 So yesterday we started talking about interval diffeomorphisms, if you remember. And in particular we were looking at two interval diffeomorphisms that have just two fixed points. So if they have just two fixed points, then these fixed points must be the end points, right? And so, and the graph cannot intersect the diagonal, so it must be something like this. So one diffeomorphism must be something like this and the other one must be something like this. So the proof, I thought I should go into the proof, but actually thinking about it is really straightforward, so I will leave it. It's exactly like the proof that they're topologically conjugate. So I stated yesterday the fact that any two such interval diffeomorphism are topologically conjugate. Notice of course that, yes, so all you do is you construct the conjugating homomorphism exactly in the same way as we constructed it the other day. Constructing a fundamental domain, fundamental domain here. Okay, because as we saw, all the points move from left to right, from one fixed point to the other, right? Because we proved that the only omega and alpha limits are fixed points, so all points have to move from one fixed point to the other. So they have to cross this fundamental domain and you can do the whole construction exactly in the linear case, okay? So rather than spending valuable lecture time doing this construction, I will let you do it as an exercise. It would be a very good exercise, yeah? There's really nothing to it. You look at the proof for the linear case and it's exactly the same in this case. So these are topologically conjugate. And what we are more interested is in the general case in which we have many fixed points, what will happen. And this also is not so difficult. So let me remind you first of all that we say that f, so that if f from i to i is an interval, is an orientation preserving interval diffeomorphism. I always mean at least the c1 diffeomorphism when I say diffeomorphism. If p is a fixed point, we say that p is hyperbolic if f prime of p is different from 1. And we say that f is hyperbolic if all fixed points are hyperbolic. So why do we insist on this hyperbolicity of the fixed point and hyperbolicity of f? So the first lemma is that if p is a hyperbolic fixed point, then it is isolated. You know what isolated means? Exactly. Okay, it's isolated as a fixed point. So there exists neighborhood u of p such that all x different from p with x in u are not fixed. So why is this term? It's a c1 diffeomorphism converges to the fixed point. I think your basic idea is right. I'm not sure however, yeah, I'm not sure if how easily you can actually prove it with that argument. But actually it's very simple. If you have, this is the diagonal. What does it mean that the fixed point is hyperbolic? It means that the derivative at the fixed point is different from 1, right? And if the derivative is different from 1, because this is c1, it means there is some neighborhood. So this is the fixed point p. It means there's some neighborhood where the derivative is also different from 1, right? So if p is hyperbolic, p hyperbolic implies f prime of p is different from 1. So there exists neighborhood u such that f prime of x is different from 1 for all x in u, right? Now suppose there was another fixed point. Suppose by contradiction there exists q in u with f of q equals q. So what would that mean? That would mean that this comes back like this. So how can this be possible? If there's a fixed point here and there's a fixed point here and the derivative here is bigger than 1, then it means by the mean value theorem that there must be some point here where the derivative is equal to 1. And this contradicts the assumption in this neighborhood. Because the only way you can map, you can have a map that maps f of p to p. f of p equals p and f of q equals q is if you have one point inside where the derivative is equal to 1. Because you have f, then f of p minus f of q equals p minus q. And so by the mean value theorem there exists some z in the interval pq such that f of p minus f of q equals f prime of z times p minus q. So this means the derivative of z is equal to 1 which contradicts the assumption. So f prime of z equals to 1 contradiction. So this is the formal argument. The intuitive geometric argument is quite clear. It's just that since the derivative is different from one, then the graph crosses the diagonal transversely and therefore there must be some neighborhood where it cannot cross it again. Otherwise it would not cross it transversely. But this is the formalization of this argument. Okay, so why is it important that this map is isolated, that these points are isolated? Well, an immediate consequence corollary is that if f is hyperbolic, so if all the fixed points are hyperbolic, then f has at most a finite number of fixed points. Why is this? Why is this? Corollary. Can you tell me? Yes. Each point is isolated. Wait, wait, wait, wait, wait. Why does this imply a finite number of fixed points? Can you? Do you know? Why? If there's an infinite number, then you would have some neighborhood of what? There would have an infinite number of fixed points. That's right. i is compact. And so? And so? I mean that's basically correct, but you need to make a link in terms of fixed points. Right? A neighborhood. Yeah, but you can have a countable number of disjoint neighborhoods. That's not a problem. You don't need to cover that. You can just take a countable number of fixed points and each one has its own neighborhood and they're all disjoint and you can still have that. I mean your argument is going in the right direction. You just need to complete it. What's? Yes. The interval is compact. Yes? I don't know. Discovering arguments, I don't think it's... Yeah. Okay. Okay. Okay. And where's the contradiction? Exactly. Okay. There you go. This is the simplest