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I'm reading @emily20's question in this thread

Let $\Omega\subset \mathbb{R}^n$ be open and bounded. Also let

$W^{1,2}(\Omega)=\left\{u \in L^{2}(\Omega)|\,\, \forall \alpha \in \mathbb{N}^{n}:|\alpha| \leq 1\,\, \exists \,D^{\alpha} u \in L^{p}(\Omega)\right\}$

denote the sobolev space.

Futher let $u_n\rightarrow u$ converge weakly in $W^{1,2}(\Omega)$. In the lecture it is sad that this implies $u_n\rightarrow u$ strongly in $L^2(\Omega)$. Does anyone have an explaination?

There is @Sub's comment that

This is Rellich-Kondrachov Theorem that says that $W^{1, p}$ is compactly embedded in $L^p$. However, your statement is not completely true. It should rather be $u_n\rightharpoonup u$ in $W^{1, p}$ implies $u_n \rightarrow u$ in $L^p$ up to a subsequence. I denote $u_n\rightharpoonup u$ for $\left(u_n\right)$ converges weakly to $u$ and $u_n \rightarrow u$ for $\left(u_n\right)$ converges strongly to $u$.

We have an exercise in Brezis' textbook Functional Analysis, Sobolev Spaces and Partial Differential Equations, i.e.,

Exercise 6.7 Let $E$ and $F$ be two Banach spaces, and let $T \in \mathcal{L}(E, F)$. Prove that $T \in \mathcal{K}(E, F)$ implies $$ \left\{\begin{array}{l} \text { For every weakly convergent sequence }\left(u_n\right) \text { in } E, \\ u_n \rightarrow u, \text { then } T u_n \rightarrow T u \text { strongly in } F . \end{array}\right. $$

So I think that @emily20's conclusion "$u_n\rightarrow u$ strongly in $L^2(\Omega)$" is correct and that @Sub's comment is wrong. Could you confirm if my understanding is correct?

Akira
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    you are right, by compact embedding you get convergence of the full sequence. Subsequence only come into play at this point when going to pointwise a.e. convergence. – daw Nov 23 '23 at 11:01
  • @daw thank you so much for your verification! – Akira Nov 23 '23 at 11:03
  • However, I think that there is another error lurking around: In order to invoke Rellich-Kondrachov, you need some regularity of $\Omega$ (e.g., a Lipschitz domain) or you have to work with $W_0^{1,p}(\Omega)$. – gerw Nov 23 '23 at 14:35
  • @gerw i guess they assume the domain to be regular enough... – Akira Nov 23 '23 at 17:02

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