# absolute continuity in trigonometric functions

I want to show the function $f:[0,1]\to \mathbb{R}$

$$f=\left\{ \begin{array}{ll} x^{3/2}\sin\left(\frac{1}{x}\right), & {x \in (0,1]} \\ 0, & x=0 \end{array} \right.$$

is absolutely continuous.

My attempt:

I broke it to functions $x^{3/2}$ and $\sin(\frac{1}{x})$. The first one is a.c. since it is increasing, for the second one I wrote the definition of absolute continuity:

$$\sin\left(\frac{1}{x_i+\delta}\right)-\sin\left(\frac{1}{x_i}\right)=2\cos\left(\frac{2x_i+\delta}{2(x_i+\delta)x_i}\right)\sin\left(\frac{-\delta}{2(x_i+\delta)x_i}\right)$$ but I don't see how it is smaller that $\epsilon\ \forall i$! Can I say for each $ϵ$ I'll find $δ=min\{ϵ,ϵ2x_i\}$ so that it converges to zero?

Another thing that I tried was using uniform integrability of ${\mbox{Diff}_\delta \ f}_{0< h\leq 1}$ but the integral results in $\Gamma$ function and imaginary number that I don't know how to handle!

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I have! But I don't understand it! At $c_k \in (\frac{1}{(k+1)\pi},\frac{1}{k\pi})$ none of $f, f', f''$ are zero! take $\frac{1}{1.5\pi}$ for example –  Anita Dec 11 '12 at 17:01
About your attempt, imagine the graph of the function $\sin(1/x)$ and determine if the total variation of this function is finite.
A natural method to prove that $f$ is absolutely continuous is to calculate $f'$ and to prove that $f'$ is Lebesgue integrable. A couple of hints for this example:
1. Suppose that $g\colon[0,1]\to\mathbb{R}$ is some function hard to integrate, but $g$ can be bounded by a simpler function $h$ (in the sense that $|g|\le h$), where $h$ is Lebesgue integrable on $[0,1]$. What can we say about $g$ in this situation?
2. For what values of $p$ is the function $x^p$ Lebesgue integrable on $[0,1]$?