Three Proofs of the Isoperimetric Inequality in R2

THREE PROOFS OF THE ISOPERIMETRIC INEQUALITY IN R2

ALEX MRAMOR

Abstract. Supplement for math 439, fall ’19.

1. Introduction The isoperimetric problem is the question of, given some fixed amount of area to enclose a domain, is there a maximal amount of area we can enclose, and what does such a configuration look like. This is a question for everyone; from people who want to make the biggest tent using the least amount of fabric (and hence saving weight/money), to founders of the city of Carthage. For us it will be an easily grasped but interesting question in “global” geometry, and will be a nice excuse to briefly introduce several fascinating ideas, which might at first seem completely unrelated.

Figure 1. Queen Dido could use as much land for founding the city of Carthage as could be encompassed by a cow hide. Here we see her making the most of the oﬀer.

1 ISOPERIMETRIC INEQUALITY IN THE PLANE 2

More precisely, we will show in three diﬀerent ways the following theorem: Theorem 1.1. (Isoperimetric inequality) Let U be a domain in R2 with smooth boundary (suppose the boundary is a regular parameterized curve), and denote by A the area of U and L the length of the boundary of U. Then L2 A ≤ (1.1) 4π

Furthermore, equality happens in the case that U is a round disc Dλ of radius λ (this is discussed at the end). Given a certain length of string/cowhide L, there are certainly better ways than others to enclose a lot of area:

Figure 2. Note that in both these cases the isoperimetric inequality holds

Naturally, similar questions can be asked in (although we haven’t discussed these in class yet) curved ambient spaces. These types of generalizations have proven to be very hard questions in fact and are still the subject of serious research, to this day.

2. The first proof: the Brunn-Minkowski inequality The ﬁrst proof we give uses the Brunn-Minkowski inequality, an inequality that has surprisingly deep connections to a number of important ﬁelds (look up the survey article of Robert Gardner for more information). In the following suppose A and B are two subsets of Rn, then the inequality is: V ol(A + B)1/n ≥ V ol(A)1/n + V ol(B)1/n (2.1)

Here, A+B is the “vector sum” of A and B i.e. A+B = ∪a∈A,b∈Ba+b. Geometrically, at least when A is about the origin, A + B is more or less given by “dragging” A along the surface of B and taking the set that is traced out. ISOPERIMETRIC INEQUALITY IN THE PLANE 3

Figure 3. The sum of a rectangle and a small disc centered at origin

The proof of this inequality can be shown by checking it directly on rectangles, and then unions of rectangles, and then general sets A and B (with appropriate assumptions on A and B). This leads us to our first exercise: Exercise 1: Prove, or look up (only after giving up), the Brunn-Minkowski inequality. With this in hand we proceed to show the isoperimetric inequality; this method is actually very robust and works in all dimensions (the next two proofs will not, however). Below we will apply the inequality by letting A be our domain U in R2, and letting B = D be a small ball of radius . First off, we see that V ol(U + D ) − V ol(U) L = lim (2.2) →0 (This is clear from a picture but honestly there are some issues with actually taking this limit that we are eliding. This is definitely fine for convex domains, which we can always reduce to using Steiner symmetrization.) Applying the Brunn-Minkowski inequality (after squaring both sides) gives: V ol(U + D ) − V ol(U) V ol(U) + V ol(D ) − 2pV ol(U)pV ol(D ) − V ol(U) ≥ (2.3) V ol(D ) 2pV ol(U)pV ol(D ) = − (2.4) ISOPERIMETRIC INEQUALITY IN THE PLANE 4

Now to unpackage this further note that V ol(D ) = π2. Hence the ﬁrst term on the √ √ right goes to zero as does, and the second term is just 2 π A. So we have √ √ L ≥ 2 π A (2.5) Squaring both sides gives the inequality. Exercise 2: Formulate and prove a higher dimensional isoperimetric inequality for domains in Rn by following the steps above.

3. The second proof: Fourier analysis This proof is due to Weil from the 1920s. First to make things a bit more natural we will make a scaling adjustment, which makes now as good a time as ever to think about the following reality check: Exercise 3: Suppose there is an inequality between area and length of the form p q A ≤ cL , with equality for some domain U1. Show that in two dimensions we must have p = 1 and q = 2. What about higher dimensions? After solving this exercise, it will be clear that it suffices to suppose the length L of the boundary α of U is precisely 2π. As before, we suppose it is a regular parameterized curve, and that it is parameterized by arclength. The point of Fourier analysis, or at least one of the points, is that it turns out we can write periodic functions (this is being less general than we could be, of course), in terms of a favorable basis, sines and cosines, and differentiation is friendlier here too. This will reduce some questions down to elementary manipulations of sequences. To be more precise, note that Z 2π sin(nt) cos(mt)dt ∀m, n = 0 (3.1) 0 and Z 2π Z 2π sin(nt) sin(mt) = πδmn, cos(nt) cos(mt) = πδmn (3.2) 0 0 So that in the infinite dimensional vector space of square integrable functions (i.e. functions where there is a definition of integral, and it is finite) on [0, 2π], sin(nt) and cos(nt) for n ∈ Z are an orthogonal set. In fact they can be seen to span this space, so in fact are an orthogonal basis for this space of functions. By Euler’s formula (this can be seen from Taylor series): eint = cos(nt) + i sin(nt) (3.3) ISOPERIMETRIC INEQUALITY IN THE PLANE 5

int So from this we can see pretty easily that in fact {e }n∈Z is a basis for the space of (complex valued) square integrable functions (this time, though, we will have R einteimtdt = 0 only if m = −n).

With this in mind, writing α(t) = (α1(t), α2(t)), represent α1 and α2 as α1(t) = P a eint, α (t) = P b eimt n∈Z n 2 m∈Z m With all this set up, we should now write L and A in terms of these sequences.

R 2π 0 L is easiest so we start with that. Here, 2π = L = 0 |α (t)|. Of course since R 2π 0 2 R 2π 0 2 0 2 our curve is arclength parameterized, this is just 0 |α (t)| = 0 α1(t) + α2(t) . We are in the space of complex valued functions apriori, but α1 and α2 and their R 2π 0 2 0 2 derivatives are real valued so we can further write this as 0 |α1(t)| + |α2(t)| It turns out that its ﬁne just to pass the derivatives through the sum, so in turn this will be equal to: Z 2π X 2 2 2 int X 2 2 2 n (|an| + |bn| )e dt = 2π n (|an| + |bn| ) (3.4) 0 n∈Z n∈Z This of course implies that

X 2 2 2 n (|an| + |bn| ) = 1 (3.5) n∈Z

To write the area A in terms of these sequences, we have to relate the functions deﬁning the boundary to the area, otherwise our work above was worth- less. This should immediately make you think back to your multivariable calculus course, because there we had various related theorems relating integrals over domains to integrals deﬁned on the boundaries of the domains. One one hand, we have R A = U 1dxdy. On the other hand, by Green’s theorem we may write: 1 Z A = | xdy − ydx| (3.6) 2 ∂U Writing this out in terms of the Fourier series like above, we see that X A = |π n(anb−n − bna−n)| (3.7) n∈Z

Again, since α1 and α2 are real valued (so they agree with their complex conjugate, one immediately sees that a−n = an and similarly for the bn. So it suﬃces to show the following: X |n(anbn − bnan)| ≤ 1 (3.8) n∈Z ISOPERIMETRIC INEQUALITY IN THE PLANE 6

This is easy though, using that 2ab ≤ a2 + b2. That’s basically the proof, with a couple small details left: Exercise 4: Fill in the following assertions • Justify (3.1), (3.2), and (3.3) • Justify (3.7) • Explain why the isoperimetric inequality follows from showing the inequality (3.8), and write out the details for why (3.8) is true.

4. Proof via the curve shortening flow Finally, we are back to differential geometry! This proof is certainly overkill, but geometric flows (very roughly speaking, deformations of shapes which deform the shapes according to their area) are a topic of intense research even now. This proof can also be used to prove stronger isoperimetric inequalities on curved surfaces, which we may (or may not) return to later in the course. Let α(s) be a closed, regular, arclength parameterized curve in the plane. The curve shortening flow of α is defined as a/the family of curves α(s, t), the ”family” parameterized by t, satisfying dα = −kn, α(s, 0) = α (4.1) dt This turns out to actually be a sort of heat equation. To compare, the (regular) heat equation starting from a function f of one variable is given by ∂f(x, t) ∂2f(x, t) − = 0, f(x, 0) = f(x) (4.2) ∂t ∂x2

The short time existence of a solution to the curve shortening flow (i.e. a family of parameterized curves solving the equation above) is actually not so simple to prove, but under reasonable assumptions a solution to the curve shortening flow exists for small times t given a curve α. Typically, an explicit solution to the curve shortening flow is not so easy to find, but under situations of high symmetry, the equation reduces to an ODE: Exercise 5: Convince yourself that a round circle will have a CSF that stays round and will ”extinguish” in finite time. You’ll note that the solution above does not exist for all time - by which we mean the flow deforms the curve into something that is no longer a regular parameterized curve, and in fact this is common for the flow of all closed curves in R2. The reason is because if two closed curves are disjoint, their flows must stay disjoint under the flow - this is an artifact of the curve shortening flow being a heat equation in fact. ISOPERIMETRIC INEQUALITY IN THE PLANE 7

