Is the map $T \mapsto T^*T$ lower-semicontinuous in the weak operator topology?

Question

Let $H$ be a separable Hilbert space and $\mathcal{L}(H)$ the set of bounded operators on $H$. If $D \in \mathcal{L}(H)$ is a positive operator, $D \geq 0$, is true that the set $$\{T \in \mathcal{L}(H)\mid T^*T \leq D\}$$ is closed in the weak operator topology (WOT) on $\mathcal{L}(H)$?

Note that the mapping $T \mapsto T^*T$ is not continuous in WOT, like the counterexample $H = \ell^2$, $T_n = S^n$ ($S$ - unilateral (right) shift) shows.

Answer 1

Suppose $T_\alpha \to T$ in WOT and $T_\alpha^*T_\alpha≤D$. What we would like to have is that $T^*T^\mathstrut≤D$, ie $$\langle x , (D-T^*T^{\mathstrut})x\rangle=\langle x,Dx\rangle - \|Tx\|^2\overset!≥0$$ for all $x$. Now we may note that $\|Tx\|=\sup_{\|y\|≤1}|\langle y, Tx\rangle|$, while $$|\langle y, Tx\rangle |= \lim_\alpha |\langle y, T_\alpha x\rangle|≤\liminf_\alpha \|T_\alpha x\|$$ if $\|y\|≤1$, giving $\|Tx\|^2≤ \liminf_\alpha\|T_\alpha x\|^2$. Now using $\langle x ,(D-T^*_\alpha T_\alpha)x\rangle ≥0$ we find: $$\langle x, Dx\rangle -\|Tx\|^2 ≥ \langle x, Dx\rangle -\liminf_\alpha \|T_\alpha x\|^2 =\limsup_\alpha \langle x, (D-T^*_\alpha T_\alpha)x\rangle$$ but for every $\alpha$ the term on the right is $≥0$, giving us the conclusion we wanted.