 Okay so let us continue with our discussion of Montel's theorem so you know so this is essentially Montel's theorem so what you do is you take so it applies so you know it is a version of Arzela Ascoli theorem adapted to the case of analytic functions okay and so as I told you Arzela Ascoli theorem everything happens on a compact set alright you need compactness on the set on which your functions are defined alright but here you know of course analytic functions are defined only on open sets on domains okay and we consider them on open connected sets namely domains. So what you have to do is you have to put all the requirements only on compact subsets okay so let me write this down so let script F be a family of analytic functions defined on a domain D okay inside the complex plane so D is an open connected set and all the functions in the family script F they are defined on this domain and they are analytic functions okay. Suppose script F is uniformly bounded on D so okay so here is again so I impulsively wrote down something that is too much to expect so you know when you want to when you want when you do the Arzela Ascoli theorem you will say you have family of functions defined on a compact set and it is uniformly bounded on the compact set okay then the Arzela Ascoli theorem is in gives you an equivalence between 2 statements once one statement is equic continuity of the family at each point and the second statement is that every sequence of functions in the family has a uniformly convergent subsequence okay. Now so you know so you know to expect a family of analytic functions to be uniformly bounded on a whole domain is too much okay so you should modify this and say that it is uniformly bounded on compact subsets of the domain okay so I will change this because this is too much on compact subsets subsets of D okay. So see a property that holds on compact subsets is called a normal property okay a normal so if you have convergence on compact subsets is called normal convergence if you have uniform boundedness on compact subsets it is called normally boundedness okay so I can state it as you know f is normally bounded on D okay. Then so you know what is so what is your Arzela Ascoli theorem the usual sense will say that you know if you have this uniform boundedness then equic continuity is equivalent to the fact that every is equivalent to the statement that every sequence has a uniformly convergent subsequence. Then the following are equivalent number one script f is equicontinuous equicontinuous at each point of D okay and the second condition will be every sequence in this family script f has a uniformly convergent subsequence but now again you should not expect uniform convergence on a domain always you should only expect normal convergence that is you should expect uniform convergence only on compact subsets. So the second statement should be written carefully you should say that every sequence fn in f has a subsequence fnk which converges normally that is uniformly on compact subsets of D okay. So you know the statement so if you compare this with the Arzela Ascoli theorem the statement is in the Arzela Ascoli theorem your family of functions is not analytic in the general ascoli theorem the family of functions is not analytic it is only continuous family of functions okay but they are complex valued of course. So the analytic condition is not there but you have a weaker condition which is just continuity okay then and in Arzela Ascoli theorem the functions are not defined on a domain they are defined on a compact subset of the complex plane and the condition on f is that it is uniformly bounded on that compact set now that is replaced now the compact set is replaced by a domain okay therefore the condition of uniform boundedness is restricted only to compact subsets of the domain alright and then you have then when you have uniform boundedness then the Arzela Ascoli theorem the philosophy is that equicontinuity is the same as the existence of subsequence that converges uniformly. So equicontinuity condition is going to anyway it is a continuity condition so it remains as it is but the existence of a subsequence which converges uniformly that also you should expect only on compact subsets so that is why we put this we say that for every sequence you get a subsequence that converges normally alright. So this is Montel's theorem and actually again you know the implication 1 implies 2 is what we are going to prove the implication 2 implies 1 is you can prove the same thing I mean if you know 2 implies 1 can be proved just in the way that we have done that one could do for the Arzela Ascoli theorem okay it is a proof by contradiction okay. So what I want to tell you is that condition 1 will always be true okay condition 1 will always be true because of Cauchy's integral formula for the derivative of an analytic function which will give a bound therefore what will happen is that 1 will always be true if the family is uniformly bounded and therefore this is always true therefore this is always true okay. So what I want to state is that this first condition is superfluous the first condition always holds okay simply because you are not working with just continuous functions you are working with analytic functions and analytic functions you know for analytic functions you have a good bound for the derivative and in fact you have a bound for all orders of derivative derivatives of every order at a point because of the Cauchy integral formula okay so 1 is always true and therefore 2 is always true so 2 is