A basic problem on bounded variation If $a > 0$ let 
$$f(x) =\left\{\begin{array}{ll}
 x^{a} \sin (x^{-a})&\text{if } 0 < x \leq 1\\
 0&\text {if }x=0
\end{array}\right.$$
Is it true that for each $0 < \alpha < 1$ the above function satisfies the Lipschitz condition of exponent $\alpha$
$$|f(x) - f(y)| \leq A|x-y|^{\alpha}$$ but which is not of bounded variation. I need some hint to start.

It is just an hint, not a complete proof. I hope it's enough for you. Consider the case $a=1$, the other are very similar. You can consider the following partition of $[0,1]$. 
$$[0,1]= \left [\frac{1}{\pi},1 \right ]\cup\ \bigcup_{k=1}^n \left [\frac{1}{(k+1)\pi},\frac{1}{k\pi} \right ]\cup \left [0,\frac{1}{n\pi} \right ].$$
Now you have that the variation on this partition satisfies
$$Var\geq\sum_{k=1}^n \left |f \left (\frac{1}{k\pi} \right )-f \left (\frac{1}{(k+1)\pi} \right ) \right |\geq \sum_{k=1}^n \left |\frac{1}{k\pi}+\frac{1}{(k+1)\pi} \right |$$
passing to the limit for $n\rightarrow\infty$ you obtain that $Var$ diverges.
Since it is a particular partition you can deduce that $f$ is not $BV$.
