 Okay, so we go on with the next talk of this morning. Jan Jan Lee will continue his lectures on second order conformal invariant elliptic equations. Please. So, we have this equation. The equation. Let's record. Oh, I didn't write that. So, so this is the, yeah, there's a function defined on gamma. And this is the conformal invariant operator. And so this is the estimate. Well, if we take the gamma equal to the half plane. So this is for the Yamabe equation. Then in that case, this equation is this equation. This is the standard equation. It's written like this. So that is this equation. So then this stays. If we have this, then we have that. So, and last time we proved the first step. So we first assume that not only there is an upper bound. We also have a lower bound. In that case, one can prove the gradient law as a bound. And it also depends on the lower bound. Step two. We can prove a holder by using this. By using this result and plus plus the Liouville for the degenerate equation. So we can prove that let's take a, take a ball. Then we can prove a bound like this. So, so this is the holder norm in a smaller ball. I didn't write as B1, but I just write B1A. It's about the same. So, so this by definition is the soup. This is the usual definition. So, so we will prove a holder, holder bound. So we define also, this is a usual definition. If we have a function, so we define in a radius this holder norm, holder quantity less than one. So we have a ball 0, 1, x will be here. So we define a quantity. Let's just, so this is for x. Well, if this, if this quantity is small, we don't, we don't really care what it is. So we, let's call it infinity. So what we care is if that quantity is bigger than one, we want to, if this holder norm is large, we want to measure, give a measurement like 1 over that. So, so we will take 1 minus x multiplied by that. So this, this, this quantity somehow is like, this quantity is like Vx alpha inverse is like, like gradient V of x. So it, this is for alpha less than one. So, so somehow for alpha equal to one, it plays a role like that. So when you rescale the holder analog of that. So then we prove this by contradiction. So suppose this is not true. Suppose holder is not true. Namely, so if not, if not, so if we do not have a holder control, then we know that, then that means for sequence of solutions, which has an upper bound, which has the upper bound and satisfy the equation. However, this thing will go to infinity, the holder. So that translates to the infimum of this will go to infinity because, yeah, this is the definition. So this will be going to infinity. So let's take a ball. They are on, say, ball of three-quarter. So we know somewhere here when there's a one-eighth, when x in here, so this inf will go to zero. We can, we can pick up a sequence. Well, we can take a maximum actually. We can just take a maximum point. So this is equal to one. Take a maximum. Maximum will not be on the boundary because on boundary it's zero, so maximum will be inside. So when x is inside my ace, so the maximum of this will go to infinity because the inf will go to zero. So this quantity will go to infinity. So x will be here, xi. So let's draw a ball of half the distance to the boundary. Let's call it sigma i. So let's say sigma i is half the distance to the boundary. So it may go to zero. It may go to zero. And let epsilon i is just log ui, xi and alpha. Let's take this. So we know that sigma i over epsilon i will go to infinity because, because of this. So this will go to, this will go to infinity because this quantity will be bigger than the quantity restricted to b1a. When restricting there, this will have a positive lower bound. And the inf goes to zero so the soup will go to infinity. So this goes to infinity. So this is twice of sigma i over epsilon i. So epsilon i over that will go to infinity. And we also know that epsilon i goes to zero because of this. This is epsilon i because of this. So if we restrict to, so left-hand side is twice of sigma i divided by epsilon i. Right-hand side is bigger than 1 half. And then I restrict this to the supremum of that ball. I can go to the supremum of that ball. I can, bigger than, let me see. This is bigger than. I want this to go to, what do I want? I want this. Sure, this goes to, goes to, so this, this is epsilon i. So this is small. So certainly this goes to, goes to the arrow. So that's very clear. So next I want to say that epsilon i will also be less than 4 log ui z alpha for every alpha in that, that ball. So when, when z is in here, so left-hand side is twice sigma i divided by epsilon i. So this is bigger than. So 2, I can take any z in that ball. So, but when I take, when I restrict z in that ball, then, then the distance to the boundary which is here will be bigger than sigma i. So it will be bigger than sigma i. So sigma i and sigma i cancels. Optional will be less than 4 times that. So this is less than 4 times this. So then, so then we can rescale v i of y equal to 1 over ui xi, enlarge this ball. So this little ball, so this little ball at the center is the maximum, almost a maximum. It's not exactly maximum, but it's like a maximum. And this ball I enlarge, so I will, I will get this. I will have that. So this enlarge to a ball of this size, bigger, bigger. And we know that log v i at zero is equal to zero because this is how we make it. So we look at log v i. So this is v i is there. At the center we choose to be the same value. And then we also know that if