 Good afternoon. So today we will complete the proof of the theorem. Let me just remind you of which theorem we started proving. So we have a linear contraction in a Banach space. So this is very general setting, infinite dimensional setting, invertible linear contraction, invertible. And we have F of the form T plus delta F, where delta F is bounded. And the Lipschitz constant satisfies this condition that is the minimum between T minus 1 minus 1 and 1 minus T. So it's a bound on the Lipschitz constant. If you remember, I said this is a way of saying that this is a small perturbation of the linear map. Then the theorem is that then there exists unique homomorphism of the form H equals identity on E plus delta H with delta H bounded, which is a conjugacy such that H composed with T equals F composed with H. So the conclusion contains two parts. It says that F and T are topologically conjugate and it also says something about the specific form of the conjugacy. It does not say that you might not be able to find a conjugacy that does not have this form here, but it says that there's a unique conjugacy of this form. So we prove two preliminary propositions and under these assumptions we show that F is a homomorphism and then we show the second proposition is that F has a unique fixed point. Okay, this is what we did yesterday. Yes, I contained this in the contraction. Yes, when I said linear contraction, that's what I meant, the norm of this. Yes, thank you. You reminded me of that several times. Yes, thank you. That's right. We need this trick to prove the propositions below. So just a remark is not so important, but these two propositions they use, one of them uses this assumption on this bound on the Lipschitz constant and the other one uses this other bound on the Lipschitz constant. Let me observe and I will come back to this later that we will apply some of these, I think, this proposition to a different Banach space and to a different function. So this holds, this is a theorem, okay, and this is a proposition that says in general if you have this setting then you have these conclusions. But in the course of the proof we will introduce a new Banach space to which we will apply this, which will have a similar setting. I will explain this as we go along. So now we are ready to start the proof of the theorem. So I will want to keep this here because I want to refer to that later. So let's start the proof. So we're actually going to prove a slightly more general statement. So suppose let G also be of the same form. So G is equal to T plus delta G with delta G bounded. And Lipschitz of delta G less than or equal to the same bound. Sorry, strictly less than T minus 1 minus 1, 1 minus. And then we will show that there exists delta H bounded such that H equals identity plus delta H is a homomorphism and it conjugates F and G. So F composed with identity E plus delta H is equal to delta H composed with G. Okay, so it's essentially the same thing except it's a slightly more general because we are taking both F and G as small perturbations of T and we are showing that they conjugate to each other. Of course, as a special case you can take delta H equal to 0 and then what you have G is exactly equal to T. If delta G is 0, G is equal to T. So the result holds for T as well as a special case. But it's just more convenient to have both sides to have a similar structure. Okay, so how are we going to show this? So we're going to do a little bit of manipulation here and then we're going to set it up as a fixed point problem in a similar way to the way we proved one of these propositions here. So first of all notice that by these propositions, let me call these propositions 1 and 2. So we know that G is a homomorphism. So by proposition 1, G is invertible. So let me write this as equation 1. I need to number these equations. So 1 is equivalent to, I'm going to compose both sides by G minus 1. So I'm going to write this and I'm also going to write F as T plus delta F so it's equivalent to T plus delta F composed with identity of E plus delta H composed with G minus 1 is equal to identity of E plus delta H. So I'm rearranging this expression here. At the moment I've just simply composed both sides by G minus 1. So I just get this and now I'm going to multiply this. I'm going to expand this composition here and this is equivalent to, equivalent to, so this is just the sum of two functions composed with this so I can just distribute it. So I get T composed with identity plus delta F composed with G minus 1 plus delta F composed with the identity plus delta H composed with G minus 1 equals identity of E plus delta H. And here I also want to just expand this as well so I'm going to write this. So this expression here is equivalent to, I'm going to take this sum out so I'm going to write this as T composed with the identity composed with G minus 1 so I just write it as T composed with G minus 1 plus T composed with delta F composed with G minus 1. Oh, sorry. Yes. Thank you. So T composed with identity composed with G minus 1 Here I have T composed with delta