TITLE: Why is Fourier analysis so handy for proving the isoperimetric inequality?
QUESTION [33 upvotes]: I have just completed an introductory course on analysis, and have been looking over my notes for the year.  For me, although it was certainly not the most powerful or important theorem which we covered, the most striking application was the Fourier analytic proof of the isoperimetric inequality.  I understand the proof, but I still have no feeling for why anyone would think to use Fourier analysis to approach this problem. I am looking for a clear reason why someone would look at this and think "a Fourier transform would simplify things here".  Even better would be a physical interpretation.  Could this somehow be related to "hearing the shape of a drum"? Is there any larger story here?

REPLY [10 votes]: I'd like to add a few words on what happens in higher dimensions. First, a convexity assumption becomes essential (as in the second proof of Hurwitz which works only for convex domains). The isoperimetric inequalities in $\mathbb R^n$, $n>2$, are much easier to deal with in case of convex bodies, and the whole problem in some sense looks most natural under the convexity assumption. Second, there are many different isoperimetric inequalities in higher dimensions. And Fourier analysis (or rather harmonic analysis on a sphere) can be successfully applied to prove at least some of them.  
There is a classical approach to isoperimetric problems based on Steiner's theorem. Let $K$ be a convex body in $\mathbb R^n$ and let $K_r$ denote the "parallel" body
$$K_r=\{x\in\mathbb R^n|\ dist(x, K)\leq r \},\quad r>0.$$ 
Then, by Steiner's theorem, there exist $n+1$ numbers $W_0^n(K),W_1^n(K),\dots,W_n^n(K)$, such that 
$$V(K_r)=Vol(K_r)=\sum\limits_{i=0}^{n}{n \choose i}W^n_i(K)r^{i}.$$
It can be shown that 
$$W^n_0(K)=V(K),\quad W^n_1(K)=\frac{S(K)}{n},\qquad(1)$$
where $S(K)$ is the surface area of $\partial K$. Moreover, $W^n_n(K)$ is equal to the volume $\pi_n$ of the unit ball in $\mathbb R^n$ and
$$W^n_{n-1}(K)=\frac{\pi_n}{2}w(K),\qquad\qquad\qquad\quad (2)$$
where $w(K)$ is the mean width of $K$. Note that for $n=2$ the perimeter $P(K)$ equals
$\pi w(K).$ The numbers $W_i^n(K)$ give some information on how the convex body $K$ is different from a ball (for the unit ball in $\mathbb R^n$, obviously $W^n_i(K)=\pi_n$ for all $i$.)
A convex body is completely determined by its support function
$$h(x)=\sup\{x\cdot y|\ y\in K\}$$
which measures the directed distance of the origin to the tangent plane of $K$ at direction $x\in S^{n-1}.$  Now, the second proof of Hurwitz deals with the Fourier decomposition of the support function of a convex 2D domain. The problem is that in dimension $n>2$ the formulas for volume and surface area in terms of the support function
cannot be expressed nicely by means of spherical harmonics. However, it is still possible
to derive an isoperimetric inequality for the numbers $W^n_{n-2}$ and $W^n_{n-1}$ via harmonic analysis, namely
$$W^n_{n-1}\geq\sqrt{\pi_n W^n_{n-2}}. \qquad\qquad\qquad(3)$$
When $n=2$ this is the standard isoperimetric inequality $P^2\geq 4\pi A$. 
If $n=3$ $(3)$ gives the isoperimetric inequality between the mean width and surface area of a convex body $$\pi [w(K)]^2\geq S(K).\qquad\qquad\qquad (4)$$ 
The proof is a straightforward extension of the second Hurwitz proof (using a decomposition of the support function into a series of spherical harmonics) and can be found here. 

Update (concerning the question in Victor's comment below). If we assume as known the inequality 
$$W_{1}^3\geq \sqrt{W_{0}^3W_{2}^3},$$
then  together with (1),(2) and (4) it implies that $S^3\geq 36\pi V^2$. ("Known" means that I don't know how to obtain the inequality using only harmonic analysis. It follows from the Alexandrov-Fenchel inequality for mixed volumes.)