we do v i, we do r for 1, v error, it is actually equal to 1. It is scaled log, this absolute i is exactly scaled like this is equal to 1. So this quantity is chosen that way. So it's equal to 1. And also for every beta bigger than 1, for y less than beta, so this log v i, log v i r for 1, this quantity is less than 4. So this is chosen because this quantity is bigger than 1 quarter. A translate will be like that. So we select it. So instead of selecting the gradient, it's a maximum. We select this holder quantity to be the maximum. So it translates to this. So then this will say that the v i will be bounded above from below by a positive constant. Because log v i is zero at one point. And on every unit bore, its holder control is less than a constant. So when you go finite distance, so it's upper and lower bounded by a constant. So translate to v i certainly it means this. So then this fall into the case that we have upper and lower positive bounds for v i. So we also want to look at the equation satisfied by v i. Well the scaling goes the right way. This scaling goes the right way. This scaling goes the right way. And the equation satisfied will be... So this v i can be written as 1 over u i x i and epsilon i n minus 2 over 2 and multiplied by epsilon i n minus i over 2. So multiply this, multiply by that. It's just like a critical exponent equation. So this is invariance. So this thing should satisfy the same equation. It's an invariant. So now this quantity will be bigger than because u i has a lower bound. It's less than b. So this is bigger than 1 over b epsilon i n minus 2 over 2. So this goes to infinity. So then this equation will multiply by a large constant because of the homogeneity. And it will be a v i gamma i. So this gamma i will go to zero. It's a constant. This is actually this raised to minus 4 over minus 2 power. So this will be going to zero. And then our earlier estimates, the estimate first step we say if something is bounded below and above by a constant and satisfying the equation, actually that proof works exactly the same if this constant is less or equal than 1. So then we will have, in that case, we will have a gradient bound. So that means in any finite range, so the gradient will be bounded. So we know that by step one, it will say that gradient of v i or log v i is the same. So it will be bounded by a constant depending on that. So now we have v i satisfying this equation. So gradient v i is trying to control on any compact set. This will be bounded by that. So we know that passing through a subsequence. So we have a v i. There's a v i. It has the following property for any gamma less than 1 in particular for gamma bigger than that. So v i will converge to some v. The convergence will be c gamma log of r n because our bounded set we control the gradient. And the limit function will be in c zero one. So and when passed through subsequence, this equation can be passed through by in viscosity sense. So we are going to see f lambda a v actually is equal to zero. Namely, this a v now belongs to the boundary of the cone. So but in that case, we know the Liouville type theorem that says v should be equal to constant. So v should be equal to constant. And well, if v is equal to constant, v i goes to v in holder gamma bigger than alpha. But we know that this holder control at zero should be one. So this can be passed to limit to log v because our convergence is higher than alpha. But a constant would not have holder measurement one. It should be holder measurement zero. So that's a contradiction. Contradicting. So therefore we have proved this gradient estimate if we assume we have proved a holder estimate under only upper bound. So by using the Liouville theorem. So steps three is we complete the proof. So now we only under this condition. So we can look at u divided by u of zero now. So let's call it u hat. So u of zero because we know u of zero is less or equal to b. So therefore f of lambda a u hat where satisfied this b should be to the minus I think it's four over n minus two power. So this is f of lambda a u. This is how the constant goes. I shouldn't be using b. Oh yeah, yeah, b. So now we are. So b, yeah, so is equal to u zero. Sorry, u zero. This is u zero. So this number is less or equal to b to the is equal to this. This is positive because this is in the denominator. This is better. So this constant is having an upper bound. But when it has an upper bound, we know holder control. So by step two, step two says that this holder is under control in b2. So this will give holder control in b2. So holder control in b2 certainly implied Hanak. So u hat will therefore be bounded and above by some constant. So because it has Hanak. So we divide by one point. So it's less than something and bigger than something else in b2. So this is what step two gives. Step two gives Hanak. So now this already have both upper and lower bound. Then use step one will give control of log u hat, which is exactly log u. This is model constant. So this is the same. This log and the gradient log u hat is the same. So that completes the proof of that. So this completes the, completes this one. So one can also, one can then prove from here, from c1 estimate to c2 estimate, using again that Bernstein type arguments. So I will not present that here. That's another use of Bernstein type arguments. So what one wants to obtain here, which is not completed yet. So is on this manifold. So the equation to look at is to look at an equation, look at