Say $f:\mathbb{R}\to(0,\infty)$ is measurable, and that
$$\forall a,b\in\mathbb{R}~~a<b\implies\log f(a)+\log f(b)\leq2\log f(\frac{a+b}{2}).$$
Why must $f$ be log-concave? (That is, why must
$$\forall a,b\in\mathbb{R}~\forall t\in(0,1)~~a<b\implies t\log f(a)+(1-t)\log f(b)\leq\log f(ta+(1-t)b)$$
hold?)


I came across this statement as I was reading the proof of Proposition 2.3.a in Saumard and Wellner. I was thinking about doing something with dyadic approximations of real numbers, but I haven't been able to make this idea rigorous yet.


*Saumard, Adrien; Wellner, Jon A.*, [**Log-concavity and strong log-concavity: a review**](http://dx.doi.org/10.1214/14-SS107), Stat. Surv. 8, 45-114 (2014). [ZBL1360.62055](https://zbmath.org/?q=an:1360.62055).



Inequality satisfied for $t=1/2$ and Measurability implies Log-Concavity
First of all, let's look at $g = -\log f$; then this is really just a question about convex functions. We want to know: if $g$ is measurable and "midpoint convex", i.e.
$$g\left(\frac{a+b}{2}\right) \le \frac{g(a)+g(b)}{2} \tag{MC}$$
 does it follow that $g$ is convex, i.e.
$$g((1-t)a+tb) \le (1-t)g(a) + t g(b), \quad 0 < t < 1 \tag{C}$$


It should be easy to show that (MC) implies that (C) holds for all dyadic rationals $t$.


It is proved in 
*Blumberg, H.*, [**On convex functions.**](http://dx.doi.org/10.2307/1988982), American M. S. Trans. 20, 40-44 (1919). [ZBL47.0232.03](https://zbmath.org/?q=an:47.0232.03) that a measurable, midpoint convex function (Blumberg just calls it "convex") is necessarily continuous. Then, since (C) holds for dyadic rationals, by continuity it holds for all real $t$.


Some other nice remarks can be found at <https://shreevatsa.wordpress.com/2010/06/29/convex-continuous-jensens/>


