Homework 4. x n x X = f(x n x) +

Transcription

1 Homework 4 1. Let X ad Y be ormed spaces, T B(X, Y ) ad {x } a sequece i X. If x x weakly, show that T x T x weakly. Solutio: We eed to show that g(t x) g(t x) g Y. It suffices to do this whe g Y = 1. Observe that Note that the we have g(t x ) g(t x) = g(t x T x) x x X = g Y T x T x Y g Y T x x X = T x x X. f(x x) sup 0 f X f X So, let ɛ > 0 be give, ad select f X with f X = 1 such that x x X < f(x x) + ɛ 2 T. Sice x x weakly, there exists a iteger N such that if > N we have This the gives, f(x ) f(x) = f(x x) < g(t x ) g(t x) T x X T ɛ 2 T. ( f(x x) + ɛ ) < ɛ. 2 T 2. If {x } is a weakly coverget sequece i a real ormed space X with x x 0. Show that there is a sequece {y m } of liear combiatios of elemets of {x } which coverge strogly to x 0. Hit: Prove that the weak closure of a covex set is the same as the strog closure. Solutio: Note that if suffices to prove that the followig statemet. If E X is a covex subset of a ormed space, the the weak closure of E, deoted by E w is the same as the strog closure of E, deoted by E. Oce we have this statemet, we ca apply it to

2 E = spa{x } i the followig way. We have that x 0 E w by hypothesis. But, E w = E by the claim. So we have that x E, which whe we upack the defiitios gives the existece of {y m } a sequece of liear combiatios of {x } that coverges strogly to x 0. So it suffices to prove the claim. Sice strog covergece implies weak covergece, we have that E E w. Suppose that x 0 E. By your exam, there exists a fuctioal f X such that f(x 0 ) < δ f(x) x E. Cosider the set {x : f(x) < δ}, this is a weak ope set of x 0 that does ot itersect E, ad so x 0 / E w. Which gives E w E. 3. Let T B(X, Y ). Show that T T uiformly if ad oly if for every ɛ > 0 there is a N, depedig oly o ɛ such that for all > N ad all x X of orm 1 we have T x T x Y < ɛ. Solutio: Suppose first that T T uiformly. The we have that for ay ɛ > 0 there exists N such that T T X Y < ɛ. The we clearly have that T x T x Y T T X Y x X = T T X Y < ɛ whe > N ad x X with x X = 1. For the other directio, suppose that ɛ > 0 is give, the we eed to show that whe > N(ɛ). Note that T T X Y < ɛ But, by the hypothesis, we have that T x T x T T X Y = sup Y 0 x X x X T x T x Y < ɛ x X whe > N(ɛ). This the clearly implies that T x T x T T X Y = sup Y 0 x X x X ɛ which proves the result. 4. If a ormed space X is reflexive, show that X is reflexive.

3 Solutio: Let h X, the for every g X there is a x X such that g = Cx sice X is reflexive. Here C is the caoical map idetifyig X ad X. Hece, h(g) = h(cx) := f(x) defies a bouded liear fuctioal f o X, ad Cf = h, where C : X X is the caoical mappig. Thus, C is surjective ad so X is reflexive. 5. Let X be a ormed space ad {f } a sequece of liear fuctioals o X. Show (a) If f f weakly, the f f weak* too; (b) Show that if X is reflexive, the the coverse holds. Solutio: Part (a) is immediate sice we have that X X where we have used the caoical mappig. Namely, each x X defies a fuctioal ˆx : X C by ˆx(f) := f(x). So, if we have weak covergece of the sequece of fuctioals, the we have weak-* covergece as well. If X is reflexive, the we have that X = X, ad so every liear fuctioal o X ca be idetified with a fuctioal of the form ˆx. 6. A weak Cauchy sequece i a ormed space is a sequece {x } i X such that for every f X the sequece {f(x )} is Cauchy. Show that a weak Cauchy sequece is bouded. Solutio: Defie ˆx : X C by ˆx (f) = f(x ). Sice {x } is a weak Cauchy sequece, the we have that {f(x )} is a Cauchy sequece i C ad hece bouded. Thus, we have that {ˆx (f)} is a bouded sequece, ad sice X is a Baach space we ca apply the Uiform Boudedess Priciple to coclude that sup ˆx < But, this the gives that sup x X <, which is what we wated. 7. A ormed space X is said to be weakly complete if each weak Cauchy sequece i X coverges weakly i X. If X is reflexive, show that X is weakly complete. Solutio: Let {x } be a weak Cauchy sequece i X. Agai, we will idetify x with ˆx via the caoical mappig. Note that by Problem 6 above, we have that sup x X <. Sice {f(x )} is Cauchy i C, it has a limit, ad we ca defie ϕ(f) = lim f(x ) = lim ˆx (f)

