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7.3 Topological Vector Spaces, the Weak and Weak⇤ Topology on Banach Spaces

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7.3 Topological Vector Spaces, the Weak and Weak⇤ Topology on Banach Spaces 138 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS 7.3 Topological Vector spaces, the weak and weak⇤ topology on Banach spaces The following generalizes normed Vector space. Definition 7.3.1. Let X be a vector space over K, K = C, or K = R, and assume that is a topology on X. we say that X is a topological vector T space (with respect to ), if (X X and K X are endowed with the T ⇥ ⇥ respective product topology) +:X X X, (x, y) x + y is continuous ⇥ ! 7! : K X (λ, x) λ x is continuous. · ⇥ 7! · A topological vector space X is called locally convex,ifeveryx X has a 2 neighborhood basis consisting of convex sets, where a set A X is called ⇢ convex if for all x, y A, and 0 <t<1, it follows that tx +1 t)y A. 2 − 2 In order to define a topology on a vector space E which turns E into a topological vector space we (only) need to define an appropriate neighbor- hood basis of 0. Proposition 7.3.2. Assume that (E, ) is a topological vector space. And T let = U , 0 U . U0 { 2T 2 } Then a) For all x E, x + = x + U : U is a neighborhood basis of x, 2 U0 { 2U0} b) for all U there is a V so that V + V U, 2U0 2U0 ⇢ c) for all U and all R>0 there is a V ,sothat 2U0 2U0 λ K : λ <R V U, { 2 | | }· ⇢ d) for all U and x E there is an ">0,sothatλx U,forall 2U0 2 2 λ K with λ <", 2 | | e) if (E, ) is Hausdor↵, then for every x E, x =0, there is a U T 2 6 2U0 with x U, 62 f) if E is locally convex, then for all U there is a convex V , 2U0 2T with V U. ⇢ 7.3. TOPOLOGICAL VECTOR SPACES, THE WEAK AND W⇤ TOPOLOGY139 Conversely, if E is a vector space over K, K = R or K = C,and U (E):0 U U0 ⇢{ 2P 2 } is non empty and is downwards directed, i.e. if for any U, V , there is 2U0 a W ,withW U V and satisfies (b), (c) and (d), then 2U0 ⇢ \ = V E : x V U : x + U V , T { ⇢ 8 2 9 2U ⇢ } defines a topological vector space for which is a neighborhood basis of 0. U0 (E, ) is Hausdor↵if also satisfies (e) and locally convex if it satisfies T U (f). Proof. Assume (E, ) is a topological vector space and is defined as T U0 above. We observe that for all x E the linear operator T : E E, z z + x 2 x ! 7! is continuous. Since also Tx T x = T x Tx = Id, it follows that Tx is an ◦ − − ◦ homeomorphism, which implies (a). Property (b) follows from the continuity of addition at 0. Indeed, we first observe that = V V : V is a U0,0 { ⇥ 2U0} neighborhood basis of (0, 0) in E E, and thus, if U ,thenthereexists ⇥ 2U0 a V so that 2U0 1 V V ( + )− (U)= (x, y) E E : x + y U , ⇥ ⇢ · · { 2 ⇥ 2 } and this translates to V + V U. ⇢ The claims (c) and (d) follow similarly from the continuity of scalar multiplication at 0. If E is Hausdor↵then clearly satisfies (e) and it U0 clearly satisfies (f) if E is locally convex. Now assume that U (E):0 U is non empty and downwards U0 ⇢{ 2P 2 } directed, that for any U, V ,thereisaW ,withW U V , and 2U0 2U0 ⇢ \ that satisfies (b), (c) and (d). Then U0 = V E : x V U : x + U V , T { ⇢ 8 2 9 2U ⇢ } is finitely intersection stable and stable by taking (arbitrary) unions. Also ,E .Thus is a topology. Also note that for x E, ; 2T T 2 = x + U : U Ux { 2U0} is a neighborhood basis of x. We need to show that addition and multiplication by scalars is continu- ous. Assume (x : i I) and (y : i I) converge in E to x E and y E, i 2 i 2 2 2 respectively, and let U .By(b)thereisaV with V + V U.We 2U0 2U0 ⇢ 140 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS can therefore choose i so that x x + V and y x + V , for i i , and, 0 i 2 i 2 ≥ 0 thus, x + y x + y + V + V x + y + U, for i i .Thisprovesthe i i 2 ⇢ ≥ 0 continuity of the addition in E. Assume (xi : i I) converges in E to x,(λi : i I) converges in K to 2 2 λ and let U . Then choose first (using property (b)) V so that 2U0 2U0 V + V U. Then, by property (c) choose W 0, so that for all ⇢ K, ⇢ 2U 2 ⇢ R := λ + 1 it follows that ⇢W V and, using (d) choose " (0, 1) | | | | ⇢ 2 so that ⇢x W , for all ⇢ K,with ⇢ ". Finally choose i0 I so that 2 2 | | 2 x x + W and λ λ <"(and thus λ <Rfor i i ), for all i i in i 2 | − i| | i| ≥ 0 ≥ 0 I (and thus also λ <Rfor i i ). 