 Ok, so let us continue with this proof of the Riemann mapping theorem. So what we have is see we have this family script F consisting of analytic functions from the given simply connected domain D which is not the whole complex plane ok and taking values in the unit disc delta alright this is the unit disc centred at the origin radius 1 right open unit disc such that of course F is analytic injective and taking F of z0 to taking z0 to 0. So z0 so z0 is a point of D alright and so we have this family script F then what we did was we had taken so the fact that each F takes values in the unit disc they told you that mod F is always strictly less than 1 for all F in the family so this tells you that the family is uniformly bounded ok. So script F is uniformly bounded and of course that we have already seen that this family is non-empty ok because you can always find a holomorphic isomorphism of D onto a sub domain of the unit disc ok that was the first step of the Riemann mapping theorem that we started to prove ok. So this family is non-empty ok and the family is uniformly bounded alright and therefore now you apply so what you do is you take so Montel's theorem will apply to say that if you have any if you take any sequence in this family then there will be a uniformly there will be a there will be a subsequence which will converge uniformly on a compact subsets namely you will always find a normally convergent subsequence normal convergence means uniform convergence of compact subsets ok. So Montel's theorem will apply but to which sequence do we apply to so what we do is we look at the we look at the supremum there is the supremum of the derivative of these functions at z0 the modulus of the derivative ok. The fact that this is the finite is because of it is basically because of Cauchy's estimates ok therefore the supremum is a finite number we called it as a alright and we proved that since all the this a is of course non-negative in fact so the fact that all these f's are all injective means that all these derivatives can never vanish ok therefore a is a positive quantity alright. And then what we did is we took sequence fn in the family f such that the if you take the corresponding sequence of derivatives and it take their moduli that converges to a and this is because a is a the supremum it has a approximation property a supremum of a set of real numbers can be gotten as the limit of a sequence from that set ok. So you that is how you get the sequence ok the sequence of functions in f such that the derivatives at z0 the modulus of the derivatives at z0 tends to a ok and then by using Montel's theorem Montel's theorem tells you that there exists a subsequence fnk which converges normally to a function f0 ok. So this is the application of Montel's theorem which says that whenever you have uniformly bounded family of analytic functions then you take any sequence in that family the sequence will admit a subsequence which converges normally ok. So normally on d which means uniformly on compact subsets of t alright. Now what we did then of course the big deal was that so then the claim was that the image of f0 under the image of d under f0 is actually the whole unit disc ok. So this f0 succeeds in mapping the given simply connected domain which is not the whole complex plane isomorphically on to the unit disc ok which is the main aim of the premon mapping theorem. So the first thing is that so there are few facts that one has to understand the first thing is as I told you first of all if you the fnk of z0 in modulus the derivatives tends to a that will tell you that f0 the derivative of f0 at z0 is also a the modulus of that is also a ok because fnk tends to f0 because it is normal convergence the derivatives will also tend to the derivative of f0 and if you apply z0 and take modulus you will get this fact. So this is limit function ok mind you a normal limit of analytic functions is analytic therefore this f0 is certainly analytic alright and it is derivative at z0 if you take the modulus that will be a alright that is because of this convergence and the definition of a and the sequence we have picked and this so the derivative is at z0 is certainly the derivative at z0 for the function f0 is non-zero so it is a non-constant function. So f0 is non-constant ok because had it been constant the derivative would be 0 alright everywhere so it is a non-constant function but then you now apply Hurwitz's theorem which tells you that whenever you have a sequence of injective analytic functions that converges normally to a function then the limit function is either constant or it is also injective but since the limit function is not constant Hurwitz's theorem will tell you that f0 is also injective f0 is also injective that is 1 to 1 or univalent in other words ok. So that is because of Hurwitz's theorem then the other thing is that you know since mod f is always less than 1 for all f in script f so that will tell you that you know of course this f the subsequence is also from there so mod fnk is always less than 1 so