 This is the last lecture, and so today I'd like to talk about some aspect of regularity of this bracket flow. So as you saw yesterday, even though you have just this one inequality, in fact, you can do quite a lot of things with this, and surprisingly, in fact, and one of the surprises of this bracket flow is that this regularity theory that I'll tell you today. So let mu t be the bracket flow. This is totally local thing, so let me say that okay, dimensional bracket flow, you know, some open set. Okay, I kept talking about bracket flow in whole Rn, but it doesn't have to be, of course, on the whole Rn, as in the definition of 1.4 or 1.5 in some open set, the Rn, okay. And then the following is 0m, 1, okay. Is that, okay, so this is just any general bracket flow, okay, dimensional bracket flow, and there's a assumption I make here is assume that this mu t is in addition is unit density flow, is unit density flow. So that means just, I remind you what this means was that for almost all time, not all time, but almost all time, I say that the mu t is of this form hk of gamma k, gamma t, where, so meaning this multiplicity function I talked about is equal to 1, okay, almost everywhere in space time. So the meaning of this is really that even if you start out with, you know, this kind of multiplicity one situation, the surface, maybe at some point, maybe some piece of the surface may come together, which may not be likely, but you know, at least you cannot, it's very difficult to exclude this kind of situation when singularity happens. And afterwards, maybe you have these two things moving together. That's the sort of picture that I'm afraid of. And I'm assuming here that that's not, that, you know, you always, you have this situation. By the way, so existence of such flow is, can be proved at least for short time, for example, you saw this existence theorem I talked about, the second and third day. And for that, if I make some, some reasonable assumption about the density of the initial data, I know that the flow stays unit density for some short time at least. So this is not totally empty assumption. Okay, so the unit density is not, well, unit density, let's see. No, it's not. Unit density is not sufficient, but I have to, okay, so let me, just a side story is if I assume that there exists some R0, such that, let's see. So your initial configuration is less than or equal to strictly less than 2 minus nu, okay, and R to, let's see, k, omega k, so there exists some nu, okay, and with this, I suppose, so for all, okay, I'm not so sure this is sufficient. I think this should be sufficient. Maybe you can relax a bit more, but just to be safe on gamma zero. Then, okay, so this kind of condition guarantees a short time, short time unit density for the flow that I constructed, okay? So unit density is not enough actually, but you need to have some more, a bit more, you know, some room. And this, this can be proved by using a few schemes of monotonicity formula. In fact, it's, it's not so difficult to do that. Okay, so just, just as a side remark, you have this. Anyway, so I haven't finished the third statement. Okay, so what is the conclusion, all right, then conclusion is that the easy way of saying the conclusion is that then almost every time and almost everywhere this flow is actually regular, see infinity flow, okay? So, but the more technical term is that for almost all time, time, and there exists a closed set, topologically closed set. GT, which is in R, which is in U, such that this GT has major zero with respect to surface measure, okay? So that's closed set, but major zero set. And such that outside of this set, okay? Outside of this set, there exists neighborhood, okay? Neighborhood of X and P, X, T in the space time, not just space, but space time, neighborhood, okay? Such that the, basically, this mu T is smooth. See infinity, mean covers the flow in this open neighborhood. I hope that this makes sense, yeah? Okay, so this GT is more like a singular set, okay? So there may be some singular time, but that's major zero. And outside of that singular time, there exists some possibly a singular set, but with respect to K-dimensional, measure is zero. And it's important that it's closed, it's closed and major zero. And outside, you take a point outside, then there's open neighborhood in space time. I hope it's clear, yeah? It's not just space, but in space time, open set. Such that this guy, well, maybe I should say support of mu T. More precisely, support of mu T is actually see infinity, mean covers the flow, very classical. Classical mean covers the flow that we've been seeing. So even though this definition is extremely sweet looking with inequality, you do still have this nice regularity. So it's almost everywhere at least, nice, smooth, okay? So it looks like kind of abstract object, but it's actually not so abstract. It's just that you have this small singularity with respect to, which is small, with respect to surface measure. And that's actually somewhat surprising. When I first heard about this, maybe more than 20 years ago, this is written actually in Brackett's famous book from 78. But it was very difficult to understand the content. And I believe that even the specialist had difficulty understanding. And this one, so it's usually called Brackett's partial regularity theorem. Partial regularity theorem, so that's from 78. But I must say that there is some original proof of this Brackett's partial regularity theorem has some serious gap in the proof, which I pointed out in my paper. And the complete proof is given by us as my student Kasai and myself. That's, let's calculate some variation, variation PDE. That is, this is actually, this is C1 alpha, regularity theory. And then I extended to C2 alpha advancing character variation. This is C2 alpha. And once you get to this, then you get to C infinity. So anyway, so these two papers really gave a clear, complete proof on this. Okay, so I don't have time to explain this whole proof. But I want to point out a few elements of this regularity theory. And, but, okay, so now, let's see, what was the things I want to say. Right, and now, some of the, the one thing that is a key of this, getting this regularity theory is, well, first you have to go from local to global. But let's see, well, one key idea maybe just for some people who are not so familiar with this type of things is that, the thing is that, one thing you can arrange is that for almost all time, let me do, to be specific, I have a subindex T0 and HK almost everywhere X0. Okay, X0. The things you can say first is that if this measure, this bracket flow is, is parabolicly magnified. Parabolicly magnified. That is, you kind of look at the point and then you kind of blow up like magnified. Parabolicly means, as you know, it's T squared X, well, you know what that is, so parabolicly magnified. Then, in fact, what you can conclude is, you're going to conclude this. It is, in fact, close to K dimensional plane. At least with respect to, well, in the sense of measure. So that's the first one observation which must be sort of familiar to you, if you have seen this kind of thing. The thing is, yeah, for almost all time and almost everywhere, if you magnify, this object is not regular at all, but still almost everywhere and almost all time, you know it's close to some K dimensional plane. So that, that can be arranged. So the key issue here is the following. So when your flow, this bracket flow is very weak sense in the sense of measure, close to a K plane, then you want to conclude that then this guy has to be regular. So I'm not looking at it anywhere, but I'm looking at the place where, at least in a weak sense, very close to K dimensional plane. Then you want to say it's regular. It's more like you're looking, you're trying to kind of do some kind of linear, linearization, if you're looking at something which is close to plane. And it's not, the regularity proof is not by linearization, but the idea is, if you're looking at things which is close to plane, then you expect that maybe the flow may be regular, okay? So the key, yes? I guess so, yes. The one that, yes, maybe you're right, yeah. I should probably say that it's close to K plus one dimensional plane in space time, yes. So the measure is supposed to be close to that. So it's kind of there in time. But it's going in time, yeah, right, right. The static, yes, static K dimensional plane, right? So the key is the, so the key is the so-called regularity theorem, which brings the bell to people who are specialized in this regularity theorem, okay, regularity theorem. So let's, let me see. So let me, just to give you the flavor of this theorem, just to make it simple, I don't try to write the statement in the strongest way, okay? So I'm using, because if I do that, then it gets kind of messy. So let's say that I do this notation. If black flow, mu t is, so just this with notation, because this is some concepts I just introduced just for this lecture. So almost that, okay. Br x0 cross, this is time variable, t0 minus r square, t0 plus r square. Okay, so this is space and time, right? With error y, if the following is true, if the following is called, okay? So what is the following? So this first one is, a is the, there exists a K dimensional plane, a through, which goes through x0, right? Such that the support of mu t inside of this ball is within the, the, the epsilon distance from this K dimensional plane, okay? So to be precise, x such that distance from x to a is less than, okay. So now, not epsilon, but to properly scale, I say epsilon times r, okay? Just, is that the right scaling? Or all t of minus t, t0 minus r square plus r square, okay? That's first condition and b is the other one. So this means, you know, this, this, this variable is staying in a, so the epsilon neighborhood of this K dimension plane for the whole time. And the next one is the closeness of the measure. So I assume that mu of t of b r x0 divided by a K dimensional measure of a unit disk minus one is less than epsilon for also, for this relevant time. And just to be precise, can I hope this makes sense? I think it's not so difficult to understand this. It's just, you know, close. This support is lying in this plane, nearby plane. And this is saying that if you compare the measure of this moving surface versus the, you know, unit disk, K dimension unit disk, the difference is small, okay, for all time. So this really is claiming sort of, well, this is a subnorm bound, right? So it's a bit, looks kind of strong, but actually it's not so because of the proposition I presented in the