Is $f(x)=x^2\cos\left(\frac{1}{x}\right)$ of bounded variation on $[-1,1]$? 
Is $f(x)=x^2\cos\left(\frac{1}{x}\right)$ with $f(0)=0$ of bounded variation on $[-1,1]$?

I know that $g(x)=x\cos\left(\frac{1}{x}\right)$ and $h(x)=x^2\cos\left(\frac{1}{x^2}\right)$ are both NOT of bounded variation, so I'm guessing that $f$ isn't either...? I just don't know which partition to take, since the proofs of the ones for $g$ and $h$ involved expressing the Variation into a harmonic series, which diverges when the number of points in the partition increases, but I don't know how to do the same for $f$. Thanks! 
 A: Hint: A function with bounded derivative is Lipschitz.
A: You can write your function as $x^2 \cos{\frac{1}{x}} = x^2 \cos{\frac{1}{x}} +3x - 3x$. This is a Jordan decomposition: $(x^2 \cos{\frac{1}{x}} +3x)' = 2x \cos{\frac{1}{x}} + \sin{\frac{1}{x}} + 3\geq -2 -1 + 3 = 0$ for all $x \in [-1,1]$, and this function must therefore be increasing. (The derivative at 0 is 3.) So we can apply Jordan's theorem and get the result.
A: I know that I am going to answer an old question, but I think the way I used to answer here is new and easy. So you can show this by using any one of the following ways:

Result$~\bf 1$: If $~f~$ is a differentiable function having bounded derivatives on $[a,b].$ Then $~f~$ is of bounded variation on $[a,b].$

Here $~f(x)=x^2\cos\left(\frac{1}{x}\right),~$ then $~f'(x)=2x\cos\left(\frac{1}{x}\right)+\sin\left(\frac{1}{x}\right)~$ for all $x\ne0~.$
So $~|f'(x)|=\left|2x\cos\left(\frac{1}{x}\right)+\sin\left(\frac{1}{x}\right)\right|\le 2\left|x\cos\left(\frac{1}{x}\right)\right|+\left|\sin\left(\frac{1}{x}\right)\right|\le 2~|x|+1~=2\cdot 1+1=3,$ as $~|\sin x|\le 1~$ and $|\cos x|\le 1~,~~\forall ~x~.$
Hence $~|f'(x)|\le 3~~,~\forall ~x\in[-1,1].$
Clearly, by the previous result, $~f~$ is of bounded variation on $[-1,1]~.$

Result$~\bf 2$: If $~f(x)=x^\alpha\cos\left(\frac{1}{x^\beta}\right),~~\forall~x\ne 0~,$ then $~f~$ is of bounded variation if and only if $~\alpha\gt\beta~.$

So in this case $~\alpha=2~$ and $~\beta=1~$. Therefore the condition  $~\alpha\gt\beta~$ satisfied and hence $~f~$ is of bounded variation on $[-1,1]~.$
