Theorem: Equivalence of Definitions of Convex Real Function

Let $f$ be a real function which is defined on a real interval $I$.
{{TFAE|def = Convex Real Function}}

Proof:
Let $f$ be convex real function on $I$ according to definition 1.
That is:
:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \le \alpha \map f x + \beta \map f y$
{{WLOG}}, 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:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map f {x_2}
      | o = \le
      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
      | c = substituting for $\alpha$ and $\beta$
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x_3 - x_1} \map f {x_2}
      | o = \le
      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x_3 - x_1} \map f {x_2}
      | o = \le
      | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}
      | o = \le
      | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
      | o = \le
      | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
      | c = 
}}
{{end-eqn}}
Again from $(1)$:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map f {x_2}
      | o = \le
      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x_3 - x_1} \map f {x_2}
      | o = \le
      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x_3 - x_1} \map f {x_2}
      | o = \le
      | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}
      | o = \le
      | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
      | o = \le
      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
      | c = 
}}
{{end-eqn}}
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.
{{qed}}