always okay. So let us look at the proof of this so what I will do is I will just I will give the proof of 1 implies 2 which is the which is a slightly technical thing okay take a point z0 in the domain and choose rho greater than 0 such that the disk mod z-z0 less than or equal to rho is contained in the domain okay. So certainly you can do this you can since the domain is an open set z0 is an interior point so there is a disk surrounding z0 which is contained in the domain and you take a slightly smaller disk its closure will also be contained in the domain and call that radius as rho okay. Then note that by Cauchy's integral formula what you will have is you see so you know my diagram is like this here is my domain D and here is my point z0 that is this I am taking this circle centered at z0 and I am going to take radius to be equal to rho-epsilon where epsilon is a very small quantity alright and what is Cauchy's integral formula Cauchy's integral formula for the derivative will tell you that for any analytic function on this closed disk alright the derivative of the function at the center of the disk is given by 1 by 2pi i integral over this boundary circle mod z-z0 is equal to rho-epsilon you will get g of zeta t zeta by zeta-z0 z0 square okay this is the Cauchy's integral formula for the derivative right this is just Cauchy's integral formula right for g analytic in this disk okay epsilon very small so this is I am just writing Cauchy's integral formula I am doing anything else right now in particular so this is you get a bound you get a bound by you know putting by parameterizing this so you will get g dash of z0 is equal to 1 by 2pi i integral from 0 to 2pi so you know the points in the z on this disk can be parameterized as zeta equal to z0 plus rho-epsilon into e power i theta this is how I can parameterize that circle okay so where theta varies from 0 to 2pi so if I transform this integral to a real integral to an integral based on a real parameter so what I will get is I will get mod I will get I will get g of so I plug in that d zeta is going to be rho-epsilon i e power i theta d theta divided by this is going to be zeta-z0 is rho-epsilon the whole square e power 2i theta okay this is what I am going to get right and now what I am going to do is I am going to take modulus and note that the modulus of the integral is less than or equal to the integral of the modulus okay I am going to use that inequality which is always used whenever you are estimating integrals it is otherwise known as the ml formula so or it leads the ml formula so mod g dash of z0 is going to be mod of this thing on the right but that will be less than or equal to you know if I take mod you must outside I will get 1 by 2pi alright and then the modulus of the integral is less than or equal to integral of the modulus so I will get 1 by 2pi integral 0 to 2pi and I will get l so let me put m into rho-epsilon this mod of i e power i theta is going to be 1 and mod d theta is just d theta because theta is increasing along this interval and here I am going to get rho-epsilon square and this is again going to be 1 and so this is what I will get where what is this m, m is bound for the modulus of g on this boundary circle okay where mod g is less than or equal to m on mod z-z0 equal to rho-epsilon okay so here this is just an inequality and you know if I calculate this if I integrate 0 to 2pi d theta I will get 2pi this 2pi will cancel this 2pi I will simply get m by rho-epsilon this is the boundary okay this is for an analytic function on this which is analytic on this closed disc now what I am going to do is I am going to apply this to all the functions in my family script F to all the functions in this family see if you take all the functions in this family alright there of course analytic therefore they are analytic on such discs alright and the point is they are all uniformly bounded so I can find a single m which will work for all the functions okay and therefore I will get this uniform bound for all the derivatives okay and that is good enough to tell me that the family is equicontinuous okay so let me make this statement for any since f is uniformly bounded on mod z-z0 less than or equal to rho you see you have the uniform boundedness on compact subsets of d okay therefore I am applying this uniform boundedness I have this uniform boundedness on this disc closed disc and this closed disc is compact because it is closed and bounded alright so this is a compact subset of d therefore this family is uniformly bounded on this okay so which is compact there exists an m such that mod f is a system of m on mod z-z0 you have this so if you so you have mod f dash of z0 is a system or equal to m by rho-epsilon as explained above so this inequality that this estimate I have got for the model of the derivative I apply it to f alright so I get this right and now what I want to say so what this tells you is that this is for what this is for every f in the family. So what you have got is you have got that m by rho-epsilon is a uniform bound for all the derivatives okay so what you have got is that all the derivatives are uniformly bounded at that point now you see now it is a fact that if derivatives you know if you have family of functions whose derivatives are uniformly bounded then that family is equicontinuous okay so I will state this as a lemma lemma 1 if so let me write g is a family of analytic functions functions on u which is domain in c and g and t and g dash which is equal to the set of all a small g dash where g belongs to g the