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Let $F$ be a set of continuous functions on $[0,1].$ Assume that every sequence in $F$ has a subsequence that converges uniformly on $[0,1].$ Prove that there is a uniform bound for all functions in $F,$ and that $F$ is equicontinuous.

This is the converse of the Arzela-Ascoli theorem. I believe I fully understand the forward proof using Bolzano-Weierstrass and a diagonalization argument to form a sequence $\{f_n\}$ from $F$ that converges on a dense subset of $[0,1],$ and then use equicontinuity to finish it off. The converse has been giving me some trouble so I have brought it here.


Edit - Thought I've been trying (6/6/16): Using "every sequence has a convergent subsequence" to show that $F$ is uniformly bounded, and "uniform convergence of subsequence" to show equicontinuity.

Since every sequence in $F$ has a convergent subsequence, $F$ must be uniformly bound because if not, one could construct a sequence that grows without bound at all points and hence would not have a convergent subsequence. (Is that valid?)

Now we wish to show that for all $n,$ $|f_n(x) - f_n(y)| < \epsilon,$ whenever $|x-y|$ is less than some $\delta.$ My instinct is to write,

$$|f_n(x) - f_n(y)| \leq |f_n(x) - f_m(x)| + |f_m(x) - f_m(y)| + |f_m(y) - f_n(y)|,$$ but this does nothing more than just change the sequence index $n \to m$ because the 1st and 3rd terms are arbitrarily small by the sequence's uniform convergence. Any hints/proofs?


Thought I've been trying (6/5/16):

Start by assuming that there exists a sequence such that $|f_n(x)| > M$ for any $M$ at some point $x \in [0,1]$ for some $n,$ then show a contradiction by showing that this sequence doesn't have a uniformly convergent subsequence. However, I am not sure if it must not have a unif. conv. subsequence, I believe it just may not have one, so I am not sure if this will work.

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  • $\begingroup$ I thought arzela-ascoli was if and only if? $\endgroup$ – Chill2Macht Jun 5 '16 at 18:28
  • $\begingroup$ Right, so normally we show if $P,$ then $Q.$ I am trying to show $Q$ implies $P.$ So I guess its a matter of how you state the original theorem, perhaps "converse" isn't appropriate if the theorem is stated as an iff. However, what I'd like to show is that if all sequences in the family of functions $F$ have a unif. conv. subsequence, then $F$ is unif. bounded and equicontinuous. $\endgroup$ – Merkh Jun 5 '16 at 18:31
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First of all note that $C[0,1]$ is a metric space and hence the hypothesis actually says that $F$ is a compact subset of $C[0,1]$ for the notion of sequential compactness and compactness coincide for metric spaces. Since a metric space if compact if and only if it is complete and totally bounded, it follows that $F$ is totally bounded and hence in particular bounded. That is, there exists a constant $M>0$ such that $$ \|f\|\leq M, $$ for all $f\in F$. This shows that $F$ is uniformly bounded.

Now, for proving the equicontinuity of $F$, first note that since $F$ is totally bounded, so for any $\epsilon>0$ there exists a finite number of elements $f_1,\ldots,f_n$ in $C[0,1]$ such that $F\subseteq \bigcup\limits_{k=1}^{n} B(f_k~;~\epsilon/3)$. We also know that every continuous function on a compact metric space is uniformly continuous. Therefore, for each $k=1,\ldots,n$, there exist $\delta_k>0$ such that $$ |f_k(x)-f_k(y)|<\epsilon/3 $$ whenever $|x-y|<\delta_k$. In particular, for any $f\in B(f_k~;~\epsilon)$ , we have $$ |f(x)-f(y)|<|f(x)-f_k(x)|+|f_k(x)-f_k(y)|+|f_k(y)-f(y)|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon $$ whenever $|x-y|<\delta_k$. It now follows easily that for $\delta=\min\limits_{1\leq k\leq n}\delta_k$, for all $f\in F$, $$ |f(x)-f(y)|<\epsilon $$ whenever $|x-y|<\delta$ and hence $F$ is equicontinuous.

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