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Let $I$ be a bounded interval and consider a sequence $(u_k)$ in $H^{1,2}(I)$ (usual Sobolev space). Suppose furthermore, that the sequence $(u_k)$ is bounded in $H^{1,2}(I)$. Then, by Rellich, we can extract a subsequence, still denoted by $u_k,$ s.t. $u_k$ converges to some $\bar{u}$ in $C^0(I)$. Furthermore, by weak compactness of bounded sets in $H^{1,2}(I)$ we can select a subsequence, s.t. $u_k$ converges to some $u$ weakly in $H^{1,2}(I)$. Thus, $u_k$ converges to $\bar{u}$ in $C^0$ and weakly to $u$ in $H^{1,2}$.

Do these limits coincide, i.e. is it true that $u=\bar{u}$?

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2 Answers 2

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Either type of convergence implies distributional convergence, among other things. So the limits must be the same.

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  • $\begingroup$ Could you elaborate on that? It just seems you are throwing another type of convergence into the box. $\endgroup$
    – Orbicular
    Commented May 19, 2011 at 10:07
  • $\begingroup$ I think this would be a right place to say: "Do your homework!" Just recall the basic definitions, and you will see that $C^0$ and $(H^1)_\sigma$ are continuously injected in $\mathcal D'_\sigma$ . Furthermore, the last space is Hausdorff. $\endgroup$
    – TaQ
    Commented May 19, 2011 at 17:48
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Both convergences imply the weak convergence in $L^2 (I)$, which has a unique limit.

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