derivatives is uniformly bounded in a neighbourhood of z0 in u then g is equicontinuous at z0 so I am setting this fact that the uniform boundedness of the derivative imply uniform boundedness of the derivatives in a family implies equicontinuity of the family uniform boundedness of the derivatives at a neighbourhood of a point in a family implies equicontinuity at that point okay and the proof of this lemma is pretty easy it is just given by simple I mean if you want to prove it for real functions it will follow I mean if you are working with real valued functions on a close bounded interval or on an interval okay then the proof will come by applying the mean value theorem okay but if you are working with complex valued functions you will have to use integration so what you do is see you know the situation is that so you know I have this u and I have this point z0 and you know I have this disc centred at z0 radius some r k mod z-z0 less than r r is greater than 0 okay this is inside u I can find such an r because after all z0 is an interior point of u and u is a open set and what I am given is that all the derivatives is uniformly bounded in the neighbourhood therefore there exists an m such that mod less of g dash is started at m in mod z-z0 less than r so this is given to me all the derivatives are bounded alright and now how do you show that the family is equicontinuous at z0 so what you do is you know you calculate g of z-g of z0 modulus okay so of course this is for all g in g okay so small g is in script g so small g dash is in script g dash and it is given that script g dash is bounded uniformly bounded in a neighbourhood of the point okay and now what is mod gz-g z0 you see what you can do see this is integral it is integral along the straight line path from z0 to z of g dash of z dz e dash zeta d zeta of course you know if you integrate g dash you will get g because after all derivative of g is g dash you can integrate g dash to get g mind you g is analytic g dash is also analytic okay and therefore this integral is actually independent of the path chosen in a simply connected neighbourhood of z0 and of course we are always considering this disc surrounding z0 which is simply connected okay. So for example you can take z to be any point here and you can take the you can simply take the straight line segment from z0 to z and you can integrate alright but then again you use the fact that modulus of the integral is less than or equal to the integral from z0 to z mod g dash of z mod d zeta okay and but what is mod g dash is you know uniformly bounded by this m so I will get this is equal to this is less than or equal to m times and integral from z0 to z mod d zeta will give you the length of the arc from z0 to z which I am considering and that is I am considering that to be a line segment so I will simply get mod z-z0 this is this is the value of integral from z0 to z mod d zeta okay normally when you integrate mod d zeta along the path you will get the arc length but now I am integrating along the straight line path so I will simply get the straight length of that straight straight line segment which is the mod z-z0 but you see so this is true for all g in script g okay so what this tells you is so given epsilon greater than 0 there exists delta which is equal to epsilon by m such that mod z-z0 less than delta which is epsilon by m implies mod g z-gz0 is less than epsilon okay for all g in g this is what you get given an epsilon I am able to find a delta okay the delta only depends on this of course this delta I fix the z0 delta depends only on epsilon okay and this z0 but it is independent of the small g in script g because this m is uniform for all g dash okay but what does this tell you this actually means that the family script g is equicontinuous at the point z0 that is the definition of equicontinuity the definition of equicontinuity is that this epsilon delta definition for continuity should hold at a point for an epsilon you should get a delta which works for all the which works simultaneously for all the functions in the family. So I have got a delta which depends only on epsilon this delta does not depend on g okay that means that the family is equicontinuous at z0 okay so this implies g is equicontinuous at z0 so that gives the lemma so what the lemma tells you is that whenever you have uniform boundedness in a neighbourhood of a point okay then you will have equicontinuity at that point okay. Now if you now look at what we have written here the uniform you have uniform boundedness of the derivative derivatives at that point okay and therefore it will work also in a small neighbourhood of this point alright and therefore you will have equi by this lemma you will have equicontinuity at every point in the domain. So you have this bound at z0 okay and it is uniform for all f alright so what will happen is it the same you can get a bound for all points in a small neighbourhood of z0 okay and therefore the derivatives are all uniformly bounded in a small neighbourhood of z0 and if you apply the lemma you will get that the family is equicontinuous at z0 so this is a statement that one is always true okay that the family will always be equicontinuous okay. So for analytic functions the derivatives will be bounded just because of Cauchy's formula okay and since derivatives are bounded equicontinuity will come automatically because that is what the lemma says that whenever derivatives are uniformly bounded you get equicontinuity alright. So by the lemma f is equicontinuous at z0 at each z0 in the in