 Okay, today we'll spend most of the time in completing the proof of the famous Riemann mapping theorem, okay, which is a kind of exercise for us, which allows us to put together the results we have encountered in this course. More than a constructive proof, we have a proof in steps in which you apply several facts, okay. There are different proofs of the same theorem, but none of them can be considered, it is, well, each of them can be considered actually an existence theorem more than a constructive one, okay, so, except for very small classes of domains in which you can find explicitly the Riemann function, the Bellomorffian with the unit disk. In general, what you know is that this function exists and it is, well, behaved with nothing more, right. So, let us start from this general fact. We are taking a domain D and C, which is simply connected and different from C. So, domain well contained in C and simply connected, which means that if you consider what is in the boundary, D is open, right. Domain means open, connected, simply connected in this case, we are assuming that also simply connectedness and different from C. So, it is contained. So, in the boundary, there are at least two points, right. So, this boundary contains at least two points. We call them A and B, right. If you want, we can actually substitute this hypothesis saying that, well, this is the case, right. At least two points are in the boundary, okay. If I remove one point from the complex plane, I don't have a simply connected domain, okay. Which point you take is not important from the logical point of view because in any case, we are considering another simply connected domain. Remember that we have defined simply connectedness in terms, in geometric terms, only for plane domains by saying that the complement in the Riemann sphere is connected, right. So, the plane is in fact simply connected, but we don't want it. But the plane minus one point, so with the punched, the punctured plane is not simply connected because in the complement you have infinite and the point, right. So, two connected components. And we have seen last time that this hypothesis is crucial also for Runge theorem, right. In the approximation, right. Okay, good. So, we have two points, not in D for sure. And we consider this function, which is one of the functions we will use in this proof, which is the following. It's a square root of z minus a over z minus b. a and b are outside of it. And this is defined in D and takes values in C, okay. What I'm saying is because a and b are not in D, this function is well defined. Are you with me? Yes. Why? Well, first b, which appears here, is the only point where the function can have some problems in the definition. The second point is that here we are taking the square root of in the complex setting. So, this number cannot be zero. Remember that zero is a branch point for any root, right. Because the definition of the squares, sorry, of the roots is related to the definition of the logarithm. So, when you are close to zero, there are problems. What we know, and I want to recall you that we proved this, that when you have a simply connected domain, and you have a function which is not vanishing in this simply connected domain, you can define the logarithm of this function, right. We proved this, right. So, in particular, this function here under the symbol of square root is a non-vanishing well-defined function in b, because it is not vanishing well-defined b is on the boundary. And so though, that we can take a single valued logarithm well-defined and this function turns out to be holomorphic in b, all right. So, I summarized what we have done here. z minus a over z minus b is well-defined in d and never vanishes, okay. So, that we have a function which is well-defined and not vanishing in a simply connected domain. We can take the logarithm, all right. And therefore, we can equivalently define as a single valued function p of z, which is a square root of z minus a over z minus b, okay. So, no singularities, no vanishing of the function under the symbol of the square root guaranteed and the fact that d is simply connected, remember that this is crucial, okay. Guarantees that you can define the logarithm, so that, okay. So, in fact, there are several ways to, so several ways. There are several functions you can define in the proof, in one of the proofs is something like the logarithm or something else. I mean, in any case, the common approach is that the function which are defined are well-defined and the fact that simply connected, that d simply connected guarantees the existence of the logarithm and the singularities and the zeros are outside of the domain. Now, what I want to prove is that no, not in d, not in d. z belongs to d and this number cannot be zero, cannot be zero. So, not, well, we have this property which is valid in the real and also in the, this is zero if and only if z is zero, right. So, the only possibility is z equal to a, but a is outside of d. So, right, okay. So, no branch point, no branch points, no singularities and what is something we can prove now is the following. The function p of z is injective, okay, as a function and as a terminology and in complex variables and function theory in particular in all textbooks, these properties for holomorphic functions which were also called regular functions, okay. In some texts we can find regular are summarizing this way, univalent. You can find many books with this terminology instead of injective and holomorphic. Well, this is just a matter of choices of terms, okay. Univalent means exactly this. It is one to one and regular or holomorphic as we are, okay. We have introduced this class of functions. So, just in case you find another book, this function is univalent or you can find out even some books there, some famous books, recent books, univalent function, the theory of univalent functions means which are holomorphic and injective, okay. And as I was telling you in before starting, so the class of univalent function in some