 Fine so let us get along with the proof of Montel's theorem see so the key point that you have to remember is that you know the difference between Montel's theorem and the Arzela-Ascoli theorem is that the Arzela-Ascoli theorem is for continuous functions okay defined on a compact metric space right continuous functions of course real valued or complex valued in fact you can even have continuous functions with values in a compact metric space whereas the Montel's theorem is for of course it is for analytic functions and then we are going to extend it to meromorphic functions and these are going to be defined on domains which are certainly not compact right. So one thing that you have to contrast and compare is that the Arzela-Ascoli theorem as it is for continuous functions defined on compact metric space whereas the Montel's theorem is for analytic functions and later on for meromorphic functions defined on a domain okay so it is not a compact set right it is an open connected set then that is so that is one difference that is actually two differences that the domains are different in the Arzela-Ascoli theorem the domain is compact metric space in the Montel theorem the domain is the domain of the function is actually a domain in the complex plane right. Then the second thing is in the Arzela-Ascoli theorem we are worried about real or complex valued continuous functions whereas in the Montel theorem you are worried about analytic functions and later on meromorphic functions okay. Then the other important thing is that what is the similarity the similarity is that both of these theorems tell you when a family of functions is compact okay. Now if you look at it from the viewpoint of the Arzela-Ascoli theorem compactness is it corresponds to sequential compactness okay and sequential compactness is just the condition that given a sequence of functions in the family you are able to extract a convergence of sequence okay and whereas in the Montel theorem what will happen is that you will have a normal version of this. So the Montel theorem will tell you that your given family of functions is compact in the following sense that given a sequence in the family you can extract a subsequence but now this subsequence is not convergent in the sense of functions that is it is not uniformly convergent on your domain but it is uniformly convergent only when restricted to compact subsets of the domain that is normal convergence okay so that is the difference. So in the Arzela-Ascoli theorem what you get is convergence with respect to uniform convergence okay and that means it is uniform convergence on the whole space whereas in the context of the Montel theorem what you will get is convergence with respect to only compact subsets that is normal convergence okay so that is the that is the subtle difference that you have to understand and the whole but the punch line is that in the Arzela-Ascoli theorem you need for this compactness or sequential compactness you need the two properties of the for these functions in the family to satisfy one is uniform boundedness okay the other one is equicontinuity alright but this is for continuous functions in the Arzela-Ascoli theorem case but if you come to the Montel case okay equicontinuity is free it comes because you are looking at analytic functions because you are looking at analytic functions you know that the derivatives of the functions are the derivatives are expressible using the Cauchy integral formula and then you can estimate the Cauchy the integrals and therefore you get the so called Cauchy estimates and the Cauchy estimates will tell you automatically that the derivatives are bounded and it is a natural fact that whenever the derivatives are bounded the original functions are themselves equicontinuous. So equicontinuity comes for free it comes automatically for if you are looking at analytic functions but if you are looking at that is in the context of Montel theorem but if you are looking at the context of the Arzela-Ascoli theorem you have to give equicontinuity as an extra condition because all you have is continuity to begin with of the functions okay. So I was so let us so the first point I want to show is that this family I am starting with the domain D in the complex plane and script F is a family of analytic functions on D and it is normally uniformly bounded so it is uniformly bounded when restricted to any compact subset of D and I have to show that given a sequence in the family I must be able to get a subsequence that converges uniformly on compact subsets okay and the first thing I will show is that equicontinuity is automatic okay. So for that pick a point is it not belonging to D okay choose rho greater than 0 so that the close disk centered at z0 radius rho the set of all points in that close disk is inside D okay you can do this because D is an open set of course I am working with a non-empty open set okay so D is an open connected set so give me a point of D by openness there is always a small disk surrounding that point which is also in D and I can take a smaller disk along with the boundary inside that disk okay so I can find this rho alright of course you know the reason