Friedrichs' Inequality

Theorem
Let $G \subset \mathbb R^n$ be bounded domain. Then for any $u \in W^{2,1}_0(G)$


 * $\|u\|_{L^2(G)} \leq diam(G) \|\nabla u \|_{L^2(G)}$,

where $diam(G) = \sup\limits_{x,y\in G} |x - y|$

Smooth functions with compact support
Let $u \in C_0^\infty(G)$. Put $u(x) = 0$ if $x=(x_1,x_2,\dots,x_n) \not\in G$. Then $u \in C_0^\infty (\mathbb R^n)$

Denote $a = \inf\limits_{x\in G} x_n, b = \sup\limits_{y \in G} x_m$

Then for any $x$


 * $|u(x_1,\dots,x_n)|^2=\left |\int\limits_a^{x_n} \frac{\partial u}{\partial x_n}(x_1,\dots,x_{n-1},t)\,dt\right|^2$

By Cauchy-Bunyakovsky-Schwarz Inequality


 * $\left|\int\limits_a^{x_n} \frac{\partial u}{\partial x_n}(x_1,\dots,x_{n-1},t)\,dt\right|^2 \leq \int\limits_a^{x_n} 1^2\,dt \int\limits_a^{x_n} \left| \frac{\partial u}{\partial x_n}(x_1,\dots,x_{n-1},t) \right|^2\,dt \leq diam(G) \int\limits_a^b |\nabla u (x_1,\dots,x_{m-1},t|^2\,dt$

Integrating this we get:


 * $ \|u\|_{L^2(G)} \leq diam(G) \int\limits_G \left ( \int\limits_a^b |\nabla u (x_1,\dots,x_{m-1},t|^2\,dt \right )\,dx$

By Fubini's Theorem


 * $\int\limits_G \left ( \int\limits_a^b |\nabla u (x_1,\dots,x_{m-1},t|^2\,dt \right )\,dx = \int\limits_a^b\,dx_m \int\limits_{\mathbb R^n} |\nabla u(x_1,\dots,x_{n-1},x_m)|^2\,dx \leq diam^2(G) \|\nabla u\|^2_{L^2(G)}$

General case
Let now $u \in W^{2,1}_0(G)$. There is a sequence $\{u_n\}_{n=1}^\infty \subset C_0^\infty(G)$ such that $\|u - u_n\|_{W^{2,1}(G)} \to 0$ as $n \to \infty$.

Then $\|u - u_n\|_{L^2(G)} \to 0$ and $\|\nabla u - \nabla u_n\|_{L^2(G)} \to 0$

As $|\|u - u_n\|| \leq \|u - u_n\|$, $\|u_n\|_{L^2(G)} \to \|u\|_{L^2(G)}$ and $\|\nabla u_n\|_{L^2(G)} \to \|\nabla u\|_{L^2(G)}$.

Since the inequality is correct for all $u_n$, it is also correct for $u$.