 Thanks Andrea. Thanks for coming. It's a pleasure to be here. Manuel, thanks for the invitation. Okay, it's nice, nice weather. Nice place and nice company to work. So, very nice stage. Okay, so I will talk about the providing these people far from the start shift. I don't know if you've heard me in any moment at any moment to ask anything because actually the good is to take something of the same kind. Today we will be relating with a space, a particular space, very special, but it's very interesting to some geometric problems and use combinatory things that are very far to the classical theory of weight. For that was the challenge, nice challenge to study this problem. To motivate the problem, the history of the weight theory and the starting point is the performance in equality related to the maximum function. So I will introduce the maximum function in the classical. Suppose that you have f function in l local integral function so we are working in Rd first in the second part of the tour we will start with that specific space. But now to present something in general, the maximum function here is the center impulse is defined like this function, like this possibility in the other direction of the board. So probably all of you saw some this operator is a classic but very important operator in harmonic analysis yes exactly Christian very very good you know a lot about this guy and of course we know that this satisfy some piece for example this operator satisfies this this that is called that measure here some constant here that's only times the number of the function this is the good type of one of the maximum function and of course there is a lot of variance of this operator instead of you can consider also you can consider queues tangles so in this context if you want tangles cubes are essentially the same but there is a standard proof that when I put w here would be a function suppose in this country locally here so but a non-linear function is not necessary but assume this and then you can change a little bit the measure instead to consider the measure to put a density w and then here it is necessary to put m of the value and this will be called for every so this is true for every of course this is the function f of this function so this is called the standard quality looks like a series of which I don't want because you can see that this value is also one so in the context of the theory and this is a learning result it is even with the method in the setting we can prove this kind of result because this is the center function so this is called covering lemma any variant of the classic covering lemma could be used to prove this so of course there is some question related with the contents here could depend on the dimension but anyway this is not the point today at the moment and ok so but in the winning theory this is one of the first result that appeared in the general point of view because this holds for every number a second seminar result that is the market result in 1972 that is the second part of the talk about but at this moment and I want to convince you in 5 minutes that this is not only a general session of the with Taiwan one of the maximum function is really this kind of quality for this operator is really good for instance suppose now we have a family of function you can consider this kind of quality is only to show one of the four of this kind of estimated so of course that is asking to generally say generalize the operator to this family in a certain way so we can consider suppose this guy we can consider the P norm of this guy the speed suppose we get a 1 and we consider this and you consider the mixed norm you ask if this is a currency so that this is essentially we can eliminate the operator we have a bold estimating this PQ estimating is this is if you know it's a here is Simon can so this here we can essentially the same here does not appear anywhere so this is a mixed PQ inequality in a vector extension of the maximum function okay so so this kind of inequality if PQ is greater than P is a general result it's inside of a general result and it's true and you don't need actually because essentially the only unit that this guy the maximum function is not linear it's linear but can be it's possible to linearize in some way this operator and it's a positive operator but if in the 70 if it was the question what about if Q is strictly bigger than P so if we understand proof this inequality true or think that inequality okay okay so we can show this because five minutes and you understand that this kind intrinsically this kind of system they have more information so imagine that this is the case Q is bigger than P let me say that in this particular case this operator the maximum function they don't know but this maximum function is all the operator okay because you can take the soup of this guy of side and then always the mfx would be less than the norm of this guy would be less than the norm of this guy okay so there is a very famous a classical result we are analysis that say that if you have interpolation result I must include that if you have some estimating in one and some estimating in infinity as that way we have only the estimating b for b between one and infinity okay so in particular and this result is very after cause for every measure a lot of things so this estimate under observation before said that actually would count that the maximum function to the power of p r maybe I will use okay so we use r this is also true dimension and here okay m of this implies this using interpolation interpolation and this result and then result of infinity okay so free this result in this okay so now with this observation of such assume that q is bigger than p so now this guy is bigger than one so I will then the number okay the conjugate number of this if you don't know is this guy q over p times q over p okay so this guy okay but don't worry it's not really right but only okay so so suppose you want to represent this the column would be q over p so this is the q over p norm of this guy and they can use quality see quality in this exponent and then we have that to calculate this norm would be considered the sum j to the power 1 because and times some g that the g would be of this property so there exists some g in this that this guy would be this norm okay maybe for this one somewhere let me show the argument how this appear and now okay