 Alright so let me just write down what I said last time, so we are trying to prove Marty's theorem which is you know it is an analog of Montel's theorem for you know Meromorphic functions okay and we have just, so what I have written here is up to one way of the proof of the theorem okay namely that if you have normal uniform boundedness of the spherical derivatives then of a family of Meromorphic functions defined on a domain in the complex plane then that family is compact in the sense that it is normally sequentially compact namely that given any sequence in that family you can find a subsequence which converges uniformly on compact subsets of your domain okay. So and I just told in words the proof that the other way of the theorem is also true namely suppose you start with a family which is normal so in the sense that a family which has this property that every sequence admits normally convergent, normally uniformly convergent subsequence okay then of course this is a family of Meromorphic functions then the claim is that if you look at the family of spherical derivatives that family has to be normally uniformly bounded namely it should be uniformly bounded on every compact subset okay and that is the way we have to write down the proof I told you in words how this can be done but now I will write it down more accurately. So let me write this here conversely so probably because I am going to the converse I will probably change colour take something else so conversely assume that script F is a family of Meromorphic functions on D such that every sequence in script F admits a normally uniformly convergent subsequence. So I am abbreviating uniformly to VFL YCGT is abbreviation of convergent and SOBSEQ is for subsequence so basically this is the right notion of saying that a family is compact okay what do I have to show we need to show that the family of spherical derivatives namely this is the set of all you take each function small f in the family script F and take its spherical derivative okay and then you get the family of spherical derivatives of this collection and what do you have to show is that this family of spherical derivatives is normally uniformly bounded okay is normally that means that you know you will have to show that it is uniformly bounded on every compact subset of D okay what does that condition mean that is for every compact subset K in D so I am using CPT to abbreviate the word compact so for every compact subset K in D okay so let me write it as a logical statement because I am going to negate it because we are going to prove it by contradiction for every K in D with K compact there exists M greater than 0 there is a uniform bound for all the spherical derivatives of the functions in your family when restricted to K that is so if I write it down it means that there exists a uniform bound which I am calling as M such that the spherical derivative of at each point is bounded by bounded above by M for all functions f is small f in the family script F and for every point z in the compact set okay this is the this is exactly the condition that the family of spherical derivatives is normally uniformly bounded mind you the spherical derivative is a is a positive is a non-negative real number okay the spherical derivative is a non-negative real number it is two times the modulus of the usual derivative divided by 1 plus modulus of the function squared okay and this is at all the points where the function is analytic and since we are considering meromorphic functions you could have points which are poles and at poles I have told you what the spherical derivative is we have evaluated it by continuity if the pole is a simple pole the spherical derivative is 2 divided by the modulus of the residue of the function at that simple pole otherwise it is 0 if it is a pole of higher order okay and of course for this exceptional function which is constantly infinity on the whole domain which is also possible in a limit of meromorphic functions or even for that matter even for analytic functions in a under normal limit for this function which is constantly in infinity equal to infinity the spherical derivative is 0 okay we should remember that so here is so what we have to show is that we have to show this condition okay and what we will do is we will prove by contradiction so what we will do is we will assume this condition is not going to hold and we will contradict the fact that you have normality of the family namely that every sequence in the family admits a subsequence which converges uniformly on compact sense so we will do that we will show a counter example to that okay. So suppose the above the above does not hold so this is proved by contradiction suppose this does not hold then what do you do so you know what you have to do is I will have to negate this okay I have to negate this logical statement okay and you know there is a very slick way of you know negating logical statements you normally replace for every with there exists and the other way around you replace there exists with for every okay that is the way you do it so you know if I negate this what I will get for every compact subset I want a certain property to hold if I negate it means that there I can find a certain compact subset where this property will not hold so this for every k in D with k compact will negate to there exists k in D with k compact so it will become there exists k subset of D with k compact okay and what is the rest of it so look at the original statement for every k in D with k compact there is this is the rest of the statement the statement is that for every f and for every z okay I can find an m such that the spherical derivatives of f and z are all bounded by m okay so this there exists m greater than 0 upon negation will become for every m greater than 0 okay and this for every small f will become there exists small f and this for every small z will become there exists a small z so the way this part negates is you will get for every capital M greater than 0 there exists an f sub m in the family okay and a point z sub m in the compact set such that the spherical derivative of f