Real Variables # 10 : Hilbert Spaces II

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1 randon ehring Real Variables # 0 : Hilbert Spaces II Exercise 20 For any sequence {f n } in H with f n = for all n, there exists f H and a subsequence {f nk } such that for all g H, one has lim (f n k, g) = (f, g). k Proof: Let g = k g ke k where e k is an orthonormal basis for H. If we can construct an f such that for all j we are done. lim (f n k, e j ) = (f, e j ). k Define the sequence a n = {(f n, e )} n=. Since (f n, e ) f n e = we have that a n is a sequence of real numbers in the compact set [, ] and thus contains a convergent subsequence a n j = {(f nj, e )} j= that converges to F. Take a 2 j = {(f n j, e 2 )} j= constructed using this subsequence and apply the same argument again to get another convergent subsequence of f nj that converges to F 2. We repeat this inductively and define with the diagonal with respect to this basis f = ak k e k. Then, by construction, for each basis element we have lim (f n k, e j ) = (f, e j ). k Exercise 2a Weak convergence does not imply strong convergence, nor does strong convergence imply convergence in the norm. Proof: Let us define the unilateral shift from l 2 to l 2 by Sx = S(x, x 2,..., ) = (0, x, x 2,...). and let us take the sequence S n. Intuitively, in some sense S n goes to 0; however, we also have that S is an isometry and that S n x = Sx = x for all x. Let us examine this further. Let f = (a, a 2,...) and g = (b, b 2,......), by Cauchy Schwarz in l 2 we have (S n f, g) = a b n+ + a 2 b n a k 2 b k 2.

2 Since g l 2, the tail of the series can be made as small as needed so (S n f, g) 0 and we have weak convergence. For strong convergence, we must have S n f = 0 for all f H; however, S n f = f. Next, take the adjoint of the unilateral shift operator S (x, x 2,..., ) = (x 2, x 3,...). We have for any x l 2 that (S ) n = (x n+, x n+2,...). Since the tail end is given by x 2 = (S ) n x 2 = x k 2 < x k 2 < which must go to 0 for the original sum to converge. Thus we have strong convergence. For convergence in the norm, we must have (S ) n 0. However, we have that (S ) n because for any element of the standard basis of l 2, e i we have (S ) n e i =. Exercise 2b For any any bounded operator T there is a sequence {T n } of bounded operators of finite rank so that T n T strongly as n. Proof: We recall that a bounded operator, A, on a separable infinite dimensional Hilbert Space, H, can be represented by an infinite matrix using an orthonormal basis, α ij = (Ae j, e i ), where Ae i = α ij e j. j= Let us define the truncation onto a finite span of {e i } by A n e i = n α ij e j. j= Let f = f ke k be an arbitrary element of H written out in this orthonormal basis. Then (A A n )f = i= j=n+ f i α ij e j 2

3 and (A A n )f = f i i= j=n+ α ij e j 0. Exercise 26 Suppose w is a measurable function on R d with 0 < w(x) < for a.e. x, and K is a measurable function on R 2d that satisfies for almost every x R d, and for almost every y R d. Then R d K(x, y) w(y)dy Aw(x) R d K(x, y) w(x)dx Aw(y) T f(x) = K(x, y)f(y)dy R d is bounded on L 2 (R d ) with T A. Proof: We have Following the hint, we have from the Cauchy-Schwarz inequality we have ( 2 ( K(x, y) f(y) dy) = [w(y) /2 K(x, y) f(y) ][w(y) ) 2 K(x, y) w(y) /2 ]dy ( ) ( ) K(x, y) w(y)dy f(y) 2 K(x, y) w(y) dy ( ) Aw(x) f(y) 2 K(x, y) w(y) dy. Using Fubini s theorem and the previous result we have T f(x) 2 = T f(x) 2 dx ( 2 = K(x, y)f(y)dy) dx ( ) Aw(x) f(y) 2 K(x, y) w(y) dy dx ( ) = A f(y) 2 w(y) K(x, y) w(x)dx dy 3

4 A f(y) 2 w(y) Aw(y)dy = A 2 f(y) 2 dy = A 2 f 2 Thus, T f A f so so T A. Exercise 27 T f(x) = π is bounded on L 2 (0, ) with norm T. 0 f(y) x + y dy Proof: We have the situation where K(x, y) = π x+y. We must apply the appropriate w from the preceding problem. Let us try for x > and 0 otherwise. Then K(x, y) w(y)dy = R w(x) = π x = x y y + x dy y y + x dy = x ln y ln(y + x) = x ln y y + x = ln( + x) = w(x) ln( + x). x Not quite what I wanted. Exercise 28 Suppose H = L 2, where is the unit ball in R d. Let K(x, y) be a measurable function on that satisfies K(x, y) A x y d+α for some α > 0, whenever x, y. Define T f(x) = K(x, y)f(y)dy. (a) T is a bounded operator on H. Proof: y a direct calculation, T f(x) 2 = T f(x) 2 dx 4

