 Today, I want to start with some results about limit functions of analytic functions. And the first result I want to present to you is the following, so take this result as Weierstrass theorem. Assume that you have a sequence of holomorphic functions defined on a domain omega, sorry, it is an open and connected subset of C, okay. And you want to investigate what are the properties of the limit and which limit will be also one of the question to use, okay, in order to have some properties. So as we know from, say, general calculus, the uniform limit is the limit to be used in order to have good properties, okay. So for instance, if you at least want to have the limit, the function f to be continuous, then it is natural to ask for uniform limit, right. So fn is holomorphic for nn, so in particular this function is also continuous, so if we are taking the uniform limit f, the limit function turns out to be continuous. And this is the minimal request. For the class of holomorphic or analytic complex analytic function, we better consider uniform limit on compact sets, okay. So we say that this convergence is taken uniformly on compact sets of omega, right. So you take a compact set and when you stick it to this compact set, you take the uniform limit, right. And well, this can be considered a bit artificial, but well uniform limit is somehow you cannot avoid. And on compact set you will see immediately why we have to consider compact sets, okay, subsets of omega, right. So I want to prove the following. So if the limit is taken this way, so we are somehow introducing a way of considering, say, a topology on these sets of functions, okay. So if this is the topology used then f is holomorphic. This is already a good result, so which means that with this topology this set is closed, okay. So the holomorphic function is closed. So the limit function is in fact again holomorphic function. How can we prove this? Well, it is in fact simple if you know the precise ingredient you need. The precise ingredient is an application of Morera theorem, remember. Morera theorem tells us that whenever you take a function which is continuous and such that the integral over any integral is zero unless it's only it is holomorphic. We have seen this in, so somehow hidden also in Gursa theorem. So we prove that this for the rectangle, but in a sense when you have this property you can always find the potential, right. And the potential turns out to be analytic and then you get that the derivative of an analytic function is in fact analytic. So take a triangle T and omega. T is a triangle, so it is a complex set, right. In other words you can always consider a neighborhood of a triangle T well contained in omega. Well contained means that well this open set is contained together with closure, right. Now what do we know about the integral, well sorry I forgot to say this. Since F is the uniform limit on triangles, okay, but on complex subsets, but in particular it is uniform, right. Then we know that the limit is as I already repeated several times it is continuous. So F for sure is continuous and in order to prove that F is holomorphic we just apply Morera theorem whenever we prove that given this we obtained that this integral is in fact zero. If we prove that this is true we are done, right. We have a continuous functions we want to evaluate the integral over any triangle and if we prove that this integral is in fact zero the value of the integral is zero then necessarily F is holomorphic because of Morera theorem. This would imply that F is holomorphic, this is by an application of Morera theorem. So the only point we have to prove is this, okay. We proceed as follows. Well remember that we have this and I put U in front of the limit to remember that we are taking the uniform limit on compact sets, all right. So Fn is holomorphic and F turns out to be the uniform limit on compact sets of these functions, right. Therefore if I calculate this is as calculating this, right. This is just substitution, okay because this is the uniform limit but one of the technical limits I show you is that well you can take out the symbol of limit from the integral whenever you have a uniform limit, right. This was already applied once and I showed you the reason why you can do this. So this is in fact the uniform limit, well it is the limit by the way. And since each of this function Fn is holomorphic then the integral over any triangle is zero because it is a closed path, right. So we have that this is the limit of a constant zero and then it is zero. So it's quite an elegant proof, right. Put it together. So now otherwise you have to struggle a lot to prove this kind of limits, limit properties. And alternative proof is as follows. Well we can also use the fact that Fn is a has a representation in terms of an integral. It has a Cauchy representation, right. For Fn the Cauchy formula can be used and this is in fact using as I said instead of a triangle a small disc contained and well contained in omega. So with this closure so we consider the value of Z to be calculated over taking the integral over C of F of C, C minus Z, right. C is a circle in omega. This is again another compact set. So we take Z and we take a small neighborhood of Z and in such a way that it is contained together with boundary. So it is well contained. So we take then the integral. Sorry, I forgot to put an N here, right. This is valid for any N. For any N you have an integral representation of the function Fn at Z. Then you take again the uniform limit here