Show $f(x) = x^2 \sin (x^{-3/2}), x\in (0,1] $, $f(0) = 0$ is of bounded variation without using improper integral

Question

Show $f(x) = x^2 \sin (x^{-3/2}), x\in (0,1] $, $f(0) = 0$ is of bounded variation.

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Try

$\forall P = \{0 = x_0 < \cdots, < x_n = 1\}$, $\exists c_k \in (x_{k-1}, x_k)$ s.t. $f(x_k) - f(x_{k-1}) = f'(c_k) (x_k - x_{k-1})$$(\because MVT)$.

Thus,

$$ V(f, P) = \sum |f(x_k) - f(x_{k-1})| = \sum |f'(c_k)| (x_k - x_{k-1}) $$

I think the next step should allow me to prove the RHS is less than some value. However, observing the following,

$$ f'(x) = \begin{cases} 2x \sin(x^{-3/2}) - \frac{3}{2} x^{-1/2} \cos(x^{-3/2}) & (x \neq 0 ) \\ 0 & (x=0) \end{cases} $$

I have $|f'(x)| \le 2x + \frac{3}{2}x^{-1/2}$, but RHS is not bounded on $[0,1]$.

Any help, including one that excludes the use of improper integral, will be appreciated.

Answer 1

Here is the more general version of the question you are asking for:

When is $F(x)=x^a\sin(x^{-b})$ with $F(0)=0$ of bounded variation on $[0,1]$?

For your example, just take $a=2$ and $b=3/2$