 So, so this is a continuation of the last lecture more or less so you see our situation is that we have taken a sequence fn of functions analytic functions defined on a domain in the complex plane okay and we have assumed that fn converges to f, fn converges to f where f is now a function again defined on the same domain okay on which each of the fn's is defined but now you are allowing f to take the value infinity okay so f is considered as a function not into the complex plane it is considered as a function into c union infinity the extended complex plane okay and the convergence is normal okay and when we say the convergence is normal we mean that it is converges it is uniform on compact subsets of the domain okay we have called the domain as d okay and since we have taken the value infinity the convergence the point wise convergence is with respect to the spherical metric okay so fn of z converges to f of z in the spherical metric this is the same as saying that the spherical distance between fn of z and f of z that goes to 0 and you want it to go to 0 normally on d okay and you know the reason why we are using this spherical metric is this is because f of z can take the value infinity then you will have to measure the distance of a point on the complex plane to the point infinity okay and that for that you have you have to do it only on the Riemann sphere so essentially use a spherical metric on the Riemann sphere okay fine so what we are trying to prove we are trying to prove this very important theorem that you know if you take a sequence of analytic function suppose it converges normally to a limit function then the limit function is either analytic that is holomorphic or completely it is infinity okay and you do not get anything in between alright so how so what do we do we try to prove this by using by applying Hurwitz's theorem and also by using the fact that the spherical metric the spherical metric is invariant with respect to inversion okay we have to use these two facts and how are we going to use it that is what we are trying to do so we have assumed that the limit function is not identically infinity okay so that means that this so if you now look at what we have written in the last lecture the last couple of lines d infinity is a set of all z and d where f takes the value infinity that is a proper subset of d and this is this is because f is not identically infinity and so d infinity is a proper set and mind you d infinity is a closed subset because you see I have already told you last time that f has to be continuous because uniform limit of continuous functions is continuous and normal limit is a locally uniform limit so it is also continuous and locally continuous is the same as continuous okay because continuity is a local property so f is certainly a continuous function and f in the inverse image of a closed set under a continuous map is a closed set the point at infinity is a closed point okay the subset consisting of only a single point is closed in the extended plane because topologically is the same as the Riemann sphere and the infinity the point at infinity corresponds to the north pole so f inverse of infinity is exactly d infinity and that is closed okay what we are trying to show is that we are trying to show d infinity is empty okay we want to show we want we want to show d sub infinity is a null set okay so which means that d is actually d so that will mean that is f is analytic on d because you know the fact is that if you show d infinity is empty then you are saying that at every point f takes only a finite complex value okay it does not take the value infinity so f is not actually a map into c union infinity is actually a map into c and then you know if you have a complex valued function which is a normal limit of analytic functions then it is analytic this is something that we have already seen this is part of for example a first course in complex analysis where you essentially have to use Cauchy's theorem and you have to use Morera's theorem okay so all you have to show is d infinity is empty and now how do you show this we exploit the fact that d is connected set see d is a domain in the complex plane in the complex plane so d is an open connected set and of course we are always worried about only non-empty sets d is on empty and remind you d since d is connected we use try to use this very important property of a connected space for a connected space the only subset that is both open and closed has to either be the null set or it has to be the full space so what you try to do is that you try to show that d infinity is open okay so d infinity is already closed suppose you show d infinity is open then d infinity has to be either so it is an open and closed subset of d which is connected so d infinity has to either be the null set or it has to be all of d but it is not all of d because I have assumed f is not identically infinity so d infinity has to become the null set and you are done okay and the theorem is proved so we because so let me write that down because of the connectedness the connectedness of d it is enough to show that d infinity is open okay that is enough to show that what does it mean it means that if you give me a point of d infinity then there is a whole neighbourhood surrounding that point which is also in d infinity that is what openness means it means that every point is an interior point