# Is $x^{3/2}\sin(\frac{1}{x})$ of bounded variation?

I'm trying to show that $f(x)$ is of bounded variation where $f(x)=x^{3/2}\sin(\frac{1}{x})$ on $[0,1]$. I think that it is but I can't show it explicitly.

Any help will be appreciated.

• Have you tried taking the derivative? If I'm not mistaken, $f$ is $C^1$ and therefore has bounded variation. Dec 3, 2014 at 20:14
• @minimalrho: It is $\mathcal C^1$ on $(0,1]$, but not on $[0,1]$. The derivative is not continuous at $0$. Dec 3, 2014 at 20:32
• @HenningMakholm Ah yes, I keep forgetting if the power has to larger than 2 or larger than 1 for the derivative to be continuous. Dec 3, 2014 at 20:47
• You can do this without integation by making an appropriate comparison with a calculus 2 $p$-series using a variation (pun intended) on the method in my answer to Curve In a Closed Interval with an Infinite Length. Note that you can get an over-estimate for the length by using (limits of) polygonal paths made up of vertical and horizontal segments. (You can get an under-estimate by summing only the vertical segments.) Dec 3, 2014 at 21:02

Hint: The total variation of $f$ on $[a,b]$ is at most $\int_a^b g(x)\,dx$, if $g(x)\ge|f'(x)|$ everywhere.
If you play a bit hard and fast you can take $[a,b]=[0,1]$ and be done quickly, but (depending on which theorems about total variation you have available) that may not be completely rigorous, because $g(x)$ will not be Riemann integrable on $[0,1]$. However, because $f$ is continuous at $0$ it actually suffices that $\int_\epsilon^1 g(x)\, dx$ exists and tends to a finite limit as $\epsilon\to 0^+$.