 Okay, so good afternoon everyone. So this is the second part of the mathematics ODE, second part that is differential equations and dynamical systems. So as you remember in the first half of the course we introduced certain fundamental concepts and we looked at the application of these concepts in several examples and in particular we concentrated on discrete time dynamical systems. So dynamical systems that you get by iterating some math on some space. So you have some space x and you have the math on the space to itself, you fix a point and you apply F repeatedly and you want to study certain properties of these systems whether it has fixed points, periodic points, dense orbits and so on. And one of the fundamental notions that we use to approach the subject is that of classifying all dynamical systems. And if you remember the core concept was that of defining some equivalence relation on the space of dynamical systems which was the notion of a conjugacy. So if we have two dynamical systems g from y to y then a conjugacy is a bijection between the two sets h and y that maps orbits to orbits. So the image of x under h will go to some point here h of x and then if you apply the map g to the point h of x you should get exactly the same point that you get by applying the map h to f of x. So this is h of f of x. So we have the definition that f and g conjugate if there exists a bijection h such that g composes with h. Please feel free to ask any questions if there's any doubts or anything you don't understand. So we discussed in the first part of the course we saw many examples. We observed that conjugacy is an equivalence relation so you can it defines an equivalence relation on the space of all dynamical systems and it defines it formalizes a way of saying that two dynamical systems are equivalent in some way. You can strengthen this condition by asking for this conjugacy for this conjugacy h to be a homeomorphism in which case we call it a topological conjugacy or in some cases h might be even a diphthalmorphism or a linear map in which case it's a stronger version of conjugacy. Okay? Everything okay with that? In a nutshell the core idea of the first part of the course. So in that we applied this to several systems mainly to linear systems in one and two dimensions and a few other simple classes of systems. In this second part of the course we're going to continue our study using this approach but we're going to look at some different techniques for constructing these conjugacies for some more interesting and sophisticated techniques in some larger and more important class of systems. Okay? This is the basic content of this course. So let me start first of all so there's going to be two main classes of systems and we are going to define them in terms of whether they are contractions or expansions and as we shall see they have very different dynamical features and very different techniques for studying them. A contraction is a map where the distance between the two points gets smaller and an expansion is a map where the distance between two points gets larger. Okay? Very basically. For the first two or three lectures we're going to concentrate on contractions. So the first section of this course we're going to talk about nonlinear contractions. So let me write first of all with the definition so let X be a metric space. X is a contraction if there exists some lambda in 0, 1 such that for all X and Y in X we have that the distance f of X minus f of Y so I will use these norms at the distance. Often our metric space will be Rn or some linear space so I will just, it's easier just to use this norm as the distance even in the abstract metric space less than or equal to lambda X minus 1. So I'm sure you've seen contractions before what is the fundamental theorem of contractions who remembers? Fixed point, exactly. Contraction mapping theorem, okay? So I'm glad that you recognize it. I imagine probably most of you have proved it before. It's a simple proof but I will prove it again for completeness, okay? Also because this will play a big role in our study of contractions more generally. So we have the contraction mapping theorem that says that X, what kind of metric space should X be? Complete metric space. That's right. X is a complete metric space, it's contraction with a constant lambda then there exists unique fixed point P in X. So most of the time when you study the contraction mapping theorem you studied it as a fixed point theorem which is partly what it is. But we are interested in dynamics, okay? So we are actually interested in the fact that this is not just the existence of a fixed point but from a dynamical point of view we have the property that a unique fixed point such that f and x converges to P for all x and x as n tends to infinity, right? So it's not just a fixed point but it's a fixed point that all the orbits converge to under iterations. So it is the attractor of the system. So there's a unique attractor. So the asymptotic behavior of all points is precisely the fixed point. And in fact this convergence is exponential, fn x minus P is less than equal to lambda n. Okay. So how do we prove this lemma? We use a couple of this theorem. So we use a couple of intermediate lemmas. We just break it down. It's a fairly simple