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Weak Topologies Weak-Type Topologies on Vector Spaces. Let X

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Weak Topologies Weak-Type Topologies on Vector Spaces. Let X Weak topologies Weak-type topologies on vector spaces. Let X be a vector space with the algebraic dual X]. Let Y ½ X] be a subspace. We want to de¯ne a topology σ on X in order to make continuous all elements of Y . Fix x0 2 X. If σ is such a topology, then the sets of the form x0 V";g := fx 2 X : jg(x) ¡ g(x0)j < "g = fx 2 X : jg(x ¡ x0)j < "g (" > 0; g 2 Y ) are open neighborhoods of x0. But this family is not a basis of σ-neighborhoods of x0, since the intersection of two of its members does not necessarily contain another member of the family. This is the reason why we instead consider the sets of the form x0 (1) V";g1;:::;gn := fx 2 X : jgi(x ¡ x0)j < "; i = 1; : : : ; ng (" > 0; n 2 N; gi 2 Y ) : x0 x0 It is easy to see that the intersection V \ V 0 of two of such sets ";g1;:::;gn " ;h1;:::;hm x0 00 0 contains V 00 where " = minf"; " g. " ;g1;:::;gn;h1;:::;hm Theorem 0.1. Let X be a vector space, and Y ½ X] a subspace which separates the points of X (that is, a so-called total subspace). 1. There exists a (unique) topology on X such that, for each x0 2 X, the sets (1) form a basis of neighborhoods of x0. This topology, denoted by σ(X; Y ), is called the weak topology determined by Y . 2. σ(X; Y ) is the weakest topology on X that makes continuous all elements of Y . 3. (X; σ(X; Y )) is a Hausdor® locally convex t.v.s. for which the sets V";g1;:::;gn = fx : jgi(x)j < "; i = 1; : : : ; ng (" > 0; gi 2 Y ) form a basis of U(0), consisting of open symmetric convex sets. 4. (X; σ(X; Y ))¤ = Y . Proof. We only sketch the proof of the ¯rst three parts. A set A ½ X will be de¯ned open if, for each a 2 A, there exist " > 0, n 2 N and g1; : : : ; gn 2 Y such a that V";g1;:::;gn ½ A. It is elementary to show that the family of such open sets satis¯es all axioms for topology. This topology is Hausdor® since Y separates the points of X. Moreover, the discussion before Theorem 0.1 implies 2. To show that the mapings (x; y) 7! x + y and (x; t) 7! tx are continuous is an easy exercise. Let us show 4. The inclusion (X; σ(X; Y ))¤ ⊃ Y is clear. To prove the reverse inclusion, we need the following algebraic lemma. ] Lemma 0.2. Let X be a vector space. Let g; f1; : : : ; fn 2 X be such that \n Ker(fi) ½ Ker(g) : i=1 Then g 2 spanff1; : : : ; fng. 1 2 Proof. Let us de¯ne a linear mapping T : X ! Rn by ¡ ¢ T x = f1(x); : : : ; fn(x) : Now, we have Ker(T ) ½ Ker(g), and this easily implies that g is constant on each set T ¡1(y)(y 2 T (X)). Let P : Rn ! T (X) be the orthogonal projection. De¯ne a functional h: Rn ! R by ¡ ¢ h(») = g T ¡1(P») : n ¤ n Then h is well-de¯ned and linear, that is h 2 (R ) . Thus there exists ® = (®1; : : : ; ®n) 2 R such that h(») = h»; ®i (» 2 Rn): Then, for each x 2 X, Xn ¡ ¡1 ¢ ®ifi(x) = hT x; ®i = h(T x) = g T (P T x) i=1 ¡ ¢ ¡ ¢ = g T ¡1(T x) = g x + Ker(T ) = g(x) : ¤ Now, let ` be a σ(X; Y )-continuous linear functional on X. There exists a neigh- borhood of 0 such that j`j · 1 on this neighborhood. This neighborhood con- tains some basic neighborhood V";g1;:::;gn which, in its turn, contains the subspace Tn L = i=1 Ker(gi). Since ` is bounded on L, we must have `jL = 0; in other words, Tn i=1 Ker(gi) ½ Ker(`). By Lemma 0.2, ` 2 spanfg1; : : : ; gng ½ Y . ¤ The topology σ(X; Y ) and the product topology on RY . Let us recall that the product topology on an arbitrary Cartesian product P = ¦i2I Ti of topological spaces is the weakest topology on P for which the canonical projections ¼k : P ! Tk ; ¼k((xi)i2I ) = xk ; are all continuous. An open basis of the product topology consists of the sets A = ¦i2I Ai where Ai ½ Ti is open (i 2 I), and Ai 6= Ti for at most ¯nitely many i's. The famous Tychono® Product Theorem asserts that a Cartesian product of compact topo- logical spaces is compact in the product topology. I A particular case of a product space is T = ¦i2I T , the space of all functions I ! T . The corresponding product topology on T I is the topology of the pointwise convergence. Now, given a vector space X and a total subspace Y ½ X], we can see every x 2 X as a function x : Y ! R, x(g) = g(x). With this identi¯cation, X can be viewed as a subset of RY , and, moreover, the topology σ(X; Y ) coincides on X with the product topology of RY (this follows easily from de¯nitions). Thus (X; σ(X; Y )) is a topological subspace of RY , and σ(X; Y ) is the topology of the pointwise convergence on Y . 