 Okay, good. So then I will introduce you now. So, okay, so welcome. Welcome everybody. Today, I'm very happy to introduce the speaker. He's doing research in harmonic and complex analysis, operative theory and interactions between analysis and probability. And today you will talk about sampling, tiling and quasi crystals. So, okay, who is yours? Okay, thank you very much, Zacharias. Well, thank you very much for your invitation for the possibility. I'm very, very happy. As Zacharias said, I want to talk about the some relationships between this classical sampling theorem, tiling and quasi crystals. So, let me begin with some prediminaries or conventions. Suppose that we have a Vorel set omega in Rd, then I will identify the space of 2 of omega with the sub space of R2 of Rd, of those elements, those representatives by Nishalmos everywhere in the complement of omega. I will use this normalization for the Fourier transform to find the exponent of the exponential. So, the inverse Fourier transform will be the same form without the minus here. Since it will appear many, many times throughout the talk, I will use this notation for the exponentials. I will note this lambda will be the frequency that appear here. So, when I consider the exponential as an element of R2 of omega, I implicitly assume that it is the exponential restricted to the set omega. So, it is multiplied by the characteristic function of omega. Well, in order to state the sampling theorem, let me recall the definition of the Bayesian spaces. So, now let me take a bounded Vorel set omega. Then the prediminary space associated to this set of omega is the sub space of those elements in R2 of Rd, whose Fourier transform, Rnishalmos, everywhere in the complement of omega. In other words, the Fourier transform of F lives in what I call L2 of omega. So, the Fourier transform, because that Fourier transform L2 is a symmetric isomorphism. So, the Fourier transform here is an isometric isomorphism between the pelibino associated to the set omega and the space L2 of omega. So, this space is the natural framework, if you want to work with signals in the nature, for instance, which do not have a really high frequencies. For instance, if you want to work with the human voice, or if you want to work with signals that the humans can listen, you don't need to work with the whole spectrum of frequencies. You can restrict your attention to the frequencies. And so, this is the natural framework, mathematical framework to work with these kind of signals. Although it is presented as a sub space of L2, inside each equivalent class there is a distinguisher representative that is very regular. Since, if you have an element in L2, the Fourier transform is causing L2, and if omega is bounded, then the Fourier transform is also in L1. And so, if you consider this integral by the dominated conversion theorem, this function here is continuous. Moreover, you can derivate this expression, and what you have at the end is that this function is not only continuous, but also it's infinite function. So, the elements of the pelibino space can be thought as very regular functions. So, you can evaluate the element of the pelibino space in different points. And it turns out that the evaluations are bounded functionals. Because if you can rewrite this integral here as in this way, what you have is that this is an linear power between this function, this function, and this other function. And so, if you use the properties of the Fourier transform created with the planchial theorem, you can transform this linear product in this other linear product. And what you get is that the value of f at x can be written as the linear product of f against this function kx here. The function kx is the inverse Fourier transform of the exponential with frequency x restricted to omega. So, it is an element in the pelibino space. The norm of kx is equal to the square root of the measure of omega. So, in this way you see that the evaluation is a continuous function, but also you see that all the evaluations are uniformly bounded. Moreover, all of them have the same, the same norm. Well, this is more or less what I wanted to say about the pelibino space. And now we're ready to recall the sampling theorem, also known as Shannon-Wittaker-Cotternikov theorem. Now, that's our trick that was mentioned to the real line, and the real line that is considered the set omega as an interval, the interval minus one-half, one-half. So, given an element in the pelibino sub i, then this theorem says that the element, the f of x can be obtained if you do the sum of the values of f of x. f at the different integer numbers combined with this function here, that is the cardinal sign. And this series converges not only in L2, but also uniformly in L3. This is a very important theorem in signal processing because it's the first step toward discretization of your function. Because observe that it's telling you that if you know the function at the integer points, you know it everywhere because you can combine these values at the integers with these functions here and you obtain the function everywhere. And for this reason, it's very, very important in applications. Let me recall very briefly the main ideas of the proof of this theorem. The first ingredient probably the most important is that something that probably you remember from the functionalized course is that the exponentials with integer frequencies constitute an orthonormal basis of L2 of i. Then if you take the Fourier of the inverse Fourier transform of this exponential, you