Verifying pointwise convergence of indicator functions

  • #1
psie
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TL;DR Summary
I'm stuck at something fairly basic I think. Let ##B(r,x)## be an open ball of radius ##r## and center ##x## in ##\mathbb R^n##. It is claimed that ##\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}## pointwise as ##r\to r_0## and ##x\to x_0## on ##\mathbb R^n\setminus S(r_0,x_0)##, where ##S(r_0,x_0)## is the sphere ##\{y:|y-x_0|=r_0\}##. I am stuck showing this.
I'm reading a proof of a lemma that $$A_rf(x)=\frac1{m(B(r,x))}\int_{B(r,x)}f(y)\,dy,$$where ##m## is Lebesgue measure, is jointly continuous in ##r## and ##x## (##A## stands for average). The claim that ##\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}## on ##\mathbb R^n\setminus S(r_0,x_0)## is made in the proof. I think there are two cases to consider. Let ##y\in \mathbb R^n\setminus S(r_0,x_0)##.
  1. ##|y-x_0|<r_0##, i.e. ##\chi_{B(r_0,x_0)}(y)=1##. Is it then also true that for some sequences ##(x_n),(r_n)## that converge to ##x_0,r_0## respectively, that ##|y-x_n|<r_n## for large enough ##n##? Why? If yes, then ##\chi_{B(r_n,x_n)}(y)=1## for large enough ##n## too.
  2. Similarly, if ##|y-x_0|>r_0##, is it then true that ##|y-x_n|>r_n##?
Also, why does ##\chi_{B(r,x)}\not\to\chi_{B(r_0,x_0)}## on the sphere ##S(r_0,x_0)##?
 
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  • #2
My intuition says 1) yes, 2) possibly, and 3) because you cannot avoid the jump from ##0## to ##1## on the sphere as there is no neighborhood completely contained in either region. However, I have to consider the triangle inequalities in detail. I think the answer to 3) is the key to the first two questions.
 
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  • #3
fresh_42 said:
My intuition says 1) yes, 2) possibly, and 3) because you cannot avoid the jump from ##0## to ##1## on the sphere as there is no neighborhood completely contained in either region. However, I have to consider the triangle inequalities in detail. I think the answer to 3) is the key to the first two questions.
Ok. I agree, I don't see how to derive ##|y-x_n|<r_n## from the triangle inequality given ##|y-x_0|<r_0## and ##x_n\to x_0,r_n\to r_0##. It doesn't lead anywhere: $$|y-x_n|\leq |y-x_0|+|x_0-x_n|<r_0+\epsilon.$$Here ##|x_0-x_n|<\epsilon## for sufficiently large ##n##. But I don't see how else to show $$\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}$$pointwise on ##\mathbb R^n\setminus S(r_0,x_0)##.
 
  • #4
Actually, I worked out 1).

Let ##\epsilon= r_0 - |y - x_0|##. For sufficiently large ##n##, ##|x_n - x_0| < \epsilon/2## and ##|r_n - r_0| < \epsilon/2 \implies r_n > r_0 - \epsilon/2##. So ##|y - x_n| < (r_0 - \epsilon) + \epsilon/2 < r_n##.
 
  • #5
psie said:
Ok. I agree, I don't see how to derive ##|y-x_n|<r_n## from the triangle inequality given ##|y-x_0|<r_0## and ##x_n\to x_0,r_n\to r_0##. It doesn't lead anywhere: $$|y-x_n|\leq |y-x_0|+|x_0-x_n|<r_0+\epsilon.$$Here ##|x_0-x_n|<\epsilon## for sufficiently large ##n##. But I don't see how else to show $$\chi_{B(r,x)}\to\chi_{B(r_0,x_0)}$$pointwise on ##\mathbb R^n\setminus S(r_0,x_0)##.
We may assume w.l.o.g. that ##x_0=0## and ##r_0=1.## We may also assume that ##
x_n=x_1/n## where ##x_1=1-c## for some constant ##c\in (0,1).## We then need the condition ##B(x_n,r_n)\stackrel{(*)}{\subset} B(0,1).## I struggle a bit to define ##r_n## appropriately such that ##\lim_{n \to \infty}r_n=1## and ##(*)## still holds, but that would be the idea.

The second is analogous with ##B(x_n,r_n)\subseteq \mathbb{R}^n\setminus B(0,1).##
 
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