Weak Topologies, Reflexivity, Adjoint operators

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1 Chapter 2 Weak Topologies, Reflexivity, Adjoint operators 2.1 Topological vector spaces and locally convex spaces Definition [Topological Vector Spaces and Locally convex Spaces] Let E be a vector space over K, with K = R or K = C and let T be a topology on E. We call (E,T ) (or simply E, if their cannot be a confusion), a topological vector space, if the addition: and the multiplication by scalars +:E E E, (x, y) x + y, : K E E, (λ, x) λx, are continuous functions. A topological vector space is called locally convex if 0 (and thus any point x E) has a neighbourhood basis consisting of convex sets. Remark. Topological vector spaces are in general not metrizable. Thus, continuity, closeness, and compactness etc, cannot be described by sequences. We will need nets. Assume that (I, ) is a directed set. This means (reflexivity) i i, for all i I, 27

2 28CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS (transitivity) if for i, jk I we have i j and j k, then i k, and (existence of upper bound) for any i, j I there is a k I, so that i k and j k). A net is a family (x i : i I) indexed over a directed set (I, ). A subnet of a net (x i : i I) is a net (y j : j J), together with a map j i j from J to I, so that x ij = y j for all j J for all i 0 I there is a j 0 J, so that i j i 0 for all j j 0. Note: A subnet of a sequence is not necessarily a subsequence. In a topological space (T,T ), we say that a net (x i : i I) converges to x, if for all open sets U with x U there is a i 0 I, so that x i U for all i i 0. If (T,T ) is Hausdorff x is unique and we denote it by lim i I x i. Using nets we can describe continuity, closeness, and compactness in arbitrary topological spaces: a) A map between two topological spaces is continuous if and only if the image of converging nets are converging. b) A subset A of a topological space S is close if and only if the limit point of every converging net in A is in A. c) A topological space S is compact if and only if every net has a converging subnet. In order to define a topology on a vector space E which turns E into a topological vector space we need (only) to define an appropriate neighborhood basis of 0. Proposition Assume that (E,T ) is a topological vector space. And let U 0 = {U T, 0 U}. Then a) For all x X, x + U 0 = {x + U : U U} is a neighborhood basis of x, b) for all U U 0 there is a V U 0 so that V + V U, c) for all U U 0 there is an ε> 0 and V U 0 so that {λ K : λ <ε} V U,

3 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 29 d) for all U U 0 and x E there is an ε> 0, so that λx U, for all λ K with λ <ε, e) if (E,T ) is Hausdorff, then for every x E, x 0, there is a U U 0 with x U, f) if E is locally convex, then there is for all U U 0 a convex V U 0, with V U, i.e. 0 has a neighborhood basis consisting of convex sets. Conversely if E is a vector space over K, K = R or K = C and U 0 {U P(E) :0 U} is non empty and is downwards directed, i.e. if for any U, V U 0, there is a W U 0, with W U V and satisfies (b) (c) and (d), then T = {V E : x V U : x + U V }, defines a topological vector space for which U 0 is the neighborhood basis of 0. (E,T ) is Hausdorff if U also satisfies (e) and locally convex if it satisfies (f). Proof. Assume (E,T ) is a topological vector space and U is defined as above. We observe that for all x E the linear operator T x : E E, z z + x is continuous. Since also T x T x = T x T x = Id, it follows that T x is an homeomorphism, which implies (a). Property (b) follows from the continuity at 0 of addition and (c) and (d) follow from the continuity of scalar multiplication at 0. If E is Hausdorff then U 0 clearly satisfies (e) and, by definition, U 0 satisfies (f) if we assume that E is locally convex. Now assume that U 0 {U P(E) :0 U} is non empty and downwards directed, that for any U, V U 0, there is a W U 0, with W U V and satisfies (b), (c) and (d). Then T = {V E : x V U : x + U V }, is finitely intersection stable and stable by taking (arbitrary) unions. Also,X T. Thus T is a topology. Also note that for x E, is a neighborhood basis of x. U x = {x + U : U U 0 }

