Equivalence of Definitions of Convex Real Function

Theorem
Let $f$ be a real function which is defined on a real interval $I$.

Proof
Let $f$ be convex real function on $I$ according to definition 1.

That is:
 * $\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \le \alpha \map f x + \beta \map f y$

, assume $x \le y$.

Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.

As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$.

Thus:
 * $\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$

So:

Again from $(1)$:

Thus:
 * $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$

So:
 * $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$

demonstrating that $f$ is convex on $I$ according to definition 3, and:


 * $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$

demonstrating that $f$ is convex on $I$ according to definition 2.

As each step is an equivalence, the argument reverses throughout.