 Good afternoon. So today we're going to continue our study of full branch maps. But we want to relax this assumption that the piecewise are fine, because that is a very strong assumption. So we're going to study an alternative property called bounded distortion. So definition, if f is a piecewise C1 full branch map has distortion if supremum over all n greater than or equal to 1 supremum over all elements omega n in pn of the supremum over all x, y in omega n of the log. So we'll take a little bit of time to read this definition. So of course, by now you understand the structure of piecewise C1 maps. So this pn is a partition. So this is the partition pn. This is one element of the partition omega n. And the element of the partition, sorry, this is not such a very good picture. But the element of the partition has the property that fn is a full branch map on pn. What is the question? Can you read my writing? fn prime. fn prime, the derivative. It's the ratio of the derivatives. So for every n, the combinatorial structure of a piecewise map independently of this definition says that for every n, you have a partition pn on which fn is a full branch map. Like each element of omega n maps to the whole thing. So what we're saying is we look for any n, we look at some n, and we look at any two points in here, and we compare the derivatives of fn prime in here. So we compare the derivative at two points, two arbitrary points here, of fn prime. So if the map is piecewise affine, what is this value here? It's always equal to 0. Because piecewise affine means that the derivative on each omega n is constant. And notice that here you're only comparing two points inside the same element omega n. So of course, from different elements, the derivative can be very different. But it's piecewise constant, piecewise affine. So this is clearly satisfied in the piecewise affine case. Because in the piecewise affine case, this is always 0. So everything here is 0. In general, it's not the case. Because in general, you start with a map which might have very, the derivative at different points might be very different. So if you take two points, x and y, the fact that they belong to omega n, if this is the original partition, means that they stay together for some amount of time. But the derivative here and here is different. Then you look at the images, and the derivative here at the images is also different. And so the ratio between the derivatives might be changing. So it could be that just because you're unlucky, the derivative at the point x is always less than the derivative at the point y. And all its image is the same. And so of course, fn prime by the chain rule is just the product of the derivatives along the orbit. So it could be that at these two points, if you take higher and higher n, this ratio could be exploding. Because it could be that this is always growing faster than this one. So even though for a fixed n, it is obvious that this is less than, oh well, it's easy to see how this can be less infinity. The crucial is that you want the supremum of all n. This is the same as saying that this is uniformly bounded. So we could put a constant here. Okay, in fact we can put some constant here. The fact that the supremum is less than infinity means that it's less than or equal to a constant for every n. So as we shall see, this is a strong property. So we're going to look at two things. First of all, we're going to show that this implies a goodicity. So it's good enough. And second, we're going to show that it actually is satisfied in certain maps. And that it holds. Because if you just look at it like this, you might say, ah, it's impossible that it holds unless the map is piecewise or fine. So let's look first at the fact that at the theorem we want to prove. So theorem, if f has bounded distortion, then the big measure is ergodic. Is it true that the big measure is invariant in general in this case? Yes, in general it is not necessarily invariant. Because in general, if you take an interval here and you look at the pre-images, it is very unlikely that they will add up exactly. But if you remember, I remarked that the definitions of invariant and ergodic are independent of each other. In general, we are interested in invariant and ergodic measures. And we shall use this fact. However, the definition of ergodic does not need the measure to be invariant. So we can define what it means for the big measure to be ergodic. It still means the same thing. And we will use the fact that the big measure is ergodic to show that certain other measures, which we will construct in certain specific situations, are themselves invariant and ergodic. So we will use this result. For the moment, we're just saying that the big measure is ergodic. OK, so how are we going to prove it? We're going to use a very similar argument to the one we used in the piecewise affine case. But first, we have to try to understand a little bit better this notion of distortion. So first of all, let me define. So let j be an interval n greater than or equal to 1. Suppose f from j into i is a c1, sorry, j to fj. So j onto its image is a c1 diffeomorphism. And then let d, sorry, fn to fn of j. So we take some iterate. So suppose we have some iterate here. This is some interval j. This is some iterate fn. This is fn of j. And then we define fn, the distortion of fn on j is equal to the supremum for all xy in j of the logarithm. So this is just this