Does left-multiplication by compact operators turn strong-convergence into norm-convergence? If $\{T_i\}_{i\in I}$ is a bounded net of operators on a Hilbert space $\mathscr H$,  converging strongly to some operator $T$, and
if $K$ is  a compact operator on $\mathscr H$, then the net
$\{T_iK\}_i$ is known to converge in norm to $TK$.
Question. Is it also true that   $\{KT_i\}_i$  converges in norm to $KT$?

PS:

*

*The present question arouse in the comments following this answer and, while I can't remember ever questioning it, neither do I remember this being discussed anywhere.  After a while I
finally figured out the answer and I thought it would be nice to record it here.

*An affirmative answer to my question is implicitly assumed in the statement of this question.

 A: This may come as a surprise but the answer is no.  In fact it is no in a very strong way!!
There exists a sequence $\{T_n\}_n$ of bounded operators, strongly converging to zero, such that $\{KT_n\}_n$ does not converge to zero in norm for every nonzero bounded operator $K$, regardless of whether $K$ is compact or even finite rank!
Let $S:\ell ^2\to \ell ^2$ be the so called unilateral shift given by $$ S(x_0, x_1, x_2, x_3, \ldots ) = (0, x_0, x_1, x_2, \ldots ).  $$ The adjoint of $S$ turns out to be given by $$ S^*(x_0, x_1, x_2, \ldots ) = (x_1, x_2, x_3, \ldots ), $$ from where one has that $S^*S$ is the identity operator.
Now let $T_n={S^*}^n$, so that $$ T_n(x_0, x_1, x_2, \ldots ) = (x_n, x_{n+1}, x_{n+2}, \ldots ), $$ and hence it is clear that $T_n\to 0$ strongly.
If $K$ is any bounded operator (compact or not) we have that $$ \|K\| = \|K{S^n}^*S^n\| \leq \|K{S^n}^*\|\|S^n\| = \|K{S^n}^*\| = \|KT_n\|, $$ so we can't have $\|KT_n\|\to 0$, unless $K=0$.


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*As mentioned in the original post, if $K$ is compact, it is well known that "$T_n\to 0$ strongly" implies "$T_nK\to 0$ in norm".


*If $T_n^*\to 0$ strongly (e.g. if $T_n\to 0$ in the $^*$-strong topology, or if $T_n$ is self-adjoint and converges to zero strongly) then $T_n^*K^*\to 0$ in norm by (1) and hence $KT_n\to 0$ in norm as well.
