When is $F(x)=x^a\sin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$? 
I'm trying to show that $F(x)=x^a\sin\left(x^{-b}\right)$ for $0<x \leq 1$ and $F(0)=0$ has bounded variation only if $a>b$. 


I know I have to show there exist an $M< \infty$ such that for any partition $0=t_0<t_1<...<t_n=1$ we have
$$\sum_{j=1}^N |F(t_j)-F(t_{j-1})|<M \iff |F(t_1)| + \sum_{j=2}^N |F(t_j)-F(t_{j-1})|<M .$$
I'm stuck here.
 A: [I'm assuming in this answer that $a,b>0$.]
Note that $f$ is continuous on $[0,1]$ and its derivative on $(0,1]$ reads as:
$$
f'(x)=ax^{a-1}\sin(x^{-b})-bx^{a-b-1}\cos(x^{-b}),\quad x\in(0,1]. 
$$
We want to study the integrability of $f$ on $[0,1]$. On the one hand, since $a-1>-1$, one has 
$$
I(x):=ax^{a-1}\sin(x^{-b})\in L^1([0,1])
$$ 
since $I$ extends continuously to $[0,1]$. Thus it suffices to study the integrability of
$$
J(x):=x^{a-b-1}\cos(x^{-b}),\quad x\in(0,1].
$$
On the other hand, according to the accepted answer to a related question   When is $\int_{0}^1|x^{a-b-1}\cos(x^{-b})|\ dx<\infty$?, one can conclude that 

$f'\in L^1([0,1])$ if and only if $a>b$. 

Now, we have the following two cases.


*

*If $0<a\leq b$, then $f'$ is not absolutely integrable on $[0,1]$, which implies that $f$ cannot be BV on $[0,1]$. 

*If $0<b<a$, then $f'$ is absolutely integrable on $[0,1]$. According to the answer to this question: Do we have $\|F\|_{TV([a,b])}=\lim_{\epsilon\to 0+}\|F\|_{TV([a+\epsilon,b])}$ if $F:[a,b]\to\mathbb{R}$ is continuous?, since $f$ is continuous, we have the total variation
$$
\|f\|_{TV([0,1])}=\lim_{\epsilon\to0+}\|f\|_{TV([\epsilon,1])}=\lim_{\epsilon\to0+}\int_\epsilon^1|f'(x)|\ dx<\infty.
$$
Hence $f$ is BV on $[0,1]$.
A: [I'm also assuming that $a,b >0$.]
Let $T_f(c,d)$ be the variation of $f$ on $[c,d]$. Suppose that $a\leq b$, we have
$$T_f(0,1)\geq \sum_{k=1}^\infty T_f(\frac{1}{\sqrt[b]{k\pi+\frac\pi2}},\frac{1}{\sqrt[b]{k\pi-\frac\pi2}})\geq \sum_{k=1}^\infty \frac{1}{(k\pi+\frac\pi2)^{\frac ab}}+\frac{1}{(k\pi-\frac\pi2)^{\frac ab}}=\infty.$$
Therefore $f$ is not bounded variation on $[0,1]$. Now we assume that $a>b$. Recalling that
$$
f'(x)=ax^{a-1}\sin(x^{-b})-bx^{a-b-1}\cos (x^{-b}),
$$ we may have $|f'(x)|\leq 2\max(a,b)x^{a-b-1}:=Mx^{a-b-1}$. By Lagrange mean value theorem, $\forall x<y\in(0,1]$, $|f(x)-f(y)|\leq Mx^{a-b-1}|x-y|$. From this we estimate
$$
T_f(\frac{1}{2^n},1)=\sum_{k=0}^{n-1} T_f(\frac{1}{2^{k+1}},\frac{1}{2^k})\leq \sum_{k=0}^{n-1}M(\frac{1}{2^{k+1}})^{a-b-1}(\frac{1}{2^{k}}-\frac{1}{2^{k+1}})<B<\infty.
$$
With $B$ a positive number independent with $n$. Hence $f$ is BV on $[0,1]$.
