It is well known that if $K$ is a compact operator on a Hilbert space, and $T_n$ is a sequence of operators converging strongly to $T$, then $K T_n$ converges in norm to $K T$.
Question : are there other operators $K$ that also satisfy this property, but are not compact ?
Thanks
The statement made by the OP that:
It is well known that if $K$ is a compact operator on a Hilbert space, and $T_n$ is a sequence of operators converging strongly to $T$, then $K T_n$ converges in norm to $K T$.
is being questioned here. Nevertheless I have no qualms with the corresponding statement in which $K T_n$ and $KT$ are replaced by $T_nK$ and $TK$, so I will read the present question with that modification.
My goal it thus to give a negative answer, namely to prove that if an operator $K$ has this property, then it is necessarily compact.
$\newcommand{\one}{{\mathbb 1}}$ Lemma. Let $T$ be a bounded operator on a Hilbert space $H$ which is not compact. Then there exists an infinite dimensional subspace $K\subseteq H$, and a constant $c>0$, such that $$ \|T\xi \|\geq c\|\xi \|, \quad \forall \xi \in K. \tag 1 $$
Proof. Let us assume first that $T$ is positive semi-definite. For each $n\in \mathbb N$, let $\one_n$ be the characteristic function of the interval $[1/n, +\infty )$, and observe that $$ x\one_n(x) \ {\buildrel {n\to\infty}\over\longrightarrow}\ x, $$ uniformly on $[0,+\infty )$, so it follows that $$ T\one_n(T) \to T, \tag 2 $$ in norm. Notice that the $\one_n(T)$ form an increasing family of self-adjoint projections and we claim that they cannot all have finite rank. This is because otherwise $T\one_n(T)$ would also be finite rank, and then $T$ would be compact by (2).
Choosing any $n$ such that $\one_n(T)$ has infinite rank, let $K$ be the range of $\one_n(T)$, hence an infinite dimensional subspace. The proof will then be concluded once we show that $$ \|T\xi \|\geq \frac1n \|\xi \|, \quad \forall \xi \in K. \tag 3 $$ To prove this notice that $$ x\one_n(x) \geq \frac{\one_n(x)}n, \quad \forall x \in [0,+\infty ), $$ so we have that $$ nT\one_n(T) \geq \one_n(T). $$ For all $\xi $ in $K$, a.k.a. the range of $\one_n(T)$, we then have that $$ \|\xi \|^2 = \langle \xi , \xi \rangle = \langle \one_n(T)\xi , \xi \rangle \leq n\langle T\one_n(T)\xi ,\xi \rangle = n\langle T\xi ,\xi \rangle \leq n\|T\xi \|\|\xi \|, $$ from where (3) follows.
Returning to the general case, in which $T$ is only assumed to be a bounded operator, let $|T|=(T^*T)^{1/2}$, which is positive semi-definite. Observing that $$ \|T(\xi)\| = \|\, |T|(\xi) \, \|,\quad \forall \xi \in H, $$ we see that any pair $(K,c)$ that works for $|T|$, also works for $T$. QED
Theorem. Let $T$ be an operator on a Hilbert space $H$ such that, whenever $\{S_n\}_n$ is a sequence of bounded operators, strongly converging to zero, one has that $S_nT\to0$ in norm. Then $T$ is compact.
Proof. Assuming by contradiction that $T$ is not compact, use the Lemma to find an infinite dimensional subspace $K\subseteq H$, and a constant $c>0$, such that (1) holds.
Setting $L=T(K)$, is is easy to see that $T$ provides a bounded, linear, invertible map $$ \check T:K\to L, $$ with bounded inverse and, in particular, that $L$ is a closed infinite dimensional subspace of $H$.
One may then easily find a sequence of operators $\{S_n\}_n\subseteq B(L)$ which converges to zero strongly, but not in norm. Extending each $S_n$ to $H$ by setting it to be zero on $L^\perp$, it is clear that the extended operators $\hat S_n$ also satisfy $\hat S_n\to0$ strongly, but not in norm.
By the assumption on $T$ we have that $\|\hat S_nT\|\to 0$, so $$ \|S_n\| = \|S_n\check T\check T^{-1}\| \leq \|S_n\check T\| \|\check T^{-1}\| = $$$$ = \|\hat S_nT|_K\| \|\check T^{-1}\| \leq \|\hat S_nT\| \|\check T^{-1}\| \to 0 $$ contradicting the choice of $S_n$. QED