4 which exists for each f X. This defies a map ϕ : X C. This map is clearly liear sice each ˆx is liear. Also, it is bouded sice Passig to the limit we have ˆx (f) = f(x ) f X x X sup x f X. ϕ(f) sup x f X. So we have that ϕ X. But, sice X is reflexive, there exists x X such that ˆx = ϕ. But, the we have f(x) = ˆx(f) = ϕ(f) = lim ˆx (f) = lim f(x ) so that x x weakly sice f(x ) f(x) for all f X. 8. Put f (t) = e it, t [ π, π]. Let L p = L p ( π, π) with respect to Lebesgue measure. If 1 p <, show that f 0 weakly i L p, but ot strogly. Solutio: Note that if p is a polyomial o [ π, π] the we clearly have that p, f L 2 0. The polyomials are dese i L p whe 1 p <, so for ay f L p, select a polyomial p such that f p L p < ɛ. The we have that f, f L 2 = f p, f L 2 + p, f L 2. The secod term goes to zero as, i particular if > N we have p, f L 2 < ɛ. For the first term, ote that we have f p, f L 2 f p p f q = f p p < ɛ where 1 p + 1 q = 1. Thus for > N we have that f, f L 2 < 2ɛ. It is immediate that f does ot coverge strogly to 0 i L p whe 1 p < sice each elemet has orm A subset of a Baach space X is weakly bouded if for all f X, sup f(x) <. Prove that S is weakly bouded if ad oly if it is strogly bouded, i.e., sup x X <.

5 Solutio: It is clear that if the set S is strogly bouded, the it must be weakly bouded sice sup f(x) sup f X x X = f X sup x X <. Suppose ow that S is weakly bouded. The we have that sup f(x) < for all f X. Defie ˆx : X C by ˆx(f) = f(x). The, we have that sup ˆx(f) = sup f(x) <. So, we have that for all x S that ˆx is a uiformly bouded family of operators from X C. Hece by the Uiform Boudedess Priciple, we have that sup ˆx X < But x X = ˆx X, ad so we have the result. 10. If {x } l 1 show that y(j)x (j) 0 for every y c 0 if ad oly if sup x l 1 < ad x (j) 0 for all j 1. Solutio: Note that we have c 0 = l 1. For otatioal simplicity, let x, y deote the ifiite sum x (j)y(j). The easy directio the follows from testig o a obvious elemet i c 0 ad the Uiform Boudedess Priciple. If we have y(j)x (j) 0 for every y c 0, the takig y = (0,..., 0, 1, 0,...) with the oe occurrig i the jth positio, we have that x (j) = y(j)x (j) 0. =0 Sice x l 1, the it is a bouded liear fuctioal o so we have that x, y x l 1 y l. The Uiform Boudedess Priciple gives that sup x l 1 <. Coversely, if we have sup x l 1 < ad x (j) 0 for all j 1, we eed to show y(j)x (j) 0. Fix y c 0 ad ote that for ay ɛ > 0 there exists J such that y(j) < ɛ.

6 Now we have that y(j)x (j) = J 1 y(j)x (j) + y(j)x (j). j=j For the secod term, we have that y(j)x (j) y(j) x (j) ɛ x l 1 ɛ sup x l 1. j=j j=j This ca be made as small as we wish by choosig ɛ appropriately. Now for the first term, we have that x (j) 0 for j = 1,..., J 1. Select N so that x (j) ɛ. The J we have that J 1 y(j)x (j) y J 1 l x (j) y l ɛ. Combiig these two estimates gives that x, y 0 as.