TransWikia.com

Conditions for pointwise convergence of indicators precomposed with uniformly continuous sequence

MathOverflow Asked by Bernard_Karkanidis on February 16, 2021

Let $X$ be a compact metric space, ${delta_n}_{n=1}^{infty}$ be a strictly monotonically decreasing sequence in $[0,1]$ converging to $0$, and ${h_n}_{n=1}^{infty}$ be a uniformly convergence sequence of continuous functions on $X$ converging to $h:Xrightarrow [0,1]$. Distinguish a non-empty compact subset $Asubseteq X$.
If:

  • $h^{-1}[0]=A$,
  • $h_n(A)subseteq [0,delta_n)$ (for every $n$),

Can we guarantee that
$$
I_{[0,delta_n)}circ h_n mbox{ converges point-wise to } I_{{0}}circ h?
$$

If not, what additional conditions am I missing?


It seems that $I_{[0,delta_n)}circ h_n(A)=1$ for all $n$; which is good. So the convergence is uniform on $A$. However, I’m most worried about the pointwise convergence on $X-A$…this I can’t manage to control…

One Answer

$newcommanddedeltanewcommandN{mathbb N}newcommandR{mathbb R}$The answer is yes. Indeed, fix any $xin X$. We need to show that $$l_n:=I{0le h_n(x)<de_n}to r:=I{h(x)=0}tag1$$ as $ntoinfty$. Let $$N:={ninNcolon0le h_n(x)<de_n}.$$

If $Nni ntoinfty$, then $l_n=1$, $h_n(x)to0$, and hence $h(x)=lim_n h_n(x)=0$, so that $r=1$ and $l_n=1to1=r$, i.e., (1) holds.

If $Nnotni ntoinfty$, then $l_n=0$ and, by the condition $h_n(A)subseteq [0,de_n)$, we have $xnotin A$, so that, by the condition $h^{-1}({0})=A$, we have $h(x)ne0$ and hence $r=0$, so that $l_n=0to0=r$, i.e., (1) again holds.

Correct answer by Iosif Pinelis on February 16, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP