# proof of monotonicity criterion

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

Suppose that $f^{\prime}(x)\geq 0$ 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 $x_{1}. 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 $f^{\prime}(x)\geq 0$ we conclude that $f(x_{2})\geq f(x_{1})$.

This proves the first claim. The other three cases can be achieved with minor modifications: replace all “$\geq$” respectively with $\leq$, $>$ 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 $\geq 0$ when $h>0$ and is $\leq 0$ when $h<0$. Anyway the ratio is $\geq 0$ 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

 $f^{\prime}(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\geq 0.$

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

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) $f^{\prime}(x)\geq 0$.

Title proof of monotonicity criterion ProofOfMonotonicityCriterion 2013-03-22 13:45:14 2013-03-22 13:45:14 paolini (1187) paolini (1187) 6 paolini (1187) Proof msc 26A06