get, you can do the computations, they are not difficult, and you get exactly these functions that appear here, this cardinal sign. Since the Fourier transform is isometric, this orthonormal basis goes to this other orthonormal basis. And so these functions constitute an orthonormal basis of your space. But also, as I said before, if you take the inverse Fourier transform of an exponential, what you get is the reproducing kernel. So this fn that you see here is exactly the inner product between f and this function. And so this sum here is just the expansion of f in terms of this specific orthonormal basis. And so it converges into, okay, so this is the first part. And the second part, the uniform conversion of this series comes from the fact that the reproducing kernels are uniformly bounded. And all of them, in this case, have no one. So using, for instance, Cauchy-Schwarz inequality, you can see that the L infinity norm is controlled by the L2 norm. So the L2 conversions of this series also implies the L infinity conversion. So you have the uniform conversion. So if you pay attention to this scheme of the proof, you see that if you want to extend this theorem to more general settings, to more general sets or even to RB, the crucial step is this one, is the existence of an orthonormal basis of exponentials in the L2 of this set. Now, the orthonormal basis in general are very, very rigid systems. So this rigidity leads us to consider some alternatives in order to be able to reach much more general sets, because otherwise we can extend this theorem, but only for some specific sets. So let me recall the definition of re-spaces and frames. Well, if you have a family of vectors in a Hilbert space, we say that this family is a re-spaces of the Hilbert space. If this operator, finding the literal tool of the set lambda to the Hilbert space, is bounded and inverted. And the operator is defined in this way. What does it mean? Okay, it means that since the operator is invertible, it starts shaping. So it tells you that any element of your space can be written in this way as a combination of the element of your basis. On the other hand, since the operator is injective, this expression is unique. The coefficients to represent the given element of your space is unique. And then this other requirement here, the boundedness, is because you have here a series. It's not like in linear algebra that you have a finite sum. Now you want that if you perturb a little bit the coefficients, you want that the result doesn't differ too much from the original result. So this is the reason why you required here the boundedness. So with the boundedness, if you perturb a little bit, a little bit in the sense of a little l2, then the sum won't change too much. This is with respect to real spaces. And what about frames? Well, look at this definition and now here is frame. These bases are frames. You see, there is only a small change and it's here. Instead of asking invertibility, the operator is injective. The idea behind frames is that you are trying to generalize a notion of system of generators that appear in linear algebra. Here, since you are asking such activity, what you have is that any element of your space can be written in this way. So in this sense, you are generalizing this idea of system of generators that appear in linear algebra. One nice fact about risk basis frames, not that any risk basis is in particular frame. The difference is that in the risk basis, this expression for given an F, this expansion of F was unique in the case of risk basis. In the case of frames, there are many different ways to write a given element of your space. But that's a unique difference. So one nice feature of frames for risk basis is that if you have a frame, then any element of your space can be written in this way. Here you have a sum and you combine these other elements in g lambda. g lambda constitute what is called a dual frame. And these elements are combined with the inner product of F against the element of your frame. So it's telling you that if you know this, all these inner products, then you can recover the element F of a given space. So if you have in some L2 of omega a frame of exponentials now, then you can see them, these special functions that appear here, dual frame. If you consider this expansion of the element F of L2 of omega, but now using the frame of exponential. Now you start to do the same as in the proof of the sampling theorem. So you start to use the Fourier Stanford. And so, since these are exponentials, designer products, they're not to be the evaluations of the functions in the pelivinar space at the points of lambda. The frequencies that appear here. In the sampling theorem what you have here where there's cardinal signs, these signs divided by x. Now what you will have is the email for your transfer of the dual frame. But still, you can see that you can recover your function everywhere just knowing the function in these points. In the point of the set lambda, capital lambda. So it's not to have a frame in order to have this kind of reconstruction scheme for functions in the pelivinar. Okay, so now let us try to study when you have in space L2 of omega, a risk basis of exponential, or a frame of exponential. For these, let me recall the two definitions. The first one is the definition of time. We say that z omega, where z omega in rd, size rd with translations in a set lambda, at level k, if this sum is equal to k, almost everywhere. What is this sum? I consider the characteristic function of omega, and I translate