Real Function is Strictly Convex iff Derivative is Strictly Increasing

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

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

Necessary Condition
Let $f$ be strictly convex 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 convex:


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

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:

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

We have:

Thus we have:

Thus:
 * $\map {f'} r < \map {f'} s$

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

Sufficient Condition
Let $f'$ be strictly increasing 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 increasing:
 * $\map {f'} \xi < \map {f'} \eta$

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 convex by definition.

Also see

 * Real Function is Convex iff Derivative is Increasing


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