4 30CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS We need to show that addition and multiplication by scalars is continuous. Assume (x i : i I) and (y i : i I) converge in E to x E and y E, and if U U 0, respectively. By (b) there is a V U 0 with V + V U. We can therefore choose i 0 so that x i x + V and y i x + V, for i i 0, and, thus, x i + y i x + y + V + V x + y + U, for i i 0. This proves the continuity of the addition in E. Assume (x i : i I) converges in E to x,(λ i : i I) converges in K to λ and let U U 0. Then choose V U 0 so that V + V U, choose ε 1 > 0 so that µx V, whenever µ ε 1, choose ε 2 > 0 and W U 0, so that {µ K : µ <ε 2 } W V, and finally choose i 0 I, so that λ λ i < min(ε 1,ε 2 ) and x i x + W, whenever i i 0. Then it follows for i i 0 λ i x i =(λ i λ)(x i x)+(λ i λ)x+λx λx+(λ i λ)w+v λx+v +V λx+u. If U 0 satisfies (e) and if x y are in E, then we can choose U U 0 so that y x U and then, using the already proven fact that addition and multiplication by scalars is continuous, there is V so that V V U. It follows that x + V and y + V are disjoint. Indeed, if x + v 1 = y + v 2, for some v 1,v 2 V it would follows that y x = v 2 v 1 U, which is a contradiction. If (f) is satisfied then E is locally convex since we observed before that U x = {x + U : U U 0 } is a neighborhood basis of x, for each x E. of Let E be a vector space over K, K = R or K = C, and let F be a subspace E # = {f : E K linear}. Assume that for each x E there is an x F so that x (x) 0, we say in that case that F is separating the elements of E from 0. Consider U 0 = { n j=1 {x E : x i (x) <ε i } : n N,x i, F, and ε i > 0 for i =1, 2ṅ }. U 0 is finitely intersection stable and it is easily checked that U 0 satisfies that for assumptions (b)-(f). It follows therefore that U 0 is the neighborhood basis of a topology which turns E into local convex Hausdorff space.

5 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 31 Definition [The Topology σ(e, F )] Let E be a vector space and let F be a separating subspace of E #. Then we denote the local convex Hausdorff topology generated by U 0 = { n j=1{x E : x i (x) <ε i } : n N,x i F, and ε i > 0 for, 2,... n }, by σ(e, F ). If E is a Banach space X and F = X we call σ(x, X ) the Weak Topology on X and denote it also by w. If E is the dualspace X of a Banach space X and we consider X (via the canonical map χ : X X ) a subspace of X we call σ(x,x) the Weak Topology on X, which is also denoted by w. Proposition Let E be a vector space and let F be a separating subspace of E #. For a net (x i ) i I E and x E lim x i = x in σ(e, F ) x F i I lim x,x i = x,x. i I An easy consequence of the geometrical version of the Hahn-Banach Theorem is the following two observation. Proposition If A is a convex subset of a Banach space X then A w = A. If a representation of the dual space of a Banach space X is not known, it might be hard to verify weak convergence of sequence directly. The following Corollary of Proposition states an equivalent criterium condition a sequence to be weakly null without using the dual space of X. Corollary For a bounded sequence (x n ) in Banach space X it follows that (x n ) is weakly null if and only if for all subsequences (z n ), all ε> 0 and there is a convex combination z = k j=1 λ jz j of (z j ) (i.e. λ i 0, for i =1, 2,... k, and l j=1 λ j =1) so that z ε. Proposition If X is a Banach space and Y is a closed subspace of X the σ(y, Y )=σ(x, X ) Y, i.e. the weak topology on Y is the the weak topology on X restricted to Y. Theorem [Theorem of Alaoglu] B X is w compact for any Banach space X.

6 32CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS Sketch of a proof. Consider the map Φ: B X x X{λ K : λ x }, x (x (x) :x X). Then we check that Φ is continuous with respect to w topology on B X and the product topology on x X {λ K : λ x }, has a closed image, and is a homeorphism from BX onto its image. Since by the Theorem of Tychanoff x X {λ K : λ x } is compact, Φ(B X ) is a compact subset, which yields (via the homeomorphism Φ 1 ) that B X is compact in the w topology. Theorem [Theorem of Goldstein] B X is (via the canonical embedding) w dense in B X. The proof follows immediately from the following Lemma. Lemma Let X be a Banach space and let x X and x 1, x 2,..., x n X Then inf x 1 x,x i x i,x 2 =0. Proof. For x X put φ(x) = n x,x i x i,x 2 and β = inf x BX φ(x), and choose a sequence (x j ) B X so that φ(x j ) β, if k. W.l.o.g we can also assume that ξ i = lim k x i,x k exists for all i =1, 2,..., n. For any t [0, 1] and any x B X we note for k N φ((1 t)x k + tx) = x,x i (1 t) x i,x k t x i,x 2 = = x,x i x i,x k + t x i,x k x 2 x,x i x i,x k 2 ( ) +2tR x,x i x i,x k x i,x k x + t 2 x i,x k x i,x 2

