Variation of $f(x)=x^\eta\sin^\varepsilon(\frac{1}{x})$ I'm asked to characterize the values of the parameters $\eta, \varepsilon$ for which the above function is of bounded variation on $[0,1]$, when we set $f(0)=0$. By "bounded variation", I mean that the following sum is bounded by some constant $c$, where $t_i$ are the boundary points of any partition $\mathcal{P}$ of the interval into finitely many segments:
$$\sum_{j=1}^n |f(t_j)-f(t_{j-1})|$$
I've made just a bit of progress: we need $\varepsilon$ rational with odd denominator, or else $\sin^{\varepsilon}(\frac{1}{x})$ won't be defined on the whole interval. Furthermore we need $\eta$ and $\varepsilon$ nonnegative, positive if we ignore the trivial cases when they're 0, or the function will be unbounded at the zeroes of $\sin(\frac{1}{x})$ or near zero, respectively, while bounded-variation functions never have essential discontinuities.
I can differentiate $f$: 
$$f'=x^{\eta-1}\sin^{\varepsilon-1}(\frac{1}{x})(\eta\sin(\frac{1}{x})-\varepsilon\cos(\frac{1}{x}))$$
This gives me critical points at the zeroes of $\sin{\frac{1}{x}}$ and at the infinitely many points where $\tan\frac{1}{x}=\frac{\varepsilon}{\eta}$. Ideally I'd estimate the variation by taking a partition at each critical point, since including all local extrema in the partition should guarantee, roughly, that I capture "all $f$'s variation". Now I'm at a loss how to proceed. Is there some nice series by which I might bound the sum I'd get in this way? Should I try a completely different approach than this using critical points? Thanks for your suggestions.
EDIT: After discussing with some other members of the course, we're pretty sure that
the value of $\varepsilon$ is immaterial and $\eta>1$ gives bounded variation while $\eta \leq 1$ does not. But the closest I have to an argument for this is to point out that the former case is just when $f'$ is absolutely integrable on $[0,1]$, which seems pertinent for satisfying Sasha's condition in the comments.
 A: Since $f$ and $|f|$ have the same total variation, we can replace $f$ by $f(x) = |x|^\eta |\sin (1/x)|^\varepsilon$ (and even allow arbitrary real $\varepsilon > 0$.) In this case, all the points where $\sin \frac1x = 0$ are minima, and all the solutions to $\tan \frac1x = \frac\varepsilon\eta$ are maxima. For every $k \in \mathbb{N}$, there is exactly one solution $x_k$ to the latter equation with ${k \pi} < \frac1{x_k} < (k+1/2) \pi$. Since $$\left|\sin \frac1{x_k}\right| = \left|\tan \frac1{x_k}\right| \left|\cos \frac1{x_k}\right| = \frac\varepsilon\eta \left|\cos \frac1{x_k}\right| \to \frac\varepsilon\eta$$ as $k \to \infty$, we get that $$\left|x_k^{-\eta}f(x_k)\right| = \left|\sin \frac1{x_k}\right|^\varepsilon \to \left(\frac\varepsilon\eta\right)^\varepsilon =:C>0 $$ as $k\to\infty$. For $k$ large, we have $C/2 \le \left|x_k^{-\eta}f(x_k)\right| \le C$, so that the total variation on the interval $\left[\frac1{k\pi}, \frac{1}{(k-1)\pi}\right]$ (whose endpoints are minima, and which contains exactly one maximum at $x_k$) is $V_k = 2f(x_k)$ and so $$\frac{C}2 ((k+1/2)\pi)^{-\eta}\le \frac{C}2 x_k^\eta \le V_k \le C x_k^\eta \le C(k\pi)^{-\eta}.$$
Summing up over $k$ and observing that both series $\sum\limits_{k=1}^\infty k^{-\eta}$ and $\sum\limits_{k=1}^\infty (k+1/2)^{-\eta}$ converge iff $\eta > 1$, you get that the total variation is finite iff $\eta>1$.
A: This hint may help you to advance with your problem: if the derivative of a function exists and is bounded on $[a,b]$, then the function is of bounded variation on $[a,b]$.
Added:
Now, the whole idea is to put conditions on $ \eta $ and $ \epsilon $, so that the derivative will exist and will be bounded on the whole interval [0,1]. See where the problem is with the derivative on [0,1].   
