Math 175 - Solutios to homework 6 Cédric De Groote November 16, 2017 Problem 1 (8.11 i the book): Let K be a compact Hermitia operator o a Hilbert space H ad let the kerel of K be {0}. Show that there is a sequece {K } of bouded liear operators o H such that K Kx x, KK x x as. Ca the K be chose so that K K I i the operator orm? Solutio: By the spectral theorem, we ca write Kx = =1 λ x, e e, for some orthoormal sequece {e } i H. This sequece has to be complete, otherwise there exists some vector x H that is orthogoal to all the e s, ad we would have Kx = =1 λ x, e e = 0; but the kerel of K is {0}. So, we kow that {e } is a complete orthoormal sequece. Furthermore, we also kow that λ 0, ad that all the λ s are ozero (otherwise the correspodig e would be i the kerel of K). Defie K by 1 K x = x, e k e k. λ k k=1 1 By the problem 1 of the homework 5, we have K = sup k< λ k < (because it s the supremum of a fiite set). We have K Kx = x, e k e k ad KK x = k=1 x, e k e k, k=1 ad both coverge to x (Theorem 4.14 i the book). We ca ot choose the K so that K K I i the operator orm. Ideed, K K is compact sice K is compact (give {x k } bouded, {Kx k } admits a coverget subsequece sice K is compact, hece {K Kx k } admits oe too sice K is bouded, hece K K is compact), ad the set of compact operators is closed i the operator orm. If K K I, that would force I to be compact, which is ot the case if H is ifiite dimesioal. 1
Problem 2 (8.13 i the book): Let K be a compact Hermitia operator o a Hilbert space H. Show that if K has ifiite rak the the rage of K is ot closed i H. Solutio: Assume that the rage of K is closed i H ad is ifiite dimesioal, ad cosider K : H im (K). The rage of K is a Hilbert space sice it is closed i H. The by the ope mappig theorem, the image of the uit ball i H cotais a ope ball aroud 0 i im (K). But sice a ope ball i a ifiite dimesioal space is ot precompact, this shows that K ca ot be compact. Problem 3 (8.14 i the book): Show that a compact Hermitia operator is the limit with respect to the operator orm of a sequece of fiite rak operators. Solutio: Diagoalize it, ad repeat Example 8.2.(iii) i the book: approximate it by the diagoal operator, but with the diagoal cut off after a certai place. Problem 4: Let M be a closed subspace of L 2 ([0, 1]) that is cotaied i C([0, 1]). (a) There exists C > 0 such that f C f L 2 for all f M. (Use the closed graph theorem). (b) For each x [0, 1] there exists g x M such that f(x) = f, g x for all f M, ad g x L 2 C. (c) The dimesio of M is at most C 2. (Hit: if {f j } is a orthoormal sequece i M, the fj (x) 2 C 2 for all x [0, 1].) Solutio: (a) Cosider F : M C([0, 1]) : f f, where M L 2 ([0, 1]) is edowed with the L 2 -orm. The graph of this operator is Γ = {(f, f) f M} M C([0, 1]). If we prove that Γ is closed i M C([0, 1]), the we will ow that F is a bouded operator by the closed graph theorem, which meas that there exists C > 0 such that f C f L 2 for all f M. Suppose that Γ (f, f ) (f, g) i M C([0, 1]). I particular, f g i, which implies that f g i L 2, sice 1 f g 2 L 2 = f g 2 dx 0 1 0 f g dx = f g. But f f i L 2 by assumptio, hece by uicity of the limit i L 2, we must have f = g, so the limit of (f, f ) is i Γ, provig that Γ is closed. (b) Cosider the fuctioal F x : M C : f f(x), where M L 2 ([0, 1]) is edowed with the L 2 -orm. This makes sese, sice M is cotaied i C([0, 1]), where we kow that we ca evaluate fuctios o poits. By the Riesz-Fréchet theorem, there exists g x M such that F x (f) = f, g x for all f M. This meas that f(x) = f, g x for all f M. Furthermore, we kow that g x = F. But F = sup f L 2 =1 f C by part (a). 2
(c) Let {f } be a orthoormal sequece i M (agai, with the L 2 ier product). By the Bessel iequality ad part (b), we have But by part (b), we also have g x, f 2 g x 2 L 2 C2. g x, f = f (x), which proves the hit. Itegratig the iequality f (x) 2 C 2 over [0, 1], we get 1 f 2 L 2 C 2 dx = C 2. 