$\newcommand{\R}{\mathbb R}$
Let $f$ be a function from $\R$ to $\R$. It is said that $f$ is midpoint-convex if for any real $x$ and $y$ we have $f(x-y)+f(x+y)\ge2f(x)$ or, equivalently, $f(x+y)-f(x)\ge f(x)-f(x-y)$. So, $f$ is midpoint-convex iff for any real $x$ and and any real $y>0$  

$$\frac{f(x+y)-f(x)}y\ge\frac{f(x)-f(x-y)}y.$$
One may call the ratio $\frac{f(x+y)-f(x)}y$ the average rate of change of the function $f$ over the interval $[x,x+y]$. Thus, $f$ is midpoint-convex iff for any real $x$ and and any real $y>0$ the average rate of change of $f$ over $[x,x+y]$ is no less than the average rate of change of $f$ over $[x-y,x]$. 


By [Sierpiński's theorem](http://matwbn.icm.edu.pl/ksiazki/fm/fm1/fm1116.pdf), every Lebesgue measurable midpoint-convex function from $\R$ to $\R$ is convex. However (assuming the axiom of choice), one can easily construct a (necessarily non-Lebesgue measurable) midpoint-convex function from $\R$ to $\R$ which is not convex -- see e.g. [this](https://www.mathcounterexamples.net/a-discontinuous-midpoint-convex-function/), where $f(i\_1)$ should be replaced by $f(b\_{i\_1})$. 


Let us now say that $f$ has non-decreasing average rate of change if 
$$\frac{f(x\_2+y)-f(x\_2)}y\ge\frac{f(x\_1+y)-f(x\_1)}y$$
for any real $x\_1,x\_2,y$ such that $x\_1\le x\_2$ and $y>0$. This is equivalent to the following: 
$$\frac{f(x-y\_2)+f(x+y\_2)}2\ge\frac{f(x-y\_1)+f(x+y\_1)}2$$
for any real $x,y\_1,y\_2$ such that $0\le y\_1\le y\_2$. 


One may note here that every function with non-decreasing average rate of change is obviously midpoint convex.


The **question** is then the one in the title: Is there a (necessarily non-Lebesgue measurable) non-convex function with non-decreasing average rate of change?



Is there a non-convex function with non-decreasing average rate of change?
The standard example of a non-convex but midpoint-convex function is additive:
given a basis $B$ of $\mathbb R$ over the rationals $\mathbb Q$, choose one member $\alpha$ of $B$ and
if $x = \sum\_{\beta \in B} c\_\beta(x) \beta$ with $c\_\beta(x) \in \mathbb Q$ (only finitely many nonzero), take $f(x) = c\_\alpha(x)$. This has non-decreasing rate of change, indeed 
$$\frac{f(x\_1 + y) - f(x\_1)}{y} = \frac{f(x\_2 + y) - f(x\_2)}{y} = \frac{f(y)}{y}$$ 