that will tell you that mod f will also be less than 1 ok. So here I mean mod f0 will also be less than 1 ok and why is this true because you know of course if all these functions are bounded by 1 therefore the limit function will also be bounded by 1 ok but I want to say it is strictly bounded by 1 I want to say the limit function cannot take the modulus of the limit function cannot be equal to 1 and why is that true because if it is equal to 1 that means there is some point in D where f0 takes a value which lies on the unit circle ok but then there is this open mapping theorem which tells you that the non-constant analytic function always the image of any open set if the image under any the image of any open set under any non-constant analytic function is an open set that means what is another way to say it the other way to say it is that an analytic the value is that analytic function takes they are all interior values analytic function cannot take a boundary value. See what is open mapping theorem say it says you take you apply you take an open set and apply the analytic function you get an image set that image set is open that means each of the values it takes is surrounded by a disk full of which are consisting of the values of the function so it cannot take a boundary value so you can also think of the open mapping theorem as saying that an analytic function cannot take a boundary value. So this f0 cannot take the boundary value 1 because you know if it takes the boundary value 1 ok then it will also take values outside a little outside the unit disc ok. It will if it takes the boundary value 1 ok if it takes a value on the unit circle ok the bound which is the boundary of the unit disc then it will also take all values in a neighbourhood surrounding that point on the unit circle and that neighbourhood will say part of that neighbourhood will be outside the unit disc okay and therefore it will take some values outside the unit disc okay. So then you will get points you will get points in D where the function limit function f capital f not is taking values outside the unit disc but it is a limit of functions all of which are taking values only inside the unit disc and that will give you a contradiction okay. A limit of a sequence of values which lie in the unit disc cannot converge to a value which is outside the unit disc okay. So that contradiction will tell you that f not also is strictly bounded by 1 in modulus okay and so use by the open mapping theorem and so and what so the moral story and of course you know if you take fnk of z not is 0 by definition and but this tends to f not of z not which will tell you that f not of z not is also 0. So all these observations tell you that this f not is actually an element of the family script f. So f not is an injective analytic function from the given simply connected domain in D which is not the whole complex domain taking values in the unit disc and taking z not to 0 okay. So f not so in other words you see you have you have for all you have the supremum a the supremum a is attained by a function that function is also in this family that is what we have proved okay the supremum a is attained by a function which is also in the same family alright on which the supremum is defined right. So this f not is in f so this is a fact that we need okay that is an observation that we need to make then now comes now I will have to tell you that you know f not takes D on to the unit disc. So I mean the moment I say this it means that you know you know f not is already injective so it is an inject and you know an injective holomorphic or analytic map is an isomorphism onto its image therefore this once you prove this you have already proved that f not is an isomorphism of D on to the unit disc okay which is the purpose of the Riemann mapping theorem alright. So now you know so the fact that f not the image of f not fills out the whole unit disc that is a fact which requires the use of hyperbolic geometry okay which is what we are going to look at. So what I am going to do is so what I am going to do is I am going to break off at this point and say a few things okay. So the first thing I want to say is that you know so we go to we change our focus for a little amount of time and look at go back to hyperbolic geometry okay. So the first thing I want to say is so here is a remark so the remark is you know that if you take a point if you take two points in the unit disc so I think maybe I should not use z not because z not is already used let me use something else theta not and theta 1 okay. What is rho sub h? Rho sub h is a hyperbolic distance the distance between these two points in the unit disc for the hyperbolic metric okay and you know that is the arc length of the arc joining these two points on the circle passing through these two points and orthogonal to the unit circle okay because that is the geodesic for the hyperbolic geometry the path of shortest length okay and of course if one of these points lies if one of these points is the origin or if both of them lie