Poincaré inequality : $\|\nabla u\|_p\geq \gamma _1\|u\|_{W^{1,p}}-\gamma _2\|u_0\|_{W^{1,p}}$

Question

Let $\Omega \subset \mathbb R^n$ a domain. I know Poincaré inequality as $$\|u\|_p\leq \gamma \|\nabla u\|_p$$ for all $u\in W_0^{1,p}$. But I have an other one (which is equivalent) that is $$\|\nabla u\|_p\geq \gamma _1\|u\|_{W^{1,p}}-\gamma _2\|u_0\|_{W^{1,p}},$$ where $u\in u_0+W_0^{1,p}$, i.e. $u=u_0$ on $\partial \Omega $. How can I prove it ?

Answer 1

First, if $u\in u_0+W_0^{1,p}$ then $u-u_0\in W_0^{1,p}$. Moreover, if $u\in W_0^{1,p}$, Poicaré says $$\|u\|_p\leq \gamma \|\nabla u\|_p\implies \|u\|_{W^{1,p}}\leq \tilde\gamma \|\nabla u\|_p.$$ Indeed, $$\|u\|_{W^{1,p}}^p=\|u\|_p^p+\|\nabla u\|_p^p\leq \gamma^p \|\nabla u\|_p^p+\|\nabla u\|_p^p=(1+\gamma ^p)\|\nabla u\|_p^p.$$

So, using this last inequality,

$$\|u-u_0\|_{W^{1,p}}\leq \gamma \|\nabla u-\nabla u_0\|_p\implies \|u\|_{W^{1,p}}-\|u_0\|_{W^{1,p}}\leq \gamma \|\nabla u\|_p+\gamma \|\nabla u_0\|_p$$ $$\implies \gamma \|u\|_{W^{1,p}}-\|u_0\|_{W^{1,p}}\leq \gamma \|\nabla u\|_{p}+\gamma \|u_0\|_{W^{1,p}}.$$ The claim follow.