 Alright so what we are discussing is the Schwartz lemma okay which where we are looking at a function f which is defined on the unit disc and it is taking values in the closure of the unit disc okay and we assume that it maps 0 to 0 okay then the Schwartz lemma says that if you take any complex number z in the unit disc then the modulus of its image cannot exceed the modulus of the the modulus of the image of that complex number under f cannot exceed the modulus of the complex number okay so and the fact is that you get equality even at one value if and only if its rotation okay and in which case you get equality everywhere okay and of course you know rotation is a bilinear transformation. And it will map the unit disc isomorphically on to the unit disc and it fixes the origin and a corollary to Schwartz lemma is that every automorphism of the unit disc every holomorphic self map of the unit disc onto itself which is an isomorphism namely which has an inverse which is also holomorphic and which fixes the origin has to be a rotation okay. So now so let us try to prove these things so the first thing is so let me prove Schwartz lemma so the idea of the proof is very easy actually you are just going to apply the maximum principle and nothing else okay so what you do is put g of z is equal to fz by z okay put g of z is equal to fz by z for z not equal to 0 okay because I am dividing by z I should put z not equal to 0 then but the beautiful thing is that of course g is analytic on the punctured unit disc namely if z is not 0 then fz by z is also analytic because numerator is fz which is analytic denominator is z which is analytic and you know the quotient of analytic functions is analytic wherever the denominator does not vanish okay therefore this g as I have defined is it as I have defined it is analytic on the punctured unit disc but the fact is it is even analytic at the origin the reason is because f of 0 equal to 0 okay so you see f of z if you write the power series if you write the power series of f of z centered at 0 namely the Taylor expansion of f of z okay so what you will get the Taylor expansion of f of z at z equal to 0 which is you know classically called the Maclaurin expansion x the Taylor expansion at 0 is called the Maclaurin expansion so the Maclaurin expansion is what it is just f of z is equal to you know f of 0 plus z f dash of 0 plus z squared by factorial 2 f double dash of 0 and so on this is what it is this is the Taylor expansion but what is f of 0 f of 0 is 0 because f is supposed to fix the origin it maps 0 to 0 so what I will get is I will get z f dash of 0 plus z squared by factorial 2 f double dash of 0 where h z is analytic on delta on the unit disc why is h z analytic because you know h z will have h z will have this power series expansion that is gotten by taking this power series expansion and you divide it by z okay namely h of z will be f dash of 0 plus z by factorial 2 f double dash of 0 and so on and that will be a convergent power series okay. So h z is given by convergent power series at the origin so it is analytic at the origin alright but on the other hand outside the origin h of z is actually f of z by z okay so what you are saying is this function g of z extends to an analytic function h z at the origin in other words g of z itself is an analytic function at the origin okay. So as h of z is equal to g of z for z not equal to 0 this shows that actually 0 is a removable singularity for g of z okay see g of z is like sin z by z g of z is like sin by z by z which a priori cannot be defined at z equal to 0 because it is a z in the denominator but actually if you write as a power series sin z by z is also defined at z equal to 0 because at 0 it has a limit okay it has a finite limit. So h removal singularity theorem says that if you have h if a function at a point where it is not defined h but h suppose there is a function which is analytic in a deleted neighbourhood of a point okay then it can be extended to analytic function at that point if one of the following three conditions satisfied or satisfied is satisfied namely the first condition is that if it tends to limit as h as you tend to that point the second condition is that if the function is bounded in a deleted neighbourhood of that point and of course the third thing is if the function has h the function can be extended h it has a power series expansion at that point okay and in fact all the three are happening here okay. So g of z is analytic on the whole unit disc so you know I will keep writing f z by h z I mean I will simply write g of z is equal