 Alright so what we are going to do now is we are going to show that if you have a normally convergent sequence of meromorphic function then the limit function is either a meromorphic function or it is identically the function which is infinity okay and the importance of this theorem is that in the limit you can only get a meromorphic function and nothing worse okay because you see meromorphic function is something that has special kind of singularities it has only poles of singularities but when you take a limit of meromorphic functions anything could have happened you could have got a function with for example essential singularity such a thing could have happened okay or you could have even got a function with non isolated singularities but all these things do not happen the nice thing happens namely if you take a normal limit of meromorphic functions you again get a meromorphic function and the worst thing that can happen is that it is the function which is identically infinity that is all okay so let us go ahead and do that the idea is similar to the idea that you know if you take a sequence of holomorphic functions then the and you take a normal limit of holomorphic functions then the limit function is either absolutely holomorphic okay it is holomorphic perfectly holomorphic or it is identically infinity okay so let me start like this let fn be a sequence of maps from D to C union infinity D inside the complex plane is a domain okay of course non-empty set non-empty open connected set and I am taking maps with values in C union infinity which is extended complex plane and the reason for that is of course you know we allow the value infinity now right and so suppose so you know suppose so let me write it here suppose fn oops suppose fn tends to f suppose fn tends to f point wise point wise on D right suppose it is a point wise convergence so that means that f is also here so your f is also here so let me write it somewhere here f is also a map from D to C union infinity and fn tends to f point wise on D you have to be when you say point wise it means for every point and when you say every point it means that fn of z tends to f of z for all z in D okay but then what does fn of z tend to f of z for all z in D mean it means that the distance between fn of z and f of z tends to 0 okay as n tends to infinity for each z in D and what distance are you going to use you just cannot use the usual Euclidean distance okay because you are allowing the value infinity you have to use a spherical metric the distance under the the spherical distance has to be used okay so let me write that down i.e fn z tends to f z as n tends to infinity for all z in D and that is supposed to mean that D the spherical distance of fn of z as a spherical distance between fn z and f of z that tends to 0 as n tends to infinity for every z in D this is what point wise convergence on D means and of course you have to use a spherical distance because the function values might be one of the points to which you are measuring from or to which you are measuring the distance could be the point at infinity so then you will have to use a spherical metric okay that we have denoted by D sub s alright so you have a sequence of functions defined on a domain D and at this moment they are only maps okay I am not even saying that they are continuous or something like that but then we have been looking at so many subsets of this collection of maps so the first important subset is those which are continuous maps from D to C union infinity these are the continuous ones so I will write CTS for continuous maps and continuous maps from D to C union infinity make sense because D is anyway a domain it is a topological space it is a subspace of the complex plane and C union infinity is a very nice topological space in fact even a complete compact metric space okay because it is basically it is isomorphic to the Riemann sphere by the stereographic projection okay so you have this and then there were smaller subsets more important subsets there was also this subset of meromorphic functions on D so this is a subset of meromorphic functions on D and so this is meromorphic so meromorphic means that you know these are functions which are analytic that is holomorphic except for a subset which is a subset of necessarily isolated points and at each of those points the function has a pole okay so analytic except for poles that is what meromorphic means alright and we have seen that this set of meromorphic functions is actually a field it is a field extension of the complex numbers algebraically it is a field and then there is the further smaller subset of holomorphic functions on D or analytic functions on D so let me write here analytic that is a smaller subset and of course these are functions which are analytic everywhere on D there are they do not have any singularities alright and what we have what we have seen is that if all the f n's are in H of D then f is either in H of D or f is identically infinity that is what we have seen okay so let me write that here we saw if all the f n's are in H of D then f is in H of D or f is identically infinity this is what we have seen in the last lectures and it is a nice thing because it says that a sequence of holomorphic functions analytic