 Alright, so we are continuing with the proof of the Arzelauskillo theorem alright, so what we have done is we have a set E which is a compact set in the plane alright and we have this family script F given to us, it is a family of continuous complex valued functions we have assumed that this family is uniformly bounded alright and we have further assumed that this family is equicontinuous at every point and we are trying to show that given any sequence in this family there is a uniformly convergent subsequence ok, so we took the points the rational points of E ok the points of E with rational coordinates ok namely points of E which are complex numbers whose both real and imaginary parts are rational numbers that is a countable density E and we have been able to extract using the diagonalization metric we are able to extract a sequence capital F sub L as a subsequence of the given sequence Fn ok with the property that this sequence the subsequence converges point wise on every at every rational point of E ok that is what we have proceeded up to so far alright. Now what I am going to do now is the time I will have to use see so far what I have used is only the I have just used the fact that the family is uniformly bounded I have just used the fact that the family is uniformly bounded and I have not used anything else whereas I still have to use the equicontinuity of the family and I will have to use the fact that everything is happening on the set E which is compact I have never used the compactness of E right we are going to use that. So how do we proceed so we do the following thing so note that so we have equicontinuity of the family script F and therefore this holds for any collection of functions in the family so it also holds for these Fls ok so let me write that down since F is equicontinuous on E so is F sub L ok after all this is just this is also a sequence in script F in fact this is a sub sequence of the given sequence Fn that we started with right. So given z0 in E ok so you have this definition of equicontinuity what it tells you is that suppose I fix an epsilon you start with a start with an epsilon given to me ok so given z0 in E and an epsilon greater than 0 ok there exist delta greater than 0 with the delta depending only on z0 and epsilon such that well the distance between you know so I should so let me write it very generally distance between F of z and F of z0 is less than epsilon whenever the distance between z and z0 is less than delta ok and here I can put for small f any member here in fact I can put it for any member in script F ok for our small f in script F in particular I can take for small f any of the capital Fls also ok this is just equicontinuity of the family script F that I have assumed so I am using that now. Now what you do is so you know for every if I fix this epsilon then you know for every point z0 so I am varying z0 then I get this disk I get this disk centered at z0 radius delta so there are 2 aspects to it first thing is as you vary the z0 ok then of course the epsilon is fixed alright but you vary the z0 then you get these disks the deltas will also vary because the deltas depend on z0 also but then you will get one disk surrounding each z0 in E ok and that will give you an open cover of E ok and now use the fact that E is compact to extract the finite sub cover ok so that is one part of the argument. So let me write that down the collection mod z-z0 less than delta delta of z0, epsilon as z0 varies over E is an open cover of E of E which is compact so here I am using the compactness of E and I am also using the I am also using the equicontinuity of the family ok. So admits a finite sub cover so you know you will get so the sub cover will be given by mod z-z1 less than delta of z1, epsilon mod z-z2 is less than delta of z2, epsilon and so on mod z-zk less than delta of zk, epsilon so I get this sub cover alright and of course you know I do not know all I know about the set E is that it is compact I do not know about anything about it is interior and so on and so forth therefore each of these sets I mean each of these disks they need not I mean this the union of all this contains E ok but it could be well bigger than E alright. So so here also whenever I say mod z-z0 less than delta I am only looking at those z for which lie in E ok I am not looking at z in this disk which does not which do not lie in E because then if z does not lie in E f of z does not make sense ok because all the functions in my family are defined only on E ok so whenever I write things like this I am assuming that the variable is also in E right. So you have this finite sub cover and let me call them zeta 1 zeta 2 because otherwise I could get into trouble zeta k so you know I have so I use z I use different symbols so that I do not confuse them with z is which are supposed to be all the points of E sub q ok. So you have you have this cover alright and then you see in each of these things in each of these disks alright of course I will be able to find a point of Eq ok I can find one point of Eq in each of these disks ok. So choose choose z I 1 in this disk mod z-zeta 1 is less than delta zeta 1, epsilon similarly z I 2 in mod z-zeta 2 less than delta of zeta 2 of epsilon and so on you choose z I there are k of them. So choose one point in each of these disks so this is zeta k ok where of course you know the zeta ij are in Eq which has been illuminated as z i z n n greater than delta