argument. So proof, i is compact. So, if there exists infinite fixed points, okay, there exists an accumulation point p which is fixed. Why is it fixed? Why is this accumulation point fixed? Exactly. Okay. Check this. So this is by continuity. Okay. You should check this. Exercise. Check this. Okay. Is p which is fixed and not isolated by definition. So it cannot be hyperbolic. Okay. So p is not hyperbolic contradiction. Okay. And this already says that this definition of hyperbolicity, it's a significant condition. Okay. So we are restricting us. It implies already quite a lot of things because, of course, there's lots of different morphisms that may have an infinite number of fixed points. But the hyperbolicity is a very natural condition and it gives already quite a lot of information about what happens. So the main proposition, we will prove here, proposition, is that if f and g are two interval differ morphisms, okay, orientation, reserving, c1 differ and hyperbolic with the same number of fixed points with the same number. Okay. Maybe I should say differently, hyperbolic, okay. Then fg topologically conjugate if and only if they have the same number of fixed points. So I will sketch the proof of this although again it is fairly simple because it will be similar to the proofs we've done before. First of all, let me just say a simple corollary. So what are the topological conjugacy classes of orientation preserving c1 hyperbolic differ morphisms? How many are they and how are they characterized? The immediate consequence of this proposition. So exactly what this proposition is saying is that if you take all the hyperbolic, all the space of hyperbolic orientation preserving c1 differ morphism, this splits exactly into a countable number of topological conjugacy classes where each conjugacy class is characterized just by the number of fixed points of that differ morphism. So each number n equals 5, 6, there is a topological conjugacy class, all the interval differ morphism c1, orientation preserving c1 hyperbolic differ morphism with 5 fixed points are all topologically conjugate to each other if and only if. So conjugacy classes hyperbolic orientation preserving c1 differs characterized by the number of fixed points. This is not even really corollary, it's just a different way to state exactly the same result. So characterized means that that is the only information you need, the number of fixed points and you got your conjugacy class. So how do we prove this proposition? So one direction is clear, right? Kevin, which direction is obvious in this proposition? There's if and only if, right? Which one is that? So we need to say if they have different number of fixed points they're not topologically conjugate. And if they have the same number of fixed points then they are topologically conjugate. Which of these is obvious? If they have the same number of fixed points then they're topologically conjugate. That's one direction. The other direction, if they do not have the same number of fixed points then they're not topologically conjugate. If and only if. Which one is obvious? Do you understand the difference between these two, right? We take two interval differ morphism. You look at them, say okay, do they have the same number of fixed points, we want to prove they're topologically conjugate. If they do not have the same number of fixed points then we need to prove that they're not topologically conjugate. If and only if. Which one is the first one? Yes, that's right, but that's the second one that I said. They should have. So anybody else can tell me? Can you tell me which is the obvious direction in this proposition? Why is that? The answer is correct, but the question is why? If they don't have some name number of fixed points they cannot be topologically conjugate. Why? What does a topological conjugacy need to, what structure does a conjugacy need to preserve? Yes, it should preserve fixed points at least, right? This is the most basic fundamental property of, not even topological conjugacy, just conjugacy. If two systems have different numbers of fixed points they cannot be conjugate. Right? Okay. So one direction here is clear. What we need to show is that if they have the same number of fixed points then they are conjugate. Okay? So let's suppose we have F from AB to itself and G from some other interval, A prime, B prime to itself. Okay? And let's suppose they have the same number of fixed points. So let's suppose that we have A equals A1, A2, AS equals B are the fixed points of A and A prime equals A prime 1, A prime 2, A prime S because it's the same number of fixed points equals B prime are the fixed points of G. Right? So we have, in this case, we have 1, 2, 3, 4, 5, 6 fixed points, right? So 1, 2, 3, 4, 5, 6 fixed points. So this is A1, A1, A2, A3, A4, A5, and A6. A1 prime, A2 prime, A3 prime, A4 prime, A5 prime, A6 prime. This is how the picture must be, okay? The only thing that can change is the number of fixed points, but this is how it needs to be. So notice, of course, that between, by definition, this is a different morphism. The finite number of fixed points between any two fixed points, the picture is either like this or like this, right? The picture is always the same between any two fixed points. If this is AI and AI plus 1, AI plus 1, this is the diagonal. So I'm kind of zooming in to this part, and then the picture will always be, say, either like this or below the diagonal, right? But it always looks in one of these two shapes. And what does this remind you of? Well, it reminds us of the previous case, where we had only two fixed points. What is the difference? It looks exactly the same. So what we have here is that this interval, AI, AI plus 1, maps also to the interval AI, AI plus 1 because AI maps to AI. AI plus 1 maps to AI plus 1, right? You can see it here. So A3 maps to this. It's