Home > Inequalities, Measure Theory, Real Analysis, shape optimization > Minkowski content and the Isoperimetric Inequality

## Minkowski content and the Isoperimetric Inequality

The proof of the isoperimetric inequality given below relies on the Brunn-Minkowski inequality and on the concept of Minkowski content.

In what follows we say that a curve parametrized by ${z(t)=(x(t),y(t)),\ t \in [a,b]}$ is simple if ${t\mapsto z(t)}$ is injective. It is a closed simple curve if ${z(a)=z(b)}$ and ${z}$ is injective on ${[a,b)}$. We say that a curve is quasi-simple if it the mapping ${z}$ is injective with perhaps finitely many exceptions.

For a compact set ${K \subset \Bbb{R}^2}$ we denote

$\displaystyle K^\delta =\{x \in \Bbb{R}^2 : d(x,K)<\delta\}.$

We say that the set ${K}$ has Minkowski content if the limit

$\displaystyle \mathcal{M}(K)=\lim_{\delta \rightarrow 0} \frac{|K^\delta|}{2\delta}$

exists. It can be proved that if ${\Gamma=\{z(t) : t \in [a,b]\}}$ is a quasi-simple curve then the Minkowski content of ${\Gamma}$ exists if and only if ${\Gamma}$ is rectifiable, and in this case ${\mathcal{M}(\Gamma)=L}$, where ${L}$ is the length of the curve.

Suppose now that ${\Omega \subset \Bbb{R}^2}$ is open, bounded and its boundary ${\partial \Omega}$ is a simple rectifiable curve ${\Gamma}$. Then we have ${4\pi |\Omega| \leq \ell(\Gamma)}$.

Proof: Denote

$\displaystyle \Omega_+(\delta) =\{x \in \Bbb{R}^2 : d(x,\Omega)<\delta\}$

and

$\displaystyle \Omega_-(\delta) = \{ x \in \Bbb{R}^2 : d(x,\Omega^c) \geq \delta\} .$

We see right away that ${\Omega_-(\delta) \subset \Omega \subset \Omega_+(\delta)}$ and ${\Omega_+(\delta)=\Omega_-(\delta) \cup \Gamma^\delta}$., and this union is disjoint. We can see that

$\displaystyle \Omega_+(\delta)=\Omega+B_\delta,\ \Omega=\Omega_-(\delta) +B_\delta,$

where the ${+}$ is the Minkowski sum of the corresponding sets and ${B_\delta}$ is the unit open disk of radius ${\delta}$. Now we apply the Brunn-Minkowski inequality and get

$\displaystyle |\Omega_+(\delta)| \geq (\sqrt{|\Omega|}+\sqrt{|B_\delta|})^2 \geq |\Omega|+2\delta\sqrt{\pi |\Omega|}$

$\displaystyle |\Omega| \geq (\sqrt{|\Omega_-(\delta)|}+\sqrt{|B_\delta|})^2 \geq |\Omega_-(\delta)|+2\delta \sqrt{\pi |\Omega_-(\delta)|}.$

Adding these two inequalities we get that

$\displaystyle |\Gamma^\delta| = |\Omega_+(\delta)|-|\Omega_-(\delta)|\geq 2\sqrt{\pi}\delta(\sqrt{|\Omega|}+\sqrt{|\Omega_-(\delta)|}).$

Divide this inequality by ${2\delta}$ and take the ${\limsup}$ as ${\delta \rightarrow 0}$ to obtain

$\displaystyle \ell(\Gamma)= \mathcal{M}(\Gamma) \geq 2\sqrt{\pi |\Omega|},$

which is just the isoperimetric inequality.