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As far as I know, a function $f$ defined on an interval $[a, b]$ is said to be of bounded variation if $$\tag{1}V_a^b(f)=\sup\left\{\sum_{P} \lvert f(x_{j+1})-f(x_j)\rvert \ :\ P\ \text{partition of }[a, b]\right\}<\infty.$$ Today I discovered that another definition is in use for a function defined in an open subset $\Omega$ of $\mathbb{R}^n$, namely (cfr. Wikipedia) we say that $f$ is of bounded variation if $$\tag{2}V(f; \Omega)=\sup\left\{ \int_{\Omega}f(x)\,\text{div}\,\phi (x)\, dx\ :\ \phi\in C^1_c(\Omega),\ \lVert \phi\rVert_{\infty}\le 1 \right\}<\infty.$$

Even if the cited Wikipedia article treats the two definitions as if they were equivalent when $\Omega=(a, b)$, this does not seem to me to be the case. The Dirichlet function $\chi_{\mathbb{Q}\cap [0, 1]}$ is not of bounded variation in $(0, 1)$ in the sense of definition (1) but it is in the sense of definition (2).

Question. What is the precise relationship between the two definitions?

Thank you for reading.

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The 2nd definition operates with equivalence classes up to equality a.e. I think in one dimension it corresponds to having a representative that is BV in the first sense. – user53153 Feb 5 '13 at 4:12
@5PM: You are probably right. If you take as $\phi$ a step function then the integral in $(2)$ reduces to a sum similar to the one in $(1)$ (but without the absolute value). Maybe playing with this observation one can show the equivalence of (1) and (2) up to a.e. equality. It would be nice to prove this thoroughly and then add it to the Wikipedia page which is a bit confusing as it is now. – Giuseppe Negro Feb 5 '13 at 19:12

1 Answer 1

up vote 14 down vote accepted

The book A First Course in Sobolev Spaces by Giovanni Leoni is a very helpful resource for calculus aspects of function spaces. The first seven chapters deal with functions of one real variable. Chapter 2 is called "Functions of Bounded Pointwise Variation" which are defined by (1) and this class is denoted $BPV(I)$, rather than $BV(I)$. Here $I$ is the interval of definition. The author comments in a footnote on pp.39-40:

Although we do not like changing standard notation, unfortunately in the literature the notation $BV(I)$ is also used for a quite different (although strictly related) function space. This book studies both spaces, so we really had to change the notation for one of them.

Section 7.1 is titled "$BV(\Omega)$ Versus $BPV(\Omega)$". Here, $\Omega$ is an open subset of $\mathbb R$ (we are still in one dimension), and $BV(\Omega)$ is defined as the class of integrable functions $u\in L^1(\Omega)$ for which there is a finite signed Radon measure $\lambda$ such that $\int u\varphi'=-\int \varphi\,d\lambda$ for all $\phi\in C_c^1(\Omega)$. This is somewhat different from, but equivalent to (2): one direction is trivial and the other is a form of Riesz representation.

Theorem 7.2. Let $\Omega\subset\mathbb R$ be an open set. If $u:\Omega\to\mathbb R$ is integrable and if it belongs to $BPV(\Omega)$, then $u\in BV(\Omega)$ and $$|Du|(\Omega)\le \operatorname{Var}u$$ Conversely, if $u\in BV(\Omega)$, then $u$ admits a right continuous representative $\bar u$ in $BPV(\Omega)$ such that $$ \operatorname{Var}\bar u = |Du|(\Omega)$$

This is proved thoroughly indeed; the proof takes three pages (pp. 216-218).

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Great! You nailed it completely. Thank you very much! – Giuseppe Negro Feb 6 '13 at 18:50
This answer deserves a lot more upvotes than just one. – Giuseppe Negro Feb 7 '13 at 1:54
@GiuseppeNegro Welcome to Math.SE! I see you are new here. :) – user53153 Feb 7 '13 at 5:09

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