Real Function is Convex iff Derivative is Increasing

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Theorem

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


Then:

$f$ is convex on $\openint a b$

if and only if:

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


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


Proof

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 $\openint a b$.

$\Box$


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.

$\blacksquare$


Also see


Sources


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