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Three Proofs of the Isoperimetric Inequality in R2

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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 = [a2A;b2Ba+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 in- equality. 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 p p right goes to zero as does, and the second term is just 2 π A. So we have p p 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 8m; 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 2 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 fe gn2Z 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 n2Z n 2 m2Z 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 jα (t)j. Of course since R 2π 0 2 R 2π 0 2 0 2 our curve is arclength parameterized, this is just 0 jα (t)j = 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 jα1(t)j + jα2(t)j 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 (janj + jbnj )e dt = 2π n (janj + jbnj ) (3.4) 0 n2Z n2Z This of course implies that X 2 2 2 n (janj + jbnj ) = 1 (3.5) n2Z To write the area A in terms of these sequences, we have to relate the func- tions 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 do- mains 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 = j xdy − ydxj (3.6) 2 @U Writing this out in terms of the Fourier series like above, we see that X A = jπ n(anb−n − bna−n)j (3.7) n2Z 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 jn(anbn − bnan)j ≤ 1 (3.8) n2Z 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.
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