continuous but not absolutely continuous function

Question

How to show by definition (without the variation of function) that $f \colon [0,1] \to \mathbb{R}$ $$ f(x) = \begin{cases} 0 \quad \mbox{if} \quad x=0 \\ x \cos \frac{\pi}{x} \quad \mbox{if} \quad x \neq 0 \end{cases} $$ is not absolutely continuous?

Answer 1

An absolutely continuous function on a bounded interval is of bounded variation. The present function $f$, defined on $[0,1]$, is not of bounded variation (it oscillates too much) and is thus not absolutely continuous. To see this, for $n\in\mathbb N$ set $x_i := \frac 1 i$, $i=1,\ldots,n$. Then $$ \sum_{i=1}^n|f(x_{i+1}) - f(x_i)| = \ldots\,\le\,\sum_{i=1}^n\frac 1 i, $$ which tends to $\infty$ as $n\to\infty$.

Answer 2

Ok, another answer not using the variation... We have to show that for every $\delta > 0$ there exist mutually disjoint open intervals $\Delta_i = (a_i,b_i)$, $i=1,\ldots,n$, such that the sum of their lengths is $<\delta$ but $\sum_1^n|f(a_i) - f(b_i)|\ge 1$. So, let $\delta > 0$ be given. Choose $N\in\mathbb N$ odd (we need this later) such that $\frac{1}{N} < \delta$. Now, let $$ \Delta_i = \left(\frac 1 {N(i+1)},\frac{1}{Ni}\right), \quad i=1,\ldots,n, $$ where $n$ is such that $\frac 1 N\sum_{i=1}^n\frac 1 i\ge 1$. Now, we have that $$ \sum_{i=1}^n\operatorname{Length}(\Delta_i) = \sum_{i=1}^n\left(\frac{1}{Ni} - \frac 1 {N(i+1)}\right) = \frac 1 N\left(1 - \frac 1 {n+1}\right) < \frac 1 N < \delta. $$ But on the other hand, \begin{align*} \sum_{i=1}^n\left|f\left(\frac 1 {Ni}\right) - f\left(\frac 1 {N(i+1)}\right)\right| &= \sum_{i=1}^n\left|\frac {\cos(Ni\pi)}{Ni} - \frac {\cos(N(i+1)\pi)}{N(i+1)}\right|\\ &= \sum_{i=1}^n\left|\frac {(-1)^{Ni}}{Ni} - \frac {(-1)^{N(i+1)}}{N(i+1)}\right|\\ &= \sum_{i=1}^n\left|\frac {1}{Ni} - \frac {(-1)^N}{N(i+1)}\right|\\ &= \sum_{i=1}^n\left(\frac {1}{Ni} + \frac {1}{N(i+1)}\right)\,\ge\,1. \end{align*} So, $f$ cannot be absolutely continuous.

Answer 3

Define $a_n= 1/(2n+1/2), b_n= 1/(2n), n =1,2,\dots $ Then the intervals $[a_n,b_n]$ are pairwise disjoint.

Let $\delta > 0.$ Choose $N$ such that $1/(2N)<\delta.$ Then each $[a_n,b_n]$ lies to the left of $\delta$ if $n\ge N.$ Thus we have

$$\sum_{n=N}^{\infty} (b_n-a_n) < \delta,$$

while

$$ \sum_{n=N}^{\infty} (f(b_n)-f(a_n)) = \sum_{n=N}^{\infty} \frac{1}{2n} = \infty.$$

This implies $f$ is not absolutely continuous on $[0,1].$