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Let $H$ be a Hilbert space and assume that $u_n$ converges weakly to $u$ in $H$. Then I want to prove that $$ \lim_{n \to \infty} \| u_n - u \|_H = 0 \;\;\Longleftrightarrow \;\; \| u \|_H \geqslant \limsup_{n \to \infty} \| u_n \|_H.$$ How can I show the equivalence of these two statements?

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$\def\norm#1{\left\|#1\right\|_H}\def\sp#1{\left\langle#1\right\rangle}$First note that we have by weak convergence \begin{align*} \norm u^2 &= \sp{u,u}\\ &= \lim \sp{u,u_n}\\ &\le \liminf \norm{u}\norm{u_n}\\ &= \norm u\cdot \liminf \norm{u_n} \end{align*} And hence $\norm u \le \liminf \norm{u_n}$. So your right hand side is equivalent to $\lim \norm{u_n} = \norm u$. Now note that \begin{align*} \norm{u_n - u}^2 &= \norm{u_n}^2 - 2\Re\sp{u_n,u} + \norm u^2\\ \end{align*} So if $\norm{u_n} \to \norm u$, the left hand side converges to $\norm u^2 - 2\norm{u}^2 + \norm u^2 = 0$. On the otherside, if $\norm{u_n - u} \to 0$, the inverse triangle inequality gives $$ \left|\norm{u_n}-\norm u\right| \le \norm{u_n-n} \to 0 $$ so $\norm{u_n} \to \norm u$.

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