 Okay, so we continue with this proof, so you know, so let me point some, let me write down something here that I had probably said but I did not write down. So you are looking at the open cover of X centered at the points X1, X2 and so on of X, so which open cover is it, it is the open cover consisting of open balls of radius delta okay that is something I did not write down, so let me write it down here. So look at the open cover of open balls of radius delta, so you look at the open cover of this space X of the metric space X which we have assumed compact okay and you are looking at the open cover which consists of open balls of radius delta that is this delta that we have chosen that we have gotten above corresponding to the epsilon or rather epsilon by 3 okay because of continuity okay. So and your for the open cover you are only looking at balls of radius delta and the centers are not of course you could have taken centers to be all points of X but then you take the centers only among the points X1, X2 and so on that is you know that is the countable dense subset of the metric space that we have cooked up okay and so this is actually in principle this is a delta net for X and you know X is compact so it is totally bounded so it every I mean so what I am saying is it is not a delta net it will give rise to a delta net because it is an open cover of open balls of radius delta okay and because of compactness this open cover will give rise to a finite sub cover. So those finite sub cover will be centered at finitely many points which are among these XIs and we label those points by XI1, XI2 etc XIM and you take the balls open balls of radius delta centered at each of these finitely many M points and you take the union you get X that is so this collection of points finite collection of points among the countable dense subset okay this is a delta net for X okay and well we will have to work with this so what we need to show is that we need to show that you know we have to show that this the whole idea is to show compactness of a family of functions. So we started with an arbitrary sequence and then we try to verify sequential compactness and we have gotten hold of this subsequence this g n okay by the diagonal argument and we have to show that this subsequence converges on all points of X at all points of X okay and uniformly and mind you the subsequence is already been cooked up in such a way that it converges on the countable dense subset okay. Now so I will have to show this for every small x in capital X I will have to show that for my given epsilon you can choose n and m sufficiently large namely greater than or equal to a certain capital N such that mod I mean the distance between g n and g m at X is can be made less than epsilon okay that is a you are just verifying that the sequence is the sequence of g n is uniformly Cauchy okay and uniformly Cauchy is means Cauchy with respect to the supremum norm the metric induced by the supremum norm okay so the point is that how do you get to an arbitrary point of X whereas the g n's are converging only on these points which are points among countable some the countable dense subset okay you have to you have to you have to interpolate with the X's with the X I's alright that is what you have to do and it is done very easily by the triangle inequality so what will happen is that see now you take any for any X in X okay there exists a j such that the distance between X and X ij is less than delta this is true that is because this X ij is X i1 through X im that is a delta net alright so there is such a j so pick that j take the corresponding X ij and you interpolate with respect to that X ij so now you can write out the triangle inequality what you will get is well well you are going to get distance between g n of X and well so I want to I want to look at g n of X minus g m of X so that is what I am going to write out oops the distance between g n of X and g m of X is that now you introduce that X ij which is to within a delta of X so you write this as this is less than or equal to g n of X minus g n of X ij plus g n of X ij minus g m of X ij plus mod g m of X ij minus g m of X so this is how you split the triangle inequality okay you extrapolate I mean you interpolate the X back to itself by this X ij and then you introduced you introduce this minus g n of X ij plus g n of X ij minus g m X ij plus g m X ij okay and now the point is that this now you see what you will have to understand is that the see because of you see in the first term it is the same function g n alright and by equicontinuity of the family okay the distance between function values for any function in the family is going to be less than epsilon by 3 if the distance between the arguments the variables is less than delta so you see this is already less than epsilon by 3 okay and so is this so the first and third term are already less than epsilon by 3 that is because of this is just because of equicontinuity due to equicontinuity mind you equicontinuity is has already been assumed for all functions so these are certain functions that family so it holds for them as well and then the so the first term and the last term are fine they are going to give me an epsilon they are going to be less than epsilon by 3 each the problem I have to worry about the middle term and but the middle term is not a big problem because actually you know the g n the sequence of g n it converges on the countable dense subset of the X i's which is labeled by the point x therefore this g n's are going to converge at X ij certainly okay so you see so how do you deal with this you see g n's converges on