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Let $f: [0,1] \rightarrow \mathbb{R}$ \begin{equation*} f(x) = \begin{cases} \hfill x^a\text{cos}(1/x^b) \hfill & \text{ if $0 < x \leq 1$ } \\ 0 \hfill & \text{ if $x = 0$ } \\ \end{cases} \end{equation*} where $a, b \in \mathbb{R}^+$. Prove: $f$ is of bounded variation on $[0,1]$ if and only if $a>b$. Do so by considering the intervals which $f$ is monotonic.

Want to show: $(\Leftarrow)$ Assuming $a > b$, $\exists M \in \mathbb{R}$ such that for all partitions on $[a,b]$ we have that $\sum_{1}^{n} |f(x_k)-f(x_{k-1})| \leq M$ or equivalently, $f$ is continuous and $f'$ is bounded on $(0,1)$. And the converse ($\Rightarrow): f \in BV[a,b]$ implies $a > b$.

We know: The amplitude of $f$ is governed by the curve $x^a$. The derivative of $f$ is $f'(x) = ax^{a-1}cos(x^{-b})+bx^{a-b-1}sin(x^{-b})$. If $a > b$, we have $a - b > 0 \Rightarrow a - b - 1 > -1$. On the other hand, if $a \leq b$ then $a - b - 1 \leq -1$.

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Not sure how to progress

user20354139
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    For the case $a\leq b$ I think it is enough to consider points $x_k$ such that $\frac1{x_k}=k\pi$. For $a>b$ I think you can show that $f$ is absolutely continuous because $f(x)=\int_a^xf'(t),dt$ then, and absolute continuity implies bounded variation. – Jonas Meyer Feb 16 '17 at 06:18

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