285K Homework #1. Sangchul Lee. April 28, 2017

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285K Homework #1 Sangchul Lee April 28, 2017 Problem 1. Suppose that X is a Banach space with respect to two norms: 1 and 2. Prove that if there is c (0, such that x 1 c x 2 for each x X, then there is also c (0, such that x 2 c x 1 for each x X. Proof. Consider the operator T : (X, 2 (X, 1 defined by T x = x. Then T is bijective and bounded by the assumption. Bounded Inverse Theorem tells that T 1 is also a bounded operator, which proves the claim with c = T 1. Problem 2. Let T be the left shift on l 2 (N and let T be its adjoint. State and prove how T acts on any particular sequence (x 1, x 2, and prove that their spectrum σ, residual spectrum σ res and set of eigenvalues σ eig are as follows: σ σ res σ eig T λ 1 λ < 1 T λ 1 λ < 1 Proof. We claim that T is the right shift: T (x 1, x 2, = (0, x 1, x 2,. Indeed, for any x = (x 1, x 2, and y = (y 1, y 2, we have y, T x = y k x k+1 = k=1 y k 1 x k = T y, x and the claim follows. Next we figure out various types of spectrum of T and T. We denote by D the open unit disc {λ C : λ < 1}. We first figure out the set of eigenvalues. k=2 1

σ eig (T = D. Indeed, the relation T x = λx is equivalent to x n = (x 1, x 1 λ, x 1 λ 2,. Now if λ σ eig (T and x is a non-zero eigenvector of λ, then we must have λ < 1 and hence σ eig (T D. Conversely, if λ D then x = (1, λ, λ 2, l 2 (N solves T x = λx. Therefore D σ eig (T. σ eig (T =. We solve T x = λx. If λ = 0, then the equation reduces to T x = 0 and hence x = 0. Otherwise, solving T x = λx gives x 1 = 0 and x n = λ 1 x n 1. Again, we have x = 0. This proves σ eig (T =. Next we investigate the residual spectrum. Since σ res (T σ eig (T, the previous computation shows that σ res (T =. Similarly, from previous computations, the relation σ res (T σ eig (T σ res (T σ eig (T is saturated and hence σ res (T = σ eig (T = D. The above four computations confirms four entries of the table. For the remaining entries, notice that σ(t and σ(t are compact subsets of C. So we have both D σ(t and D σ(t. But since the spectral radius r(t of T satisfies r(t T 1, we must have σ(t D and hence the equality holds. Similar argument applied to T 1 gives the identity σ(t = D as well. Problem 3. Construct sequences of operators {T n } that: (a converge strongly but not in norm, (b converge weakly but not strongly. Proof. Notice that in each case, convergence in the stronger sense implies convergence to the same limit in the weaker sense. So it suffices to construct a sequence (T n that converges to T in the weaker sense but does not converge to T in the stronger sense. In both examples, we will utilize an orthonormal subset {ψ n } of H. (a Define T n by T n x = n k=1 ψ k, x ψ k. Then the Bessel s inequality k=1 ψ k, x 2 x 2 tells that the linear operator T given by T x = k=1 ψ k, x ψ k converges for each x H and defines a bounded operator. So T n converges to T strongly. On the other hand, (T T n ψ n+1 = 1 shows that T T n 1 for all n and therefore T n does not converge in norm. (b Let T n = ψ 1, x ψ n. Then for any x, y H we have y, T n x = y, ψ n ψ 1, x 0 as n and hence T n converges to 0 weakly. On the other hand, T n x = ψ 1, x does not converge to 0 unless we already have ψ 1, x = 0. So T n does not converge strongly. 2

