 time constraints. So the second thought and the second thought, actually, is that we conduct the university, normally by the integration, which is, of course, an area of research where I don't know about it. Yeah, it's okay. Okay. So thanks. I will, so I will, what I will do is for this lecture, I will rather quickly run through some slides to describe some, this type of problems and I will describe some problem which I, which people have been trying to prove. And then I will start with some estimates in this direction. So at the beginning, maybe it runs pretty fast. Okay. So my, so here it is a classical problem where, so we have a compact remaining manifold. So the question is whether or not this remaining metric you can find a positive function so that the new metric, the scalar curvature is constant. So this u, then the problem is in terms of u, it's just this semi-linear partial differential equation because the left-hand side divided by this power is the new scalar curvature. It's just a computation from definition. So this is the equation to be solved. So this problem actually is a variational problem. So fix up manifold if you look at the remaining metric. So there is a functional. You integrate the scalar curvature divided by volume of certain power. So then it is a functional. So there is a critical point. Euler Lagrange equation. So if you look at a critical point of a functional, so that's actually an equation. So this is actually the Einstein equation. So, and if one, because we are looking for new metric of this form, it's a variational problem. It's a conformal metric. So we only look at functional of positive function multiplied by this g. So you run through your, this positive function. So one can think, so this is one can think of, if the total metric space is r3, you can say this is a line in the r3. So then you can, one can look at this functional and restrict it to this conformal class. Then the critical point will be given by new scalar curvature equal to constant. So, namely this equation actually has a variational formulation. So actually there are three mutually exclusive cases to, depending on the first eigenvalue of this operator. This is the Laplacian operator. This is called conformal Laplacian. So one should solve this equation with this constant should be the first eigenvalue of this operator. So this problem was solved through many years work. So after that, after this problem was solved, people look at solution space, try to understand the space of solutions. So there are three cases. One is first eigenvalue is zero. Well, then that's a linear problem. So this is a second order elliptic operator equal to zero now. So then it is known that the eigenspace for the first eigenvalue should be one dimensional. So zero, so existence is also clear. So this is very clear. So another case is the first eigenvalue of this linear operator is negative. So one needs to solve an equation with a negative sign. So in that case, the uniqueness is actually clear. And it follows from standard maximum principle. And existence in this case actually is also known, can be proved easily. So because if you take a small positive epsilon as a constant, so it's going to be a sub solution. And the other one will be a super solution. So once there's a sub solution less than a super solution for such an elliptic equation, then there will be a solution in between. So that's clear. But in this case, if the first eigenvalue is positive, it's much more complex. So if this manifold is our standard sphere, then all solutions are known. So it's one known. And it's unique modular conformative morphism of the standard sphere. So if it's not SN, solution space is more complicated. So one question would be is say whether all solutions are bounded. So if all solutions are bounded, you know the solution space is compact with very high norm. So there's a lot of work on this problem. So starting from Shane and then for many works. And eventually the result is the answer is no if dimension n is greater or equal to 25. And dimension below 25, then the answer is yes. And actually in this case the yes was depending on positive mass theorem. So this part is the part what I will describe analysis about is try to solve a fully nonlinear version. Yeah, might be problem. So for this problem, there is a remaining manifold. There's a tensor called shout and tensor. So this is the Ricci curvature, scalar curvature and the metric. And the shout and tensor is one can think of it as ambiance matrix, symmetric matrix, let's say. So let lambda to be the eigenvalue of this tensor with respect to the metric g. Then when we take a trace, we actually get scalar curvature, modular, a harmless constant. So therefore the amabi problem in this case can be rephrased as follows. On a remaining manifold, if we assume the sum of the eigenvalues is positive, whether or not one can find a conformal metric to make it equal to one. And a more general question would be, let's say for some other function, you start from it's positive, whether or not you can make it equal to one. So this is for function equal to the sum of lambda, the lambdas. So one can look at other function. So this equation will be a second order fully nonlinear elliptic equation because this H e tilde in terms of, yeah. So it's written in terms of u is like