example of Lipschitz condition

Statement 1.

Let f(x)=x2. Then f satisfies the Lipschitz conditionMathworldPlanetmath on [a,b] for finite real numbers a<b.


We want to show that for some real constant L, and for all x,y[a,b],


Let x,y[a,b]. Clearly if x=y, the above inequalityMathworldPlanetmath holds, so assume xy. Since x and y are interchangable in the above equation, it can be assumed without loss of generality that x<y.

Since f is differentiableMathworldPlanetmathPlanetmath on (a,b), by the mean-value theorem, there is a z(x,y) such that


that is,


Taking the modulus of both sides gives


Finally, to find L it is necessary to consider all possible values of z:

|x2-y2||x-y| =2|z|

Thus, for all x,y[a,b],


as required. ∎

Statement 2.

Additionally, L=2max{|a|,|b|} is the Lipschitz constant of f.


Assume |b||a|, since if |b|<|a|, it is possible to consider -f instead of f. This also implies that b>0. Let ε>0 be sufficiently small that a<b-ε and that higher powers of ε can be ignored. Now,

|f(b)-f(b-ε)||b-(b-ε)| =b2-(b-ε)2b-(b-ε)

By the assumptionPlanetmathPlanetmath above, b=max{|a|,|b|}. Thus, since b,b-ε[a,b] and by the definition of the Lipschitz condition,


However, the result from the previous proof gives


Combining these inequalities provides


and the result follows by trichotomy. ∎

Title example of Lipschitz condition
Canonical name ExampleOfLipschitzCondition
Date of creation 2013-03-22 17:14:16
Last modified on 2013-03-22 17:14:16
Owner me_and (17092)
Last modified by me_and (17092)
Numerical id 10
Author me_and (17092)
Entry type Example
Classification msc 26A16