the countable dense subset consisting of X 1, X 2 and so on so in particular it will converge on you know this finite subset okay so in particular on the finite subset which is given by X i1 etc up to X i m and of course you know the X ij that I have chosen is inside this finite set okay but then the point is that you know by definition of convergence of a sequence I can certainly find an n a capital N such that small n and small m greater than capital M will make sure that this quantity is always less than epsilon by 3 irrespective of the X ij that I choose that is because there are only finitely many you may get you will get one bound one subscript for each i each j and then you take probably the maximum amongst all those okay so there exists n such that n, m greater than n implies that the distance between g n of X i l and g m of X i l can be made less than epsilon by 3 if yeah for all for all for all l varying from 1 to m okay so this can be done because it can be done for each X i it can be done for X i1 it can be done for X i2 and so on and for each you get a you get an integer and then you take the maximum amongst all those okay there only finitely may this can be managed so that is it then you are done so what this will tell you is that if m and n are greater than n okay then what you will get is that the left hand side which is g n X minus g m X is less than epsilon okay and mind you this is independent of X okay this has got nothing to do with X X did not matter right so this is independent of X so this sequence g n is uniformly Cauchy and that is the proof of theorem that ends the proof of theorem okay so we have demonstrated sequential compactness so if you to just to complete the proof accurately what we have shown is that we have shown that g n is uniformly Cauchy okay but the g ns come from this subset which is already a close subset okay so but it is a close subset of a complete metric space therefore it is also complete and therefore showing that is the g ns is Cauchy is the same as showing that the g ns are convergent the sequence of g ns is convergent and that is what you wanted we started with an arbitrary family arbitrary sequence of functions from the subset and we have produced a subsequence which converges okay so that is the Arzela Ascoli theorem right now what we need to do is that we need to now go back to complex analysis and look at our the kind of functions that we are interested in we are interested in you know analytic functions and then we are interested in of course Meromorphic functions that is our final aim so what does it mean for Meromorphic functions so the beautiful thing is that you know I mean for to begin with at least let us say analytic functions the beautiful thing is that you know see the main point is the following what does Arzela Ascoli theorem say it says that if you are for example looking at continuous bounded real valued functions on a compact metric space then if you want the compact if you want a compactness of a subspace that is a collection of functions okay then that is equivalent of course that subspace has to be closed and bounded because compactness always implies closed and bounded okay but what you need to extra put extra is equicontinuity okay so basically you need boundedness you need a closed subs for a closed subset to be compact you need it to be bounded and it has to be equicontinuous that is what you want okay and mind you this is a nice thing because it is easy to equicontinuity is more friendlier to check rather than checking something like count total boundedness which is very very difficult to check okay for a family of functions it is not so easy okay so now when you go to the context of complex analysis and if you are looking at analytic functions what happens is something very beautiful happens see the what you get is you know analytic functions for analytic functions you have the Cauchy integral formula you have the integral formulas see these integral formulas if you if you apply the so called integral inequality the ML inequality which says that the modulus of an integral is bounded by M times L where M is the maximum modulus of the integrand on the contour of integration and L is the length of the contour of integration okay this is the ML inequality if you apply this ML inequalities to the Cauchy integral formula what you get is what you get are called the Cauchy estimates okay so the beautiful thing is that for analytic functions you have Cauchy estimates okay and what these Cauchy estimates will tell you that is that if you are working on a compact set okay it will tell you that the derivatives are bounded okay because mind you the derivatives of an analytic function are given by the Cauchy integral formulas the general Cauchy integral formula will give you the nth derivative okay so if you therefore you know the bound the bounds for the derivatives are given by applying the ML inequalities to the Cauchy integral formula so what happens is that if you have an analytic function on a compact set for example on a closed disc if you want then all the derivatives are all bounded okay all the derivatives are bounded so what happens is that in some sense you get boundedness of derivatives but the beautiful thing is once you have boundedness of derivatives that always implies something stronger for the original functions it implies equicontinuity so what happens is if you are working with analytic functions equicontinuity is automatic okay equicontinuity is just automatic so you know therefore what happens is that you know just uniform boundedness will give you sequential compactness that is the big deal the