# Arzelą-Ascoli theorem

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• Apr 5th 2006, 03:04 PM
miss_lolitta
Arzelą-Ascoli theorem
Hi

Please, can some one give me lecture and exercises in Arzelą-Ascoli theorem ??

Thanks in advance!!

Miss_Lolitta
• Apr 5th 2006, 06:40 PM
ThePerfectHacker
Quote:

Originally Posted by miss_lolitta
Hi

Please, can some one give me lecture and exercises in Arzelą-Ascoli theorem ??

Thanks in advance!!

Miss_Lolitta

Do not click here!!!
• Apr 6th 2006, 02:56 AM
Rebesques
Why would a miss-"lolitta" be interested in such messy stuff? :eek:

Anyway. Here's one: Let $C^{0,a}$ be the set of Hölder continuous functions on $[0,1]$, with exponent $0. Define the norm

$||f||_a={\rm max}_{x\in [0,1]}|f(x)|+{\rm sup}_{x,y\in [0,1]}\frac{|f(x)-f(y)|}{|x-y|^a}$.

Show that the closed unit ball $||f||_a\leq 1$ of $C^{0,a}$ is compact in the space of all continuous functions on $[0,1]$.

(Enjoy :rolleyes:.
Taken from Royden's "Real Analysis" )
• Apr 6th 2006, 06:17 AM
miss_lolitta
Really this is messy stuff!! :eek:

I wanted lecture in it because I have an assignment how to write a lecture with explaining by examples then represent this lecture in class..

But I have no ideas about this theorem?? :confused:

and How I perform this a simply.. without complexity?? :cool:

ThePerfectHacker & Rebesques....

Thanks so much :)
• Apr 6th 2006, 10:22 AM
Rebesques
Well... If you want, you might say something like this. The Bolzano-Weirstrass theorem, sais that a bounded sequence converges. What can we say about a sequence of functions?

A problem is, we need to be precise about the mode of convergence - so consider the best possible, uniform convergence - the one implied by the sup-norm: $||f||=\sup|f(x)|$

So, if we simply have a sequence of functions $(f_n)$, and it is bounded (i.e., the sup-norm is finite for all), do we have uniform convergence? That is, is there a function $f$ with $||f_n-f||=\sup_{x}|f_n(x)-f(x)|\rightarrow 0$ ???

No. The domain could be unbounded, and so our functions may not approach a limit function (though they are bounded).

We fix this (and alike problems) by demanding the domain be compact, say the interval $[0,1]$

So, are we now to say uniform convergence is guaranteed? No, again. It could happen we only have pointwise convergence alone (not in the sup-norm, but only that for each $x$, $\lim_n f_n(x)=f(x)$.)

The additional hypothesis to resolve this, is (uniform) equicontinuity; That is, the $\delta$ in the $\epsilon-\delta$ definition of continuity, holds globally for all $x$ and $f_n$.

(Notice that that the Holder continuous functions in the last exercise do not satisfy this uniformly - but a relaxed form of the theorem can be used.)

So, every bounded and (uniformly) equicontinuous sequence of functions defined on $[0,1]$, necessarily has a (uniformly) convergent subsequence - this is what the theorem says. :confused: :eek:

Example. Take $\sin(nx)=:f_n(x), \ x\in [0,1]$. This is a bounded and equicontinuous sequence (show!). Note that for every $x, \ \lim_n f_n(x)$ does not exist (besides trivial cases).

But, by the Ascoli-Arzela theorem, there exists some subsequence $( \sin(k_nx))$ and a function $f(x), \ x\in [0,1]$ with

${\sup}_{x\in [0,1]} |{\rm sin}(k_n x)-f(x)|\rightarrow 0, \ n\rightarrow\infty$ .

Hope that helps.
• Apr 6th 2006, 12:45 PM
miss_lolitta
That's perfect!!
Well done!! ;)

Please,,Can you prove that :
Let K be a compact subset of Rp and let F be a collection of functions which continuous on K and have values in Rq. The following properties are equivalent:

1- The family F is bounded and uniformly equicontinuous on K.
2- Every sequence from F has a subsequence which is uniformly convergent on K.( this is another state for Arzelą-Ascoli theorem ) :confused:

Thanks :) soooo00S00oooo :) much
• Apr 7th 2006, 11:18 AM
Rebesques
Quote:

Thanks soooo00S00oooo much
no need, love. i am just dead bored at work.

Quote:

Let K be a compact subset of Rp and let F be a collection of functions which continuous on K and have values in Rq. The following properties are equivalent:

1- The family F is bounded and uniformly equicontinuous on K.
2- Every sequence from F has a subsequence which is uniformly convergent on K.( this is another state for Arzelą-Ascoli theorem)

1 implies 2: This is the proof of the Arzela-Ascoli theorem. It is kind of tedious, but at least there's a diagonal argument.

a) Consider all points of $K$ with rational coefficients. Call this (numerable) set $Q$, and denote its elements by $x_m, \ m=1,2\ldots$.

b) Let $(f_n)$ be any sequence of elements of $F$. We show it has a pointwise convergent subsequence on $Q$.