so in other words what does it mean it means that if you take a point z not in d infinity namely a point z not where f takes the value infinity then there is a small neighbourhood there is a small disc surrounding z not where also f takes the value infinity that is what you have to show if f is infinity at a point then f is infinity at all points in a small disc surrounding that point this is what will mean to say that d infinity is an open set okay so that is what we are going to do that is exactly what we are going to do alright and for that we are going to use could be this theorem we are going to use the the invariance of the spherical metric with respect to inversion okay and how we are good you will see this so start with so let us start with and z not in d infinity okay and of course I should tell you that of course I am assuming d infinity is non-empty okay to begin with if you want to be very logical you can say that let us assume d infinity is non-empty because if d infinity is empty we are anyway done okay and so you assume d infinity is non-empty and then prove and get a contradiction okay so if you want to be very logical you can say like that so in any case I am assuming I start with a point so you know this is this is often a feature of mathematics see finally d infinity is empty that is what you want to show you want to show that the set is empty you want to show that there is no point in that set but then you know it is round about the way you do it what you do is you assume it is non-empty and then you try to once it is non-empty you try to get some properties of the set which will give you a contradiction so in this case you assume d infinity is non-empty okay and then you get the fact that d infinity has to be everything and that is not true because f is not identically infinity okay so this often happens in mathematics so you start with the point z0 in d infinity so f of z0 is infinity this is what it is and now but mind you f is a continuous map okay f is a continuous map into c union infinity and it is therefore it is continuous at z0 also and what I want to say is that since it is continuous at z0 okay you can find a sufficiently small neighbourhood of z0 such that all the function values on that neighbourhood okay are close to infinity to within whatever epsilon distance you want and mind you you have to now use the spherical metric okay so I am using the continuity of f at z0 okay so since f is continuous the given epsilon greater than 0 there exists a delta greater than 0 such that well mod z minus z0 less than delta in d will imply that I should write this with respect to the spherical metric d sub s f of z comma f of z0 which by the way is infinity this can be made less than epsilon okay so you see I have to use the spherical metric okay so I am just saying that since f takes the value infinity at z0 in a sufficiently small neighbourhood of z0 f has to take values close to infinity so the distance between function values fz and infinity which is f of z0 that can be made as small as I want if I choose sufficiently small neighbourhood delta neighbourhood of z0 and of course I have to choose it in d of course okay because I want f of z to make sense because f of z is make sense only for z in d okay fine now you see mind you what this means you see try to understand what this means it means that if you take this small disc centered at z0 radius delta okay then the image of that disc lies in a neighbourhood of infinity because the distance between f of z and f of z0 which is which is equal to infinity is less than epsilon means f of z lies in a neighbourhood of infinity that means that these small disc centered at z0 radius delta is mapped to the exterior of a sufficiently large circle okay so you must think that as epsilon becomes smaller and smaller you are getting the exterior of sufficiently larger and larger circles okay so if you are thinking in terms of radii of circles you should think of 1 by epsilon or 1 by epsilon squared or something like that as epsilon goes to 0 because then 1 by a positive power of epsilon greater than 1 will go to infinity okay fine so this is what it means now you see so you know so let me tell you basically what the philosophy is all about see the idea is very very simple let me give you the idea of the proof you see f is so there is this neighbourhood of z0 okay at z0 f is infinity okay and then there is a small neighbourhood of z0 where f is close to infinity okay so you know if I draw a diagram it is going to be something like this so let me draw a diagram it helps to draw a diagram so here is z0 and there is a small disc surrounding z0 radius delta and what f is doing is that it is mapping this onto neighbourhood of infinity which is you know the exterior of sufficiently large disc this is what is happening so the interior of this okay is going to the exterior of a sufficiently large disc and so this is this disc is centre at z0 is radius delta sufficiently small and the image of this disc I am not saying the image is all of the you know exterior of that large circle but it is a subset of that okay so yeah so this is the situation see what this tells you therefore is that f is certainly bounded away from 0 in a neighbourhood of z0 right because at z0 f is taking the value infinity alright and in a neighbourhood of z0 it has to take values close to infinity and values close to infinity