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Are all monotonic of bounded variation, or would there exist a counterexample of a function that is monotonic but not of bounded variation?

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    There is even a characterization: a function on a closed interval is of bounded variation if and only if it is a difference of monotonic increasing functions. – Thomas Andrews Apr 25 '21 at 03:45
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    The only such examples would be functions on an infinite domain, like $f(x)=x$ for $-\infty \lt x \lt \infty$. – herb steinberg Apr 25 '21 at 03:48
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    On an unbounded region, like $[0,\infty)$ you need the function to be bounded for it to be of bounded variation. Same for open intervals. $f(x)=-1/x$ is monotonic increasing on $(0,1)$ but is not of bounded variation. – Thomas Andrews Apr 25 '21 at 03:50
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    @herbsteinberg There are also examples defined only on open intervals. In any non-compact interval, you need to add the condition that $f$ is bounded. – Thomas Andrews Apr 25 '21 at 03:52

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A function defined on a closed interval which is monotonic is indeed of bounded variation. If $f$ is monotonic increasing on $[a,b]$ then the total variation is always $f(b)-f(a).$

Indeed, there is a characterization that a function on $[a,b]$ is of bounded variation if and only if it can be written as a difference of two monotonic increasing functions.

For other intervals, you need to add the condition that the function is bounded.

For example, on $(0,1)$ the function $g(x)=-\frac1x$ is monotonic increasing, but not of bounded variation.

On $\mathbb R,$ $h(x)=x$ is increasing and not of bounded variation.

Thomas Andrews
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