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 $x\in X$.
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Sign up to join this communityLet $X$ 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 $x\in X$.
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.
No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consider the subspace $Z_k$ of $B(X)$ comprising operators mapping ${\rm span}\{z_1, \ldots, z_k\}$ to itself that are 0 on some fixed complement of ${\rm span}\{z_1, \ldots, z_k\}$ (so that $\dim Z_k = k^2$). We may then choose a finite $\tfrac{1}{k}$-net $T_{k,j}$ of the sphere $\{T\in Z_k\colon \|T\|=k\}$. Let $S$ be the union of all the nets picked above. We claim that 0 is in the SOT-closure of $S$.
Indeed, suppose that $U$ is a SOT-open neighbourhood of 0. Let $x_1, \ldots, x_n\in X$ and let $\varepsilon > 0$ be such that $$\{T\in B(X)\colon \max_{i\leqslant n} \|Tx_i\| < 2\varepsilon \}\subseteq U.$$ Take $k$ with $1/k <\varepsilon$. When $k$ is large enough, by the density of $(z_n)_{n=1}^\infty$ there must be $T_k\in Z_k$ with $\|T_k\|=k$ such that $\|T_kx_i\|<\varepsilon$ for all $i$. Pick $j$ so that $\|T_{k,j} - T_k\|\leqslant\tfrac{1}{k}$. Thus, $$\|T_{k,j} x_i\| = \|T_{k,j} x_i - T_kx_i + T_k x_i\|\leqslant \tfrac{1}{k}+\varepsilon<2\varepsilon, $$ that is, $T_{k,j}\in U$.
Consequently, the set $S$ is not SOT-closed however it is SOT-sequentially closed.
Not sure whether this is equivalent to former answers. For maps between metric spaces, continuous is equivalent to sequentially continuous (see this wiki page). Consider the composition of operators in $B(X)$, it is a map from $B(X) \times B(X) \to B(X)$. It can be shown that this map is sequentially continuous (by playing with the definitions), but it is not continuous (See a counterexample here). So $B(X)$ is not metrizable.