to f z by z and just remember that this is the expression for g when z is not equal to 0 when z is equal to 0 it is actually h okay and h is actually the power series expansion of g at the origin at and the origin is a removable singularity so it is a it is a point to which g can be extended analytically okay. So well fine so after that remark what we do next is now we are in a position to apply maximum principle okay so what you will do what we will do is the following thing so what you are going to do is you are going to take this disc is the unit disc and so I have this function g which is going from the unit disc what I am going to do is I am going to take a circle centre at the origin radius r small r okay so and I am going to look at the function g of z which is f z by z alright and I am going to look at what its modulus is on the circle okay so on mod z is equal to r which is less than 1 okay so I am looking at all the points on a circle centre at the origin radius 1 radius r small r where r is a fraction okay what is mod g z mod g z is mod f z by z because mod z is equal to r and r is positive okay so certainly z is not 0 and g has a expression f of z by z when z is not 0 and if I calculate mod g I am going to get mod f z by mod z and that is well that is less than or equal to 1 by r because mod f z is always less than or equal to 1 is something that is given to me that is just analytic expression for the geometric fact that f takes values in the closed unit okay and mod z is equal to r is already assumed because I am estimating the mod g z on the circle where mod z is equal to r so I get this if you take mod z less than or equal to r mod z less than or equal to r if you take this closed disc centre at 0 radius r including the boundary okay and if you look at the function g z it is analytic there okay the maximum principle will tell you that it is modulus will be maximum on the boundary so here is where I am applying the maximum principle so on mod z less than or equal to r mod g z has maximum on mod z is equal to r okay but on mod z is equal to r mod g z is less than or equal to 1 by r and therefore maximum principle will tell you that on this whole disc mod g z is less than or equal to 1 by r so the upshot is on this whole closed disc on here what you are getting is mod g z is equal to 1 by r I mean see the point is you make an estimate of mod g z on the boundary of that closed disc which is mod z equal to r and that will also be an upper bound for the values inside because this is the maximum principle the maximum principle tells you that the mod will attain its maximum only on the boundary so if you know a bound for the function on the boundary that bound will also be a bound for the function values on the interior and of course the function in this case is the modulus of the analytic function okay so by the maximum principle okay so you know when in applying maximum the maximum principle I am using the fact that you see g is analytic and therefore harmonic because g is analytic means that both its real and imaginary parts are harmonic and therefore g is also harmonic and I am using the maximum principle for harmonic functions okay so it applies to g so mod g has a maximum on mod z equal to r but on mod z is equal to r it is bounded by 1 by r therefore 1 by r is a bound for g on the whole closed disc okay so for mod z less than or equal to r mod g z is less than or equal to 1 by r this is what you get now what you do is that you take the limit as r tends to 1 minus okay if you take the limit as r tends to 1 minus we get mod g z is strictly less than is less than or equal to 1 for mod z less than so and mod g z less than or equal to 1 will tell you that mod fz by z is less than or equal to 1 and that translates to mod fz less than or equal to z for mod z less than 1 which is the first assertion in the shorts lemma okay so this implies so you get this of course you know there is a little bit of trouble at z equal to 0 but at z equal to 0 this holds because mod f mod f0 is 0 and of course here I have to put mod mod f0 is 0 which is equal to mod 0 further let z0 be such that mod z0 is less than 1 I will of course need z0 not equal to 0 and mod of f of z0 is equal to mod z0 okay so in this statement I should correct myself so z0 should be non-zero because if I do not put that condition then this is always true for z0 equal to 0 okay so z0 should be non-zero so suppose there is an z0 which is not 0 where mod fz0 is equal to mod z0 okay so then what you are going to get is that you are going to get then mod gz0 is going to become equal to 1 okay