why I am including the boundary is because I want a compact set okay I if you take a close disk it is both close and bounded so it is compact and then I can use the hypothesis that the family of functions is then uniformly bounded on this compact set okay. So let M be a uniform bound for all let M be a uniform bound for all f belonging to script f inside mod z minus z0 less than or equal to rho okay so what does this mean this means that mod f of z is less than or equal to M for all f in script f and for all z with mod z minus z0 less than or equal to rho okay. And now what I would like to show is that I would like to show that in a smaller disk what happens is if you take a smaller close disk then the family is actually equicontinuous so what I do is I choose a smaller close disk so basically you know if I draw a diagram so it is going to be like this I am going to have this I am going to have this disk centered at z0 radius rho and then you know now what you do is choose a rho 1 which is less than rho rho 1 which is positive and less than rho okay so that I can consider the smaller close disk of radius rho 1 centered at z0 okay and look at z in mod z minus z0 less than or equal to rho 1 okay so the situation is like this I take a smaller disk and this length is rho 1 okay I am looking at the smaller disk alright. Now for z prime in this smaller close disk by the Cauchy integral formula well if you want second Cauchy integral formula what do you have if I write it out I will get mod well first let me write out the formula without worrying about estimating it so f dash of f dash of z prime is 1 by 2 pi i integral over mod z minus z0 is equal to rho f of z dz by z minus z prime the whole square so this is the Cauchy integral formula okay and the mind you the contour of integration is this outer circle taken with the positive sense right and now the point is that you have this but then I want to basically what I am trying to show is that I am trying to show that in the smaller close disk all the derivatives are bounded okay and because all those derivatives are bounded in the smaller close disk the family is equicontinuous okay so all I am saying is that pick any point there is a small close disk around that point where the family is equicontinuous and since the point that you picked was arbitrary this will tell you that the family is equicontinuous on all of D okay so you get equicontinuity for free okay and of course the reason is because you have the Cauchy integral formula okay so what is this there is a modulus of this will be modulus of this and you know the modulus of the integral is less than or equal to the integral of the modulus it is less than or equal to the maximum of the modulus of the integrand as the argument a variable of integration varies over the region of integration multiplied by the length of the contour so what I am going to get is I am going to get this is less than or equal to 1 by 2 pi that is what I will get for this term outside 1 by 2 pi i and the length of this contour is going to be 2 pi rho it is just a circle radius rho alright and then what about the integrand the modulus mod f is going to be bounded by m because mod f is always bounded by m for all f in the family script f okay so mod f is going to be bounded by m so I can put this in here and as far as the denominator is concerned you see look at this see the z is lying on the contour of integration okay and I am my z prime is see my z prime is lying in the smaller disc so my z prime is here okay and the distance between z and z prime is therefore at least rho minus rho 1 okay so mod z minus z prime is greater than or equal to rho minus rho 1 for mod z minus z not is equal to rho okay if you take z on the outer circle then the distance from a point to the inner circle has to be at least rho minus rho 1 which is the difference of the radii okay the closest z prime can get to z is along this radial line from z not to z and that closest distance is rho minus rho 1 okay so you know why I need this inequality because now I want to I have to invert the inequality and then the direction of the inequality will change so I will get 1 by mod z minus z prime the whole squared mod will be less than or equal to 1 by rho minus rho 1 the whole squared so I can put a rho minus rho 1 the whole squared here okay so finally I will end up with just m rho by rho minus rho 1 the whole squared okay and this is a bound this is a bound so this is true this is true for all f all functions f in script f and for all z prime with lying in the inner disc mod z prime minus z not less than or equal to rho 1 okay so this happens so in some what have we shown we have just shown that the modulus of the derivatives of all your functions inside close in a smaller closed disc they are all uniformly bounded by this constant okay and this constant has got nothing to do with any particular point see this constant that I have got on the right side is m rho by rho minus rho in the whole squared that has got nothing to do with the point z prime except that the z prime should lie inside this in smaller disc and it has got nothing to do with also the function f that I chose from the family so