If $\|x_n-u_n\|\to 0$ and $x_{n_{i}}\rightharpoonup w,$ then $u_{n_{i}} \rightharpoonup w.$

Question

Assuming that $\{u_n\}$ and $\{x_n\}$ are bounded sequences in a Hilbert space, $H$. Also, if $\|x_n-u_n\|\to 0$ and $x_{n_{i}}\rightharpoonup w,$ how does $u_{n_{i}} \rightharpoonup w?$

My trial By Eberlein-Smul'yan Theorem, there exists a subsequence $\{u_{n_{k}}\}$ of $\{u_n\}$ such that $u_{n_{i}} \rightharpoonup u^*\in H$.

Question: How do I show that $u^*=w$ or from $\|x_{n_{i}}-u_{n_{i}}\|\to 0,$ that $u_{n_{i}} \rightharpoonup w?$

Answer 1

We have $\Vert x_n - u_n \Vert \rightarrow 0$ and thus $x_n - u_n \rightharpoonup 0$ (strong convergence implies weak convergence, this follows directly from Cauchy-Schwarz) and therefore $x_{n_i} - u_{n_i} \rightharpoonup 0$. However, if also $x_{n_i} \rightharpoonup w$, then we obtain $u_{n_i} = (u_{n_i} - x_{n_i}) + x_{n_i} \rightharpoonup 0 + w =w$.