A point in the weak closure but not in the weak sequential closure - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:06:14Z http://mathoverflow.net/feeds/question/91217 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91217/a-point-in-the-weak-closure-but-not-in-the-weak-sequential-closure A point in the weak closure but not in the weak sequential closure Kale 2012-03-14T20:36:46Z 2012-03-15T21:26:01Z <p>I'm trying to find a proof of this counterexample by von Neumann:</p> <p>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 <code>$S=\{ x_{mn} : m, n\geq 1\}$</code>. Von Neumann shows that $0$ is in the weak closure of this set but no sequence in $S$ convergess weakly to $0$.</p> http://mathoverflow.net/questions/91217/a-point-in-the-weak-closure-but-not-in-the-weak-sequential-closure/91276#91276 Answer by Aaron Tikuisis for A point in the weak closure but not in the weak sequential closure Aaron Tikuisis 2012-03-15T13:10:13Z 2012-03-15T13:10:13Z <p>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 <code>$$\{x \in \ell^2: |\langle x, y\rangle| &lt; 1\}$$</code> of $0$ does not intersect $S$.</p> http://mathoverflow.net/questions/91217/a-point-in-the-weak-closure-but-not-in-the-weak-sequential-closure/91321#91321 Answer by Bill Johnson for A point in the weak closure but not in the weak sequential closure Bill Johnson 2012-03-15T19:02:44Z 2012-03-15T21:26:01Z <p>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.</p> <p>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.</p>