Weak topology is not metrizable

functional-analysismetric-spacesmetrizabilityweak-topology

I'm now reading some properties of weak topology, I have some problems which may related to the topology property in non-metrizable space ($E $ is a Banach Space):

  1. I know that $E^*$ with weak* topology is not a metrizable space, but the classical results shows that $B_{E^*} = \{f:\|f\|_{E^*} \leq 1 \}$ is metrizable in weak* topology provided $E$ is separable. How can we explain this contradiction?

  2. Let $S = \{x: \|x\| = 1, x \in E\}$, then the closure of $S$ in weak topology $\sigma(E,E^*)$ is $B_E = \{x:\|x\| \leq 1, x \in E\}$. However a result shows that If we want to find a sequence $\{x_n\}$ s.t. $x_n \rightharpoonup 0$,$\|x_n\| = 1$, we often need the extra condition that $E^*$ be separable or reflexive. This two results seem a contradiction to me.

  3. Prove the following result: if $M \subset E$ is a linear subspace, $f_0 \in E^*$, then there exists some $g_0 \in M^\perp$ such that:
    \begin{equation*}
    \inf_{g \in M^\perp} \|f_0 – g\| = \|f_0 – g_0\|
    \end{equation*}

Best Answer

  1. That the unit ball, $ B_{E^*} $ is metrizable, does not mean that $E^*$ is. Actually, all balls in $E^*$ are metrizable, but you cannot define a global metric, that generates the weak* topology. So there is really no contradiction here.

  2. Remember that $E$ endowed with the weak topology is not first countable, and thus it is not sufficient to consider sequences when finding closures. In general, there only needs to be a net converging weakly to $0$.

  3. Let $(g_n)_{n\geq 1}\subset M^\perp$ be a sequence such that $$ \lim_{n\to\infty}\lVert f_0-g_n \rVert=\inf_{g\in M^\perp}\lVert f_0-g \rVert. $$ Clearly $g_n$ is norm bounded, and we may assume, by possibly passing to a subnet, $(g_\alpha)_{\alpha\in A}$, that $(g_\alpha)_{\alpha\in A}$ converges to some $ g_0\in M^\perp$ in the weak* topology. Now using that the norm is weak* lower semicontinuous we have $$ \inf_{g\in M^\perp}\lVert f_0-g \rVert=\lim_{\alpha\in A}\lVert f_0-g_\alpha \rVert=\liminf_{\alpha\in A}\lVert f_0-g_\alpha \rVert\geq \lVert f_0-g_0 \rVert. $$ On the other hand, it is obvious that $\inf_{g\in M^\perp}\lVert f_0-g \rVert\leq \lVert f_0-g_0 \rVert$, since $g_0\in M^\perp$. So $ \inf_{g\in M^\perp}\lVert f_0-g \rVert= \lVert f_0-g_0 \rVert. $

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