 Come back to our third lecture on the area minimizing graphs. So now for this lecture, first of all, quickly, I will prove the Georgian theorem on the C1 alpha regularity of area minimizing graphs in the interior when they have small excess. OK, so recall our hypothesis. So we have a function u, which is going from the unit disk into our n. And graph of u is area minimizing. OK, and then we know that the excess, which is what we call in the following way. So the excess of the graph of u on the cylinder of radius 1 centered at 0 with respect to the horizontal plane is very small, so it's less than, say, some epsilon. So now fix a point p, which is equal to some x u of x. And this is inside the cylinder of radius 1 half. And pick a ball of radius 1 half centered on this point. And now, in particular, compute the excess in the ball of radius 1 half of the graph of u, of course. And since the ball of radius 1 half centered at p is contained in the cylinder, when I compute the excess with respect to this horizontal plane, I get something which is relatively small. So I only get a factor times this, because this excess has a normalizing factor. And then when I minimize overall planes, of course, I get something possibly smaller. So this is going to be less than a constant times epsilon. So now pick some plane, pi, which optimizes this excess. By the elementary lemma that we have seen a couple of lectures ago, the difference between this plane pi and the plane pi 0, so the angle between the two of them, has actually to be comparable to the square root of this excess. Pi minus pi 0 is actually pretty small. So now inside this ball, we know that if we have epsilon sufficiently small, we can actually pick a cylinder whose base is parallel to this plane pi. And it has slightly smaller radius. And all the graph inside the cylinder is contained inside my ball. So this is a cylinder of radius 1 half, say minus eta. This is center at the point P. And then it has base pi. And I mean, when we are using this notation over here, what we mean is that we take a ball of radius 1 half minus eta centered at the point pi. We intersect it with the cylinder pi. And then we add pi perp to this. So this is pi. And this is pi perp. OK, very good. And then in the cylinder, what we know is that the excess is very small. And that exists. So if we call E this excess over here, so that exists a function V from the ball of radius 1 half minus eta centered at P. So this is the disk, actually, in parallel to pi. So this is a notation that we will use for this object over here, B 1 half minus eta. And then P is the center point. And pi is the plane in which the disk is actually contained into pi perp. And that's actually two things. So first of all, the Lipschitz constant of V is less or equal than a constant times E to a power gamma. And the set where V and U are different, well, let's say V and U prime, where U prime is the new function which is giving you the graph in these coordinates. OK, so the area of this is actually less or equal than a constant and then E to the power 1 plus gamma. OK, so if we make now what we have already done before. So first of all, from the proof of the Lipschitz approximation theorem that we had last time, you actually see that since I can patch V and U prime, so although V is not a minimizer, I cannot find the competitor for V which gains in area better than E to the power 1 plus something. So recall from the Lipschitz approximation, so if Z is a competitor for V, then the graph, so the m-dimensional volume of the graph of Z must be bigger or equal than the m-dimensional volume of the graph of U prime minus an error which was the patching between V and U prime because V and U prime are not exactly the same function and this was like 1 plus, say, some gamma. And of course, then we have also a similar estimate with the graph of V. On the other hand, we have the Taylor expansion for our excess and for our area. So we know that we know two things. So we know, first of all, that the m-dimensional volume of the graph of V, it's approximately the area of the unit disk plus 1 half, the integral over this disk of the energy plus an error which is like E to the power 1 plus gamma. And then we know something else also. We know that similarly, the volume of the graph of U prime minus the area of the disk down is also equal to 1 half, the integral of the U prime squared plus big O of E to the 1 plus gamma. So this means essentially that since this is almost E, so this is comparable to E, so this means essentially that this indirectly energy is comparable to E. And if you renormalize the function V by dividing by E to the power 1 half, OK, so this is comparable to 1. Since, however, you minimize the area over there and the error that you have over here is a constant E to the power 1 plus gamma, you actually can conclude something which is very interesting that if you have a competitor for this V divided by E to the 1 half, indirectly energy such a competitor can only gain something like E to the power gamma. So if V prime, say, or if G is a competitor for V divided by E to the 1 half, so then the directly energy of G has to be bigger or equal than the directly energy of this V divided by E to the power 1 half plus E to the gamma. OK, so this is actually telling you that this function V divided by E to the 1 half is very close to be an harmonic function, OK? And in particular, out of these considerations, it's not difficult to see that there is an harmonic function which is W12 close to this normalized approximating function V with an error which is actually proportional to this E over gamma. But in fact, we don't know that. I mean, we don't need that. What we really need is that there exists an harmonic function H such that the W12 distance between H and V divided by E to the 1 half, OK? It's a little o of 1. So what this one means is just I treat this as, I mean, thinking that I take the excess very small. Of course, this translates into the fact that there exists an harmonic function H prime such that H prime minus V in W12 is a little o of your excess E, OK? So now, here we have a very simple lemma which I leave to you as an exercise. I mean, this is a semester or three weeks activity on harmonic analysis. And this is actually a purely harmonic analysis lemma. So if you have an harmonic function H, so if H is an harmonic function, say, in the bowl of radius, so in the disc of radius sigma, OK, and you pick a tau which is less than sigma, OK? Then you can expand your harmonic function H in spherical harmonics. And you will actually find the following interesting fact. So if I integrate over the bowl of radius tau dH minus the average over the bowl of radius sigma of dH, and I square it, OK? So the energy of this is going to be less or equal than tau divided by sigma to the power m, which is the dimension, plus 2 times the integral of dH squared, OK? So now, I apply this to the H prime. So I apply this to the H prime. And then I apply it actually to, I mean, I use the fact that H prime is close to v. And I use the fact that v is close to u. And then I will actually derive the following inequality. So in the bowl of radius 1 quarter, if I take du minus the average of du squared, OK? So with respect to the bowl of radius 1 half minus eta, I will have something like this. So this is going to be less or equal than 1 half. OK, so this is going to be less or equal than 1 quarter to the power m plus 2, OK? And then I'm dividing by 1 half minus eta to the power m plus 2. And here, I will have the integral of du squared, OK? And then here, I have something which is little o of dxs, OK? And now, it's not too difficult to imagine the following. So if the integral, sorry, du prime squared here. So if the integral of du prime squared was the xs, and here I have b of 1 half minus eta. So if the integral of du prime squared was comparable to the xs with respect to the horizontal plane, so now there's another tailed expansion that you have to do, which is telling you that this energy over here is close also to an xs. But in excess, where instead of comparing the tangent planes to the horizontal plane, you are comparing the tangent plane to the plane whose graph is given by the average of du prime, OK? So take the plane pi 1, and which is the graph of the linear function, say, x prime gets mapped into this average on the Bolo radius 1 half minus eta of du prime times x prime. So it's a slightly tilted plane. And then conclude that the xs of the graph of u prime, but which is just the graph of u on the cylinder of radius 1 quarter centered at your point p compared to this new plane pi 1, OK? So this is going to be less or equal than something which is up to this constant eta, essentially 1 half to the power m plus 2, OK? So modulo errors. So it's something like 1 half to the power m plus 2. And then there's something like a constant times eta, essentially, OK? And then you have the integral of du prime squared, and that's actually the excess of the graph of u in c 1 half minus eta, and then compare to pi. And then here there is a little o of e, OK? So now remember that the cylindrical excess is not normalized, but the spherical excess is normalized. So when I compute actually the spherical excess, I have to divide by a power m of the radius. So here this power, I mean, if I pass to the spherical excess, this power m is actually eaten by denormalization, but there will be a gain which remains, which is this extra power 2, OK? So if you pass to spherical excesses, you will then find that the excess of the graph of