Midpoint convexity

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous and midpoint convex function, which means that

$$\forall x,y\in \mathbb{R}\;,\;\;\displaystyle{ f\left(\frac{x+y}{2}\right)\leq \frac{f(x)+f(y)}{2} .}$$

Assume there exists $a,b\in \mathbb{R}$ such that $f(a)=f(b)=0$.

Show that $f\leq 0$ on $[a,b]$. And in a second time, deduce that $f$ is convex.

Any hint would appreciated.

Answer 1

Hint: To make it simple, look at the case $a = 0, b = 1$; any other case reduces to this via $$ g(x) = f(\frac{x-a}{b-a}). $$ Can you show that if $f(x), f(y) \le 0$, then $f(p) \le 0$ where $p$ is the average of $x$ and $y$?

If so, then you know that $f$ is negative at every point of the form $$ \frac{n}{2^k} $$ where $n$ is an integer and $0 \le \frac{n}{2^k} \le 1$.

Now: in what way have we used the continuity of $f$ so far?

Answer 2

John Hughes gave a very nice path to the first part. For the second part, use the transformation $$ g(x)=f(x)-mx-k $$ where $y=mx+k$ is the line through $(a,f(a))$ and $(b,f(b))$. Then $g(a)=g(b)=0$.