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Let $\Omega\subset\mathbb{R}^3$ be a bounded, open domain with a nice boundary. By the Rellich-Kondrakov theorem, $H^1(\Omega)$ is compact in $L^2(\Omega)$. Suppose $u_n\rightharpoonup u$ in $H^1$, and that $\|u_n\|_2=1$ for all $n$. It follows that the convergence is strong in $L^2$ by taking a subsequence. Call this limit $v$.

Question: Is $u = v$? I suspect not, but would like a counterexample!

Suppose $u\neq v$. Then $u_n$ cannot converge to $v$ weakly, since weak limits are unique. Then there is at least one $f\in H^{-1}$ such that $\langle u_n-v, f\rangle\nrightarrow 0$.

Considering that $\langle u_n-v, g\rangle\rightarrow 0$ for all $g\in L^2\subset H^{-1}$, one must find $f$ outside $L^2$.

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up vote 2 down vote accepted

$\def\weakto{\rightharpoonup}$ We have $u = v$: Since $u_n \weakto u$ in $H^1$, so $u_n \weakto u$ in $L^2$. As $u_n \to v$ in $L^2$, we have - as norm convergence implies weak convergence - that $u_n \weakto v$ in $L^2$. Weak limits are unique, hance $u = v$.

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Thanks! It was actually that simple ... –  Simen K. Sep 19 '12 at 11:06

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