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From Peter Lax Functional analysis page 104:

Show that a weakly sequentially compact set is bounded.

Definition. A subset $C$ of a Banach space $X$ is called weakly sequentially compact if any sequence of points in $C$ has a subsequence weakly convergent to a point of $C$.

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I did not down vote but perhaps you might want to show some of your thoughts. –  Matt N. Nov 13 '12 at 16:07
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, some people on this site consider the use of imperatives (show, prove, do etc.) rude. Consider editing you question. –  Dennis Gulko Nov 13 '12 at 16:32
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What do you mean by weakly sequential compact? Is it in $\Bbb R^n$ or in any metric space? –  Berci Nov 13 '12 at 16:48
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@Berci: Sequentially compact in the weak topology on $X$. –  Brian M. Scott Nov 13 '12 at 19:39

1 Answer 1

up vote 2 down vote accepted

The key is the following result:

If $X$ is a normed space, and $\{x_n\}\subset X$ a sequence which converges weakly to $X$, then $\sup_{n\in\Bbb N}\lVert x_n\rVert<\infty$.

It's a consequence of Baire's categories theorem applied to $X^*$, the topological dual of $X$.

We show that an unbounded set cannot be weakly compact. If $S$ is not bounded, let $\{x_n\}\subset S$ such that $\lVert x_n\rVert\geqslant n$. If $\{x_{n_k}\}$ is a subsequence of $\{x_n\}$, then for each $k$, $\lVert x_{n_k}\rVert\geqslant n_k$. In particular, $\{x_{n_k}\}$ is not bounded hence cannot be weakly convergent.

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