previous lecture. In fact, some of the weak closeness does imply actually the L infinity closeness, in fact. So this one's not so strong. So, but I state in this way for this, for the simplicity. This one is really closeness by measure, right? I mean, you are comparing the measure, your measure in two with the disk. Okay, so these are the definition of almost flat, right? Okay, so then the conclusion is, if it's like that, then it's regular, basically, yeah, it's not, right? Okay, so the theorem is, theorem is that this is, as I said, this is cosine t and t equals 14. Is that supposed, oh, no, sorry, let's see. Okay, maybe I should state this way, okay, so suppose we have a mu t. It's a local result, but just to be simple, I think I'll just state it in the whole entire space. It's a unit density k-dimensional graph at all in Rn. It doesn't have to be Rn actually, but just let me write this way. There's some, certain relevant constant which I don't want to deal with, so. So if you just fix delta, then there exists some coefficient, zero, which depends on dimension and how far you are away from t equals zero. And also, let's say initial measure, total measure, okay? Which is small, okay? Which is actually small, this equals zero, with the following property, okay? So this is totally independent of the flow itself, except that, well, you're assuming that this is, this, it may depend on the size, as well as how far you are away from time zero, okay? Right, and then the conclusion is that whenever this flow is almost flat, then it's regular, okay? So let's suppose mu t is almost flat. Almost flat in pr x0 cross t0 minus r square t0 plus r square. Which is in the, away from zero, with error, error, y0, then the support of mu t in this smaller ball. So this is more like interior regularity result. So I need to go in a bit inside, t0 minus r square t0 plus r square is c infinity, okay? So as I said, if it's a unit density flow, almost everywhere and almost all time, you can arrange so that your flow is satisfying this condition, okay? And then so you have this partial regularity. And I notice that, of course, this almost flatness condition is not satisfied around, for example, junction point. Your junction point is not going to satisfy this almost flatness condition. So I'm not claiming any regularity at the junction point, for example, okay? So away from junction point is what we are looking at, okay? And also it's important that we have an estimate also, okay? So now moreover, because this is more, having an estimate is sometimes more very important. Let's look at this cross. In this, in here, the support of mu t is represented as a graph, as a graph of function. Graph, graph, graph over this plane A that I talked about there. Over A, okay? So it's not just, I'm claiming it's smooth, but also there's an estimate as a graph, okay? And just for the notational convenience, let us do a change of variable. Do a change of variable and also appropriate also one of rotation. Let us change the variable, okay? So that just to write this as a graph, it's a bit messy. So that this plane A is just rk cross 0, okay? 0, m minus k, and x0 is 0, t0 is 0. And also by doing a stretching by factor of r, I can assume that r to be equal to 1, okay? So I am just, you know, you can arrange so that this is true, okay? Just can't, this A plane is moved back to the origin and then rotate, just stretch, okay? I get you there. And then this support can be represented as a graph over this rk, and so then there exists a f, which is from k dimension ball of radius one-half, and time is minus one-half to one-half to rn minus k, okay? So that's a function which is going to represent the moving surface, such that this in curvature flow is just nothing but just this graph, okay? So the support of mu t, well, this, I have to look at the k-dimensional ball cross cylinder and intersecting with say b1, okay? So that's, is equal to just a graph, okay? So this is x1, xk, f1, x1, xk, and so forth, fn minus k, x1, xk. So it's really a graph. So x1 is where this x1, xk is in the ball radius one-half, okay? So that's, that really is a precise language that says, you know, this support is really presented as a graph, okay, of f, and also you have estimate, okay? And also the CL norm, let's say, just to be very casual in this, so this usual else or the derivative in space and time, okay? This is bounded by constant depending on L, n, k, and so forth, and maybe some other constant that I wrote there, let's just skip that. And times the, okay, here is, instead of defining precisely, I just say L infinity of, of height of the support of mu t in this ball, say. So this is like L infinity, you know, or you can say, well, in this case, maybe y is the one that we have there as a height, but, and you can actually replace this, if you, you can replace this by something weaker, something I'd say minus one to one dt of e1 of distance function from x to a square, yeah, maybe I should, something like this, okay? Or, maybe I should say, or, let's sign code. So this is some quantity that is, if rocket flow is close to displaying k, this is something that you, you know, it's going to be small, okay? It's, there's some argument, but I can't be there, but you can say that, okay? See, I think that's it. Any question at this point? I hope that this makes sense, okay? And note that this is more like the property of a heated question, if you know what it is. In a heated question, you can, you know, you have the estimate of all derivatives in terms of some weeks norm like such as L2 of function itself, right? Actually heated question, it's better, but actually here it's, it's not as good, but you can make it better than this, in fact, which, yes, e1 alpha, yes, right, right, yeah, I was going to tell you this, but okay, maybe, well, actually, maybe I can, yeah, okay, no, just, I have this note comment, yeah. So first you do c1 alpha, and then in fact, to get to c2 alpha, c1 alpha still doesn't give you the equality, you see, you see, black is, formulation is by inequality. So even if you have c1 alpha, you still don't have equality yet. So now, so that means you don't have PDE yet, even in a week form, right? Well, actually, it's not that, in fact, that's, that's not the case, in fact, but yes, you have to get, once you get c1 alpha, you actually have to do another sudden kind of broad argument, and which involve, in fact, very interestingly new monotonicity type formula, which is available only for main coverage of flow, in fact. So this is something that I actually found just by chance, and which I don't have time to talk about this, but it's in the paper of this second one, if you're interested. It's actually very interesting, somehow. You use, okay, so maybe I should say a bit more. So this monotonicity formula is the one that you make a measure, you look at the distance from the paraboloid, which is a solution of the heat equation. So you have some kind of second order approximation. So the way you do it, you look at the solution of the heat equation, you look at the distance from that, and then you show that, that distance function squared times the, this backward heat kernel is monotone decreasing. In fact, that's, then it's a, on not only that, you have extra term, which give you some kind of differential energy control of the graph. And then, that's a key estimate, and using that, you get c2 alpha. And then, once you get that, you prove that the velocity is, in fact, equal to mean coverage of vector. Yeah, then once you have that, then that's classical after that, yeah. So, okay, that's clear. Okay, so let's see. So for the rest of the time, I would like to give you some kind of indication of idea of the proof. Okay, so maybe I should first point out the following. So, for the people who've never seen this kind of thing, I just want to, but with some background in PD, maybe, in the case of a minimal surface, if, okay, so just let's consider the, as a comparison, the static case. Okay, so that is the case where mean curvature is not moving. So that means it's a minimal surface. Okay, what do you do in the case of a minimal surface? And, by the way, this regularity theory corresponds to so-called other regularity theorem if it's a time independent, okay? So, my proof should reduce to another regularity proof, okay, basically. But, okay, so now, recall that in the case of a minimal surface, that is, in the case that this mean curvature, okay, is equal to zero, okay? So that is really a time independent situation. You know, this velocity was mean curvature, so it's time independent. Now, what is the equation in this case? So, equation is this one equation, okay? So, let's say mu is equal to hk, okay? So, in this case, the equation we have is this, as we, as I explained this. Just let me, like, this way. Is equal to minus of, use this many times, h, but now this is equal to zero. So, this is true for any vector field, okay? So that's equal to, that's the equation that you, you know, you have. And if you think about the sort of linearization of this very rough language, but just to give you idea, linearizing, assuming that everything is nice and smooth, okay? The linearized equation of this is roughly speaking when, when, when say, when you're looking at a problem in Rn plus 1 and k is equal to n, okay? So just, just to, just to make it simple, okay? Just a hypersurface case, the linearization of this equation is going to be just this one, basically, okay? So, if, if you think this gamma is a graph, is a graph of function f, okay? So, assuming that this, this guy is represented as a single, as a scalar function f, then the linearizing of this is going to be this, so this, this equation, okay? This is equal to 0 for any test function, okay? So that's a very rough language, but it is in some sense true that this, this equation reduces to this relationship, which is really, you know, weak form of f being Laplacian, you know, or harmonic function, right? That's, if you linearize this, this is what you have. So, from this, well, well, you know it's, it's harmonic function, nice and smooth, but the point is, okay, now if you have this, what you can do is you, you use a suitable test function to show that, well, in this case, you multiply, you, you use phi to be, phi to be f times another test function or something, and then you do integration by parts, and you get this, this equation is less than equal to some constant times. Well, let's say maybe just this casual, something like this, right? That is the gradient, this energy is bounded by cell 2, no? And so this is something so called