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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.

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  • $\begingroup$ Have you tried taking the derivative? If I'm not mistaken, $f$ is $C^1$ and therefore has bounded variation. $\endgroup$
    – minimalrho
    Dec 3, 2014 at 20:14
  • $\begingroup$ @minimalrho: It is $\mathcal C^1$ on $(0,1]$, but not on $[0,1]$. The derivative is not continuous at $0$. $\endgroup$ Dec 3, 2014 at 20:32
  • $\begingroup$ @HenningMakholm Ah yes, I keep forgetting if the power has to larger than 2 or larger than 1 for the derivative to be continuous. $\endgroup$
    – minimalrho
    Dec 3, 2014 at 20:47
  • $\begingroup$ 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.) $\endgroup$ Dec 3, 2014 at 21:02

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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 $\epsilon\to 0^+$.

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