 Alright, so let us continue with the proof of Zalkman's lemma, okay. So basically what this lemma is about, it is about characterizing a non-normality of a family, okay. So you know so I begin with a family script f of Meromorphic functions, okay and I am assuming the family is defined on this domain D, alright and I assume that the family is not normal, okay. Then as the non-normality will manifest at some point, okay, the family is not normal if it is not normal at least one point in the domain and how do you get that non-normal point? That is exactly what Zalkman's lemma is all about. So you see what it says is that you can find the sequence of points in the domain converging to this point Z0 which is the so called which is the point of non-normality and you can find these decreasing sequence of radii, positive radii so that you know if you take and you will be able to find see the fact that the family is not normal means what? It means that the family is not sequentially compact, normally sequentially compact. I mean our definition of normal is normally sequentially compact and that is the correct version of compactness for us, okay. When you are looking at a family of analytic functions or meromorphic functions the correct version of compactness is normal sequential compactness that is given every sequence you should find a normally convergent subsequence and when you say a family is not normal what you are saying is that you are just saying that there is a sequence for which you cannot find any normally convergent subsequence, okay and Zalkman's lemma tells you that you can find such a sequence and that is the sequence here fn. You can find this sequence it is a non-normal family in fact the sequence itself forms a non-normal family so that if you take the members of the sequence and then you take the corresponding zoomed functions, okay. So gn is the zooming of fn centered at zn and with the magnification factor 1 by epsilon n, okay. Then this zoomed family converges normally on the whole complex plane to a non-constant meromorphic function g, okay and the point is that the non-constancy of the meromorphic function reflects in the fact that its spherical derivative is not 0 because you know the moment the spherical derivative of a meromorphic function is 0 it means it has to be constant, alright. So this non-constant now the non-constancy of g as a meromorphic function is further fixed by this fact that the spherical derivative at the origin is 1 and all the spherical derivatives are bounded by 1, okay. So this is Zalkman's lemma so the point about this lemma is that if a family of meromorphic functions on a domain is not normal it gives you a non-normal point Z0 and it gives you a non-normal sequence in the family which violates normality in a neighbourhood of Z0 that is the whole point, alright. And I have explained to you that I have explained to you what happens if the family were normal, if the family were normal what would happen is that no matter what kind of what Z0 you choose and the sequence Zn you choose like this and you choose these any radii epsilon going to 0, okay. The zoom functions will always converge normally to a constant meromorphic function I mean to a constant function, okay. So the normality of the family will tell you that always the zoom functions will be constant and the normal that the non-normality of the family is reflected by being able to find if a sequence for which the zoom functions do not converge to normally to a constant function but actually they converge normally to a non-constant meromorphic function, okay. So it is the limit function that matters if it is normal the zoom functions will always converge to a constant if it is not normal I can find the situation where the zoom functions converge to a non-constant meromorphic function that is the whole point, okay. So let us see a proof of this the proof is tricky so as I have mentioned in the you know the reference material the textbook that I am following is that of Gamelin, okay and the proof as it is mentioned that is tricky as you will see. So we will make a couple of reductions the first thing is that you know what is given to me is that the family is not normal, okay and now you know we have already proved Marty's theorem which is a characterization of normality namely it says that a family is normal if and only if you know if you take the family of spherical derivatives that is normally uniformly bounded, okay. So recall if you take a family of analytic functions okay the condition that that family is normal that is normally sequentially compact is by Montel's theorem equivalent to the family being normally uniformly bounded and if you consider meromorphic functions you get the analogous theorem which is Marty's theorem which says that the condition for normality is that the family of spherical derivatives is normally uniformly bounded. So the spherical derivatives being normally uniformly bounded is equivalent to normality of the family so the family is not normal you have a violation of the bounded normally uniform boundedness of spherical derivatives and what does that mean it means that there is a compact set on which you know the spherical derivatives can become unbounded. So this means by Marty's theorem you can actually find a sequence of points okay and functions such that the corresponding functions at those points the spherical derivatives go to infinity plus infinity, okay. So this is so that is the first step so let me write this by Marty's theorem there exists a sequence Wn tending to W0 in D and functions fn in the family f such that fn hash of Wn goes to plus infinity and Wn are in a compact subset, okay. So I can find this just because of Marty's