domains are very well studied and are related for instance to problems like the Bieberbach conjecture, right. So, the functions which are considered are functions which are holomorphic in a unit disk, injective there, so they are univalent and up to a normalization you can assume that they fix the origin and have the derivative at one equal to one. So, the famous Bieberbach conjecture and now it is a theorem that gives sharp estimates of the coefficients of in the power expansion of zero of this class of functions, injective and holomorphic, all right. Good. So, how do you prove that this function is like well you don't have to do very much but say assume that P of Z1 is equal to P of Z2. If you prove that this implies that Z1 is equal to Z2 then we are done, right. So, this is the classical proof that this is nothing ready to complex or not complex, this is because we are using the definition of functions, right. So, but the function is explicitly written here, so we can apply and this number here is the probability that its square is the same, right. So, that you have that well this implies that and well this implies that Z1 equal to Z2, right. Well this is obvious, okay. This is the kind of very simple exercise but let me also observe that assume that P of Z1 is equal to alpha. Then it is not possible that the value minus alpha is taken by any Z in D. Assume that alpha is a value taken by P at a sum Z1, at only one Z1. We can say this because the function is now injective, right. Take one value, call it alpha. Take the opposite of this value. This is not taken by any Z by the by any Z in D which means that well if you want to have this is alpha and this is minus alpha, this is 0, okay. Well somehow P of D cannot into 6 minus alpha, right. This is just to have an idea. Why? Well the reason is somehow related to what we have we have already shown for injectivity because if you assume that P of Z1 is alpha but there is a say Z2 equal to minus alpha, right. So that you can have this of this so I am saying this P Z1 is alpha and assume that there exist by contradictions there exist one Z2 such that P of D such that P of Z2 is minus alpha then you have this, right. And then again taking the square of the the amounts we are considering and these qualities we obtained that Z1 minus A over Z1 minus B is Z2 minus A over Z2 minus B which implies that Z1 is equal to Z2 which is contradiction, all right. So same calculation shows us also this important fact. What can be also we also have that P of D is open, okay. This is because P is holomorphic, right and P is not constant, right. So the image of an open set is open. How can I say that P is not constant? Well we prove that P is in fact injective so cannot be, right. It is even more, okay. So P of D is open, right and I can say that call D1 P of D which is open and take W0 in D1, right. Then what I I know that well I can find a small neighborhood W0 contained in D1 which is in fact the consequence of the fact that the D1 is open, okay. So if Z0 if W0 sorry is in D1 that exists the positive C such that this small neighborhood is contained in D1 because of the fact that W0 is a point in an open set. So neighborhood, open neighborhood is contained. But if I consider minus W0 and I consider this inequality the set of all W in C such W minus minus W0 smaller than C. This is not contained in D1, correct. And I can show you this as I said before. This is W0. This is a small neighborhood centered at W0 of radius C in D1. And here is minus W0 and here is a small neighborhood and the opposite. And we prove that the opposite of the values cannot be taken by any point, right. All right. And this is nice because well when you have something which in general in this course when you have to prove something well assume that the function is not vanishing. We have used it several times not vanishing there. So you define immediately the function one over, okay. And use this fact to obtain a contradiction. Now we are taking, okay. We already know that the function is injective, okay. And I am assuming that P of Z1 is equal to alpha. Now I want to say that there is no extra point in D such that any point in D in the domain D such that the value associated to this point is minus alpha. If you assume this you obtain that Z1 is equal to Z2. And Z2 was there by contradiction the point such that P of Z2 is equal to minus alpha which means that Z1 is mapped by P into alpha and minus alpha. So it wouldn't be even a function, right. It is a single value. One point, one image, one value for the function P. So it cannot be that Z1 is mapped into alpha and minus alpha, right. And it cannot be, well we might say no, there is one point, one alpha. Alpha equal to 0 is the only, right. If alpha is, it cannot be 0 because as I said P of Z is never 0. Pardon me, here? Yes. It means that Z1 is equal to Z2. This is the conclusion, right. If I am assuming that this is, this is the contradiction true then I obtain that Z1 is equal to Z2, right. This is what? That P of Z1 is alpha. P of Z1 is also minus alpha, right. And this is possible only if alpha is equal to minus alpha because it is function. But alpha is minus alpha only if, if and only if alpha is 0. But 0 cannot be taken as a value. It's not in D and it's not in P of D because the square root of Z is 0 if and only if Z is 0, right. So putting together all the stuff you have in the value function and P of D does not intersect minus P of D. What, what do I mean with minus P of D? Minus P of D is nothing, okay. It's the opposite of any, of any P of D, of P element in P of D, right. Good. So I have that, I can define now this. So I take a special Z, W0. This is, so I take W0 in D1 which is P of D. I use this notation. I notice that this, sorry, that P of Z, okay, which was called W is not vanishing when Z varies in D. And I put the same C in front, it's in front, it's an enumerator. And I call it F0 of Z. This function is well defined and holomorphic in D. Why? Well, because P of Z plus W0 is never 0. As Z varies in D, right. Why