Isoperimetric-like inequality for non-convex sets - MathOverflow

Isoperimetric-like inequality for non-convex sets

2011-10-10T09:58:57Z

The classical isoperimetric inequality can be stated as follows: if $A$ and $B$ are sets in the plane with the same area, and if $B$ is a disk, then the perimeter of $A$ is larger than the perimeter of $B$.

There are several ways to define the perimeter. Here is a unusual one: if $A \subset \mathbb{R}^2$ is a convex set, the Cauchy-Crofton formula says that the perimeter of $A$ equals the measure of the set of lines that hit $A$, or

$$ p(A) = \frac{1}{2} \int_{S^1} \lambda(P_{\theta} A) d\theta, $$

where $P_\theta$ is the orthogonal projection in the direction $\theta \in S^1$, and $\lambda$ the Lebesgue measure on any line.

Now, this definition of $p(A)$ makes sense for non-necessarily convex sets, excepts that it is not the usual notion of perimeter, so let's call it rather "mean shadow". My question if whether the isoperimetric inequality holds for the mean shadow instead of perimeter: if $A,B$ are (open, say) subsets of the plane with equal area, and if $B$ is a disk, is the mean shadow of $A$ larger that the mean shadow of $B$ ?

The inequality is true if $A$ is convex, and we can assume that $A$ is a disjoint union of convex sets (since taking the convex hull of a connected set does not change the mean shadow).

Answer by Anton Petrunin for Isoperimetric-like inequality for non-convex sets

2011-10-11T03:40:34Z

As you noticed, it is sufficient to consider the case $$F=\bigcup_{i=1}^n F_i$$ where $F_1$, $F_2,\dots, F_n$ are disjoint convex figures with nonempty interior. Let $s$ be mean shadow of $F$. Denote by $K$ the convex hull of all $F$. Note that
$$\mathop{\rm length}(\partial K\cap F)\le s.$$

We will prove the following claim: one can bite from $F$ some arbitrary small area $a$ so that mean shadow decrease by amount almost $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s$ (say $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s{\cdot}(1-\tfrac{a}{s^2})$ will do). Once it is proved, we can bite whole $F$ by very small pieces, when nothing remains you will add things up and get the inequality you need.

The claim is easy to prove in case if $\partial F$ has a corner (i.e., the curvature of $\partial F$ has an atom at some point). Note that the total curvature of $\partial K$ is $2{\cdot}\pi$, therefore there is a point $p\in \partial K$ with curvature $\ge 2{\cdot}\pi{\cdot}\tfrac1s$. The point $p$ has to lie on $\partial F$ since $\partial K\backslash \partial F$ is a collection of line segments. Moreover, if there are no corners, we can assume that $p$ is not an end of segment of $\partial K\cap F$.

This proof is a bit technical to formalize, but this is possible. (If I would have to write it down, I would better find an other one.)

Answer by Jean-Marc Schlenker for Isoperimetric-like inequality for non-convex sets

2011-10-11T13:59:51Z

I believe that the answer is positive. If $A$ is connected, then it has the same mean shadow as its convex hull $CH(A)$ so the isoperimetric inequality for $CH(A)$ shows that the mean shadow of $A$ is larger than the mean shadow of $B$.

If $A$ is not connected I believe the same inequality holds. I'll sketch a proof when $A$ has finitely many connected components $A_1, \cdots, A_n$, the general case then follows by an approximation argument. Choose a point $x\in CH(A)$ and define a 1-parameter family of deformations of $A$ by making a parallel translate of each connected component $A_i$ so that its barycenter moves towards $x$, say at constant speed to reach it at time $t=1$. Stop this 1-parameter family of deformation as soon as a contact occurs between two connected components, then merge those two connected components and repeat.

The point is that the area of $A$ does not change under this deformation, however the mean shadow is non-increasing -- actually the size of the shadow is non-increasing in every direction. At the end of this deformation one obtains a connected set to which the usual isoperimetric inequality can be applied, and the same inequality then also applies to $A$.