 You have seen yesterday with Kadim a proof that this map on the torus proved that this map is ergodic, remember? But this proof used Fourier series and it strongly relied on the fact that this is linear, well, homogeneous, let's say linear. Today we are going to see another proof which is also for this system, but it is also useful for the nonlinear case because in the future we will be focused on the nonlinear case. So let me first introduce you some tools we are going to use and at the end of the class I would like to identify the tools that we have used so that we know how to use them in the future. Okay, so remember we have seen the mean ergodic theorem. Let me remind you, sorry. This is the von Neumann version. Is this size okay? Okay, thanks. Okay, so we have seen that for every, let's put L2, we had go to zero. Okay, and in particular I don't know probably you have seen it in exercises. This F is ergodic, if and only if this map is constant almost everywhere. Let me put it like this. Okay, and I don't know if you have seen it as an exercise, but for all three I will make a remark in a minute. Let me put it here. I don't know if you have seen it in an exercise, but if this is constant, if F is ergodic and F, this is constant, then this constant equals this. Okay, I will leave you the details of what I am going to say next as an exercise, but let me just do a little remark. You have seen with Stefano the point-wise version of this, which is the bulk of ergodic theorem, and you have also seen it for L1, wasn't it? Stated F. In fact, let me state it like this. If mu is invariant, you have that almost everywhere converges to a function. Okay, so now we have that if this converges in L2, this implies that there is a subsequence that converges almost everywhere, but since this sequence converges almost everywhere, this implies that this map and this map coincide almost everywhere. Yes, not necessarily. This is for any F, let's say for our purposes, just put it continuous. No, here also not, because this is a priori, this is not constant. Okay, if F is ergodic, this will be constant and vice versa, but this holds for our purposes of today, let's say that it's a homeomorphism, but it's more general than that. Okay, I will say that it's a homeomorphism because I want... Yes, I have said somewhere that was mu invariant here, it's also mu is F invariant. Both theorem have the same. I don't know if Stefano has told it, but we can iterate forward or backwards, and we would obtain the limit exists both in the future and in the past, okay? I want to do this comment because we have seen in the von Neumann theorem, we have seen that if F is measurable and invertible and its inverse is measurable, then we have that E number, the conditional expectation for F and for F to the minus one was the same. So in particular, this limit in the future and in the past will be the same. But since this also holds, this will have the consequence, the consequence, these two limits will coincide almost everywhere. I will leave this as an exercise, but the idea is this. Remember that this implies that it has a convergent subsequence, but we know that it's convergent mu almost everywhere. So these two limits will coincide almost everywhere. So we will use a couple of tricks in this proof of today. One trick was given as an exercise. In order to check that F is ergodic, we have to check that this limit is constant for all phi in L2. But in fact, it is enough to prove it for all phi continues, okay? This is trick one. This is a very general trick. Why is this useful for us? Because we are going to use in this method, we are going to strongly use the structure of the stable and stable manifolds that Amy described today. Remember the stable manifolds were the set of points that approach are asymptotic in the future. Manifold and the unstable manifold, the set of points. So we are going to use this. And what's here that comes trick two? Yes, here we are going to, now we are going to focus on the torus, and so let me change this. This is going to be the torus, and this mu is going to be volume, okay? And so let me put it here. So here it comes trick two. Trick two is an exercise, but think a little bit. If two points are in the same stable manifold, and we iterate forward, these points get as near as we wish. So if phi is continuous, these values will be as close as we wish. And so this will have the same limit, okay? So trick two is that if y belongs to the same stable manifold as x, then they will have the same bulk of forward limit. And the same way, if y belongs to the stable manifold of x, then phi minus y, x equals phi minus y. If they don't exist, we don't care about them. Or another trick could be to take the lim soup of this and just focus on the points where it exists. But we are going to focus on the points where this limit exists, okay? Well, if it exists here, it will exist in the whole stable manifold. And if it doesn't exist here, it will not exist on the whole stable manifold, okay? So we will focus on the points where this limit exists. So this is trick two. Trick one is very general. It doesn't depend on the structure of the torus or anything. Trick two is more particular to our case. And we have trick three that we haven't proved it, but trick three is that you can go anywhere in the torus by following a stable and a stable path, okay? You can choose any two points in the torus, and then you can go along a stable manifold and stable manifold, stable manifold, and eventually you will reach to the second point, okay? It's like streets, okay? Stable and stable manifolds are like streets on the torus. They allow you to reach from any point to any other point, okay? So this is trick three. So let me write it somewhere. Let me write it like this. Okay, so we will go further on trick two in a minute. But in the torus, Amy has already told you the stable leaves and the unstable leaves have a rational slope. So if we want to take a look at these streets, they will look like this. Let me draw it in different colors. So we will have here the red ones will be the unstable manifold and the green ones are the stable manifold, okay? But now we are, oh yes, where's x? Let's put to be safer x and y. So y will be any point here and x will be any point here, okay? In fact, it would be better if they are inside the square I will draw in a minute. Okay, so now we can consider square and just because I like it that it's not necessary, we can consider a square such that it's delimited by stable and unstable manifolds, okay? I chose it to be like this, but it's not necessary because I like it. This yellow x will be a square, okay? So let's put y inside, okay? So then we will focus, we will call local unstable manifold and local stable manifold. The pieces of manifolds that are inside the square, okay? This is just a small comment. Stable manifolds and unstable manifolds are dense in the torus. Each stable manifold is dense in the torus, okay? So this, no, we're going, it's just a small remark. So this is a piece of local stable manifold, but in fact this, the same stable manifold will appear here infinitely many times as different local stable manifolds. It's just a comment, doesn't have to do with the proof, just for you to know. Okay, so we're going to focus in this square and what we are going to prove is that the map, the bulk of average is constant on this square. So if it is constant for every five continuous, then this will hold for any square so that we can cover the torus with squares that overlap so it will be constant on the torus, okay? So we are now reduced to proving that the bulk of average is constant on the square. Yes, let me write it. Goal is to prove that this converges to a constant on S, okay? In other words, phi x, phi plus x is constant on S. Since, yes, we are going to do this with any square and squares overlap, you can cover the torus with squares and so they are constant so you get a global constant for this. So our goal, let me write the goal in fewer words, our goal is to prove that phi plus, phi minus are constant on S and this will hold for all S and then we are going to be done. So what's the idea of the proof? Well, Amy gave the idea of the proof in her lecture. The idea of the proof is to show that phi plus is almost everywhere constant on stable leaves and then we will have that, they are, for the points that exist the limit, they are equal on unstable leaves, we will see that because we have seen, I have erased it now, we have seen this, this coincides almost everywhere so we first move along stable leaves and prove that this is almost everywhere constant then we move along unstable leaves and these are constant then we will have constant all over the square but let me go into details because devil is in the details. Okay, so I will move first on unstable leaves, sorry but this will be the same. So let me define the good points, the good points are going to be these points, are going to be the points where the bulk of limit exists and also it's the same for the future and for the past, okay? So we are going to define the good points for X in S, this already implies that the limit exists, okay? So it exists, it's unique and they coincide, okay? So this is the set of good points. You will prove it in the problem session today that this has measure one on the charts but in particular the density, this is volume measure, okay? The density of G on S is one. Almost every point on S is a good point. This meaning that this limit exists, this limit exists and they both coincide, okay? By our previous remark we know that this holds and so you have to fill in the details in the problem sessions today, okay? So the bad points are going to be the complement and so naturally the measure of the bad points is going to be zero, okay? So in our case the stable manifold and the unstable manifolds are straight lines, okay? If you haven't seen it today you will see it in the problem session. For the 2111, the stable manifold and the unstable manifold is a foliation as Amy has said but it's a foliation made of straight lines, just straight lines, okay? In particular, Fubini holds, okay? This will be a key concept, okay? So since we have that this has measure one and this has measure zero, we will have, let me tell you what this is in a minute. This is the Lebesgue measure restricted to the unstable leaves, to the unstable local leaves, okay? But by Fubini we know that the measure can be decomposed. We can, in order to get the measure on this we can move along one stable line and integrate along the unstable lines over the stable line, okay? You just integrate this over the base and you take the measure over each of the straight lines. This holds because these are straight lines so it's just the Fubini you have learned in calculus, okay? So in particular, this is the Lebesgue measure over the unstable leaves so in particular you will have for almost every x and s this will hold, okay? Some exceptional points but we won't care because they will have measure zero, okay? Okay, and so this will imply or this is equivalent to say the measure of good points will be one, so the density of good points is one along unstable leaves, okay? Okay, now let's focus on what we want to do. So remember our goal was to prove that five plus and five minus are constant, okay? So let's pick some point. This is, let me erase here because I will need this. So let's pick a good point. Let's pick a good point but a point which is also a density point. You can also choose a good point that is a density point of G, okay? So for this point, it's a density point of G in the local unstable manifold. This is one of this. For almost every x you have this but we can choose also x to belong in G. This accessory, we don't need it but it's more elegant to choose a point in