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Chaper 3 Weak Topologies. Reflexive Space .Separabe Space. Uniform Convex Spaces. III.1 The weakest Topology Recall on the weakest topology which renders a family of mapping continuous i : X Yi iI arbitary set topological space To define the weakest topology on X such that i for each Let Yi is continuous from X to iI 1 wi F (Yi ), ( wi ) must be open in X For any finite set (*) 1 ( wi ), iF FI wi : open in The family of the sets of the form (*) form a base of a topology F of X The topology is the weakest topology that renders all i continuous Yi Proposition III.1 Let xn F be a sequence in X, then F(Yi ) xn x i ( xn ) i ( x) i I " " For each i I i ( x ) w and w is open in Yi 1 1 x i ( w) and i ( w) is open in X N s.t. 1 n N xn i ( w ) i ( xn ) w Hence i ( xn ) i ( x ) " " It is sufficient to consider U 1 i ( wi ) iF 1 where x i ( wi ) i F , For each i, i ( x ) wi N i s.t. n N i i ( xn ) wi Let N max N i i F , then n N i ( xn ) wi 1 i F xn i ( wi ) iF Proposition III.2 Let Z be a topological space and i Z X Yi i I Then is continuous is continuous from Z to i Yi i I " " For each i I If w is open in Yi , then i 1 ( w) is open in X , 1 1 ( i ( w)) is open in Z sin ce is continuous Hence i is continuous " " Let U 1 i ( wi ) iF 1 U 1 i iF iF 1 1 ( wi ) 1 i ( wi ) 1 ( ) ( wi ) is open in Z i iF III.2 Definition and properties of the weak topology σ(E,E´) Definition σ(E,E´) E: Banach space E´: topological dual of E f ( x) f , x x E, f E see next page ( E , E ) Definition : The weak topology is the weakest topology on E such that f :E R is continuous for each f E Proposition III.3 The topology is Hausdorff ( E , E ) on E Let x y E By sec ond geometric form of Hahn Banach Thm f E and R s.t. f , x and f ,y Let O1 z E : f , z , and O2 z E : f , z , ( , ) is ( E , E ) open in E , 1 O2 f ( , ) is ( E , E ) open in E , then O1 f 1 x O1 , y O2 , and O1 O2 Hence the topo log y ( E , E ) on E is Hausderff . Proposition III.4 Let x0 E neighborhood of ; we obtain a base of x0 by consider sets of the form V x E : fi , x x0 , i F 0, fi E , and F is finite where Proposition III.5 Let (i) (ii) if xn be a sequence in E. Then xn x weakly f , xn f , x f E xn x xn x strongly, then weakly. (iii) if xn x xn weakly, then is bounded and x lim inf xn n (iv) if xn x weakly and fn f strongly in E´, then f n , xn f , x (i ) By Pr oposition III .1 (ii) Assume xn x strongly. Since f , xn f , x f , xn x f xn x f , xn f , x By Pr oposition III .1, xn x weakly (iii) If xn x weakly, then by (i ) f , xn f , x f E By Corollary II .2, sup xn n and x lim inf xn n f n , xn f , x (iv) f n , xn f , xn f , xn f , x f n f , xn f , xn f , x f n f , xn f , xn f , x f n f xn f , x n f , x By (i ) By (iii) f , xn f , x 0 xn is Since f n f , Hence i.e. bound fn f 0 f n , xn f , x 0 f n , xn f , x Exercise Let E , F be real normed vector space consider on E and F the topologies ( E , E ) and ( F , F ) respectively. Then the product topology on E X F is ( E F , E F ) Proposition III.6 If dim E ,then ( E , E ) is strong topology on E. To show that a strong open set is ( E , E ) open. It is sufficient to show that for x0 E and for an open nhd . U of x, there is a ( E , E ) open nhd . V of