this. This is the formula. You define, we have the definition of the shout and tensor one compute. So after this conformal change, it's going to be like this, involving second derivative. So then this is a nonlinear expression. So if we look at eigenvalues, so it's nonlinear. So we want that equal to one. So it's a nonlinear second order equation and we will only look at the case. This equation is elliptic. So more precisely because here we have a function f, a general function f. So more precisely we look at gamma. So we have a, in R n, this is where the eigenvalues live. So there's a gamma. This is the domain of definition of this f. So this gamma is a comb. So this gamma is an open convex symmetric comb. Symmetric meaning if lambda one, lambda n in it, we make a permutation, it's still in it. So it's a vertex at origin. It contains the positive comb. It's like the first quadrant. All coordinates are positive. And gamma one is half space. Gamma one is on this side. It's a half space. So, and f is a function, c one, and f lambda i, partial derivative in each lambda i is positive. This means the equation is elliptic. f is positive inside. f equal to zero on the boundary. So these are important examples. We take this case elementary symmetric function. Gamma k is the component where sigma k is positive containing the positive comb. So then, this is a, these are important examples. So then, so the fully nonlinear yamabe problem is to assume the initially eigenvalues of this Schouten tensor is in this gamma. So find a u so that to make this equal to one. So when you look at this equal to one, one, yeah, it's, yeah, okay. So we solve this. So the yamabe problem is actually a special case for f equal to sigma one. So we take the sum of all eigenvalues. Well, one, this is, this looks rather general. So can one really solve something like that? So well, the answer is yes. If this f, if the domain is locally conformally flat. So then it's very general. So general f gamma like this. You can always solve it. So let's, let's look at the important case for elementary symmetric functions. Then if this k is greater or equal to n over two, then one can solve it. For k equal to two, one can also solve it because this case, it's a variational problem. One has a variational, one have a functional. One can look for gradient point of that functional. So there's a lot of work on it. So still there are open problem. So the main open problem here is look at a special case for sigma k and k between three and less than n over two. So that's, so the answer would be yes if one can prove a priori estimates. So, namely if one have a solution, then one show solution is upper bounded by a constant. So, so this is the main remaining open problem now. So in this, so well, why, why it is sufficient if one prove this a priori bound, then the problem is solved. Well, so this is a fully nonlinear equation. But for this equation, if we have upper bound with C, then so through more than 10 years work of people, one already know that there will be a lower bound also. And one also know that the gradient of U will be upper bounded. So this is C1 estimates. And then one also know C2 estimates. Once you have C0, C1, you will have C2. And once it's C2, then if this is a f is concave, like in this case, then one classical theorem like Krilov's Evans will give C2 alpha, then shoulder estimates will give all derivative estimates. So, namely, so using already known results, one have all derivative estimates. So if one have all derivative estimates, then one can deal this problem by, by, by, by a homotopy to a subcritical problem. And then one can solve it. So one can make a homotopy to a subcritical equation. So, so anyway, so the, so this is the remaining is to prove C0 estimate. So what I will describe here will be concerning the estimates to go say from C0 to up. So already if there's time I will talk about some cases C0 estimates. So, so in particular, we will not worry about remaining manifold for the time being. We, we want to establish estimates for Euclidean space. And then how, and then to, to go from Euclidean to remaining one have another step because that equation will have low order terms because of the remaining metric. So here I will only describe the Euclidean case. And this Euclidean equation will also occur when one tries to prove this C0 estimate. So, so usually when one treats C0 estimates, one can say suppose it's not true, then C0 estimates fail, then one rescale that sequence of solutions, one will end up with a solution in the whole space. And then, then this will be the equation if one rescale appropriately. So, so this will be an AU is a second order operator like this. So this operator looks, looks complex, but this operator is the only conformally invariant second order operator. So it's a nitro object. And also many of the proofs can be made using this conformal invariance property. And it doesn't, yeah, one can use it that way. So, well, so this type of equations, there's a classic work of Louise and Nuremberg and Sprach. So they studied a second order fully nonlinear elliptic equations, they are kind of problem. So this, the equation we look at is kind of resembles that, that equation. It has some low order terms. And, and it has some additional feature is there's a conformal invariance of the