Strict topology and $*$-strong toppology on $B(H)$ coincide In the paper Woronowicz - $C^*$-algebras generated by unbounded elements,  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.
 A: 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).