this constant C appears here? Because I also know that, well, what I've done here is that W plus W0, sorry, smaller than C is not contained in D1, right. So this guarantees the good definition when I assume that W is P of Z, okay. This number is never 0. But furthermore, this is also true. It is normalized such a way that this number cannot exceed C in the modules of this complex number so that this ratio cannot exceed 1. So that have a function, so if not, finally from D into the closure of the unit disk. So after these two or three steps, elementary steps, we have chosen from the domain D a function well defined, holomorphic, injective, as we'll see, mapping the unit disk into the closure, mapping the domain D into the closure of the unit disk, okay. The function F0 is also injective. This is quite obvious, but let us go to the details of this. Well, you can see it either directly, you prove it well, you assume that F of Z0, Z1 is equal to F of Z2 and this implies that Z1 equal to Z2 and invite you to verify it. You can write it explicitly and using the fact that P is injective you conclude almost immediately. But if you want to be more sophisticated and avoid calculations, which is always the case, avoid calculations, well you can observe that F0 of Z is defined in the following way, C over P of Z plus W0 and so that F0 is the composition of what? You have a P of Z and you add W0, right, so you have a translation. Then you take an inversion and then you take a dilation. Okay, so let me write this. T of W is W plus W0. I of W is 1 over W and D of W is CW, right. Take P of Z, apply, so compose T to P of D, you obtain P of Z plus W0, right. Then apply the inversion 1 over P of Z plus W0, then apply the dilation, you obtain C over, so you obtain as I said F0. This functions here are the basic tools, okay, in what we have found in the Mabius transformation. These are linear fractional transformations. They are all invertible, so they are all injective. If P is injective and you are composing P with the injective function, the compose function is in fact injective, right. So finally we have an injective function from the simple connected domain D, which can be as odd as you want, so you have only one hypothesis added that D is not the complex plane, because otherwise the function, if not, would be immediately constant. An entire function which is bounded is constant, your real theorem. So there is no hope to find, okay, a biomorphic from the complex plane and the unit disk. So in fact we are considering an injective function in order to avoid constant functions, right. And that's why when I showed you why we're starting the automorphism of the unit disk, I said well the unit disk is in some sense the model of any bounded simple connected domain, okay. So at the end of the story we'll find that the domain D, and this is the statement of the Riemann mapping theorem, is biomorphic to the unit disk. So at the time being we have just one injective function from the domain D and to the closure of the unit disk, okay. And here comes the part of the steps which are less constructive. Up to now you can say well we can do something, right. We can write, you can take a and b properly and then imagine that you can have something up to now, yes. But now we consider the set of functions which are like they're not holomorphic and injective, right. And as I said this can be found to be also simply univalent, summarized with the term univalent, okay. Well this is a matter of time. So this class of functions, so in principle you might wonder if this class, if this family of function exists and is not empty, right. Say well I have no idea, D is just a simple connected domain which is not C, right. That's the only hypothesis I have. Well I show that if not belongs to this class. So it is not empty. So we're not talking about the empty set which is always a good hypothesis to have, right. Otherwise you can prove many properties, right. The empty set is very good, okay, for many results. So good. What can I say about this class of functions? This family of function turns out to be a family of function holomorphic and say uniformly bounded. They are all in modules smaller or equal to one, right. Because they map the domain D holomorphically and injectively one-to-one way into the closure of the unit disk. So the modules of any F, since you have this, then you can say that for any Z and D this is true and for any F, right. So uniformly bounded. So maybe I put it on a new slide. I have that F is a family of uniformly bounded holomorphic functions, right. And by Montel theorem this guarantees that the family F is also normal, right. It is to say is a normal family, from any sequence of functions in F you can extract a subsequence which converges, right. The point is that what we know from Weisler theorem is the function, the limit function turns out to be holomorphic because you are considering sequences of holomorphic functions. What we don't know is whether the function, the limit function is in F, is still in F or not. So we want to know if F is closed with respect to this convergence. So it is bounded as a family of functions. If it were closed it would be somehow compact, right. And unfortunately the answer is no, it is not the case. So what you have is that as an application of as we will see an application of the Horowitz theorem, well proved so far, the limit function of an injective function of a sequence of injective functions is not necessarily injective, okay. So let us collect the ideas up to now. So F is normal since it is uniformly bounded because we have Montel theorem, right. And this is okay. So which means that for any sequence in F, so we are taking a sequence of holomorphic and injective function from the into the closure of the unit task that exists a subsequence, okay. That exists, okay. This is contained in here, right. So such that so uniformly, right. I use G, okay. Sorry, I use G because then F would be used later, right. So there is a limit function which is obtained as a uniform