Example. We show that the function \begin{eqnarray*} f\!:\; x\mapsto\! & \left\{ \begin {array}{ll} x\cos\frac{\pi}{x} & \mbox{when}\,\,x \neq 0,\\ 0 & \mbox{when}\,\, x = 0, \end{array}. 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 $\frac{1}{m} < a$ 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) $$f\left(\frac{1}{\nu}\right) = \frac{1}{\nu}\cos\nu\pi = \frac{(-1)^\nu}{\nu}.$$ 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.

Bibliography

1

E. LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset III. Toinen osa. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).