OPERATOR THEORY ON HILBERT SPACE. Class notes. John Petrovic

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1 OPERATOR THEORY ON HILBERT SPACE Class notes John Petrovic

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3 Contents Chapter 1. Hilbert space Definition and Properties Orthogonality Subspaces Weak topology 9 Chapter 2. Operators on Hilbert Space Definition and Examples Adjoint Operator topologies Invariant and Reducing Subspaces Finite rank operators Compact Operators Normal operators 27 Chapter 3. Spectrum Invertibility Spectrum Parts of the spectrum Spectrum of a compact operator Spectrum of a normal operator 43 iii

4 iv CONTENTS Chapter 4. Invariant subspaces Compact operators Line integrals Invariant subspaces for compact operators Normal operators 56 Chapter 5. Spectral radius algebras Compact operators 64

5 CHAPTER 1 Hilbert space 1.1. Definition and Properties In order to define Hilbert space H we need to specify several of its features. First, it is a complex vector space the field of scalars is C (complex numbers). [See Royden, p. 217.] Second, it is an inner product space. This means that there is a complex valued function x, y defined on H H with the properties that, for all x, y, z H and α, β C: (a) αx + βy, z = α x, z + β y, z ; it is linear in the first argument; (b) x, y = y, x ; it is Hermitian symmetric; (c) x, x 0; it is non-negative; (d) x, x = 0 iff x = 0; it is positive. In every inner product space it is possible to define a norm as x = x, x 1/2. Exercise Prove that this is indeed a norm. Finally, Hilbert space is complete in this norm (meaning: in the topology induced by this norm). Example C n is an inner product space with x, y = n k=1 x ky k and, consequently, the norm x = n k=1 x k 2. Completeness: if {x (k) } k=1 is a Cauchy sequence in Cn (here x (k) = (x (k) 1, x(k) 2,..., x(k) n )) then so is x (k) m for any fixed m, 1 m n, and C is complete. Example Let H 0 denote the collection of all complex sequences, i.e. functions a : N C, characterized by the fact that a n 0 for a finite number of positive integers n. Define the inner product on H 0 by a, b = n=0 a nb n. The space H 0 is not complete in the induced norm. Indeed, the sequence {a (k) } k N, defined by a (k) n = 1/2 n if n k and a (k) n = 0 if n > k is a Cauchy sequence, but not convergent. 1

6 2 1. HILBERT SPACE Example Let l 2 denote the collection of all complex sequences a = {a n } n=1 such that n=1 a n 2 converges. Define the inner product on l 2 by a, b = n=1 a nb n. Suppose that {a (k) } k=1 is a Cauchy sequence in l 2. Then so is {a (k) n } k=1 for each n, hence there exists a n = lim k a (k) n. First we show that a l 2. Indeed, choose K so that for k K we have a (k) a (K) 1. Then, using Minkowski s Inequality for sequences (see Royden, p. 122), for any N N, { N } 1/2 { N } 1/2 a n 2 a n a (K) n 2 + n=1 n=1 { N 1=0 a (K) n 2 } 1/2 = lim k { N n=1 a (k) n lim sup a (k) a (K) + a (K) 1 + a (K). k a (K) n 2 } 1/2 + { N n=1 } 1/2 a (K) n 2 Thus a = {a n } l 2. Moreover, {a (k) } converges to a, i.e. lim k a a (k) = 0. Let ɛ > 0 and choose M so that k, j M implies that a (k) a (j) < ɛ. For such k M and any N, we have N n=1 a n a (k) n 2 = lim N j n=1 a (j) n a (k) n 2 lim sup a (j) a (k) 2 ɛ 2. j Since N is arbitrary, it follows that a a (k) ɛ and, therefore, l 2 is Hilbert space. Example The space L 2 of functions f : X C, such that X f 2 dµ < (where X is usually [0, 1] and µ Lebesgue measure). The inner product is defined by f, g = X fg dµ and L2 is complete by the Riesz Fisher Theorem (see Royden, p. 125). Example The space H 2. Let X = T (the unit circle) and µ the normalized Lebesgue measure on T. The Hardy space H 2 consists of those functions in L 2 (T) such that f, e int = 0 for n = 1, 2,.... Some important facts. Proposition (Parallelogram Law). x + y 2 + x y 2 = 2 x y 2. Proposition (Polarization Identity). 4 x, y = x + y, x + y x y, x y + i x + iy, x + iy i x iy, x iy. Exercise Prove Propositions and

7 1.2. ORTHOGONALITY 3 Problem 1. Let be a norm on Banach space X, and define x, y as in Polarization Identity. Assuming that the norm satisfies the Parallelogram Law, prove that x, y defines an inner product Orthogonality In Linear Algebra a basis of a vector space is defined as a minimal spanning set. In Hilbert space such a definition is not very practical. It is hard to speak of minimality when a basis can be infinite. In fact, a basis can be uncountable, so if {e i } i I is such a basis, what is the meaning of i I x ie i? Definition An orthonormal subset of Hilbert space H is a set E such that (a) e = 1, for all e E; (b) if e 1, e 2 E and e 1 e 2 then e 1, e 2 = 0. An orthonormal basis in H is a maximal orthonormal set. We use abbreviations o.n.s. and o.n.b. for orthonormal set and orthonormal basis, respectively. Theorem Every Hilbert space has an orthonormal basis. Proof. Let e be a unit vector in H. Then E = {e} is an orthonormal set. Let M be the collection of all orthonormal sets in H that contain E. By the Hausdorff Maximal Principle (Royden, p.25) there exists a maximal chain C of such orthonormal sets, partially ordered by inclusion. Let N be the union of all elements of C. Then N is a maximal orthonormal set, hence a basis of H. If the set {e} is replaced by any orthonormal set, the same proof yields a stronger result. Theorem Every orthonormal set in Hilbert space can be extended to an orthonormal basis. Example For k N, let e k denote the sequence with only one non-zero entry, lying in the kth position and equal to 1. The set {e k } k N is an o.n.b. for l 2. (If a vector x l 2 is orthogonal to all e k, then each of its components is zero, so x = 0.) Example The set {e 1, e 3, e 5,... } is an orthonormal set in l 2 but not a basis. Example The set { 1 2π, cos t π, sin t π, cos 2t π, } sin 2t,... is an o.n.b. in L 2 ( π, π). π

