# Real Function is Strictly Convex iff Derivative is Strictly Increasing

## Theorem

Let $f$ be a real function which is differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Then $f$ is strictly convex on $\left({a \,.\,.\, b}\right)$ iff its derivative $f'$ is strictly increasing on $\left({a \,.\,.\, b}\right)$.

## Proof

### Necessary Condition

Let $f$ be strictly convex on $\left({a \,.\,.\, b}\right)$.

Let $x_1, x_2, x_3, x_4 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3 < x_4$.

By the definition of strictly convex:

$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} < \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2} < \dfrac {f \left({x_4}\right) - f \left({x_3}\right)} {x_4 - x_3}$

Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$.

Thus:

$f' \left({x_1}\right) < f' \left({x_4}\right)$

Hence $f'$ is strictly increasing on $\left({a \,.\,.\, b}\right)$.

$\Box$

### Sufficient Condition

Let $f'$ be strictly increasing on $\left({a \,.\,.\, b}\right)$.

Let $x_1, x_2, x_3 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3$.

By the Mean Value Theorem:

$\exists \xi: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} = f' \left({\xi}\right)$
$\exists \eta: \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2} = f' \left({\eta}\right)$

where $x_1 < \xi < x_2 < \eta < x_3$.

Since $f'$ is strictly increasing:

$f' \left({\xi}\right) < f' \left({\eta}\right)$

Thus:

$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} < \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

Hence $f$ is strictly convex by definition.

$\blacksquare$