H composed with G minus 1 with delta H composed with g minus 1 plus this expression I'm going to leave as it is, delta F composed with identity of E plus delta H composed with g minus 1 equals identity E plus delta H. And finally, OK, so finally I'm going to rewrite that as so this is equivalent to I'm just going to reorder it a little bit in a way that's better. So t composed with delta H composed with g minus 1 plus delta F composed with ID of E plus delta H composed with g minus 1 plus t composed with g minus 1. And then here I'm just going to subtract the identity on both sides because I want to have the identity on this side and I'm going to write equals delta H. So my conclusion is that this expression here is equivalent to this expression here. All I've done is some basic algebraic rearrangements. And what's the difference between these two ways of writing this expression? Can anyone guess why? I went through all this trouble to rewrite this like that. Exactly, exactly. All this because I want to formulate this as a fixed point problem. This is going to be a fixed point in a different space because it's going to be a fixed point in the space of functions because I want to find delta H as a fixed point. So now I would like to say that this is a kind of function where delta H is the variable of this function and I want a fixed point. I want delta H to be the fixed point of this function. And then I'm going to use the contraction mapping theorem to get these fixed points. So how do I do that? Well, first of all, let me define the space of bounded functions. So let E equals the space C0, B, E, E. And to understand this notation, this is the space of continuous C0 bounded functions from E to E. And equals space of phi from E to E such that phi is continuous and bounded. Because this is the space in which I want delta H to live. Because what I'm looking for is a function delta H, which is continuous and bounded, to get a homomorphism. And then for phi in E, I can define a map. I can define a new function for your phi exactly like this. So I define it like this. It's going to be t composed with phi composed with g minus 1 plus delta F composed with the identity plus phi composed with g minus 1 plus t composed with g minus 1 minus the identity on E. So you see, of course, F and G are fixed. F and G are fixed in the theorem because we fix an F, then we fix another G. What we're trying to do is show the conjugacy between F and G. So F and G are given. So I'm allowed to use F and G in the definition of this function. And indeed, it's obvious that we need to use F and G because the function delta H will depend on F and G because it's going to be a conjugacy between F and G. So it makes sense. So this is a new function on the space E. We need to show that this new function is still continuous and bounded. So we need to show that this is a well-defined function from E to itself. We haven't done that yet. At the moment, it's just well-defined in the sense that it gives a function. You plug in a point x, and these are all functions on the space E. And so it gives you some other point. We need to show that this gives a function that still belongs to E. And that's going to be our first lemma here. So lemma of phi in E, we have that F of phi also belongs to E. But this is fairly easy, actually. I mean, this looks quite complicated. But if you break it down and you look at it slowly, if you look at it bit by bit, it's fairly straightforward. So this is just a sum of various functions. So we just need to show that each one of these is continuous and bounded, right? So the first two terms. So this term here is bounded because G minus 1 is a homomorphism. Phi is bounded, right? And so you take the composition of G minus 1. It doesn't matter what G minus 1 is. Phi is bounded, which means that the image of the composition phi and G minus 1 is bounded. And then you just apply this linear map to the bounded function. So it's clearly bounded. So yeah? Bounded just means that the image is contained in some, the supremum of the images of the points is bounded, right? So because phi is bounded, then if you look at the supremum of the images of all the points and this function is bounded. So T composed with phi composed with G minus 1 bounded because G is bounded. The same thing here. So delta F is bounded, right? So whatever this is, the result of this composition is bounded. So delta F composed with identity of E plus phi composed with G minus 1 is bounded because delta F is bounded. The last terms is not completely obvious because the identity is not bounded. But we also have here that these are not bounded either. And fortunately, they cancel out exactly because we have that notice that T, so T composed with G minus 1 minus the identity of E, we can write as T minus G composed with G minus 1, right? We just compose both sides by G. And T minus G is delta G. Exactly. So this is exactly equal to delta G, which is bounded. OK, so that was a close one. OK, so this shows that the function that we get out of this is bounded. We just need to show that it's continuous, but that is obvious because everything is continuous here. All these functions are continuous, so phi is continuous. OK, so excuse me. Ah, yes, thank you. Equals delta G composed with G minus 1 is bounded because delta G minus delta G. Thank you. Still bounded. OK, so we have a map from curly E to itself, right? From the space of all bounded continuous functions to itself. And this map is designed precisely so that if we can find the fixed points for this map, then that fixed point is exactly the delta H that we're looking for, and we have solved our problem. At least we've shown the existence of the delta H, bounded and continuous, which gives us the conjugacy. So we want to apply some fixed point theorem, right? So what do we need to do to apply the fixed point theorem? First of all, we need, yes? Exactly, exactly. So we need to have that this is a Banach space, which it is, and I will not prove it because it's a fairly standard fact. So we write what the norm is. So E is a Banach space in particular complete, in particular it's a complete metric space with the norm of phi. So the norm of phi is just equal to the supremum over all x in E of the norm of phi of x with the norm, the E-norm. This is the obvious norm. It's the sub-norm. It's just a standard sub-norm for continuous functions. Uniform norm sometimes is called. OK, so this is a standard fact. I'm sure you probably will have done it. This is a complete metric space. This is a Banach space. So we have our map on a complete metric space. We want the fixed point. We just want to show it's a contraction. Or we want to show, sorry, yes. So we need to show that it's got a fixed point. So how are we going to show that it's a fixed point? Well, in fact, going to show not necessarily that it's a contraction, but we're going to use this proposition here, but apply to this map F, big F, on big E. OK, instead of on little F, on little E. Because this applies in a general setting. That's what I was telling you before. So we're going to use this proposition that have F as a unique fixed point. And how are we going to do that? By showing that this map itself can be written as a perturbation of a linear map on this Banach space. This is where this is the key subtle point. This is the we've come to the heart in some sense of this whole argument. And I'm trying to go slowly step by step so you don't get lost and you can follow each bit. So we're going to write it like this. So let me raise this. So this map F is a map from E to E. It's given by this. And I claim that this is the linear part, that this is linear. And I will call it curly T. And this here is the perturbation of that linear part. And I will call it. So we're not going to show directly that it's a contraction. It's a little bit more sophisticated argument. I am now going to show that this is a linear map from E to E and that this linear map is a contraction. And then I'm going to show that this delta F satisfies the assumptions here. So it's bounded. And it satisfies the Lipschitz condition. I think it was this one that we need to apply for position 2. And then we will apply for position 2 that says that this map has a unique fixed point. So lemma 1 or lemma 2 for this purpose, for this class. T from E to E is linear. And the norm of T, the operator norm of T, like the norm of T as a linear map, so the supremum of all unit vectors of the image of T is less than or equal to the norm of the original T in the original Banach space, which is less than 1. So this T is a linear contraction. So proof. Let's show first of all that it's linear. So we just check. We just use a definition of linearity and check. There's nothing. So we take two functions, phi and psi in E. And we take two scalars, alpha and beta in R. And you remember the definition of linearity just says that curly T of alpha phi plus beta psi. We're going to have to show that this is equal to alpha T of phi plus beta T of psi. That's just the definition of linearity. So we just apply the definition of curly T, or big T. Let me call it big T, or curly T, because this is already a big T, curly T. So this is OK. So what I mean by this is this is still a function of phi, T of phi, delta f of phi. This is what these are. So this map here sends phi to this composition. So this is our new function. So this map, curly T, maps this to that composition, which means it maps this to this composition, alpha phi plus beta psi composed with g minus 1. This is the definition of curly T. This is just a sum of functions applied to g minus 1. So I can multiply this out. So I can write T composed with alpha phi composed with g minus 1 plus beta psi composed with g minus 1. And now we use the linearity of T. And so this is just equal 2. So by linearity, I can multiply this out. So T applied to this plus T applied to this. T applied to this is just equal to alpha T composed with phi composed with g minus 1 plus beta T composed with psi composed with g minus 1. And this is just exactly this. So this is just exactly equal to alpha tau, sorry, curly T of phi plus beta curly T of psi. So this shows linearity. Now we need to show that this is a contraction. It's a linear contraction. So for the norm, we have that the norm of curly T on E is by definition equal to the supremum over all phi with curly norm 1 of tau of phi in curly E banach space. And by the definition of tau, this is just equal to the supremum phi equal to 1 of this function here, which is T composed with phi composed with g minus 1. OK, so we take the supremum over all functions phi with norm 1. And so all we need to get is a bound for this. So for all phi with phi of epsilon equals 1, we have that T composed with phi composed with g minus 1 is less than or equal to. So this norm here is just the maximum. So now here we're taking the norm of this as a function in this banach space. This is just the supremum of the images of all the points. And so this is bounded. Because phi is bounded, this is bounded. We apply the linear map to these points that are bounded. So this is less than or equal to the norm of the linear map in E multiplied by the maximum value that we can get here. But phi is bounded, and it has norm 1. So the maximum value that this function can take is 1. So this is just 1. I can write this. So this is the norm on phi composed with g minus 1 in E. But this is just 1. And so this is less than or equal to T in E, which is what we wanted to show, which is less than 1. So this proves the first part of the lemma, the fact that this function from curly E to curly E is a linear contraction. Now what we need to show is that this is a small Lipschitz perturbation. So let me create a bit more space, actually. I just wanted to leave this, because I wanted to remind you that we're going to use this proposition here, which, if you remember, uses this assumption. The Lipschitz of delta f is less than 1 minus T. So this is what we're going to show. So lemma 3 is Lipschitz with Lipschitz delta f less than or equal to, sorry, delta f less than or equal to Lipschitz of delta f, which is less than or equal to 1 minus T. OK, so I'm just going to do this calculation. Then we'll take a couple of minutes break so you can relax your minds before we complete the proof. Let me give this proof first. So this is delta f. So first of all, let me remind we've got to do some estimates on the Lipschitz constant here. So let me remind you how that is defined. So the Lipschitz recall that Lipschitz of delta curly f is equal to the supremum over all functions phi and psi in E of delta f of phi minus delta f of psi over phi minus psi A. Yes, OK. And the same for the Lipschitz constant of delta small f, because we're going to compare these two is the same thing but just with the appropriate norm, right? So x different from y in E. And here we take delta f of x minus delta f of y in this norm E over x minus y in E. So these are the two quantities which we need to compare, OK? So we just plug in. This is a classic proof where you just plug in the definitions and you get what you want. So let's write delta f of phi minus delta f of psi is equal to. So what is the definition of delta f of phi? So this is just equal to delta little f composed with the identity plus phi composed with g minus 1 minus delta little f composed with the identity E plus psi composed with g minus 1, because in the definition of delta f there's another term, but notice that the other term does not, it cancels out. The other term is just that t composed with g minus 1 minus the identity. So it's the same term on both sides, so it just cancels out. So here, again, we just multiply everything out here. So we get delta f composed with the identity of E composed with g minus 1 plus delta f composed with phi composed with g minus 1 minus delta f composed with the identity composed with g minus 1 minus delta f composed with psi composed with g minus 1. And these two cancel. This and this cancel. So we have delta f composed with phi composed with g minus 1 minus delta f composed with psi composed with g minus 1. And I'm going to write this as this is equal to delta f composed with phi minus delta f composed with psi all composed with g minus 1. I can just add them like that. And the reason I want to add them is because here we have a bounded function. Here we have a bounded function. g minus 1 is a homomorphism. So that means that the norm of this delta f phi minus delta f psi in the curly Banach space is just equal to the norm delta f delta little f composed with phi minus delta little f composed with psi in the normal Banach space. So once we have these norms are the same, we just have from the definitions delta curly f of phi minus delta curly f of psi over phi minus psi is equal to. Sorry, this should be this is still a function in E. So this is still the curly E. So this is just equal to delta f composed with phi minus delta little f composed psi over phi minus psi. And now we use the definitions of these norms. So this is equal to on the numerator we have the supremum over all z in E, the supremum norm. This is a function in curly E. This is a supremum norm. It's a definition. It's just a supremum over all z in E of delta f composed with phi of z minus delta f composed with psi of z E. And on the bottom we have also the supremum over z in E of phi of z minus psi of z E. This is the place where we switch from one norm to the other. And so now if you think a little bit, you realize that we have the supremum here. So with a less than or equal time we can take the supremum of the ratio, supremum over z in E of this ratio delta f composed with phi of z minus delta f composed with psi of z over because here we put something that is smaller than this. So this is greater than or equal to. And that's it. And this is exactly the definition of the Lipschitz constant of delta little f. Well, this is less than the Lipschitz constant of. So this shows, so let me just say what we have shown so far. So we have shown on curly E to curly E is of the form curly f is equal to curly t plus delta curly f where t is linear. The norm of t is less than 1. And the Lipschitz constant of delta f, actually the Lipschitz this is less than or equal to the norm of t of E less than 1. And the Lipschitz constant of delta f is less than or equal to the Lipschitz constant of delta f. And so the assumptions of the position 2 hold and delta f has a unique fixed point which we shall call delta H that belongs to E. Moreover, the conjugacy condition is satisfied. So f plus delta H is equal to IDE plus delta H composed with G. Delta H in E, so it's bounded and continuous. There's still one more thing we haven't shown. So what is it that we haven't shown? Exactly, exactly. So let's take now just a couple of minutes break, refresh your brains. It's very easy, it will just be a few lines. But I think now is a good time to take a break for the first part and then we'll come back just in a couple of minutes. So after the break now we can complete the proof of the theorem. As we said, we have shown that there exists a unique bounded and continuous delta H. So that the conjugacy equation is satisfied. But this does not automatically imply that this is a homeomorphism. So to show that it's a homeomorphism, we need to show that it's invertible. And what we're going to do is actually find the inverse of this function. And the way we're going to find the inverse is by applying, in principle, the whole argument we had before to the equation with f and g swapped. So by the same argument used above, there exists unique delta H bar such that g composed with ide plus delta H bar is equal to ide plus delta H bar composed with f. In other words, of course we don't have to do it because we already did it. It's exactly symmetric. And that's one of the reasons why we wanted to work with g instead of t so that we could swap them around and it would be the same thing. So if you start with wanting to prove this equation, you will get everything exactly the same as we did before. But of course, you will end up with a different delta H bar than delta H, because the role that g and f play in that functional equation is not exactly the same. So you will get all the same estimates, all the same bound. You will get a delta H bar, but it's a different one and it will satisfy this. So we have two. We have one that's a conjugacy between f and g like this and we have another function that's a conjugacy between g and f like this. And then we just do a little bit of manipulation. So let's call this equation 2, this equation 3. So starting with 3, so what we can do is we can compose both sides on the right by id plus h. So composing on the right by the continuous function id plus delta H, we get g composed with the identity of e plus delta H bar composed with identity of e plus delta H equals id plus delta H composed with f composed with id e plus delta H. Yes, sorry, bar here, yes. Now we can use 2 here. We can use the conjugacy in 2. So this is equal by 2 to identity of e plus delta H bar composed with identity of e plus delta H composed with g, which means that this composition is the identity. So g is conjugate to g itself just by the identity. So we've just proved it. We've just proved it. We have this, we've proved this, and we've proved this. We've proved both. So we can use them. So what we do is we start with this expression here and we compose on the right by this. To compose on the right, this doesn't need to be homomorphism. It's just a continuous function. We compose like this and like this. But now here, we have the conjugacy above, which means that this is equal to this. Well, it means, OK, so we haven't completely proved that it's the identity yet, because for two functions to be the inverse of each other, they need to be the identity whichever way you compose them. We have uniqueness of delta H. Exactly, exactly. So by applying the same, the fact that we need to make an observation, which is that this function here is also of the form identity of e plus a continuous bounded function. So observe, notice that if we have identity plus phi composed with identity plus psi, this is just, so we sum this. So this is equal to the identity, identity composed with identity plus psi plus phi composed with the identity plus psi. And this is just equal to the identity composed with the identity. So identity plus psi. And this is phi composed with the identity is phi. And phi composed with psi is just phi composed with psi. So this is equal to the identity plus a function, which is psi plus phi plus phi composed with psi, which is a bounded and continuous function. So this means that this is a composition of identity plus a bounded continuous function, and identity plus a bounded continuous function. So that means that this itself, this composition, is a bounded continuous function. So if we write this function here, we write it in the form ide plus phi plus delta h, big h, then g composed with identity of e plus delta h is equal to identity of e. Sorry, so this is the composition the other way around. So I need both. So I need both. So I need to do the same thing with this. So here I need to compose on the left, compose by identity of e plus delta h bar. And here I do a similar thing. So here I'm going to get identity of e plus delta h bar composed with f, composed with identity of e plus delta h is equal to identity of e plus delta h bar composed with identity of e plus delta h, composed with g. And let me just see which way do I want to know. I've got it the wrong way around here. So composing both sides on the left by id plus delta h bar. Yes, the problem is, this is the same thing. Yes, yes, OK. I don't know why. I suddenly thought that these were switched. So this implies g composed with identity of e plus delta h is equal to identity of e plus delta h composed with g. That's right. Sorry about that. Exactly. And because of uniqueness, delta h is equal to g. So by uniqueness of delta h, we must have delta h equals 0 since g composed with the identity equals identity composed with g. That's right. Yeah, so strictly speaking, the reason why I was doing it the other way, strictly speaking, you do need to show that the composition of these two, the other way around, is also equal to the identity. Because by the same, exactly the same calculation. Because the definition of the inverse is a function such that f composed of f minus 1 equals f minus 1 composed of f equals the identity. So you do exactly the same calculation there, and then you get exactly the same thing. OK, so we've proved that if you have a linear contraction, so let's just remind. So we've proved that if you have a map f equals t as delta f, where t is less than 1 and Lipschitz of delta f is small, then f is topologically conjugate to t. This is the model of the result. So two comments. One, so both of them to do with the fact that how much can we generalize this? So does really t need to be a contraction for this to be true. So if you remember the first semester, we discussed linear maps a lot. And we discussed the topological conjugacy of linear maps. And we realized that for many results, the key property was hyperbolicity of linear maps. And a linear map is hyperbolic if it has some part of the eigenvalues which are contracting and some part of the eigenvalues which are expanding. So this result is actually true in much more generality if you assume that t is a hyperbolic linear map. Contracting linear map is one example of a hyperbolic linear map. And in the hyperbolic case, this is called the Grobman-Hartman theorem. You will see it in some of the books that I gave you in the bibliography, some of the standard dynamical systems book. The proof is similar, but there is a significant additional complication because of the direction that is expanding. And you need to change it. It's a similar kind of construction. And it reduces it to a fixed point theorem, but it's quite a bit more involved. So I decided to give here just this case in which is a contraction which shows the basic idea of the argument in a slightly simpler setting. This is one remark. So the second remark is that, however, as you remember in the finite dimensional setting, there are some linear maps which are not contracting in that sense, but still all the orbits converge to the fixed point. Do you remember? What is the difference between having a contracting linear map in R2, for example? What's the difference between having a contracting linear map and a linear map where everything converges to the origin? Is it the same thing? Does every map like that? Is it contracting? Is that it? So all of this, we've proved it in a very general setting of Banach spaces, which are infinite dimensional spaces. Last semester, we looked at linear maps in finite dimensions. In particular, we looked at two dimensional linear map. So now let's suppose T from R2 to R2 is a linear map. This is much more familiar territory. So we studied extensively all kinds of two dimensional linear maps. And we looked at their dynamics and their behavior. And you know that, of course, if the norm of T is less than 1, what does the norm of T being less than 1 mean? The norm of T being less than 1 means that if you look at the action of the linear map, for example, if you look at the image of the unit circle, it will map inside itself, because T less than 