limit uniformly on compact set. Viestra theorem, theorem G is holomorphic and G maps the unit, sorry, the disk, sorry, the domain D into the closure, right. This is what we know. The limit exists because of Montel theorem. So the limit in the sense of uniform convergence on compact sets and the function, the limit function G is holomorphic. But I cannot say that G is in F in general because the additional request is that G has to be injective. I would say yes if G is injective. But is it true that from an injective class, an injective, sorry, sequence of holomorphic function, you obtain a limit which is an injective holomorphic function? The answer is no, unfortunately, as I anticipated. This is 7, right. So equivalently, is it true in general that the limit function of sequence of injective holomorphic functions is still an injective holomorphic function? So what we know is that the limit function is holomorphic by bias of the theorem, as I already told you. What I want to show you is that not necessarily this limit function turns out to be injective. In fact, consider, well, Z naught and D and consider the sequence Gn of Z to be Fn of Z minus Fn of Z naught, right. So we start for a sequence. Without passing through the subsequence, we already assume that this is convergent to a function. If you want to pass through the subsequence, then you consider the directory subsequence. And you construct a new sequence, Gn of Z to be Fn of Z minus Fn of Z naught. Z naught is a point in D, right. This is, again, a sequence of holomorphic functions. This Gn of Z is holomorphic, right. Fn of Z is holomorphic. This is a constant term, though, okay. This is holomorphic. What we know is Gn of Z naught is equal to zero. I know that Fn is a family. Since F is normal, Fn is a sequence which converges, as I said, to a function G, right. So I have passing through the limit I obtain, well, this tends to G of Z naught. This tends to G of Z. And this tends to another function, H of Z, right. We have this equality. Now what we have is that, well, the limit for F exists because F is, the family F is normal. So we are taking up to a subsequence, a sequence of function in a normal class, in a normal family. So this function here, by Weisstrass theorem, is holomorphic as well. But what we know from Hurwitz theorem is that this function is vanishing at Z naught because we have a sequence of functions which vanishes at Z naught for any n, right. Or it is identically zero. So if the second, sorry, what I am saying is that, well, G of Z naught is equal to zero. But Gn of Z is not zero if Z is different from Z naught, right. Because F is supposed to be an injective function, right. So it vanishes only Z naught. But the limit function can be identically zero. If it is the case that H is identically zero, it is constant. And G of Z, the limit function G is constant, equal to G, to G of Z naught, right. So it is not injective. Because we have studied the zeros of limit functions, right, using the argument principle, we have shown that the limit function can be either with the same zeros or so without zeros or identically zero. So if F of Z is identically zero, equivalently G of Z is identically to G of Z naught. So it is constant. Then G is not injective. So this fact shows you that you cannot take the limit of univalent functions and be in the class of univalent function for sure. There are some obstructions. In principle, you can have a function, the limit function, which is constant. You can make examples of this, right. Take one over n times Z, you have a sequence, right, which tends to zero. For instance, okay. So now the idea is consider another class. This is seven, right. So consider F prime in F. And F prime is defined as follows. Such that the derivative of F at C in modulus is greater or equal to F prime zero of X C, X C S and D. So you pick one point in the domain D. You evaluate the function F naught with constructors so far, not in X C, but its derivative. Take the modulus and say that, what you are considering here is the set of all functions in F whose derivative at C have modulus greater or equal to this number, F prime, the modulus of F prime zero at C. So of course, F naught belongs to F prime. So again, we are talking about a non-empty family. So F prime. What do we have? We have that F prime is normal since it is a subfamily of a normal family, right. And is finally compact. So let me say what I mean. So we have seen that the only risky situation is that the limit function might be constant. Otherwise, it is injective, right. It vanishes only at Z naught, right. No other zeros, right. Good. All right. So assume G prime, well, sorry, the derivative, the limit function sequence of functions in F prime is constant. Then the derivative of a constant function is zero and here we have something which is positive, okay. So you have a contradiction, right. So F naught as derivative C positive whose modulus is positive and cannot be, right, that the limit function is by continuity. This in fact depends on the fact that when you take F in F prime and you take this, evaluate the derivative of F at C, this function is continuous. In fact, well, how do you prove this? Okay. In general, you have this. What we will use it in fact is that, well, this is but something very reasonable. If F tends to, well, this can be calculated using standard notation like this, right. Everything here is composition of continuous operations over complex valued and injective in particular holomorphic functions. So everything is continuous. And what we want to see here is that, well, we want to prove actually that this is something reasonable. You take the soup of F prime of C as F varies in F prime. I want to prove that this is in fact a max, okay. So take F in a sequence in F called this soup to be M, right. The soup always exists, right. If it is taken for a value which belongs to the class F prime, then this soup becomes a max, right. The soup always exists. I take this as a, but you take a continuous functional of something which