I have been doing some excercises on total variation when the following questions came up to my mind:

(1) Let $f$ be continuous on the interval $[0,1]$ and be of bounded variation. Is it true that its total vatiation function $TV(f_{[0,x]})$ is uniformly continuous? i.e. Is it true that for $\forall\space\epsilon>0$, $\exists\space\delta>0$, such that for arbitrary interval $[a,b]$ with $|b-a|<\delta$, we have $TV(f_{[a,b]})<\epsilon$?

(2) Alternatively, if it is not uniformly continuous, can I say that for $\forall\space{}x\in[0,1]\text{ and } \forall\space\epsilon>0$ , $\exists\space\text{ nondegenerate interval }I \text{ such that }x\in{}I\text{ and } TV(f_{I})<\epsilon$?

Thank you!

  • 1
    $\begingroup$ Hint: What is the total variation function of a monotonous function? $\endgroup$
    – Did
    May 12, 2012 at 11:46
  • $\begingroup$ @Didier I think it should be the absolute value of the difference of end point values? But I still don't see how I can use that on this question... $\endgroup$
    – Vokram
    May 12, 2012 at 12:16
  • $\begingroup$ Hence, for example, if $f(0)=0$ and $f$ is nondecreasing and if $g(x)=$TV$(f_{[0,x]})$, then $g=\ldots$ $\endgroup$
    – Did
    May 12, 2012 at 12:53
  • $\begingroup$ @Didier yeah in that case $g$ would be continuous. Hmm but sorry if I'm getting your point really slow here, I did not assume f is monotone in the question. Are you saying that I should use Jordan's decomposition and express it as the difference of two such functions? But I think that way I will get an inequality instead of an equality and there's no guarantee they are continuous any more $\endgroup$
    – Vokram
    May 12, 2012 at 13:10
  • $\begingroup$ This shows is that the regularity of $x\mapsto$TV$(f_{[0,x]})$ is at most the regularity of $f$, in general. $\endgroup$
    – Did
    May 12, 2012 at 13:31

1 Answer 1


Since a continuous function on a compact set is uniformly continous, you're actually asking whether the total variation of a continuous function is continuous.

Let $\epsilon\gt0$ and $x_0\in(0,1]$ be given. Since the total variation is non-decreasing, it suffices to find a point $x\lt x_0$ with $TV(f_{[x,x_0]})\lt\epsilon$ to show that the total variation is left-continuous, and thus by symmetry also right-continuous and hence continuous.

Pick some point $x_1\lt x_0$. By definition there is a partition of $[x_1,x_0]$ such that the sum of absolute differences of function values over the partition is within $\epsilon/2$ of $TV(f_{[x_1,x_0]})$. Since $f$ is continuous, we can find a point $x$ between the last intermediate point of the partition and $x_0$ such that $|f(x)-f(x_0)|\lt\epsilon/2$. Refining the partition with this point doesn't decrease its sum of absolute differences. Now the sum of absolute differences in the partition up to $x$ is within $\epsilon/2+\epsilon/2=\epsilon$ of $TV(f_{[x_1,x_0]})$, and thus so is $TV(f_{[x_1,x]})$; hence $TV(f_{[x,x_0]})\lt\epsilon$ as required.

  • 6
    $\begingroup$ Nice! However, I had hard time understanding the last sentence, so maybe it is worth to supply a few more details as follows: Let $V$ denote the variation over the partition chosen within $\epsilon/2$ of $TV(f_{[x_1,x_0]})$ as described above. Then $$TV(f_{[x_1,x_0]})<V+\epsilon/2.$$ Also, $V-|f(x)-f(x_0)|$ is some variation over interval $[x_1,x]$, so we have: $$TV(f_{[x_1,x]})\ge V-|f(x)-f(x_0)|.$$ Finally, $TV(f_{[x_1,x_0]})=TV(f_{[x_1,x]})+TV(f_{[x,x_0]})$, so that: $$TV(f_{[x,x_0]}) = TV(f_{[x_1,x_0]}) - TV(f_{[x_1,x]})<(V+\epsilon/2) - (V-|f(x)-f(x_0)|)<\epsilon/2+\epsilon/2=\epsilon.$$ $\endgroup$
    – mathreader
    Jul 7, 2013 at 0:09
  • 1
    $\begingroup$ Perhaps the downvoter would care to explain the downvote? $\endgroup$
    – joriki
    Jul 27, 2016 at 22:55
  • 1
    $\begingroup$ @mathreader actually $V-|f(x)-f(x_0)|$ is no more than some variation over $[x_1, x]$. But the proof follows the way you said. +1 for the comment $\endgroup$
    – Filburt
    Feb 19, 2017 at 20:31
  • 1
    $\begingroup$ @Brofessor, Fiburt: Now I see that there is a little mistake in my comment of 6 years ago. The sentence after the first formula should read: "Also, $V+|f(x)-f(x_0)|$ is some variation over interval $[x_1,x]$...". Indeed, $V$ is some variation over interval $[x_1,x_0]$, and we add to it the difference of values of $f$ on $x_0,x$ and so we get some variation over $[x_1,x]$. So the next formula should read: $$TV(f_{[x_1,x]})\ge V+|f(x)-f(x_0)|.$$ But of course, $V+|f(x)-f(x_0)|\ge V-|f(x)-f(x_0)|$, so the second inequality is still valid, as is everything after it. Thanks for noticing! $\endgroup$
    – mathreader
    Jun 22, 2019 at 14:36
  • 2
    $\begingroup$ @Brofessor You are right, I mixed up the notation. Let me start anew. Let $P'=\{x_1,\dots,x_k,x_0\}$ be a partition such that its variation $V'=\sum_{j=1}^{k-1}|f(x_j)-f(x_{j+1})|+|f(x_k)-f(x_0)|$ has the property that $TV(f_{[x_1,x_0]})<V'+\epsilon/2$. We refine partition $P'$ by introducing a new point $x$ in between $x_k$ and $x_0$. The new variation $V=V'-|f(x_k)-f(x_0)|+|f(x_k)-f(x)|+|f(x)-f(x_0)|\ge V'$ by the triangle inequality. But now $V-|f(x_0)-f(x)|$ is some variation over $[x_1,x]$, so it is no bigger than the total variation on that interval. Thus, the second inequality holds. $\endgroup$
    – mathreader
    Jun 23, 2019 at 7:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .