Friedrichs' Inequality

Theorem
Let $G \subset \R^n$ be bounded domain.

Then for any $u \in W^{2, 1}_0 \left({G}\right)$:


 * $\left\Vert{u}\right\Vert_{L^2 \left({G}\right)} \le \operatorname {diam} \left({G}\right) \left\Vert{\nabla u}\right\Vert_{L^2 \left({G}\right)}$

where:
 * $\operatorname {diam} \left({G}\right) := \sup\limits_{x, y \mathop \in G} |x - y|$

Smooth functions with compact support
Let $u \in C_0^\infty \left({G}\right)$.

Put $u(x) = 0$ if $x=(x_1,x_2,\dots,x_n) \notin G$.

Then:
 * $u \in C_0^\infty (\R^n)$

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

Then for any $x$:


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

By Cauchy-Bunyakovsky-Schwarz Inequality


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

Integrating this we get:


 * $\displaystyle \left\Vert{u}\right\Vert_{L^2 \left({G}\right)} \le \operatorname {diam} \left({G}\right) \int\limits_G \left ( \int\limits_a^b |\nabla u (x_1,\dots,x_{m-1},t|^2 \ \mathrm d t \right) \ \mathrm d x$

By Fubini's Theorem


 * $\displaystyle \int\limits_G \left ( \int\limits_a^b |\nabla u (x_1,\dots,x_{m-1},t|^2 \ \mathrm d t \right ) \ \mathrm d x = \int\limits_a^b \ \mathrm d x_m \int\limits_{\mathbb R^n} |\nabla u(x_1,\dots,x_{n-1},x_m)|^2 \ \mathrm d x \le \operatorname {diam}^2 \left({G}\right) \left\Vert{\nabla u}\right\Vert^2_{L^2 \left({G}\right)}$

General case
Let now $u \in W^{2, 1}_0 \left({G}\right)$.

There is a sequence $\displaystyle \{u_n\}_{n \mathop = 1}^\infty \subset C_0^\infty \left({G}\right)$ such that:
 * $\left\Vert{u - u_n}\right\Vert_{W^{2,1} \left({G}\right)} \to 0$

as $n \to \infty$.

Then:
 * $\left\Vert{u - u_n}\right\Vert_{L^2 \left({G}\right)} \to 0$

and:
 * $\left\Vert{\nabla u - \nabla u_n}\right\Vert_{L^2 \left({G}\right)} \to 0$

As $|\left\Vert{u - u_n}\right\Vert| \le \left\Vert{u - u_n}\right\Vert$, it follows that:
 * $\left\Vert{u_n}\right\Vert_{L^2 \left({G}\right)} \to \left\Vert{u}\right\Vert_{L^2 \left({G}\right)}$

and:
 * $\left\Vert{\nabla u_n}\right\Vert_{L^2 \left({G}\right)} \to \left\Vert{\nabla u}\right\Vert_{L^2 \left({G}\right)}$

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