 Recording in progress. Thank you all for coming this afternoon. Our analysis is a pleasure to have Professor Diego Moreira from the Universidade Federal do Ciara in Brazil. Visit us for what the second or third time? Second time. Okay, we'll be talking today about up to the boundary grade investments in Bernoulli type 3 boundary problems. It's a pleasure to have you again here, Diego. May this continue your string of visits. All right, thanks a lot. So I'd like to thank everybody for coming. It's a nice opportunity to be here again and see all of you from last year. Okay, so what I plan to do today as you talk about this is a kind of a recent work that I did in collaboration with Ederson Braga. Ederson was my first PhD student in Brazil and he is a professor at Universidade Federal do Ciara, the University of what I am at. So let me give you a pretty quick with the mouse a little bit sometimes just to or click on the video. So here is a quick outline of almost certainly I don't have the time to go all the way to the end so I will stop somewhere in between but my goal is here at least to present you the statements of the results and at least give you some hints why the result should be true when eventually provide some of the ingredients of the constructions of the elements that I did for the program. So the outline is a cluster of results in LPD versus unbounded coefficients. So then I'll talk a little bit about gradient up to the boundary, the players and pearls. Then I'll try to state the main result and then I will revisit at least the interior case so that to give you some feedback on what it is and in fact I will be finishing somewhere talking about the barriers because I don't think that I will have time to talk about 7 and 8 which is the strategy of the group for the boundary estimate and the trace estimate. So I'll start very slowly. So when my perspective is the following my understanding is that there has been an effort in the last 20 years for where many people are interested to do an extension of the theory of the regularity of PDEs for PDEs that have bounded coefficients for the PDE that have unbounded coefficients. When you do this change that you go from bounded coefficients to unbounded coefficients eventually this is a delicate issue because some of the classical theorems if you are not very careful on the regularity of the coefficients that you are putting in the equation, the classical results may fail. So let me show you a couple of them. So the first example is due to Safano. There's a very nice and simple example in 2010. Consider this function which is u of x 1 minus x square. So you see that this function satisfies this very simple linear elite PDE, Laplacian of u plus where this drift term, the drift coefficient is given by this vector field here and x over the x square and you can see pretty clearly that the sub of u is not attained on the boundary. And why this happens? Maximum principle fails and the maximum principle fails. The reason is just because of the low regularity of the coefficient. The coefficient is in ln minus epsilon for every epsilon. So the coefficient is not in ln. So this b here is not in ln. That's why the maximum principle fails. We know that if b is 0 or if b is in an infinity, the theorem is true. But if b is lightly below ln, the theorem fails. This is not a property of the maximum principle. If you go to the Hopf, the Hopf principle also, the same thing happens. You see Hopf-Fellenic and Nietzsche's up to the boundary. So this also due to Safanoff, right? So you consider this cube, right? And in this paper, Safanoff constructed two functions, right, which are continuous up to the boundary. They vanish in one of the faces of the cube and they are solutions to this elite PDE again, the same one as before, where the b now is in ln, right? So these are the, I'm just summarizing the details of his construction. However, the Hopf-Lemma fails, right? The infimum of u1 over xn is 0. So the solution reads the boundary of the cube in a tangential fashion. Not only that, the solution u2 is not continuous, is not Lipschitz also. See, u2 over xn goes to infinity as you approach the boundary. So the Hopf-Fellenic-Lemma fails, the Lipschitz up to the boundary. This is the Hopf-Fellenic-Lemma that fails, the Lipschitz up to the boundary fails, and the boundary harness also fails because the boundary harness, if the boundary harness were true, those two ingredients, those two quantities should be fine and they are not, right? So why is this as the case? So Hopf-Fellenic-Lemma and Lipschitz are like both fails, reason, the b is in ln, right? So it's like, it's something like that, that's just the first n minus 1 coordinate. Yeah, the rest is the n minus 1 coordinate here, and here is the xn. So here is our n minus 1. This is the notation, okay? Okay, now, just a personal happy mention, the Hopf boundary regularity. If you have, let's say, u of x to be the very simple function fx times y2 in the first quadrant, right? So consider this function in the first quadrant here. This is my domain, right? x times y. So this is a harmonic polynomial, right? And the normal derivative vanishes at the origin, right? So you have a positive harmonic function that vanishes on the boundary, but the normal derivative is there, also vanish, and so this side of it Hopf-Fellenic. And the reason is there is a coordinate on the boundary, right? This has nothing to do with the coefficients, but I just want to emphasize that the fact that there is a coordinate here makes the Hopf-Fellenic also fails. Okay, so let me summarize this. So let me go to the classic theory and then I'll try to point out some of the difficulties. So this theorem is going to be in the framework of PDEs. I want to transfer these to free bounded problems, but before I do it, I need to see what is the danger when I walk into the road of trying to extend the theorems from PDEs to free bounded. So the relevant theory of PDEs has a follow. If you give me a U as a bounded solution and some function in i q r q is bigger than n and you consider this equation, Laplace of u equals to f in b1 plus. b1 plus for me is half a ball. Let me set