so we have the possibility to put the integral inside and now we can apply p is between one and we can apply this result okay and then we pass the operator to the other function we have and of course the parameters and estimate there and we have here p and here we have mg okay and now we can use we can I put the integral inside okay so now we can remove the integral and use here and we have the sum here here one over p we have m of p but now this is sorry here at least and now we use here to the power p over p q over p and then we have the integral of this guy q over p one p over q and this we have here the integral of mg to the power q over p then one over p so this this is what we want to take here if we put one over p that is necessary to put it inside but here observe these guys these guys appear here the maximum function of g and this was because we use this estimate but now if you want if you know that this guy is working it is the gays because the maximum function is bounded we don't we don't measure in this exponent bigger than one and we can't eliminate this guy and this is a sensitive one okay so okay so only a calculation to show for instance from this guy we can obtain vector value estimate by 3 using the maximum function so this is one of the application the paper even this appears essentially this kind of calculation appearing in the paper but there is this idea I will not speak about extrapolation because it is a specific thing but it is the starting point in extrapolation extrapolation is essentially something like this there is a class of way called it suppose you I will say only this word but maybe I recommend of some okay so this is a one way okay so satisfied that the maximum function is controlled in general by differentiation the next time we know that the function will be controlled the maximum the function will be controlled by differentiation if we have the other side in equality this is a class of way it is nice there is no trigger there is no only constant there is a kind of dimension this kind of guy okay so don't worry if you don't know about that but this the important thing is that it is true because it is a consequence of this inequality I can put here this guy this constant here and here the same way so in fact this is a characterization this kind of way to satisfy this and if this inequality calls the value must be satisfied that condition so this is the a one way to satisfy this condition extrapolation in particular extrapolation said that if you have this inequality for any way in a one then you can pass to every P from that way with every way in a P and in particular the very measure is this is not necessary to have the other end point okay to extrapolation extrapolation okay but you need this for every way and then you can pass this is a very very nice result to prove your difference but the starting point sorry prove your difference to be positive you require the weights to be positive or you said all the operator the weight yes the weight must be positive no negative function yes yes in LP so not all you are saying yes yes yes yes no negative weight because our positive operator no negative one the measure must be no negative okay okay so this result is by doing to prove your difference but in my point of view the starting point of this result in this paper by the person listen okay so this is why what it would be nice and of course we can we can share with you yes yes you can correct guys using like other kinds of operator yes maximum operators the difference to some nice is the same class or not for every convex essentially you can use cube convex symmetric would be essentially the same okay few points of completely different it's really stronger the fact that you know no no it's completely different but if you fix reasonable in the sense that the set is in between two cubes or two walls it comes on depending on the dimension but it's essentially the same okay there and this is two possibilities but there is a lot but there is at least two possibilities of this quality we want to try this around to understand that this guy is no okay no maybe okay this guy this is quite hidden okay so so understand here this is so there are at least two one of them is okay white here that's not that this guy must be if you want to put the same way you need a condition okay so this is the result you can put everywhere but if you want to recover the same measure here you need a condition that is a strong condition but okay so now we're going to this is going to wait because wait here and appear the other way so okay so we can reduce to one white estimate if you have some condition so very natural extension of this kind of result is why is always if you put here instead of what about a similar more similar a human okay we'll put the square function or another and point estimate square function different operator around the classical stuff are possible to same was okay so okay but always even here the operator to put another positive a positive eventually bigger operator and there is a lot of variant of this kind of problem sorry but it's supposed to mention there is a lot of why we mentioned some variant that I was working with we learned Carlos Perez a long time ago and was calling the who put the Hebrew function the Hebrew operator but this is not the don't worry I will not cut any calculation about this so forget this if you don't know the definition of the chemical don't worry it's a more singular so if you ask this in equality okay this in equality was disproved by reggaeton in 2010 so this is called the mac and hog within in a meeting of time ago mac and hog proposed this question and it's not true positive result is put some bigger operator the maximum function or integration there is a lot of result related with the N5 if you know about the 5 5 will be placed 5d will be the maximum function is to put 5 something a little bigger this is like young but there is a lot of result the best what result was in the 90 was by Carlos Perez put LL to log L to some power epsilon bigger than 0 but this was improved in 2015 by Ray Lacy Michael Lacy Ray and Domingo Salsales and it's possible to put P log log P to about 1 plus epsilon okay so this is a funny history but okay so there is a variant of the A1 A1 looks very easy but this is true for the maximum function I have proved it yet but okay so imagine that this inequality is weaker