sub m at z sub m is going to be greater than m this is how it negates okay this is the negation of that statement. So in particular what I can do is that you know basically I want to show that if I assume if that this happens I want to show that I am going to get a contradiction and contradiction to what contradiction to what I have assumed namely that the family whenever you have a sequence you can extract normally convergent subsequence okay so I will have to cook up a sequence and the way I do it is since this negated statement is true for k you start putting m equal to 1, 2, 3, 4 so you make m larger and larger so that m goes to infinity and then you know by going to infinity you expect that the spherical what you will get is you will get a sequence of points and a sequence of functions which where the spherical derivatives at those corresponding points is going to infinity but then this cannot be this cannot come from the original family if it were normal that is because normal convergence of a family also implies normal convergence of the spherical derivatives and mind you the spherical derivative is a finite quantity okay it cannot be it is not an infinite quantity okay so that is how we will get the contradiction so what I will do is so let me write it down these are gory details but sometimes you should know how to write down things that is very important and sometimes you should also be able to just say it in words for example you would find in several textbooks they probably the textbook would become very voluminous if they write down every detail so they might just qualitatively say it in a few words okay but then it is your duty to write it down you have to translate it okay so this is part of this exercise whenever you read a book okay but since this is a lecture I am bound to explain as many details as I can so I will do it so put m equal to 1 2 and so on get f 1 f 2 and so on this is a these are all functions in the family script f and you get these points z 1 z 2 and so on which these are all points in k with the spherical derivative of f i at z i greater than i okay I get this this is when I when I put m equal to 1 2 and so on okay now you know how the proof will go on the one hand I have this sequence of functions so I can always extract a normally convergent subsequence because that is part of my assumption okay on the other hand I also have the sequence of points okay it is a sequence of points but where does it lie it lies in a compact set therefore there is a convergent subsequence because you know any compact this compact set is a compact metric space and for a compact metric space you know that means compactness means is equal to sequential compactness which means that from every sequence you can extract a convergent subsequence so what I can do is that I can do it either way so but so let me do it like this since k is compact so sequentially compact because compactness and sequential compactness are equivalent from metric spaces there exists subsequence z let me call this as z n i okay there is the subsequence which converges okay to it will converge to a point and that point has to be in k because you know a compact set is closed okay because it is a set is closed so if a set is closed it will contain limits okay so z n, z sub n i is a subsequence and it is a convergent subsequence so the limit should also be in k because k is closed okay fine so you have this so you have these z n i is going to z not on the one hand okay this is a compactness of k where I am that is being used and then I will also use the fact that the family is normal okay then it is compact okay so well by hypothesis there exists a subsequence so you know well there exists a subsequence of f sub n i okay so you see I already have the original sequence okay and I have the original sequence of points when I take a subsequence of points which converges I have gotten a subsequence of the original sequence and I am considering the corresponding functions in the family of the in the sequence of functions and I am that is already a subsequence and I am extracting a further subsequence from that because that is what the normal sequential compactness is all about from give me any sequence I can always extract a normally convergent subsequence so I am not trying to I am not extracting a convergent normally convergent subsequence from f 1, f 2 and so on but I am extracting a normally convergent subsequence from the subsequence given by f n 1, f n 2 and so on okay so there is a subsequence of f n i say let me call this as well the notation gets a little bit bad but does not matter so I get this subsequence I get a subsequence of this which converges uniformly on k okay in fact it will be a subsequence which will converge uniformly on all compact subsets of D okay that is what the theorem that is what the hypothesis promises but I am restricting I am just worried about k for the moment because k is a compact set when I am working now but here is the issue is that this f n i l okay this will go to certain f not okay because you see a normal limit of if you take a normal limit of meromorphic functions that is also meromorphic that is something that we prove okay see a normal limit of meromorphic functions is either identically infinity okay or it is a meromorphic function it can even be holomorphic it can even be holomorphic okay this is what we have already seen okay so this f not what can happen with f not is that f not can either be the function is identically infinity or f not can be a honest meromorphic function it could even be a holomorphic function okay so note that either f not is identically infinity okay or well f not is meromorphic function on D okay so this is there alright and what is it that I want you to understand see this convergence is uniform on k this f i n f n i l converges to f not is uniform on k it is a uniform convergence so what it means is given epsilon greater than 0 okay that you can find an n such that if your i sub l is greater than n alright then the distance between these two functions the distance between the function values at any point of k can be made less than