is calculated starting from a normal family and you repeat the same argument where I'm going to show you and you see that it is equivalent. So you have a soup, you can consider a sequence of value of a function in F prime such that this number tends to M by definition of soup, right. But this function, this sequence of functions here are contained in a normal family. So you extract some subsequence, make it converge, and the value, so the limit function F is such that this is true because of bias stress theorem again, okay. So this limit exists by normality and this is true by bias stress, all right. So normality guarantees the distance of the limit and the continuity of the modulus and the fact that the limit of the derivative of a sequence is equal to the derivative of the limit function that is to say one of the consequences of bias stress theorem guarantees that this is in fact true and then F prime is in fact compact, right. So modulus of F prime of C, the max, okay. In our setting we can say that so for our purposes we can say the following, so we are number 10, okay. So there exists F tilde and F prime such that the modulus of F tilde of F prime, so the derivative of F tilde at C is greater or equal to F prime of C for any F, so there is this maximum, yes. Now I want to prove that the following facts are true, so F tilde, well I don't know anything about this F tilde, okay. I can only say that well this F tilde exists once again, okay. But I can show you and surprisingly that this is true, that the function which is extremal for this problem, take the maximum of the modulus of the derivative at C, guarantees that this is true and how can you prove it? Well it is quite simple, we use what, maybe strong formation again. Assume that this is not, all right. And take this new function which is the composition of F tilde with a suitable Mabius transformation, right. So if you want a transformation I considering is this. The Mabius transformation I consider is, okay. Here is a bar, right, it's a conjugation. So F tilde at C is not 0, this is a Mabius transformation different from the identity. If it is 0 it is the identity. You said this, this vanishes, this vanishes as well, and so, okay. But then I have to have that, when I have to calculate the derivative of F star, well first let me observe that this function here still belongs to F prime, why? Well it is allomorphic in the simply connected domain D, takes values into the closed unit disk, why? Because we are then applying a Mabius transformation maps the unit disk into itself and the boundary into itself, so no problem. It is injective, yes it is, because it is a composition of a function which belongs to F prime with a Mabius transformation. Once again the Mabius transformation is injective, so you are composing injective functions. And what do we have? Well we have that this is the extremal function for this inequality. So in particular it is true that this is satisfied, right. So this is for any F, in particular it is greater than the modulus of F prime, of F not prime, which is one of the functions in F prime, all right. So the F star I'm considering here is should be in F prime. But what do we have? Well calculate, when we calculate the derivative of F star at Xe we obtain, well what? I have tilde prime of Xe, right, times 1 minus modulus of F of Xe squared, correct. And then 1 minus modulus of F of Xe squared, squared and then minus what? The derivative of the denominator times the numerator, right. But when it is evaluated at Xe it is 0, right. So this, cancel this, are you with me? Good. And what I obtained from this very elementary calculation, we are applying the Leibniz rule, right. We obtained that the modulus of this F star I constructed of the modulus, sorry of the derivative of F star at Xe is modulus of F tilde prime over 1 minus modulus of F tilde Xe squared, which is, since this is smaller or equal to 1, this is greater or equal to modulus. And this is a contradiction because F star would be a function in F prime, whose derivative at Xe is modulus greater or equal of the function which realizes the maximum of the modulus, right. So this is a contradiction. So it turns out that when you have the extremar problem solved, automatically you also know that Xe is mapped by the extremar map into the origin. Something mysterious, okay. In geometry this is the case, okay. Good. Now what is left to prove? We have an injected function from the unit disc, sorry again, from the simplified domain D into the closure of the unit disc, right. F wave proof and it is injected. F wave proof that the function is also onto a unit disc, we are done. We have a biomorphan from D, the simplified domain, not the plane, entire plane into the complex unit disc, right. So what is left to prove is, well that F tilde here is also onto, which means that if I equivalently or for any alpha in D, unit discs or modulus of alpha is smaller than 1, that exists as Z in D such that F tilde of Z is alpha. And I can say that this is unique because of injectivity, right. Well once again, assume not such as Z in D and then we construct another class, another pair of functions and we show that we get a contradiction. Using the fact that F tilde is the extremar function from the problem of modulus of F prime, of the derivative of F prime is maximum, solve this. So the first function I want to consider is the following. So F by contradiction, F of Z, F tilde of Z is not alpha for any Z in D, take this function to be alpha minus F tilde of Z over 1 minus, sorry 1 plus F tilde of Z, right, 1 plus so on. So it is a mebius transformation up to a sign, right, composed to F tilde, composed to a mebius transformation then the square root of this. And this function is well defined again in single valid, why? Because you see this by the hypothesis, why we are taking this as an hypothesis, so this never vanish, right. So there is no Z in D such that phi of Z is 0. So there are no branch points and this never vanish as well because alpha has modulus more than 1. So