${f_n}$ is a sequence of continuous functions on $\Bbb R$, and $f_n \rightarrow f$ uniformly on every finite interval $[a,b]$. If each $f_n$ is bounded, is it true that $f$ must be bounded?
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Here is a counterexample:
$f_n(x) = |x|$ if $|x| < n$
$f_n(x) = n$ otherwise
TonyK
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Could you explain what you mean by smoothing out? – kiwifruit Mar 15 '14 at 20:29
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@David: Thanks $-$ I missed the continuity requirement. I've fixed it now. – TonyK Mar 15 '14 at 20:30
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Maybe I'm confused, but this doesn't converge uniformly – Marc Mar 15 '14 at 20:31
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@Marc: It converges uniformly on every bounded interval. – TonyK Mar 15 '14 at 20:31
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Ah, I didn't read carefully enough... – Marc Mar 15 '14 at 20:32
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@Marc: That makes two of us :-) – TonyK Mar 15 '14 at 20:32
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Note that it is true if $f_n\rightarrow f$ uniformly. – Marc Mar 15 '14 at 20:32
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Could you explain why this converges on every bounded interval, but not uniformly? – kiwifruit Mar 15 '14 at 20:38
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@kiwifruit: $(f_n)$ converges uniformly on every bound interval, but not on $\mathbb R$. What exactly is your question? – TonyK Mar 15 '14 at 20:39
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@Marc: Yes of course. – TonyK Mar 15 '14 at 20:40
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Hmm, I don't quite understand the difference between the two. Can't we say R is a combination of many bounded intervals? – kiwifruit Mar 15 '14 at 20:40
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@kiwifruit: You think that if $(f_n)$ converges uniformly on every bounded interval, then it must converge uniformly on $\mathbb R$? Then you must sit down and do some thinking, perhaps with my counterexample in mind. – TonyK Mar 15 '14 at 20:43
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It is quite straightforward to prove if the convergence is uniformly. Use the definition that in that case for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\sup_{x\in\mathbb{R}}|f_n(x)-f(x)|<\epsilon$ for all $n\ge N$. – Marc Mar 15 '14 at 20:44
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@Marc: Yes, we know. – TonyK Mar 15 '14 at 20:45
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It was intended for Kiwifruit;) – Marc Mar 15 '14 at 20:45
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Hmm, I'm having a hard time picturing the counterexample graphically, so it's a bit difficult for me to see what the difference is. – kiwifruit Mar 15 '14 at 20:47
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@kiwifruit: I don't see what more we can usefully say. Sooner or later, you have to think things out for yourself. – TonyK Mar 15 '14 at 20:49
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Answer is NO. For example, Let $f(x)=x$, $f_n(x)=x$ if $-n\leq x\leq n$, $f_n(x)=-n$ if $ x\leq -n$, and $f_n(x)=n$ if $ x\geq n$.
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