It is known that a function with bounded variation is differentiable almost everywhere.
There are also functions with unbounded variation that are differentiable almost everywhere (e.g. take $f:[0,1]\rightarrow\mathbb{R}$ with $f(0)=0$ and $f(x)=1/x$ otherwise).
But does there exist a function on $[0,1]$ with unbounded variation that is everywhere differentiable?
Best Answer
Consider $f:[0,1]\to\mathbb{R}$ by $f(x)=x^{2}\sin(\frac{1}{x^{2}})$ for $x\neq0$ and $f(0)=0$.
$\int_{0}^{1}\lvert f'(x)\rvert dx=\int_{0}^{1}\lvert2x\sin(\frac{1}{x^{2}})-\frac{2}{x}\cos(\frac{1}{x^{2}})\rvert\ge\int_{0}^{1}\frac{2}{x}\lvert\cos(\frac{1}{x^{2}})\rvert dx-\int_{0}^{1}2x\vert\sin(\frac{1}{x^{2}})\rvert$
$\ge\int_{0}^{1}\frac{2}{x}\lvert\cos(\frac{1}{x^{2}})\rvert dx-1$.