proof but it's nice to see. It's a nice proof. So I will give it. If x is a metric space and f is a contraction, then for all x in x, the forward orbit of x, right, which we write as a sequence xn, n equals 0 to infinity, x0 is equal to x, x1 is equal to f of x, and so on, is a Cauchy sequence. So remember what a Cauchy sequence is? So recall that a sequence is Cauchy if for all epsilon there exists some n which depends on epsilon, such that xn minus xm is less than equal to epsilon for all nm greater than equal to n epsilon. You agree? This is the definition of Cauchy sequence. No, Cauchy sequence is a kind of a way of saying that the sequence is converging without actually saying that it's converging because the word, the notation to say that something is converging implies that it's converging to a point in the space, whereas this is not necessarily the case. The notion of Cauchy is just saying that the points as you move forward in the sequence they're getting closer and closer to each other. So why is this a Cauchy sequence? So let x equals x0 in x, then we have that xn plus 1 minus xn. This is precisely equal to f, sorry then for all n greater than equal to 0, xn plus 1 minus xn is just equal to f of xn minus f xn minus 1 because this sequence is the orbit of a point. So xn plus 1 is the image of xn and xn is the image of xn plus 1. So this is just exactly the same point and by the definition of contraction this is less than equal to lambda xn minus xn minus 1. And then of course I can do the same thing. So these points here are the image of xn minus 1 and xn minus 2. So here I get this is less than equal to lambda squared xn minus 1 minus xn minus 2 and so on. And I end up with lambda n times x1 minus x0. So this is not or is not yet Cauchy but if we take so for n greater than equal to n greater than equal to n we can just write xn minus xn. This is what we need to estimate for the Cauchy property and we just write this by the triangle inequality as xn minus xn minus 1 plus xn minus 1 minus xn minus 2 and so on up to xn plus 1 minus xn, okay, triangle inequality. And then each of these we estimate by this inequality here. So this is less than or equal to lambda n minus 1 in this case times x1 minus x0 plus lambda n minus 2 x1 minus x0 and so on plus lambda m x1 minus x0, okay. And then we can just take out the x1 minus x0. So this is less than or equal to x1 minus x0 times the sum from j equals m to n minus 1 of lambda j and of course this is just less than or equal to x1 minus x0 times the sum all the way from j equals big n to infinity of lambda j. Because m and n are both bigger than or equal to big n so we just bound it above by this, okay. But of course you know that this sequence here is converging to zero so when you take n are very large. This is very small. This is a geometric decreasing sequence so it's bounded by some constant times lambda to the n say and so by taking n sufficiently large you can make this less than or equal to epsilon if n is sufficiently large and this shows that it's a Cauchy sequence. Any questions? Okay, so to get that this is a Cauchy sequence we didn't need to assume that x was a complete metric space but now the assumptions of the theorem are that they are a complete metric space. The definition of a complete metric space is that every Cauchy sequence converges and what does it mean that it converges? It means that there is a point in the space to which it converges, okay. So since x is complete there exists p in x such that xn converges to p where xn is the sequence of this point x. So now we want to show that this point is a fixed point. So let me write that also as a separate lemma, you know it's a very small lemma, lemma if again a general lemma so x is a metric space, f is a continuous map. For this lemma we don't need f to be a contraction and then suppose that x in x, the orbit of x is given by xn, n equals 0 to infinity and xn converges to some point p, then p is a fixed point, okay. So this is very simple, we have the sequence xn, we can define a new sequence yn to be equal to xn plus 1 which is equal to f of xn, okay. Then clearly yn also converges to p because it's the same sequence, we just index it differently, okay. And then we just have that f of p is equal to f of the limit as n tends to infinity of xn by continuity, the image of the limit is the limit of the images, right here we just use continuity, limit as n tends to infinity of f of xn which is equal to the limit as n tends to infinity of yn which is equal to p, so p is a fixed point. So this is enough to prove the existence of a fixed point, right because we've shown that if you take any initial condition, notice that this fixed point is constructed dynamically in some sense, you choose an arbitrary point, you look at its sequence, you look at where it converges, that's where your fixed point is. We need to show uniqueness of the fixed point, why do we have uniqueness of fixed points? Yeah, because otherwise it will contradict that it's a contraction, right. So this gives, so this gives existence of the fixed point, okay, if p and p tilde in x are two fixed points, suppose by contradiction p is different from p tilde then this would mean that f of p minus f of p tilde which because the both fixed point is equal to