a fixed point, by definition. So this is also A3. This here is A4. So this is also A4. So really what you have is the interval A3. This interval here, A3, A4, mapping to the interval A3, A4, right? So as a different morphism. So this itself is an interval different morphism that I can draw like this, right? If I want, I can draw it. I could draw it really... Here I try to draw as an example of how it is as part of the bigger picture. But if you just restrict yourself to this, what you get is a map of this interval into itself that looks exactly like this and it's got two fixed points that are exactly at the end points. So the reason I'm telling you this is because this is the clue to the construction of the homomorphism. So how do you suggest we construct this homomorphism? Here. Well, we can do it step by step, right? So let's look at this interval here, A1, A2, which is an interval different morphism. And let's look at the corresponding interval here, also A1, A2, which is also an interval different morphism. Both of them have just two fixed points at the end points. So what did we just prove before? That they're topologically conjugate. Okay? The map is topologically conjugate. So it means we can define a homomorphism from A1, A2 to A'1, A'2, which conjugates the dynamics inside these two inside these two intervals. Okay? And notice, of course, that this is made of pieces that are all independent of each other. So basically these different morphisms can be split up. A1, A2 maps to A1, A2. So when you start in here, you stay in here forever, right? And when you start inside this interval, you stay inside this interval forever. Indeed, we saw exactly what happens, right? What happens is that every point in here will converge under iteration either to one fixed point or the other fixed point. So it's very simple. It's just like the dynamics when you have just two fixed points, like this. So we construct a homomorphism between this interval and this interval. Then we construct a homomorphism between A2 and A3 that maps to A2, A3, and so on. And we do it for each of these intervals. And because these homomorphisms coincide on the end points, this homomorphism maps A2 to A2' and this homomorphism also maps A2 to A2'. Then they come together to construct, to become one big homomorphism from the whole interval to itself. So it's very straightforward, really. So define for each interval Ai, Ai plus 1 and Ai prime, Ai plus 1 we can define a conjugacy H from Ai, Ai plus 1 to A prime, Ai, A prime, Ai plus 1. Let's call this Ai and then let H from AB on the whole interval to A prime B prime be defined by H restricted to Ai, Ai plus 1 to be simply equal to Ai. Any questions? Okay? No? What are your doubts? Did I go too fast? I see some puzzled faces and I'm not sure which part of this argument is puzzling. So do you agree that on each interval between two fixed points the dynamics is just like the dynamics of intervals with two fixed points that we studied before? So on this map we have an interval that maps to itself with only two fixed points. And on this interval we have also an interval to itself with these two points with only two endpoints and so there's a conjugacy. There's a map, a homomorphism between the interval A to A3 which I call Ai to Ai prime, Ai plus 1 which conjugates the dynamics restricted to these two intervals. It exactly has the conjugacy property. If you take a point in here and you map F it maps to the image of H of this point here. It exactly is a conjugacy between the dynamics restricted to these two intervals. And then you can define because all these maps H, Ai they're defined independently by their coincides so H1 maps this interval to this interval H2 maps this interval to this interval they both are defined at the point A2 but they both have the same image as the point A2 so they're all consistent with each other so you can just define a global homomorphism by defining this homomorphism to be equal to H1 restricted to this H2 restricted to this and so on and then it's not difficult to check that this is a conjugacy, okay? Let me maybe write it as an exercise just to check this but exercise H is a conjugacy is a topological conjugacy between F and G. The pasting lemma What's the pasting lemma? Yes, you're right, you're right, yes. Right, you're right, so yes I kind of assume that you're right but in principle we'd have to show that H defining this way is continuous, yes. But that's why I was remarking that it has the same Yes, because the continuity from the left comes from the fact that it's a homomorphism here the continuity from the right comes from the fact that it's a homomorphism for the interval, so it's continuous from the left and from the right because they have the same image. You're right, this is an observation that should be made, thank you. And the fact that it's a conjugacy follows immediately from the fact that it's a conjugacy on each interval independently, right? And these intervals they do not mix. Points that are in this interval stays in this interval in forward and backward time forever and so on, so it's a conjugacy this technique of fundamental domains that we introduced in the linear case has already proved quite powerful in general cases, right? We've already used it, we looked at it in detail in the case of linear systems and then we used it I did not repeat the argument because it's exactly the same we used it in the special case in which we have an interval with just two fixed points and then interestingly that was sufficient in the general case because the general case reduces to a finite number of intervals each of which is exactly that property of having just two fixed points at the end points. All right, so now we want to now that we have really classified this is part of the goal at each of these steps you take a class of systems and you classify them up to