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Let $f_n : (a,b) \to \mathbb{R}$ be functions that have finite number of maxima and minima, for $n = 1,2,3...$. Let D be a countable dense subset of $(a,b)$. If sequence $\{f_n\}$ converges to $f$ such that the convergence is non-uniform at the points $x \in D$ and uniform at the points $x \in (a,b)\setminus D$, then does it imply that $f(x) = 0 \forall x \in (a,b)\setminus D$ ?

EDIT : The question didn't come out as what i expected. So I am posting a new one. sorry for the inconvinience.

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"uniform at the points $x \in (a,b)\setminus D$" seems to be a bit of an unfortunate formulation. Do you mean to say "uniformly on $(a,b) \setminus D$"? –  t.b. May 13 '11 at 8:25
    
@Theo : there seems to be a problem –  Rajesh D May 13 '11 at 8:27
    
@Rajesh: Does non-uniform mean pointwise convergence –  user9413 May 13 '11 at 8:28
    
Seems I posted my comment as an answer. Anyway, @Theo, do you understand "non-uniform" here? Pointwise but not uniform? –  Glen Wheeler May 13 '11 at 8:29
    
@Theo, @Glen : I will reformulate with your permission...in few minutes –  Rajesh D May 13 '11 at 8:29

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Certainly not. Perhaps you might want to further quantify what you mean by "non-uniform". Note that the sequence of functions $f_i(x) = c$ where $c$ is a fixed non-zero number converges pointwise (as well as uniformly, of course), and is a counterexample.

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