Prove that $x^2|\sin(\frac1x)|$ is absolutely continuous on $[0,1]$ Could you help me prove that $f = x^2|\sin(\frac1x)|$ is absolutely continuous on $[0,1]$?
I tried to prove this with this:
f has a derivative f′ almost everywhere, the derivative is Lebesgue integrable, and
$f(x)=f(a)+\int _{a}^{x}f'(t)\,dt$
for all x on [a,b].
The derivative of the function is: $2x \mid (\sin(1/x)) \mid - \sin(\frac2x)/(2 \mid \sin\frac1x\mid )$. But I am not sure how to proceed, since there are many pointswhere $f$ is not differentiable. Thank you in advance!
 A: It suffices to prove that $g(x) = x^2 \sin(\frac1x)$ is absolutely continuous on $[0, 1]$, because that implies that $f = |g|$ is absolutely continuous as well.
$g$ is differentiable on $[0, 1]$ with a bounded derivative:
$$
 |g'(x)| = |2x \sin(\frac1x) - \cos(\frac1x) | \le 3
$$
for $0 < x \le 1$, and $g'(0) = 0$.
Therefore $g$ is Lipschitz continuous. On a compact interval, Lipschitz continuity implies absolute continuity.
A: Hints: It is clear that $f'$ exists at al but countably many points and $|f'|
 \leq 3$ a.e. Let $g(x)=\int_0^{x}f'(t)dt$. Then $g$ is absolutely continuous on $[0,1]$ and $g'=f'$ a.e.. Now consider $[a,1]$ where $0<a<1$. On this interval there are only a finite number of points where $f$ is not differentiable. These points give a partition on $[a,1]$. On each subinterval of this partition $f-g$ is a constant. By continuity of $f$ and $g$ conclude that $f-g$ is a constant on $[a,1]$ for each $a$. Now use continuity again to see  that this constant does not depend on $a$. This proves that there is a constant $c$ with $f(x)=c+\int_0^{x}f'(t)dt$ for $0<x\leq 1$.
