# Weak*-norm continuous operators on von Neumann algebras

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?

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.

• Dear Mateusz, thank you for the thorough answer. This satisfyingly settles the case of "internal" weak*-norm-continuous operators (i.e., thus given by mutliplication by an element of $M$.) The situation might be different if one considers maps $T\colon M\to M$. In particular, if $F\subseteq M$ is finite-dimensional, and if $T\colon M\to F$ is normal, then $T$ is also weak*-norm-continuous. Thus, there tend to be many such maps, and I guess that if $M$ is hyperfinite then there are enough to approximate the identity on $M$ in some sense. – Hannes Thiel Jan 31 '17 at 9:56
• Yes, that's correct. Such a map has to have finite rank because it is, in particular, weak-norm continuous, where I really mean the weak topology in the Banach space meaning; it is known that only finite rank maps have this continuity property. – Mateusz Wasilewski Jan 31 '17 at 13:02

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:

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

1. 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. ]