s.t. V U Suppose B ( x0, r ) U . Choose a basis e1 , e2 ,, en of E s.t ei 1 i 1,, n n For x E , x f i ( x )ei , f i E i 1,, n i 1 n x x0 f i , x x0 ei i 1 n f i , x x0 ei i 1 n f i , x x0 i 1 n if f i , x x0 i 1,, n Hence if we let V x E ; f i , x x0 i 1,, n then V B ( x0 , n ) Choose s.t. n r, then V B ( x0 , r ) U Remark If dim E ,then ( E , E ) is strictly weaker then the strong topology. Let S x E ; x 1 S is strongly closed . Claim : S ( E , E ) B (0,1) pf : Let x0 B (0,1) and let V x E ; f i , x x0 i 1,, n where 0 and f1 ,, f n E There is y0 E s.t. y0 0 and f i , y0 0 i 1,, n because co dim x E ; f i , x 0 n i 1 [ If there is no such y0 , then the map : E R n defined by x f1 ( x ),, f ( xn ) is 1 1 then dim R ( ) n dim E n ] Consider the function g (t ) x0 ty0 g is continuous fun. of x, g (0) x0 1 and lim g (t ) , then there is t0 0 s.t t x0 t0 y0 1 i.e x0 t0 y0 S But f i , x0 t0 y0 x0 0 i 1,, n then x0 t0 y0 V x0 t0 y0 V S x0 S ( E , E ) III.3 Weak topology, convex set and linear operators Theorem III.7 Let CE be convex, then C is weakly closed if and only if C is strongly closed. To show : If C is strongly closed , then C is weakly closed . Let x0 C , there is 0 f E and R c f , x0 f , x s.t. x C Let V x E ; f , x , then V is a nhd . of x0 and V C then V C c C is weakly open C is weakly closed c Remark The proof actually show that every every strongly closed convex set is an intersection of closed half spaces H f , x E f , x C Corollary III.8 If : E ( , ] is convex l.s.c. w.r.t. strongly topology then In particular, if then is l.s.c. w.r.t. ( E , E ) xn x weakly ( ( E, E)) ( x ) lim inf ( xn ) n Let then then then A x E ; ( x ) , R A is strongly closed and convex. A is ( E , E ) closed is ( E , E ) l.s.c Observation : Let ( x ) x , then is strongly continuous and convex is strongly l.s.c and convex then is ( E , E ) l.s.c Hence if xn x weakly ( ( E , E )), then x lim inf xn n Theorem III.9 Let E and F be Banach spaces and let T :EF be linear continuous (strongly) , then T is linear continuous on E with to F with ( F , F ) And conversely. ( E , E ) By Pr oposition III .2 it is sufficient to show that x f , Tx is continuous from ( E , ( E , E )) x f , Tx is strongly continuous x f , Tx is ( E , E ) l.s.c x f , Tx is ( E , E ) l.s.c x f , Tx is ( E , E ) u.s.c x f , Tx is ( E , E ) continuous Conversely, sup pose that T is linear continuous from ( E , ( E , E )) to ( F , ( F , F )) then G (T ) is closed in E F w.r.t product topo log y ( E , E ) ( F , F ) which is ( E F , E F ) i.e. G (T ) is closed w.r.t. ( E F , E F ) G (T ) is convex in E F then G (T ) is closed in E F w.r.t strong topo log y Closed Graph Theorem T is continuou from ( E , E ) to ( F , F ) On Remark E, ( E, E) is weak topology j : E E by j ( x ), f f , x f E j( x) x In genernal j is not surjective If j ( E ) E E is called reflexive III.4 The weak* topology σ(E′,E) The weak* topology ( E , E ) is the weakest topology on E´ such that f f ,x is continuous for all xE Proposition III.10 The weak* topology on E´ is Hausdorff Proposition III.11 One obtains a base of a nhds for a f0 E by considering sets of the form V f E ; f f 0 , xi , i 1,, n 0 , x1, x2 ,, xn in E Proposition III.12 Let (i) f n be a sequence in E´, then