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 norm of the function minus its average value (this is known as a BMO norm) is controlled by the norm of its derivative.

Anyway, here goes.

Theorem. Suppose that is a simple (sufficiently smooth) closed curve in with length and enclosing an area . Then .

Proof #1. Suppose that is a parametrization for . By scaling, we may assume that — scaling points to changes arclength by a factor of (by the arclength formula) and the area by a factor of (by linear algebra). Furthermore, we shall parametrize by arclength so that for every .

Let and denote the coordinate functions for , i.e. Observe that both are periodic, and so we may think of them as living on the circle. We can then compute some Fourier series:

Since both and are real-valued functions, we know, for instance, that , and so Parseval’s identity says that

We know the arclength formula is given by

Since , we have that

Hence

Green’s Theorem tells us that

Thinking of as the inner product between and , using the fact that and are real-valued and hence , and the fact that integrates to zero unless , we see that

But we now have the foolish inequalities

This gives

and hence

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 when , 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:

where denotes . Since gives the constant term for the Fourier series of , this Poincare inequality follows immediately by the Fourier series formula for the derivative, since

In fact, since the terms coincide exactly, one sees that

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

Assume that the orientation of our parametrization makes the inside of the absolute value positive. Use this formula and the fact that to deduce that

Using the fact that , we get that

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

Thus, again, The argument that equality only holds for circles is just as easy as before.

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.

July 31, 2008 at 11:22 pm |

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.

July 31, 2008 at 11:35 pm |

Your proofs can be slightly simplified by using complex numbers, i.e., etc.

December 11, 2008 at 7:43 pm |

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.

March 19, 2011 at 9:51 pm |

[...] [2] http://cornellmath.wordpress.com/2008/05/16/two-cute-proofs-of-the-isoperimetric-inequality/ [...]

November 12, 2011 at 7:06 am |

It is beautiful proof.

Is there any proof using Fourier analysis for the n-dimensional case?

April 6, 2012 at 1:08 pm |

[...] 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 [...]

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