it is a uniform constant okay so what you have shown is that the because of analyticity the functions the derivatives are bounded now because the derivatives are bounded you get equicontinuity and how is that you get that by actually integrating these functions so well you know if I had taken so far if I take 2 points z 1 and z 2 with mod z 1 mod z i minus z not less than or equal to rho 1 that is if you take 2 points in the smaller closed disc okay you take 2 points z 1 and z 2 in this smaller closed disc then if you f z 1 minus f z 2 is nothing but the result of integrating from z 1 to z 2 of f dash of z dz okay this is because of just because of this fundamental theorem of integral calculus the moment the function has an antiderivative then the integral is just the difference of evaluation of the function at the final and initial points okay so you have this but of course the path of integration here really does not matter so long as the path lies inside this closed disc and well you know I am taking the path to be the straight line segment from z 1 to z 2 if I take that straight line segment path then what will happen is that well if I put a mod to this I will get this is less than or equal to mod f dash of z okay which is now I have got a bound for that and mind you this z is lying on the path of integration and therefore it is inside this smaller closed disc so this bound applies so I will get this m rho by rho minus rho 1 the whole squared this is the bound for f dash of z okay and the length of this contour from z 1 to z 2 is just the length of the line segment from z 1 to z 2 that I am choosing and so it will be a mod z 1 minus z 2 this is what I will get okay so now that is it this gives me equicontinuity because so for epsilon greater than 0 choose delta to be less than you know what to do you have to choose it to be less than rho minus rho 1 the whole squared by m rho times epsilon you choose this okay then mod z 1 minus z 2 less than delta will imply that mod f z 1 minus f z 2 will be less than you look at this quantity I have to replace this mod z 1 minus z 2 by the bigger quantity rho minus rho 1 the whole squared by m rho epsilon okay and I will get less than epsilon okay so basically this is true for all z 1 and z 2 inside mod z minus z 1 mod z minus z 1 not less than or equal to rho 1 okay. So what you have shown is that given any two points the moment two points inside the smaller closed disc are lesser than delta the function values are lesser than epsilon and this works for any two points and for any function so you have shown a kind of you have shown a uniform equicontinuity of the family inside the smaller closed disc okay so thus script f is equicontinuous in inside mod z minus z not in the smaller closed disc since z not was arbitrary f is equicontinuous on D so I am writing eq cts for equicontinuous on D so this is the important point I mean this is the distinguishing feature between Montel's theorem and Arzela-Ascoli theorem in philosophy while in the Arzela-Ascoli theorem you need equicontinuity in the Montel theorem you get equicontinuity for free because you are already working with analytic functions and so the moral of the story is if you have a uniform bound for your functions that gives rise to a uniform bound for the derivatives okay that is because of the Cauchy integral formula and estimation and the uniform bound for the derivatives gives you equicontinuity for the original functions. So just the uniform boundedness of the original functions for analytic functions is enough to give you equicontinuity also okay so you are now in good shape now you see now I can actually apply the Arzela-Ascoli theorem for example on this if I take any you take any compact subset of D okay then these continuous functions will restrict to continuous functions on that compact set of course okay any continuous function on a topological space if you restrict it to a subspace it will remain continuous if you take for the subspace the subspace topology okay so if you take a compact subset of D if you restrict these family of functions you are going to get continuous complex valued functions on the compact set and they are also going to be equicontinuous because you already checked equicontinuity alright so and mind you all you all already assume that the family is normally bounded normally uniformly bounded therefore they are also going to be uniformly bounded on the compact set therefore the usual Arzela-Ascoli theorem applies and given any sequence I can extract a subsequence okay so I have brought myself into the purview of applying the usual Arzela-Ascoli theorem okay so let me write that down thus for any compact subset K in D the Arzela-Ascoli theorem applies to give a convergence of sequence for any given sequence in script F okay so you are able to all apply the Arzela-Ascoli theorem but you are still but we are still not we are we I must say we are only halfway through the Montel theorem see there is a small set again there is a small subtlety that you have to notice what have we achieved so far give me a compact set K suppose I start with a sequence in the family F give me a compact set K I will be able to get