0 But f 2 L 2 is exactly the umber of f s that we have, sice each term i this sum is 1 by orthoormality of {f }. This proves that the dimesio of M ca be at maximum C 2. Problem 5: I this problem, we defie weak covergece o a ormed liear space. Let E be a ormed liear space ad {x } =1 E. We say that x coverges weakly to x if for every y E, we write this as x x. y(x ) y(x); (a) Let E be a Baach space ad suppose {x } =1 with x x. Show that sup x <. You may use that if x E, the sup y(x) = x. y E, y =1 (b) Suppose that E is separable ad (E ) = E. Show that if {y } =1 E with y 1, the there exists a elemet y E ad a subsequece {y k } so that y k y. That is, show that the closed uit ball i E is weakly compact. [Hit: see homework 2 problem 8.] Solutio: Notice that i this questio, E is merely a ormed liear space. Therefore, we ca ot assume existece of a ier product, or completeess of that space. (a) For all, defie T : E C : y y(x ). We have T = sup T (y) = sup y(x ) = x, y E, y =1 y E, y =1 where we used the fact stated i the questio at the last equality. Now, sice x x, we have y(x ) y(x) for every y E. I particular, the sequece { y(x ) } must be bouded: sup T (y) < y E. 3
By the priciple of uiform boudedess, we therefore have that which meas that sice T = x. sup T <, sup x < (b) We use a diagoal argumet. Sice E is separable, let S = {x m } be a coutable set of poits that is dese i E. By a diagoal argumet like i homework 2 problem 8, we ca fid a subsequece {y k } of {y } such that y k (x m ) coverges (say, to α m ) for every m. Defie y : S C by y(x m ) = α m. We prove that y is uiformly cotiuous o S. Let x r, x s S. The y(x r ) y(x s ) = lim y k (x r ) y k (x s ) k = lim y k (x r x s ) lim y k x r x s x r x s k k sice y 1. So y : S C is uiformly cotiuous, hece it exteds to a cotiuous fuctio y : E = S C, by a geeral topology theorem (ot hard to prove). Notice that we eed uiform cotiuity; for example, f : R \ {0} R : x 1 x is cotiuous, but ot uiformly cotiuous, ad does ot exted cotiuously to f : R R. We ow show that y is liear. Let v, v E, ad let x v ad x v with {x } S ad {x } S. The αx + βx αv + βv. By cotiuity of y, y(αv + βv ) = lim y(αx + βx ) = lim lim k y k (αx + βx ) = α lim lim k y k (x ) + β lim lim k y k (x ) = α lim y(x ) + β lim y(x ) = αy(v) + βy(v ) We used the cotiuity of y at the first lie, the defiitio of y at the secod lie, the cotiuity of each y k at the third lie, the defiitio of y at the fourth lie ad the cotiuity of y at the fifth lie. This proves that y is liear. The two previous paragraphs prove that y E. Now, we eed to check that y k y. Let x E = (E ). We eed to check that x(y k ) = y k (x) y(x) = x(y), where we use the idetificatio betwee E ad (E ) at the two equalities. By costructio, we ow that this is true if x S. If ot, let {x } S such that x x. We have y k (x) y(x) y k (x) y k (x ) + y k (x ) y(x ) + y(x ) y(x) x x + y k (x ) y(x ) + y x x, where we used y 1 to boud the first term. Now, take large eough so that the first ad third terms are small, ad the choose k large eough so that the secod oe is small as well. This proves that y k (x) y(x) for every x E, ad hece that y k y. 4
Problem 6: Suppose that E ad F are Baach spaces so that E is separable with (E ) = E. (a) Suppose that T : E F is a liear map so that if {x } =1 has x x implies T x T x. Show that T is bouded. (b) Suppose that T : E F is a liear map so that if {x } =1 has x x implies T x T x (i.e. i the orm topology). Show that T is compact. You may use without proof that if B is a Baach space ad B is separable, the B is separable. Solutio: (a) We wat to use the closed graph theorem. Take (x, T x ) Γ(T ) (the graph of T ), such that (x, y ) (x, y) i E F. I particular, x x, so x x. By assumptio, T x T x. But T x y, so T x y. By uicity of weak limit, we must have y = T x, so Γ(T ) is closed ad T is bouded by the closed graph theorem. (b) Take a sequece {x } E that is bouded. We have to show that its image uder T admits a coverget subsequece. Sice E = (E ) is separable, we kow that E is separable as well by the fact stated i the problem. So we ca apply problem 5 (b), ad coclude that {x } has a weakly coverget subsequece {x k }. By the assumptio, {T x k } coverges i F, hece we proved that T is compact. Problem 7: Show that there is o slowest rate of decay of the terms of a absolutely coverget series; that is, there is o sequece {a } =1 of positive umbers such that a c < if ad oly if {c } is bouded. (Hit: Cosider the map T : l l 1 give by T f() = a f(). The set of f such that f() = 0 for all but fiitely may is dese i l 1 but ot l.) Solutio: Suppose that such a sequece {a } exists, ad