epsilon and mind you now because you are working with meromorphic functions you could very well they could very well take the value infinity and you should not use the usual distance you should use the spherical distance so you will have so I have to write it like this d spherical of f of n i l f sub n i l of z comma f not of z this can be made less than epsilon okay the spherical distance between the value of these two this the functions in the sequence beyond a certain stage and the value at the limit that can be made as small as I want this is uniform convergence and the point is that this n has got nothing to do with z so this is for all z okay this is the uniformity it works for all z okay so in particular you know you can now see what I am getting at you see I have these z n i's these z n i's are going to z not okay and therefore the f if you give me any function f in the family by continuity f of z n i will go to f of z not alright and but the point is for f if I had taken f n i okay what will happen if you take the original sequence of functions f 1 f 2 and so if you take f i and evaluate it at z i not the function but I mean the spherical derivative okay then it is greater than i okay we have this so if I take this f n i and evaluate and take its spherical derivative and then evaluate at z n i I am going to get something that is greater than n i okay and that is suppose to go to f n i of z not and now if I let n i tend to infinity it will go to f not of z not so essentially you know f not of z not is going is coming very close to a sequence of quantities which are becoming larger and larger then how can it be because the I mean not f not of z not but I mean spherical derivative of f not of z not so let me write it out yeah so the first statement I need to make is so this was the lemma that I had I had rubbed out last time so this lemma is that if g n converges to g normally in suppose this is a normal convergence then the spherical derivatives also respect that so then the spherical derivatives of the g n will go to the spherical derivatives of g normally okay so this is a lemma that I rubbed out which we need to use at this point if you have a sequence of meromorphic function which converges to a limit function and suppose all this is happening normally namely it is happening uniformly on compact subsets of your domain D then of course we have already seen that the limit function is going to be meromorphic or it is identically infinity okay so this g can be either meromorphic or it is identically infinity and the point is no matter what that is the if you take the corresponding sequence of spherical derivatives that will converge also to the spherical derivative of the limit and that will happen normally okay and remember that if g is the function which is identically infinity there is no problem with the spherical derivative to 0 it is not something that is strange okay it is not an undefined quantity or something like that okay. So you have this so I am using this so if I use this what I will get is that you see I have this f n i l going to f not so this will tell me that f n i l the spherical derivatives will go to f not the spherical derivative of f not okay and this is happening mind you this is going to happen normally so it is going to be this all this is going to happen uniformly on k okay so this is uniformly on k because k is compact alright and what does this mean this means that you know the if you calculate mind you when I am comparing when I am comparing values of usual meromorphic functions I will have to use this the spherical metric okay but when I am comparing values of the spherical derivative okay I have to use the usual distance function on r because spherical derivatives are non-negative real numbers. What this means is that given an epsilon prime greater than 0 there exists an n prime such that if the index is greater than n prime then the distance between f n i l hash and f not hash for at any z is can be made less than epsilon prime okay I can do this for all z in k this is what it means and I am using the modulus function here because these values are all real values okay you see I will now interpolate so what I will do is that I will use the triangle inequality and write this as f n i j of z i j and then I will add f not of z i j okay I can do this plus I then I will write this f not of z i j minus f not and of course everything is a spherical derivative is not just the original functions so and then I will have f hash of z not. So this is just by good old triangle inequality on the real line right I have just added and subtract the spherical derivative z i j okay and of f not okay so now you know what I am going to do so this will be n i j that is what this quantity is alright and now you know by now you know what is happening you see this quantity is n i j and as i j tends to infinity this quantity this is going to go to infinity okay this is going to go to infinity as i j tends to infinity okay you know that alright so what you are saying is what is there on the left is the distance from f not hash of z not okay to a point to a value which is going bigger and bigger and you are saying that can be bounded by the sum of two quantities okay see the first quantity you see the first quantity can be made less than epsilon that is because of the uniform convergence of f n i j to f not okay see we have just written it down above the f n i j is the converge uniformly to f not on k so the lemma says that therefore the spherical derivative is also converge so the distance between f n i j hash and f not hash at any point can be made as small as I want so the moral story is that this quantity here can be made less than epsilon okay and this is okay for any z okay so I plug in z i j so I can make this less than epsilon okay and look at this quantity here what is it so this is less than epsilon of course I will have to put in some condition and the condition is that if this n i j is chosen greater than n I think it was so it is less than epsilon prime if you chose the i j greater than n prime so let me write that so it is this can be made less than epsilon