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 Convex iff Derivative is Increasing
- Real Function is Strictly Convex iff Derivative is Strictly Increasing
Sources
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- 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.): $\S 3.3, \S 3.4$