this notation that I'm going to be using for quite a while. So b1 plus again like this, this is bxm again. Okay, so if I have Laplace of u equals to f and I prescribe some c1 alpha boundary data here, I want to see what is the regularity of the solution and we know from the boundary regularity of PDEs that the solution is c1 alpha all the way up to the boundary. So if I don't say to half a ball like this, the solution is c1 alpha all the way up to the boundary. Of course, of course, if the solution is c1 alpha all the way up to the boundary, the solution is also leapshitz. Not all the solution is c1 alpha, but I have an estimate for the solution. The c1 alpha norm is bounded by a universal constant times the L e 50 norm of u plus the L q norm of the right hand side plus the c1 alpha norm of the boundary data. Right? Of course, if I am c1 alpha, of course, I am leapshitz too. This is a triviality. It is a triviality, but I am pointing this out because for a free boundary problem, there is no hope that this estimate will be true, but this one survives. So I emphasize in the leapshitz just because this is what the best way of life that I can hope for free boundary problems. This estimate also works for if I replace the Laplace of the Laplacian by some full nonlinear or for the Laplacian, if I do a little correction on the right hand side. In fact, for the Laplace here, it appears a power here, 1 over p minus 1. For the full nonlinear, the estimate is exactly that one. No change. Okay. So let me go back and say that if I have the solution where the coefficients of the equation are nice. So the operator is very nice. The right hand side is in L q with q bigger than n. The boundary data is very nice because it's c1 alpha and the boundary is also nice because it's flat. So if everything is okay, the equation is okay, coefficients are okay, boundary data is okay, boundary is okay, then the gradient survives and the solution goes c1 alpha all the way up to the boundary. This is not always the case. The existence of the gradient is very sensitive to this ingredient, the regularity of the coefficients, the regularity of the boundary data, the regularity of the boundary. In other words, if any of those things goes bad, you may fail to have the gradient, the estimate of the gradient up to the boundary. Let's see some examples. So this is an example that I constructed with Lihe 1 a couple of years ago. So we constructed a function Laplace of u equals to f, this is in R2. Laplace of u equals to f, u s continues all the way up to the boundary, vanishes on the boundary on the flat boundary, but this f is in L2. It's not in Lq for q bigger than n, it's exactly where q is n. So f is in L2, right? Then what happens? Then solution is not even leap sheets in any tiny ball around the origin, let alone differentiable, let alone c1 alpha. In other words, if the coefficients is bad in the sense that the coefficients is Ln and mets the dimension, the gradient may fail to exist around in any point of the boundary. So the solution is not even leap sheets and the normal derivative blows up as you approach the origin. So here we have the regularity of the boundary is okay because the boundary is flat. The regularity of the boundary data is also okay, but the data is zero, right? But the regularity of the coefficient is bad because the coefficient is in Ln and this is enough to break the existence of the gradient on the boundary. Okay, here is another simple interesting example. I cannot press this thing twice otherwise it goes out. So here's an example that you can take. This is an exercise from Evans book, actually. So you take a harmonic function in the half space, continuous all the way up to the boundary, but on the boundary, the function is this cone in the x prime variable, right? So that is a representation formula for using the Poisson kernel, et cetera. If you do this computation, you can see that this first order differential coefficient mimicking the first derivative, the normal derivative, it blows up, right? So again, we have the regularity of the boundary is okay because the boundary is flat, there's a plane, right? There are a lot of the coefficients. They're spectacular because the function is harmonic, right? But the regularity of the boundary data is terrible. The boundary data is leap sheets, right? There is a kink at the origin and so the gradient does not exist and the normal derivative blows up just because it is very sensitive to the similarity of the boundary data on the order. Finally, let's consider this example. So you have a harmonic, you have a, this is a sector, a circular sector where you take this alpha to be bigger than pi. Then you take a harmonic function which is one around here and zero at the edges of this cone and you make a smooth transition here the way you like from one to zero just to glue it in a continuous fashion, right? If this is, you know, true, you have a formula for that. This is like the imaginary or the real part of z to the alpha, let's say, right? Well, the same way when you do this computation here, the regularity of the boundary data is okay because the boundary data is zero in the neighborhood of this kink. The regularity of the coefficient is also okay because the function is harmonic, right? But the regularity of the boundary is the, is the, is the villain here. The boundary has a kink and also the gradient didn't exist there. So in other words, if I am in a PD, if I am, if I want to start the regularity of a solution of a uniformly PD all the way up to the boundary, I have to be careful with these three ingredients. The regularity of the boundary, the regularity of the boundary data, and the regularity of the coefficients. Because if any of those guys goes bad, the gradient may fail, the gradient actually may blow up on the boundary. Okay, towards the main result. So this is not to be afraid of this slide, let me just put it, explain simple words, what is that? So I will consider a fully nonlinear operator, right? And fully nonlinear, and I will consider the case where this fully nonlinear operator is uniformly elliptic. This means what? Uniform elliptic, ellipticity just means that