C.6 Adjoints for Operators on Hilbert Spaces

C.6 Adjoints for Operators on Hilbert Spaces 317 Additional Problems C.11. Let E R be measurable. Given 1 p and a measurable weight function w: E (0, ), the weighted L p space L p s (R) consists of all measurable functions f : E C for which f p,w = fw p <. Prove that L p w (R) is a Banach space with respect to the norm p,w. Assuming Theorem C.38, prove that if 1 p < then L p w(e) = L p 1/w (E). C.6 Adjoints for Operators on Hilbert Spaces If A is an m n complex matrix and x y is the ordinary dot product on C d, then Ax y = x A y, x C n, y C m, where A = A T = A H is the Hermitian, or conjugate transpose, of A. As an application of the Riesz Representation Theorem, we will show that there is an analogue of the Hermitian matrix for linear operators on Hilbert spaces. In Section C.10 we will see that this extends further to operators on Banach spaces, but in that setting we need to appeal to the Hahn Banach Theorem in order to construct the adjoint. Throughout this section, we will let H and K denote Hilbert spaces. C.6.1 Adjoints of Bounded Operators Exercise C.40. Let L: H K be a bounded linear operator. (a) For each g K, define a functional µ g : H C by f, µ g = Lf, g, f H. Note that, following Notation C.36, f, µ g denotes the action of the functional µ g on the vector f, while Lf, g denotes the inner product of the vectors Lf, g K. Show that µ g H. Consequently, by the Riesz Representation Theorem, there exists a unique element g H such that f, µ g = f, g, f H. (b) Define L : K H by L g = g. Show that L is a bounded linear map, that (L ) = L, and that L = L. We formalize this as a definition. Definition C.41 (Adjoint). The adjoint of L B(H, K) is the unique operator L : K H that satisfies Lf, g = f, L g, f H, g K.

318 C Functional Analysis and Operator Theory When H = K, we use the following additional terminology. Definition C.42 (Self-Adjoint and Normal Operators). Let L B(H) be given. (a) We say that L is self-adjoint or Hermitian if L = L. Equivalently, L is self-adjoint if f, g H, Lf, g = f, Lg. (b) We say that L is normal if L commutes with its adjoint, i.e., if LL = L L. All self-adjoint operators are normal, but not all normal operators are self-adjoint (Problem C.19). Remark C.43. Each complex m n matrix A determines a linear map of C n to C m. The adjoint A of this map is determined by the conjugate transpose of the matrix A. That is, identifying matrices with maps, we have A = A T = A H. This conjugate transpose matrix A H is called the Hermitian of A. For matrices it is customary to instead say that A is Hermitian if A = A H, instead of saying that A is self-adjoint. If L B(H, K), then L is a linear map of H into K, but the mapping of B(H, K) to B(K, H) given by L L is antilinear, as (αa + βb) = ᾱa + βb. Exercise C.40 shows that L = L, and hence the map L L is an isometry. We will need some facts about the relationship between invariant subspaces and adjoints. Definition C.44. Let A B(H) be given. A closed subspace M H is said to be invariant under A if A(M) M, where A(M) = {Af : f M}. Note that we do not require that A(M) be equal to M. The simplest example of an invariant subspace is M = span{f} where f is an eigenvector of A. Exercise C.45. Show that if a closed subspace M H is invariant under A B(H), then M is invariant under A. C.6.2 Adjoints of Unbounded Operators Adjoints can also be defined for unbounded operators, although now we must be careful with domains. For example, consider the differentiation operator Df = f. This operator is not defined on all of L 2 (R), but instead is densely defined in the sense of Notation C.3. For example, D maps the dense subspace S = { f C 1 0 (R) : f, f L 2 (R) } into L 2 (R). Although D: S L 2 (R) is unbounded, if f, g S then we have fg C 0 (R), so integration by parts yields

Df, g = C.6 Adjoints for Operators on Hilbert Spaces 319 f (x)g(x) dx = f(x)g (x) dx = f, Dg. Hence D = D as an operator mapping S into L 2 (R). The important point in the preceding example is that if g S is fixed, then f Df, g is a bounded linear functional on S, even though D is an unbounded operator. The following exercise extends this to general operators. Exercise C.46. Suppose that S is a dense subspace of H, and L: S K is a linear, but not necessarily bounded, operator. Let S = {g K : f Lf, g is a bounded linear functional on S}. Show that there is an operator L : S H such that Lf, g = f, L g, f S, g S. If necessary, we can always restrict a densely defined operator to a smaller but still dense domain. Given an operator L mapping some dense subspace of H into H, if we can find some dense subspace S on which L is defined and such that Lf, g = f, Lg for f, g S, then we say that L is self-adjoint. C.6.3 Bounded Self-Adjoint Operators on Hilbert Spaces We now focus in more detail on bounded self-adjoint operators on Hilbert spaces, which have many useful properties and appear often throughout this volume. Some of the results that we describe next hold only for complex Hilbert spaces; we will indicate the changes needed for real Hilbert spaces. Recall that, following the notational conventions laid out in Section A.1, we always assume in this volume that a vector space is over the complex field unless specifically stated otherwise. For complex Hilbert spaces, we have the following characterization of selfadjoint operators. This theorem does not have an analogue for real Hilbert spaces. Theorem C.47. Let H be a (complex) Hilbert space, and let A B(H) be given. Then: A is self-adjoint Af, f R for all f H. Proof.. Assume that Af, f is real for all f. Choose any f, g H. Then A(f + g), f + g = Af, f + Af, g + Ag, f + Ag, g. Since A(f + g), f + g, Af, f, and Ag, g are all real, we conclude that Af, g + Ag, f is real. Hence it equals its own complex conjugate: Af, g + Ag, f = Af, g + Ag, f = g, Af + f, Ag. (C.11)

