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

Title | proof of monotonicity criterion |
---|---|

Canonical name | ProofOfMonotonicityCriterion |

Date of creation | 2013-03-22 13:45:14 |

Last modified on | 2013-03-22 13:45:14 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 6 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 26A06 |