 into small cubes equal ones say k to the power of d small cubes q and k's large compared to my constant small k q. Then the doubling index function h at 1 over k q is larger than the minimum of the doubling index on small cubes times a constant say k over 10 small k here. Now let us assume that the I don't think we need it let us let us prove it I don't think we need any any assumption on this one right right now. But the conclusion is if the index is large for each small q and it's really huge for for the large q. We take this cube one over k q and we will go from this one to the large cube q in this way. So I take this small this cube one over k q and take the maximum point it could be inside or outside but take the maximum point there. Take it there I look at the doubling constant for this cube yes we will assume that all the doublings are larger than one. Then there is a point around this one where the value is larger than this maximum point because when I double the cube I get constant larger than one it means that the maximum of the larger cube is big. So I take this cube here not double it by multiplied by this constant k small and I'll find a point where the maximum is larger than before. So we can find take x0 in one over k q such that h over x0 is equal to the maximum of h over this cube. Then we can find point x1 in small cube q multiplied by our constant k such that the value of x1 is at least e to the power this minimum sorry yes and we assume that x0 is small q is the small cube that contains our point x0 where the maximum is obtained. So we took a maximum point in this cube and took cube from our partition that contains this point x0. Then we're going to find point x1 in this cube where the maximum is at least e to the power n times the maximum that we had before and we're going to repeat the steps and see how many times we're going to repeat it and originally we were in the cube one over k times q think about size of side of the q as one then side of the cube q is one over k and each time we go small k over large k steps in the distance. So we can repeat this procedure until we reach the end of our large cube and we will find point s where the value is at least e to the power sn value at this point and if you think about how many this shelves would go in one step we go not more than k shelves and we have from the middle one to the large one approximately large k so we can have s that is larger than constant k over k there and it gives you immediately the doubling index of the cube one over kq is at least s and this was the minimum this is a constant k over k minimum of the doubling indices of small cubes for the moment there was no assumption of function h whatsoever h was any nice bounded function you take the maximums and you go from one scale to another the only thing that we assumed is that the mac the doubling in each cube is larger than one so that we do our step from maximum point find even larger value because now I will end up in a new find x1 say it's over here I take a small cube x1 that contains this one it's a new cube but we assume that for all cubes we have this property that the doubling index is larger than one at least n so from this one I can still find a point in there where the value is even larger for arbitrary function it could happen that you go back here if you have solution of a polyptic PD you have maximum principle assume that you'll go to the boundary have can repeat the steps and you see that you can have many of them and give you this this constant over that over here for the moment we use nothing about our function and we want to talk about solutions of elliptic PDs and there we have the idea that doubling should be increasing function we talked about doubling of on the balls now we have a double in the cubes but up to some constants it's still kind of an increasing function so true doubling index of a cube should look like that so we have function H think of it as a solution of elliptic equation now and define a beautiful and H of Q as superman over doubling indexes of H let me put large Q here small Q there over all Q so we have a cube we'll look at all small cubes you know Q Q if look at the doubling index for this one and takes a supreme I will assume that L H is equal to zero in a logical if you do it for arbitrary function you can end up with infinity here but when we're looking at solutions of elliptic PDs we know that we have traces of monotonicity of the doubling and it tells you that this supreme is finite moreover one can check that for this one the old doubling index of a cube is larger than constant times this modified index of the cube minus another constant so if this supremum is large if you can find a Cuban side with a large doublet index then doubling is for your large cube is also large there is a control of the usual doubling through this supreme doubling and this function is nice because this is monotone by the definition if Q 1 is a Cuban Q 2 there was a question say if you have two cubes then by the definition you have this property that they increase so the question was if I want to put a constant here and say that if okay so one more thing that I want to do today is to iterate the observation here it told you that we have can find one Cuban side where the doubling index is not as large as the global one and remember I want to find many cubes where the doubling is not as large as the global one so now I'm going to talk about solutions of elliptic PDs as I assume that that this is true in a big cube and we partition Q into b to the power d small equal cubes Q for each of them this modified doubling index is bounded by the doubling index of the large cube by this inequality but we claim that there are many cubes where the doubling index is smaller so then the number of cubes Q where the doubling is larger than one half of the doubling of the initial one is small the total number of cubes is b to the power d and this number is less than b to the power d minus constant so there exists c and b not such that and those constant depend on the elliptistic constant and Lipschitz constant of the equation if you have this solution of this equation in such cube and you divide into small cubes with b larger than this number then the number of cubes with big doubling is small compared to the total number of cubes total number is b to the power d and here we win some some power and there is nothing new in this statement except for comparison of this modified doubling and the old one and this lemur I will not go into technical details but what you do roughly speaking you take your cube Q you divide it into k to the power d small cubes then you know that at least one of them will have usual doubling that is less than the doubling on this scale divided by a constant so let us skip technicalities and think about the initial doubling as increasing one I have a cube Q I divide this one into k to the power d small squares and then I have at least one square where the doubling is less than n over 2 and all the other squares where the doubling is still less than or equal to n and then I repeat this procedure take each small cube and divide it further into k to the power d cubes here this one the doubling there is already less than n over 2 so for each smaller cube it still be less than n over 2 for those ones for each one I'll get one u cube where the doubling is then n over 2 so after two steps if you think a little bit you will see that we have k to the power d minus one squared cubes where the doubling is less than n and all others this is kd plus kd minus one where the doubling is less than or equal to n over 2 now let us iterate it many times then we will get k to the power ld cubes in total and the number of ones with big doubling will be this one and in all the rest we will already know that doubling is less than or equal to n over 2 so the number of cubes where the doubling is larger than one half of the initial one is this quantity so the number of cube where the doubling and the thing that I'm cheating I'm confusing this doubling index with that one and technical thing that you can work out this is bounded by kd minus one to the power l where the total number of cube cubes is k to the power d if you do know this by b to the power d this is b to smaller power and it gives you this estimate so this is the basic thing for induction that we will do tomorrow and see how we can use this simple observation to prove the estimate on the size where the solution of elliptic equation is small I think you have to stop now and thank you for your attention