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In Wikipedia, the Ascoli theorem requires the functions to be continuous on the closed and bounded interval $[a,b]$. However, in the proof given in the book "Theory of Ordinary Differential Equations" by Coddington and Levinson (Tata MaGraw-Hill Edition 1972) (given at page 5, Ascoli Lemma) the authors only require that the interval be bounded. The lemma is:-

On a bounded interval $I$, let $F=\{f\}$ be an infinite, uniformly bounded, equicontinuous set of functions. Then $F$ contains a sequence $\{f_n\}$, $n = 1,2,\cdots,$ which is uniformly convergent on $I$.

(The authors mention in the text before the lemma that $I$ denotes an open interval.)

Reading through the article on Compact Space on wikipedia, I get a feeling that the interval should be compact. However, the proof in the above book also seems to be correct.

If someone has access to this book (as it is not available on google books), can you please clarify.

Thanks in advance.

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Perhaps you could provide a quotation of the theorem, as it appears in "Theory of Ordinary Differential Equations", in your post. If you did, you would be more likely to get an answer. –  matt Jan 4 '12 at 9:53
    
The statement you wrote talks only about convergence - but you didn't state which convergence. What you will not get without compactness is uniform convergence (and it is easy to find a counter example). But there are other definitions - you can get point convergence, or - a definition which is popular for the use of ODEs - "uniform convergence on every compact subset". –  yaakov Jan 4 '12 at 10:23
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1 Answer

up vote 4 down vote accepted

The authors define "equicontinuous on $I$" to mean "uniformly equicontinuous on $I$". That is, for any $\epsilon>0$ there is a $\delta>0$ such that for any $f\in \cal F$ and for any $x$, $y$ in $I$,

$$|f(x)-f(y)|<\epsilon\quad\text{ whenever }\quad|x-y|<\delta.$$

The theorem you are referring to does give uniform convergence (you omitted this in your post). And, indeed, with the assumption of uniform equicontinuty and uniform boundedness, the assumption that $I$ be closed is not needed.

What is needed is that the interval be totally bounded. If you examine the proof in Coddington and Levinson , you'll see that's what they are using.

See here for a different proof.


Incidentally, some authors say $\cal F$ is equicontinuous over $I$ to mean that given $x\in I$ and $\epsilon>0$, there is a $\delta_{x,\epsilon}$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta_{x,\epsilon}$ and $f\in\cal F$ . One can show that if $I$ is compact, then this notion of equicontinuity implies the notion of uniform equicontinuity above.

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Thanks for finding the precise mistake and the help. I have not understood the definition of equicontinuity and uniform equicontinuity clearly before. –  jpv Jan 5 '12 at 6:17
    
The Wikipedia article on this theorem assumes uniform equicontinuity (as the dependence of $\delta$ on $x$ has not been made clear) and a closed and bounded interval. From your answer I do not think that both these assumptions are necessary. Is it a good idea to edit the Wiki page? –  jpv Jan 5 '12 at 6:27
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