on a diameter then that geodesic turns out to be just a straight line segment okay and otherwise it is not a straight line it is always a curved path it will be an arc of a circle which is which passes through these two points and which is perpendicular to the unit circle which is the boundary of the open unit disc right. Now what is the formula for this? The formula for this is you see you know you know given any two points in the unit disc you know you can always find a mobius transformation and automorphism in the unit disc that maps one of them to 0 and maps the other one onto if you want a point on the real axis okay you can always do that okay. So you know situations like this I have so this is my unit disc and you know I have two points z0 I mean zeta0 and let us say zeta1 okay then you know I can write this mobius transformation if you want let me give it a name capital H okay and you know I can find a mobius transformation which will map the unit disc to back to the unit disc that is it will be an automorphism unit disc and it will map the point zeta0 to 0 I can make zeta0 go to 0 and I can make zeta1 go to a point on the real axis I can do that okay this can always be done how because you know you have to just define H of z H of zeta to be you know if you put zeta-zeta0 by 1-zeta0 bar zeta okay this will map zeta0 to 0 and this is certainly of the form of the general form of an automorphism of the unit disc okay and then the only thing is when I put zeta1 I will get H zeta1 is zeta1-zeta0 by 1-zeta0 bar zeta1 that is a point in the unit disc it may not lie on the positive real axis but you know if I whatever its argument is if I multiply by you know e power i beta okay so that beta is the negative of the argument of this when I substitute zeta equal to zeta1 then H of zeta1 will lie on the real axis. So you know so this is something that this is an adjustment we can always do so what we will end up with is I think this so you know H of zeta is equal to e power i beta beta times this you put this okay if you multiply by e power i beta it is still an automorphism of unit disc because e power i beta is just rotation about the origin okay and you know general automorphism unit disc looks like this that is also something that we have seen already okay where you put beta to be negative of principle argument of you substitute zeta1 without the e power i beta tau okay. So now if you calculate H of zeta1 and you calculate the argument of H of zeta1 argument of H of zeta1 will be argument of beta I mean argument of e power i beta which is beta plus this okay so you will get 0. So the fact that its principle argument is 0 tells you that it is on the real axis so it is a positive real number it is a fraction and it is inside the unit disc so it is a fraction. So you can do this now but what is the you know what is the advantage of doing this see the advantage of doing this is you know that the hyperbolic you know that the automorphisms of the unit disc are isometries for the hyperbolic metric every for the distance given by the hyperbolic metric the every automorphism unit disc is an isometric okay. In fact in a way the you know the original version I mean the pix lemma for example says that that you know if you have an analytic function which maps unit disc into the unit disc if it is not an automorphism then it will be a strict contraction in terms of the hyperbolic metric otherwise it will be it will preserve the hyperbolic metric and it will be an automorphism okay. So in this case this H is an automorphism so it will preserve the hyperbolic metric so this is the same as hyperbolic distance between H of zeta not and H of zeta 1 and this is just the hyperbolic distance between 0 because H zeta not is 0 and H zeta 1 is so H zeta 1 is whatever it is of course it is a real number okay. And you know we derived a formula for this is this is half lawn of half lawn of 1 minus H zeta 1 by 1 plus H zeta I mean 1 plus H zeta 1 by 1 minus H zeta okay. So this is a formula for the hyperbolic distance between any two points of unit disc because this is something we already calculated if you had real number R here okay where R is a fraction you will get half lawn 1 plus R by 1 minus R right that is just the radial distance from the point from the origin to the point with point lying on the real axis with coordinate R okay. So this is a calculation we have already done I need you to remember this therefore take g from delta to delta analytic not an isomorphism okay take a function take an analytic map from the unit disc to the unit disc which is not an isomorphism alright take such a map. Now what this will tell you is that you know I mean what does pix lemma tell you pix lemma tells you that g is a strict contraction g is a strict contraction with respect to the hyperbolic metric okay. So that is what pix lemma tells you alright what does that mean it means that if you take 2 points z and zeta 0 and zeta 1 in the unit disc okay you apply the points to g you take there that is you take the images under g and you apply you