topological conjugacy we have essentially done this whereas again as we did in the linear case the question of structural stability remember what structural stability means you perturb the system a little bit and does it remain inside the same conjugacy class so to do this we need to introduce a very important interesting concept which is what do we mean by a small perturbation of this map in the linear case it was obvious because the linear maps are parameterized in the real line a of x equals ax so you just change a a little bit and you get another linear map here we don't have such a parameterization here we just have interval different morphism and we want to perturb this a little bit so we need to give a meaning to what we mean by perturbing this a little bit any suggestions for what we mean by small perturbation of this so okay let me draw another picture because this one so let's think of intuitively how you would go about defining a small perturbation of this so let's suppose we have a map an interval different morphism that looks therefore say like this so it's got two fixed points sorry it's got four fixed points what is a small perturbation of this well that is something that we will have to yes that's what you would like from you would kind of like if you can find a way of defining a small perturbation to guarantee that it doesn't change the number of fixed points you would automatically get structural stability as long as they're still hyperbolic so it should not change the number of fixed points and they should remain hyperbolic and then you guarantee structural stability right so what kind of perturbation can we want to make sure that that happens excuse me so okay so you could just shift everything up and down you have a little bit of a problem because these are fixed these two points so you cannot because it's a different morphism of the interval this will always map here and this will always map there like you cannot change that so you can't really shift the whole thing up although the intuition is natural to say okay can we shift it up or down a little bit but maybe we can just shift it up or down in different places right sorry point wise yes so we could say okay this remains closed we could just say the graph is close to this the new graph is close to the old graph epsilon very good so let's define what we call the c0 distance so let we define fg from i to i the continuous maps then we define what's called the c0 distance between f and g is equal to the supremum for all x in i of f of x minus g of x supremum which in this case is also because i is compact so I could write maximum here so what does this mean this means that for every point x so we have two suppose we have two maps for each point x you look at f of x and g of x and you take the distance between them and you take the supremum of all of this this is sometimes called the uniform norm or c0 c0 distance c0 distance or uniform norm sometimes it's also called uniform norm ok so if so d0 fg is equal to epsilon if f of x minus g of x is less than equal to epsilon for all x in i this is what this means so what can we change within an epsilon neighborhood so suppose we let me just throw this picture again suppose we start with one map ok you choose an epsilon and you say ok let's look at all the maps within this epsilon neighborhood how much can the situation change so what does choosing an epsilon mean it means that for each point for each x the new map is allowed to be within an epsilon plus or minus epsilon distance the new graph is contained if this is f of x then g of x is allowed to be within this epsilon neighborhood of f of x so really you have a kind of neighborhood of the graph an epsilon neighborhood I mean I've drawn it quite big but of course you can imagine it very small so the new graph needs to be contained in this so following your suggestion here can we change the number of fix points with a graph that lies inside this epsilon neighborhood of f we cannot maybe so for example maybe I can draw something like this doesn't even need to oscillate that much can I avoid this by making epsilon smaller yeah I mean I drew this picture with this size but of course I can scale down the whole picture and I can draw it with arbitrarily small epsilon so what does this suggest that with this notion of distance even hyperbolic maps will not be structurally stable because however small the epsilon I choose I will be able to find inside I will be able to find another map g which is close enough to f but has a different number of fix points and therefore it belongs to different topological conjugacy class so the system is not structurally stable but I can also have different notions of distance here and this is why it's very interesting to think of different topologies so this distance this is a metric and you should check this I will put it in the exercise to check this is actually a metric on the space of continuous maps just also a small observation is that this is actually a metric that's defined on all continuous maps not necessarily invertible but of course all homeomorphism is a subset of this and so this defines a metric also on the space of all homeomorphisms this is a metric as a metric it uses a topology on the space of continuous maps this is the topology the topology says that these two points are close if they are within their graphs are within epsilon of each other we can define a different metric on this we can define for example there's many many different metrics on this we can define what we call d1 of fg and we will define it like this to supremum over all x in i of f of x minus g of x plus f prime of x minus g prime of x what is the difference between these two distance functions the slope must also be close so at each point what you do is you look at every x again and you look at the value of the functions and you take