1 means that every unit vector is contracted. And in particular, it means that all the points converge to the origin. But there's also some linear maps where all the points converge to the origin, but it's not contracting in this sense. What linear maps are those? Do you remember? Right. What are the maps where you know that the origin is a globally attracting fixed point? What's the characterization of these linear maps? If you bring it into diagonal form, this is the key point. So what is the distinction between the original linear map and the map in diagonal form? So remember that when you bring them into diagonal, by bringing into diagonal form, you find a linear conjugacy between your original map and a map that has some kind of diagonal form if the eigenvalues are real or some kind of normal form that we studied in the case in which the eigenvalues are complex. But when you do this linear conjugacy, the eigenvalues stay the same. Remember that. The eigenvalues stay the same. So in the linear case, you might have this situation if you remember. For example, if you have complex eigenvalues, you might have that you have a linear map with complex eigenvalues. And the phase space, the orbits of the linear map, they lie on some spirals. But the spirals look like this. You recognize such a picture? Yes? Is this map contracting? Not in this sense, right? Because you can see there are points, for example, this point here might be mapped to this point here. Eventually, it goes to zero. But it doesn't go monotonically to zero. This point gets mapped to this. So the unit circle gets mapped to an ellipse, but it doesn't get mapped inside the unit circle. So it's not contracting in this case. So natural question is, well, does our theorem apply to this map as well as to the map if we take this, if we do the linear conjugacy, remember to the normal form, what does the linear conjugacy do? The linear conjugacy by some matrix P precisely maps these to much more symmetric spirals in which the point is monotonically shrinking towards the origin. In this normal form, this matrix is contracting. So the operator norm of that matrix is less than 1. So the norm of the matrix changes. The eigenvalues of the matrix don't change, but the norm of the matrix changes in this case. In this case, the norm is bigger than 1. In this case, it's less than 1. This is not a satisfactory situation because these two are linearly conjugate. They are really, according to our philosophy, they are essentially the same dynamical system in a very strong sense. Much more than topological conjugacy. It's a linear conjugacy. We would like to say that if this theorem, this Lipschitz perturbation theorem applies to some map like this, which it does, because this map satisfies the linear map, satisfies the assumptions of that it's contracting, if we take a small Lipschitz perturbation of this, we do not destroy the topological conjugacy class. What about this? What if we start with this linear map and we take a small Lipschitz perturbation? Will we still not destroy the topological conjugacy class? It would make sense. This is what we would like to have in this case. So this is, in fact, true. And the way we show it is that really this question about the norm is much more flexible than we want, than we think. So really, there's a different way to look at this linear conjugacy. And in some sense, this linear conjugacy, you can think of this linear conjugacy as a change of coordinates or a change of metric or a change of norms in some ways. So another way to think of this linear conjugacy is that there is a different norm on your space, such that in that norm, this looks like this. You can just think of like that. And this is simply the norm that measures this distance here as this distance here. It's a measure that is, it's a norm that makes these distances a little bit shorter and these distances a little bit longer and it looks the same. So if we use that norm, then the matrix is contacting in that norm. And then as long as we define the Lipschitz perturbation, because the Lipschitz perturbation is defined also in terms of the norm, as long as we define the Lipschitz perturbation to be small in terms of that norm, then everything should apply to this situation as long as we look at it in the right norm. So we are out of time now, so I'm not going to do that now, but when we, the next lecture, we will just do a simple calculation to show that if you have in the two-dimensional case, if you have a system that has all the eigenvalues less than one, which is the condition that we had to construct this linear conjugacy, or just more general in arbitrary dimension, if you know that your system is linearly conjugate to one that is contracting, then you can use that to actually construct a contracting norm for your original system, and then you can apply this theorem. So for the moment, I just wanted to present the problem, so the next lecture, we can go straight into and prove this change of norms that we'll apply in this case. So I think we've done enough for today. Thank you very much.