Egalitarian measures

Question

A question I got asked I while ago:

If $T$ is a triangle in $\mathbb R^2$, is there a function $f:T\to\mathbb R$ such that the integral of $f$ over each straight segment connecting two points in the boundary of $T$ not on the same side is always $1$?

(Of course, you can change $T$ for your favorite convex set... and the problem should really be seen as asking for what sets is the answer affirmative, mostly)

f had better not be continuous, right? Becuase there will be problems at each corner. — Kevin Buzzard, Nov 27 '09 at 17:03
Isn't your question answered by the basic properties of the Radon transform? ie your function transforms to a characteristic function of the lines that intersect your set non-trivially. Apply inverse radon transform. http://en.wikipedia.org/wiki/Radon_transform — Ryan Budney, Nov 27 '09 at 18:18
@Ryan: Sure, the question really is about the Radon transform (and thank you for pointing it out), but it doesn't seem to me that this observation makes it easy. — Harald Hanche-Olsen, Nov 27 '09 at 18:58
Indeed, one way to understand the question is: for what convex sets does that anti-transform exist? This makes it clear that the answer depends on the regularity of the boundary, for example (but that is quite apparent just by drawing a few pictures!) but there are global obstacles too, like the one observed by Harald, modulo well-behavedness of the function, as Greg observes. — Mariano Suárez-Álvarez, Nov 27 '09 at 19:05
BTW, the original question about triangle is answered. So you might want to mark it or reformulate the question for all convex figures... — Anton Petrunin, Dec 11 '09 at 17:26

Harald Hanche-Olsen · Accepted Answer · 2015-06-29T13:47:33.690

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In general, no. For the double integral $\iint_T f(x,y)\,dx\,dy$ will be the height on any side, as is seen by turning the triangle with one side parallel to an axis and performing the integral. So at least, $T$ has to be equilateral. I don't know the answer in that case.

Edit: Wait, wait – the same trick works even if I turn the triangle at any angle, hence all heights (defined as $\sup_{RT} y-\inf_{RT} y$ where $R$ is a rotation) must be the same. That is never true for a triangle, and limits the number of convex sets seriously – but there are still non-circles that might satisfy the criterion. Once more, I don't know the answer, but for circles at least, it should in principle be straightforward to check if a radially symmetric function will do. And of course, if there is a solution, there is a radially symmetric one, as can be seen by rotating the solution and taking the average of all its rotated variants.

Edit²: For the unit disk (and radially symmetric $f$) the answer should in principle be obtainable by the Abel transform (which is really nothing but the Radon transform on radially symmetric functions). The required Abel transform $\Phi$ should be the characteristic function of the interval $[0,1]$ (we only use positive $x$ due to symmetry), and the inverse Abel transform provides the answer: $$f(r)=\frac{-1}{\pi}\int_r^\infty \frac{\Phi'(x)\,dx}{\sqrt{x^2-r^2}}=\frac{1}{\pi\sqrt{1-r^2}}$$ when $0\le r<1$. Being lazy, I checked the answer using Maple, and it seems right.

Addendum: Anton Petrunin pointed out in a comment that the above measure is the push-forward of the surface measure on the unit sphere on the unit disk under projection. It is well-known that the surface area of the portion of a disk between two parallel planes depends only on the distance between the planes (and is proportional to said distance), which ties in nicely with the desired property of $f$ on the disk.

edited Jun 29 '15 at 13:47

answered Nov 27 '09 at 17:21

Harald Hanche-Olsen

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While the original question was for triangles, my best guess is that only interiors of constant width curves were the candidates with most chances; one can do this for a circle, for example. – Mariano Suárez-Álvarez Nov 27 '09 at 17:28
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Stupid comment: if you turn the triangle so that no sides are parallel to an axis and compute the double integral then, do you get an immediate contradiction? – Kevin Buzzard Nov 27 '09 at 17:30
@buzzard: You're right. I edited my answer without having seen your comment. – Harald Hanche-Olsen Nov 27 '09 at 17:34
And reverse the order of integration, now you get that integral will be the length of a side. Or, indeed, use any orientation, then that integral is the "width" in the direction of the y-axis. So: triangles don't work but we get a new question: can this be done in a convex curve with width 1 in every direction? In particular, can it be done for a circle? – Gerald Edgar Nov 27 '09 at 17:35
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This argument works provided that the integrand satisfies Fubini's theorem. I don't know if possible wilder, conditionally convergent solutions are part of the original question. – Greg Kuperberg Nov 27 '09 at 18:55
@Harald: indeed, that is the solution I had for the disc. – Mariano Suárez-Álvarez Nov 27 '09 at 19:35
@Greg: Ummm … yeah, that's gonna make the problem real hard. – Harald Hanche-Olsen Nov 27 '09 at 19:39
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BTW, in case of disc, your measure is push-forward measure for projection from unit shpere to unit disc. – Anton Petrunin Nov 27 '09 at 21:04
@Anton: Brilliant. And of course we all know, from teaching calculus if nothing else, that the surface area of the portion of a sphere between two parallel planes depends only on the distance between the planes. – Harald Hanche-Olsen Nov 27 '09 at 21:09
How is it the push forward? The measure of a little piece cut out by a small chord of the disc is pretty large, yet its preimage under the projection is small... – Mariano Suárez-Álvarez Nov 27 '09 at 21:24
Take two parallel lines with a distance δ, both intersecting the disk. The integral over the disk between the two lines is δ (which is not actually “pretty large”). Similarly, take two parallel planes with a distance δ, both intersecting the unit sphere. The area of the sphere between the two planes is 2πδ. Apart from the multiplicative constant 2π, that is the same thing. – Harald Hanche-Olsen Nov 27 '09 at 22:23
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Do I understand it right: For a figure with constant width one can apply inverse of Radon transform and ONLY need to check that support is inside of the original figure? – Anton Petrunin Nov 28 '09 at 19:47
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@Anton: No, I don't think so, though I am not an expert on this. Apparently, deciding what functions are in the range of the Radon transform is not altogether trivial. Helgason has some results on this (http://www.ams.org/mathscinet-getitem?mr=692054), but I don't have this reference. He also has a theorem showing you don't need to check: If f decays fast and Rf vanishes on all lines not intersecting a compact convex set K, then supp f is contained in K (http://www.ams.org/mathscinet-getitem?mr=172311). – Harald Hanche-Olsen Dec 05 '09 at 20:36

Egalitarian measures

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