Let the function $f_a:[0,1] \to \Bbb R$ be defined by
$$f_a(x)=\begin{cases} x^a \cdot \cos(\frac{1}{x}) & 0 < x \leq 1 ;\\ 0 & x=0.\end{cases}$$
Find all values $a\ge 0$ such that
(a) $f_a$ is continous on $[0,1].$
(b) $f_a$ is of bounded variations on $[0, 1]$
(c) $f_a$ is absolutely continuous on $[0, 1]$
Do I have to show that the function is Lipschitz continuous for some values of $a$? How can one find these values of $a$?
For (a), the only possible continuity issues that can occur are at $x=0$. So make sure $\lim_{x\rightarrow 0^+}fa(x)=fa(0)=0$. Not all values of $a$ work for this.
For $(b)$, use the fact that a Bounded variation function is differentiable almost everywhere and that the total variation equals $\int_0^1 |fa'(x)|dx$, which needs to be finite.
For $(c)$, an absolutely continuous function must satisfy the fundamental theorem of calculus: $fa(x)=\int_0^x fa'(t)dt$.