ok. So you choose one rational point from each of these disks ok I have finitely many disks which cover E ok and in each of these disks you choose a rational point alright choose a rational point. Now what you do is that see what you should now notice is that if you give me any point of E give me any point of E that point has to lie in one of this ok and corresponding to that you have a certain z ij ok. So you know so that is how I actually extend all my results to for example convergence to any point of E ok. So you know if you want if you think of so if z belongs to E then z is then z-zeta zeta I let me use some notation then mod z-zeta is less than delta zeta sub l, epsilon for some l with 1 less than or equal to l less than or equal to k this will happen ok and z i l is also in mod z-zeta l less than delta zeta l commands ok. So I have this right now you know basically what I am trying to show I am trying to show that my aim is you know starting with a sequence fn which I have picked in this family f for which I have extracted the subsequence which converges on the rational points I am trying to actually show that this subsequence converges uniformly not just converges but converges uniformly and that too not only at the rational points but on all of E I am trying to achieve both in one stroke. So here what I have is that this subsequence converges only on the rational points ok and converges on the rational points means point wise converges it does in uniform convergence alright but what I am trying to show is that not only does it converge at the rational points I am trying to say it converges everywhere and not only I am trying to say it converges everywhere I am trying to say it converges everywhere uniformly ok that is what I am trying to do. So how do I check that it converges uniformly so you know I will have to just check that you know I just have to check that at any if I substitute any z here in this sequence of functions I get this sequence of complex numbers complex values I have to show that that sequence is uniformly caution ok I have to show that that is uniformly caution right if I show that then I will get uniform convergence of that sequence. So what I will have to do is that I will have to compare fm of z-fn of z for m and n sufficiently left this is what I will have to compare right now what you use is you use the fact that you see this z is in this which contains the point zeta l alright and which also contains the zil so what I do is I add and subtract the function values at these two points. So what I do is I write it as fm z-fm of you know if you want zeta l then plus fm of zeta l to undo the effect of adding it then I put minus fn of zeta minus fn of now I introduce the point zil then I undo by adding fn of zil and then I write the final fn of z then I use a triangle inequality to and group these use the triangle inequality to get mod fm of z-fm of zeta l plus this is less than or equal to this plus modulus of fm of zeta l minus fn of zil plus modulus of fn of zil minus fn of z okay I use so you know I am so basically the so basically you know the diagram is like this so here is if you want this is E okay and there is this point zeta l so there is this point zeta l which is a point of E okay and there is this disk centered at zeta l radius delta okay and this intersects E where this intersects E that is where I have chosen my point z okay my chosen point z is there plus this also contains one point of the set of rational coordinates which I have taken as zil zi sub l so that is also here so this is zil okay so this is my situation right so maybe I will so this is E and this is zeta l so this is my situation right now I mean basically if I show that for m and n sufficiently large if I can make this less than epsilon independent of z okay then that is enough to show that the sequence capital F or sequence of capital F else they converge uniformly on E alright so that is my aim my aim is I want to find m a bound beyond which for all for all values of m and n beyond which for all values of z okay I can make this less than epsilon alright so what do I how do I do that so you know the so the first thing that you must notice is that you see this the first expression in the last expression are the difference in the function values the same function involving the same function so this involves fm and this involves fn alright and of course all these F's are all part of the family script F which has equicontinuity alright so I will have to make an adjustment so that you know this term works out to lesser than epsilon by 3 this term works out to less than epsilon by 3 and this term also works out less than epsilon by 3 so that when I add all the 3 it works out to less than epsilon which is what I want so how do I make this and this less than epsilon by 3 it is very simple because I will use I will use equicontinuity in the neighbourhood of zeta L this delta of zeta L, epsilon such that fz-fzeta L in modulus can be made less than epsilon okay but I can choose that epsilon to be less than I can make that epsilon epsilon by 3 so what I do is that you know here I do I make the I go back and make this change so what I do given epsilon greater than 0 alright you choose a delta alright which depends only on z0 and epsilon and not on the function f in the family such that this is not less than epsilon but let me make it epsilon by 3 okay suppose I do this alright then would I be done so