Two Cute Proofs of the Isoperimetric Inequality

The blog has been pretty quiet the last few weeks with the usual end-of-term business, research, and A-exams (mine is coming up quite soon). I was looking through some of my notes recently and came upon two very short Fourier analysis proofs of the isoperimetric inequality. Both proofs are among my all-time favorites; the result is of general interest (though it is subsumed in more general and useful facts), and the proofs are quick and elegant. The proofs are similar, but the second generates a Poincare inequality which is one of the fundamental tools of analysis — basically, the inequality says that for a function with a derivative, the $L^2$ norm of the function minus its average value (this is known as a BMO norm) is controlled by the $L^2$ norm of its derivative.

Anyway, here goes.

Theorem. Suppose that $\Gamma$ is a simple (sufficiently smooth) closed curve in $\mathbb{R}^2$ with length $L$ and enclosing an area $A$ . Then $A\le L^2/4\pi$ .

Proof #1. Suppose that $\gamma:[0,2\pi]\rightarrow\mathbb{R}^2$ is a parametrization for $\Gamma$ . By scaling, we may assume that $L=2\pi$ — scaling points $(x,y)$ to $(\delta x,\delta y)$ changes arclength by a factor of $\delta$ (by the arclength formula) and the area by a factor of $\delta^2$ (by linear algebra). Furthermore, we shall parametrize $\Gamma$ by arclength so that $|\gamma '(s)|=1$ for every $s$ .

Let $x(s)$ and $y(s)$ denote the coordinate functions for $\gamma$ , i.e. $\gamma(s) = (x(s),y(s)).$ Observe that both are periodic, and so we may think of them as living on the circle. We can then compute some Fourier series:

$\displaystyle{x(s) =\sum_{n\in\mathbb{Z}}a_n e^{ins}.}$

$\displaystyle{x'(s) =\sum_{n\in\mathbb{Z}}ina_n e^{ins}.}$

$\displaystyle{y(s) =\sum_{n\in\mathbb{Z}}b_n e^{ins}.}$

$\displaystyle{y'(s) =\sum_{n\in\mathbb{Z}}inb_n e^{ins}.}$

Since both $x$ and $y$ are real-valued functions, we know, for instance, that $x^2=|x|^2$ , and so Parseval’s identity says that

$\displaystyle{\frac{1}{2\pi}\int_0^{2\pi}[x'(s)]^2=\sum_{n\in\mathbb{Z}}|na_n|^2.}$

We know the arclength formula is given by

$\displaystyle{L=\int_0^{2\pi}|\gamma '(s)|ds.}$

Since $|\gamma '(s)|^2=|\gamma '(s)|=1$ , we have that

$\displaystyle{2\pi = L = \int_0^{2\pi}[x'(s)]^2+[y'(s)]^2ds.}$

Hence

$\displaystyle{\sum_{n\in\mathbb{Z}}|n|^2(|a_n|^2+|b_n|^2) = 1.}$

Green’s Theorem tells us that

$\displaystyle{A = \frac{1}{2}\left|\int_\Gamma xdy - ydx\right| = \frac{1}{2}\left|\int_0^{2\pi}x(s)y'(s) - y(s)x'(s)ds\right|.}$

Thinking of $\int_0^{2\pi}f(s)\bar{g}(s)ds$ as the inner product between $f$ and $g$ , using the fact that $x$ and $y$ are real-valued and hence $x=\bar{x}$ , and the fact that $e^{inx}e^{imx}$ integrates to zero unless $n=-m$ , we see that

$\displaystyle{A = \pi \left|\sum_{n\in\mathbb{Z}}n(a_n\bar{b}_n - b_n\bar{a}_n)\right|.}$

But we now have the foolish inequalities

$\displaystyle{|a_n\bar{b}_n - b_n\bar{a}_n| \le 2|a_n||b_n|\le |a_n|^2 + |b_n|^2.}$

This gives

$\displaystyle{\frac{A}{\pi}=\left|\sum_{n\in\mathbb{Z}}n(a_n\bar{b}_n - b_n\bar{a}_n)\right|}$

$\displaystyle{\le \sum_{n\in\mathbb{Z}}|n|(|a_n|^2 + |b_n|^2)}$

$\displaystyle{\le \sum_{n\in\mathbb{Z}}n^2(|a_n|^2 + |b_n|^2)=1,}$

and hence $A\le\pi.$

The fact that the circle is the only curve for which equality between the two is possible follows by the above with the facts that $|n|<|n|^2$ when $|n|>1$ , some high school algebra, and knowing what a circle is.

Proof #2. Another fairly easy way to prove the isoperimetric inequality is to use a Poincare inequality:

$\displaystyle{\int_0^{2\pi}|f-c_f|^2\le\int_0^{2\pi}|f'|^2.}$

where $c_f$ denotes $\frac{1}{2\pi} \int_0^{2\pi}f$ . Since $c_f$ gives the constant term for the Fourier series of $f$ , this Poincare inequality follows immediately by the Fourier series formula for the derivative, since

$\displaystyle{\int_0^{2\pi}|f'|^2=\sum_{n\in\mathbb{Z}}n^2|a_n|^2\ge\sum_{n\neq 0}|a_n|^2=\int_0^{2\pi}|f-c_f|^2.}$

In fact, since the $n=1$ terms coincide exactly, one sees that

$\displaystyle{\int_0^{2\pi}|f'|^2-\int_0^{2\pi}|f-c_f|^2=\sum_{|n|\ge2}(n^2-1)|a_n|^2.}$

Hence the only way one can get equality is if $a_n=0$ for all $|n|\ge2$ . To get the isoperimetric inequality, we use Green’s Theorem again, to get a slightly different (but more familiar) area formula:

$\displaystyle{A=\left|\int_0^{2\pi}x(s)y'(s)\right|.}$

Assume that the orientation of our parametrization makes the inside of the absolute value positive. Use this formula and the fact that $\int_0^{2\pi}y'=0$ to deduce that

$\displaystyle{2A=2\int_0^{2\pi}xy'=2\int_0^{2\pi}(x-c_x)y'.}$

Using the fact that $2ab=a^2+b^2-(a-b)^2$ , we get that

$\displaystyle{2A=\int_0^{2\pi}(x-c_x)^2+(y')^2-(c-c_x-y')^2}$

$\displaystyle{\le\int_0^{2\pi}(x-c_x)^2+(y')^2.}$

The last inequality follows since we’re dropping a negative term. Applying the Poincare inequality, we get that

$\displaystyle{2A\le\int_0^{2\pi}(x')^2+(y')^2=\int_0^{2\pi}1=2\pi.}$

Thus, again, $A\le\pi.$ The argument that equality only holds for circles is just as easy as before.

This entry was posted on May 16, 2008 at 6:54 am and is filed under Basic Grad Student, Guests, Undergraduate. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Two Cute Proofs of the Isoperimetric Inequality”

Aaron F. Says:
June 13, 2008 at 3:05 pm
I like the first proof a lot! It’s the first pure-mathematical application of Fourier decomposition that I’ve seen.
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Two Cute Proofs of the Isoperimetric Inequality

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