# WOT convergence to SOT convergence

Let $H$ be a Hilbertspace and $T_n \in B(H)$ a sequence of operators with $T_{n+1} \geq T_{n}$. I want to to show that if there is a self-adjoint $T\in B(H)$ with $T_n \stackrel{WOT}{\rightarrow}T$ then $T_n \stackrel{SOT}{\rightarrow}T$ ($T$ possibly different?). There is a hint that should first show that $(T-T_n)^2\leq ||T-T_n||(T-T_n)$ which I havbe done. Then I get for $x \in H$ $$||T-T_n(x)||^2=|\langle (T-T_n)^2(x),x\rangle |\leq ||T-T_n|| |\langle (T-T_n)(x),x \rangle|$$ Now I know that

$|\langle (T-T_n)(x),x \rangle|\rightarrow 0$, what I do not know is that $||T-T_n||$ is bounded. Am I on the right track here or am I misinterpreting the hint? Thank you.

• $0 \leqslant T-T_n \leqslant T$ – Daniel Fischer Nov 5 '14 at 10:47
• Thank you for your reply. But wouldn't that mean that $T_n \geq 0$ for all $n \in \mathbb{N}$? – Zolf69 Nov 5 '14 at 11:04
• Oh, right, $0 \leqslant T-T_n \leqslant T - T_0$. – Daniel Fischer Nov 5 '14 at 11:05

You can bound $\|T - T_n\|$ using the uniform boundedness principle.
Fix $x \in H$ and let $f_n(y) = \langle T_n x, y \rangle$. Clearly $f_n$ is a bounded linear functional on $H$. Moreover, for each $y$, since $f_n(y) \to \langle Tx, y\rangle$ by WOT convergence, in particular $|f_n(y)|$ is bounded. Then the uniform boundedness principle asserts that $\|f_n\|$ is bounded. But $\|f_n\| = \|Tx\|$, so $\|Tx\|$ is bounded for each $x$. Using the uniform boundedness principle again, $\|T_n\|$ is bounded. To conclude, $\|T-T_n\| \le \|T\| + \|T_n\|$ by the triangle inequality.