functions can either go to infinity or it will go to a analytic function that is all you do not get in you do not get something weird in between for example you do not even get a meromorphic function which is not holomorphic okay fine and you remember we used basically we used 2 important things we used the invariance of the spherical metric with respect to inversion and the other thing that we used was the Hurwitz's theorem okay theorem of Hurwitz alright. Now what I want to say is that the similar theorem is true if if you take f n's to be a meromorphic okay so if you take what we will show today is that if all the f n's are in M of D then either f is in M of D or f is identically infinity so you are just extending this from the holomorphic case to the meromorphic case that is what we want to do okay and that is also that is also a very nice thing see the point is that you know why all these theorems are important is that you see on the one hand they are very believable I mean you expect them to happen because under normal convergence good properties are preserved okay you know from basic analysis if you have normal convergence which is actually locally uniform convergence so essentially it is uniform convergence locally okay and under uniform convergence all nice things happen the if you take a uniform limit of continuous functions you get a continuous function if you take a uniform limit of analytic functions you get an analytic function okay if you take a uniform limit of integrals okay then that is the same as integrating the uniform limit you can interchange limit and integral you can interchange you can do term wise differentiation if you are working with a series which is uniformly converging okay so uniform convergence is a very nice thing so under uniform convergence you expect everything to go smoothly so it is correct to expect that if you take a sequence of holomorphic functions if it converges normally that is locally uniformly then the limit function is also holomorphic that is correct to expect and that is what happens similarly if you have sequence of meromorphic functions if it converges normally to a limit function the limit function is also meromorphic the only thing that you will have to worry about is the other extreme possibility which is that the function becomes identically infinity that is a possibility you cannot ignore okay but that is the worst thing that will happen and nothing in between these two happens so it is important that this is very very important because you know it should not happen that I start with a series of meromorphic functions or holomorphic functions if they are holomorphic functions they do not have any singularities if they are meromorphic functions they have poles okay but in the limit suddenly I should not get a function which is crazy enough to have say essential singularities or I should not get a function which has a singularity which is non-isolated I mean such horrible things should not happen and you can expect such horrible things should not happen because of normal convergence which is locally uniform convergence and this is exactly what we are trying to show in these lectures I mean this is so you know it is this is always part of mathematics you guess that something nice will happen but then to prove it you will have to do some work you will have to use some theorems okay. So well so let us see so first of all I want to say that see if you take so first thing I want you to understand is that if all these f n's were not just maps but suppose they were continuous okay then f also becomes continuous because a uniform limit of continuous functions is continuous and a normal limit is a locally uniform limit alright therefore so let me write that down if so let me use a different colour see if all f n are continuous functions from D to C union infinity okay and the f n's converge to f normally normally on D normally on D and this always means with respect to spherical metric so you know here I must tell you that in the previous slide here when I say if f n is if the f n are all holomorphic then f is holomorphic or f is identically infinity here of course I am assuming that the convergence is not just point wise but it is in fact normal f n converges to f normally this is very important this is very very important of course normal convergence I did not mention it there but it is important okay so point wise convergence is useless it point wise limits can not be as good as you want them to be okay but normal convergence is very very important here right so suppose you take all the f n's are continuous and assume that the convergence is normal on D then of course f is also continuous and this is just the very well known fact that uniform limit of continuous functions is continuous alright you know that and the point is that normal limit is something that is locally a uniform limit okay so it means that see given any point okay because you are working on an open set you can always find a closed disc surrounding centered at that point which is inside your open set okay and such a closed disc is of course compact and normal convergence promises you that the convergence will be uniform on that closed disc in particular it will be uniform on the on the open disc which