Real Function is Strictly Concave iff Derivative is Strictly Decreasing

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Theorem

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


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


Proof

Necessary Condition

Let $f$ be strictly concave on $\openint a b$.

Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$.

We are to show that:

$\map {f'} r > \map {f'} s$


Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that:

$r < x_1 < x_2 < x_3 < s$

By the definition of strictly concave:

$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} > \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} > \dfrac {\map f s - \map f {x_3} } {s - x_3}$

and:

$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} > \dfrac {\map f s - \map f {x_2} } {s - x_2}$


Let $x_1 \to r^+$.

Then:

\(\ds \dfrac {\map f {x_1} - \map f r} {x_1 - r}\) \(>\) \(\ds \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}\) \(\ge\) \(\ds \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}\) Limits Preserve Inequalities
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \map {f'} r\) \(\ge\) \(\ds \dfrac {\map f {x_2} - \map f r} {x_2 - r}\) Definition of Derivative of Real Function at Point


Similarly, let $x_3 \to s^-$.

We have:

\(\ds \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}\) \(>\) \(\ds \dfrac {\map f s - \map f {x_3} } {s - x_3}\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}\) \(\ge\) \(\ds \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}\) Limits Preserve Inequalities
\(\text {(4)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {\map f s - \map f {x_2} } {s - x_2}\) \(\ge\) \(\ds \map {f'} s\) Definition of Derivative of Real Function at Point


Thus we have:

\(\ds \map {f'} r\) \(\ge\) \(\ds \dfrac {\map f {x_2} - \map f r} {x_2 - r}\) from $(3)$
\(\ds \) \(>\) \(\ds \dfrac {\map f s - \map f {x_2} } {s - x_2}\) from $(2)$
\(\ds \) \(\ge\) \(\ds \map {f'} s\) from $(4)$


Thus:

$\map {f'} r > \map {f'} s$

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


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$\Box$


Sufficient Condition

Let $f'$ be strictly 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 strictly decreasing:

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

Thus:

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

Hence $f$ is strictly concave by definition.

$\blacksquare$


Also see


Sources

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