Chapter 2
Weak Topologies and Reflexivity
2.1 Topological Vector Spaces and Locally Convex Spaces
Definition 2.1.1. [Topological Vector Spaces and Locally Convex Spaces] Let E be a vector space over K,withK = R or K = C and let be a T topology on E. We call (E, ) (or simply E, if there cannot be a confusion), T a topological vector space, if the addition:
+:E E E, (x, y) x + y, ⇥ ! 7! and the multiplication by scalars
: K E E, ( , x) x, · ⇥ ! 7! 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 2 convex sets. Remark. Topological vector spaces are in general not metrizable. Thus, continuity, closedeness, and compactness etc, cannot be described by se- quences. We will need nets. Assume that (I, )isadirected set . This means (reflexivity) i i, for all i I, • 2 (transitivity) if for i, j, k I we have i j and j k,theni k, and • 2 27 28 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY
(existence of upper bound) for any i, j I there is a k I, so that • 2 2 i k and j k. A net is a family (x : i I)indexedoveradirectedset(I, ). i 2 A subnet of a net (x : i I)isanet(y : j J), together with a map i 2 j 2 j i from J to I, so that x = y , for all j J, and for all i I there is 7! j ij j 2 0 2 a j J, so that i i for all j j . 0 2 j 0 0 Note: A subnet of a sequence is not necessarily a subsequence.
Definition 2.1.2. In a topological space (T, ), we say that a net (x : i I) T i 2 converges to x, if for all open sets U with x U there is an i I, so that 2 0 2 x U for all i i .If(T, ) is Hausdor↵ x is unique and we denote it by i 2 0 T limi I xi. 2 Using nets we can describe continuity, closedness, 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 closed 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 con- vergent subnet.
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 2.1.3. 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 : d) for all U and x E there is an ">0,sothat x U,forall 2U0 2 2 K with <", 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. ⇢ 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 ⇢ } 30 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 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 ⇢ 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 Definition 2.1.4. If E is a topological vector space over K, we call E⇤ = f : E K : f linear and continuous . { ! } Definition 2.1.5. [The Topology (E,F)] Let E be a vector space and let F be a separating subspace of E#. Then we denote the locally convex Hausdor↵topology generated by n 0 = x E : x⇤(x) <"i : n N,x⇤ F, and "i > 0 ,i=1,...,n , U { 2 | i | } 2 i 2 j=1 \ by (E,F). If E is a locally convex space we call (E,E⇤), as in the case of Banach spaces, the Weak Topology on E and denote it also by w.IfE,sayE = F ⇤, for some locally convex space F , we call (F ⇤,F) the weak⇤ topology and denote it by w⇤ (if no confusion can happen). From the Hahn Banach Theorem for Banach spaces it follows that the weak topology turns a Banach space X into a Hausdor↵space, and we can see X, (X, X⇤) as a locally convex space. Similarly X⇤, (X⇤,X) is a locally convex space which is Hausdor↵. Proposition 2.1.6. Assume that X is a Banach space and that X⇤ denotes its dual with respect to the norm. Then X, (X, X⇤) ⇤ = X⇤ and X⇤, (X⇤,X) ⇤ = X. Proposition 2.1.6 follows from a more general principle. Proposition 2.1.7. Let E be a locally convex space and E⇤ its dual space. Equip E⇤ with the topology (E⇤,E). Then E⇤, (E⇤,E) ,and E, (E,E⇤) are locally convex spaces whose duals are E and E , respectively (where we ⇤ identify e E in the canonical way with a map defined on E ). 2 ⇤ Proof. It is clear that E belongs to E⇤, (E⇤,E) ⇤ in the following sense: If e E and if (e) is the function on E which assigns to f E the scalar 2 ⇤ 2 ⇤ f,e, ,then (e)isin E , (E ,E) ⇤. From now on we identify e with h i ⇤ ⇤ (e) and simply write e instead of (e). Assume : E⇤ K is linear and (E⇤,E)-continuous. ! 1 U = f E⇤ : , f < 1 = ( 1, 1) { 2 |h i| } 32 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY is then an (E,E )-open neighborhood and thus there are e ,e ,...,e E ⇤ 1 2 n 2 and ">0 so that n f E⇤ e ,f <" U. { 2 ||h j i| }⇢ j\=1 It follows from this that n ker(e ) ker( ). j ⇢ j\=1 Indeed, n n ker(e )= f E⇤ e ,f < " j { 2 ||h j i| } j\=1 \>0 j\=1 n = f E⇤ e ,f <" · { 2 ||h j i| } \>0 j\=1 U ⇢ · \>0 = f E⇤ : , f < 1 ·{ 2 |h i| } \>0 = f E⇤ : , f < =ker( ). { 2 |h i| } \>0 Now an easy linear algebra argument implies that is a linear combination of e ,e ,...,e which yields that E. 1 2 n 2 Proposition 2.1.8. Let E be a vector space and let F be a separating sub- space of E#. For a net (xi)i I E and x E 2 ⇢ 2 lim xi = x in (E,F) x⇤ F lim x⇤,xi = x⇤,x . i I () 8 2 i I h i h i 2 2 Exercises 1. Show that in a topological space (T, )asetA is closed if and and T only if for a net (xi)i I A which converges to some x it follows that 2 ⇢ x A. 2 2.1. TOPOLOGICAL AND LOCALLY CONVEX VECTOR SPACES 33 2. Assume that E is a locally convex vectors space. Show that (E, (E,E⇤)) is also a local convex vector space and that (E, (E,E⇤))⇤ = E⇤. 3. Prove Proposition 2.1.8. 34 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 2.2 Geometric Version of the Hahn-Banach The- orem for locally convex spaces We want to formulate a geometric version of the Hahn-Banach Theorem. Definition 2.2.1. AsubsetA of a vector space V over K is called convex if for all a, b A and all [0, 1] also a +(1 )b A. 2 2 2 If A V we define the convex hull of A by ⇢ conv(A)= C : A C V,C convex ⇢ ⇢ n n \ = jaj : n N, j [0, 1],ai A, for i=1,...,n, and j =1 . 