when you change the operator from NP to MQ, this difference is controlled above and below by the putchi operators with some, with some the gradient parts, and here eventually made a super linear part. So ellipticity, in general, ellipticity means, elliptic equation means that this F is monotone in the matrix. Uniform ellipticity means that it's quantitatively monotone. Quantitatively monotone in this sense, in the sense that this difference is controlled above and below by the difference of the matrix and the difference of the gradients. This F, as I have seen by previous examples, this F I will consider in LQ, where Q is bigger than N, because we know that if we are in LN, we have got an example, right? And this drift also, this drift coefficient here, which is undivided, I'm going to consider also bigger than N and bigger than Q. And this M are the putchi extreme operators, which are the two extreme operators that we use in the definition to define what that uniform ellipticity means. Okay, so let me just give a quick description of what is going on here. The origin of this theory goes back to Kaffareli, Kranbel, Gokhan, and SVF for full non-equations with measurable ingredients, a famous paper in CPAM in 1996. The existence as LBP, how do I connect quality with unbounded coefficients? There's also super linear growth has been done for many people, Koik, SVF, Bojansi, Rakovi, Rewan, Kril, Opsafana, Dome, many, many, many, there are many very interesting results on the work of equations with unbounded coefficients. And great data estimates in theory on the boundary. There are a couple of people that have been working on that, and Gabriele Norbin, Izvyek, myself, Ihe, Edison, João Vitor, a bunch of other people, and there's a type of estimate that we are searching for. Like the solution, they see the c1-alpha norm of the solution, although we are to the boundary, the solution is not only c1-alpha, but we have an estimate where the c1-alpha norm is controlled again by the LFN norm of U, by the LQ norm of the right-hand side and the c1-alpha of the boundary there. Okay, here is the goal now. There was the following. Instead of having a PDE, right? I don't have a PDE anymore. I have a function U, let's say which is continuous, and the function is positive somewhere. So this is the positivity set of my function U. I have no clue about the geometry, the shape, whatever of the set. And I'm going to impose the same problem. So I have a solution of this PDE, not in the entire bowl, but eventually just inside this set which may be a priori a crazy set, right? I will prescribe a boundary data, right? U equals to phi here on the fixed boundary, right? And I'm going to add this condition here. So the free boundary for me is exactly the interface that separates when U is positive from when U is negative. And on the free boundary here, right, I will have that the gradient only along the free boundary is bounded by some function C, let's say bounded function, right? The question is, can I recover, can I recover the same type of result that I have before where I have a PDE with where there is no free boundary? How about if I replace the fully-dominated one by a pila plus? Can I say so that's the, let me convince you, using the classical theory of free boundary problems. So remember that what happens in the PDE, when the PDE does not hold only in this set, which may be crazy a priori, right? But it holds in the whole bowl. And I have a, let's say a C1 alpha boundary data prescribed here. The solution is C1 alpha all the way up to the boundary for free boundary problems. So this is now a free boundary problem where I have, I have a solution just holds here. I have a C1 alpha boundary data. And I know that the gradient is bounded just along the free boundary. So the goal is to propagate these boundaries of the gradient, not only on the free boundary, I want to propagate it inside. This is what I want to do. Solutions for this problem can never be better than lip sheets. Let me convince you of that. This is a classical free boundary problem. So let's say that you have a negative harmonic function whenever it's positive. And let's say that the gradient is one along the free boundary. How do I get a function satisfying these conditions? I just minimize this function. Right? So the reasoning here is the following. If I don't have this part, so let me try to use this. If I don't have the characteristic part and my function is just gradient of you, if I minimize, I become harmonic. But if I have this guy here, this guy is, so the guy is collecting the contribution of the positivity set. So in order to minimize, there is a competition between these two terms. I may need to vanish somewhere in order to make the contribution of this term as little as possible. So my function is no negative, but since I'm competing, I am minimizing this competition, the function has to bloom on the bottom somewhere in order to try to minimize energy. So the function glues here, let's say, let's say that I prescribe a boundary data here, you can think as a wire, right? And then I am going to minimize this function among all the functions that I have this wire prescribed on the boundary. So the solution glues somewhere in the bottom, function is no negative and glues somewhere here. You can easily prove, right, that whenever the function is positive in the positivity set, which is the set outside here, right, this term may become irrelevant because the function is already positive. So minimizing locally the function in the positivity set, it is as if this term has no contribution. So the function becomes harmonic. Okay, suppose for a second, right, that the free boundary here, which is the boundary of the positivity set becomes a smooth set. This in fact happens. We know that this free boundary is smooth as analytic in dimensions two, three and four. It is open for five and six. And we know that in after dimension seven, the similar set may exist. But let's take in our tool. In our tool, this free boundary is smooth. Then what do I have? I have a harmonic function that vanishes on the boundary of a situ set. So what Lemma tells me that the gradient gets the