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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!

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    I agree with you: probably my suggestion works only if $p=2$. Generally speaking, the closure of $C_0^\infty(\mathbb{R}^N)$ with respect to the norm $\left( \int |\nabla u|^2 \right)^{1/2}$ gives you $\{u \in L^{2^*} \mid \nabla u \in L^2\}$. There is no hope, in general, to interpolate and get $u \in L^p$ for other values of $p$. However, I can't find a counter-example.2012-05-06

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I do not if this hint really works, I'm a bit confused. Anyway, I propose it. By the Gagliardo inequality, there exists a constant $C>0$ such that $\|\varphi_n\|_{L^{2^*}} \leq C \|\nabla \varphi_n\|_{L^{2}}.$ Hence, by assumption, $\varphi_n \to 0$ strongly in $L^{2^*}$. Up to subsequences, $\varphi_n \to 0$ almost everywhere. Actually, by elementary interpolation, $\varphi_n \to 0$ strongly in $L^q$ for every $q \in (p,2^*)$. Question: can we conclude now that $\varphi=0$?

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    Can we move on like this: $\varphi_n\rightharpoonup \varphi$ in $ L^p(\Omega)$ implies $\varphi_n \to \varphi$ in $\mathcal{D}'(\Omega)$ (in the sense of distribution), $\varphi_n \to 0$ in $L^q(\mathbb{R}^n)$ is actually $\varphi_n \to 0$ in $L^q(\Omega)$, which implies $\varphi_n \to 0$ in $\mathcal{D}'(\Omega)$. So we obtain $\varphi=0$ a.e. This is something like my former question [link](http://math.stackexchange.com/questions/140811/do-the-two-limits-coincide). But I'm not so sure about my "proof".2012-05-06