big deal is equicontinuity comes for free if you are going to work not just with continuous functions but if you are working with analytic functions okay that is the philosophy okay that is the direction in which I am going to explain how these things work so let me make this so in that generality the theorem the version of the as a last called theorem is a very important theorem it is called Montel's theorem okay so we go on to Montel's theorem which is the key ingredient for proofs of many important theorems in complex analysis including of course Ray the including of course the Rayman mapping theorem and you know the Picard theorems and so on so let me put this Montel's theorem so here is Montel's theorem so let D in the complex plane be a domain of course I am always assuming it is an open connected set and it is non-empty of course okay let so let script F so let script F be a family you can call it as family or collection whatever you want be a family of analytic functions on D which is normally uniformly bounded on D okay so what is this normally uniformly bounded whenever you say for a certain property if you say normally that property it means that that property is to be verified not for all sets but it is to be verified only for compact subsets okay so when I say normally uniformly bounded on D it means that it is uniformly bounded on compact subsets of D okay so let me write that that is uniformly bounded on compact subsets so I am using abbreviations UFLY for uniformly BDD for bounded CPT for compact okay so suppose it is normally uniformly bounded okay then you have sequential compactness okay then you have compactness basically okay but the only thing is that you should see compactness as sequential compactness okay and the sequential compactness normally what does it mean it means that given any sequence you have a convergence of sequence but the only thing is here it is not just convergent on the whole it will be you will get a convergence okay which first of all it is a convergence of sequence of functions so it will be normal it will be a uniform convergence okay mind you whenever you are talking about convergence for functions it is always a kind of uniform convergence alright for example that is how it is if you are looking at the continuous complex valued or real valued functions on a compact metric space okay so it is uniform convergence but it is not again just uniform convergence but it is uniform convergence restricted only to compact subsets so it is uniform convergence in the you know normal sense okay so that is the kind of sequential compactness that you will get so that is the that is the result okay then every sequence in F has a subsequence that converges uniformly on compact subsets okay there are several little subtleties in this in the statement of this theorem which I will try to explain okay now so let me give you let us go to the proof of this okay so what I want to tell you is that you see in how do you contrast this with respect to the usual Arzela-Ascoli theorem so the usual Arzela-Ascoli theorem is for functions defined on a compact metric space okay that is the first thing then the second thing is in the usual Arzela-Ascoli theorem whenever you are talking about convergence it is uniform convergence okay it is just uniform convergence it means uniform convergence on the whole space right and the third thing is that the Arzela-Ascoli theorem there is that if you want compactness which is the same as sequential compactness okay that is every sequence has a convergence of sequence that for that you will have to give boundedness which is actually uniform boundedness okay plus you have to give you continuity okay now the big deal is when you come to complex analysis when you come to analytic functions I have already told you that the problem with analytic functions is that the convergence always is never uniform on a on a on a domain it is only uniform when you restrict it to compact sets so you have to change the convergence to normal convergence so you must not require always convergence you should require convergence only on compact subsets okay that is the first change you have to make the second change is that you can get rid of you do not need unique equicontinuity okay and so you just get uniform in a sense you are you are saying that uniform boundedness implies sequential compactness that is what you are saying okay but the beautiful thing is that the sequential compactness is with respect to normal convergence okay that is one important thing the other thing is the functions are not defined on a compact set they are defined on a domain okay that is another difference in the usual Arzela Ascoli theorem you are looking at functions they are defined on a compact metric space whereas here you are looking at analytic functions which are certainly continuous but they are defined on not a compact set they are defined on an open set open connected set okay that is the difference these are the differences you so now you but you can see that the point is that you are able to you when you come from the real top from the topological side to the complex analysis side okay you replace convergence uniform convergence by normal convergence you replace you forget equicontinuity because it comes for free okay so how does one prove this? So the proof is very very simple the first thing I want to tell you is that if you are see if you are looking at a compact subset of the domain then there is nothing great because you can directly apply Arzela Ascoli theorem okay and you have to use the bounded the Cauchy estimates okay so for example let me tell you so