# Functions of bounded variation on $\mathbb{R}^n$

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?

• If you apply the second formula and use integration by parts you recover the original definition ($n=1$). Dec 9, 2022 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. Dec 9, 2022 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. Dec 9, 2022 at 16:55
• Does this answer your question? Bounded variation on $\mathbb{R}$ Dec 9, 2022 at 19:29
• 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. Jan 3, 2023 at 9:11