Prove or disprove: if fnfnf_n converges uniformly on AAA, then fnfnf_n converges uniformly on ¯A\overline{A}.

I am interested in the statement

Let AA be a set and fnf_n be a sequence of functions defined on ¯A\overline{A}. Assume that fnf_n pointwise converges to a function ff on ¯A\overline{A}. If fnf_n converges uniformly on AA to ff, then fnf_n converges unifomly on ¯A\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 22-fold:

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

I assume no a priori condition on fnf_n and ff.

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Maybe you need to add continuity into the mix?
– copper.hat
Oct 21 at 2:08

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1 Answer
1

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False.

Let A=QA=\mathbb Q and define fn(x)=f(x)=0f_n(x)=f(x)=0آ for each x∈Qx\in\mathbb Q and n∈Nn\in\mathbb N. If x∈R∖Qx\in\mathbb R\setminus \mathbb Q, then let fn(x)=x/nf_n(x)=x/n and f(x)=0f(x)=0 for each n∈Nn\in\mathbb N. Clearly, (fn)n∈N(f_n)_{n\in\mathbb N} converges pointwise to ffآ on ¯A=R\overline A=\mathbb R. However, convergence is not uniform on ¯A\overline A, even though it is on AA.

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

  

 

@JackyChong Trivially. Both fnf_n (for every n∈Nn\in\mathbb N) and ff identically vanish on AA.
– triple_sec
Oct 21 at 2:07

  

 

I see. Cute example.
– Jacky Chong
Oct 21 at 2:07