Midpoint-Convex Function is Rational Convex

Theorem
Let $I$ be a non-empty real interval.

Let $f: I \to \R$ be a real function.

If $f$ is midpoint-convex, then $f$ is rational-convex.

Proof
It suffices to show that for each $n \in \N$ and for any choice of $n$ elements $x_1,\dots,x_n \in I$, we have that:

via forward-backward induction.

The statement holds for $n=0$ vacuously and $n=1$ as $f(x/1)=f(x)/1$ for each $x\in I$.

If $x_1,x_2$ are two points in $I$, then as $f$ is midpoint convex,

Suppose that if $n=2^k$, then for any choice $x_1,\dots,x_n$ of $n$ elements in $I$, we have

Let $n=2^{k+1}$. Let $x_1,\dots,x_n\in I$.

which completes the induction for integers of the form $2^k$.

If $2^k\leq n\leq 2^{k+1}$ for some integer $k$, then let $x_1,\dots,x_n\in I$ and set $\overline{x}=\frac{x_1+\dots+x_n}{n}$.

It follows that

from which we obtain

completing the proof.