 So, hello, thank you for coming for the last lecture. I will try to prove the main result today, and I hope you will be surprised how simple it is. I'm not joking. So let me repeat the result that we formulated last time. We have a solution of this equation in some, say, ball. B, we have a subset of a smaller ball here, or depends on the dimension. And we assume that the measure of E is positive. Then what we want to prove is that the maximum over the ball is bounded of our solution. U is bounded by the maximum over this set of positive measure times the maximum over the big ball or unit ball there. E is some quite well said here of positive measure, and the measure is fixed. The constants depend on the measure of E, my parameter R, and constants of the operator, ellipticity and Lipschitz constant of the operator. You can think about it as a propagation of smallness result if you know that your function is bounded a prory by 1 and is small on the set E. Then I'm saying that this function is bounded by constant E to the power of gamma on this middle ball. As in this Rebault's theorem, you can repeat these steps many times. Instead of the middle ball, you can think about any compact set inside. But what we're saying is that the propagation of smallness in this like that. I have a bound epsilon on set of positive measure. Then on any compact set, I have bound like that C times epsilon to the power gamma. The dependence on this small parameter epsilon is like that. So it's a generalization of the three ball theorem. And as far as I know, this question was asked by Landis in the 60s if this is true for set of positive measure instead of the ball inside. And there was a nice work in the Rashfili in the 80s who gave a bound here that would go to 0 when epsilon goes to 0, but not as nice as the exponential bound that you would expect from three spheres. We also reformulated it in the following way. So that it will look like cubes better than balls and replace this estimate by the following one. I have solution of this equation in unit cube or say constant time this cube. I define doubling index of u over any cube as the log of the maximum. And I look at the global doubling, that is the supremum q in q of these doubling indices. And then we said that our estimate would follow from the following one that if you look at points in your cube, q is any cube in q0 such that it's less than e2 minus a maximum of u over q. Then the measure of this set is bounded by constant e2 minus a over n times beta n is the doubling constant over the cube and beta and c are some parameters that depend on the nopticity and Lipschitz constants of our operator. And this estimate that we are going to prove. Before we do that, let me comment on this inequality. I think it's a trivial comment for many of you. But if you look at it and you haven't seen it carefully before, I want to say that this is nothing else but estimate of b more norm of the log of your solution. What is written here is that measure of the points in the cube where maximum over q of the log minus g of x is larger than, I'm taking the logarithm of this inequality is larger than a. This measure is bounded by exponential minus a beta over n. And it's nearly an inequality that says you that the b more norm of the log of u is bounded by constant times the frequency there. And to prove this, I will look at it in the following way. So let me denote by m of u a q the ratio of the measure of the points in q where u of x is less than e to minus a maximum of a q over the q. I guess I forgot to normalize my inequality here. If I don't assume that q is a unit cube, I have to put the measure of q in there. So this is the quantity I'm looking for. And I want to estimate it in this way. So we want to know that this is bounded by constant e to minus a beta, but they want to find c and beta side of this inequality holds. I believe in this inequality, so I believe that the maximum of those is finite if I maximize over all possible cubes all possible solutions of elliptic equations with coefficients that satisfy some estimates. So I will denote by m, a, and n the maximum of all this ratios where the maximum is taken over all cubes and all operators of the form and all solutions to this one. I fix the constants for a, ellipticity constant, and Lipschitz constant, and I look at the maximum of this ratio for all possible cubes solutions there. You may ask why maximum exists, and we will explain that in a minute. Let me first rewrite the estimates and tell you what is the idea of the proof. So now what we want to know is that m of a, n is bounded by. Look at it as a function of two parameters a and n. For some parameters, this is trivial. If the ratio a over n is small, I can make this constant large so that this inequality holds. It's always bounded by 1. This ratio is always bounded by 1. So I will choose my constant c such that when a is less than n, we have this inequality in a trivial way. Then we will prove that this inequality holds in a strip where n is bounded. We will bound the n and improve the inequality here. And from here, we will do double induction. I will go to a point there and compare the value of my function m of a, n at this point with some of the values with some coefficients at two points, 1 below in n, 1 to the left in a. So that the double induction go this way. First, my base will be n from 0 to n, 0. The next step, I will double n and prove it for all a induction in a. Then go up again, prove for all n on the next level, for all a's in this way. And so this will be the double induction argument. But let us start with the base of the induction when n is bounded. And we will see that we have some non-trivial estimate there. The main idea is to use a oscillation theorem. It tells you that if you have a solution to elliptic PDE and some square, and you look at the half of this square, then the oscillation of your function u over the half of the square is bounded by constant time the oscillation over the square. Without this, it will be trivial estimate. But this holds with some coefficient tau that is less than 1 uniformly over all elliptic operators with known estimates on the coefficients. And by oscillation, I mean the difference between supremum and the infimum without absolute values over u over q. So how we will find set where our function is not very small? How about q over q? Suppose that the maximum of u over q is. Don't have nice letters. All of them are taken. OK, suppose this one. And divide this cube into small ones. First of all, on each small cube, the supremum of the absolute value is controlled because we control the doubling index. I assume that I divide it into many pieces. I can double that one or multiply