Sequential continuity and continuity of WOT/SOT

Question

Let $H$ be a Hilbert space and $L(H)$ denote the space of bounded linear operators. Let $\phi:L(H)\rightarrow \mathbb{C}$ be a sequentially continuous map with respect to (wrt) WOT/SOT, then is it also continuous wrt WOT/SOT?

Answer 1

Not necesarily. Any ultraweakly continuous state is of the form $\phi=\sum\limits_{k=1}^\infty \langle \cdot x_k, x_k \rangle$, where the $x_n$s can be assumed to be mutually orthogonal and $\sum\limits_{k=1}^\infty ||x_k||^2=1$, and will be sequentially WOT/SOT-continuous, but not weakly/strongly continuous if all $x_k$s are non-zero.

Indeed, if $T_n \rightarrow T$ weakly then $\langle T_n x, y\rangle \rightarrow \langle Tx, y\rangle$ for all $x, y\in H$. In particular, the set $\{ \langle T_n x, y\rangle: n \in \mathbb{N}\}$ is bounded for each $x$ and $y$ in $H$. Fixing $x\in H$ and letting $y$ range across $H$ this means that the collection of bounded linear functionals $\{\langle T_n x, \cdot\rangle : n\in \mathbb{N}\} \subset H^*$ is pointwise bounded. By the uniform boundedness principle (UBP), this means that the $\langle T_n x, \cdot\rangle$s are uniformly bounded in norm, i.e. $\{||T_nx||: n\in \mathbb{N}\}$ is bounded for each $x\in H$. Applying the UBP again, this means that $\{||T_n||: n\in \mathbb{N}\}$ is bounded. A simple approximation argument then shows that $\phi(T_n) \rightarrow \phi(T)$.

The above argument shows that $\phi$ is weakly/strongly sequentially continuous. It is easy to see that $\phi=\sum\limits_{k=1}^\infty \langle \cdot x_k, x_k \rangle$ cannot be written in the form $\sum\limits_{n=1}^N \langle \cdot y_n, y_n' \rangle$ so is not weakly/strongly continuous.