2 2 2 n Xj=1 Xj=1 o AsubsetA V is called absorbing if for all x V there is an 0 (2.1) v V : µ (v) < 1 C v V : µ (v) 1 . { 2 C }⇢ ⇢{ 2 C } If V is a locally convex space space and if 0 is in the open kernel of C, then µC is continuous at 0. Proof. Since C is absorbing 0 C and µ (0) = 0. If u, v V and ">0 2 C 2 is arbitrary, we find 0 < u <µC (u)+" and 0 < v <µC (v)+", so that u/ C and v/ C and thus u 2 v 2 u + v u v = u + v C, u + v u + v u u + v v 2 which implies that µ (u + v) + µ (u)+µ (v)+2", and, since, C u v C C ">0 is arbitrary, µ (u + v) µ (u)+µ (v). C C C Finally for >0 and v V 2 v r v µ ( v)=inf r>0: C = inf : C = µ (v). C r 2 r 2 C n o n o 2.2. GEOMETRIC VERSION OF THE HAHN-BANACH THEOREM 35 To show the first inclusion in (2.1) assume v V with µ (v) < 1, there 2 C is a 0 < <1 so that v/ C, and, thus, 2 v v = +(1 )0 C. 2 The second inclusion is clear since for v C it follows that v = v C. 2 1 2 If V is a locally convex space and and 0 C0, then there is a an open 2 convex neighborhood U of 0, so that 0 U C. Now let (x )beanet 2 ⇢ i which converges in V to 0. Since with U also "U is a neighborhood of 0, for ">0, we obtain for any ">0 an i I, so that for all i i in I it follows 0 2 0 that x "U. i 2 µ (x ) µ (x ) "µ (x ) ". C i U i "U i Theorem 2.2.3. (The Geometric Hahn-Banach Theorem for locally convex spaces) Let C be a non empty, closed convex subset of a locally convex and Hausdor↵space E and let x E C. 0 2 \ Then there is an x E so that ⇤ 2 ⇤ sup ( x⇤,x ) < ( x⇤,x0 ). x C < h i < h i 2 Proof. We first assume that K = R and we also assume w.l.o.g. that 0 C 2 (otherwise pass to C x and x x for some x C). Let U be convex 0 2 open neighborhood of 0 so that C (x + U)= ,thenletV be an open \ 0 ; neighborhood of 0 so that V V U and let D = C + V .It follows that ⇢ also (x + V ) D = . 0 \ ; From Lemma 2.2.2 it follows that µD is a sublinear functional on E, which is continuous at 0. On the one dimensional space Y = span(x0)define f : Y R,↵x0 ↵µD(x0). ! 7! Then f(y) µ (y) for all y Y (if y = ↵x ,with↵>0 this follows from D 2 0 the positive homogeneity of µD, and if ↵<0 this is clear). By Theorem 1.4.2 we can extend f to a linear function F , defined on all of E,with F (x) µ (x) for all x E.Sinceµ is continous at 0 it follows F is D 2 D continuous at 0 and thus in E⇤. Moreover, if x C it follows that F (x0) > 1 supx C µD(x)) 1. If 2 2 K = C we first choose F , by considering E to be a real locally convex space, and then put f(x)=F (x) iF (ix). It is then easily checked that F is a complex linear bounded functional on E. 36 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Corollary 2.2.4. Assume that A and B are two convex closed subsets of of a locally convex space E, with for which there is an open neighborhood U of 0 with (A + U) (B + U)= . \ ; Then there is an x⇤ E⇤ and ↵ R so that 2 2 ( x⇤,x ) ↵ ( x⇤,y ), for all x A and y B. < h i C = A B = x y : x A and y B . 2 2 we note that 0 C is convex and that 62 Applying Theorem 2.2.3 we obtain an x X so that ⇤ 2 ⇤ sup ( x⇤,x ) < ( x⇤, 0 )=0. x C < h i < h i 2 But this means that for all x A and all y B ( x ,x y ) < 0 and thus 2 2 < h ⇤ i ( x⇤,x ) < ( x⇤,y ). < h i < h i An easy consequence of the geometrical version of the Hahn-Banach Theorem 2.2.3 is the following two observation. Proposition 2.2.5. If A is a convex subset of a Banach space X then w A = Ak·k. If a representation of the dual space of a Banach space X is not known, it might be hard to verify weak convergence of a sequence directly. The following Corollary of Proposition 2.2.5 states an equivalent criterium for a sequence to be weakly null without using the dual space of X. Corollary 2.2.6. For a bounded sequence (xn) in Banach space X it follows that (xn) is weakly null if and only if for all subsequences (zn),all">0 there is a convex combination z = k z of (z ) (i.e. 0,for j=1 j j j i i =1, 2,...,k,and l =1)sothat z " . j=1 j P k k P 2.2. GEOMETRIC VERSION OF THE HAHN-BANACH THEOREM 37 Exercises 1. Prove Proposition 2.2.5. 2. Let X be a Banach space with norm . Show that µ = . k·k BX k·k 3. Prove Corollary 2.2.6. 4. Show that `1 is not isomorphic to a subspace of c0. 5.⇤ Show that there is an x⇤ `⇤ so that 2 1 a) x = 1, k ⇤k b) x⇤,x =limi xi, for x =(xi) c = (⇠i):limi ⇠i exists h i !1 2 { !1 } c) If x =(⇠i) ` , and ⇠i 0, for i N,then x⇤,x 0, and 2 1 2 h i d) If x =(⇠i) ` and x0 =(⇠2,⇠3,...)then x⇤,x0 = x⇤,x 2 1 h i h i 38 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 2.3 Reflexivity and Weak Topology Proposition 2.3.1. If X is a Banach space and Y is a closed subspace of X, then (Y,Y )= (X, X ) Y , i.e. the weak topology on Y is the weak ⇤ ⇤ \ topology on X restricted to Y . Theorem 2.3.2. (Theorem of Alaoglu, c.f. [Fol, Theorem 5.18] ) BX⇤ is w⇤ compact for any Banach space X. Sketch of a proof. Consider the map :B⇤ K : x ,x⇤ (x⇤(x):x X). X ! { 2 | |k k} 7! 2 x X Y2 Then we check that is continuous with respect to w⇤ topology on BX⇤ and the product topology on K : x , has a closed image, x X { 2 | |k k} and is a homeorphism from B onto2 its image. QX⇤ Since by the Theorem of Tychano↵ K : x is compact, x X { 2 | |k k} (B ) is a compact subset, which yields2 (via the homeomorphism 1) X⇤ Q that BX⇤ is compact in the w⇤ topology. Theorem 2.3.3. (Theorem of Goldstein) BX is (via the canonical embedding) w⇤ dense in BX⇤⇤ . The proof follows easily from the following Lemma (see Problem 1). Lemma 2.3.4. Let X be a Banach space and let x X ,withx 1, ⇤⇤ 2 ⇤⇤ ⇤⇤ and x , x ,...,x X . Then 1⇤ 2⇤ n⇤ 2 ⇤ n 2 inf x⇤⇤,xi⇤ xi⇤,x =0. x 1 |h i h i| k k Xi=1 n 2 Proof. For x X put (x)= i=1 x⇤⇤,xi⇤ xi⇤,x and =infx BX (x), 2 |h i h i| 2 and choose a sequence (xj) B so that (xj) ,ifj . ⇢P X & %1 W.l.o.g we can also assume that ⇠i =limk xi⇤,xk exists for all i = !1h i 1, 2,...,n. For any t [0, 1] and any x BX we note for k N 2 2 2 ((1 t)x + tx) k n 2 = x⇤⇤,x⇤ (1 t) x⇤,x t x⇤,x |h i i h i ki h i i| Xi=1 2.3. REFLEXIVITY AND WEAK TOPOLOGY 39 n 2 = x⇤⇤,x⇤ x⇤,x + t x⇤,x x |h i i h i ki h i k i| i=1 Xn 2 = x⇤⇤,x⇤ x⇤,x |h i i h i ki| i=1 X n n 2 2 +2t x ,x x ,x x⇤,x x + t x⇤,x x⇤,x < h ⇤⇤ i⇤i h i⇤ kih i k i |h i ki h i i| ⇣ i=1 ⌘ i=1 X n X n 2 2 k +2t ( x⇤⇤,xi⇤ ⇠i)(⇠i xi⇤,x ) + t ⇠i xi⇤,x . ! !1 < h i h i | h i| i=1 i=1 ⇣ X =: i ⌘ X From the minimality of |it follows{z that} for all x B , that for all t>0 2 X the second term needs to be non negative, and, thus, n n ⇠ ( x⇤,x )withx⇤ := x⇤, < i i =lim (xk) k !1 40 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY n 2 =lim x⇤⇤,xi⇤ xi⇤,xk k |h i h i| !1 i=1 n X 2 = x⇤⇤,x⇤ ⇠ |h i i i| i=1 Xn = 2 | i| i=1 Xn = i i i=1 Xn = ( x⇤⇤,x⇤ ⇠ ) i h i i i i=1 X n = i( x⇤⇤,x⇤ ⇠i) (since R) < h i i 2 ⇣ i=1 ⌘ X n = x⇤⇤,x⇤ ⇠ Theorem 2.3.5. Let X be a Banach space. Then X is reflexive if and only if BX is compact in the weak topology. Proof. Let : X, X be the canonical embedding. ! ⇤⇤ “ ”IfX is reflexive and thus is onto it follows that is an home- ) omorphism between (BX , (X, X⇤)) and (BX⇤⇤ , (X⇤⇤,X⇤)). But by the Theorem of Alaoglu 2.3.2 (BX⇤⇤ , (X⇤⇤,X⇤)) is compact. “ ” Assume that B , (X, X ) is compact, and assume that x B ( X ⇤ ⇤⇤ 2 X⇤⇤ we need to show that there is an x B so that (x)=x , or equivalently 2 X ⇤⇤ that x ,x = x ,x for all x X . h ⇤ i h ⇤⇤ ⇤i ⇤ 2 ⇤ For any finite set A = x ,...,x X and for any ">0 we can, { 1⇤ n⇤ }⇢ ⇤ according to Lemma 2.3.4, choose an x B so that (A,") 2 X n 2 x⇤⇤ x ,x⇤ ". |h (A,") i i| Xi=1 The set I = (A,"):A X⇤ finite and ">0 , { ⇢ } 2.3. REFLEXIVITY AND WEAK TOPOLOGY 41 is directed via (A,") (A ,"): A A and " ".Thus,by 0 0 () ⇢ 0 0 compactness, the net x :(A,") I must have a subnet (z : j J) (A,") 2 j 2 which converges weakly to some element x B . 2 X We claim that x ,x = x ,x , for all x X .Indeed,letj i h ⇤ i h ⇤⇤ ⇤i ⇤ 2 ⇤ 7! j be the map from J to I, so that z = x , for all j J, and so that, for j ij 2 any i I there is a j with i i , for j j . Let x X and ">0. 0 2 0 j 0 0 ⇤ 2 ⇤ Put i =( x ,") I, choose j , so that i i , for all j j , and choose 0 { ⇤} 2 0 j 0 0 j J, j j , so that x z ,x <", for all j j . It follows therefore 1 2 1 0 |h j ⇤i| 1 that (note that for i =(A," ) it follows that x A and " ") j1 0 ⇤ 2 0 x⇤⇤ x, x⇤ x⇤⇤ xi ,x⇤ + zj x, x⇤ " + " =2". h i h j1 i h 1 i Since ">0 and x⇤ X ⇤ were arbitrary we deduce our claim. 2 Theorem 2.3.6. For a Banach space X the following are equivalent. 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 2.2.5 ) ⇢ yields that B = B Y is a (X, X )-closed and, thus, (X, X )-compact Y X \ ⇤ ⇤ subset of BX . 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 (Y,Y ) ⇤ 2 ⇤ ⇤ ⇤ is the restriction of (X, X⇤) to the subspace Y .Thus,BY is (Y,Y ⇤)- compact, which implies, by Theorem 2.3.5 that Y is reflexive. “(a) (b)” If X is reflexive then (X ,X )= (X ,X). Since by the ) ⇤ ⇤⇤ ⇤ Theorem of Alaoglu 2.3.2 BX⇤ is (X⇤,X)-compact the claim follows from Theorem 2.3.5. “(c) (a)” clear. ) “(b) (a)” If X is reflexive, then, by “(a) (b) ” X is also reflexive and ) ⇤ ) ⇤⇤ thus, the implication “(a) (c)” yields that X is reflexive. ) Similar ideas as in the proof of Theorem 2.3.3 are used to show the following result which characterizes when a Banach space X is a dual space of another space. Theorem 2.3.7. Assume that X is a Banach space and Z is a closed sub- space of X⇤,sothatBX is compact in the topology (X, Z),andsothat x =supz B z(x) . k || 2 Z | | 42 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Then Z⇤ is isometrically isomorphic to X and the map T : X Z⇤,x f , with f (z)= z,x ,forx X and z Z, ! 7! x x h i 2 2 is an isometrical isomorphism onto Z⇤. Proof. We first note that T (BX )is (Z⇤,Z)denseinBZ⇤ .Indeed,ifthis is not true we can apply the Geometric Hahn Banach Theorem for lo- cally convex spaces (Theorem 2.2.3) applied to the locally convex space Z⇤, (Z⇤,Z) whose dual is by Proposition 2.1.6 Z, (Z, Z⇤) , and obtain elements z S and z S so that ⇤ 2 Z⇤ 2 Z 1= z =supx, z < z⇤,z =1, k k x B h i h i 2 X which is a contradiction. Secondly, our assumption says that T (BX )is (Z⇤,Z)-compact. To see that note that if (xi)i I is a net in X and z⇤ Z⇤,then 2 2 (fx )i I converges to z⇤ with respect to (Z⇤,Z) i 2 lim xi,z = z⇤,z for all z Z () i I h i h i 2 2 z⇤ T (BX ) and (X, Z⇤) lim xi,z = z⇤ (By assumption). () 2 i I h i 2 Exercises 1. Show Theorem 2.3.3 using Lemma 2.3.4. 2. Prove Proposition 2.3.1. 3. Show that B`⇤ is not sequentially compact in the w⇤-topology. 1 Hint: Consider the unit vector basis of `1 seen as subsequence of B`⇤ . 1 4. 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⇤ 2 i I there is a j i,with x c,wheneverj j . 2 0 k jk 0 5. Show that in each infinite dimensional Banach space X there is a weakly null net in SX . 2.3. REFLEXIVITY AND WEAK TOPOLOGY 43 6.⇤ Prove that every weakly null sequence in `1 is norm null. Hint: Assume that (x ) S is weakly null. Then there is a subse- n ⇢ `1 quence xn and a block sequence (zk) so that limk xn zk = 0. k !1 k k k Here we mean by a block sequence a sequence (zn)in`1 of the form zn =(0, 0, 0 ...0,⇣(nk 1 + 1),⇣(n, nk 1 + 2)),...,⇣(n, nk), 0, 0,...), nk 1times | {z } where 1 = n1 2.4 Annihilators, Complemented Subspaces Definition 2.4.1. (Annihilators, Pre-Annihilators) Assume X is a Banach space. Let M X and N X . We call ⇢ ⇢ ⇤ M ? = x⇤ X⇤ : x M x⇤,x =0 X⇤, { 2 8 2 h i }⇢ the annihilator of M and N = x X : x⇤ N x⇤,x =0 X, ? { 2 8 2 h i }⇢ the pre-annihilator of N. Proposition 2.4.2. Let X be a Banach space, and assume M X and ⇢ N X . ⇢ ⇤ a) M ? is a closed subspace of X⇤, M ? =(span(M))?,and(M ?) = span(M), ? b) N is a closed subspace of X, N = (span(N)) ,andspan(N) ? ? ? ⇢ (N )?. ? c) span(M)=X M = 0 . () ? { } Proposition 2.4.3. If X is Banach space and Y X is a closed subspace ⇢ then (X/Y )⇤ is isometrically isomorphic to Y ? via the operator :(X/Y )⇤ Y ?, with (z⇤)(x)=z⇤(x). ! (recall x := x + Y X/Y for x X). 2 2 Proof. Let Q : X X/Y be the quotient map. ! For z (X/Y ) , (z ), as defined above, can be written as (z )= ⇤ 2 ⇤ ⇤ ⇤ z Q.Thus (z ) X .SinceQ(Y )= 0 it follows that (z ) Y . ⇤ ⇤ 2 ⇤ { } ⇤ 2 ? For z (X/Y ) we have ⇤ 2 ⇤ (z⇤) =supz⇤,Q(x) =supz⇤, x = z⇤ (X/Y )⇤ , k k x B h i x B h i k k 2 X 2 X/Y where the second equality follows on the one hand from the fact that Q(x) k k x , for x X, and on the other hand, from the fact that for any x = x+Y k k 2 2 X/Y there is a sequence (yn) Y so that lim supn x + yn = x . ⇢ !1 k k k k Thus is an isometric embedding. If x Y X ,wedefine ⇤ 2 ? ⇢ ⇤ z⇤ : X/Y K,x+ Y x⇤,x . ! 