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
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
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.