# How to show that the function $x^\alpha \sin\left(\frac{1}{x}\right)$ ($\alpha > 1$) is of bounded variation on $(0,1]$?

I am given the function $$f(x) = \begin{cases} x^\alpha \sin\left(\frac{1}{x}\right) &\text{on }(0,1], \\ 0 & \text{if }x = 0. \end{cases}$$ how do I show that this function is of bounded variation on $[0,1]$ if $\alpha >1$?

The variation is given by $$Vf=\sup\{\Sigma_{n=1}^N|f(x_n)-f(x_{n-1})|: 0=x_0 <x_1,..<x_N=b\}.$$

What I managed to do was to show that it was of bounded variation when $\alpha \ge 2$, because then the derivative is bounded, so the result follow from the mean value theorem. But what about the case $1<\alpha<2$?

Any tips?

$f$ derivative is $$f^\prime(x) = \begin{cases} \alpha x^{\alpha-1} \sin\left(\frac{1}{x}\right) - \frac{1}{x^{2-\alpha}} \cos\left(\frac{1}{x}\right) &\text{on }(0,1], \\ 0 & \text{if }x = 0. \end{cases}$$ Hence $$\vert f^\prime(x) \vert \le \alpha x^{\alpha-1} + x^{\alpha-2}$$ The integrals $\int_0^1 x^{\alpha-1}dx$ and $\int_0^1 x^{\alpha-2}dx$ both converge for $1 < \alpha < 2$. Hence $$V_0^1(f) \le \int_0^1 \vert f^\prime(x) \vert dx$$ and $f$ is of bounded variation on $[0,1]$ as the RHS of the inequality is finite.
• @user119615 Because $f^\prime$ is continuous on $(0,1]$. So for $0 < u < v \le 1$, you have $f(v)-f(u)=\int_u^v f^\prime(t) dt$. Hence $\vert f(v)-f(u) \vert=\vert \int_u^v f^\prime(t) dt \vert \le \int_u^v \vert f^\prime(t) \vert dt$. – mathcounterexamples.net Dec 10 '15 at 16:57
• I see that this gives that $V_\epsilon^1f\le\int_\epsilon^1|f'|dt\le\int_0^1|f'|dt$. But how do you go from this to the case $V_0^1f$? – user119615 Dec 10 '15 at 17:31
• As $f$ is continuous on $[0,1]$, you can prove that $\lim\limits_{\epsilon \to 0} V_\epsilon^1(f) = V_0^1(f)$ which is a finite number as $V_\epsilon^1(f) \le \int_0^1 \vert f^\prime(t) \vert dt$ for all $\epsilon$. – mathcounterexamples.net Dec 10 '15 at 17:58