 Okay so what we need to do now as I was studying in the last lecture is to work with domains which probably include the point at infinity okay so you know so this is a this is an important turning point a little technical but it is very easily understandable. So all this time you know so let me say a few things when you probably do a first course in complex analysis you are working with a domain you are working with a function which is defined on a domain in the complex plane okay which means it is defined on an open connected set of course non-empty okay and the function is taking complex values okay but then we want to work not just with analytic functions we want to work with meromorphic functions and the problem with the meromorphic function is at a pole the function goes to infinity and therefore at a pole you do not get continuity okay so you are forced to include the point at infinity in the co-domain of the function okay and that leads you to look at functions with values in the extended complex plane okay so that is what we have been doing so far what we have been doing so far is we have been looking at functions which are defined on a domain in the complex plane okay but taking values in the extended complex plane alright and in for such functions we have proved lot of theorems so for example we have proved that you know we have proved that if you have a family of functions which converges normally okay then if all the functions are analytic that is holomorphic then the limit function is either analytic or it is identically infinity similarly if all the functions are meromorphic okay then the limit function is either meromorphic or it is identically infinity and then we have proved of course Montel's theorem that if all the functions are or all analytic if you have a family of analytic functions okay then it every sequence admits a normally convergent subsequence if and only if the family of that family is normally uniformly bounded okay that is uniformly bounded on complex subsets that was Montel's theorem and then we also proved Marty's theorem which is much more stronger namely you take a family of meromorphic functions and assume that then the condition that any sequence in that admits a normally convergent subsequence is equivalent to the spherical derivatives being normally bounded and we resort to spherical derivatives because for meromorphic functions usual derivatives will not be defined at the poles okay. So in all these statements the domain of the functions was always a subset of the complex plane and but of course the co-domain we took it to be the extended complex plane and mind you therefore what is happening is in the domain the metric you are looking at is the usual Euclidean metric because after all it is a subset of the complex plane where you have the Euclidean metric whereas in the co-domain the metric you are looking at is the spherical metric because the co-domain is the extended complex plane it is identified by the stereographic projection with the Riemann sphere and you actually take the spherical metric on the Riemann sphere alright which is exact which is the distance between the distance between two points being given by the length of the minor arc of the greater circle on the Riemann sphere passing through those two points okay. So this is the setup of things now what we want to do is we want to extend all these theorems okay to the domain not only a domain in the complex plane but you want extended to a domain in the extended complex plane okay that is the next step that means you are now allowing infinity as a value of the variable you are allowing infinity in the domain okay and you want to write out the same you want the same results again and the fact is it is true it will work all these results will be true for a domain even in the extended complex plane but then we need to fix it okay. So I will tell you where the problem lies the problem lies in let us try to be naive and usually the see the advantage of being naive is that you will think naturally okay and the disadvantages it may not work but the advantage is that you know where you go wrong and then you can correct yourself okay so there is always an advantage to being naive in the first place so suppose D is a domain in the extended complex plane okay and certainly I am looking at a domain which contains a point at infinity okay because if it does not contain a point at infinity then it is a usual domain and I have all the theorems for usual domains in the complex plane I have already proved okay. So let me look at a domain in the extended complex plane which contains a point at infinity okay now I can just say a sequence of functions converges normally on the domain if it converges on compact subsets this is usual definition okay but the problem is that this definition will not work why it will not work is because if you take a compact subset of a domain even if you take a domain in the extended complex plane that compact subset cannot contain the point at infinity because any name root of infinity is unbounded you understand so therefore the problem is that if you say a family of functions or let us suppose you are looking at a sequence of functions and if you say if you say it normally converges on a domain which contains the point at infinity actually you are not taking care of the normal convergence at infinity because when you say it normally converges what does it mean it means that it is converging on compact subsets uniformly convergent on compact subsets but what are compact subsets of a domain even in the extended complex plane see if you look at it with respect to the usual Euclidean plane okay no neighbourhood of infinity okay can be a compact subset of the usual plane okay so what is happening is that even if you naively define that a sequence of functions is converging normally on a domain containing the point at infinity what you are actually defining is only that it is that the sequence of functions is converging normally only on the domain the punctured domain with the point in infinity removed okay so you are not able to take care of normal convergence at infinity that is the problem okay so how do you tackle this so the way of tackling this is the is again the same old philosophy