TITLE: Examples of differentiable functions that are not of bounded variation
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: Sticking with the example $f(x) = x^2\sin (1/x^2),$ which is everywhere differentiable, I think it's easier to deal with its total variation directly from the definition. Let $a_n = 1/\sqrt {\pi/2+2n\pi},\, b_n = 1/\sqrt {2n\pi}.$ Then
$$\sum_{n=1}^{\infty}|f(b_n)-f(a_n)| = \sum_{n=1}^{\infty}|f(a_n)|= \sum_{n=1}^{\infty}\frac{1}{\pi/2+2n\pi}.$$
Apply the limit comparsion test, using the terms $1/n$ for comparison, to see this series diverges. Since the total variation of $f$ on $[0,1]$ is at least the $\sup$ of all partial sums of this infinite series, we see $f$ has unbounded total variation on $[0,1].$