 Ok, zelo je to moja zeločna lekova, ozvrčenja izgleda za tezvrčenje z geometričnej zeločne. Zeločno, da se vse zeločno vse zeločno. Zeločno, da se zeločno vse zeločno. Vse zeločno, da se zeločno vse zeločno. Zeločno, da se zeločno. Zeločno, da se zeločno. Mislim, da imaš štah, ki je veliko pozena, na brano za sve. V začnem stavodnem komputetih, da je stavodna zelo, zato igravim ta nekaj ječen visi全erati tko, ki je evakovati tron. In nekaj, ko stavim, omrečizavate, Prvno, je to nekaj strategij, da se vsega rezačne rezačne za superemetrge in svarju. I pa svoj taj problem nekaj tega. Vse rezačne rezačne rezačne in vsega rezačne rezačne nekaj pravno taj pravno, da pravno svašte aplikacije, ali nekaj naše vsev problem, vse do vsega sajte v Klaudiju pravno zelo, da se vsega naše vsega rezačne aplikacije. v vseh razrednih razrednih, z kakorah razrednih. Pa vseh bi z�apravili, da je to početne, da lahko se je vseh vseh, da je to početne. Pa vseh tačno, da je pravdala, začnaš, da je to početne. Poznih z vsem. Zelo izgledam, skupaj. Zelo izgledam, da nekaj začne. Zelo v Bruminkovski, da se početne. Zelo vseh zelo izglede, da se početne. in svoj skupaj, ko je, da je vzlokal. Načo jazelje v izglednih spiričnih, bo vzlokal stabilitivi za sobolev in kualitiv. In maybe Gallardo Nivenberg. Te vzlokal je funkcional načo vsej vsej, da jazelje je, da je je vzlokal k svoj zestavljaj, s katerim, za sobolev. Načo, stabilitivi in kvalitivi, za kajšči vsej vsej vsej vsej vsej, in asyntotic behavior. So, for applications, these are used in asyntotic behavior of solutions to parabolic. My have supreme metric in quality. For instance, they appear naturally in... Let's say, in general, in variational problems. So, in the notes, the technique to Claudio, there is some discussion in one of them about applications of this stability and equality in variational problems. Problem coming actually from... to study the shape of crystals, while in another note, there is... there are some possibilities for... quality, which can be used to study a long time behavior of solutions to parabolic PDs. So, this is just to mention some possible applications. Something which happens, I can tell you, as an experience, in these problems is that, unfortunately, there is no universal method to prove stability. Like, whenever you have a new problem, if you are lucky, some variant of non-methods work, but sometimes you just have to start from scratch and find a completely new approach, which is going to be exactly the point now for the Bruminkowski. So, just very briefly, just in ten minutes, I would like to mention an approach for a supreme metric in equality, which works for the classical supreme metric in equality, wouldn't work for the anisotropic case. And this kind of connects also with what you saw in Manuela's lecture, because he used regularity theory for minimal surfaces. So, let's say... So, another approach for stability. These approach have been started and then improved with time, but I think the first one we use these in this context has been... have been Cikalese and Leonardo. So, the idea is the following. The idea is that you want to prove, that something like the deficit of a set controls up to a constant, the symmetry, square. So, this is what you want to prove. So, you say... Okay, suppose that by contradiction, there exists a sequence ek, such that... So, we have the volume fixed, ek equal to the volume b1, delta ek over asymmetry of ek square is going to zero. So, you have a sequence, which contradicts your inequality. Then you say, okay, what's the idea? The idea is that I know full-grader result. Full-grader tells me that if I am a smooth, nice smooth perturbation of the sphere, the result is true. So, if I could show that this sequence, ek, was at some moment of this form, I would get a contradiction. The problem is that, a priori, I have no information on this sequence. They are completely random set. It's easy to see that they are not of that form, because if you give me a set ek, no matter how it looks like, I can always, for instance, get a very small tentacle. And so, even if this was ek, I get a rather small tentacle. This would be... I can do this in such a way that deficit is almost the same, the asymmetry is almost the same. And so, you see that this new set, ek would still be a set which satisfies its property, but it's very rough. So, I'm saying a set which satisfies something like this, a priori has no regularity. It could be extremely rough. On the other hand, I could do the following. I would say, OK, I have a family like this, which is... I mean, I have a family ek, which is doing this. Let me replace ek by a new set fk, where fk is a minimizer of the following function. Minimum of volume of f equal 1. So, I have the volume constraint. Sorry, b1. So, I have the volume constraint. And then I say, let me fix the asymmetry of fk to be equal to the asymmetry of ek. So, asymmetry of f equal asymmetry of ek, and you minimize delta f. So, you say, among all sets which have this denominator fixed, let me minimize the denominator. OK? I can do this. And you have... So, this function is what? This is nothing else that the perimeter of f minus the perimeter of the unit board. So, really what I'm minimizing is kind of the perimeter with an asymmetry constraint. So, I'm minimizing the perimeter among all sets which have the same volume, maybe 1, and some fixed asymmetry. And so, this fk, but of course, have a deficit, which is less than delta fk, is less or equal to delta ek. So, in particular, since the asymmetries are the same, I have this, and this was going to zero. So, also these sets have this ratio going to zero. But now these are less random sets because fk is constructed as a minimizer of some variational problem. So, maybe I could hope that these sets now are smoother because they are minimizer of the perimeter. The problem is that I have some constraints which are kind of annoying. So, if it was just minimizing the perimeter with volume constraint, well, the minimizer, okay, would be a ball, but in some sense, you can hope that, okay, this constraint doesn't mess up too much with the regularity. It's kind of difficult to work directly with that constraint. So, instead of putting on constraint, there is a... So, the difficulty here of this approach is to prove that fk is smooth. So, when you have this kind of problems where you have a constraint like this and you don't really know how to deal with it, you can try to say, okay, let me not force the constraint completely. Let me just force it. Let me just put it as a penalization term. So, try another weaker approach, in some sense. You say, okay, this constraint really kind of bothers me. I don't know how to deal with it. So, instead of minimizing that, I'm going to minimize delta of f plus a of f minus a of k, a of ek, where I just put the volume constraint now. So, instead of putting this, imposing that this is zero, I kind of remove it and I put it as an additional term in the function, which means that whenever af is not the same as ek, I pay something. Okay? So, why this? So, define, let now fk be a minimizer of this. So, I started with fk, minimizer of this and I discovered that this is not bad, but I don't know how to prove regularly. So, let's forget this. Let's try this one. Let's say fk is a minimizer of this. What do I need? I need to make sure that this is still satisfied. But, so, if fk is a minimizer, so what do I have? I have that delta of fk plus a of fk minus a of ek is less or equal than delta of ek, which remember that delta is just perimeter of the perimeter of the unit bowl. I can add the perimeter of the unit bowl of both sides and you get this, actually. Just because delta is the same as the perimeter of the constant. So, on the right, so you get this by minimality. And now perimeter of ek, I'm gonna, so I'm just gonna, so this is delta of ek plus perimeter of the unit bowl and this it's what? It's a little, actually, since it is going to zero, I know that this is bounded by perimeter of the unit bowl plus a constant, which is gonna be small, but I don't even care, a symmetry of ek square. This come from this. Right? The delta is controlled by ek square. And now here, the perimeter of fk by the superimetric inequality, for sure is greater than the perimeter of the unit bowl. Fk is a guy which has measure equal to unit bowl