of you know how to tackle the point at infinity you tackle the point at infinity by you know inverting the variable and looking at the point at 0 okay so you know whenever you want to look study f of z at z equal to infinity what was the original idea we studied f of 1 by w at w equal to 0 and when the moment so the neighbourhood of infinity will translate to a neighbourhood of 0 and a neighbourhood of 0 is again now a good old neighbourhood and in the good old complex plane and you can do work with it which you have already proved theorems there okay so here is the definition so the definition is suppose d is a domain in the extended complex plane containing the point infinity when do I say a sequence of functions converges normally on d infinity I say that I have to say it in two pieces I have to first say that the convergence is normal on d minus infinity okay that is throw away infinity use your throwing away one point okay so it is still an open set okay mind you infinity is a closed point if you look at in the extended complex plane which is identified with the Riemann sphere the infinity is identified with the north pole on the Riemann sphere okay so you take d minus infinity that is an open that is a nice domain in the complex plane okay it is going to just by removing a point you cannot make it disconnected okay and because it is open okay so it is still going to be open connected so it is a domain in the complex plane you require that on that domain punctured at infinity the given family converges normally okay which means you are requiring that there is uniform convergence on compact subsets of domain on compact subsets of the plane which intersect d that is all okay that is one condition the second condition is you take the same sequence of functions change the variable from z to 1 by w and say that that converges normally in a neighborhood of the origin okay so you give a definition outside the outside infinity and you give a definition in a neighborhood of infinity by translating it to a neighborhood of 0 and this is the definition that we will make and this is the definition that works okay you will see that this with this definition you can translate all the theorems that we have proved you can just use all the theorems that you have proved so far to get theorems for the case when the domain includes the includes a point at infinity okay so let me write this down so the next thing is let me use a different color normal convergence at infinity so let f n of z be a sequence of continuous functions I do not even need continuous let me just say the sequence of functions on a domain d of the extended complex plane with infinity in the domain okay and taking values in the extended complex plane okay so basically you can think of f n as a sequence defined on an you know an open on a domain on the Riemann sphere and taking values on the Riemann sphere if you think of the extended complex plane as a Riemann sphere you can think that your source is a domain it is an open connected set on the Riemann sphere okay your d is on the Riemann sphere it is an open connected set that contains the north pole which is supposed to correspond to the point at infinity and you have these functions each of these functions are defined on that d and they are taking values again in the Riemann sphere which means they can take the value infinity okay so you take a sequence of functions so when do we say that this sequence converges normally on d so let me write that down we say that f n converges normally on d on d if number 1 f n so the sequence f n converges normally on d minus infinity which is d minus infinity is a domain in the good old complex plane okay and you know defining convergence on a domain in the complex plane is something that we have already done it is just uniform it is just uniform convergence on compact subsets okay and this and is very very important number 2 okay f n of 1 by w converges normally in a neighborhood of w equal to 0 so here is the so this second statement is what actually takes care of normal convergence at infinity and you know it is very beautiful because normal convergence at infinity means you should ensure uniform convergence on compact subsets of infinity okay but the problem is there is no compact subset of infinity that you can think of in the usual complex plane you can think of it only on the Riemann sphere so if you want to translate it back to the usual complex plane you have to translate from neighborhoods of infinity to neighborhoods of 0 by making this inversion z going to 1 z going to 1 by z which is w okay so you replace z by 1 by w right so this is the definition alright now comes now you see this definition is piece wise what you have done is you have got normal convergence outside infinity that is the first statement because outside infinity in the domain is again a domain in the usual complex plane you have no problems with that and the second part of the definition says you have convergence normal convergence at infinity okay that is what the second one says because you know after setting f n of 1 by w at w equal to 0 a neighborhood of w equal to 0 is the same as studying f of z in a neighborhood of infinity alright but because it is a neighborhood of 0 I know what converges normally means okay fine now the so this is the right definition and this definition works like I will put this as D E F for definition okay so this definition works but then there are certain remarks that needs to be made so that you know you realize that you always make a definition if a definition has to suit a particular condition you make a twist in the definition then you have to check whether your definition is consistent so one of the things that you can ask is the following since I have defined the normal convergence of sequence in two pieces can it happen that on each piece I get a different limit function you can ask that and the answer is no you cannot if you are working with continuous functions this cannot happen because of continuity okay that is so this definition will work properly with continuous functions okay so let me write this down suppose so here is a remark so the remark is suppose each f n of z is continuous on D okay okay then f n converges to a continuous