A point in the weak closure but not in the weak sequential closure

I'm trying to find a proof of this counterexample by von Neumann:

Let $x_{mn}\in \ell^2$ be defined by $$x_{mn}(m)=n \quad,\quad x_{mn}(n)=m \quad,\quad x_{mn}(k)=0 \hbox{ otherwise,}$$ and let $S=\{ x_{mn} : m, n\geq 1\}$. Von Neumann shows that $0$ is in the weak closure of this set but no sequence in $S$ convergess weakly to $0$.

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I have taken the liberty of rewriting and retitling the question, hopefully preserving the original sense. That said, I am not sure if the question really belongs on MO – Yemon Choi Mar 14 '12 at 21:30

As Aaron pointed out, "von Neumann's" example is really a non example. To salvage the problem, restate it as: construct a sequence in $\ell_2$ which has $0$ in its weak closure, but no subsequence converges weakly to $0$. First note that such a sequence must be unbounded (by Eberlein-Smulian). Secondly, observe that it is enough to have for each $\epsilon > 0$ a (necessarily bounded) subsequence that converges weakly to a point whose norm is at most $\epsilon$ (and, of course, no subsequence that converges weakly to $0$). With these "hints", it is easy to construct an example: Let $x_{nm}(k)$ be $1/n$ if $k=1$, $n$ if $k=m>1$, and $0$ otherwise. With the "obvious" definition, $0$ is in the $2$-weak sequential closure of $x_{nm}$ but not in the $1$-weak sequential closure. From this beginning it is natural to define for each countable ordinal $\alpha$ the $\alpha$-weak sequential closure and to state an obvious problem. Another (not very difficult once you understand the example above) problem is to build a sequence in $\ell_2$ whose norms tend to infinity and yet $0$ is in the weak closure of the sequence.

Another striking example of the phenomena sought by the OP is the following. Take a dense sequence in the unit sphere of $\ell_1$. Then $0$ is in the weak closure of the sequence but no subsequence converges weakly to $0$ because $\ell_1$ has the Shur property.

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Did you mean "unit sphere of $\ell_1$"? – Philip Brooker Mar 15 '12 at 20:03
Thanks, Philip. I corrected the typo. – Bill Johnson Mar 15 '12 at 20:18
Another small typo: I think in the definition of $x_{nm}(k)$, you want $n=m > 1$ to read $k=m > 1$. – Aaron Tikuisis Mar 15 '12 at 21:21
Thanks, Aaron; I corrected it. – Bill Johnson Mar 15 '12 at 21:26

I don't believe that $0$ is a weak cluster point of this set. For example, consider $y \in \ell^2$ defined by $$y(k) = 1/k.$$ Then we have, for any $m,n$ that $$\langle x_{m,n}, y \rangle = m/n + n/m \geq 2.$$ Therefore, the weak neighbourhood $$\{x \in \ell^2: |\langle x, y\rangle| < 1\}$$ of $0$ does not intersect $S$.

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It is possible that I made an error in transcription when reformatting the original question, but unless the OP turns up thete's no way of knowing – Yemon Choi Mar 15 '12 at 18:25