Let $f:\mathbb{R} \to \mathbb{R}$ be a function. Define the total variation function $V_f:[0,T] \times [0,T] \to \mathbb{R}$ over an interval $[s,s’]$ as usual: $$V_f(s,s’) = \sup_{D \subset [s,s’]} \sum_{j \in D}\lvert f(t_j)-f(t_{j-1}) \rvert$$

for all $s,s’$ in $[0,T]$. Here, $D = \{t_0, t_1, \ldots, t_n\}$, with $t_0=0$, $t_n=T$ and $t_j > t_{j-1}$ is referred to as a partition of the interval $[0,T] $.

We refer to $f$ as being of bounded variation over $[0,T]$ if $V_f(0,T)<\infty$. Moreover, we say that $V_f$ is Lipschitz if there exists a constant $C$ such that $V(s,s’) \leq C \lvert s-s’\rvert $ for all $s,s’$ in $[0,T]$.

Question: for $f$ continuous of bounded variation, is the total variation function $V_f$ Lipschitz?

I have two clues. First, a paper I saw shows that any uniformly continuous function $g$ is Lipschitz plus an epsilon, i.e. for all $\epsilon$ there exists a $C_\epsilon$ such that:

$g(x,y) \leq C_\epsilon \lvert x-y\rvert + \epsilon$

Since the total variation function of a continuous function is itself continuous (I do not know where I know this fact from, but I am quite sure it is true) on a compact, it is uniformly continuous. This is not quite what I need for my research, but it might be nice-enough that I can make things work.

The second clue is trying to make things work directly. Note that the total variation behaves, in some sense, like the arc length. This particular sense is that, if the arc length of the function is finite, then so is the total variation (an easy proof). Moreover, for any given partition $D$, the arc length of a function over any subinterval is finite. Hence, the bounded variation is finite. Hence, for that fixed partition it is easy to find a constant such that $V(t_j, t_{j-1}) \leq C_D \lvert t_j - t_{j-1} \rvert$ holds for all $j$. In fact, The following $C_D$ will do the job:

$$C_D= \sup_{j \in D} \frac{V(t_j, t_{j-1})}{ \lvert t_j - t_{j-1} \rvert} $$

The question I am asking would be thus reduced to: does the $C_D$ constant explode when we take supremum over all $D$? That is, whether the following statement is true:

$$\sup_{D \subset [0,T]} C_D < \infty$$

PS: if there was any value to the arc length detour, so to speak, is that, personally, it helps me visualise what is going on. This is because the arc length is the length of the path described by the function, and the arc length is closely related to the total variation.

EDIT: in the original question, I had forgotten to add that the function $f$ should be continuous.


1 Answer 1


$f(x)=\sqrt x$ for $0<x<1$, $f(x)=0$ for $x \leq 0$ and $f(x)=1$ for $x \geq 1$ defines a function of bounded variation whose total variation is not Lipschitz. Since $f$ is increasing on $(0,1)$ it is its own total variation on this interval and $f$ is not Lipschitz.

  • $\begingroup$ Thanks, note that your function is not continuous at $x=1 = T$. I forgot to add function should be continuous of bounded total variation, in regards to the question I am interested to answer. $\endgroup$ Sep 28, 2021 at 0:03
  • $\begingroup$ It is not proper to change a question after an answer is posted. Anyway, there was a trivial modification needed and I have edited my answer. @MartinGeller $\endgroup$ Sep 28, 2021 at 0:12
  • $\begingroup$ Yes, sorry about that. It felt like reposting the whole question was a bit too much . It would be good if you could briefly explain why the total variation of your function is not Lipschitz. $\endgroup$ Sep 28, 2021 at 0:17
  • $\begingroup$ @MartinGeller No problem. I have edited again to say why the example works. $\endgroup$ Sep 28, 2021 at 0:21

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