is a subset of the closed disc okay so whenever you have uniform convergence on a set you will always have you will automatically have uniform convergence on any subset okay alright so you get locally uniform convergence and therefore the locally the limit function f will become continuous and if you and continuity is a local property so if a function is locally continuous then it is continuous if it is if you have global function okay so that is why f is continuous now what is very very important is the following thing if you add this condition that now notice that since you are working with values in the extended plane this is very very important not only f n makes sense 1 by f n also makes sense and not only f makes sense 1 by f also makes sense okay so you see you see so let me say this note then that 1 by f n also makes sense and 1 by f also makes sense so let me say so makes sense so let me let me put something more general 1 by f n certainly makes sense as a map from D to C union infinity okay 1 by f n of z is very well defined if f n of z is infinity then 1 by f n of z is 0 if f n of z is 0 then 1 by f n of z is infinity and if f n of z is a non-zero complex number then 1 by f n of z is the inverse of that complex number so it is very well defined so this 1 by f n is also these 1 by f n also makes sense okay and not only that similarly 1 by f also makes sense okay any function on the domain which has values in the extended plane has an inverse okay and for example if you take the worst case which is the function which is identically 0 you do not have to worry its inverse is the function which is identically infinity okay but it is not an inverse in the algebraic sense that you know the function multiplied by the inverse will not give you 1 okay you should not go ahead and write the function 0 multiplied by the function infinity is equal to 1 that is not correct okay but this is a convention it is a convention okay so what we say is if you take the function which is identically 0 then its inverse function is defined to be the function which is identically infinity but it is not an inverse in the algebraic sense you have to be careful about that okay fine so but what is the advantage of this advantage of this is that you know if you are looking at meromorphic functions then the inverses of meromorphic functions are also meromorphic functions that is the advantage okay. So now what I want to tell you is that you see if suppose I assume that f n converges to f normally okay on D with respect to spherical metric then it also happens that 1 by f n converges to 1 by f normally on D with respect to the spherical metric that will also happen that is just because of the fact that the spherical metric is invariant with respect to inversion okay so let me write that note also that 1 by f n converges to 1 by f normally on D with respect to the spherical metric because the distance spherical distance between f n of z and f of z is the same as the spherical distance between 1 by f n of z and 1 by f of z this is just the invariance of the spherical metric under inversion okay. I told you that the inversion on the complex plane if you transport it via the stereographic projection to the Riemann sphere what it will give you is it will give you a rotation of the Riemann sphere it will give you a rotation of the Riemann sphere by 180 degrees okay so under rotation of a sphere the distance the spherical distance between two points on the sphere will not change obviously okay so this is something that we have seen before. So the point is that it is beautiful if f n converges to f normally then 1 by f n converges to also 1 by f normally and these are equivalent mind you f n converges to f if and only if 1 by f n converges to 1 by f okay it is a simultaneous statement these are two simultaneous statements and both of them are equivalent okay. So the point is that therefore what I am trying to tell you is the philosophy is as follows whenever you are looking at functions with values in the extended plane always think automatically of the reciprocal functions also the inverse of those functions okay they also make sense okay that is the advantage okay. Now what I want to tell you is that you see suppose now let me do the following thing suppose all the functions f n were all meromorphic okay what I want to actually say is that f will be meromorphic okay and that is the important result that we want to see okay. So now let me assume that suppose f n are all meromorphic functions on D okay then the theorem is that f is also meromorphic on D so here is a theorem then f is meromorphic or the other extreme case is that f is identically infinity okay so this is the theorem if f n is meromorphic if each f n is meromorphic then the normal limit of f n namely f that is meromorphic or it is identically infinity what it means is that the limit function will have if at all it has singularities they will only be isolated they will be poles and you would not get anything worse than that okay. Now what I want you to understand is that I see first and foremost f n are meromorphic okay and that will imply that 1 by f n are also meromorphic okay see because see the reciprocal of a meromorphic function is also a meromorphic function right and the you