An elementary proof of James' characterization of weak compactness - PDF document

See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/231992716 An elementary proof of James' characterization of weak compactness (Lecture slides) Conference Paper in Bulletin of the Australian Mathematical Society · May 2010 CITATIONS READS 4 114 1 author: Warren B. Moors University of Auckland 89 PUBLICATIONS 956 CITATIONS SEE PROFILE All content following this page was uploaded by Warren B. Moors on 08 June 2014. The user has requested enhancement of the downloaded file.

An Elementary Proof of James’ Characterisation of weak Compactness Warren B. Moors Department of Mathematics The University of Auckland Auckland New Zealand

Background The purpose of this talk is to give a self-contained proof of James’ characterisation of weak compactness (in the case of separable Banach spaces). The proof is completely elemen- tary and does not require recourse to integral representations nor Simons’ inequality. It only requires results from linear topology (in particular the Krein-Milman Theorem) and Eke- land’s variational principle (Bishop-Phelps Theorem). The idea of the proof is due to V. Fonf, J. Lindenstrauss and R. Phelps. Proposition 1 Let { K j : 1 ≤ j ≤ n } be convex subsets of a vector space V . Then � n n � � co K j = λ j k j : ( λ j , k j ) ∈ [0 , 1] × K j j =1 j =1 � n � for all 1 ≤ j ≤ n and λ j = 1 . j =1

From this we may easily obtain the following result. Theorem 1 Let { K j : 1 ≤ j ≤ n } be weak ∗ compact con- n � vex subsets of the dual of a Banach space X . Then co K j j =1 is weak ∗ compact. We say that a subset E of a set K in a vector space V is an extremal subset of K if x, y ∈ E whenever λx + (1 − λ ) y ∈ E, x, y ∈ K and 0 < λ < 1 . A point x is called an extreme point if the set { x } is an extremal subset of K . For a set K in a vector space X we will denote the set of all extreme points of K by Ext ( K ) . Proposition 2 Let K be a nonempty subset of a vector space V . Suppose that E ∗ ⊆ E ⊆ K . If E ∗ is an extremal subset of E and E is an extremal subset of K then E ∗ is an extremal subset of K . In particular, Ext ( E ) ⊆ Ext ( K ) .

We may now present our first key result. Theorem 2 (Milman’s Theorem) Let E be a nonempty sub- set of the dual of a Banach space X . If K := co weak ∗ ( E ) is weak ∗ weak ∗ compact then Ext ( K ) ⊆ E . Proof: Let e ∗ be any element of Ext ( K ) and let N be any weak ∗ closed and convex weak ∗ neighbourhood of 0 ∈ X ∗ . Then E ∗ ⊆ � weak ∗ Let E ∗ := E x ∗ ∈ E ∗ ( x ∗ + N ) . . So by n in E ∗ such compactness there exist a finite set y ∗ 1 , y ∗ 2 , . . . , y ∗ that E ∗ ⊆ � n j =1 ( y ∗ j + N ) . For each 1 ≤ j ≤ n , let K j := j + N ) ∩ K . Then each K j is weak ∗ compact and convex ( y ∗ and E ⊆ E ∗ ⊆ � n j =1 K j . Therefore, n n � � e ∗ ∈ K = co weak ∗ ( E ) ⊆ co weak ∗ K j = co K j . j =1 j =1 Thus, e ∗ = � n j =1 λ j k j for some ( λ j , k j ) ∈ [0 , 1] × K j with � n j =1 λ j = 1 . Since e ∗ ∈ Ext ( K ) , there exists an i ∈ { 1 , 2 , . . . , n }

such that λ i = 1 (and λ j = 0 for all j ∈ { 1 , 2 , . . . n } \ { i } ). Therefore, e ∗ = k i ∈ K i ⊆ y ∗ i + N ⊆ E ∗ + N . Since N was an arbitrary weak ∗ closed convex weak ∗ neighbourhood of 0 , e ∗ ∈ E ∗ . ❦ ✂✁ Theorem 3 Let X be a Banach space. Then every nonempty weak ∗ compact convex subset of X ∗ has an extreme point. Proof: Let K be a nonempty weak ∗ compact convex subset of X ∗ and let X ⊆ 2 K \{ ∅ } be the set of all nonempty weak ∗ compact convex extremal subsets of K . Then X � = ∅ since K ∈ X . Now, ( X, ⊆ ) is a nonempty partially ordered set. We will use Zorn’s lemma to show that ( X, ⊆ ) has a minimal element. To this end, let T ⊆ X be a totally ordered subset of X (i.e., ( T, ⊆ ) is a totally ordered set). Let K ∞ := � C ∈ T C . Then ∅ � = K ∞ is a weak ∗ compact convex subset of K . Moreover, K ∞ is an extremal subset of K since if x ∗ , y ∗ ∈ K and 0 < λ < 1 and λx ∗ + (1 − λ ) y ∗ ∈ K ∞ then for each

C ∈ T , λx ∗ + (1 − λ ) y ∗ ∈ C ; which implies that x ∗ , y ∗ ∈ C . That is, x ∗ , y ∗ ∈ K ∞ . Therefore, K ∞ ∈ X and K ∞ ⊆ C for every C ∈ T , i.e., T has a lower bound in X . Thus, by Zorn’s Lemma, ( X ⊆ ) has a minimal element K M . Claim: K M is a singleton. Supppose, in order to obtain a contradiction, that K M is not a singleton. Then there exist x ∗ , y ∗ ∈ K M such that x ∗ � = y ∗ . Choose x ∈ X such that x ∗ ( x ) � = y ∗ ( x ) . Let K ∗ := { z ∗ ∈ K M : � x ( z ∗ ) = max x ( w ∗ ) } . w ∗ ∈ K M � Then ∅ � = K ∗ ⊆ K M and K ∗ ∈ X . Thus, K ∗ = K M ; which implies that x ∗ ( x ) = y ∗ ( x ) . Thus, we have obtained a contradiction and so K M is indeed a singleton. It now follows from the definition of an extreme point that the only member ❦ of K M is an extreme point of K . ✂✁ In order to prove the well-known consequence of this result we need a separation result (which we will not prove here).