term here, basically, where the interval j, in this case, is this integral omega n. So the bounded distortion says that the supremum of all n and of all integrals j equals omega n in pn, this value here is uniformly bounded away from infinity. So that's just a definition. And what is the geometric meaning of the distortion? So I claim that this is the distortion of fn on j. As we said before, if fn is affine on j, in other words, if the derivative is constant, then we get 0 for this distortion. And if you remember, what does constant derivative mean? How did we use geometrically the property of constant derivative? We took some set a here of certain measure, or some set j prime. And then here we had the image of j prime, fn of j prime. And what could we say in the piecewise affine case about these intervals? That's right. So if you have piecewise affine, well, in general, you can use the mean value theorem to say something about these lengths. But without knowing anything about what the actual derivative was, the fact that the map was affine told us what? It told us that the ratio between these intervals is preserved, whatever the length is or the derivative is. Because the derivative is constant, it means that the stretching or contraction that goes on is uniform here. So if j prime is half the size of j, then fn of j prime would be half the size of fn of j. The ratios is preserved. And that's what we use in the piecewise affine case. And what we're going to show in the following lemma is that this gives you exactly a bound to how these ratios can change when you apply the map. So lemma for all j prime and j subintervals, we have the ratio between fn of j prime and fn of j is less than or equal to j prime over j times e to the power d and greater than or equal j prime over j times e to the power minus d, where d is exactly that. It's equal to d fn of j. So here we have the ratio between the images, the Lebesgue measure of the images. Here we have the ratio of the original ratio. And we're saying that the quotient between the ratios is uniformly bounded above by e to the d and below by e to the minus d. And you can see that this coincides in the piecewise affine case. Does this work in the piecewise affine case? What did we say that this was in the piecewise affine case? Zero. And so what happens here? It works, OK? Because you get one here, one here, which means you get equality because it's squeezed on both sides. So how do we prove this? Well, this is, as you kind of suggest, is just a simple application of the mean value theorem. So by the mean value theorem, there exists x in j prime and y in j. Such that fn prime of x is equal, fn prime, sorry, fn prime of j is equal to, sorry, let me write it directly. Such that, yes. Sorry, that was fine. fn prime of x is equal to the image fn, fn of j prime over j prime, right? And fn prime of y is equal to fn of j over f, right? The mean value theorem. So the image fnj prime is equal to fn prime of x times j prime for some x in j prime. And the same thing for j. And so we just take the ratios. So we have that fnj. So we take the quotient between these two things. So j prime divided, sorry, this should be j. fnj prime over j times j prime over j is equal to prime of j prime equals over j. And this is equal to, OK? This actually is kind of going in that direction, right? So from that, I have that the ratio between these derivatives is equal to this, which is equal to this, OK? And so from the definition of distortion, what do you know about this? This is the assumption we have, right? So the assumption is that this, so the maximum amount of this ratio is the distortion. And so we have that this is less than or equal to. So by definition of d, we have that fn prime of x over fn prime of y is less than or equal to e to the d, e to the d, and greater than or equal to the minus d. And so we get the result, OK? And so, OK, you plug it into here, and you get exactly the result. So it's really just the application of the mean value theorem, and you get exactly the. OK. So what was the second ingredient we used to prove the codicity or Lebesgue measure for piecewise affine maps? Do you remember? I pointed out there were two ingredients. One was the fact there was affine, which here we're hoping to replace by the boundary distortion. And there was a second ingredient that we used. Remember, we used the argument with Lebesgue density points, right? We found the Lebesgue density points, and we iterated the Lebesgue density points, and that's how we showed our codicity. So what did we need to use to be able to use the argument for Lebesgue density points of the measure? We needed to use the fact that maximum of the Lebesgue measure of omega n overall omega npn goes to 0. In other words, that the intervals of the partition element shrink uniformly. And what ingredient did we use here? In the piecewise affine case, do you remember? That's bigger than 1. Bigger than 1. That's right. But now what do we know about the derivative? Do we know that it's bigger than 1? We've made no assumptions on the derivative here. So this little exercise or this little proof is going to this little calculation. We'll show you how powerful the boundary distortion is. Because remember, also in the piecewise affine case, we did not assume that the derivative was bigger than 1. How did we conclude that the derivative would be bigger than 1 in the piecewise affine case? Exactly, exactly. So