# Mod-10 Lec-38 Arzela-Ascoli Theorem: Under Uniform Boundedness, Equicontinuity and Uniform

Arzela-Ascoli Theorem_Under Uniform Boundedness,

Equicontinuity and Uniform Sequential Compactness are Equivalent Ok so ahh so the context is the proof of the

Riemann mapping theorem ok and what you done is in the last we have looked at hyperbolic

geometry ok. Now there is one more ahh technical ahh ahh deto that we have to take to be able

to complete the proof of Riemann mapping theorem and that is that so called ahh complex version

of Arzela Ascoli theorem and the so called Montel theorem ok. So that is what I am going to discuss alright.

So basically the the Arzela Ascoli theorem and the Montel theorem, they are all theorem

which ahh ahh you know guarantee that you are given a family of functions ok on a compact

domain ok ahh that any sequence in that family has a uniformly convergent subsequence ok.

So so that so let me tell you the general idea, general ideas is you see I have some

a family function ok a family of functions defined on a on a domain alright. And let us assume that the family is defined

on ahh I mean for example if you are thinking of the simplest case of real valued function

ok then you think when you assume that is a real valued functions are all define on

a on a closed bounded interval ok which is the compact subset connected subset of real

value alright or more generally if you are thing of function on the plane ok. Then you think of functions which are define

on a domain on the plane ok and in fact you assume ahh at lease to begin to assume that

you ahh with all they also extend continuously to the boundary of the domain that the domain

is bounded and you know ahh ahh the domain is bounded and then you add the boundary which

is compact ok. so you have family of complex function complex valued functions defined

on the domain. Of course all functions here interesting setting

continuous ok, then the question is ahh ahh if you take a sequence of functions from this

family ok to expect that the sequence will converge any sequence of functions will converge

too much ok, we expect that any given sequence of functions will converge is too much ok,

but what you can always expect is that exactly is a subsequence which converges. So ahh ahh the idea of the Arzela Ascoli theorem

and Montel theorem is there under good conditions ok, you can always ensure that you give me

any sequence of functions in a family satisfying of course that on a compact set ok, it will

always have a subsequence which converges uniformly ok. So the general point is that

you want you have family functions ok and the domain effectively where you are studying

is this compact alright or in the or if you are looking at for example analytic functions

you are looking at analytic functions. The property that are you you look at the

analytic functions on close disc in your domain which are a compact closed and bounded discs

ok and the result that you want to set you want to you want the given any sequence from

this family there is a subsequence which converges uniformly ok. Now ahh why is this so important

this is important because you see the moment you say there is a subsequence which converges. It tells you that at least that is the limit

function for subsequent though the whole the given sequence of functions need not converge,

but at least a subsequence converges, but the fact that the converges uniform they will

see all the properties of the functions will also carry over to the limit. For example

if you are looking at a uniform convergence and uniform limit of analytic functions uniform

limit of continuous functions continues because of this. So you just do not want you know just convergence

of functions that does not help because if you just a ordinary convergence of functions

you can go wrong in the sense that the limit function may not have the good properties

original functions you can have a sequence of continuous functions which converts to

limit functions is not contagious you may have the limit function which continuity ok. You are not such things happen so that is

when you are studying convergent the functions is natural that you at least demand uniform

convergence. So that all good properties pass on to the limiting function ok. So the both

the Arzela Ascoli theorem and Montel theorem they are basically situations which guarantee

that you can always find a subsequence ahh of uniformly convergent functions ok subsequence

converges uniform the sequence of function that converges uniform ok, that is generally

idea. Now let us get the technicalities. So ahh you know so here is a ahh so here is

an situation like let F the function a family or collection ahh so let me do the following

thing ahh let me ahh put the title as ahh Arzela Ascoli and Montel theorems. So these

these are very ahh technical theorem but they are basically easy to prove and they are very

powerful ok. So let F be the family family of collection

of ahh function ahh define on and continues and complex valued

on ahh a compact subsets D into complex plane ok. So I have a compact subset of complex plane

