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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 $a>0$.2012-06-08
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    @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!?2012-06-08
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    You are correct.2012-06-08
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    I excluded the possibility of having zero functions (because this will not effect my problem). Does it change anything now?2012-06-08

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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!2012-06-08
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    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 $n$!2012-06-08
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    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!2012-06-08
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    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 $|f_n(x)|$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. – 2012-06-08
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    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!2012-06-08
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    Yes, but when talking about convergence of functions make sure to specify in which sense they converge: here $|f_{n_k}|$ converges uniformly to 0 (and therefore also in the pointwise sense).2012-06-08
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    Can I ask why the convergence is uniformly here?2012-06-09
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    $f_n$ converges to $f$ uniformly iff the scalar sequence $\sup_x |f_n-f|$ converges to 0. Here of course $f=0$.2012-06-09
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    Thanks Generic Human, this discussion was very helpful for me.2012-06-09
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    I have this question for you, you said in your first comment above "If no subsequence ..., that is if $\liminf_{n\to\infty}\sup|f_{n}|=l>0$", do we knowe that this limit exists?2012-06-09
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    Yes we do: the $\inf_{m\ge n} \sup |f_m|$ is well-defined because the numbers are non-negative, and this forms a non-decreasing sequence: therefore a (possibly infinite) limit always exists, but because the sequence is bounded from above by $M$ we always have a finite limit.2012-06-11