we had a different morphism from small to big. And because the derivative is constant, it has to be bigger than 1. Here, the derivative is not constant, so the derivative could change. So you could have from small to big would be sufficient for the derivative to be very large in some small part, and it could be very small in some other part. And so it could be some regions where the derivative is very small. And then when you iterate the map, there might be some intervals that always are unlucky and they always stay in the small region where the derivative is small. And so the omega n's in the partition, they do not shrink there. Because we're not guaranteed that everywhere the derivative is bigger than 1. But we have bounded distortion. So even though it's not piecewise affine, we have bounded distortion. So we cannot have a situation where if you look at the little interval and maps to the whole thing, the ratio between the derivatives cannot be too big. So we cannot have a region where the derivative is really big and a region where the derivative is really small. So morally, we still have that the derivative is not piecewise affine, is not constant, but it's almost constant, okay? In this way. And we will show that this is sufficient, in fact, to get this property. So how do we do it? So first of all, let delta equals the maximum of all omega in P of the size of omega. So we have our map. This is the original map, okay? So we have finite or maybe infinite partition elements here. We take the biggest interval. The biggest interval is well defined. The biggest interval of the original partition. And then what we have is that for all n greater than equal to one for all omega n in Pn, we have that f n minus one of omega n. What do we know about the n minus one image of omega n? So f n minus one belongs to P. It's one of the elements of this partition, right? Because then at the next stage, it gets mapped f n of omega n is everything. And the way it's everything is because at time n plus one, it maps to one of these and then it maps to everything. This is the combinatorial structure we've got, okay? So this means that we know that the size of f on n minus one omega n is less than or equal to what? It's of course less than or equal to one, but we can do better than that. Why did I introduce this delta here? It's less than or equal to delta. Why is that? Exactly, okay? So I mean delta might not be small. Maybe delta is not the best symbol. Delta usually you think of as something small, okay? Sorry, maybe delta I should have used some other element, but delta is just the maximum size of these partition elements. Because this belongs to one of these partition elements, it has to be less than or equal to delta, whatever delta. One minus, I could have written this one minus delta, so on. And therefore, now omega n, recall that omega n belongs to some omega n minus one, where f n minus one of omega n minus one equals all the interval i. So the picture we have here is the following, is that we have an interval omega n minus one. Here we have an interval inside here, which is the form omega n, right? And after n minus one iterations, i f n minus one of omega n minus one maps onto everything, and f n minus one of omega n maps to some element of the original partition p. Okay, this is just the combinatorial structure. So so far, I've not used any assumptions or have anything. I've just written some notation that comes from the properties of the combinatorial structure of full branch maps. So we're going to use the bounded distortion property. Here we have that this is a C1 different morphism, because f n minus one is a piecewise C1 different morphism on element of p n minus one, omega n minus one. And so when we apply the bounded distortion property here, so we have that omega n minus one minus omega n over omega n minus one. And in other words, now it just makes, it's a little bit simpler. We are comparing the ratio between the complement of omega n and omega n minus one. If we wanted to, we could compare the ratio between omega n and this, but it doesn't make any difference, okay? So what the bounded distortion says is that the quotient between this and the full interval is comparable to the quotient between this and the full interval on omega n minus one. Or of course, equivalently, the quotient between the complement of this interval and the quotient between the complement on this interval and this is also bounded by the same constants, because they themselves are intervals, okay? So what we write is that this quotient is greater than or equal to e to the minus d times the ratio in the image, and the image I will write as f n minus one omega n minus one minus f n minus one omega n over minus one. So this is just these two intervals here, quotient between the full interval, okay? And now this, so this is greater than or equal to what? The numerator here, so the total value of this, sorry? One minus delta, exactly, right? So this is greater than or equal to e minus d, one minus delta over, what is this? One, okay, over one. So what is this number here? This is bigger or less than one. This is a positive number, yeah, because delta is less than one, and this is some number between zero and one. This is some number between zero and one. The product is some number between zero and one, okay? So for simplicity, let me just call it, let me define it one minus tau. I define tau in this way, because it's easier