function f on D okay so you get only one function okay and what is the similar statements hold similar statements hold okay so let me not say that because I need to expand on that okay so what is the point what is the proof the proof is that well you see on D minus infinity okay f n of z will converge to well let us say f of z alright and f z is continuous this is because of normal due to normal convergence you know normal limit of continuous functions is continuous and mind you you should consider the limit function f also to be a function with values in the extended complex plane okay and this so this is because of part one of the definition now part two of the definition will tell you that in a neighbourhood of the origin f n of 1 by w will also converge to something and I will call that as g of w okay so also on an neighbourhood of w equal to 0 f n of 1 by w converges to g n of w sorry g of w so I will in principle I should expect to get a function g and again g is continuous at 0 and in fact g is continuous everywhere again because again I am using just the fact that normal limit of continuous function is continuous okay so there is only one thing that for w not equal to 0 okay z is equal to 1 by w is not infinity okay so z lies in D minus infinity okay and therefore what will happen is by the uniqueness of limits of a sequence point wise you will get that f of z will be equal to g of what is this g of 1 by z okay so this implies for w not equal to 0 okay f of 1 by w is the same as g of so maybe I should call this as you know here I have to say the following thing I should call this as g n of w okay I should call g n of w as f n of 1 by w and g n of w tends to g of w okay and the point is that this g of w is f of 1 by w is g of w that is what I am getting okay if I put z equal to 1 by w right so but then you see and both are continuous at 0 okay both are continuous at 0 mind you because of continuity therefore limit w tends to 0 okay so what this will tell you is that you know the so this will tell you that f of infinity will be g of 0 okay so this implies that the limit function the limit function is f f of z even at infinity so you get a unique limit okay that is the whole point so there is normal convergence defining it piecewise that is one outside infinity and the other one in an a board of infinity by translating to an a board of 0 okay though it is a 2 piece definition normal convergence will still give rise to only one function but mind you all functions are being taken with values in the extended complex plane and that in the target the metric you are using is always a spherical mention okay so you know unless you give a argument like this okay things could go wrong see after all I have defined normal normality separately in 2 pieces normality I mean normal convergence in 2 pieces okay and then it could happen that on each piece I could get different limits at least what this tells you is that it will not happen if your functions are continuous okay and that is going to be the case because we are going to deal only with analytic functions or monomorphic functions and you know analytic functions are of course continuous and even if you if you recall even if you take an analytic function at infinity by definition it is a function which is continuous at infinity you know and by a version of the Rayman's removable singularities theorem saying that a function is analytic at infinity means that it should be bounded at infinity okay so it means that the function value at infinity is a finite complex number okay so analytic function at infinity makes sense so analytic functions on a domain in the extended complex plane containing the point at infinity makes sense for us okay and similarly monomorphic functions also make sense because what is a monomorphic function on a domain which contains a point at infinity it is see it is supposed to be a monomorphic function on the punctured domain with the punctured at infinity and you invert the variable and that should give you a monomorphic function at the origin okay always you go to saying that a function is monomorphic at infinity is a saying f of z is monomorphic at infinity okay in a neighborhood of infinity is the same as f of 1 by z is monomorphic at z equal to 0 so you always translate back to at infinity you always translate back to neighborhood at 0 so it makes sense okay so monomorphic functions on an extent on a domain in the extended complex plane containing the point at infinity makes sense and all these functions with values in the extended complex plane also makes sense so it is a we are in a perfect situation and all these functions are all continuous monomorphic functions are continuous because you allowed at the pole you define the function value to be infinity and you allowed infinity to be in the co-domain of the function okay so we are in the right setup now what we need to know is we need to check each of those theorems that we proved we need to deduce that those theorems also work for such a domain okay so here is the first one so theorem. Let D be a domain in c union infinity and f n a sequence so a sequence of holomorphic functions or analytic functions on D taking values in c considered as subset of c union infinity okay. If f n converges to f normally on D then either f is analytic on D or f is identically infinity on D okay so you see the point is that we have already proved this theorem when D does not contain the point infinity when D is a subset of the usual complex plane we have already proved this theorem that is you take a sequence of analytic functions and you assume that it converges normally then the limit function has to be either analytic or it will be identically infinity okay and you cannot get anything in between for example you cannot get a strictly monomorphic function in between okay and the reason is a pole cannot pop up at infinity I mean in the limit and if you remember this was because if you invert the variable a 0 cannot pop up at the limit because of Hurwitz's theorem so Hurwitz's theorem is working in the background okay. So we will have to only worry about the case when D contains infinity that is the extension we are particularly interested in so let me write that down we have already proved