The Isoperimetric Inequality

This will be a cool application of the concepts introduced in the last post. We call a curve rectifiable if there exists finite M such that for any partition a=t_0<t_1<\cdots<t_N=b, \sum^N_{j=1}|z(t_j)-z(t_{j-1})\le M, i.e. the curve has some notion of “finite length,” where length is the supremum of this quantity taken over all partitions or equivalently the infimum over all upper bounds M. It follows by definition that if the x– and y-parametrizing functions of the curve are of bounded variation, the curve is rectifiable. It is not, however, true that the classical arc length formula holds for all curves of bounded variation: just consider a curve where the x and y functions are the Cantor-Lebesgue function. We get a straight line from the origin to (1,1), but the derivative of the function is zero almost everywhere.

Result 4: The arc length formula does work if we assume absolute continuity of x(t) and y(t).

Proof: We will prove that the total variation of a complex valued function over [a,b] is \int^b_a|F'(t)|dt, because then we can just substitute F(t) = x(t)+iy(t). As usual, we will prove that there is inequality in both directions. But recall that for absolutely continuous functions, the fundamental theorem of calculus holds, so pick a partition a = t_0<t_1<\cdots<t_N=b so that \sum^N_{j=1}|F(t_j)-F(t_{j-1}| = \sum^N_{j=1}\left|\int^{t_j}_{t_{j-1}}F'(t)\right|dt\le\int^b_a|F'(t)|dt. In the other direction, recall that step functions are dense in L^1, so we can find an approximating step function g to F' so that h = F'-g has integral arbitrarily small. By the triangle inequality, T_F(a,b)\ge T_G(a,b)-T_H(a,b)>T_G(a,b)-\epsilon, where G(x) = \int^x_a g(t)dt and H(x) = \int^x_ah(t)dt. We can bound T_G(a,b) by taking a partition where the adjacent intervals are over constant parts of the step function g so that T_G(a,b)\ge \sum^N_{j=1}\left|\int^{t_j}_{t_{j-1}}g(t)\right|dt=\int^b_a|g(t)|dt. But recall that we picked g to be extremely close to F' so that \int^b_a|g(t)|dt\ge\int^b_a|F'(t)dt|-\epsilon, so T_F(a,b)\ge\int^b_a|F'(t)|dt - 2\epsilon and we get inequality in the other direction.

Now before we proceed to state and prove the isoperimetric inequality, let’s get some vocabulary under our belt. Define the one-dimensional Minkowski content M(K) of a curve K to be \lim_{\delta\to 0}\frac{m(K^{\delta}}{2\delta}, where K^{\delta} denotes the set of points which are at most \delta away from any point in K. Define a simple curve to be a curve that doesn’t double over on itself, a quasi-simple curve to be a curve such that t\to z(t) is injective except for finitely many points, and a closed curve to be one that starts where it ends. As the name suggests, the Minkowski content of a curve turns out to be precisely its length if the curve is rectifiable and quasi-simple.

Proof: Define M^*(K) and M_*(K) to be the limit superior and limit inferior of \frac{m(K^{\delta})}{2\delta}, respectively. To prove this, we prove two results on M^* and M_*, respectively: i) if \Gamma is quasi-simple and M_* is finite, then the curve is rectifiable and the length is bounded above by M_*, ii) if the curve is rectifiable then M^* is bounded above by the length.

Begin with any partition a = t_0<t_1<\cdots<t_N=b and let L_P denote the length of this polygonal line. Then by absolute continuity, we can pick large enough subintervals I_j = [a_j,b_j] of each of the (t_{j-1},t_j) so that \sum^N_{j=1}|z(b_j)-z(a_j)|\ge L_P-\epsilon. Now consider the sub-curves \Gamma_j over these intervals and their padded versions \Gamma^{\delta}_j, where we pick \delta so that they are all still disjoint. Their union is contained within \Gamma^{\delta} obviously, so m(\Gamma^{delta})\ge \sum^N_{j=1}m(\Gamma^{\delta}_j). What we will next prove will allow us to take the step \sum^N_{j=1}m(\Gamma^{\delta}_j)\ge 2\delta\sum^N_{j=1}|z(b_j)-z(a_j)|, after which we are done.

