Theorem 1   Suppose $f \colon (a,b) \to \mathbb{R}$ is continuous and that, for all $x,y \in (a,b)$ , $$ f \left( {x + y \over 2} \right) \le {f(x) + f(y) \over 2} . $$ Then $f$ is convex.

Proof. We begin by showing that, for any natural numbers $n$ and $m \le 2^n$ , $$ f \left( {m x + (2^n - m) y \over 2^n} \right) \le {m f(x) + (2^n - m) f(y) \over 2^n} $$ by induction. When $n=1$ , there are three possibilities: $m=1$ , $m=0$ , and $m=2$ . The first possibility is a hypothesis of the theorem being proven and the other two possibilities are trivial.

Assume that $$ f \left( {m' x + (2^n - m') y \over 2^n} \right) \le {m' f(x) + (2^n - m') f(y) \over 2^n} $$ for some $n$ and all $m' \le 2^n$ . Let $m$ be a number less than or equal to $2^{n+1}$ . Then either $m \le 2^n$ or $2^{n+1} - m \le 2^n$ . In the former case we have

In the other case, we can reverse the roles of $x$ and $y$ .

Now, every real number $s$ has a binary expansion; in other words, there exists a sequence $\{m_n\}$ of integers such that $\lim_{n \to \infty} m_n / 2^n = s$ . If $0 \le s \le 1$ , then we also have $0 \le m_n \le 2^n$ so, by what we proved above, $$ f \left( {m_n x + (2^n - m_n) y \over 2^n} \right) \le {m_n f(x) + (2^n - m_n) f(y) \over 2^n} . $$ Since $f$ is assumed to be continuous, we may take the limit of both sides and conclude $$ f( s x + (1 - s) y ) \le s f (x) + (1 - s) f (y) , $$ which implies that $f$ is convex.