transversal. If the gradient gets here transversal, right? So if I go, if I come from positive side, the gradient has transversal. If I come from the zero side, the gradient is zero. There is a jump on the grade along the free boundary. So there is no chance for the for the gradient to be similar because it's discontinuous on the free boundary. So the best I could hope for that solution is Lipschitz. So if I take minimize, in other words, if I take minimizer of this problem, this free boundary problem is the Euler Lagrange equation of this function. So if I am interested to prove the regularity of solutions for this equation, if I take this as an example, I know that I can never expect my solution to be above Lipschitz. Okay, so then let me go back to the problem. Here I'll put the whole assumptions. I have a full and only a PDE. The right hand side is in LQ with Q is bigger than M. The drift term that measures the ellipticity of that is also in some LQ zero. I prescribed a C1 alpha boundary data. This huge condition here is just saying that F is elliptic. And I put some continuity of the coefficient, which is natural for this type of theory, right? We know from the D-Georgia or Creosafano that if you get even linear equations with bounded measurable coefficients, solutions are only C alpha. Since I am interested in proving C1 alpha theory, I have to put some control of some sort of the coefficients. So I am asking the coefficients to be continuous, right? Then what happens? So here is a theorem, the theorem which is the substitute of the theorem that I discussed above. So there is actually a picture of that which is better, but let me try to explain this. So suppose my function is continuous, it is a viscosity solution of a free boundary problem when the boundary data is C1 alpha. So let me go to the picture. I think this picture is better. It is better to see what is going on. So I have, this is my free boundary problem. I have this PDE inside the positivity set. I prescribed a C1 alpha boundary data down here, and I have also this free boundary condition. The question is, can I prove or not that the solution is elliptic all the way up to the boundary? Can I ever be better than elliptic? Elliptic is the best that I could hope for. So my question is if I can prove that the solution is elliptic all the way up to the boundary. And then let me go back here to show you, yes, I can under two situations. First of all, P condition. What is the DPD condition? DPD condition means that the function phi, which is C1 alpha, vanishes anywhere where the function vanishes, the gradient also vanishes. So if I have this condition, which is very natural from the physics perspective, from the physics perspective, it's very natural. So I can prove that you're under this condition. Let me explain why this is reasonable. What does DPD stand for? The Generate Phase Transition. Sorry, I forgot to say that. So let me prove this really easily that so what I want to prove, I want to prove an estimate for u plus. I'm unable at this moment to prove this is like I want phase problem. I'm unable to prove that the gradient is bounded if the solution change phase. So I just want my theorem only works if I am bounded the gradient of the positive part. Okay, let's forget the free boundary. Let's see what happens even for harmonic functions. Suppose that I have a C1 alpha boundary data here and I want to prove that the gradient of u plus is bounded. Right? Then when I look at this, I look, look, I want to prove that the gradient of u plus is bounded. What is the boundary data of u plus, the boundary data of u plus is phi plus. But if phi is C1 alpha, phi plus destroys a priority the regularity of phi. Right? So phi plus, so you are composing a C1 alpha function with a Lipschitz function, which is the x plus. So phi plus is at most Lipschitz. So in this case, right? So I would have say a harmonic or a sub harmonic function here, u plus, right? Where the boundary data is only Lipschitz. I have a sub harmonic function where the boundary data is Lipschitz. We know from harmonic analysis, for instance, the harmonic functions with Lipschitz boundary data are never Lipschitz. They are in the same class at most. This is not true for C alpha. If I give a harmonic function with a C alpha boundary data for alpha between 0 and 1 is straight, so we can see alpha all the way up to the boundary. But it fails when alpha is equal to 1. So I need to resolve this issue, right? So because I have u plus and the boundary data is only Lipschitz. How do you avoid the boundary data to be Lipschitz? What I ask is that phi plus is also C1 alpha. Otherwise, there is no chance I can prove the theorem. Because if the boundary data is only Lipschitz, there is no chance that I'm going to prove that the gradient is bound. And let me tell you, if phi is C1 alpha, phi plus is C1 alpha if and only if the DPT holds. So DPT becomes a natural condition because of that. Okay, so then the theorem says the following. If the boundary data satisfies DPT, then my solution is Lipschitz, right? I have a bound for the gradient, the solution is Lipschitz. And what is the estimate? The estimate is exactly the estimate that I had before. In other words, the gradient is bound by the LQ norm of u plus the LQ norm of the right-hand side by the C1 alpha norm of the boundary data plus this gradient, which is the free boundary condition. After all, this is a free boundary problem. So the free boundary condition has to play a role somewhere. What is the other condition of the theorem? I can completely disregard the regularity of the boundary data. So boundary data is C1 alpha. Okay, I can disregard DPT if I ask also that the solution becomes a sub-solution across the free boundary. In other words, I do not care about DPT if I ask that when I cross the solution, cross the free boundary in a super solution fashion. It is not artificial. There are many examples in the similar perturbation theory of free boundary problems where this kind of thing you get for free. There are many examples in the literature. Okay, so difficulties in the linear setting. The difficulties here are to overcome the use of potential