to write this tau, sorry, one, yes. Okay, so really I'm just going to rewrite this in a way that is a little bit easier. So I write one over omega n over omega n minus one is equal to omega n minus one minus omega n over minus one, and this is just equal to n over minus one, and this is greater than or equal to one minus tau, okay? So omega n, okay? So for all n, now tau, notice of course tau does not depend on n, and it does not depend on the specific interval that we've chosen, right? Because this is a bound that depends only on delta, which is universal, depends on the map and the partition, and d, which is also constant, that depends completely on the map. So this tau only depends on these two. So for every n, and for all intervals omega n in omega n minus one, omega n then equal to tau for some tau, right? Because here I just have one minus this is greater than or equal to this, so I move them around and I get omega n over omega n minus one is less than or equal to tau. And what do I do with this? Exactly, okay? I just repeat this, so this will give me tau squared omega n minus two, okay, and so on, and I just get tau to the n here. So this is, I think, really interesting, this calculation that we did here, because it shows that the bounded distortion, at least in this one-dimensional setting, together with the full branch, implies some kind of, well, not some kind, implies exponential shrinking of these components, just like in the uniformly expanding case. Okay, so this was all we needed. So for the, to complete the proof of the ergodicity is actually exactly the same as the piecewise affine case, because we have all the ingredients, except if you remember, I brought your attention to a place where we had an equality, and we only needed an inequality. If you look back at the proof of the piecewise affine case, so if we now complete the proof of theorem, exactly piecewise affine case we get. So I won't repeat it, because you can just look at the notes. You use Lebesgue density theorem, you take a set of positive measures, you assume it's got full measure, and you want to show that it has, sorry, you assume it has positive measure, you want to show it's got full measure, so we want to show that the complement has zero measure, right, and by taking a Lebesgue density point and iterating using the piecewise affine map, we get that the iterate between the complement and the whole original interval is given by Fn of omega n intersection AC over Fn of omega n, okay? And in this case, we have, this is less than equal e to the d of omega n intersection AC over I, which is therefore less than or equal to e d times epsilon. So if you remember, in the piecewise affine case, here we had an equality, because we've shown that these two sides were equal because of the affine property, then these quotients were defined, here we just have this bound, but that is quite enough because the d is independent of n. This is the crucial point, we use the fact that d is independent of n, so there's always a d, so we can take epsilon as small as we want, and then using the Lebesgue density theme, we can take n as large as we want, and we get exactly the same estimate. And this implies that AC is equal to zero. This implies AC equals zero, so Lebesgue. So when you study this, you should go back and look at the piecewise affine case, make sure you understand how the argument is really essentially the same. So this shows that Lebesgue measures the logic. So now the question is, how do you check that a map has bounded distortion? Because if I give you a map, the bounded distortion property says that the supremum overall n, you know, it has a lot of supremums. You need to study all eight sets of the map, and you need to look at all points and all intervals and so on. So it's not at all obvious how to check that the map has bounded distortion. So we're going to prove the following theorem. So question, bounded distortion. So let me make first definition. So I'm going to prove a slightly weaker version of expansivity than we had before. So F is uniformly expanding. So until now I think we've said that uniformly expanding means the derivative is bigger than one everywhere. But if you remember, I made one remark in the previous course saying that for all the results you actually need a weaker form of uniform expansivity, which is the following. There exists constant c and lambda greater than zero, such that for all x in i, and for all n greater than equal to one, fn prime of x is greater than equal c e to the lambda n. So e to the lambda is bigger than one, right? So if the derivative is bigger than one everywhere, this is clearly satisfied, even for c equals one. The point is that there are many situations in which the map is not uniformly expanding, but some iterates is uniformly expanding. So you might have a point that has a small derivative for a few iterates or even for many iterates, but then after a while it starts expanding uniformly. And so this constant c allows that possibility in this definition. Because if you take c very close, this allows to have for n equals one, two, three, it allows this to be small, less than one, so this condition is still satisfied. But as long as for every point, eventually it starts going exponentially fast, and the c does not depend on the point x. So this is a uniform constant c, then this is also called uniformly expanding. This condition is exactly equivalent to saying there exists some big n, such that the iterate f begin is uniformly expanding in the previous case. So the derivative is bigger than one. So theorem c one full branch map, piecewise c one, suppose f is uniformly expanding, there exists some constant k greater than zero, such that the supremum overall omega in p the supremum, sorry, I will need c two in this case actually. So supremum overall x, y in omega, double prime of x over f prime of x is less than or equal to k. Then there exists k tilde greater than zero such that for all n greater than or equal to one for all omega n in pn and for all x, y in omega n. Yes, excuse me. The y. Ah, sorry, yes, thank you. No, sorry, sorry, I do want the y. Yeah, now you take the second and the first derivative at all possible points in the interval and you look at the maximum ratio between the second and the first derivative. Sorry, and I also want a squared here. I'm trying to go a little bit too fast. So then there exists a k tilde such that log of fn prime of x over fn prime of y is less than or equal to k tilde is less than or equal to k tilde times fn x minus fn y which in particular is less than or equal to k tilde. So in particular, so why do you think this is useful? So we wanted to check whether f has bound the distortion and we have replaced, we've put some different conditions that say that these imply that f has bound the distortion. Why are these conditions any better than having the bound the distortion? Exactly, exactly, so the definition of bound the distortion involves all iterates of the map whereas these conditions, the uniformly expanding and this just depend on some finite iteration of the map in particular and for the first iterative. So this is just the first iterative of the map, the first and second derivative and this for example, if the derivative is bigger than one everywhere then it's satisfied automatically for the first iterative of the map. So this is the kind of condition you like to have on a map, right? You like to say if a map is like this then it has certain properties. You don't like to have a definition saying if a map satisfies for all iterates something then it has certain properties is not really very useful. Although it is useful in this case because we're able to check using this theorem, these properties. So how do we show this? Yes, you need C2. For the theorem, you're right. For, well, for no, the previous theorem just said that if you have bound the distortion then you have the big measures of garlic. And now we're saying that to check bound the distortion we won't have to be C2 and to satisfy these properties. Yeah, so for this theorem we need C2. So if you want to apply this theorem to a map it needs to be C2 and satisfy these properties and then it has bound the distortion. Okay, so let's start by proving the first part. Now, this first part is a very nice. Wait, so what are we going to prove, want to prove? No, sorry, so we're going to prove in two in three separate lemmas. So let me give a name to these, let me call this star. Let me call this double star. So lemma one says that star implies that for all omega in P and all x, y in omega we have f prime of x over f prime of y minus one is less than equal to k, f of x minus. So I know that at the moment this is very technical, all these bounded distortion and all these formula and all these equations, okay. But there is some geometric intuition behind it. So I want to try to help you organize this material so it makes sense and it's not just a bunch of complicated formulas. So we're going to use, so the result is bounded distortion has a very concrete geometrical interpretation that I gave you before. So it's significant. The two assumptions we're going to use here one is that f is uniformly expanding and the other one is this condition here, okay. So in the course of the proof I want to be clear about when we're using these two conditions. So in this lemma we're just going to use the first condition star. And what does the first condition star say? In some saying it's putting a bound on the second derivative in terms of the first derivative. And so it's saying that the first derivative cannot change too much because after all that's what bounded distortion is about, right. We said bounded distortion is the fact that the first derivative cannot be really big here and really small here, okay. And I'm saying that the way to control this is essentially boils down to this. Using the second derivative we control the fact that the first one is not changing too much. And more precisely this is how we're going to, this is how it turns out that it's useful to formulate this. So that condition gives you this which says that if you take any two points inside one partition element, the ratio between the two, there's a kind of Lipschitz condition for this. So this ratio's close to one and how close it is to one depends Lipschitz on the distance not between X and Y but on the images of X and Y. That's just turns out to be the right formulation for the calculations that we need. So again, this is just a mean value argument. So by the mean value theorem, we have that F of X minus F of Y is equal to some F prime of Xi times X minus Y for some Xi in X, Y and that F prime of X minus F prime of Y. So we can use the mean value theorem also for the derivatives. So the derivative is a function. We measure this derivative at these two points and this is F double prime of some, so let me