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.

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