because after all z0 is not 0 so mod g I mean gz0 is just fz0 by z0 and mod gz0 will be mod fz0 by mod z0 and that will be 1 okay but then again you apply the maximum principle to mod g okay that will tell you that mod g will always be equal it will tell you that g has to be constant okay so because mod g can attain its maximum only on the boundary of the unit disc it can provided extends to the boundary of the unit disc if it does not extend to the boundary of the unit disc it can never attain the maximum so whereas mod g is bounded by 1 mod g is bounded by 1 maximum value 1 cannot be attained in the interior if it is attained in the interior then g has to be a constant this is the maximum principle okay so by the maximum principle g is a constant so and with g is a constant and mod gz0 constant with modulus 1 so I will get g of z is equal to e power i alpha because a constant a complex number with constant with modulus 1 has to be a form e to the i alpha and that means that because g is f of z by z it will tell you that f of z is just e to the i alpha z so it is a rotation okay so that finishes the proof of the Schwarz lemma okay that finishes the proof of the Schwarz lemma and I must again point out that I had made earlier the mistake of not saying that z0 is non-zero okay that is important because if I did not say that then I already have mod f0 equal to 0 equal to mod 0 right so this is the proof of the Schwarz lemma I mean what you must understand is that the moment f is a rotation then f becomes a bilinear transformation therefore it is 1 to 1 also it becomes a conformal isomorphism okay so what your Schwarz lemma actually says is that if you take a analytic function from the unit disk into the unit disk the closure of the unit disk which takes the origin to the origin then either it is strict contraction in terms of length okay that is mod fz is strictly less than mod z for all z with mod z less than 1 or it is a rotation okay so let us look at this corollary let us try to prove this corollary proof of corollary so what I have to do is I will have to take a holomorphic automorphism of this which fixes the origin and I have to say that it is a rotation so let f from delta to delta be a holomorphic that is analytic isomorphism such with f of 0 equal to 0 okay so you take a holomorphic isomorphism self isomorphism of the unit disk right now of course to f I can apply Schwarz lemma because condition of conditions for Schwarz lemma is that f should be analytic on the unit disk it should take values inside the closed unit disk in this case it is taking values in the unit disk itself and it is defined on the unit disk and it takes 0 to 0 so all the conditions Schwarz lemma are satisfied so I can apply Schwarz lemma and I will get by Schwarz lemma f of z mod fz is less than or equal to mod z for all z with mod z strictly less than 1 I will get this by applying Schwarz lemma to f but mind you what is given is that f is a holomorphic isomorphism which means f inverse is also like f f inverse is also a holomorphic map from delta to delta and f inverse will also take 0 to 0 because f of 0 is 0 f inverse 0 will also be 0 so I can also apply Schwarz lemma to f inverse ok since Schwarz lemma also applies to f inverse f inverse mod f inverse of w is less than or equal to mod w for all w with mod w less than 1 I can apply Schwarz lemma to f inverse which is the inverse of f which is given to me to exist because f is given to be a holomorphic isomorphism which means f has an inverse and that inverse is also holomorphic ok but then you put w equal to fz in this put w equal to fz in this will give you that mod z is less than or equal to mod fz ok and then you compare these two opposite inequalities mod fz less than or equal to mod z mod z less than or equal to mod fz and you will get mod fz is equal to mod z ok and that is it that should tell you that f is a rotation. So mod fz is equal to mod z for all z with mod z less than 1 but we already seen the proof of the Schwarz lemma that whenever mod fz is equal to mod z holds for a single z not different from the origin then f has to be a rotation. So here it is what you are getting is that it holds for all the points of the unit disk ok you have more than what you need ok. So this will imply again by Schwarz lemma that f is a rotation. So that proves the fact that proves the corollary it says that the only holomorphic automorphisms of unit disk that fix the origin they are all rotations ok fine. So having done this what I am going to embark upon is to go into discussion of the Rayman mapping theorem which is something that I want to whose proof which is what I would like to discuss in the coming lectures. It is a very deep theorem and the proof is not easy it involves several facts and so but to make a preliminary discussion about it I wanted this fact about automorphisms of the unit disk. So let me start with the Rayman mapping theorem which is what our long term goal in the next few lectures is proof of the Rayman mapping theorem ok. So Rayman mapping theorem so this is a theorem which says that you take any domain in the complex plane which is simply connected and assume that it is not the whole complex plane then that domain can be mapped by a holomorphic isomorphism or to the unit disk. So in other words if you take simply connected domains in the complex plane and you go modulo holomorphic isomorphism you will get only a set containing two elements one will be the isomorphism class of the whole complex plane and the other will be the isomorphism class of the unit disk and mind you the unit disk is isomorphic to the upper half plane in fact it is any disk is any disk is any open disk is holomorphically isomorphic into any open half plane because you can always find a mobius transformation that will map the interior of a disk to any half plane ok. So geometrically up to holomorphic isomorphism unit disk is the same as a half plane any disk is like the unit disk alright any finite disk it looks like a half plane that is one holomorphic isomorphism class the other holomorphic isomorphism class is a isomorphism class of the whole complex plane and these are the only two holomorphic isomorphism classes of simply connected domains in the complex on the complex plane ok and that is the statement of the Rayman mapping thing. So let me state that any simply connected domain not equal to C I am writing it also in words I also put it in symbols D is simply connected the proper domain in the complex plane is holomorphically isomorphic isomorphic to the unit disk ok so this is the Rayman this is the celebrated famous Rayman mapping theorem right and so in other words what you are saying is that if you give me a simply connected domain which is different from the complex numbers then there is a holomorphic isomorphism from D to delta which is unit disk ok and that holomorphic isomorphism can be made in fact unique uhhh in the following way. So let me explain that in fact given z0 point of D a real number lambda greater than 0 there exists a unique holomorphic holomorphic is the same as analytic or conformal isomorphism f from D to delta with f of z0 is equal to 0 f dash of z0 is equal to lambda. So uhhh so the Rayman mapping theorem says that uhhh any simply connected domain uhhh which is not the whole complex plane can be holomorphically mapped on to the unit disk and you can make that mapping unique if you fix a point if you if you require that a point of D a fixed point of D goes to the origin under this map and also that at that point the derivative of the map at that point is a fixed real number ok. So these conditions uhhh make the map f uhhh unique ok and it is uhhh uhhh so so you know there are two parts to this one part is to find a map f ok then which is uhhh which is rather the hard part the easier part is to say that once you have a map f like this it is it is unique and it is the uniqueness part that will use Schwarz's lemma ok or the corollary of Schwarz's lemma. So what I will do is uhhh uhhh I will I will first try to uhhh apply Schwarz's lemma which we have just seen to show that you know if a map like that exists it is unique such a map uhhh a map like that is called uhhh uhhh is given a special name it is called the Riemann map it is called the Riemann map of the domain D ok and the Riemann map is unique if you fix the value of a point on the domain and the derivative at that of the map at that point ok uhhh I of course you I have fixed the value of the point z0 to be 0 for convenience and most people uhhh are in several textbooks you would you will see that lambda is taken to be equal to 1 ok. But in principle you could take lambda to be any positive uhhh real value ok. Now uhhh so let us let us first prove uniqueness ok. Suppose f1 and f2 f1, f2 for R uhhh from D to delta are holomorphic isomorphisms with these conditions f1 of z0 is 0 is also equal to f2 of z0 and derivative of f1 at z0 is the given lambda which is equal to the derivative of f2 at z1. I just want to show that f1 and f2 are one and the same map right. So what I do is that I compose f1 with uhhh one of the f i's with the inverse of the other and realize that I get a conformal