS 1.3. THE HAHN-BANACH THEOREM find some element y Y \ M. By corollary to the Hahn-Banach Theorem, there exists f Y such that f(y) and f =. Then M (T f)(x) = (f T )(x) = f(t x) = (1.75) since T x M and f = on M. Since this holds for all x X we conclude that f is a non-trivial element in the kernel of T, which is a contradiction since T is injective and as such has trivial kernel. Exercise Let X be a real vector space and let P be a subset of X such that (i) If x, y P then x + y P, (ii) If x P and λ then λx P, (iii) If x P and x P then x =. a. Show that the relation defined by x y if and only if y x P is a partial ordering on X. b. Let M is a subspace of X such that for each x X there exists y M with x y. If f is a linear functional on M such that f(x) for x M P then there is a linear functional F on X such that F (x) for x P and F = f. M a. To show that this is a partial ordering, we need to show that it is reflexive, antisymmetric and transtive. (a) Reflexive: Since P is non-empty, let y P and note that λy P for all λ. In particular, by choosing λ = we have that P. Now x x since x x = P and we conclude that is reflexive. (b) Anti-Symmetric: Let x, y P be such that x y and y x. By definition, we then have that x y P and y x = (x y) P. By the third property of P, we conclude that x y = so x = y as required. (c) Transitivity: Let x, y, z P be such that x y and y z; then y x P and z y P. The first property then implies that z x = (z y) + (y z) P, so is transitive. 25

28 1.3. THE HAHN-BANACH THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS b. Define the function p : X R by p(x) = inf {f(y) : y M and x y}. (1.76) We claim that p is a sublinear functional, for which we need to show that p is subadditive and homogeneous. (a) Sub-Additivity: Let x 1, x 2 X so that p(x 1 + x 2 ) = inf {f(y) : y M, x 1 + x 2 y} = inf {f(y) : y M, y x 1 x 2 P } inf {f(y) : y M, y x 1 P } + inf {f(y) : y M, y x 2 P } = p(x 1 ) + p(x 2 ) where we have used the fact that taking the infimum over each component separately is less restrictive and hence yields a greater bound on the value of the function f(y). (b) Homogeneity: Let λ and x X so that p(λx) = inf {f(y) : y M, y λx P } = inf {f(y) : y M, 1λ } y x P = inf {f(λz) : z x P } = inf {λf(z) : z x P } = λp(x) where we have used the fact that f is a linear functional to pull out the scalar component. Fix x M so that if x y for some y M then f(y) = f(y x) + f(x) f(x) and so x y implies that f(x) f(y). Since p(x) infimizes over all possible y M such that x y, it will attain its minimal value at x and so f(x) = p(x) for all x M. By the Hahn-Banach theorem, we then know there exists a linear functional F : X R such that F (x) p(x) for all x X and F = f. All that remains to be shown is M that F (x) for all x P. For the sake of contradiction, assume x P such that F (x) <. By linearity, we then have that F ( x) = F (x) >. (1.77) On the other hand, since M is a subspace we know P. ( x) = x P and so Hence x since p( x) = inf {f(y) : y M, y ( x) P } = f() = (1.78) since f is a linear map and hence must take zero to zero. We have shown that F ( x) p( x), but this is a contradiction, hence F (x) > for all x P. 26

29 CHAPTER 1. FUNCTIONAL ANALYSIS 1.4. THE BAIRE CATEGORY THEOREM Exercise Let X and Y be Banach spaces, T L(X, Y ), N(T ) = {x : T x = } and M = T (X). Show that X/N(T ) is isomorphic to M if and only if M is closed. Let us begin by showing that if X/N(T ) is isomorphism to M then M is closed. By Exercise 15 of Folland we have that N(T ) is a closed proper subspace, so by Exercise 12 of Folland we know that since X is complete then X/N(T ) is also complete. Since X/N(T ) = M we then have that M is a complete subspace of Y, which is also a Banach space, and hence M must be closed. On the other hand, let us assume that M is closed and show that X/N(T ) = M. By Exercise 15 of Folland, we know there exists a unique S L (X/N(T ), Y ) such that T = S π and T = S. We immediately conclude that S is bounded since T is bounded, and S is surjective on the range of T since if y = T x then S(π(x)) = T x = y. Now S is injective since ker(s) = {x + N(T ) : S(x + N(t)) = } = {x X : S(π(x)) = } = {x X : T (x) = } = N(T ). By Corollary to the Open Mapping Theorem, S is a bijective, bounded, linear map from X/N(T ) onto M and hence is an isomorphism. 1.4 The Baire Category Theorem Definition We say that a set F X is nowhere dense if the interior of its closure is empty; that is, if ( F ) =. Theorem (Baire Category). Let (X, d) be a complete metric space. 1. If U n X is a collection of open and dense sets, then n U n is dense. 2. X cannot be expressed as the countable union of nowhere dense sets. 27

