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Let $f(x)=x^2\sin^2(\pi/x)$ for $x\in (0,1]$ and $f(0)=0$. Show that $f(x)$ is absolutely continuous in $[0,1]$, but $|f|^{1/2}(x)$ is not.

Attempt. Note that $f'(x)=2x\sin^2(\pi/x)-\pi\sin(2\pi/x)$ implies that $|f'(x)|\leq 2x+\pi$ for $x\in (0,1]$. So we have $$\int_0^1|f'(x)|dx\leq \int_0^1 (2x+\pi)dx<\infty$$ Hence, $f$ is absolutely continuous.

For the second part, I claim that $x|\sin(\pi/x)|$ is not of bounded variation to conclude that it is not absolutely continuous. But, I couldn't prove it. In fact, I am able to prove that the function $x\sin(\pi/x)$ is not of bounded variation by using the sequence $x_n=\frac{2}{2n+1}$.

I would be glad if someone could help me and check my attempt. Thanks!

zhw.
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Ergin Süer
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  • If $g$ is continuous and $|g|$ is of bounded variation, then $g$ is of bounded variation. See https://math.stackexchange.com/questions/2299679/f-is-of-bounded-variation-rightarrow-f-is-of-bounded-variation – Feng Jan 14 '20 at 01:57
  • you still need to show that $f(x)=\int_0^xf'(t)dt+f(0)$ – operatorerror Jan 14 '20 at 02:09

1 Answers1

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Your estimate works to prove that the almost everywhere derivative is $L^1$, and your proof sketch is how I would prove the derivative is not BV and so not AC.

However, you still must show that the fundamental theorem of calculus holds. This is readily done using the usual calculus FTC though, indeed fix $\delta>0$. Then, $$ x^2\sin^2(\pi/x)-\delta^2\sin^2(\pi/\delta)=\int_\delta^x f'(t)\mathrm dt=\int_0^xf'(t)\mathrm dt-\int_0^\delta f'(t)\mathrm dt $$ Now taking $\delta\downarrow 0$, $$ x^2\sin^2(\pi/x)=\int_0^xf'(t)\mathrm dt-\lim_{\delta\downarrow 0}\int_0^\delta f'(t)\mathrm dt $$ but using your estimate, $$ \left|\int_0^\delta f'(t)\mathrm dt \right|\leq \delta^2+\pi\delta\to 0 $$

operatorerror
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