Following the proof of the existence theorem in chapter 1 of Hofer-Zehnder, Symplectic Invariants and Hamiltonian Dynamics, I find:
We have used the well-known fact that the closed unit ball of a Hilbert space is weakly sequentially compact.
I googled a bit, and found this. I have also managed to put together a fully understandable proof of the same result from Wikipedia, where I specify fully understandable (to me) because I know it is, whereas that pdf is something I haven't read through yet. But the point is the pdf assumes the Hilbert space is separable. Now, HZ uses this to deduce a bounded sequence in $H_1(S^1)$, the space of absolutely continuous $2\pi$-periodic functions $\mathbb{R}\to\mathbb{R}^{2n}$ with square-integrable derivatives (i.e. derivatives in $L_2(S^1)$). I can justify this usage of the result only in two ways:
- The result holds for all Hilbert spaces, not only separable ones, which is what the form of the statement suggests;
- That space is separable.
Which is true? And how would I prove it?
Best Answer
The result holds for all Hilbert spaces. The generalization is surprisingly simple.
Let us assume, you know the proof for separable Hilbert spaces.
Now, let $X$ be any Hilbert space and $\{x_n\} \subset X$ be a bounded sequence. Now, the space $\tilde X = \overline{\operatorname{span}(\{x_n\})}$ is separable and you can invoke the result in this space. Hence, a subsequence converges towards $x \in \tilde X$. Since this space inherits the norm of $X$, the subsequence also converges in $X$.