theoretical arguments like harmonic measure plus some kind of influence function. The theorem, some of the theorems are proven for linear equations with very smooth coefficients. But there, the harmonic analysis is a second order LEP, and the proof of that is using this potential theoretical. But if I work with the linear, there is no potential for this type of equation. You also see the use of some reflection principles. Reflection principles do not get along very well with the viscosity theory for fully linear PDEs. So we have to somehow try to overcome the use of reflection principles. And third, to overcome the presence of the unbounded right hand side, which is kind of new here for free boundary problems. The passage from the bounded to the unbounded coefficients, I mean, at least to my understanding, in Bernoulli type free bounded problems, I think this is the first result. So what is the novel? What is the novel of work? All the estimates are consequence of geometry of barriers. So I need to construct barriers for unbounded coefficient equations. And then with the use of these barriers, right, I can replace all the proofs that are new proofs when I cannot use the potential theoretical arguments. So with this barriers, we can use the geometry of the barriers to produce new proofs of these ingredients and eventually the theorem at the end. So here is how the pieces of the proof got together. It took quite a while to discover this diagram and what implies the other. So essentially everything follows from the geometry of the barriers. So the geometry of the barriers implies a lindes lemma, which measures the expansion of the level sets from the boundary to the interior. It also controls the solution by the negative distance function for the free boundary. It allows me to prove the trace estimate along tangential points. And here is where the ideas from harmonica have disappeared. This is due to Carlos Koenig in the study of the maximum operators for the solutions for the directly problem when you have some LP boundary data. So you have the trace estimates. And then with some continuation, a sub-solution lemmas across the free boundary, you can use the control by the distance, then regular grade NPD estimates change in the result. Let me try to, I mean, so this is how the pieces of the how the pieces of the, how the pieces of the proof glued together. And you can say, so the proof is very involved. It's a very geometric proof. The lindes lemma, I mean, this is a name that you can find in the literature with many, many meanings, right? In the meaning that I'm using here is you have some information on the boundary. So you have a solution or a sub or a super solution, let's say, of a PDUF, some information on the boundary. So some quote, let's say solution is bigger than M on the boundary. Then you can prove that this solution becomes bigger than a multiple of M inside the domain. So it's like if the information from the boundary propagates inside the domain. So the proof is very intricate, especially the boundary when you have to relate with the snow potentiality issues about the points. So let me try to give you an idea, at least in the interior, which the proof is more elementary, why barriers are important and why barriers are fundamental to the proof. So let me talk about the interior elliptics regularity, right? So if you have a free boundary problem to prove that the solution is elliptics, the most crucial step that you have to do is to control the function U, suppose that you have a solution in your basic free boundary problem in B1. What you have to do is to control the function by the distance to the free boundary. If you can prove that you can control your solution U by the distance to the free boundary, this implies elliptics regularity. Let me show you how. So let me go back and revisit the basic theory of harmonic function. Everything follows from there. Suppose that you have a negative C1, but you can take C1 up to the boundary, harmonic function, right? And I want to bound the gradient at the origin, that's it, right? Then the gradient estimate says that the gradient is bounded by the L infinity norm of U over R, right? This is the classical gradient estimate. But since the function is not negative, the L infinity norm of U in Br over 2 becomes the soup of U in Br over 2. And you can control by the infimum by harmonic. But if you can control by the infimum, you can control by the value of the solution at the origin. So the gradient of the solution in the center of the ball is controlled by U0 over the radius. Suppose now that additionally to this setting, I give you a point on the boundary of this ball where the function is on negative, the function vanishes there and the normal derivative exists there also. Then you have a quantified way of the whole dilemma. I think this, my impression is this first appears in the work of Bristitka, Fadere, Nielenberg in late 80s, I think, 89 or something, right? And so if you have a harmonic function, non-negative, the quantitative version of the whole dilemma says that because the whole dilemma says what? If you have a harmonic function that vanishes on the boundary, the normal derivative hits there with an angle. But the quantitative version quantifies the angle, the angle as U0 over R, right? In other words, if you glue those two estimates together, the gradient that is controlled by the value of the center at over R, by hope, the value of the center over R is bounded by normal derivative. Then what do you get? You get that the gradient of U, right, is bounded by the normal derivative. I like to see this as saying that the Hopf-Flema allows the propagation of the information from the boundary to the interior because I have a, I have an information on the boundary, say let's say about the normal derivative, far away from the center. But the Hopf-Flema allows me to transfer the information from the boundary to the interior. Let me prove this realistically that this proves that the solutions of free boundary problems are elliptical. Suppose that I have the following situation. I have a harmonic function here, here U is positive, here negative, there. Suppose