3 The weak topology of a normed space. Let X be a normed space. Since X¤ (which is a subspace of X]) separates points by the Hahn-Banach theorem, the topology σ(X; X¤) makes X a Hausdor® locally convex t.v.s. This topology is called the weak topology of X, and is denoted by w or wX . Let us list some of its basic properties. (By ¿k¢k we denote the norm topology.) (a) A basis of U(0) in the w-topology consists of the sets ¤ ¤ ¤ V ¤ ¤ = fx 2 X : jy (x)j < "; i = 1; : : : ; ng (" > 0; n 2 N; y 2 X ): ";y1 ;:::;yn i i (b) (X; w)¤ = X¤ (see Theorem 0.1). (c) w · ¿k¢k (since the basic neighborhoods from (a) are clearly ¿k¢k-open). (d) w = ¿k¢k if and only if dim(X) < 1. (Indeed, if X is ini¯nite-dimensional, then each basic neighborhood V ¤ ¤ contains the nontrivial subspace ";y1 ;:::;yn Tn ¤ i=1 Ker(yi ) which is not contained in the open unit ball. If X is ¯nite- dimensional, then w = ¿k¢k since all vector topologies on X coincide.) w ¤ ¤ ¤ ¤ (e) xn ! x if and only if y (xn) ! y (x) for each y 2 X . (This holds since w is the topology of the pointwise convergence on X¤.) The weak¤ topology of a dual of a normed space. Let X be a normed space. Then X can be considered as a subspace of X¤¤ (by the canonical isometric im- mersion) and hence also as a subspace of (X¤)]. Moreover, X clearly separates the points of X¤. Thus the topology σ(X¤;X) is a Hausdor® locally convex vector topology on X¤. This topology is called the weak¤ topology of X¤ and is denoted by ¤ ¤ w or wX¤ . Let us remark that the w¤ topology of X¤ does not depend only on the normed space X¤ but also on its predual X. There exist examples of distinct Banach spaces ¤ ¤ ¤ ¤ X1 and X2, for which X1 and X2 are isometrically isomorphic but (X1 ; w ) and ¤ ¤ (X2 ; w ) are not isomorphic as topological vector spaces. It can also happen that, for a normed space X, the topologies σ(X¤;X) and σ(X¤; X) (where X is the completion of X) are di®erent. Let us list some of the basic properties of the w¤-topology. (a) A basis of U(0) in the w¤-topology consists of the sets ¤ ¤ ¤ W";y1;:::;yn = fx 2 X : jx (yi)j < "; i = 1; : : : ; ng (" > 0; n 2 N; yi 2 X): (b) (X¤; w¤)¤ = X. ¤ (c) w · wX¤ · ¿k¢k. ¤ (d) w = ¿k¢k if and only if X is ¯nite-dimensional (from the same reson as for the weak topology). ¤ ¤ (e) w = wX¤ if and only if X is reflexive. (If X is reflexive, then w = ¤ ¤ ¤¤ ¤ ¤ ¤ ¤ σ(X ;X) = σ(X ;X ) = wX¤ . If w = wX¤ , then X = (X ; w ) = ¤ ¤ ¤¤ (X ; wX¤ ) = X , that is, X is reflexive.) ¤ ¤ w ¤ ¤ ¤ (f) xn ! x if and only if xn(y) ! x (y) for each y 2 Y . 4 Some results concerning weak topologies. ¤ ¤ Theorem 0.3. Let X be a normed space, fxng ½ X, and fxng ½ X . (a) If fxng is w-convergent, then fxng is bounded. ¤ ¤ ¤ (b) If X is a Banach space and fxng is w -convergent, then fxng is bounded. Proof. (a) The points xn can be viewed as continuous linear functionals on the Ba- nach space X¤. Since they are pointwise bounded, the Banach-Steinhaus Uniform Boundedness Principle implies that their norms are equi-bounded. (b) The func- ¤ tionals xn are pointwise bounded on the Banach space X. Use again the Uniform Boundedness Principle. ¤ The following proposition is simple but useful. (The simple arrow \!" always denotes the convergence in the norm topology.) ¤ ¤ Proposition 0.4. Let X be a normed space, fxng ½ X, and fxng ½ X , x0 2 X, ¤ ¤ x0 2 X . Assume that at least one of the following conditions hold: ¤ ¤ ¤ (a) xn ! x0, xn(x0) ! x0(x0), and fxng is bounded; ¤ ¤ w ¤ ¤ (b) xn ! x0, xn ! x0, and either X is a Banach space or fxng is bounded; ¤ ¤ ¤ ¤ (c) xn ! x0 and x0(xn) ! x0(x0), and fxng is bounded; ¤ ¤ w (d) xn ! x0 and xn ! x0.
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