# *-homomorphism between concrete von Neumann algebras is SOT-SOT continuous iff it is WOT-WOT continuous

Let $\mathcal H, \mathcal K$ be Hilbert spaces and $M \subseteq B(\mathcal H)$ a (concrete) von Neumann algebra (Here, $B(\mathcal H)$ denotes the algebra of bounded operators on $\mathcal H$). Furthermore, let $$\pi \colon M \to B(\mathcal K)$$ be a *-Homomorphism. Note that $\pi$ is automatically a contractive, i.e. especially bounded, operator.

I want to prove that $\pi$ is strong-strong (SOT-SOT) continuous if and only if $\pi$ is weak-weak (WOT-WOT) continuous.

strong-strong means that $B(\mathcal H)$ as well as $B(\mathcal K)$ carry the strong operator topology (SOT). Same goes for the term weak-weak and the weak operator topology (WOT).

I know that the topological duals of $B(\mathcal H)$ w.r.t the strong/weak operator topologies coincide, i.e. a linear functional on $B(\mathcal H)$ is strongly continuous if and only if it is weakly continuous. The same property is inherited by the sub-von Neumann Algebra $M$.

Do you know how to prove this or can you share a reference on this matter?

EDIT. I have asked another more general question which could yield an answer to this concrete problem. See here. However, I am not conviced that there is a closed-graph theorem for the strong and weak operator topologies.

2nd EDIT (06.04.2017)

I think one can prove WOT-WOT implies SOT-SOT in a straigthforward way:

It suffices to prove that if $x_i \overset{\mathrm{SOT}}{\to} 0$ then $\pi(x_i) \overset{\mathrm{SOT}}{\to} 0$.

We know that $$x_i \overset{\mathrm{SOT}}{\to} 0 \iff x_i^* x_i \overset{\mathrm{WOT}}{\to} 0 \tag{\ast}.$$

Now if $x_i \overset{\mathrm{SOT}}{\to} 0$, since $\pi$ is WOT-WOT continuous, ($\ast$) gives that $$\pi(x_i)^* \pi(x_i) = \pi(x_i^* x_i) \overset{\mathrm{WOT}}{\to} 0$$ which, again by ($\ast$) implies that $\pi(x_i) \overset{\mathrm{SOT}}{\to} 0$.