that my function is harmonic here. And suppose that the free boundary is smooth, C1 let's say, and that the normal derivative on the green here, the normal derivative here is bounded. So I have a harmonic function, negative, positive here, negative there in the boundary, in the normal derivative is bounded here. I want to prove that the solution is linked in, say, half a ball. But it's kind of very clear, right? This is kind of clear. Why? Because what do we know from here that the gradient at the center of the ball is controlled by the normal derivative? This is what they're saying. If you have a negative harmonic function, negative harmonic function if the function is C1, if the normal derivative exists on the boundary, the gradient on the center of the ball is bounded by the normal derivative. Let me apply here, take a point on the positivity set and compute the distance to the free boundary and take this ball and apply that lemma to this ball. Then the gradient here is bounded by the normal derivative there. But the normal derivative there is bounded because it's a data of the problem, right? The gradient is bounded there. So the gradient is uniformly bounded anywhere you take here, so the solution is linked, right? But then, so the point is, I think I heard the order of these slides, the problem is realized in, right? And the free boundary may not be smooth anymore. In order to use the reason, the proof that I gave, I assume that the free boundary is C1. But remember that in a free boundary problem, my goal at the end is to prove that the free boundary is regular. So I cannot use that the free boundary is regular at all problems. Free boundary may not be smooth. The normal derivative may not exist, right? And the free boundary may not become regular. So what should I do then? I should try to search for an intermediate quantity to do the intermediate point that will do the job that the normal derivative will do, but I don't have the normal derivative now. And the state is u0 over r, you see? If I have a non-negative harmonic function and a point on the boundary where the solution vanishes, u of 0 is bounded by the normal derivative, which in general case does not exist, right? But in other words, so u0 over r may be controlled by the normal derivative, but this quantity also controls the gradient. So the normal derivative controls this guy, the normal derivative controls this guy, and this guy controls the gradient, but the normal derivative may not exist. So I have to find a substitute for the normal derivative eventually. So the u0 over r is the key quantity here, right? Okay, so what are the ideas? The idea is to formulate the probability of viscosity sense. Whenever you don't have some differential ingredient in the problem, you change this for test functions. I don't have it, but any time that I test I get whoever is smooth, this guy will satisfy the property. So in order to overcome this difficulty, the idea is to formulate the price, the formulate the problem in the viscosity sense, the price that I have to pay, I have to solve good barriers because otherwise I cannot test the definition accordingly. So here is the definition of the viscosity solution. I will not spend too much time on that, but you have a viscosity solution of this problem, right? A harmonic function that satisfies this condition on the free boundary. If whenever you have a touching function, touching it from below, you transfer the bound of the gradient there to that function. So I cannot talk about the gradient of u because u is continuous, but whenever I touch u from below what I smooth function, I can talk about the gradient, so I impose the estimate on the gradient of the testing function. Let me just mention this. This is a barrier, a very basic barrier. So if I give you a ring of radius out and out over two, it's very easy to, you can construct by hand this function. It's very easy to construct a harmonic function that vanishes on the outer boundary and the s of some height in the inner boundary. Let's see if I have a picture for that. Here is a picture, much better than that. So it's very easy, suppose that this, I have a ring of radius one-half and one. It's very easy, you can construct by hand, a harmonic function which says the value of the solution here is some quantity, like say m, right, and vanishes there. I have a formula for that, but I don't care about the formula. What is important to me is the geometry. The geometry of this barrier suppose that this radius here out and out over two, whenever you solve this by hand, you can see that the normal derivative, the inner normal derivative of this function, along with the free boundary here, along the outer boundary, is proportional to the height over the radius. Basically, as intuitively, right, if you push a tangent line here, eventually this is smaller than the height over the radius. In fact, it can prove much more. You can prove not only the normal derivative on the boundary, it's proportional to the height of the area, but the whole gradient in the most proportional to that. Okay. Doing that, let me conclude. I gave a heuristic proof why solutions to free boundary problem of one phase should be lipped if the free boundary is C1. Not suppose that the free boundary is not C1 and I will take a viscosity solution and I will prove for you that the solution is Lipschitz. Let me do it by hand here. So take a point on the free boundary, take a point in the positivity side, and then you take the ball that touches the free boundary there. Since the function is non-negative and the harmonic here, it satisfies harmonic inequality. This means what? This means that all the solution in this tiny ball is comparable to the value of the solution at the center, meaning that u of x or every x here is bigger than some multiple of x0. In other words, you construct the cylinder, the solution is on the top. But if the solution is on the top, then you create this lily flower, the piece of the fundamental solution, the barrier that I just discussed a minute ago. And then by comparison principle, u is harmonic, right? It is above my barrier, which is also harmonic in this sphere. In the