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I know what it means for a function $f$ on a bounded interval $[a,b]$ to be of bounded variation. But what does it mean for $f$ to be of BV on $\mathbb{R}$? Does it mean that $\lim_{N\rightarrow\infty} V[-N,N] < \infty$? Here $V$ is the total variation of $f$.

On an interval $[a,b]$, $f$ is of bounded variation if $\sup_\Gamma \sum_{i=1}^m |f(x_i) - f(x_{i-1})| < \infty$ where the supremum is over partitions $\Gamma=\{x_0 < x_1<...<x_m \}$ of $[a,b]$.

user394438
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  • What definition are you using for being BV on $[a.b]$? – πr8 May 06 '17 at 23:59
  • Added the definition to the question – user394438 May 07 '17 at 00:02
  • Okay. There's an alternative characterisation of BV that says, for $\Omega \subset \mathbb{R}$, the total variation of $f$ on $\Omega$ is given by $|f|{BV} := \sup { \int{\Omega} f(x)g'(x) : |g| \le 1 \text{ on } \Omega, g \text{ is smooth} }$, and then the BV space on the domain $\Omega$ is the set of $f$ such that $|f|_{BV}$ is finite. This definition is good in that it generalises instantly to your case ($\Omega = \mathbb{R}$) and, with a bit of careful consideration, to $\Omega \subset \mathbb{R}^n$. – πr8 May 07 '17 at 00:08

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It means that $$\sup\sum_{i=1}^n|f(x_i)-f(x_{i-1})|<\infty,$$ where the supremum is taken over all possible $x_0<\dots<x_n$. And yes it is equivalent to the limit condition you wrote.

Gio67
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  • Is there any place that proves this (the equivalence), or follows directly by the definition? – Math May 01 '19 at 01:11
  • I don’t understand the question. What I wrote is the definition of a function of bounded variation defined on the real line – Gio67 May 01 '19 at 14:21
  • Yes, I agree with you. What I wanted to know is how to prove that your definition is equivalent to $\lim_{N \rightarrow \infty}V[-N,N]<\infty$. – Math May 01 '19 at 14:26
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    If you take a partition for the real line $x_0<\cdots<x_n$ you can find $N$ so large that $[-N,N]$ contains all the points in the partition. So $\sum |f(x_i)-f(x_{i-1})|\le V[-N,N]$. Since $V$ is increasing, $\sum |f(x_i)-f(x_{i-1})|\le \lim_NV[-N,N]$. Take the sup over all partitions and you get one inequality. The converse comes from the fact that any partition in $[-N,N]$ is a partition for the real line – Gio67 May 02 '19 at 11:14