7! h i 2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES 45 First note that this map is well defined (since x ,x+ y = x ,x+ y for h ⇤ 1i h ⇤ 2i y ,y Y ). Since x is linear, z is also linear, and z , x = x ,x , for 1 2 2 ⇤ ⇤ |h ⇤ i| |h ⇤ i| all x X, and thus z = x . Finally, since 2 k ⇤k(X/Y )⇤ k ⇤k (z⇤),x = z⇤,Q(x) = x⇤,x , h i h i h i it follows that (z⇤)=x⇤, and thus that is surjective. Proposition 2.4.4. Assume X and Y are Banach spaces and T L(X, Y ). 2 Then (2.2) T (X)? = (T ⇤) and T (Y ) (T )? N ⇤ ⇤ ⇢N (2.3) T (X)= (T ⇤) and T ⇤(Y ⇤) = (T ). N ? ? N Proof. We only prove (2.2). The verification of (2.3) is similar. For y Y ⇤ 2 ⇤ y⇤ T (X)? x X y⇤,T(x) =0 2 () 8 2 h i x X T ⇤(y⇤),x =0 () 8 2 h i T ⇤(y⇤)=0 y⇤ (T ⇤), () () 2N which proves the first part of (2.2), and for y Y and all x (T ), ⇤ 2 ⇤ 2N it follows that T (y ),x = y ,T(x) = 0, which implies that T (Y ) h ⇤ ⇤ i h ⇤ i ⇤ ⇤ ⇢ (T ) , and, thus, T (X ) (T ) since , (T ) is closed. N ? ⇤ ⇤ ⇢N ? N ? Definition 2.4.5. Let X be a Banach space and let U and V be two closed subspaces of X. We say that X is the complemented sum of U and V and we write X = U V , if for every x X there are u U and v V , so that 2 2 2 x = u + v and so that this representation of x as sum of an element of U and an element of V is unique. We say that a closed subspace Y of X is complemented in X if there is a closed subspace Z of X so that X = Y Z. Remark. Assume that the Banach space X is the complemented sum of the two closed subspaces U and V . We note that this implies that U V = 0 . \ { } We can define two maps P : X U and Q : X V ! ! where we define P (x) U and Q(x) V by the equation x = P (x)+Q(y), 2 2 with P (x) U and Q(x) V (which, by assumption, has a unique solution). 2 2 Note that P and Q are linear. Indeed if P (x1)=u1, P (x2)=u2, Q(x1)=v1, 46 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Q(x2)=v2, then for , µ K we have x1 +µx2 = u1 +µu2 + v1 +µv2, and 2 thus, by uniqueness P ( x1+µx2)= u1+µu2, and Q( x1+µx2)= v1+µv2. Secondly it follows that P P = P , and Q Q = Q. Indeed, for any x X we we write P (x)=P (x)+0 U + V , and since this representation 2 2 of P (x) is unique it follows that P (P (x)) = P (x). The argument for Q is the same. Finally it follows that, again using the uniqueness argument, that P is the identity on U and Q is the identity on V . We therefore proved that a) P is linear, b) the image of P is U c) P is idempotent, i.e. P 2 = P We say in that case that P is a linear projection onto U. Similarly Q is a a linear projection onto V , and P and Q are complementary to each other, meaning that P (X) Q(X)= 0 and P +Q = Id. A linear map P : X X \ { } ! with the properties (a) and (c) is called projection. The next Proposition will show that P and Q as defined in above remark are actually bounded. Lemma 2.4.6. Assume that X is the complemented sum of two closed sub- spaces U and V . Then the projections P and Q as defined in above remark are bounded. Proof. Consider the norm on X defined by ||| · ||| x = P (x) + Q(x) , for x X. ||| ||| k k k k 2 We claim that (X, ) is also a Banach space. Indeed if (x ) X with ||| · ||| n ⇢ 1 1 1 x = P (x ) + Q(x ) < . ||| n||| k n k k n k 1 n=1 n=1 n=1 X X X Then u = 1 P (x ) U, v = 1 Q(x ) V (U and V are assumed n=1 n 2 n=1 n 2 to be closed) converge in U and V with respect to ,respectively.Since P P k·k also x = 1 x converges with respect to and k·k|||·||| n=1 n k·k n n n 1 P x = xn =lim P (xj)+Q(xj) =lim P (xj)+ lim Q(xj)=u+v, n n n n=1 !1 j=1 !1 j=1 !1 j=1 X X X X 2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES 47 and n n n x x = u P (x )+v Q(x ) n n n Xj=1 Xj=1 Xj=1 n n = u P (xn) + v Q(xn) n 0, ! !1 j=1 j=1 X X (here all series are meant to converge with respect to ) which proves that k·k (X, ) is complete. ||| · ||| Since the identity is a bijective linear bounded operator from (X, )to ||| · ||| (X, ) it has by Corollary 1.3.6 of the Closed Graph Theorem a continuous k·k inverse and is thus an isomorphy. Since P (x) x and Q(x) x k k||| ||| k k||| ||| we deduce our claim. Proposition 2.4.7. Assume that X is a Banach space and that P : X X, ! is a bounded projection onto a closed subspace of X. Then X = P (X) (P ). N Theorem 2.4.8. There is no linear bounded operator T : ` ` so that 1 ! 1 the kernel of T equals to c0. Corollary 2.4.9. c0 is not complemented in ` . 1 Proof of Theorem 2.4.8. For n N we let e⇤ be the n-th coordinate func- 2 n tional on ` ,i.e. 1 en⇤ : ` K,x=(xj) xn. 1 ! 7! Step 1. If T : ` ` is bounded and linear, then 1 ! 1 1 (T )= (e⇤ T ). N N n n=1 \ Indeed, note that x (T ) n N e⇤ (T (x)) = e⇤ ,T(x) =0. 2N () 8 2 n h n i In order to prove our claim we will show that c0 cannot be the intersection of the kernel of countably many functionals in `⇤ . 1 Step 2. There is an uncountable family (N↵ : ↵ I) of infinite subsets of N 2 for which N N is finite whenever ↵ = are in I. ↵ \ 6 48 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Write the rational numbers Q as a sequence (qj : j N), and choose for 2 each r R asequence(nk(r):k N), so that (qn (r) : k N) converges to 2 2 k 2 r. Then, for r R let Nr = nk(r):k N . The family (Nr : r R)then 2 { 2 } 2 satisfies the claim in Step 2. For i I,putx↵ =1N↵ ` ,i.e. 2 2 1 (↵) (↵) 1ifk N↵ x↵ =(⇠k : k N)with⇠k = 2 2 (0ifk N↵. 62 Step 3. If f `⇤ and c0 (f)then ↵ I : f(x↵) =0 is countable. 