In my functional analysis course we proved the Banach-Alaoglu theorem and this corollary :
Banach-Alaoglu theorem :Let $X$ be a separable Banach spaceLet
$\{f_n\}$ be a bounded sequence in $X'$, then there exists a
weakly*-convergent subsequence $\{f_{n_k}\}$ to some $f \in X'$.Corollary :Let $X$ be a reflexive Banach space such that the dual $X'$
is separable. Let $\{x_n\}$ be a bounded sequence in $X$, then there
exists a weakly-convergent subsequence $\{x_{n_k}\}$ to some $x\in$
$ X$.
The proof of the corollary consisted in applying the Banach-Alaoglu theorem to the spaces $X'$ and $X''$ and for that we needed $X'$ to be separable.
Now in my Calculus of variations course as a summary of important results from functional analysis I have this theorem withouth proof:
Theorem :Let $X$ be a reflexive Banach Then every bounded sequence
has a weakly convergent subsequence.
which I think it should be the corollary above so I wonder, is this wrong? The hypothesis that the dual $X'$ is separable is missing. Or maybe my version of the corollary was an easier-to-prove one?
Best Answer
People add separability because in that case the proof is easier and is known as Helly's Selection Theorem. One can give a proof independent of the usual proof of Banach Alaoglu which uses Tykonov's Theorem. One proceeds by a diagonalization argument by exploiting separability. Actually, a Banach Space is separable if and only if the unit ball in weak$^{*}$ topology is metrizable. Thus showing sequential compactness suffices.
Suppose you have proved the statement:
Consider now $X$ to be a reflexive banach space and and let $x_{n}$ to be a bounded sequence. Then consider $V=\overline{\text{span}}\{x_{1},..x_{n},...\}$. Then, $V$ itself is a reflexive and separable Banach space and hence has a subsequence $x_{n_{k}}$ that converges weakly to some $x_{0}\in V$.
Now let $f\in V^{*}$ be a linear functional, then $f(x_{n_{k}})\to f(x_{0})$. By Hahn-Banach Theorem, $f$ has an extension $f'\in X^{*}$ such that $f'|_{V}=f$. Then we have $f'(x_{n_{k}})\to f(x_{0})$ . Also for any linear functional $F\in X^{*}$, we have that $F|_{V}$ is a bounded linear functional in $V^{*}$. Thus $F(x_{n_{k}})\to F(x_{0})$ .
Thus we have that $f(x_{n_{k}})\to f(x_{0})$ for all linear functionals $f\in X^{*}$. Thus $x_{n_{k}}$ converges weakly to $x_{0}$ .
Thus, extending from the separable case to the non-separable case is not really difficult.