is F of Z and what this here is a uniform limit. This arrow this vertical arrow is a uniform limit but since we are applying on the right hand side a uniform limit then we can say well this is which tells us that the two on the right hand side on the left hand side we have an equality, right. This tends to F the integral of F of C and then since F is continuous remember that when we when we used the fact that from the integral representation formula then we obtained that the function is analytic which shows something more general. It suffices to have a continuous function on a path, right. And what you obtain from here is an analytic function. So it's a kind of magic box whenever you put inside a continuous function over the curve of course you are integrating. This the result is an analytic function. It is analytic, right. So in this case we are applying two facts. This is the limit of continuous function in particular so uniform limit so it is continuous. Here we have that this uniform limit pass the symbol of integral and then you obtain that this integral here is this but then the function which is continuous in this integration guarantees that output is in fact complex analytic. So again it is allomorphic. This is just another proof which also leads to something else we can say about the convergence of uniform convergence of classes of allomorphic functions. In fact it is also true that sorry for this three well we also know from the Cauchy integral formula that the case derivative of a function of which is allomorphic so take for instance the element of the sequence can be expressed using once again the Cauchy integral representation, right. It is what this is k plus 1, right. And maybe there is a k factorial here, right in front probably but this is not important I mean. So this can be done for any k, right. k means case derivative. So on the right hand side you have only something different from the previous expression in the denominator nothing here, right. So you can express the case derivative changing the degree of the denominator in this ratio but the function here is still continuous, right. So take the kth derivative as a function, right. And well take the uniform limit again, okay. Here we have I don't know what we have but on the right hand side I have this is k factorial over 2 pi i f of xc, xc minus z, k plus 1, xc which is fql, f koz. So what we also have as a side effect result if you want in this second approach is that not only the uniform limit on compact set is a holomore phi but each derivative of the, so if you consider the sequences of case derivative of the sequence considered, so we have infinitely many sequences from the sequence, right. Take the kth derivative of each function and consider this as a sequence. Well this converges uniformly to what? To the kth derivative of the function of the limit function. So this is very important, okay. It's a very good result and this is known as bias result. Okay. Another application is well for instance it's a corollary. You have that, assume that f, like they take this notation, left f, yeah as I said, fn, fn is a sequence of holomorphic functions. And now consider this n equal to 1 to k, right. Assume this converges uniformly compact sets of j to something, g of z. So you can take also the sum, right. This is another sequence, the finite sum and the limit, assume that the limit is uniform on compact set. So then then we also have that the kth derivative of j is the uniform limit of the sums of the kth derivative of fn, okay. So it's very, very precise. So then the kth derivative of g is in fact the sum and this sum I'm taking, okay. Well this is the uniform limit, right. If you want because it is a limit inside in this, in this, what do you mean? So what I'm taking here is as a sequence, consider s of k, right. And I take the s of k converges to uniform to g of z, n is from 1 to k. Yes, exactly. Well what I'm writing here is I have to say this way, right. If you want, the uniform limit is okay. This is maybe it's not as precise as I want. This is a uniform limit as k tends to infinity of summation, oh right, k is not precise, okay. I understand, I understand. Not k, maybe I have to use some extra letter, okay. So from k to, from m equal to 1 to m, right. Sorry, sorry for this, okay. So the kth derivative of fn, yes, sorry. Then yeah, double index. Wow, this is another good. So let me continue by telling you that in fact from this theorem, we have many, many other important facts. To take for instance this fact, to take fn, a sequence, well define g, this is holomorphic. This result is known as Hurwitz theorem, right. Assume that fn tends to f uniformly on compact sets of g. So then we know that the f is in compact holomorphic, right, from the previous theorem. Assume that fn of z is not 0 for any n and for n is z and g. So assume that all these functions here in the sequence do not vanish on g. Then either f is identically 0 in g or f of z is not 0 for any z in g. And the statement here is known as Hurwitz theorem. How can we prove this? Well it's not difficult. It's just, and again, a proper application of standard tools we have so far developed. In particular, when we're talking about zeros, we have the argument principle, right, which tells us, we are dealing with holomorphic functions. So no poles can appear, right. So we have the argument principle which tells us something, right, about zeros. So as a proof, let me briefly recall you that, well, assume that f is not identically 0, right. First, if f is