catch-operative type equation, inequality, okay? So this linear equation gives you this, but also this nonlinear equation also gives something similar. I don't have time to talk about this, but you get something like this, okay, in principle, in a, sort of, in the language of countable rate percent, okay? So, now, using this type of things, well, that's, that's a sort of starting point of the regularity theory, even for other regularity theory, okay? Now, what we have is quite different situation, right, for this bracket flow. I mean, we have this inequality, which is, you can, which you can use only for positive test function, even. It's not even any test function, but only for positive test function, okay? So, how, what, what is the sort of procedure that, that you can think of? So, now, let's look at equation and see what is the structure of this. So the proof is actually very different from elliptic case, MSA. It's, it's very different. In often, parabolic problem, sometimes it's very similar to elliptic problem, right? Sometimes you just, it's, it's probably not, it's a bit too much to say that it's just modification, right? But, but sometimes it's like that, right? You do elliptic and then you can do parabolic in a similar way. But here is extremely different, I'd say, very different proof. An idea in both is also very different. Okay, so remember, remember, the, the blackest inequality is, is this guy, this one, which are, in a, in a, in a distribution sense. And let's just consider the test function, which is just a space dependent, okay? Just for simplicity. But remember, this is, has to be positive, no negative function. Is less than or equal to nabura phi dot h mu minus phi times h mu square d mu t. That, that was the inequality. And remember that, that a priori bound we have is, this mean curvature is bounded in spacetime in L2, okay? Not, it's, it's, it's a, this guy is just a bounded in L2 in spacetime integration. That's, I guess that's about it. We don't have much. Now, now, so let's, let's go to the idea. So now, okay, again, just consider the situation of hypersurface case. Just let k go to n. And presumably, we are looking at the place where things is supposed to be flat, right? So in that case, co-dimension, one case, this quantity, let's see. Let, let phi to be, okay, also. Let phi x to be more or less like a characteristic function, of this set, rn, okay, rn plus 1, and x1 square. So this is, assume that this, this guy is a cylinder. I guess I can use that, yeah, okay. I take this function, which is like a characteristic function of cylinder, okay? Or radius 1. So xn plus 1, I'm looking at cylinder. And this function is like really a characteristic function. So, and also, I think I might as well take a homogeneously, homogeneous one that's independent in this direction. Phi is, Phi is really, Phi is really depending only on n, okay? So, no dependence on xn plus 1. So it's like 1 and 0, okay? That's a, that's a sort of the image of Phi. If you like, you can think this is more like a characteristic function itself. But since I can flagging a characteristic function here, so I'm just doing a little bit of smoothing, okay? So I just think that this is a smoothed out characteristic function. Then, what is this? This function, this, this right hand side is just, you know, if you just think this is more like a characteristic function of the cylinder, this is nothing, of course, by this surface graph. I mean, this is after all the measure of the surface, so that's what you have in the cylinder. Okay, let's call the cylinder c or something, okay? Inside, yeah, okay. Okay, so this, or this, this one. If there's any hole or anything, that's what you have. And so that means that if, if everything is very nice, what you're looking at is really this surface area, okay? Now I might, it's, it's, I can just subtract constant, okay? I can do this because just, it's a constant, okay? So it's sitting there, right this way, okay? I hope this makes sense. Okay, I can suddenly take one inside. And if, if, if all, you know, everything is fine, then this, and if you, your surface is a nice sky, then you can do a theta expansion. And this comes out to be about one-half number of f square. You know, if, if everything is nice, I mean, this is really nice, if you can do this, but it can, of course, but just to give you an idea, yeah? So, so this is roughly, you're looking at the change of the judicial energy, in some sense, if everything is nice, okay? So this left-hand side is really like a change of the judicial energy. The gradient square, you know, L2 is the one that you're looking at. And this part, well, just very roughly speaking, this is mean curvature. So if, if it's a mean curvature, and if you linearize it, this comes out to be about laplacian of f, right, of square, okay? So you're dumping by this term, this laplacian of f square. You're trying to reduce this judicial energy by this laplacian square. Well, this is, you have this term, but just for a little bit. This is the, driving the energy down, okay, in time. So that, this, you see this kind of structure in this? Okay, so, that you see, okay, so now I want to explain a bit more technical key point, okay, and just to explain this proposition, which may be, technical looking, but geometrically sort of