Real Function is Convex iff Derivative is Increasing

Theorem
Let $f$ be a real function which is differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Then:
 * $f$ is convex on $\left({a \,.\,.\, b}\right)$ iff its derivative $f'$ is increasing on $\left({a \,.\,.\, b}\right)$
 * $f$ is concave on $\left({a \,.\,.\, b}\right)$ iff its derivative $f'$ is decreasing on $\left({a \,.\,.\, b}\right)$.

The strict version of this theorem also holds. That is:
 * $f$ is strictly convex on $\left({a \,.\,.\, b}\right)$ iff its derivative $f'$ is strictly increasing on $\left({a \,.\,.\, b}\right)$
 * $f$ is strictly concave on $\left({a \,.\,.\, b}\right)$ iff its derivative $f'$ is strictly decreasing on $\left({a \,.\,.\, b}\right)$.

Corollary

 * $f$ is convex on $\left({a \,.\,.\, b}\right)$ iff its second derivative $D^2 f$ is $\ge 0$ on $\left({a \,.\,.\, b}\right)$
 * $f$ is concave on $\left({a \,.\,.\, b}\right)$ iff its second derivative $D^2 f$ is $\le 0$ on $\left({a \,.\,.\, b}\right)$.

Proof
We will prove the convex case. The concave case follows by a similar argument.

First, suppose $f'$ is increasing on $\left({a \,.\,.\, b}\right)$.

Let $x_1, x_2, x_3 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3$.

By the Mean Value Theorem, we have:


 * $\displaystyle \exists \xi: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} = f^{\prime} \left({\xi}\right)$
 * $\displaystyle \exists \eta: \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2} = f^{\prime} \left({\eta}\right)$.

where $x_1 < \xi < x_2 < \eta < x_3$.

Since $f'$ is increasing, $f^{\prime} \left({\xi}\right) \le f^{\prime} \left({\eta}\right)$.

Thus we have that $\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$.

Hence $f$ is convex by definition.

The proof of the strict version is identical.

Now suppose that $f$ is convex on $\left({a \,.\,.\, b}\right)$.

Let $x_1, x_2, x_3, x_4 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3 < x_4$.

By the definition of convex function, we have:


 * $\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2} \le \frac {f \left({x_4}\right) - f \left({x_3}\right)} {x_4 - x_3}$

Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$.

In the case that $f$ is strictly convex on $\left({a \,.\,.\, b}\right)$, one would use the fact that the left-hand side as a function of $x_2$ is strictly increasing. This follows directly from the definition of a strictly convex function.

Thus we find that $f' \left({x_1}\right) \le f' \left({x_4}\right)$.

Hence it follows that $f'$ is increasing on $\left({a \,.\,.\, b}\right)$.

Proof of Corollary
This result follows directly from Derivative of Monotone Function.

Note
Thus is proved the intuitive result that a convex function "gets steeper".