Real Function is Convex iff Derivative is Increasing

Theorem
Let $f$ be a real function which is differentiable on the open interval $\openint a b$.

Then:
 * $f$ is convex on $\openint a b$


 * its derivative $f'$ is increasing on $\openint a b$.
 * its derivative $f'$ is increasing on $\openint a b$.

Thus the intuitive result that a convex function "gets steeper".

Necessary Condition
Let $f$ be convex on $\openint a b$.

Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that:
 * $x_1 < x_2 < x_3 < x_4$

By the definition of convex function:


 * $\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \le \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$

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

Thus:
 * $\map {f'} {x_1} \le \map {f'} {x_4}$

Hence $f'$ is increasing on $\openint a b$.

Sufficient Condition
Let $f'$ be increasing on $\openint a b$.

Let $x_1, x_2, x_3 \in \openint a b$ such that $x_1 < x_2 < x_3$.

By the Mean Value Theorem:


 * $\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
 * $\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$

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

Since $f'$ is increasing:
 * $\map {f'} \xi \le \map {f'} \eta$

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

Hence $f$ is convex by definition.

Also see

 * Real Function is Strictly Convex iff Derivative is Strictly Increasing


 * Real Function is Concave iff Derivative is Decreasing
 * Real Function is Strictly Concave iff Derivative is Strictly Decreasing