 this video will be about the Riemann mapping theorem. So what I'm going to do is state the theorem and then sketch the proof of it. So first of all state what it is. Suppose you take any complicated open subset D of the complex plane. The Riemann mapping theorem says that this is isomorphic to the open unit disk. So this is just the complex number z with z having absolute value less than 1. The only conditions are that D should be simply connected and open as I said. And next condition is that D should not be the whole of the complex plane. It fails for for this one case. So Riemann sort of gave an argument for it in his thesis in about 1851. However his argument was notoriously incomplete. It used something Riemann called the Dirichlet principle and it wasn't really clear under what circumstances this was true and it took about 50 years to sort out what was actually going on. So the first complete proofs of it were given about 50 or 60 years later by Osgood and Carotheodori and Coebi. So the theorem really ought to be named after them as well. Notice by the way that this open subset D can be pretty wild. So for example if I take an open square then I can remove bits of it looking like this. So I can just remove lines and you find that for the boundary of the square there's actually no arc connecting a point on this boundary to a point in here because it would have to oscillate up and down infinite distance. And moreover we can make it even weirder. So I can I can remove some sort of weird fractal like object so I can take some sort of infinitely branching fractal in some very complicated way. So this open set can really be extremely hairy and it's sort of amazing that you can always make an isomorphic to that. That means you should have a holomorphic function from F from D to the unit disk whose inverse is also holomorphic. Then we have this funny exception that D can't be the whole of C. This follows because of Leuville's theorem which says that there is no there any bounded map on the complex numbers must be constant. So if you've got a map from the whole complex numbers to the unit disk it must just be constant so it can't be an isomorphism. So now I'm going to sketch the main idea of the proof. So the proof comes in four steps. So the idea is we're going to look at maps from D to the open unit disk. Let's call this U. So here's going to be the map F and here we've got some funny set D. I'm going to choose a point in D so I'm going to choose a special point Z0 and I'm going to choose the point nought in the unit disk and I want the matter of the property that F of Z0 is 0. So it's going to take that point to that point and we're going to choose F so that it's holomorphic and injective and we want to maximize the derivative of F at Z0 and what we want to show is that there is we need to show there's a unique such function and it satisfies the conditions of Riemann's mapping theorem. So suppose let's call the space of all such maps F. So we need to show four things. First of all we need to show that F is non-empty rather obviously if there aren't any maps of if there aren't any injective maps at all we certainly can't find an isomorphism. Secondly we need to show that F prime of Z0 is bounded above and if it's bounded above then it's and and that there is at least one such F then these two maps together say that the values of F the derivative of F at Z0 have a supremum. Thirdly we want to show there exists sorry there exists F with F prime of Z0 maximal. This is a rather subtle point I mean we can find functions with F prime Z0 arbitrarily close to the maximum but it's not clear that we can really take a limit of them and this is actually something like this was the problem with Riemann's original proof he kind of assumed you could you could take a limit of functions and this is rather subtle problem sometimes you can and sometimes you can't and finally we've got to show that F is injective and surjective where where this F is the thing with the derivative as large as possible. So I'll go through these four steps first of all we want to find we need to find one F mapping D to the unit bore and we notice this actually fails for D the complex numbers as I mentioned before by Lieuville's theorem here I want F to be injective and holomorphic and this is obvious if the domain D is bounded because we can just translate it and multiply it by constant and that will put it inside the unit bore and so what we want to do is to reduce the case when D is bounded and because D is not equal to C so D omits some point and we may as well assume at this point is the origin because we can just translate it and D is simply connected and it omits zero and what this means is we can define the square root function on D so you know normally there's a bit of a problem defining the square root function because it is a branch point at the origin and so on but if we've got a simply connected region not containing zero then we can actually define the square root function and then we can check the image of D under the square root is not dense in the complex numbers notice that D might have been dense for instance it might be the complex numbers minus the minus the positive real axis or something like that but by by doing the square root trick we can make sure it's not dense so we can assume again if it's not dense we can translate and assume it omits a small disc with center at zero and then if we if we map D if we map it by taking z to one over z this will give us a bounded domain in the complex numbers because D omits a small disc around zero so we've reduced the bounded case so there's at least one such function f as I said this is the point at which we need to use the fact that D is not equal to the whole of the complex numbers next we're going to show that f prime c zero is bounded what this means