When is $F(x)=x^a\sin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?

Question

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 $a>b$.

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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.

Answer 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 $a>0$, one has $$ I(x):=ax^{a-1}\sin(x^{-b})\in L^1([0,1]) $$ since $\int_0^1|I(x)|\,dx\le\int_0^1ax^{a-1}\,dx=1.$ 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 $[0,1]$.

Answer 2

[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, $\forall x<y\in(0,1]$, $|f(x)-f(y)|\leq M\left(\max_{t\in[x,y]}t^{a-b-1}\right)(y-x)$. From this we estimate $$ \begin{align}T_f\left(\frac1{2^n},1\right)&=\sum_{k=0}^{n-1} T_f\left(\frac1{2^{k+1}},\frac1{2^k}\right)\\ &\leq \sum_{k=0}^{n-1}M\max\left(\frac1{2^{(k+1)(a-b-1)}},\frac1{2^{k(a-b-1)}}\right)\left(\frac1{2^{k}}-\frac1{2^{k+1}}\right)\\&<B<\infty,\end{align} $$ with $B$ a positive number independent of $n$. Hence $f$ is BV on $[0,1]$.