30 1.4. THE BAIRE CATEGORY THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS Proof. 1. It suffices to show that for any open W X we have that ( ) U n W. (1.79) n=1 To do this we will construct a Cauchy sequence x n x such that x W and x U n for each n. Since U 1 is dense we know that U 1 W is open and non-empty, so we may find a point x 1 and r 1 > such that B(x 1, r 1 ) U 1 W. It suffices to find a nested sequence of closed balls B(x n, r n ) such that B(x n, r n ) B(x n 1, r n 1 ) U n (1.8) n and r n. Assume that we have iteratively created a nested sequence of closed balls for some number n 1. Consider the set B(x n 1, r n 1 ) U n which is not empty since U n is dense. Hence we can find another open ball B(x n, ρ n ) such that B ( ) x n, ρn 2 B(x n 1, r n 1 ) U n so we set r n = ρn. 2 The sequence {x n } n=1 is Cauchy and each x n B(x N, r N ) provided that n > N. Thus x n x and x B(x N, r N ) U N for all N which gives the desired result. 2. For the sake of contradiction, assume that we can write X = F n (1.81) n=1 where the F n are a collection of nowhere dense sets. It is then easy to see that ( F n ) c is open (since it is the complement of a closed set), and dense (since F n has empty interior). Hence = X c = ( F n ) c (1.82) is an intersection of open dense sets, but is a contradiction to part (a) since the empty set is not dense in X. n=1 If (X, d) is not complete but is homeomorphic to a complete metric space ( X, d) then the Baire Category theorem still holds. For example, the set (, 1) with the Euclidean metric is not complete, but is homeomorphic to R with the Euclidean metric, and hence the theorem applies Meager Sets Definition A set U X is meager if U can be written as the countable union of nowhere dense sets. 28

31 CHAPTER 1. FUNCTIONAL ANALYSIS 1.4. THE BAIRE CATEGORY THEOREM The primary purpose of Baire Category is to show existence. In particular, if we want to show there exists an element with a certain property, we show the set of objects that do not have this property is meager. Definition If X, Y are Banach spaces then a map T L(X, Y ) is said to be open if T (B 1 ) contains an open ball B(, r) Y for some r >. For metric spaces X, Y, we say that T is open if for any U X open we have that T (U) Y is open. Theorem (Open Mapping Theorem). If X, Y are Banach spaces, T L(X, Y ) and T is surjective then T is open. Proof. Via the definition of an open map, it is sufficient to show that T (B 1 ) contains an open ball. Denote by B n X the open ball of radius n. Since T is surjective, we know that we can cover Y by the image of these open balls of radius n; that is, Y = T (B n ). (1.83) n=1 Since Y is complete (it s a Banach space) we know that not all of the T (B n ) are nowhere dense so in particular there is a T (B n ) that is not nowhere dense. Since the map x nx is a homeomorphism, we conclude that T (B 1 ) in particular cannot be nowhere dense. That is, there exists a y Y and r > such that B(y, 4r) T (B 1 ). (1.84) Let y 1 be some element such that y 1 y < 2r and take x 1 B 1 to be an element in the preimage of y 1 so that T x 1 = y 1. Thus B(y 1, 2r) T (B 1 ). Choose y such that y < 2r and y = T x 1 + (y T x 1 ) T (B 2 ) (1.85) since T x 1 T (B n ) and y 1 T x 1 T (B n ). Hence B Y (r, ) = { y < r} is an open ball in Y and we get that B Y (r, ) T (B 1 ). ( We claim then that B r Y, ) T (B 2 1 ). By an appropriate rescaling, consider the set { y < r2 n } so that y T (B 2 n). Pick y < r so that there exists x 2 1 B r such that 2 y T x 1 < r. By induction, there exists x 4 n B 1 such that 2n n y tx j < j=1 29 r. (1.86) 2n 1