the difference between the value of the functions but you also take the difference between the slopes of the functions the derivative of the functions at this point so for example here the slope is quite different even though the value of the functions is very close here for example at this fixed point the slope is different even if they have the same fixed point and the distance is zero the slopes are different so the distance is non-zero so there's a very interesting to appreciate the difference between these two this is also metric on the space of continuous maps so this also induces a topology but these are different topologies which one of these topologies is stronger in the sense which one you might have two functions which are very close in one of these metrics but very far in the other metric d1 is stronger in what sense that you can be very far in the d1 metric even though you're very close in the d0 metric for example these functions if you take very small epsilon any two functions that are inside this neighborhood will be close in the c0 metric but if like in this case you put some slopes and you change the slope quite a bit they will be far away in the c1 metric let me give you a more explicit example which I think helps very much to illustrate this so remark d0 and d1 induce a very different topologies on the space ok so in this case you need the map to be differentiable right so in this case fg fg should be c1 for this to make sense so of course for c1 functions they're also continuous so in the space of c1 functions you can define both of these metrics on the space just of continuous functions you can only define this one but on the space of c1 functions you can define both so just on the space c1 i of c1 maps so example let so example suppose i is equal to the interval 01 and let f of x be the constant function 1 half so interval 01 1 half so f of x is just constant function 1 half right and g of x is equal to 1 half plus epsilon sin x over epsilon what does this look like so as you know of course the sin of anything oscillates between 1 and minus 1 right so it remains inside the epsilon neighborhood here so we have here we have 1 half plus epsilon 1 half minus epsilon the only thing that this epsilon does is increase the number of oscillations right because if epsilon is small then really as x varies between 0 and 1 this becomes very can go from 0 to becoming quite big so there's more oscillations right so it looks like something like this with the number of oscillations increases with epsilon so what is the c0 distance between these so d 0 between f and g is equal to epsilon let's parameterize this by epsilon for the c1 distance notice that the derivative exactly so the derivative of this is just cosine of x okay when you differentiate is cosine of x that oscillates also between 1 and minus 1 and the derivative of this of course is just equal to 0 so what is the c1 distance is actually equal to 1 plus epsilon right 1 plus epsilon sorry yes maybe maybe yes so as epsilon goes to 0 okay so the crucial observation here is what happens you can take arbitrarily small epsilon and what happens the distance in the c0 distance between f and g is going to 0 the distance between f and g epsilon is not going to 0 it's bounded yes okay let me yes you're right okay so let me also use this opportunity to remark that I chose to put a plus here there's different ways in which you can write this if you want you can just take the maximum between these two there's some slightly different ways of writing the norm they're all absolutely equivalent I just decided to write plus but you're right it's not completely clear that's allowed a little bit it's 0 you're right so this is probably maybe equal to 1 anyway the important observation is that it's always greater than equal to 1 this distance okay for me this is a really interesting application so you study in abstract you study topology you study function spaces you study different topology on the same space so what we have here topology is the same space okay but what does it mean to have two different topologies on the same space it means that you take a sequence of functions it's the same sequence of functions okay in one case this sequence of functions is converging to this case in one topology in the sense that the distance is going to 0 in the other topology it is not converging to this function it's different ways of measuring distance between within this space okay I think this is a very simple but really excellent example to illustrate this so it also shows how much stronger this C1 topology is because to converge in the C1 topology you need to converge in a much stronger sense the map needs to converge and the derivatives need to converge okay so what is our conclusion of all of this about structural stability let me state the result here so position at f hyperbolic C1 interval differential okay always orientation preserving okay so maybe I'll present this in a little table it might be useful so we're interested in structural stability and as you remember there's two facts that affect the structural stability that's the type of conjugacy that you define so we can have either topological conjugacy or we can have C1 conjugacy and the other one is the topology that we have on our space so we can have the C0 topology and we can have the C1 topology and now we want to fill in in each of these boxes whether this map is structurally stable in this setting or in this setting or in this setting so what is the result what do you think are the results going to be so three of these we are almost immediate okay the last one will be a little bit delicate so first of all and the C1 conjugacy what does C1 conjugacy mean do you remember so hyperbolic C1 interval differential morphism yes so we suppose we have f is a C1 hyperbolic interval differential morphism so I didn't discuss