# 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.

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Could you use the fact that $f(x)$ is of bounded variation if $\int g(x) \mathrm{d}f(x)$ exists for a continuous function $g(x)$. – Sasha Aug 28 '12 at 5:38
Thanks, Sasha. I suppose this will lead to the solution, though I don't yet see what $g$s to attempt for the cases when $f'$ is unbounded. – Kevin Carlson Aug 28 '12 at 9:01
I believe you can avoid integration and differentiation, and simply work directly from the definition of "bounded variation". Try using comparison tests with appropriate $p$-series (from elementary calculus) and partitions like what I did in my answer at Curve In a Closed Interval with an Infinite Length. In that answer I established unbounded variation. In a bounded variation case, you'll want to find an over-estimate of the length that converges. – Dave L. Renfro Sep 5 '12 at 19:02

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 $\eta>1$.

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$\varepsilon$ begins as an arbitrary real, but ends up being an arbitrary rational with odd denominator...which is precisely the content of my second paragraph. Thanks very much for the answer, though! I've got to go for now, and will read it carefully later tonight. – Kevin Carlson Oct 28 '12 at 18:40
Oh, I missed that, but I think it does not really make a difference. You can always pass to the absolute value $|f|$, and as long as $\varepsilon>0$ is real, the above proof would apply to $F(x) = |x|^\eta |\sin (1/x)|^\varepsilon$, even for irrational $\varepsilon$, I think. – Lukas Geyer Oct 28 '12 at 20:44
Edited to make it work for general $\varepsilon > 0$. – Lukas Geyer Oct 28 '12 at 20:52
Wonderful, thanks very much! – Kevin Carlson Oct 28 '12 at 22:05

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 $[a,b]$.

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].

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Thanks, but now what to do when the derivative is unbounded? – Kevin Carlson Aug 28 '12 at 9:00
@KevinCarlson: See the added. – Mhenni Benghorbal Aug 28 '12 at 11:20
yes, but sometimes a function may be BV with unbounded derivative. – Kevin Carlson Aug 28 '12 at 12:01
@KevinCarlson:First find the conditions on $\eta$ and $\epsilon$ so that the derivative exists on [0,1]. Note that, boundedness of the derivative is a sufficient condition. – Mhenni Benghorbal Aug 28 '12 at 12:36
I agree that it's sufficient. I was pointing out that it's not necessary: $\eta=\frac{3}{2}, \varepsilon=2$ gives a counterexample. I'm hoping to give a complete classification of this family. – Kevin Carlson Aug 28 '12 at 12:50