Determine for which parameters $\alpha, \beta \in [0, +\infty[$ the function
$$f_{\alpha,\beta}:[0,1] \to \mathbb{R}, \quad f_{\alpha, \beta}(x) :=
\begin{cases}
x^{\alpha}\ \text{sin}(\frac{1}{x^{\beta}}) & \text{für} \ x \in \ ]0,1]\\
0 & \text{für} \ x = 0 \\
\end{cases}$$
$a)$ has bounded variation,
$b)$ is absolutely continuous.
I know that any absolutely continuous function has bounded variation. So I wanted to solve b) first in order to get an impression what choices of $\alpha$ and $\beta$ I could make.
So let's look at any decomposition of $[0,1]$ such that $\sum_{k=1}^{m}(b_k - a_{k}) < \delta$ for $\delta > 0$. Then we consider the decomposition of $f$ into those subintervals
$$\sum_{k=1}^{m} |f(b_k) - f(a_k)| = \sum_{k=1}^{m} \biggl|b_k^{\alpha}\ \text{sin}\biggl(\frac{1}{b_k^{\beta}}\biggr) - a_k^{\alpha}\ \text{sin}\biggl(\frac{1}{a_k^{\beta}}\biggr)\biggr| \leq \sum_{k=1}^{m} |b_k^{\alpha} + a_k^{\alpha}| < \sum_{k=1}^m |b^{\alpha}_k - a^{\alpha}_{k}| + 2|a^{\alpha}_{k}|$$
So, based on this estimate, the parameter $b$ has no influence on the absolute continuity. We see that if $\alpha = 0$, then $\sum_{k=1}^m |1-1| + 2|1| = 2m$ can't be made arbitrarily small.
However, I am struggling to go any further than that. I know certain results for specific $\alpha$ and $\beta$ (e.g. $\alpha = \beta = 1$ has no bounded variation) but I would like to develop a systematic approach to solve that question. Can anybody please help me?
Below I will give a (very detailed) sketch of one possible way to approach this problem.
Since $f$ is continuous on $[0,1]$ and differentiable on $(0,1]$, the variation $V[f;0,1]$ is bounded above by $\int_0^1 |f'(t)|\,dt$. Thus a sufficient condition for $f$ to be of bounded variation is that that $f' \in L^1[0,1]$. Since $f'(x) = x^{a-b-1}(ax^b\sin(x^{-b}) - b\cos(x^{-b}))$, it follows that $f' \in L^1$ if and only if $a> b$; to see this, try and compute the integral--- you'll see that the problem area is around zero.
Thus after you fill in the details (I've made two big claims in the above paragraph), you have proved that if $a>b$, then $f$ is of bounded variation. This is in fact an if and only if; when $a \leq b$, the variation of $f$ on $[0,1]$ is infinite. To see this, let $c_n$ be the $n$th critical point of $\sin(x^{-b})$ from the origin (i.e., $(\sin(x^{-b}))'|_{c_n} =0$), and let $P_k$ be the partition given by the first $k$ elements of the sequence: $$P_k = \{c_0, \dots, c_k\}$$
You can show that the variation $\Gamma_{P_k}$ of $f$ over $P_k$ goes to infinity by the comparison test: there is a divergent series $\sum_{i=1}^\infty z_n = \infty$ so that for each $k$ we have $\sum_{i=1}^k z_n \leq \Gamma_{P_k}$, thus $\lim_{k \to \infty} \Gamma_{P_k} = \infty$. But as $V[f;0,1] = \sup\{\Gamma_P : P \text{ is a partition}\}$, this means that $V[f;0,1] = \infty$ when $a \leq b$.
This covers the case of bounded variation. We get for free that $f$ is not absolutely continuous when $a \leq b$, since it isn't even of bounded variation! If $a > b$, we're also done: the argument we used to show $(a >b \Longrightarrow f \in BV)$ actually proved that $f$ is absolutely continuous, as for any $L^1$ function $g$, the function $G(x) = \int_0^x g(t)\,dt$ is absolutely continuous (why?)