Analyzing the convexity of a superadditive function

Question

Let $f:[0,1] \to [0,1]$ be a continuous and increasing function with $f(0)=0$, $f(1)=1$, such that for all $0 \leq x<y \leq 1$, $\frac{f(y)-f(x)}{y-x} \leq \frac{f(1)-f(1-(y-x))}{y-x}$. Given that $f$ is superadditive, I want to analyze if $f$ is convex.

Answer 1

I think $f$ may be non-convex. Consider the following example: Let $\alpha >1$. Define \begin{equation*} f(x) = \begin{cases} \alpha x, & 0\leq x\leq \frac{1}{2\alpha};\\ 1/2, & \frac{1}{2\alpha} < x \leq 1-\frac{1}{2\alpha};\\ \alpha(x-1+\frac{1}{2\alpha})+\frac{1}{2}, &1-\frac{1}{2\alpha}<x\leq 1. \end{cases} \end{equation*}