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?
Best Answer
$f$ 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.