very interesting. This is, I make all this geometric software of inequality. So that's, it's in my book, it's in my paper. But, okay, so, let's see. So again, let me, well, let me choose a different data. Eighta, one, okay, and B, smooth approximation. Okay, so this is, I do general dimension then. Approximation of characteristic function of the k-dimensional cylinder. Okay, so this is x square, x1 square plus xk square is less than one. Okay, so this is k-dimensional sphere. I mean, you can think of hypersurface case. They're not just n in Rn plus one, but let's do this. And let omega k to that to be the, say, t. Okay, I'll explain what this is. Eighta square, hkx, where t is just rk cross zero, okay? Well, I'll explain this now. Okay, so the picture is what we saw here in case of hypersurface k or n case. But, not that this is, you know, if this is really characteristic function, not just approximation, this is going to be precisely omega sub k. It's really, you know, integrating this t, t is really this k-dimensional plane. So, you know, since this, if this is characteristic function, it's really exactly omega k. But, you know, it's slightly different. So, I just mentioned that this number is almost omega k. Okay, so these are fixed things, you know? Eighta and omega k to that is fixed. Now, the things I'm going to write has nothing to do with time. Okay, so this is time-independence statement. Now, let mu to be a k-dimensional house of measure restricted to gamma, where gamma is the one that I'm going to talk about. Well, you can think this as nice smooth surface if you like. Actually, even for smooth surface, this guy's, what I'm going to talk about is non-trivial. So, maybe you can think this as smooth surface if you like, okay? Now, this, right, and then I define alpha square to be the mean curvature square times eta square mu. So, alpha is, okay, maybe I should, okay? Alpha is this L2 mean curvature square, L2 mean curvature, with eta multiplication. And let's see to be also these things. This is a distance function of x to t squared, and this is, let's see. Okay, so x1k square, so this is again a cylinder, but twice as big. I want to take a little bit bigger one, maybe take one-half, okay? And also assume, I need a little bit of control of this density ratio. We are x over rk. This is for all. Okay, let's call this as a cylinder c, maybe c2, okay, sorry. Or this is a cylinder of radius 2, so let's see. We are x is in c2. This is bounded by say e1, okay? So, this is a bit technical point, but I need some kind of density ratio. By the way, Brachy fluid satisfies this type of, you know, bound from yesterday's talk. So, this is not a problem for us. Now, then the statement is following. So then, there exists three constants, c1, alpha1, and c1, which as follows. These are the constants which depend, okay, maybe I should, which depend, which only depend on dimension and e1, okay, and with the following, okay? So, what is the statement? So, there are two scenarios, two case. Okay, so that if, okay, so after I write, I will explain this, but eta square d mu, let's see, I think, minus omega k. If this is less than omega k to the, okay, so these numbers are written so that I exaggerate the estimate, but it's, of course, doesn't have to be 100. Some number, okay, and alpha is less than alpha1, then the following holds, okay? So then, in fact, what we have is this omega square d mu minus omega k. This is bounded by the two case, c1 times alpha k, 2k over k minus 2. So these are not, so this exponent I will explain in a moment, but let me write this, c square. If k is bigger than 3, and if k is 1 or 2, you have some of the better c1, alpha2 over 3, c1 half plus c square. Okay, so that means you don't have this term, lacking this term. And the b is, okay, let's see. If this is not true, if this is in the range of, and let's see, this is another exaggeration, but 99 k to the, okay, in this range, then the conclusion is alpha has a lower bound. Okay, so that's a statement. Okay, so this requires a bit of explanation. So this is really an estimate, which has nothing to do with time, okay? But this is very essential to get the estimate or bracket flow. So this is saying the following. So you see, these constants are sort of independent of surface at all, just depend on the dimension, and e1, that's this one. Now, the claim is the following. So if this case is 1, this is the one that this mu, which is the surface measure, this is the case where your surface is very close to a disc. Okay, because you see, if this mu happened to be just a plane, t itself, then this is 0, of course. Okay, you see, remember this definition, you know, this definition, if this is t, that's 0, okay? So that means this a is a situation where the surface measure is very close to, you know, flat plane. Then also, I'm assuming that L2 mean curvature is small, okay? So curvature, L2 is small. So these two assumptions, actually the difference is, in fact, not just this, but actually you get this estimate, that is, you have a mean curvature raised by this far, and actually these are the lower order term in some sense. So the main term is this guy. Okay, so this is the most important term. Okay, maybe I just skip this part because they are like something that's not so important after all. Well, actually, this