12
$\begingroup$

Could one give an example of a strictly increasing, continuous but not absolutely continuous, function on $[0,1]$ into $[0,1]$ or on $[0,1)$ into $R$ or any of the related combinations of 1-d domain and range?

$\endgroup$
10
  • 2
    $\begingroup$ the Cantor function is not strictly increasing. $\endgroup$ Dec 17, 2013 at 18:10
  • $\begingroup$ The Cantor function is the most classical example. See math.stackexchange.com/questions/125345/… $\endgroup$ Dec 17, 2013 at 18:17
  • $\begingroup$ Wouldn't $x \mapsto 1/(1-x)$ on $[0,1)$ work for this? $\endgroup$ Dec 17, 2013 at 18:19
  • 1
    $\begingroup$ When the interval is not compact, any continuous function which is not uniformly continuous will do (since absolute continuity implies uniform continuity). $\endgroup$
    – user113529
    Dec 17, 2013 at 18:47
  • 1
    $\begingroup$ I say that's one possible definition, @user43687. Whether it's a good definition depends on what you want to do with it. Sometimes, the really important property is being an integral of an $L^1_{\text{loc}}$ function, then it's convenient to just say AC. Other times, you're interested in a function mapping sets of small measure to sets of small measure, then it's not a good definition. $\endgroup$ Dec 17, 2013 at 21:11

3 Answers 3

13
$\begingroup$

Just add the identity function, $\text{id}(x) = x$, to the Cantor function, $\text{c}$. The sum of continuous functions are continuous, and the sum of an increasing function with a strictly increasing one is strictly increasing.

As in the proof that $\text{c}$ is not absolutely continuous choose $\epsilon < 1$. For every $\delta > 0$ there is a finite pairwise disjoint sequence of intervals $(x_k,y_k)$ covering the zero-measure Cantor set with

$$ \sum_{k} |y_{k} - x_{k}| < \delta $$

And since the $\text{c}$ only changes on the Cantor set

$$\sum_{k} |\text{c}(y_{k}) - \text{c}(x_{k})| = 1$$

But

$$\begin{align} (\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k})) &= (\text{id}(y_{k}) - \text{id}(x_{k})) + (c(y_{k}) - c(x_{k})) \\ &\ge c(y_{k}) - c(x_{k}) \end{align}$$

So a fortiori

$$\sum_{k} |(\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k}))| \ge 1$$

$\endgroup$
2
  • 2
    $\begingroup$ This maps to $[0,2]$ but scale by $1/2$ and choose $\epsilon < 1/2$ as well if $[0,1]$ is desired. $\endgroup$
    – A. Webb
    Dec 17, 2013 at 19:13
  • $\begingroup$ Excellent, thank you, A. Webb. $\endgroup$
    – Hans
    Dec 17, 2013 at 19:48
7
$\begingroup$

Let $f:[0,1)\to \mathbb{R}$, $f(x)=\tan(\pi x/2)$. This function is continuous and strictly increasing but not absolutely continuos.

Just to show that this function is in fact not absolutely continuous. Take $\epsilon=1$, and suppose there is a $\delta>0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_{k},y_{k})$ of $[0,1)$ satisfies $\sum_{k}|x_k - y_k| < \delta$ then we have $\sum_{k} |f(y_k)-f(x_k)| < 1$.

Since $\lim_{x \to 1-}f(x) = +\infty$ and $f$ is continuous, let $x_0 \in [1-\delta,1)$ so we can get $y_0 \in [1-\delta,1)$ such that $f(y_0)-f(x_0)>1$.

Using only the interval $(x_0,y_0)$ to test the definition of absolute continuity we have then that $|y_0 - x_0|<\delta$ but $|f(y_0) - f(x_0)|>1$. Therefore, $f$ is not absolutely continuous.

$\endgroup$
3
$\begingroup$

Counterexample number $8.30$ of "Counterexamples in Analysis" by Gelbaum and Olmsted (which can be found here) provides a continuous, strictly increasing function on $[0,1]$ which is singular. Since it is not constant, it can't also be absolutely continuous.

$\endgroup$
4
  • $\begingroup$ Could you please explain in more detail your last sentence "Since it is not constant, it can't also be absolutely continuous."? $\endgroup$
    – Hans
    Dec 17, 2013 at 19:12
  • 1
    $\begingroup$ @Hansen See this. $\endgroup$ Dec 17, 2013 at 19:19
  • $\begingroup$ @Hansen On compact intervals being absolutely continuous means the function has a derivative almost everywhere and that fundamental theorem of Lebesgue calculus applies. Singular functions have a zero derivative almost everywhere. $\endgroup$
    – A. Webb
    Dec 17, 2013 at 19:26
  • $\begingroup$ @DavidMitra, A.Webb and yoknapatawpha, thank you all. $\endgroup$
    – Hans
    Dec 17, 2013 at 19:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .