# Closed unit ball of Hilbert space: sequentially compact in weak topology?

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:

1. The result holds for all Hilbert spaces, not only separable ones, which is what the form of the statement suggests;
2. That space is separable.

Which is true? And how would I prove it?

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 $X$.

• It's about weak convergence, not norm convergence. In my view, the solution lies in the fact that $X= \tilde X + \tilde X ^{\perp}$. Jul 20, 2019 at 17:36

Theorem A By "1.&4." of [1], the closed unit ball is weakly sequentially compact in any reflexive Banach space.
Fact B Hilbert spaces are reflexive.

References to Theorem A: [1] https://en.wikipedia.org/wiki/Reflexive_space#Properties "Let X be a Banach space. The following are equivalent.
1. The space X is reflexive.
2. The continuous dual of X is reflexive.
3. The closed unit ball of X is compact in the weak topology. (This is known as Kakutani's Theorem.)
4. Every bounded sequence in X has a weakly convergent subsequence.
5. Every continuous linear functional on X attains its maximum on the closed unit ball in X. (James' theorem)"

Another reference for 1.<=>4. is Theorem 6.2, p. 174 in Robert C. McOwen, Partial Differential Equations: Methods and Applications, 1996 (420 pp.): https://books.google.fi/books?id=TuNHsNC1Yf0C&pg=PA174&lpg=PA174&dq=the+unit+ball+of+finite+sequences+is+%22weakly+sequentially+compact%22&source=bl&ots=WRLojyHsH9&sig=nPQfhL-XPoDxXZSaKa4pmy_uX2k&hl=fi&sa=X&ved=0ahUKEwjfz_ufnJbQAhWD3CwKHU_CBqA4ChDoAQg9MAY#v=onepage&q=the%20unit%20ball%20of%20finite%20sequences%20is%20%22weakly%20sequentially%20compact%22&f=false

Another proof of the Hilbert case of "Theorem A" is here including "all results it cites, with explanations".

Non-generalization. 1. The set $A:=\{1_n\}_{n\in\mathbb N}$ is closed and bounded but not weakly sequentially compact (nor compact) in $\ell^1(\mathbb N)$.*
2. Therefore, no set containing $A$ is weakly sequentially compact (or compact).
3. In particular, the unit ball of $\ell^1(\mathbb N)$ is not weakly sequentially compact (nor compact).

*) Proof: $(\ell^1)^*=\ell^\infty$ and $1_k 1_n\rightarrow 0$, as $n\rightarrow\infty$, so no subsequence of $(1_n)_{n\in\mathbb N}$ can converge to anything but $x=0\in A$, but $f 1_n=1\rightarrow 1\ne f0$ when $f=(1,1,1,\ldots)\in\ell^\infty$, so $x=0$ is not the limit of any such subsequence. So $A$ is not w.s.c. (hence not weakly compact, by the Eberlein–Šmulian theorem). But clearly $A^c$ is open, QED.