# Is there a function whose total variation is unbounded but its restricted variation (mesh converging to zero) is bounded?

Suppose $$f$$ is a bounded function on $$[a,b]$$, its total variation is defined to be $$\mathrm{Var}(f) = \sup_{\tau} \sum_{i=1}^{n} |f(x_{i})-f(x_{i-1})|.$$ Furthermore, if $$f$$ is continuous on $$[a,b]$$, it can be proved that $$\mathrm{Var}(f) = \lim_{|\tau|\to 0}\sum_{i=1}^{n} |f(x_{i})-f(x_{i-1})|,$$ where $$|\tau|$$ denotes the mesh of the partition $$\tau$$.

When $$f$$ is discontinuous, can anyone give a counterexample to the above equality, that is the total variation cannot always be calculated by letting the mesh approaching zero?

• If the limit exists then it is equal to the total variation. I don't know if there is an example where the limits does not exist. Apr 5, 2019 at 12:07
• The question asked in the title is different that the one in the main formulation.
– Hayk
Apr 17, 2019 at 15:59

To see that the limit and supremum in question might produce difference results define $$f:[0,1]\to[0,1]$$ as $$f(x) = \begin{cases} 0, &\text{ if } \ x \neq 1/2, \\ 1, &\text{ if } \ x = 1/2. \end{cases}$$ Clearly $$\mathrm{Var}(f) = 2$$, but any partition of $$[0,1]$$, with diameter however small, will result in $$0$$ variance-sum unless it includes the point $$1/2$$ in the partition of $$[0,1]$$.
• @C.Davide, if I understand you correctly, there cannot be a counterexample that is partition independent. This is because the $\sup$ itself is realized on a partition with diameter approaching to $0$. Indeed, notice that $\sup$ is a limit of some sequence of partitions. But for any partition of $[a,b]$ adding more points to it can only increase the variation-sum. Hence, WLOG, we may assume that the sequence of partitions realizing the $\sup$ has diameter converging to $0$. Thus, there is at least one sequence of partitions (meshes) with decaying diameter, that coincides with the $\sup$.