Let $H$ be a Hilbert space. Let $T:dom(T) \rightarrow H$ be a densely defined operator. Suppose there is a sequence $\{u_j \}$ in the domain of $T$ such that $u_j \rightarrow u$, and $||Tu_j||$ is bounded, prove that exists elements $w_k \rightarrow u$, such that $Tw_k \rightarrow \alpha$ for some $\alpha \in H$.
The proof I am given is as follows:
- we take a subsequence $u_{j_k}$ such that $\{ Tu_{j_k} \}$ is weakly convergent in $H$.
- Observe weak and norm closure of convex sets coincide.
- By consdiering the convex hull of $\{Tu_{j_k}\}$ there is a suitable sequence sequence $w_k$ each of which is convex combination of $\{u_{j_k}\}$ such that $Tw_k$ converges in norm.
I could see how we can choose $w_k$ such that $Tw_k$ converges in norm. But I don't see why we can choose such that $w_k \rightarrow u$.
The sequence $(Tu_j)$ is bounded, so it has a weakly convergent subsequence, say, $(Tu_{j_k})$. By a theorem of Banach-Saks, a weakly convergent sequence in a Hilbert space has a subsequence whose arithmetic means converge in norm. Therefore, there is a further subsequence, denoted for simplicity by $(Tu_l)$, such that the sequence $$\frac{1}{n}\sum_{l=1}^nTu_l$$ converges in norm. Put $w_n=\frac{1}{n}\sum_{l=1}^nu_l$. Since $u_k$ converges to $u$, so does its subsequence $u_l$, and so does the sequence of means $w_n$. Thus, $Tw_n$ converges in norm, and $w_n$ converges to $u$, as required.