a conformal metric. So we change to a new metric and we want the equation equal to one. So this is the equation we wrote to get this gradient estimate. In the, when this metric is our end, we write f equal to one. So eventually what we really want is somehow we have such an equation on the manifold. And we want to prove that the solutions will be universally bounded. We want to prove this. So if you, we want to prove this. This c depends on this m and g and depends on this f. So we want to prove an a priori estimate. So whenever we can prove such an a priori estimates, we will be able to produce a solution. We will be able to produce a solution. So using our estimates show like gradient estimates and also the secondary estimates and so on. So using available estimates, we will be able to prove existence. If one can prove a priori estimates. So this, this, this is open when this f, which is the case elementary symmetric function for k greater or equal to two or less than n over two. So in this range, it is not known. If k is greater or equal to n over two, it is known. But of course, I should say that this is not conformally equivalent to standard sphere, wrong sphere. In that case, this estimate is not true because all solutions are classified and and the explicit and the conformal difumofism group is non-compact. So it's not true. So actually the only manifold with non-compact conformal difumofism group is standard SN. So namely, there is no obvious way to make, make, make you to lose upper bound. If it's not SN. So this is not known. So for SN existence is true. It's true. Existing through just all solutions are completely classified. Right. So but, but no upper bound in that case. So, so well one way of proving existence in the case k greater or equal to three and less than that is to prove an upper estimate. So this is what we describe some estimates now. So now, so what one would, what one, what one would say that. So if there is a sequence of solution such that the maximum of the sequence of solution goes to infinity, then satisfying the equation. Then once, one wants to, to say something about the sequence, say something about the sequence and eventually to say this is not possible. So, so one wants to analyze the behavior of this UI. One wants to analyze that then hoping that with more and more information one say it's not possible to have such a blow up. So before that one needs to get, get hold of this UI somehow. So what we, so, so in this case, in this case. And one, one simple question is whether this UI can only have finitely many blow up points on the manifold. So, so to go further it will be good if one can show that otherwise this behavior of UI is very complex. This is not easy to handle. And also even after proving say one has finitely many blow up points. One would like to say focusing on one blow up point whether or not it's a standard way of blowing up. It's just like a rescaling of, of a solid extremum. So whether one can prove that. So proving that will also help further analysis improving this is not possible. So one wants to analyze this behavior of UI as much as I can for to begin with. So, so for example, whether or not the H1 norm of this sequence is bounded. So, next I want to provide analysis of such a sequence again on RN. So, so I would, so it has not been known how to extend that to the manifold yet. Even though efforts are being made for that. So I want to describe the estimates for that in some detail for the Euclidean case only. So before giving the proof, so I want to describe the result on Euclidean space. So which is known so far. So on Euclidean space. So here is what is known so far. So, so then we take a sequence. So now we are in Euclidean domain now. So we take this equation in Euclidean domain. So in a ball of radius two, we take a ball of radius two and we have a sequence of equations. So suppose there is a blow up. So soup in unit ball goes to infinity. So then for any absent positive, we have the following description of the blow up. So solution leaves a ball of radius two. So then we know that we will have finitely points. So denoted by x1k and x2k maybe x3k. So we have a number of points and they are the number is so they are universally bounded away from each other. So they are universally distance away. This capital K depends only on this data. So and xik is a local maximum point. So we can draw universal radius of ball. So here center is the max. Center is the max. And all the max are comparable. So all the max are comparable. And inside each ball, inside each ball, this uk comparing to a standard function. And the standard function is the solving of extremum. So one half, this is less than epsilon u xik, and x inside the ball. This capital U is the standard solving function. So this is the capital U. Capital U is here. This is the solving of extremum. It's a fixed function. And those represent a shift of center. This is mu is the nitro scaling to this solving function. So it's a standard function there. So we know that this uk in c zero norm, it is being measured like that. And away from this ball, so we have all these balls, and away from this ball, we know the rate, this uk i of x, this centers. So away outside this little ball, away, they decay as one over the maximum rate. Which is exactly the rate of the standard bubble. So it's the rate of that. This is an estimate which can be achieved for any sequence of blow-up solutions of this equation, locally. And it would be very good if