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.

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  1. June 11, 2012 at 8:15 pm

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