proof of monotonicity criterion

Let us start from the implications “$\Rightarrow$”.

Suppose that $f^{\prime }(x)\ge 0$ for all $x\in (a,b)$. We want to prove that therefore $f$ is increasing. So take $x_1,x_2\in [a,b]$ with . Applying the mean-value Theorem on the interval $[x_1,x_2]$ we know that there exists a point $x\in (x_1,x_2)$ such that

$$f(x_2)-f(x_1)=f^{\prime }(x)(x_2-x_1)$$

and being $f^{\prime }(x)\ge 0$ we conclude that $f(x_2)\ge f(x_1)$.

This proves the first claim. The other three cases can be achieved with minor modifications: replace all “$\ge$” respectively with $\le$, $>$ and .

Let us now prove the implication “$\Leftarrow$” for the first and second statement.

Given $x\in (a,b)$ consider the ratio

$$\frac{f(x+h)-f(x)}{h}.$$

If $f$ is increasing the numerator of this ratio is $\ge 0$ when $h>0$ and is $\le 0$ when . Anyway the ratio is $\ge 0$ since the denominator has the same sign of the numerator. Since we know by hypothesis that the function $f$ is differentiable in $x$ we can pass to the limit to conclude that

$$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\ge 0.$$

If $f$ is decreasing the ratio considered turns out to be $\le 0$ hence the conclusion $f^{\prime }(x)\le 0$.

Notice that if we suppose that $f$ is strictly increasing we obtain the this ratio is $>0$, but passing to the limit as $h\to 0$ we cannot conclude that $f^{\prime }(x)>0$ but only (again) $f^{\prime }(x)\ge 0$.