this for infinity not belonging to D so the essential thing that so let the essential thing is we allow infinity to belong to D okay now go back to the definition of normal convergence for a domain containing the point infinity what is the definition first thing is you throw out infinity and on the remaining thing which is a domain in the complex plane there is normal convergence okay and the other thing is that you take a neighborhood of 0 and look at the functions of the variable inverted in a neighborhood of 0 okay so now so let me write that down for now for D minus infinity f n will converge to f normally so either f is identically infinity on D minus infinity or f is analytic on D minus infinity okay this is because of the fact that D minus infinity is a usual domain the usual complex plane and for the usual complex one we have proved such theorem okay whenever you have normal convergence of analytic functions the limit is either identically infinity or it is uniformly an analytic function okay but you know if f is identically infinity in D minus on D minus infinity it will also be infinity at infinity because f is continuous okay so we have to only take care of the situation when f is not identically infinity and prove that f is analytic on D even at infinity so infinity is the only problem okay so let me write that down if f is identically infinity on D minus infinity then by continuity of f on D f is identically infinity on all of D because f of infinity will become infinity okay so if f is not identically infinity infinity becomes an isolated singular point of f okay because mind you f is analytic on D minus infinity D minus infinity is a neighborhood of infinity it is a deleted neighborhood of infinity D minus infinity is a deleted neighborhood of infinity okay and f is analytic on that that means f is analytic in a neighborhood of infinity that means infinity is an isolated singular point for f and I to check that f is analytic at infinity I have to only check f is bounded at infinity but why is that true that is because Because I have normal convergence at infinity okay which is normal convergence if you change the variable to 1 by W and look at W equal to 0 okay now you will use a second part of the definition of normal convergence which is compact convergence I mean normal convergence at infinity you use that and then you are done okay. So now we also have we have also fn of 1 by W converges to f of 1 by W in normally normally in a neighbourhood of 0 of W equal to 0 this is the second part of the definition okay and mind you that is you could have taken a you could have taken a small enough open disk at the origin which will be a domain it will be connected okay but it would even be simply connected okay and now on that domain okay you look at the fns okay the point is that each of the fns is also analytic at infinity you see what look at what was given to me what we what was given to us we have started with sequence of holomorphic functions on D and we are looking at infinity and we have assumed infinity belongs to D it means that each fn is already analytic at infinity each f in is already analytic at infinity okay that means fn of z is analytic at z equal to infinity that means fn of z is bounded at z equal to infinity that means f1 of 1 by W is bounded at W equal to 0 and because the f is the normal limit of the fns f of 1 by W will also be bounded at W is equal to 0 and that is the same as saying that f is analytic at infinity and you have done okay so that is it. So let me write this down now since each fn is analytic at infinity each fn of w is analytic at 0 so f of 1 by w is bounded at 0 which means f is analytic at infinity thus if f is not identically infinity then f is also holomorphic on D and that is the proof okay. So you see you are able to extend the theorem that we already proved that a normal limit of analytic functions can either be analytic or it will be identically infinity even if your domain contains infinity you are able to do that okay and the technical point was to deal with normal convergence at infinity and that is cleverly done by translating a neighbourhood of infinity into a neighbourhood of 0 by inverting the variable which is the usual philosophy that we have always been using to study the point at infinity okay fine so the same kind of argument will give you the corresponding theorem for you know it will give you the corresponding theorem for meromorphic functions okay so let me write that down theorem. Let fn be a sequence of meromorphic functions on a domain D in the external complex plane suppose fn converges normally on D then either the limit f equal to limit fn is identically infinity on D or it is meromorphic on D okay so this is just extending the previous theorem into the meromorphic case okay but now the point is that essentially you want to deal with the domain which contains the point at infinity and what is the proof? Proof is exactly the same okay the proof is exactly the same let me write it out if D does not contain infinity this has already been proved so assume D contains infinity okay we have that f is meromorphic on D minus infinity okay so again let me stop and say a couple of things please remember that when you say normal convergence now okay it is uniform convergence on compact subsets alright but either you must look at compact subsets of D minus infinity or you should look at compact subsets of a neighbourhood of the origin with the variable inverted that is the point okay and in both cases for the variable you are using only the Euclidean metric but for the function values you are using the spherical metric okay you have to remember that that is a big difference and the other important thing is that you know you are trying to make use of the theorems that you have already proved for a domain in the usual complex plane and trying to deduce the corresponding theorems when the domain contains the point at infinity okay so these are things that you should highlight in your mind okay so you see so let me repeat what I said if infinity is not in D then D is a usual domain and you know for a limit of meromorphic functions the limit can either be identically infinity or it can be meromorphic okay and mind you this is a very important