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?

  • $\begingroup$ So your question is how to show that the unit ball is weakly sequentially compact without using the axiom of choice? $\endgroup$
    – Potato
    Commented Aug 22, 2013 at 4:38
  • $\begingroup$ Yes, using separability and Hilbert space. It really shouldn't be too bad... The question could also be rephrased to (complete the proof in wikipedia's article about Banach Alaoglu in separable case) $\endgroup$
    – Evan
    Commented Aug 22, 2013 at 4:41
  • $\begingroup$ Oh! I found proofwiki.org/wiki/Banach-Alaoglu_Theorem as well. It looks like the way I should prove that $x_\infty$ is in $l^2$ is via duality. That certainly makes sense... well, I guess that's that... So the proof is simple even if we talk about separable Banach spaces. $\endgroup$
    – Evan
    Commented Aug 22, 2013 at 5:32

1 Answer 1




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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