Let $H$ be a Hilbert space and $T_n\in\mathcal{L}(H)$. Suppose that for every $x\in H$ we have that $T_nx\rightharpoonup Tx$ where "$\rightharpoonup$" denotes weak convergence. Prove that $T\in\mathcal{L}(H)$ and $\|T\|\leq \liminf_{n\rightarrow \infty} \|T_n\|$.
I can't prove the part $T\in\mathcal{L}(H)$ (although I suppose it needs to use the uniform boundedness principle) and I'm not sure about my solution in the second part. Here it is: assuming $T\in\mathcal{L}(H)$, without loss of generality, we can assume $\|T\|=1$ (otherwise set $U=\frac{T}{\|T\|}$). Then we have $$\|Tx\|=\langle Tx, Tx \rangle = \lim_{n\rightarrow \infty} \langle Tx, T_nx\rangle $$ since $T_nx\rightharpoonup x$ and we bound $\langle Tx, T_nx\rangle$: $$\langle Tx, T_nx \rangle \leq \|Tx\|\cdot \|T_nx\|\leq \|T\|\cdot \|x\|\cdot \|T_n\|\cdot \|x\|\overset{\|T\|=1}{\leq}\|T_n\|\|x\|^2$$ so $$\|Tx\|=\lim_{n\rightarrow \infty}\langle Tx,T_nx\rangle\leq \liminf_{n\rightarrow \infty} \|T_n\|\|x\|^2$$ and by taking supremum over $\|x\|\leq 1$ we have $$\|T\|=\sup_{\|x\|\leq 1}\|Tx\|\leq \sup_{x\leq 1} \liminf_{n\rightarrow \infty} \|T_n\|\|x\|^2 \leq \liminf_{n\rightarrow \infty} \|T_n\|$$ as required.
I would appreciate it if someone could tell me how to prove $T\in\mathcal{L}(H)$ and whether my solution for the other part is correct!
If $T_n x \rightharpoonup Tx $ then $T_n x $ is weakly bounded and hence norm bounded. Therefore for all $x\in H$ the sequence $ ||T_n x||$ is bounded and therefore from uniform boundeness principle the sequence $||T_n ||$ is bounded thus $T$ is a bounded linear operator.