# Help with Arzela-Ascoli theorem proof

• Feb 28th 2010, 05:02 PM
Diego
Help with Arzela-Ascoli theorem proof
Our teacher gaves a few days ago the standard proof of this theorem which can be found in many text books, dealing with finding a dense numerable set of the set in which the functions are defined, diagonilization, etc.

The thing is in this proof I can not understand why equicontinuity is necessary and not just simple uniform continuity for each function of the diagonal.

Any help would be appreciated.
• Feb 28th 2010, 05:13 PM
Drexel28
Quote:

Originally Posted by Diego
Our teacher gaves a few days ago the standard proof of this theorem which can be found in many text books, dealing with finding a dense numerable set of the set in which the functions are defined, diagonilization, etc.

The thing is in this proof I can not understand why equicontinuity is necessary and not just simple uniform continuity for each function of the diagonal.

Any help would be appreciated.

Assuming this really is the most famous proof...

At one point we need to say "...there exists a $\displaystyle \delta>0$ such that $\displaystyle d(x,x')<\delta\implies |f_n(x)-f_n(x')|<\frac{\varepsilon}{3}$ for all $\displaystyle f_n\in \Delta$". If all the $\displaystyle f_n\in \Delta$ were merely uniformly continuous we would certainly have a corresponding $\displaystyle \delta_n>0$ which satisfies the right conditions. But, to make sure it works for all $\displaystyle f_n$'s we would need to take $\displaystyle \inf_{n\in\mathbb{N}}\delta_n$ and there is a very real possibility that this is zero.
• Feb 28th 2010, 06:05 PM
Diego
The thing is that I don't see why the same delta is needed
Your absolutely right, if the same $\displaystyle \delta$ were needed then then inf of the deltas could be zero.

But the thing is I don't see why the same delta is needed. i have the following proof ( which I'll summarize in some parts to avoid a long text):

The proof follows much on the lines of the proof given in Elements of Real Analysis, Bartle

Given $\displaystyle C(K,\mathbb{R}^{q})$( the set of continuous function from K to $\displaystyle \mathbb{R}^{q}),$ where K is compact and a subset of $\displaystyle \mathbb{R}^{p}$ we want to show that if $\displaystyle \Im$ is a subset of $\displaystyle C(K,\mathbb{R}^{q}$)and $\displaystyle \Im$ is bounded and equicontinuous then there exists a subsequence for every sequence of functions in $\displaystyle \Im$ that converges.

We can find a set $\displaystyle A := \left\lbrace x_{1},x_{2},...\right\rbrace$ which is numerable and dense in K, then we can find a sequence of functions which i'll lable $\displaystyle \left\lbrace g_{n}\right\rbrace$ which for every point in A the sequence $\displaystyle \left\lbrace g_{n}(x_{i})\right\rbrace$ converges in $\displaystyle \mathbb{R}^{q}$.

The thing is that then from equicontinuity we can find a $\displaystyle \delta$ which makes every function $\displaystyle \|g_{n}(x)-g_{n}(y)\|<\epsilon/3$, if $\displaystyle \|x-y\|<\delta$ and then take open balls with center in each $\displaystyle x_{i}$ and radius $\displaystyle \delta$ the union of these balls complete covers K, and because K is compact we can find a finite open subcover such that it covers K,using the fact that these functions are Cauchy convergent in $\displaystyle \mathbb{R}^{q}$. we find that for $\displaystyle x \in K$, $\displaystyle \|g_{n}(x)-g_{m}(x)\|<\epsilon$

Why can't we take open balls with different radius delta (i.e., use uniform continuity)?
• Feb 28th 2010, 06:12 PM
Drexel28
Quote:

Originally Posted by Diego
Your absolutely right, if the same $\displaystyle \delta$ were needed then then inf of the deltas could be zero.

But the thing is I don't see why the same delta is needed. i have the following proof ( which I'll summarize in some parts to avoid a long text):

The proof follows much on the lines of the proof given in Elements of Real Analysis, Bartle

Given $\displaystyle C(K,\mathbb{R}^{q})$( the set of continuous function from K to $\displaystyle \mathbb{R}^{q}),$ where K is compact and a subset of $\displaystyle \mathbb{R}^{p}$ we want to show that if $\displaystyle \Im$ is a subset of $\displaystyle C(K,\mathbb{R}^{q}$)and $\displaystyle \Im$ is bounded and equicontinuous then every subsequence of a sequence of functions in $\displaystyle \Im$.

We can find a set $\displaystyle A := \left\lbrace x_{1},x_{2},...\right\rbrace$ which is numerable and dense in K, then we can find a sequence of functions which i'll lable $\displaystyle \left\lbrace g_{n}\right\rbrace$ which for every point in A the sequence $\displaystyle \left\lbrace g_{n}(x_{i})\right\rbrace$ converges in $\displaystyle \mathbb{R}^{q}$.

The thing is that then from equicontinuity we can find a $\displaystyle \delta$ which makes every function $\displaystyle \|g_{n}(x)-g_{n}(y)\|<\epsilon/3$, if $\displaystyle \|x-y\|<\delta$ and then take open balls with center in each $\displaystyle x_{i}$ and radius $\displaystyle \delta$ the union of these balls complete covers K, and because K is compact we can find an open subcover such that it covers K,using the fact that these functions are Cauchy convergent in $\displaystyle \mathbb{R}^{q}$. we find that for [tex]x \in K[\math], $\displaystyle \|g_{n}(x)-g_{m}(x)\|<\epsilon$

Why can't we take open balls with different radius delta (i.e., use uniform continuity)?

Because, if memory serves, we later need to make a statement that has to do with $\displaystyle \f_m(x)-f_n(x)|$ and if we don't have $\displaystyle \delta$ that works for all of them we're S.O.L.

If my memory doesnt serve tell me and I'll go look ath the proof.
• Feb 28th 2010, 06:41 PM
Diego
most probably but I don't see it.
Continuing with the proof:
Then $\displaystyle \exists \left\lbrace z_{1},z_{2},...,z_{r} \right\rbrace$ such that K is a subset of the open balls with radius delta and center in each of these points. Then we can find an N (because there is a finite number of points) such that the functions $\displaystyle g_{n}(z_{i})$ are smaller (using the norm of $\displaystyle \mathbb{R}_{q}$) than epsilon / 3 then you just add up things using the triangle inequality and you get the result.
• Feb 28th 2010, 06:44 PM
Drexel28
Quote:

Originally Posted by Diego
Continuing with the proof:
Then $\displaystyle \exists \left\lbrace z_{1},z_{2},...,z_{r} \right\rbrace$ such that K is a subset of the open balls with radius delta and center in each of these points. Then we can find an N (because there is a finite number of points) such that the functions $\displaystyle g_{n}(z_{i})$ are smaller using the norm than epsilon / 3 then you just add up things using the triangle inequality and you get the result.

I need to go back and look at the proof, I'll get back to you (it's a pretty long proof if I remember correctly).

I'm sure someone will answer it before I do though.
• Mar 29th 2010, 10:23 AM
Diego
I now know why
We can not take different deltas because, then the union of the open balls for these deltas might not cover K, for instance in [0, 1], if our numerable set does not include 1 we could have open balls that get tinnier as we choose x closer and closer to 1, but they never include 1.