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I have two questions:
1. I know that the uniform limit of a continuous functions is continuous. But I'm wondering whether this is true if the convergence is locally uniform. That is the uniform convergence is true within any bounded interval.
2. Also I'm confused with the uniqueness of the limit.
For example:
Define
$f_n:[0,1]\rightarrow[0,1]$ such that
$$f_n(x) = \begin{cases} 1-nx &, 0 \leq x\leq \frac{1}{n} \\ \\0 &, \frac{1}{n} \leq x\leq 1 \end{cases}$$ Define $f:[0,1]\rightarrow [0,1]$ to be zero function.
Define $h:[0,1]\rightarrow [0,1]$ to be $$h(x)=\begin{cases} 1 &, x=0\\ \\ 0 &, x\neq 0 \end{cases}$$
and we can prove that: $\forall x,y\in (0,1]$, $f_n$ converges uniformly to $f$ as well as to $h$ within $[x,y]$ (That is locally uniformly). Please point out the mistake I have done

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1 Answer 1

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Since continuity is a local property (that is, in order to determine whether $f$ is continuous at $a$, all that matters is how $f$ behaves near $a$), yes, local uniform convergence preserves continuity.

Concerning your other question, $(f_n)_{n\in\mathbb N}$ neither uniformly to no function and converges pointwise to $h$.

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  • $\begingroup$ Thank you for the answer for the continuity. But for the other part, my concern is about Locally uniformly convergence. $\endgroup$
    – Charith
    Dec 8, 2018 at 14:57
  • $\begingroup$ But I wrote in answer that local uniform convergence preserves continuity. Wasn't that the question? $\endgroup$ Dec 8, 2018 at 14:59
  • $\begingroup$ Yes I agree with you about the continuity. But how about the uniqueness of the limit $\endgroup$
    – Charith
    Dec 8, 2018 at 15:02
  • $\begingroup$ The limit is unique (if it exists) because, for each $x$ in the domain of the $f_n$'s, $f(x)=\lim_{n\to\infty}f_n(x)$ and the limit of a sequence of numbers is unique (again, if it exists). $\endgroup$ Dec 8, 2018 at 15:06

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