Friedrichs' Inequality

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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:

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

Proof

Smooth functions with compact support

This article, or a section of it, needs explaining. In particular: In several places in the below, this construct was seen: $\vert \nabla u (x_1, \dots, x_{m - 1}, t\vert^2$ where it appears that the close of the round bracket parenthesis "$)$" was been omitted. It has been assmued in the corrected version that it should be this: $\vert\nabla u (x_1,\dots,x_{m-1},t)\vert^2$ but it is not obvious and may require someone knowledgeable to check this out. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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:

\(\ds \size {\int \limits_a^{x_n} \map {\frac {\partial u} {\partial x_n} } {x_1, \ldots, x_{n - 1}, t} \rd t}^2\)

\(\le\)

\(\ds \int \limits_a^{x_n} 1^2 \rd t \int \limits_a^{x_n} \size {\map {\frac {\partial u} {\partial x_n} } {x_1, \ldots, x_{n - 1}, t} }^2 \rd t\)

\(\ds \)

\(\le\)

\(\ds \map {\operatorname {diam} } G \int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t\)

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:

\(\ds \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x\)

\(=\)

\(\ds \int \limits_a^b \rd x_m \int \limits_{\mathbb R^n} \size {\map {\nabla u} {x_1, \ldots, x_{n - 1}, x_m} }^2 \rd x\)

\(\ds \)

\(\le\)

\(\ds \map {\operatorname {diam}^2} G \norm {\nabla u}^2_{\map {L^2} G}\)

$\blacksquare$

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 $u$.

$\blacksquare$

Source of Name

This entry was named for Kurt Otto Friedrichs.