201B, Winter 11, Professor John Hunter Homework 9 Solutions

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1 2B, Winter, Professor John Hunter Homework 9 Solutions. If A : H H is a bounded, self-adjoint linear operator, show that A n = A n for every n N. (You can use the results proved in class.) Proof. Let A : H H be a bounded, self-adjoint linear operator. Then we can prove that A k is also a bounded self-adjoint operator for any k N. Why? The answer is because of the following argument. Assume that A = M, where M is some constant greater than zero. Then by induction we can conlude the following: and A n = A n A A n A A n = M n (A n ) = (A n A) = A (A n ) = A(A n ) = A n. Hence, we got that indeed A n is also a bounded, self-adjoint linear operator. The linearity of A k follows right away from the linearity of A. The spectral radius of a bounded self-adjoint operator is given by its norm; more precisely, remember that A k = r(a k ) = lim A kn /n. Relabelling the indices so that u = nk, we have that A k = lim A u k/u u ( ) k = lim u Au /u =r(a) k = A k. Hence, we are done proving our problem.

2 2. Define m : (, ) (, ) be if < x < /4 m(x) = 2(x /4) if /4 x 3/4 if 3/4 < x < and let M : L 2 (, ) L 2 (, ) be the multiplication operator Mf = mf. Determine the spectrum of M and classify it into point, continuous and residual spectrum. Describe the eigenspace of any eigenvalues in the point spectrum. Proof. due to Amanda. Fisrt let s establish some properties about the operator M. Clearly, M is a self-adjoint operator since we have the following: f, Mg = f(x)m(x)g(x)dx = m(x)f(x)g(x)dx = Mf, g. Note that here the fact that m(x) is a real valued function, i.e., m(x) = m(x), helped us to conclude that M is self-adjoint. M is self-adjoint, implies that σ r (M) = and σ(m) [ M, M ]. This results that I have just stated can be found in Prop.9.8 and Lemma 9.3 form your textbook. Looking at the definition of m(x), we can see that the maximum value that m can attain is and we can also observe that m 2 (x) for all x [, ]. Hence, applying the definition of the norm of M we get: M = Mf(x) f = ( = f = f = = ( ) /2 m 2 (x) f(x) 2 dx ) /2 f(x) 2 dx If we take any f(x) L 2 (, ) with port on (3/4, ) such that f(x) =, we see that Mf(x) =, so M =. Hence, σ(m) [, ]. Notice that for any nonzero function f(x) with port on (, /4) that Mf(x) =, so these are functions with eigenvalue. Furthermore, we know such functions exist. Similarly, any nonzero function f(x) with port on (3/4, ) satisfies Mf(x) = f(x) so these are functions

3 with eigenvalue. So {, } σ p (M). Consider any λ [, ). Then since m(x) for all x, it follows that m(x) λ λ > for all x. Since (M λi)f(x) = (m(x) λ)f(x), if (M λi)f(x) = then f(x) = a.e., so it follows that M λi is one-to-one. Also, if g(x) L 2 (, ) then m(x) λ g(x) is well-defined and in L2 (, ), so So M λi is also onto. (M λi) g(x) = g(x). m(x) λ Hence, for λ [, ) we see that λ ρ(m), so σ(m) [, ]. If λ (, ), then we still have M λi is one-to-one since m(x) λ is nonzero for all but one value of x so (M λi)f(x) = (m(x) λ)f(x) = implies f(x) =. However, M λi is not onto. If it were onto, then ran(m λi). Then there would be an f(x) L 2 (, ) such that (M λi)f(x) = (m(x) λ)f(x) = so f(x) = m(x) λ However, there is some x (/4, 3/4) such that m(x ) = 2(x /4) = λ. It follows that Since f(x) 2 2 = /4 3/4 /4 3/4 λ dx + 2 /4 dx. (2(x /4) λ) 2 (2(x /4) λ) dx + 2 3/4 we see that f(x) / L 2 (, ). Hence, M λi is not onto. ( λ) 2 dx Therefore, λ σ(m) for λ (, ). Particularly, since M λi is one-to-one, we know that λ σ c (M) or λ σ r (M). From class we know that σ r (M) = since M is self-adjoint. Hence, it must be that λ σ c (M), Therefore, the spectrum σ(m) = [, ] where σ p (M) = {, } and σ c (M) = (, ).

4 3 Suppose that {λ n } is a sequence of complex numbers such that λ n as n and define the operator K : l 2 (N) l 2 (N) by K(x, x 2,..., x n,...) = (λ x, λ 2 x 2,..., λ n x n,...) (a) Prove that K is a compact operator. (Recall that a set is precompact iff it is totally bounded) Proof. Let {λ n } and let K : l 2 (N) l 2 (N) be defined by K(x, x 2,...) = (λ x, λ 2 x 2,...). We want to show that K is a compact operator. We know that finite-rank operators are compact operators (you have proved this as a homework problem in 2A) and we also know that the uniform limit of compact operators is a compact operator. Let K n : l 2 (N) l 2 (N) be given by K n (x, x 2,...) = (λ x, λ 2 x 2,..., λ n x n,,,...). Then K n is finite-rank operator and hence, a compact operator. Note that K K n = (K K n )x x = = (,...,, λ n+ x n+, λ n+2 x n+2,...) x = = λ i x i 2 x = i=n+ x = i=n+ x = j=n+ x = j=n+ x = j=n+ λ j 2, j=n+ λ i 2 x i 2 λ j 2 i=n+ x i 2 λ j 2 x i 2 i= λ j 2 x 2 which can be made arbitrarily small for sufficiently large n, i.e, as n. Therefore we can conclude that K n K uniformly and K is compact.

