Real Function is Concave iff Derivative is Decreasing

Theorem

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

Then $f$ is concave on $\openint a b$ if and only if its derivative $f'$ is decreasing on $\openint a b$.

Thus the intuitive result that a concave function "gets less steep".

Proof

Necessary Condition

Let $f$ be concave 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 concave:

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

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

Thus:

$\map {f'} {x_1} \ge \map {f'} {x_4}$

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

$\Box$

Sufficient Condition

Let $f'$ be decreasing on $\openint a b$.

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

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 decreasing:

$\map {f'} \xi \ge \map {f'} \eta$

Thus:

$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$

Hence $f$ is concave by definition.

$\blacksquare$

Also see

Real Function is Strictly Concave iff Derivative is Strictly Decreasing

Sources

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2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.): $\S 3.3, \S 3.4$