absolutely continuous on [0,1] versus absolutely continuous on [ε,1] for every ε>0
Lemma.
Define f:R→R by
f(x)={0 if x=0xsin(1x) if x≠0. |
Then f is absolutely continuous (http://planetmath.org/AbsolutelyContinuousFunction2) on [ε,1] for every ε>0 but is not absolutely continuous on [0,1].
Proof.
Note that f is continuous on [0,1] and differentiable
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+1ε⋅1=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 x1≠x2, |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 Lipschitz 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 variation 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 |
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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 |