call this Xi one and this Xi two times X minus Y for some Xi one, Xi two and Xi. And so taking the ratio of these two, we just get that this ratio is equal to this ratio, right, for some Xi one, Xi two and so we write that F prime of X minus F prime of X times X minus F prime of Y is equal to F double prime Xi two over F prime Xi one times F. And what do we do with this? What's the assumption I said we're using in the lemma? We're using this assumption, right, almost less than K, there's a square here, right? So this is less than or equal to F prime of X Xi one times K because we have the square here times K times F of X minus F of Y times, sorry, wait, what do I want to write here? So I know that this is less than or equal to K times F prime of Xi. So yeah, so this is less than or equal to K F prime of Xi one times F of X minus F of Y and this is basically what we want. So yeah, so actually, sorry, I want to say it just a little bit different for complete. Okay, I'm sorry, let me just say. So we need to use the fact that we have a supremum here, right? So by star, by star, we have that F double prime Xi two over F prime Xi one is less than or equal to K times F prime of Xi for N Xi in omega because we have a supremum here. And so substituting this into here, we have that F prime X minus F prime of Y is less than or equal to K times F prime of Xi times F of X minus F of Y, okay? And since this is true for every Xi, we can take Xi equals Y. So letting Xi equals Y, we divide through by F prime of Y, okay? And we get exactly what we want, which is F prime of X over F prime of Y minus one less than or equal to K F of X. Let's see, because this is true for every combination of X and Y, right? So in particular, it's true if I take, so just apply, so clearly it's true for Xi one here, right? So the question is why is it true for any Xi? Because, no, no, I think it's just a simple consequence of this because we have that, let's see if I made a note in the notes here. Question, let's see. Yeah, I guess if I wanted to be strictly, perhaps I need to change the constant K for this to be true. Yes, maybe I need to change the constant K, which is fine, but so maybe you're right. So maybe I should, okay, maybe I need a little bit of an, a little extra argument. So this is Xi one, okay? And this is less than or equal to some K hat of F prime of Xi for all Xi in omega because indeed, because F is uniformly expanding, so the derivative needs to be bounded away from zero, at least for it to be uniformly expanding. And if the derivative is bounded away from zero, then it means that the ratio, it means that I can always bound this derivative at some point is less than or equal to some constant times the derivative at any other point. No, that's not completely true, actually. If the derivative is unbounded, this is not completely true. Okay, it's a good point. I will figure this out and let you know tomorrow. Okay, it's a good point. It's a small point, but it's, it should be correct. It should be written in a correct way, yes, you're right. I will check that or you can try to work it out as an exercise. Okay, let's assume that for the moment and let's continue. Can you see the solution? Excuse me. If you use another K, how do you get it? By what I thought just before? Well, the problem is that derivative might be unbounded. The derivative could be unbounded. We're not assuming the derivative is bounded. So I'm not completely sure. Okay, I will come back to it. But you can think about it. It's an interesting subtle point, actually. Let's assume that it works. It doesn't look like it should be impossible. You know, it's just a question of understanding how to formulate it. And let's go on to the next lemma. So let's assume now that we have this lemma here and now we will prove the main fundamental distortion calculation, which is a very interesting calculation. As far as the calculation can be interesting. So lemma two. So now we just suppose this property here. So let me call this, okay, three stars since I used two stars earlier. This equation here implies for all n greater than one, for all omega n in pn, we have that the distortion of fn on omega n is less than equal to k times the sum i equals one to n of f i x minus f i y. For all x, y, yes, let me write it like this actually. So this is a very interesting statement. And it says that to measure the distortion, remember the distortion is the log of the ratios. It's the supremum, right? This is the supremum over all x, y in omega n of the log of fn prime of x over, sorry, I shouldn't write it. So remember that this is equal to the supremum over all x, y omega n of the log fn prime of x over, and what does this say? This says that all you need to do is sum, so this is our omega n, and you know that up to time n, it maps here. This is fn omega n, and you need to sum the iterates, the length of the iterates. This is f omega n, this is f i. So you look at the size of the iterates, and you sum them all up, and that's the distortion. So what is the calculation here? It's a one-line proof, but it will be a slightly long line. So by the chain rule, we call that fn prime of x is just equal to the derivative along the orbit, right? fn minus one of x. So fn prime is just equal to the derivative along the orbit, right? So what we want to estimate is this, for any two points x and y, supremum. So here we take log over fn prime of y and we write this as the log of this product