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 $u \left({x}\right) = 0$ if $x = \left({x_1, x_2, \ldots, x_n}\right) \notin G$.

Then:
 * $u \in C_0^\infty \left({\R^n}\right)$

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

Then for any $x$:


 * $\displaystyle \left\vert{u \left({x_1, x_2, \ldots, x_n}\right)}\right\vert^2 = \left\vert{\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \left({x_1, x_2, \ldots, x_{n-1}, t}\right) \ \mathrm d t}\right\vert^2$

By Cauchy-Bunyakovsky-Schwarz Inequality:

Integrating this we get:


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

By Fubini's Theorem:

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

There exists a sequence $\displaystyle \left\langle{u_n}\right\rangle_{n \mathop = 1}^\infty \subset C_0^\infty \left({G}\right)$ 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 $\left\vert{\,\left\Vert{u - u_n}\right\Vert\,}\right\vert \le \left\Vert{u - u_n}\right\Vert$, 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 $u$.