it cannot be that this number is 0 as well. So this function phi is well defined, so it is well defined, holomorphic, and furthermore we have that phi of Z has modulus more than equal to 1 for any Z. And this is because once again we are considering F tilde to be a function from D into the unit disk, then we compose with a mebius transformation, take this square root so, okay, good. And phi is injective, right. While this can be proven directly because if you assume that phi of Z 1 which is alpha minus F tilde Z 1 over 1 plus alpha bar F tilde Z 1 is equal to phi of Z 2 alpha minus F tilde of Z 2 over 1 plus alpha bar F tilde of Z 2, then again taking the square you obtain an equality of mebius composed with the function F tilde and F tilde is injective. So that this is equivalent to alpha minus F tilde Z 1 over 1 minus 1 plus alpha bar F tilde of Z 1, F tilde of Z 2. While this is mebius composed to F at Z 1, same mebius composed to F tilde at Z 2, mebius and F tilde are both injective, then we conclude that this is possible only when Z 1 is equal to Z 2. Now starting on phi we construct another function, this is just somehow technical but we require, so we compose once again with a mebius transformation. So we take phi composed with another mebius transformation, we obtain psi. So psi has a holomorphic well defined, here there are no problems, it can vanish, there are no roots. What is important that this denominator never vanishes, in fact never vanishes because this number here is in the unit disk and it is injective because it is the composition of mebius and phi, phi is injective as I showed you. So psi is for free say in some sense injective, it is for free and I also have that since this is true then psi of Z is also such that modulus of psi of Z is smaller or equal to 1. Right? This is also obvious, this is a composition with a notomorphism of the unit disk with a function which takes its values into the unit disk. So it just a rotation into the unit disk or something or good transformation. So psi is eventually in F prime, it is injective, holomorphic well defined and maps the disk sorry the domain D into the closure of the unit disk. So it is in what we have now that if we calculate the derivative of psi at Xe and take the modulus and we discover that by the assumption we have that is that F tilde of Z is not alpha for any Z, then you obtain the P C which is defined using this hypothesis in several steps right here, here right we apply this to have a good definition. We obtain a function which has derivative at Xe whose modulus is greater than the modulus of the derivative of tilde Xe and this is a contradiction. Well how to calculate? Well the derivative of Xe is once again F prime Xe times 1 minus modulus of V Xe squared and then I have minus and then it is plus right plus V tilde Xe times F prime of Xe times this evaluated at Xe which is 0 as we did before. So the numerator going calculated at Xe is 0 over 1 minus phi Xe squared. So once again this counts as this, this is C prime Xe. Now what is this? Okay I recall and this new slide sorry I recall this new slide that we have these two functions P of Z is square root of alpha minus F tilde of Z 1 plus alpha bar F tilde of Z right and P C of Z is P of Z minus phi Xe over 1 minus phi of Z bar phi of Z. These are the two functions we are considering that in order to define psi I need phi. In order to define phi I have to know that F tilde of Z is never alpha okay with this hypothesis I can construct these two functions and they turn out to be good same objective holomorphic well defined and mapping the domain D into the closure of the unit disc and well C prime of Xe is F prime of Xe over 1 I calculated this minus square right that's what I calculated. So this is just what I have so far. Now let us evaluate F of Xe onto this is F Xe right. F of Xe is well square root of alpha minus well F tilde of Xe is 0 that's what we have proved okay just before considering the function of phi and psi. So well F of Xe is square root of alpha that's it. F prime of Xe is well F prime of Xe is 1 over 2 square root of phi of Xe right times what times the derivative of this right evaluated Xe which is minus F tilde prime of Xe right 1 plus modulus of alpha squared because no sorry I'm sorry I'm sorry this has to be calculated. So the derivative of F prime is the using the chain rule the derivative of a compositional function right so 1 over 2 the function evaluated Xe which is correct then I have to evaluate the derivative of this function under the symbol of the root right which is the derivative of the numerator at Xe minus because it is a minus in front times the denominator evaluated Xe but F tilde of Xe is 0 so it is 1 right and then I have the derivative of the denominator with a minus in front so minus and then I have alpha bar F prime F tilde prime of Xe times this evaluated Xe which is only alpha because F of Xe is 0 so if I'm not mistaken that much this is 1 over 2 square root of alpha times square root of sorry over 1 over I'm sorry well maybe and then I have here 1 minus well 1 minus 1 plus this the minus in front and F tilde prime of Xe is it pardon me the minus is in front right minus minus well the minus comes from from the denominator right is it so minus alpha tilde times the derivative of F tilde Xe times the numerator which is alpha right and then F tilde of Xe is 0 so we have this so minus is this good so I'm wrong because this is not is it 1 over 2 F Xe right so sorry it's just a chain rule this calculates 0 right now I didn't want to make all the calculations it's very boring right in any case though the idea is that I have to calculate the derivative of us of a root of a composition of functions so I calculate this and then I have 1 over 2 the function here see evaluated Xe right so I have that this number is here given and then I calculated the derivative of this composition of F tilde and the maybes a suitable maybes which gives us this is it correct or not and probably what I forgot to consider here is something else