identically 0, there is nothing to say. I mean, well it is one of the possibilities. If f is not identically 0, then f is analytic, is holomorphic, right. So it eventually vanishes in a discrete set, right, of points, and points are isolated, right. So the zero set of f, the zeros of f are isolated. Since f is analytic, right, cannot accumulate. Otherwise, it's constantly zero, but we are assuming that f is not constantly zero. Now, good. Now, take z naught zero of f and g. If f of z naught is equal to zero, then for sure, f of z is different from zero in neighborhood, right. That is because the zeros of f are isolated, right. So we have this situation. This is possible zero and in a neighborhood, the function f is not vanishing. Now, consider a smaller c of rho to be the set of z such that z minus z naught is equal to rho smaller than r. So with the radius rho smaller than r. So this is a compact set. We are taking just the circle, right. And therefore, I have that the continuous function modulus of f has a minimum on c, rho, and this minimum, call it m, is positive. Because we know that the value of the function f is never zero inside of this puncture disk. In particular, when we restrict our consideration to c rho, we are considering something which is outside of z naught. So this minimum exists and it is positive. Good. Now, on this set which is compact once again, I can certainly consider, this is c rho, this is not a disk. It is a circle. On here, I can also consider one over f of z because z is n of c rho. I am sure that this number here is well defined for the function, no problem. And this can be written as the limit, uniform limit over c r of one over f n of z. Correct? This can be done because we are dealing with uniform limit over a compact set. And f n is a sequence of allomorphy functions which are not vanishing. So no problem. And we also seen that for any derivative we have that f prime of z is in fact a u limit of f prime n of z, right. We have seen this for any derivative in particular for the first derivative. So now it is clear what I am looking for. I am looking for what? For this f prime of z over f of z to be as it is expected the limit of f prime n over f n uniform, right, on this compact set. And then I can also say well the integral over c rho is minus, why? Because the limit is uniform, right. But this number is 0 because the function is not vanishing in entire, right. This is the argument principle applied to f n, to each f n. It tells you that this number on the right hand side is 0. Are you with me? So it means that there is no 0 inside the circle. So in particular this is not 0. So sorry, this is sorry, this is 0. And so f cannot vanish. So this consideration can be repeated for any point so the function cannot vanish. So actually what we have here is something more that well the number of zeros are the same. Say for a certain point, okay, from n greater to n naught. If you assume this then you can say that the number of zeros of this function is the same as the functions, the number of zeros of f n for n greater than or not. This is somehow more specific result. If you're not assuming that f n is a sequence of vanishing or not vanishing holomorphic function then what you discover from here is the number of zero of the limit function is the same of infinitely many elements of the sequence in the same way. This is another possible extension of who it's feeling. Good. Now as I said, if you're dealing with classes of, with sets of functions and you put a topology on it, it's not true to ask whether this this set turns out to be or not closed. And now we have shown that using the the uniform uniform convergence then we are certainly dealing with something which is then closed, okay. So the limit function is again holomorphic. And now the next step is to consider properties in order to be sure that from a class, from a sequence you can extract a subsequence which is convergent uniformly on the set. So this is the notion which we'll introduce now. In fact we say that the family, this is the definition, a family f of holomorphic function, okay. This is a family. So each function is defined in omega and it takes value in to see is normal given any sequence of holomorphic functions f. So you take a sequence from the, from the, from the family f there exist a subsequence which converges uniformly sets of omega to a holomorphic function. So the idea is that while we want to be sure that while this is in fact a condition which reminds compactness, right. Have a sequence up to a subsequence you have a convergence sequence, right. Well this is just a definition and as it often occurs in, in mathematics well definition is not well it can be enlightening but not useful somehow, okay. Because this, this is what we want, but we need something, some equivalent or some other conditions to be sure that this condition is, is fulfilled, right. And in fact we, we give an also this definition here seven say that a function, sorry a family of holomorphic functions is locally bounded. So all the functions are defined in omega and, and take values are in to see, right. It's locally bounded if for any a in omega there exists an m and r positive such that for any f in f we have that f of z is smaller than m if z minus a is smaller than r, right. Locally bounded, okay. This is definition. So now the result which gives an equivalent condition for normality is famous result and complex analysis due to Montel, okay. We'll just give a sketch of the proof and in fact it, it depends on a similar