see so this makes sense for so this all this is okay if you know with a small modification alright this is provided f n is not identically infinity okay because a function which is identically infinity we do not I mean we do not you know it is an extra function that we add alright we do not consider it to be meromorphic we consider it as an extreme case the function which is identically infinity okay but it is inverse will be the function which is identically 0 and that is a nice constant function it is a holomorphic function also okay. So when I write this I am assuming that the f n that I am looking at is not identically infinity and when you may ask what if that f n is identically infinity in that case 1 by f n is defined to be the function is identically 0 and that is by definition it is a constant function it is holomorphic okay so that case is always you know you keep it separate when a function is identically infinity okay and also this case that if f n is identically 0 then 1 by f n is defined to be identically infinity alright and so you know you should also I also throw out the case when f n is not identically 0 when f n is identically 0 I throw out that case okay because then 1 by f n becomes the function is identically infinity and the function is which is identically infinity is not considered as a meromorphic function okay because it has it is a meromorphic function should have only poles only at you know it has only poles and they should be an isolated set of points at the poles it goes to infinity but it cannot go to infinity everywhere right. Now what I want you to understand is that see I want you to see that therefore since 1 by f n are all meromorphic 1 by f n are including the case of 1 by f n being infinity okay all these 1 by f n are of course continuous mind you because this set of meromorphic functions is actually contained in the set of all continuous maps because it is continuous because mind you at a pole you are allowing the value infinity for a function. So what you must understand is that all these 1 by f n are in fact continuous functions and therefore if you take if f n tends normally to f then because of the invariance of the spherical metric under inversion 1 by f n will tend normally to 1 by f and 1 by f will certainly be continuous that is because each 1 by f n is certainly continuous even if 1 by f n is the function which is identically infinity it is continuous mind you even the extreme case. So what I want you to understand is that if f n tends to f normally implies 1 by f n tends to 1 by f normally this happens and this implies that this 1 by f is certainly a continuous function from D to C union infinity and this includes the case when 1 by f n is I mean when 1 by f is even identically the function which is infinity okay mind you the function which is identically infinity is here but it is not here it is not in the meromorphic functions by definition right fine. So what we want to show is that we want to show this 1 by f is meromorphic that is what we want to show now I want you to so let me say another thing that I wanted to say but I did not and that is that how do you see 1 by f n is meromorphic it is very very simple you see look at the places where the function f n has poles this is an isolated set of points and 1 by f n will have zeros at those points okay if a function has a pole at a point then its reciprocal will have a zero at that point and the order of the zero will be exactly equal to the order of the pole okay and if a function has a zero at a point then its reciprocal will have a pole. So you see if you take f n to be a meromorphic function then there is it is analytic outside an isolated set of points which are poles okay and outside an isolated set of points you will get an open subset of D and inside that again the set of zeros will again be an isolated set of points because you see the set of zeros of an analytic function are always isolated and this isolated set of points which are the zeros of f n they will be the isolated set of points which are the poles of 1 by f n. So 1 by f n has only poles as singularities so 1 by f n is meromorphic you have to understand that okay so now you see what I want to say is that so we had defined this set D infinity of f is D infinity of f is a set of points in D where f takes the value infinity so D infinity of f is so let me write that this set of all z belonging to D such that f of z is equal to infinity and this is just f inverse of infinity and this is a close set it is a close subset of D okay and well and it is close because the point infinity is a closed point in the extended plane because it corresponds to the north pole on the Riemann sphere under the stereographic projection and the inverse image of a closed set under continuous function is close f is continuous and so D infinity of f is a closed set okay and the point is that if you assume that f is not identically infinity okay then you are just saying that D this is the same as saying that D infinity of f is a proper subset of D okay D infinity of f is a proper subset okay because if D infinity of f is all of D then it is the same as saying f is identically infinity okay so D infinity of f is a proper subset of D what we want to show is that we want to show f