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Suppose $f$ is a function defined on some interval $\mathrm{J}$ such that $$ f \left( \dfrac{t_1 + t_2}{2} \right) \leq \dfrac{f(t_1) + f(t_2)}{2}, $$ for all pairs $t_1, t_2$ in $\mathrm{J};$ this is sometimes referred to as "midpoint convex." Furthermore, assume $f$ is a Baire function (this is equivalent to $f$ being Borel measurable). Then, $f$ is convex, meaning that $f(\alpha_1 t_1 + \alpha_2 t_2) \leq \alpha_1 f(t_1) + \alpha_2 f(t_2)$ for all $t_1, t_2$ as before and $\alpha_1 + \alpha_2 = 1$ both non-negative.

I read this result while reading the book of William Feller "An Introduction to Probability Theory and Its Applications" without a hint on how to proceed.

I know how to prove the result when $f$ is continuous (e.g. Midpoint-Convex and Continuous Implies Convex). After a few attempts, I don't see a way around to the measurable case. Any hints or references appreciated.

William M.
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  • Here it is proven that midpoint convex and bounded is continous, therefore convex https://math.stackexchange.com/questions/3637220/midpoint-convex-bounded-function-is-continuous so you just need to prove that your function is locally bounded. This reminds me of the proof that an additive measurable function is continuous, where you also start by proving that it is locally bounded, using the SteinHaus Theorem https://en.wikipedia.org/wiki/Steinhaus_theorem . Maybe your claim can be proven this way :) – Aitor Iribar Lopez Jan 12 '22 at 00:46
  • Also maybe you can try to have a look at this article "On the Continuity of Mid-Point Convex Functions" by J. D. Maitland Wright – Aitor Iribar Lopez Jan 12 '22 at 01:03

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First step: We prove that if $f$ is midpoint-convex and bounded near a point $x$ then $f$ is continuous in $x$.

To do so, we assume WLOG that $f$ is bounded by $M$ in $(a,b)$, $f(0)=0$ and $f$ is not continuous at $0$ (because $f$ is midpoint-convex/continuous iff $f-c$ is). Let $\epsilon > 0$ be such that there is a sequence $x_n$ converging to $0$ with $|f(x_n)|>\epsilon$. Note that, due to the inequality $f(x) + f(-x) \geq 0$ we may assume, changing $x_n$ by $-x_n$, that $f(x_n)>0$ so $f(x_n) \geq \epsilon$. By midpoint-convexity, and since $f(0)=0$, $f(2x)>2f(x)$, so $$ f(2^k x_n) \geq 2^k \epsilon $$ which leads to a contradiction if we pick $k$ big enough such that $2^k \epsilon >M$ and $n$ big enough such that $2^k x_n \in (a,b)$.

Second step: Now suppose $f$ is measurable on an interval $(a,b)$ but it is not locally bounded near $x$, with $f(x)=0$. After restricting the interval we can assume that $(a,b) = (x-2 \delta , x+ 2 \delta)$ for some $\delta >0$. For each $k>0$ pick $y_k$ in $(x_0-\delta, x_0+\delta)$ such that $f(y_k)>k$ (we can take positive values as before, using $f(y_k)+f(2x-y_k) \geq 0$).

Then for any point $z \in (y_k-\delta, y_k+\delta) \subset (x-2 \delta , x+ 2 \delta)$ we have $f(z)+f(2y_k-z) \geq 2k$. Note that $2y_k-z$ is the symmetric of $z$ with respect to $y_k$ so it is contained in $(y_k-\delta, y_k+\delta)$. Therefore, since $f$ is measurable, the set of points in $(y_k-\delta, y_k+\delta)$ with $f(z)>k$ has measure at least $\delta$. In particular, the set $\{z \in (a,b): f(z) >k\}$is measurable and has measure at least $\delta$ for all $k$. This is a contradiction because taking $k$ to be natural numbers, you would get that

$$0<\delta \leq \lim_{k \to \infty}m\left(\{ z \in (a,b) : f(x) >k \}\right) = m \left(\bigcap_{k \in \mathbb N} \{ z \in (a,b) : f(x) >k \}\right)=m(\emptyset)=0$$

Aitor Iribar Lopez
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