4
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kavi Rama Murthy
    Apr 5, 2019 at 12:07
  • 2
    $\begingroup$ The question asked in the title is different that the one in the main formulation. $\endgroup$
    – Hayk
    Apr 17, 2019 at 15:59

1 Answer 1

Reset to default
4
$\begingroup$

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 $[0,1]$.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for this simple example. But the limit here is mesh-dependent, how about mesh-independent counterexample? Is there such a function : (1) for any mesh appproaching zero, its restricted variation exists. (2) But its true total variation is quite different from the restricted variaton in (1). $\endgroup$
    – C. Davide
    Apr 20, 2019 at 6:58
  • $\begingroup$ @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 $\sup$. $\endgroup$
    – Hayk
    Apr 20, 2019 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.