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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).

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I don't quite understand the question. You can always compare an open set to its convex hull, it will have the same mean shadow, but a smaller area. So the isoperimetric inequality for open subsets should follow from the convex case? –  Jean-Marc Schlenker Oct 10 '11 at 10:03
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For connected sets, taking the convex hull does not change the mean shadow, but for non-connected sets it increases ... –  Guillaume Aubrun Oct 10 '11 at 10:09
    
I'm not sure whether considering open sets changes the following argument substantially: Take two closed balls with volume one half and radii $r$. Put them with canters apart by a distance of $L$ in the plane. Then for $L\to\infty$ the mean shadow will be $4\pi r$. Further the mean shadow is monotone in $L$ and decreases for decreasing $L$ as the overlap in the projection becomes larger if the balls are nearer. Eventually, for $L=2r$ the balls will touch and you have a connected set, where the mean shadow conincides with the perimeter of the convexification. –  André Schlichting Oct 11 '11 at 9:40

2 Answers 2

up vote 4 down vote accepted

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.)

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I don't understand why the total length of $\partial K \cap F$ is at least as big as the mean shadow of $F$. It seems to me that $\partial K \cap F$ can be arbitrary small for a given mean shadow (consider the case when $F_1,F_2,F_3$ are very small balls around the vertices of a large equilateral triangle, and $F_4$ anything inside the triangle). Did I miss something ? –  Guillaume Aubrun Oct 12 '11 at 12:00
    
@Guillaume, now it is fixed (if you would read further you would see that this is a misprint). –  Anton Petrunin Oct 12 '11 at 17:19
    
@Anton, what if the curvature is concentrated at corners where $\partial K$ meets $F$? –  Sergei Ivanov Oct 12 '11 at 21:34
    
OK, I got it now, that's indeed a nice argument although it may be hard to write down. Thanks ! This gives also a proof of the usual isoperimetric inequality which I didn't know. Is this proof standard, or written somewhere ? –  Guillaume Aubrun Oct 13 '11 at 8:58
    
@Guillaume. I learned this idea from Kliener's proof of isoperimetric inequality for 3-dimensional Hadamard space. –  Anton Petrunin Oct 13 '11 at 15:52

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 $A$.

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That's a very interesting idea, but it's not clear to me that it works, and it does not seem true that the shadow in every direction is non-increasing when the sets move as you describe. For example take 2 (disjoint) congruent equilateral triangles, one pointing to the right, and one pointing to the left, such that their vertical shadows coincide. When you move them along the line passing though their barycenters, the vertical shadow is actually increasing ! –  Guillaume Aubrun Oct 11 '11 at 14:21
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You're right -- I think that the mean shadow probably decreases, but the argument I gave is not correct. I'll look into this later, unless some else can make it work... –  Jean-Marc Schlenker Oct 11 '11 at 15:34
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Note that your argument works (shadows in every directions decrease) in the case when $A_1,\dots,A_n$ have a center of symmetry, basically by the one-dimensional version of the Hadwiger-Kneser-Poulsen conjecture. –  Guillaume Aubrun Oct 11 '11 at 17:15

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