p minus p tilde, okay, contradicts the contraction, contradicts contraction, okay. So we have that p is equal to p tilde, so this gives uniqueness. And finally the exponential convergence to the fixed point is obvious, again it just uses the contraction, finally we have that f n of x minus p is just equal to f of, f n minus 1 of x minus f of p, okay. And now we use the fact that this is a contraction, so this is less than equal to lambda, f n minus 1 of x minus p, okay, and then you repeat the same argument here, okay, and you just less than equal to lambda n x minus p. So this gives exponential convergence, okay. So this was familiar territory for you. Now let's see what interesting things we can say using the contraction mapping theorem. So first simple application is just a question of which systems are contracting. How do you know that a system is contracting? And one of the simplest ways to do this is just to look at the derivative of the map, because the derivative tells you something about the contraction. So very simple application, proposition, let f r n to r n be a C1 map, suppose the derivative at the point x is less than equal to lambda less than 1 for all x in r n. So then f is a contraction, and so has a unique globally-attracting fixed point, globally-attracting because every point in r n will converge to that fixed point. So you know how to prove this? Who tells me how to prove this? Simple proof. How do we show that it's a contraction? Exactly, okay. So let me write it out just for completeness again. But it's very simple application of the mean value theorem. Let me first do it in one dimension when it's particularly simple, all right. So proof for n equals 1, we just have, like you said, the mean value theorem shows that f of x minus f of y is equal to f prime of z times x minus y for some z between x and y, okay. But if we assume the derivative is always less than equal to lambda, then this is just less than equal to lambda x minus y, which is exactly a contraction. So in higher dimension, it's also essentially exactly the same thing. You just need to use a slightly more kind of sophisticated version. So in general, if you take your two points in r n, x and y, then how can you use the information about the derivative? Well, you can just take the line joining these two points. This is the distance is given by the length of this line. And you look at the image of this line. And if the image of this line is less than lambda times the image of the line, then they must have come closer, right? So it's a contraction. So how do we do that? So if we let this line be called gamma, so gamma 0 tau to r n is a straight line, straight segment, straight line segment joining x and y. And we suppose it's parametrized by arc length. It's a parametrized arc length. So this is exactly the length so that gamma prime of t is equal to 1 for all t in 0 tau. Then we can just f of x minus f of y is just equal to the length is less than or equal to the length of the image of f of gamma. And this we can just write as equal to integral from 0 to tau of the derivative in gamma of t of gamma prime of t, which is the tangent direction of gamma prime, right? This is basically the chain rule in higher dimensions, right? And now this is just less than or equal to the integral between 0 and tau of the norm of df in gamma t dt. And this is less than or equal to lambda tau, which is equal to lambda. And so this is a contraction. So your question is that the matrix is a n by n matrix. And we're looking at the norm of, so this is what norm we're considering on the derivative. OK, good question. So you're right, because we introduced several norms. So in this case, we use the operator norm on the derivative. Let me, it's precisely the norm that we need for this to be true. So this is equal to the supremum over all v of norm 1 of df x of f. And this is the Euclidean norm in Rn. So the usual norm, you're right. In the first part of the course, we introduce a different norm on matrices. All the norms are equivalent. But in this case, we want specifically the norm that has to do with the expansion and contraction of the derivative. So this is the maximum expansion of the derivative. So this means the fact that this is less than equal to lambda means that if you take any unit vector, its image will be less than equal to lambda under this derivative. Excuse me? Yes, the norm is equal to 1. Thank you. OK, so this was a simple application. We will have some more interesting applications of the contraction mapping theorem. First, let's prove one more lemma about the fact that in certain situations, if you have a family of contractions, then the fixed point changes continuously as you change the contraction. So let me state this. So lemma that omega and x in metric spaces and x is complete. f omega cross x to x be a continuous map. So what does this mean that it's a continuous map? So when we write it in this, what this really means, this is a product of two metric spaces. It maps to one metric space. So the way to think about this is that this metric space is parameterizing a family of maps from x to x. Because for each omega in omega, what you get is a map from x to x. So the fact that this is continuous means that this family varies continuously with omega. So let me write it. So i.e. the maps f omega x equals f omega x is continuous in both x and omega. So suppose that for all these omegas, each f omega is a contraction. So suppose there exists lambda in 0, 1 such that f omega x minus f omega y is less than or equal to lambda x minus y for all omega in omega and for all x, y. So all this is saying is that we have a family of maps from x to x. And these families change continuously depending on some parameter omega. But all of them are uniformly contracting in this way. And since x is a complete metric space by the contraction mapping theorem for each omega, we will have a fixed point of the map f omega. And the conclusion is then the fixed point p of omega depends continuously on omega. Yes. OK, so you're not clear about this dependence on x and omega. All I'm saying in this sentence is I'm clarifying or maybe making it more difficult for you to understand what this means. It's exactly the same statement. So I'm saying that this map is continuous. This is a map from one metric space to another metric space. It's continuous. But all I'm saying is that the best way to think about this continuity is to think of the fact that for each omega you have a map from x to x that I write like this. But this is just the same as this. In this notation, these are the two variables, omega and x. This is a point in x. Yes, it's continuous in both x and omega. So this is a continuous map, which means the image depends continuously whether you change omega or whether you change x in both variables. OK, so the proof again is very simple. So we use the contraction mapping theorem and the triangle inequality. We get x minus p omega is less than or equal to x minus f omega of x plus f omega of x minus p omega. Just a triangle inequality for any x and any and the fixed point p omega. We have this. And this is less than or equal to x minus f omega of x plus. So what is this? This is just saying that this is the fixed point of f. This is f omega of x. So the distance between f omega of x and the fixed point is this point is converging to the fixed point at the rate lambda. So this is just lambda times x minus p omega. So we now have x minus p omega on both sides. So taking this to this side, we get a factor 1 minus lambda x minus p omega. And so this gives x minus p omega is less than or equal to x minus f omega of x over 1 minus lambda. So in particular, choosing. So this is true for any x. This is just a general inequality for any x. We have this. Now I'm going to choose x equal to p omega tilde for some arbitrary omega tilde in omega for different parameter value. And then we get p of omega tilde minus p omega, which is what we want to estimate, because we want to show that p depends continuously on omega is less than or equal to 1 over 1 minus lambda times p omega tilde minus f omega p omega tilde. So here I haven't done anything. I've just substituted the point p omega tilde for x here. So just by this, I get this inequality. So what does this inequality mean? So what happens as omega converges to omega tilde? So this is the fixed point for omega tilde. It's not the fixed point for omega. So f omega p omega tilde is not p omega tilde. But because f is continuous in omega, then when omega is very close to omega tilde, f omega of p omega tilde will be very close to omega tilde. And in fact, they will converge. So by the continuity of f with respect to omega, we have that f omega of p omega tilde converges to f omega tilde p of omega tilde. This is just a continuity with respect to omega tilde. But of course, since this is the fixed point, this is exactly equal to p omega tilde. And that's it, because what we're saying is that as omega converges to omega tilde, this converges to p of omega tilde. Here we've got omega tilde. So that means this is converging to 0. And because this is bound for this, this means this is converging to 0. So as omega comes close to omega tilde, this is converging to 0, which means these depend continuously on each other. So you need to look at it a little bit. It's quite elementary, but you need to think about it just because of this substitution of p omega tilde and exactly what this means. You need to spend a few minutes looking at it, and then you will see. Sorry? We only use the completeness to say that for each omega there's a fixed point. We apply the contraction mapping thing to say that there is a fixed point. And then we don't use the fact that omega is complete. Omega is just a metric space that parameterizes this family of maps. So it's very general. Family of maps parameterized by omega. And then the fixed point depends continuously on the map, which is what you expect, of course. There's nothing unusual about that. OK, so these are the first preliminary lemmas on contractions. So as I said at the beginning, our basic tool, our basic goal is to classify dynamical systems using the notion of conjugacy. At the moment we have said that if you have a contraction, basically the dynamics is very simple in some sense, at least in the sense that the omega limits are just a fixed point. Everything converges to a fixed point. However, that doesn't mean that two contractions are necessarily conjugate, topologically conjugate to each other because the way that the points converge to the fixed point might be quite different. So if you really want to understand the finer structure of contractions, we would like to talk about the conjugacy, topological conjugacy, and the construction behind contractions. Unfortunately, in a complete generality, we cannot make any statement about the fact that any two contractions are topologically conjugate, for example. That is not generally true. So we're going to prove a more limited theorem, which however is very important, about the fact that if you have a linear contraction, then any small perturbation of that is topologically conjugate to the linear contraction. So this is going to be the main theorem of this section. So let me state it carefully here. We're going to be looking at perturbations of linear contractions. So I'm going to prove this in a fairly general setting. So let E be a Banach space. You know what a Banach space is? It's a vector space, which might be infinite dimensional instead of finite dimensional. There's nothing really particularly mysterious about that. And it has a norm. It's a complete metric space. It's a metric space, right? Complete norm vector space. So we have a norm, which we write like this. And we suppose that T is a linear invertible linear contraction. So T is an invertible linear transformation. And by a contraction, I mean that the norm contracts. So the norm of T is equal to the supremum. Again, the same norm I gave before, the supremum over all unit vectors of T of v is less than or equal to lambda less than 1. This is what I mean by a linear contraction. OK, so what's the theorem? So theorem. Suppose we have another map F from E to E of the form F equals T plus delta F, where delta F. So delta F from E to E is bounded. Bounded means that the maximum norm, right? So this means that the supremum over all v in E of delta F of v is finite. This is what I mean by bounded. So I'm going to define two conditions. This is one condition. And then the other condition is that it's a Lipschitz. And that the Lipschitz constant of delta F is less than or equal to the minimum of the inverse of the norm of the inverse of T and 1 minus the norm of T. OK, so you need to think about this a little bit. But the norm, the inverse of the norm of the inverse, if you think about it for a linear map, is simply the infimum. This is just equal to the infimum over v equals 1 of T of v. So this is the supremum means because every vector is contracted, the supremum is the one that is contracted the least. Whereas this is the one that is contracted the most. So I should put here also that, yes, invertible linear contraction. OK, so these are the assumptions. And then the conclusion is that then T and F are topologically conjugate. And in fact, we'll make a slightly more precise statement. So more precisely, there exists unique homomorphism of the form H equals the identity plus delta H with delta H bounded, which is a conjugate. It's such that H composed with T equals F composed with H. So it's not too hard to believe this result, but the proof will be quite involved. What this is saying is that if you take a linear map that's contracting and you take a nonlinear perturbation of this map that is sufficiently small perturbation, the sufficiently small is contained in these two conditions. A small perturbation of the linear map, then the new map that you get, this perturbation, does not destroy the topological conjugacy class. It's a bit like saying that this system is structurally stable as long as you take perturbations of this particular kind. And in fact, we will show that the conjugating homomorphism is close to the identity also in this sense here. If the perturbation, of course, is zero, the identity conjugates the linear map to itself. If it's a small perturbation, the conjugacy small is close to the identity. So there's various reasons for proving this theorem. First of all, it has its intrinsic interest in the statement, but another reason is that we will now introduce a new, very interesting techniques for proving this conjugacy. So in the first part of the course, basically all the conjugacies we proved were using this technique of fundamental domains in several situations. In this course, we will introduce at least two other very important kinds of techniques to construct this conjugacy. And this will be one of these new techniques. And this is a kind of technique that is also fairly applicable and interesting in different contexts and different settings and different areas of mathematics here. OK, so we will give a complete proof of this. Let me start today with just a couple of propositions that we will use. So the first proposition is this proposition. We have the same setting. We have E is a Banach space. T is an invertible linear contraction. We have f e to e is of the form f equals t plus delta f with lip sheets of delta f less than or equal to t minus 1 minus 1. So this is a preliminary proposition and we'd use just one of these various conditions. Then f is a homomorphism. So notice, of course, that the initial linear map is, of course, a homomorphism because it's invertible. It's an invertible linear map. So it's a homomorphism. It's a bijection. What this is claiming. So if you make a nonlinear perturbation, there is nothing to guarantee that it