Arzelà-Ascoli theorem

**miss_lolitta** · April 5th 2006, 03:04 PM

Hi

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

Thanks in advance!!

Miss_Lolitta

**ThePerfectHacker** · April 5th 2006, 06:40 PM

Originally Posted by miss_lolitta

Hi

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

Thanks in advance!!

Miss_Lolitta

Do not click here!!!

**Rebesques** · April 6th 2006, 02:56 AM

Why would a miss-"lolitta" be interested in such messy stuff?

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

.
Taken from Royden's "Real Analysis" )

**miss_lolitta** · April 6th 2006, 06:17 AM

Really this is messy stuff!!

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??

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

ThePerfectHacker & Rebesques....

Thanks so much

**Rebesques** · April 6th 2006, 10:22 AM

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.

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.

**miss_lolitta** · April 6th 2006, 12:45 PM

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 Arzelà-Ascoli theorem )

Thanks

soooo00S00oooo

much

**Rebesques** · April 7th 2006, 11:18 AM

Thanks soooo00S00oooo much

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

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 Arzelà-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.

**miss_lolitta** · April 7th 2006, 03:38 PM

Really I don't know how can I thank ..

I'm grateful for you

I repeat...

Thanks

soooo00S00oooo

much

**miss_lolitta** · April 7th 2006, 10:54 PM

Originally Posted by Rebesques

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

I'm so sorry

**Rebesques** · April 10th 2006, 10:57 AM

I'm so sorry

Not as sorry as I am.

Do come back with more theorems like this.

**miss_lolitta** · April 10th 2006, 12:58 PM

ok

click here if u sure

good luck

**Rebesques** · April 10th 2006, 03:02 PM

That's nothing like the other one.

**miss_lolitta** · April 11th 2006, 10:53 AM

I know that

but you understand it on the fly

try again to take bonus!!

**Rebesques** · April 12th 2006, 03:07 PM

try again to take bonus!!

Sorry... I can't be bribed.

(...unless it's really much.)

**miss_lolitta** · April 13th 2006, 03:12 PM

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!!

just because Rebesques helped me without any bribe..

regard..

lolitta

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