function of not bounded variation

Example.  We show that the function

$f:x\mapsto$

which is continuous in the whole $ℝ$, is not of bounded variation on any interval containing the zero.

Let us take e.g. the interval  $[0,a]$.  Chose a positive integer $m$ such that   and the partition of the interval with the points   $\frac{1}{m},\frac{1}{m+1},\frac{1}{m+2},\mathrm{\dots },\frac{1}{n}$  into the subintervals$[0,\frac{1}{n}],[\frac{1}{n},\frac{1}{n-1}],\mathrm{\dots },[\frac{1}{m+1},\frac{1}{m}],[\frac{1}{m},a]$.  For each positive integer $\nu$ we have (see this (http://planetmath.org/CosineAtMultiplesOfStraightAngle))

$$f\left(\frac{1}{\nu }\right)=\frac{1}{\nu }\cos \nu \pi =\frac{{(-1)}^{\nu }}{\nu }.$$

Thus we see that the total variation of $f$ in all partitions of  $[0,a]$  is at least

$$\frac{1}{n}+\left(\frac{1}{n}+\frac{1}{n-1}\right)+\mathrm{\dots }+\left(\frac{1}{m+1}+\frac{1}{m}\right)=\frac{1}{m}+2\sum _{\nu =m+1}^n\frac{1}{\nu }.$$

Since the harmonic series diverges, the above sum increases to $\mathrm{\infty }$ as  $n\to \mathrm{\infty }$.  Accordingly, the total variation must be infinite, and the function $f$ is not of bounded variation on  $[0,a]$.

It is not difficult to justify that $f$ is of bounded variation on any finite interval that does not contain 0.