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Let $H$ be a Hilbert space and let $B(H)$ denote the Banach space of bounded operators on $H$. Then there are several topologies we can endow on $B(H)$. I am interested in the case of the strong operator topology and and the ultrastrong operator topology (sometimes also called $\sigma$-strong).

Note that the ultrastrong topology is finer than the strong topology. If a set $X$ is ultrastrongly dense, then it is of course also strongly dense. I am having trouble coming up with a counter-example for the converse.

Does anyone have a simple example of a set which is strongly dense, but fails to be ultrastrongly dense? I would also be interested in similar examples for weak/ultraweakly dense sets.

From Martin Argerami's answer below, it seems like no such example exists at least when $X$ is a $*$-algebra. I am still interested in the case where $X$ is merely a set with no additional structure.


For the sake of completeness, the two topologies can be defined as follows. For each $\xi \in H$, let $\|\cdot \|_\xi$ be the semi-norm on $B(H)$ defined by $$\|T\|_\xi = \|T\xi\|.$$ Then the strong operator topology is the topology induced by the collection of semi-norms $\{\|\cdot \|_\xi\}_{\xi\in H}$. Likewise, if we are given a sequence of vectors $s=(\xi_n)_{n=1}^\infty$, such that $$\sum_{n=1}^\infty \|\xi_n\|^2 < \infty,$$ we can define the semi-norm $\|\cdot \|_s$ by $$\|T\|_s = \left(\sum_{n=1}^\infty \|T\xi_n\|^2\right)^{1/2}.$$ The ultrastrong topology is then the topology induced by the collection $\{\|\cdot \|_s\}$, where $s$ is an arbitrary $L^2$-convergent sequence.

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No such example exists, since the strong and ultra-strong topologies agree on bounded sets. Suppose that $X\subset B(H)$ is strong dense, and fix $a\in B(H)$. Then there exists $\{x_j\}\subset X$ with $x_j\to a$ strongly. By Kaplansky's Density Theorem, we may assume that $\{x_j\}$ is bounded by $\|a\|$. On the ball $B_{\|a\|}(0)$ the two topologies agree, so $x_j\to a$ ultrastrongly. Thus $X$ is ultrastrong dense.

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  • $\begingroup$ This is great. I suspected this to be true, but couldn't prove it, so I started looking for counter-examples. I was missing the usage of Kaplanski's. Thank you! $\endgroup$ – EuYu Nov 18 '19 at 23:05
  • $\begingroup$ A follow-up question. Kaplanski's density theorem seems to require that $X$ is a $C^*$-algebra. How do I conclude the existence of a bounded net when $X$ is just a set? $\endgroup$ – EuYu Nov 18 '19 at 23:20
  • $\begingroup$ For Kaplansky you don't need a C$^*$-algebra, just a $*$-algebra. But you are right, if $X$ is just a set my argument does not apply. $\endgroup$ – Martin Argerami Nov 19 '19 at 0:46
  • $\begingroup$ Do you know if the result continues to hold for a set? $\endgroup$ – EuYu Nov 19 '19 at 7:25

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