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 repeated dim ker(k µ n ) times, then µ n 2 <. Observe that for every f, g L 2 (X, Ω, µ) we have Kf, g (Kf)(x)g(x)dx Å k(x, y)f(y)dy g(x)dx Å k(x, y)f(y)g(x)dy dx. (We shortly write dx for dµ(x)) By applying Tonelli s theorem we see that f(y)g(x) H : L 2 (X X, Ω Ω, µ µ) therefore k(x, y)f(y)g(x) L (X X, Ω Ω, µ µ) and hence by Fubini s theorem Å Kf, g k(x, y)f(y)g(x)dx dy Å f(y) k(x, y)g(x)dx dy Therefore K is self-adjoint. Å f(y) k(y, x)g(x)dx dy f(y)(kg)(y)dy f, Kg. For the second part, note that by Corollary 5.4, (ker K) has an orthonormal basis {e n } n N such that each e n is an eigenvector associated to the eigenvalue µ n. Note that for fixed n, we have {m N : µ m µ n } {m N : e m ker(k µ n I)}. And since {e m ker(k µ n I)} is an (orthonormal) basis for ker(k µ n I), we get {m N : µ m µ n } {e m : e m ker(k µ n I)} dim ker(k µ n I). By Lemma 4.8, the set {φ ij : i, j N} defined by φ ij (x, y) e i (x)e j (y)
REAL ANALYSIS II HOMEWORK 3 2 is an orthonormal subset of H. And using Fubini, we have k, φ ij H k(x, y)φ ij (x, y)dxdy Å k(x, y)e i (x)dx e j (y)dy (Ke i )(y)e j (y)dy Ke i, e j L 2 (X,Ω,µ) µ i e i, e j L 2 (X,Ω,µ) µ i δ ij where δ ij is the Kronecker delta. Therefore by Bessel s inequality (and Tonelli s theorem for series) we get k 2 H k, φ ij 2 µ i δ ij 2 µ i 2. i,j N i N j N i N 4. If T is a compact self-adjoint operator and {e n } and {µ n } are as in (5.4) and if h is a given vector in H, show that there is a vector f in H such that T f h if and only if h ker T and n µ 2 n h, e n 2 <. Find the form of the general vector such that T f h. Suppose h T f for some f H. Then for every g ker T we have h, g T f, g f, T g f, 0 0 so h ker T. And by Corollary 5.4, we have h, e n T f, e n µ n f, e n and hence f, e n µ n h, e n. Therefore since {e n } is an orthonormal set, By Bessel s inequality we have f 2 n n n f, e n 2 µ n 2 h, e n 2 µ 2 n h, e n 2 as µ n s are real. Conversely, suppose h ker T and n µ 2 n h, e n 2 <. Recall that since {e n } is an orthonormal basis for (ker T ), there exists a unitary mapping ϕ : (ker T ) l 2 (N) g { g, e n } n N. Let f ϕ ({µ n h, e n } n N ). More concretely, f is the limit of the Cauchy sequence given by the partial sums N µ n h, e n e n in (ker T ), or shortly f n µ n h, e n e n.
