Is Eberlein-Smulian necessary to prove the fact that every bounded sequence of a Hilbert space has a weakly convergent subsequence?

Question

The standard answer I have seen on stackexchange is something like this: Hilbert space is the dual of itself, so applying Banach-Alaoglu, we see that the bounded sequence is contained in a closed ball which is weakly compact. Thus the sequence contains a weakly convergent subsequence.

However, weak compactness does NOT necessarily imply sequential weak compactness. This turns out to be true, by the Eberlein-Smulian Theorem, which states thtat weak compactness and sequential weak compactness are equivalent for Banach spaces. My question is: Is there any proof of the fact not using Eberlein-Smulian? Otherwise the urban legend that you only need Banach-Alaoglu to show this fact needs to be corrected!

Answer 1

It suffices to prove this for separable Hilbert spaces since we can restrict our attention to $\overline{L(x_n)}$ if $x_n$ is the given sequence.

Then we can do this completely by hand. Let $\{ e_j\}$ be an ONB. Since $\langle e_1, x_n\rangle$ is a bounded sequence of complex numbers, we can select a convergent subsequence, and then a sub-subsequence of this that makes $\langle e_2, x_n\rangle$ convergent etc. Then a diagonal argument produces a subsequence for which $\langle e_j, x_{n_k}\rangle$ converges for all $j$.

It's now easy to see that this sequence converges weakly (use one more time that $\|x_n\|\le C$).

Answer 2

The result you are looking for is that, in a reflexive space, every bounded sequence has a weakly convergent subsequence. If I'm not mistaken, Eberlein-Smulian is the converse of that result (that this property characterizes reflexivity)

In your case, it is a culmination of two facts:

1. If $E$ is a separable Banach space, then closed unit ball $B^{\ast}$ in $E^{\ast}$ is metrizable in the weak-$\ast$ topology. In fact, if $\{x_n\}$ is a countable dense subset of $E$, then $$ d(f,g) = \sum_{n=1}^{\infty} \frac{1}{2^n\|x_n\|}|(f-g)(x_n)| $$ defines a metric on $B^{\ast}$. It then takes some work to prove that this metric induces the weak-$\ast$ topology.

2. Now if $E$ is reflexive, then any closed subspace is also reflexive. Hence if $\{x_n\}$ is bounded, you can replace $E$ by the separable space $F = \overline{\text{span}\{x_n\}}$, and then apply (1).