# Inequality satisfied for $t=1/2$ and Measurability implies Log-Concavity

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, Stat. Surv. 8, 45-114 (2014). ZBL1360.62055.

• It is not very natural to consider pointwise values of a measurable function. As for me, a more natural question (the answer is also positive) is whether this inequality for almost all pairs $a,b$ does imply that the function $\log f$ is equivalent to a concave function. – Fedor Petrov Dec 26 '17 at 18:21

## 1 Answer

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., American M. S. Trans. 20, 40-44 (1919). ZBL47.0232.03 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/