prime if so let me write it correctly i j can be made greater than n prime okay I can do this right and again I think I have missed the subscript needs to be corrected here so let me do it this is an issue with double subscripts or triple subscripts okay so this quantity this can be made less than epsilon okay so I have this and then what about the second quantity the second quantity is the spherical derivative at so there is a zero missing here the spherical derivative of f not at z i j minus the spherical derivative of f not at z not okay but mind you this also I can make sufficiently small that is because of the continuity of the spherical derivative this spherical derivative is anyway continuous okay so the moral of the story is that I can make this also this also can be made less than if you want epsilon double prime if you choose some yeah again I have this problem cropping up that I will have to worry about this double subscript let me write this n i j if n sub i j if i j is sufficiently large is greater than say n double prime I can do this and this is just what am I using here I am just using I do not have space to write but maybe I will come to continue here continuity use continuity of f not hash at z not and remember that the z n i j is also tend to z not the z not was a limit point okay so that is it you have a quantity on the left okay which is very close to very large number okay and this large number is becoming larger that is very close to a large number is because it is bounded by a quantity on the right which is very small okay the sum of epsilon and epsilon prime I can make it as small as I want so it means that the quantity on the left which is the spherical derivative at z not of f not that can that comes arbitrarily close to any large number okay and that is impossible because that is a contradiction because this quantity is a finite quantity the spherical derivative at any point is a finite quantity so it cannot go it cannot be within an epsilon distance of an increasing sequence of numbers just cannot happen so that is the contradiction so let me put this here so this is finite quantity and the totality of all is that well that the whole thing that you have written down managed to get is observed okay so let me write that down so this is observed all this is observed cannot happen so we get a contradiction okay and that finishes the proof okay so spherical derivatives need to be bounded okay and the point I want you to remember is that of course you know sometimes writing down these gory details is a pain you will have to worry about subscripts and things like that but then once in a while you should do this because then you get a feel of having written down something very accurately okay and of course you also should try to be very elegant and say without any notations you must be able to just say in words so if you want to say all this is crazy thing in words you know what you would say is that well I have sequence of points and I have sequence of functions which if that is if I want to prove my contradiction I will find a compact set on which the uniform boundedness of the derivatives will be will not be true on that compact subset I can get a sequence of functions and a sequence of points such that the corresponding spherical derivatives are going to be unbounded okay and but then the points have to converge because it is a compact subset at least the subsequence of points have to converge because it is a compact subset okay so and also the functions must allow a convergent subsequence if you put these two together what will happen is that you will get a sequence of spherical derivatives becoming unbounded okay but coming arbitrarily close to the spherical derivative of the limit function at that given point and that is not possible because the spherical derivative of a function at a given point is a finite quantity okay so that is how you can say it elegantly in words okay fine so that finishes Marty's theorem so mind you Marty's theorem is very powerful because you see it is more powerful than it is more powerful than the usual Montel theorem because you see Montel's theorem the original Montel's theorem says that you know it is only for analytic functions and what you had required there was well you needed normal uniform boundedness of the family of analytic functions alright whereas in Marty's theorem you generalize from analytic functions to meromorphic functions okay and you do not require bounded normal uniform boundedness of the functions but you require normal uniform boundedness of the spherical derivatives of the functions okay so this is a difference so if you so suppose I have a family of analytic functions okay and suppose I know that their derivatives are uniformly bounded normally uniformly bounded okay I can still conclude something from I cannot apply the original Montel theorem okay but I can apply Marty's theorem and say that I can still conclude that this family of analytic functions will converge to a limit and since it is normal convergence okay I mean if you take a sequence of functions from this family I can extract a subsequence that will converge to a limit and that limit will either be holomorphic or it will be identically infinity so you see Marty's theorem is more powerful that is what I want to tell you okay. So let me again go back to that slide that I wrote out here it is so here is so this is let me go back to this slide you see there was this thing that I have circled in violet surrounded by a small I have written it in red the family of derivatives is normally uniformly bounded is an intermediate step okay so if I start with the family which is normally uniformly bounded and suppose it is a family of analytic functions because of the Cauchy integral formula okay I get that the family of derivatives is normally uniformly bounded okay and from that I get equicontinuity and once I have equicontinuity and uniform boundedness I can apply Arzela's theorem okay and then whatever I want I get by doing a diagonalization argument but now you see suppose I am given a family of analytic functions okay on a domain suppose I am