320 C Functional Analysis and Operator Theory Similarly, after examining the equation A(f + ig), f + ig = Af, f i Af, g + i Ag, f + Ag, g, we conclude that Af, g Ag, f = g, Af + f, Ag. (C.12) Adding (C.11) and (C.12) together, we obtain 2 Af, g = 2 f, Ag = 2 A f, g. Since this is true for every f and g, we conclude that A = A. The next result gives an alternative formula for the operator norm of a self-adjoint operator. This theorem holds for both real and complex Hilbert spaces. Theorem C.48. If A B(H) is self-adjoint, then A = sup Af, f. f =1 Proof. Set M = sup f =1 Af, f. By Cauchy Bunyakowski Schwarz and the definition of operator norm, it follows that M A. Choose any unit vectors f, g H. Then, by expanding the inner products, canceling terms, and using the fact that A = A, we see that A(f + g), f + g A(f g), f g = 2 Af, g + 2 Ag, f = 2 Af, g + 2 g, Af = 4 Re Af, g. Therefore, applying the definition of M and using the Parallelogram Law, we have 4 Re Af, g A(f + g), f + g + A(f g), f g M f + g 2 + M f g 2 = 2M ( f 2 + g 2) = 4M. That is, Re Af, g M for every choice of unit vectors f and g. Write Af, g = α Af, g where α C satisfies α = 1. Then ᾱg is another unit vector, so Af, g = α Af, g = Af, ᾱg M. Hence and therefore A M. Af = sup Af, g M, g =1

C.6 Adjoints for Operators on Hilbert Spaces 321 As a corollary, we obtain the following useful fact for operators on a complex Hilbert space. This result also holds for operators on a real Hilbert space if we include the assumption A = A as a hypothesis. Exercise C.49. Let H be a (complex) Hilbert space. If A B(H) and Af, f = 0 for every f, then A = 0. C.6.4 Positive and Positive Definite Operators on Hilbert Spaces Among the self-adjoint operators, we distinguish the special class of positive operators. Definition C.50 (Positive and Positive Definite Operators). Let A B(H) be given. (a) We say that A is positive or nonnegative, denoted A 0, if Af, f 0 for every f H. (b) We say that A is positive definite or strictly positive, denoted A > 0, if Af, f > 0 for every nonzero vector f H. By Theorem C.47, since we are dealing with complex Hilbert spaces, all positive and positive definite operators are self-adjoint. When dealing with real Hilbert spaces, the assumption of self-adjointness should be added in the definition of positive and positive definite operators. Exercise C.51. Show that if A B(H, K), then A A B(H) and AA B(K) are both positive operators. Determine conditions on A that imply that A A or AA is positive definite. Additional Problems C.12. Let H 1, H 2, H 3 be Hilbert spaces. Show that if A B(H 1, H 2 ) and B B(H 2, H 3 ), then (A ) = A and (BA) = A B. C.13. Show that if A B(H, K) is a topological isomorphism then A B(K, H) is also a topological isomorphism, and (A 1 ) = (A ) 1. C.14. Show that if A B(H, K), then A = A = A A 1/2 = AA 1/2. C.15. Show that A A defines an involution on B(H) in the sense of Definition C.32. Remark: In the language of operator theory, the fact that this involution satisfies A A = A 2 means that B(H) is an example of a C -algebra. C.16. Given A B(H), show that A is normal if and only if Af = A f for every f H.

322 C Functional Analysis and Operator Theory C.17. Show that if A B(H, K) then the following statements hold. (a) ker(a) = range(a ). (b) ker(a) = range(a ). (c) A is injective if and only if range(a ) is dense in H. C.18. Given A B(H), show that ker(a) = ker(a A) and range(a A) = range(a ). Use this to show that if A is normal, then ker(a) = ker(a ) and range(a) = range(a ). C.19. Fix λ l, and let M λ be the multiplication operator defined in Exercise C.14. Find M λ, and show that M λ is normal. Determine when M λ is self-adjoint, positive, or positive definite. C.20. Fix φ L (R), and let M φ be the multiplication operator defined in Exercise C.15. Find M φ, and show that M φ is normal. Determine when M φ is self-adjoint, positive, or positive definite. C.21. Given k L 2 (R 2 ), let L k be the integral operator with kernel k. Show that the adjoint operator (L k ) is the integral operator L k whose kernel is k (x, y) = k(y, x). Characterize those kernels k L 2 (R 2 ) corresponding to self-adjoint operators L k. C.22. Let L and R be the left- and right-shift operators from Problem C.2. Show that R = L, and conclude that L and R are not normal. C.23. Let {e n } n N be an orthonormal basis for a separable Hilbert space H. Let T : H l 2 (N) be the analysis operator Tf = { f, e n } n N. Find a formula for the synthesis operator T : l 2 (N) H. C.24. Let M be a closed subspace of H, and let P B(H) be given. Show that P is the orthogonal projection of H onto M if and only if P 2 = P, P = P, and range(p) = M. C.25. Suppose that A, B B(H) are self-adjoint. Show that ABA and BAB are self-adjoint, but AB is self-adjoint if and only if AB = BA. Exhibit selfadjoint operators A, B that do not commute. C.26. Let L: H H be a self-adjoint operator on H, either bounded or unbounded and densely defined. Show that all eigenvalues of L are real, and that eigenvectors of L corresponding to distinct eigenvalues are orthogonal. C.27. Let L B(H) be normal. Show that if λ is an eigenvalue of L, then λ is an eigenvalue of L, and the λ-eigenspace of L equals the λ-eigenspace of L. Also show that eigenvectors of L corresponding to distinct eigenvalues are orthogonal.

C.7 Compact Operators on Hilbert Spaces 323 C.28. Let A B(H) be given. Show that if A B(H) is a positive operator, then all eigenvalues of A are real and nonnegative, and if A B(H) is a positive definite operator, then all eigenvalues of A are real and strictly positive. C.7 Compact Operators on Hilbert Spaces In many senses, compact operators on a Hilbert space are the ones that are most similar to linear operators on finite-dimensional spaces. Throughout this section, H and K will denote Hilbert spaces. C.7.1 Definition and Basic Properties The closed unit ball makes frequent appearances in any discussion of compact operators, so we introduce a notation for it. Notation C.52. Ball H will denote the closed unit ball in a Hilbert space H: Ball H = { f H : f 1 }. By Problem A.25, Ball H is compact if and only if H is finite-dimensional. If H and K are finite-dimensional and T : H K is linear, then T will map the closed unit ball in H to a closed ellipsoid in K, which is necessarily compact. Even if K is infinite-dimensional, if H is finite-dimensional then T(Ball H ) will be a finite-dimensional ellipsoid in K, and hence will still be compact. However, this need not be the case if H is infinite-dimensional. For example, if H is infinite-dimensional and I : H H is the identity map, then I(Ball H ) = Ball H is not compact. In general, even if T : H K is bounded and linear, T(Ball H ) need not be closed in K (see Problem C.32). Definition C.53 (Compact Operators). A linear operator T : H K is compact if T(Ball H ) has compact closure in K. We define and set B 0 (H) = B 0 (H, H). B 0 (H, K) = { T : H K : T is compact }, The next exercise gives some useful reformulations of the definition of compact operator. Exercise C.54. Given a linear operator T : H K, show that the following statements are equivalent. (a) T is compact. (b) T(Ball H ) is totally bounded.