theorem because Marty's theorem says that you know normality is equivalent to the spherical derivatives being uniformly bounded on compact subsets, okay. Fine now we will make a couple of reductions what we will do is for convenience we will assume that you know W0 is actually the origin okay you assume W0 is the origin and how can you do that you can do that by simply translating the domain to so that you make W0 the origin. So you translate the domain by minus W0 you will get a new domain and you look at the functions there the translated functions. So without loss of generality what you can do is you can assume that W0 is the origin okay and that is one thing and the second thing is that you can also assume that the moment you assume W0 is the origin so origin is a point of D okay then of course there is a small disc surrounding the origin which is also in D because after all D is an open set and you by using a scaling transformation you assume that the unit disc along with the boundary is also in D okay. So these are you scale the domain I mean you scale the domain and you translate the domain so that you can assume without loss of generality that the compact set you are looking at where you got this sequence Wn is actually the unit disc okay along with the boundary unit circle and the sequence actually converges to the origin okay. So we will make these assumptions without any loss of generality so let me write that down without loss of generality we may assume W0 equal to 0 and mod Z less than or equal to 1 is in D okay so for this all you have to do is that you have to translate D by minus W0 and then you have to scale D suitably so that the unit disc which is a neighborhood of W0 equal to 0 is inside D alright fine. So you can do this so you see my picture is now like this so I have so here is my I think I will have to okay so let me go down so here is my picture so I have so this is a complex plane and I have this is the origin so this is the unit disc and well this is inside D. So you know your domain D contains the unit disc so this is D and well this is a z plane and of course there are these so there is the sequence of points Wn Wn plus 1 that is tending towards the origin okay and what you are given is that okay fine so I have this now what you do is that you so the tricky thing is that so you know what I am looking for I am looking for I have to extract this sequence Zn okay which goes to Z0 and I have to extract this sequence of functions which you know which have this property that the zoomed functions okay they converge to a non-gosson meromorphic function so you see the trick is the following so what you do is you put Rn to be the maximum over mod Z less than or equal to 1 of fn hash of Z and then you multiplied by 1 minus mod Z okay so you do this so this is the tricky part the trick is you see what is given to you is these functions fn hash the spherical derivatives of those fn hash is just the spherical derivative of fn okay and of course you know spherical derivative is continuous mind you the spherical derivative of meromorphic function is a continuous function it is a continuous non-negative real valued function okay it is a positive function okay at worst it can be 0 right which is what happens if the function is a constant okay but the point is that you mind you the spherical derivative has no problems at poles when the meromorphic function has poles there is no problem with the spherical derivative unlike the usual derivative the usual derivative is not defined at a pole because it is a singular point but a spherical derivative is defined at a pole and we have already seen that if it is a pole of higher order then the spherical derivative is 0 if it is a pole of order 1 namely a simple pole then the spherical derivative is 2 divided by the modulus of the residue at that pole okay so the spherical derivative is a nice continuous function okay a non-negative real valued function and you are looking at this function on this domain mod z less than or equal to 1 which is a compact set. So if you are looking at a continuous function on a compact set continuous real valued function on a compact set you know the function is of course you know it will be uniformly continuous and it will attain its maximum and minimum therefore this maximum is well defined okay and the point is that you see what is given to me is that these f n hash they become larger and larger okay at points which are getting closer and closer to the origin you see as n tends to infinity w n converges to 0 okay that means as n tends to infinity w n goes closer and closer to 0 and what is f n hash of w n that is going to infinity that means f n hash attains large values closer and closer to the origin as n becomes large alright therefore so you know what one does is that it could happen that the maximum values of f n hash could also be taken close to the boundary okay but if you go close to the boundary this quantity becomes very small. If you go close to the boundary of the unit disc the quantity 1 minus mod z will become very small and that will offset this the value of f n hash at that point okay that is heuristically this is the reason for multiplying by 1 minus mod z okay instead of just considering the maximum of f n hash 1 minus f n hash of z. So mind you 1 minus mod z is also continuous it is also continuous you know real valued function non-negative real valued function inside the unit disc so there is no problem about it okay so the product is of course continuous real valued function so that is it has a maximum okay now you have to make a series of observations the first thing is suppose Z n is such that mod z n is less than or equal to 1 and R n is attained at z n so R n is