Let $X$ be a separable Banach space. Is the strong operator topology metrizable on $B(X)$, the space of all bounded operators on $X$?


SOT-$\lim T\_i=0~$ if and only if $~\lim \|T\_ix\|=0$ for every $x\in X$. 



Is the strong operator topology metrizable?
A quick proof using the open mapping theorem: It follows easily from the uniform boundedness principle that $(B(X),SOT)$ is sequentially complete. If it were metrizable it would thus be a Fréchet space and the open mapping theorem implies that the continuous identity $(B(X),\|\cdot\|\_{op})\to (B(X),SOT)$ would be open, i.e., $SOT$ coincides with the operator norm topology which is not true if $X$ is infinite dimensional.


EDIT. You do not need a *sledgehammer* to show that $SOT$ differs from the operator norm topology: Fixing a non-zero element $y\in X$ we have an embedding $X^\*\to B(X)$, $f\mapsto f\otimes y$ where $f\otimes y$ maps $x$ to $f(x)y$. The induced topologies on $X^\*$ are the weak$^\*$ topology and the dual norm topology. They differ because every continuous semi-norm of the weak$^\*$ topology has a huge kernel.