324 C Functional Analysis and Operator Theory (c) If {f n } n N is a bounded sequence in H, then {Tf n } n N contains a convergent subsequence in K. We show now that all compact operators are bounded. Theorem C.55. B 0 (H, K) B(H, K). Proof. Assume that T : H K is linear but unbounded. Then there exist vectors f n H such that f n = 1 but Tf n n. Therefore every subsequence of {Tf n } n N is unbounded, and hence cannot converge. Exercise C.54 therefore implies that T is not compact. Now we consider the structure of the space of compact operators. Theorem C.56 (Limits of Compact Operators). B 0 (H, K) is a closed subspace of B(H, K). Specifically, if T n : H K are compact, T : H K is linear and bounded, and T T n 0, then T is compact. Proof. Exercise: Show that B 0 (H, K) is a subspace of B(H, K). We must show that B 0 (H, K) is closed. Assume that T n are compact operators and that T n T in operator norm. By Exercise C.54, to show that T is compact, it suffices to show that T(Ball H ) is a totally bounded subset of K. Choose any ε > 0. Then there exists an n such that T T n < ε. Since T n is compact, we know that T n (Ball H ) is totally bounded. This implies that there exist finitely many points h 1,...,h m Ball H such that T n (Ball H ) m j=1 B ε (T n h j ). (C.13) We will show that T(Ball H ) is totally bounded by showing that T(Ball H ) m j=1 B 3ε (Th j ). (C.14) Choose any element of T(Ball H ), i.e., any vector Tf with f 1. Then T n f T n (Ball H ), so by equation (C.13) there must be some j such that T n f T n h j < ε. Consequently, Tf Th j Tf T n f + T n f T n h j + T n h j Th j < T T n f + ε + T n T h j < ε 1 + ε + ε 1 = 3ε. Hence equation (C.14) follows, so T is compact. Compact operators are well-behaved with respect to compositions. Exercise C.57. Let H 1, H 2, H 3 be Hilbert spaces.

C.7 Compact Operators on Hilbert Spaces 325 (a) Show that if A: H 1 H 2 is bounded and linear and T : H 2 H 3 is compact, then TA: H 1 H 3 is compact. (b) Show that if T : H 1 H 2 is compact and A: H 2 H 3 is bounded and linear, then AT : H 1 H 3 is compact. In particular, B 0 (H) is a closed, two-sided ideal in B(H) under composition of operators. It is possible for the product of two noncompact operators to be compact. An example is given in Exercise 2.52, which shows that the product of a noncompact time-limiting operator with a noncompact frequency-limiting operator can be compact. C.7.2 Finite-Rank Operators Recall that the rank of an operator T : H K is the vector space dimension of range(t). We say that T is a finite-rank operator if range(t) is finitedimensional. We set B 00 (H, K) = { T B(H, K) : T is finite-rank }, and let B 00 (H) = B 00 (H, H). Problem C.8 shows that a linear, finite-rank operator need not be bounded (that is why we include the assumption of boundedness in the definition of B 00 (H, K) above). On the other hand, every finite-rank operator that is bounded must be compact. Exercise C.58. Show that if T : H K is bounded, linear, and has finite rank, then T is compact. Thus, B 00 (H, K) B 0 (H, K). Combining the preceding exercise with Theorem C.56, we obtain a useful technique for showing that a given operator is compact. Specifically, if we are given T : H K, and if we can find bounded linear finite-rank operators T n such that T T n 0, then T must be compact. Moreover, the converse is also true: Every compact operator T : H K can be realized as the operator norm limit of a sequence of bounded finite-rank operators. Theorem C.59. B 00 (H, K) is a dense subspace of B 0 (H, K). In particular, if T : H K is compact, then there exist bounded linear finite-rank operators T n : H K such that T n T in operator norm. Proof. Assume that T is compact, and set R = range(t). If R is finitedimensional, then T is finite-rank, and we are done. So, assume that R is infinite-dimensional. Problem C.34 shows that R is separable. Therefore,

326 C Functional Analysis and Operator Theory since R is closed, we can find a countable orthonormal basis {e n } n N for R. For any f H we have Tf R, so Tf = Tf, e n e n, f H. Define T N f = N Tf, e n e n, f H, and note that T N = P N T where P N is the orthogonal projection of K onto the closed subspace span{e 1,...,e N }. By definition, we have that T N converges to T in the strong operator topology, i.e., T N f Tf for every f. Our goal is to show that T N T in operator norm. Choose any ε > 0. Since T(Ball H ) is totally bounded, it is covered by finitely many ε-balls centered at points in T(Ball H ). Hence, there exist h 1,..., h m Ball H such that T(Ball H ) m k=1 B ε (Th k ). Since lim N Th k T N h k = 0 for k = 1,...,m, we can find an N 0 such that N > N 0, Th k T N h k < ε, k = 1,...,m. Choose any f with f = 1 and any N > N 0. Then Tf B ε (Th k ) for some k, i.e., Tf Th k < ε. Therefore so T N f T N h k = P N Tf P N Th k P N Tf Th k < 1 ε, Tf T N f Tf Th k + Th k T N h k + T N h k T N f < ε + ε + ε = 3ε. This is true for every unit vector, so T T N 3ε for all N > N 0, and thus T T N 0. The next exercise characterizes the finite-rank operators, and uses this characterization to show that compactness is preserved under adjoints. Exercise C.60. (a) Show that T B(H, K) has finite rank if and only if there exist ϕ 1,...,ϕ N H and ψ 1,..., ψ N K such that N Tf = f, ϕ k ψ k, f H. (C.15) k=1 In case this holds, find T and show that T is also finite-rank. (b) Show that if T B(H, K), then T is compact if and only if T is compact. Following Notation C.18, an alternative way to write the finite-rank operator given in equation (C.15) is T = N k=1 (ψ k ϕ k ).