The sequence $f_n(x_1)$ is bounded; By the Bolzano-Weierstrass theorem, it has a convergent subsequence, say $f_{1,n}(x_1), \ n\in {\bf N}.$ Apply the same argument for $f_{1,n}(x_2)$, yielding a convergent subsequence $f_{2,n}(x_2), \ n\in {\bf N}.$ (note: the sequence $(f_{2,n})$ is thus convergent at $x_1$ and $x_2$). Proceed in this fashion, to obtain a subsequence $(f_{k,n})_{k,n\in {\bf N}}$ of the original sequence $(f_n)$. Then, the sequence $(f_{n,n})$ converges for all elements of $Q$.

c) We now show that $(f_{n,n})$ converges uniformly on $K$. That is, $f_{n,n}(x)$ is Cauchy for all $x$, in the uniform convergence norm.

Let $x\in K, \ m,n\in {\bf N}$ (these will be fixed later) and $\epsilon >0$. There exists some $x_i\in Q$ with

$|f_{n,n}(x)-f_{n,n}(x_i)| \ , \ |f_{m,m}(x)-f_{m,m}(x_i)|\leq \epsilon/3 \ (1)$

due to equicontinuity of the functions and the density of $Q$ in $K$. Also, we have

$|f_{n,n}(x_i)-f_{m,m}(x_i)|\leq \epsilon/3 \ (2)$

for sufficiently large $n,m$.

Therefore

$|f_{n,n}(x)-f_{m,m}(x)|\leq|f_{n,n}(x)-f_{n,n}(x_i)|+|f_{n,n}(x_i)-f_{m,m}(x_i)|
$

$+|f_{m,m}(x_i)-f_{m,m}(x)|\leq \epsilon,
$

by (1) and (2). So

$\sup_{x\in K}|f_{n,n}(x)-f_{m,m}(x)|\leq \epsilon,
$

for all sufficiently large $n,m$; this means the subsequence is uniformly convergent on $K$.

2 implies 1: Let $C(K)$ be the metric space of continuous functions on $K$ endowed with the uniform convergence norm.

The given condition implies that $F$ is compact in $C(K)$, so it is bounded and closed in its topology.

We now show it is equicontinuous. Let $f$ belong to $F$. Then there is a sequence $(f_n)$ in $F$ converging uniformly to $f$.

Let $\epsilon>0, \ x,\ y \in K$ (to be fixed later). Consider $n$ so large that

$|f_n(x)-f(x)|, \ |f_n(y)-f(y)|\leq \epsilon/3 \ (3)$

and, from the uniform continuity of $f_m$, there exists
$\delta=\delta(\epsilon)$ such that

$|f_n(x)-f_n(y)|\leq \epsilon/3 \ (4)$

for $|x-y|\leq \delta$. We thus obtain

$|f(x)-f(y)|\leq
|f_n(x)-f(x)|+|f_n(x)-f_n(y)|
$

$+|f_n(y)-f(y)|\leq \epsilon$

due to (3) and (4). So $F$ is uniformly equicontinuous.
• Apr 7th 2006, 03:38 PM
miss_lolitta
Really I don't know how can I thank ..

I'm grateful for you

I repeat...

Thanks :) soooo00S00oooo :) much
• Apr 7th 2006, 10:54 PM
miss_lolitta
Quote:

Originally Posted by Rebesques
no need, love. i am just dead bored at work.

I'm so sorry :(
• Apr 10th 2006, 10:57 AM
Rebesques
Quote:

I'm so sorry

Not as sorry as I am.

Do come back with more theorems like this.
• Apr 10th 2006, 12:58 PM
miss_lolitta
ok

click here if u sure :p

good luck :rolleyes:
• Apr 10th 2006, 03:02 PM
Rebesques
That's nothing like the other one.
• Apr 11th 2006, 10:53 AM
miss_lolitta
I know that

but you understand it on the fly

try again to take bonus!! :D
• Apr 12th 2006, 03:07 PM
Rebesques
Quote:

try again to take bonus!!
Sorry... I can't be bribed. ;)

(...unless it's really much.)
• Apr 13th 2006, 03:12 PM
miss_lolitta
oh,you can't be bribed..

but bonus was candy..Do you like this bribe?? ;)

Rebesques,,I would like to tell you that presentation I presented it to my professor who will say the result is it good or no on next Monday ,,

but I'm sure it's excellent!! :cool:

just because Rebesques helped me without any bribe..

regard..

lolitta