is that no matter which injective f you take there's a universal upper bound for them and for this we recall the Schwartz lemma most difficult part of the Schwartz lemma is trying to remember that Schwartz doesn't have a t in it because there's a some Schwartz is with a t and some without it's kind of like Lorentz there are lots of physicists called Lorentz some with a t and some without a t anyway um the Schwartz lemma says that if you've got a map from the unit disc to the unit disc then f prime of zero has absolute value less than 1 and it's less than 1 unless um f is a rotation and we're going to use this later and this is quite easy to prove all you do is you look at fc over z and we can check this is bounded by 1 by by looking at its values on the circle very close to the unit circle and since this is bounded by 1 this easily implies the derivative is at most 1 at 0 and if this is equal to 1 at some point then f of z over z is constant which implies that f must actually be a rotation anyway um what this means is um we can now easily bound all these functions because we take our domain d and we're mapping it to the unit disc um well what we do is we um we take our point z zero and we take just take some small disc around z zero and a function f is bounded on this disc it maps this disc into this disc so by Schwartz is lemma we have a universal um upper bound on all such on the derivatives of all such functions at z zero um so um what we can find is we can find a sequence of functions f1 f2 and so on with fi of z zero the derivative of fi at z zero tends to m which is the maximum value of um um f of z zero prime not the maximum value we haven't shown the maximum exist so let's say it should be called the supremum because we've shown that there is such a finite value m and the problem is do these fi's converge to some limit f and this is in general this is a rather tricky problem um so if you've got a sequence of functions obviously you can't always that that doesn't necessarily converge but you can ask if the functions are bounded can we find a convergent subsequence and this sounds quite plausible because if you've got a convergent sequence of real numbers we can find a convergent subsequence if it's if it's bounded so sorry if we've got a bounded sequence of real numbers we can find a convergent subsequence and the answer is in general no for instance if we look at functions f from the unit interval to the unit interval we can find the sequence that has no convergent subsequence and the first one will sort of oscillate a little bit and the second one will oscillate more and the third one will oscillate even more and so on so we can have a sequence of bounded functions with no convergent subsequence um and the problem here is that the derivatives get large or if f isn't differentiable it's it's the differences between points get very large and there's a basic theorem called the ozala-ascoli theorem which says that we can find a convergent subsequence if um well the ozala theorem says we can find a convergent subsequence if the if the sequence is equicontinuous and the trouble with this is that no one ever quite remember what equicontinuous means but roughly what it means is that if we have a good bound or the derivatives of fi and what we mean by good bound i'm not going to worry about too much but the point is if we can bound derivatives of these functions then we can always find a convergent subsequence and the point is that if we're working with real functions there's no easy way to bound the derivatives if we're working with complex polymorphic functions then we've got a good bound so the derivative of a function at a is given by 1 over 2 pi i times the integral of f of z over z minus a squared dz and what you notice here is that if we have a bound on f of z then we get a bound on this integral here this is a we're integrating a little circle around a so we get a bound on the derivative there's one slight catch we need to be able to integrate around this little circle so if we have a bound of f on some region we get a bound on the derivative in a slightly smaller region and this turns out to be good enough to show that we get a convergent subsequence well you need to think about what convergence means and and it turns out to mean uniformly convergence on compact subsets and i'm not going to worry too much about what that means and the point is that if we've got a bounded sequence of holomorphic functions in the complex plane then there's this very general theorem saying you can nearly always find a nice convergent subsequence for some in some sense of the word convergent um by the way i can give an example of um some an example of why you need to be a bit careful what we're trying to do is we're trying to maximize um the derivative of a function at some point given some boundary conditions um like it has to take z zero to zero um if you try and maximize um the integral from zero to one of f of z dz given that f of z um is bounded by one and f of zero equals f of one equals zero then there's actually no solution to this um because we can find functions we can find a function like that or we could have another function that looks like this and so on so you can make the the the supremum of this is one but there's no function that actually takes the supremum so you've got to be a bit careful about um saying that if you've got a sequence of functions then you can find some function maximizing some value because it doesn't always hold um the point is the arsella ascoli theorem says we can actually find a function maximizing the derivative of of f prime of zero um but that's that that's because we can we've got a good bound for the derivative of these functions notice by the way that the derivatives here are getting very large so so as usual if if you can't bound the derivatives