Problem 4. Let D C be a domain and f : D B(H be a function that is weakly holomorphic, i.e., with the complex derivative defined in the weak operator topology. Prove that f is (strongly holomorphic. Do the same for analytic functions. Proof. The following lemma is useful for our purpose. Lemma. For a non-empty subset F B(H, the followings are equivalent. (a F is bounded, i.e., sup T F T B(H < (b F is weakly bounded, i.e., sup T F ϕ, Tψ < for any ϕ, ψ H. The proof is almost trivial. Indeed, the only non-trivial implication is (. Fix ϕ and notice that sup T F Tϕ, ψ < for each ψ H. Applying the uniform boundedness principle (UBP, it follows that sup T F Tϕ H = sup T F Tϕ, H <. Therefore the claim follows by applying UBP again. Assume first that f : D B(H is weakly holomorphic. This amounts to saying that the function f ϕ,ψ (z = ϕ, f (zψ is holomorphic for each ϕ, ψ H. We first prove that there exists a function f : D B(H satisfying f ϕ,ψ (z = ϕ, f (zψ for any ϕ, ψ H and z D. To this end, let z 0 D and choose r > 0 such that B(z 0, r D. Then for each z B(z 0, r \ {z 0 }, the map (ϕ, ψ f ϕ,ψ(z f ϕ,ψ (z 0 z z 0 = ϕ, f (w f (z ψ w z defines a sesquilinear form. Moreover, since this converges as z z 0 for each ϕ, ψ H, the lemma above shows that C := sup f (z f (z 0 z z 0 <. z B(z 0,r\{z 0 } Since (ϕ, ψ f ϕ,ψ (z 0 is a pointwise limit of a sesquilinear form, it is also a sesquilinear form. Moreover, f ϕ,ψ (z 0 = lim z z0 ϕ, f (z f (z 0 ψ C ϕ ψ z z 0 and hence (ϕ, ψ f ϕ,ψ (z 0 is bounded. Therefore such f (z 0 exists by the Riesz representation theorem for bounded sesquilinear forms. Next, we prove that f is holomorphic in the norm topology with the derivative f constructed as above. For each z 0 D, the function z f ϕ,ψ(z f ϕ,ψ (z 0 f ϕ,ψ (z 0(z z 0 (z z 0 2 = ϕ, f (z f (z 0 f (z 0 (z z 0 (z z 0 2 ψ 3

extends to a holomorphic function on D. If we choose r > 0 so that B(z 0, r D, then this observation shows that sup ϕ, f (z f (z 0 f (z 0 (z z 0 (z z 0 2 ψ < z B(z 0,r\{z 0 } By the lemma above, this yields sup z B(z 0,r\{z 0 } f (z f (z 0 f (z 0 (z z 0 (z z 0 2 < This bound immediately implies that f is holomorphic in the norm topology with the derivative f as required. Finally, equivalence of weak analyticity and strong analyticity is an immediate consequence from the following equivalent together with the proof above. Proposition. Let B be a Banach space and D C be a domain. Then f : D B is holomorphic if and only if it is analytic. We do not show this claim here, but the proof is essentially the same as in the classical case B = C with due modification. Problem 5. Prove that if T is self-adjoint (and, by our convention, bounded, then the spectral radius of T equals T. Proof. We begin by noting that T 2 = T T is true for any bounded operator T. Indeed, this follows from T 2 = sup T x H 2 = sup x, T T x H T T T T = T 2. x H =1 x H =1 In particular, if T is self-adjoint then T 2 = T 2. Next, if T is self-adjoint then T n is also self-adjoint for n 1. Then a repeated application of the previous identity gives T 2n = T 2n. Therefore T = lim n T 2n 1/2n = r(t. The last equality follows from the Gelfand s formula, which is already proved in the class. 4

