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(Am writing this post in a rush, out of office, so cannot give adequate links etc right now.)

There is a classical and well-understood definition of what it means for a continuous function $f:[a,b]\to{\mathbb R}$ to be absolutely continuous: we should have $$ \sup \{ \sum_{i=1}^n (t_i-t_{i-1}) \vert f(t_i)-f(t_{i-1}) \vert \} < \infty $$ where the supremum is over all partitions $a=t_0 < t_1 < \dots < t_n=b$.

From my limited reading on Wikipedia and various other online searches, it transpires that there are several different notions of "functions of bounded variation" defined on higher-dimensional "domains", with the quotes indicating that I am uncertain of the precise technical qualifiers. (The setting given in Wikipedia is for open subsets of ${\mathbb R}^n$; googling has also shown links to papers defining BV for Riemannian manifolds.)

It is not clear to me how these definitions in higher dimensions could be approached in a naive way via approximating sums, as in one variable. So my rather naive question is: can such a definition be made to work on, say, the $2$-sphere? What about higher spheres?

Also, since BV/AC are metric notions and not just topological ones, would I get definitions closer to the one-dimensional case if I worked with a suitable polyhedron or a triangulation of the $2$-sphere? (and if so, does such an approach obviously generalise to higher spheres?)

(The motivation for this question is slightly indirect, and I may try to give more context when I am next in the office with free time and working brain.)

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  • $\begingroup$ I tend to prefer to think of AC as a fact about measures and BV as a fact about functions and by stretching the definitions a bit you get them to coincide on intervals. Reading your question it feels like you are asking about something different still. A motivation perhaps will be very much in order. (Also, I guess since you are tagging metric geometry, the naive way of BV along all geodesics is out.) Lastly, intuitively I would think that using triangulations, the scaling would be wrong. (If you take differences between adjacent vertices and add them up, the sup would be $\infty$ always.) $\endgroup$
    – Willie Wong
    Commented Mar 7, 2012 at 10:12
  • $\begingroup$ Yemon, you talk about metric and topological notions, but what about measure-theoretic, à la Radon-Nikodym? $\endgroup$
    – Tom Leinster
    Commented Mar 7, 2012 at 13:43
  • $\begingroup$ To me BV consists of absolutely integrable functions on a finite dimensional domain whose distributional partial derivatives exist and are measures with total bounded variation. BV spaces are interesting but difficult to work with. Pelczynski and various collaborators (including me one time) have studied BV this millennium--just do an author search on mathscinet for Pelczynski, A*. $\endgroup$
    – Bill Johnson
    Commented Mar 7, 2012 at 21:57
  • $\begingroup$ Thanks for the comments - I will try to edit the post to clarify my own muddled thinking and come up with something better-posed. $\endgroup$
    – Yemon Choi
    Commented Mar 7, 2012 at 23:46

1 Answer 1

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There is a difference between BV and AC. AC functions on an interval are integrable functions whose distributional derivatives are representable by integrable functions. BV function on an interval are integrable functions whose distributional derivatives are distributions representable by (signed) measures.

We can then define the space AC functions on a Riemann manifold to be the Sobolev space $W^{1,1}$ (one derivative in $L^1$), i.e., integrable functions with integrable gradients. These functions are definitely not continuous so they may not be as appealing to you.

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