Friedrichs' Inequality

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

Then for any $u \in \map {W^{2, 1}_0} G$:


 * $\norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \norm {\nabla u}_{\map {L^2} G}$

where:
 * $\map {\operatorname {diam} } G := \sup \limits_{x, y \mathop \in G} \size {x - y}$

Smooth functions with compact support
Let $u \in \map {C_0^\infty} G$.

Put $\map u x = 0$ if $x = \tuple {x_1, x_2, \ldots, x_n} \notin G$.

Then:
 * $u \in \map {C_0^\infty} {\R^n}$

Denote by $a,b$ the extremes of the last coordinate: $a = \inf \limits_{x \mathop \in G} x_n$ and $b = \sup \limits_{y \mathop \in G} x_n$

Then for any $x$:


 * $\ds \size {\map u {x_1, x_2, \ldots, x_n} }^2 = \size {\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \tuple {x_1, x_2, \ldots, x_{n - 1}, t} \rd t}^2$

By Cauchy-Bunyakovsky-Schwarz Inequality:

Integrating this we get:


 * $\ds \norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x$

By Fubini's Theorem:

General case
Let now $u \in \map {W^{2, 1}_0} G$.

There exists a sequence $\ds \sequence {u_n}_{n \mathop = 1}^\infty \subset \map {C_0^\infty} G$ such that:
 * $\norm {u - u_n}_{\map {W^{2, 1} } G} \to 0$

as $n \to \infty$.

Then:
 * $\norm {u - u_n}_{\map {L^2} G} \to 0$

and:
 * $\norm {\nabla u - \nabla u_n}_{\map {L^2} G} \to 0$

As $\size {\, \norm {u - u_n} \,} \le \norm {u - u_n}$, it follows that:
 * $\norm {u_n}_{\map {L^2} G} \to \norm u_{\map {L^2} G}$

and:
 * $\norm {\nabla u_n}_{\map {L^2} G} \to \norm {\nabla u}_{\map {L^2} G}$

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