equation. So in, in the talks, I will describe some estimates. So for this Euclidean space and I will mention some open problems. So the first problem I will describe a Liouville type theorem. So, so this theorem says that if we have a positive continuous function in Rn minus zero, so and satisfied in viscosity sense, the eigenvalues of this Au, we have n eigenvalues, they lie on the boundary here. So the equation is the eigenvalue lies there. So we, let's just look at U is C2, then it has clear meaning what that means. Otherwise there's a weak meaning. So, and we can think of a C2. So if the equation is satisfied in Rn minus zero in the space minus a point, then the solution must be radially symmetric about a point. And corollary is if the equation is satisfied in the whole space, then it's a constant. So, so I will describe if gamma is equal to gamma one. So where gamma one is here is the half space. So then this statement is actually the classical Liouville theorem. The reason is this lambda Au belongs to here. It's the same as lambda U equal to zero. So if we take all, if we take the trace, these two terms will disappear. So it just means that. So then the classical Liouville theorem says that the corollary one is the classical Liouville theorem. So it says that a positive harmonic function has to be a constant. So I will describe a proof of this. So let's say U is C2. So I will only, I think I will describe a proof of U is C2. And I will indicate some, at some key point, some proof depends only on the C0 property of U. And that's a key reason why the whole thing can go to C0. So, and let's prove for theorem one. So of course, if one needs to prove, well, this Liouville theorem is needed when studying the problem I described before. So when there's a need to establish something like that. And naturally, one can think of those ways to prove the classical Liouville theorem. So which one can be made going through to this nonlinear set. Anyway, so actually the way we describe here. Okay, yeah. So this, if we look at the harmonic function, we know when we make a Kelvin transformation it's still a harmonic function. So here we have this invariance also. So it's invariant under Kelvin transform, Kelvin transformation. So what is a Kelvin transformation? So if we take a ball of radius lambda, sorry that I use lambda for the radius to denote radius and here to denote eigenvalues. So I will just keep this. So there are two lambda here going on. And here is a ball of radius lambda and centered at x. And if we have a function U, whenever we have a function U, we can cook up a function. It's called a Kelvin transformation. So it's lambda minus x. So it's a Kelvin transformation. And the U 0 1 is the one which may look more, certainly look more familiar. It's a Kelvin transformation. This is with respect, unit ball centered at 0. Here we just change the ball naturally. So that's the Kelvin transformation. And with this expression, one can calculate that actually the eigenvalue of this is actually the same as that. So the eigenvalue of this conformal fashion A U, when we make a Kelvin transformation, it's actually the same. Essentially this is the only operator which can give you this properly. That's a defining property for the operator. So because of that, then the equation, so this is an equation. So lambda, this is our equation. So this equation is invariant. So that when you have a solution of this equation, the equation we look at, then when we make a Kelvin transformation, it's always a solution. So our equation is invariant under Kelvin transformation. We know this property for harmonic functions. Harmonic functions has this property. So we are going to prove this theorem. We are going to prove theorem one by proving the following. So we'll prove for every x in our n, but not equal to 0. So this is the origin. So we'll prove for every x which is not equal to 0. And when we draw a ball of radius lambda for lambda, bigger than zero, less than x. So for every point not equal to 0, so then we'll prove that U x lambda y, the Kelvin transformation is greater or equal to U of y. For every y belongs to this ball minus this point x and also p. I'm going to call this point p. So p is the Kelvin transformation maps p through. So y goes to x plus lambda squared y minus x over y minus x squared. So if I take p here, I'm going to be mapped to zero and vice versa. So this is a symmetry point of zero with respect to the boundary of this sphere. So I denote that as p. So after the Kelvin transformation when y is inside, if y takes p, then this expression, U x lambda expression, you can see a singularity because you may have a singularity as zero. And also here, if I take y equal to the center, then if you look at the expression here when y is at x, this will be infinity. So one may have another singularity. So there are two possible singularities here. So that's why we write this in equality for ball minus these two points. So we are going to prove this. So once proving this, we will see symmetry. The reason is, so if one can establish this, then we are going to see symmetry because first from here, you will see that if you have zero and you have any x, if you draw a ball like this, so we have U x and we take the radius as absolute value, then we are going to have this order. Then this is true for every x not equal to zero. This is just by limit. You can go to there. So then we will see that we want to prove symmetry. So now I take x1 axis here. I take x prime as the remaining. And if we take every y, for every y, which y1 is positive, if I take any point y on this half plane, then we see that for x large, for r large, if I take this point as x equal to r zero, zero, if I take r very large, this ball will contain y. So for large y, we know that y will belong to b, r, x for large r. Now we know this inequality. We know from there, this has an order U of y for large r because this y is inside now. So then I send r to infinity. Send r to infinity. So this will converge to U of y hat. So y hat is a symmetry point here with respect to the plane. So this is minus 1, minus y1, y prime. So minus y1 and y prime. Ah, cannot read now. Right, sorry. So this one actually goes to U of minus y1, y prime. And this is y1 and y prime. So we have an order then. So this point, the value of U is greater or equal than that point. We can prove the other way. We put the ball on this side. So we showed that they are equal. So then, and this is true for any, this x1 axis can be chosen arbitrarily. So this function U will be symmetric with any, with respect to any plane going through the origin and we have the symmetry. So the main thing is to prove this. So once we have a way to establish this, we'll get the symmetry. So we will prove this result by proving the following theorem. So we know that, so here we know that this U, we know U in this setting will be equal to U on the boundary of this region, of this ball. Okay, we know that. Because this Kelvin transformation is a reflection with respect to that. We can look at the formula. So they are the same. They are the same. And we also know that lambda of AU satisfy the equation inside. And we know that the reflected one satisfy the equation may miss two points. We want our conclusion is we have this order. So this is a maximum principle statement. So we need a maximum principle statement like this. We have a bounded domain. This omega will be this ball. And we have two, we have some singular points. So here are two singular points, x and p. One solution which is, this is more general which is on one half space. This should be minus gamma. So I shouldn't have a bar there. Okay. I shouldn't have a bar there. No bar there. So our minus gamma. And the V is belongs to this belongs to gamma bar. So then we have V greater than U on the boundary. And we want to have this order inside. So this is the maximum principle statement. And this, yeah. So if this is harmonic function, and we have that. So we first describe a proof of this without singularity. Namely, this statement. So suppose these two functions, again there's no bar here. So two functions, they satisfy, and for the time being we just think of this boundary gamma. We can think of that. So if we have an order on the boundary, then we have an order inside. But on the ball. Yeah. Yeah. They are equal because of this. Yeah. So this will be lambda. So this cancels, so y minus x. So this becomes y. Infinity. U x lambda at x. Yeah. Wait, wait. This will be bigger than that. Yes. Oh, we are going to prove that. We prove that. Yeah. That has to be proved. Yeah. Yeah. Yes. Yeah. That needs to be proved. Yes. Yes. Yes. That needs to be proved. And so I would say that's the main harder part to prove. Yeah. That's the only part in a way the tougher one to deal with. Well, I may not. But anyway, that state may not be proved. But when proving this, both require some thoughts. Yeah. But that's not obvious, this order. Not obvious. Yeah. So this is one step. One step. One step is to see the idea is assuming everything is smooth. Yeah. Then there's a separate step to address this singularity. So lambda of Au belongs to Rn minus gamma bar. So the original, I should not have that. So or lambda of Av belongs to gamma. So this means this is a strict subsolution. So for the gamma one case, this is just say Laplacian U is strictly positive. So this is a strict super solution. So in the Laplacian case, this is Laplacian V should be less than zero. So if one of them satisfies the equation super strict. So then this statement is very simple. So this is simple. So the proof is straightforward. So let's describe the proof. This proof. So then, so then what we prove by contradiction. So if not, one can, maybe one can draw something like this. This is going to be a V and maybe U would be something like this. This would be U. And here I will, this is omega. So V on the boundary is greater equal than U. But inside somewhere V, U is bigger than V. So there will be a picture like this. So then I use blue, blue can run, blue is okay. So then we multiply this V by, we multiply this V by a number. We can find a V bigger than one to put it in the following position. So, so namely, so this V hat is greater than or equal to U in omega. And V hat is bigger than U on boundary omega. And V hat actually equal to U at one interior point. So it's quite clear from the picture. So, and this equation, if we look at the expression of this AU, so lambda of AV hat is simply equal to a positive multiple of lambda of AV because of the expression. This is not readable anymore. So, so therefore this is still satisfying the same, same equation. So satisfying the same equation. So this is belongs to gamma bond. It's still satisfying the same equation because gamma is a cone, or definition gamma is a cone. So then at the touching point, so this is the expert. So at touching point, we see the gradient should be the same because there's a minimum point of the difference. So, and the Hessian should be greater than that. So of course, we already know that this is the same. Then if we look at the expression of AU, we see immediately that AV hat at X bar should be less than AU X bar. And certainly, eigenvalues will also have that order. So property, one property of one property of this gamma is if lambda is in gamma, mu greater or equal to zero, namely mu is every coordinates of mu is greater or equal to zero. Then the sum will still be in gamma. This follows from the convexity of gamma and the fact this gamma contains the positive cone. So now, for instance, we said if either, for instance, if we assume this, the other one is the same. If we assume this, then this will imply lambda of AU would also belong to gamma and would violate the assumption violating lambda of AU should be in Rn minus gamma. As I said, this should be taken out of that gamma bar. So it's a contradiction. So, so namely, if we have one of these conditions is strict, then the proof just follows from this very quickly. So then the question is, in this case, what, how to prove? Well, actually this proof may remind us of the, a proof which one may, maybe the first proof of many people have seen to prove the maximum principle for Laplacian. It's a similar proof. So then what do we need to do? So we only need to prove the following. We only need to say there exists a sequence of solution of, in C2 and Vi goes to V in C0 and lambda of AVI is in, in gamma. So we only need to approximate our V so that we have C0 convergence, but we are able to make this eigenvalue go to gamma. For example, you know previously this, this eigenvalue of V may very well be lying here. So you want to perturb your V in C0 norm controlling this eigenvalues inside. So, so this depends on the equation. So if, if you can do that, then it works. Our proof just works. So the main thing is to make this work. The main thing is to achieve that. You want to be able to perturb, to make up approximation of V, but to move this eigenvalues inside. So from, from super solution to strict super solution. So to do that, it's easier to work with this expression. It's harder to, well, well, for this it doesn't matter that much if we don't keep track of the dependence that carefully for this we don't need to. So anyway, so let's make a change anyway. So then we write it as in that. So we just rewrite this in terms of W. So that's the expression. When we compute, it looks like that. So let's, so we are, instead of perturbing V, we are going to perturb W. W expression looks neater. So, so let's take, define a function. Let's take this function. So we take some small data and this is just a exponential function. So then a calculation, a calculation will give the following. So for epsilon and delta small, we are going to prove this, we are able to prove this inequality. Here, we need the structure of this operator. So namely, this is equation dependent. We are working with this equation. It has this nice property. And moreover, the dependence of this, as in the proof one can see that the dependence of this only depends on C zero information. We'll see that. Just look at the proof. We'll see that the smallness depends only on a lower bound, a positive lower bound of W. So that's the reason all these theorems will work going to C zero sense. And to C, we'll of course, if one can have this inequality, that means from a W, where AW is say in the closure, when we make a small perturbation in C zero norm, we are pushing that inside by a strict positive amount of lambda. We are pushing it that way. So then after that, we can just take, say, we can take epsilon i equal to delta i goes to zero, and we define our V i through this W. And this V i will have the property because clearly this will go to that C zero, and this will... So how to make this calculation? So we can... So how to prove this step? Here it just says that if we can do this, so this approximation, this property will just follow immediately. So to prove this step, one calculate W plus epsilon because this AW is quadratic. So we have AW plus linear term in epsilon and plus quadratic in epsilon. So if we replace in this expression, hash and W by that, because this is the definition of AW. So then if you insert that, the second derivative will be replaced by AW itself plus first derivative information. Then we have that. So this is harmless because this is a cone structure says if you change, this is still in gamma bar. And that term, the gradient term actually is a good term. So this will help later also. So this is a good term actually. And then the explicit expression, if we compute, we will see that it's bigger than that because this term raised to a square can be absorbed by that and left with a delta square term. So for terms as small, one will have this. So the linear term is actually gaining that. So then we squeeze epsilon, we will have that. So this proves this inequality. So we have proved this maximum principle statement for smooth in the domain. So maybe I'll stop here.