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

Note that by using the polarization identity it follows that $\langle T_nx,y\rangle\to \langle Tx,y\rangle$ for all $x,y\in H$.

Now fix a constant $x\in H$. By letting $y=Tx$ we have $\langle T_nx, Tx\rangle\to \langle Tx,Tx\rangle$ and $\langle Tx, T_nx\rangle=\overline{\langle T_nx, Tx\rangle}\to \overline{\langle Tx,Tx\rangle}=\langle Tx,Tx\rangle$. Combine this with the attempt you made, and you get the result.