are certainly non-zero values because values close to infinity are supposed to be values which lie in the exterior of a large on the exterior of a large circle alright so f is f is going to be non-zero in particular okay and now imagine your fn converges to f uniformly on compact subsets of D that is given to you that is given to us because we have assumed fn converges to f normally on D alright so because of so so in particular you know if I choose this delta sufficiently small so that even the boundary of that circle mod z-z0 equal to delta is also inside D I can do that if I make delta a little bit smaller if you want okay then mod z-z0 less than or equal to delta becomes a compact set because you know it is now closed and bounded it is a compact subset of D and therefore fn will converge to f normally on that right because I in fact uniformly on that because it is a compact set so if I choose delta sufficiently small I can make sure that on this close small close this centred at delta centred at z0 radius delta the convergence of fn to f is uniform okay but then f is never 0 there f is not 0 on that on that closed disc on that small disc because it is f values are in neighborhood of infinity so f is not 0 so that means because of uniform convergence fn are also non-zero beyond a certain stage in that in the closed disc okay and if fn are non-zero they are non-zero analytic functions so 1 by fn will become holomorphic they will become analytic okay and you know fn converges to f in the spherical metric because of the property of the invariance of the spherical metric with respect to inversion fn converges to f in the spherical metric will tell you that 1 by fn converges to 1 by fn 1 by f in the spherical metric okay so 1 by fn will converge to 1 by f and these 1 by fn are all analytic in the disc and 1 by f therefore will become analytic in the disc but what is 1 by f of z0 it is 1 by infinity it is 0 so z0 becomes a 0 for 1 by f okay and 1 by fn is a sequence of analytic functions that is converging normally to 1 by f in the disc apply Hurwitz's theorem what it will tell you is that all the 1 by fn's beyond a certain stage they will have zeros as many zeros with multiplicities as the 0 z0 of 1 by f okay but then you see if 1 by fn has zeros that means fn will have poles alright but fn are analytic how can they have poles that is a contradiction therefore the moral of the story is that you get a contradiction okay and therefore so you can see you can either see this is a contradiction or you can go one step further and say that see where you can look at where Hurwitz's theorem will go wrong see Hurwitz's theorem can go wrong in the following sense if you take a sequence of analytic functions if they are converging normally to a limit function okay then Hurwitz's theorem says that a 0 of the limit function comes from the zeros of the functions which converge to that limit okay it comes as a limit point of zeros of functions that converge to that limit function but then there is one extreme possibility the limit function itself could have been identically 0 okay if the limit function is identically 0 okay then that is a case which we do not deal with in the actual Hurwitz's theorem see in the Hurwitz's theorem we always assume that you have an isolated 0 of the limit function so the only way this Hurwitz's theorem can fail to apply here is that 1 by f becomes identically 0 but then if you say that 1 by f becomes identically 0 you are just saying that f is identically infinity in a neighbourhood of z0 and that means that wherever f is infinity if you take a point where f is infinity then there is a neighbourhood surrounding the point where again f is infinity so d infinity is open and we are done okay so there are both so there are these ways of looking at it which will give you the proof of the theorem okay so now let me write down things in words okay so let me write this we can choose delta small enough so that mod z minus z0 less than delta is in D okay and now you see you also have this uniform convergence since fn converges to f uniformly I should say normally oops normally fn converges to f uniformly I am abbreviating uniformly to u of ly on mod z minus z0 less than or equal to delta okay and well note that so what does this mean this means that if for the spherical metric if you take fn of z and f of z this see this distance can be made lesser I mean this distance goes to 0 uniformly on mod z minus z0 less than or equal to delta because of course mod minus z mod z minus z0 less than or equal to delta is compact okay and what does uniform convergence means uniform convergence means that you can choose an index capital N large enough such that for all small and greater than capital N you know you can make this spherical distance lesser than epsilon if you want okay so mind you we already started with some epsilon okay let us keep that epsilon so there exists an N sufficiently large so that N greater than or equal to capital N implies that the distance spherical distance between fn of z and f of z can be made less than epsilon but here is the uniform here is the uniformity of the convergence for all z in mod z minus z0 less than or equal to delta independent of N see the point is that you can choose this capital N in a way that has got nothing to do with this z that is the uniformity of the convergence okay. In general if it is point wise convergence