one can achieve something similar to that, or even at the beginning, even weaker than that. Or just extend this to remain manifold. So I want to describe a few ingredients in proving this. Well, the first ingredient is a Liouville type theorem. And actually there's an open problem here. So it says that if there is a sequence of solutions to this f, so first for this equation, if we write this equation, and if we write in the case this f and gamma is equal to sigma one, which is the trace, and gamma one, and this is lambda one, as lambda n. So in that case, this equation will be modular, harmless constant. It is this equation. Well, I think I can even write the constant in. It is exactly this equation. That equation is this equation. So for this equation, it is known that all solutions of that can be written down explicitly. So this is the theorem of Luiz. Later work extending this to the general fully nonlinear conforming embarrassment situation, and solutions are all given by that. And that's a Liouville type theorem. So here to prove the previously mentioned result, we need a strengthened version of that. So, namely, we have a sequence of solutions which lives in larger and larger balls, but we know that it goes to a continuous function. And then we would like to classify them as the same. That is an ingredient in proving that blow-up behavior. So, well, of course, if one has such a sequence of solutions living in larger balls, and the limit will satisfy the equation in viscous distance. So if we know how to classify all viscosity solutions, so then this result will just follow. But right now, somehow we don't know how to prove this thing. For even take V in C1 alpha, we still don't know how to classify these viscosity solutions. So this equation is elliptic, but not uniformly elliptic. Until when should my lectures stop? 5 minutes. So we have a solution in larger and larger balls. So we know that this Vk will go to V, let's say, C0 log of Rn. We know this V is super harmonic. We know V is super harmonic. So therefore V will be bigger than a fundamental solution power. Then, because in the proof, this Rk, as long as Rk goes to infinity, it's enough for the proof. So we can shrink this Rk appropriately to make Vk and V very, very small. So we can make it as small as we want, comparing to this radius. So in particular, we can insist that this Vk has this property. So this can be less or equal. Because Vk, on compact set, converges uniformly. So I can make that radius be smaller, but still maintaining going to infinity. So this is just take that ball smaller. You know, fixed ball, certainly this can be achieved. Then passing through a subsequence somehow. One can always achieve something like that. So then what we will prove is, so we have a ball here. We are going to take this X. So we are going to take this X. And we know that for very small ball with radius lambda, take a small ball, and we make a Kelvin transformation. So when we make a Kelvin transformation, this Kelvin transformation is written there. This one will be less or equal to V in the ball, the complement of the ball. So if lambda positive is small. So this smallness depends on K, depends on K and depends on X. So we take very small ball. When we make a Kelvin transformation, we are going to see an order outside. And that's not hard at all. That's quite straightforward. So this statement. And then we can enlarge this radius and to the maximum position, and we call it lambda K bar of X. We make it lambda K bar of X. So this lambda bar K of X, one can show that it has a lower bound, which is independent of K. And this statement uses the gradient estimates because this estimate is uniform. So we need this estimate. We have a lower bound, which is independent of K. And the upper bound, by definition, we just... So mainly the lower bound will follow from here. So that part, I did not write, spell out the details. Essentially, it doesn't really use much information. Just look at the function and do it carefully and you will see that. So then, I will quickly say two minutes about this. Next time I will start from here with more details. So why we'll then take limit of this lambda bar K of X. So you enlarge the sphere to the maximum position. You take the limf to see in the limit. And one can prove by maximum principle and half lemma, either this lambda bar is always infinity or lambda bar of X is always finite. So either or. If this lambda bar of X is infinity, then this V will have a property just from the definition that for any ball in the space, the Kelvin transformation will be less or equal than the function itself. That will simply say V is a constant. But constant is not a solution. So then lambda bar of X should be less than infinity for all X. And also, this means this process somehow stops. This lambda bar of X, it stops. So one can actually show that at infinity, this V has this property. They all have this property. At infinity, this is equal to alpha. And then that means we have arrived at the following situation. We have a super harmonic function in our N. And for every X in our N, there exists a positive number so that the Kelvin transformation is staying below. And at infinity, all this Kelvin transformation takes the same value. So in fact, such a function will have a specific form. It will actually have a specific form. So this property will give this specific form. Once we have this specific form, it's basically proved the classification. So I think I will stop today.