2 1 ⇢N { 2 6 } In order to verify Step 3 let An = ↵ : f(x↵) 1/n , for n N. { | | } 2 It is enough to show that for n N the set An is finite. To do so, let 2 k ↵1,↵2,...,↵k be distinct elements of An and put x = j=1 sign f(x↵j ) x↵j (for a C we put sign(a)=a/ a ) and deduce that f(x) k/n. Now 2 | | P consider M = N N .ThenN M is infinite, and thus it follows j ↵j \ i=j ↵i ↵j \ j for 6 S k x˜ = sign(f(x↵j ))1Mj Xj=1 that f(x)=f(˜x)(sincex x˜ c ). Since the M , j =1, 2,...,k are 2 0 j pairwise disjoint, it follows that x˜ = 1, and thus k k1 k f(x)=f(˜x) f . n k k Which implies that that A can have at most n f elements. n k k Step 4. If c0 n1=1 (fn), for a sequence (fn) `⇤ , then there is an ↵ I ⇢ N ⇢ 1 2 so that x↵ 1 (fn). In particular this implies that c0 = (fn). 2 Tn=1 N 6 n N N Indeed, Step 3 yields that 2 T T C = ↵ I : fn(x↵) = 0 for some n N = ↵ I : fn(x↵) =0 , { 2 6 2 } { 2 6 } n N [2 is countable, and thus I C is not empty. \ Remark. Assume that Z is any subspace of ` which is isomorphic to c0, 1 then Z is not complemented. The proof of that statement is a bit harder. Theorem 2.4.10. [So] Assume Y is a subspace of a separable Banach space X and T : Y c is linear and bounded. Then T can be extended to a ! 0 linear and bounded operator T˜ : X c . Moreover, T˜ can be chosen so that ! 0 T˜ 2 T . k k k k 2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES 49 Corollary 2.4.11. Assume that X is a separable Banach space which con- tains a subspace Y which is isomorphic to c0. Then Y is complemented in X. Proof. Let T : Y c be an isomorphism. Then extend T to T˜ : X c ! 0 ! 0 and put P = T 1 T˜. Proof of Theorem 2.4.10. Note that an operator T : Y c is defined by a ! 0 (Y ,Y)nullsequence(y ) Y ,i.e. ⇤ n⇤ ⇢ ⇤ T : Y c0,y y⇤,y : n N . ! 7! h n i 2 We would like to use the Hahn Banach theorem and extend each yn⇤ to an element x X ,with y = x , and define n⇤ 2 n⇤ k n⇤k k n⇤ k T˜(x):=(x⇤ ,x : n N),xX. h n i 2 2 But the problem is that (xn⇤ ) might not be (X⇤,X) convergent to 0, and thus we can only say that ( xn⇤ ,x : n N) ` , but not necessarily in c0. h i 2 2 1 Thus we will need to change the xn⇤ somehow so that they are still extensions of the yn⇤ but also (X⇤,X)null. Let B = T B . B is (X ,X)-compact and metrizable (since X k k X⇤ ⇤ is separable). Denote the metric which generates the (X⇤,X)-topology by d( , ). Put K = B Y .SinceY X is (X ,X)-closed, K is (X ,X)- · · \ ? ? ⇢ ⇤ ⇤ ⇤ compact . Also note that every (X⇤,X)-accumulation point of (xn⇤ )liesin K. Indeed, this follows from the fact that xn⇤ (y)=yn⇤(y) n 0, for all ! !1 y Y . This implies that limn d(xn⇤ ,K) = 0, thus we can choose (zn⇤) K 2 !1 ⇢ so that limn d(xn⇤ ,zn⇤) = 0, and thus (xn⇤ zn⇤)is (X⇤,X)-null and for !1 y Y it follows that x⇤ z⇤,y = x⇤ ,y , n N. Choosing therefore 2 h n n i h n i 2 ˜ T : X c0,x( x⇤ z⇤,x : n N), ! 7! h n n i 2 yields our claim. Remark. Zippin [Zi] proved the converse of Theorem: if Z is an infinite- dimensional separable Banach space admitting a projection from any sepa- rable Banach space X containing it, then Z is isomorphic to c0. Exercises 1. Prove Proposition 2.4.2. 2. a) Assume that ` isomorphic to a subspace Y of some Banach 1 space X,thenY is complemented in X. 50 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY b) Assume Z is a closed subspace of a Banach space X, and T : Z ! ` is linear and bounded. Then T can be extended to a linear 1 and bounded operator T˜ : X ` ,with T˜ = T . ! 1 k k k k 3. Show that for a Banach space X, the dual space X⇤ is isometrically isomorphic to a complemented subspace of X⇤⇤⇤, via the canonical embedding. 4. Prove Proposition 2.4.7. 5. Prove (2.3) in Proposition 2.4.4. 2.5. THE THEOREM OF EBERLEIN SMULIAN 51 2.5 The Theorem of Eberlein Smulian For infinite dimensional Banach spaces the weak topology is not metrizable (see Exercise 1). Nevertheless compactness in the weak topology can be characterized by sequences. Theorem 2.5.1. (The Theorem of Eberlein- Smulian) Let X be a Banach space. For subset K the following are equivalent. (X,X⇤) a) K is relatively (X, X⇤) compact, i.e. K is compact. b) Every sequence in K contains a (X, X⇤)-convergent subsequence. c) Every sequence in K has a (X, X⇤)-accumulation point. We will need the following Lemma. Lemma 2.5.2. Let X be a Banach space and assume that there is a count- able set C = xn⇤ : n N BX ,sothatC = 0 .Inthatcasewesay { 2 }⇢ ⇤ ? { } that C is total for X. Consider for x, y 1 n d(x, y)= 2 x⇤ ,x y . |h n i| n=1 X Then d is a metric on X,andforany (X, X⇤)-compact set K, (X, X⇤) coincides on K with the metric generated by d. The proof of Lemma 2.5.2 goes along the lines of Exercise 1 in this section. Lemma 2.5.3. Assume that X is separable. Then there is a countable total set C X . ⇢ ⇤ Proof. Let D X be dense, and choose by the Corollary 1.4.6 of the The- ⇢ orem of Hahn Banach for each element x D, an element y S so 2 x⇤ 2 X⇤ that y ,x = x .PutC = y : x D .Ifx X, x = 0, is arbi- h x⇤ i k k { x⇤ 2 } 2 6 trary then there is a sequence (xk) D, so that limk xk = x, and thus ⇢ !1 limk yx⇤ ,x = x > 0. Thus there is an x⇤ C so that x⇤,x = 0, !1h k i k k 2 h i6 which implies that C is total. 52 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Proof of Theorem 2.5.1. “(a) (b)” Assume that K is (X, X )-compact ) ⇤ (if necessary, pass to the closure) and let (x ) K be a sequence, and put n ⇢ X0 = span(xn : n N). X0 is a separable Banach space. By Proposition 2 2.2.5 the topology (X0,X0⇤) coincides with the restriction of (X, X⇤)to X .Thus,K = K X is (X ,X )-compact. Since X is separable, by 0 0 \ 0 0 0⇤ 0 Lemma 2.5.3 there exists a countable set C BX , so that C = 0 . ⇢ 0⇤ ? { } It follows therefore from Lemma 2.5.2 that (K , (X ,X ) K )is 0 0 0⇤ \ 0 metrizable and thus (xn) has a convergent subsequence in K0. Again, using the fact that on X0 the weak topology coincides with the weak topology on X, we deduce our claim. “(b) (c)” clear. ) “(c) (a)” Assume K X satisfies (c). We first observe that K is (norm) ) ⇢ bounded. Indeed, for x⇤ X⇤,thesetAx = x⇤,x : x K K is 2 ⇤ {h i 2 }⇢ the continuous image of A (under x⇤) and thus has the property that every sequence has an accumulation point