suppose suppose Z0 is a point of D okay let rho be greater than 0 so that the disc mod Z minus Z0 less than or equal to rho is in D okay you choose a sufficiently small radius so that the close disc is inside D alright of course I can always find an since Z0 is a point of D of D which is an open set I can always it is an interior point so there is always a disc open disc surrounding Z0 is also in D now you take disc of slightly shorter smaller radius okay and that close disc will also be in D you can take that as your rho okay so you the reason for taking the boundary also is you know pretty well because then I get a compact set because a close and bounded set so it is compact and once it is compact I can apply all the hypothesis I have so now what happens is watch that you know if I take this if you take the family if you look at a family of analytic functions on D and you restrict it to this close disc okay what you are getting is a family of continuous functions complex valued functions mind you analytic functions are continuous of course okay they are complex valued continuous functions and you are restricting them to a closed disc which is a compact subset it is also compact metric space so actually you know Arzela as you are in the situation of Arzela Ascoli theorem the Arzela Ascoli theorem actually needs only continuous real or complex valued functions defined on a compact metric space okay so you are in that situation alright so since you are in that situation you are already given that this family is normally uniformly bounded it means that it is uniformly bounded on compact subsets therefore this family is bounded on this close disc because it is a compact subset so you already have boundedness you have boundedness of the family which is actually uniform boundedness okay you have that already. Now what more do you require for extracting a convergence subsequence from a given sequence what you require is that you require equicontinuity okay but the point is that because of analyticity and the Cauchy estimates equicontinuity is automatic so let us see why that is true you see that so let me say the following thing well if Z prime is in the set of all Z such that mod Z minus Z0 is less than or equal to rho okay you take a point here and then you see what will happen is so you know if you want let us take something that that is right so yeah fine so what you do so let me draw a diagram so that it is easier to visualize so here is so here is my Z0 and here is my close disc centre at Z0 radius rho and here is my Z prime okay now what you do is that well so notice so you can see you know you can choose a delta such that the close disc centre at Z prime radius delta lies inside this close disc okay so I can choose a delta like this okay choose delta greater than 0 so that mod Z minus Z prime less than or equal to delta lies in mod Z minus Z0 less than or equal to rho you can do this okay and then now you do the following thing you look at what is look at the Cauchy integral formula see by the Cauchy integral formula well this is the well this is the second Cauchy integral formula which is for the derivative the first derivative f dash of Z prime okay is what it is 1 by 2 pi i integral over integral in the positive sense over this circle mod Z minus Z prime is equal to rho delta okay of f Z t Z by Z minus Z prime alright and I am going to get the whole square okay this is the first Cauchy integral formula right or rather second Cauchy integral formula okay the first Cauchy integral formula is for the function itself which is a 0 think of it as 0 derivative okay so this is a Cauchy integral formula this is true for all functions f in f this is fine this is because after all f is a family of analytic functions so this is true now you now apply this I am just trying to write out the Cauchy estimate so mod f of mod f prime of Z prime is what this is going to be modulus of this integral but that is less than or equal to the maximum value of the integrand multiplied by the length of the contour which in this case is the circle of radius delta okay centered at Z prime so what I am going to get is I am going to get so this is less than or equal to I am going to get 1 by 2 pi is what I am going to get if I put a mod here and the length of the contour is well it is no if you put a square then it is an f prime if you if you put so what is the Cauchy what is see f not is if you put this is the first formula so if you want the nth derivative you have to put n plus 1 okay so what is this so I will get see if I calculate the modulus of this I will get mod f Z okay so you know let me put let me put m here so I am so for the modulus of f Z I am going to get an m okay and then for the mod Z minus Z prime the whole square see that is mod Z minus Z prime is delta because the variable of integration is Z and the variable of integration lies on the region of integration which in this case is this contour which is positively oriented this is the orientation the usual positive orientation so this is m by delta squared okay and I am going to get and I am going to get the the length of the contour and that is going to be 2 pi delta okay so basically I am going to get m by delta and what is this m see this m is the common bound for all the functions in your family on this closed disc on this big closed disc that is because that is given you see you are given that look at go back go above and look at this see you are given you are given that this family f is a family of analytic functions it is normally bounded normally uniformly bounded on D so it means that it is uniformly bounded on every compact set on