right no it is correct this is just the denominator this is what this is just the numerator right of this expression because I want to show a contradiction by showing that F C prime of Xe has a modulus greater than F tilde prime of Xe so to complete this is this over 1 minus modulus of F of Xe square which is this that is to say this is let me say 1 over 2 square root of alpha then a minus in front of 1 plus modulus of alpha square times F tilde prime of Xe over 1 minus modulus of alpha there is there is something which is which I'm not convinced about there should be a plus here somewhere right I'm sorry I'm just it's probably here the mistake right here is that what you pointed out so with this so what I have after this calculation should be that F C prime of Xe is 1 over 2 square root of alpha 1 minus modulus of alpha times minus what minus 1 plus modulus of alpha square times F tilde prime of Xe and this can be simplified in this notation modulus of alpha minus 1 times modulus of alpha plus 1 over 1 minus modulus of alpha right so this cancel this you have a minus in front so C prime of Xe is minus 1 over 2 square root of alpha this cancel this right so 1 plus modulus of alpha times F tilde prime of Xe and believe me or not this cannot be because we have this right we can write this and this way is that take 1 minus 2 times this and then this right and this is greater well this is 2 modulus of alpha right that is to say that that the derivative of psi prime Xe is greater than the derivative of F tilde prime Xe and this is a contradiction so if you assume that a value alpha in the unit disk is not taken by the function of tilde you can construct a function at psi which has derivative at Xe in modules greater than F tilde which is the extremal function for this for this calculation so we have finally the following so F tilde 16 we have that F tilde must be on 2 and therefore F tilde from D D is pi holomorphism what we don't have up to now is that well this pi holomorph which exists because we have shown that it cannot it can be constructed in some way has a limit function but it can be obtained well that F tilde is not unique and this is because well remember that if we compose a mobius to F tilde we obtain again transformation from the disk to from the domain into the disk which is good with the good properties for instance if you take this function here well G tilde is a biohomomorphism and you can also observe that the derivative is the same right in modules at Xe so there is no contradiction the fact that there are some extremal functions not only one so it can be uniquely determined if you assume that for instance that the derivative well what do you what do you what is the effect of multiplying ae to i theta 2 to F it's sensitive to rotate right the values and then the derivative is multiplied by preferred direction e i theta similarly so if you if you for instance say well I want not just one biomorphism but the biomorphism which maps Xe into 0 right so you decide that a certain point Xe is mapping to 0 and then the derivative Xe of the function is for instance positive in the sense that this real and positive means that it has to be oriented in such a way that this product it turns out to be a real number then you have just one possibility of choosing theta there is no way to avoid this so the general statement is so this is the Riemann mapping theorem given a domain the simply connected in C and not C there exist a bihologomorphism furthermore bihologomorphism is unique F we assume that we assume that the value a certain value Xe is mapping to 0 and the derivative of the biomorphic Xe is positive so it is a real number and positive this depends essentially on the fact that you have the freedom of choosing a composition with an automorphism the automorphism of the unidisk act acts transitively so you can move it to the origin but then you have also the freedom of choosing a rotation and if you require that the derivative at C is a positive real number then you fix the direction of the rotation and you have only one possibility not the freedom of choosing whatever you want as a composition as I said this can be an instructive exercise to collect the ideas of several theorem statements we have encountered in the course but it's not constructive in the sense that you don't have a function you know that the function exists but in several cases okay this is enough just to know that you can well for instance if I give you a very odd simply connected domain whatever I like so for instance well okay this is the contour of a simply connected domain in the plane I have no idea of what are the functions which are holomorphic in this domain to itself but I can say well I can perfectly describe the automorphism well there is a function which maps everything into the unidisk call it raw okay by Riemann okay raw I compose and then I come back take w here move it here take a maybe transformation which I know characterize a maybe transformation is it's an automorphism any automorphism of the unidisk is in fact a maybe transformation then come back and I define a function which is an automorphism which is invertible in this odd domain D using this raw so in some sense this conjugates everything to the disk and that's why I said that the disk is the model which has to be started instead of considering all other simply connected domains but in some cases other models are much better and in some cases you can explicitly have the Riemann function so the the biomorphism is also called the Riemann function because of the Riemann mapping theorem so this is in fact an instructive example and it is related to one of the exercises I gave you for the for the next assignment and it is the following assume that you consider as a domain D the half plane this half plane the right half plane as it is normally considered so this is a simply connected domain right and it is not a plane it is unbounded but doesn't really mean anything because you just want to know that it is not the plane and well I invite