in which both close and bounded and I ahve collection of function ahh defined on points

of this capital E taking values in complex numbers that we meet I am assuming that all

continues ok and they are complex plane aright, now you we all know ahh what the definition

of continuity at a point, the definition of continuity at a point is that you know given

epsilon ok ahh. The function values at the point ahh the function

value enable the point can be drop with an Epsilon which is epsilon distance for the

function value at that point. If you choose points a delta neighbourhood of the given

point that is the continuity of the given point and so let me just write that down recall

that f the family is continuous at point ahh the z if ahh given epsilon=0 or x is delta

greater than zero sets that ahh whenever the distance between z and z0 is less than delta. And course z is point at e then distance between

Fz and fz0 can be less than 0. SO this is an ordinary Epsilon delta definition of continuity

of the functions small f in this family at the point z0 ok. Now what you should notice

is that you know this delta ahh this delta depends on ahh of course is delta depends

on epsilon ok and this delta also depends on the point z0 and it also depends on the

point ahh f. It also depends on the function f ok, so delta

delta is actually delta of f, ahh f,z0, epsilon, then if you change for you know if you keep

the function f the same, if you keep the epsilon the same but if you change z0 the delta will

change ok that is the dependence of delta on z0 ok and of course if you change epsilon

of f also the delta will chose ok. So this delta depends on these 3 things alright. Now you see ahh suppose that you know you

are able to find a delta that is independent of this f okj suppose you find delta that

does not depend on f ok that means the same delta for the given epsilon the same delta

will work for every f ok for a given epsilon and given z0. The same delta will work for

a every f small if in the family script f, if that happens we say that the family is

equicontinuous at the point z ok. If delta is independent ok ahh if if a delta

if a delta independent of f and depending only on ahh z0 and epsilon ahh can be found

for every epsilon greater than zero we will say ahh f the family f is equicontinuous at

z0. So this is the most of equicontinuity ok. So for all the functions in the family

you know you are saying that ahh the function values can be made ahh the function values

near the function the function values at points near to the point Z0 can be made to within

epsilon distance of a function value at z0. If you choose a sufficiently small neighbourhood

of the point z0, but the same neighbourhood works for all functions ok, it works uniformly

for all functions right, so you know so the point is that you are able to find there is

no dependence on the particular member of the smallest f of the family script f that

is the whole point. Of course ahh that is clear that if a family is if a family is if

your family is equicontinuity at the point. Then it should be continuous at that point

ok, because equicontinuity is stronger than ordinary continuity alright

and ahh and the about this equicontinuity is that you know this is one of the ingredients

ahh one of the hypothesis that in the context of the Arzela Ascoli theorem or the Montel’s

theorems, it will ensure that you know ahh it always extracts as a sequence of functions

which converges uniformly ok. So this is equicontinuity and so that is one

of the technical ingredients. So you know if you if you take any two points if you take

any 2 points in this disc ok then the distance between them is certainly a 2 delta aright

and the distance between the function values by a triangle inequality is less than 2 epsilon

alright. So what this also tells you is that it tells you that ahh ahh it tells you that

each f will be kind of uniformly continuous ahh on each of this discs ok. So but anyway see the other thing that one

wants to worry about is ahh so uniform boundedness, so we say script f is uniformly bounded on

e if of course you know ahh in all these in this argument I have not have still not use

the compactness of the subsidy right, but I could have defined it for any subsidy of

the complex plane right is definition makes sense for any subsidy of complex plane. But the point is that the compactness is one

of the ingredients for theorem ok. So ahh of course in all these things I do not have

assumed these compact but I am keeping e compact in view of these terms right. So so we say

f is uniformly bounded on e if ahh modfz is less than equal to m or all z in e and for

all small f, so this is uniform bounded ok, so of course boundedness of a function composite

function values means that is modulus bounded ok. So for all values of the functions you take

the modulus all this module i they do they are bounded above by some positive real number

ok, you are able to find some positive real number m such fact that modz ahh is always

