Let $(E,\|\cdot \|)$ a normed Banach space . Let $\sigma $ the weak topology on $E$. Let $B=\{x\in E\mid \|x\|\leq 1\}$. Suppose $(B,\sigma )$ is metrizable.
1) Will all balls be metrizable ?
2) Why $E$ won't be necessarily metrizable ?
Attempts
1) I would say yes using the following argument : Let $B_k=\{x\in E\mid \|x\|\leq k\}$. I define $$d_k(x,y)=d\left(\frac{x}{k},\frac{y}{k}\right)$$ where $d$ is the metric on $B$. Does it work ?
2) Since all ball are metrizable (if 1) is true), why can't we define a metric on $E$ as follow : if $x,y\in E$, let $B'$ a ball that contain $x$. It's metrizable, therefore, there is a metric $d$ on $B'$. Then the distance of $x,y$ is $d(x,y)$. Doesn't work ?
3) If 2) doesn't work, does $E$ is at least locally metrizable ? i.e. I can define a distance on all open that are bounded ? (or maybe it need to be convex ?)
Best Answer
(1) Works obviously.
(2) Fails because $d(x,y)$ isn't uniquely defined.
(3) The weak topology in a infinite dimensional Banach space is never metrizable: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2008&task=show_msg&msg=2476.0001
In general, topological vector spaces are metrizable iff first countable (Topological Vector Spaces IV: Completeness and Metrizability).