8 4 1. HILBERT SPACE Example The set { } 1 e int : n Z is another o.n.b. in L 2 ( π, π). 2π In Linear Algebra, if {e i } i I is an o.n.b. then every vector x can be written as i I x, e i e i. In Hilbert space our first task is to make sense of this sum since the index set I need not be countable. x 2. Theorem (Bessel s Inequality). Let {e i } k i=1 be an o.n.s. in H, and let x H. Then k i=1 x, e i 2 Proof. If we write x i = x, e i, then k k k k k k 0 x x i e i 2 = x x i e i, x x i e i = x 2 2Re x, x i e i + x i e i, x j e j i=1 i=1 i=1 i=1 i=1 j=1 k k k k k k = x 2 2Re x i x, e i + x i x j e i, e j = x 2 2Re x i x i + x i x i = x 2 x i 2. i=1 i=1 j=1 i=1 i=1 i=1 Corollary Let E = {e i } i I be an o.n.s. in H, and let x H. Then x, e i 0 for at most a countable number of i I. Proof. Let x H be fixed and let E n = {e i : x i 1/n}. If e i1, e i2,..., e ik E n then k x 2 x ij 2 k(1/n 2 ). j=1 So, for each n N, E n is a finite set, and E = n E n. In view of Corollary the expressions like x, e i e i turn out to be the usual infinite series. Our next task is to establish their convergence. The following Lemma will be helpful in this direction. Lemma If {x i } i N is a sequence of complex numbers and {e i } i N is an o.n.s. in H, then the series i N x ie i and i N x i 2 are equiconvergent.

9 1.2. ORTHOGONALITY 5 Proof. Let s m and σ m denote the partials sums of i N x ie i and i N x i 2, respectively. Then s m s n 2 = m i=n+1 x ie i 2 = m i=n+1 x ie i, m j=n+1 x je j = m i=n+1 x i 2 = σ m σ n so the series are equiconvergent. Now we can establish the convergence of i I x, e i e i. We will use notation x i = x, e i for the Fourier coefficients of x H relative to the fixed basis {e i } i I. Corollary (Parseval s Identity). Let {e i } i I be an o.n.s. in H, and let x H. Then the series i I x ie i and i I x i 2 converge and i I x ie i 2 = i I x i 2. Proof. Since only a countable number of terms in each series is non-zero, we can rearrange them and consider the series i=1 x ie i and i=1 x i 2. The latter series converges by the Bessel s Inequality and Lemma implies that the former series converges too. Moreover, their partial sums s m and σ m satisfy s m = σ m, so the last assertion of the corollary follows by letting m go to. Now we are in the position to show that, in Hilbert space, every o.n.b. indeed spans H. Of course, the minimality is a direct consequence of the definition. Theorem Let E = {e i } i I be an o.n.b. in H. Then, for each x H, x = i I x ie i, where x i = x, e i. Proof. Let x i = x, e i and y = x i I x ie i. (Well defined since the series converges.) Then y, e k = x, e k i I x ie i, e k = 0, for each k I, so y E. If y 0, then E {y/ y } is an o.n.s., contradiciting the maximality of E, so y = 0. The following is the analogue of a well known Linear Algebra fact. We use notation card I for the cardinal number of the set I. Theorem Any two orthonormal bases {e i } i I and {f j } j J in H have the same cardinal number. Proof. We will assume that both cardinal numbers are infinite. If either of them is finite, one knows from Linear Algebra that the other one is finite and equal to the first. Let j J be fixed and let I j = {i I : f j, e i

10 6 1. HILBERT SPACE 0}. By Corollary 1.2.4, I j is at most countable. Further, j J I j = I. Indeed, if i 0 I \ j J I j then f j, e i0 = 0 for all j J so it would follow that e i0 = 0. Since card I j ℵ 0 we see that card I card J ℵ 0 = card J. Similarly, card J card I. By Cantor Bernstein Theorem, (see, e.g., Proofs from the book, p.90) card I = card J. of H. Definition The dimension of Hilbert space H, denoted by dim H, is the cardinal number of a basis In this course we will assume that dim H ℵ 0. Exercise If H is an infinite dimensional Hilbert space, then H is separable iff dim H = ℵ 0. [Given a countable basis, use rational coefficients. Given a countable dense set, approximate each element of a basis close enough to exclude all other basis elements.] Next, we want to address the question: when can we identify two Hilbert spaces? We need a vector space isomorphism (i.e., a linear bijection) that preserves the inner product. Definition If H and K are Hilbert spaces, an isomorphism is a linear surjection U : H K such that, for all x, y H, Ux, Uy = x, y. In this situation we say that H and K are isomorphic. Exercise Prove that Ux, Uy = x, y for all x, y H iff Ux = x for all x H. Conclude that a Hilbert space isomorphism is injective. Theorem Every separable Hilbert space of infinite dimension is isomorphic to l 2. Every Hilbert space of finite dimension n is isomorphic to C n. Proof. We will assume that H is an infinite dimensional Hilbert space and leave the finite dimensional case as an exercise. Since H is separable, there exists an o.n.b. {e n } n=1. For x H, let x i = x, e i and U(x) = (x 1, x 2, x 3,... ). By Parseval s Identity, the series i=1 x i 2 converges, so the sequence (x 1, x 2, x 3,... ) belongs to l 2. Thus U is well-defined, linear (because the inner product is linear in the first argument), and isometric:

11 1.3. SUBSPACES 7 Ux 2 = i=1 x i 2 = x 2. Finally, if (y 1, y 2, y 3,... ) l 2 then i=1 y i 2 converges so, by Lemma 1.2.5, n=1 y ne n converges and U( n=1 y ne n ) = (y 1, y 1, y 1,... ). Thus, U is surjective and the theorem is proved. Exercise Prove that every Hilbert space of finite dimension n is isomorphic to C n. Problem 2. Let H ne a separable Hilbert space and M a subspace of H. Prove that M is a separable Hilbert space. Problem 3. The Haar system {ϕ m,n }, m N, 1 n 2 m, is defined as: Prove that this system is an o.n.b. of L 2 [0, 1]. 2 m/2, if n 1 2 m x n 1/2 2 m, ϕ m,n (x) = 2 m/2, if n 1/2 2 m x n 2 m, [ ) n 1 0, if x / 2 m, n 2 m Subspaces Example Let H = L 2 [0, 1] and let G be a measurable subset of [0, 1]. Denote by L 2 (G) the set of functions in L 2 that vanish outside of G. Then L 2 (G) is a closed subspace of H. Further, if f L 2 (G) and g L 2 (G c ), then f, g = 0. Definition If M is a closed subspace of the Hilbert space H, then the orthogonal complement of M, denoted M, is the set of vectors in H orthogonal to every vector in M. Exercise Prove that M is a closed subspace of H. Theorem Let M be a closed subspace of Hilbert space H, and let x H. Then there exist unique vectors y in M and z in M so that x = y + z.

12 8 1. HILBERT SPACE Proof. Let {e i } i I and {f j } j J be orthonormal bases for M and M, respectively. Their union is an o.n.b. of H so x = i I x, e i e i + j J x, f j f j and we define y = i I x, e i e i, z = j J x, f j f j. Then y M, z M, and x = y + z. Suppose now that x = y 1 + z 1 = y 2 + z 2, where y 1, y 2 M and z 1, z 2 M. Then y 1 y 2 = z 2 z 1 belongs to both M and M, so y 1 y 2, y 1 y 2 = 0 and it follows that y 1 = y 2, and consequently z 1 = z 2. Definition In the situation described in Theorem we say that H is the orthogonal direct sum of M and M, and we write H = M M. When z = x + y with x M and y M we often write z = x y. Theorem allows us to define a map P : H M by P x = y. It is called the orthogonal projection of H onto M, and it is denoted by P M. Here are some of its properties. Theorem Let M be a closed subspace of Hilbert space H and let P be the orthogonal projection on M. Then: (a) P is a linear transformation; (b) P x x, for all x H; (c) P 2 = P ; (d) Ker P = M and Ran P = M. Proof. Let {e i } i I and {f j } j J be orthonormal bases for M and M, respectively, and let Q = I P be the orthonormal projection on M. If x, x H and α, α C, then P (α x +α x ) = i I α x +α x, e i = α P x + α P x, so (a) holds. (b) If x H, then x = P x + Qx and P x Qx. Therefore, x 2 = P x 2 + Qx 2 P x 2. (c) If y M then P y = y. Now, for any x H, P x M so P 2 x = P (P x) = P x. (d) If P x = 0 then x = Qx M. If x M then Qx = x by (c), so P x = 0. The other assertion is obvious. Problem 4. Prove that P M x is the unique point in M that is nearest to x, meaning that x P M x = inf{ x h : h M}.

13 1.4. WEAK TOPOLOGY 9 Problem 5. In L 2 [0, 1] find the orthogonal complement to the subspace consisting of: (a) all polynomials in x; (b) all polynomials in x 2 ; (c) all polynomials in x with the free term equal to 0; (d) all polynomials in x with the sum of coefficients equal to 0. Problem 6. If M and N are subspaces of Hilbert space that are orthogonal to each other, then the sum M + N = {x + y : x M, y N } is a subspace. Show that the theorem is not true if M and N are either: closed but not orthogonal or orthogonal but not closed Weak topology Read Royden, page Example Consider the sequence of functions {cos nt} n N in L 1 [0, 2π]. It is easy to see that this sequence is not convergent. However, for any function f L, 1 0 f(t) cos nt dt 0 as n. Since L is the dual space of L 1, we say that cos nt 0 weakly, and we write w lim n cos nt = 0. Example Consider the sequence of functions {cos nt} n N in L [0, 2π]. Notice that, while not a convergent sequence, if f L 1 then 1 0 f(t) cos nt dt 0 as n. Since L is the dual space of L 1, we say that cos nt 0 in the weak topology. In a Banach space X it is useful to consider three topologies: the norm topology, induced by the norm; weak topology the smallest topology in which all bounded linear functionals on X are continuous; weak topology (meaningful when X is the dual space of Y so that Y X ) the smallest topology in which some bounded linear functionals on X are continuous (those that can be identified as elements of Y). In order to dicuss these topologies (and understand their role), we need to find out what bounded linear functionals on Hilbert space H look like.

14 10 1. HILBERT SPACE Theorem (Riesz Representation Theorem). If L is a bounded linear functional on H, then there is a unique vector y H such that L(x) = x, y for every x H. Moreover, L = y. Proof. Assuming that such y exists, we can write it as y = i N y ie i relative to a fixed o.n.b. {e i } i N. Then y i = y, e i = e i, y = L(e i ). Therefore, we define y = i N L(e i) e i, and all it remains to prove is the convergence of the series. Let s n = n i=1 L(e i) e i. Then L(s n ) = n i=1 L(e i) Le i = s n 2, so s n 2 L s n from which it follows that s n L. Lemma Thus the series n i=1 L(e i) e i converges and the result follows from We see that if L H, the dual space of H, then L = L y. The mapping Φ : H H defined by Φ(y) = L y is a norm preserving surjection. It is conjugate linear: Φ(α 1 y 1 + α 2 y 2 ) = α 1 y 1 + α 2 y 2. Nevertheless, we identify H with H. Consequently, H is reflexive (i.e., H = H) so the weak and weak topologies on H coincide. Therefore, we will work with 2 topologies: weak and norm induced. The absence of a qualifier will always mean that it is the latter. Exercise Prove that the weak topology is weaker than the norm toplogy, i.e., if G is a weakly open set then G is an open set. Example If {e n } n N is an orthonormal sequence in H then w lim e n = 0 but the sequence is not convergent. Exercise Prove that the Hilbert space norm is continuous but not weakly continuous. The following result shows why weak topology is important. [See Royden, p. 237] Theorem (Banach-Alaoglu). The unit ball {x H : x 1} in Hilbert space H is weakly compact. Remark The unit ball B 1 of H is NOT compact (assuming that H is infinite dimensional). Reason: if {e n } n N is an o.n.b. then the set {e 1, e 2, e 3,... } is closed but not totally bounded, hence not compact. Exercise Prove that if a bounded set in H is weakly closed then it is weakly compact.

15 1.4. WEAK TOPOLOGY 11 In spite of the fact that the weak topology is weaker then the norm topology, some of the standard results remain true. Theorem A weakly convergent sequence is bounded. Proof. Suppose that x n is a weakly convergent sequence. Then, for any y H, the sequence x n, y is a convergent sequence of complex numbers, which implies that it is bounded. In other words, for any y H there exists C = C(y) > 0 such that x n, y C. This means that, for each n N, x n can be viewed as a bounded linear functional on H. By the Uniform Bounded Principle (Royden, p. 232), these functionals are uniformly bounded, i.e., there exists M > 0 such that, for all n N, x n M. Although weakly convergent sequence need not be convergent there are situation when it does. Theorem If {x n } n N is a weakly convergent sequence in a compact set K then it is convergent. Proof. Since {x n } n N K, it has an accumulation point x and a subsequence x n converging to z. If {x n } had another accumulation point x, then there would be another subsequence x n converging to w. It would follow that w lim x n = x and w lim x n = x. Since {x n } is weakly convergent this implies that x = x, so it has only one accumulation point, namely the limit. By definition, the weak topology W is the smallest one in which every bounded linear functional L on H is continuous. This means that, for any such L and any open set G in the complex plane, L 1 (G) W. Since open disks form a base of the usual topology in C it suffices to require that L 1 (G) W for each open disk G. Notice that x L 1 (G) iff L(x) G, so if G = {z : z z 0 < r} and z 0 = L(x 0 ) then x L 1 (G) iff L(x x 0 ) < r. Now Riesz Representation Theorem implies that L 1 (G) = {x H : x x 0, y < r} for some y H. We conclude that a subbase of W consists of the sets W = W (x 0 ; y, r) = {x H : x x 0, y < r}. Exercise Prove that a bounded linear functional L is continuous in a topology T iff L 1 (G) T for every open disk G.