in K. This implies that Ax⇤ is bounded in K for all x⇤ X⇤, but this implies by the Banach Steinhaus Theorem 2 1.3.8 that A X must be bounded. ⇢ Let : X, X⇤⇤ be the canonical embedding. By the Theorem of ! (X⇤⇤,X⇤) Alaoglu 2.3.2, it follows that (K) is (X⇤⇤,X⇤)-compact. There- (X ,X ) fore it will be enough to show that (K) ⇤⇤ ⇤ (X) (because this ⇢ would imply that every net ( (x ):i I) (K) has a subnet which i 2 ⇢ ( (X),X⇤)) converges to some element (x) (X)). (X⇤⇤,X⇤) 2 So, fix x⇤⇤ (K) . Recursively we will choose for each k N, 0 2 2 xk K, and for each k N afinitesetA⇤ SX , so that 2 2 k ⇢ ⇤ 1 x⇤⇤ (x ),x⇤ < for all x⇤ A⇤, if k 1,(2.4) h 0 k i k 2 j 0 j Assuming that x1,x2,...,xk 1 and A0⇤,A1⇤,...,Ak⇤ 1 have been chosen for some k>1, we can first choose x K so that (2.4) is satisfied (since k 2 A is finite for j =1, 2,...,k 1), and then, since span(x , (x ),j k)is j⇤ 0⇤⇤ j a finite dimensional space we can choose A S so that (2.5) holds. k⇤ ⇢ X⇤ By our assumption (c)thesequence(xk) has an (X, X⇤)- accumulation point x0. By Proposition 2.3.1 it follows that (X,X⇤) x0 Y = span(xk : k N)k·k = span(xk : k N) . 2 2 2 2.5. THE THEOREM OF EBERLEIN SMULIAN 53 We will show that x0⇤⇤ = (x0) (which will finish the proof ). First note that for any x⇤ j N Aj⇤ 2 2 S x0⇤⇤ (x0),x⇤ lim inf x0⇤⇤ (xk),x⇤ + x⇤,xk x0 =0. h i k h i h i !1 Secondly consider the space Z = span(x⇤⇤, (xk),k N)k·k X⇤⇤ it follows 0 2 ⇢ from (2.5) that the set of restrictions of elements of k1=1 Ak⇤ to Y is total in Z and thus that S 1 x⇤⇤ (x ) Z A⇤ = 0 , 0 0 2 \ k { } ⇣ k[=1 ⌘? which implies our claim. Exercises 1. Prove that if X is a separable Banach space (BX⇤ , (X⇤,X)) is metriz- able. 2. For an infinite dimensional Banach space prove that (X, (X, X⇤)) is not metrizable. Hint: Exercise 4 in Section 2.3. 3. Prove that for two Banach spaces X and Y , the adjoint of a lin- ear bounded operator T : X Y is w -continuous (i.e (Y ,Y) ! ⇤ ⇤ (X⇤,X)-continuous). 4. Show that `1 isometric to a subspace of C[0, 1]. 5.⇤ Show that `1 is not complemented in C[0, 1]. You might need to prove the following two things: a) There is no separable complemented subspace X of ` which con- 1 tains c0 (this follows from the fact that c0 is not complemented in ` 1 but complemented in each separable super space) b) Every separable sub space X of M[0, 1] sits inside another separable subspace Y of M[0, 1] which is complemented in M[0, 1]. Idea for b) take dense sequence (µ )in S take ⌫ = 2 n mu then n X | n| prove that L1(⌫) is complemented in X. P 54 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 2.6 Characterizations of Reflexivity by Pt´ak We present several characterization of the reflexivity of a Banach space, due to Pt´ak[Ptak]. We assume in this section that our Banach spaces are defined over the real field R. Theorem 2.6.1. The following conditions for a Banach space X are equiv- alent 1. X is not reflexive. 2. For each ✓ (0, 1) there are sequences (x ) B and (x ) 2 i i1=1 ⇢ X i⇤ i1=1 ⇢ BX⇤ ,sothat ✓ if j i,and (2.6) xj⇤(xi)= (0 if j>i. 3. For some ✓>0 there are sequences (x ) B and (x ) B , i i1=1 ⇢ X i⇤ i1=1 ⇢ X⇤ for which (2.6) holds. 4. For each ✓ (0, 1) there is a sequence (x ) B ,sothat 2 i i1=1 ⇢ X (2.7) dist conv(x ,x ,...,x ), conv(x ,x ,...) ✓. 1 2 n n+1 n+2 5. For some ✓>0 there is a sequence (x ) B ,sothat(2.7) holds. i i1=1 ⇢ X For the proof we will need Helly’s Lemma. Lemma 2.6.2. Let Y be an infinite-dimensional normed linear space y1⇤, y ,...,y Y , M>0 and let c ,c ,...,c be scalars. 2⇤ n⇤ 2 ⇤ 1 2 n The following are equivalent (M) The Moment Condition For all ">0 there exists y Y with 2 y = M + " and y⇤(y)=c for k =1, 2,...,n. k k k k (H) Helly’s Condition n n n a c M a y⇤ for any sequence (a ) of scalars. j j j j j j=1 j=1 j=1 X X 2.6. CHARACTERIZATIONS OF REFLEXIVITY BY PTAK´ 55 Proof. “(M) (H)”. Let ">0 and assume y Y satisfies the condition ) 2 in (M). Then n n n n a c = a y⇤(y) y a y⇤ =(M + ") a y⇤ , j j j j k k· j j j j j=1 j=1 j=1 j=1 X X X X which implies (H), since ">0 was arbitrary. “(H) (M)” Assume (H) and let ">0. We can assume that not all the ) yk⇤ are vanishing (otherwise also all the ck have to be equal to 0, and any y y ,y ,...y with y = M + " will satisfy the conditions in (M)). 2{ 1⇤ 2⇤ n⇤}? k k We can also, for the same reason assume that not all cj’s vanish. Secondly, we can assume, after reordering the y , that for some k 1, 2,...,n the j⇤ 2{ } sequence (y )k is linear independent and y ,y ,...,y span(y : j⇤ j=1 k⇤+1 k⇤+2 n⇤ 2 j⇤ j =1, 2,...,k). This implies that if we have a y Y ,with y = M + " 2 k k and yj⇤(y)=cj, for j =1, 2,...,k, then it also follows that yj⇤(y)=cj, for (j) j = k+1,k+2,...,n. Indeed, for j = k+1,k+2,...,n, choose scalars (ai : k (j) i =1, 2,...,k) so that yj⇤ = i=1 ai yi⇤, for j = k+1,k+2,...,n. Now, the k (j) inequality in (H) implies thatP cj = i=1 ai ci, for j = k +1,k+2,...,n. Indeed, choose a = 1, a = 0, if i k+1,k+2,...,n j and a = a(j), j i 2P{ }\{ } i i if i 1, 2,...,k , which implies that the right hand of the equation in (M) 2{ } vanishes. This yields that yj⇤(y)=cj, for j = k +1,k+2,...,n. We can therefore restrict ourselves to satisfy the second condition in (H) for all j =1, 2,...,k. Define for j =1, 2,...,k the a ne subspace H = y Y : y (y)=c .ThenG = k H is an a ne subspace of j { 2 j⇤ j} j=1 j codimension k. Note, that if we pick y G,thenG = y + G0,whereG0 is 2 T the closed subspace k G = y Y : y⇤(y)=0 . 