so it's perimeter is greater or equal. So, when you compare this, you see that this plus this term is less or equal than this, you discover, you simplify the perimeters and then you discover that afk minus a ek is less or equal to constant a ek square, which tells you that when ek, the symmetry is very small, you see that the symmetry of fk is kind of comparable. So, you have to think that the symmetry is very small. So, this is a lower order, much smaller than this. Right? So, this is telling me, if you have small numbers, this is telling you that up to constants like something of this for, you have that, so for k large enough, I will have something like this. Right? Give me three numbers and you have that this is small. Then, as this goes to zero, at some moment, you will have something like this. So, it means that the symmetry, okay, it's not the same, but they're comparable. So, of course, I can just go back here, right? I will not have something like this. I will have maybe a factor four in front because the symmetry now, the ratio between these two is bounded by factor four. So, the symmetry are not the same. Right? So, I have, of course, delta of this is less than delta of this. It's just this inequality. Right? So, I have that now still. So, delta fk over afk squared is going to zero. But now, for this functional, what am I really doing? So, let me rewrite it. So, since the deficit, again, what is the deficit? Perimeter minus perimeter of the ball. Perimeter of the ball is just a constant. So, it doesn't count. So, fk is a minimizer of this, we say. Perimeter of f plus af minus afk with volume of fx. Okay, so this, this is like you are minimizing the perimeter plus some other term with a volume constraint. So, the volume constraint is really not that difficult to handle. Like, let's forget for a second. So, forget for a second that and think that you are minimizing something like this. You are minimizing perimeter plus this term. What's the key point? The point is that this term, so this is a perimeter term and this is volume type. These are measures. It's about measures. The symmetry is a measure. So, what you can, in some sense, you are the graph. So, if you take, like the boundary of f locally and you take a ball of radius r and then you say, okay, let me compare the perimeter of f with the perimeter of some set. So, let me use a different call. So, you take this set, you call it g. So, this is the boundary of g. So, you take, so let g be such that g delta f is containing some ball br. So, this is a ball of radius r. So, you have a set which is with a perturbation of f only inside the ball. Then, you can use g as a test here and you get that the perimeter, you know that the perimeter of fk, sorry, this was fk, fk. Plus the symmetry of fk minus the symmetry of ek is less or equal than the perimeter of g plus symmetry of g minus the symmetry of ek and then you take this guy here and the point is that since f and g differ only inside this ball you get an information of this form. Perimeter of fk is less or equal than perimeter of g minus the difference between these two guys but, so this symmetry is something like this is like fk delta, this is the volume of the difference between fk and b1 and this is, roughly speaking, the difference between g and b1. So, if you do this minus this you can control this with f delta g the difference between the volume of f and g this term minus this term and this is like is controlled by r to the n because they differ only in a ball of radius r. So, this is, let me keep going here so what we discover is that the perimeter of fk is less or equal than perimeter of g plus a constant r to the n for every g such that f delta g is contained in some ball of radius r around some point x dot. So, when you perturb your graph, your set it doesn't matter if it's a graph this set could be completely rough, it doesn't matter but whatever perturbation you do the perimeter of f is minimal up to r to the n and r to the n is kind of lower order respect to the perimeter because the perimeter inside a ball of radius r roughly speaking is r to the n minus 1 the perimeter of a set inside a ball of radius r. So, here I'm saying this set is almost a minimizer of the perimeter function up to a lower, so this is lower order term with respect to the perimeter. So, this is not what do I want to say, I'm saying what I want to say, the boundary of k is not a minimal surface, it's not the minimal surface that you saw in Emanuele's class in this case he had co-dimension I mean this is co-dimension 1 Emanuele had arbitrary co-dimension but he had minimal surfaces which minimized the perimeter with fixed boundary condition he had a surface such as whenever you perturb it the area has to increase where the perimeter has to increase up to an error but the error being lower order is not a regularity theory. So, being lower order so this means that fk is what is called an almost minimal surface and since we are in co-dimension 1 now I'm not going to enter into the theory but so boundary of k is n minus 1 dimensional and also we know since the symmetry is very small that fk it's very close to the unit ball so it means that my set so here I have the unit ball and my set fk a priori could be a very rough perturbation of the ball this a priori what fk may look like but because it's an almost minimal surface so it satisfies some kind of equation and here it doesn't matter that we are in co-dimension 1 but the fact is that almost minimal surface which is close to a ball actually implies by the regularity theory boundary of fk is a c1 alpha graph over boundary of v1 so this cannot happen actually your set is not like this the regularity theory tells you no your boundary of fk is this regularity for minimal surface which are close to some things there is a regularity theory developed for this so here I am using something deep and high in one trivial but what's the point is that now my sets satisfy full grade assumption my set now for k large enough large enough fk satisfy full grade hypothesis which tells me that delta of fk is greater or equal in a dimensional constant the symmetry of fk square but it's a contradiction because we are assuming that remember delta of fk over symmetry or fk square was going to zero so in some sense I suppose that my sets were not good for that I replaced them so I started from sets in k which were completely random I didn't know what they looked like I replaced them by new sets which still satisfy the fact that this ratio was going to zero but they were smoother and then for the particular cut and then I could show that these sets and then for smooth perturbation result is true so I get a contradiction so this is kind of approach which is very efficient it gives you in some sense it tells you the only thing you have to check when you want to prove stability is when you are a smooth perturbation on the sphere in some sense full grade case is the only one to check because all the rest you can get by this kind of abstract machinery compactness argument so it is deep like the regularity theory and so it is more direct but for some aspects it is less elementary because you use deeper results but it works in several other problems now to show you that the theory is not over in some sense it is not that the methods you saw like this one developing around the minimizer or using an optimal transport map it is more or less once you have this one way or another you solve a problem give you another problem where any approach that I I mean that I describe to you but also anyone else I knew up to now just fail unfortunately or fortunately because then you have to think something new which is interesting so so let's start with a different problem which is Bruminkowski so Bruminkowski is about volumes of sets so the idea is the following you have two sets A and B in RM so actually last week you already saw what the Bruminkowski inequality is because Guido gave a proof of Bruminkowski inequality using optimal transport let's say so I have two sets A and B I am going to assume that A and B are compact and I am going to define their sum A plus B to be the set of points of the form X in A, sorry, X plus Y where X is in A and Y is in B so you start to take all possible sums of points in A and B so what Bruminkowski is about how can I relate the volume of this guy to the volume of these guys so it's about relating volumes so what's the inequality in one dimension so in the one decase the Bruminkowski inequality is very easy and it tells you that the measure of A plus B is greater or equal to the measure of A plus the measure of B to give you the proof, the very elementary so up to a translation you can assume that the minimum so all these inequalities are translation invariant you can always translate the sets nothing will change in the volumes so you can assume that the minimum of A is zero and the max of B is zero so you have this picture you have the line you have the origin A is somewhere here is some compact set which lives on this half line and B lives on the left you see that A plus B contains well, in A plus B I can take always A plus zero because zero is in B and then I can also take B, any point in B with zero because zero is in A so you take points of some of this form this is just A itself this is just B itself and this is a disjoint union this is equal to A union B and you have a disjoint union so the volume is greater equal than the volume of the union but this is because they are disjoint the sum of the volumes so one dimension is just this simple trick now what happens in higher dimension in higher dimension you will try to see if this is the right inequality and I tell you no and actually a simple way to guess what the right inequality is is to test it with balls because it's an easy example so what about n greater equal than 2 in higher dimension let's do an example take A to be the ball centered at zero with radius rho and B to be the ball centered at zero with radius r then exercise so this is example this is exercise A plus B is nothing else than the ball of radius r plus rho so if you take a ball and use some other ball you just have the sum of the radii but then what is the volume so the volume of A plus B is a dimensional constant time r plus rho to the n and the same dimensional constant r to the n and the volume of B is the same dimensional constant sorry A was B rho so this is rho to the n this is r to the n so if you want an inequality which is kind of sharp at least you would like maybe I mean you hope that you are not doing something completely being not sharp is to get powers 1 over n because then this implies that if I take powers 1 over n I get this because you take powers 1 over n and then you have r plus rho rho and r so this is inequality in this special case you see that balls which usually should be the best sets usually the minimizer for every inequality especially with so many three symmetries if you want a sharp inequality on balls you have to get power 1 over n otherwise you don't have equality so if there is a sharp inequality it should be something like this and in fact the Bruminkowski inequality is this we just put a greater or equal so theorem Bruminkowski which is called Bruminkowski but they proved it only for convex sets I think Brum proved it and Bruminkowski did the quality case but only for convex sets A and B compact and non-empty then the volume of A plus B 1 over n greater or equal than A 1 over n let me give you a proof of this so Bruminkowski inequality is one of the inequalities that you can't prove by induction so it's kind of interesting actually just a comment for people who know that Bruminkowski is like an equality of sets there is analogous of Bruminkowski for functions which is called Precopalindler and then Precopalindler has the nice feature that it goes up by induction so in some sense you can prove Bruminkowski precopalindler in one dimension precopalindler in one dimension Bruminkowski in one dimension but if you remain in the space of sets you cannot go from one dimension to one dimension there is no induction argument for Bruminkowski which I am going to show you is kind of neat and cute the proof of this is by approximation in induction so you start from the case so let's do the case where A and B are boxes with parallel sides parallel sides so just to do a picture this is my A and this is my B boxes with parallel sides and let's say that the sides so sides of A they have length A1 AM let me call this the length of the different sides in this case would be A1 and A2 B1 and B2 so sides of B they have length B1 and Bn so you have two boxes and this is the length of the sides so what do we want to prove we want to prove that we want to prove this in equality so I am going to look what is this so I want to prove that this quantity is equal to 1 what is this quantity this is equal so the volume of A is the product of the I and the volume of A plus B other exercise so if you do A plus B that would be a box with sides A1 plus B1 AM plus Bn so if you have this A plus B A1 plus B1 and AM plus Bn boxes, I mean these are cubes, parallel people this is a box for me this is not a box in box, in very few box means like this this picture and so the sum of course the side just adds so I get AI plus BI and I have 1 over N plus the same product of BI product AI plus BI 1 over N and the good thing about products is that if you have products on the top and products on the bottom you can just put the product in front so this is the same as product of AI AI plus BI 1 over N plus product BI AI plus BI 1 over N and now we already saw it we used it last time arithmetic geometric inequality so arithmetic mean geometric mean inequality tells me that the product to the power of 1 over N is less than the average so this is less than 1 over N sum over AI AI plus BI and I get the same here plus 1 over N sum over AI of BI AI plus BI so apply to each of them arithmetic geometric inequality now just put the sum together 1 over N sum over AI whatever, AI divided by AI plus BI plus BI AI plus BI so I have AI plus BI over AI plus BI which is just 1 and the sum goes from 1 to N so you have 1 over N sum from 1 to N of 1 so you get 1 which was the inequality I wanted so Bruminkowski inequality for boxes is nothing else that arithmetic geometric inequality so this is now how do I do the general case the general case is done by induction on the total number of boxes so in 2 we do induction over the number of boxings and b they say A as 100 boxes and B as 200 so the total number of boxes is 300 and they do induction on the total number of boxes F so I did the case where there are two boxes A is 1, B is 1 and now I do induction over the total so so the next case would be 3 boxes so if I have 3 boxes at least one of the two must have 2 boxes so let's say without loss of generality A as at least 2 boxes so number of boxes are all disjoint so induction over the number of boxes all boxes are disjoint respectively for A and B so A cannot be this this is not admissible A can be something like this so this is not this is okay so these are disjoint boxes so if A is at least 2 boxes since these are disjoint again I didn't write down but boxes and my boxes are always parallel to each other these are all parallel all boxes are disjoint and parallel so something like this the axes are all parallel so since the boxes are disjoint I can always find so I can find one coordinate hyperplane which splits as the union as two sets A1 disjoint union where these are again union of boxes so let me draw a picture to show you what I mean it's easier to draw them to write and to see what's going on so I'm just saying that because this joint I can I can always find an hyperplane which could be the vertical one or the horizontal one but one of the coordinate hyperplane let's say this one A is here and another number of boxes is always here so this hyperplane is not allowed to cut any box A minus I can find an hyperplane which splits my set A so this is my set A which splits A minus and A plus and I have at least one box on the left and one box on the right so at this level I didn't cut so here the hyperplane touch no boxes in the assumption they are non-empty so I am doing number of boxes I mean also because when you do this set with an empty set you get empty set so it's not very interesting so the hyperplane does not intersect A so I am finding an hyperplane which does not intersect A and split my A in two sets now let me draw B I am going to take another hyperplane so I should find an hyperplane parallel to the previous one so that the following happens so let me call let me use colors to so maybe this hyperplane I take an hyperplane vertical