Let $\{f_n\}$ be a sequence of functions converging pointwise to $f$ such that $\lim_{n \rightarrow \infty } f_n (x_n)=f(x)$ for every sequence $\{x_n \}$ converging to $x$. Then is it true $f_n \rightarrow f$ uniformly?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Probably no: math.stackexchange.com/questions/1574573/… $\endgroup$– SiminoreNov 15, 2016 at 9:32
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
This is in general false. Consider the sequence of functions $f_n(x) = x/n$. Then $f_n(x_n)$ converges to $0$ for every convergent sequence $x_n$, but the sequence $f_n$ does not converge uniformly on $\mathbb{R}$ to $0$.
-
$\begingroup$ Are there any (weak) conditions we can add for it to hold? $\endgroup$ Nov 15, 2016 at 9:38
-
1$\begingroup$ If the domain of $f$ is compact, the theorem holds. I don't think there are any other sensible conditions. For example, uniform boundedness is also too weak, since $f(x) = \min\{|x|/n, 1\}$ is a conterexample. $\endgroup$– DominikNov 15, 2016 at 9:40