it by a constant. I'm in a larger piece. The maximum here is 1, but the maximum on this one is also bounded because I know the doubling index. Yes, it depends on the doubling index and the size of the small cube. So let me say that I divided my cube into cubes and depends on the number of these cubes. And the other hand, we know that the oscillation of u on a small cube is not very large. It's bounded by some constant. I can take tau to a large power if I think about cube's q is very small. I repeat this oscillation lemma several times. I can make the oscillation of this one to be less than very small constant a time the oscillation of the initial cube that is bounded by 2 because the maximum was equal to all at me. A that depends on my j times constant that depends on. Yes, let me denote it still by tau. So this will be tau 0 that we get from the oscillation lemma when we do it once. And when we do it many times, I have a new tau that is very small tau goes to 0 as j goes to infinity. This is true for all small q and this is true for all small q. I still control my function on the big cube, right? I control the doubling index for this q. I know that the maximum is less than 1 and the applied at the maximum over k q is bounded by e to the power n, right? So that's why I write this constant here instead of 2. So we found 1 divided q into cubes. Look at the maximum everywhere is not very small and the oscillation could be done small if we take the number of cubes large enough. Remember n is fixed. In particular, if you take one cube where your function attained the maximum value on the cube q, I can make the oscillation less than 1. And then I'll know that on this cube my function is never 0. If I take oscillation less than 1 half, I will know that on the cube where the maximum is attained, my function is at most 1 half there. So I'll find one piece where the function is not small. I'll leave this and that probably. And it gives me the first estimate. If I divide it into j cubes, one of them is not in this set, doesn't intersect this set. So this is bounded by j minus 1 over j times cube. There is one cube with a maximum value where oscillation is small and we have good portion. So now I'm going to iterate this one. I know that on each cube the maximum is not very small. It depends on this fixed parameter j that I take and n. And if you iterate it, you will see that set where u of x is bounded by 1 half b to the power l is bounded by this constant to the power l times q, just iterating this estimate. And I claim that this gives you the basis of induction. I have the right decay of the measure of the set where the function is small, right decay in a. I forget about n now, but this, when you take a power of something less than 1 and get a power of something less than 1 here and get the right decay. If you'll try to write down how my parameters depend on this frequency in this basis of induction, I guess the best thing that you will get is exponential in n here, not linear thing. So this is the first step that tells you that there is some estimate if you fix the frequency, the doubling index here. Then there is decay of this measure of the set of the function is small with the right dependence on a. And now I want to go from this base of induction to induction step. And I'll do induction from n to 2n. And at the same time, use the fact that here I have some base of induction in a. So I will do induction in a as well. Yes, it's just arbitrary. Oh, beta will depend on this one. So we choose a beta set it for this n less than n0. We have this estimate. And we'll see if we can extend this estimate to other n's. Or we'll have to adjust this beta to make our induction step to work as well. And this is what I said. I said it's a very simple proof. It's a surprisingly simple thing to do. I'm going to write an inequality for my function m of a n by taking a cube or q and dividing it into, once again, many pieces as we did yesterday. We know that if we take a cube divided into many pieces, there is one piece where the doubling is less than for the large cube. Just need one cube with doubling that is less. So I have this supremum. I take a function u for which a solution for which the supremum is nearly obtained. Think of it as a solution of elliptic equation in this cube. Divide cube into many pieces. Call them q's once again. Say k. This is no. Size it for at least one of them. The doubling is less than or equal to the doubling of q and q over 2. I can do it if my doubling is large enough. I have to compare doubling to the usual doubling. If I choose my n0 large, then I can do this step. Now, let us look at our function in each of these squares. It solves an equation with the same elliptistic constants and Lipschitz constants if you want to renormalize and think about your cube as unit cube when you renormalize your Lipschitz constants are even better. So what we have is that this u is, sorry, let me write it this way. So I will sum over all cubes, but one, the one here, and then add this cube where I know that the doubling is less than it was. And I want to estimate my measure by some of the measures. I have similar things for small cubes, but I have to adjust the sink here. I'm still comparing the value of my function to the value of the maximum on the big cube. And I want to compare it to the maximum in this particular cube. But I know the doubling. So I know that the maximum of my function here is not much smaller than the maximum in this cube because I control the doubling n. So we probably can take this away now and write the maximum on any cube cube of u is at least e to minus constant times n maximum of a cube. And then my inequality is the following. Taking supremum there, I have k minus 1 over k terms like that. What changed? a is now mu constant. a is less than it was. It's a minus cn. And the doubling is the same. But I have one term where doubling is just half of this one. If you look at my picture here, I was taking a point, a m, and where I want to have an estimate. I got some cubes where a is smaller and one cube where n is 1 half of those. And I know that the value here is bounded by a combination of those two. And this one is with k is less than 1. And here I already know that my estimate from induction of what is this. Here I know this estimate because I'm doing conduction in a. And I have my base in a. And now you have to write down what we have and see that this will be k minus 1 over k times e minus some constant, c0, should write beta times the ratio of those two, a over n minus c, but with constant less than 1 plus e minus beta 2a over n minus. It's not difficult to convince yourself that this one, if you choose the beta in the right way, is.