Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
By the way, there are several different proofs of isoperimetric inequality using Fourier series (Hurwitz alone gave two). –  Victor Protsak Jun 5 '10 at 22:43
1  
The proposed connection with hearing the shape of the drum is very nice. It is known that you can hear the volume of a drum (but, in general, not the shape). –  Gil Kalai Jun 8 '10 at 14:52

5 Answers 5

up vote 43 down vote accepted

Experience with Fourier analysis and representation theory has shown that every time a problem is invariant with respect to a group symmetry, the representation theory of that group is likely to be relevant. If the group is abelian, the representation theory is given by the Fourier transform on that group.

In this case, the relevant symmetry group is that of reparameterising the arclength parameterisation of the perimeter by translation. This operation does not change the area or the perimeter. When combined with the observation (from Stokes theorem) that both the area and perimeter of a body can be easily recovered from the arclength parameterisation, this naturally suggests to use Fourier analysis in the arclength variable.

share|improve this answer

It seems to me that it is worth mentioning that only the 2-d isoperimetric inequality is easily proved using Fourier analysis but not higher dimensional geometric inequalities. In addition to the group invariance property cited by Terry Tao, there is the simple fact that a closed curve can be represented by a pair of periodic functions, and, if the curve is parameterized properly, its length and area are nicely represented by integrals of quadratic polynomials of the periodic functions and their derivatives. All in all, a nice setup for Fourier series. If the integrands were higher degree polynomials, a proof might still be possible but I'm not sure it would be as easy. And is there an analogous proof in higher dimensions?

share|improve this answer
1  
Just about the only way in higher dimensions is via Brunn-Minkowski inequality. Does it have any Fourier analytic meaning? –  Victor Protsak Jun 6 '10 at 1:29
1  
Arguably, the Brunn-Minkowski inequality is itself an isoperimetric inequality, so it's best to talk about how it is proved. Every proof I know involves some form of symmetrization via rearrangement (the fashionable terms is mass transportation). There are definitely deep connections between Brunn-Minkowski and Fourier analysis, but I will let people who understand this much better than me explain this. –  Deane Yang Jun 6 '10 at 1:35
    
I will also be very interested to learn a Fourier proof of the isoperimetric inequality for d>2.I was always very curious with the question why Fourier analysis is NOT handy forproving isoperimetric inequalities in dimension 3 and higher given the beautiful d=2 proof. –  Gil Kalai Jun 8 '10 at 14:49

This doesn't answer the question of why Fourier analysis works, but it certainly is an answer to how one might think "Hmm ... perhaps we're in the domain of Fourier analysis here." It's that the surface area of a shape X is defined in terms of the volume of X+B when B is a small ball. There is a close relationship between sumsets and convolutions (the sumset is precisely the set of points where the convolution of the characteristic functions of the two sets is non-zero), and every time you have a convolution, thoughts of Fourier analysis should be triggered.

The reason I say this doesn't answer the question of why Fourier analysis works is that there is a difference between the convolution and the support of the convolution, and the latter does not transform nicely. But that shouldn't stop one thinking of Fourier analysis and attempting to find some way of relating surface area to convolutions.

share|improve this answer
    
That's a nice answer. For some reason I had never heard that interpretation of surface area before, and it's nice to hear. –  Peter Samuelson Jun 9 '10 at 15:46

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.)

share|improve this answer
3  
Awesome! Do you know whether $S^3\geq 36\pi V^2$ has $\textit{any}$ proof based on harmonic analysis (not necessarily of Hurwitz type)? –  Victor Protsak Jun 7 '10 at 3:49

My two cents. It is known since Antiquity that the circle is the curve of minimal length maximizing inscribed area. So you know that you are looking for the circle, from the start.

A series in cos(x) and sin(x) is certainly a good way to represent circles, and composition of circles. Circles turning around circles, turning around circles, were called epicycles in the terminology of Middle Ages and Renaissance. They were used to approximate the trajectories of planets, which are close to circles. So using trigonometric functions to approximate curves close to circle is an idea which is quite old, and from the historical viewpoint, quite natural.

Now in modern courses in Fourier analysis, justifications for introducing these series usually come from the heat equation, faithful to the original application Fourier had in mind, or from purely mathematical considerations about the nice properties of orthogonal basis. So the link with curves close to circles is somehow lost.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.