Is $x^2\cos\left(\frac{1}{x}\right)$ of bounded variation on $[-1,1]$?

Question

Define $$g(x) = \begin{cases} x^2\cos\left(\frac{1}{x}\right) &\mbox{if } x \neq 0\\ 0 & \mbox{if } x = 0. \end{cases} $$
Is g of bounded variation on $[-1,1]$.

My attempt:-To show that $g$ is bounded variation its enough to show that its Lipschitz since class of Lipschitz is inside class of bounded variation . its clear that $g$ is continuous on $[-1,1]$ and we have $g^{\prime}(x)=2x\cos(1/x)+\sin(1/x)$. in addition we have $|g^{\prime}(x)|\le 3$ for $-1<x<1$. that is the upper and lower derivatives are bounded on $(-1,1)$ hence $g$ is Lipschitz hence its bounded variation.

My Concern:- I am not sure that the derivatives of the function g(x) is exist at 0. any help and suggestion ?

Answer 1

The derivative $g'$ exists on $[-1,1]$, $\lim\limits_{h \to 0}\dfrac{g(0+h)-g(0)}{h}=\lim\limits_{h \to 0} h\cos(\frac{1}{h}) = 0$.