5 = ( 2 K(x, y)f(y)dy) dx ( 2 A f(y) x y dy) d+α dx However, Cauchy Schwarz tells us ( 2 f(y) x y dy) d+α = = M f(y) 2 x y d+α dy f(y) 2 x y d+α dy x y d+α dy The last integral converges (to say, M) because the numerator goes as r d and the denominator goes as r d α. ( 2 T f(x) 2 A f(y) x y dy) d+α dx AM f(y) 2 x y d+α dxdy AM 2 f(y) 2 dy = AM 2 f 2 (b) T is compact. Proof: Let us take the operators T n given by T n f(x) = K n (x, y)f(y)dy where K n (x, y) = K(x, y) if x y /n and 0 otherwise. The truncation K n is still in L 2 ( ) so it is still a Hilber-Schmidt operator and therefore compact. Also by the absolute continuity of the integral K n (x, y) K(x, y) dy A x y d+α dy = A z d+α dy 0. x y /n z /n For the last step we used Prop.2 from Chapter 2. (c) T is a Hilbert-Schmidt operator if and only if α > d/2. Proof: According to our text, an operator is Hilbert-Schmidt if and only if K L 2 ( ), that is 5

6 x y 2d+2α dydx < The numerator must go as r d and the denominator goes as r 2d 2α. We need or a > d/2. d (2d 2α) = d + 2α < Exercise 29 Let T be a compact operator on a Hilbert space H, and assume λ 0. (a) The range of λi T defined by {g H : g = (λi T )f, for some f H} is closed. Proof: Let g j = (λi T )f j and V λ = ker(λi T ) which is of finite rank and thus closed. Since V λ is closed, f n has on orthogonal decomposition f n = x n + y n where x n V λ and y n Vλ. We can ignore the x n as g j = (λi T )f j = (λi T )y j. So only take f n Vλ. Assume f n is not bounded, then we can take a subsequence such that f n 0 for all n and lim f n =. Normalizing h n = f n / f n then h i Vλ, h n = and (λi T )h n = g j / f j 0 since {f n } is bounded, y compactness of T we have that a subsequence of {T h n } converges. Thus, h n converges to h with h = and (λi T )h = lim(λi T )h n 0, so h n V λ. However, h n V λ so h h n 2 = 2 which is a contradiction. Thus f j is bounded. Since f j is bounded, we have that a subsequence of T f j converges. Then λf j = (λi T )f j + T f j must also converge to say f V λ. Thus, g = lim(λi T )f j = (λi T )f so we have that the image of λi T is closed 6

7 (b) This may fail when λ = 0. Proof: Take any operator whose image is not closed. (c) Show that the range of λi T is all of H if and only if the null-space of λi T is trivial. Proof: I will quote two results contained in almost every textbook (except ours, which I am starting to really dislike). If X is a linear subspace of H then X = X and if T is a bounded operator ker T = (ImT ). Then, since Im(λI T ) is closed, Im(λI T ) = Im(λI T ) = [ker(λi T ) ] = ker(λi T ) Thus if the range of λi T is all of H if and only if ker(λi T ) is all of H if and only if ker(λi T ) = {0}. Exercise 30 Let H = L 2 (S ). Fix a bounded sequence {λ n } of complex numbers, and define an operator T f by T f(x) λ n a n e inx whenever f(x) a n e inx. (a) T is a bounded operator on H and T = sup n λ n. Proof: As we know that e inx is an orthornormal basis, we have from Parseval s identity that and Let λ = sup n λ n, then f 2 = T f 2 = a n 2 λ n 2 a n 2. T f 2 = λ n 2 a n 2 7

8 λ 2 a n 2 = λ 2 a n 2 = λ 2 a n 2 = λ 2 f 2 so we have T λ. We must have some sequence λ n whose sup λ n and take x n = e iλnx. Then since T is bounded (and therefore continuous) we have T x = lim T x n = lim λ n e iλnx = lim λ n = λ so T λ. Thus, T = λ. (b) T commutes with translations, that is, we define τ h f(x) = f(x h) then T τ h = τ h T for every h R. Proof: We have = = where b n = a n e inh. Then τ h f(x) = f(x h) = a n e in(x h) a n e inh e inx b n e inx T τ h = λ n b n e inx = λ n a n e inh e inx and τ h T f(x) =. = λ n a n e in(x h) λ n a n e inh e inx 8

