It looks very easy but I must admit I am struggling with this problem.


Okay, let $M$ be a von Neumann algebra acting on a Hilbert space $H$ and let $K$ be another Hilbert space. Suppose $h\colon M\to B(K)$ is a \*-homomorphism which is continuous in the strong operator topology. Is the image $N=h(M)$ a von Neumann algebra?



Is the image of a vNA under a SOT-continuous morphism still a vNA?
If you consider $M / \mathrm{ker}h$ instead of $M$ ($\mathrm{ker}h$ is SOT-closed, so you end up with a von Neumann algebra), you may assume that $h$ is injective, hence isometric, since it as a $\ast$-homomorphism between $C^{\ast}$-algebras. We now wish to show that its image is SOT-closed. This is probabliy not very wise of me, but now I use the fact that strong continuity implies complete additivity (very easy) and this implies ultraweak (i.e. weak$^{\ast}$) continuity (somewhat technical)$^{1}$. Now pick $x$ in ultraweak closure of $h(M)$; we may assume that it is self-adjoint. By Kaplansky density theorem, you may approximate it in strong operator topology by elements from $h(M)$, which have norm not greater than $\|x\|$. So, we have a bounded net $(x\_{i})$ converging to $x$ in strong operator topology, hence also ultraweakly. Since $h$ is isometric, we can go back to $M$ and there we have a bounded net $(h^{-1}(x\_{i}))$. Since balls are ultraweakly compact, there exists a subnet, which converges in ultraweak topology, and ultraweak continuity implies that $x$ is, in fact, contained in the image of $h$. 


Well, this turned out to be not very pleasant method but I hope it is quite clear what is happening.


$^{1}$Starting from this moment, one can also use Krein-Smulian theorem and it is probably a better way.