your domain okay so for the point of the story is that you know equicontinuity is automatic just because the derivatives are all automatically bounded because of Cauchy's integral formula right. So everything comes from just uniform boundedness uniform boundedness of your family of analytic functions on compact subsets will automatically give you uniform boundedness on compact subsets of derivatives you will get uniform boundedness of derivatives in compact neighbourhoods of each point and that will give you equicontinuity at that point and in this way you can cover all the points so you get equicontinuity everywhere right. So this part one this is always true okay and so I will have to now do this I will have to show that so this is always true so therefore this is always true so the only thing I will have to do is I will have to show that give me a sequence here I will have to show that I can cook up a subsequence which converges uniformly on compact subsets okay. So what I will do is okay so let me retain this so start with a sequence enough okay I already know that the whole family script f is equicontinuous at each point of the domain alright start with a sequence now what you do is so we do a clever construction okay so the construction is you know see I want to basically apply Arzela-Ascoli theorem but you know Arzela-Ascoli theorem the usual Arzela-Ascoli theorem that I want to apply will only work on compact subsets therefore you know what I have to do is I have to find I have to chop down this domain into a union an increasing union of compact subsets and apply repeatedly Arzela-Ascoli theorem on each member of the union and then apply a diagonalization argument okay. So what you do is you do the following thing so this is the trick this is the trick of chopping up a non-compact set okay into a union of compact subsets which cover it so what you do is for every n greater than or equal to 1 let En be so you know you look at mod z less than or equal to n this is the disk this is the close this centered at the origin radius n okay. So you know if so if I so let me draw a diagram so that you can think a little bit just for motivation so you know you have suppose my domain is like this okay of course the way I have drawn it the domain is already the closure of the domain is already compact okay but let me do the following thing let me just remove this so that you know you can think of the domain as being probably unbounded this is part of the boundary of the domain alright. So this is my domain D and this is the boundary of D okay now what you do is you look at mod z less than or equal to n alright so that is going to be so you know if I take n I am going to get a disk like this okay and if I take n plus 1 I will get a bigger disk so this is n and this is n plus 1 okay as n increases these close disks they cover the whole complex plane okay the union of all these disks as n goes from 1 to infinity is the whole complex plane alright and now what I will do is you see I intersect it with the set of all points in the boundary the set of all points in the domain D such that the distance of that point to the boundary is greater than or equal to 1 by n look at this rather funny condition so the condition is I am looking at all the points in the domain okay which lie inside this disk and whose distance from the boundary by distance of course I mean perpendicular shortest distance okay the shortest distance from the boundary is at least 1 by n okay that means I am avoiding points whose I am avoiding points in the domain whose distance from the boundary is less than 1 by n okay. So if you if you think of it like this if you think of it like that then what will happen is you know so here is this is the portion of the boundary this is the portion of the boundary alright and what I am doing is I am avoiding all the points whose distance is so if I take this smaller disk okay if I intersect the smaller disk with which is mod z less than or equal to n with the domain alright what I will get is I will get this okay this is what I will get alright this is the intersection of the smaller disk mod z less than or equal to n with the domain alright and of course this is the portion of the boundary this is the portion of the boundary that intersects the disk mod z less than or equal to n right. Now what you do in this boundary you throw out all those points in the domain okay you take do not take all the do not take all of the shaded region but throw out all the points in the domain whose distance is less than 1 by n okay. So it means that you know I am throwing out all points here now I am just I am just avoiding all points close enough whose distance is so you know this is what I am throwing out okay I am throwing out this because all points here the distance with the boundary is less than 1 by n I am throwing that out I am just throwing out a piece of the domain which is close to the boundary okay and it is this shaded set which is e1 I mean this is en this shaded set this is en okay. If you take en plus 1 what will happen is that I will get this whole intersection minus I throw out all points whose distance from the boundary is less than 1 by n plus 1 which is smaller distance than this okay so you see as n becomes larger see you can see something that is happening as n becomes larger I am covering more and more of the domain because after all as n goes to infinity these discs will cover the whole complex plane therefore as n becomes larger I am covering more and more of the domain and I am what I am throwing out is lesser