 We start with the first lecture by Professor Li. Thanks. So I will give some ingredients in the proof of this blow-up analysis. So the result says that if we have a solution in a ball of radius 2 and if the supremum in a unit ball goes to infinity. So we would like to have a good description of these solutions. So eventually we want to have such a description on many of those. So here we so far have it for in Euclidean domain. So what it says is there is a finite number of blow-up points universal distance away. And the heights near each blow-up points are all comparable by universal constants. And each blow-up profile is like a rescale of the sublif extremal function. And the closeness is measured in c0 close to this rescaling of the sublif extremal function. And so the first statement somehow, this proposition ensures that the blow-up profile is the same. It's the same b. And it is not known whether or not if a solution has weaker regularity to this intact. If it's c2, it is known. This problem, if solution is a classical solution, classification is known. If it's c1, alpha, it is not known. And so I will describe a proof of this proposition. So from which one can see also some of the arguments used somehow need to be so far to not work for solving this open problem. So to start a proof of that, so we first make some simplifications. So first, because our solution is actually superharmonic. So v is superharmonic. So because of the problem, so because of this cone structure. So it is always, we only deal with superharmonic functions. So that will imply by the maximum principle, the zero will be positive. And this is for all y greater or equal to 1. So this follows from maximum principle to compare with the harmonic function. This is a harmonic function on the boundary of the unit ball. If you take c0 less than the minimum of v, then we have this inequality. So then we can pass to subsequence. So we may assume without loss of generality by passing to subsequence and by shrinking this rk smaller and making this c0 smaller. So we may assume that vk of y minus v of y is very, very close to this v. And so this is bigger than. So we just make this rk smaller but still goes to infinity and maybe make this c0 a little bit smaller. So we will have this. We will have this. So then this is a ball of rk. And we take x sitting here. And we will see that. So we will see that when we make a Kelvin transformation, we take a ball as lambda small. And we take vk x lambda. We make a Kelvin transformation. We compare in two solutions outside. We are going to see that this is less or equal than v of y for y in v rk minus v lambda of x. So first, if lambda positive is small, this can be seen from the expression of this Kelvin transformation. A Kelvin transformation, a Kelvin transformation. So if you make lambda very, very small and outside here, one can see quite clearly it's going to be very small. So there's no problem. And here it is also OK because actually this inequality is holds in y minus x less than, for example, any number of a. If you want such an inequality to hold up to a, and even only if this r to the n minus 2 over 2, this vk, if you write in local coordinates, and it is monitored. It's the same thing. So this is just a rewriting of this inequality. So if you want this inequality to hold from 0 to here up to a, you just need to check the monotonously of this. This is just an equivalent statement. So this is certainly monotone because you have any positive c1 function. This will be positive when r is very small. So there's no problem with this to get started, this procedure. And then as usual, like in moving plane argument or moving sphere, here moving sphere just the same as moving plane, but using conformal invariance. So sometimes because there are more spheres in our n. So planes, if you look at the stereographic projection, it's on SN. So planes are spheres going through the North Pole. If you use freely all spheres in our n, you are actually using all spheres on SN. Sometimes give you advantage. And OK, so then we enlarged this radius lambda to the largest position. That's the lambda k of x. So lambda k of x, k bar of x. So this number is we keep this inequality enlarging the radius of the ball until you cannot do that anymore. So that defines a number. And then actually one can prove that this lambda k bar of x is bigger than 1 over c of x positive. And of course, this is we defined. We have an upper bound. So the lower bound is actually independent of k. So here you need to make some estimates on the gradient of this function, which we actually proved such gradient estimates. Next, so we have a sequence of solution vk. It goes to v on any compact subsets. So we define this the limit of this lambda k bar. We define it as a function of. So I actually can prove that this, so first for v, if this is finite, so we know because we have this order for vk, we can pass for any lambda less than this lambda bar, we can pass this inequality to the limit. So we know that, so whether or not it's infinity for any lambda radius less than that, we know that vx lambda, we have this order. Because we can pass this inequality