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As a generalization of Prove that $x^\alpha \cdot\sin(1/x)$ is absolutely continuous on $(0,1)$ :

Let $f : (0, 1] \to \mathbb{R}$ be the function denoted by $f(x) = x^a \sin(1/x^b)$.

Determine for which $a,b$ the function $f$ is absolutely continuous.

So at least we know what happens when $b=1$ according to the link.

In addition, for what $a,b$ is $f$ uniformly continuous but not absolutely continuous? (cf Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$ )

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  • Question asked yesterday: http://math.stackexchange.com/questions/525307/absolute-continuity-of-xa-sinx-b/525405#525405 – Jonas Meyer Oct 15 '13 at 03:39
  • i think this is not correct when both $a<0$ and $b<04 right ? i don't want to ask another question but i think we need to say something more – Airbag May 12 '14 at 15:18

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If $a>b$, then $$ f'(x) = ax^{a-1}\sin(1/x^b) -bx^{a-b-1}\cos(1/x^b) $$ and so $$ \int_0^1 |f'(x)|dx \leq a\int_0^1 x^{a-1}dx + b \int_0^1 x^{a-b-1}dx < \infty $$ and hence $f$ is absolutely continuous.

If $a\leq b$, then consider a partition $$ 0 < x_0 < x_1 < x_2 < \ldots < x_{2N} < 1 $$ given by $$ x_{2n} = (2\pi n + \pi/2)^{-1/b}, \quad\text{ and } x_{2n+1} = (2n\pi)^{-1/b} $$ Then $$ \sum_{n=1}^N |f(x_{2n+1}) - f(x_{2n})| \geq \sum_{n=1}^N \frac{1}{(2\pi n + \pi/2)^{a/b}} $$ which diverges since $a/b \leq 1$, and so $f$ is not of bounded variation.

Prahlad Vaidyanathan
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  • i think this is not correct when both $a<0$ and $b<0$ right ? i don't want to ask another question but i think we need to say something more. – Airbag May 12 '14 at 14:39