continues to be a bijection because you could make a perturbation where the injectivity fails, for example. You could perturb it so the two nearby points now cross, so map to the same point. So this is the statement. The statement is that by imposing this condition, which is a way of saying that this is a small perturbation, you cannot lose the injectivity, basically, is the key point. OK, so this is a first example of an interesting application of the contraction mapping theorem in a non-trivial setting. The one that we did about the derivative is a kind of almost trivial application. Now we'll actually prove this using the contraction mapping theorem. But to do that, we need to set things up in a special way. So how do we do it? Well, first of all, what do we need to prove? We need to prove that f is a homomorphism, which means we need to prove that it's a bijection, it's continuous, and f minus 1 is continuous. So the fact that f is continuous is obvious because t is a linear map, which is continuous. Delta f is Lipschitz, so in particular it's continuous, so it's continuous. So the continuity of f is obvious. So f is continuous by definition. So what do we need to show? We need to show that it's a bijection. So we need to show that f is a bijection and that f minus 1 is continuous. And that will show that it's a homomorphism. So to show that f is a bijection, we need to show that it's a surjective and injective, as usual. So to show that f is surjective, we need to show surjective, just a definition of surjective. So we need to show that for all y in E, there exists x in E such that f of x equals y. This is surjective. If we show that this x is unique, then we will have shown that it is also injective. If we show if x is unique, then f is also injective. And if we show that this x depends continuously on y, if, moreover, x depends continuously on y, then x, if we show that it's surjective and it's injective, then x is equal to f minus 1 of y. So the fact that x depends continuously on y is exactly the same thing as saying that x equals f minus 1 of y is continuous. Then we will have shown that f is a homomorphous. So we fix a y, if you want, in the image. We show that there exists some point that maps to it, gives surjectivity. This is a unique point, gives injectivity. And this point depends continuously on y, continuity of f minus 1. And this already starts reminding us of something. We're trying to show that for each y, there exists a unique x that depends continuously on y. So we're starting to set it up in this way. So how do we implement this? Well, what does it mean that f of x is equal to y? Notice, so f of x equals to y is exactly the same thing as saying, so f of x is of the form t plus delta f of x equals y, because that's just how f is given. This is equivalent to t of x plus delta f of x equals y, because this is just the sum of two functions. This is equivalent to writing y minus delta f of x equals t of x. I've just subtracted delta f of x from both sides. And now, because t is invertible, I can just apply t minus 1 to both sides. So this is the same as t minus 1 of y minus t minus 1 composed with delta f of x equals x. So this looks like I just pulled out of my hat these equalities there, all very elementary. But the reason why I've done this, why I like this form, even though this form looks very, doesn't look particularly nice, is because now I can write this as a fixed point problem. So now let for y in E, we define a map, theta of y from e to e as theta y of x equals t minus 1 of y minus t minus 1 composed with delta f of x. So you see, I'm now using the set e itself to parameterize this family of maps. For each y, I've defined a map that obviously depends on y, because I have y as a parameter for this family of maps. But I think of it as a map on x, from e to e, where the variable is x, and I plug the x in here. And what I want is for this to be satisfied, which means I'm looking exactly for a fixed point of this map. If I get a fixed point of this map, that means theta y of x equals x. That means this is equal to x, which means this is satisfied, which means this is satisfied. So it is sufficient to show for every y in E, the map theta y e to e has a unique fixed point, which depends continuously on y, because that will give us exactly what I was asking for before. And as I said before, all we need is to show that for every y, it takes this unique fixed point x that satisfies this, and that depends continuously on y. And that's exactly what we would get, because this fixed point of theta y gives value of x that satisfies this, which is equivalent to satisfying this. And we need to show it's unique and depends continuously on y. And the nice thing is we have set it up exactly in the context of the lemma we had before, which is a family of contraction mappings. So all we need to show is that each of these phi of theta y is a contraction by some constant lambda uniformly in y, and then we have automatically all, exactly all these properties. So how do we know that this is a contraction? So it is sufficient to prove this to show that theta y is a family of contractions. OK, it's clear that this depends continuously on y because of the definition here. Everything is continuous, family of contractions, uniformly