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A function $f:[a,b]\to\mathbb{R}$ is said to be of bounded variation if $$ \sup_I \sum_{i=1}^n |f(x_i)-f(x_{i-1})| \le V $$ for some finite $V>0$, where the supremum is over all finite partitions $I=(a=x_0<x_1<\ldots<x_n=b)$ of $[a,b]$. I am looking for metric-space analogues of this notion.

One natural idea is to consider a function $f:X\to\mathbb{R}$, where $(X,d)$ is a bounded metric space. We can define $f$ to be of bounded variation if $$ \sup_{\mathcal{P}} \sum_{A\in\mathcal{P}} (\sup_A f-\inf_A f) \le V,$$ where $\mathcal{P}$ is the collection of all finite partitions of $X$. Has this notion been studied somewhere?

EDIT: Following @August Cleaner's comment: of course something is missing from the above: the metric! I should have defined it as something like $$ \sup_{\epsilon>0}\sup_{\mathcal{P}_\epsilon} \sum_{A\in\mathcal{P}_\epsilon} (\sup_A f-\inf_A f) \le V,$$ where $\mathcal{P}_\epsilon$ is a finite partition where each block has diameter at most $\epsilon$. That at least agrees with the standard definition on $[a,b]$, I think.

EDIT II: Following Martin Hairer's example, here's a third attempt. For $x_1,\ldots,x_n\in X$, we can define the Voronoi partition induced by this finite set: every $x\in X$ is associated with the closest $x_i$. [There's the issue of how to break ties, but we're going to consider all such partitions and hence all possible tie-breaking schemes.] Let $\mathcal{P}_n$ be the collection of all Voronoi partitions induced by $n$ points, and say that $f:X\to\mathbb{R}$ has bounded variation if $$ \sup_{n>0}\sup_{\mathcal{P}_n} \sum_{A\in\mathcal{P}_n} (\sup_A f-\inf_A f) \le V.$$ Any comments on this definition?

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    $\begingroup$ I turn your attention to the fact that the suggested notion is quite different from the classical, e.g. the function $y=x$ on $[0,1]$ has infinite variation. Also this notion does not depend on the metric. $\endgroup$
    – August Cleaner
    Commented Mar 22, 2016 at 10:14
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    $\begingroup$ Blocks in a partition don't need to be connected. In particular, one can easily have a partition of $[0,1]$ into $1/\epsilon^2$ blocks of diameter $\epsilon$, so the identity map still has infinite variation with the new definition. There is a notion of BV for functions of several real variables (see Wikipedia), but I am not sure that it generalises well. One naive generalisation would be to ask that $f\circ g$ is BV, uniformly over all $g \colon \mathbb{R} \to X$ having Lipschitz constant $1$. $\endgroup$
    – Martin Hairer
    Commented Mar 22, 2016 at 13:02
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    $\begingroup$ There is some notion of BV functions for metric measure spaces, which seems more natural when looking at the definition for BV in several real variables. Google gives a bunch of results, two of which are cvgmt.sns.it/paper/2738 and sciencedirect.com/science/article/pii/S0021782403000369 $\endgroup$
    – Willie Wong
    Commented Mar 22, 2016 at 15:13
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    $\begingroup$ For your EDIT II: let your metric space be $[0,1]\times[0,1]$ and let $f(x,y) = x$. Let $n > 1$. Take $(x_i,y_i) = (0, i / (n-1))$ for $i < n$. Then for the $\mathcal{P}_n$ adapted to $\{(x_i,y_i)\}_{i = 0, \ldots, n-1}$ you have that $$ \sum_{A\in \mathcal{P}_n} (\sup_A f - \inf_A f) = n $$ So after taking the sup in $n$ you get that this function has infinite total variation by your definition. $\endgroup$
    – Willie Wong
    Commented Mar 22, 2016 at 15:17
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    $\begingroup$ Maybe a better way to go about it is to recall that $f\in BV$ if and only if (the distribution) $f'$ is a measure, and then generalize this condition. $\endgroup$
    – Christian Remling
    Commented Mar 22, 2016 at 20:11

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