over f prime of y, right? This is simply the chain rule, so I take the product of all of these to get fn prime, and I write this as a sum, so let's write this as a sum i equals zero to n minus one of the logs of the same thing, f prime in f i of x, f prime in f y of y. And now I'm going to do a little trick, the usual trick, so I'm going to write this as plus or minus, so I'm going to write this as the sum. Okay, I will skip one step because you can easily see it. This is the sum i equals zero n minus one of log of f prime f i of x minus f prime of a i, f i of y over f prime of f i of y, plus one, okay? So I just added and subtracted f prime f i of y. And now I'm just going to take this absolute value inside and so I put a less than or equal to the sum i equals zero n minus one, and I'm going to write this as the log, and then I'm going to write f prime f i of x minus f prime f i of y over f prime f i of y plus one, sorry. I still have the parenthesis here, which is plus one, everything parenthesis. And then this is less than or equal to the sum i equals zero n minus one of just this without the log. So f prime f i of x minus f prime f i of y over f prime f i of y. And why is that? And now why is this? Exactly, because log of one plus x is always less than or equal to x for all positive x. And this is less than or equal to, okay? Well, here I just have the sum i equals zero n minus one of prime f i of x minus f prime f i of y minus one. And what bound do I have for that? So notice that I'm not iterating the derivative. I'm just taking the derivative in two different points and notice that x and y are chosen in omega n. So what does that mean about f i x and f i y? That's right, in particular they always belong to one element of the original partition, to the same element of the original partition, right? Because omega n, if you look at the units of omega n, it always stays inside one element of the original partition until after time n it maps onto the whole thing. But before that it always maps to an element like you say of omega n minus one, omega n minus two, omega n minus three and so on, which are all the refinements of the original partition. So that means that this applies because we're applying this to points that are inside some element of the original partition and therefore this gives less than or equal to constant k times the sum i equals zero to n minus one of f i of x minus f i of y Okay, but of course since x and y belong to omega n, this is just less than or equal to k times the sum of f i omega n. One line proof, sorry, sorry, sorry, sorry. Ah, ah, ah, you're right, you're right, you're right. Because this is f prime in x and y is less than or equal to this in the image. Thank you, thank you. So you're right, so here you have i, that's why actually I was wondering why, I couldn't remember why, this was from y to n. So you look at the images, which are exactly to this. Yes, thank you very much. Good, so we only have a few minutes left but we want to prove the last lemma. So we still haven't proved, so we've proved two lemmas, right, we've started with star, which was in the assumptions of the theorem and we use this to prove this. Then we use this to prove this property here. Now, remember what the statement of the lemma says, the lemma of the, the statement of the lemma was a bound on this, on the supremum overall n. What did it say, what did the statement of the lemma say? Can I erase? Okay, so what did the statement of the lemma say? Of the theorem, sorry. The theorem said that this was uniformly bounded for any n and for any omega n. This is less than equal to a constant. So what we've done is we've reduced this to a bound on this. So what we need to show is that this is less than equal to a constant, independent of n. Now this is a little bit subtle here. So what we mean is we need to find a bound for the sum that is independent of n. How can we find a bound for the sum that is independent of n? Yes, but what is happening to this interval here when you iterate it? Is it shrinking exponentially fast? Is this a geometric series? What happens to this? What's the size of these intervals as you increase? These are shrinking. F i of omega n, omega n is very small. It's growing geometrically. So if they're growing geometrically, how are we going to get a bound that is independent of n? Let me state the lemma in the meanwhile. Here we are going to use the uniformly expansion. Expansion. F uniformly expanding, okay? Then there exists some k tilde depending on c and lambda such that for all n greater than equal to one and for all omega n in pn and for all x and y in omega n, we have that sum i equals one to n of f ix minus fi y is less than or equal to k tilde fnx minus fny. So in particular, if we take x and y to be the end points of omega n here, then we get that the sum of these, this is just equal to one. So the sum of these will be less than equal to k tilde that depends only on c lambda not on n and therefore we get the theorem, okay? So this lemma is the final lemma that we need to get the theorem. This is a stronger statement because it implies that if you take x and y very close, small it's a kind of leaps its condition but we're particularly interested just in the conclusions when x and y are the end points so we get a uniform bound. But this is the lemma but let me go back to how are we going to hope to prove this lemma, right? Because when you start with these two points