f n hash of z n times 1 minus mod z n okay so R n which is the maximum is attained at some z n okay so look at that z n so and this is a z n that I actually want or probably a subsequence of that as you will see. See the first thing is note that you see R n is greater than or equal to you know f n hash of w n into 1 minus mod w n this happens because R n mind you is the maximum of f n hash into 1 f n hash of z into 1 minus mod z so if you put z equal to w n so the maximum value will always be greater than any of the other values so I will get this but then you see as n tends to infinity you see this goes to 1 okay because w n tends to 0 and this fellow goes to infinity okay because that is the original assumption the f n hash the spherical derivatives go to infinity okay that is how we pick the sequence w n because it was violating normality violating the conditions of Marty's theorem okay so you see so what is happening is that this will tell you that you know R n will tend to infinity so these R ns are becoming bigger and bigger and bigger okay that is something that you have to understand first. Now you look at this so you know if you look at this definition of z n okay what it will tell you is that the f n hash of z n will also go to infinity because you see if you take R n this is f n hash of z n times 1 minus mod z n and this is certainly you know greater than this is less than or equal to f n hash z n because you know after all 1 minus mod z n is less than or equal to 1 okay so this going to plus infinity as n tends to infinity will imply that the f n hash of z n will also go to plus infinity okay so this implies that this goes to infinity plus infinity as n tends to infinity okay so what you have done is you have got this from the sequence w n which goes to 0 to w naught you have cooked up this other sequence z n okay and the point is that the spherical derivatives at the z n also go to infinity plus infinity just like the spherical derivatives at the w n's go to plus infinity okay but the point is that of course this is the sequence z n that you have got that need not be convergent it is just a sequence okay but anyway it is a sequence inside the unit disk and you know unit disk is compact sequentially compact therefore there is a convergence of sequence and therefore without loss of generality can assume that you know this sequence z n is actually convergent okay so we will make that assumption without loss of generality we assume z n converges to z naught with you know z naught also of course in the unit disk because a unit disk is closed okay of course when I say unit disk I am I am also including the boundary it is not the open unit disk okay fine so I have gotten we have gotten hold of this sequence actually alright and now the point is that we have so you know what is our aim our aim is you have to get this sequence of functions and you have to get these sequence of points such that and then you have to get a certain sequence of radii okay such that the zoom functions they converge to a non-constant meromorphic function so where do you get those sequence of decreasing radii okay and that is that comes very simply so what you do is you put epsilon n to be 1 by f n hash of z n okay and then this will go to this will of course go to 0 as it will go to 0 plus as n tends to infinity that is because the f n hash of z n is going to plus infinity right so this will serve as the zooming radii so now everything is in place we have gotten what we want okay and so see so let me write this down since R n is f n hash of z n times 1 minus mod z n what you will get is that you will get epsilon n R n is equal to 1 minus mod z n okay because epsilon n is just defined to be 1 by f n hash of z n right and now what you do is that you do the following thing you put g you take the zoom functions okay so you take g n of zeta to be well f n so it is just you zoom the function f n you zoom the function f n centered at z n with the magnification factor 1 by epsilon n and use the variable zeta so this is going to be f n of z n plus epsilon n times zeta so this is a zoom function this is the family of zoom functions okay mind you we have to find a family of zoom functions which converge normally to non-constant Meromorphic function this is this will be that family okay and where is this defined you see you know this is defined for mod zeta lesser than R n you see what is happening is that you have so the diagram is like this you have this unit disk this is origin this is this is 1 and then you have z n somewhere here okay then you see if you take the smaller disk centered at z n its radius will be 1 minus mod z n this radius will be 1 minus mod z n but this 1 minus mod z n is as I have written above it is just epsilon n R n okay so if I think of a zeta here variable zeta here okay then you know the maximum distance of z n to zeta can be epsilon n R n okay and that means that the maximum value of zeta can be up to R n because I have reduced you know I have actually used the scaling factor 1 by epsilon n okay so but the point is look at these functions g n the zoom functions these zoom functions are defined on mod zeta less than R n and mind you R n tends to plus infinity so what it means is that as before you know the zoom functions are eventually defined on any compact subset of the plane okay so g n is defined for n sufficiently large on any compact subset subset of the complex plane and here of course the complex plane you are looking at is a zeta plane mind you you have brought in this new variable the zooming variable zeta okay so now the now the fact is that g n you know g n does the job that is all you have to verify okay and how does one do that see mind you we want to show