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We know that $W^{1,p}(\Omega)\hookrightarrow L^q(\Omega)$ if $\Omega\subset\mathbb R^N$ is a bounded open and $\partial \Omega$ is $C^1$, $1\leq p<N$ and $1\leq q<p^*:=\frac{Np}{N-p}$. Moreover, for every bounded sequence $(u_n)$ in $W^{1,p}(\Omega)$ (unless subsequence), $u_n\rightarrow u_0$ in $L^q(\Omega)$. I want to $u_0\in W^{1,p}(\Omega)$, but I only know that $u_0\in L^q(\Omega)$. Is it true that $u_0\in W^{1,p}(\Omega)$? and how can I prove it?

My attempt:

Since $W^{1,p}(\Omega)$ is reflexive, we obtain $u_n\rightharpoonup u_0$ in $W^{1,p}(\Omega)$. We conclue with compact embedding.

The problem in my attempt is that $W^{1,p}(\Omega)$ is not reflexive for $p=1$.

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  • $\begingroup$ For $p=1$, try to construct a sequence for which $u'_n$ converges (in some sense) to a Dirac delta functional. $\endgroup$
    – daw
    Commented Jul 21, 2018 at 19:44
  • $\begingroup$ Even with this hint, I did not make it. $\endgroup$ Commented Jul 22, 2018 at 18:22
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    $\begingroup$ @daw I don't think that gives a valid counterexample as you seem to suggest taking $N=1$; the questions is considering the case $N > p =1.$ $\endgroup$
    – ktoi
    Commented Jul 23, 2018 at 11:59

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