and I define B plus to be what is on the left so here I can cut boxes when I call B minus what is on the right this is B minus and I choose it previous one so that the fraction of B which is on the left is the same as the fraction of A which was on the left so the volume of A plus over A is equal to the volume of B plus over B so what I did is the following I cut first and let's say that I have 40% of the mass here and 60% of the volume here then I take an hyperplane and start to move it and I look how much fraction of volume I am on the left with respect to the right I stop when I have 40% here and 60% here this is my hyperplane it's a ratio it's just a matter of percentage I want to make sure that the ratio on the left here is the same as the ratio on the left here so we have also generality up to a translation now I redo the argument that I did in one dimension so up to a translation we can assume coincide with let's say x1 equals 0 so these are both hyperplanes x1 equals 0 and then what happens you see immediately then A plus plus B plus A minus from A minus plus B minus in a sens that if I take a point here and I point here the sum will have coordinate less or equal to 0 if I take a point here and I point here the sum will have coordinate greater or equal to 0 so sums of points which are both on the left, we remain on the left sum of points which are both on the right remain to the right so in part of the sets are disjoint I mean this is always true that they are disjoint I didn't need to do a translation to say that they were disjoint it was just easier to visualize if I tell you that they are both x1 equals 0 so in this way x1 is less or equal to 0 here so this is contained x1 less or equal to 0 and this is contained in x1 greater or equal to 0 ok so now I say then the volume of A plus B is greater or equal then the volume of A plus plus B plus plus the volume of A minus plus B minus just because these sets are disjoint and of course each of these sets is contained here this is contained here, this is contained here so the sum is contained and same here these are disjoint and they are both contained in plus B and now just apply so now the point is that since I split A without counting boxes so you see that the number of boxes of A plus is strictly less than the number of boxes in A number of boxes of A minus is strictly less than the number of A the number of boxes in B plus is just less or equal I don't know if they decrease it because I cut so maybe when I cut I created new boxes so in fact it is possible that I didn't the number of boxes is the same for B because this then with the cut I would have still one box and one box so it didn't increase at least and same for B minus so I can apply the inductive hypothesis because now the sum of the boxes of A plus plus B plus is strictly less than this same for this because here it has strictly equalities so by induction I know that the result is true for A plus B plus so by induction A plus plus B plus is greater or equal than A plus one over N plus B plus one over N power N and now here is the moment where it is crucial that I chose the ratios in this way because now this is what A plus which multiplies one plus B plus over A plus one over N power N but this by this choice this is the same as A over B so this is exactly the same if I replace here A plus one over N to the power N and now the same is true for A minus and B minus because of course these ratio like these implies the same if I put A minus and B minus if it was 1440 on the left it's going to be 660 on the right so the proportion on the left implies the proportion on the right is the same so A minus I get so here I am using that A minus over A is also equal to B minus over B which is just an immediate consequence of that and so you get here A minus one plus B one over N A one over N so the same inequality but with the minus and now you just add inequalities and you are done so let's let A plus B is greater equal than the sum so A plus plus A minus one plus B one over N A one over N power N and that's so this proves that this proves that the result is true whenever A and B are finite union of boxes there is as large as you want because by induction you have as many boxes and by approximation you get the general compact case general compact set there is an argument by approximation you approximate compact sets with finite union of a lot of boxes you apply the result and you pass to the limit I am going to stop in a second and then next hour we are going to discuss the stability but let me mention two interesting facts one is is a interesting remark that if you have never seen it is useful to know so Bruminkowski is one of the strongest inequalities for this kind of geometric setting for instance Bruminkowski implies that superimetric inequality also the anisotropic one so remark one is that Bruminkowski implies isoperimetric inequality also anisotropic how do you show this at least for most sets you say ok let me do directly the anisotropic case if you remember the definition of yes what was the anisotropic perimeter that's the limit for epsilon going to zero of the volume of e plus epsilon k minus e over epsilon for a perimeter remember if the classical perimeter corresponds to k equal to unit ball ok classical perimeter means k equal to b1 ok now what do we have we have a so this was the epsilon neighborhood if you remember yesterday was the epsilon neighborhood if you look at the notes and now we have a sum Minkowski sum and the Minkowski inequality tells me that the volume of this is greater or equal than the volume of e 1 over n plus the volume of this 1 over n what is the volume of this the volume of epsilon k 1 over n power n minus e epsilon limit as epsilon goes to zero but then this is a volume to power 1 over n so it's one homogeneous the volume scale is power n but then you have 1 over n outside so the epsilon goes out so this is limit epsilon goes to zero e 1 over n plus epsilon k 1 over n power n minus volume of e over epsilon this is n measure of k 1 over n e n minus 1 over n so this is the superimetric inequality which have constant so this is the first remark so you see also why Bruminkowski is interesting it implies our superimetric inequality in one line a second remark is the quality case that is not obvious from the proof but I'm going to mention to you and then we're going to see more later in the stability so if a plus b power 1 over n is equal to a 1 over n plus b 1 over n then this means that a and b are convex and they are homotetic so this means that they are a is equal to the form lambda b plus x dot for some lambda positive and x dot in a b so the only way actually this is a leaf and a leaf so a quality in Bruminkowski happens if and a leaf both sets are convex and actually they are the same convex set up to deletion OK so let me stop here we can take so up to 07 and then we start again OK so what I would like to discuss now is the stability for Bruminkowski so it's not very easy from whenever you have a proof which is not done directly so you have to argue by approximation, induction, whatever it's very difficult to even understand the quality case so to understand the quality case I would just like to remember 5 minutes almost the proof by optimal transport and which kind of make you understand a bit why things work but it's gonna be useless for stability and we'll see why so let me stop of Bruminkowski by optimal transport it's not gonna be useless but let's say it's gonna be useless for the general case so the widow gave a proof last week based on some displacement convexity and quality let me just give a direct proof which is more or less a copy of what I already did the last time for the superimetric inequality so you have your set you have your set b and you also say two probability measures to a and b so uniform density in a uniform density in b now you take the optimal transport map t equal grad phi which sends mu on to nu so t push forward mu equal nu phi is convex you have that the determinant of grad t its b over a now the observation that you can make is the following that so t is sending a on to b so t of a is equal to b which implies that if I take the set so let's consider this set identity plus t of a so the image of a through the map identity plus t which is the set of points of the form x plus t of x where x is in a so this set this is in a, this is in b because points in a are sent to b so this is contained in a plus b so what I can say is that the volume of a plus b is greater or equal than the volume of this set which is what is the integral over a of the Jacobian of this map so the determinant of identity plus grad t but now grad t is remember the action of phi because t is grad phi and now the observation which is always based on some other geometric inequality that's all you have is the following if you want exercise if you give me non-negative numbers let's say a1, an non-negative then the product of 1 plus ai is greater or equal i from 1 to n so get ready at the 1 plus the product of the i 1 over n power n so let's apply this inequality there so what is here the determinant so the determinant of identity plus action of phi is the product if I call lambda i the eigenvalues of the action of phi is the product of 1 plus lambda i because you just diagonalize the matrix and when it's diagonal it's just the product then here this is greater or equal by the inequality 1 plus the product of the lambda i 1 over n power n but this is what this is the determinant of action of phi which is the determinant of grad t and this is b over a because that's the remember the determinant of grad t was b over a this is the determinant of the transport map and so here what we get is so I go back here and I say this is greater or equal than the integral over a of 1 plus determinant power 1 over n so b over a 1 over n power n outside and this is a constant function so you get volume of a 1 plus b over a 1 over n where there are only two inequalities and that's cute but what about now if I want to use it for stability so just 2 days ago we saw that optimal transport proof for the superimaging equality was very efficient to prove stability so stability issue so the question is the following I'm gonna assume let's say that the volume of a and b is bounded by two constants that I'm gonna call little lambda and capital lambda so I just want my sets not to be too small or too big so I trap their volume in some range so Bruminkowski is not scale invariant so this is not a free assumption I'm gonna make it as an assumption I'm gonna assume my sets are neither too small nor too big and for every once I put a bound of the volumes I want to show how far they are from each other so I define delta of a and b my deficit to be the error in Bruminkowski I would like to measure how far they are from each other but in fact there are two things that I want to measure because what is the quality case the quality case tells me that they are homothetic but also they are convex so I would like delta to measure not only how far they are from each other in shape but also how far they are from being convex so let me define the following quantity so for simplicity anyhow we will not have time to look at general case so for simplicity let's assume so this is a simplification it's not without loss of generality I'm just doing for simplicity let's assume that they have the same volume so I'm gonna define first of all the standard one distance so alpha on the a and b a and b is the infimum among all translation of a delta x plus b so how far they are in l1 up to a translation so this is what measure how far they are from be homothetic this is something we proved with Francesco and Aldo in l9 was that if you already assume that a and b are convex then alpha a and b is controlled by a constant square root of delta eb so if in some sense you get rid of the convexity part you say I don't want to worry about proving that Bruminkowski implies convexity so I'm gonna assume already their convex then I can get a stability in equality and proof uses the optimal transport proof so it's like, it's actually even easier than the one I gave you two days ago it's the same kind of strategy that the one I gave you two days ago with a superimetric inequality but in some sense it's easier because the sets are convex you don't have that problem of last time where I had sets which could have something like this because now sets are convex so this disappear for free so this is not the point of what I want to say so the kind of stability result for this part is the same strategy so that's a case where a strategy works for a superimetric inequality works for Bruminkowski but actually that's um let's now wonder what can we say when the set is not convex so I want to treat the general case now so I don't want to assume convexity and as we'll see the problem is already complicated in very simple cases so let's assume again to simplify a equal b so we simplify even more life now the sets are already the same so if the sets are the same I don't have to show that they are homotetic that part is taken care of but now I want to show that they are convex so what is now delta so delta of a now it's volume of a plus a one over n minus twice the volume of a one over n so this is my deficit so some of it itself is not the same as a because so a plus a is the set of points of the form x plus x prime where x is prime in a so in some sense you are taking some different points and in general you get more this set is bigger than just deleting so what is the goal so goal I would like to use delta of a to control the measure of the the distance of a from its convex cell so kove isn't my convex cell of a so I would like to use the deficit in Bruminkowski to show that my set is almost convex okay so let me tell you already just a philosophical comment why the optimal transport proof cannot give you anything just because the optimal transport proof it's an empty proof when a is equal to b because if you're sending a on to a the transport map is the identity and you're not doing anything you're just transporting a set to itself and you're just doing trivial inequalities so the proof I gave which had a meaning for different sets in some sense for same sets there is no nothing deep behind it so when we tried when we were working on this problem without the Francesco we tried for several months actually to try to understand the general case and we realized that we couldn't do it then we said okay let's try the stupid case because you always think that this is a stupid case dimension one and at least just to feel better with yourself you say let me do a simple case at least if I can do it feel a bit better after wasting last two months proving nothing and we tried that for probably another mount and was still stuck then we said okay let's do one dimension a equal b the easiest possible case you can think of we tried probably another mount at least and then we just gave up for us we had no clue how to do that we were just getting crazy and it was at that moment where I was in Los Angeles that I was lucky to meet an analyst who knows arithmetic and so I was chatting with Terence Thau who told me that the one dimensional case is known and is a theorem in additive combinatorics from the sixties good take the book look at the theorem like 12 pages proof you get the one dimensional stability result use a lot of machinery, your subgroups greater common divisor and stuff which is not like standard when you work with these kind of problems because the point is that the result is not I mean the result in additive combinatorics is not a result about sets in R the result in additive combinatorics is a result about sets in Z but then you think okay if I have a set in R and discretize it with intervals and then to inch interval and then I get a set in Z and then the problem becomes a problem in Z but the problem in Z has much more structure than just the problem in R because in Z you have to deal with the fact that points have also in between what do I mean take an interval in R multiply by 2 you get an interval take a set full in Z I mean 0, 1 until 10 multiply by 2 you get 0, 2, 4, 6, 8 so in some sense being full is not which is stable in Z what is stable in Z is to be an arithmetic progression that is something that you preserve when you multiply so in Z the family of convex sets of intervals becomes all the arithmetic progressions but the arithmetic progressions you have to kind of quotient every time if you want to have like a canonical guy by the greater common divisor divide by the greater common divisor you remove the holes so this presence of the greater common divisor everywhere makes the proof much more complicated anyhow so let me show you now the one dimensional proof is not 12 pages because if you only want the result in R you can actually simplify it so we want to prove something which is less so let me make let me start with an example so we are dimension n equal 1 and I am going to take the following example so I take a set which is union of two intervals or length one-half so one-half ah ok sorry delta v, I am going to assume here let me normalize the volume otherwise this is not scale invariant so let me normalize the volume so let me take a to be union of two intervals then if you do ah let's consider a plus a over wow ok done in this way a plus a over two so this means that you take x prime plus x plus x prime over two so you have to take converse combination so you take points you take points here you take points here you do converse combination you take a point here and a point here you do converse combination you get another interval so you get one-half length one-half so the measure of a plus a is twice and this is three-halves so this is three so delta of a which is here n is one so delta of a which is a plus a minus two the measure of a in this case is one and you see that the convex hollow of a minus a is as large as you want because you can always take these intervals them very far apart so this is as large as you want while this is fixed and is one so that's just to say that you cannot prove just delta controls this quantity because you have a counter example so you need delta to be small enough if you want to prove something and amazing thing of these are theorem which is called Freiman theorem it's so 59 says the following so this is a corollari of Freiman theorem Freiman theorem is about sets in z so if a is a set in r with volume one let's say and if delta of a is strictly less than one counter example then the measure of the convex hollow of a minus a is less or equal than delta e so once you kill that counter example you have a very sharp stability in equality it tells you that the distance from your convex hollow is controlled constant one, power one from the deficit see there is no square root, there is no constant it's a very strong result ok so let me show you the argument so personally we tried to do something like that by end I remember we never thought about I mean I don't know how to prove this result without passing through the integer that's what I want to claim I mean without using the really the result in z and passing through the limit I don't have a proof directly in r which would be interesting because I don't know I don't know if what I'm giving you is the simplest possible proof but that's what you have as of now so let's prove Freiman so we start with Alema which is called Koshi Davenport inequality so these we are I think 1817 with Koshi in 1923 with Davenport I mean Koshi proved and Davenport just reproved it they both took out the name so the lemma said the following let a and b be two sets non-empty always two sets in z over pz so take the integers modulus pz and p is a prime then you have a Bruminkowski type inequality which tells you the cardinality the volume means cardinality cardinality of a plus b so you still have a plus b in z plus b z just do the sum models z so this is the sum in z over pz and you do the cardinality this is greater or equal then the minimum between the volume the cardinality of a plus b minus one and p so what this inequality is telling you so forget about the minus one either the volume of a plus b is greater or equal then volume of a plus the volume of b which is Bruminkowski or it can be false in the case where this is everything when the cardinality of a plus b is p so it's the whole group so either a plus b is the whole group or you have a Bruminkowski inequality so these are the two possible cases the proof it's in this proof is crucial that you have p prime so proof is by induction I can give you a reference if you are interested in looking at this by induction on the cardinality of b so you start with cardinality one on cardinality of b and uses that z over pz as non-trivial subgroups so the fact is that the subgroups of z over pz is only zero or everything that's what he uses and there is an induction where the cardinality of b it's not completely trivial there is a trick here it's not a trivial induction but it's half a page then once you find the right idea let me give this for granted let me show you how this lemma gives or less what we want so the corollary of this result is this other effect so now we go to z let's say proposition let a be a subset of z now where the minimum of a is zero and the max of a is some number p where p is prime we wanna discuss in a second if the cardinality of a plus a is less than three cardinality of a minus three then zero p minus a is less or equal to two minus a no a plus a minus two a plus one ok, it may look ugly but if you think one minute with me follow me for one minute you will see that it's not that bad so this is z and I have a bunch of points and I'm assuming the following so up to a translation I can always assume that my set starts at zero so I assume that the first point in a is at zero so here I have my first point in a and then maybe I have all the points and I'm gonna assume that the last point in z is p and p is a prime so my set goes from zero to p why I'm doing this remember what I said before I have, in zed I have arithmetic progressions which are all the analysis of common sets but there is only one arithmetic progression which starts from zero and finishes at a prime so the fact that the last point is a prime ensure me that there is no other arithmetic progression besides zero p so that's the reason so this is like zero, the full set is the only convex set which can go from zero to a prime that's the reason why p prime is important and then what am I saying so let's say to me where we are going we want to prove something in r so when you want to prove something in r what do you do, you take your set you discretize using a billion of intervals so you cut in a lot of very small intervals you approximate the intervals with points and then you transport the point in the problem in zed but once I'm in zed I will have still billions of points so my sets are huge so this means that in this assumption we can forget about minus three minus three is just a small number with respect to billions and here plus one also is a small number so morally what am I saying here here I'm saying a plus a is strictly less than 3a which is so a plus a strictly less than 3a is the same then delta of a strictly less than a which was my assumption and here zero piece of the points so this is like convex hollow of a minus a less or equal then delta of a so if you forget about the minus three and the one I'm just this is exactly the theorem I stated if the deficit is smaller than the volume of a then the convex hollow of a minus a is less than the deficit which is Fremont's theorem in r so this is exactly the same plus one on the other side so this is really the result in r I mean this is almost the result in r so once we get this the result in r is very easy so let's prove it so let phi p from z over z over pz be the canonical projection so you take an integer and you just send it modulus p inside z over pz then the observation is that when you have a plus a so what happens in a plus a a plus a contains so zero is in a so it contains zero plus a and it contains also p plus a because p is also in a but the point is that so this is nothing other than a but the point is that these two sets when you send them in z over pz they get to the same because p is equal to zero so phi p of a plus p it's equal to phi p of a which means that when I take a plus a and I map it down to zp there is a cancellation one these two sets collapse to each other so this fact implies that the cardinality of phi p decreases because there is a cancellation so these two copies in z they are almost rejoined, they only have an intersection there is only one point in common so you have here zero p, let's say 2p here you add a somewhere and here you have p plus a you see in z these sets have only one point of intersection but when you map them down so the cardinality of these decreases with respect to a plus a by cardinality of a I may miss somewhere plus 1 or minus 1 but this should be correct