is analytic except for poles. So you want to show that this D infinity of f which is a closed set these are certainly points where f takes the value infinity okay. So you want to show that they are all poles and in particular you want to show that this D infinity of f mind you it is isolated that is a see that is a important point like it could be connected it could be a curve after all it is a subset of a domain in the complex plane a subset of domain is an open set and open set can contain curves it can even contain a sequence of points which has a limit it can contain so many things. So it is not even clear that D infinity is isolated set of points okay so that we have to prove okay so that see that is the important part okay that you have to observe. So of course you know the way you handle it is by looking at 1 by f okay because 1 by f is now there for you to play with because 1 by f is already continuous you see that is the point. So you have it here 1 by f is here it is already continuous alright and you can use it right and 1 by fn converges normally to 1 by f so you can use that. So let me say the following thing see let us take this D minus D infinity if you take this D minus D infinity this if you take f of D minus D infinity this will go into the complex plane okay because you have thrown out D infinity which is a set of points where f takes the value infinity. So at the other points which is in the points of D minus D infinity f will take value other than infinity so the only values other than infinity in the external complex plane are complex values. So f so that means this and mind you D minus D infinity is open D minus D infinity is open because D infinity is closed alright and now you have the function f you have a function f which is defined on this open set D minus D infinity and it is taking complex values and what is this f this f is again a normal limit of fn. See f is a normal limit of fn on all of D so it is also a normal limit of fn on a subset of D okay if f fn tends to f normally on D itself so fn will tend to f normally on any subset of D okay. So since D minus D infinity is a subset of D fn will tend to f normally on D minus D infinity and what is this f on D minus D infinity is a complex valued function alright just let me think for a moment I am trying to look at I am trying to look at a point where f has a value takes the value infinity so I am trying to look at a point Z not in D infinity so I want to say it is a pole so I want to say if Z not belongs to D infinity I want to say f of f has a pole at Z not but the way to verify that f has a pole at Z not is to verify is equivalent to verifying that 1 by f has a 0 at Z not. So I will have to show that 1 by f has a 0 isolated 0 at Z not okay but then what I will have to show is that 1 by f is analytic in a neighborhood of Z not in a deleted neighborhood of Z not I will have to show that first okay and I must make sure and then I will get that Z not is an isolated 0 of 1 by f 1 by f being analytic in a neighborhood of Z not will tell me that f is analytic in a deleted neighborhood of Z not and Z not is a pole okay and if and now this will tell me that f is meromorphic okay that is what I will have to do so how do I show that see the ideas of the proof are similar to the ideas that we used in the earlier lectures okay so let me do the following thing. So D infinity of f is actually D0 of 1 by f okay so D0 of 1 by f is the set of all Z belonging to D such that f 1 by f of Z is 0 okay mind you 1 by f is make sense it is a continuous map okay and therefore 1 by f of Z equal to 0 will be the locus of points where 1 by f takes the value 0 it is the inverse image of 0 under 1 by f which is continuous so it is a mind you it is a closed set okay because the single point 0 is a closed set in CUNA infinity any single point is a closed set so this is so if you want let me write this this is just 1 by f inverse of 0 okay so take a point Z not in D infinity of f okay then take a point Z not in D infinity of f then what happens this is a point where f takes the value infinity so 1 by f takes the value 0 okay since 1 by f takes the value 0 and 1 by f is continuous okay there is a small disc surrounding Z not in fact even a closed disc surrounding Z not where 1 by f is bounded okay because it is just continuity see 1 by f is a map from it is a continuous map from D to CUNA infinity okay and if you use continuous so in particular it means it is continuous at Z not also okay that means that and 1 by f of Z not is 0 okay so what it means what does continue to what will continue to tell you tell you that given an epsilon greater than 0 there exists a delta greater than 0 such that if you make the distance between so I can use the usual distance because I am now going to because I am on D if you make the distance between Z and Z not less than delta okay then the distance the spherical metric between f of 1 by f of Z and 1 by f of Z not by the way which is 0 this quantity is 0 that can be made less than epsilon this is just continue the definition of continuity of 1 by f at Z not given an epsilon okay I can make 1 by f Z to be as close as I want to 0 to within an epsilon distance of 0 if I choose Z sufficiently close to Z not and how sufficiently close that is what