before going here I didn't discuss in the case with several fixed points the problem of the C1 conjugacy but you remember what are the obstructions to C1 conjugacy when are two maps C1 conjugate two maps with the same number of fixed points so that topologically conjugate when will they be also when do we know for sure that at least they are not C1 conjugate they have a different number of fixed points they are clearly not conjugate but if they have the same number of fixed points so that they are topologically conjugate what do we know about whether they might or might not be C1 conjugate exactly the derivatives of the fixed points have to be the same right so if you have two maps both of which have the same number of fixed points so that these are topologically conjugate but the derivative are different at the fixed points then they cannot be C1 conjugate so what is the chance of getting structural stability for C1 conjugacy if you perturb a little bit this map either in the C0 or in C1 topology it's clear that you can make small perturbations and you can change the derivative a little bit at the fixed point even if you fix epsilon in the C1 topology you're allowed the derivative to be a little bit different the C1 topology just says the map should be closed and the derivative should be closed but not exactly the same so clearly you can change a little bit at this fixed point so on the C1 conjugacy there is unstable there's no chance in both cases structurally unstable I'm not giving a proof I'm giving the statement of the proposition here in the C0 topology what do we think we saw the example before it means if you take so you're looking at topological conjugacy now and you're saying okay I just take an epsilon neighborhood of the graph and I make a perturbation do I stay inside the same topological conjugacy class? No so structurally unstable so the only hope we have left is this one here when you make a small perturbation in the C1 which means that you assume that the graph is inside inside this neighborhood and the derivative is also inside this neighborhood and the derivative is also epsilon close to the derivative okay so in fact we will get this structurally stable in this case this is the only case in which we have structural stability okay so I will give the proof it's not difficult but maybe let's just take a couple of minutes break and then we'll come back to it okay so let's prove so in the C1 conjugate we look at it in these different cases so for the C1 conjugacy it's very easy C1 conjugacy we need to show that a map F is structurally unstable hyperbolic linear map if if P is a fixed point of F then we have this picture here then it is essentially obvious that you can find a nearby map that has a slightly different derivative at this point you can even take this map to be almost the same until here then you just change the derivative a little bit and you take another map that has a smaller change of derivative so this map is no longer C1 conjugate to the previous one okay then clearly we can find another map G with D1 FG less than equal to epsilon such that and even the same fixed point with the same fixed point but G prime of P different from F prime of P and so G is not C1 conjugate to F and so F is not structurally stable so of course D1 less than equal to epsilon is stronger than D0 FG is equal to epsilon so obviously exactly the same thing applies even more if you take this in the C0 topology of both of these cases in terms of the C1 conjugacy this is yet another example of what I keep insisting of how strong the C1 conjugacy is so you basically have very hard to have structural stability with C1 conjugacy although I should mention that there are certain situations certain systems that have special symmetry so for example you could say okay I want to look at all the interval the C1 interval, hyperbolic interval that have some special property that is due caused maybe by some model some phenomena that you're trying to model or whatever reason there might be some symmetry or some restriction on the kind of maps you can consider which means that you can only perturb in a certain direction in a certain way and it could be that within that subfamily you do have C1 structural stability right? so I want to emphasize both of these things so in this particular family of maps which is all C1 interval, hyperbolic difference of it with the C1 conjugacy you have structural instability but if you take a subfamily of this it could be that this family in fact fixes the derivative at the fixed points even within if you fix the derivative at these four fixed points you still have a lot of perturbations you can make because you can change the derivative and the map everywhere else and it might still be C1 conjugate if you fix the derivative at the fixed points right? so I'm not saying that there are no other maps that are C1 conjugate to this but just in this particular topology that we've chosen there's quite a lot of scope for perturbations and therefore it's structurally unstable if you limit the kind of perturbations you can make then you might get structural stability so again this depends always very much not only implicit in the topology is the specific space that we've chosen right? so this topology comes with a space you could have the same topology but on a subspace of the full space and you could have structural stability here so when we say topology it comes with a space so this is the space of all C0 maps of the interval and this is the space of all C1 maps of the interval okay so this is an important remark okay so what about the topological conjugacy so let's look at the case in which we have topological conjugacy and C0 topology C0 topology topology so again we have our fixed point which looks something like this okay we have some epsilon neighborhood we have some C0 neighborhood locally of this