Prove that $x^\alpha \cdot\sin(1/x)$ is absolutely continuous on $(0,1)$ I would appreciate any help with my HW exercise:

Prove that $f(x) = x^\alpha \cdot \sin(1/x)$ is absolutely continuous on $(0,1)$, when $\alpha>1$.

It's easy to find the derivative of $f$:
$$f'(x) = \alpha x^{\alpha-1} \sin(1/x) - x^{\alpha-2} \cos(1/x).$$
So, when $1<\alpha<2$, the function is not Lipschitz, and that is the main problem.
I searched for similar questions and found this:
Examples of absolutely continuous functions that are not Lipschitz.
and this:
http://mathdl.maa.org/images/cms_upload/0002989049585.di021349.02p00072.pdf
But I couldn't understand the PDF file, which merely handles the case $\alpha=3/2$.
Thanks!
 A: I assume that $1<\alpha<2$.  If not, as remarked by the questioner, $f(x)$ is Lipschitz on $(0,1)$ and the problem is simpler.
You can use the following approach.  Take a small $\delta>0$.  You wish to bound $V=\sum_i |f(x_i)-f(y_i)|$ whenever $S=\{(x_1,y_1),\dots,(x_n,y_n)\}$ is a finite set of pairwise disjoint intervals contained in $(0,1)$ with $\sum_i |x_i-y_i|<\delta$.
Let $N$ be the smallest integer such that $N>(2\pi \delta)^{-1}$, and split the interval into two pieces, $(0,1/(2\pi N))$ and $(1/(2\pi N),1)$. By splitting an interval in $S$ if necessary, which does not decrease $V$, you can assume that each member of $S$ is in either $(0,1/(2\pi N))$ or $(1/(2\pi N),1)$.  
Let $V_1$ be the portion of $V$ coming from the intervals in $(0, 1/(2\pi N))$.  Here, the function $f(x)$ has alternating local maxima and minima.  The maxima are close to the values $x=2/((4n+1)\pi)$; let $M_n$ be the value of $x$ close to $x=2/((4n+1)\pi)$ where $f(x)$ has a local maximum.  Similarly, the minima are close to the values $x=2/((4n+3)\pi)$; let $m_n$ be the value of $x$ close to $x=2/((4n+3)\pi)$ where $f(x)$ has a local minimum.
Argue that
$$V_1\le |f(M_N)|+|f(M_N)-f(m_N)|+|f(m_N)-f(M_{N+1})|+\cdots\ \ \ (1)$$
and that
$$
f(M_n)=f(m_n)=O(n^{-\alpha}).\qquad (2)
$$
Then, combining $(1)$ and $(2)$, conclude that $V_1=O(N^{1-\alpha})$.
Let $V_2$ be the portion of $V$ coming from the intervals in $(1/(2\pi N),1)$.  In this portion of $(0,1)$, $|f'(x)|$ is bounded above, so the function $f(x)$ is Lipschitz.  Argue that the Lipschitz constant is $O(\delta^{\alpha-2})$.  Therefore, $V_2$ is $O(\delta^{\alpha-1})$.
Since $V=V_1+V_2$, adding the above estimates together should prove that $V\to 0$ as $\delta\to 0$.
A: For any $ \delta > 0 $, the function $ f(x) $ is Lipschitz on $ [\delta,1] $, since 
$$ |f'(x)| = |a x^{a-1} \sin \frac{1}{x} - x^{a-2} \cos \frac{1}{x}|  \leq  |a x^{a-1} \sin \frac{1}{x}| +  |x^{a-2} \cos \frac{1}{x}| \leq a |x|^{a-1}  +  |x|^{a-2}  < a + \delta^{a-2} < \infty \ \ \ (**)$$
Therefore, for every $ x,y \in [\delta,1] $ I can find a number $ c $ such that $ \frac{|f(x)-f(y)|}{|x-y|} < c $
(since $ |f'(x)| < \infty $ on $ [\delta , 1] $) and therefore $ f(x) $ is AC on $ [\delta , 1] $.\
$$ f(x)=f(\delta)+\int^{x}_{\delta} f'(t)dt \Rightarrow f(x)=f(\delta)+\int^{1}_{0} f'(t) \chi_{[\delta , x]} dt $$
 Choose $ \delta = \frac{1}{n} $ and take the limit as $ n $ approaches infinity
$$ \lim_{n \rightarrow \infty} f(x)= \lim_{n \rightarrow \infty} f(\frac{1}{n})+ \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]} dt \ \ \  (*)$$
Moreover, $ f(x) $ is continuous at $x=0$. Because 
$$-1 \leq  \sin \frac{1}{x} \leq 1 \Rightarrow -x^a \leq x^a \sin \frac{1}{x} \leq x^a \Rightarrow \lim_{x\rightarrow 0^+} -x^a = 0 \leq \lim_{x\rightarrow 0^+} x^a \sin \frac{1}{x} \leq \lim_{x\rightarrow 0^+} x^a = 0  \Rightarrow \lim_{x\rightarrow 0^+} x^a \sin \frac{1}{x}=0 $$ And because $ f(0)=0 $ we conclude that $ f(x) $ is continuous at $ x=0 $.
$$ (*) \Rightarrow f(x)=f(0) + \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]} dt $$
By $ (**) $ we know that $ |f'(t) \chi_{[\frac{1}{n} , x]}| \leq |f'(t)| \leq ax^{a-1} + x^{a-2} = g(x)$, if $ a>1 $ then $ g(x) $ is Riemann integrable over $ [0,1] $ (and therefore is Lebesgue-integrable) since 
$$ \int^{1}_{0} ax^{a-1} + x^{a-2} = a + \frac{1}{a-1} < \infty $$
 And because $ f'(t) \chi_{[\frac{1}{n} , x]}  $ converges pointwise a.e to $ f'(t) \chi_{[0 , x]} $, using the Lebesgue Dominated Convergence theorem, we would have 
$$ f(x)=f(0) + \lim_{n \rightarrow \infty} \int^{1}_{0} f'(t) \chi_{[\frac{1}{n} , x]}dt \Longrightarrow_{L.D.C} f(x)=f(0) + \int^{1}_{0} f'(t) \chi_{[0 , x]}dt  \\
  \Rightarrow 
f(x)=f(0) + \int^{x}_{0} f'(t)dt \Longrightarrow  f(x) \  is \ AC \ if  a > 1 \ \ \ \square$$ 
