[Math] Separable Hilbert space weak sequential compactness

functional-analysishilbert-spaces

To preface, Banach-Alaoglu shows weak* sequential compactness of the unit ball, and in Hilbert spaces weak* and weak convergence is the same. So I already know that the unit ball of a Hilbert space is weakly-sequentially-compact.

However, for separable Hilbert spaces, and let's just focus on $l^2$ as we have an inner product isomorphism, there should be an easy constructive proof, but I don't quite see it.

Let $x_n$ be a bounded sequence in $l^2$. By diagonalization we can obtain a subsequence for which $x_{n_k} \to x_\infty$ pointwise. For that matter, we can obtain a subsequence for which $\langle x_{n_k} – x_\infty, g_i \rangle \to 0$ for a dense countable subset $g_i$ of $l^2$.

So, the question is, can we show that $x_\infty \in l^2$ ?

Alternatively, perhaps there is a different route. But the idea is that we should not have to appeal to the axiom of choice (which Banach Alaoglu does).
Ok, so as I type this, I see http://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem mentions defining a metric using a dense set. But for some reason, I still don't see how to complete the proof. What is missing? It appears that $x_{n_k}$ converges to $x_\infty$ in the metric defined in wikipedia.. hmm… but why is that metric complete? Any ideas?

Best Answer

From

http://www.proofwiki.org/wiki/Banach-Alaoglu_Theorem

there were just a few more steps:

  1. Let $l(g_i) = \lim_{n_k} \langle x_{n_k} , g_i \rangle = \langle x_\infty, g_i \rangle$. (We can obtain $x_\infty$ by including the standard basis in the dense $g_i$).
  2. Extend to all $g \in l^2$ via $l(g) = \lim_i l(g_i)$. This makes $l$ a linear functional, and can be bounded based on $x_{n_k}$. $|l(g)| \leq |l(g)-l(g_i)| + |l(g_i) - l_{n_k}(g_i)| + |l_{n_k}(g_i)|$.
  3. Riesz representation and uniqueness shows that $l(x) = \langle x_\infty, x \rangle$.

Then we have the weak convergence as desired.

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