and the ball goes to infinity. So v lives on the whole space. So it is actually for this. So before taking the limit, if we look at this vk, so here is something we cannot really apply strong maximum principle and half lambda to this v. So that's the main reason we cannot settle that open problem at the beginning. So some maximum principle argument on v, we could not perform directly. So we have to, we can only perform on vk. And somehow there is a limit process we are using. So here anyway, so we are able to prove by maximum principle argument and half lambda performed on this vk solution. So we know that either this is identically infinity or finite. So this we can prove. What do I mean by making maximum principle and half lambda argument? What I mean is like this. So if this is zero, this is a very large ball radius rk. This is a boundary. So if we have a function, something like vk, so then we take a ball somehow and we reflect it. So this is b, let's put it x, b lambda of x. So when we reflect, the function is going to be like this. If at the beginning, lambda very small, it goes like this. This is like vk x lambda y. So that's the function being reflected. You can see the formula from the beginning. If you reflect, it will go like this. Then you enlarge the ball. There are some possibilities which make this inequality can be violated, start to be violated. So what can happen clearly is, for example, when this function somehow goes like this, touch like this, and go like that. So then at some, for enlarged lambda, you hit it from below and touch inside. So next moment, it can go out. But this is not possible because for vk, it satisfies an elliptic equation. And this reflection also satisfies the same equation. So then when we take a difference, it's going to, the difference will satisfy a second order elliptic equation if we look at the difference. This is greater than 0 and equal to 0 at one point. Then one can have the strong maximum principle to say they have to be identically equal to 0. So then they will all the way stick together. So another possibility is near here. At some moments, this becomes tangent. So this becomes tangent. So here, become tangent. So then two solutions, greater or equal than, then we assume inside is still greater than the error. One greater than the other, only here become tangent. If in that case, then it violates the half lemma. Because inside positive, a half lemma says the derivative should be positive. So they should go with different derivative here. Cannot become tangent. So therefore, for this moving sphere process to stop, actually, unless you stick together, in that case, this statement is also true. At the moment when it stops at this lambda k bar, if it is finite, that means on the boundary, they hit together. So you always have a boundary touch. So you always, boundary will touch. So the statement is whenever this moving sphere procedure stops at this radius, then this function, when you reflect it, the reflected outside the piece will touch on the boundary on this circle, on this sphere. You will touch it. So that's the only possibility that this procedure can stop. So this says what you can make use of this stopping radius. So they touch. So then you can argue that after going to limit, then your v. So this case is easier to handle. Because in this case, last time we once proved that. So this case, meaning this v will have the property for any sphere. You make a reflection. The reflection will be below v. So namely, in that case, by definition, you take any ball and then you take v, x and lambda. This one will always be below this v of y outside. Because that's the meaning this lambda bar of x is infinity. Because lambda bar of x, meaning you will get for any lambda less than that, the reflection stays below. But this, we showed that this will imply, so for example, I take any plane in the space. So we can make a balls like that. And we can prove symmetry with respect to this plane. So therefore, this v is symmetric with respect to any plane. Therefore, v is constant. So in this case, it can be ruled out very quickly. So we only need to worry about this case. So this case occurs when this lambda k is always also finite. So that means for vk, you always have boundary touch on a ball of rk. So that you always carry a touching point. And from there, you can argue the limit function actually has this property. The limit function has the property that at infinity, so this is the Kelvin transformation value at infinity. That value will touch the value of v at infinity. So the touching of vk and its Kelvin transformation on the boundary of rk at a point will translate to this. So it will mean this v will make a Kelvin transformation with touch at infinity. So that means this function v has the following property. So v will be locally leptrist because we proved this gradient estimate. And then we know that this v has this property. It's super harmonic. And for every x in the space, there is a radius so that the Kelvin transformation cannot move anymore. So they stick together at infinity. So this means there's a radius outside this ball. The Kelvin transformation stays