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

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.

• Do you already know that the norm on a Hilbert space is weakly lower semicontinuous? Commented Jul 23, 2020 at 16:27
• @MaoWao Yes, this is well known Commented Jul 23, 2020 at 16:59

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.
• I don't understand how $T \mapsto T^*T$ is continuous here: If $S$ is the right shift on $\ell^2$ and $T_n = S^n$, then certainly $T_n$ is bounded, as $||T_n|| \leq 1$ and $T_n \to 0$ in WOT, but $T_n^*T_n = I$, so clearly $T_n^* T_n \not \to 0$ here Commented Jul 23, 2020 at 18:25