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?
[Math] Example of continuous but not absolutely continuous strictly increasing function
absolute-continuityexamples-counterexamplesfunctionsreal-analysis
Related Solutions
$$f(x)=\begin{cases} 1/\log x \quad &\text{if } x\in (0,1/2] \\ 0 &\text{if }x=0\end{cases}$$ Consider the behavior of $f(x)/x^\alpha$ at zero.
Also notice that $f'$ is bounded on $[1/n,1/2]$ for all $n=3,4,\dots$.
Here is a way to construct such a function.
- Draw $f(x)=x^3+x^2+x$.
- At every integer $n$, draw $g(x)$ between $n$ and $n+1$ as the secant from $f(n)$ to $f(n+1)$.
- $g(x)$ is now continuous, differentiable at every point between the integers, but not at the integers.
EDIT v2 I've edited to use a strictly increasing function.
Here is a graphical depiction of the concept:
Best Answer
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
$$\sum_{k} |(\text{id}(y_{k}) + c(y_{k})) - (\text{id}(x_{k}) + c(x_{k}))| \ge 1$$