16 12 1. HILBERT SPACE Problem 7. Prove that a subspace of Hilbert space is closed iff it is weakly closed. Problem 8. Prove that Hilbert space is weakly complete. Problem 9. Let {x n } n N be a sequence in Hilbert space with the property that x n = 1, for all n, and x m, x n = c, if m n. Prove that {x n } n N is weakly convergent. Problem 10. Find the weak closure of the unit sphere in Hilbert space.

17 CHAPTER 2 Operators on Hilbert Space Nobody, except topologists, is interested in problems about Hilbert space; the people who work in Hilbert space are interested in problems about operators. Paul Halmos 2.1. Definition and Examples Read Section 10.2 in Royden s book. Operator always means linear and bounded. The algebra of all bounded linear operators on H is denoted by L(H). Example Let H = C n and A = [a ij ] an n n matrix. The operator of multiplication by A is linear and bounded. Indeed, for x = (x 1, x 2,..., x n ) and M = sup 1 i n ( n j=1 a ij 2 ) 1/2, so A M. Ax = sup 1 i n n j=1 a ij x j sup 1 i n n a ij 2 j=1 1/2 n x j 2 j=1 1/2 = M x Example Let H = l 2 and A = [a ij ] i,j=1, where a ij = c i if i = j and a ij = 0 if i j. We call such matrix diagonal and denote it by diag(c 1, c 2,... ), or diag(c n ). The operator A (or, more precisely, the operator of multiplication by A) is bounded iff c = (c 1, c 2,... ) l (i.e., when c is a bounded sequence). Indeed, let x = (x 1, x 2,... ) l 2, so Ax = (c 1 x 1, c 2 x 2,... ) and Ax 2 = i=1 c ix i 2. If c i M, i N, then Ax 2 M 2 i=1 x i 2 = x 2 so A is bounded. On the other hand, if c / l, then for each n there exists i n so that c in n. Then Ae in = c in e in n and A is unbounded. Remark It is extremely hard to decide, in general, whether an operator A is bounded just by studying its matrix [ Ae j, e i ] i,j=1. 13

18 14 2. OPERATORS ON HILBERT SPACE Example Let H = l 2 and let S be the unilateral shift, defined by S(x 1, x 2,... ) = (0, x 1, x 2,... ). Notice that S(x 1, x 2,... ) 2 = x x = x 2 so S = 1. In fact, S is an isometry, hence injective, but it is not surjective! Example (Multiplication on L 2 ). Let h be a measurable function and define M h f, for f L 2, by (M h f)(t) = h(t)f(t). If h L (essentially bounded functions see Royden, p. 118), then M h f 2 = hf 2 h 2 f 2 = h 2 f 2 so M h is a bounded operator on L 2 and M h h. On the other hand, for ɛ > 0, there exists a set C [0, 1] of positive measure so that h(t) h ɛ for t C. If f = χ C then M h f 2 = hf 2 = h 2 ( h ɛ) 2 µ(c) = ( h ɛ) 2 f 2, C and it follows that M h h ɛ. We conclude that M h = h and M h is bounded iff h L. Example (Integral operators on L 2 ). Let K : [0, 1] [0, 1] C be measurable and square integrable with respect to planar Lebesgue measure. We define the operator T K by (T k f)(x) = 1 K(x, y)f(y) dy. Now 0 T K f 2 = T k f(x) 2 dx = { 1 } { 1 K(x, y) 2 dy Therefore, T K is bounded and T K 0 0 K(x, y)f(y) dy 0 } f(y) 2 dy { K(x, y) 2 dydx 2 dx 1 0 dx = f 2 1 } 1/2. ( 1 2 K(x, y)f(y) dy) dx K(x, y) 2 dydx. Example (Weighted shifts). Let H = l 2 and let {c n } n N be a bounded sequence of complex numbers. A weighted shift W on l 2 is defined by W (x 1, x 2,... ) = (0, c 1 x 1, c 2 x 2,... ). It can be written as W = S diag(c n ) so it is a bounded operator and W = diag(c n ). In some situations it is useful to have an alternate formula for the operator norm. In what follows we will use notation B 1 for the closed unit ball of H, i.e. B 1 = {x H : x 1}. Proposition Let T be linear operator on Hilbert space. Then T = sup{ T x, y : x, y B 1 }.

19 2.2. ADJOINT 15 Proof. Let α denote the supremum above, and let us assume that T 0 (otherwise there is nothing to prove). Clearly, for x, y B 1, T x, y T, so α T. In the other direction, α sup{ T x, y : x, y B 1, T x 0, y = = sup{ T x, T x T x : x B 1, T x 0} = sup{ T x : x B 1, T x 0} T x T x } = T, and the proof is complete Adjoint In Linear Algebra we learn that the column space of matrix A = [a ij ] n i,j=1 and the null space of its transpose A T are orthogonal complements in R n. In C n, A T needs to be replaced by A = [a ji ] n i,j=1. In this situation, (2.1) Ax, y = x, A y. Exercise Prove that, if A is an n n matrix and x, y C n, then Ax, y = x, A y. Example Let h L and let M h be the operator of multiplication on L 2. Then (M h ) = M h. The following result will show that a relation (2.1) is available for any operator. Proposition If T is an operator on H then there exists a unique operator S on H such that T x, y = x, Sy, for all x, y H. Proof. Let y H be fixed. Then ϕ(x) = T x, y is a bounded linear functional on H. By Riesz Representation Theorem there exists a unique z H such that ϕ(x) = x, z, for all x H. Define Sy = z. Then