the characteristic function of omega using the element of lambda. I sum all this translation, and this sum have to be equal to k, almost everywhere. In this case, we say that omega dials the space, rd in this case, at level k, the level is given here. When the level is one, we simply say that the set dials the space, we are specifying the level. On the other hand, let me recall that in rd, four lattice, simply a lattice in this talk, is the image of set b through an inverted matrix. So in particular, a lattice is subgroup of the additive group rd. Even a lattice, there is notion of dual lattice, the dual lattice consists of those elements in rd, such that the exponential with frequency mu in this case are lambda periodic. Another way to say the same is that the dual lattice consists of those elements mu in rd, such that if you take a product of mu and lambda, this belongs to the integers for any lambda in capital lambda, in the original lattice. You can also prove that the dual lattice of this expression is, as in the case of the lattice, so it's a lattice and the inverse in the invertible matrix associated to the dual lattice is this one. This is the inverse of the transpose of the matrix associated to the original lattice lambda. By the way, I never put lambda here, this lattice here is lambda, or this is the dual of this one. With this definition, let me mention some some results about the existence of this basis or our basis related with times. So consider a lattice in rd and a Borel set in rd. Our first result is so-called Fugleder theorem, and it tells you that you have the exponentials with frequencies in the lattice lambda constituted and also an orthogonal basis for L2 of omega if and only if the set omega times the space with the dual lattice. So you have a very nice relationship between the existence of orthogonal basis of exponentials and time of this late existence of orthogonal basis of exponential with some properties, geometric properties of the set omega. More recently, there was appeared this beautiful theorem of Fugleder theorem that relates also existence now of free spaces with k times. More precisely, what they proved was that if you have a bounded set, then if this set k times the space rd with a lattice lambda, then there is a risk basis of exponentials for the space L2 of omega. This is very nice. The first result appears this relation explicitly between tiling of k tiling and the existence of free spaces of exponential. Before this theorem, Schroding-Marsow has proved a special case of the theorem where the set omega was the finite union of this finite union of qs of equal side events. After this theorem appeared, Mihailis Konosakis also give an alternative proof of this result and with his method he was able to say something more about the structure. If you see in both results, we are dealing with lattices. This is taking you that you have a group acting in both situations. When you have a group action on your problem, you know that you can have a reduction in some sense of your problem. So with Elona Vor and Carlos Cabrilli we studied these problems from this point of view, trying to use this reduction that comes from the fact that you have a lattice acting on your problem. So we made this version of Graysstatt and Leibniz there. There exist elements a1, ak in rd such that for any bounded or set omega in rd such that in k-tiles rd with Elona, then the exponentials with frequencies in this shift of the dual lattice. The dual lattice here is shifted by a1 and here a2 and a3 and until ak. These system exponentials constitute a risk basis for L2 of omega. So here what you have is that the same k-tiles elements work for any omega that k-tiles rd. And you have here that you can say something about the structure of the risk basis that you get in these spaces. This was also the structure that was found in Michalis in his paper. On the other hand, this k-tapel here are Scheneck in the sense that if you pick them at random in the unique yield, for instance, with probability one, the k-tapel that you have chosen will work for this yarn. Using again this structure provided by the lattices, you can prove also a kind of converse of this result. Which is this one that tells you that if you have a Borel set, a bounded Borel set in a v such that L2 of omega meets a risk basis of this particular form, then the set omega had to k-tile the space rd with lambda. Here you use the dual. And so in this way you close the circle as in the case of Fugler's yarn. In the case of Fugler's yarn you should remember you had an if and only if. You had the existence of a normal basis, if and only if you have the tiling with the dual lattice. Here you again have something similar, you have that you have risk basis with this structure if and only if your set k-tiles the space with the dual lattice. In particular, it's telling you that for instance in the space L2 of a triangle, you cannot find risk basis with this kind of structure, because the triangle cannot k-tile the plane. There is one important difference between the Fugler's yarn and this yarn here, that is this hypothesis of boundedness. Here we are requiring the boundedness of the Borel set and this condition cannot be dropped. You can find counterexamples. Although, but then there are some results from boundedness sets formed by Yana Karvaha and her PhD thesis. You had to ask some extra medical requirements to the set in order to have a risk basis of exponentials like this. Well, this is about risk basis. In all these yarns you have a lot of structure. What happens