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Given a function $f:[a,b]\rightarrow\mathbb{R}$, we define the variation of $f$ on $[a,b]$ as follows:

\begin{equation*} V_a^b(f)=\sup\left\{\sum_{i=1}^{n}{|f(t_i)-f(t_{i-1})|}\middle|P=\{t_i\}_{i=0}^n\in\mathscr{P}([a,b])\right\}. \end{equation*}

where $\mathscr{P}([a,b])$ denotes the set of partitions of the interval $[a,b]$. It's very easy to generalised this definition when the codomain is a metric space $(X,d)$. It's enough to substitute $|f(t_i)-f(t_{i-1})|$ by $d(f(t_i),f(t_{i-1}))$. We say that $f$ is a function of bounded variation if $V_a^b(f)<+\infty$.

I need to learn about functions of bounded variations defined on an open set of $\mathbb{R}^n$. Given a function $f\in\mathcal{L}^1(U,\mathbb{R})$, we define the variation of $f$ on $U$ as follows:

\begin{equation*} V_U(f)=\sup\left\{\int_{U}{f(x)\cdot\text{div}(\varphi)(x)\,dx}\middle|\varphi\in\mathcal{C}_c^1(U,\mathbb{R}),|\varphi|\leq 1\text{ on $U$}\right\}. \end{equation*}

I don't understand why this is the natural definition. Could anyone explain how to get to this definition in a natural way?

Mittens
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Raúl Filigrana Villalba
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  • If you apply the second formula and use integration by parts you recover the original definition ($n=1$). – Mittens Dec 09 '22 at 16:32
  • $f$ doesn't have to be differentiable by the definition. Anyway, what I want is like a motivation that brings me from the first definition to the second one, not the other way around. – Raúl Filigrana Villalba Dec 09 '22 at 16:44
  • It is actually going from 2nd to first that tells you why the second definition is a indeed a generalization of the usual variation in the real line. Notice that functions of finite variation in the real line are almost surely differentiable. – Mittens Dec 09 '22 at 16:55
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    Does this answer your question? Bounded variation on $\mathbb{R}$ – Mittens Dec 09 '22 at 19:29
  • The variation is not that set. The variation is the supremum of that set. – Paul Sinclair Dec 10 '22 at 17:05
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    This Q&A may shed some light on your doubts, Raúl. Intuitively, the advantage the definition of Lamberto Cesari has on the one of Camille Jordan (and its $n$-dimensional analogues) is that the former one is independent on the value of the function $f$ takes on any zero Lebesgue measure on its domain of definition: this also implies the invariance of the variation respect to coordinate changes. – Daniele Tampieri Jan 03 '23 at 09:11

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