20 16 2. OPERATORS ON HILBERT SPACE T x, y = x, Sy. To show that S is linear, let Sy 1 = z 1, Sy 2 = z 2, and let x H. Then x, S(α 1 y 1 + α 2 y 2 ) = T x, α 1 y 1 + α 2 y 2 = α 1 T x, y 1 + α 2 T x, y 2 = α 1 x, Sy 1 + α 2 x, Sy 2 = x, α 1 Sy 1 + α 2 Sy 2. By the uniqueness part of Riesz Representation Theorem S is linear. That S is unique can be deduced by contradiction: if x, Sy = x, S y for all x, y H then x, Sy S y = 0 for all x which implies that Sy S y = 0 for all y, hence S = S. Finally, S is bounded: Sy 2 = Sy, Sy = T Sy, y T Sy y T Sy y so Sy T y and S T. Definition If T L(H) then the adjoint of T, denoted T, is the unique operator on H satisfying T x, y = x, T y, for all x, y H. Here are some of the basic properties of the involution T T. Proposition (a) I = I (b) T = (T ) = T ; (c) T = T ; (d) (α 1 T 1 + α 2 T 2 ) = α 1 T1 + α 2 T2 ; (e) (T 1 T 2 ) = T2 T1 ; (f) if T is invertible then so is T and (T ) 1 = (T 1 ) ; (g) T 2 = T T. Proof. The assertion (a) is obvious and (b) follows from x, T y = T x, y = y, T x = T y, x = x, T y. It was shown in the proof of Proposition that T T so T T T and (c) follows from (b). We leave (d) as an exercise and notice that x, (T 1 T 2 ) y = T 1 T 2 x, y = T 2 x, (T 1 ) y = x, (T 2 ) (T 1 ) y establishes (e). As a consequence of (a) and (e), T (T 1 ) = (T 1 T ) = I and (T 1 ) T = (T T 1 ) = I which

21 2.3. OPERATOR TOPOLOGIES 17 is (f). Finally, T T T T = T 2 and to prove the opposite inequality let ɛ > 0 and let x be a unit vector such that T x T ɛ. Then T T T T x T T x, x = T x 2 > ( T ɛ) 2, and (g) is proved. Example Let H = l 2 and the let S be the unilateral shift (see Example 2.1.3). Then S (x 1, x 2,... ) = (x 2, x 3,... ). The operator S is called the backward shift. Example Let T K be the integral operator on L 2 (see Example 2.1.5). Then (T K ) = T K, where K (x, y) = K(y, x). We now give the Hilbert space formulation of the relation with which we have opened this section. Theorem If T is an operator on Hilbert space H then Ker T = (Ran T ). Proof. Let x Ker T and let y Ran T. Then there exists z H such that y = T z. Therefore x, y = x, T z = T x, z = 0 so x (Ran T ). In the other direction, if x (Ran T ) and z H, then T x, z = x, T z = 0. Taking z = T x we see that T x = 0, and the proof is complete. We notice that, for T L(H) and x, y H, the expression T x, y is a form that is linear in the first and conjugate linear in the second argument. It turns out that this is sufficient for a polarization identity. Proposition (Second Polarization Identity). 4 T x, y = T (x + y), x + y T (x y), x y + i T (x + iy), x + iy i T (x iy), x iy. Exercise Prove Second Polarization Identity Operator topologies In this section we take a look at the algebra L(H). It has three useful topologies which lead to 3 different types of convergence. Definition A sequence of operators T n L(H) converges uniformly (or in norm) to an operator T if T n T 0, n. A sequence of operators T n L(H) converges strongly to an operator T if T n x T x 0,

22 18 2. OPERATORS ON HILBERT SPACE n, for all x H. A sequence of operators T n L(H) converges weakly to an operator T if T n x T x, y 0, n, for any x, y H. It follows from the definition that the weak topology is the weakest of the three, while then norm topology (a.k.a. the uniform topology) is the strongest. Are they different? Proposition The operator norm is continuous with respect to the uniform topology but discontinuous with respect to the strong and weak topologies. Proof. The first assertion is a consequence of the inequality A B A B. To prove the other two, let {e n } n N be an o.n.b. of H, H n = k=n e k, P n = P Hn. Then P n 0 strongly (hence weakly) since P n x 2 = k=n+1 x k 2 0. However, P n = 1 which does not converge to 0. Example We say that an operator T is a rank one operator if there exist u, v H so that T x = x, v u. We use the notation T = u v. Let T n = e n e 1. Then T n x, y = x 1 y n 0 while T n x = x 1 e n is not a convergent sequence. Thus, the weak and strong topologies are different. Example The involution T T is continuous in uniform topology. ( T n T = T n T ). Also, it is continuous in the weak topology, because (T n T )x, y = x, (T n T )y = (T n T )y, x. However, it is not continuous in the strong topologies. Counterexample: let S be the unilateral shift, and T n = (S ) n. Then T n 0 strongly but {T n} is not a strongly convergent sequence. Indeed, for any x = (x 1, x 2,... ) H, T n x 2 = (x n+1, x n+2,... ) 2 = k=n x k 2 0, as n. On the other hand, for x = e 1, T nx = S n e 1 = e n, which is not a convergent sequence. An operator T L(H) is a continuous mapping when H is given the strong topology. We will write, following Halmos, (s s). One may ask about the other types of continuity. Theorem The three types of continuity (s s), (w w), and (s w) are all equivalent.

23 2.3. OPERATOR TOPOLOGIES 19 Proof. Suppose that T is continuous, and let W be a weakly open neighborhood of T x 0 in H. We will show that T 1 (W ) is weakly open. It suffices to prove this assertion in the case when W belongs to the subbase of the weak topology. To that end, let W = W (T x 0, y, r) = {x H : x T x 0, y < r}. Then z T 1 (W ) T z W T z T x 0, y < ɛ z x 0, T y < ɛ. We see that z T 1 (W ) iff z V (x 0, T y, ɛ) so T 1 (W ) = V which is a weakly open set. The implication (w w) (s w) is trivial, so we concentrate on the implication (s w) (s s). To that end, suppose that T is not continuous. Then it is unbounded, so there exists a sequence {x n } n N of unit vectors such that T x n n 2, n N. Clearly, x n /n 0 and the assumption (s w) implies that T x n /n weakly converges to 0. By Theorem the sequence {T x n /n} is bounded which contradicts the fact that T x n /n n. The fact that every operator in L(H) is weakly continuous has an interesting consequence. Corollary If T is a linear operator on H then T (B 1 ) is closed. Proof. Banach-Alaoglu Theorem established that B 1 is weakly compact so, by Theorem 2.3.2, T (B 1 ) is weakly compact, hence weakly closed, hence norm closed. Exercise Prove that if F is a closed and bounded set in H then T (F ) is closed. At the end of this section we consider a situation that occurs quite frequently. Theorem Let M be a linear manifold that is dense in Hilbert space H. Every bounded linear transformation T : M H can be uniquely extended to a bounded linear transformation ˆT : H H. In addition, the operator norm of T equals ˆT. Proof. Let x H. Then there exists a sequence {x n } n N M converging to x. Since {x n } n N is also a Cauchy sequence, for every ɛ > 0 there exists N N such that, m, n N x m x n < ɛ/ T. It follows that, for m, n N, T x m T x n < ɛ, so {T x n } n N is a Cauchy sequence, hence convergent, and there exists y = lim n T x n. We will define ˆT x = y, i.e., ˆT (lim x n ) = lim T x n.