fixed point we just need to do a local analysis and we need to show that we can create some fixed points in here fairly clear so really I just need to say it's clear that you can perturb this to get some fixed points so I will just write it but so let P a fixed point of F fixed point of F then for all epsilon there exists a neighborhood UP of P such that F of UP is contained in P minus epsilon P plus epsilon okay so the way I'm formulating this so this is P this maps to P because it's a fixed point so here is P plus epsilon here is P minus epsilon so neighborhood that maps some small neighborhood that maps inside UP so this is UP that maps inside this neighborhood right and then it's clear that we can change this interval the way this interval maps to this interval we can change it so that we create a different morphism and it has additional fixed points I'm not exactly sure how more what more to say because it's clear so we can make a small local perturbation of F in the C0 topology C0 topology such that perturbation G of F of F in C0 topology such that G has additional fixed points in UP and that way you lose the structural stability okay so these are all really the structural instability are fairly clear to see the less trivial part of structural stability under the C1 topology so let's finish with that so now we look at topological conjugacy and C1 topology okay so here we need to show that we cannot create any new fixed points and we cannot destroy any new fixed points so okay we need to show that any sufficiently small C1 perturbation will have the same number of hyperbolic fixed points these four hyperbolic fixed points in this case so we need to show that we cannot destroy the hyperbolicity of these fixed points why can we not destroy the hyperbolicity of these fixed points exactly exactly because since geometrically exactly geometrically it's clear to see that the derivative not equal to 1 means that this is transverse to the diagonal so you can change the slope a little bit and still stays transverse to the diagonal because in the C1 topology you're allowed to change the derivative only by epsilon so if epsilon is sufficiently small it will not become tangent to the diagonal and intuitively why can you not destroy any of these fixed points well if you shift the map a little bit you cannot destroy these fixed points right you can change it a little bit but you cannot move it only if this if you have the graph very close to the diagonal then you could shift this part of graph up a little bit to destroy these fixed points but of course you need to shift it up by more than some epsilon sufficiently small if epsilon is sufficiently small this bit will always be below the diagonal right if epsilon if you take a point that's below the diagonal for sufficiently small epsilon even in the C0 topology this will remain below the diagonal and whereas this will remain above the diagonal so you have this point is above the diagonal this point is below the diagonal this point is above the diagonal so you must have at least these two fixed points in between so you cannot really destroy the fixed points right let me try to write down these arguments for me and then we need to show that you cannot create any more fixed points which is the most crucial thing right so how do we do it so we need to split we're going to have two separate arguments we're going to look so these are the fixed points these are the fixed points these are also fixed points we're going to take some delta neighborhoods of the fixed points here here here and here and then we're going to look outside these neighborhoods we're going to say things cannot change because here you're far away from the diagonal and so when you make a small perturbation you cannot intersect the diagonal and then here we're going to look at the situation separately so there exists delta so let so let epsilon greater than 0 and g i to i d1 of fg less than equal to epsilon so there exists delta and delta prime greater than 0 neighborhoods up equals p minus delta p plus delta for each fixed point p for f such that all this joint such that up uq is empty if p is different from q and f of p minus f of p plus or minus delta is greater than equal to delta prime for every fixed p so what I'm doing is I'm trying to isolate these two neighborhoods so here of course this is one sided in the case of the two fixed points here they're one sided in the other case they're two sided neighborhoods and we have this property that outside these neighborhoods the graph is far away from the diagonal these are these neighborhoods so first of all pairwise this joint and second what I'm saying here is that f of p minus f of p plus or minus delta so you take f of p and you take f of p minus delta then this distance here is greater than equal to delta prime and this distance here is greater than equal to delta prime and this distance here and also and also x minus x is greater than equal to delta prime for all x that does not belong to the union of the up for all fixed p so I can choose delta prime sufficiently small that all the points outside these neighborhoods have a distance from the diagonal of greater than delta prime and then I'm going to consider these two situations separately so as long as g so if epsilon is small enough for example epsilon less than delta prime then any map that is even g0 close to f satisfies g of x is different from x for all x that does not belong to the union overall to the union let me call this just u does not belong to u okay this is the first statement this is for points outside these neighborhoods okay so this is important to understand what the argument I'm saying I'm saying okay I'm dividing the argument into two parts I define these neighborhoods and I say outside these neighborhoods the graph is always more than delta prime from the diagonal so that means as long