The upshot one may encompass a given closed curve C with a huge circle, who’s flow we know doesn’t exist for all time, to see the same must be true for the flow of C. So, the flow will eventually approach something which is no longer a regular parameterized curve. It could shrink to a point, or for all we know it could develop a neckpinch as the figure below suggests:

Figure 4. By neckjpinch in the above we mean where the ﬂow makes points come very close together while some points stay far apart

It turns out however, by a remarkable theorem of Grayson shown in the mid 1980s (you know people who were alive before this theorem was proved!) that neckpinches actually never occur, so that curves only become ”unflowable” if they shrink to points. Now onto the isoperimetric inequality. It should be reasonable that, since the curve moves by speed κ, the area of a domain U as you evolve its boundary ∂U by the CSF changes by dA Z = − κ = −2π (4.3) dt ∂U The last inequality perhaps comes as bit of a shock - the mean curvature flow pre- serves embeddedness of a closed curve (or so that, if a closed curve doesn’t cross itself initially, it won’t later on), and it is an important theorem that the integral of curvature of a plane curve measures how much it ”winds.” To sell you on this, inte- grating the curvature tells you how many radians the tangent vector sweeps out as it spins around as you go along the closed curve, since the curvature is the derivative of the direction the points as you move along the curve - this angle must be an integer ISOPERIMETRIC INEQUALITY IN THE PLANE 8 multiple of 2π since the curve is closed. Since the curve doesn’t intersect itself, you can see at least from a picture it must be 2π precisely. As for the length of the boundary, it evolves as follows: dL Z = − κ2 (4.4) dt To see this, parameterize the flow of the ∂U by α = α(s, t) so that L = R |dα/ds|; in this case it is legal to pass the d/dt through the integral, and you can calculate to see the formula using the Serret formulas Exercise 6: Fill in the details to what was written above to get (4.4). Now at last we are ready to prove the isoperimetric inequality using the curve shortening flow - or to kill a flea with a hammer. In one line we can relate the evolution of area and length under the CSF; all integrals are on ∂U: dA Z Z Z dL dL2 −4π t = −4π κ = 2( κ)2 ≤ 2L κ2 = −2L = − (4.5) dt t t dt dt Since by Grayson we know every curve will go extinct in finite time, say by time T , and at that time length and area will be zero, we get the isoperimetric inequality by integration.

5. What about equality? To wrap things up let’s briefly discuss the case of equality in the isoperimetric inequality - this section like the one before will be skip some details, and could be handled in a more elementary way, but this approach is reminiscent of the “varational method” commonplace in modern day geometry so is worthwhile. First note that round circles (and the discs they bound) satisfy the equality L2 = 4πA (indeed, L = 2πr and A = πr2). There is still the question as to whether there are regions besides discs satisfying L2 = 4πA however. So, consider a region U bounded by a unit speed parameterized curve α satisfying L2 = 4πA (of course, imaginably there is a region U with a really pathological boundary satisfying the equality, but by ”geometric measure theory” our assumption is fine under some very weak assumptions which we sweep under the rug). It is easy to see it suffices to show that the curvature of α, κ, is constant. Exercise 7: Show that a closed, unit speed parameterized plane curve with constant curvature traces out a round circle. With this in mind and continuing to (slightly) suspend rigour note that if we perturb the trace of our curve α by tfν, where f is a function on α and t is small ISOPERIMETRIC INEQUALITY IN THE PLANE 9 enough to ensure the perturbation of the trace is still embedded (for our case, take this to mean it bounds a region), then we will have Z Z A0(0) = f, L0(0) = κf (5.1) where A and L are the area of the enclosed region and length of its boundary (compare with the section above). Restricting ourselves to perturbations such that R f = 0, we see it will suffice to show that we can arrange such a perturbation so that R κf < 0 when κ is not constant; if equality for L2 = 4πA was already achieved this would imply there were domains violating the isoperimetric inequality! A picture (which perhaps could be more convincing!!) which gives the basic idea is the following:

Figure 5. If α has nonconstant curvature, we can ﬁnd perturbations which don’t change the area but which ”round” the curve

Of course, the reader should try to make this rigorous: Exercise 8: Generalize the picture above to all noncircular curves and, in doing so, complete the proof of the rigidity case of the isoperimetric inequality.