so we have a cancellation of the cardinality at least one copy of a collapse and OK so now my assumption was that I had the assumption so this implies that cp of a plus a is less or equal so a plus a minus a and my assumption this was less than 3 a minus 3 minus a and maybe I you know that's correct so this is 2a minus 3 and now this is the same as 2 5p of a minus 1 because when you take a and you map z into zp zero and p collapse to the same point so you lose one point so the cardinality of these is one less than the cardinality of these so here you are using that OK so we have this and this is nothing else the same as 5p of a plus 5p of a in zp so what I discovered that since cardinality of 5p of a plus 5p of a in zp is less than twice the cardinality of 5p of a minus 1 you apply Koshidavenport Koshidavenport inequality tells me the following this is either greater or equal than this or greater than p so since it is strictly less it must be greater than p so this implies that the cardinality of 5p of a plus a which is the same as this is greater or equal than p and once you have this you are done because you say this and use that one this tells me that p less or equal a plus a minus a this is exactly equivalent as the cardinality of 0p minus a is less or equal than a plus a minus 2a plus 1 which was my statement so actually it is very elementary and also the other one I told you is not that complicated so this proves the result for z how do you do the case of r so now that we have the lem and this proposition we can do the general case so proof of Freiman proof of Freiman is the following so let's say without loss of generality I can assume that the convex solid A is let's say the interval now we are back in r 0n I am going to take a sequence pk of numbers converging to plus infinity primes and then I define ak to be the union of intervals ijk all the intervals ijk so I choose only the intervals which intersect so union over j so the ijk intersection a is non-empty and ijk I take them in form I will draw a picture in a second so jm pk plus 1 j plus 1 m pk plus 1 so what am I doing I am taking the interval 0m the convex solid without loss of generality my convex solid I say it is 0m just up to translation it is an interval and assuming it is 0m so what I am saying is that for you take 0m and you divide it in intervals all the same length and these are k plus 1 intervals and then approximate my set whatever it was with a union of intervals of this form so this is my set ek now I take my set ak which is an approximation of A and I transport it to z in the sense that whenever there is an interval I get a point and in this way so if you define bk the set of j such that ijk is in A in ak so you take the indices to the interval then this set bk now is a subset of z to this set you apply the last inequality and you discover that 0 pk minus bk is less or equal than bk plus bk minus twice bk plus 1 for this one you have just to check that the fact that the deficit was less than 1 implies that you can apply the last proposition to bk and then this is exactly the same so this is in terms of cardinality but if you go back to ak this is telling you in terms of ak that 0m minus ak so this is equivalent less or equal than ak plus ak so delta ak plus 1 plus 1 over pk plus 1 that's what happens it's exactly the same and then you just let k maybe a bit fast you should have all the elements you can try to redo by yourself this part it's not the fuel, just to check that all the details are there but there is no surprise in this proof so you see this stability for grooming cost in one dimension with this method as I told you I have no clue how to get a proof without passing through zp just as a comment there are a lot of theorems of Freiman type so it's something for which there is no obvious analog for in R I wouldn't know because so the kind of theorem so this is just really a one minute parenthesis the Freiman theorem is about stuff of the following form it tells you okay take a set for instance in so Freiman type theorems you take a set in z and you look at the garden idea of A plus A so what do you expect of the garden idea of A plus A roughly speaking the garden idea of A plus A if you take a random set should be A square if A is kind of random because all the possible points you have it's order A square because you take all sums so you have A choices here and garden idea of A choices for this point and garden idea of A choices for this one so all the sums gives you something A square so we have an extreme case because we are really assuming something very strong that A plus A is almost is not much more than twice the garden idea of A but already if you assume so the assumption that the garden idea of A plus A is bounded by a constant even large thus the garden idea of A is non trivial if the garden idea of A is large is already a very trivial assumption if you give me a large set with billions of points and you know that A plus A is less or equal than KAA then I can tell you something about A for instance A contains big chunk of arithmetic progressions it's not a random set it has some order inside so this is a non trivial hypothesis we tell you something about the geometry of these sets but I wouldn't know how to I mean I don't have a real analysis of this in you can do this not only in Z but in ZD you can also have higher dimension also in ZD now let's go back to our problem we saw how to do stability for Bruminkowski in one dimension if you wonder what about higher dimension actually we saw only the case A equal B but you can do also so the case in one dimension there is a sharp stability result also when A is equal to B so theorem if you take two sets in R non empty and if delta of A and B is strictly less than the minimum between the measure of A and the volume of B so this is now A plus B minus A minus B then delta of A minus A is less than or equal to delta of B is less than or equal to delta of B and delta of B minus B is less than or equal to delta of B so there is an analogous which is passed again through this machinery but there is an analogous to be the result very sharp also for different sets so the question is what about higher dimension so there is no result at least to my knowledge and to Terenstav knowledge which is probably more important about higher dimensional case in Z so Z2, Z3, there is nothing and the issue is that as I mentioned to you in the proof of the Koshidavan protein equality you use that Zp as no non trivial subgroups once you go into dimension you would have Zp, Zq and there you always have non trivial subgroups so that's the issue so this is why there is no generalization in higher dimension so let me state a theorem so I'm gonna state just one of the two theorems we have the easiest one for the case equal B so if I have A in Rn non-empty actually directly we put volume of A equal 1 then there exist constants delta of N and Cn positive such that if delta of A is less than less or equal than delta of N then convex hollow A this is this is with Derrick Gerrison with this theorem in this year less or equal than constant delta of A alpha of N where alpha of N is some expression number alpha of N is something like 1 over 2 A to the N minus 1 N factorial N minus 1 factorial something like that very nice exponent nice was ironic so and we have a result so what's the main problem to prove this result so this result so I should mention remark in 2012 Mike Christ proved the following fact delta of A goes to 0 implies convex hollow A minus A goes to 0 and something analysis for this is non trivial and the main issue that you have with respect to all problems where you have isoparimetric inequalities so what I mentioned isoparimetric quality the key point was the compactness perimeter bounded implies precompactness that was the first class this compactness is crucial when you want to do argument by precontradiction the kind of strategy to prove this you would say ok let me assume that is false I have a sequence where this goes to 0 but this doesn't go to 0 you would like to start up a sequence prove that it converts to something and prove that in the limit you get a contradiction but it's very difficult to show that if this goes to 0 your sequence is compact in some sense the sets are compact in L1 L1 lock whatever is very difficult because this is not a like a