one's sort of important, but this is, you can interpret, but okay, so maybe I should say these are important, okay? And so this is a dimension bigger than three k's, okay? And if you don't have the, oh, three should be three also. Three is included here, so, yeah, sorry. So then in low dimension case, you don't have this term, okay? So the main term is going to be this one, okay? Okay, so you have sort of sovereign space like, you know, exponent, right? But it's not, I say it's geometric just because it's not a linear dependence. You see, as I told you, this is like deletion energy if everything is linearized. But this one, you know, this is not, it's not, if it's a linear theory, you should have this bounded by something like this, right? I mean, something which is low that is bounded by a higher order term. But you see, you have this power, so in fact, you're raising this by basically k minus two, okay? So in that sense, it's kind of geometric. It's not like, comes from linear theory, okay? And the second case is also, I hope this makes sense, this is a case where your measure is a bit away from this. You see, this is, you see, you're not this case, but you're away. You know, this guy is bigger than this number, but less than this number, okay? So less than this number means it's not close to two omega t tilde, okay? Not close to that or not close to zero, okay? You know, if this is zero, this is violated. If it's two omega k tilde, it's violated. So you are in between. So that means, maybe just to be more specific, you have this way. Here's omega k tilde, which is, which is the case where you have a flat plane, okay? And you have a, if you have two disks, you have two k tilde. But if there's nothing, it's zero. So this case B is the one that you are away from this one disk, but you are also away from this zero. So you are here. That makes sense, yeah? Yes? No, no. I think, so meaning, yeah, this, I think, well, maybe you should believe this picture more. Yeah, you are in this range. That's what it is. Yeah. Okay. Sorry, am I missing something? I wonder. I think this should be right. I'm not so sure. Okay. Yeah. Okay. Please take this picture as the right statement, if you like. Okay. You are, that's the one that you're excluding. Yeah. Yeah. That's right. Yeah. Yeah. In extreme case, suppose this mu is just flat two plane extremely close to each other. You know, then this side is going to be all zero, right? So, yeah, you don't want that. So this is a condition I'm excluding two flat plane or empty set. Okay. Okay. So the conclusion is you have this lower bound for L2 mean curvature. This means that either you have some kind of hole, maybe, you know, to have these kind of conditions, then mean coverage cannot be extremely small. Okay. There's some low bound, which is, you know, independent of the surface. Okay. So there's these two scenarios in this range. And the proof of this is fairly actually tricky. You have to, in fact, that there's a similar statement in, well, for specialist, I just mentioned that there's a, this estimate analogous to this in alert paper. But alert paper only gives actually square here, in fact, instead of this power. And having this power is, in fact, turned out to be essential in the argument that I'm going to talk about in a moment. So you have to do this very carefully, some very careful covering argument and some estimate called a cylindrical gross estimate. And also you have to do some kind of Lipschitz approximation at this level already to get this. This one, I feel like I'm written a very careful proof for this in my paper. So if you're interested, please take a look. Okay. So why do we have this? Let's be clear in a moment. So this has to do with the estimate that we want. Okay. So now you just write E of t to be, now I back to the time dependent situation where a square d mu t, that's a bracket flow and minus omega k tilde. Okay. And now I think this as a time dependent quantity, which is moving in time. Right. And then this proposition 4.1 gives you the, okay, so it is not a precise statement, but let me just say very roughly that this gives, just to give you the idea. Now d dt of E t is less than or equal to minus minimum of alpha 1 and E t of k minus 2 over k. Okay. So I think maybe I should restrict myself to k bigger than 3k. The law dimension case is easier to deal with. I think c t is the time dependent c now. I think c t is time dependent t. Okay. Okay. So why we have this? You see the thing is, you see the bracket inequality tells you that if you use 8 square as a test function, bracket inequality number 6 tells you that d dt of 8 square d mu t is less than or equal to alpha and number 8 dot h minus h, 8 square, 8 square. Now note here is alpha here. This guy is exactly the, here's alpha, you know, maybe I should say now alpha t. It's time dependent quantity. So this one is some term, first, this term which is the first term, okay, this minus alpha t square. Okay. So now as I told you, you know, this is just a constant. So this left-hand side is nothing but just this is d dt of e t. This is d dt of e t. Now forget this one just for simplicity. Now this is saying that this is less than or equal to alpha t. Okay. Now we are close to the, we are close to the, our measure is sort of close to one, okay. So