calculate the hyperbolic distance okay the distance between the image points will be strictly less than the distance between the original points. So it will be strictly less than the hyperbolic distance between the original points so this will happen this is pix lemma that any analytic map or self map of the unit disc which is not a isomorphism which is not an automorphism it has to contract okay you take 2 points and then you take their images the distance between the images under the hyperbolic metric is smaller than strictly smaller than the original distance distance between the original points under the hyperbolic metric okay. If you know if zeta 0 is fixed and zeta 1 tends to zeta 0 okay what will happen you see as zeta 1 as the points come closer and closer the obviously the distance between the 2 points will come closer and closer and it will tend to 0. So what will happen is that the hyperbolic distance between zeta 0 and zeta 1 that tends to 0 okay and as zeta 1 tends to zeta 0 g zeta 1 will also tend to g zeta 0 because of continuity of g okay g zeta 1 tends to g zeta 0 it will imply that the hyperbolic distance between g zeta 0 and g zeta 1 will also tend to 0 okay. So I am just stating the obvious thing that as the points come closer their images also come closer okay and both sides will tend to 0 okay. See if 2 quantities go to 0 okay it is not necessary that their ratio goes to 0 okay it can happen the ratio can go to infinity or it can go to 0 or it can go to some finite value. So these 2 go to 0 but the fact is that if you take the ratio rho if you take the ratio of the distance between the images and the distance between the 2 given points okay this quantity can be approximated by the following number 2r by 1 plus r squared okay where zeta 0, zeta 1 belong to mod z less than or equal to small r which is strictly less than. So in fact I want to say in fact I do not even want to say this I want to say that this is always less than or equal to okay this is a bounded quantity and this is the bound okay and this bound is and in fact this bound is even attained okay and the bound is attained okay. So there are you can find point zeta 0 and zeta 1 in this disc for which the ratio of the distance is exactly equal to this so that bound is also attained. So for so this is very important this is not for any g this is for g zeta is equal to zeta square this is for this this is for the square function okay. So I mean if g is any analytic self map from the unit disc to the unit disc which is not an isomorphism it is a contraction okay and the point about both sides is both sides goes to go to 0 as the points approach each other okay but if you take the ratio the ratio is a 0 by 0 form okay and a 0 by 0 form can behave in any way that you want but it will go to 0 because the numerator is strictly lesser than the denominator alright this is strictly less than this okay. So you can expect it to go to 0 alright but then you can get a very special expression for what this ratio is for zeta 1 very close to zeta 0 okay and in fact if you take both zeta 0 and zeta 1 in this close disc okay then this is bounded by this number 2r by 1 plus r square where and that is for the particular case of the function g which is zeta square okay the square function. The square function is of course an analytic map from the unit disc to the unit disc but you know it is not it is not it is not an isomorphism because it is not injective because you have 2 square roots going to the same it is 2 to 1 in a deleted I mean in the deleted disc I mean from the disc if you remove the origin then both z and minus z will go to the same value so it is a 2 to 1 map alright. So this is not an injective map so it is not a holomorphic isomorphism therefore by pix lemma it will be a contraction but the point is when you calculate this ratio for zeta 1 and zeta 0 inside this close disc this is the bound for this ratio okay but the question is how do you get this number okay for that you know you have these points zeta 0 and zeta 1 you can move zeta 0 to the origin you move zeta 1 to this r by using a map h like this okay and then you calculate now you calculate this ratio you calculate this ratio this ratio will be rho h of you know this ratio it will be rho h of 0 r square divided by rho h of 0 r okay I am calculating it for I am simply calculating it for the for the for these two points and there the effect of the function zeta going to zeta square for those two points okay instead of calculating it for two general points zeta 1 and zeta 0 because I can move them to these two points alright. So you know for example if you do this calculation okay you will get you will get half lawn you will get half lawn of 1 plus r square by 1 minus r square by half lawn of 1 plus r by 1 minus r okay this is what it will be alright and then you know if you try to apply if you try to study it as r tends to 0 okay this is a you know as r tends to 0 this is lawn this is lawn 1 this is a 0 by 0 form so you can apply Lopital's rule okay. So applying Lopital's rule