result in analysis which is known as Ascoli-Alzela theorem. Have you ever heard of Ascoli-Alzela theorem? Well it's a diagonal process you take, this is normally studied in, in functional analysis, okay. Okay so just, just mention what I, I want to say is that f is normal if and only if f is locally, locally bounded. Well to see one fact it is simple and the other saying particularly so assume that f is normal and if f were not locally bounded means that exist assume that f is normal and f not locally bounded. So the, the proof goes by contradiction which means that exists a complex set k in omega k not empty such that we have that the soup z in k and f in f is plus infinity. It's not locally bounded so it cannot be bounded, right. So in particular, in, in a, in particular it is possible to consider a sequence f n in f such that when I consider the, this is max actually as the belongs to k is a, k is a compact this is greater or equal to m, right. So I, I know that this supremum is plus, no I'm assuming by contradiction that the, the f is not locally bounded so this property is not true so that I have that this supremum is plus infinity that in particular I take a sequence of, of functions we have the property of, of determining this, this soup as plus infinity so in particular I assume this, right. So, but we are assuming that f is normal, f is normal so out from f n I consider f and k a subsequence, okay because it is normal so from a sequence I extract a subsequence which converges uniform converges on compact sets, right. Therefore, I have that the soup of f and k z minus f of z is smaller than epsilon, right. We are taking a compact set over this compact set we have this because the convergence is uniform on compact set, right. And on the other hand since f and k was extracted from f n, f and k is smaller equal to sorry soup k z as z is in capital K, right. Whereas on the other hand f is a uniform limit of continuous limits so f is continuous and therefore this number here is bounded when z is the compact set, okay. That is where the contradiction which follows from this so nk is smaller or equal to f modules of f and k of z, right. Smaller or equal to k of f of z plus well this is f of z, right. Take the soup, soup then this is smaller than epsilon plus m, right. Because we split and divide this, right. This number is made can be made smaller than epsilon because of this and this number can be made smaller than m because of this. So we have that this and bounded and k which goes to infinity is actually bounded by a real number it's a contradiction, right. So normality guarantees that the class f the family f is locally bounded. Well on the other hand if you consider f as I said this is a sketch of the proof not precise proof, right. f locally bounded is to say each f is each f each function f is locally not as modules which is bounded, okay. Essentially, good. So from locally bounded and you use Cauchy estimates and this proves that this is smaller or equal to a constant times z minus z naught. Whatever f you take in f you have this, okay. Why how can I obtain this? Well look f of z is 1 over 2 pi i integral of dixie. f of z naught is 1 over 2 pi i that put c right constant f of c c minus z naught dixie. So that difference f of z sorry minus f of z naught is look 1 over 2 pi i then I put in the same integral numerator is the same the denominator okay is well c minus z naught then minus c plus c right is it. So the z minus naught comes from this and is independent from c so it can be taken out from the integral and this is the contribution from the other hand this number here is bounded and this number here since we are integrating can be also controlled. So by the Cauchy estimates we have this which tells us the following so for any f f of z minus f of z naught is smaller equal to one constant z minus in modules of z minus you know so that f is functions are equicontinuous right equicontinuity is the key ingredient with the locally boundness bounded and equicontinuous using ascoliola guarantees that for any from any sequence of functions then you extract a subsequence which is convergent uniform and complex set this is ascoliola hypothesis fulfilled okay so I'm not saying that this is a complete proof but then it's what we can say up to now okay so then from here so f equicontinuous plus f locally bounded then using ascoliola guarantees normality as I said in the ascoliola theorem the process you're doing is something a diagonal process because any complex set can be in fact covered by a finite number of this you repeat because this consideration to and say you approach so you in some sense use approximation stuff and you repeat this this consideration for any complex set and so on and so forth okay so and this is for analytic functions what you will need so in some sense normality can be replaced by locally bounded property this is what we need right so the next task is to somehow have the similar description for the class of meromorphic functions because we all consider what we consider today is something related exclusively for analytic complex analytic functions right so for instance Holby's theorem required a function to be followed by otherwise the argument principle wouldn't so the fact that this is zero in the in the case of a meromorphic function the the integral of f prime over f it doesn't guarantee that is no zero at all we can have a zero and the poles of the the values of the integral cancel right so for the meromorphic functions we will not enter the details of the