less than equal to m, so this m is a bound for f or modf ok and you want the same bound

to work for every smallest f encrypted, if that happens you say the family is uniformly

bounded on e ok. So you have so you have these 2 facts and

now comes the ahh now I can say this Arzela Ascoli theorem, so I am stating only one version

of the theorem which is the version that we need but there are versions of the theorem

for defined on compact whole work spaces with functions taking values in ahh metric spaces

and so on so for and they are very general questions ok. But this is the version that period that is

the version that I am going that I am define ahh that is the version I am going to state

and improve, so ahh so here is with ahh let script f the family of continuous complex

valued functions on the compact subset e of the complex plane

ok ahh suppose ahh f is uniformly bounded on e ok then the following are equivalent

number 1 f is equicontinuous at each point of e. Number 2 every sequence of f has uniformly

convergent subsequence. So this is the this is the version of the Arzela Ascoli theorem

that you know ok. So you have so again let me explain you have this compact subset e

in the complex plane, so the compactness is very very important alright and you have script

f in the family of continuous functions defined on this complex set e and taking complex values

alright. And you put the condition that this family

is uniformly bounded on e ok, so there is there is an there is a positive M which is

an upper bound for the modulus of Fz for all ze and for all small f in script f ok. Then

the the Arzela Ascoli theorem actually tells you that the condition for being able to extract

a uniformly convergent subsequence from any sequence in the family is equivalent just

demanding that the family is equal continuous at every point of e ok. So each pic for a family of when you have

uniformly when you have uniform boundedness ok, then equicontinuity is equivalent to be

able to extract a uniformly convergent subsequence ok, this is all you can state a elegant, if

you are having functions defined on a compact set ok which are uniformly bounded, we have

family of functions which are defined on a compact set and suppose a family is uniformly

bounded. Then what is the condition that is equivalent

to being able to extract a uniformly convergent subsequence from any given sequence of functions

the condition is simply the equicontinuity of the family at each point of the compact

set ok. So this is the Arzela Ascoli theorem right and ahh what I am going to do is and

what I am going do is I am going to go ahh ahh I am going to next one to the Montel theorem

which I need to ahh which I need to use in the proof of Riemann mapping theorem. But the Montel theorem is ahh but the true

the Montel theorem I need only ahh the implication that one implies to I do not need the other

part of the currency is 2 implies 1. So what I will do is I will just indicate how to one

implies 2 and 2 implies 1 is a reasonably easy exercise ok and in fact even the proof

that one implies two just parallel the proof ahh that you would have seen in the real case

in a first course in real analysis ok. So so let me do that, so proof of ahh 1 implies

2 so you see ahh so you know this is a ahh this is a standard technique of diagonalization

that is used to prove this implication Arzela Ascoli theorem even the even for ahh real

value function defined on you know a close bounded interval on the real line ok the same

proof we got ok. So ahh how does I begin so what is given you are given a compact subset

e in the complex plane. You have given a family script f after a composite

value functions on e and you are given that this family uniformly bounded so that is this

constant m which bounce the modulus of the function values at each of e and for every

function the family uniformly is the same constant works regardless the point and regardless

of the function ok and what is given to me is it is equicontinuity is given to me ok. So so what you do is you you make for the

fact you make use of the fact that ahh ahh if you you know on the real line if you take

the rational numbers take the points a rational that countable and the distance ok. So similarly

if you take the plane which is it ok to ok if you take all the points with rational coordinates

then that is accountable ok, and it is also dense ok, so this existence of a countable

dense subset is what is used ok. So what we do the you do the following thing

like ahh E subQ D the set of points of D with rational points ok, so ah of course you know

here ahh in other words I am looking at ahh points that complex plane as points on points

of r2 and I am going to rational I mean both the real and imaginary parts are rational

ok. So that is each of Q is actually ahh set of all ahh x+iy which so that xiy ok. Then of course you know that ahh then you

know that piece of Q is countable ok and it is and it is dense in EQ E because its closure

will be its closure will be equal ok then E sub Q is countable and E sub Q closure will

be triumph, because E sub Q is just E sub Q is just ahh you know E intersection with

q cross q, Q crossed you if you think of CS r cross r the points in the complex plane

every complex numbers as rational coordinate should be Q cross Q. And how do you take EQ is you will just EQ

by intersecting E with Q cross and you know the subset up a countable set count you know

