# 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 $\openint a b$ if and only if its derivative $f'$ is strictly increasing on $\openint a b$.

## Proof

### 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:

 $\ds \dfrac {\map f {x_1} - \map f r} {x_1 - r}$ $<$ $\ds \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ $\ds \leadsto \ \$ $\ds \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}$ $\le$ $\ds \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}$ Limits Preserve Inequalities $\text {(3)}: \quad$ $\ds \leadsto \ \$ $\ds \map {f'} r$ $\le$ $\ds \dfrac {\map f {x_2} - \map f r} {x_2 - r}$ 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 \ \$ $\ds \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$ $\le$ $\ds \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}$ Limits Preserve Inequalities $\text {(4)}: \quad$ $\ds \leadsto \ \$ $\ds \dfrac {\map f s - \map f {x_2} } {s - x_2}$ $\le$ $\ds \map {f'} s$ Definition of Derivative of Real Function at Point

Thus we have:

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

Thus:

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

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

$\Box$

### 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.

$\blacksquare$