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

$f(x) = x^\alpha \sin \frac 1 {x^\beta}$ , whenver $x \in (0,1]$ and $f(0) = 0$ where $\alpha > \beta$. Then for what values $\alpha$ and $\beta$, $f$ is a function of bounded variation on $[0,1]$?

I think the theorem which states that "If $f : [a,b] \longrightarrow \mathbb R$ be a continuous function on $[a,b]$ and if $f'$ exists and is bounded on $(a,b)$ then $f$ is a function of bounded variation on $[a,b]$" can be applied to solve this kind of problem. But we also know that there exist functions which are of BV but do not satisfy the above theorem. So, my question is what is the actual way to proceed?

Please help me. Thank you in advance.


marked as duplicate by clark, Alex M., Daniel W. Farlow, Davide Giraudo real-analysis Nov 8 '16 at 21:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @clark well, yes, but that question does not have an (accepted) answer. $\endgroup$ – Thomas Nov 8 '16 at 18:30
  • $\begingroup$ No,I don't find anything worth in the replies of the question for which my question was considered to be a complete duplicate.But what I have analysed so far it has come that if $\alpha = 0$ then $\beta < 0$ yield BV functions on $[0,1]$ and for $\alpha \geq 1$ then for $\beta \leq {\alpha} - 1 $ again yield BV functions on $[0,1]$.But if $\alpha < 0$ or $0< \alpha < 1$ I can't make any satisfactory conclusion.Precisely when $\alpha = \frac {1} {2}$ and $\beta = \frac {1} {3}$ or if $\alpha = - \frac {1} {2}$ and $\beta = -1$ then what will happen?Please give me some suggestions. $\endgroup$ – user386602 Nov 9 '16 at 8:36