not given that the family is normally uniformly bounded suppose I am not given that okay I cannot apply the original Marty's theorem suppose I am given instead of being given that the original family of analytic functions is bounded suppose I am given that the family of derivatives usual derivatives which makes sense for analytic functions suppose I am given that that family is uniformly normally uniformly bounded okay I cannot apply Montel's theorem because I do not have the uniform boundedness normal uniform boundedness of that family itself I have only the normal uniform boundedness of the derivatives of the family I cannot apply the usual model theorem but then the uniform boundedness of the ordinary the boundedness of the ordinary derivative will give rise will imply the boundedness of the spherical derivative because of this inequality you see if you go down because that is the so this is what I have circled in violet the you see the spherical derivative is bounded by 2 times the bound for the normal derivative the usual derivative. So if I am given a family of analytic functions such that the usual derivatives are normally uniformly bounded it follows that the family of spherical derivatives is normally uniformly bounded okay so I can apply Marty's theorem Marty's theorem will tell me now therefore that any sequence in that family admits a convergent subsequence a normally convergent subsequence okay and that normally convergent subsequence is what it is a it is a convergent sequence of holomorphic functions but the only thing is that the convergence is with respect to the spherical metric and then you know what happens my limit function can will either be holomorphic that is analytic or it will be identically infinity okay. So what is the upshot of all this the upshot of all this is let me state it if you have a family of analytic functions on a domain and suppose you know that the usual derivatives of the analytic functions are normally uniformly bounded okay then that family is compact in the sense that you may any sequence in that family I can find a subsequence of convergence normally convergent subsequence and the limit function will either be again analytic or it will be identically infinity this is what I will get because of Marty's theorem and mind you I cannot apply the original model theorem because for the original model theorem I need boundedness uniform boundedness on compact subsets of the original family that is not given to me what is given to me is only uniform boundedness on compact subsets of the derivatives okay so that so in that sense Marty's theorem is very very strong than Montel's theorem okay so with that you know I have more or less we have more or less come to one point here in our discussion I will tell you what we need to do next okay. So if you see all these things that we have proved we have been looking at a domain in the complex plane okay so what are the things that we proved first we proved that so for domain in the complex plane we defined normal convergence okay that is uniform convergence on compact subsets and then we proved that if you take a normal limit of analytic functions okay then the limit is either identically infinity okay if you use a spherical metric or it will be again an analytic function okay and if you take same thing happens for meromorphic functions if you take a family of meromorphic functions okay if you if it is if it is if the family for example if you take a sequence of meromorphic functions which is normally uniformly convergent then the limit function is again going to be either meromorphic it or it will be identically infinity you do not get any strange situations okay the moral of the story is when you take a normal limit of analytic functions we will get an analytic function and the extreme cases you will get the function is identically infinity same for meromorphic functions okay and we have seen that for example if you take a sequence of meromorphic functions is not going to go to a sequence of analytic functions is not going to go to strictly meromorphic function okay a pole cannot pop up at the limit okay such things cannot happen so it is all very well behaved and then we have proved the Montel's theorem and we have proved Mati's theorem okay now all this is for usual domains what about a domain which contains the point at infinity that is the next we want to we want to include all extend all these results to a domain in the extended plane what is the problem with the domain in the extended plane the problem with the domain in the extended plane is the point at infinity at for the point at infinity okay you cannot find a compact neighbourhood at least on the usual plane okay so if you take suppose D so what I am trying to say is that you know if you say that if I have a if I am taking a domain in the extended plane that means it is actually a domain in the usual plane it is in the exterior of large enough disk and it includes a point at infinity okay and whenever I take compact subset of that it will not include the point at infinity the point at infinity cannot come into any compact subset okay so there is a problem if you take a domain which contains a point at infinity okay you are never able to get a compact neighbourhood of infinity in the usual sense okay so what is the meaning of normal convergence at infinity that has to be made sense of you have to understand what normal convergence at infinity means you will have to understand that in that context you have to you have to extend all these terms okay so what I will do in the coming lecture is coming lectures is you know try to look at normal convergence at infinity and extend all these results to the case when your domain is a domain in the extended plane which can possibly contain the even the point at infinity okay so this is a very very important step okay and then we have enough tools to go ahead with the with another theorem called Zaltman's lemma and that will lead directly to stronger version of Montel's theorem on omission of values and that will lead to finally to the Picard theorems okay so we will see that in the forthcoming lecture.