then you have a lot trouble showing that there's a convergent sum sequence um so now we come to point four we have found f um from d to u maximizing f prime of z zero um here where f is injective and holomorphic and now what we want to do is is we need to show f is injective and surjective and showing it's injective is fairly easy and showing it as surjective is kind of tricky um so so for showing it's injective it's not very hard to see this so suppose the limit isn't injective so suppose the um so as we've got d here and um under f it sort of maps to something that's not injective so so it might sort of overlap itself a bit there and then since the image of this is open you can see if if you deform it slightly it's still not injective um but since f is a limit of some sequence f1 f2 f3 and so where all these f i's are injective f must also be injective and this depends on the fact that d is open by the way if d is closed then the the limit of injective maps from d to something need not to be injective um so um we've got the the tricky part of the whole proof is proving this limit is is surjective so how do we do that um well let's first of all we we need to show um that if we've got some map f going from d to the unit disc taking um f take taking z0 to zero we've got to show that if f is not onto we can find a new um uh function g with g prime z0 strictly greater than f prime of z0 um and to do this I first need to quickly review mobius transformations so we recall that we've got some mobius transformations that go from the unit disc to the unit disc and these take z to a um um z minus a over 1 minus a bar z and here we can choose any a in in the unit disc and we can even multiply this by e to the i theta where where theta is some real number um and these form a group of holomorphic maps from the unit disc to itself and it acts transitively um that means it can take any point in the unit disc to any other point in the unit disc by by some some group element um furthermore it's you can easily check that the only elements of this group fixing mapping the mapping z the only elements of the group of all maps from the unit disc to itself mapping zero to zero are rotations that follows easily from Schwarz's lemma that I mentioned a bit earlier so so this is the full group of all maps from the unit disc to itself um so now let's try and construct this function g so let's think what we've got so we've got this domain d and we've got a map f of from d to the unit disc and it maps z zero to the point zero and we're going to suppose that it emits some point d so it emits some point that I'll write in blue so I think I'll write it in pink because blue doesn't seem to show up very well um so um if if we think of d as being some sort of purple region then we can its image in purple might be something like that and then what I'm going to do is I'm going to apply mobius transformation and move this points that it doesn't take move this special point to zero and put a blob in the middle so it looks the same um and this purple region will will end up looking something like that and not the image of the point zero will now be somewhere here so this is a mobius transformation and now I'm going to apply the square root and this will map the um point zero to itself and it will do something funny to the image of this purple region and map it to something else and finally I'm going to apply another mobius transformation in order to move this point back to zero so um things will now look like this um here I'll have this the image of d will look something like this and there will again be another pink point there and um what I'm going to do is I'm going to take this map my new map to be the composition of all these maps so this is my new map g and what we want to we need to show that g prime of z zero is absolute value greater than f prime of z zero um and and this will finish the proof um so how do we do that well um what we do is we look at this composition let's call it h um which goes from this purple region to this purple region here h isn't defined on the whole unit this because we kind of took a square root at this point and this this will follow from if we can show that um h prime of c zero is has absolute value less than one by the rule for composition of derivatives so we've just got to prove this and this in turn will follow from the fact that h um if we take the inverse of h and take its derivative at sorry that should be a zero not a z zero that zero then this is absolute value less than one by the taking the inverse of a derivative and there are two ways to do this first of all we could write down an explicit formula for h it's not too difficult it's a Mobius transformation followed by a square root followed by Mobius transformation and we could differentiate explicitly and after about a page of calculation we could show that it was actually an absolute value um greater than one so that should have been a greater than one um so uh um however there's an easy way to see it without any calculation which is to observe that h is a holomorphic map from the unit disc to the unit disc and now Schwarz's lemma implies that its derivative at zero is less than one if h is the minus one is not a rotation and it's not a rotation because we stuck a square root sign in there I mean I mean if we just had Mobius transformations it might have been a rotation um so this completes the proof of Riemann's mapping theorem notice by the way there's nothing particularly special about the square root sign here um all we need is that the inverse of this maps the unit disc to itself in a nice way so instead of taking square roots we could also use nth roots provided n is greater than or equal to two um we can't take n equals one because then we would run into this problem that this map might actually be a rotation