automorphism of the unit disc which by Schwarz lemma is a rotation because it will fix the origin ok. So you see so the situation is like this. So I have I have D I have delta I have f1 then I have f2 then I have delta here ok. So if I if I go like this I get the map which is f2 inverse followed by f1 ok. And this f2 inverse followed by f1 uhhh what is the property of this map this is holomorphic it is a holomorphic uhhh map. It is a holomorphic isomorphism because it is a composition of two holomorphic isomorphisms f1 is a holomorphic isomorphism and f2 is a holomorphic isomorphism therefore f2 inverse is also a holomorphic isomorphism. The inverse of a holomorphic I the inverse of an isomorphism is always an isomorphism ok. So this is a composition of isomorphism therefore this is also an isomorphism and uhhh so this is uhhh this is an isomorphism is a holomorphic isomorphism and uhhh where does 0 go to 0 goes to 0. Because you see z0 under this map goes to 0 and under this map also goes to 0. So if I compose it I will get 0 goes to 0 alright. So it is a holomorphic isomorphism from delta to delta which fixes the origin and as we have seen just now seen as a corollary of Schwarz's lemma uhhh this has to be a rotation ok. So by uhhh corollary to Schwarz's lemma f1 circle f2 inverse of w is equal to uhhh e power i alpha w it has to be a rotation of course whenever I write e power i alpha of course I am assuming alpha is real ok. Because if I write e power i alpha with alpha complex then it is uhhh no longer uhhh your rotation ok whenever I write e to the i alpha I am always assuming alpha is real. So uhhh so this is because of the corollary to Schwarz's lemma ok. And uhhh you see you know if I calculate f1 circle f2 inverse uhhh uhhh uhhh so if I if I take the derivative of this ok if I take the derivative of this what I will get is that I will get f1 circle f2 inverse derivative at w is equal to derivative of e power i alpha at w which is e power i alpha ok. The derivative with respect to w of e to the i alpha w is just e to the i alpha ok. And on this side I will get derivative of this expression but then for this you apply the chain rule ok. So what you will get is you will get f1 dash of f2 inverse of w into f2 inverse f2 inverse dash of w is equal to e to the i alpha ok. This is just uhhh applying the chain rule of differentiation alright. And uhhh and mind you uhhh in particular you know I I can put I can put any value for w here. So uhhh because it is an identity for all w. So I can put w equal to 0 ok. Put w equal to 0 and what I will get I will get f1 dash of f2 inverse of 0 times f2 inverse dash of 0 is e to the i alpha this is what I will get ok. But then uhhh you have to uhhh remember that f2 inverse of 0 is actually uhhh z0 because f2 of z0 is 0 ok. And therefore uhhh the uhhh so what you will get is you will get f1 dash of z0 f1 dash of z0 right into f2 inverse derivative at 0 is e to the i alpha ok. Now you see f1 dash z0 is given to be lambda. So I will get lambda and the fact is uhhh so let me write that separately ok. Now I claim that f2 inverse derivative at 0 is also uhhh is is equal to 1 by lambda ok. That is just because uhhh again uhhh chain rule applied to f2 and f2 inverse ok. So f2 uhhh circle f2 inverse of uhhh uhhh w is w ok. And if you apply the chain rule you will get f2 dash of f2 inverse of w times f2 inverse dash w is equal to 1 ok. You put w equal to 0 and you will get f2 dash of z0 into f2 inverse dash fdup of 0 is equal to 1. And this will tell you that f2 inverse dash of 0 is 1 by lambda ok. Mind you lambda is a lambda is a non is a it is nonzero it is positive. And therefore you know well uhhh if you if you look at both uhhh what you will get is that you will get e to the i alpha is equal to 1 ok. And once e to the i alpha is 1 uhhh you will get f2 f1 circle f2 inverse w is equal to w. And this will actually tell you that f1 equal to f2 ok. So uhhh uhhh so that so by using the uhhh shots is lemma or rather the color of the shots lemma namely that every automorphism of the unit disc that fixes the origin is a rotation. You are able to show that a Rayman map uhhh if it exists which is specified at a point of the domain at the simply connected domain which is not equal to the complex plane. And it is uhhh uhhh if if it is if it is image is specified at one point and it is derivative at that point is specified then the map is unique. The uniqueness comes from shots lemma actually ok. Now uhhh so this is the easier part this is the uniqueness of uhhh this Rayman map. But now we have to prove the existence of the Rayman