Pretty straightforward question. Let $\mathscr{H}$ be a Hilbert space and $\mathscr{L}(\mathscr{H})$ the algebra of bounded operators on $\mathscr{H}$. Define the strong-$*$ operator topology to be the topology on $\mathscr{L}(\mathscr{H})$ generated by the seminorms $\{ T \mapsto \| T v \|, T \mapsto \|T^* v \| : v \in \mathscr{H} \}$. Is the map $F : x \mapsto x^* x$ continuous in the strong-* operator topology?
Here's what I know so far. The adjoint map $x \mapsto x^*$ is continuous in the strong-* operator topology. Multiplication of operators is generally separately continuous (i.e. the maps $x \mapsto x y, x \mapsto yx$ are continuous for fixed choices of $y$) in the SOT, but is not separately continuous (i.e. the map $(x, y) \mapsto x y$ is not continuous).
My guess (and it's just a guess) is that the map $F(x) = x^* x$ should be strong-* operator topology-continuous, but I don't know how to show this or find a counterexample.
I know the claim is false when we look at the weak operator topology. Let $S : \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ be the shift operator $S (a_1, a_2, a_3, \ldots) = (0, a_1, a_2, a_3, \ldots)$. Then $S^*(a_1, a_2, a_3, \ldots) = (a_2, a_3, a_4, \ldots)$, so $S^* S = I$, and $S S^*$ just replaces the first coordinate with $0$. In this scenario, we have that the sequences $(S^n)_{n \in \mathbb{N}}, ((S^*)^n)_{n \in \mathbb{N}}$ both converge weakly to $0$, but $(S^*)^n S^n = I$ for all $n \in \mathbb{N}$. This is not, however, a counterexample for the strong-* operator topology, since though $(S^*)^n$ converges strongly to $0$, the sequence $(S^n)$ only converges weakly (not strongly) to $0$.