for the capital N will depend on epsilon of course but it will also depend on the z particular z you are looking at the point z but the uniform convergence is that now you have this capital N depending only on epsilon and not depending on z could have been anything so let us look at this d sub s of I am trying to compare fn of z and f of z0 which is infinity and this is by triangle inequality it is d sub s fn of z and now you put f of z okay and then put d sub s f of z and f of z0 this I can do put fz okay you introduce this fz as a third point the triangle to apply the triangle inequality okay now if you do this see this is certainly less than see the ds fn z fz that is less than epsilon as I have underlined above so I will get an epsilon plus the ds fz fz0 is also less than epsilon that is because of continuity of f at z0 so I will get epsilon plus epsilon is 2 epsilon okay so what this will tell you is that this will tell you what we want and you know this is for this is for n greater than or equal to capital N. So what this will tell you is that beyond a certain stage all the fns they are a neighborhood of infinity so they do not vanish and that is what I want I want all the fns not to vanish beyond a certain stage why because then I can invert them and say that they will the inverses will also be analytic okay so thus so this is what I want thus fn for n greater than or equal to small n greater than or capital N do not vanish on mod z minus z0 less than or equal to delta and hence 1 by fn n greater than or equal to capital N or analytic there analytic or holomorphic there okay I wanted to invert the fns and why I wanted to invert the fns because you see I want to use this you know the invariance of the spherical metric with respect to inversion. So now ds of f of fn of z comma f of z this is the same as ds of 1 by fn of z comma 1 by f of z you have this okay here is where I am using the invariance of the spherical metric with respect to inversion and mind you what is it that we have what is our basic assumption our basic assumption is that this fellow on the left goes to 0 normally on d okay this is our assumption original assumption but that quantity is equal to the quantity on the right side okay and you see but there is a small thing this is the quantity on the left side goes to 0 normally on d it goes to 0 uniformly on mod z minus z0 less than or equal to delta because it is a compact subset of d okay. So uniformly on mod z minus z0 less than or equal to delta that let me write that but you see this equality that I have written that is valid everywhere but the only problem is you know I want see I want 1 by fn to be analytic okay and that does not happen everywhere that happens anyway on mod z minus z0 less than or equal to delta so see I am worried about the equality only in mod z minus z0 less than or equal to delta because the 1 by fns are analytic there for small n greater than or equal to capital N okay 1 by fn converges to 1 by f okay and these in see this is uniformly in mod z minus z0 less than or equal to delta alright this is uniform convergence and the 1 by fns these 1 by fns are all analytic in mod z minus z0 less than or equal to delta okay so you have sequence of analytic functions that is converging to a limit function okay and this is this convergence is actually uniform convergence and mind you the limit function is also complex value that is the big deal see notice f is f f f takes values in the neighborhood of infinity right f at z0 is infinity and in that small disc surrounding z0 f takes values in the neighborhood of infinity that is the reason why the distance spherical distance between f of z and f of z0 which is infinity is less than epsilon that is what it means so f takes values also in the neighborhood of infinity there and if f takes values in the neighborhood of infinity f is not 0 of course okay and 1 by f will become very small because f is very large so 1 by f is bounded so the moral of the story is that this is a 1 by f is a not in that neighborhood mod z minus z0 less than or equal to delta 1 by f is a complex valued function that is it is a bounded complex valued function that is the big deal there okay it does not take the value infinity 1 by f never takes the value infinity okay because f is f takes values in a neighborhood of infinity okay so 1 by f this is bounded in mod z minus z0 less than or equal to delta okay now you know now we are again using this standard theorem from a first course in complex analysis that you are normal limit of analytic functions if the limit function is also a complex valued function then the limit function is continuous and in fact it is analytic. So the moral of the story is see all this pain is to say that 1 by f makes sense and 1 by f is a complex value of course 1 by f 1 by f z0 is going to be 0 because 1 by f z0 is 1 by infinity which is 0 according to our conventions alright so this 1 by f is in a neighborhood of 0 actually it takes values in a neighborhood of 0 because f takes values in a neighborhood of infinity okay so let me write that here takes values in a neighborhood of 0 which is actually 1 by f z0 okay now what does all this tell you this all these things will now tell you that 1 by f is analytic at z0 okay and of course 1 and at z0 what happens at 1 at z0 1 by f is 0 because it is infinity f of z is infinity f of z0 is infinity therefore z0 becomes a 