C.7 Compact Operators on Hilbert Spaces 327 C.7.3 Integral Operators with Square-Integrable Kernels The next exercise shows that integral operators that have a square-integrable kernel (which we call the Hilbert Schmidt integral operators) are compact. We will refine this yet further in Theorem C.79 below. Exercise C.61 (Hilbert Schmidt Integral Operators). Fix k L 2 (R 2 ), and let L k be the integral operator with kernel k. (a) Let {e mn } m,n N be an orthonormal basis for L 2 (R 2 ) of the form e mn = e m e n, as constructed in Problem C.29. Define k N = N m=1 N k, e mn e mn, and show that the corresponding integral operator L kn is bounded and finite-rank. (b) Show that L k L kn is the integral operator with kernel k k N, and that L k L kn k k N 2. Conclude that L k is a compact mapping of L 2 (R) into itself. Additional Problems C.29. Let {e n } n N be any orthonormal basis for L 2 (R). Define e mn (x, y) = (e m e n )(x, y) = e m (x)e n (y), x, y R. Show that {e mn } m,n N is an orthonormal basis for L 2 (R 2 ). C.30. Suppose that T : H K is compact and {e n } n N is an orthonormal sequence in H. Show that Te n 0. C.31. Let H be an infinite-dimensional Hilbert space. Show that if T : H K is compact and injective, then T 1 : range(t) H is unbounded. C.32. Fix λ l, and let M λ be the multiplication operator defined in Exercise C.14. Show that M λ is compact if and only if λ c 0 (i.e., λ n 0 as n ). Show that if λ c 0 and λ n 0 for every n, then T(Ball H ) is not closed in l 2. C.33. Fix φ L (R), and let M φ be the multiplication operator defined in Exercise C.15, where we take p = 2. Show that M φ is compact if and only if φ = 0 a.e. C.34. Prove that if T B 0 (H, K), then range(t) is a separable subspace of K. C.35. Show that if T : H H is compact and λ 0 is an eigenvalue of T, then the corresponding eigenspace ker(t λi) is finite-dimensional.

328 C Functional Analysis and Operator Theory C.8 The Spectral Theorem for Compact Self-Adjoint Operators Throughout this section, H will denote a Hilbert space. The Spectral Theorem provides a fundamental decomposition for selfadjoint operators on a Hilbert space. The version of the Spectral Theorem that we will need in this volume asserts that if T is a compact self-adjoint operator on H, then there is an orthonormal basis for H that consists of eigenvectors of T, and T has a simple representation with respect to this orthonormal basis. This is only a basic version of the Spectral Theorem. The theorem generalizes to compact normal operators with little change, allowing eigenvalues to be complex instead of real. Much more deeply, however, there is a Spectral Theorem for self-adjoint operators that are not compact, and even for some that are unbounded. We refer to texts on operator theory for such extensions, e.g., see Conway [Con00]. C.8.1 Existence of an Eigenvalue In the finite-dimensional setting, the first step in proving the Spectral Theorem is to note that every n n matrix always has at least one (complex) eigenvalue. Unfortunately, this is not true in infinite dimensions, even for compact operators. Exercise C.62. (a) The Volterra operator on L 2 [0, 1] is the integral operator V f(x) = x 0 f(y)dy. Show that V is compact, not self-adjoint, and has no eigenvalues. (b) Show that the multiplication operator M : L 2 [0, 1] L 2 [0, 1] given by M f(x) = xf(x) is self-adjoint (in fact, positive definite), but is not compact and has no eigenvalues. On the other hand, we will show that an operator that is both compact and self-adjoint must have an eigenvalue. In order to do this, we first need the following sufficient condition for the existence of an eigenvalue of a compact operator. Lemma C.63. If T : H H is compact and λ 0, then inf Tf λf = 0 = λ is an eigenvalue of T. f =1 Proof. Suppose we can find unit vectors f n such that Tf n λf n 0. Since T is compact, {Tf n } n N has a convergent subsequence, say Tf nk g H. Then λf nk = ( λf nk Tf nk ) + Tfnk 0 + g = g. Since the f nk are unit vectors and λ 0, we conclude that g 0. Moreover, since T is continuous it follows that λtf nk Tg. But we also know that λtf nk λg, so we conclude that Tg = λg.

C.8 The Spectral Theorem for Compact Self-Adjoint Operators 329 Now we show that compactness combined with self-adjointness implies the existence of an eigenvalue. Since the eigenvalues of a self-adjoint operator are real and are bounded in absolute value by the operator norm, we know that any eigenvalue λ must lie in the range T λ T. Theorem C.64. If T : H H is compact and self-adjoint, then either T or T is an eigenvalue of T. Proof. Since T is self-adjoint, we have T = sup f =1 Tf, f by Theorem C.48. Hence, there must exist unit vectors f n such that Tf n, f n T. Since T is self-adjoint, every inner product Tf n, f n is real, so we can find a subsequence { Tg n, g n } n N that converges either to T or to T. Let λ be either T or T, as appropriate. Then we have g n = 1 for every n and Tg n, g n λ. Hence, since both λ and Tg n, g n are real, Tg n λg n 2 = Tg n 2 2λ Tg n, g n + λ 2 g n 2 T 2 g n 2 2λ Tg n, g n + λ 2 g n 2 = λ 2 2λ Tg n, g n + λ 2 λ 2 2λ 2 + λ 2 = 0. It therefore follows from Lemma C.63 that λ is an eigenvalue of T. C.8.2 The Spectral Theorem We obtain the Spectral Theorem by iterating Theorem C.64. The index set J in the following theorem is either J = {1,...,N} with N = dim(range(t)) if T has finite rank, or J = N otherwise. Theorem C.65 (Spectral Theorem for Compact Self-Adjoint Operators). Let T : H H be compact and self-adjoint. Then, there exist: (a) countably many nonzero real numbers {λ n } n J, either finitely many or with λ n 0 if infinitely many, and (b) an orthonormal basis {e n } n J of range(t), such that Tf = n J λ n f, e n e n, f H. (C.16) Each λ n is an eigenvalue of T, and each e n is a corresponding eigenvector. Proof. If T = 0 then we can take J =, so assume that T is not the zero operator. Since T is compact, Problem C.34 implies that range(t) is separable, and hence has either a finite or a countably infinite orthonormal basis. Define H 1 = H and T 1 = T. By Theorem C.64, T 1 has an eigenvalue λ 1 that satisfies λ 1 = T 1 > 0. Let e 1 be a corresponding eigenvector, normalized so that e 1 = 1.

