Von Neumann Algebras and Convergence in the Strong$^\star$ Topology Let $\mathbb{A}$ be a von Neumann algebra, and let $f : \mathbb{A} \rightarrow \mathbb{A}$ be an ultra-continuous mapping (normal mapping) and let $(x_n)_{n\in \mathbb{N}}$ be a bounded sequence of $\mathbb{A}$ such that  $x_n \rightarrow x$ in the strong$^\star$ topology.
In this case, can it be inferred that the sequence $(f(x_n))_{n\in \mathbb{N}}$ is convergent in the strong$^\star$ topology to $f(x)$ ?
 A: Yes, provided that the map $f$ is completely positive. Here is a proof following the approach presented in Blackadar's encyclopedia; I will assume that $f$ is also contractive, but this can be assumed without loss of generality.
Let $f$ be completely positive and assume that $f$ is normal, i.e. continuous for the ultraweak topologies. Take a bounded net $(x_i)$ in $A$ so that $x_i\to0$ in the strong-star topology. Then we have that $x_i\to0$ and $x_i^*\to0$ strongly, so $x_i^*x_i\to0$ strongly (because multiplication is strongly continuous on bounded sets), so $x_i^*x_i\to0$ weakly, thus $x_i^*x_i\to0$ ultraweakly (because the weak and ultraweak topologies agree on bounded sets). But then since $f$ is ultraweakly continuous (this is what normal means) we have that $f(x_i^*x_i)\to0$ ultraweakly. Now since $f$ is completely positive, we have the inequality
$$0\leq f(x_i)^*f(x_i)\leq f(x_i^*x_i)\to0 $$
so $f(x_i)^*f(x_i)\to0$ ultraweakly, thus $f(x_i)^*f(x_i)\to0$ weakly; but then if $\xi\in H$ where $A\subset B(H)$ then
$$\|f(x_i)\xi\|^2=\langle f(x_i)\xi,f(x_i)\xi\rangle=\langle f(x_i)^*f(x_i)\xi,\xi\rangle\to0$$
which proves that $f(x_i)\to0$ strongly.
Similarly, $x_ix_i^*\to0$ ultraweakly, hence $f(x_ix_i^*)\to0$ ultraweakly, hence $f(x_i)f(x_i)^*\to0$ ultraweakly, hence $f(x_i)^*\to0$ strongly and this shows the strong-star convergence of the net $\{f(x_i)\}_{i}$ to $0$, which is what we wanted to prove.