32 1.4. THE BAIRE CATEGORY THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS Since the sum is absolutely convergent, x < 1, we know that x converges and x n < 1 (1.87) n and so ( ) T x n = y. (1.88) n Corollary If X and Y are Banach spaces and T L(X, Y ) is bijective, then T is an isomorphism; that is, T 1 L(X, Y ). Proof. Note that T (B 1 ) B y (, ɛ) implies that T 1 1 and so T is bounded as required. ɛ Equivalently, since T is surjective it is open, and T open is the same as T 1 being continuous. Injectivity gives the isomorphism. Definition Let X and Y be Banach spaces. Then T : X Y is a linear map then we say that T is a closed operator if its graph Γ = {(x, y) X Y ; y = T (x)} is closed in the product topology. Theorem (Closed Graph Theorem). Let X and Y be Banach spaces, and T : X Y be linear. If T is closed, then T is bounded. Proof. The product space X Y is still a Banach space, and since T is linear the graph Γ of T is a closed linear subspace of X Y, and hence is also complete. Consider the two projection maps π 1 : X Y X and π 2 : X Y Y acting as π 1 (x, y) = x and π 2 (x, y) = y. It is then clear that π 1 L(Γ, X) and π 2 L(Γ, Y ). Consider the restriction of these projection maps to Γ, for which we can rewrite T as T = π 2 π 1 1. (1.89) To show that T is bounded it is then sufficient to show that both π 2 and π1 1 are bounded. Clearly π 2 is bounded and we see that π 1 is a bijection between Γ and X. Since Γ is a Banach space, by Corollary we then have that π1 1 is bounded, and hence T is bounded. 3

33 CHAPTER 1. FUNCTIONAL ANALYSIS 1.4. THE BAIRE CATEGORY THEOREM The utility of the closed graph theorem can be closely related to the notion of continuity. In particular, we know that T is a continuous function if whenever x n x then T x n T x. What the Closed Graph Theorem tells us is that if x n x and T x n y then y = T x. Theorem (Uniform Boundedness Theorem). If X, Y are normed vector spaces and A L(X, Y ). 1. If there exists a non-meager subset S X such that for all x S we have that sup T A T x <, then sup T <. (1.9) T A 2. If X is a Banach space and if for each x X separately we have sup T A T x < then sup T <. (1.91) T A Proof. 1. Let E n be the set of points for which the supremum is bounded by n; that is { } E n = x S : sup T x n = {x X : T x n}. (1.92) T A T A Since E n is the intersection of a collection of closed sets, it is necessarily closed itself. We can write S = n E n, and so the Baire Category Theorem since S is non-meager we know that there is at least one E n which is not nowhere dense; that is, E n contains an open ball E n B(x, ρ) B(x, r). (1.93) We want to show that there exists M > such that T x M for all x 1 and for all T A. We claim that E 2n B(, r). Note that this claim implies the theorem since for any x X with x r we have that T x 2n for all T A, which implies that for all x 1 we have T x 2n. To see this, note that for all x B(, r) we have that r x = (x + x ) x, and that x + x, x B(x, r) and hence T x T (x + x ) + T x n + n = 2n (1.94) which is precisely what we wanted to show. 2. Note that X is non-meager by the Baire Category Theorem, and hence part (1) implies the necessary theorem. 31

34 1.4. THE BAIRE CATEGORY THEOREM CHAPTER 1. FUNCTIONAL ANALYSIS Exercise Let X and Y be Banach spaces and let {T n } n=1 be a sequence in L(X, Y ) such that lim T n x exists for every x X. If we define T x = lim T n x then show that T L(X, Y ). For any fixed x, y X and a, b R we have T (ax + by) = lim n T n (ax + by) = a lim n T n x + b lim n T n y = at x + bt y and so T is linear. Define A = {T n : n N} L(X, Y ). Since each T n is individually bounded, we know that sup T n x, x X (1.95) T A and so we can apply part (b) of the Uniform Boundedness Principle to guarantee the existence of M > such that T n M for all n N (in particular, T n x M for all x X and n N). Then T = sup lim T nx M (1.96) n and so T is also bounded. x =1 Exercise Let X, Y, Z be Banach spaces and let B : X Y Z be a separately continuous bilinear map. Show that B is jointly continuous. Denote by B x : Y Z the function B x (y) = B(x, y) so that by assumption B x L(Y, Z) for each x X, and similarly B y : X Z so that B y L(X, Z) for each y Y. Let C x, C y > be such that B x y C x y and B y x C y x. Define A X = {B x : x X}, A Y = {B y : y Y }. (1.97) If B 1 (X) = {x X : x 1} then on this set, we have that for any y Y sup B x y = sup B(x, y) B x A X x B 1 (X) = sup B y x x B 1 (X) sup C y x = C y. x B 1 (X) By the Uniform Boundedness Principle, we know there exists M such that B x M for all x X. The argument is precisely the same in the case of B y and so we know there 32