if you want just a simple risk basis and you don't care about the structure. Well, we don't know too much about risk basis of exponentials in spaces L2. And here is a list of what we don't know and we know shortly. Just to have an idea, look at, we don't know for instance if the L2 of the Borel meter is a basis of exponential or not. We don't know if the L2 of the triangle is a basis of exponential. I said that we know that it does not admit a basis of exponentials with that specific structure but without structure we don't know if we have a risk basis of exponential. Actually, we don't know if there exists a bounded set without risk basis of exponentials. So there are many open problems here. What we know is, for instance, if you take a finite union of the shown cubes and then the corresponding L2 of this set, admit a risk basis of exponential. And this is very deep result due to Gadi-Kosman-Shaharnitsa. And another theorem about places where we know the existence of risk basis is this one about centralized symmetric polytopes with centralized symmetric phases. This is a quite new result here last year, I think, due to... But you see, we don't know too much about the existence of risk basis of exponential. So let us move to Frank's and see what can say about that. In the finite dimensional setting, we know that if you have a system generator, it has more elements or at least the same amount of elements as basis in your space. In this setting, the intuition say that if you have a frame, you have to have more elements than in the case of a basis. But how to formalize this idea? Because both of them have to be accountable. One way to formalize this idea is through this very intensive. Let me recall briefly the definition. If you have a set lambda in RB, let me write in this way qr of x. In this way, I denote the q of center at x and with side lengths r. And then the upper density is defined in this way. You count how many elements of lambda you have in this qr of x and then you divide by the measure of the q. And you repeat this with all the cubes in the space and all the points in RB. In order to define the upper density, you take the supremum of these quantities and then you take the limit when the side lengths of the cube tend to infinity. And to define the lower density, you take the infimum of these quantities over all the x in RB and then you take the limit. You can prove that the limit exists in both cases. And these are called the lower density and the upper value density. And using these densities, you can get the following result. This is due to lambda. It tells you that if you have a re-spaces of exponentials for your space delta of omega, then it is a re-spaces then both the upper and lower density coincides. It has uniform density and we denoted only by b without the plus or minus. And these two densities have to be equal to the measure of your set omega. In the case of Fourier frames, in the case of frames of exponentials, the density of the frequencies, the lower density had to be greater or equal than the measure of omega. In this way, you can formalize this envision that in the frames, you have to have more elements in the basis. But a natural question that appeared here is, okay, in a frame, I need more elements. You can sense this lower bound here is the kind of critical value for your density. But can I find frames of exponentials, of frequencies, have density as close as I want to this critical value measure of omega? This is very important in applications because remember that I told you that when you go to the sampling zone, when you go to the Bayesian space, the frequencies that appear here are then formed in the evaluations that you have seen. And so the more points that you have, the more frequencies that you have in your frame, the more points that you have to sample from your scene. So the more complicated that will be your device to sample the signal. So you want frames with the lower density, with very small density, the smallest that you can have. So this is a natural question here. It was proved by Shorby, Shorby Marzo, that this is actually true, that you can find, given a bounded word set of omega, given epsilon, rate of zero, then you can find a set of frequencies whose density is the measure of omega. This is lower than, lower or equal than the measure of omega plus epsilon. The exponential with these frequencies constitutes a frame for L2 of omega. So you can find arbitrary frames with frequencies arbitrary close to this critical value measure of omega. The proof of the result is very nice. Let me say a few words about the proof. If you have your original set omega, this yellow one, you can approximate the set omega in measure by a set omega epsilon that consists on a union of equal silence cubes. And so this set omega multitized RD with some lattice. And therefore, by the previous terms of three spaces, you know that for the L2 of this omega epsilon, you have a risk base of exponential. And the frequencies, by the Landau theorem, have to have density equal to the measure of omega epsilon. Now, if you take this risk basis of exponential from omega epsilon to omega, what you get is a frame for L2 of omega. And the density of the frequencies is equal to the measure of omega epsilon can be taken as close as you want to the measure of omega. With Elona, Agora and Carlos, and really we could try to mimic this idea and motion settings that are locally compact. And using more of this idea of shorty, we could prove that awesome that setting we can find in four-year frames close to the