So we want to prove in general that m(\Gamma^{\delta})\ge 2\delta|z(b)-z(a)|. Simply rotate the curve so that the endpoints lie on the x-axis, with x-coordinates of A and B. Take slices \Gamma^{\delta}_x at fixed x. These will contain intervals of length 2\delta so that by Fubini’s Theorem, m(\Gamma^{\delta}) = \int_{\mathbb{R}}m(\Gamma^{\delta}_x)dx\ge\int^B_Am(\Gamma^{\delta}_x)\ge 2\delta(B-A)=2\delta|z(b)-z(a)|.

To prove our second claim, we introduce another natural way of parametrizing curves, by distance rather than time. If L(a,b) is the length of the curve from start point t = a to end point t = b, let s(t) = L(a,t). Redefine our curve z(t) = x(t)+iy(t) as \tilde{z}(s) = \tilde{x}(s)+\tilde{y}(s) so that \tilde{z}(s) = z(t) when s = s(t). Observe that |\tilde{z}(s_1)-\tilde{z}(s_2)|\le |s_1-s_2| in general, so the components \tilde{x}(s) and \tilde{y}(s) are absolutely continuous and |\tilde{z}(s)|=1 for almost all s. Furthermore, the formula for arc length still works.

Returning to our proof that M^*(\Gamma)\le L for \Gamma rectifiable, let’s use this new parametrization z(s). Consider the sequence of functions F_n(s) = \sup_{0<|h|<1/n}\left|\frac{z(s+h)-z(s)}{h}-z'(s)\right|, where we have extended z to equal z(0) to the left of 0 and z(L) to the right of L. By continuity of z, we only need to worry about taking the supremum over countably many h, so F_n is the supremum of countably many measurable functions and thus itself measurable. The sequence tends to zero almost everywhere, and by Egorov’s theorem (a result in measure theory that I’ll prove in a later post about Littlewood’s three principles), we have uniform convergence outside of a set E_{\epsilon} of measure less than \epsilon. With this in mind, for fixed \epsilon, also pick an r_{\epsilon}=1/n with n large enough that F_n(s)<\epsilon.

Now partition the interval into intervals I_1,...,I_N each of length \rho less than r_{\epsilon}, except possibly the last one. Call the intervals which do not lie entirely in E_{\epsilon} the “good intervals” and those which do the “bad intervals.” Denote \Gamma_j again to be the sub-curve on \Gamma corresponding to I_j. What follows is quite bashy, but the basic point is to find bounds for each \Gamma^{\delta}_j that can be made arbitrarily small.

In the case of good intervals, pick some s_0\in I_j=[a_j,b_j] which does not lie in E_{\epsilon}. Rotate and translate \Gamma_j so that z(s_0) = 0 and z'(s_0) = 1. We can check that because |z(s_0+h)-h|<\epsilon|h| by our choice of s_0, and because |h|\le\rho<r_{\epsilon}, that the rectangle [a_j-s_0-\epsilon\rho,b_j-s_0+\epsilon\rho]\times[-\epsilon\rho,\epsilon\rho] contains the curve and the rectangle containing \Gamma^{\delta}_j has measure 2\delta\rho+O(\epsilon\delta\rho+\delta^2+\epsilon\rho^2).

In the case of bad intervals, we still have that |z(s)-z(s')|\le |s-s'| so that the bad sub-curves are in balls of radius \rho and thus m(\Gamma^{\delta}_j)\le O(\delta^2+\rho^2). So in total, summing up to get a bound on m(\Gamma^{\delta}) and then dividing by 2\delta and rearranging, we eventually get that \frac{m(\Gamma^{\delta})}{2\delta}\le L+O\left(\rho+\epsilon+\frac{\delta}{rho}+\frac{\epsilon\rho}{\delta}+\frac{\rho^2}{\delta}\right).

