# 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:

\(\displaystyle \dfrac {\map f {x_1} - \map f r} {x_1 - r}\) | \(<\) | \(\displaystyle \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}\) | \(\le\) | \(\displaystyle \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}\) | Limits Preserve Inequalities | |||||||||

\(\text {(3)}: \quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {f'} r\) | \(\le\) | \(\displaystyle \dfrac {\map f {x_2} - \map f r} {x_2 - r}\) | Definition of Derivative of Real Function at Point |

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

We have:

\(\displaystyle \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}\) | \(<\) | \(\displaystyle \dfrac {\map f s - \map f {x_3} } {s - x_3}\) | from $(1)$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}\) | \(\le\) | \(\displaystyle \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}\) | Limits Preserve Inequalities | |||||||||

\(\text {(4)}: \quad\) | \(\displaystyle \leadsto \ \ \) | \(\displaystyle \dfrac {\map f s - \map f {x_2} } {s - x_2}\) | \(\le\) | \(\displaystyle \map {f'} s\) | Definition of Derivative of Real Function at Point |

Thus we have:

\(\displaystyle \map {f'} r\) | \(\le\) | \(\displaystyle \dfrac {\map f {x_2} - \map f r} {x_2 - r}\) | from $(3)$ | ||||||||||

\(\displaystyle \) | \(<\) | \(\displaystyle \dfrac {\map f s - \map f {x_2} } {s - x_2}\) | from $(2)$ | ||||||||||

\(\displaystyle \) | \(\le\) | \(\displaystyle \map {f'} s\) | 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.

$\blacksquare$