Q is countable therefore Q cross Q is also countable alright and therefore a subset of

countable that worries of Q also count and rational numbers are dense ahh and therefore

E subQ will be dense E and E is the closure of E sub Q and the complex plane and you get

back Q ok. So read this first right now the whole ahh

you see this is something of a mystery for example you know when this all facts that

you keep using all the time but if you really want road the marital deeply in a certain

way when you are only perfect. For example you know rational numbers countable ok which

means that you know all the rational numbers can be put in a significant sequence ok. So I can write rational numbers as a sequence

xn ok and that is very ahh that something that you cannot imagine okay because given

any rational number ahh your your your enumerating rational numbers in some order alright, but

then the usual order that you know of the usual order that you know of on the real line

you can tell what is the immediate next rational number two given rational number simply because

how worth goes you can always find the rational number close to it ok. So you cannot see what is the next version

number but here is very or using some you know very abstract settle to say that the

countability allows you to index and you know enumerate all the rationales ok, so this si

the high set a abstracting that you use alright that something that in practice only we really

cannot do it, you will got expect to do it right, so ahh in some sense that will be connected

to the axiom of choice ok which is as you know see it well ordering principle and Jones

Lamar. And these are all and Jones Lamar is not a

limit it is it is actually a result that is actually an axiom which you accept and you

cannot do it only you can prove it only if you are seeing each other equal informs memory

Johns Avenue reaction of choice as well ordering alright, so this is the depth of section that

is involved but if you say its bread and butter when you do analysis ok. So you therefore you know I had written E

sub Q as zi and ok this is the reading complex subset, you write all that all points in E

you any write them alright and you all you write this completely existential you really

do not know what is it one is rz it to it is all you know is that you can write all

this points like this that is used and I will use that as you say. So so you write like

this, so this corresponds to what this corresponds to the fact that this is countable ok. Or it also corresponds the fact that this

is countable and ahh therefore we can order it using real numbers in using natural numbers

therefore if you choose some ordering with respect to write the natural numbers you get

a secret and that’ is the sequence I am writing ok, so this si something we obstruct right

but what you do is now what you do is you do the following thing, you take. And of course you know what I am supposed

to do if I am supposed to take a sequence in the family and I am suppose that produces

subsequent which is converges uniformly ok not just converges but I wanted to convert

uniformly right. So so start with start with sequence fn in this family ok consider ahh

so take the first ahh point ok and apply fn quit considers and what will happen is since

modules of all this fellows is less than equal to M which by the theorem. We have a convergent subsequence

which we write as fn1 of z1 ok, so this is fn 0f z1 is a sequence ahh of all numbers

which is bounded, so it convergence of sequences and called converges sequences f n1 of z ok.

So this fn1 is the subsequent fn fn1 is the subsequence from fn right, now what you do

if you repeat this process you repeat this process with z2 adn with the functions in

the subsequent ok. Now consider what you do is you take this

fn1 you take applied to set and ahh let anyone verify and wanted, so I get sequence of function

values of z2 ok and again the same argument works again modulus of fn1 of z2 is less than

or equal to M ahh will tell you ahh that theorem will tell you there exists a subsequence ahh

fn2 of z2 of fnof z2 which converges ok. So you see in the first step I am applying all

the members of the field sequence ahh to the point z1. And from that I get a subsequence, in the

second step what I do if I forget that the points and once but I only look at this subsequence

fn1 and apply z2 to do it and again apply ahh theorem and give the uniform boundedness

to show that subsequence of a subsequence ok which converges at z2 ok and notice see

fn1 already converges Z1 and fn2 is a subsequence of a fn1. Therefore fn2 will not only converge

z2 it is also converge z1 ok. So note that fn12 of z2 also converge ok,

so now you go by induction ok by induction the we get for every n greater than equal

to 1 a subsequence. So let use M or k a subsequence fnk of fnk-1 such that fnk of z0 converges

for j less than or equal to j ok. So you know ahh so this is this is what is happening right.