Problem 6. For a bounded self-adjoint operator T on H, let φ T : C(σ(T B(H be the (unique map such that φ T (P = P(T for P a polynomial and φ T ( f B(H = f C(σ(T, f C(σ(T Given ψ H, let µ ψ be the Borel measure on σ(t such that, for each f C(σ(T, ψ, φ T ( f ψ = f (λ µ ψ (dλ (1 Let B(σ(T be the class of bounded measurable functions on σ(t. Construct an extension φ T : B(σ(T B(H of φ T such that (1 holds for all f B(σ(T and all ψ H. Proof. Notice that for any f C(σ(T, we have φ T ( f ψ 2 H = φ T ( f ψ, φ T ( f ψ = ψ, φ T ( f f ψ = f (λ 2 µ ψ (dλ = f L 2 (µ ψ. (2 Then for each ψ H, the linear map Phi T,ψ : C(σ(T H defined by Phi T,ψ ( f = φ T ( f ψ is an L 2 -isometry. Since C(σ(T is a dense subset of L 2 (µ ψ, the identity (2 allows us to uniquely extends Phi T,ψ to all of L 2 (µ ψ. We denote this extension by PhiExt T,ψ. Before moving to construct the map φ T of our interest, we check some properties of PhiExt T,ψ. The usual approximation argument shows that PhiExt T,ψ is linear. The identity (1 extends to PhiExt T,ψ in an appropriate sense. Indeed, for each f L 2 (µ ψ, we can find an approximating sequence ( f n C(σ(T that converges to f in L 2 (µ ψ. Then φ T ( f n ψ PhiExt T,ψ ( f in H, and f n f in L 1 (µ ψ, where the latter follows from the Cauchy-Schwarz inequality f n f L 1 (µ ψ ψ H f n f L 2 (µ ψ. This shows ψ, PhiExt T,ψ ( f = lim ψ, φ T ( f ψ = lim f n (λ µ ψ (dλ = f (λ µ ψ (dλ. (3 n n We may identify B(σ(T as a subset of L 2 (µ ψ, since µ ψ is a finite measure. This way, we find that PhiExt T,ψ ( f is well-defined for each f B(σ(T. Now we would like to define the map φ T : B(σ(T B(H. In view of the previous construction, it seems natural to define φ T ( f for any f B(σ(T by φ T ( f : ψ PhiExt T,ψ ( f. 5

Indeed, the previous construction shows that φ T coincides with φ T on C(σ(T and that (1 extends to this map. So it may sound like everything is settled. In fact, that is far from the truth. We have not proved that φ T ( f B(H. Since the notion of convergence in L 2 (µ ψ need not (and in general, does not coincide for different vectors ψ H, it seems very hard to work directly with the extension PhiExt. Instead of sticking to this pesky L 2 -business, we appeal to a monotone class argument. Define the family F by F = { f B(σ(T : φ T ( f B(H}. In order to invoke a monotone class argument, we check the following properties: (a Since φ T extends φ T, we have C(σ(T F. (b We check that F is a vector space. Indeed, assume that f 1,, f n F and c 1,, c n C. Since PhiExt T,ψ is linear for each ψ H, we readily check that φ T ( f (ψ = n j=1 c j φ T ( f j ψ holds for any ψ H. Then by the assumption, it follows that φ T ( f = n j=1 c j φ T ( f j B(H and therefore f H. (c Let ( f n be a uniformly bounded sequence in F such that f n converges pointwise to a function f on σ(t. Then we know that f B(σ(T. Moreover, by (2 and the bounded convergence theorem, we have φ T ( f n ψ φ T ( f (ψ H = f n f L 2 (µ ψ n 0 This proves that φ T ( f n converges strongly and the limit function is φ T ( f. Therefore φ T ( f is also a bounded linear operator and hence f F. This three properties are enough to initiate a monotone class argument to show that F = B(σ(T. Therefore φ T is a well-defined map B(σ(T B(H that extends φ T as expected. Problem 7. For the map φ T from the previous exercise, prove that σ(φ T ( f = { f (λ : λ σ(t} holds for all f C(σ(T. Give an example of T and f B(σ(T for which this fails (with φ T ( f replacing φ T ( f, of course. Proof. We first prove that σ(φ T ( f f (σ(t. If λ f (σ(t, then it follows from the compactness of f (σ(t that dist(λ, f (σ(t > 0. So we have (λ f 1 C(σ(T. By the continuous functional calculus, this implies that λ φ T ( f has a bounded inverse. Therefore λ σ(φ T ( f. Before moving to the proof of the reverse direction, we introduce a simple lemma. 6

