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Let's consider positive real numbers $\alpha,\beta>0$. Then let's define the function:

$ f(x)= x^{\alpha} sin(\frac{1}{x^\beta})$ if $x\in (0,1]$.


Prove that if $\alpha>\beta>0$ then $f$ is absolutely continuous. I was Reading the solution of Kaczor, he said that it's because: $$ f\left( x \right) = \int\limits_0^x {\alpha t^{\alpha - 1} \sin \left( {\frac{1} {{t^\beta }}} \right) - \beta t^{\alpha - \beta - 1} \cos \left( {\frac{1} {{t^\beta }}} \right)} $$ and the integrand is integrable on $[0,1]$. But I don't know how to prove that :S. At least in the case $\alpha - \beta \ge 1$ , then it's easy to see that is in fact continuous, but in a general case I don't know if it's true :S.

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up vote 1 down vote accepted

Each of the two parts of the sum of the integral has the form $x^{-r}$ times a bounded function, where $r$ is less than 1. Then the absolute value of each part is bounded above by $Cx^{-r}$, which is integrable. Since the absolute value has a bounded integral, each part is integrable. Since each is integrable, their difference is integrable.

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