0 | i| ≥ 0 λ x = λ (x x)+(λ λ)x+λx λx + λ W + V λx + V + V λx + U. i i i i − i− 2 i ⇢ ⇢ If satisfies (e) and if x = y are in E, then we can choose U U0 6 2U0 so that y x U and then, using the already proven fact that addition − 62 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 + v1 = y + v2, for some v ,v V it would follows that y x = v v U,whichisa 1 2 2 − 2 − 1 2 contradiction. If (f) is satisfied then E is locally convex since we observed before that = x + U : U is a neighborhood basis of x, for each x E. Ux { 2U0} 2 Theorem 7.3.3. (The Hyperplane Separation Theorem) Let X be a locally convex topological space. Let K X be nonempty, closed and convex, and z X K. ⇢ 2 \ Then there exists a continuous linear functional F : X R for which ! F (z) < inf F (x). x K 2 We will need the following two propositions Proposition 7.3.4. Every element of an open subset U of X is an internal point. (there is some λ for which x + λ x E whenever λ λ ). 0 0 · 2 | | 0 Proof. Exercise Proposition 7.3.5. AlinearmapF : X K is continuous if and only ! there is a neighborhood of 0 on which F is bounded Proof. Exercise. 7.3. TOPOLOGICAL VECTOR SPACES, THE WEAK AND W⇤ TOPOLOGY141 Proof. We can choose a convex 0-neighborhood U so that (z + U) K = . \ ; We can assume that U = U (after replacing U by U ( U)). − \ − By the Hyperplane Separation Lemma we can find a linear functional F : X R so that ! sup F (x) inf F (x). x z+U x K 2 2 Now since K is not empty, let x0 K and thus M =infx K F (x) F (x0) < 2 2 , And thus 1 sup F (x) F (z) +sup F (x) F (z) + M, x U | || | x U+z | || | 2 2 so by 7.3.5 F is continuous. Since F = 0 we can choose y X,withF (y) = 0 After multiplying y 6 2 6 with a small enough but number (Multiplication by scalars is continuous) we can assume that y U, and after multiplying y by 1 if necessary, we 2 − can assume that F (y) > 0. Thus F (z) <F(y)+F (z)=F (y + z) sup F (x) inf F (x), x z+U x K 2 2 which proves our claim. Theorem 7.3.6. Let (p↵)↵ A be a family of seminorms on a (real or com- 2 plex) vectorspace X. For ↵ A, x X and ">0,put 2 2 U = y X : p (x y) <" . x,↵," 2 ↵ − Let be the topology generated by the sets U . T x,↵," Then a) For each x X, the finite intersections of the sets U , ↵ A, 2 x,↵," 2 ">0, form a neighborhood basis of x. b) For a net (xi)i I X and x X, (xi) converges to x if and only if 2 ⇢ 2 for all ↵ A, limi I p↵(xi x)=0. 2 2 − c) (X, ) is a local convex topological space. T Proof. (a) By Proposition 6.1.7 the finite intersection of a generating system of a topology is a basis of that topology (recall that is called a B⇢T basis if = U : x U is a neighborhood basis). Secondly if Bx { 2B 2 } 142 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS x n U ,weletδ = " p (x x ) it follows from the triangular 2 j=1 xj ,↵j ,"j j j − ↵j − j inequality that T n n x U U , 2 x,↵j ,δj ⇢ xj ,↵j ,"j j\=1 j\=1 which proves our claim. (b) It follows from (a) that xi i I x ">0,↵ A i0 I i i0 xi Ux,↵," ! 2 () 8 2 9 2 8 ≥ 2 ↵ Ap↵(xi x) i I 0. () 8 2 − ! 2 (c) the continuity of the vector operations follows from Proposition 6.3.4 (which characterizes continuity via nets) and (b): If xi i I x and yi i I y, then for all ↵ A, p↵(xi x) i I 0 and ! 2 ! 2 2 − ! 2 p↵(yi y) i I 0, and thus by the triangular inequality p↵(xi+yi x y) i I − ! 2 − − ! 2 0, and thus, again by (b) xi + yi i I x + y.
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