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 $$Df$$ is increasing on $$\left({a \, . \, . \, b}\right)$$;
 * $$f$$ is concave on $$\left({a \, . \, . \, b}\right)$$ iff its derivative $$Df$$ is decreasing on $$\left({a \, . \, . \, b}\right)$$.

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

First, suppose $$Df$$ 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:


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

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

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

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

Hence $$f$$ is convex by definition.

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:

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

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

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

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

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