instead of taking the deltas you make the change of taking delta by 2 okay that still continues to be an open cover alright and so I will have to change to e to delta by 2 everywhere okay so I get a finite sub cover and then I choose one rational point in each of these finitely many open discs which cover e alright and then I note that if z is a point of e then this is going to lie in one of the sets because they cover e okay so it lies in one of the one of these discs I choose the I choose that one and call the center zeta sub l alright and I also have a I can also get a rational point of e in that set okay right and now I write this but when I write this expression what I do is I do not I try not to bring in this center okay but I just work with this rational this point with rational coordinate that I have chosen so instead of this I put zil here and also zil here okay so then I am in good shape so you see I get zil here I get zil here and now I have the same zil in the central term so you know if I look at the distance in z and zil that will be less than that will be less than delta by the triangle inequality because you know mod zil is going to be less than or equal to mod z minus zil plus mod of zil minus zil and which is less than delta of zil comma epsilon because each one is less than half delta alright and then I get this inequality I get this and of course this is zil okay and then if I look at the third one I am comparing zil and z the function values at these points of the function fn so again I do not mess up with zil but I compare with z okay and again I have the same I have the same inequality and that will tell me that that is why that will tell me that this is still less than epsilon by 3 okay so I am in good shape so I have to only theory about this fellow and here is where I use the fact that the capital F's the subsequence of capital F's will converge at every rational point okay. So since the fl converts so let me put fm converge on eq and you know this zil is a point of eq for the given epsilon there exist an n which will depend on you know epsilon and it will depend on epsilon and it will also depend on zil such that the distance between fm at zil and fn at zil can be made less than epsilon for m and n greater than or equal to this n of epsilon, zil okay this is just the Cauchy nature of the sequence fm of zil because fm of zil converges so it is Cauchy so this is the Cauchy condition okay and but then I get all these n's alright and the you know now if I start with any z okay this for that z this l may change okay if I start with any z in e this index l that comes in all this that will depend on z okay but there are only finitely many else the l can only range from 1 to k okay and therefore there are only finitely many such numbers capital N okay and so if I take m and n to be larger than the maximum of all these then the modulus of this is always less than epsilon independent of z and that gives me uniform convergence of the subsequence of capital F's okay and that finishes the proof okay so for n, m greater than or equal to maximum of all these n epsilon, zil 1 less than or equal to l less than or equal to k we have mod fm of z minus fn of z is less than epsilon okay independent of z of z in e and this independence of z in e tells you that the sequence of capital F's converges uniformly on e. So this implies fn converges uniformly this is yeah you are right I want you are right all the 3 should add up to epsilon this is already epsilon by 3 this is epsilon by 3 so this one has to be also adjusted to be epsilon by 3 so I will put epsilon by 3 here so you choose n so that instead of getting this less than epsilon you get epsilon by 3 yeah thank you for pointing that out only then you will get this sum less than epsilon otherwise you will get something more than that okay. So the moral of the story is so you know basically it is a game of adjusting these epsilons and deltas carefully alright and even if you go wrong a little you know how you know what to do to adjust it okay so it is something this epsilon delta adjustment is something that you always I mean do in any analysis course alright. So the idea is therefore you know you use the uniform boundedness so you know let me some let us summarize what the proof did you see you have this family script F from that family you have taken a subsequence and what you want to show that this subsequence I mean from this family you have taken a sequence not subsequence you have taken a sequence and what you want to show is that it has a subsequence which converges uniformly. So how do you get that subsequence what you do is you first get hold of a subsequence which converges on the rational points okay that uses the uniform boundedness and the diagonalization trick okay once you have a subsequence which converges point wise on the rational points then you use the compactness of e and the equicontinuity to show that it converges uniformly not only on rational points but at every point okay and what it means is therefore the if you take the limit of the subsequence you will get a continuous function you will certainly get a continuous function so in particular you know this applies to for example if your functions were all analytic functions in the interior of e okay then you will end up with an analytic function in the interior of e the limit function will also be analytic in the interior of e right. So that is the so that is one way