proofs as we start from now okay started to to to be a bit more sketchy and I collected my my my what we need for our result in these slides which I'm going to show you if I if I can okay so some remarks on meromorphic functions so the class of meromorphic function on a on a domain g will be denoted by mg is it okay for you because on the on the on the screen the okay well you will receive it but in any case you can read it okay mg so is the set of meromorphic functions so function with poles only poles are allowed as singularities right functions which are meromorphic can be extended continuously as functions from the domain g to the Riemann sphere by putting the value at the at the pole to be infinite right so in this way in this way can you can you read it so if f is meromorphic and you add the value infinite to hz which is a pole then the the extending function turns out to be in fact continuous right but on c hat said on the Riemann sphere we introduced the distance of two points right and this is what you've done in one of the previous lessons right since we since we are dealing in the in topology uh consideration for function from g into the Riemann sphere we have to deal with this with this distance so this is the distance of two points and then we also have the similar expression for this from infinity remember right so there is something which changes the the shape of balls from the euclidean topology right we have a denominator which appears right however if we take this distance which is the distance we introduce using the stereographic projection and it works fine the class of meromorphic function is not complete because you take the sequence of functions which are constant okay so this is a z set of x okay anyway uh this this sequence of constant function whose values are respectively n okay or any n which is not to be cosci but the convergence is the convert so the the limit function is the function which is constantly equal to infinite which strictly speaking is not meromorphic because it would would mean that the poles are the entire set we said that the poles like the zeros are isolated but well with the some artificial uh considered well we add the the function which is constantly equal to infinite as a possible say extremal case for meromorphic function and this way we have that the space is complete which is important to now the first important result is the following which extends in some sense to meromorphic case the to the meromorphic setting what we had for analytic functions of the bison theorem so take a sequence of meromorphic function and take the uniform limit on compact set so in principle you can expect to have something very worse as a final result so you know but the good result is this that either the limit function is meromorphic itself or it is the function which is constantly equal to infinite right assume so i give you just the idea of the proof assume that the function is not identically equal to infinite right the limit function all right and take a value which is not so take a z which is not a pole so if you are far away if you are not considering a pole the value of the function is bounded right and this is the same in a neighborhood of the point right so if you stay away from the from the poles this can be considered the restriction can be considered as a meromorph as a meromorphic actually as analytic function and you repeat everything whereas when the value is infinite right and the function is not identically equal to infinite then you take the puncture disc around the pole and you repeat the consideration of bias of the theorem and everything works fine because well the function like for zero so the the poles are isolated and what is outside say in a neighborhood is something where you can make the consideration as an analytic case okay so that you have that the sequence of meromorphic function convert to a meromorphic function or to the function which has all the values equal to infinite now speaking of normality for meromorphic function requires to introduce this somehow index okay to take mu f of z to be two f prime modules of frame z plus over one plus modules of f z squared it's just a function mu f all right from g into r so there are infinitely many functions like this real value functions because on the right side you have a okay and of course when the the the value is taken for a pole you have to take the limit okay so notice that this mu depends on f and then z you take f in the class of meromorphic function calculate this for any z right what can be proven is not difficult the mu f is a continuous function and well just a matter of calculation to prove that mu f is the same as mu or one over f remember that if f is allomorphic one over f is not necessarily allomorphic but if f is meromorphic one over f is meromorphic right so this is just because of calculations and this this i have to say this new function introduce associated to the class of meromorphic function is important because well this is the definition for normality and that's all definitions that they quite have also equivalent conditions so we take a sequence in in a family of meromorphic function and if there exists a subsequence which uniformly converges on complex set to either a meromorphic function or to the function infinity so this is an extended respect case normality in the analytic case then necessarily the function the family of function is called normal and the equivalent condition we have an equivalent condition okay for normality for a class of meromorphic