thing you know it tells you that normal limits are good because after all what is a meromorphic function a meromorphic function is an analytic except for poles okay but when a function goes to a limit the limit function could be horrible see the limit function could have been an analytic function with non pole singularities it may it could have been an analytic function with essential singularities or even worse limit could have been an analytic function with non isolated singularities such horrible things could have happened but the fact is normal convergence prevents that see normal convergence you know is always a it is locally uniform convergence okay because every point has a neighbourhood which is compact okay so at every point you can find a neighbourhood compact neighbourhood where you will have uniform convergence and therefore you on that neighbourhood also you will have uniform convergence so it is locally uniform convergence normal convergence is and therefore because of your local uniformness everything nice happens you know the moment you have uniform convergence limits of continuous functions are continuous limits of analytic functions are analytic and so on but so this is happening globally right and there is one more thing that I have to tell you that here the moment I assume f n is a sequence of meromorphic functions on the domain which contains a point at infinity and I assume that it converges in mind you a unique limit function is already defined that is because you know the f n are meromorphic and therefore they are continuous and I told you that whenever you take a continuous limit of continuous functions if you take a normal limit then the limit function is continuous even though your definition of normal convergence has been split into two pieces one for the domain with infinity punctured and the other for neighbourhood of infinity thought of as a neighbourhood of the origin okay. So even this existence of f a uniform function f a single function f is because of the continuity of all these functions because meromorphic functions are continuous functions considered as functions into the excellent complex plane that is something that you should not forget okay. So all right so if you take the domain which contains a point at infinity then d minus infinity is a usual domain in the usual complex plane and you already proved a theorem for that so the function f is meromorphic on d infinity or f is identically infinity on d minus infinity okay. So this is something that we have already proved we have that f is meromorphic on d minus infinity or f is identically infinity on d minus infinity all right and again the same old argument if f is identically infinity on d minus infinity then it has to be infinity at infinity because of continuity of f okay. So we have to only deal with the condition when f is not identically infinity right so let me write that down so if f is identically infinity on d minus infinity then f is identically infinity on d as f is continuous at in continuous on d in fact continuous at infinity okay. So if f is not identically infinity we have f is meromorphic on d minus infinity okay and therefore infinity becomes a singular point okay infinity is certainly a singular point because the problem now is slightly more complicated as it stands it looks you can say infinity is a singular point but you cannot immediately say that infinity is an isolated singular point that is the point that is the issue. So infinity is a singular point because if you take d minus infinity there are only poles on d minus infinity f is you know it is meromorphic so on d minus infinity there are poles and poles are of course isolated so as far as d minus infinity is concerned all the singular points are isolated but infinity itself may be a non-isolated singular point it could happen poles could accumulate at infinity if you have a sequence of singular points going to a point then that point is not an cannot be an isolated singular point so you have this problem but then that will not happen because of the normal convergence at infinity because what is happening at infinity is being controlled by what is happening when the variable is inverted in a neighbourhood of 0 okay so let me say this now in neighbourhood of W equal to 0 f n of 1 by W converges to f of 1 by W normally and what I want you to understand is that you change the variable from z to 1 by W alright and if you are looking at z not equal to infinity you are looking at W not equal to 0 okay and the point I want to make is that even at W equal to 0 I want to say that the f n's okay they continue to be meromorphic that is because it is already given to you see what is given to you is that each of these functions is meromorphic on D so infinity the point infinity which is in D can either be a pole by itself for each of the f n's or it can be a point of analyticity infinity cannot be any worse okay so that means each of these f n's of 1 by W are meromorphic in a neighbourhood of W equal to 0 okay but this neighbourhood of W equal to 0 on that neighbourhood that is a usual neighbourhood of the complex plane and you have normal convergence of a sequence of meromorphic functions therefore the limit function can be either identically infinity or it can be meromorphic but the limit function is not identically infinity because f is not identically infinity okay so f of 1 by W has to be you know a meromorphic function at W equal to 0 that means you are saying f of z is meromorphic at infinity and you are done okay so you escape and you get the proof of the statement that you want so let me write that down so let me write this and since each f n is each f n z is meromorphic on D f n of 1 by W is meromorphic neighbourhood of W equal to 0 so f of 1 by W is meromorphic at 0 at W equal to 0 as f is not identically infinity so f is so f of z is also meromorphic at infinity and we are done okay and so you this horrible thing of infinity being a non-isolated singular point for f does not happen so you see we are still in very nice in a very nice situation okay so what I need to do is what we need to do next is try to generalize a Montel theorem and Marty's theorem for the case of meromorphic functions and we will do that in the next lecture.