24 20 2. OPERATORS ON HILBERT SPACE First we need to establish that the definition is independent of the sequence {x n } n N. If {x n} n N is another sequence converging to x, we form the sequence (x 1, x 1, x 2, x 2,... ) which also converges to x. By the previous, the sequence (T x 1, T x 1, T x 2, T x 2,... ) must converge, and therefore, both of the subsequences {T x n } n N and {T x n} n N must have the same limit. Notice that, if x n x, the continuity of the norm implies that ˆT x = lim T x n = lim T x n lim T x n = T x so ˆT T. Since the other inequality is obvious we see that ˆT = T. In particular, ˆT is a bounded operator. Also, ˆT (αx + βy) = ˆT (α lim x n + β lim y n ) = ˆT (lim(αx n + βy n )) = lim T (αx n + βy n ) = lim(αt x n + βt y n ) = α lim T x n + β lim T y n = α ˆT x + β ˆT y, so ˆT is linear. Finally, suppose that T 1 and T 2 are two continuous extensions of T, and let x H. If x n x, the continuity implies that both T 1 x n T 1 x and T 2 x n T 2 x. If x n M then T 1 x n = T 2 x n, so T 1 x = T 2 x. Therefore, the extension is unique, and the proof is complete. Need an example 2.4. Invariant and Reducing Subspaces When M is a closed subspace of H, we can always write H = M M. Relative to this decomposition, any operator T acting on H can be written as a 2 2 matrix with operator entries (2.2) T = X Z Y. W It is sometimes convenient to consider only the initial space or the target space as a direct sum. In such a situation [ ] we will use a 1 2 or 2 1 matrix. Thus X Y will describe an operator T : M M H; if f M and g M then [ X Y ] [ ] f g = Xf + Y g. A subspace M is invariant for T if, for any x M, T x M. It is reducing for T if both M and M are invariant for T.

25 2.4. INVARIANT AND REDUCING SUBSPACES 21 Example The subspace (0) consisting of zero vector only is an invariant subspace for any operator T. Also, H is an invariant subspace for any operator T. Because they are invariant for every operator they are called trivial. A big open problem in Operator theory is whether every operator has a non-trivial invariant subspace. Example If M is a closed subspace of H and T 1 is an operator on M with values in M, then the operator T = T 1 0, defined by T x = T 1 x if x M and T x = 0 if x M is an operator in L(H). However, if M is not invariant for T 1, the same definition (T x = T 1 x for x M, T x = 0 for x M ) describes the operator [ ] T 1 0. Proposition If T is an operator on Hilbert space H, and P = P M is the projection onto the closed subspace M, then the following are equivalent: (a) M is invariant for T ; (b) P T P = T P ; (c) Z = 0 in (2.2). Proof. It is not hard to see that the matrix for P is [ I ] so P T P T P = [ ] 0 0 Z 0. This establishes (b) (c). Since [ ] [ f g M iff g = 0, we see that T f ] [ ] 0 = Xf Zf M for all x H iff Z = 0 so (a) (c). Example Let S be the unilateral shift, n N, and M = k n e k. Then SM = k n+1 e k M. Proposition If T is an operator on Hilbert space H, and P = P M then the following are equivalent: (a) M is reducing for T ; (b) P T = T P ; (c) Y, Z = 0 in (2.2); (d) M is invariant for T and T. Proof. Since P T T P = [ 0 Y Z 0 ] we see that (b) (c). Further, the matrix for T is [ X Z Y W ] so, by Proposition 2.4.1, M is invariant for T and T iff Z = Y = 0 and (c) (d). In order to prove that (a) (d)

26 22 2. OPERATORS ON HILBERT SPACE it suffices to show that M is invariant for T iff M is invariant for T. By Proposition 2.4.1, M is invariant for T iff Y = 0 (iff Y = 0). On the other hand T [ ] [ ] 0 g = Y g W g M iff Y g = 0 for all g. Exercise Prove that the matrix for T is [ ] X Z Y W. Example Let T = M h, let E [0, 1], m(e) > 0, and let M = L 2 (E). If f M then T f = hf M. Also, T = M h and T f = hf M, so M is reducing for T. Example Let S be the unilateral shift, n N, and M = k n e k. Then M is invariant for S but not reducing, since e n M but S e n = e n 1 / M Finite rank operators The closest relatives of finite matrices are the finite rank operators. Definition An operator T is a finite rank operator if its range is finite dimensional. We denote the set of finite rank operators by F. Example If T is a rank one operator u v (see Example 2.3.1) then the range of u v is the one dimensional subspace spanned by u, so u v F. The rank one operators turn out to be the building blocks out of which finite rank operators are made. Proposition If T is a linear operator on H then T belongs to F iff there exist vectors u 1,u 2,...,u n, and v 1,v 2,...,v n such that T x = n i=1 x, v i u i. Proof. Suppose that Ran T is of finite dimension n, and let e 1, e 2,..., e n be an o.n.b. of Ran T. Then T x = n i=1 T x, e i e i = n i=1 x, T e i e i. We leave the converse as an exercise. Exercise Prove that if there exist vectors u 1, u 2,..., u n, v 1, v 2,..., v n such that T x = n i=1 x, v i u i, for all x H, then Ran T is of dimension at most n.

27 2.6. COMPACT OPERATORS 23 Exercise Prove that if T = u i v i then T = v i u i. The next theorem summarizes some very important properties of the class F. Theorem The set F is a minimal -ideal in L(H). Here the star means that F is closed under the operation of taking adjoints. Proof. It is obvious that F is a subspace of L(H). Furthermore, if T F and A L(H), then Ran T A Ran T so T A F. Also, if T is of finite rank, then according to Proposition 2.5.1, T = n i=1 u i v i so T = n i=1 v i u i. It follows that T F and the same is true of T A, for any A L(H). Consequently, AT is of finite rank, and F is a -ideal. To see that it is minimal, it suffices to show that, if J is a non-zero ideal, then J contains all rank one operators. Let T J, T 0. Then there exists vectors x, y, such that y = 1 and y = T x. Let u v be a rank one operator. Since J is an ideal, it contains the product (u y)t (x v) which equals u v. A finite rank operator is a generalization of a finite matrix. What happens when we take the closure of F in some topology? Exercise Prove that the strong closure of F is L(H). [Hint: Prove that P n I strongly.] Conclude that the weak closure of F is also L(H) Compact Operators Exercise established that the strong closure of F is L(H). Therefore, we consider the norm topology. Definition An operator T in L(H) is compact if it is the limit of a sequence of finite rank operators. We denote the set of compact operators by K. Example Let T = diag(c n ) as in Example 2.1.2, with lim n c n = 0. Then T is compact. Reason: take T n = diag(c 1, c 2,..., c n, 0, 0,... ). Then T n F and T T n = sup{ c k : k n + 1} 0. It follows that T is compact.

