Basic Question about absolutely continuous functions

  • #1

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My question is maybe elementary but I don't know the answer. I have a function f absolutely continuous in (a,c) and in (c,b), f continuous in c. Is f absolutely continuous in (a,b)?
I think the answer is negative but I can't find a counterexample. I really apreciatte your help.
 

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  • #2
mathman
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Possible counterexample. two semicircles (diameters on top), touch at c. The derivative is plus infinite to the left and minus infinite to the right (of c).
 
  • #3
mathwonk
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i think this looks true. given epsilon you have to find a delta so that on any finite sequence of non overlapping intervals of total length less than delta, the total variation of the function is less than epsilon. try considering two cases, how small does delta have to be if none of the intervals contain c and how small does it have to be if some interval does contain c, then try to satisfy both conditions at once.
 
  • #4
Svein
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An absolutely continuous function can be expressed as the difference between two non-decreasing continuous functions (and vice versa). Apply that.
 
  • #5
S.G. Janssens
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An absolutely continuous function can be expressed as the difference between two non-decreasing continuous functions (and vice versa). Apply that.
This is not true.
 
  • #6
Svein
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This is not true.
G. Ye. Shilov: Mathematical Analysis (A Special Course). Pergamon Press 1965, page 306.
 
  • #7
S.G. Janssens
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G. Ye. Shilov: Mathematical Analysis (A Special Course). Pergamon Press 1965, page 306.
The "vice versa" is not true, because there exist continuous, non-decreasing functions that are not absolutely continuous.

Let ##f : [0,1] \to \mathbb{R}## be the Cantor function. Then ##f## is continuous, non-decreasing, but not absolutely continuous. The zero function is continuous and non-decreasing. The difference ##f - 0## is not absolutely continuous.
 
  • #8
FactChecker
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That dang Cantor. If it wasn't for him, math would be much simpler. ;>)
 
  • #9
mathwonk
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i believe Svein is thinking of (continuous) functions of bounded variation, a larger class than absolutely continuous ones.
 
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  • #10
Delta2
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The "vice versa" is not true, because there exist continuous, non-decreasing functions that are not absolutely continuous.

Let ##f : [0,1] \to \mathbb{R}## be the Cantor function. Then ##f## is continuous, non-decreasing, but not absolutely continuous. The zero function is continuous and non-decreasing. The difference ##f - 0## is not absolutely continuous.
I don't have the book Svein suggests to check it myself but perhaps the theorem/lemma is about non-decreasing AND non-identically zero functions.
 
  • #11
S.G. Janssens
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I don't have the book Svein suggests to check it myself but perhaps the theorem/lemma is about non-decreasing AND non-identically zero functions.
I don't think so.

Just replace the ##f## from before by the sum ##g + h##, where ##g## is the Cantor function and ##h## is any absolutely continuous, non-decreasing function that is not identically zero. Then ##g + h## is continuous, non-decreasing but not absolutely continuous. The difference ##(g + h) - h## is not absolutely continuous.

The forward implication in post #4 is true, because absolutely continuous functions are of bounded variation. It is the "vice versa" there that is not true.
 

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