is decided by the delta there is a delta okay and you know I can even put less than or equal to that is because you know I can choose a smaller disk such that smaller disk centered at Z not radius delta such that the close disk including the boundary circle that also lies inside D because D is after all an open set I can always choose this disk sufficiently small in D okay because D is an open set and then I can make sure that even the boundary of the disk is inside D I can do that right and you see what is see what does this tell you this tells you that you see the distance between 1 by f and 0 1 by f is very close to 0 okay and you see you also have this fact that the 1 by f n they converge to 1 by f mind you normally. So the first thing I want you to understand is that since 1 by f n converges to 1 by f normally the 1 by f n will converge to 1 by f uniformly on this close disk that is because this close disk is compact and because of that all the 1 by f ends beyond a certain stage they will also take values in a neighbourhood of 0 okay and that means that in a neighbourhood of 0 all the 1 by in a neighbourhood of Z not okay all the 1 by f n beyond a certain stage they are all going to be bounded and that means that they are all going to be analytic because mind you our originally our f n's were all meromorphic. So 1 by f n's are also meromorphic but you know meromorphic function if it is bounded at a point then that has to be a good point if a meromorphic function is bounded in the neighbourhood of a point okay then you know in it cannot it can assume in that point has to be a good point by the Riemann's removable singularity theorem. What can happen is in the neighbourhood of that point it can have some poles because it is a meromorphic function but it cannot have a pole because at a pole it will take the value infinity I am putting the restriction that it is taking values close to 0 so it means that in a neighbourhood of Z not all these f n's 1 by f n's they are all going to be analytic and since all the 1 by f n's are analytic and they converge uniformly to 1 by f in this disk centre at Z not 1 by f becomes analytic at Z not and once 1 by f becomes analytic at Z not Z not becomes a 0 of 1 by f which is now an analytic function therefore it is isolated. So Z not becomes an isolated 0 of 1 by f and that is that will tell you that Z not will be an isolated pole for f so what this argument tells you is that every Z not where f takes the value infinity is actually a pole of f and therefore f is a meromorphic and that is the that ends the proof okay so let me write that down so let me write these things in words note that 1 by f is bounded near 0 in mod Z minus Z not Z not equal to delta also 1 by f n converges to 1 by f uniformly so I am abbreviating it as u of l y on mod Z minus Z not less than or equal to delta because of normal convergence and because mod Z minus Z not less than or equal to delta is compact so for n sufficiently large the 1 by f n's are bounded near 0 in mod Z minus Z not less than or equal to delta okay and hence by if you want Riemann's removable singularities theorem Z not is 0 of 1 by f and of course here I am saying that 1 by f is 1 by f is analytic because 1 by f n's are all analytic beyond a certain stage all the 1 by f n's are analytic and 1 by f n's take to 1 by f and it is uniform convergence therefore 1 by f is analytic so 1 by f is analytic and it has a 0 at Z not so it is an isolated 0 okay note that 1 by f n n sufficiently large are analytic so 1 by f is analytic being a normal limit being a uniform limit so u f is abbreviation for uniform so all this happens in mod Z minus Z not less than or equal to delta you may need to make delta smaller if you want but that is not a problem the point is that E Z not is thus 0 of the analytic function 1 by f in mod Z minus Z not in mod Z minus Z not less than or equal to delta so it is isolated thus f of Z has a pole at Z not so f is actually a meromorphic function and that is the end of the proof so it is a very nice fact that if you take a normal limit of meromorphic functions barring the extreme case that the normal limit is identically infinity the limit is again a meromorphic function okay so that ends the proof of this fact okay and what we saw last class was that if you put the additional condition that the f n's are all holomorphic and you assume that the limit function is not identically infinity then the limit function is also holomorphic and why is that true now you can see that if you take the f n's additional to be holomorphic you know and if it is not and the limit is not identically infinity you know by whatever we have proved now you know that the limit is meromorphic okay but if it is really meromorphic at a certain point namely if it has a pole at a certain point then what will happen is that its reciprocal will have a 0 and Hurwitz's theorem will say that if 1 by f has a 0 then 1 by f n's will start having 0's as n becomes large but the moment 1 by f n start having 0's f n will start having poles and that will that is not possible if you assume f n's to be holomorphic therefore you see that all the if all the f n's are holomorphic then f also has to be holomorphic or it has to be identically infinity okay so I will stop here.