Let $H$ a Complex Hilbert space, $T, T_n \in B(H)$. Prove that $T_n x \to Tx$.

Question

I need help for the next exercise:

Let $H$ a complex Hilbert space and $T, T_n \in B(H)$ for $n \geq 1$ and $\|T_{n} x\| \to \|T x\|$ and $\langle T_n x,x \rangle \to \langle Tx,x \rangle$ for all $x \in H$. Prove that $T_n x \to Tx$ for all $x \in H$.

I tried to do the following: $\| T_n x - Tx \|^2 = \langle T_n x - Tx, T_n x - Tx \rangle = \langle T_n x, T_n x \rangle - \langle T_n x, Tx \rangle - \langle Tx, T_n x \rangle + \langle Tx, Tx \rangle = \|T_n x\|^2 + \| Tx \|^2 - (\langle T_n x, Tx \rangle + \langle Tx, T_n x \rangle)$ I think that $\langle T_n x, Tx \rangle + \langle Tx, T_n x \rangle \to \langle Tx,Tx \rangle + \langle Tx, Tx \rangle$

but I really couldn't prove this using the fact that $\langle T_n x,x \rangle \to \langle Tx,x \rangle$.

I appreciate if someone can help me. Buen día a todos.

Answer 1

You cannot prove it, since it is false. Let $H$ be $\Bbb R^2$, with its standard inner product. Take $T_n(a,b)=(-1)^n(b,-a)$, and $T(a,b)=(-b,a)$. Then:

• each $T_n$ has norm $1$, and so does $T$, which implies that $\lim_{n\to\infty}\|T_n\|=\|T\|$;
• for each $n\in\Bbb N$ and each $x\in\Bbb R^2$, $\langle T_nx,x\rangle=\langle Tx,x\rangle=0$, and therefore $\lim_{n\to\infty}\langle T_nx,x\rangle=\langle Tx,x\rangle$ .

But you don't have $\lim_{n\to\infty}T_n=T$.