defie T : l l 1 : (c 1, c 2, ) (a 1 c 1, a 2 c 2, ). The if coditio o {a } says that T maps l to l 1, ad the oly if says that T is surjective. Sice T is clearly ijective, we coclude that T is bijective. To prove that T is cotiuous: T (c 1, c 2, ) l 1 = (a 1 c 1, a 2 c 2, ) l 1 = ( ) ( ) a c sup c a = a (c 1, c 2, ) l. This proves that T is cotiuous, sice a < by assumptio (takig c = 1 for every ). By the bouded iverse theorem, T 1 is also cotiuous. Therefore, T defies a homeomorphism betwee the topological spaces l 1 ad l. Homeomorphisms take dese sets to dese sets. The map T above clearly takes the set of sequeces with oly fiitely may ozero terms i l to the set of sequeces with oly fiitely may ozeroterms i l 1 bijectively. Therefore, we would fid a cotradictio if we prove that this set is dese i l 1 but ot i l. 5
It is clearly ot dese i l : takig the sequece (1, 1, 1, ), ay sequece (c 1, c 2, ) with oly fiitely may ozero terms would be at distace at least 1 of (1, 1, 1, ) i the l orm. To see that it is dese i l 1, take ay x = (x 1, x 2, ) l 1. The, defie x (k) = { x if k 0 if > k. For every k, this sequece is i the set metioed above. Furthermore, (x (k) ) x = x l 1, =k+1 which coverges to 0 as k because the series x coverges (as x l 1 ). This proves that (x (k) ) coverges to x i l 1, which proves the claim. Problem 8: Let H be a Hilbert space ad T L(H, H) compact ad self adjoit. Deote by σ p (T ) := {λ C λ is a eigevalue of T }, σ ap (T ) := {λ C λ is a approximate eigevalue of T }, where we say λ is a approximate eigevalue if there exists {x } =1 H so that x H = 1 ad (λi T )x 0. (a) Show that σ(t ) σ p (T ) {0}. Moreover, if H is ifiite dimesioal, the σ(t ) = σ p (T ) {0}. (b) Show that σ(t ) = σ ap (T ). Solutio: Sice T is compact ad self-adjoit, we ca diagoalize it. Call λ 1, λ 2, its eigevalues (some of them may be 0), with correspodig complete orthoormal basis {e }. Furthermore, λ 0 (if H is ifiite dimesioal). (a) Take λ / σ p (T ) {0}. We wat to show that λ / σ(t ), i.e. that λi T is ivertible. By the bouded iverse theorem, it suffices to prove that it is ijective ad surjective. It is ijective, sice we assume that λ is ot a eigevalue. To prove that it is surjective, we costruct a bouded iverse. Defie S L(H, H) by Se = 1 λ λ e (this makes sese sice λ λ, sice λ / σ p (T ) while λ σ p (T )). Sice (λi T )e = (λ λ )e, this is ideed a (set-theoretic) iverse to λi T. Let us prove that S is cotiuous. By the problem 1 of homework 5, we kow that S = sup λ λ. Sice λ 0, the oly accumulatio poit of the sequeces (λ ) is 0. But we 1 chose λ 0, hece there exists ε > 0 such that λ λ > ε for every. Therefore λ λ < 1 ε, provig that S is bouded. Together, the previous two paragraphs show that λ / σ(t ). To show the moreover, sice it is clear that σ p (T ) σ(t ), it suffices to prove that 0 σ(t ) i the case where 0 / σ p (T ). If 0 / σ p (T ), the T is ijective. If 0 / σ(t ), the T is also surjective. But compact operators ca ot be surjective i a ifiite dimesioal space: the bouded iverse theorem would give a bouded iverse T 1 to T, ad the we ca write I = T T 1. But T is compact, hece T T 1 is as well, ad I is ot compact whe H is ifiite dimesioal. 6
(a) (Shorter solutio) By problem 3 of homework 5 (ad by the spectral theorem), σ(t ) is the closure of {λ }. So if H is fiite dimesioal, σ(t ) = σ p (T ). If H is ifiite dimesioal, the sice the oly accumulatio poit of (λ ) is 0 (because the λ s coverge to 0 by the spectral theorem), we have that σ(t ) = σ p (T ) {0}. (b) Suppose first that λ / σ(t ), i.e. λi T has a bouded iverse (λi T ) 1. Suppose that λ σ ap (T ), i.e. there exists (x ) H such that x = 1 ad (λi T )x 0. The (λi T )x 0, hece x = (λi T ) 1 (λi T )x 0 as well by cotiuity of (λi T ) 1. But x = 1; cotradictio. This proves that σ ap (T ) σ(t ). Coversely, let us show that σ p (T ) {0} σ ap (T ); this will imply the result sice σ(t ) σ p (T ) {0} by part (a). First, 0 σ ap (T ); ideed, take x = e (the orthoormal basis we got from applyig the spectral theorem to T ); this satisfies x = 1, but (0 I T )(x ) = λ x = λ 0. Now, take λ σ p (T ), say λ = λ k for some k. The, take the costat sequece x = e k for every. This satisfies x = 1 for every, but (λi T )(x ) = 0 = 0, provig that λ is a approximate eigevalue. 7