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f:RdR≥0 is log-concave if log(f) is concave (and the domain of log(f) is convex).

Theorem: For all σ on the sphere Sd-1 and r∈R, gσ(r) := ∫σ.x=rf(x)dS(x) is a log-concave function of r. (Note: g, as a function of σ and r, is the Radon transform of f.)

Question: does this characterize log-concavity? That is, if gσ(r) is log-concave as a function of r for all σ, is f log-concave?

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2 Answers 2

up vote 4 down vote accepted

If I understand the question correctly, I think the answer is no.

Start with the following : if f is the indicator function of the unit ball, then the function g(r) is strictly log-concave close to 0 (this function does not depend on theta).

Now, let h be the indicator function of the ball of radius r<1. Then f-epsilon.h is never log-concave for any epsilon>0, and its Radon transform (which again is independent of theta) remains log-concave is epsilon is small enough.

(this is especially easy to see in dimension 2, in which case the Radon transforms of both f and h are second-degree polynomials on their support)

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Great! Such a simple example, too. –  Darsh Ranjan Oct 23 '09 at 1:28

The following answer exploits a wider context, where non-negative functions $f$, viewed as densities of absolutely continuous measures $f(x)dx$, are replaced by non-negative measures $\mu$. Because a concave function unbounded from above is $\equiv+\infty$, a log-concave measure $\mu$ is naturally an absolutely continuous measure with log-concave density.

Therefore the Lebesgue measure over the $2$-dimensional sphere $S^2$ is not log-concave over ${\mathbb R}^3$. Nevertheless, its Radon transform in a direction $\sigma$ is $$g_\sigma=2\pi\chi_{(-1,1)},$$ which is log-concave for every $\sigma$.

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I like this example a lot, since a uniform measure on a sphere is far from log-concave. (And apparently I had never computed its Radon transform.) –  Darsh Ranjan Sep 15 '10 at 0:18

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