Prove $f : [0,1] \to \mathbb{R}$, $f(x) = x^2 \sin(1/x^2) \mathbf{1}_{\{ 0 < x \le 1\}}$ is of bounded variation.

Question

I need to show the following function is of bounded variation.

$f : [0,1] \to \mathbb{R}, x \mapsto x^2 \sin(1/x^2) \mathbf{1}_{\{ 0 < x \le 1\}}$

What I tried:

Since $f$ : differentiable on $[0,1]$, we have

$$
f'(x) = \begin{cases} 2x \sin(1/x^2) – \frac{2}{x} \cos(1/x^2) & 0<x\le1 \\[7pt] 0 & x=0\end{cases}
$$

Clearly $f'$ is not bounded, so with this method I cannot say if $f$ is BV or not. Is there any help?

Answer 1

HInt: Let

$$a_n = \frac{1}{\sqrt {\pi/2+2n\pi}},\,\, b_n =\frac{1}{\sqrt {2n\pi}}.$$

Consider the sums $\sum_{n=1}^{N}|f(b_n)-f(a_n)|.$