28 24 2. OPERATORS ON HILBERT SPACE Example Let T = T K as in Example If K L 2 ([0, 1] [0, 1]) then T K is compact. We will point out at several different sequences in F that all converge to T K We start with a function theoretic approach: simple functions are dense in L 2 (Royden, p. 128), and a similar proof establishes that simple functions are dense in L 2 ([0, 1] [0, 1]). Since a simple function is a linear combination of the characteristic functions of rectangles χ [a,b] [c,d] (x, y) = χ [a,b] (x)χ [c,d] (y) it follows that K(x, y) is the L 2 limit of functions of the form K n (x, y) = n i=1 f i(x)g i (y), so T K is the norm limit of T Kn, which are all finite rank operators. Exercise Verify that T Kn F, if K n (x, y) = n i=1 f i(x)g i (y). Our second approach is exploiting the fact that L 2 is Hilbert space. If {e j } j N is an o.n.b. of L 2 we can, for a fixed y, write K(x, y) = j=1 k j(y)e j (x). Now define K N (x, y) = N j=1 k j(y)e j (x) and notice that T KN T K as N. Exercise Verify that T KN F and that lim N T KN = T K, if K N (x, y) is as above. Our last method is based on the matrix for T K. Let k ij = T K e j, e i, with {e n } n N an o.n.b. of L 2 ([0, 1]). First we notice that, for any f L 2, k f, e k 2 = k f, e k e k 2 = f 2. Therefore, T K e j, e i 2 = e j, TKe i 2 = TKe i, e j, 2 = TKe i 2 j=1 j=1 j=1 = 1 It follows that, for any n N, n k ij 2 = 0 1 i=1 j=1 i=1 0 0 K (y, x)e i (x) dx n i=1 0 2 dy = K(x, y)e i (x) dx K(x, y)e i (x) dx dy = K(x, y)e i (x) dx 1 n 1 0 i= dy. K(x, y)e i (x) dx dy = K(x, y) 2 dxdy dy

29 2.6. COMPACT OPERATORS 25 so the series k ij 2 converges. Operators whose matrices satisfy this condition are called Hilbert Schmidt i=1 j=1 operators. The Hilbert-Scmidt norm is defined as T K 2 = { i=1 j=1 k ij 2 } 1/2, and it satisfies the inequality A A 2. Hilbert-Scmidt operators are compact because we can define T n to be the matrix consisting of the first n rows of the matrix of T K and having the remaining entries 0. Then each T n F and T n T K 0. Indeed, Ran T n {e 1, e 2,..., e n }, and T K T n 2 T K T n 2 2 = k ij 2 0, n. i=n+1 j=1 Exercise Prove that the Hilbert-Scmidt norm is indeed a norm and, for any T L(H), T T 2. Next we consider some of the properties of compact operators. The first one follows directly from the definition. Theorem The set K is the smallest closed -ideal in L(H). The following result reveals the motivation for calling these operators compact. Theorem An operator T in L(H) is compact iff it maps the closed unit ball of H into a compact set. Proof. Suppose that K is compact and let {y n } n N be a sequence in K(B 1 ). We will show that there exists a subsequence of {y n } that converges to an element of K(B 1 ). Notice that, for every n N, y n = Kx n, and x n belongs to the weakly compact set B 1. Thus, there exists a subsequence {x nk } converging weakly to x B 1. Thus, it suffices to show that Kx nk converges to Kx. Let {K n } be a sequence in F thaty converges to K. For any m N, K m (B 1 ) is a bounded and closed set (by Corollary 2.3.3) that is contained in a finite dimensional subspace of H, so it is compact. By Theorem 1.4.4, {K m x nk } k N converges to K m x. Now, let ɛ > 0. Then there exists N N such that K K N < ɛ/3. Further, with N fixed, there exists k 0 N so that, for k k 0, K N x nk K N x < ɛ/3. Therefore, for k k 0, Kx nk Kx (K K N )x nk + K N (x nk x) + (K N K)x < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Thus, y nk = Kx nk is a convergent subsequence converging to Kx K(B 1 ) so K(B 1 ) is a compact set.

30 26 2. OPERATORS ON HILBERT SPACE Suppose now that K(B 1 ) is compact and let n N. Notice that y K(B1)B(y, 1/n) is an open covering of the compact set K(B 1 ), so there exist vectors x (n) 1, x(n) 2,..., x(n) k H so that k i=1 B(Kx(n) i, 1/n) is a covering of K(B 1 ). Let H n be the span of Kx (n) 1, Kx(n) 2,..., Kx(n) k and P n the orthogonal projection on H n. Finally, let K n = P n K. Clearly, K n F. Let ɛ > 0, and choose N > 1/ɛ. If n N, and x 1, then Kx K n x = Kx P n Kx. Since P n Kx is the point in H n closest to Kx, it follows that Kx K n x inf 1 i n Kx Kx (n) i < 1/n < ɛ. Thus K n K and the proof is complete. Remark In many texts the characterization of compact operators, established in Theorem 2.6.2, is taken to be the definition of a compact operator. Exercise Prove that if F is a closed and bounded set in H and T is a compact operator in L(H) then T (F ) is a compact set. There is another characterization of compact operators: Proposition If T is a linear operator on H then T is compact iff it maps every weakly convergent sequence into a convergent sequence. In this situation, if w lim x n = x then lim T x n = T x. Proof. Suppose first that T is compact and let w lim x n = x. By Proposition 1.4.3, there exists M > 0 such that, for all n N, x n M. Therefore, T x n /M T (B 1 ), which is compact by Theorem Now Theorem implies that lim x n = x. In order to establish the converse, we will demonstrate that T (B 1 ) is compact by showing that every sequence in T (B 1 ) has a convergent subsequence. Let {y n } n N T (B 1 ). Then y n = T x n, for x n B 1, so the Banach Alaoglu Theorem implies that {x n } has a weakly convergent subsequence {x nk } and, by assumption, {T x nk } is a (strongly) convergent subsequence of {T x n }. Example We have seen in Example that if T diag(c n ) and c n 0, then T is compact. The converse is also true: if {e n } is the o.n.b. which makes T diagonal, then T e n 0 (because w lim e n = 0 and T is compact) so c n e n 0.

