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 $x\in X$.


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.

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    $\begingroup$ Is there an easy argument to see that the SOT topology is different from the norm topology on a general Banach space without using, for example, the Josefson-Nissenzweig theorem? $\endgroup$ – Tomasz Kania Apr 26 '18 at 16:47
  • $\begingroup$ @ Tomek Kania Probably it works: the product on $B(X)$ is jointly norm continuous but this point is not valid concerning SOT. $\endgroup$ – Ali Bagheri Apr 27 '18 at 6:03
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    $\begingroup$ The second paragraph is also a proof all by itself, since the weak-* topology on $X^*$ is not metrizable, so it cannot embed in any metrizable space. $\endgroup$ – Nate Eldredge Nov 17 '18 at 0:55

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.

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    $\begingroup$ But note that its restriction to the unit ball is metrizable, by $d(S,T) = \sum 2^{-n} \|(S - T)x_n\|$ where $(x_n)$ is a dense sequence in the unit ball of $X$. That's something I've used once or twice. $\endgroup$ – Nik Weaver Apr 26 '18 at 11:51
  • $\begingroup$ @NikWeaver, you are completely right. The OP may want to see: math.stackexchange.com/questions/2515140/… $\endgroup$ – Tomasz Kania Apr 26 '18 at 13:04

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.


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