to the critical density. In other locally compact gradient groups, you don't have cubes, you don't have dilation so you have some technical problems to exactly the same, but you can try to follow this scheme and using this idea you can get a similar result in locally compact gradient groups too. Well, now, if you see these schemes, if you have a set, if you want to construct a frame for that set, you can use this scheme of shorty, but what happened if you set change with the time? Suppose that your set evolves with the time. Well, it presents the measure, but it becomes a longer and longer, so the diameter of your set starts to increase. So this scheme here will give you for the different set, different frames of exponentials. So a question that could appear in this setting is, okay, if I have this domain that evolves with the time in such a way it presents a measure about the diameter increases, increases. Can I find a frame exponential for all these domains in the final time? Can I find a kind of universal frame? Maybe a little bit more precise. It's possible to find a set of frequencies such that the exponential with these frequencies constitute the frame for L2 of omega, for every omega such that the measure of omega is lower than the lower density of lambda. This is a requirement that appeared in Landau's theorem, perhaps with an extra equality in the case of Landau, but it's possible to obtain this. Well, if you don't add any structure to your set or any other condition, the answer is no. It was provided by Oleski and Lianoski. But if you add some topological condition to omega, like a compartments, it turns out that the answer is yes. This is very simple reason. Oleski and Lianoski proved that there exists a set of lambda in Rd density is uniform and equal to one such that the exponential with frequencies in lambda constitute a normal basis for L2 of omega, for every compact set of omega in Rd such that the measure of omega is smaller than one. One here is the density of one. Moreover, they prove that this lambda can be taken as a small perturbation of this regular lattice set D. This is a very impressive result because you have many, many compact sets with very different structures and it's telling you that the same frame can be used to all of them. Another surprise that appeared in this universal problem is that the intuition says that this set lambda cannot have a structure. You have to per tool set D in some random way, cannot have some structure because it has to work for many different compact sets. So one, we expect to find a set without a very rigid structure, but it turns out that you can find universal sets of frequencies. So the frequencies such as this and the corresponding potentials are frames for every compact set corresponding to a measure with a lot of structure. And this is, here is where appeared the connection with quasi-critics. Let me recall the definition of model sets. I want to construct a model set in Rd. Okay, so I expand Rd in n other dimensions. I add n extra dimensions to Rd. And so in the Cartesian product of Rd times Rn, I consider a lattice gamma. And I say that gamma is in generic position if, when I consider the two canonical projections, Rd, sorry, this should be Rn. So both projections restricted to the lattice are injected. And both projections, the image of the lattice to both projections have to be dense in Rd and Rn. A simple way to construct a lattice in generic position is to make this post that you have R and you add one extra dimension with the Cartesian product with R. And then in R2, you consider the lattice step two, and then you rotate it in a rational portion of two parts. Then the lattice that you get is a lattice in generic position. And satisfy these four conditions, two injectivities and two density conditions. Well, when you have a lattice in generic position, you can define what's called the model set. So take a lattice in generic position in Rd times Rn. In Rn, take a set omega that's usually called a window, and the corresponding model set to this lattice and in this window is this set here. And this is the projections onto Rd of all the elements in the lattice whose second projection is in the window. Here is an example. Here, the window is this interval here. And so in this light blue part, you have all the elements of the lattice that belongs, whose second projection belongs to this window. So these are the elements of the lattice that you project to the first space Rd. And the projection is what's called this model set lambda omega. If you have worked with the almost periodic functions, you will see that this construction has this flavor of almost periodic functions. It's more precise in this, but I don't want to do these details. Let me give you an example. Suppose that you take a new rational number, alpha, and you take this matrix, one plus alpha minus one, alpha minus one. And so you consider this lattice. You can prove that this lattice is in genetic position, too. So a typical element in this lattice gamma of this form, one plus alpha m minus m and alpha minus m, where m and m are integers. And then you consider the window as a window, the interval zero. So when the construction say, okay, you have to consider those elements in the lattice, the second projection lies in the window. So the second projection has to be in the zero one. So these part here have to be in zero one. Since n and m are integers, it's obliged that m had to be the integer part of n alpha. And so if I denote in this way the fractional part of an alpha, what you have