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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!

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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.

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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 $0<x\leq 1$.

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