Lemma. Let A, B B(H. Then for any λ σ(a, we have dist(λ, σ(b R λ (A 1 B A. Proof of Lemma. We may assume B A R λ (A < 1 without losing generality. Under this condition, the series R λ (A n=0 ((B AR λ (A n converges in B(H to the resolvent R λ (B. Moreover, this gives R λ (B R λ (A 1 B A R λ (A. Similar consideration shows that if µ λ R λ (B < 1 then the resolvent R µ (B exists and hence µ σ(b. So it follows that This completes the proof of lemma. //// dist(λ, σ(b R λ (B 1 R λ (A 1 B A. Now we return to the original proof and show that f (σ(t σ(φ T ( f. Assume that λ σ(φ T ( f. Then for any polynomial p, dist(λ, σ(φ T (p = dist(λ, p(σ(t = inf λ p(µ µ σ(t inf µ σ(t ( λ f (µ + p f C(σ(T = dist(λ, f (σ(t + p f C(σ(T. Here, the first equality follows from what is already proved in the class. Then dist(λ, f (σ(t dist(λ, σ(φ T (p p f C(σ(T R λ (φ T ( f 1 φ T (p φ T ( f p f C(σ(T = R λ (φ T ( f 1 2 p f C(σ(T. Taking p f in C(σ(T, we find that dist(λ, f (σ(t R λ (φ T ( f 1 > 0. Therefore λ f (σ(t and the claim follows. Finally, we construct a counter-example when f is allowed to be in B(σ(T. Pick a bounded sequence a = (a n in R such that the set of limit points of a is [ 1, 1]. (For instance, a n = cos n does the job. Then let H = l 2 (N and consider the operator T : l 2 (N l 2 (N which acts by multiplication by a, i.e., T(x n n N = (a n x n n N. It is straightforward to check that T is a bounded self-adjoint operator T x 2 a l (N x l 2 (N, x, T y = x n a n y n = a n x n y n = T x, y n N n N 7

and that p(t(x n n N = (p(a n x n n N. This relation continues to hold when p is replaced by bounded Borel-measurable functions. This follows from our construction of the extension φ T together with the monotone convergence theorem applied to x, p(tx l 2 (N = p(a n x n 2 = n=1 p(λ µ x (dλ where µ x = x n 2 δ an In order to see how this operator serves as a counter-example, we first check that a m σ(t for any m. This follows from Te m = a m e m where e m = (δ mn n N. Then by the compactness, [ 1, 1] σ(t. Now choose α [ 1, 1] \ {a n : n N} and define a function on R by 1, if λ > α, f (λ = 1/2, if λ = α, 0, if λ < α. Then f B(σ(T and we have f (σ(t = {0, 1 2, 1}. On the other hand, f (a n {0, 1} for all n N. This shows that φ T ( f is an orthogonal projection and hence σ( φ T ( f {0, 1}. Therefore f (σ(t σ( φ T ( f as desired. n=1 Problem 8. Given two finite Borel measures µ, ν on R with bounded support, define T µ f (λ := λ f (λ for f L 2 (µ and T ν f (λ := λ f (λ for f L 2 (ν. Show that T µ, T ν unitarily equivalent µ, ν equivalent (We recall that two measures are equivalent if they have the same null sets. Proof. ( Define the operator U : L 2 (µ L 2 (ν by U f = (dµ/dν 1/2 f, where dµ/dν is the Radon-Nikodym derivative of µ with respect to ν. 1 Then for any f L 2 (µ, it follows that U f 2 dν = f 2 dµ dν dν = f 2 dµ R R and thus U is an isometry. Moreover, since dµ dν dν dµ = 1 both µ-a.e. and ν-a.e., it follows that U 1 exists and is an isometry given by U 1 f = (dν/dµ 1/2 f. So U is a unitary transformation. The identity T µ = U 1 T ν U is straightforward to check. R 1 dµ/dν is defined in ν-a.e. sense, but any non-negative version will work for the argument. Similar remark applies to dν/dµ. 8

( Let U : L 2 (µ L 2 (ν be a unitary transformation such that T µ = U 1 T ν U holds. Then for any polynomial p, we have p(t µ = U 1 p(t ν U. Now let ϕ = U1 2 L 1 (ν, where 1 L 2 (µ denotes the constant function with value 1. Then p(λ µ(dλ = 1, p(t µ 1 L 2 (µ = U1, p(t ν U1 L 2 (ν = p(λϕ(λ ν(dλ. R In view of the bounded convergence theorem and the monotone class theorem, this identity remains valid for any bounded Borel-measurable functions p on R. Thus it follows that µ ν with dµ/dν = ϕ. Finally, interchanging the rule of µ and ν proves that ν µ also holds. Therefore the claim is true. R Problem 9. Prove Weyl s criterion: For T bounded and self-adjoint, { } σ(t = λ R : {ψ n } n N H, ψ n = 1, lim (λ Tψ n = 0 n Use it to conclude that σ(t = σ ac (T σ sc (T σ pp (T. Proof. (Part 1 For the first part of the claim, notice that [ ] {ψ n } n N H, ψ n = 1, lim (λ Tψ n = 0 n So it suffices to show that inf (λ Tψ = 0. ψ =1 ρ(t = {λ C : inf (λ Tψ > 0}. (4 ψ =1 We begin by noting that σ res (T = for a bounded self-adjoint operator T. Indeed, this follows from the relation σ res (T σ eig (T. Next, denote by ρ(t the right-hand side of (4. If λ ρ(t, then ψ H = R λ (λ Tψ H R λ (λ Tψ H shows that inf ψ =1 (λ Tψ R λ 1 > 0 and hence λ ρ(t. This proves ρ(t ρ(t. Conversely, assume that λ ρ(t. The assumption λ ρ(t implies that λ T is injective. Also, the observation σ res (T = forces that D = Ran(λ T is a dense subset of H. (We are not excluding, and in fact will prove, that D = H. Now let c = inf ψ =1 (λ Tψ. Then ψ D : c 1 ψ H = c 1 (λ T(λ T 1 ψ H (λ T 1 ψ H 9

shows that (λ T 1 : D H uniquely extends to a bounded operator on H, which we denote by R. This satisfies R(λ T = id H since R D = (λ T 1. Moreover, applying the density argument to ψ D : (λ TRψ = (λ T(λ T 1 ψ = ψ proves that (λ TR = id H as well. Therefore λ T is invertible with the bounded inverse R and hence λ ρ(t. (Part 2 Let us see how Weyl s criterion leads to the identity. If λ σ (T for some type {ac, sc, pp}, then by Weyl s criterion there exists a sequence (ψ n H such that ψ n = 1 and (λ Tψ n 0 as n. Again by Weyl s criterion, this implies λ σ(t. For the reverse direction, assume that λ σ(t and choose a sequence (ψ n in H according to Weyl s criterion. We utilize the orthogonal decomposition H = H ac H sc H pp to write ψ n = ψ ac,n + ψ sc,n + ψ pp,n, where φ,n H for each type. Now from the identity 1 = ψ n 2 = ψ ac,n 2 + ψ sc,n 2 + ψ pp,n 2, we must have ψ,n 1/ 3 for some type. Passing to a subsequence if necessary, it follows that there exists a type for which ψ,n 1/ 3 for all n. Then for the normalized version ψ n = ψ,n / ψ,n, we have (λ T ψ n 3 (λ Tψ,n 3 (λ Tψ n n 0. In the second inequality, we exploited the fact that the decomposition H = H ac H sc H pp is T-invariant. Therefore λ σ (T as required and the desired equality follows. Problem 10. Prove Stone s formula: If T is bounded and self-adjoint, R λ = (λ T 1 for λ ρ(t is the resolvent, and Π A is the spectral projection associated with a Borel set A R, then 1 b s- lim (R λ iε R λ+iε dλ = 1 ( Π[a,b] + Π (a,b ε 0 2πi a 2 Here the limit is in the strong operator topology; the integral is in Bochner sense. Proof. Let F : [a, b] C(σ(T be continuous. Then adopting the usual Riemann-sum business shows the following identity between Bochner integrals b a φ T (F(λ dλ = φ T ( b a F(λ dλ. 10

We apply this observation to F(λ = (λ iε 1 (λ + iε 1. Then we have R λ iε R λ+iε = φ T (F(λ and hence 1 b ( 1 b ( 1 (R λ iε R λ+iε dλ = φ T 2πi a 2πi a λ iε 1 λ + iε ( 1 b 1 = φ T π a (λ 2 + ε 2 dλ ( ( 1 b = φ T π arctan 1 ( a ε π arctan ε } {{ } =:g ε ( Notice that g ε (x 1 uniformly in x R and in ε > 0. Moreover, g ε converges pointwise to dλ. lim g ε (x = 1 ε 2 1 [a,b](x + 1 1, x (a, b 2 1 (a,b(x = 1/2, x {a, b} Therefore the integral converges strongly to as desired. 0, otherwise x R 1 b ( 1 s- lim (R λ iε R λ+iε dλ = φ T ε 0 2πi a 2 1 [a,b] + 1 2 1 (a,b = 1 2 Π [a,b] + 1 2 Π (a,b 11

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