second sphere, there's a triviality because u is non-negative and my barrier is 0. So u is harmonic in this ball, my barrier is harmonic in this ball, and on the boundary of the ring, u is above the boundary. So maximum principle says the solution is above all over. In other words, I patched my solution in blue by a smooth function on the free boundary point. The viscosity theory tells me then what? That the gradient of the solution here is bounded by the gradient of the solution here is bounded by the free boundary condition. By the gradient of the solution here is proportional to the height of the radius. The height is the value of the solution here and the radius is the distance to the free boundary. So I just controlled in a universal way a solution by the distance. So here is the formal rule of everything I said, right? By the end of the day, let me see if I have a by the end of the day, I control u by the distance to the free boundary and then since the function is harmonic and classical everything becomes. Okay, now what happens? So this is for two phase problems. Let me skip that. I can reduce the two phase problem to one phase problem by this very powerful ingredient called the outcalf-relimon atomistic form. If the problem now, if the function is not only positive, it becomes negative eventually and also, right? I have a way to control the product of the global derivatives on the free boundary once I once I use the argument that I used before by the outcalf-relimon atomistic formula, it is as if the problem becomes one phase and then I can return it. Let me skip that. So this is the outcalf-relimon atomistic formula. So here's the fundamental ingredient that we constructed. I want to substitute the barriers that we constructed by hand and in a table for the harmonic functions. And we did that using the putsch-extremal operators for fully nonlinear, you see? Instead of having Laplace equals to zero, we take this fully nonlinear operators, both of them, right? And the same scenario is zero in the, let's concentrate on this one. The zero in the outer sphere and some value of the inner sphere. And then we prove that the solution of this problem exists and that satisfies the same geometry that we had before. In other words, the height is proportional to the gradient, which is proportional to the function over the distance provided what? Provided the right-hand side is much smaller than that. This is the key point. In other words, let's think on that. If f is zero, morally dysfunction, it is like the previous barrier if f is zero. The fact that there is a fully nonlinear operator is not a big deal if this is zero. So if the function is zero, the barrier is exactly that one that I just wrote, right? But if there is a right-hand side, yeah, it interferes in the geometry of the barrier. But it says that if the size, if the size of the right-hand side is not to be incompatible to the height, the geometry is preserved. And here there is another interesting point, which is the following. So this thing here is just to say that the geometry of this barrier is exactly the same we did before. In fact, if the right-hand side is very tiny like this condition here, human eye cannot distinguish between the solution for the Laplace and the solution for that one because the geometry is the same. I don't care about how do I get to, if I have a formula or not for the solution, what is relevant for me is the geometry. It's the only thing that I use to prove the Bipschitz estimate. But here there is an interesting thing also. Since I have also a bound for the gradient, I also have the gradient is also comparable to the height. Then my solution becomes also a good solution for quasi-linear operators like Pilaplace because you know that the Pilaplace, the Gilaplace, they have a part of the gradient in front. If I lose the control of the gradient, I cannot handle this operator. But if the gradient is bounded above and below, which is happening here, the gradient here is bounded above and below because the gradient is universally comparable to the height, then I can use this barrier for quasi-linear operators. Here is the geometry of this barrier. If the right hand side is very small, so if the right hand side is the LQ norm of F is very tiny in comparison with the height, the geometry is like this as it was before. Then what happens now? Then I repeat the proof. How do I repeat the proof? So if I am in that old scenario, same old scenario, right? What I do is I apply Harman's inequality. It is as before. No, because I have a right hand side. The right hand side has a contribution into Harman. But then I use this dichotomy. This dichotomy says if the contribution of the right hand side is much smaller than the height, then Harman will work as before. Not only that, the geometry of the barrier also works as before. So under condition 1, I can reproduce the proof that I did before, word by word. In other words, if the right hand side is much smaller than the height, the geometry of the barrier does the job as it did before. In the other case, where the height of the solution is not bigger than the right hand side is smaller, I get for free, because the solution is controlled by the distance. This power here is bigger than one, so I can replace it by one. So the function is controlled by the distance trivially in this case. In the previous case, I repeat the proof that I did before, and I also get the solution. So in the first case, I get that. In the second case, I get that. So the crucial point are the barriers. If I had the barriers, I can reproduce word by word what happens in the interior of life. For the boundary of life, that's another story. For the boundary of life, I don't have time to talk. I still have two minutes to finish. I don't have time to talk. The proof is, in fact, the proofs are new, because the proofs for the boundary depends on potential theoretical arguments. So we need to come up with new proofs. But all of the proofs are consequences of the geometry. And one of the consequences of the barriers is to prove the hope flamer. So if I have a non-negative solution of a super solution of a fully nonlinear PDE like