Since T is continuous and linear, we get REAL ANALYSIS II HOMEWORK 3 3 T f n µ n h, e n T (e n ) n n µ n h, e n µ n e n h, e n e n h where the last equality holds because we are assuming h (ker T ) (in general it would be only P (h) where P is the projection of H onto (ker T ) ). Conway, Page 53 2. In Theorem 6.2 show that n λ 2 n < (see Exercise 5.3). Here is the setup: L : D L 2 [a, b] is the Sturm-Liouville operator which is assumed to be injective. Then by Theorem 6.9 there is a Hilbert-Schmidt operator G on L 2 [a, b] such that Im G D and the restriction of G to L 2 [a, b] D is the inverse of L. Moreover G is self-adjoint, with eigenvalues λ, λ 2,... in Theorem 6.2 s notation, which are all real. Therefore by Exercise 5.3 we have n λ 2 n <. 3. In Theorem 6.2 show that h D if and only if h L 2 [a, b] and n λ 2 n h, e n 2 <. If h D, show that h(x) n h, e n e n (x), where this series converges uniformly and absolutely on [a, b]. By Exercise 5.4, as G is compact self-adjoint operator, we get that h D Im G if and only if h (ker G) L 2 [a, b] and (λ n ) 2 h, e n 2 λ 2 n h, e n 2 <. n n The repetitions of λ n s is not an issue because all of the eigenspaces of G are onedimensional by Theorem 6.2 (a) and (b). Assume h D. As {e n } is an orthonormal basis for L 2 [a, b], we have h h, e n e n n That is, if we define h N N n h, e n e n, then the sequence {h N } converges to h in the L 2 -norm. Stein & Shakarchi, Chapter 6 29. Let T be a compact operator on a Hilbert space H, and assume λ 0. (a) Show that the range of λi T defined by {g H : g (λi T )f, for some f H} is closed. [Hint: Suppose that g j g, where g j (λi T )f j. Let V λ denote the eigenspace of T corresponding to λ, that is, the kernel of λi T. Why
REAL ANALYSIS II HOMEWORK 3 4 can one assume that f j V λ? Under this assumption prove that {f j } is a bounded sequence.] (b) Show by example that this may fail when λ 0. (c) Show that the range of λi T is all of H if and only if the null-space of λi T is trivial. (a) Start as said in the hint. Since λ 0, V λ is finite-dimensional and in particular a closed subspace. So f j u j + v j where u j V λ and v j V λ. Then g j (λi T )(u j + v j ) (λi T )v j so by replacing f j with v j, we may assume f j V λ. If {f j } has a subsequence identically zero, that implies g 0 so there is nothing to show. So we may assume {f j } has no such subsequence and then by throwing away finitely many terms, we may assume that f j 0 for every j. Now let e j f j / f j. Since {e j } is a sequence of unit vectors and T is compact, there is a convergent subsequence {T e jk } of {T e j }. By replacing {f j } with {f jk }, we may assume that {T e j } is convergent, say to T e j u. Now observe that e j λ ((λi T )e j + T e j ) Ç å λ gj f j + T e j Suppose that f j s are not bounded. Then by passing to a subsequence we may assume that f j. Consequently g j / f j 0 so e j λ u. Then we have the following: Since e j V λ, the limit λ u and hence u is in V λ. Since e j s are unit vectors, λ u so u λ 0. So u 0. Since T is continuous, T λ u u. So T u λu, which means u V λ. This is a contradiction. Thus {f j } is bounded. Hence as T is compact, {T f j } has a convergent subsequence {T f jk }. By replacing f j with f jk we may assume that {T f j } is convergent, say f j f. Finally this yields that g j (λi T )f j (λi T )f because (λi T ) is continuous; so (λi T )f g and g Im(λI T ). (b) Note that if λ 0, the range of λi T is simply im T. So we want a compact operator whose image is not closed. Consider the operator T defined in Exercise 33. In its solution we show that it is compact. Also note that because Æ T ϕ k, ϕ l k ϕ k+, ϕ l k δ k+,l l δ k,l Æ ϕ k, l ϕ l T is defined by T ϕ k ϕ k k. Now since k 2 ψ N N k ϕ k+, <, if we let
REAL ANALYSIS II HOMEWORK 3 5 {ψ N } is a convergent sequence in H, say with the limit ψ. By the definition of T, we have ( N ) ψ N T ϕ k Im T. However we claim that ψ / im T. For the sake of contradiction, suppose T ϕ ψ. Then for k 2 we have ϕ, ϕ k (k ) ϕ, ϕ k k (k ) ϕ, T ϕ k (k ) ψ, ϕ k (k ) k which contradicts Bessel s inequality. (c) In general we claim that for a bounded linear operator S, we have Indeed, if h ker S then (im S) ker S. h, Sf S h, f 0, f 0 for every f H. Hence ker S (im S). And if h (im S) then for every f H we have S h, f h, Sf 0. Hence S h 0, so we get the reverse inclusion. In our case, letting S λi T, by (a) we have The claim is immediate from here. im S im S ((im S) ) (ker S ) (ker(λi T )). 33. Let H be a Hilbert space with basis {ϕ k }. Verify that the operator T defined by is compact, but has no eigenvectors. T (ϕ k ) k ϕ k+ Suppose T (ψ) λψ for some λ C and ψ H. So there exists scalars {c k } such that ψ c k ϕ k. Then λc k ϕ k λψ T (ψ) c k k ϕ k+. Since {ϕ k } is an (orthonormal) basis, this means λc 0 and λc k+ c k /k for k. There are two cases: λ 0. Then for every k, we have c k /k 0 and hence c k 0.
