proof of monotonicity criterion


Let us start from the implicationsMathworldPlanetmath”.

Suppose that f(x)0 for all x(a,b). We want to prove that therefore f is increasing. So take x1,x2[a,b] with x1<x2. Applying the mean-value Theorem on the interval [x1,x2] we know that there exists a point x(x1,x2) such that

f(x2)-f(x1)=f(x)(x2-x1)

and being f(x)0 we conclude that f(x2)f(x1).

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

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

Given x(a,b) consider the ratio

f(x+h)-f(x)h.

If f is increasing the numerator of this ratio is 0 when h>0 and is 0 when h<0. Anyway the ratio is 0 since the denominator has the same sign of the numerator. Since we know by hypothesisMathworldPlanetmath that the function f is differentiableMathworldPlanetmathPlanetmath in x we can pass to the limit to conclude that

f(x)=limh0f(x+h)-f(x)h0.

If f is decreasing the ratio considered turns out to be 0 hence the conclusionMathworldPlanetmath f(x)0.

Notice that if we suppose that f is strictly increasing we obtain the this ratio is >0, but passing to the limit as h0 we cannot conclude that f(x)>0 but only (again) f(x)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