330 C Functional Analysis and Operator Theory Let H 2 = {e 1 } and let T 2 = T H2. If T 2 = 0, then we stop at this point. Otherwise, we continue as follows. Since span{e 1 } is invariant under T 1, we know from Exercise C.45 that H 2 is invariant under T 1 = T 1. Exercise: Show that T 2 : H 2 H 2 is compact and self-adjoint. Therefore T 2 has an eigenvalue λ 2 such that λ 2 = T 2 > 0, and since T 2 is a restriction of T 1, we have λ 2 = T 2 T 1 = λ 1. Let e 2 be a corresponding unit eigenvector. By definition of H 2, we have e 2 e 1. Further, λ 2 is an eigenvalue of T (not just T 2 ), and e 2 is the corresponding eigenvector of T. Now set H 3 = {e 1, e 2 } and T 3 = T H3. If T 3 = 0, then we stop at this point. Otherwise, we continue as before to construct an eigenvalue λ 3 and unit eigenvector e 3 (which will be orthogonal to both e 1 and e 2 ). As we continue in this process, there are two possibilities. Case 1: T N+1 = 0 for some N. In this case, since H N+1 = {e 1,...,e N }, we have H = span{e 1,..., e N } H N+1. Therefore, each f H can be written uniquely as f = N f, e n e n + v f where v f H N+1. Since Tv f = T N+1 v f = 0, we have Tf = N f, e n Te n + Tv f = N λ n f, e n e n. In this case T is finite-rank and has the required form. Case 2: T N 0 for any N. In this case we obtain countably many eigenvalues λ n and corresponding orthonormal eigenvectors e n. Since T is compact, Problem C.30 implies that Te n 0. Since Te n = λ n e n and e n = 1, we conclude that λ n 0. Let M = span{e n } n N. Then {e n } n N is an orthonormal basis for M, and H = M M. Hence, each f H can be written uniquely as f = f, e n e n + v f for some v f M. Therefore Tf = f, e n Te n + Tv f = λ n f, e n e n + Tv f. Note that since span{e 1,...,e N } M, we have for any N > 0 that

C.8 The Spectral Theorem for Compact Self-Adjoint Operators 331 v f M span{e 1,...,e N } = H N. Since T HN = T N, we therefore have Tv f = T N v f T N v f = λ N v f 0 as N. Consequently Tv f = 0, so T has the required form. Exercise: Complete the proof by showing that, in either case, {e n } n J is an orthonormal basis for range(t). Since the nonzero eigenvalues of a compact self-adjoint operator converge to zero (if there are infinitely many), we usually order them in decreasing order according to absolute value: λ 1 λ 2. The multiplicity of a given eigenvalue λ is the number of times it is repeated in the list of eigenvalues given above, or, equivalently, it is the dimension of the λ-eigenspace ker(t λi). The multiplicity of any given eigenvalue is finite. The Spectral Theorem is stated above in terms of the nonzero eigenvalues of T. The zero eigenspace is ker(t), and the vectors in this space are eigenvectors of T for λ = 0. If H is separable, then ker(t) is separable, so by combining an orthonormal basis for ker(t) with the orthonormal basis for range(t) given by Theorem C.65, we obtain the following reformulation of the Spectral Theorem, which states that every compact, self-adjoint operator on a separable Hilbert space is actually a multiplication operator M λ of the form introduced in Exercise C.14. Exercise C.66. Let H be a separable Hilbert space. Let J = {1,..., dim(h)} if H is finite-dimensional, or J = N if H is infinite-dimensional. Given T B(H), show that the following statements are equivalent. (a) T is compact and self-adjoint. (b) There exists a real sequence λ = {λ n } n J, with λ c 0 if J = N, and an orthonormal basis {e n } n J for H such that Tf = M λ f = n J λ n f, e n e n, f H. For the finite-dimensional setting, the Spectral Theorem corresponds to the diagonalization of Hermitian matrices. Exercise C.67. Let A be an n n complex matrix, viewed as a mapping of C n into itself. Show that A is a Hermitian matrix if and only if A = UΛU 1 where U is unitary and Λ is real and diagonal. Also show that, in this case, the diagonal entries of Λ are the eigenvalues of A, and the columns of U form an orthonormal basis for C n consisting of eigenvectors of A.

332 C Functional Analysis and Operator Theory As an application of the Spectral Theorem, we construct a square root of a compact positive operator. Exercise C.68. Assume that T : H H is both compact and positive, and let equation (C.16) be the representation of T given by the Spectral Theorem. Define T 1/2 f = λ 1/2 n f, e n e n, f H. n J Show that T 1/2 is well-defined, compact, and positive, and that (T 1/2 ) 2 = T 1/2 T 1/2 = T. Additional Problems C.36. Define the spectral radius ρ(t) of an m m matrix T to be ρ(t) = max{ λ : λ is an eigenvalue of T }. Let A be an m n complex matrix. Show directly that if we place the l 2 norm on both C n and C m, then the operator norm of A as a mapping of C n into C m is A = ρ(a A) 1/2. Compare this to Problem C.7. C.9 Hilbert Schmidt Operators The Hilbert Schmidt operators are a special subclass of the compact operators on a Hilbert space. Although the definition of a Hilbert Schmidt operator makes sense when H is nonseparable (by using a complete orthonormal system instead of a countable orthonormal basis), in this volume we will only need to consider Hilbert Schmidt operators on separable spaces. Therefore, throughout this section, H will denote a separable Hilbert space. C.9.1 Definition and Basic Properties Definition C.69. An operator T B(H) is Hilbert Schmidt if there exists an orthonormal basis {e n } n N for H such that Te n 2 <. The next exercise shows that the choice of orthonormal basis is irrelevant. Exercise C.70. Fix T B(H). Given an orthonormal basis E = {e n } n N for H, set ( ) 1/2 S(E) = Te n 2. Show that S(E) is independent of the choice of orthonormal basis E. That is, if S(E) is finite for one orthonormal basis then it is finite for all and always takes the same value, and if S(E) is infinite for one orthonormal basis then it is infinite for all.

