A continuous function on $[0,1]$ not of bounded variation

Question

I'm looking for a continuous function $f$ defined on the compact interval $[0,1]$ which is not of bounded variation.

I think such function might exist. Any idea?

Of course the function $f$ such that $$ f(x) = \begin{cases} 1 & \text{if $x \in [0,1] \cap \mathbb{Q}$} \\\\ 0 & \text{if $x \notin [0,1] \cap \mathbb{Q}$} \end{cases} $$ is not of bounded variation on $[0,1]$, but it is not continuous on $[0,1]$.

Answer 1

Consider any continuous function passing through the points $(\frac1{2n},\frac1n)$ and $(\frac1{2n+1},0)$, e.g. composed of linear segments. It must have infinite variation because $\sum\frac1n=\infty$.

Answer 2

For example $x\sin(1/x)$ (and 0 for $x=0$).

The variation is unbounded, because it is bounded ;-) from below by the sum of absolute values of extrema, but it is the harmonic-like series.

Answer 3

The Weierstrass Function restricted to the interval $[0,1]$

Answer 4

Here is an general idea. A function of bounded variation on $[0,1]$ is necessary differentiable a.e.. In another word, any continuous function which is not differentiable on a positive measure set is not of bounded variation. So, Weierstrass Function is not of bounded variation for sure.

Also, some function which oscillate too much can not be bounded variation as well, for example $$ u(x)=x^a\sin(\frac{1}{x^b}) $$ is not of bounded variation as long as $1 \le a\leq b$.