below. But the value of the reflection touches at infinity of this v. So in fact, this statement is enough to deduce that v should be of this form. So what this v does, so let's recall, we have a lemma. So the lemma says that if we take a little ball, so suppose we have a function which is super harmonic outside this little ball. And u can be, u is actually for this lemma, u is l1 is enough. So l1 minus 0. And if we have two functions, let's say c1 and n. So u of y is greater or equal than two functions. Then if u of y is equal to w1, 0, and w2, 0, let's say w1, w2 is differentiable as a point. Let's just say c1. So in that case, we must have gradient w1, 0, equal to gradient w2, 0. So namely, the picture when draws in one dimension should not happen. So one dimension, you would draw u like this, which is harmonic outside 0. And this would be w1. And this would be w2. So they can stick together there, but having different gradients. So as soon as you go one dimension up, this should not happen. So whenever you are able to make two functions touch there at one point, you know these two functions must have the same gradients there. It's quite, so the way to use it, which I have actually used this a number of times, which can give me some results, is to create many such w. So to make them stick at one point, then you will have information that all the gradients should be the same. And often, one can extract some useful information from there. So here, the situation we have created is, so it happens at infinity in this case. So here, these functions after, so this function is like, so v is like u and 0 is like infinity. So these family of functions stay below this function, but they touch, this is the value at infinity. They are the same. They touch the value of v at infinity. So that it is like we created a lot of w here and stayed below a function. And they touch at infinity, not 0 though. So then from here, from this lemma, and this v is superharmonic. And from this lemma, one knows that the gradient of this function, the value at infinity should be the same. So for every x, the gradient of this function at infinity will be the same vector. So there's a lot of information buried there. So the infinity is the same as 0. It's because this is a conformal invariant. You make a Kelvin transformation, then infinity becomes 0. It falls into this lemma. So this is the lemma. And we apply the lemma with, for w1, w2, we cook up many w. So let's call it wx. So this is the Kelvin transformation of this. So we turned. The Kelvin transformation is performed with the unit ball centered at 0, which turns infinity to the origin. And then we can apply this lemma. So that means this function, wx, for every x, this function, we take gradient at 0. It should be a fixed vector. It should be a fixed vector. So when calculated, it's easy to calculate the gradient of that function at 0. So when you calculate it, you see this function. You see that this is equal to a constant. This alpha is the value of v at infinity. So this is a linear function. This is v of x. This is a gradient v of x. This is a number. So this thing should be a constant. Should be a constant. Well, then you rewrite it. We are saying all this gradient should be 0. Now, gradient is 0. That means all this inside should be a constant. So this is a v to some power plus a quadratic polynomial should be 0. So then you can just write down what v is. So once you have this expression, and then it should satisfy the equation, so everything is very clear. So this v is being classified. So after this is being classified, so what we want to know is, so what we want to know is, if we have a sequence of solutions to this confirmed invariant equation, you know, bo of v2. And if in b1, so the supremum goes to infinity. We want to know what happened to that. Well, one situation would be, you have a maximum sub-haul value blow up like this. All the other is smaller than that, somehow like this picture. So then you analyze that first, and then put the picture together. So in that case, if that's the case, let's say this 0, something like this. So I would rescale our function by dividing it by uk of 0. And the right way to rescale is, so this will keep everything invariant. So this will be vk of. So this will make the solution will be in large bore. And the vk will be 0, will be 1. And because we are assuming we are in a position that all these other are slower height, so vk will actually be less than 1. So this will be a situation to analyze first. So then we want to know what happens in this situation. So in all the other situations, one can somewhat reduce to there. So when we look at the situation, it's this situation. So we have a vk, a sequence of solutions, which is in a bore with enlarging radius. So this will be the radius rk. So this will be rk goes to infinity. And the vk at 0 is 1, which is the maximum. And we want to know what happens to this solution, very large, enlarging bore, what can one say? Well, in that case, one can prove that. So you shrink this bore with delta prime. But that's a universal size. So you shrink that. Then you know that this vk is close to this u by