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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?)


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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
up vote 3 down vote accepted

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 $f_n(x_n)\not\to f(x)$.

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Excellent, thank you :) – user938272 May 12 '11 at 0:20

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