map ok. You have to show that there is a map from the uhhh from the given simply connected domain which is not the whole complex plane to the unit disc ok. So, let me let me recall the the the the fact that the domain is simply connected is uhhh by definition it means that the any two uhhh any any curve uhhh any path in the domain starting at any point can be continuously shrunk to a point to that point. That means that there are no holes in that domain ok. If you think of it uhhh as uhhh in terms of the region that in that is enclosed by any closed curve in the domain then you do not want any holes there. The domain should not contain any holes ok. Because if there is a hole then you cannot continuously shrink a closed curve to a point ok fine. So, uhhh it is very very important that you have the simple connectedness. So, the question is how do I produce uhhh how do I produce a uhhh holomorphic map from this domain to the unit disc ok which is an isomorphism. So, the first step is you try to get hold of some holomorphic map which maps the domain into a sub domain of the unit disc ok. So, let me explain to you the the first step towards the proof of the Riemann mapping theorem namely the first step towards the proof of the existence of the Riemann map. The first step what you do is you show that the domain D can be conformally that is isomorphically it can be mapped in onto a sub domain of the unit disc. First of all you do that then you modify that map so that it can fill out the whole unit disc ok. And I am I am saying it loosely modify means it is not just modify you will have to do a lot of things ok. But first of all given any simply connected domain how do I land at least into the unit disc. So, uhhh the the beautiful point here is uhhh uhhh existence of uhhh a logarithm for a non vanishing holomorphic function on uhhh on a simply connected domain that is uhhh essentially used. So, so let me explain that. So, what I am going to do is uhhh step 1 so this is existence of the Riemann map. So, I am going on with step 1 uhhh find holomorphic isomorphism h from D to h of D which is a subset of uhhh the unit disc ok. So, the first step is to map first of all D into holomorphically isomorphically into a sub domain of the unit disc. So, what you do what we do is a following thing. So, so here we exploit all the uhhh we exploit the fact that the domain is not the whole complex plane and the fact that the domain is simply connected ok. So, what we do is since uhhh the domain is not the whole complex plane there exist need or not complex number which is in uhhh which is outside the domain I can find such a complex number because uhhh it is not uhhh the domain is not the whole complex plane. So, there is something outside the domain which is in the complex plane ok. You take this we have uhhh now now the next thing I am going to use is the fact that this domain is simply connected ok. The function z- uhhh uhhh g of z is equal to z-neta not is non vanishing on uhhh on D that is obvious because that because the point neta not is not in D. So, z-neta not with z varying in D can never be 0 ok. And since it is a non vanishing uhhh function and it is of course analytic ok it is it is an analytic function after all it is a translation. It is translation by minus neta not ok it is an analytic function. This analytic function uhhh is non vanishing on this domain D which is simply connected therefore it has a logarithm ok. Since D is simply connected we have an analytic branch of log z-neta not on D ok and therefore the moment I have log z-neta not I will have analytic branch of the square root of z-neta not on D and my claim is that function will do the job of mapping D uhhh I can use that function carefully to map D onto a sub domain of the unit disc ok. So, let me write that and stop thus we have an analytic branch of root of z-neta not as exponential of half log z-neta not on D. We can use this branch branch of root of z-neta not to get the holomorphic isomorphism. So, I will stop here we will continue in the next lecture. So, I am just trying to say that you are using a square root of z-neta not to get to map cleverly to map D first into a sub domain of the unit disc ok. And the fact is that you are able to write this z-neta not because there is a netonaut which is outside your simply connected domain your simply that uses the fact the simply connected domain is not the whole complex domain. And the and the fact that you have a analytic branch of the square root is because uses it uses the fact that the domain D is simply connected ok. So, I will stop here.