Operators that improve strong convergence to norm convergence? 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 
 A: 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
