TITLE: Weak*-norm continuous operators on von Neumann algebras
QUESTION [8 upvotes]: Let $M$ be a von Neumann algebra with predual $M_*$, and let $T\colon M\to M$ be a bounded, linear map. Let us say that $T$ is (sequentially) weak*-norm continuous if for every net (sequence) $(a_j)_j$ in $M$ that weak* converges to $a$ in $M$ we have $\lim_j \| T(a_j) - T(a) \| = 0$.
In particular, given $h\in M$, we may ask when the multiplication operator $L_h\colon M\to M$, $a\mapsto ha$, is (sequentially) weak*-norm continuous. 
I suppose such operators on $M$ and elements of $M$ have been considered before. Is there an established terminology or some papers where this has been used?
Is it known when $M$ has an approximate unit $(h_j)_j$ such that multiplication by each $h_j$ is weak*-norm continuous?

REPLY [3 votes]: Your example of a multiplication operator has the property of being weak$^*$-continuous, and in that setting, we can say quite a lot, with an elementary proof.
Claim: Let $E$ be a Banach space and $T:E^*\rightarrow E^*$ be a weak$^*$-continuous operator (that is, the adjoint of some $S:E\rightarrow E$).  Then the following are equivalent:

For any bounded net $(a_j)$ in $E^*$ which is weak$^*$-convergent to 0, we have that $\|T(a_j)\| \rightarrow 0$.
$S$ is a compact operator.
$T$ is a compact operator.

Proof: That $S$ is compact if and only if $T$ is compact is Schauder's theorem.  Set $X:=\{S(\omega):\omega\in E, \|\omega\|\leq 1\}$ and notice that (1) is equivalent to:

For any bounded net $(a_j)$ in $E^*$ which is weak$^*$-convergent to 0, we have that $\langle a_j,\omega\rangle\rightarrow 0$ uniformly on $X$.

Then $S$ is compact if and only if $X$ is compact.  If $X$ is compact, then a simple $\epsilon$-net argument shows that (4) holds (using that the net $(a_j)$ is assumed bounded).  Conversely, suppose that $X$ is not compact, so there is $\epsilon>0$ such that $X$ admits no $2\epsilon$-net.  For any finite-dimensional subspace $N\subseteq E$, any closed and bounded subset of $N$ is compact, and so the distance from $N$ to $X$ must be at least $\epsilon$, say.  By Hahn-Banach we can find $a_N\in E^*$ and $\omega_N\in X$ with $\langle a_N,\omega \rangle=0$ for all $\omega\in N$, with $|\langle a_N, \omega_N \rangle| \geq \epsilon$ and with $\|a_N\|\leq 1$.  The (bounded) net $(a_N)$ hence shows that (4) does not hold.
If $E$ is separable then you can consider (bounded) sequences in place of nets.
Thus, I believe, your original question is equivalent to asking when the (left) multiplication operator given by $h\in M$ is compact.
[ I stressed the word "bounded".  If you allow unbounded nets then the condition becomes that $X$ has finite-dimensional span, i.e. that $S$ (equivalently $T$) is a finite-rank operator.  In the sequence case, the Principle of Uniform Boundedness says you have to be bounded anyway. ]