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)

  • $\begingroup$ f had better not be continuous, right? Becuase there will be problems at each corner. $\endgroup$ – Kevin Buzzard Nov 27 '09 at 17:03
  • $\begingroup$ $f$ should be nice enough in the interior of $T$ (I guess) $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '09 at 17:04
  • $\begingroup$ I guess we should just imagine f defined on the interior. $\endgroup$ – Kevin Buzzard Nov 27 '09 at 17:29
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    $\begingroup$ 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. en.wikipedia.org/wiki/Radon_transform $\endgroup$ – Ryan Budney Nov 27 '09 at 18:18
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    $\begingroup$ BTW, the original question about triangle is answered. So you might want to mark it or reformulate the question for all convex figures... $\endgroup$ – Anton Petrunin Dec 11 '09 at 17:26

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.

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    $\begingroup$ 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? $\endgroup$ – Kevin Buzzard Nov 27 '09 at 17:30
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    $\begingroup$ 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. $\endgroup$ – Greg Kuperberg Nov 27 '09 at 18:55
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    $\begingroup$ BTW, in case of disc, your measure is push-forward measure for projection from unit shpere to unit disc. $\endgroup$ – Anton Petrunin Nov 27 '09 at 21:04
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    $\begingroup$ 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? $\endgroup$ – Anton Petrunin Nov 28 '09 at 19:47
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    $\begingroup$ @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 (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 (ams.org/mathscinet-getitem?mr=172311). $\endgroup$ – Harald Hanche-Olsen Dec 5 '09 at 20:36

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