## 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. $\Box$