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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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up vote 1 down vote accepted

Hint: Denote the inner product on $H$ by $\langle,\rangle$. Then $\lim_{k\to\infty}\langle u,u_k\rangle=\langle u,u\rangle$.

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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
Thank you for the comment. – Ann Nov 12 '12 at 12:12
@Ann: You are welcome. – 23rd Nov 12 '12 at 12:37

For a general normed space it is enough to use that the norm is weakly sequentially lower semicontinuous, then $$ ||u|| \leq lim\ inf\ ||u_k|| \leq lim\ sup\ ||u_k|| \leq R.$$

Also notice (it is not clear from the statement of the problem that this fact is known to you) that every weakly convergent sequence is bounded. To see this you may use the canonical embedding $J:X \rightarrow X'',J(u)(f)=f(u)$. Because the sets $\{J(u_k)(f)\}$ are bounded for every $f \in X'$, the set $\{J(u_k)\}$ is bounded by the uniform boundedness principle. But then you use that $||J(u_k)||=||u_k||$ and you are done.

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