G that satisfies this, okay? So in particular, since this is a point of G in the unstable leaf, we will have because this is a density point of G and G is the set of points such that this holds and so this will hold for almost every y. So let me draw it, let me draw it here. We will change. So we have here, here is x. X is a good point and so for almost all x, y in x in w u local of x the forward limit and the backward limit will coincide, okay? For almost all y in this local and stable leaf of x. You follow me up to here. So we pick a good point. This is a good point. So this is not only this belongs here so this will hold for x but not only will hold for x but for x it will hold for almost every y in the unstable leaf of x, okay? Okay, but so we have that phi plus y equals phi minus y. So this is almost, on one hand this is almost everywhere constant. Let me, so on one hand y belongs to w u log of x by the exercise, by the trick two we had phi minus y equals phi minus x, okay? And it was phi plus x because this is good and let's call this a constant t zero, okay? Just not to get confused. So for all, this is not for almost everywhere. For all y in the unstable manifold of x we will have this. This because of trick two is exist because x is a good point and this equals this because x is a good point and this is because I named it, okay? Now let's erase this. Now let's take almost every y in w u log x is a good point. Why? Because we chose it to be a density point of t. And so almost every y will satisfy that this equals this, okay? This is because almost every y is a good point. We have this point y we take the stable leaves of y and this phi plus will be constant on this stable leaf because of trick two again. So this will continue with orange. So this is going to be phi plus z for all z w s log of y, okay? So we will have... Let me draw it straight. We will have x here. We will have almost every y will have the same value of phi plus and phi minus. And then by trick two we can go up, okay? Because of the definition of g and of the fact that it is a density point of fubini, we have this almost everywhere constant here both for the future and on the past and then we take almost every point here and then all points here will be constant. All points here will be phi plus will be the same, okay? So phi plus z will equal c0 for all in w s log of y for almost every y in w u log of x, okay? Now we apply fubini again. We integrate this and we will get the whole square. Okay, so let's... We have this, let's do fubini again. So fubini again, we will have the measure we integrate over the unstable leaf intersection g and over each of this y we integrate and the density of this set on s will be one, okay? So this implies that phi... Yes, no, here m is the big measure. Yes, yes. Because here we are integrating we are taking w u log x and then we are taking the union of all the stable manifolds on x that belong here. So this will be... contained in the tutorials. Okay, so this implies that almost every set will be constant. So by Birkoff... No, not yet by Birkoff. This will be almost everywhere constant on s, but now we can tile the whole torus with squares. Not tile, we can cover because we want them to overlap and since they overlap and they coincide the constant will be constant over the whole torus. Okay, so we will have that this is constant but then since this is constant by Birkoff theorem it will be ergodic. Okay, so we have five minutes and I would like to remind you what are the elements. Yes, x is a density point of g. In fact, it's just for elegance because I want this to happen. I use that x is a density point of g in Fubini to prove that this holds for almost every y in w, x, u of x. No, it's... Yes, that's Fubini. We are using that we want that almost every y in w, u log of x, y belongs to g. Okay, so if g is a density point the measure of g around x is one and so we want... Oh, I saw almost one. So we want this to happen because this implies this. Yes. Yes, no, no. We are using just the measure of g on the square is one. Yes, we are using in this sense an x is one point of this. Okay? Well, let me just recall the elements we have used here. We have strongly used that the function is continuous because if it weren't continuous then we wouldn't have that this is constant along the stable manifolds. This is an element that the map is continuous but this is more or less standard so I will not recall it as an element. So what we are using, we are using that, okay, we are using that five plus constant. Okay, we are using this. This is one element. Other element is that this w, e, s, that's good foliations. What do I mean by good foliations? Because we didn't use that the map is linear in fact. We used that we could apply Fubini. If we can apply Fubini then the linear part is not relevant but in order to do that we need that these stable and unstable manifolds are good foliations. Okay, what do we mean by good foliations? I don't want to get into technical details but it's just enough that we can apply Fubini. Good that we can apply Fubini. Fubini to decompose and then Fubini to compose again. We use it twice. And the third element that we have used is that while this square covers the whole manifold so we can reach from each point, this is, let me call it like this, we can reach from each point to each other point in the torus by following stable and unstable paths. It's like a street system. By a path made by stable and unstable leaves. There are more technical details but this is essentially what I would like you to remember. The three essential points are that this is almost constant, that we could decompose the measure into stable and unstable leaves and we could compose them again. And we needed that we can go from any point in the torus to any other point of the torus by a path made of U.S. leaves. Okay, and this is it. We are finished today. Thank you.