this, then I have the hope. If I had u, which is a solution of those two differential inequalities, I have hope. You see u over r, remember the previous hope, the normal derivative was bounded below by u0 over r. Since my function now is in LQ, I have now the contribution of the right-hand side. This was proven, I did this with Ederson, but at the very same time, Boyan Cirocone had got a completely different proof. And for fully nonlinear cases, his theorem is stronger than ours. They come up at the same time, but his theorem is stronger. But the fact is our proof also works for the Gila-plus, because our barrier, our argument is geometric, our barriers have control on the gradient. So the same proof that we did here by a little adaptation also works for the hope flamer for the homogeneous case. So here is where I enter in the discussions of the strategy of the proof, which is very delicate and I will skip. Here is the crucial part of the proof, where you have some trace estimates. This is a very nice idea that came from Kenick to control the behavior of the solution on non-tongential cones and transfer the regularity inside the cones to the regularity on the boundary. This is what hope flamer does. And here is the latest boundary proof, and I can show you later, Giovanni, I don't have time to do it now. But it essentially says the following, the solution has some height on the boundary, and there is a super solution of some fully-knowing MPDE. If you enter in the domain, the solution may decays a bit, but it is still proportional to that value. That's what I mean with these line details, which it has known in many contexts, but they take in the context of fully-knowing MPDs with unbounded measurable coefficients. I don't know any other, I think maybe this is new, at least my impression. And here I have all the iteration schemes that we have to do, and the final ingredients of the proof. They are very technical, so thank you very much. For the very nice talk, any comments or questions from the audience? Some questions in the beginning. You showed these nice examples. You showed these nice examples where you, if you have the data of your MPDE in LN, the result fails, right? So it seems natural to assume, then, some of these results that your function f, on the right-hand side, or part of the question, is in LQ for Q bigger than N. Right. And things work. But, you know, I mean, there are some spaces in between, right? Sure. LN and LQ. You could take an LN log square, true log Q. Very nice. Do people consider these spaces? Some of these examples, can you actually assume less than LQ? Can you assume LN log, a power of log? Let me tell you what is the, what is not, yes, let me tell you. That's a very nice question. Let me give you a kind of answer that it is true. We know it is true. Some of, I mean, the whole frame is not written in that context, but the regularity, some of the regularity unions written, and the point is the following. Let's say that you get LN, let's say you have a domain, or any ball inside the domain. Let's say, yeah, you consider, you consider the LN norm here. What do we know? We know that if, if the LN norm in these balls are universally controlled by a model of continuity, which is genicontinuous, it is out holds. Then there is a work, a very interesting work of Giuseppe Mignoni, which has to be, there are some results of Agnid-Bundler-Geo, so for full and linear, for full and linear equations where they write these Lorentz spaces, where this, this Lorentz space like this, where the distribution function has some integrability conditions. So they still want to have a regularity theory, it works for right-hand side of this space also. So this is Lihe 1, Lihe 1, and I think this is just LNN in the Lorentz space notation. Yes, right, and then you're going to move a little bit to LN1. Yes, right. So here is, this result is Giuseppe Mignoni and Agnid-Bundler-Geo. So this is done for at the level of the, so there are some results from the regularity theory that works in this case, or this, let's say C1, suppose that I give of Laplace of U equals to a half, C1 half a boundary data, the right-hand side there are in none of those space. Solutions has a gradient with some models of continuity all the way up to the boundary. So this is a fact, but this has to be converted into the Ho-Plemma and the Bayer, here's a lot of things, and this is one, one of the ideas that I have to be working in the next, with some students, we are, we started working on that. Yeah, right, between, so in other words, some of the regularity theory up to the boundary, the gradient survives, but to apply for free boundary problems, you need to use, there is a piece of information to prove Ho-Plemma, to construct barriers and to apply for free boundaries. And this is one of the projects that I'm doing with some students, very nice question. And I think, at least as far as I know, those are the, the sharp, let's say, the sharp spaces where you cannot improve. I mean, unless you move to another type of space for it. Another type of space like you can make. If you just consider LB spaces, LN is the better. Then if you enter into the business of Florian space, then you start with LNN and you say that, okay, I can go up to LN1, but not better. Right, not better. Then you can. And then another, for instance, there is a student in Fortaleza, just finished his PhD last year, and we he proved the same theorem for divergence equations. But for diverse equations, we change the right hand side and we put the right hand side in the more spaces. Right? So in the more spaces, I think you can go to more spaces too. And there, I mean, you have to play a little bit with the exponents to see what is going on. So then I put the IQ a few bigger than M, because I don't want the, I just want to find the argument, at least to show that in classical cases, you have the estimates for free balance. Now there is this whole business of finding the optimality of the exponents, which is a very, very interesting question, very interesting question. Very nice, very nice. Do you have questions or comments? No, let's thank Diego again. We can stop the recording.