f f * n for ( E , E ) fn , x f , x x E (ii) If fn f fn f strongly, then for ( E, E) (iii) If then fn f fn f for ( E, E) for ( E, E ) fn f * (iv) If fn then is bounded and f lim inf f n n (v) If fn f * and xn x strongly, then f n , xn f , x Lemma III.2 Let X be a v.s. and ,1,,n are linear functionals´on X such that i (v) 0 i 1,, n (v) 0 n 1 ,, n R s.t. i i i 1 Let F : X R n 1 be defined by F (u ) (u ),1 (u ),, n (u ) u X Then F ( X ) R( F ) is closed convex set in R Since (1,0,,0) F ( X ) , , 1 ,, n R not all zero and R s.t. n (u ) ii (u ) u X i 1 n (u ) i i (u ) 0 u X i 1 0 i (u ) i (u ) i 1 n n 1 Proposition III.13 If : E R is linear continuous´w.r.t ( E , E ) then there is x E s.t. ( f ) f , x f E There is a nhd . V of 0 for ( E , E ) s.t. ( f ) 1 may take V as follows V f E ; f , xi i 1,, n In particular, if f , xi 0 i 1,, n then ( f ) 0 n Lemma III .2 ( f ) i f , xi i 1 n f , i xi i 1 n Take x i xi i 1 then ( f ) f , x f E f E Corollary III.14 If H is a hyperplane in E´ closed w.r.t ( E , E ) Then H is of the form H f E ; f , x ,0 x E , R H is of the form * H f E ; ( f ) , where 0 E , R Suppose f 0 H , then there is V H c where V f E ; f f 0 , xi i 1,, n x1 , x2 ,, xn E , 0 V is convex (13) Either ( f ) f V (13) or ( f ) f V From (13) ( g ) ( f 0 ) g W V f 0 From(13) ( g ) ( f 0 ) g W V f 0 W is a ( E , E ) nhd . of 0 then is ( E , E ) continuous By Theorem III .13 x E s.t. ( f ) f , x f E then H f E ; f , x III.5 Reflexive spaces On Remark E, ( E, E) is weak topology j : E E by j ( x ), f f , x f E j( x) x j is isometry j(E) is closed vector subspace of E In genernal j is not surjective If j ( E ) E E is called reflexive Lemma 1 (Helly) p.1 Let E be a Banach space, f1, f 2 ,, f n E 1,2 ,,n R and are fixed. Then following statements are equivalent Lemma 1 (Helly) (i) 0, x BE s.t. fi , x i (ii) p.2 n n i 1 i 1 ii i fi where i 1,2,, n 1 , 2 ,, n R BE x E ; x 1 n (i ) (ii) Let 1 , 2 ,, n R and s i i 1 n n i 1 i 1 (i ) i f i , x i i s n n i 1 i 1 i i i f i , x s n i f i s i 1 By letting 0, we have n n i 1 i 1 ii i f i (ii) (i ) Let 1 , 2 ,, n R define the map : E R n n and by ( x ) f1 , x , f 2 , x ,, f n , x Then (i ) ( BE ) Suppose that (i ) doesn' t hold . then ( BE ) Since ( BE ) is a closed convex set in R , n by strictly separation principle there is a 1 , 2 ,, n R ( x) n i 1 i 1 i fi , x i i n i 1 i 1 s.t. x BE n n n x BE i f i i i which contradicts (ii) Lemma 2 (Goldstine) Let E be a Banach space. Then JB E is dense in w.r.t the weak* topology ( E, E) on E BE Let BE and V be a nhd of in the form V E ; , f i i 1,, n Let i , f i 1 , 2 ,, n R i 1,, n and note that we have n n n i 1 i 1 i 1 ii , i f i i f i Helly Lemma 0, x BE s.t. f i , x i i 1,, n Jx , f i , f i i 1,, n Jx , f i i 1,, n Jx V Theorem (Banach Alaoglu-Bornbaki) BE f E ; f 1 is compact w.r.t. ( E , E ) Theorem A Banach space E is reflexive if and only if BE is compact w.r.t weak topology ( E , E ) " " Assume that E is reflexive. Then JBE BE , sin ce J is isometry Claim : J : E , ( E , E E , ( E , E For any f E 1 ( f J 1 )( ) f , J 1 ( ) , f E then f J 1 is continuous from E , ( E , E to R By Pr oposition III .2, J 1 is continuous from E , ( E , E to E , ( E , E Since BE is compact w.r.t ( E , E ) and by Claim, BE