function is given by proving a locally boundness of mu f okay so family of meromorphic function meromorphic there are some mistakes sorry misprints meromorphic functions is normal if and only if mu f and mu f means the set of mu the functions okay mu f from g into r as f varies in the family of meromorphic function turns out to be locally bounded this is not impossible to prove it's not you know so it's beyond the interest of the course okay just to know that there is a similar result so locally boundness of the functions guarantees normality of families of holomorphic functions and local boundness of something related to families of meromorphic function guarantees normality for meromorphic and where the two notions are somehow extend because meromorphic functions are form a larger class of function okay and let me also put here what we also have for meromorphic this room the theorem is quite an important one assume that g is an open set take e a set which has this problem is in the complement of g in the Riemann sphere and such that its closure intersects it oh sorry its closure intersects each connected component of this complement in the Riemann sphere then given any holomorphic function f in g and any compact set k well you can approximate this holomorphic function f with a rational function with a meromorphic function whose poles are in e okay this is the room the theorem so any holomorphic function can be approximated by a rational function over complex set so and in some sense you also can control the poles the poles can be okay only in the in the in the complement of the set g in particular and this is the theorem you probably already know assume that g is simply connected so the complement is connected right there is only one connective component of the complement in the Riemann sphere this is one of the characterization of simply connectedness we can assume to take e to be infinite so the function which have which are meromorphic and poles are only in at infinity are in fact polynomials right so that putting together the two results we know is that in fact a function which is holomorphic in in a simply connected domain can be approximated uniformly on compact set by polynomials this is probably an extension to holomorphic setting of what is known as stone bias plus theorem or something like this right right so if you ever follow so it depends the geometry of the domain as you can see here forces the function to be approximated or not by a class of polynomials which are of course also can be considered as meromorphic function but in general if you don't know anything you have to say well rational functions or so portions of polynomial okay in general approximates okay and before continuing let me just come back to yes to the hypothesis of Runge theorem so let me come back so I show you that well you you have to consider e and the complement of g and the component in the Riemann sphere right and such that this cloud its closure intersects each connected component of c hat of the component of g right so this apparently seems to be quite weird as the condition though means but it cannot be relaxed this condition cannot and here is a counter example so take as g c minus the origin this is not simply connected right because the complement of g is as two points and assume that I want to relax a little bit so I just say well e is one over two points not the other so there are two connected components right two points separated points right so I take e to be well just infinite and I take the function f of z one over z which is okay holomorphic and g right right so assume that the well the theorem so the the corollary to runge theorem is still valid with this relaxed conditions on e is just intersecting one or two components or two components intersects only one right which means that one over z can be approximated right by polynomials but this number can be very smaller if z is in a complex set over complex set so take as a complex set this number here this this is a compact set compact in g yeah g is the puncture plane and this is an annulus right so what I'm saying is that if I apply a runge theorem I would have that one over z which is holomorphic right is approximated by this function sorry this is approximated is the limit of polynomials right so that I can make this distance smaller than one over n right or from this I obtained it one minus z p n z is smaller than modulus of z over n and this is true for z in k right I'm saying I approximate a function which is holomorphic by polynomials because I'm considering a rational function functions whose poles are at infinity so polynomials but this convergence is made on compact set so I take this compact set k in particular good now I have that when z well if z as modulus equal to to n so it is in k here so this number here is one one in particular because modulus of z is n I've taken this I have the freedom of choosing any compact set I'm doing this just to show you that we have a contradiction right so I taking this on purpose so that I have this inequality therefore I have that p and z the modulus of p and z can be written as one over n z p and z because modulus of z is n right and this is all right this is smaller or equal to one over n one minus well p sorry z p n z minus one and plus one this is the same sorry so that this is smaller equal to one over n plus one over n and this is smaller than one right because of the previous consideration so that this number is two over n let me show you again and the previous consideration