Posts Tagged ‘Eric Carlen’

Stable isoperimetric inequality

October 10, 2011

Eric Carlen from Rutgers gave a colloquium last week in which he bought up some curious questions and facts regarding the ‘stability’ of standard geometric inequalities such as the isoperimetric and Brunn-Minkowski inequality. To prevent myself from forgetting it, I’m dropping a short note on this matter here. Interestingly I was unable to locate any reference to this nor did I take any notes, hence this post is completely based on my recollection of a lunch five days ago.

–Many thanks to Marco Barchies, serval very high-quality references are located now. It turns out that starting with Fusco-Maggi-Pratelli ’06  which contains a full proof of the sharp bound, there has been a collective progress on shorter/different proofs and variations of the theorem made. See comments below!

As we all know, for sets in \mathbb{R}^n, the isoperimetric inequality is sharp only when the set is a round ball. Now what if it’s ‘almost sharp’? Do we always have to have a set that’s ‘close’ to a round sphere? What’s the appropriate sense of ‘closeness’ to use?

One might first attempt to use the Hausdorff distance:

D(S, B_1(\bar{0})) = \inf_{t \in \mathbb{R}^n}\{D_H(S+t, B_1(\bar{0})).

However, we can easily see that, in dimension 3 or higher, a ball of radius slightly small than 1 with a long and thin finger sticking out would have volume 1, surface volume \varepsilon larger than that of the unit ball, but huge Hausdorff distance:

In the plane, however it’s a classical theorem that any region S of area \pi and perimeter m_1(\partial S) \leq 2\pi +\varepsilon as D(S, B_1(\bar{0})) \leq  f(\varepsilon) where f(\varepsilon) \rightarrow 0 as \varepsilon \rightarrow 0 (well, that f is because I forgot the exact bound, but should be linear in \varepsilon).

So what distance should we consider in higher dimensions? Turns out the nature thing is the L^1 norm:

D(S, B_1(\bar{0})) = \inf_{t\in \mathbb{R}^n} \mbox{vol}((S+t)\Delta B_1(\bar{0}))

where \Delta is the symmetric difference.

First we can see that this clearly solves our problem with the thin finger:

To simplify notation, let’s normalize our set S \subseteq \mathbb{R}^n to have volume 1. Let B_n denote the ball with n-dimensional volume 1 in \mathbb{R}^n (note: not the unit ball). p_n = \mbox{vol}_{n-1}(\partial B_n) be the (n-1 dimensional) measure of the boundary of B_n.

Now we have a relation D(S, B_n)^2 \leq C_n (\mbox{vol}_{n-1}(\partial S) - p_n)

As said in the talk (and I can’t find any source to verify), there was a result in the 90’s that D(S, B_n)^4 \leq C_n (\mbox{vol}_{n-1}(\partial S) - p_n) and the square one is fairly recent. The sharp constant C_n is still unknown (not that I care much about the actual constant).

At the first glance I find the square quite curious (I thought it should depend on the dimension maybe like n/(n-1) or something, since we are comparing some n-dimensional volume with (n-1) dimensional volume), let’s see why we should expect square here:

Take the simplest case, if we have a n-dimensional unit cube C_n, how does the left and right hand side change when we perturbe it to a rectangle with small eccentricity?

As we can see, D(R_\varepsilon, C_n) is roughly p_n \varepsilon. The new boundary consists of two faces with measure 1+\varepsilon, two faces of measure 1-\varepsilon and 2 \times (n-2) faces with volume 1-\epsilon^2. Hence the linear term cancels out and we are left with a change in the order of \varepsilon^2! (well, actually to keep the volume 1, we need to have 1-\varepsilon/(1+\varepsilon) instead of 1-\varepsilon, but it would still give \varepsilon^2)

It’s not hard to see that ellipses with small eccentricity behaves like rectangles.

Hence the square here is actually sharp. One can check that no matter how many of the n side-length you perturbe, as long as the volume stay the same (up to O(\varepsilon^2)) the linear term of the change in boundary measure always cancels out.

There is an analoge of this stability theorem for the Brunn-Minkowski inequality, i.e. Given two sets of volume V_1, V_2, if the sum set has volume only a little bit larger than that of two round balls with those volumes, are the sets L^1 close to round balls? I believe it’s said this is only known under some restrictions on the sets (such as convex), which is strange to me since non-convex sets would only make the inequality worse (meaning the sum set has larger volume), don’t they?

I just can’t think of what could possibly go wrong for non-convex sets…(Really hope to find some reference on that!)

Anyways, speaking of sum sets, the following question recently caught my imagination (pointed out to me by Percy Wong, thank him~ and I shall quote him ‘this might sound like probability, but it’s really geometry!’):

Given a set T \subseteq l^2 (or \mathbb{R}^n), we define two quantities:

G(T)=\mathbb{E}(\sup_{p \in T} \Sigma p_i g_i) and

B(T) = \mathbb{E}(\sup_{p \in T} \Sigma p_i \epsilon_i)

where \mathbb{E} is the expected value, \{g_i\} are independent random variables with a standard normal distribution (mean 0, variance 1) and \{\epsilon_i\} are independent Bernoulli random variables.

Question: Given any T, can we always find T' such that

T \subseteq T' + B_{l^1}(\bar{0}, c B(T)) and

G(T') \leq c B(T)

To find out more about the question, see Chapter 4 of this book. By the way, I should mention that there is a $5000 prize for this :-P