absolutely continuous on [0,1] versus absolutely continuous on [ε,1] for every ε>0


Lemma.

Define f:RR by

f(x)={0 if x=0xsin(1x) if x0.

Then f is absolutely continuousMathworldPlanetmath (http://planetmath.org/AbsolutelyContinuousFunction2) on [ε,1] for every ε>0 but is not absolutely continuous on [0,1].

Proof.

Note that f is continuousMathworldPlanetmath on [0,1] and differentiableMathworldPlanetmathPlanetmath on (0,1] with f(x)=sin(1x)-1xcos(1x).

Let ε>0. Then for all x[ε,1]:

|f(x)|=|sin(1x)-1xcos(1x)||sin(1x)|+|1x||cos(1x)|1+1ε1=1+1ε

Since f is continuous on [ε,1] and differentiable on (ε,1), the mean value theorem (http://planetmath.org/MeanValueTheorem) can be applied to f. Thus, for every x1,x2(ε,1) with x1x2, |f(x2)-f(x1)x2-x1|1+1ε. This yields |f(x2)-f(x1)|(1+1ε)|x2-x1|, which also holds when x1=x2. Thus, f is LipschitzPlanetmathPlanetmath on (ε,1). It follows that f is absolutely continuous on [ε,1].

On the other hand, it can be verified that f is not of bounded variationMathworldPlanetmath on [0,1] and thus cannot be absolutely continuous on [0,1]. ∎

Title absolutely continuous on [0,1] versus absolutely continuous on [ε,1] for every ε>0
Canonical name AbsolutelyContinuousOn01VersusAbsolutelyContinuousOnvarepsilon1ForEveryvarepsilon0
Date of creation 2013-03-22 16:12:19
Last modified on 2013-03-22 16:12:19
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 11
Author Wkbj79 (1863)
Entry type Example
Classification msc 26A46
Classification msc 26B30