In the paper [Woronowicz - $C^\*$-algebras generated by unbounded elements](https://www.fuw.edu.pl/%7Eslworono/PDF-y/GENER.pdf), I read that the $\*$-strong operator topology on $B(H)$ and the strict topology on $B(H)$ coincide. I believe this means the following:


Consider the multiplier algebra $M(B\_0(H)) =B(H)$. Via this canonical identification, $B(H)$ obtains the strict topology. Explicitly, a net $(x\_i)$ in $B(H)$ converges strictly to $x$ iff $\|x\_i y -x y\|\to 0$ and $\|yx\_i \to yx\| \to 0$ for all compact operators $y \in B\_0(H)$.


A net $(x\_i)$ in $B(H)$ converges strongly iff $\|x\_i\xi-x\xi\| \to0$ and $\|x\_i^\*\xi-x^\*\xi\| \to 0$ for all $\xi \in H$.



> 
> **Question**: How to show that $x\_i \to x$ in the $\*$-strong topology implies $x\_i \to x$ strictly?
> 
> 
> 


Here is my attempt:


Using rank-one operators, I managed to show that $x\_i \to x$ strictly implies $x\_i \to x$ in the $\*$-strong topology. Conversely, I managed to show that if $x\_i \to x$ in the $\*$-strong topology, then $\|x\_iy -xy\| \to 0$ and $\|yx\_i-yx\| \to 0$ for all **finite-rank** operators $y$ on $H$. If $y,y' \in B\_0(H)$, we can make the estimate (similarly for the other component)
$$\|yx\_i -yx\| \leq \|yx\_i- y'x\_i\| + \|y'x\_i-y' x \| + \|y'x-yx\|$$
and choosing $y'$ a finite rank operator that approximates $y$, it would follow that the left hand side converges to $0$, if we can show that $(x\_i)$ is norm-bounded.



Strict topology and $*$-strong toppology on $B(H)$ coincide
Yeah, I don't think this is true. The two topologies do agree on bounded sets, as you have already observed.


For a counterexample, let $H$ be a separable, infinite-dimensional Hilbert space and let $\mathcal{I}$ be the set of all finite-dimensional subspaces of $H$, ordered by inclusion. For $E \in \mathcal{I}$ let $x\_E$ be $d\cdot(I-P\_E)$ where $d = {\rm dim}(E)$ and $I-P\_E$ is the orthogonal projection onto $E^\perp$. This net converges $\ast$-strongly to zero because for any $v \in H$, any finite-dimensional subspace $E$ which contains $v$ will satisfy $x\_Ev = 0$. So $x\_Ev$ is eventually zero, for any $v \in H$. This shows strong convergence to zero, and then $\ast$-strong convergence is immediate because each $x\_E$ is self-adjoint.


However, I claim that $(x\_E)$ does not converge strictly to zero. Specifically, fix an orthonormal basis $(e\_n)$ and let $T$ be the operator $Te\_n = \frac{1}{n}e\_n$; I claim that $||x\_ET|| \to \infty$. This is because, given any $E$, we can find a value of $n$ for which $(I-P\_E)e\_n \neq 0$, and then for any finite-dimensional subspace $F$ contained in the orthocomplement of $E \vee {\rm span}(e\_n)$ we have $P\_{E \vee F}e\_n = (P\_E + P\_F)e\_n = P\_Ee\_n$. So $$x\_{E \vee F}Te\_n = \frac{1}{n}x\_{E \vee F}e\_n = \frac{d}{n}(I - P\_{E \vee F})e\_n = \frac{d}{n}(I - P\_E)e\_n,$$ and since $(I-P\_E)e\_n \neq 0$ and $d = {\rm dim}(E \vee F)$ can be arbitrarily large, this shows that $||x\_{E \vee F}T||$ can be arbitrarily large (as $F$ varies, keeping $E$ fixed).