of the Arzela-Ascoli theorem which is what we need okay there is the other statement in Arzela-Ascoli theorem which says that if you have this property that given any sequence in the family f which is uniformly bounded you are able to extract a uniformly convergent subsequence then you have to the other statement in the theorem will tell you that you will have to show that this family is equicontinuous okay and that is a proof that I am not going to go into okay it is a proof by contradiction similar to the proof that you would have seen in real analysis okay so I am not going to go into that but what I need to do is to go to Montel's theorem okay. So let me explain Montel's theorem so let me state Montel's theorem at least so this is the theorem that we actually need alright and what is the theorem well so now you see now you want a statement for analytic functions okay you want a statement for analytic functions so you must understand how to modify the statement you see when you are looking at analytic functions you know analytic functions are defined only on open sets they are not defined on closed sets in particular you cannot think of them as being defined on compact sets okay because analyticity at a point means it means there is a whole neighbourhood of the point where the function is defined so it has to be an interior point unless a point is an interior point of the domain of a function you cannot talk about its analyticity there so I cannot simply talk about you know analytic functions on a compact set does not make any sense right therefore but therefore you know how will you adapt the Arzela Ascoli theorem to the complex analysis setting where you have analytic functions. So the idea is what you do is whatever conditions you put they should be in the Arzela Ascoli theorem all conditions are conditions of functions on a compact set alright whatever conditions you have they are all conditions on a compact set the hypothesis are all conditions of functions on a compact set and the conclusion is also about functions on a compact set okay and how do you adapt it to the complex analysis setting where you want to apply to analytic functions what you do is you do take analytic functions on a domain but put all the conditions on compact subsets of the domain okay so you are working with analytic functions but whatever continuity I mean whatever conditions you put try to put them on compact subsets of the domain so this is the so called normal conditions okay for example if a sequence of functions converges on compact subsets it is called normal convergence okay if a sequence of functions is uniformly bounded on compact subsets it is said to be normally bounded okay then the natural version of Arzela Ascoli theorem for analytic functions that you can expect is if you have a family of analytic functions on a domain okay suppose they are normally bounded okay that is they are bounded on compact subsets right then the equicontinuity is equivalent to normal convergence namely normal convergence of a subsequence of any given sequence okay so the statement will be that you have you know a family of analytic functions on a domain okay and instead of assuming them to be uniformly bounded on the domain you will assume them only to be uniformly bounded on compact subsets of the domain which will anyway follow actually because of the Cauchy estimates just because of analyticity so you do not have to assume that but it is already there okay then the model theorem will tell you that if these functions are equicontinuous if this family is equicontinuous then given any sequence in this family you can extract a subsequence which converges normally namely which converges on compact subsets of the domain okay and this version of the theorem this version of Arzela Ascoli theorem is called Montel's theorem okay so you must understand the philosophical shift from going from you know Arzela Ascoli theorem to the Montel theorem. In the Arzela Ascoli theorem all conditions all hypotheses all conclusions are for functions defined on complexes okay but when you come to Montel's theorem you are working with analytic functions and analytic functions are not defined on compact sets they are defined basically they are defined only on open sets therefore whatever conditions you put you do not put on the functions themselves on the whole domain which is open but you put it on compact subsets of the domain so you change everything to fit into compact subsets of the domain okay so instead of uniform bounded you will get normally bounded so instead of saying they are uniformly bounded on the whole domain you will assume that they are uniformly bounded on compact subsets of the domain and instead of getting uniform convergence on the domain which is too strong which you usually do not get you will get uniform convergence on compact subsets of the domain which is called normal convergence and if you adapt it to this situation in this way then you get Montel's theorem and the proof is surprisingly Arzela Ascoli theorem plus another application of diagonalization so already the Arzela Ascoli theorem has used one level of diagonalization as we saw in the proof but when you adapt it to complex analysis as Montel's theorem you have another you have to do one more diagonalization where at each stage you have to apply the Arzela Ascoli theorem and then that will give you the Montel theorem okay so we will discuss that in the next lecture.