# Prove that $x^\frac{3}{2}\sin (\frac{1}{x})$ is a function of bounded variation for $x\in(0,1]$. [closed]

Prove that $x^\frac{3}{2}\sin (\frac{1}{x})$ is a function of bounded variation for $x\in(0,1]$.

## closed as off-topic by user223391, Peter Woolfitt, user91500, Claude Leibovici, Jyrki LahtonenJul 15 '15 at 8:03

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• I tried to show that the derivative is bounded, with no luck... – Ran Kashtan Jul 15 '15 at 6:17
• Undoubtedly you have seen other variants of topologists sine curve, so you know what to expect, when you zoom in a neighbordhood of the origin - countably infinitely many peaks and valleys packed denser and denser. But here the $x^{3/2}$ tends to zero, so those peaks and valleys also become shallower as we go. Their altitude differences form a series, and your task is to check the resulting series converges absolutely. Of course, you can also follow the route in Marco's or avid19's answer :-) – Jyrki Lahtonen Jul 15 '15 at 7:08

Hint: if $f$ is differentiable on $(0,1)$ then $\operatorname{Var}(f)=\int_0^1 |f'(x)|\mathrm{d}x$.
$$\int_{0}^{1}\left|\left(x^{3/2}\sin\left(1/x\right)\right)'\right|dx\leq\frac{3}{2}\int_{0}^{1}\left|\sqrt{x}\sin\left(\frac{1}{x}\right)\right|dx+\int_{0}^{1}\frac{\left|\cos\left(\frac{1}{x}\right)\right|}{\sqrt{x}}dx\leq$$ $$\leq\frac{3}{2}\int_{0}^{1}\sqrt{x}dx+\int_{0}^{1}\frac{1}{\sqrt{x}}dx=3.$$