C.9 Hilbert Schmidt Operators 333 Definition C.71. The space of Hilbert Schmidt operators on H is B 2 (H) = { T B(H) : T is Hilbert Schmidt }. The Hilbert Schmidt norm of T B 2 (H) is T HS = T B2 = ( ) 1/2 Te n 2, where {e n } n N is any orthonormal basis for H. The notation B 2 (H) emphasizes that the space of Hilbert Schmidt operators is the central (p = 2) space in the family of Schatten classes B p (H) described below. We will use the notations HS and B2 interchangeably. Note that if we write T 2 B 2 = Te n 2 = Te n, e m 2, (C.17) m=1 then we see that the Hilbert Schmidt norm is simply the l 2 -norm of the matrix representation of T with respect to the orthonormal basis {e n } n N. The fact that B2 is indeed a norm on B 2 (H), and some of the other basic properties of Hilbert Schmidt operators, are developed in the next exercise. Exercise C.72. Prove the following facts. (a) T T B2 for all T B 2 (H). (b) B2 is a norm on B 2 (H), and B 2 (H) is complete in this norm. (c) B 2 (H) is closed under adjoints, and T B2 = T B2 for all T B 2 (H). (d) If T B 2 (H) and A B(H), then AT, TA B 2 (H), and we have AT B2 A T B2, TA B2 A T B2. Consequently, B 2 (H) is a two-sided ideal in B(H). (e) All finite-rank operators are Hilbert Schmidt, i.e., B 00 (H) B 2 (H). (f) All Hilbert Schmidt operators are compact, i.e., B 2 (H) B 0 (H). (g) The space B 00 (H) of finite-rank operators is dense in B 2 (H) with respect to both the operator norm and the Hilbert Schmidt norm. C.9.2 Singular Numbers and Schatten Classes We will give an equivalent formulation of Hilbert Schmidt operators in terms of their singular numbers. Singular numbers can be defined for any compact operator T : H H. Although a compact operator need not have any eigenvalues, the operator T T is both compact and self-adjoint (in fact, positive). Therefore, we can apply the Spectral Theorem to T T to deduce that

334 C Functional Analysis and Operator Theory there exists an orthonormal sequence {e n } n J and corresponding real numbers {µ n } n J such that T Tf = n J µ n f, e n e n, f H. The µ n are the nonzero eigenvalues of T T, and since T T is positive, these µ n are strictly positive. There are either finitely many nonzero eigenvalues of T T, or if infinitely many then we have µ n 0. The index set J is either J = {1,...,N} with N = dim(range(t T)) if T T is finite-rank (which by Problem C.18 happens if and only if T is finite-rank), or J = N otherwise. Definition C.73 (Singular Numbers). Let T : H H be compact, and let {µ n } n J and {e n } n J be as constructed above. Then the singular numbers or singular values of T are taken in decreasing order: s n (T) = µ 1/2 n, n J, s 1 (T) s 2 (T) > 0. The vectors e n are corresponding singular vectors of T. In particular, a bounded finite-rank operator has only finitely many nonzero singular numbers. The singular numbers of a self-adjoint operator can be written directly in terms of its eigenvalues. Exercise C.74. Show that if T is compact and self-adjoint and {λ n (T)} n J are its nonzero eigenvalues taken in decreasing order, then s n (T) = λ n (T). Consequently, if T is compact and positive then s n (T) = λ n (T). Exercise C.68 shows how to construct the square root of a compact positive operator. If T is an arbitrary compact operator, then T T is positive and compact, and we have a special name for its square root. Definition C.75 (Absolute Value). If T : H H is compact, then the positive compact operator is the absolute value of T. T = (T T) 1/2 In this terminology, the singular numbers of a compact operator T are the nonzero eigenvalues of T = (T T) 1/2, so we have the relations s n (T) = λ n ( T ) = λ n (T T) 1/2. In particular, since T T is positive, its largest eigenvalue is its operator norm. Therefore, s 1 (T) = λ 1 (T T) 1/2 = T T 1/2 = T, (C.18) the final equality following from Problem C.14.

C.9 Hilbert Schmidt Operators 335 Remark C.76. It can be shown that every bounded operator A on H can be written in the form A = UP where P is positive and U is a partial isometry (meaning that Uf = f for all f (keru) ), and this polar decomposition is unique if we impose the condition that ker(u) = ker(p). In particular, if A is compact then P coincides with the operator A defined above. Thus, absolute value can be defined for arbitrary bounded operators, although we will only need it for compact operators in this volume. Now we can reformulate the definition of Hilbert Schmidt operators in terms of singular numbers. Exercise C.77. Let T : H H be compact, and let s = {s n (T)} n J be the singular numbers of T. Show that T is Hilbert Schmidt if and only if s l 2 (J), and, in this case, T B2 = s 2 = ( ) 1/2 s n (T) 2. n J More generally, the Schatten class B p (H) is the space of compact operators that have l p singular numbers. Definition C.78 (Schatten Class). Given 1 p <, the Schatten class B p (H) consists of all compact operators T : H H whose corresponding singular numbers {s n (T)} n J satisfy T Bp = ( ) 1/p s n (T) p <. n J Note that since a bounded finite-rank operator has only finitely many singular numbers, we have B 00 (H) B p (H), for each 1 p <. In fact, it can be shown that the finite-rank operators are dense in B p (H) for each p. Since the l -norm of the singular values is s = sup s n (T) = s 1 (T) = T, n J we usually define the p = Schatten class to be B (H) = B 0 (H), the space of all compact operators on H. As in the discussion leading up to Exercise C.66, it is often convenient to extend the singular vectors so that we obtain an orthonormal basis for all of H. The singular vectors {e n } n J as we have defined them form an orthonormal basis for range(t). Since T T is self-adjoint, we have range(t) = ker(t T). Hence we can form an orthonormal basis for all of H by combining {e n } n N with an orthonormal basis {f m } m I for ker(t T). We can regard the vectors

336 C Functional Analysis and Operator Theory f m as being singular vectors of T corresponding to zero singular numbers. Each vector f H can be written and we have f = n J T Tf = n J f, e n e n + n I f, e n T Te n + n I f, f m f m, f, f m T Tf m = n J s n (T) 2 f, e n e n. C.9.3 Hilbert Schmidt Integral Operators We now specialize to the case of integral operators on L 2 (R). Theorem C.19 showed that if k L 2 (R 2 ), then the corresponding integral operator L k is a bounded operator on L 2 (R), with operator norm L k k 2. Exercise C.61 further showed that L k is compact. The next result shows that L k is actually Hilbert Schmidt when k L 2 (R 2 ), and further that all Hilbert Schmidt operators on L 2 (R) can be written as integral operators with square-integrable kernels. Theorem C.79 (Hilbert Schmidt Kernel Theorem). (a) If k L 2 (R 2 ), then the integral operator L k with kernel k is Hilbert Schmidt, and L k B2 = k 2. (b) If T is a Hilbert Schmidt operator on L 2 (R), then there exists k L 2 (R 2 ) such that T = L k. (c) The mapping k L k is an isometric isomorphism of L 2 (R 2 ) onto B 2 (L 2 (R)). Proof. (a) Assume that k L 2 (R 2 ), and choose any orthonormal basis {e n } n N for L 2 (R). Problem C.29 shows that if we set e mn (x, y) = (e m e n )(x, y) = e m (x)e n (y), then {e mn } m,n N is an orthonormal basis for L 2 (R 2 ). Since k e mn L 1 (R 2 ), Fubini s Theorem allows us to interchange integrals in the following calculation: k, e mn = k(x, y)e m (x) e n (y)dxdy = = ( ) k(x, y)e n (y)dy e m (x) dx L k e n (x)e m (x) dx = L k e n, e m.

