# Sequence of real valued functions converges pointwise but not uniformly to a continuous function

Suppose $f_n : [a,b] \to \mathbb{R}$ continuous, and that $(f_n)$ converges pointwise but not uniformly to continuous $f$. Show there is a sequence $x_m \to x \in [a,b]$ such that $f_n(x_n)$ does not converge to $f(x)$.

My thoughts:

I think Bolzano-Weierstrass could be useful here. If $f_n$ doesn't converge uniformly, then there exists an $\epsilon_0 > 0$ such that for any $N$ there exists an $n \geq N$ and an $x_k \in [a,b]$ with $|f_n(x_k) - f(x_k)| \geq \epsilon_o$. So letting $N$ vary from 1 (say) upwards, we generate a sequence $(x_k)$ which is obviously bounded. By B-W, this has a convergent subsequence, say $x_m \to x$. Now I've thought about considering $|f_n(x_n) - f(x)| = |f_n(x_n) - f(x_n) + f(x_n) - f(x)| \geq |f_n(x_n) - f(x_n)| - |f(x_n) - f(x)|$. The first of these terms is greater than $\epsilon_o$ for large enough $n$, and the second tends to 0 and so we can make it as small as we want by choosing large enough $n$. This feels like the right sort of idea, but how do I turn it into a rigorous proof? (or if I'm wrong, why?)

Thanks

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I think there will be a confusion between the indexing of functions given originally and the sequence given later. The question if asked like " Show that there is a sequence $x_{k} \to x \in [a,b]$ such that $f_{n_{k} }(x_{k})$ does not converge to $f(x)$" would be clearer. – Kasun Fernando Jul 4 '12 at 5:57

I think what you have so far is fine. Maybe all you're missing is: for $n$ sufficiently large, $|f(x_n)-f(x)|\leq\epsilon_0/2$, so $|f_n(x_n)-f(x)|\geq\epsilon_0/2$ by your estimate. Hence $f_n(x_n)\not\to f(x)$.