xn and omega n, then the images are growing and so you're summing intervals that are growing all the time. So how can we hope to get a bound that is independent of n which is the number of terms that we are summing and why do we specifically need the uniform expansion? It would be better if we had uniform contraction for this, right? Then it would be obvious. But how does it depend on the uniform expanding? When in doubt always use the mean value theorem. How do we use the mean value theorem here? So how can we get a bound when we are summing a series of terms that are increasing? How can we get a uniform bound? The bigger n is and the more they grow. Is that true? What I'm saying? Yes and no. Okay, I'm not sure if I know and the sum what you're saying but I think you're along the right lines is the creature lines is I am purposefully trying to mislead you, okay? Because I'm emphasizing the facts that these are growing which is true, okay? But how much can they grow? There's a maximum amount that they can grow, right? How big is this at time n? fn of omega n is how big? Is one. So whatever the value of n is they're growing exponentially fast, geometrically but the maximum size is always one. So if you go backwards you actually get a geometrically decreasing sequence. So this independence of n which I keep emphasizing and it's very important is not completely true that it's independent of n because there's something else in this calculation that does depend on n and what is it? It's not the final bound but it's something else does depend on n. It's the initial interval that you choose right here. If you take n very large it's true that you're summing up many more terms but how big is the interval that you start from? It's very small, okay? So let's try to formalize this calculation a little bit. So here we have omega n, okay? Here we have two points x and y, x and y which could be the end point so it could be something smaller. So let's call this interval omega and then we look at the images. So we look at omega n of omega n equals i and fn of omega is something, yeah? And how do we apply the mean value theorem? Well, we don't apply the mean value theorem going forward. If this is some, so here we have some intermediate step. So this is f i of omega n and this here is f i of omega, right? Then what we're going to apply the mean value theorem is to this step here. This is fn minus i. So what we know, so here we have f i, here we have fn minus i and so by the mean value theorem we have that fn of x minus fn of y equals fn of omega which is equal to fn minus i of f i of omega, right? Which is equal to fn minus i, okay? fn minus i in some point psi i, psi, let me call this n minus i times f i of omega. So there's some point here psi n minus i so that if you look at the derivative from here to here this element here is bigger than this factor here by this derivative here and what do we know about this derivative here? This is greater than or equal to c e to the lambda n minus i times f i of omega, okay? So what we're comparing is this image to this image, okay? And the way we do it is we're now going to show that therefore this is exponentially small compared to this going backwards, okay? So we have that the sum, let me write it here or the sum i equals 1 to n of f i of x, okay? This is of course, this is equal to, I can write this as c e lambda n minus i f i w so the interval i equals 1 to n of f i of x minus f i of y is less than or equal to 1 over c e to the minus lambda n minus i f sum. And this, so what is the sum here from i equals 1 to n? Well I can rewrite it as, so notice that when i equals 1 here I'm going to get n minus 1 and as I increase i when i is n, this gives 0, right? So I can actually write this as 1 over c the sum i equals 0 to n minus 1 of e to the minus lambda i or maybe j let's change the index just so not to create confusion j minus f n of y and this is of course less than or equal to 1 over c how do we make this bound independent of n? This is the key step when we want to make it independent of n so far it's always a bound this bound here is less than something that increases as you increase n but we want a bound that is independent of n so what's the very last step to make this bound independent of n? Exactly, I can just sum for all j's and then I get a bound independent of n so this is less than or equal to 1 over c the sum j equals 0 to infinity of e to the minus lambda j of f n of x minus f n of n and this is equal to so I can write this as k tilde f n of x minus f n of y where this k tilde is just 1 over c times the sum so as we saw here it depends only on c and lambda so it does not depend on any other constants related to the map okay so I'm sorry we took a few minutes longer than usual now but what we've just to summarize remember what we've done today is we've shown that if a map has bounded distortion then that is enough to prove that Lebesgue measure is ergodic theorem number 1 theorem number 2 to check that a map has bounded distortion it just needs to be uniformly expanding and it needs to satisfy that condition on the second derivative over the square of the first derivative so some finite number of iterates are sufficient to show that it's bounded distortion okay so we've given some checkable conditions to show that Lebesgue measure is ergodic and in the last lecture tomorrow we will show a nice class of examples of maps that satisfy this condition thank you