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Let $u\in W^{1,p}(\Omega )$ where $\Omega= B_2\backslash B_1$ (and $B_r=\{x\in \mathbb R^n\mid \|x\|_2<r\}$). We suppose that $u|_{\partial B_1}=0$. Show that there is a constant $\gamma >0$ s.t. $$\|u\|_{L^p}\leq \gamma \|\nabla u\|_{L^p}.$$

It looks to be almost Poincaré inequality, but to use poincaré we must have $u|_{\partial B_2}=0$ too. So how can I solve this problem ?

MSE
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  • What is $| \cdot |$? If you trace the proof by contradiction of the Poincaré inequality, it doesn't require $u$ to vanish on $\partial B_2$. – Michał Miśkiewicz Apr 18 '17 at 22:14
  • @MichałMiśkiewicz: Thanks, I edited my answer. I tried as you did, but not conclusive. – MSE Apr 19 '17 at 09:46
  • In the lecture we once had the following generalized Poincare-Friedrichs inequality: Let $f \in H^1(\Omega)'$ and $f(1) \neq 0$. Then there exists a constant $C < \infty$ such that $$||v||^2_{H^1} \leq C \big(||\nabla v||^2_{L^2}+f^2(v)\big)$$ – Cahn Apr 19 '17 at 10:51
  • The well-known inequality is taking this one and setting $f(v)=\int_{\partial\Omega} \text{tr} \ u ~\text{d}s$. In the same way set $f(v)=\int_{\partial B_1} \text{tr} \ u ~\text{d}s$ so by the Poincare-Friedrichs inequality (continuity follows from the trace theorem) one gets $||v||^2_{L^2} \leq C ||\nabla v||_{L^2}^2$. So it is enough that the trace operator of $u$ vansishes on part of the boundary. But right now I don't know if there is such an inequality also for $L^p$. – Cahn Apr 19 '17 at 10:51

1 Answers1

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Assume that such $\gamma$ does not exist. Then for each $k = 1,2,\ldots$ there is some $u_k \in W^{1,p}(\Omega)$ such that $u_k = 0$ on $\partial B_1$, $\|u_k\|_{L^p}(\Omega) = 1$ and $\|\nabla u_k\|_{L^p}(\Omega) \le 1/k$ (this can be obtained by scaling).

By Rellich-Kondrashov, we can choose a subsequence such that $u_k \to u$ in $L^p(\Omega)$. Since $\nabla u_k \to 0$ in $L^p(\Omega)$, we conclude that $\nabla u = 0$ and the convergence $u_k \to u$ is also in $W^{1,p}(\Omega)$. By continuity of trace we have $u = 0$ on $\partial B_1$, which together with $\nabla u = 0$ gives us $u = 0$. On the other hand, $\| u \|_{L^p(\Omega)} = \lim_{k \to \infty} \| u_k \|_{L^p(\Omega)} = 1$. This contradiction shows the existence of $\gamma$ satisfying the inequality.

Michał Miśkiewicz
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  • I'm sorry, I don't really see the contradiction. Why the fact that $u=0$ on $\partial B_1$ with $\nabla u=0$ gives $u=0$ on $\Omega $ ? Indeed, $\nabla u=0$ tell us that $u$ is constant a.e. since the boundary has measure $0$, it can be $0$ on the boundary, but be not $0$ on $\Omega $. There is something I don't get here. Thank you. – idm Jan 19 '18 at 14:40
  • If you already know $u \equiv c$ a.e. in $\Omega$, then the trace of $u$ is also the constant function $c$. As you noted, restricting to sets of measure zero doesn't make sense for functions defined a.e., but look up traces of Sobolev functions. – Michał Miśkiewicz Jan 19 '18 at 23:47
  • Could you please give me more details because I'm not totally convinced of the fact that $u=0$ on $\Omega $ (but of course I trust you :-)) – idm Jan 20 '18 at 10:09
  • Sure, but which step is problematic for you? – Michał Miśkiewicz Jan 20 '18 at 10:13
  • Thank you. My problem is from $u=C$ a.e. in $\Omega $ and $u|_{\partial \Omega }=0$, therefore $u=0$ in $\Omega $ – idm Jan 20 '18 at 10:36
  • I already addressed that in the previous comment. If $u = C$ a.e. in $\Omega$, then $u|{\partial \Omega} = C$ in the sense of trace, so (since also $u|{\partial \Omega} = 0$) $C$ has to be zero. I feel that reading about traces should help. – Michał Miśkiewicz Jan 20 '18 at 10:59
  • Thanks a lot for your answer. – idm Jan 20 '18 at 11:34