# Stable isoperimetric inequality

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