you are either this guy or this guy. Either, you know, one of the, one of the others, right. Now if you are a bit away, note that you have a low bound of the mean curvature. So that means, you see, this is going to be minus constant. So it just keeps reducing the area at constant speed, right, if you are away. Now if you are not, so now after a while, okay, so suppose at t equal to zero, your, your surface area was, was in this range. But after a while you know for sure that the measure is going to be less than, you know, say, you are going to enter, you are going to enter this range. Okay. Now once you are here, now note that e, here's e t is going to be bounded by this quantity, but this means roughly speaking, if you invert or take a, this means e t is, you know, this, if you raise this by, let's see, k, k minus. k minus two is alpha. I mean, if you forget about all the, all the other terms just for simplicity. But you see, you have a, there you have minus sign, so it should be like this, right. Okay. Maybe I just have an absolute value or something. Okay. In some sense you get this, this term. And now, so, and note that this is exponent is smaller than one, so if you solve the all the e, in finite time, this e has to be equal to zero. Okay. In finite, the fact that this is bigger than, this is less than one is extremely important. If it's not, if it's one, it takes, you know, even a long time. But since this is less than one, strictly, it becomes zero, but actually it doesn't become zero because of this error term. Okay. So it doesn't, but the point is you can bound this e in terms of this c in some finite time for sure. And actually this argument can do also, you can do this argument backward in time too. I don't have time to talk about it. But the point is with this kind of all the argument, you get the following type of estimate. That is dt in the intermediate time where t is let's say minus one half and one half. This is bounded by some constant times of ct where t is say minus one to one. Okay. Something like this. I mean conceptually, not exactly, but conceptually by this all the argument. And then note that this, as I told you, is more like a deletion energy. And this is more like L2 energy. L2, you know, L2 norm. So you're bounding the, you know, soup norm of the deletion energy in terms of the, well, I guess this is not L2 exactly, but you see this is like L2 norm. You have to take a soup over time, but again you can actually bound this by something weaker. Okay. But anyway, you see analog, right? Because you see this is more like couchopoly type of inequality at this point. You're bounding the deletion energy in terms of L2 norm, right? So that's exactly like the couchopoly. And once you have this, the deviation of this surface is roughly corresponding to the mean curvature square too. Because the change of this is basically controlled by the mean curvature square. So this kind of quantity now is controlled by L2. Okay. So you are sort of in business now. You can do a bunch of this type of estimate. Then you can actually do this blow up argument of the deletion energy. And there are many technicality which is difficult. Well, one somewhat peculiar difference from elliptic cases. Well, this is one. But the other point is when you do a blow up argument, you need to know precisely the height of your support of the muti. And there you do need the super norm bound of somehow the height of the support. Okay. Maybe I should just say you need the, what was the number? The last estimate I had in the previous lecture, I told you that you need to have a sharp linear dependence on the, I guess, 3.2. You need 3.2 to do a blow up argument. Which was not a case for other case. I said, other regularity theory, when you do blow up argument, you don't really need to have a sharp sup bound at all. Actually, you don't use it at all. But for here, because of the fact that we have a test function which needs to have a sign. You know, it has to be positive. You have to be very careful. And you need to have sup bound in terms of the blow up argument. Okay. So that's just a very brief idea. And as I told you, C to alpha involves some other new estimate, but I don't have time. Now, just to conclude, I just mentioned that, so this gives you some kind of at least generic regularity of this bracket flow. But the other things which is known in this kind of framework, I mean, strictly speaking, in this framework, is we know a little bit about 1D triple junction regularity. That is, if your flow is close to one-dimensional triple junction in a weak sense of measure, then you have certain, we have some regularity theory for that. But that's about all we know at this point. And so, and this is, of course, assuming the unit density as well. But if you don't have these unit density assumptions, even the static case or stationary case, it's known it's very, very difficult to do anything. But maybe if you try to do bracket flow with high amount of pressure, maybe there's some hint. Actually, maybe if you do something more general, maybe you're lucky. I don't know. That's possible. And I hope that some of the estimates that we gained during this estimate, during this result, may be useful for other geometric analysis programs. So with this, I'd like to conclude my lecture. Thank you.