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?



Weak*-norm continuous operators on von Neumann algebras
Since taking the adjoint is continuous in both topologies, the set of $h$ such that the multiplication map has the said continuity property is $\ast$-closed, so we may assume that $h$ is self-adjoint -- let's take a look at its spectral decomposition. Suppose that the spectrum has a continous bit, say that $b\neq 0$ is an element of the spectrum that is not an eigenvalue. Then the sequence of spectral projections $E\_{\mathbb{R} \setminus (b-\frac{1}{n}, b + \frac{1}{n})}$ converges to identity in weak\*-topology but the images do not converge in norm. Therefore any non-zero element of the spectrum has to be an eigenvalue. Let's take a look at eigenspaces. If the orthogonal projection $P$ onto the eigenspace can be approximated from below by a sequence of projections in our von Neumann algebra that are not equal to $P$ then we get a contradiction again; this means that $P$ has to be a finite sum of minimal projections. If any set of the form $\mathbb{R} \setminus (-\varepsilon, \varepsilon)$ contains infinitely many eigenvalues then we can also construct an appropriate sequence of projections showing that the multiplication map is not weak\*-norm continuous. Therefore $h$ has to be "compact", where we think of minimal projections as being one-dimensional. It is now important to check whether multiplication by a minimal projection is weak\*-norm continuous.


Since $h$ is built from minimal projections, if we consider the type decomposition of $M$, $h$ only lives in the type I part, we can therefore disregard type II and type III parts. The type I algebras are always direct sums of algebras of the form $B(\mathsf{H}) \otimes L^{\infty}(X,\mu)$, where in our case the measures that appear have to be discrete. 


Let's consider the factor case for a while. If the factor is finite dimensional then of course it works, so consider $B(\ell\_2)$. A minimal projection is just a rank one projection if we consider the natural representation, say $p= |e\_1\rangle\langle e\_1|$, where $(e\_n)\_{n \in \mathbb{N}}$ is the canonical orthonormal basis. The sequence of partial isometries $v\_n := |e\_1\rangle \langle e\_n|$ converges in weak\*-topology to $0$ but the sequence $pv\_n=v\_n$ does not converge to $0$ in norm, so only the finite dimensional factors remain.


The last important case to consider is an algebra of the form $M\_k \otimes \ell\_{\infty}$ (the measure has to be discrete and the finite case is trivial). A minimal projection in this case is of the form $p=|e\_1\rangle\langle e\_1| \otimes \delta\_1$. Let $(A\_n)$ be a sequence of matrices that converges in weak\*-topology. Each $A\_n$ is of the form $A\_n= (f\_{ij}^{n})$, where $f\_{ij} \in \ell\_{\infty}$. We have that $pA\_n$ is only the first row of $A\_n$ multiplied by $\delta\_1$, so it is the following vector $(\delta\_1 f\_{11}^n,\dots, \delta\_1 f\_{1k}^n)$. Its norm is equal to $\| \sum\_{i=1}^{k} |\delta\_1 f\_{1i}|^2\|\_{\infty}$. Because of the $\delta\_1$ factor this function is supported at just one point, so the norm is $\sum\_{i=1}^{k} |f\_{1i}^{n}(1)|^2$ -- this surely converges to $0$ if the sequence $A\_n$ converged to $0$ in weak\*-topology.


The conclusion is that in the type decompostion of $M$ one just needs to consider bits of the form $M\_n \otimes \ell\_{\infty}^{k}$, where $k$ may be infinite.


