Home > shape optimization > Proof of the isoperimetric inequality 2

## Proof of the isoperimetric inequality 2

I will continue the series of proofs for the isoperimetric inequality in the two dimensional case, i.e. if a simple closed curve ${ \Gamma}$ (which we suppose for simplicity that it consists of piecewise ${ C^1}$ curves) of length ${ L}$ encloses an area ${ A}$ then ${ L^2 \geq 4\pi A}$ and the equality is attained only if ${ \Gamma}$ is the boundary of a circle.

The proof I give here assumes again the existence of the optimal shape, a non-trivial fact proved here. Let’s make a few remarks about the optimal shape. Like in the proof presented here, it is reasonable to assume that the plane region enclosed by ${L}$ denoted by ${\Omega}$ is convex.

One of the basic applications of continuity to plane geometry to see that for any direction ${\overrightarrow{d}}$ in the plane there exists a line parallel to this direction which splits ${\Omega}$ in two region of equal areas. To do this, take a ‘sweep’ of the plane by a line parallel to ${\overrightarrow{d}}$, i.e. choose a coordinate system ${xOy}$ in the plane such that ${\Omega}$ is included in the halfplane ${x>0}$, and denote ${f(x)=|\Omega \cap \{ y \in \Bbb{R}^2 : y, where ${|\cdot|}$ is the Lebesgue measure. It can be proved that ${f}$ is in fact Lipschitz continuous (note that ${\Omega}$ is considered to be bounded). Since ${f(0)=0}$ and ${\lim_{x \rightarrow \infty}f(x)=|\Omega|}$, the continuity of ${f}$ implies that there exists ${x_d >0}$ such that ${f(x_d)=|\Omega|/2}$, and the line ${x=x_d}$ has the desired property.

Using the fact presented above and another continuity argument we can prove that there exists a line ${\ell}$ which splits ${\Omega}$ into two regions which have the same area and the same perimeter. For every direction which we can characterize by the angle ${\theta \in (0,2\pi)}$ which this direction makes with the positive side of the axis ${Ox}$ there exists a line ${\ell_\theta}$ which splits ${\Omega}$ in two regions of equal areas. For such a line ${\ell_\theta}$ we denote by ${P_+(\theta)}$ and ${P_-(\theta)}$ the perimeter of ${\Omega}$ on one side and the other of ${\ell_\theta}$. We choose the ${+}$ and ${-}$ signs such that the orientation is consistent with the case ${\theta=0}$. Then we can define ${f:[0,2\pi]}$ by

$\displaystyle f(\theta)= P_+(\theta)-P_-(\theta),$

a function which is again continuous. We note that ${f(\theta+\pi)=-f(\theta)}$ since after a rotation by ${\pi}$ the line support of the direction is the same, the direction points the other way. Therefore we can find a ${\theta \in [0,\pi]}$ such that ${f(\theta)=0}$ and we are done.

Using this trick we can choose this line which splits ${\Omega}$ into two parts with equal areas and equal perimeters and reduce the problem to proving that half the optimal shape is half a circle. For this denote ${A,B}$ the intersection of ${\ell}$ with ${\Gamma}$ and pick one half of ${\Omega}$. Suppose that this half ${\Omega'}$ is not a half-circle. Then we can find a point ${C}$ on the boundary (not on ${\ell}$) such that the angle ${ \angle ACB}$ is less or more than ${\pi/2}$. The area of ${ACB}$ is given by the formula

$\displaystyle \text{Area} ABC=\frac{AC\cdot BC \cdot \sin \angle ACB}{2},$

so if we ‘glue’ the portions of ${\Omega' \setminus \Delta ABC}$ to the sides ${AC,BC}$, and consider a hinge in ${C}$ such that these regions are mobile (see the figure). If we move the considered regions until ${\angle C=\pi/2}$ then we have increased the area preserving the perimeter, and by the symmetry by ${\ell}$ we get a contradiction to the fact that ${\Omega}$ is optimal.

This contradiction shows that half of ${\Omega}$ is half a circle, and therefore ${\Omega}$ itself is a circle.