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If $\{f_{n}(x)\}$ is a sequence of continuous functions on $\mathbb R$, with $|f_{n}(x)|\leq C, \forall n$, and $\lim_{n\to\infty}f_{n}(x)=0$ uniformly on $\mathbb R$, does there exist a subsequence of $\{f_{n}\}$ which is decreasing on $\mathbb R$?

Edit: What if $\{\sup_{\mathbb R}|f_{n}(x)|\}$ is also converges to 0, and all $f_{n}$ are positive continuous functions with the above properties?

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$f_n(x) =\frac{\sin(nx)}{n}$ –  N. S. Jun 11 '12 at 17:20
    
Take for $f_n$ a bump of height $1/n$ and width $1$ centered at $n$. –  Did Jun 11 '12 at 17:34
    
Will any type of "bump" work? –  Adam Rubinson Jun 11 '12 at 17:36
1  
@AdamRubinson: Yes. Draw a picture. –  Harald Hanche-Olsen Jun 11 '12 at 17:43
    
@Adam: And please use the @ thing. –  Did Jun 11 '12 at 17:51

1 Answer 1

No. Consider $f_n(x) \equiv -1/n$. Then $|f_n(x)|\le 1 \ \forall n $, $f_n \rightarrow 0$ uniformily on $\mathbb{R}$ and All subsequence of $(f_n)$ is strictly increasing.

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your functions are not bounded (as Mathblue requires) –  Adam Rubinson Jun 11 '12 at 17:47
    
It is bounded by one. –  user29999 Jun 11 '12 at 17:54
    
Not bounded? $|f_n(x)| \le 1/n$. –  Robert Israel Jun 11 '12 at 17:55
    
oooh yeah I read it as 1/x. My bad –  Adam Rubinson Jun 11 '12 at 17:55

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