is that this model set is n plus the integer, the fractional part of n alpha. And so this is the corresponding model set to this, this window and this lattice. And here is a picture of the dots, the elements also of lambda omega. And I wanted to show you this picture because in some sense, you have here another type of stylus. And instead of using only one type, you have two different types of styles, the light blue one and the orange one. You combine them in an aperiodic way. And this model sets are also related with aperiodic diamonds. It's related with one known pair of aperiodic diamonds. Here the model sets are the points, the vertices of these polymers that we are here. And they are also used to describe the diffraction patterns of some new material in the 80s. And that's the reason why this model set sometimes are also called classic crystals. What is the connection with our problem? Well, the connection comes from this result of Mateo Meier. First of all, consider a lattice generic position in rd times r. Consider as a window interval. And then you can prove that the density of the corresponding model set is uniform and is given by the measure of the window divided by the measure of the lattice. The measure of the lattice is the measure of any fundamental domain of the lattice. For instance, this one, if the lattice has its own. This calls in general not only for this particular model set, always the density is given by the measure of the window divided by the measure of the momentum. If you take this particular model sets where you have increases on one dimension and you take this window interval, then what Mateo Meier proved is very impressive result. For every compact set, such that the measure of the compact set is lower than the density of this model set, then the freak, the exponential with frequencies in this model set is a frame for L12K. So this set has this universal property that I mentioned in the theorem of Holley's theorem, and this set is a very structured set. It's not so grand. And so this is very impressive for me. Now to finish, one more, and we say that with Mateo and Lola Vora and Carlos Caballi, we try to extend this kind of result to local compact Aurelian groups. The theory of something can be done in these groups and some of these groups in the condition that the dual group to be compactly generated. These special groups are basically due to the structure of theorems isomorphic to this type of group. This group of isomorphic to D copies of R times L copies of the torus, and here a discrete and countable Aurelian group. And so many, many results of something can be extended to this setting, in particular land also can be extended to this setting. But the difficulties in these groups comes from this group, from B. Because this part is very nice, it's a big group, a lot of analysis here, but this group here is really the source of all the difficulties. This group is just a countable Aurelian group. So it can be, for instance, the rational numbers with a discrete of polish. Or it can be this rougher group, the B and roots of the unity with a prime number with a discrete of polish. So it can be very weird. So I wanted to study this quasi-crystals and the model sets and the applications to sampling theory. The first problem that appears is, okay, is there any lattice in generic position in this group? At that time, we asked, in a year, if he knew I was asking about this problem, he said that it would be nice to know if every group G has this property, if you extend it with R and then inside this Cartesian product you have a lattice in generic position. It turns out that it's not always possible. With Lona, Carlos and Bazarab, we proved that in order to have here a lattice in generic position, then what you need is an if and only if condition that the part here, this discrete part in your group, cannot contain a copy of set P to the power B plus M plus 1 or any prime P. Okay, let me very briefly say a few words about why this very strange condition here. The reason why this condition appears is that if you want a lattice in a generic position, the possession has to satisfy the four conditions, the two conditions of dance range. So these conditions implies that there is some more system between part of your group D and part of some R to the power R to D plus M. And the problem comes from the elements of D with finite order because in R, you don't have an element of finite order. So the problems that arise from different prime numbers can be split in different places using the fact that the dimension of R as Q vector space is infinite. So using this fact, you can split the problems of different piece. But when you work with the same P, then you have to pay a higher price to absorb each element of finite order of order P. You have to pay the price of one dimension of one copy of R. And so here you have exactly D plus M copies of R. So until D plus M, you don't have any problem, but you cannot absorb any more, any other copy of any other element of order P. So that's the reason why this condition here. Well, I think that's finished here. Thank you. Thank you very much for your attention. Thank you very much. If there are any questions, you can either raise your hand in the participants list or let me know in the chat that you have a question and I will take you in order. Let's see. Any questions. I want to ask you the first question then. In this, you mentioned some some sorts of results, especially the one that you had the omega epsilon covering the omega. And then you get, you get a frame