every compact subset of D so on this closed disc of center at Z not and radius rho which is of course a compact subset of D it is bounded okay all the functions are bounded and I am taking that bound to be m so let me write this where m is uniform is the uniform bound for all f in the functions f in this family script f on this disc center at Z not radius close disc with radius rho okay you can put this bound independent of delta also okay so you can you can so this delta that I chose seem to depend on the Z prime all right but then you can get rid of this delta so that I can get a you know uniform a uniform bound for the derivative f dash is as follows see the first thing that you can notice is that you know I can change this this contour integration which is the this smaller circle center at Z prime radius delta to the larger circle which is mod Z minus Z not is equal to rho and I can do this that because of the fact that f is analytic in the bigger closed disc and also the point Z prime is also enclosed by the by the bigger circle okay so you know the the therefore this formula for f dash of Z prime is valid okay so so in this way I have gotten rid of the delta in the integrating contour okay then the other thing is I will have to get of the delta here appearing the bound here and so for that what I will do is that I will just have to restrict Z prime to be with inside a you know a circle of radius rho by 2 okay from from I mean centered at Z not so restrict you know is at prime to mod Z minus Z not is not equal to rho by 2 okay if you do this then you see this effectively makes this delta which is supposed to be the you know the distance between the point Z which is on the contour of integration and the point Z prime which is inside the contour this distance from Z to Z prime what will it be you see the Z is now going to lie on the outer contour the other circle and the Z prime inside okay and you see the minimum distance between Z and Z prime is rho by 2 and therefore you know so this delta here will essentially be you know this delta is supposed to be the distance between should be modulus of Z minus Z prime okay and that that distance is at least rho by 2 alright and of course therefore you know 1 by delta squared will become less than or equal to 4 by rho squared so basically instead of this delta squared and you will get a rho squared and I get a 4 here okay the inequality will get reversed if you take reciprocals and of course this this 2 pi delta will become 2 pi rho because that is the length of the the contour the which is the length of the outer circle so in effect this bound will become 4 m by rho okay and that will become a bound that has got nothing to do with delta alright now you see of course m is the uniform bound for all the functions in the family on the on the bigger closed disc centered at Z dot and radius rho but what does this give you you see now you calculate what is f of z1 minus f of z2 okay what will this will what will this be if you take z1 and z2 inside this to be two specific instances of Z prime so so here so here is z1 if you want and here is z2 okay and of course you know this integral is going to be independent of the path so long as the path is you know inside this closed disc that is because f dash is analytic there and what will happen is that you know you are going to get this is by the m l inequality this is less than or equal to integral from z1 to z2 mod f dash of z dz mod dz and now you know this now you can apply this bound this mod f dash of z is is less than or equal to 4 m by rho times this mod dz is going to give you mod z1 minus z2 and that is for example if you take the straight line segment from z1 to z2 okay and this is valid for all z1 and z2 in this closed disc mod z minus z0 less than or equal to rho by 2 okay so you see what does this tell you this tells you that the you know the I can make the distance between fz1 and fz2 you know small the moment small enough the moment I can make the distance between z1 and z2 small enough okay and this is and in a way that is independent of the particular choice of z1 and z2 okay and also in a way that has got nothing to do with the function f because this m is a uniform bound for all the functions okay and that is exactly saying that f all the functions f the whole family functions script f okay that is equicontinuous on on this disc centered at z0 radius rho by 2 okay and that is how you get equicontinuity for free okay. See the last inequality tells you mod fz1 minus fz2 is less than or equal to some constant times z1 minus z2 okay so what that gives equicontinuity that is like a Lipschitz condition you see given an epsilon you choose carefully the delta and the way you choose delta is independent of z1 and z2 so independent of z1 and z2 you are saying the distance between f of z1 and f of z2 can be made lesser than epsilon whenever z1 and z2 are within a delta and independent of z1 and z2 that is equicontinuity you see it is actually it is beautiful it is uniform it is a kind of uniform continuity because you are it does not you do not worry about whether it is z1 or which z1 or z2 it is and you it also works for all functions f so it is a kind of uniform equicontinuity that is what you get I mean that is the power of this of the Cauchy integral formula that you are using okay and this is what you get if you assume analyticity you get this you get equicontinuity just like that okay so the only thing that is left is uniform boundedness okay but that is already assumed so you get sequential compactness but in the normal sense that is the that is the point that is modulus 0 okay all right