that you know g n converges normally to a non-constant meromorphic function okay that is what we want to show that is the whole point now again use what is theorem see to after all g n are also meromorphic because g n are just you know obtained from f n by translation and scaling okay g n is just f n translated see you take the variable of the f n okay and you know you translate that variable by minus z n and then you divide by a scale it by 1 by epsilon and you will get f n okay so f n have been obtained from g n by translation and scaling so g n are also meromorphic okay and well and what am I trying to show I am trying to show that the g n converge normally but again I can apply Marty's theorem to show that the g n converges normally I will have to only show that the g n are normally uniformly bounded I mean the spherical derivatives of the g n are normally uniformly bounded okay so that is all that is what I check okay and that is very that is just an estimate so how do I check that see you will see that g n hash so you know what will happen is g n hash of if you calculate g n hash of zeta this is spherical derivative of g n of zeta mind you this is so I will have to take the spherical derivative of f n of z n plus epsilon n zeta so this is the spherical derivative I have to take okay but then taking the spherical derivative you know will be the same as taking the spherical derivative of f n and then I will get a multiplication factor epsilon n you know the spherical derivative will become smaller for the zoom function the spherical derivative will become smaller by the inverse of the zooming factor the zooming factor is 1 by epsilon n the inverse of the zooming factor is epsilon n okay and so you know this is just chain rule of differentiation so this is epsilon n times f n hash of z n plus epsilon n times zeta this is what you get alright and now you see what you must understand is that now I have a I have this inequality because you know f n hash z n times 1 minus mod z n is R n and that is the maximum value okay so recall that we have this definition of R n here R n is f n hash z n times 1 minus mod z n and mind you that is the maximum value of this quantity R n is actually the maximum value of f n hash multiplied f n hash of z multiplied by 1 minus mod z okay therefore what you can see is that R n is certainly going to be greater than or equal to the value of f n hash times 1 minus mod z for any mod z for any z in the unit disk okay so I will put this for z I will put this so I can put z n plus epsilon n zeta and here I will get z n plus epsilon n zeta so this is correct okay because in fact when you put zeta equal to 0 this it is the value on the right is actually R n R n is a maximum value okay so you have this but then but you see now I can use this to get now I you know so this is the quantity here and this is the quantity that is appearing here okay which is multiplied by epsilon is g n hash of zeta so I can use this to get a bound for g n hash of zeta so what will I get I will get g n hash of zeta is equal to epsilon n times this but this rectangle this thing is less than or equal to R n by 1 minus mod z n plus epsilon n zeta so I will get this is less than or equal to epsilon n R n by 1 minus mod z n plus epsilon n zeta all right but then you see triangle inequality mod z n plus epsilon n zeta is less than or equal to mod z n plus epsilon n mod zeta okay and therefore what I will is that this is also less than or equal to epsilon n Rn by 1 minus mod Zn minus epsilon n mod Zeta I will get this right and now mind you go back so here is why this proof is tricky this 1 minus mod Zn mind you is epsilon n Rn okay that has to be trickly used so this 1 minus mod Zn I can put epsilon n Rn then you can see this epsilon n is coming out both the numerator and denominator and get cancelled so you see I get epsilon n Rn by epsilon n Rn minus epsilon n mod Zeta and this becomes that is less than or equal to Rn by Rn minus mod Zeta okay and you see now so this is what this is the estimate for the spherical derivative of Gn right and see the point is that if mod Zeta is less than say some R okay there exists an n large enough such that Rn is going to be greater than R okay because after all the Rn's tend to plus infinity okay so that is beyond a certain stage all the Rn's are greater than R so that means that you know you can so you know I can divide by Rn and let me put equal to here I will get 1 by 1 minus mod Zeta by Rn okay and if mod Zeta is less than R and Rn is greater than R okay then Gn is defined on mod Zeta less than R okay then Gn hash Zeta is defined on mod Zeta less than R okay because mod Zeta less than R contains is contained in mod Zeta less than Rn and mod Zeta less than Rn is the domain of Gn okay so Gn hash is defined and in fact it is not only for n it is also for higher values of n okay so you know let me write Gn hash of n plus m is equal to 1, 0, 1, 2 and so on so all these Gn's are defined okay and the point is in any case this quantity you get this estimate Gn hash of Zeta is bounded by 1 by 1 minus R by Rn which is bounded by 1 by 1 minus R okay so and finally I have gotten this 1 by 1 minus R without any condition on the subscript n small n okay and that is a uniform bound for Gn beyond a certain stage for Gn hash beyond a certain stage alright and that is it now Marty's theorem will tell you that Gn hash has to therefore you know it has to converge normally okay so by Marty's theorem Gn admits a subsequence that converges normally on the whole complex plane okay and without loss of generality we may assume that this subsequence is Gn itself okay so after all I am interested in a convergent subsequence and what I have got is a sequence which I