Let $f:\mathbb R\to\mathbb R$ be a function such that it exists an $\varepsilon>0$ such that for all $x,y\in \mathbb R$ with $|x-y|>\varepsilon$, one has that $f(\frac{x+y}{2})\leq\frac{f(x)+f(y)}{2}$. Recall that a function $g:\mathbb R\to\mathbb R$ is said mid-convex if $g(\frac{x+y}{2})\leq\frac{g(x)+g(y)}{2}$ for all $x,y\in \mathbb R$.
My question is: Is $f$ mid-convex?
Thank you for your help.
Consider the map $f(x)=\vert \vert x \vert -1\vert$. You’ll verify that $f$ satisfies your criteria for $\epsilon =100$. However, $f$ is not mid convex.
A too small bump...