Theorem 4 Let K be a nonempty weak ∗ compact convex subset of the dual of a Banach space X . If x ∗ ∈ X ∗ is not a member of K then there exists an x ∈ X such that x ( x ∗ ) > max x ( y ∗ ) . � y ∗ ∈ K � Theorem 5 (Krein-Milman Theorem) Let K be a nonempty weak ∗ compact convex subset of the dual of a Banach space X . Then K = co weak ∗ Ext ( K ) . Proof: Suppose, in order to obtain a contradiction, that co weak ∗ Ext ( K ) � K. Then there exists x ∗ ∈ K \ co weak ∗ Ext ( K ) . Choose x ∈ X x ( y ∗ ) : y ∗ ∈ co weak ∗ Ext ( K ) } . Let x ( x ∗ ) > max { � such that � K ∗ := { z ∗ ∈ K : � x ( z ∗ ) = max x ( y ∗ ) } . y ∗ ∈ K �

Now, K ∗ is a nonempty weak ∗ compact convex extremal sub- set of K . Therefore, by Theorem 3, there exists an e ∗ ∈ Ext ( K ∗ ) ⊆ Ext ( K ) . However, e ∗ �∈ co weak ∗ Ext ( K ) . Thus, we have obtained a contradiction. Hence the state- ❦ ment of the Krein-Milman theorem holds ✂✁ This concludes the necessary linear topology required in order to prove James’ Theorem. Our next goal is to prove the Bishop-Phelps Theorem. To do this we start will some convex analysis. Let f : X → R be a continuous convex function defined on a Banach space X . Then for each x 0 ∈ X we define the subdifferential of f at x 0 to be: ∂f ( x 0 ) := { x ∗ ∈ X ∗ x ∗ ( x ) + [ f ( x 0 ) − x ∗ ( x 0 )] ≤ f ( x ) : for all x ∈ X } . Then for each x ∈ X , ∂f ( x ) , is a nonempty weak ∗ compact

convex subset of X ∗ . We will require two facts about the subdifferential: (a) If f ( x ∞ ) = min x ∈ X f ( x ) then 0 ∈ ∂f ( x ∞ ) (this follows directly from the definition); (b) If h : X → R is also a continuous convex function then ∂ ( h + f )( x ) = ∂h ( x ) + ∂f ( x ) for all x ∈ X. Next, we prove Ekeland’s variational principle. Theorem 6 (E.V.P.) Suppose that f : X → R is a bounded below lower semi-continuous function defined on a Banach space X . If ε > 0 , x 0 ∈ X and f ( x 0 ) ≤ inf y ∈ X f ( y ) + ε 2 then there exists x ∞ ∈ X such that � x ∞ − x 0 � ≤ ε and the function f + ε � · − x ∞ � attains its minimum value at x ∞ . Moreover, if f is continuous and convex then 0 ∈ ∂f ( x ∞ ) + εB X ∗ .

Proof: We shall inductively define a sequence ( x n : n ∈ N ) in X and a sequence ( D n : n ∈ N ) of closed subsets of X such that (i) D n := { x ∈ D n − 1 : f ( x ) ≤ f ( x n − 1 ) − ε � x − x n − 1 �} ; (ii) x n ∈ D n ; (iii) f ( x n ) ≤ inf x ∈ D n f ( x ) + ε 2 / ( n + 1) . Set D 0 := X . In the base step we let D 1 := { x ∈ D 0 : f ( x ) ≤ f ( x 0 ) − ε � x − x 0 �} and choose x 1 ∈ D 1 so that f ( x 1 ) ≤ inf x ∈ D 1 f ( x ) + ε 2 / 2 . Then at the ( n + 1) th -step we let D n +1 := { x ∈ D n : f ( x ) ≤ f ( x n ) − ε � x − x n �} and we choose x n +1 ∈ D n +1 such that x ∈ D n +1 f ( x ) + ε 2 / ( n + 2) . f ( x n +1 ) ≤ inf

This completes the induction. Now, by construction, ∅ � = D n +1 ⊆ D n for all n ∈ N . It is also easy to see that sup {� x − x n � : x ∈ D n +1 } ≤ ε/ ( n +1) . Indeed, if x ∈ D n +1 and � x − x n � > ε/ ( n + 1) then � � = f ( x n ) − ε 2 / ( n + 1) f ( x ) < f ( x n ) − ε ( ε/ ( n + 1)) � � y ∈ D n f ( y ) + ε 2 / ( n + 1) − ε 2 / ( n + 1) = inf ≤ inf y ∈ D n f ( y ); which contradicts the fact that x ∈ D n +1 ⊆ D n . Let { x ∞ } := � ∞ n =1 D n . Fix x ∈ X \ { x ∞ } and let n be the first natural number such that x �∈ D n , i.e., x ∈ D n − 1 \ D n . Then, f ( x ∞ ) − ε � x − x ∞ � ≤ f ( x n − 1 ) − ε � x − x n − 1 � < f ( x ) since f ( x ∞ ) ≤ f ( x n − 1 ) − ε � x n − 1 − x ∞ � since x ∞ ∈ D n � � ≤ f ( x n − 1 ) − ε � x − x n − 1 � − � x − x ∞ � .