you to verify that this function here which is holomorphic and D as such that oh sorry I put the Q but I wanted to use the Z right that's why because I'm I'm working with material sometimes sorry Z okay this is holomorphic in D well in fact it is well defined in D D is the open unit disk so the only problem is that this function is in fact a meromorphic function in the plane and has a pole of order one at one but one is on the boundary of the unit this so it is well defined and well we can also write the power expansion right because modulus of Z is more than one right so it is holomorphic that that shows you that the function is definitely holomorphic well defined and it is also injective and this is really seen as I take 1 plus C over 1 minus Z 1 is equal to 1 plus Z 2 over 1 minus Z 2 right and this is f and all f Z 1 is equal to Z 2 so it is injective it is a holomorphic and in the last few minutes let me just say that if you prove this right but this number is positive as you can easily see in so this is one half then I multiply and then I have 1 plus Z minus module Z square minus Z bar then I have plus 1 plus Z bar minus Z and then I have minus module of Z square right and then I have 1 minus Z module square so Z cancel this Z bar cancel this Z bar and then have 2 minus 2 modules of Z square right over 2 so it is 1 minus module of Z square over which is positive is Z in the unit disk so what we have is that this function here maps the unit disk holomorphically into the half plane what you are asked to prove and that that this part of the exercise probably you have the upper half plane right inside of the right half plane which is just a matter of rotating half planes well what is not difficult to see is that well you can always find a Z which solves this equation when W belongs to H plus that is to say the function is also subjective it's on 2 so it is the bilomorphic this is one of the few cases this function is called the Cayley transformation and so the half plane is another model of the unit disk and it has been heavily studied for several reasons so let me just give you one motivation we have studied the automorphism unit disk and as I said from this we can in principle obtain the information of the automorphism of any simply connected or made different from the plane because when you use this conjugation and you study everything there in this case we have an explicit reman this is one of the few cases an explicit and reasonable reman mapping all right and the half plane we have considered so the H plus here we can see quite easily what are the automorphies without fixed points for instance if you remember when counting the fixed points of the automorphism of the unit disk we have shown that either the fixed points are one inside and one outside or both are on the boundary they can coincide but can be also different right and the unit disk this is not easy to see it is easy to see when it has one fixed point it's essentially you have the subgroup which is the isotropic group of the origin right the rotations this is the model right you move the fixed point to the origin and say okay this is nothing but well by Schwartz lemma e i theta times z right rotation good but if I ask you well can you please show me an explicit example of an automorphism without fixed points and with only one fixed point coincident with the other at the boundary this is called the parabolic case well this is somehow hard or if you ask you well no well let us start with the easy case two fixed points in the boundary well you have to somehow struggle a little bit to write a function which maps the unit disk into itself it is injected with two points which are fixed on the boundary here it is much easier and I show why well you take z well take w okay in the half plane in the right half plane and map it into lambda w with lambda a real number well take it different from zero well this is an automorphism as you can see right you can invert it and you're still in the half plane right it maps the real axing into itself zero into zero but it is this is the boundary you have to imagine that the boundary of the the unit disk is now the vertical line the vertical y axis including infinite which is associated with point one right and this correspondence so zero is a point on the boundary the other point which is mapping to itself is the point of infinity exactly and you can see that if lambda is greater than one then one point is how do you say attractive and the other is repelling take the well you study this in dynamics okay one well if lambda is greater than one you take the orbits of any point points tend to go to infinity if lambda is smaller than one well of course the opposite tends to go to the origin if lambda is equal to one well this is the identity everything is fixed okay no dynamics nothing interesting another example is the following take w plus i which means that you start from a point and you move along the vertical line this is this is nothing but a translation right in the plane in the half plane it is a an automorphism of course you can take a two i or something like that with an imaginary translation contribution on the translation only imaginary so that this function maps the half plane into itself it is invertible as you can easily see well this function has no fixed points on the boundary except the point of infinity which is the only fixed point in this case and this can be another so another surrogate model in case you don't like the unit disk all it's difficult here is to describe the case with a fixed point so there is a balance so if you want to say the fixed point case automorphism description is much better and in the unit disk whereas and so uh oh i run out of my time so so the theorem tells you that you can use this or half plane if you like more as the model for any simply connected domain different from the domain plane but for the for the plane we have the description of automorphism so in some sense we cover all simply connected cases right and that's all for today thank you