9 (c) Conversely, if T is any bounded operator on H that commutes with translations, then T is a Fourier multiplier operator. Proof: Apply this to the basis element e inx and let T e inx = k a ke ikx and τe ins = e inh e inx. Assuming this operators commute τ h T e inx = k a k e ikh e ikx is equivalent to T τ h e inx = e inh k a k e ikx. This only holds if a k is 0 except for k = n then we have (say) a n = λ n. Thus every basis element is an eigenvector and we have a fourier multiplier sequence. Exercise 33 Let H be a Hilbert space with basis {e k }. operator T defined by Verify that the is compact, but has no eigenvectors. T (e k ) = k e k+ Proof: Let us verify that T has no non-zero eigenvectors. Assume λ is an eigenvalue of T, then for an arbitrary element x = (x, e k)e k we have (x, e k ) e k+ = T x = λx = k (x, e k )e k. If λ = 0, we have (x, e k ) = 0 for all k, a contradiction to the assumption that x is a non-zero eigenvector. Assume λ 0. As this is a basis, equating coefficients for e we have λ(x, e ) = 0, thus (x, e ) = 0 as we assumed that λ 0. Next, for e 2 we have λ(x, e 2 ) = (x, e ) = 0, thus (x, e 2 ) = 0. For e 3, we have λ(x, e 3 ) = (x, e 2 )/2 = 0, and (x, e 3 ) = 0. Similarly, we find (x, e k ) = 0 for all k thus x = 0. Again we have a contradiction, thus T has no eigenvectors. Define T k by ( n ) T n x = T (x, e k )e k = Then, T n is of finite rank and thus compact. n (x, e k )T e k 9

10 Proposition 6. (ii) tells us that if we show that T n T 0 then T is compact. Taking our arbitrary element x H we have using Cauchy- Schwarz (and remembering that in this basis it behaves like l 2 ) (T n T ) = x x (x, e k )T e k (x, e k ) T e k (x, e k ) 2 k 2 e k k2 T e k 2 Exercise 34 Let K be a Hilbert-Schmidt kernel which is real and symmetric. Then, as we saw, the operator T whose kernel is K is compact and symmetric. Let {φ k (x)} be the eigenvectors ( with eigenvalues λ k ) that diagonalize T. Then: (a) k λ k 2 < Proof: We have T φ k = λ k φ k and that φ k is a basis, so by Parseval s identity we have T φ k 2 = λ k φ k 2 = λ k 2 However, since φ k is in the Hilbert space, it must have finite norm, thus k λ k 2 <. (b) K(x, y) k λ kφ k (x)φ k (y) is the expansion of K in the basis {φ k (x)φ k (y)}. Proof: We know that in general for a Hilbert-Schmidt operator we have K(x, y) k,l a kl φ k (x)φ l (y). T φ n (x) = K(x, y)φ n (y)dy 0

11 = = k = k = k a kl φ k (x)φ l (y)φ n (y)dy k,l l a kl φ k (x) a kl φ k (x)δ ln l a kn φ k (x). φ l (y)φ n (y)dy However, T φ n = λ n φ n, so a kn φ k (x) = λ n φ n k so by comparing coefficients of basis elements we have that a kn = λ n δ kn thus K(x, y) k,l = k,l = k a kl φ k (x)φ l (y) λ n δ kn φ k (x)φ l (y) λ k φ k (x)φ k (y). We can again see that n λ 2 < by the conditions that K(x, y) L 2. (c) Suppose T is a compact operator which is symmetric. Then T is of Hilbert-Schmidt type if and only if n λ n 2 <, where {λ n } are the eigenvalues of T counted according to their multiplicities. Proof: We must only show the converse as we have already shown the previous implications. Since T is a compact symmetric operator, the Spectral theorem gives us a basis {φ k } of H that consists of eigenvectors T φ k = λ k φ k. We can then define an L 2 by (because n λ n 2 < ) by K(x, y) = k λ k φ k (x)φ k (y) Defining T f(x) = K(x, y)f(y)dy

12 we have that T φ m (x) = λ k φ k (x)φ k (y)φ n (y)dy = λφ m (x). k Since this agrees for all basis elements and thus an arbitrary function f, we must have that T is Hilbert-Schmidt. 2

REAL ANALYSIS II HOMEWORK 3. Conway, Page 49

REAL ANALYSIS II HOMEWORK 3. Conway, Page 49 REAL ANALYSIS II HOMEWORK 3 CİHAN BAHRAN Conway, Page 49 3. Let K and k be as in Proposition 4.7 and suppose that k(x, y) k(y, x). Show that K is self-adjoint and if {µ n } are the eigenvalues of K, each