J 1 ( BE ) is compact w.r.t ( E , E ) " " Suppose that BE is compact w.r.t ( E , E ) J is continuous with norm By Thm III .9, J is continuous from ( E , ( E , E )) to ( E , ( E , E )) J is continuous from ( E , ( E , E )) to ( E , ( E , E )), because ( E , E ) is wea ker than ( E , E ) JBE is compact w.r.t ( E , E ) By Goldstine Lemma , JBE BE then JE E Hence E is reflexive. Exercise Suppose that E is a reflexive Banach space . Show that evere closed vector subspace M of E is reflexive. Since E is reflexive, BE is compact w.r.t. ( E , E ) M is strongly closed then M is closed w.r.t. ( E , E ) BM BE M is compact w.r.t. ( E , E ) For any x0 M and a ( M , M ) nhb. V of x0 in the form V x M ; f i , x x0 , i 1,, n where f i M i 1,, n Claim : V M c is a ( E , E ) nhb. of x0 fi ( x) ˆ [ Let f i ( x ) f ( x0 ) if x M if x M Since M is strongly closed , fˆi E V M c x E ; fˆ , x x0 , i 1,, n is a ( E , E ) nhb. of x0 ] If V I is a family of ( M , M ) open sets such that V BM V M I then V M c BM and by Claim I c I is a family of ( E , E ) open sets Since BM is compact w.r.t. ( E , E ), n 1 ,, n I s.t. V M c BM i 1 i n V BM i 1 i Hence BM is compact w.r.t ( M , M ) Therefore M is reflexive Corollary 1 Let E be a Banach space. Then E is reflexive if and only if E is reflexive (1) Suppose that E is reflexive. Then ( E , E ) ( E , E ) By Alaoglu Bornbaki Thm BE is compact w.r.t ( E , E ) BE is compact w.r.t ( E , E ) Hence E is reflexive. ( 2) Suppose that E is reflexive. then by (1), E is reflexive. Since JE is closed subspace of E , by Exercise, JE is reflexive Since J is isometry between E and JE , E is reflexive. Corollary 2 Let E be a reflexive Banach space. Suppose that if K is closed convex and bounded subset of E . Then K is compact w.r.t ( E , E ) Since K is strongly closed and convex , by Thm III .7 K is closed w.r.t ( E , E ) Since K is bounded , K mBE for some m 0 Since E is reflexive, mBE is compact w.r.t ( E , E ) K is compact w.r.t ( E , E ). Uniformly Convex A Banach space is called uniform convex if for all ε>0 , there is δ>0 such that if x, y BE with x y x y then 1 2 x y 2 x y Counter Example for Uniformly Convex Consider (R , ) 2 ( x1 , x2 ) x1 x2 is not uniform convex. see next page y (0,1) x y 2 (-1,0) x (1,0) (0,-1) Example for Uniformly Convex Consider (R , ) ( x1,, xn ) n 2 x1 2 x2 2 xn is uniform convex. see next page For any 0 For any x, y B(0,1) and 2 2 2 x y x y 2 x y 2 2 x y 2 x y 2 x y 2 ( Parallelog ram x y 2 42 x y 2 2 1 2 4 2 x y 2 1 1 1 1 2 4 4 Take 1 1 x y 1 2 2 4 , we have Thm) Theorem A uniformly convex Banach space E is reflexive. Let E , 1 To show that JBE Since JBE is closed in E strongly, it is sufficient to show that for 0, x BE s.t. Jx Given 0, let 0 be as in the definition of uniform convexity f E with f 1 s.t. , f 1 2 Let V E ; , f , 2 V is a nhb. of w.r.t E , E Since JBE is dense in BE w.r.t E , E , x BE s.t. Jx V Claim : Jx BE [ Suppose not. Then W Jx BE c Since BE is compact w.r.t E , E , BE is closed w.r.t E , E Jx BE is closed w.r.t E , E W is open w.r.t E , E xˆ BE s.t. Jxˆ V W We have then (sin ce Jx, Jxˆ V ) f , x , f and f , xˆ , f 2 2 , f f , x xˆ x xˆ 2 x xˆ , f (1 ) 1 2 2 2 2 x xˆ But x xˆ J ( x xˆ ) Jx Jxˆ , sin ce Jxˆ W a contradiction.