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I have this problem which I think the Mean Value Theorem for continuous functions may apply.

Let $\{f_{n}\}_{n\geq 1}$ be a sequence of non-zero continuous real functions on $\mathbb R$, with the following properties:

(1) $$\sup_{x\in \mathbb R}|f_{n}(x)|\leq M$$ for some $M>0$, i.e., the sequence is uniformly bounded on $\mathbb R$.

(2) There exists a countable set $W \subset\mathbb R$ such that $$\sup_{w\in W}|f_{n}(w)|\to 0$$ as $n\to \infty$

Question: Is there an $a\in (0,M)$ on the $y$-axis, such that -for every $n$ - we can find a point, say $x_{n}$ on the $x$-axis with $|f_{n}(x_{n})|=a$?

Note: The set $W$ has no accumulation (limit) point.

Edit: I aded that the functions are nonzero.

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Perhaps the $\sup$ in (1) should be attained? Otherwise, take $f_n = 0$, this satisfies all the conditions, but obviously there is no point for which any $|f_n|$ equals $a>0$. –  copper.hat Jun 8 '12 at 22:45
    
@copper.hat: Thank you for your comment, but I'm not given this information. In fact, I'm given that $|f_{n}(x)|\leq M$ for all $x\in \mathbb R$, and all $n\geq 1$, and I think this means that $\sup_{\mathbb R}|f_{n}|\leq M$, I'm right!? –  Alpha Jun 8 '12 at 22:53
    
You are correct. –  copper.hat Jun 8 '12 at 22:58
    
I excluded the possibility of having zero functions (because this will not effect my problem). Does it change anything now? –  Alpha Jun 8 '12 at 23:02

1 Answer 1

Take $f_n(x)=\frac{1}{n}\sin \pi x$ and $W=\mathbb Z$. For any non-zero $a$, $f_n(x)\ne a$ for all large enough $n$.

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Ok, so in this case $f_{n}\to 0$ as $n\to \infty$, that make sense. Is it the only case where the result fail? I mean, if we also exclude this possibility! –  Alpha Jun 8 '12 at 23:15
    
If no subsequence of $\sup |f_n(x)|$ converges to 0, that is if $\liminf\limits_{n\to\infty} \sup\limits_x |f_n(x)| = l>0$, then you can take any $a\in (0,l]$: for all large enough $n$, there is $u_n$ s.t. $|f_n(u_n)|\ge l$ (definition of $l$) and $v_n$ s.t. $|f_n(v_n)|\le a$ (definition of $W$), so you can apply the MVT. Note however that the result is only true for large $n$, not for every $n$! –  Generic Human Jun 8 '12 at 23:22
    
Do we need that "no subsequence of $\sup|f_{n}(x)|$ converges to 0"?? I think if we just assume that $|f_{n}(x)|$ does not converge to 0, which will imply that $\sup_{\mathbb R}|f_{n}(x)|$ does not converges to 0, is enough! –  Alpha Jun 8 '12 at 23:31
    
No, it's not enough: if any subsequence of $\sup |f_n(x)|$ converges to 0, then for any $a>0$, from some point onward in that subsequence we'll have $|f_n(x)|<a$ for all $x$, so the result fails. Saying a sequence does not converge to 0 is much weaker than saying it converges to a non-zero limit. –  Generic Human Jun 8 '12 at 23:38
    
Ohh I see. One last thing: If there is a subsequence of $\sup|f_{n}(x)|$ converges to 0, say $\sup|f_{n_{k}}|$, then this will imply that there exists a subsequence $|f_{n_{k}}|$ of $|f_{n}|$ such that $|f_{n_{k}}|$ converges to 0, right! –  Alpha Jun 8 '12 at 23:43

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