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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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    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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    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