Therefore, L k 2 B 2 = L k e n 2 = = C.9 Hilbert Schmidt Operators 337 ( ) L k e n, e m 2 m=1 m=1 k, e mn 2 = k 2 2 <, so L k is Hilbert Schmidt and the mapping k L k is isometric. (b) Suppose that T is Hilbert Schmidt on L 2 (R). Choose any orthonormal basis {e n } n N for L 2 (R), and let e mn be as above. Equation (C.17) shows that { Te m, e n } m,n N l 2 (N 2 ). Therefore, the series k = m=1 Te n, e m e mn converges unconditionally in L 2 (R 2 ). Hence k L 2 (R 2 ), so the integral operator L k is Hilbert Schmidt. Since part (a) shows that k L k is an isometry, we conclude that L k = Te n, e m L emn, (C.19) m=1 with unconditional convergence of this series in Hilbert Schmidt norm. Since the Hilbert Schmidt norm dominates the operator norm, the series in equation (C.19) also converges unconditionally in operator norm. Hence, given f L 2 (R), we have that L k f = m=1 Te n, e m L emn f, where the series converges unconditionally in L 2 (R). Now, Example C.17 shows that L emn is the rank one operator L emn f = f, e n e m. Therefore, using the unconditionality of the convergence to reorder the summations, we have L k f = Te n, e m f, e n e m = = m=1 ( ) f, e n Te n, e m e m m=1 f, e n Te n ( ) = T f, e n e n = Tf,

338 C Functional Analysis and Operator Theory where the continuity of T is used to obtain the penultimate equality. Hence T = L k. Thus B 2 (L 2 (R)) is isometrically isomorphic to L 2 (R 2 ), and therefore inherits a Hilbert space structure from L 2 (R 2 ). In terms of kernels, the inner product on B 2 (L 2 (R)) is L k, L h = k, h. B 2 (L 2 (R)) is a separable Hilbert space, for if {e n } n N is any orthonormal basis for L 2 (R) and we set e mn = e m e n, then {L emn } m,n N is an orthonormal basis for B 2 (L 2 (R)). By Example C.17, each operator L emn is a bounded, rank one operator whose range is span{e m }. As discussed in Notation C.18, we often identify this operator with the function e m e n, and write L emn = e m e n. Thus {e m e n } m,n N is an orthonormal basis for B 2 (L 2 (R)) consisting of bounded rank one operators, and equation (C.19) is the orthonormal basis representation of a Hilbert Schmidt operator as a superposition of rank one operators, which we can also write as T = Te n, e m (e m e n ). m=1 This series converges both in Hilbert Schmidt norm and in operator norm. C.9.4 Trace-Class Operators Definition C.80 (Trace-Class Operators). A compact operator T B(H) is trace-class if its singular numbers {s n (T)} n J satisfy s n (T) <. n The space of trace-class operators on H is the Schatten class B 1 (H). Since the largest singular number of a compact operator T is s 1 (T) = T, the operator norm is dominated by the trace-class norm, or indeed by any Schatten-class norm: T T Bp, 1 p. The next exercise provides an equivalent definition of the trace-class norm reminiscent of Definition C.71. Exercise C.81. Suppose that T : H H is compact, and prove the following facts.

C.10 The Hahn Banach Theorem 339 (a) The quantity T en, e n is independent of the choice of orthonormal basis {e n } n N. (b) T B 1 (H) if and only if T en, e n < for some orthonormal basis {e n } n N. (c) If T B 1 (H), then T B1 = T en, e n. (d) T B 1 (H) if and only if T 1/2 B 2 (H). The trace class B 1 (H) is a Banach space with respect to the norm B1, and is also a two-sided ideal in B(H). All finite-rank operators are trace-class, and every trace-class operator is the limit in trace-class norm of a sequence of finite-rank operators. The trace of a trace-class operator T B 1 (H) is trace(t) = Ten, e n, where {e n } n N is an orthonormal basis for H. This series converges and is independent of the choice of orthonormal basis. Thus the trace is the sum of the diagonal elements of the matrix representation of T with respect to an orthonormal basis. The trace induces an inner product on the space of Hilbert Schmidt operators. Specifically, if A, B B 2 (H), then AB is trace-class, and A, B = trace(ab ) defines an inner product on B 2 (H) whose induced norm is exactly B2. Additional Problems C.37. Given g, h H and 1 p <, show that the Schatten-class norm of g h is g h Bp = g h, where (g h)(f) = f, h g. C.10 The Hahn Banach Theorem The Hahn Banach Theorem is a powerful result dealing with extension of linear functionals defined on subspaces of a space. C.10.1 Abstract Statement of the Hahn Banach Theorem We concentrate in this volume on complex vector spaces. For such spaces, the abstract version of the Hahn Banach Theorem is as follows.

340 C Functional Analysis and Operator Theory Theorem C.82 (Hahn Banach Theorem). Let X be a complex vector space and let ρ be a seminorm on X. If M is a subspace of X and λ: M C is a linear functional on M satisfying f, λ ρ(f), f M, then there exists a linear functional Λ: X C such that Λ M = λ and f, Λ ρ(f), f X. For real vector spaces, the dominating seminorm ρ can be replaced by a sublinear function q: X R that satisfies q(f + g) q(f) + q(g) and q(cf) = q(f) for all f, g X and scalars c 0. The proof of the Hahn Banach Theorem takes some preparation, and therefore we will omit it. The proof for arbitrary vector spaces is not constructive, as it relies on the Axiom of Choice in the form of Zorn s Lemma. The Hahn Banach Theorem is intimately tied to convexity, and there are generalizations to more abstract settings such as locally convex topological vector spaces. C.10.2 Implications of the Hahn Banach Theorem In practice, it is usually not the Hahn Banach Theorem itself but rather its corollaries that are applied. Therefore we will concentrate on the implications of the Hahn Banach Theorem. Since these corollaries are so important, when invoking any one of them it is customary to write by the Hahn Banach Theorem instead of by a corollary to the Hahn Banach Theorem. Our first corollary states that any bounded linear functional λ on a subspace M of a normed space X has an extension to the entire space whose operator norm on X equals the operator norm of λ on M. This is easy to prove when the space is a Hilbert space, but it is far from obvious that such an extension should be be possible on non-inner product spaces. Note that we do not require the subspace M to be closed. Corollary C.83 (Hahn Banach). Let X be a normed space and M a subspace of X. If λ M, then there exists Λ X such that Λ M = λ and Λ = λ. Proof. Set ρ(f) = λ f for f X. Note that ρ is defined on all of X, and is a seminorm on X (in fact, it is a norm if λ 0). Further, f M, f, λ f λ = ρ(f). Hence, Theorem C.82 implies that there exists a linear functional Λ: X C such that Λ M = λ (which implies Λ λ ) and f X, which implies that Λ λ. f, Λ ρ(f) = λ f,

