# Boundedness of a weakly convergent sequence of functions

Let $u_k \in X$ be a sequence of functions such that $\| u_k \|_X \leqslant R < \infty$ for all $k$. Let $X$ be Hilbert space. If there exists $u \in X$ such that $u_k$ converges weakly to $u$ in $X$, i.e. $$u^k \longrightarrow u \;\;(weakly) \;\;\;\text{in} \;\;\; X \;\;\;(k \to \infty)$$ then how can I show that $\| u \|_X \leqslant R$ ?

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Hint: Denote the inner product on $H$ by $\langle,\rangle$. Then $\lim_{k\to\infty}\langle u,u_k\rangle=\langle u,u\rangle$.
Thank you very much! Then is the argument below right? " $\| u \|^2 \leqslant \lim_{k \to \infty} | < u_k, u> | \leqslant \lim_{k \to \infty} \| u_k \| \| u \| \leqslant R \| u \|$. –  Ann Nov 12 '12 at 11:52
@Ann: Yes, you are right. It would be better to write $\limsup_{k\to\infty}\|u_k\|$, because the limit may not exist. –  23rd Nov 12 '12 at 12:06