0 of an analytic function and what is the property of a 0 of an analytic function it is isolated. So the moral of the story is that 1 by f is analytic at z0 z0 is a 0 of 1 by f now appeal to Hurwitz's theorem now appeal to Hurwitz's theorem and what will Hurwitz's theorem say Hurwitz's theorem will say that if 1 by f is not identically 0 then z0 will be an isolated 0 for 1 by f and suppose it has a certain order L then all the f n's all the 1 by f n's for n sufficiently large will also have L0's in a neighborhood of z0 but 0's of 1 by f n are the same as poles of f n and that is not allowed because f n's are all analytic how can they have poles so you cannot apply Hurwitz's theorem you should not be able to apply Hurwitz's theorem the only way out is that 1 by f should be identically 0 in that neighborhood and 1 by f being identically 0 in that neighborhood is the same as f being identically infinity in that neighborhood and that is all we want okay. So you see here is where Hurwitz's theorem comes in alright so let me write that down so let me write this here by Hurwitz's theorem either 1 by f is identically 0 in mod z minus z0 less than or equal to delta which means so I should say which means f is identically infinity in mod z minus z0 less than delta alright or e z0 is an isolated of 1 by f say of order L and for n sufficiently large 1 by f n also have L0's in mod z minus z0 less than delta which is impossible which is impossible as f n will then have L poles in mod z minus z0 less than delta less than delta for n sufficiently large okay. So what this tells you is that the only way out so f is identically infinity that is the only way out for mod z minus z0 less than delta okay and this means that mod z minus z0 less than delta is contained in d infinity. So you start with the z0 in d infinity I am able to find a whole disc surrounding z0 which is also in d infinity so d infinity is open so d infinity is open but already we know d infinity is closed so d infinity is both an open and closed subset of it is both open and closed subset of d which is connected so there is no other choice it has to either be d or empty but it is not d because f is not identically infinity so it has to be m just the empty set and we are done okay thus d infinity is both open and closed in d which is connected I should remove the comma here so d infinity is not the null set or d but f not identically infinity implies that d infinity is not d so d infinity is the null set and this means f is analytic on d and that finishes the proof okay. So this is a very very nice theorem okay so if a sequence of holomorphic functions on a domain converges in the spherical metric normally then the limit function is either holomorphic that is analytic or you go to the other extreme the limit function is identically infinity you do not get something in between and the idea is what could you get in between well if you want something mild you can have a meromorphic function which means that you get some points which are isolated where the limit function develops poles okay but what this theorem says is that it simply cannot develop a pole somewhere okay and you know now the reason philosophically why f cannot develop a pole at a point because you see if f n converges to f in the spherical metric and f develops a pole at a point then because of the invariance of the spherical metric with respect to inversion 1 by f n will converge to 1 by f okay but then 1 by f n will converge to 1 by f and if f has a pole at a point 1 by f will have a 0 there so the 1 by f n will start having zeros by Hurwitz's theorem and they will give rise to poles of f n which is not possible okay of course there is a much versa thing that you can expect that the f n converges to f, f of course you know this f if you look at the locus where f is not infinity that is of course an open set because that is a compliment of D infinity and on that open set f is going to be analytic there is no problem it is a honest complex valued function which is a uniform limit of analytic function so it is analytic there is no problem but what kind of a set D infinity is is very very you know it is very mysterious what could have happened is that this D what is what prevents D infinity from being set with non-empty interior okay why should be an isolated set of points if D infinity is an isolated set of points it means that f is a meromorphic function but why should it be an isolated set of points why cannot D infinity be a curve okay if it is a curve then D infinity then it means that your f has non-isolated singularities okay why should D infinity you know if you want f to be meromorphic the condition is that D infinity must be an isolated set of points. So the next theorem that we are going to prove in the next lecture is that if you drop the assumption that the f n's are holomorphic assume that they are meromorphic then also the limit function f will be meromorphic it would not be any worse that is D infinity will only be a set of isolated points and it cannot be it cannot contain non-isolated points so you cannot have a sequence of a sequence of meromorphic function it goes to a limit function then that limit function you know it cannot have horrible non-isolated singularities it can have only isolated singularities and they have to only be poles okay this is again a very good thing and we will prove this in the next lecture okay.