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I am interested in the statement

Let $A$ be a set and $f_n$ be a sequence of functions defined on $\overline{A}$. Assume that $f_n$ pointwise converges to a function $f$ on $\overline{A}$. If $f_n$ converges uniformly on $A$ to $f$, then $f_n$ converges unifomly on $\overline{A}$.

I initially wanted to use the contraposition of this statement to answer this question. But trying to prove it lead me to think that it may false. But cannot find counter-examples.

My question is then $2$-fold:

  1. If the statement is false, what is a simple counter-example?
  2. Under what conditions is the statement known to be true? It is true if $\overline{A}\setminus A$ is finite for example.

I assume no a priori condition on $f_n$ and $f$.

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  • $\begingroup$ Maybe you need to add continuity into the mix? $\endgroup$ – copper.hat Oct 21 '16 at 2:08
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False.

Let $A=\mathbb Q$ and define $f_n(x)=f(x)=0$ for each $x\in\mathbb Q$ and $n\in\mathbb N$. If $x\in\mathbb R\setminus \mathbb Q$, then let $f_n(x)=x/n$ for each $n\in\mathbb N$ and $f(x)=0$. Clearly, $(f_n)_{n\in\mathbb N}$ converges pointwise to $f$ on $\overline A=\mathbb R$. However, convergence is not uniform on $\overline A$, even though it is on $A$.


Incidentally, this counterexample reveals that the additional assumption that the limit function $f$ is continuous (even uniformly continuous) on $\overline A$ is still not strong enough to make the claim true.

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  • $\begingroup$ @JackyChong Trivially. Both $f_n$ (for every $n\in\mathbb N$) and $f$ identically vanish on $A$. $\endgroup$ – triple_sec Oct 21 '16 at 2:07
  • $\begingroup$ I see. Cute example. $\endgroup$ – Jacky Chong Oct 21 '16 at 2:07

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