and lesser I am throwing out points very very close to the boundary of the domain okay therefore in this way I will cover the whole domain so what you must understand is that and of course you know this is a compact set okay this is a compact set and this is a close set alright this is a compact set and this is a close set and therefore the intersection is continues to be compact so the moral of the story is that en is compact for every n and union n equal to 1 to infinity en is your domain your domain has been chopped up into compact sets so this is the I mean this is the clever trick that one needs to make use of to be able to apply Arzela Ascoli theorem okay so en is compact for everyone for every n union of all the en's is d and of course you know en you can see that you know en plus 1 will contain en plus 1 will contain en alright en plus 1 will contain en so it is increasing it is an increasing sequence alright and another beautiful thing is you take any compact subset of d any compact subset of d will be contained in a sufficiently large en okay any compact subset of any compact subset subset of d is contained in a sufficiently large in en for en sufficiently large so you see these are the these are the properties of this of the en's the en's are all compact their union is d they are increasing and any compact subset of d is contained in en for en sufficiently large okay and this is essentially you can think of this as chopping the domain out into I think instead of even saying chopping I should say you know your it is more of a filling out the domain in terms of an increasing sequence of compact subsets the en's fill out the domain the union is the whole domain okay now since each en is compact okay I can apply Arzela Ascoli theorem because you know each en is compact and each en is a compact subset of d but on compact subset sub d I have uniform boundedness because it is normally uniformly bounded okay and of course Iq continuity is already there okay so I can apply Arzela Ascoli theorem on each compact subset okay so what I do is I do the following thing I again I cleverly use again a diagonalization argument and if you recall already in proving the Arzela Ascoli theorem we use a diagonalization argument okay so we again use a diagonalization argument apply Arzela Ascoli theorem to e1 to the sequence on e1 okay e1 is a compact subset of d and I have sequence of functions on e1 that sequence is uniformly bounded why because the sequence is part of the family script f which is uniformly bounded on compact subsets therefore it is also uniformly bounded on e1 alright now Arzela Ascoli theorem will tell you that there is a subsequence which will converge uniformly on e1 okay so to get a subsequence fn1 of fn that converges uniformly on e1 okay now what I do is I take the subsequence and apply Arzela Ascoli theorem to it on e2 okay so I go to the next bigger set okay now apply Arzela Ascoli to this fn1 on e2 okay to get fn2 a further subsequence which converges on e2 uniformly okay now what I will do is I will take this fn2 and apply Arzela Ascoli theorem to it on e3 and I will proceed like this okay by induction we get a subsequence fnk of fnk-1 which converges uniformly on e for all k greater than okay alright and now comes the big deal now what you do is you from this seek from all this you take the diagonal subsequence that will give you a subsequence of the original sequence that will converge uniformly on compact subsets of t and that is the end of the proof okay so again you know I will again draw that the same kind of diagram that we drew for the diagonalization argument in the Arzela Ascoli theorem the proof of Arzela Ascoli theorem see so you know you have the situation like this you have you have you know you have fn so you have fn1 and then I have the subsequence fn2 then I have the subsequence fn3 and so on and so this is you know this is some fi1 fi2 fi3 and this is this is the sequence it converges uniformly on e1 okay then here I have fj1 fj2 fj3 that converges uniformly on e2 then I have fk1 fk2 fk3 that converges uniformly on e3 okay and it goes on like this and now what I am going to do is I am going to take this diagonal subsequence okay you define fm where fm is equal to mth member of fnm okay then fm is contained in the intersection of all these subsequences which is in of course everything is a subsequence of fn converges uniformly on compact subsets of t and that finishes the proof and why does it converge uniformly on compact subset of d because you take any compact subset of d any compact subset of d is in some en and okay and so you take any compact subset of d it will be in some ek but on ek capital fk capital fk plus 1 capital fk plus 2 etc they all converge uniformly on ek okay so this sequence of functions eventually converges uniformly on every compact subset of d therefore it converges after all convergence itself is an eventual thing it has to happen only beyond a certain finite stage alright therefore you get the last statement that this sequence converges uniformly on compact subset of d okay and that gives the proof of 1 implies 2 okay where 1 is already true we have seen so we have proved 2 is true okay and that gives the proof of Montel's theorem okay so you see it is a so the clever thing is to chop the domain up into these increasing sequence of compact subsets and then to repeatedly apply Arzela's theorem and again get a apply a diagonalization argument okay so I will stop here and we will continue with the proof of Riemann mapping theorem in the next lecture.