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 $x_1, \dots, x_n \in I$:

via Backwards Induction.

Basis for the Induction
The statement holds:
 * for $n = 0$ vacuously
 * for $n = 1$ as $\map f {\dfrac x 1} = \dfrac {\map f x} 1$ for each $x \in I$.

Let $x_1, x_2$ be two points in $I$.

Then as $f$ is midpoint-convex:

This is the basis for the induction.

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

This is the induction hypothesis.

Induction Step
This is the induction step:

Let $n = 2^{k + 1}$.

Let $x_1, \dots, x_n \in I$.

Then:

which completes the Proof by Mathematical Induction for integers of the form $2^k$.

Backwards Step
Let $2^k \le n \le 2^{k + 1}$ for some integer $k$.

Let $x_1, \dots, x_n \in I$.

Defined $\overline x := \dfrac {x_1 + \dots + x_n} n$.

Then:

from which we obtain:

completing the proof.