So you know if you if you ahh write it in a pictorial ahh, so you have n1 which is ahh

with the property that fn1 fn1 convergys at Z1 alright. Therefore if I write it I will get f I will

get fi z1 fi2 of z2 and let me so let me write fi3 of z3 again z1 and so on. So this is this

is converges at z0 alright then I will get so this is sequence fn1 then for this I get

subsequence fn2 ok, this fn2 is subsequence of a fn1 that means all the integers that

occurs here they are among the indices among the indices, but still I write it only in

this order. So now I will write it as fj1(z2) fj2(z2)

fj3(z2) and so on and this will converge at both z1 and z2 ok and mind j1 and j2 j3 is

subsequence of i1, i2, i3 and i1, i2, i3 is subsequence of rational numbers ok. Then if

I again repeat the process once more I get a fn3 I get this subsequence which is a further

subsequence of this and if I write the indices as fk1 of z3 fk2 of z3 and fk3 of z3 and so

on. Then I get the subsequence of this subsequence

which converges at z1, z2 and z3 ok, I get this ahh situation like this aright and now

what you do you know you take the diagonal sequence of functions and you take this you

take this you take this you take this is called diagnologist, you take the diagonals of subsequence

ok. So consider consider the subsequence ahh ahh so you know I wll give it up. I give it a special symbol I called capital

F FL is actually Lf member of fnl, this is how it bigin. So F1 is first member of fn1,

F2 is second member of a fn2, F3 is third member of fn3 that is your different ok. So

this is so this so this lie here is F1 this lie here is F2, this lie here is F3 and and

so on, so this diagonal sequence of little ok. Now the beautiful thing about design a

sequence is that converge at every point of EQ ok. So that is the and that is the power of the

diagonalization process, you are able to extract this, after all you want a sequence which

converges on all of E ok, but then you know that because everything is continuous ok ahh

if you can get uniform continuity and dense open subset of E ok, then you will get everything

alright and what is the dense of open subset E is this open subset and what you are what

helps in the diagonalization process is the fact that this count ok. That is what allows you to enumerate and then

expect this diagram alright. So what is the point. So the the fact that lies is that fn

ahh Converges are F I think are useless L l greater han 1 converges on EQ, so this is

the bit of it, why because you see why is that true because you see F, what is fl, (FL)

is actually lth member of Fnl where if you take the sequence fn ahh ahh l converges at

z1 etc of zl ok. See if this si the lf member of fnl alright

and the but if you look at fnl that sequence converges at all points up to zl alright and

therefore you know if you give me any point of ahh EQ that point will be because of this

is a numeration ok which is as far as I told you any point of Eq is some zl any point of

Eq some zl and but you know fnl the sequence fnl will converge zl alright. Therefore all ahh fl is greater than l which

apart come from subsequence of this, you will also converge that and therefore this itself

will converge at zl ok. So let me write that if z belongs to Eq then z is zl and since

fnl of zlconverges we have that the subsequence ft of zl t greate than equal to l converges

ok and you for a sequence of function it converge a point it is enough to converge belongs of

the state. So what I am saying is that the sequence fl

will you know converge ahh at the point zl at least aftet zl but and this is, so this

will prove this one alright. So the moral of the story is you are able to extract here

subsequence which converges point wise on this set of point ok. Now from this and equicontinuity

we can show ahh that you have ahh that this that this seqwunce is equals the diagonal

equal you extract it converge is actually uniform and that will give to proof of it.

So I will continue with that in next video.

## 2 Replies to “Mod-10 Lec-38 Arzela-Ascoli Theorem: Under Uniform Boundedness, Equicontinuity and Uniform”

nice explanation

very nicely explained .