Solution. Let \(V\) be the total variation of \(f\) on \([a,b]\). For any \(x\in [a,b]\) the partition \(\{a,x,b\}\) gives \[\left |f(x)-f(a)\right | \leq \left |f(x)-f(a)\right | + \left |f(b)-f(x)\right | \leq V,\] so \(\left |f(x)\right | \leq \left |f(a)\right | + V\) for every \(x\), and \(f\) is bounded.

For the example take \[f(x) = \begin {cases} x\sin \left (\pi /x\right ), & 0<x\leq 1,\\ 0, & x=0,\end {cases}\] which is continuous on \([0,1]\), including at \(0\) since \(\left |f(x)\right |\leq x\). At the points \(x_k = 2/(2k+1)\) the function takes the values \(\pm x_k\) alternately, so a partition through those points contributes variation at least \(\sum _k 2/(2k+1)\), which diverges. Hence the total variation is infinite.

Continuity does not imply bounded variation, and bounded variation does not imply continuity — a step function has variation equal to its jump. The two conditions are independent, which is why the class of functions of bounded variation is introduced separately.