from a race basis from the mega epsilon. My question is a philosophical one in these sorts of results. Is there a way that you extract when you send epsilon to zero some sort of pre compactness of your objects, you know to to to have the hope at the end to get the race basis at the end or maybe get the better information on the on the frame that you get for omega or these are generally very hard to quantify. Sorry, sorry. Could be repeat the question you know that you have a race basis for a little bit bigger. And from that you want to extract some information about the most more philosophical if there are any if people if there is any sort of result or in the line that you want to when you when you send epsilon to zero you can extract some you have some compactness argument for me. It is usually very hard because when when you approximate your you will say, when epsilon tends to zero, the problem is that if you have a race basis, this operator that appeared in the definition of a race basis was invertible. And the problem when when you let the science epsilon tends to zero is that the enormous the inverse of this of this operator tends to infinity. So, as when you go to your set, your your risk basis are less and less risk basis sometimes the operators are they go to infinity or they may go to infinity. I mean, for example, if you know a priority that omega has a race basis, and if you apply this process with omega epsilon, do you recover the risk basis of your original omega from this omega epsilon risk basis approximations, would you recover. Well, in that case, I don't know, I don't know. I was thinking, for instance, in the case of the of the ball that the natural we don't know if there is any risk basis. And so you can, you can try to approximate the ball by this, this sense of some union of the squares. But in that case, you have this this problem. But if you already know that you have a race basis in your space, I don't know if you can recover. I don't know. I don't know what happened in that case. For example, you mentioned a result where you have a risk basis for a polytope which is centered centered symmetrically with each face centered symmetric. Right. So this one would be an example that you already know you have a risk basis and you could approximate by cubes in this color on type the composition. And then probably, I don't know, maybe interesting to see if some subsequence along when epsilon goes to zero would converge back to the basis that you had. I mean, this is a philosophical question. Yes. I like very much. Okay, thank you. Thank you. So are there any other questions? Can I ask something? Yes, please. Yeah, absolutely. So I would just like to ask you about this, whether all these problems about sets, which are more singular know somehow like this kind of sets. These things work there somehow. What's the situation now. Weird results now I guess. There are some results in the relations. Well, it's still an open program to know if, for instance, in the term I can talk to the space to associated to the measure. If in that set in that space, you have a frame of exponentials. As far as I know, for some kind of counter set, you know that there exists even also normal basis of exponentials, but they are very special counter sets where there is a medical relationship between the number of cuts that you do, how many intervals that you keep in each step. For instance, in the counter set, you divide the interval in three parts and take two parts. And so two and three are co-primes. And so this made the situation very difficult. Another example is you divide the interval zero one in four pieces and you take the first on the third. You repeat this construction. This is the and in this counter set of the L2 of the measure associated to this counter procedure, you can find also normal basis of exponentials. But in the program here, one of the problems, many problems, but one of the problems is that if you have a basis of if you have an optimal basis there is no good notion of density to deal with them. You don't have a base like you don't have a land outside. You have a kind of density that's associated to a half of measure, but it doesn't work very, very well. So, I don't know. At least as far as I know, there is no many things in this direction. You can replace, for instance, frames by other types of four-year expansions like effective sequences. Yes, that effective sequences exist in any L2 of a singular measure. They have good properties. But about frames, I don't know. The limit procedures are difficult because you are changing in some sense the dimensions. If you want, for instance, to go from in the ternary context, you want to, okay, I constructed this basis for the L2 of the you divide the 0, 1 in three intervals and you construct the risk basis for the L2 of the first and third interval. Then you construct a risk basis. You go one step in the construction of the count of set and you construct the risk basis for the first, second, third and fourth interval. And you go in this way. In each step, what you have are risk basis for the first half of the dimension one. But in the limit, you go to a set that has another half of the dimension. And so this creates a lot of problems. So it's quite difficult to do this. Okay, thank you. Thank you very much. Thank you very much. Any last question? Okay, so if not, I would like to thank the speaker again. Thank you very much. And next week, we will have Ivaldo Nunes as our speaker. So looking forward to that. 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