Suppose we have a sequence of smooth functions with compact support: $\{\varphi_n\}\subseteq C_c^{\infty}(\Omega)$, here $\Omega\subseteq \mathbb{R}^n,n\geqslant 2$ is open. Suppose additionally we have $ \varphi_n\rightharpoonup \varphi \text{ in } L^p(\Omega) \text{ (weakly convergence)} $ for some $p< 2^*=\frac{2n}{n-2}$. And $ \nabla\varphi_n\to 0 \text{ in } L^2(\Omega). $ Then do we have $\varphi=0$ a.e.?
If $\Omega$ is bounded, I think I can say yes.
Since we can set a ball $B\supseteq \Omega$ and then $\{\varphi_n\}\subseteq H_0^1(B)$. Poincare inequality implies $\varphi_n\to 0$ in $H_0^1(\Omega)$. By Rellich-Kondrachov theorem there is a subsequence, still denotes by $\{\varphi_n\}$, converges to some $u$ in $L^p(\Omega)$ and hence $u=\varphi$ a.e. That is $\varphi_n\to 0$ in $L^2(\Omega)$ and $\varphi_n\to \varphi$ in $L^p(\Omega)$ and hence $\varphi=0$ a.e.
But what happens if $\Omega$ is unbounded?
In this case I can't use Poincare inequality and Rellich-Kondrachov theorem. Is there an another way to prove it or a counterexample exists?
Any advice will be appreciated!