REAL ANALYSIS II HOMEWORK 3 6 λ 0. Then λc 0 yields c 0. But then c 2 /2 0 and so c 2 0. Inductively we get c k 0 for every k. So in any case, we get ψ 0. Thus T has no eigenvectors. To see that T is compact, let {ψ n } be a sequence in H such that ψ n for every n. We will show that the sequence {T (ψ n )} has a convergent subsequence. Writing ψ n c n,k ϕ k, we have c n,k 2 for every n. Given ε > 0, there exists N N such that kn+ k 2 < ε/4. Now for each n let x n (c,, c,2,..., c,n ) C N. So {x n } is a sequence in C N such that x n. Therefore by Bolzano-Weierstrass theorem {x n } has a convergent subsequence {x nj }. So there exists M such that p, q > M implies that x np x nq 2 < ε/2. Then for every p, q > M we have T ψ np T ψ 2 c nq np,k c nq,k 2 H k N c np,k c nq,k 2 c np,k c nq,k 2 + k 2 kn+ k 2 N c np,k c nq,k 2 2 + kn+ k 2 ε 2 + 2 ε 4 ε. So {T ψ nj } is a Cauchy sequence in H, therefore it is convergent. 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 <. (b) K(x, y) λ k ϕ k (x)ϕ k (y) is the expansion of K in the basis {ϕ k (x)ϕ k (y)}. (c) Suppose that 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. (a) and (b) By the spectral theorem we know that {ϕ k } is an orthonormal basis for L 2 (R d ). Then by what we have shown in problem 3 in page 49 of Conway, K, ψ jk L 2 (R d R d ) δ jkλ k where {ψ jk } s are defined by ψ jk (x, y) ϕ j (x)ϕ k (x). So the only thing we need to verify is that the eigenfunctions {ϕ k } can be taken to be real. Write u k + iv k for the
REAL ANALYSIS II HOMEWORK 3 7 decomposition of the function φ k to the real and imaginary parts. Then since K is real-valued, both T u k and T v k are real functions where T u k + it v k T ϕ k λ k ϕ k λ k u k + iλ k v k. As λ k is real, by comparing the real and imaginary parts above we get T u k λ k u k and T v k λ k v k. So both u k and v k belong to the eigenspace of λ k. By applying Gram- Schmidt to {u k, v k } on each eigenspace, we get an orthonormal basis of real-valued eigenfunctions. (c) One implication is by (a). For the reverse, assume n λ n 2 <. We may assume by the spectral theorem that λ > λ 2 > Then if we define n K n (x, y) λ k ϕ k (x)ϕ k (y), the sequence {K n } in L 2 (R d R d ) is Cauchy and hence has a limit K. Note that given any f L 2 (R d ), we have n K n (x, y)f(y)dy λ k ϕ k (x)ϕ k (y)f(y)dy n n ( n λ k ϕ k (x) ϕ k (y)f(y)dy λ k ϕ k, f ϕ k (x) ) λ k ϕ k, f ϕ k (x). Thus if we write P n for the projection of L 2 (R d ) onto the subspace spanned by {ϕ,..., ϕ n } the above computation shows that the finite rank operator T n P n T is an integral operator with kernel K n. So being the difference of two compact symmetric operators, T T n is compact and symmetric with the eigenvalues {λ n+, λ n+2,... }. Thus T T n λ n+ and hence T n T in operator norm since λ n 0 by the spectral theorem. Now let T be the integral operator with kernel K. Then T T n is an integral operator with kernel K K n ; therefore T T n K Kn L 2 (R d R d ) by Proposition 5.5. But K n K in L 2 (R d R d ), hence T n T in the operator norm. Thus we conclude T T is an integral operator.