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Let $u_n\in H^1(\Omega),\ \Omega\subset\mathbb{R}^N, N\geq 1$ be functions such that $u_n\to u\in L^2(\Omega)$ and the convergence takes place in $L^2(\Omega)$. Is it true that $u\in H^1(\Omega)$?

So, in general, are Sobolev spaces $W^{k,p}(\Omega)$ closed subsets in $L^p(\Omega)$ with the norm $\Vert\cdot\Vert_p$?

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No, not in general. Take some $u \in L^p(\Omega)$ that is not in $W^{k, p}(\Omega)$. I do not know your $k, p, \Omega$, but sometimes you can easily construct those with characteristic functions on (half-)lines. E.g. $\chi_{[0, 1]}(x)$ has no weak derivative on $[-1, 1]$.

It is well known that then there is some sequence $(u_n)_{n \in \mathbb{N}} \subseteq C_0^\infty(\Omega) \subseteq W^{k, p}(\Omega)$ that converges to $u$ in $L^p$. You can research this e.g. in Alt's linear functional analysis.

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  • $\begingroup$ In general, I think the characteristic function of a half space, or, more precisely, of $\{x\in\Omega\mid x_1\leq a\}$, has weak partial derivative in $x_1$-direction given by the surface measure of $\{x\in\Omega\mid x_1=a\}$, which is not a function. Of course one has to choose $a$ appropriately to guarantee a "large enough" intersection. $\endgroup$
    – MaoWao
    Feb 9, 2022 at 15:07
  • $\begingroup$ Yes. That is why I said "sometimes". Sometimes also $\log \log \lvert x \rvert$ works. Depends on your situation $\endgroup$ Feb 9, 2022 at 17:36

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