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.

Solution 1:

[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 $[0,1]$.