0 { 2 j } j\=1 We need to show that (2.8) N := inf y : y G M. k k 2 Then our claim would follow, since the Intermediate Value implies that there must be some y in G for which N (2.9) span(G)= ry : y G and r R . { 2 2 } We note that g 1. Indeed, otherwise choose a sequence (y ) G, k ⇤k n ⇢ with limn yn = N, and note that !1 k k N = g⇤(yn) g⇤ yn n g⇤ N Proof of Theorem 2.6.1. “(i) (ii)” Assume that X is not reflexive and that ) ✓ (0, 1). Since X is a closed subspace of X (as usual we are identifying X 2 ⇤⇤ with its image under the canonical embedding into X⇤⇤), there is a functional x X that x = 1, x 0 and x (x ) >✓for some ⇤⇤⇤ 2 ⇤⇤⇤ k ⇤⇤⇤k ⇤⇤⇤|X ⌘ ⇤⇤⇤ ⇤⇤ x X ,with x < 1 (see Exercice 1). ⇤⇤ 2 ⇤⇤ k ⇤⇤k Now we will choose inductively xn BX and x⇤ BX , n N, at each 2 n 2 ⇤ 2 step assuming that the condition (2.6) holds up to n, and additionally, that x⇤⇤(xn⇤ )=✓. For n = 1 we simply choose x S so that x (x )=✓ and then we 1⇤ 2 X⇤ ⇤⇤ 1⇤ choose x B so that x (x )=✓. Assuming we have chosen x ,x ,...,x 1 2 X 1⇤ 1 1 2 n 2.6. CHARACTERIZATIONS OF REFLEXIVITY BY PTAK´ 57 and x1⇤,x2⇤,...,xn⇤ so that ✓ if j i n, and (2.10) xj⇤(xi)= (0ifi a ✓ = M a x⇤⇤⇤(x⇤⇤) | n+1| | n+1 | n n = M x⇤⇤⇤ a x + a x⇤⇤ M a x⇤⇤⇤(x )+a x⇤⇤ . j j n+1 j j n+1 ⇣ j=1 ⌘ j=1 X X We can therefore choose an x X , x 1 so that x (x ) = 0 for n⇤+1 2 ⇤ k n⇤+1k n⇤+1 j all j =1, 2,...,n and x⇤⇤(xn⇤+1)=✓. Secondly, we note that the functionals x1⇤,x2⇤,...,xn⇤+1,thenumbers ✓,✓,...,✓, and the number M = x < 1 satisfy Helly’s condition. Indeed, k ⇤⇤k for scalars a1,...,an+1 we have n+1 n+1 n+1 a ✓ = a x⇤⇤(x⇤) M a x⇤ . j j j j j j=1 j=1 j=1 X X X We can therefore find x B , so that x (x )= ✓, for all j = n+1 2 X j⇤ n+1 1, 2,...,n. “(ii) (iv)” and “(iii) (v)” Fix a ✓ (0, 1) for which there are sequences ) ) 2 (x ) B and (x ) B for which (2.6) holds. Let x = n a x j ⇢ X j⇤ ⇢ X⇤ j=1 j j 2 conv(x1,x2,...,xn) and z = 1 bjxj conv(xn+1,xn+2,...)then j=n+1 2 P z x Px⇤ (z x)=x⇤ (y)=✓, k k n+1 n+1 which implies our claim. “(iv) (v)” obvious. ) “(v) (i)” Assume that for ✓>0 and the sequence (x ) B satisfies (2.7). ) j ⇢ X Now assume that our claim is false and X is reflexive. Define Cn = conv(xj : j n + 1), for n N,thenthesetsCn, n N, are 2 2 weakly compact, C1 C2 .... Thus there is an element v n N Cn.We 2 2 can approximate v by some u conv(xj : j N), with u v <✓/2. There 2 2 k k T is some n so that v conv(x ,...,x ). But now it follows, since u C , 2 1 n 2 n+1 that dist conv(x ,...,x ), conv(x ,x ,...) v u <✓/2, which is 1 n n+1 n+2 k k a contradiction and finishes the proof. 58 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Exercises 1. Assume that X is not reflexive and 0 < ⇥ < 1. Then there exists an x X , x = 1, and x X , x < 1, so that ⇤⇤⇤ 2 ⇤⇤⇤ k ⇤⇤⇤k ⇤⇤ 2 ⇤⇤ k ⇤⇤k x (x ) > ⇥, and X 0. ⇤⇤⇤ ⇤⇤ ⇤⇤⇤|X ⌘ 2.7 The Principle of Local Reflexivity In this section we proof a result by J. Lindenstrauss and H. Rosenthal [LR] which states that for a Banach space X the finite dimensional subspaces of the bidual X⇤⇤ are in a certain sense have “similar” finite dimensional subspaces of X. Theorem 2.7.1. [LR] [The Principle of Local Reflexivity] Let X be a Banach space and let F X and G X be finite dimensional ⇢ ⇤⇤ ⇢ ⇤ subspaces of X⇤⇤ and X⇤ respectively. Then, given ">0, there is a subspace E of X containing F X (we \ identify X with its image under the canonical embedding) with dim E = dim F and an isomorphism T : F E with T T 1 1+" such that ! k k·k k (2.11) T (x)=x if x F X and 2 \ (2.12) x⇤,T(x⇤⇤) = x⇤⇤,x⇤ if x⇤ G, x⇤⇤ F. h i h i 2 2 We need several Lemmas before we can prove Theorem 2.7.1. The first one is a corollary of the Geometric Hahn-Banach Theorem Proposition 2.7.2. (Variation of the Geometric Version of the Theorem of Hahn Banach) Assume that X is a Banach space and C X is convex with C = and ⇢ 6 ; let x X C (so x could be in the boundary of C). Then there exists an 2 \ x⇤ X⇤ so that 2 0 x⇤,z < x⇤,x for all z C , Proof. We first show that there is an x X so that T (x)=y. Assume this 2 where not true, then we could find by the Hahn-Banach Theorem (Corollary 1.4.5) an element y Y , so that y (z) = 0, for all z T (X) and y ,y =1 ⇤ 2 ⇤ ⇤ 2 h ⇤ i (T (X) is closed). But this yields T (y ),x = y ,T(x) = 0, for all x X, h ⇤ ⇤ i h ⇤ i 2 and, thus, T ⇤(y⇤) = 0. Thus 0= x⇤⇤,T⇤(y⇤) = T ⇤⇤(x⇤⇤),y⇤ = y, y⇤ =1, h i h i h i which is a contradiction. Secondly, assume that y T (X) T (B ). Since T is surjective onto 2 \ X its (closed) image Z = T (X) it follows from the Open Mapping Theorem that T (BX ) is open in Z, and we can use variation of the geometric version of the Hahn-Banach Theorem, Proposition (2.7.2), and chose z Z ,so ⇤ 2 ⇤ that z ,T(x) < 1= z ,y for all x B . Again by the Theorem of h ⇤ i h ⇤ i 2 X Hahn-Banach (Corollary 1.4.4) we can extend z⇤ to an element y⇤ in Y ⇤.It follows that T ⇤(y⇤) =supT ⇤(y⇤),x =supz⇤,T(x) 1, k k x B h i x B h i 2 X 2 X and thus, since x < 1, it follows that k ⇤⇤k y⇤,y = y⇤,T⇤⇤(x⇤⇤) = x⇤⇤,T⇤(y⇤) < 1, |h i| |h i| |h i| which is a contradiction. Lemma 2.7.4. Let T : X Y be a bounded linear operator between two ! Banach spaces X and Y with closed range, and assume that F : X Y has ! finite rank. Then T + F also has closed range. Proof. Assume the claim is not true. Put S = T + K and consider the map S : X/ (S) Y, x + (S) S(x) N ! N ! which is a well defined linear bounded Operator, and which by Proposition 1.3.11 cannot be an isomorphism onto its image. Therefore we can choose sequence (z )inX/ (S), with z = 1 and n N k nk xn zn,with1 xn 2, for n N, so that 2 k k 2 lim S(zn)= lim S(xn) = 0 and dist(xn, (S)) 1. n n !1 !1 N 60 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Since the sequence (F (xn):n N) is a bounded sequence in a finite 2 dimensional space, we can, after passing to a subsequence, assume that (F (xn):n N) converges to some y Y and, hence, 2 2 lim T (xn)= y. n !1 Since T has closed range there is an x X, so that T (x)= y. Using again 2 the equivalences in Proposition 1.3.11 and the fact that T (x ) y = T (x), n ! if n , it follows for some constant C>0 that %1 lim dist x xn, (T ) lim C T (x xn) =0, n n !1 N !1 and, thus, y F (x)= lim F (xn) F (x) F ( (T )), n !1 2 N so we can write y F (x) as y F (x)=F (u), where u (T ). 