I was wondering if $f_n$ converges pointwise to $f$ on a closed interval, $E$, must there be closed interval $E' \subseteq E$ such that $f_n|_{E'}$ converges uniformly to $f|_{E'}$.
Currently, with my intuition, I can't think of an counter-example.
Asked
Active
Viewed 222 times
4
-
The actual theorem along these lines is called Egorov's theorem. – Ian Feb 16 '20 at 17:54
1 Answers
7
Let $(q_n)_{n\in\mathbb N}$ be an enumeration of $\mathbb Q\cap[0,1]$. For each $n\in\mathbb N$, let $f_n\colon[0,1]\longrightarrow\mathbb R$ be equal to $\chi_{\{q_n\}}$; in other words, $f_n(x)=0$, unless $x=q_n$, in which case $f_n(x)=1$. Then $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function. However, there is no interval $[a,b]\subset[0,1]$ (with $a<b$) such that the convergence is uniform on $[a,b]$, since $[a,b]$ contains infinitely many $q_n$'s.
José Carlos Santos
- 440,053
-
Interesting! Are you aware of an example when $f_n$ are all continuous? I foolishly forgot to include that in my original question. – Zach Hunter Feb 16 '20 at 18:17
-
@ZacharyHunter I'm pretty sure you can just imitate this example in the continuous case: take $f_{m,n}$ continuous with $\lim_{m \to \infty} f_{m,n} = \chi_{{ q_n }}$ pointwise, then enumerate that 2D array into a sequence. – Ian Feb 16 '20 at 20:01
-
@ZacharyHunter : Just put a bump of small width, say $n^{-2}$ centered at $q_n$ in $f_n$. This is a generic method of making a sequence of point-indicator-like functions into a sequence of continuous functions. – Eric Towers Feb 17 '20 at 03:08