than the previous one and now I am saying that we have the condition A1 because if not this is infinite what about this question okay so looks like what is the question here is that the dependency of this content will be linear this is a question okay so why this question is interesting why in 2008 2009 was working with Carlos and Andre and in that moment was motivated with this question this is related with the problem that is called A2 concept that was proven by Johnson that said that this guy actually no this guy was proven by any kind of same with regularity the linear growth in the actual estimate that also holds is linear and this was related with we just know about the tram equation regularity is related with a lot of things but okay so if we prove that by extrapolation and working a little bit with we could obtain the A2 concept and if we really prove this guy we will prove that that in that moment was also the question okay so after I get empty the proof that this guy was okay so in that moment we obtain lower lower this guy okay so this is a little bit starting so this was proven by Carlos and then Johnson okay so in that moment we saw that this result would be not sure okay so we can remove this and I think it was earlier I think it was earlier okay okay so later aside of both basing actually for the non-necessary one over three will be necessary so the week was open sometime if it's really possible to put something smaller than one and it's not possible because this is true and this was proven more recently by Carlos okay so Lucas okay okay so but this is there is a very interesting problem related to change the operator here Israel also today it's not all there is several problems related to change the operator and also change the operator okay so this is a considerable level of state inequality and the motivation today is not to change the operator it's changing the operator so what's all again constant it's an equality space space is not now when we have even the we don't we don't quite so in some metric space but our interest will be exactly this if all the guys X is this one okay so and first the metric the measure in this set I will mention why was the study that problem let me see the time okay okay certainly okay okay so define this guy okay I would say here okay suppose you have you a metric pressure space see we can define this guy okay so as I said you need some you need some geometric property to prove this kind of things even we don't it's not so easy even it's very easy to construct even in that example this okay so let me motivate us that one is a result right now something here because x metric pressure space satisfies what is calling the n micro normally condition if you never mean that we have in that space the measure for radius controlled by some constant measure condition because it's possible to cover the lemma so when condition means okay but depending on n is controlled by a constant that don't depend actually I would say something because strong it would here any y original the same so this is a strong micro this condition is this and this is interesting of course it's very interesting result but this for instance I cover there is a result of understanding long time ago that said this is the quality for the here supposed to be dimension is big and then this is true okay so if you consider more I think that also in the same paper close to that paper this but if you consider other norms no no no sorry this is an electronic measure this is okay so but this is so in particular if you consider this result recovered but actually they prove okay so to say something really more for instance if you know about if you know for instance this this function micro w no no no no because because you have this okay but this is true depends here will depend on it you can iterate so no it's not right you fix it yes I know so fix and then okay and then you have the ball radius r less than the universal constant times that you apply this twice you get four thirds square less than the constant universal square square so much something bigger than two maybe three times four thirds you get something bigger than two and you also get a cq and that but c was universal so okay I will answer suppose that you have r4 and I meant to explain that the measure of the similar to forward this guy certified was proven by now and some of these guys the measure is not complete because the constant okay so would be some constant that depend on n that certified but this is the point then here estimate will be will be changed okay so we will know again this is the point you this is possible that you will take something some constant here depend on there of course we choose yes you're right but this this is not enough to say that we have this sharpest okay this is the point okay so but in particular this kind of space this kind of space so this can be applied to this guy and then if you have this this is not okay so this could be a very small set inside of our deal but they also prove that if you fix in here there is a little bit there is a little bit that the result is sharp in the sense that they prove that bigger than if I show then there exists a one week so this mean that actually it's impossible to obtain with this condition it's impossible to obtain something okay so and they mentioned that there actually seems that when you you're doubling or micro doubling or weak doubling condition seems a dynamic space the week could be could be however they recover the following result that this was interesting result in the today that's what we call this guy suppose here k equal to dk this is the the root k 3 mean this allowed a root k equal to 2 children each children 2 children take k with k children each children k children this is associated with the 3 groups they all then k plus 1 okay so here you can very simple measure and distance the distance will be take this guy and and the distance will be the number will be a natural 0 natural number of 0 that will be the number of issues to go to from x there is a unique but in that case so in that case is 4 so this is the distance and the measure will be the content measure density 1 in each vertex so this is our metric measure space then now I don't prove that okay I will not hear that way because the graph people in graph 30 say the measurement is content measurement so they prove that I will get here but I