Home > Fourier Series, shape optimization > Proof of the Isoperimetric Inequality 5

## Proof of the Isoperimetric Inequality 5

Suppose ${\Gamma}$ is a simple, sufficiently smooth closed curve in ${\Bbb{R}^2}$. If ${L}$ is its length and ${A}$ is the area of the region it encloses then the following inequality holds

$\displaystyle 4 \pi A \leq L^2.$

I am going to present another proof of the isoperimetric inequality. This time with Fourier series.

Proof: Suppose ${\gamma : [0,2\pi] \rightarrow \Bbb{R}^2}$ is a parametrization for ${\Gamma}$, and we may assume that ${\gamma}$ is parametrized with constant speed, i.e. ${|\gamma'(t)|=L/2\pi}$ on ${[0,2\pi]}$. Denote by ${x,y}$ the coordinate functions, i.e. ${\gamma(t)=(x(t),y(t))}$. Then ${x,y}$ can be extended periodically to the whole ${\Bbb{R}}$ and we have

$\displaystyle x(t)=\sum_{n \in \Bbb{Z}} a_n e^{int}$

$\displaystyle y(t)=\sum_{n \in \Bbb{Z}} b_n e^{int}$

$\displaystyle x'(t)=\sum_{n \in \Bbb{Z}} ina_n e^{int}$

$\displaystyle y'(t)=\sum_{n \in \Bbb{Z}} ina_n e^{int}$

Note that all these function are real valued. Parseval’s identity gives

$\displaystyle \frac{L^2}{2\pi} =\int_0^{2\pi} (x'(s))^2+(y'(s))^2 ds=2\pi\sum_{n \in \Bbb{Z}}n^2(|a_n|^2 +|b_n|^2).$

By Green’s formula we have

$\displaystyle A=\frac{1}{2} \left| \int_0^{2\pi}(x(s)y'(s)-y(s)x'(s))ds \right|=\pi \left| \sum_{n \in \Bbb{Z}} n(a_n\overline{b}_n-\overline{a}_nb_n) \right|,$

and this is because ${x,y}$ are real valued, so ${a_{-n}=\overline a_n,\ b_{-n}=\overline{b}_n}$ and

$\displaystyle \int_0^{2\pi} e^{i(m+n)t}dt=\begin{cases} 2 \pi & m+n=0 \\ 0 & \text{otherwise} \end{cases}.$

Then we have

$\displaystyle 4 \pi A = 4\pi^2 \left| \sum_{n \in \Bbb{Z}} n(a_n\overline{b}_n-\overline{a}_nb_n) \right| \leq 4\pi^2 \sum_{n \in \Bbb{Z}}|n||a_n\overline{b}_n-\overline{a}_nb_n|\leq$

$\displaystyle \leq 4 \pi^2\sum_{n \in \Bbb{Z}} |n| (|a_n|^2+|b_n|^2)\leq 4\pi^2 \sum_{n \in \Bbb{Z}} n^2 (|a_n|^2+|b_n|^2)=L^2.$

Therefore the inequality is proved. For the equality to hold we first need that ${a_n=b_n=0}$ when ${|n| \geq 2}$. Then we also have

$\displaystyle |a_n\overline{b}_n-\overline{a}_nb_n| \leq 2|a_n||b_n| \leq |a_n|^2+|b_n|^2,$

so ${|a_n|=|b_n|}$ for ${|n|= 1}$. And we see right away that the equality holds only for the circle.