Re: Weak convergence

From: G. A. Edgar (edgar_at_math.ohio-state.edu.invalid)
Date: 10/30/04 

Date: Sat, 30 Oct 2004 07:55:42 -0400

In article <41836b94$0$3665$8fcfb975@news.wanadoo.fr>, Julien Santini <santini.julien@wanadoo.fr> wrote:

> Hello,
>
> In the following, C denotes the complex numbers.
> Let E be a complex Banach space, and A a subset of E. My book calls A
> "weakly closed" whenever for any sequence (a_n) of elements of A, if lim
> x*(a_n)=x*(a) for any bounded linear functional x*: E->C, then a is in A.
>
> A natural question -that is not mentioned in my book- would be to determine
> whether a weakly closed set A is closed for the weak topology, whose
> subbasis is given by {x*^(-1)(U); x*: E->C bounded, U open set of C}.
>
> I fail when trying to prove that whenever a is in Adh(A)-A, then there
> exists a sequence (a_n) of elements of A weakly converging to a. The fact is
> I can control the behaviour of a_n according to a finite number of x* only.
> Any suggestion ? - a one word answer would suffice, I'd just want to know if
> I should try proving it or not -.
>
> --
> Julien Santini
>
>

If you use weak convergence of nets (generalized sequences), then weakly
closed described that way is equivalent to closed in the weak topology.
But in general it is not. If your subset A is convex, then sequences are
enough. But in general, sequences are not enough.

In the Banach space l^1 of summable sequences, it is surprising that
a squence converges weakly if an only if if converges in norm.
(a gliding hump argument...) But the weak and norm topologies are
different, so there are sets A that are norm closed but not weakly
closed. Such as the set of all x with norm >=1 . 

-- 
G. A. Edgar                               http://www.math.ohio-state.edu/~edgar/

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