derivative, it's not a perimeter and in particular you have to think that so remark if A is convex delta of A is 0 but the set of convex sets of volume 1 is not so I can have a sequence of convex sets which become they will not converge to anything they will just collapse so this is in some sense if you want to prove something like that by contradiction you have to get rid of this degeneracy for people who know this this is handled by a John type lem argument but it's not trivial because we don't know that the sets are convex and then even if you know that your sets are bounded still you have to show that they are it's not trivial, I mean they're just bounded sets, they could oscillate a lot and so that's something that you have to prevent as well and so you have to start the information from delta let me just very briefly in this compactness it proves that the family of sets where this goes to 0 is pre-compact in L1 up to an affine transformation which gets rid of this degeneracy and then it gets exactly do by contradiction but it's so the proof, it does directly the proof for the boom in Koski case and it's done, it has a paper for the two-dimensional case and for the n-dimensional case it has another paper because it's much more complicated in n-dimensional than in two-dimensional so what's the strategy of the proof so the strategy of proof if you want to prove a theorem like that either you give me a proof which it proves inside the proof you also have Freiman's argument or something which substitutes Freiman or you use Freiman as an inductive step so since I have no clue how to prove Freiman directly we use Freiman as a starting point so we do by induction over the dimension at least in this way the case n equal 1 has been done so strategy of proof is by induction so you are in a random and you assume that you already proved the result up to n-1 so you start with a set let's say a and you say ok I start to split my space as r and rn-1 and I start to look at the intersection of a with lines I get something here I call this set Ay so Ay is a intersection y cross r the line passing through y so y is a point in rn-1 and the the first observation that you can make is the following the observation is that the volume of a plus a minus twice the volume of a let me write in this way this is nothing I mean the deficit I wrote before was I mean these are all comparable if the volume is 1 I mean the deficit before was this but the volume of a is 1 so if this is small this is smaller so let me start with this so this is by Fubini I can say that the volume of a is the length of each of this section so it's the length of Ay so here I have minus so h1 means the length of the set Ay so this is the 1d out of measure and the same I can do here I can do the h1 of a plus a over 2 y so you take the slice for the other set now you can observe easy exercise that a plus a over 2y contains Ay plus Ay over 2 so if you take your set you slice it with y and you do sum of points here these are going to be a set which is contained in the slice which is contained a plus over 2 with coordinate y so here you can say that this is h1 of Ay plus Ay over 2 minus h1 of Ay so to this you apply the 1d case this is the deficit in one dimension the 1d case which tells you that either this is large or controls the convex hull so either this is greater or equal than h1 over y let's say over 2 or it controls convex hull of Ay minus Ay so what you discover is something like this that you intersect with lines and there are only two possibilities either most of the time you get almost an interval so this is small or you have a lot some mass so this guy is greater than this so you get some yeah I finished the watch out so that's why I'm doing this color so this is greater or equal than the integral over some, so this is like the bad case and this is the good case so I have when y is bad I can say that this is greater than h1 of Ay over 2 Ay and when y is good and partition and partition y is good and bad I get that this controls kaj y minus ay h1 so you discover and this is delta so this is my function this is delta so delta controls this so it tells me that most of the segments either are intervals or this is the starting point of all the rest which goes on for a while where sometimes you have to start to get so you have to kill the generasy and so that you have to get rid of these sets very long and thin so there is let's say a bunch of work to show the following that up to a linear transformation of determinant equal to determinant 1 we can assume a to be bounded so this is a sense how to do an affine transformation of determinant 1 the point is that the determinant 1 transformation do not change the deficit nor the statement of the theorem the determinant transformation to everything and the statement is invariant so you can always do it yeah universally bounded in some VR R universe so everything is contained in a huge box and once you are there so in some sense you get rid of these so you start to know that your set has some segments so in some sense your set is not too it doesn't have many holes but still you are not that close to convexity you still you could think that your set could be just to give you to be kind of weakly and still maybe it doesn't have many holes so the fact that there are segments is very difficult still from there to the duty that you are close to convex but what you know is that the measure of A plus A over 2 is very close to the measure of A which tells you something like that that whenever I take a point in A x and another point in A y and I do the convex combination I take a point and I can take points all over and you see that when taking a convex combination you start to discover these points should not be too far I mean if the measure doesn't increase too much you cannot have a picture like this because convex hollow points will fill up a lot of mass inside and so you end up to prove something like this and I am done just two minutes to write this and then stop you would like to prove a statement like this so lemma so let's say that I have a function f from set E into R f is bounded let's say by capital M and it satisfies two properties so that f of x is lessor equal than f of x plus x prime over 2 plus eta for every x x prime x plus x prime over 2 in E so you have a set which satisfies a kind of concavity inequality with a small error eta is small and the points are inside the set where the function is defined and you know that the sets where the function is defined it's also almost convex and let's say it's non-trivial convex hollow V it's contained into B, R and the measure of V is 1 so it's a non-trivial set so you have a set with measure 1 and you know that for almost all points for all points inside you are almost concave you satisfy a kind of concavity inequality with a small error then there exists a function capital F from the convex hollow V into R concave such that the integral is constant eta to some alpha so you want to prove that if you are almost concavity inequality in a set which is almost full measure inside a convex set then you are very close to a concave function so this is one of the key step in the proof where I don't have an elementary proof for it that I can tell you in two lines it's kind of technical and pretty involved even long and this is one of the key points which are some moment you are going to apply to show that your profile is very close to a concave profile and then you apply for the bottom part to say that the lower profile is also close to a convex profile to minus and so your set is close to convex so if you apply the lemma the lemma tells you it's going to imply that A is close to some K where K is and from this you conclude so I'm out of time so I stop here are you at least with this but I didn't enter too much also because it's mostly technical and I think it's much easier to understand the other proofs I gave to you the point of these problems especially of this one is that you see that techniques even if the final statement is always the same it can be very different it can involve different ideas I found a very interesting person when I saw this connection with the attitude combinatorics and I think it would be nice if there could be a direct proof which kind of avoids this attitude world but at the moment is unknown so thank you for the attention