Let $f: [-1, 1]$ be defined by $f(x) = x\sin(\frac{1}{x})$ when $ x \neq 0$ and $f(0) = 0$. Prove that there do not exist increasing functions $g$ and $h$, defined on $[-1, 1]$ such that $f(x)=g(x)-h(x)$ for every $x$.
So I have tried applying criteria for differentiability and integrability for $f$ and found that it is not differentiable at $x=0$, however the function is Riemann integrable. I know that monotone functions are always Riemann integrable, but not too sure whether this is the correct path in proving this statement.
Also, I have tried reasoning by noting that $x_n = \frac{1}{n \pi}$ always yields $f(x_n) = 0$, but it seems possible to construct increasing functions such that they intersect infinitely many times but perhaps $f$ doesn't behave too well that this can't be done? I am not too sure about this approach and have not gotten far with it.
Any help would be greatly appreciated - thanks!