will tell you that you can differentiate both the numerator and the denominator okay it is a 0 by 0 form if you take limit as r tends to 1 I mean r tends to 0 okay it is a it is an infinitesimal I mean it is a 0 by 0 indeterminate form so if you apply Lopital's rule this can be what you will get is you know if you calculate it you will get 2r by 1 plus r square just apply Lopital's rule for or very very small you will get this okay and that is very simple you just differentiate the numerator and then you differentiate the denominator and divide you will get this expression okay. So the moral of the story is that I need to know how this quantity looks like you know when the point zeta 1 comes very close to the point zeta 0 I need that fact right and in particular of course you know you must remember that this is this is less than 1 is certainly less than 1 because if you cross multiply and move things to the other side I will get 1 minus r of the whole square which is strictly greater than 0 right. So this I mean the fact that this is less than 1 is the statement that I mean it again reinforces the fact that it is a contraction it is a strict contraction alright. Now anyway actually probably even this expression is not so important for me I want I want to say that there is a constant okay this ratio is less than or equal to a constant which is less than 1 okay I need that fact for the square function right. Now what you do is now we do this very nice lemma so here is a lemma which is kind of one shouldn't call it a lemma in fact it is more than a theorem because it uses lot of stuff okay and which is it is a critical lemma that we need to complete the proof of the Riemann mapping theorem. So the lemma is the following the lemma is if let me use d not in delta is a simply connected domain such that d not is not equal to the whole unit disc suppose you have simply connected domain in the unit disc which is not the whole unit disc okay and if 0 is a point of d not then there exists an analytic map or holomorphic map psi from d not to delta such that psi take 0 to 0 and the derivative of psi at the origin as modulus greater than 1 okay. So this is a rather technical lemma okay it is a rather technical lemma which uses hyperbolic geometry but this is what we need for completing the proof of the Riemann mapping theorem. So what this tells is you see the situation is like this the situation is I have I have this I have this unit disc I will draw a bigger one so that I can fill in with other things so and I have I have a domain I have domain d not which is not equal to the whole unit disc okay and such that it contains the point it contains the origin okay. Then what this lemma says is that I can find you know a holomorphic map that maps d not again into another domain in the unit disc okay such that 0 will go to 0 alright but the derivative at this at 0 can be made greater than 1 okay. So this is like you know you see the point I want you I mean the way I want you to think about it is like this see what is go back to our Schwarz lemma okay go back to Schwarz lemma go back to actually infinitesimal version of Schwarz lemma what does it say from if you are having an analytic map from the unit disc to unit disc okay then the derivative at the origin in modulus has to be bounded above by 1 the derivative at the origin cannot exceed 1 it has to be less than or equal to 1 and it will be equal to 1 exactly when there is an automorphism if it is strictly less than 1 it is certainly not an automorphism okay now that lemma has this is like you know the effect of statements similar to that lemma not for maps from the unit disc to the unit disc but for maps from a smaller simply connected sub domain of the unit disc to the unit disc. So what that lemma says is that if your map is from the unit disc to the unit disc then the derivative at the origin modulus is bounded above by 1 but if you try to map a smaller domain a smaller simply connected domain to the unit disc by an analytic map okay you can always succeed in exceeding the you can always succeed in exceeding that bound for the derivative at the origin okay see this will not happen if d not is equal to delta if d not is equal to delta then Schwarz lemma the infinitesimal version of Schwarz lemma will tell you that the derivative is less than or equal to 1 and in fact if it is not an automorphism it will be strictly less than 1 you can never exceed 1 and you will get 1 only when it is an automorphism okay but the moment your smaller simply connected domain you can manage to find a analytic map which whose derivative at the origin modulus is greater than 1 this is like you know this tells you what you can this tells you about analytic maps from a simply connected proper sub domain of the unit disc to the unit disc it also always tells you that you can get you can always find one thing that that you know that is opposite to what you will get for the Schwarz lemma had d not be in the whole unit disc right. So you should see it in that you should see it in that point of view okay.