C.10 The Hahn Banach Theorem 341 Given a normed space X and given µ X, the operator norm of µ is µ = sup f, µ. f X, f =1 Thus, we obtain the operator norm of µ by looking back at its action on X. The next corollary provides a complementary viewpoint: The norm of f X can be obtained by looking forward to its action on X. Corollary C.84 (Hahn Banach). If X is a normed space and f X, then f = sup f, µ. µ X, µ =1 (C.20) Further, the supremum is achieved. Proof. Fix f X, and let α denote the supremum on the right-hand side of equation (C.20). Since f, µ f µ, we have α f. Let M = span{x}, and define λ: M C by cf, λ = c f. Then λ M and λ = 1. Corollary C.83 therefore implies that there exists some Λ X with Λ M = λ and Λ = λ = 1. In particular, since f M, we have α f, Λ = f, λ = f, and therefore the supremum in equation (C.20) is achieved. Now we can give one of the most powerful and often-used implications of the Hahn Banach Theorem. It states that we can find a bounded linear functional that separates a point from a closed subspace of a normed space. This is easy to prove constructively for the case of a Hilbert space, but it is quite amazing that we can do this in arbitrary normed spaces. Corollary C.85 (Hahn Banach). Let X be a normed linear space. Suppose that: (a) M is a closed subspace of X, (b) f 0 X\M, and (c) d = dist(f 0, M) = inf { f 0 m : m M }. Then there exists µ X such that f 0, µ = 1, µ M = 0, and µ = 1 d. Proof. Note that d > 0 since M is closed. Define M 1 = span{m, x 0 }. Since f 0 / M, each f M 1 can be written uniquely as f = m f + t f f 0 for some m f M and t f C. Define λ: M 1 C by f, λ = t f. Then λ is linear, λ M = 0, and f 0, λ = 1. If f M 1 and t f 0, then we have m f /t f M, so f = t f f 0 + m f = t f f 0 ( mf t f ) tf d.

342 C Functional Analysis and Operator Theory If t f = 0 (so f M), this is still true. Hence, f, λ = t f f /d for all f M 1. Therefore λ 1/d. On the other hand, there exist m n M such that f 0 m n d. Since λ vanishes on M, we therefore have 1 = f 0, λ = f 0 m n, λ f 0 m n λ d λ. Therefore λ 1/d. Applying Corollary C.83, there exists a µ X such that µ M1 = λ and µ = λ. This functional µ has all of the required properties. C.10.3 Orthogonal Complements in Normed Spaces We can generalize the notion of orthogonal complements of subsets of Hilbert spaces to normed spaces by allowing the orthogonal complement of S X to be a subset of X rather than X. Definition C.86 (Orthogonal Complement). If S is a subset of a normed space X, then its orthogonal complement S is the subset of X defined by S = { µ X : f, µ = 0 for all f S }. Exercise C.87. Show that S is a closed subspace of X. The next exercise gives a useful characterization of complete sequences in a normed space. The analogous characterization for the case of Hilbert spaces appeared as Exercise A.101. Exercise C.88. Let S be a subset of a normed space X. Prove that S is complete S = {0}. Consequently, a subspace M of X is dense in X if and only if M = {0}. C.10.4 X and Reflexivity Let X be a normed linear space. Following our notational conventions, if f X and µ X then we write f, µ to denote the action of µ on f, and we take this notation to be linear in f and antilinear in µ. Holding µ fixed, the mapping f f, µ is a bounded linear functional on X. On the other hand, if we hold f fixed, then µ f, µ is an antilinear functional on X. We can make a linear mapping by considering µ f, µ instead. Thus, each element f X determines a linear functional f : X C by the formula µ, f = f, µ for µ X. If this linear functional is bounded, then it is a bounded linear functional on X and hence belongs to the dual space of X, which we denote by X. The next result shows that f is indeed bounded and has operator norm f = f. This gives us a a natural identification of X with a subset of X.

C.10 The Hahn Banach Theorem 343 Theorem C.89. Let X be a normed space. Given f X, define f : X C by µ, f = f, µ, µ X. Then f is a bounded linear functional on X, with operator norm f = f. Consequently, the mapping is a linear isometry of X into X. T : X X f Proof. By definition of the operator norm, we have f = f, sup µ, f. µ X, µ =1 On the other hand, by the Hahn Banach Theorem in the form of Corollary C.84, we have f = sup µ X, µ =1 f, µ. Since f, µ = µ, f, the result follows. Definition C.90 (Natural Embedding of X into X ). Let X be a normed space. (a) The mapping T : f f defined in Theorem C.89 is called the natural embedding or the canonical embedding of X into X. (b) If the natural embedding of X into X is surjective, then we say that X is reflexive. Remark C.91. We emphasize that in order for X to be called reflexive, the natural embedding must be a surjective isometry. There exist Banach spaces X such that X = X even though X is not reflexive [Jam51]. Exercise C.92. Let E R d be Lebesgue measurable. Show that l p and L p (E) are reflexive for each 1 < p <. C.10.5 Adjoints of Operators on Banach Spaces We saw in Section C.6 that if H, K are Hilbert spaces and A B(H, K), then there exists an adjoint operator A B(K, H) uniquely defined by the condition f H, g H, Af, g K = f, A g H. (C.21) Now we will consider the case where X, Y are Banach spaces and A B(X, Y ). We will see that, as a consequence of the Hahn Banach Theorem, there exists a unique adjoint A B(Y, X ) defined by an equation analogous to equation (C.21). However, while we have A: X Y, the adjoint will be a map A : Y X. In particular, unlike the Hilbert space case, we cannot consider compositions of A with A.