okay this can be made smaller than one when z as modulus equal to n so in particular I'm choosing such a okay good I'm saying that well I repeat this holomorphic function in G can be approximated by polynomials so that this distance is smaller than one over n right I can make it as small as I want in any compact set so I take this specific compact set subset of G and show you that from this you obtain this inequality and from this inequality you then this so that p n of z is smaller equal to over n and this is contradiction so as I said the the hypothesis of intersection of the connected components by the closure of E cannot be relaxed quite surprisingly and well before finishing let me let me just tell you that there is another theorem I want to discuss with you the result I'm not going to tell you anything about the proof of this result but it is important to know that there is there is this result so from Runge theorem we know that we can approximate on compact set any holomorphic function with rational functions and we have also detailed the description of the of the position of the poles right and this is Runge on the other hand if you want if you start from singularities which are poles it's always the case that f a meromorphic function exists with these singularities prescribe singularities prescribe order prescribe principle parts the answer is yes and this is the theorem which gives you the description so take an open set u and c and the subset of point without limit points this is important otherwise the function is said I would be this function we are looking for would be would be not meromorphic in a strict sense so it would have infinitely many poles and accumulating at some point right so you take sk of z to be as a rational function okay and then the poles of s a go z are all concentrating ak right so mk represents the order of of the pole of s a f s k of z right so for each point in u in a sorry in this subset of points without limit points you consider sk of z and MetaClaffler proved that in fact well starting with this hypothesis there is a meromorphic function half in u so it is function with poles but whose poles are precisely the ak is prescribed and with principle parts sk sk of z so this solves the problem starting from a generic but so reasonable choice of points and prescribed to each point on order and the principle parts of the residue and all the other coefficients in the power in the Loran expansion right we have a finite tail with negative coefficients with negative indexes in the coefficients well what you can find is a global meromorphic function with prescribed poles and prescribed principle parts well just to complete there is one extra page here and this is what is known as Vitaly theorem and Vitaly theorem is not well known I think but tells you something about the possibility to conclude that sequence of holomorphic function converges uniformly on compact sets so you start from a sequence of holomorphic functions and g is a domain of c okay which is uniformly bounded on compact sets if a is any subset with the limit point such that at this at any z okay of a you have point wise convergence of the sequence so you have you show you make a choice of z in a and then you obtain a sequence of complex numbers associated to the sequence of values of the sequence of functions and if this is a sequence with which converges for any z in a then f n converges uniformly on compact sets well the proof and this is this concludes the today's lesson is as follows assume by contradiction so by contradiction take k in the u compact and I consider f n k cos e k minus f m k cos e k is greater or equal to delta for any k this is the compact k this is for any k in n okay and the z k is assume by contradiction that convergence is not uniformly on the that f n does not converge uniformly on compact so I can take f and k right z k and m such a well this is in the sequence such that this is the the possibility but however from the locally bounded property we know that from n and k we obtain we can extract a subsequence convergent because this is the application of montel theorem and from m and k we obtain another function g which is also all about by montel theorem and similarly since the z k is a sequence of points right in a compact set we can extract a subsequence which converges okay to z naught and this forms okay this is true and this z k is a in our consideration because it has a limit point so what we know is that okay f of n k z tends to f and n k of z naught right as the sorry tends to f of z naught right and f of m k z tends to g of z naught however because of this hypothesis right this would be the only possibility so I have g of z naught minus f of z naught and I write it g of z naught minus f m k of z plus f m k of z then plus minus f n k of z plus f n k of z then minus f of f of z naught so that modules of g of z naught minus f of z naught is is equal to g of z naught this is smaller equal right well this tends to 0 right and this tends to 0 as well right so that this is what you can see remains actually greater or equal to delta because of our assumption assumption contradiction so the two functions instead of agreeing have to be distance and this is a contradiction of our hypothesis so the two functions are in fact the same function okay and z naught so this Vitaly theorem is the last result we see about about a sequence of holomorphic functions and the meromorphic function and then we apply this for the next consideration okay so I think it does stop here yes