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## Generalized Poincare Inequality

Consider $\Omega \subset \Bbb{R}^N$ a bounded domain with Lipschitz boundary. If $H$ is a non-zero, closed subspace in $H^1(\Omega)$, which does not contain the non-zero constant functions, then there is a constant $C>0$, depending on $\Omega$, such that $\|u \|_{L^2}\leq C \| |\nabla u | \|_{L^2},\ \forall u \in H$.

Note that this generalizes the usual Poincare inequality, which says that the above inequality holds for some $C>0$ on the space $H_0^1(\Omega)$, a space which does not contain the non-zero constant functions.

Proof: Suppose that the given inequality is not true. Then for every $n \in \Bbb{N}^*$ there exists $v_n \in H$ such that $\displaystyle \frac{1}{n} > \frac{ \| | \nabla v_n | \|_{L^2}}{\|v_n \|_{L^2}}$. Consider the sequence $(u_n) \subset H$ defined by $u_n=v_n / \|v_n \|_{L^2}$. From the previous inequality it follows that $\| | \nabla u_n | \|_{L^2} < 1/n$, therefore $\| |\nabla u_n | \|_{L^2} \to 0$ as $n \to \infty$.

On the other hand we have $\|u_n\|_{H^1}^2=1+\| | \nabla u_n | \|_{L^2}^2$, which implies that $(u_n)$ is bounded in $H^1(\Omega)$. Then there is a subsequence denoted without loss of generality $(u_n)$ which converges weakly to $u \in H^1(\Omega)$. Since $H$ is closed, it follows that $u \in H$. Since the inclusion $H^1(\Omega) \subset L^2(\Omega)$ is compact, it follows that $(u_n)$ has a subsequence which converges strongly in $L^2(\Omega)$ to the same $u$. Without loss of generality, denote this sequence by $(u_n)$. Then, using the weak-sequential lower semicontinuity of $\|\cdot \|_{H^1}$ we have

$\displaystyle \|u\|_{L^2}+\| |\nabla u| \|_{L^2}^2=\|u\|_{H^1}\leq \liminf_{n \to \infty} \|u_n\|_{H_1}^2=$

$\displaystyle =\liminf_{n \to \infty}(\|u_n\|_{L^2}^2+\| |\nabla u_n|\|_{L^2}^2)=\|u\|_{L^2}^2$.

From the above, we get that $\| |\nabla u| \|_{L^2}=0$ and therefore $u$ is constant. Moreover, $\|u\|=\lim \|u_n\|=1$, and we have found a non-zero constant $u\in H$. This contradicts the definition of $H$.