a convergence of sequence but you see if I change the compact set I may get a different convergence of sequence okay whereas what does the Montel theorem say the Montel theorem says you start with a sequence you can get a subsequence the same subsequence which will work on every compact set you see that is the subtle difference so we are halfway through we are for given a sequence in F in the family script F you can always extract a subsequence which converges on that compact set but if you change the compact set the subsequence can change what Montel theorem promises is one subsequence that will work for all compact sets okay and the way to get that is it is a very very clever thing it is again a diagonalization argument okay if you go back to the Arzela Ascoli theorem we use a diagonal diagonalization argument actually in the proof of the Arzela Ascoli theorem one way where we assume where we try to show that you know if a family is equicontinuous and uniformly bounded then it is sequentially compact or compact so what we did was we started we we took a family of continuous functions in your family and we try to extract the convergent subsequence but since we were in a close space we just contented ourselves with extracting a Cauchy subsequence okay but the point is how did you get this Cauchy subsequence we knew that the matrix space was on which the functions were defined was compact so we knew it is separable so we took a separable dense subset we took a countable subset of points x1, x2, x3 and so on which is dense and then what we did was we took the sequence first applied it to x1 okay and by boundedness this gave you a subsequence of functions which converged at x1 and then we applied that subsequence of functions to x2 okay and again being bounded we got a further subsequence which converged at x2 and then we did this process at infinitum okay we got a matrix of functions and then we took the diagonal sequence and that diagonal sequence was a sequence of functions that converged on this countable dense subset and then we used equicontinuity to interpolate and to conclude that therefore this diagonal sequence converges on all of x and uniformly okay that is so you see the same kind of diagonalization argument we will use now okay and the so and the key to that is that is a special construction what you do is that you construct a sequence of increasing sequence of compact sets which fill out your domain D okay so it is more like you know if your domain D for example is a whole complex plane okay an increasing sequence of compact sets which fills out the complex plane will be just sequence of disks close disks of increasing radii okay so it is just that but you will have to take care about the boundary okay so we will do that so let me write this down note that this convergent subsequence could change if k were changed but Montel's theorem promises a single subsequence that converges uniformly on any compact subset okay so that is the subtle difference so what we will do is that so I am going to look at these sets Dn so here is my set Dn here is how I define it it is a set of all points in D with the following property mod Z is less than or equal to n okay so it is a subset of the close disk centered at the origin radius n alright so it is a anyway that this first condition tells you that each Dn is automatically bounded because there is a bound alright and then and here is a second condition which is the distance from Z to the boundary is at least 1 by n okay so it is a it is a rather you know it is a rather nice condition okay so what I want you to note is that you know there are only two essentially two cases you have to understand see if dou D is empty what does it mean if dou D is empty it means D is the whole space okay dou D can be empty only if and only if D is the whole space because you know after all dou D put together dou D is the boundary of D dou D and that is a close set you know if you take dou D and take the union with D you get D closure okay therefore dou D being empty is the same as D equal to D closure but if D is equal to D closure you are saying D is closed but D is already open D is both open and closed okay therefore and if it is non-empty it has to be the whole space because the whole space is connected okay so the only so the situation that dou D is empty corresponds to D is the whole complex plane okay that means actually you are looking at a family of entire functions okay and in that case this Dn is just the sequence of discs of increasing radius okay and this condition will become useless the distance between D, Z and dou D is greater than or equal to 1 by n is superfluous okay you write this condition only when it makes sense only when dou D is non-empty okay if dou D is empty if you want define the distance to be infinity so that infinity is always greater than or equal to 1 by n okay if you want but right think of this condition only when dou D is non-empty okay so you know what is it that you are doing the reason why we put this condition is because you know you cannot include everything in D okay if you try to include everything in D you will have issues if you try to because D is open okay because D is open basically I want all these Dn's to be compact okay so that the fact is that D is actually the union of all these Dn's D is the union of all these Dn's this is one the second thing rather facts the second thing is Dn is compact for every for each n that is the second condition and the third this is the very important condition any compact subset is contained in some Dn and any compact subset K of D is contained in some Dn so this is the beauty about this collection Dn and you see it is this compactness that I am worried about see each Dn is automatically bounded there is no problem about that each Dn is automatically bounded so the only thing that prevents it from being compact is it not being closed okay if each Dn if you check that each Dn is also bound is also closed then each Dn will become both closed and bounded so it will become compact okay now the big deal is it is for making sure that each Dn is closed that I have put this extra condition that D distance of Z to do D is greater than or equal to 1 by n that is the reason I have put it put that is because you see suppose my region is just unit disc suppose D is just the unit disc okay suppose D is just set of all Z such that mod Z less than 1 okay then what is D1? D1 is D what is D1 is set of all points which is whose modulus is less than or equal to 1 okay that is if I forget the other condition okay forget the second condition that the distance from the point to the boundary of D is greater than or equal to 1 by n suppose I do not put that condition and if D is the unit disc then D1 is D so D1 becomes the unit disc and it is not compact because it is not closed okay so therefore it is for the closeness of each Dn that you put this extra condition the distance you are throwing away part of the boundary okay you are taking points from certain particular distance 1 by n from that boundary and you are not allowing points to go very close to the boundary you allow up to a certain distance minimum distance which is 1 by n and that condition makes each Dn closed I mean that is the significance of this condition that I have circled okay so if you draw a diagram you know it is more like this it is like you know so here so let me draw somewhere here so you know suppose this is the plane and suppose this is your domain D alright then you know what is D1? D1 is all those points which are lying in the so let me draw this this is too big get rid of this oops so you know so these are all the points in the unit circle okay this is mod z less than or equal to 1 and so I will also include the boundary because I have allowed less than or equal to 1 okay and if I and then I am looking at all those points of D whose distance from I think I will have to slightly modify this diagram so that I get a non-empty set so let me draw something like this okay so suppose this is my so this is my suppose this is my suppose this is my D okay and mind you D does not contain the boundary so you know I should redraw this I should draw it like this so this is my dou D this is the boundary of D okay it is like it is like a half plane except that the boundary is not aligned if you want okay. Now what is D1? D1 is all those points lying in the closed unit disk okay whose distance from this boundary is at least 1 so if you actually calculate it you see what you will get is you will get this so with this distance 1 okay this will be D1 this is what D1 will be okay and this boundary is included okay because these are the points whose distance from the boundary dou D is at least 1 and points less than with distance less than 1 are all not included okay so this is D1 if I next you know if you take D2 okay then what will you get so I have this disk now this is a bigger disk radius 2 alright and what am I going to get I am going to look at all those points inside this closed disk also for which the distance from the boundary is greater than or equal to half now okay so what I will get is I will get this I will get this so basically I will get this region that I have shaded with vertical lines that is going to be D2 okay so like this you can see that these on the one hand the D ends are becoming bigger and bigger so as to cover D on the other hand they are coming closer and closer to the boundary of D so that you do not lose any points very close to the boundary okay this is what is happening but keeping all the D ends away from the boundary by distance of 1 by n is the condition to make them a closed subset make each of these closed okay so that is the reason each Dn is therefore closed and bounded so each Dn is compact and the union of all the Dn is D okay so you can make this you can write out all the details basically using properties of the fact basically using the fact that you know the distance function the metric function is a continuous function that is what you have to use if you write down everything in detail okay the distance function is always a continuous function so you have to use that alright and you can see several things it is very clear that all the Dn will cover D okay because you me any point of D okay it has to lie in some big enough closed disk some mod z less than or equal to n