Find the values of $\alpha$ and $\beta$ for which the function $f$ is of bounded variation on $[0,1]$?

Question

The question is :

Let $f : [0,1] \longrightarrow \mathbb {R}$ be defined by

$f(x) = x^{\alpha} \sin {\frac {1} {x^{\beta}}}$ , whenever $x \in (0,1]$ and $=0$ , whenever $x = 0$.

Then find the values of $\alpha$ and $\beta$ for which $f$ becomes a function of bounded variation on $[0,1]$.

Please give me a right way to proceed.Thank you in advance.

I have just obtained the result which is $\alpha > 2$ and $\beta < \alpha - 2$.Is it correct at all?Please verify it.

Answer 1

Hint: If $f$ is continuously differentiable on $[a,b],$ then the total variation of $f$ on $[a,b]$ is $\int_a^b|f'(x)|\, dx.$

Answer 2

Claim. Let $\alpha>0$, $\beta > 0$ and let $f$ be as given. Then $f \in \operatorname{BV}(\left[0,1\right])$ if and only if $\alpha > \beta$.

Proof. Suppose that $\alpha>\beta$. Let $h:\left[0,1\right]\to\mathbf{R}$ be defined by $$ h(x)= \begin{cases} \displaystyle{x^{\alpha/(\alpha-\beta)}\sin\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)}&,\text{ if } 0<x\le 1\\ 0&,\text{ if } x = 0. \end{cases} $$ Then $h$ is differentiable at every $x\in \left[0,1\right]$. In fact, we have $$ h^{\prime}(x)= \begin{cases} \displaystyle{\frac{\alpha}{\alpha-\beta}x^{\beta/(\alpha-\beta)}\sin\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)-\frac{\beta}{\alpha-\beta}\cos\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)}&,\text{ if }0<x\le1\\ 0&,\text{ if }x=0. \end{cases} $$ In particular, $h^{\prime}$ is bounded, hence $h$ is Lipschitzian of order $1$, i.e., there is some $C>0$ such that for all $x\in\left[0,1\right]$ and all $y\in\left[0,1\right]$, we have $$ \left|h(x)-h(y)\right|\le C\left|x-y\right|. $$

Substituting $x\mapsto x^{\alpha-\beta}$, $y\mapsto y^{\alpha-\beta}$, we finally get $$ \left|f(x)-f(y)\right|\le C\left|x^{\alpha-\beta}-y^{\alpha-\beta}\right|, $$ therefore the sums appearing in the definition of total variation are bounded above by $C(t_{n}^{\alpha-\beta} - t_{0}^{\alpha-\beta})=C(1-0)=C$, which implies $f\in\operatorname{BV}(\left[0,1\right])$.

Suppose that $\alpha\le\beta$ and consider the following partition $$ \left\lbrace 0, \left(\frac{2}{\pi(2n+1)}\right)^{1/\beta}, \left(\frac{2}{\pi(2n)}\right)^{1/\beta},\left(\frac{2}{\pi(2n-1)}\right)^{1/\beta},\cdots,\left(\frac{2}{\pi}\right)^{1/\beta},1 \right\rbrace. $$ The following holds: $$ \begin{aligned} \sum_{k=1}^{2n+2}\left|f(t_{k})-f(t_{k-1})\right|&\ge\sum_{k=2}^{2n+1}\left|f(t_{k})-f(t_{k-1})\right|\\ &=\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=2}^{2n+1}\left|\frac{1}{(2n+2-k)^{\alpha/\beta}}\sin\left(\frac{k\pi}{2}\right)-\frac{1}{(2n+2-(k-1))^{\alpha/\beta}}\sin\left(\frac{(k-1)\pi}{2}\right)\right| \\ &\ge\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=1}^{n}\frac{1}{\left(2(n-k)+1\right)^{\alpha/\beta}}\\ &=\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=0}^{n-1}\frac{1}{\left(2k+1\right)^{\alpha/\beta}}. \end{aligned} $$ Because $0<\alpha\le\beta$, the last sum can be made arbitrarily large. Consequently, $f\notin \operatorname{BV}(\left[0,1\right])$ and the claim is proved.