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 ( , x) 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 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 K : 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,( i : i I) converges in K to 2 2 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 := + 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 <"(and thus Proposition 7.3.4. Every element of an open subset U of X is an internal point. (there is some for which x + x E whenever ). 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) 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 ,welet = " 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 , j ⇢ 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. ! 2 Similarly we can show the continuity of the multiplication. The triangular inequality also proves that for x X the sets of the form n 2 j=1 Ux,↵j , j are convex and by (a) they form a neighborhood basis of x. T Proposition 7.3.7. Suppose that X and Y are two locally convex topological spaces whose topologies are generated by the families of seminors (p↵)↵ A 2 and (q ) B, respectively. 2 Then a linear map T : X Y is continuous if and only if for every ! B, there are n N, ↵1,↵2,...↵n A,andC>0 such that for all 2 2 2 x X 2 n q (Tx) C p (x) for all x X. aj 2 Xj=1 Proof. “ ”IfT is continuous and B there is a neighborhood U of 0 ) 1 2 in X such that U T y Y : q (y) < 1 . By 7.3.6 we can assume that n ⇢ { 2 } U = j=1 Ux,↵j,"j for some n N, ↵1,↵2,...↵n A, and "1,"2,..."n 2 1 2 2 (0 ). Put" =min1 j n "j, and C = " .Thusq (x) < 1whenever 1 S p (x) <", for all j =1, 2,...,n.Soletx X.Eitherp (x) > 0 for ↵j 2 ↵j some j 1, 2,...,n .Putz = "x/ n p (x). Thus p (z) <"for all 2{ } i=1 ↵i ↵i i =1, 2,...,n and therefore P n n 1 q (T (x)) = " p (x)q (T (z)) C p (x). ↵j ↵j Xi=1 Xi=1 7.3. TOPOLOGICAL VECTOR SPACES, THE WEAK AND W⇤ TOPOLOGY143 Or p (x) = 0 for all j 1, 2,...,n ,thenp (rx) = 0 for all j ↵j 2{ } ↵j 2 1, 2,...,n , for all r>0, and thus rq (T (x)) = q (T (rx)) < 1, for all { } r>0 which means that q (T (x)) = 0 which also implies that q (T (X)) Cp (x) ↵j and finishes the proof of our claim. On a normed linear space X and its dual X⇤ we define the following weak and weak⇤ topology. Definition 7.3.8. Let X be a normed linear space and X⇤ it’s dual space. Then the weak topology on X is generated by the following semi norms: For x X ,put ⇤ 2 ⇤ px : X R,px (x)= x⇤(x) , for x X ⇤ ! ⇤ | | 2 A neighborhood basis of x X of the weak topology is formed therefore by 2 the sets n U = Ux,x ,"j ,n N,x⇤ X⇤,"j > 0, for 1 j n,where j⇤ 2 j 2 j\=1 U = z X : x⇤(x z) <", for x⇤ X⇤, and ">0. x,x⇤," { 2 | | 2 and by Theorem 7.3.6 (b) a net (x ) weakly converges to some x if and i i ⇢ only if lim x⇤(xi)=x⇤(x) for all x⇤ X⇤. i I 2 2 The weak or w -topology on X is is generated by the following semi ⇤ ⇤ ⇤ norms For x X,put 2 qx : X⇤ R,qx(x⇤)= x⇤(x) , for x⇤ X⇤ ! | | 2 A neighborhood basis of x X of the weak topology is formed therefore ⇤ 2 ⇤ by the sets n U = Ux ,xj ,"j ,n N,xj X,"j > 0, for 1 j n,where ⇤ 2 2 j\=1 U = z⇤ X⇤ : (x⇤ z⇤)(x) <", for x X, and ">0. x⇤,x," { 2 | | 2 144 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS and by Theorem 7.3.6 (b) a net (x ) weakly converges to some x if and i⇤ i ⇢ ⇤ only if lim xi⇤(x)=x⇤(x) for all x X. i I 2 2 Let X and Y be Banach spaces. The strong operator topology on L(X, Y ) (which is still weaker, i.e., coarser, than the topology generated by the operator norm) is the locally convex topology generated by the semi norms r : L(X, Y ) [0, ),T T (x) . x ! 1 7! k k Thus a net (Ti)i I L(X, Y ) converges to T L(X, Y ) in the strong 2 ⇢ 2 operator topology if and only if limi I (Ti T )(x) = 0 for all x X 2 k k 2 The weak operator topology on L(X, Y ) is the locally convex topology generated by the semi norms s : L(X, Y ) [0, ),T y⇤(T (x)). x,y⇤ ! 1 7! | (Note that for S, T L(X, Y ): y ((T + S)(x)) = y (T (x)+S(x)) 2 | ⇤ | | ⇤ | y⇤(T (x)) + y⇤(S(x)) ). Thus a net (Ti)i I L(X, Y ) converges to T | | | | 2 ⇢ 2 L(X, Y ) in the weak operators topology if and only if limi I y⇤((Ti 2 k T )(x)) = 0 for all x X and y Y . k 2 ⇤ 2 ⇤ Theorem 7.3.9. (Theorem of Alaoglu) (often also called Theorem of Alaoglu Bourbaki) B = x X : x 1 is compact with respect to the weak -topology. X⇤ { ⇤ 2 ⇤ k ⇤k } ⇤ Proof. For x X put 2 Dx = ⇠ K : ⇠ x . 