11
$\begingroup$

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 $x\in X$.

$\endgroup$

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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? $\endgroup$ Apr 26, 2018 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. $\endgroup$
    – ABB
    Apr 27, 2018 at 6:03
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 17, 2018 at 0:55
16
$\begingroup$

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.

$\endgroup$
2
  • 7
    $\begingroup$ 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, 2018 at 11:51
  • $\begingroup$ @NikWeaver, you are completely right. The OP may want to see: math.stackexchange.com/questions/2515140/… $\endgroup$ Apr 26, 2018 at 13:04
3
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.