okay and if the point is and it has to be at some distance away from the boundary because D does not contain its boundary it is an open set so every point in D has to be away from the boundary so you can always find every point of D in some Dn okay so the union of all the Dn is covered D okay and all the Dn are compact and what is more beautiful is you give me a compact subset of D if you take a compact subset of D such a compact subset will be in some Dn certainly because it will be contained in some big enough closed disk okay and it will have because it is compact its distance from the boundary of D will be some finite quantity okay so it will be contained in Dn for n sufficiently large so given any compact subset of D it has to be contained in Dn for some sufficiently large n okay therefore the beauty is that you have constructed this increasing sequence of compact subsets which cover the whole of D and now comes the trick what you do is each of these for each of these subsets you can apply the Arzela Ascoli theorem okay and then use a diagonalization argument and pick out a diagonal subsequence which will now converge on all compact subsets of D and that is the sequence that is promised by Montel's theorem so that is what I am going to write down so let me write this down let us go okay so let me write it here let f 1, f 2 etc be a sequence in script f and apply the Arzela Ascoli theorem to extract a convergent subsequence on D1 okay mind you D1 is a compact set okay it is a compact subset of D so on D1 it is a compact subset so it is a compact matrix space and therefore you know I can apply the Arzela Ascoli theorem and on D1 I can get a convergent subsequence by and here when I say convergent subsequence is uniformly convergent okay because it is convergence in the space of functions so extract so here let me put uniformly okay so I will call that subsequence as f 1, 1, f 1, 2, f 1, 3 now you see that you are in the diagonalization business okay now what will you do now you take this subsequence f 1, 1, f 1, 2, f 1, 3 and again apply the Arzela Ascoli theorem to get another subsequence which will converge uniformly on D2 and call that as f 2, 1, f 2, 2, f 2, 3 okay and proceed in this manner and then you take the diagonal sequence the diagonal sequence will be a subsequence of the original sequence which will converge on all D which will converge compactly which will converge uniformly on all Dn's and since any compact subset of D is contained in some Dn therefore it will also converge uniformly on all compact subsets of D and I am done okay so that is the trick so let me write that down. Then apply the Arzela Ascoli theorem to extract a subsequence f 2, 2, f 2, 1 that converges uniformly on D2 okay continue this way to get f n 1, f n 2 that converges uniformly on I should say D1, D2, etc Dn so in fact I should not put comma I mean instead of putting comma I can actually put union d i, i equal to 1 to n okay and that is it so we are in the following situation so you have this original sequence f 1, f 2 and so on then you have this f 1, 1, 2 and so on this is subsequence and the point is converges uniformly on D1 and then you have f 2, 1, f 2, 2 and so on this is also a subsequence this that converges uniformly on D2 and D1 so that is D1 union D2 okay and you go like this add infinitum you end up with f n 1, f n 2 that converges uniformly on union i equal to 1 to n D i okay then the subsequence the diagonal subsequence so this is a diagonal trick the diagonal subsequence f 1, 1, f 2, 2, f 3, 3 converges uniformly on compact subsets of D that is and that finishes the proof of the of course I must tell you that D1 union D2 is just D2 okay because D2 is bigger than and this union i equal to 1 to n D i is just D n actually okay each D i includes the DJs for j less than i because it is an increasing collection of compact subsets okay and any compact subset of D is contained in one of the DJs okay and therefore this diagonal subsequence will converge uniformly on that compact set so starting with starting with a sequence of analytic functions defined on the domain D okay just putting the condition that this analytic this sequence is uniformly bounded on compact subsets of D namely that is normally uniformly bounded just that condition is good enough to guarantee that you get compactness in the sense that any sequence from this family will admit a convergence subsequence and that is the point of Montoff's theorem so what we will need to do is that we will have to you see somehow the derivatives are entering into the picture and the trick is that you want to extend this to Meromorphic functions but you know for Meromorphic functions derivatives are not defined at the poles but then we have replacement for that we have the spherical derivatives so the trick is that you try to get another version of this Montoff's theorem which will work for Meromorphic functions and you try to use a spherical derivative and still everything works okay so you get a Montoff's theorem for Meromorphic functions okay and that is what we are going to do next because finally we have to worry about Meromorphic functions for the proof of the Picard theorem okay.