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 ?
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.