7 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 33 ( ) k β +2tR ( x,x i }{{ ξ i) (ξ } i x i,x ) + t 2 ξ i x i,x 2. =:λ i From the minimality of β it follows that for all x B X ( ) R λ i ξ i R( x,x ) with x := λ i x i, and thus Indeed, write x,x = re ia, then ( x R λ i ξ i ). ( ) x,x = e ia x,x = x,e ia x R λ i ξ. On the other hand lim k x,x k = n λ iξ i and thus which implies that So x β = lim φ(x k ) k = λ i 2 = = λ i λ i ( λ i ξ i R λ i ξ i ), ( x = R λ i ξ i ). λ i ( x,x i ξ i ) ( ) = R λ i ( x,x i ξ i )

8 34CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS ( ) = R x,x R λ i ξ i Thus β = 0 which proves our claim. x x x x x =0. Theorem Let X be a Banach space. Then X is reflexive if and only if B X is compact in the weak topology. Proof. Let χ : X X be the canonical embedding. If X is reflexive and thus χ is onto it follows that χ is an homeomorphism between (B X,σ(X, X )) and (B X,σ(X,X )). But by the Theorem of Alaoglu Bourbaki (B X,σ(X,X )) is compact. Assume that (B X,σ(X, X ) is compact, and assume that x B X we need to show that there is an x B X so that χ(x) =x, or equivalently that x,x = x,x for all x X. For any finite set A = {x 1,... x n} X and for any ε> 0 we can, according to Lemma , choose an x (A,ε) B X so that The set x x (A,ε),x i 2 ε. I = {(A, ε) :A X finite and ε> 0}, is directed via (A, ε) (A,ε ): A A and ε ε. Thus, by compactness, the net ( x (A,ε) :(A, ε) I ) must have a subnet (z j : j J) which converges weakly to some element x B X. We claim that x,x = x,x, for all x X. Indeed, let j i j be the map from J to I, so that z j = x ij, for all j J, and so that, for any j 0 there is a i 0 with j i j 0, for i i 0. Let x X and ε> 0. Put i 0 =({x },ε) I, choose j 0, so that i j i 0, for all j j 0, and choose j 1 J, j 1 j 0, so that x zj,x <ε, for all j j 1. It follows therefore that (note that for i j1 =(A, ε ) it follows that x A and ε ε) x x, x x x ij1,x + zj1 x, x ε + ε =2ε. Since ε> 0 and x X where arbitrary we deduce our claim. Theorem For a Banach space X the following are equivalent.

9 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 35 a) X is reflexive, b) X is reflexive, c) every closed subspace of X is reflexive Proof. (a) (c) Assume Y X is a closed subspace. Proposition yields that B Y = B X Y is a σ(x, X )-closed and, thus, σ(x, X )-compact subset of B X. Since, by the Theorem of Hahn-Banach (Corollary 1.4.4), every y Y can be extended to an element in X, it follows that σ(y, Y ) is the restriction of σ(x, X ) to the subspace Y. Thus, B Y is σ(y, Y )-compact, which implies, by Theorem that Y is reflexive. (a) (b) If X is reflexive then σ(x,x )=σ(x,x). Since by the Theorem of Alaoglu Bourbaki B X is σ(x,x)-compact the claim follows from Theorem (c) (a) clear. (b) (a) If X is reflexive, then, (a) (b) X is also reflexive and thus, the implication (a) (c) yields that X is reflexive. An important consequence of the Uniform Boundedness Principle is the following Theorem [Theorem of Banach-Steinhaus] a) If A X, and sup x A x,x <, for all x X, then A is (norm) bounded. b) If A X, and sup x A x,x <, for all x X, then A is (norm) bounded. In particular weak compact subsets of X and weak compact subsets of X are norm bounded. Exercises 1. Show Theorem using Lemma Prove Proposition and Corollary

10 36CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS 3. Show that B l is not sequentially compact in the w -topology. Hint: Consider the unit vector basis of l 1 seen as subsequence of B l Prove that for a Banach space X every w converging sequence in X is bounded, but that if X is infinite dimensional, X contains nets (x i : i ) which converge to 0, but so that for every c>0 and all i I there is a j 0 i, with x j c, whenever j j Show that in each infinite dimensional Banach space X there is a weakly null net in S X. 6. Prove that every weakly null sequence in l 1 is norm null. Hint: Assume that (x n ) S l1 is weakly null. Then there is a subsequence x nk and a block sequence (z k ) so that lim k x nk z k = 0.