Finally, we have to be careful what we choose for the length \rho of the sub-intervals, because we want it to be close to \delta; it turns out \frac{\delta}{\sqrt{\epsilon}} is a good choice, because then for small enough \delta we still have that \rho<r_{\epsilon} and the above inequality becomes \frac{m(\Gamma^{\delta})}{2\delta}\le L+O\left(\frac{\delta}{\sqrt{\epsilon}}+\epsilon+\sqrt{\epsilon}+\frac{\delta}{\epsilon}\right). Taking the limit superior as \delta goes to zero, we get an upper bound of L+O(\epsilon+\sqrt{\epsilon}) so that for \epsilon arbitrarily small, we get that M^*(\Gamma)\le L as desired.

Now we are finally ready to state and prove the isoperimetric theorem. Our hypotheses are that \Omega is a bounded open subset in \mathbb{R}^2 and its boundary \Gamma is rectifiable.

Result 5: 4\pi m(\Gamma)\le L(\Gamma)^2. In other words, of all closed curves with the same length, the circle encloses the largest area.

Proof: This proof makes use of the following cool result:

Brunn-Minkowski Inequality: If A and B are measurable sets in \mathbb{R}^d and their sum is measurable (the sum of two sets is simply the set of all sums of points), then m(A+B)^{1/d}\ge m(A)^{1/d}+m(B)^{1/d}.

Proof: We progress in order of “fineness” of the sets involved. In the case where A and B are rectangles, this turns out simply to be AM-GM. Now if they are the union of finitely many almost disjoint rectangles, the key is to induct on the number of rectangles in A and B. Pick R_1 and R_2, rectangles in A, and split them by a hyperplane. Denote the half of A on one side of this hyperplane to be A_1 and the other half to be A_2 so that both have one fewer rectangle than A. Then translate B so that its two halves as split by the hyperplane are in the same proportions as the halves of A. Note that the union of A_1+B_1 and A_2+B_2, disjoint by construction, lies in A+B. By our induction hypothesis and a bit of calculation, we get the desired result.

To prove the case in which A and B are any open sets of finite measure, invoke the fact that every open set can be written as a countable union of almost disjoint closed cubes. Now if A and B are arbitrary compact sets, note that A+B is compact; we can easily construct “buffered” sets A^{\epsilon} and B^{\epsilon} which are open and defined to e points at most \epsilon away from A and B, respectively, and from our result on open sets the inequality is evident for compact sets as well. But because we can approximate measurable sets in general from the inside using compact sets, we get Brunn-Minkowski for all measurable sets.

Returning to the isoperimetric theorem, let’s construct “rings” \Omega_- and \Omega_+, defined respectively to be the points at least \delta away from the complement of \Omega and the points at most \delta away from \Omega. Denote the disc centered at the origin and of radius \delta to be D(\delta). Then D(\delta)+\Omega_-\subset \Omega and \Omega+D(\delta)\subset\Omega_+, so by Brunn-Minkowski and some rearranging, we get that

m(\Omega_+)\ge\left(m(\Omega)^{1/2}+m(D(\delta))^{1/2}\right)^2\ge m(\Omega)+2\pi^{1/2}\delta m(\Omega)^{1/2}

as well as

-m(\Omega_-)\ge -m(\Omega)+2\pi^{1/2}\delta m(\Omega_-)^{1/2}.

Note additionally that m(\Gamma^{\delta})=m(\Omega_+)-m(\Omega_-), so from our inequalities, m(\Omega^{\delta})\ge 2\pi^{1/2}\delta(m(\Omega)^{1/2}+m(\Omega_-)^{1/2}). Dividing by 2\delta to get the upper Minkowski content when \delta\to 0 and using a previous result, we conclude that L(\Gamma)\ge \pi^{1/2}(2m(\Omega)^{1/2}). Rearranging gives the isoperimetric theorem.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s