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 Aug 22 '13 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 Aug 22 '13 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 Aug 22 '13 at 5:32



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.