My aim is to prove that in a Hilbert space, any sequence has a weakly convergent subsequence. To prove this, I'm trying to prove that:
In a Hilbert space, every bounded and closed set is weakly relatively compact
I have tried it via Banach-Alaoglu theorem but I find it difficult as this theorem applies to the dual space with the weak-* topology and not the weak one. Anyway I have the feeling that there is an easier way to prove it.
Any help would be highly appreciated. Thank you.
Best Answer
Banach-Alaoglu and Riesz theorem are an overkilling, even if I really like it as a proof. Moreover, your statement is true in general in reflexive spaces and it is usually known as the easy part of Eberlein-Smulian theorem. Recalling the following definitions,
A fundamental result is
From Kakutani's theorem follow
Now we can state
Proof. Set $M = \overline{\mbox{span}_{\mathbb K}}\{x_n\}$. By Theorem 2, $M$ is reflexive and obviously is separable. Hence, by Corollary 2, $M'$ is reflexive and separable. Then $J(B_M) = B_{M''}$ is compact and metrizable with respect to the $\sigma(M'', M')$ topology, so there exists $\{x_{n_k}\}$ subsequence and $F = J(x)$ s.t. $J(x_{n_k}) \rightharpoonup J(x)$ when $k\to \infty$ with respect to the $\sigma(M'', M')$, i.e., by definition, $\langle J(x_{n_k}), f \langle_{M'',M'} \to \langle J(x_n), f \rangle_{M'',M'}$. Since $M$ is closed, $x \in M$ and $\langle f, x_{n_k} \rangle_{M',M} \to \langle f, x_n \rangle_{M',M}$ when $k\to\infty$. Thanks to the Hahn-Banach theorem, there exists $L \in X'$ s.t. $f := L_{|M}$ and hence we have $\langle L, x_{n_k} \rangle_{X',X} \to \langle L, x_{n_k} \rangle_{X',X}.$ $\square$
Now consider a Hilbert space $H$. The scalar product on $H$ induces a norm which satisfies the parallelogramm identity (trivial):
$$\| x + y \|^2 + \| x - y \|^2 = 2( \|x\|^2 + \|y\|^2)$$
for all $x,y\in H$. Such a norm is uniformly convex and hence, by
we have $H$ is reflexive. So the previous Theorem 4 holds and your claim follows as a corollary of it.
I know your method was quicker, but I listed a few of results you would probably know, provided you can apply Banach-Alaoglou theorem. By the way, in the previous work there is no needing of knowledge of Hilbert spaces theory, except for the definition. In this sense, this is the easiest proof, even is longer.
References.