# function of not bounded variation

Example.  We show that the function

 $\displaystyle f\!:\;x\mapsto$ $\displaystyle\left\{\begin{array}[]{ll}x\cos\frac{\pi}{x}&\mbox{when}\,\,x\neq 0% ,\\ 0&\mbox{when}\,\,x=0,\end{array}\right.$

which is continuous  in the whole $\mathbb{R}$, 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   $\frac{1}{m} and the partition of the interval with the points   $\frac{1}{m},\,\frac{1}{m+1},\,\frac{1}{m+2},\,\ldots,\,\frac{1}{n}$  into the subintervals $[0,\,\frac{1}{n}],\;[\frac{1}{n},\,\frac{1}{n-1}],\;\ldots,\;[\frac{1}{m+1},\,% \frac{1}{m}],\;[\frac{1}{m},\,a]$.  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

 $\frac{1}{n}\!+\!\left(\frac{1}{n}\!+\!\frac{1}{n\!-\!1}\right)\!+\ldots+\!% \left(\frac{1}{m\!+\!1}+\frac{1}{m}\right)=\frac{1}{m}+2\!\sum_{\nu=m+1}^{n}% \frac{1}{\nu}.$

Since the harmonic series diverges, the above sum increases to $\infty$ as  $n\to\infty$.  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.

## References

• 1 E. Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset III. Toinen osa.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).
Title function of not bounded variation FunctionOfNotBoundedVariation 2013-03-22 17:56:29 2013-03-22 17:56:29 pahio (2872) pahio (2872) 6 pahio (2872) Example msc 26A45 example of unbounded variation function of unbounded variation