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Suppose $f$ is a continuous function on the compact set $[0,1]$. Is true that given $\epsilon>0$, there exist $\delta>0$ such that for any partition $P=\{x=x_0<x_1,...,<x_{i-1}<y=x_i\}$ of the interval $[x,y]$, with $|x-y|<\delta$, we have that $$\sum_{n=1}^i|f(x_n)-f(x_{n-1})|<\epsilon$$

Tomás
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  • Interesting, I don't knew this property, I was trying to prove it only by using the uniformly continuity of $f$. – Tomás Feb 22 '13 at 16:49
  • What you described is related to bounded variation. http://en.wikipedia.org/wiki/Bounded_variation#BV_functions_of_one_variable – tom Feb 22 '13 at 16:53
  • In fact, I was trying to prove that if $f$ is continuous, then $V_0^x f$ is continuous. I realized that if $f$ satisfies this property, then it is easy to show the continuity of $V_0^x f$, but now I saw that my argument is wrong... – Tomás Feb 22 '13 at 16:55

4 Answers4

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Consider the function

$$f(x)=\begin{cases} 0,&\text{if }x=0\\ x\sin\frac1x,&\text{if }0<x\le 1\;; \end{cases}$$

clearly $f$ is continuous, but you can make

$$\sum_{n=1}^i|f(x_n)-f(x_{n-1})|$$

as large as you like when $x=0$, even if $y$ is very small: if you choose the right partition, you’re essentially adding up terms of the harmonic series.

Brian M. Scott
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Not in general. Every continuous function that is not of bounded variation provides a counterexample.

If $f$ is not of bounded variation, then for all $\delta>0$, there exists an interval $[x,y]\subset [0,1]$ with $|x-y|<\delta$ and there exists a partition $P=\{x=x_0<x_1,...,<x_{i-1}<y=x_i\}$ such that $\sum_{n=1}^i|f(x_n)-f(x_{n-1})|>1$. This can be seen by breaking up $[0,1]$ into finitely many intervals of length less than $\delta$, because $f$ cannot have bounded variation on all of these without having bounded variation on the entire interval.

This property is implied by absolute continuity, but not conversely. E.g., every monotone continuous function satisfies this property, but need not be absolutely continuous (for example, the Cantor-Lebesgue function).

Jonas Meyer
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No, the function $f(x) = x \sin(1/x)$, $f(0)=0$ is not absolutely continuous

Emanuele Paolini
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    I think that argument that $xsin(1/x)$ is not absolutely continuous is not right. Becouse cantor function, which is not absolutely continuous, satisfy desired propertie(because it is continuous and monotone). – tom Feb 22 '13 at 16:49
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But function $x\sin\left(\frac{1}{x}\right)$ does not have bounded variation and it is continuous on $[0,1]$.

tom
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