in y. So we get a continuous family of contractions that contracts uniformly in y, so we have all the conditions. So now we just plug in the conditions. So let x and x prime be two points in E, and then we have theta y of x minus theta y of x prime is equal to, OK, I raised the definition, but you have it of theta y. And you can see that what you get here is just t minus 1 composed with delta f of x minus t minus 1 composed with delta f of x prime. Because the term t minus 1 of y is the same term in both, so it just cancels out. So t minus 1 is linear, so we can just write t minus 1 composed with delta f of x minus delta f of x prime, and this is t minus 1, and this is bounded above. This is a linear map, so we take the norm of the linear map. This is less than or equal to norm of the linear map because this is the norm of this map is the maximum expansion of this map times the norm of this difference here, delta f of x minus delta f of x prime. And what is our assumption? Our assumption is precisely that delta f is Lipschitz. So we just have that this is less than or equal to t minus 1 times the Lipschitz constant of delta f times x minus x prime. And the assumption was precisely the Lipschitz constant is less than 1 over this, so this means this is less than 1. This is equal to some lambda less than 1, independently of y. So theta y is a contraction. Yes, 1 over t minus 1, 1 over the right. So assumption was that maybe I wrote it wrong. Lipschitz of f is less than or equal to t minus 1 minus 1. Is that what I wrote? Right. So the product of these two is less than 1, right? This is? Sorry? Ah, sorry. Sorry, sorry, sorry. Let it be strict. Yes, thank you. Yes. OK, I know you must be a bit tired, but let me just prove one more lemma, because then after that we'll be able to start the proof of the theorem in the next class. So one other preliminary lemma. OK, so this proves that if a small perturbation of the linear map, then it's still a homeomorphism. And I just want to emphasize again that this is a first example in this argument where we use the contraction mapping theorem in a non-trivial way, right? We have a problem that when you look at it, it looks nothing like any contraction mapping theorem, but we convert it into a fixed point problem, OK? So the second proposition which we will want to use in the proof of the theorem. So proposition, again, we have the same. I shouldn't have deleted it. Banach space t e to e is an invertible linear contraction. F is of the form t plus delta f. And this time we use the other assumption on the Lipschitz less than equal 1 minus t, right? If you remember, this was the second assumption. And I'm writing these propositions just to clarify exactly where we will use these assumptions in the theorem to get these two conditions. So then f has a unique fixed point. So this is a little bit different. But again, it's of the same kind of result. It's saying we have a linear map and then we have a small perturbation. It still has a fixed point, OK? Again, this is fairly easy to believe this result. And the proof is also fairly straightforward in this case. So proof is just one calculation. So for x and y in e, we have f of x, OK? So here we will just show that f itself is a contraction. And then we get a unique fixed point. So because we know that t is contracting and because we know that the Lipschitz constant of this is small, in this case we will just conclude that f is still a contraction and therefore has a unique fixed point. So f of x minus f of y is equal 2. So we write f as t plus delta f. So first we write just we use a little kind of triangle inequality trick. So first I write this as f of x minus t of x plus t of x plus t of y minus t of y minus f of y, OK? I just add and subtract t of x and t of y. And then I use this to write it as f of x minus t of x minus f of y minus t of y. I have another brackets here plus t of x minus t of y. And this is just equal to f minus t of x minus f minus t of y plus so yes, so plus t. This is a linear map so I can just write it as t of x minus y. Now f minus t is exactly delta f. So this is less than or equal to delta f of x minus delta f of y plus less than or equal so I can take the norm of t out here. I bound this t of x minus y by the norm of t times the distance between x and y. And then I use the Lipschitz bound on this. This is less than or equal to the Lipschitz of delta f times x minus y plus the norm of t times x minus y. And this, of course, is just equal to the Lipschitz constant of delta f plus the norm of t all multiplied by x minus y. And then we have exactly our assumption was that Lipschitz delta of f is smaller than 1 minus t. So this is less than 1. And so f is a contraction. Sorry, I just write less than or equal to 1. So notice, of course, that this is not what we're trying to prove. What this says is that if you take a small perturbation of the linear map, it has a fixed point. But we're trying to prove much more. We're trying to prove that f is topologically conjugate to t under those conditions. So we're going to use that. We still have a non-trivial part of the argument left. But we will use these two technical lemmas as part of the argument, which we can carry out next time. So I think for today we've done enough work. OK, thank you.