3
$\begingroup$

Let ${C_n}$ be a family of curves (on the unit square $M$) such that $C_\infty$ is a space filling curve and $f(x,y)$ a function of two variables. Does the following identity hold? $$lim_{n\to\infty}\int_{C_n}f(x,y)\,ds=\iint_Mf(x,y)\,dx\,dy$$

Can it even be computed? The above identity is surely incorrect but the idea is that if we interpret the line integral {double integral} as the area {volume} between the curve {unit square} and the function ant take the limit, when $C_n$ "approaches" the area of the unit square, the line integral should approach the volume over it

Thanks in advance

  • 0
    I'm not sure, but I think the first limit is $+\infty$ at least in some cases. Take, as an example, the Peano curve and take $f=1$. The length of the curves approximating the Peano curve should converge to the area of the square. Though, at any step the Peano approximations take more twists, gaining more length and going eventually to $\infty$. In fact, if I remember correctly, any two points on the Peano curve have infinite "distance" (i.e. the piece of the curve between those two points has infinite length): When dealing with fractals, things go pretty crazy. – 2017-04-29
  • 0
    @Harnak thanks for the comment, I suspected something like that but still does it make sense to ask this kind of question? Maybe some modified identity involving the integrals works, like "area" to "volume" (integrals) just as the step from "line/length" to "area" (space-filling curve) – 2017-04-29
  • 0
    Well, the problem with this is that you're "summing" lengths without any weight. For example, if you take a unit square and take all the vertical lines in the square, those will have length $1$, but a straight summation (well, you'd also have to specifiy in which sense you're summing) will always turn out to be infinite. To perform a proper summation you should integrate all vertical lines. Euristically this means that each length is multiplied by an infinitesimal $dx$ which allows you to step from lengths to ares. – 2017-04-29
  • 0
    @Harnak do you mean infinitesimal "width" given to the line? – 2017-04-29
  • 0
    To add to previous comment: if you "multiply" the lengths of the vertical lines by $dx$ you have now rectangles with infinitesimal sides, which you can sum through integration. That's the different with a straight summation, which will just take the same number (the length) and sum it over and over, everntually going to infinity. So even if your Peano curve had finite length, you'd have to think of a way to sum things properly there, because you can't hope to get areas out of lengths, without an integration. I hope I made clear what I mean. – 2017-04-29
  • 0
    yes, I mean exactly that. – 2017-04-29
  • 0
    @Harnak I don't see how that can get to the relation I want. It would give you a volume but how do you deal with corners in the curve? – 2017-04-29
  • 0
    @Harnak nevermind, corners are not a problem :) – 2017-04-29
  • 0
    Well I'm not sure myself on how to pose this problem. I was just pointing out that "summing straight lengths" is not a good strategy to try obtain an area. Though, if you come up with some idea, I'm curious about it! :) – 2017-04-29

0 Answers 0