know admits a convergent subsequence so without loss of generality I replace a sequence by the convergent subsequence okay if I do not do this then I will have to use a double subscript okay but it really does not matter but now this Gn does the job because you see what happens is that first of all this tells you this bound on Gn hash so what you will do is now you let Rn to you let Rn to tend to infinity okay then R by Rn will go to 0 okay so you let n tend to infinity then Rn goes to infinity R by Rn goes to 0 and this quantity goes to 1 okay and that will tell you that all the Gn hash they are all bounded by 1 okay so clearly you get all the Gn hash are all bounded by 1 that is one condition and what is the other thing what about Gn hash of 0 if you calculate Gn hash of 0 Gn hash of 0 is going to be what so let us go back to what we have here so go to this formula here you put Zeta equal to 0 Gn hash of 0 is epsilon n Fn hash of Zn but you see epsilon n Fn hash of Zn is 1 because epsilon n is actually 1 by Fn hash of Zn so Gn hash of 0 is actually 1 okay so let us see that all these are all little tricks that have I mean they are all there okay you have to look at them okay that is the reason why this proof is tricky so this is actually 1 okay so if the Gn converge to G normally on C then the Gn hash of course converge to G hash normally on C so what this means is that if the since the Gn hash are all bounded by 1 the limit G hash is also bounded by 1 and since all the Gn hash at 0 is equal to or equal to 1 G hash also at 0 will be equal to 1 just by properties of limits and you have done with the proof of Zalkman's lemma okay and we have used Marty's theorem that is the whole point right. Now what I want you to understand is that in this Zalkman's lemma you have basically you have a condition for non-normality of a family okay and the fact is that the converse of Zalkman's lemma is also true namely if you have a family script F such that you are able to find a sequence is it attending to Z0 and a sequence of radii epsilon n and also a sequence of functions such that the zoomed family converges normally to a non-constant meromorphic function then the original family has to be not normal it has to be not in fact normality will be actually you know normality will be violated at the point Z0 Z0 is a point where normality of the family is violated okay and in what sense this the point is that at Z0 the spherical derivative is become as you approach Z0 through Zn spherical derivative is become unbounded and the unboundedness of the spherical derivatives is the same as non-normality because that is Marty's theorem okay. So let me probably say give this as a note that note the converse of Zalkman's lemma is true is true and why is that so for if F is a family such that there exists a sequence Zn going to Z0 with and there are the sequence of radii going to 0 with the zoomed functions and you have family of functions and there exists a sequence Fn family of functions in F such that the zoomed functions Gn which is a zooming of Fn centered at Zn with the magnification factor 1 by epsilon n in the new variable Zeta suppose this goes normally converges to G Zeta on a complex plane with G hash of 0 equal to 1 and G hash it is always less than or equal to 1 suppose this happens okay then the family cannot be normal and why is that true that is very very simple because you see then script F is not normal why is that true it is very simple because you know Gn hash of Zeta as I as we just calculated we have seen this is epsilon n times Fn hash of you know Zn plus epsilon n Zeta you know this and put Zeta equal to 0 what you will get is Gn hash of 0 is equal to epsilon n into Fn hash of Zn okay but you see Gn hash of 0 G hash of 0 is 1 okay and Gn hash take goes to G okay so Gn hash goes to G hash of 0 which is 1 okay so if you take the and mind you all these epsilon n's are going to 0 okay so this happens as n tends to infinity okay so what will it mean see the you have a product of two quantities one of them is going to 0 but the product is bounded that means the other has to go to infinity so what you will get is that so this implies that this has to go to plus infinity okay and what does that mean it means that you have violated the family the sequence Fn has violated the conditions of Marty's theorem you have found functions which whose spherical derivatives are going to plus infinity the spherical derivatives are not bounded and where is this happening see Fn hash spherical derivative of Fn at Zn is becoming larger and larger and larger going to plus infinity and the Zn's are approaching Z0 okay mind you Zn's are all approaching Z0 so what is happening is that if you look at a compact neighbourhood of Z0 an open disc close disc containing Z0 you see that on that compact neighbourhood okay these Fn hash is are not going to be bounded uniformly because you are going to plus infinity and now Marty's theorem will tell you therefore that this even the Fn's that sequence itself as a family is not normal okay so that implies not normality so the converse of Zalkman's lemma is also true so Zalkman's lemma is actually an if and only if condition okay but the beautiful thing is that beautiful thing about the lemma is that you if you get a non norm if a family is not normal your the lemma is able to guarantee the existence of a non normal point and a non normal sequence at that point okay you get both the point and a sequence that violates normality okay so what is now left is that I will have to use this to prove Picard's theorems and we will do that in the coming lectures. So let me write this here so this implies by Marty's theorem that Fn hence F is not normal okay so I will stop here.