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