Faber-Krahn inequality

I record a proof of the Faber-Krahn inequality here, mainly for my own benefit.

Let {M} be one of the standard space forms: the Euclidean space {\mathbb R^n}, the unit sphere {\mathbb S^n}, or the hyperbolic space {\mathbb H^n}. Suppose {\Omega } is a bounded domain in {M} with smooth boundary {\partial \Omega} which is a closed hypersurface. Then we can define {\Omega^\#} to be the geodesic ball in {M} which has the same volume as {\Omega}.

Let {\lambda_1(\Omega)} be the first Laplacian eigenvalue of {\Omega} under the Dirichlet boundary condition. i.e. {\lambda_1(\Omega)>0} is the smallest number {\lambda} such that there exists a non-zero smooth function {f} with

\displaystyle \begin{array}{rl} \displaystyle   \Delta f =-\lambda f\;  \displaystyle \textrm{in }\Omega, \quad f =0 \;\textrm{on }\partial \Omega. \end{array}

Theorem 1 (Faber-Krahn inequality) Suppose {\Omega } is a bounded domain in {M} with smooth boundary {\partial \Omega}. Then

\displaystyle \begin{array}{rl} \displaystyle   \lambda_1(\Omega)\ge \lambda_1(\Omega^\#).  \end{array}

The equality holds if and only if {\Omega} is a geodesic ball.

Proof: Let {|B_R|=|\Omega|}. Suppose {f} is the first eigenfunction on {\Omega}. Symmetrize {f} by the following procedure: define {g: B_R\rightarrow \mathbb R^+} to be the radial function such that for any {t},

\displaystyle \begin{array}{rl} \displaystyle \mathrm{Vol} \{f\ge t\} = \mathrm{Vol} \{g\ge t\}. \end{array}

By Fubini’s theorem,

\displaystyle \begin{array}{rl} \displaystyle   \int_\Omega f^2 =\int_0^\infty \mathrm{Vol}(f^2\ge t)dt =\int_0^\infty \mathrm{Vol}(g^2\ge t)dt =\int_{B_R} g^2.  \ \ \ \ \ (1)\end{array}

It is easy to see that {|\nabla g|=| g'(r)| } is constant on the level set of {g}, so we have

\displaystyle \begin{array}{rl} \displaystyle   \mathrm{Area}\{g=t\} =\int_{\{g=t\}}1dS =\left(\int_{\{g=t\}}|\nabla g|dS \int_{\{g=t\}}\frac{1}{|\nabla g|}dS \right)^{\frac{1}{2}}. \end{array}

By the isoperimetric inequality,

\displaystyle \begin{array}{rl} \displaystyle   \left(\int_{\{f=t\}}|\nabla f|dS \int_{\{f=t\}}\frac{1}{|\nabla f|}dS \right)^{\frac{1}{2}} \ge& \displaystyle  \left(\int_{\{f=t\}}1dS\right)^2\\ =& \displaystyle  \left(\mathrm{Area}\{ f=t \}\right)^2\\ \ge & \displaystyle  \left(\mathrm{Area} \{g=t\} \right)^2\\ =& \displaystyle \left(\int_{\{g=t\}}|\nabla g|dS \int_{\{g=t\}}\frac{1}{|\nabla g|}dS \right)^{\frac{1}{2}}.  \ \ \ \ \ (2)\end{array}

By the co-area formula,

\displaystyle \begin{array}{rl} \displaystyle   \frac{d}{dt}\left(\mathrm{Vol}\{f\ge t\}\right)=-\int_{\{f=t\}}\frac{1}{|\nabla f|}dS \quad \textrm{ and }\quad \frac{d}{dt}\left(\mathrm{Vol}\{g\ge t\}\right)=-\int_{\{g=t\}}\frac{1}{|\nabla g|}dS. \end{array}

As

\displaystyle \begin{array}{rl} \displaystyle   \frac{d}{dt}\left(\mathrm{Vol}\{f\ge t\}\right)= \frac{d}{dt}\left(\mathrm{Vol}\{g\ge t\}\right) \end{array}

by the construction {g}, so in view of (2), we must have

\displaystyle \begin{array}{rl} \displaystyle   \int_{\{f=t\}}|\nabla f|dS\ge \int_{\{f=t\}}|\nabla g|dS. \end{array}

By coarea formula again,

\displaystyle \begin{array}{rl} \displaystyle   \int_{\Omega} |\nabla f|^2 =& \displaystyle \int_{0}^{\infty}\left(\int_{\{f=t\}}|\nabla f|dS\right)dt\\ \ge& \displaystyle \int_{0}^{\infty}\left(\int_{\{g=t\}}|\nabla g|dS\right)dt\\ =& \displaystyle \int_{B_R} |\nabla g|^2.  \ \ \ \ \ (3)\end{array}

By (1), (3) and minmax principle,

\displaystyle \begin{array}{rl} \displaystyle   \lambda_1(\Omega) =\frac{\int_\Omega|\nabla f|^2}{\int_\Omega f^2} \ge\frac{\int_\Omega|\nabla g|^2}{\int_\Omega g^2} \ge\lambda_1(B_R). \end{array}

By the isoperimetric inequality, the equality holds if and only if {\Omega} is a geodesic ball. \Box

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This entry was posted in Analysis, Differential equations, Functional analysis, Geometry, Inequalities. Bookmark the permalink.

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