# 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!

-
 See A14. in this article mathdl.maa.org/images/cms_upload/… – Deven Ware Dec 11 '12 at 16:57 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