forget okay this is very important that k is bigger than equal to if not it is very easy this space because it is 1 it is a dummy so they prove that maximum function is here but in that way that I consider okay so what is the interesting point this guy does not depend we can take every day and consider for so as I said the measure here suppose this is a natural number because this is essentially depending on the root of the guy x will be k over r so imagine you move in a way so it is completely of and so it is very exponential okay so this is in that point of view is very far kind of measure so okay let me say because this paper is and to show an example he was he was thinking before was not correct in that sense that is not could be a very but the behavior of the respect to the dummy the dummy behavior of the measure the dummy property however could be true so so I saw this result and I say okay people my university that is over there or not sleeping Israel and Martin that is working graphs and Israel now is not thinking okay so we wasn't interested to say okay what's hello to bro I think I'm going to say in a quantity in a frame okay it will be true so okay so it's not possible okay so I'm sorry that it's important that this result can be obtained by I no so we don't know this paper but simultaneously independently now and from a geometry proof to the paper that actually result okay so really we know the proof by now that it's combinatorial nice very nice proof combinatorial symmetry of this guy okay so as I said the beginning was a nice challenge to okay there is no possible to use any I mean who could try to to change the measure and change the density and it's possible to obtain this or okay so the result will be a weaker result this aberration okay so it's if you consider it's bigger so okay so this guy you know what proof but actually it must be the case only depend on this not depend on k but go to infinity when this is going to you understand that the MS is you win a little bit of integrability so it's bigger so it's still bigger that iteration in fact we also proof that this is essentially short because it's not possible to put 6n it's not possible to put x is a sequence of functions go to infinity infinity this way it's not possible to put w mn of w for any duration of okay so essentially it's necessary to have a little bit more integrability okay so okay so this is very singular the maximum function this show okay so it's very surprising that all people that I mentioned before we consider independent on k but in any case the space is very singular because the maximum functions it's not possible to function in the classical case as I said a lot of singular operators it's possible to put 2n of m of m but here for the maximum function it's not possible to put m 32 okay so this is completely okay we were lucky that we were lucky that actually we can find the same because in that case because maybe it would be a problem of our proof that appears but no okay so there is also a result in the case p bigger than 1 compared to the weighted theory but okay so in fact I can't mention because each p is different this is the same the same but of course it's not a p condition the typical the classical a p condition is completely far to characterize but there are some conditions if I have time it is also not far okay so let me say only I will not talk about the a p but there is a some results for p bigger than 1 the paper that we have with israel but let me say only some things if I mean it's there is also I don't want to forget that there is a recent result of fractional operator in this content by cosine and secret and there is also some ambitious student Manuel is standing the linear this guy in this context okay so to mention a very nice estimate that suppose that p and p I will forget about k but we are stopping to take k and this belongs to me we have a restriction that you guys there we have the problems measured here with x and a with natural number and then I will now put this significant identity k 2 so this this is the part of that is x and f and the distance between them is okay so fix one second so who is the possibility to all point here are two over k because the complete wall will be people that for every point is distance in general it is very easy to check that this measure is less than this guy and in the same way okay so it is very easy to check this however this is sharpness Manuel sharpness you know they mentioned you are sharp so we can edit this part no okay so this is a this is very nice in the sense that here this guy here second this is k 10 this guy r is pretty big but r over 2 is pretty much smaller than this k so in fact if you have two walls essentially of the same measure and you have same measure here we have one of these guys similar and k to the power r over 2 and here we have k to the power r so really this is in fact it is sharp and this is the main this is the main this is related but it is related with the which is the situation in the restrict with that estimate for set for set e suppose that you are treating the average the average this is the one and then the point is that in general this guy will be because the maximum function is type 11 then will be also type 11 however this kind of estimate you have a good decay in the constant respect to r even and this is the key so our main work was to hold true because our both follow the general idea but the point is hold to introduce and we don't have the symmetry in the way ok so when you mention this finish that is possible ok so I would denote this guy to prevent in the second variable ok and then it is supposed to prove can be estimate constant k over r here is prime plus 1 and this guy is supposed to mention how we can do this in 4 minutes but essentially the point is that this guy always at this time is prime is less than ok this essentially will have a good decay to repeat the two completely as in the case a weighted case ok so this estimate could be maybe it's possible to improve but no it's not the case because if you can obtain of course this prime is infinity this will be 1 but this is not say nothing and actually it's impossible to say something because it's not true to put w and because it's not possible so in some moment it seems that we lost information ok so I think that I'm sorry so I can finish here so thank you very nice talk so any questions or comments