31 2.7. NORMAL OPERATORS 27 It is useful to know that compactness is inherited by the parts of an operator. Theorem Suppose that T is a compact operator on Hilbert space H = M M and that, relative to this decomposition, T = [ X Z W Y ]. Then each of the operators X, Y, Z, W is compact. Proof. Let {T n } be a sequence of finite rank operators that converges to T. Write, for each n N, Y n T n = [ X n ] Z n W n. Then all the operators Xn, Y n, Z n, W n F and they converge to X, Y, Z, W, respectively. Exercise Prove that X n, Y n, Z n, W n F and that they converge to X, Y, Z, W, respectively. [Consider the projections P 1 = P M and P 2 = P M and notice that, for example P 1 T P 2 = [ 0 Y 0 0 ], so Y n Y T n T and Ran Y n Ran P 1 T n P 2 the later being finite dimensional.] 2.7. Normal operators Definition If T is an operator on Hilbert space H then: (a) T is normal if T T = T T ; (b) T is self-adjoint (or Hermitian) if T = T ; (c) T is positive if T x, x 0 for all x H; (d) T is unitary if T T = T T = I. Example Let T = diag(c n ). Then T = diag(c n ) so T is normal. Also, T = T iff c n R, n N, and T is positive iff c n 0, n N. Finally, T T = diag( c n 2 ) so T is unitary iff c n = 1, n N. Exercise Let T = M h on L 2. Prove that T is normal and that it is: self-adjoint iff h(x) R, a.e.; positive iff h(x) 0 a.e.; unitary iff h(x) = 1 a.e.. The relationship between T and T that defines each of these classes allows us to establish some of their significant properties. Proposition An operator T on Hilbert space H is self-adjoint iff T x, x is real for any x H.

32 28 2. OPERATORS ON HILBERT SPACE Proof. If T = T then T x, x = x, T x = x, T x = T x, x so T x, x R. On the other hand, if T x, x is real for any x H then Second Polarization Identity implies that T x, y = T y, x so T = T. Exercise Prove that T x, x R implies that T x, y = T y, x. Corollary If P is a positive operator on Hilbert space H then P is self-adjoint. Example If P is the orthogonal projection on a subspace M of Hilbert space H, then P is a positive operator. Indeed, if z H write z = x + y relative to H = M M. By Theorem 1.3.2, P z = x and P y = 0, so P z, z = x, x + y = x 2 0. Combining Theorem and Example we see that every projection is a positive idempotent. In fact, the converse is also true. Theorem If T is an idempotent self-adjoint operator then T is a projection on M = {x H : T x = x}. Proof. Let z H and write it as z = T z + (z T z). Now T (T z) = T z so T z M. Also, z T z M. Indeed, if x M, then x, z T z = x, z x, T z = x, z T x, z = 0. By Proposition 2.1.1, the norm of every operator T in L(H) can be computed by considering the supremum of the values of its bilinear form T x, y. The next result shows that, when T is self adjoint, it suffices to consider only some pairs of x, y B 1. Proposition If T is a self-adjoint operator on Hilbert space H then T = sup{ T x, x : x = 1}. Proof. Clearly, T x, x T x 2, so if we denote by α the supremum above, we have that α T. To prove that α = T, we use the Second Polarization Identity, and we notice that, in view of the assumption T = T and Proposition 2.7.1, 4Re T x, y = T (x + y), x+y T (x y), x y. Moreover, using Parallelogram Law, and assuming that x and y are unit vectors, we obtain that 4Re T x, y α x+y 2 +α x y 2 = α(2 x 2 +2 y 2 ) = 4α. When x is selected so that T x 0, y = T x/ T x we obtain Re T x α so T α.

33 2.7. NORMAL OPERATORS 29 Exercise Prove that two product of two self-adjoint operators is self-adjoint iff the operators commute. Remark If we write A = (T + T )/2 and B = (T T )/2i then the operators A, B are self-adjoint and T = A + ib. We call them the real part and the imaginary part of T. Proposition If T is an operator on Hilbert space H then the following are equivalent. (a) T is a normal operator; (b) T x = T x for all x H; (c) the real and imaginary part of T commute. Proof. Notice that T x 2 T x 2 = (T T T T )x, x. If T is normal then the right side is 0, so (a) implies (b). If (b) is true, then the left side is 0, for all x. Since T T T T is self-adjoint, Proposition implies that its norm is 0, so (b) implies (a). A calculation shows that, if A and B are the real and imaginary part of T, resp., then AB BA = (T T T T )/2i so (a) is equivalent to (c). In Definition we have introduced the concept of the Hilbert space isomorphsim. Since it preserves the inner product ( Ux, Uy = x, y ), it preserves the norm, and hence both weak and strong toplogies. Therefore, if U : H K, we do not distinguish between an operator T L(H) and UT U 1 L(H), and we say that they are unitarily equivalent. Since, by Definition 2.7.1, an operator T is unitary iff T T = T T = I, we should check that UU = U U = I. Exercise Verify that UU = I K and U U = I H. Notice that both equalities need to be verified, because it is quite possible for one to hold but not the other. Example: the unilateral shift S satisfies S S = I SS. Exercise Prove that T is an isometry iff T T = I. Exercise asserts that the product of two self-adjoint operators is itself self-adjoint iff the operators commute. What if self-adjoint is replaced by normal? If M, N are commuting normal operators, their product

34 30 2. OPERATORS ON HILBERT SPACE is normal if MN commutes with N M and it looks like we need the additional assumption that M commutes with N (which also gives that M commutes with N). When an operator T commutes with both N and N we say that T doubly commutes with N. When N is normal we can establish even a stronger result. Theorem (Fuglede Putnam Theorem). Suppose that M, N are normal operators and T L(H) intertwines M and N: MT = T N. Then M T = T N. Proof. Let λ be a complex number, and denote A = λm, B = λn. Notice that AT = T B, so A 2 T = A(AT ) = A(T B) = (AT )B = (T B)B = T B 2, and inductively A k T = T B K, for k N. It follows that, if we denote the exponential function by exp(z), exp(a)t = T exp(b). It is not hard to see that exp( A) exp(a) = I so T = exp( A)T exp(b). If we denote by U 1 = exp(a A), U 2 = exp(b B ), then both U 1, U 2 are unitary operators. Indeed, U 1 = [ (A A) n /n!] = (A A ) n /n! = exp(a A ) = U 1 1, and similarly for U 2. Now we have that exp(a )T exp( B ) = U 1 T U 2 and exp(a )T exp( B ) = T. We conclude that exp(λm )T exp( λn ) = T for all λ C. Now f(λ) = exp(λm )T exp( λn ) is an entire bounded function, hence a constant. Therefore, f (0) = 0. On the other hand, f (λ) = M exp(λm )T exp( λn ) + exp(λm )T exp( λn )( N ) so f (0) = M T T N, and the theorem is proved. Exercise Prove that exp( T ) exp(t ) = I for any operator T L(H). Corollary The product of two normal operators is itself normal iff the operators commute. Exercise Prove Corollary