5 (b) Let P n : l 2 (N) l 2 (N) be the orthogonal projection onto the nth component, P n (x, x 2,..., x n,...) = (,,...,, x n,,...) In what sense (uniformly, strongly, weakly) does the sum n N λ np n converge to K? Does your answer change in λ as n Proof. Since P n (x, x 2,...) = (,...,, x n,,...), using the same K n as the one we defined above, we have K n = n λ n P n. k= We have probed above that K n K uniformly. Now we discuss the case when λ n does not converge to, but it is still bounded. In this case we can see that K n doesn t converge uniformly to K, and the reason is the following: lim K K n = lim lim x = i=n+ x = i=n+ λ i x i 2 λ i 2 x i 2 ; and since we assumed that λ n then there exists an ɛ > such that for every M N we can find m > M such that λ m 2 > ɛ. Picking x = e m, we get that lim K K n > ɛ. This proves that indeed K n doesn t converge uniformly to K. In the same context, meaning the case when λ n does not converge to, but it is still bounded, we want to see if K n converges strongly to K. For this, let s fix x l 2 (N). Assume also that λ i G for every i N, where G is a positive constant. Therefore lim (K K n)x = lim lim G 2 x i 2. i=n+ i=n+ λ i x i 2

6 But, this last term converges to as n, since x l 2 (N), by definition implies x i 2 x i 2 as n. i=n+ i= Hence, K n K strongly. This clearly implies that K n K weakly. If we assume that the sequence {λ n } n is not bounded either, then from the work we did above, we can see that K n K weakly. Implicitly this tells you that K n doesn t converge strongly or uniformly to K. 4. Determine the spectra of the left and right shift operators on l 2 (N) S(x, x 2, x 3,...) = (, x, x 2,...), T (x, x 2, x 3,...) = (x 2, x 3, x 4,...), and classify them into point, continuous, or residual spectrum. Proof. due to Amanda. We have previously shown that S = = T. So by a theorem if λ C such that λ >, then λ ρ(s) and λ ρ(t ). Now I will prove the following four claims: (a) Claim: S λi is one-to-one for all λ C such that λ. Proof. I will treat this in two cases. If λ = then Sx = implies x i = for all i so x =. Hence, S λi is one-to-one. Suppose that < λ. Then (S λi)x =, implies (, x, x 2,...) = (λx, λx 2, λx 3,...) Since λ, = λx implies x =. Since λx n = x n, a simple induction shows that x n = for all n. Hence, x =, so S λi is one-to-one for all λ such that λ. (b) Claim: S λi is not onto for λ. Proof. Note that if λ = that e is clearly not in ran(s λi) = ran(s). Suppose that λ. Then (S λi)x = e implies that λx = and x n λx n = for n 2. An induction argument shows that x = /λ, and x n = /λ n. However, ( ) n x = λ 2 n=

7 Note the above sum is a geometric series with r =. Since < λ 2, λ 2 it follows that. Hence the above sum diverges, so x / l 2 (N). λ 2 Therefore, e / ran(s λi) so this is not onto. (c) Claim: T λi is one-to-one for λ =. Proof. Suppose that it was not one-to-one. Then there would be some nonzero x l 2 (N) such that (T λi)x = or (x 2, x 3, x 4,...) = (λx, λx 2, λx 3,...) Hence, x 2 = λx and a simple induction shows that x n = λ n x. Therefore, x implies x, and since λ = we see that x = x 2 λ 2n = x 2 n= n= a contradiction so it must be that x =, and it follows that T λi is one-to-one. (d) Claim: T λi is not one-to-one for λ <. Proof. By a similar argument as above, we see that any nonzero x that satisfies (T λi)x = is of the form Choose x =. Then x = (x, λx, λ 2 x, λ 3 x,...) x = ( λ 2 ) n n= Since the above is a geometric series with r = λ 2 <, we see that the sum converges, so x l 2 (N) is nonzero and satisfies that (T λi)x =. Hence, T λi is not one-to-one. Note that claims (a) and (b) show that σ(s) = {λ C : λ }. Recall that ran(s λi) is dense iff ker((s λi) ) = ker(t λi) = {}. Since λ = λ, claims (c) and (d) show that ran(s λi) is dense iff λ =. Therefore, σ c (S) = {λ C : λ = } and σ r (S) = {λ C : λ < }. Also, from the above work we get σ p (S) =. Note that claim (d) shows us that σ p (T ) = {λ C : λ < }. Since σ(t ) is a closed set, we know that σ(t ) = {λ C : λ } (as from above we know that λ > implies λ ρ(t )). For λ = we know that ker((t λi) ) = ker(s λi) = {} by claim (a). Hence, we know that ran(t λi) is dense. Therefore, it must be that σ c (T ) = {λ C : λ = } and σ r (T ) =.