2 | |k k Then Dx K, is compact for all x X, and thus by Tychano↵’s Theorem ⇢ 2 6.5.1 D = Dx x X Y2 is compact with respect to the product topology. Note that D is exactly the set of all functions f : X K,with f(x) x . Moreover the ! | |kk product topology D coincides with the topology of pointwise convergence. BX⇤ consists exactly of the elements in D which are linera, and the w⇤ topology is the restriction of the product topology on D to BX⇤ . It is therefore enough to show that B is closed in D.Solet(x ) B so that f D X⇤ i⇤ ⇢ X⇤ 2 exists so that f(x)=limi I xi⇤(x) for all x X. Thus for a, b K, x, z X 2 2 2 2 f(ax + by)=limxi⇤(ax + by)=limaxi⇤(x)+limbxi⇤(y)=af(x)+bf(z), i I i I i I 2 2 2 Thus f is linear. Moreover f(x) x , for x X, and thus f B . | |k k 2 2 X⇤ 7.4. THE KREIN-MILMAN THEOREM 145 7.4 The Krein-Milman Theorem Throughout this section X will be a locally convex vector spaces Definition 7.4.1. Let K X be convex. ⇢ A nonempty subset E of K is called an extreme subset of K if it is convex and closed, and has the property that if x E and there are u, v K so 2 2 that x = u +(1 )v, for some 0 < <1, then both u and v are in E. x K is called an extreme point of K if x is an extreme set for K. 2 { } This means that if we we write x as x = u +(1 )v, for some 0 < <1 and u, v K,thenu = v = x. 2 Proposition 7.4.2. Let K X convex. ⇢ 1. If the intersection of a collection of extreme subsets of a convex set K X is nonempty, then the intersection is an extreme subset of K ⇢ and, moreover, if A is an extreme subset of B and B is an extreme subset of K, then A is an extreme subset of K. 2. An extremal subset of an extremal subset of K, is an extremal subset of K. Proof. Exercise. Lemma 7.4.3. Let K be a nonempty, compact, convex subset of a locally convex topological vector space X and :X R be linear and continuous. ! Then the set of point in K at which takes its maximum/minimum value on K is an extreme subset of K. Proof. Since K is compact takes its maximal value m on a compact set E. Since is linear E is also convex. Assume that x E and that x = 2 u +(1 )v,withu, v K, and 0 < <1. It follows that u and v are 2 both in E. Indeed, assume that for example u E,then 62 m = (x)= (u)+(1 ) (v) (u)+(1 )m < m, which is a contradiction. Lemma 7.4.4. Let K be a nonempty, compact, convex subset of a locally convex topological vector space X. Then K has an extreme point. Proof. The strategy of the proof is first to apply Zorn’s Lemma to find an extreme subset E of K that contains no proper subset which also is an extreme subset of K. 146 CHAPTER 7. ELEMENTS OF FUNCTIONAL ANALYSIS Let = E K : E = is an extreme subset of K . E { ⇢ 6 ; } Let be linearly ordered by “ ”. Then L⇢E E0 = E E \ 2L is also in . Since the elements of are compact and have the finite inter- E L section property E =. By 7.4.2 E is an extremal set for K. 0 6 0 By the Zorn’s Lemma there is a minimal extremal set F for K. We need to show that F is a singleton. Assume not, and let x, y F , 2 x = y. By the Separation Theorem 7.3.3 there is a continuous linear : 6 X R, so that (x) < (y). From Lemma 7.4.3 it follows that the subset ! G ( F for which assumes its minimum of F is extremal for F .SinceF is extremal for K it follows that G is extremal for K. For a set S we call the convex hull of S the set ⇢ conh(S)= K : K convex and S K ⇢ ⇢ \ F S finite, 0 s 1,s F, = ss : ⇢ 2 . s =1 (s F s F ) X2 2 P Theorem 7.4.5. Let K X, non empty, convex and compact. Let Ext(K) ⇢ be the set of extreme points of K,the K = conh(Ext(K)). Proof. It is clear that conh(Ext(K)) K. Assume that conh(Ext(K)) ( K, ⇢ and pick z K conh(Ext(K)). By the Hyperplane Separation Theorem 0 2 \ 7.3.3 there is a continuous linear functional : X R so that ! (7.2) (z0) > max (x) sup (x). x conh(Ext(K)) x Ext(K) 2 2 Let m = maxx K (x). By Lemma 7.4.4 the set 2 M = x K : (x)=m , { 2 } is an extremal set of K.SinceM is nonempty, compact and convex it has an extreme point x0 by Lemma 7.4.4. By Proposition 7.4.2 x0 is an extreme point of K. But this means that (z ) m = (x ) which contradicts 0 0 (7.2).