Characterizing the Radon transforms of log-concave functions

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 $\sigma$.

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 $g_\sigma(r)$ 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 if 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)