# In compact embedding Theorem, $u_0$ lies in $W^{1,p}(\Omega)$?

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 $p=1$.

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