# Characterization of sphere.

I'm editing the question because I think the previous formulation was leaving a key element of the problem out and that was making it impossible to answer the question. I tried to update/improve the notations, but if there is something wrong with them, feel free to let me know in the comments.

let $\mathcal{A}$ is the unit square in $\mathbb{R}^2$ and $m_{\mathcal{A}}$ is the area of $\mathcal{A}$.

Next, consider the set $\mathcal{H}$ of all sub-sets of $\mathcal{A}$ that have area $m_{\mathcal{A}}/2$.

Now, given $H\in\mathcal{H}$, define the set $\mathcal{B}_H$ as the set of all lines having non empty intersection with $H$.

Now consider the problem of finding $H^*\in\mathcal{H}$ that minimizes:

$$\int_{\pmb a\in\mathcal{B}_H}\underset{\pmb x\in H}{\max}\quad M^2(\pmb x,\pmb a)d\pmb a$$

where

$$M^2(\pmb x,\pmb a)=(\pmb x'\pmb a)^2/||\pmb a||^2$$

is the squared orthogonal distance of $\pmb x$ to $\pmb a$.

I think $H^*$ is the circle with area $m_{\mathcal{A}}/2$. This is based partly on intuition, partly on trying various simple an randomly shaped sets $H$ on a computer. Now I m looking for a proof that this intuition is true or false an I don t really know how to proceed from here.

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Most of your notation is fine. But what distance is meant by squared orthogonal? Do you just mean $(x_1-a_1)^2+(x_2-a_2)^2$? The subscripts mean coordinates: $x=(x_1,x_2)$ and $a=(a_1, a_2)$. –  Eric Stucky Jun 27 '14 at 23:41
Yeah, I wasn't going to say anything but I don't know what $a'x$ means either. My best guess is dot product: $a_1x_1+a_2x_2$? (Also: definitely not "inf", because that gives a number, not a set. I would just say "I am seeking an $H^*\in\mathcal H$ which minimizes [the integral]".) –  Eric Stucky Jun 28 '14 at 20:43
Maybe I'm a bit thicker than Eric, but I don't understand your notation. The dot product of $\mathbf a$ and $\mathbf x$ is not the same as the distance between $\mathbf a$ and $\mathbf x$. And in neither case is the set of all $\mathbf a$ such that $\exists\mathbf x\in H:\mathbf a'\mathbf x=0$ the set of all lines intersecting $H$. –  Rahul Jul 2 '14 at 9:41
It's nothing to be embarrassed about if you have a conceptually clear question but are unfamiliar with the proper notation; you should just write your question in conceptual terms that you are comfortable with. It's perfectly possible to write a good mathematical question with little or no formal notation, and anyway an excess of notation can be just as hard to read as it is hard to write. –  Rahul Jul 2 '14 at 12:35
@Rahul: thanks! I was effraid the question would be rejected if it was too worldly for this site. I've removed all notations that are unnecessary. I felt that I still needed the integral though and hence the $M(x,a)$ that is causing so much confusion. –  user42397 Jul 2 '14 at 13:01

I think this is a nice question. If I'm not mistaken, you're looking for the shape $H$ of area $1/2$ which minimizes $$f(H) = \int{\rm dist}(a,H)^2\,\mathrm da$$ where ${\rm dist}(a,H)$ is the distance from the line $a$ to the farthest point in $H$, and the integral is taken over all lines $a$ which intersect $H$.

But what does it mean to integrate over a set of lines? We have to define what $\mathrm da$ actually means; that is, we have to choose a measure over the space of lines in the plane. A natural choice is the kinematic measure, which is invariant to rigid motions: If we describe a line $a$ by its distance from the origin, $p\in\mathbb R$, and the direction to the closest point, $\theta\in[0,\pi)$, then the kinematic measure is simply $\mathrm da = \mathrm dp\,\mathrm d\theta.$

Figure from "Integral Geometry & Geometric Probability" by Andrejs Treibergs.

For any given $\theta$, let $p^-$ and $p^+$ be the smallest and largest values of $p$ for which the line $a = (p,\theta)$ intersects $H$. That is, $H$ is sandwiched between the two parallel lines $(p^-,\theta)$ and $(p^+,\theta)$. Assuming $H$ is connected, the set of lines with orientation $\theta$ that intersect $H$ is precisely the ones for which $p$ lies between $p^-$ and $p^+$. So our integral becomes $$f(H) = \int_0^{\pi}\int_{p^-}^{p^+}{\rm dist}(a,H)^2\,\mathrm dp\,\mathrm d\theta.$$

The point on $H$ farthest from the line $a = (p,\theta)$ lies on either $(p^-,\theta)$ or $(p^+,\theta)$, so $${\rm dist}(a,H) = \max(p-p^-,p^+-p)$$ and \begin{align} \int_{p^-}^{p^+}{\rm dist}^2(a,H)\,\mathrm dp &= \int_{p^-}^{p^+}\max(p-p^-,p^+-p)^2\,\mathrm dp \\ &= \frac7{12}(p^+-p^-)^3 \\ &=\frac7{12}{\rm width}_H(\theta)^3, \end{align} where ${\rm width}_H(\theta) = p^+-p^-$ is the "width" of $H$ in the direction $\theta$. Consequently, we have $$f(H) = \frac7{12}\int_0^{\pi}{\rm width}_H(\theta)^3\,\mathrm d\theta$$ which we can interpret as $7\pi/12$ times the "mean cubed width" of $H$.

Figure from the Wikipedia article "Mean width".

Thus, at least if we restrict ourselves to connected shapes, your problem is equivalent to finding the shape of constant area $1/2$ with the smallest mean cubed width. It certainly seems likely that such a shape must be a circle, but I don't know how to prove it. Two thoughts:

1. If you were minimizing the mean width (without the cube inside the integral), then the minimal shape would indeed be a circle. This follows from the isoperimetric inequality and the fact that the mean width is $1/\pi$ times the perimeter of the convex hull of $H$.

2. It seems natural to rule out disconnected shapes by arguing that one could always move the connected components towards each other without increasing the width, but I'm having trouble justifying that rigorously.

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Thanks. This answer provides a lot of elements. I will take the time this week end to read thoroughly the linked references. It certainly enhance my understanding of the problem. –  user42397 Jul 4 '14 at 9:04
Can you confirm that the kinematic measure also equivariant to affine transformations? –  user1963 Jul 8 '14 at 9:38
@user1963: It's not. If you squash the plane down in the $y$ direction, lines parallel to the $x$-axis are now much more likely than lines parallel to the $y$-axis. –  Rahul Jul 8 '14 at 9:41