# Example of a strongly dense set which is not ultrastrongly dense.

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.

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.
• 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? – EuYu Nov 18 '19 at 23:20
• 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. – Martin Argerami Nov 19 '19 at 0:46