2N Thus lim dist(xn x u, (T )) = 0 and lim F (xn) F (x) F (u) =0. n n !1 N !1 k k F (T ) has also closed range, Proposition 1.3.11 yields (C being some posi- |N tive constant) lim sup dist(xn x u, (F ) (T )) lim sup C F (xn) F (x) F (u) =0. n N \N n k k !1 !1 Since T (x+u)= y = F (x+u) (by choice of u), and thus (T +F )(x+u)= 0 which means that x + u (T + F ). Therefore 2N lim sup dist(xn, (T + F )) = lim sup dist(xn x u, (T + F )) n N n N !1 !1 lim sup dist(xn x u, (T ) (F )) = 0. n N \N !1 But this contradicts our assumption on the sequence (xn). Lemma 2.7.5. Let X be a Banach space, A =(ai,j)i m,j n an m be n matrix and B =(bi,j)i p,j n a p by n matrix, and assume that B has only real entries (even if K = C). 2.7. THE PRINCIPLE OF LOCAL REFLEXIVITY 61 Suppose that y1,...,ym X, y⇤,...,y⇤ X⇤, ⇠1,...,⇠p R,and 2 1 p 2 2 x⇤⇤,...,x⇤⇤ B satisfy the following equations: 1 n 2 X⇤⇤ n (2.13) ai,jxj⇤⇤ = yi, for all i =1, 2,...,m,and Xj=1 n (2.14) yi⇤, bi,jxj⇤⇤ = ⇠i, for all i =1, 2,...,p. D Xj=1 E Then there are vectors x ,...,x B satisfying: 1 n 2 X n (2.15) ai,jxj = yi, for all i =1, 2,...,m,and Xj=1 n (2.16) yi⇤, bi,jxj = ⇠i, for all i =1, 2,...,p. D Xj=1 E Proof. Recall from Linear Algebra that we can write the matrix A as a product A = U P V ,whereU and V are invertible and P is of the form I 0 P = r , 00 ✓ ◆ r where r is the rank of A and Ir the identity on K . For a general s by t matrix C =(ci,j)i s,j t consider the operator t t s TC : ` (X) ` (X), (x1,x2,...,xt) ci,jxj : i =1, 2,...,m . 1 ! 1 7! ⇣ Xj=1 ⌘ (1) If s = t and if C is invertible then TC is an isomorphism. Also if C and C(2) are two matrices so that the number of columns of C(1) is equal to the (2) number of rows of C one easily computes that TC(1) C(2) = TC(1) TC(2) . Secondly it is clear that TP is a closed operator (P defined as above), since n TP is simply the projection onto the first r coordinates in ` (X). It follows therefore that T = T T T is an operator1 with closed A U P V range. Secondly define the operator n m p SA : ` (X) ` (X) ` , 1 ! 1 1 n p (x1,...xn) TA(x1,...xn), yi⇤, bi,jxj . 7! i=1! ⇣D Xj=1 E⌘ 62 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY SA can be written as the sum of TA and a finite rank operator and has therefore also closed range by Lemma 2.7.4. Since the second adjoint of SA⇤⇤ is the operator n m p SA⇤⇤ : ` (X⇤⇤) ` (X⇤⇤) ` , 1 ! 1 1 n p (x1⇤⇤,...xn⇤⇤) TA⇤⇤(x1⇤⇤,...xn⇤⇤), yi⇤, bi,jxj⇤⇤ 7! i=1! ⇣D Xj=1 E⌘ with t n m TA⇤⇤ :` (X⇤⇤) ` (X⇤⇤), (x1⇤⇤,x2⇤⇤,...xn⇤⇤) ai,jxj⇤⇤ : i=1, 2,...,m , 1 ! 1 7! ⇣ Xj=1 ⌘ our claim follows from Lemma 2.7.3. N Lemma 2.7.6. Let E be a finite dimensional space and (xi)i=1 is an "-net of S for some 0 <"<1/3.IfT : E E is a linear map so that E ! (1 ") T (x ) (1 + "), for all j =1, 2,...N. k j k Then 1 3" 1+" x T (x) x , for all x E, 1 " k kk k1 "k k 2 and thus 2 1 (1 + ") T T . k k·k k(1 ")(1 3") We are now ready to proof Theorem 2.7.1. Proof of Theorem 2.7.1. Let F X⇤⇤ and G X⇤ be finite dimensional ⇢ ⇢ (1+ )2 subspaces, and let 0 <"<1. Choose >0, so that (1 )(1 3 ) <", and a N N -net (xj⇤⇤)j=1 of SF . It can be shown that (xj⇤⇤)j=1 span all of F ,butwe can also simply assume without loss of generality, that it does, since we can add a basis of F . Let N N S : R F, (⇠1,⇠2,...,⇠N ) ⇠jx⇤⇤, ! 7! j Xj=1 and note that S is surjective. 2.7. THE PRINCIPLE OF LOCAL REFLEXIVITY 63 1 (i) Put H = S (F X), and let (a : i =1, 2,...,m) be a basis of H, (i) (i) \ write a as a =(ai,1,ai,2,...ai,N ), and define A to be the m by N matrix A =(ai,j)i m,j N . For i =1, 2,...,m put N (i) y = S(a )= a x⇤⇤ F X, i i,j j 2 \ Xj=1 choose x ,x ,...,x S so that x ,x > 1 , and pick a basis 1⇤ 2⇤ N⇤ 2 X⇤ h j⇤⇤ j⇤i g ,g ,...g of G. { 1⇤ 2⇤ `⇤} Consider the following system of equations in N unknowns z1⇤⇤, z2⇤⇤, ...,zN⇤⇤ in X⇤⇤: N ai,jzj⇤⇤ = yi for i =1, 2,...,m Xj=1 z⇤⇤,x⇤ = x⇤⇤,x⇤ for j =1, 2,...,N and h j j i h j j i z⇤⇤,g = x⇤⇤,g⇤ for j =1, 2,...,N and k =1, 2,...,`. h j ki h j ki By construction zj⇤⇤ = xj⇤⇤, j =1, 2,...,N, is a solution to these equations. Since x =1< 1+ , for j =1, 2,...,N, we can use Lemma 2.7.5 and k j⇤⇤k find x ,x ,...x X,with x =1< 1+ , for j =1, 2,...,N,which 1 2 N 2 k jk solve above equations. Define N N S1 : R X, (⇠1,⇠2,...,⇠N ) ⇠jxj. ! 7! Xj=1 We claim that the null space of S is contained in the null space of S1.Indeed N N N if we assumed that ⇠ K , and j=1 ⇠jxj = 0, but j=1 xj⇤⇤ = 0, then, 2 N 6 Lemma 2.7.6 (consider the operator F R , x⇤⇤ x⇤⇤,x⇤ ) there is an P ! 7! hP j i i 1, 2,...N so that 2{ } N x⇤, x⇤⇤ =0, i j 6 D Xj=1 E but since x ,x = x ,x this is a contradiction. h j⇤⇤ i⇤i h j i⇤i It follows therefore that we can find a linear map T : F X so that N ! S1 = TS. Denoting the standard basis of R by (ei)i N we deduce that x = S (e )=T S(e )=T (x ), and thus i 1 i i i⇤⇤ 1+ > x = T (x⇤⇤) x⇤,x = x⇤⇤,x⇤ > 1 . k ik k i k |h i ii| h i i i| 64 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY By Lemma 2.7.6 and the choice of it follows therefore that T T 1 k k·k k 1+". Note that for ⇠ H = S 1(F X), say ⇠ = m a(i), we compute 2 \ i=1 i m m N P (i) S1(⇠)= iS1(a )= i ai,jxj Xi=1 Xi=1 Xj=1 m N m (i) = i ai,jxj⇤⇤ = iS(a )=S(⇠). Xi=1 Xj=1 Xi=1 We deduce therefore for x F X, that T (x)=x. 2 \ Finally from the third part of the system of equations it follows, that x⇤,T(x⇤⇤) = x⇤,x = x⇤,x , for all j =1, 2,...,N and x⇤ G, h j i h ji h ji 2 and, thus (since the xj⇤⇤ span all of F ), that x⇤,T(x⇤⇤) = x⇤⇤,x⇤ , for all x⇤⇤ F and x⇤ G. h i h i 2 2 Exercises 1. Prove Proposition 2.7.2. 2. Prove Lemma 2.7.6. 3. A Banach space X is said to have the Bounded Approximation Property (BAP), if there is a sequence of finite rank operators (T ), T : X X, n n ! so that for all x X, x =limn Tn(x). 2 !1 a) Show that ` ,1 p< has (BAP). p 1 And if you know a bit of probability theory: L [0, 1], 1 p< , p 1 has also (BAP). b) Show that if X⇤ is separable and has (BAP), so does X. Hint: There are finite rank operators Tn : X⇤ X⇤, n N, ! 2 which point wise converge to the identity. Consider the adjoints, but notice that they might not map X to X.