# Why $\frac{1}{2}\|\nabla u_n\|^2-\|h\|_{L^2}\|u\|_{L^2}\geq \gamma _3\|u_n\|^2_{W^{1,2}}-\gamma _4$

Let $\Omega \subset \mathbb R^n$ open, bounded with Lipschitz boundary. Let $(u_n)\subset W^{1,2}(\Omega )$ s.t. $u_n^\Omega :=\frac{1}{|\Omega |}\int_\Omega u_n=0.$ Let $h\in L^2(\Omega )$. Then, by Poincaré inequality, we have that $$\|u_n-u_n^\Omega \|_{L^2}\leq C\|\nabla u_n\|_{L^2},$$ for a certain $C>0$. We want to bound $\|u_n\|_{W^{1,2}}$. In my course it's written : There are $\gamma _1,\gamma _2,\gamma _3,\gamma _4>0$ s.t.

$$1\leq \frac{1}{2}\|\nabla u_n\|_{L^2}^2-\|h\|_{L^2}\|u_n\|_{L^2}\underset{(1)}{\geq} \frac{1}{4}\|\nabla u_n\|^2_{L^2}+\gamma _1 \|u_n\|_{L^2}^2-\gamma _2\|u_n\|_{L^2}\underset{(2)}{\geq} \gamma _3\|u_n\|_{W^{1,2}}-\gamma _4.$$

I think that if we have the inequality $(1)$, the inequality $(2)$ follow since we have (for suitable constants $A,B,C,D>0$) that $$\frac{1}{4}\|\nabla u_n\|^2_{L^2}+\gamma _1 \|u_n\|_{L^2}^2-\gamma _2\|u_n\|_{L^2}\geq A\|u_n\|_{W^{1,2}}^2-B\|u_n\|_{W^{1,2}}\geq C\|u\|_{W^{1,2}}^2+D.$$

My question

My problem is that I can't get (1) using Poincaré. Any idea ?

We have that $$\frac{1}{2}\|\nabla u_n\|_2^2 = \frac{1}{4}\|\nabla u_n\|_2^2 + \frac{1}{4}\|\nabla u_n\|_2^2 \geq \frac{1}{4}\|\nabla u_n\|_2^2 + \frac{1}{4C^2}\|u_n\|_2^2,$$ hence (1) follows by choosing $\gamma_1 := 1/(4C^2)$ and $\gamma_2 := \|h\|_2$.
• Ok, but why didn't they do directly $\frac{1}{2}\|\nabla u_n\|_2^2\geq C\|u_n\|_2^2$ by poincaré and conclude ? – MSE Jun 28 '17 at 16:07
• If you want to have $\|u_n\|_{W^{1,2}}^2$ at r.h.s. you need a term containing the norm of the gradient. – Rigel Jun 28 '17 at 16:38