## 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.

### 9 Responses to “Two Cute Proofs of the Isoperimetric Inequality”

1. Aaron F. Says:

I like the first proof a lot! It’s the first pure-mathematical application of Fourier decomposition that I’ve seen.

2. misha Says:

There is an even cuter and totally elementary proof. It follows from the fact that the maximal area polygon with a given number of sides and a given perimeter must be regular. To see it, we first notice that it has to be convex (otherwise its convex hull will have bigger area and smaller perimeter). By cutting a triangle off this polygon, we can see that the area of this triangle is maximized when it is isosceles, and we conclude that the maximal area polygon is equilateral, I.e., the lengths of all sides are the same. And finally, by cutting a quadrilateral off the maximal area polygon, we can see that the area of this quadrilateral will be maximized when the 2 adjacent angles of our polygon (that are also the 2 angles of the quadrilateral that we have cut off) are equal to each other. So our polygon must be regular.

For a region bounded by a (piecewise) smooth curve, we can see again that it must be convex (by the same argument as for the polygons). To prove that the boundary must be a circle, we inscribe any equilateral polygon into it and observe that it must be regular, otherwise the area of the region can be increased by making this polygon regular by changing its angles and keeping the segments of the region adjacent to the sides of the polygon the same, because the area of the whole region is the area of the inscribed polygon + the total area of the segments cut off from this region by the sides of the polygon. Q.E.D.

3. misha Says:

Your proofs can be slightly simplified by using complex numbers, i.e., $z=x+iy$ etc.

4. Scott Says:

Xavier Cabre recently found the most beautiful proof of the isoperimetric inequality I have ever seen. It brought tears to my eyes.

Cabre, X., “Elliptic PDE’s in probabilities in geometry and symmetry and regularity of solutions,” Disc. Cont. Dynamical Systems 20 (3), 425-457 (2008).

Don’t be turned off by the title if you are scared of PDEs, it is elementary.

5. The Isoperimetric Inequality « Chaitanya's Random Pages Says:
6. Tung Says:

It is beautiful proof.
Is there any proof using Fourier analysis for the n-dimensional case?

7. Isoperimetric problem (Dido’s Problem) | kReese.net Says:

[...] 5, 2011]. [6] Luthy, P. “Two Cute Proofs of the Isoperimetric Inequality” Internet: http://cornellmath.wordpress.com/2008/05/16/two-cute-proofs-of-the-isoperimetric-inequality/, May 16, 2008 [September 5, 2011]. [7]Hoory, S., Linial, N., and Wigderson, A. “Expander [...]