an epsilon. This will be an epsilon. So first, we know that this vk, like so now we start with the picture. This is a bore of rk, bore picture of capital U. So this is vk. So that will go to u in some, for example. So it's lower. So it will converge to that u, capital U. But that doesn't mean one do not have far away. We have some crazy things. Because in the limit, they all go. So that one wants to rule out that kind of possibility. So first, one shows that the minimum on a bore of radius r will not be too high. Will not be too high. So the reason is that if it is very high, this standard solution, you have some property. So the standard solution, u is this solution. When we make a transformation for this solution, Kelvin transformation for this function, if lambda is less than 1, it's always stay below. If lambda is a bore, it's bigger than 1, it goes up. If lambda is 1, they become the same. So this is for the standard solution, it has these properties. So if this happens, then one can get a contradiction by doing this. One can then reflect this moving sphere. This procedure will be able to go all the way to lambda bigger than that. So this will be possible. So because when this reflection, we have CDRO control of vk to u inside. So we're making a Kelvin transformation outside. CDRO is under control. So that will rule out the boundary touching. So one can go all the way. So then after sending k to infinity, this moving sphere procedure will be performed for u all the way up to 1 plus epsilon square. So this violates the picture of this function capital U. So this shows that this vk, we want to prove that this vk is very close to u. So first, we prove that for any radius br, so the minimum on this boundary cannot be too high. At least there's one point, which is not very high. And then this vk can be bigger than this, 1 minus epsilon of that. This comes from the fact that this vk is super harmonic. So somehow one can compare using super harmonicity to compare this. So we have a lower bound. So to rule out this, so this comes from super harmonicity. To rule out that vk cannot go too high, so one can estimate the energy, which is small. So I think I'm running out of this. So one will have energy very small. So one will control that for fixed point on to outside, the energy is small. So for this equation, if one control the energy, then one can control L infinity norm. So it's like small energy imply regularity arguments. Because when small energy, if there's no bound, then when we scale that, one will lead to an entire solution. But there's a Liouville theorem, then the energy will be too big. So that's the kind of small energy imply regularity argument. And then this lemma, then one can show that this vk will be upper bounded by this u. So we first show that the minimum on each sphere will not be too high, compared to the standard profile. So this one shows the upper is also bounded by the sublif function profile. And so this argument, so this actually uses our local gradient estimates to achieve this. So next, one wants to show that everything will be within 1 plus 2 epsilon of the standard profile. So we know that the minimum cannot be, the minimum cannot be too high. We know the maximum cannot be, we know this cannot be like CUR. So this we proved. But it's possible that we want to rule out that the maximum is, say, we want to rule out the maximum is actually bigger than 1 plus 2 epsilon UR. So then there will be an oscillation somehow. So this is smaller than that, the maximum is higher. So on this ball of radius R on the boundary, one will get oscillation. So one wants to rule out that picture. So the picture to be ruled out is we have a ball very, very large Br. So here there is one point which is, let's call it somewhere here, let's call it zk and zk hat. So this function vk and zk hat, this is smaller than 1 plus epsilon U zk. But on the other one, we have a bigger than 1 plus 2 epsilon. So there's an oscillation on this sphere. So that's the picture to be ruled out. But in that case, we have an upper bound. So we will have estimates. So when we make a rescale, we call this vk hat defined like this. We know this is bounded, so we prove this bounded. So the vk is bounded. And it's satisfying equation, the right hand side will go to 0. So we have proved estimates for that. So then this vk hat will have a limit called v star. So this v star will be a solution which we discussed before. This v star will satisfy the degenerate equation we looked at before. The convergence is at least 0. So we have a convergence. And this picture will transform to vk hat. This surface will become b1 because we did a rescaling. So that means for vk hat on the unit ball boundary, you have one point which is bigger than 1 plus epsilon, while the other one will be smaller than that. So this is the vk hat will have this oscillation. But the limit is a radial function. So we prove that the limit function should be radial. So this should imply v hat star radial. But then this is a contradiction. It is not possible because after this, so then the rest, one can put things together using previous estimates. So I think I will stop here.