# Do we have $\varphi=0$ a.e.?

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?

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Can't you conclude that $\varphi_n \to 0$ strongly in $L^p_{\mathrm{loc}}$? –  Siminore May 6 '12 at 8:22
@Siminore: Forgive me if I misunderstand your comment. Do you mean deal with the sequence in some compact subset $K$ of $\Omega$? To my knowledge, since each $\varphi_n$ has different support, I can't use Poincare inequality in H^1(K), and neither Rellich-Kandrachov theorem. If I misunderstand your comment, please tell me more details. Thanks! –  Y.Z May 6 '12 at 8:30
It seems to me that Rellich's theorem states that $H^1(\mathbb{R}^N)$ is compactly embedded into $L^p_{\mathrm{loc}}$ for subcritical values of $p$. –  Siminore May 6 '12 at 8:52
@I think I got your point: $\{\varphi_n\}$ bounded in $H^1(\mathbb{R}^n)$ implies bounded in $H^1(B)$ for any ball, by Rellich's theorem have a strongly convergent subsequence, say, $\{\varphi_n\}$ in $L^p(B)$, which means compactly embedded into $L_{\mathrm{loc}}^p$. But I still can't conclude $\varphi_n \to 0$ strongly in $L_{\mathrm{loc}}^p$. I can't even show $\{\varphi_n\}$ is bounded in $H^1(B)$! Would you please tell me more details? Or post an anwser? Thanks! –  Y.Z May 6 '12 at 10:07
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. –  Siminore May 6 '12 at 10:12

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 $\varphi=0$?
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. But I'm not so sure about my "proof". –  Y.Z May 6 '12 at 12:30