Equivalence of Definitions of Convex Real Function

Theorem

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

The following definitions of the concept of Convex Real Function are equivalent:

Definition 1

$f$ is convex on $I$ if and only if:

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

Definition 2

$f$ is convex on $I$ if and only if:

$\forall x_1, x_2, x_3 \in I: x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

Definition 3

$f$ is convex on $I$ if and only if:

$\forall x_1, x_2, x_3 \in I: x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$

Proof

That is:

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

Without loss of generality, 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:

 $\text {(1)}: \quad$ $\ds \map f {x_2}$ $\le$ $\ds \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}$ substituting for $\alpha$ and $\beta$ $\ds \leadstoandfrom \ \$ $\ds \paren {x_3 - x_1} \map f {x_2}$ $\le$ $\ds \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}$ $\ds \leadstoandfrom \ \$ $\ds \paren {x_3 - x_1} \map f {x_2}$ $\le$ $\ds \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}$ $\ds \leadstoandfrom \ \$ $\ds \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}$ $\le$ $\ds \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}$ $\ds \leadstoandfrom \ \$ $\ds \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $\le$ $\ds \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$

Again from $(1)$:

 $\text {(1)}: \quad$ $\ds \map f {x_2}$ $\le$ $\ds \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}$ $\ds \leadstoandfrom \ \$ $\ds \paren {x_3 - x_1} \map f {x_2}$ $\le$ $\ds \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}$ $\ds \leadstoandfrom \ \$ $\ds \paren {x_3 - x_1} \map f {x_2}$ $\le$ $\ds \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}$ $\ds \leadstoandfrom \ \$ $\ds \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}$ $\le$ $\ds \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}$ $\ds \leadstoandfrom \ \$ $\ds \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$ $\le$ $\ds \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$

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.

$\blacksquare$