as I take epsilon smaller than delta prime then any epsilon neighborhood even in the C0 topology cannot intersect the diagonal so I cannot create any new fixed points in this region outside these neighborhoods right so this guarantees I cannot create any things any new fixed points out here so all I need to do is inside these neighborhoods so what do I do inside these neighbors now I need to so that inside these neighbors I do not create and I do not get rid of any fixed points so for each inside this neighborhood inside each neighborhood up so what does the neighborhood u p so by this property here so inside each u p we have that f f of p plus or minus f of p minus delta and f of p plus delta lie on opposite sides of the diagonal so if d0 f of g is less than delta then g p minus delta and g of p plus delta still lie on opposite sides of the diagonal so here I know that this distance at these end points is bigger than delta sorry bigger than delta prime this is bigger than delta prime so if I take a small perturbation even in the c0 topology and I take a small perturbation then I know that this point still lies above the diagonal this other point still lies below the diagonal so there must exist at least one fixed point in there so this shows that I cannot destroy any fixed point right so there exists at least one fixed point for g in this neighborhood up and I almost finished so because in this neighborhood of this fixed point I have this picture here so I have this distance here is greater than delta prime this distance here is also greater than delta prime okay so if I look at the image so this is the point this is the neighborhood u of p so the end points what I just said there is the what I'm trying to show is it clear I've dealt with what happens outside these neighbors now I need to show that inside this neighborhood any small perturbation we still have a unique hyperbolic fixed point okay so I need to show that I cannot lose this fixed point here and the only way to lose this fixed point is if the whole graph is either below or above the diagonal but this cannot be because this distance is delta prime this distance is delta prime as long as the c is zero distance of g is less than delta prime then this point will remain above this point will remain below and I need to cross the diagonal okay so this shows that I cannot lose that fixed point the only thing that's left to show is that I cannot create any new fixed points and this is the only place where we use the fact that the perturbation is in the c1 metric both of these properties just use the fact that these maps are close in the c0 metric as we saw in the c0 metric you could create some new fixed points okay but in the c1 metric again is very simple so if g so if delta is sufficiently small then we can guarantee that f prime of x is different from 1 for all x in U p right so I can assume in fact this so since the derivative here p is different from 1 I can assume we have a small enough neighborhood so that the derivative everywhere in f of p is different from 1 so if g if epsilon is small and of g is less than equal to epsilon okay then also g prime of x should be different from 1 for all x in U p so I assume this is very small the derivative is different from 1 also for g therefore it's different from 1 if I take epsilon less than 1 minus f prime of x basically and therefore I can use the same argument by contradiction that I used before right so assume assume by contradiction that there exists q in U p with g of q equals q so that there exists another fixed point which is what we're trying to disprove so we have now our map g there is another fixed point in U p and then this means exactly by before so by the mean value theorem by the mean value theorem f g of p minus right and we suppose so we suppose that there are two fixed points because we don't know that fixed so we suppose that there exists q and q prime in U p such that g of q equals q and g of q prime equals q prime sorry my handwriting so we want to show that there's a unique fixed point assume by contradiction there exists two then you have that g I'm sorry this notation is really gq minus g of q prime is equal to f prime of x x prime of z times q minus q prime for some z in this interval q q prime sorry thank you g prime and this implies because these are fixed points this implies that g prime of z is equal to 1 ok so here I guess it should have well doesn't matter because the orientation preserving so the positive but g prime of z is equal to 1 which is a contradiction which is a contradiction because we've taken the neighborhood small so that this is different so this completes the proof right so we've shown make sure you look at this part of the proof ok this is not a very difficult proof but it's very important we we divide it into certain ways we use kind of several bits of information several definitions and in particular in this last part we really you really use various aspects of you use the c0 and the c1 topology in various situations and we look at the inside these neighbors outside the neighbors so this is a very important proof to make sure you review ok so this basically completes our section on interval diffeumophism after linear maps we've looked at interval diffeumophism we've looked at the notion of hyperbolic interval diffeumophism we've done essentially a complete classification of topological conjugacy and we've looked at the problem of structural stability of the interval diffeumophism so this also more or less for the moment completes our study of one-dimensional systems which are the systems in one dimension in which we've introduced and experimented a lot of concepts and techniques and from the next lecture we're going to start looking at two-dimensional or higher-dimensional systems starting from linear maps which again are the simplest cases in higher dimensions