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The original prompt of this question is an homework of mine. The homework is about the sets of finite perimeter in $\mathbb{R}$.
So I took the definition of set of finite perimeter form my notes, which is (in its general form for measurable subsets of $\mathbb{R}^n$):

$E\subset\mathbb{R}^n$ measurable set, has finite perimeter in $A\subset\mathbb{R}^n$ open, iff $\chi_{E}\in BV(A).$

Thus, I went back to the definition of function of bounded variation which is:

Let $A\subset\mathbb{R}^n$ be a fixed open set. A function $f:A\rightarrow\mathbb{R}$ is of bounded variation on $A$ iff $f\in L^{1}(A)$ and $$\infty>\vert\vert Df\vert\vert(A):=sup\{\int_{A}fdiv(\phi)dx,\ \phi\in C^{1}_c(A,\mathbb{R}^n),\ \ \vert\vert\phi\vert\vert_{\infty}\}.$$ We call $\vert\vert Df\vert\vert(A)$ the variation of $f$ in $A$.

Here is my question: why do we take $f\in L^{1}(A)$?
Observations:

  • The variation of $f$ makes sense even if $f\in L^{1}_{loc}(A)$, the restriction $f\in L^1(A)$ seems unnecessary (at least at this point).
  • I'm aware of the definition of function of local bounded variation, but this doesn't really answer the question.
  • In this way we cannot define the perimeter for all unbounded measurable sets, e.g. $\mathbb{R}$ doesn't have perimeter (in the above sense) since $\chi_{\mathbb{R}}\notin L^1(\mathbb{R})$.
  • I know the notion of reduced boundary and I know that in one dimension the perimeter of a set is the number of elements of set of the jumps of its characteristic function (of course this fact is true when the perimeter is defined). Now, given this geometric interpretetation (jumps of the characteristic function), it seems strange that the perimeter is undefined for sets like $[a,\infty)$, since their characteristic function jumps only at $a$.

Several books give this definition (or something equivalent), thus this is not a choice of my professor. There must be some reason (maybe technical) for this fact.

P.s. I tagged this question with 'bounded variation' even if it is formally wrong, because there isn't an analogue tag for functions of several variables. If it is a problem I'll remove it.

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Generally, one can't answer the question "why is the definition made this way" on the basis of the definition alone. The decision is made on the basis of what appears after the definition, possibly 100 pages after it. Some possible reasons to require $L^1$ in the definition of $BV$:

  1. It makes $BV$ a normed space, with the norm being the total variation of $Df$. If locally integrable functions were allowed, then constant functions would be included but their variation it zero.

  2. The Sobolev embedding $BV(\mathbb{R}^n)\subset L^{n/(n-1)}(\mathbb{R}^n)$ would not hold if $L^1_{\rm loc}$ functions were allowed.

You make the point that this choice excludes some sets from having finite perimeter. But this is mostly in one dimension, where the notion of perimeter (and other geometric concepts) are not of much interest. Outside of the one-dimensional case, we hardly gain anything at all. The plane $\mathbb{R}^2$ would have infinite perimeter, as would any subset of $\mathbb{R}^2$ with bounded complement.

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    $\begingroup$ I've talked also with a classmate and we agreed that the "problem" of something being excluded is only one dimensional. Thank for pointing out other interesting issues. $\endgroup$
    – Uskebasi
    May 31, 2017 at 15:09

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