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Suppose ${f_n}$ are uniformly bounded and equicontinuous on some closed interval $[a,b]$. Therefore, by Arzela-Ascoli we know that $f_n$ has a uniformly convergent subsequence.

But we can also apply Arzela-Ascoli to the subsequences of $f_n$ to get that every subsequence of $f_n$ has a subsequence that converges uniformly. And we have some extra information that says each of the subsequences of subsequences converges uniformly to the same thing.

How do you conclude that $f_n$ converges uniformly from this?

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Wait, so every (strict) subsequence of $f_n$ converges uniformly? – Alex Becker Sep 18 '12 at 0:05
Every subsequence has a subsequence which converges uniformly. So that isn't immediate. – StuartHa Sep 18 '12 at 0:14
False even if the functions $f_n$ are constant. – GEdgar Sep 18 '12 at 0:17
To be clear: you are imposing that there is a function $g\in C[0,1]$ so that every subsequence of $(f_n)$ has a further subsequence that converges to $g$ in $C[0,1]$? If so, consider the contrapositive of your claim. – David Mitra Sep 18 '12 at 0:27
In any topological space if you have a sequence $f_n$ with the property that every subsequence has a further subsequence $f_{n_k}$ which converges to a fixed $f$ in the space, then $f_n$ converges to that $f$. The proof is to observe that if $f_n$ does not converge to $f$ then there is a subsequence which does not ever get 'close' to $f$. But then this subsequence has a subsequence which converges to $f$, which is your contradiction. – Chris Janjigian Sep 18 '12 at 0:47
up vote 3 down vote accepted

Let $f$ be the common limit of the uniformly convergent subsequence. Suppose $f_n$ does not converge uniformly to $f$, so we have some $\epsilon>0$ and a collection of points $x_{n_k}\in [a,b]$ such that $n_k\to \infty$ and $\|f_{n_k}(x_{n_k})-f(x_{n_k})\|>\epsilon$. But then no subsequence of $f_{n_k}$ can converge uniformly to $f$, a contradiction. Hence $f_n$ converges uniformly to $f$.

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Ohhh, you are too smart! – StuartHa Sep 18 '12 at 6:10

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