u in b 1 half p, with respect to the plane pi y, but then if I optimize, I get even something smaller. So this is less or equal than 1 quarter plus my eta. And then here there's going to be the excess. Instead of having it in the cylinder of radius 1 half minus eta, I use the lemma, and I compare it to the ball of radius 1 half, which is what I started with, OK? And then I have a little o of e, which is controlled by this over here. And so maybe here I just have to put twice eta for every eta that you give me. I will choose epsilon sufficiently small, so that, OK? Is that b 1 fourth on the left? Sorry? On the left is that b 1 fourth? b 1 fourth, yes, sorry. OK, so now what you see over here is a decay estimate, right? So you started that you had a small excess over here, and actually you find that in the ball of radius 1 quarter, you had excess, which has decayed by a factor, almost 1 quarter. So I could actually, if I choose my constant correctly, so here I could just say this is 1 divided by 2 to the power 2 minus, say, 2 alpha, OK? So almost 1 quarter. Of course, now you started with a small excess, and you actually find that in a smaller boiler you have even a smaller excess. Well, this means that I could actually iterate this idea even on the ball of radius 1 quarter, OK? And what I, in fact, discover out of this is what is called the Georgian's excess decay. So the Georgian's excess decay actually tells me that under the assumption that I have a small excess for every p in the ball of radius 1 half, and for every r, which is less than 1 half, I actually discover that my excess in the ball of radius r centered at p is actually decaying like a constant times the initial excess in the cylinder of radius 1. So let us call it e. And then I have r to the power 2 minus 2 alpha, OK? So now we are essentially done with the C1 alpha theorem of the Georgian. If you know something which is called Moray's type, decay estimates or Moray lemma or Moray campanato lemma depending whether you come from Italy, for instance. So we have already seen that this excess, I mean, we have already seen that if we stick, OK, we have seen sometimes ago that if we stick the average of the function, right, and we compare the excess, I mean, we compute the excess with respect to the plane, which is given by the average of the function, right? So the L2 distance to this average is actually comparable to the excess. So you do the same computation over here. And what you discover is that on the ball of radius r p down in the horizontal plane, if I actually compute the u and I subtract the average of u squared, this quantity, remember, this is actually normalized by a factor r to the power m. So here I can take the average. So this quantity over here is comparable to the excess, say, in twice the radius because this would be like the excess in the cylinder. And so this one also decays like a constant e r to the power 2 minus 2 alpha. OK, and now Moray's lemma gives you. And I mean, if you don't know what Moray's lemma is, you can actually get a complete proof in the lecture notes. So Moray's lemma now gives you that out of this decay estimate, du is actually c0 alpha. I mean, not only it's c0 alpha, so somehow it's maybe suggestive to take a square root of this. Sorry, it's c1 minus alpha. So maybe I should have called actually this 1 minus alpha. I should have called alpha. OK, no, too late. So you see somehow that this 1 minus alpha is related to this exponent. And then you can simply guess that the size of the c1 minus alpha norm is this e to the power 1 half, which is the linear estimate which I promised you. And of course, this you're going to have it on the Bolo radius 1 half. I'm very good. Yeah, this average. Yeah, absolutely. Exactly. So this is telling you the L2 oscillation of the function in the Bolo radius r is comparable to a certain power of r. And out of this, actually, you, I mean, the idea of more Islam is that, for instance, out of this estimate, you just discover that the averages of the derivatives is actually a convergent series. And it converges with the rate, which is exactly this r to the power of 1 minus alpha. I mean, you're summing geometric series, which is, for instance, is telling you right away that exists a limit of the average at every point. So every point is, for instance, a back point for the function du. And then you get continuity by similar tricks. Good. So we are right on schedule. So the idea was to spend the first lecture giving you the introduction, then one hour and a half on the Georges theorem, and then one hour and a half on Amgren's theorem. So now for the remaining part of this lecture and for the lecture of tomorrow, I'm going to give you actually an idea on how you can improve this c1 1 minus alpha estimates to some c1 beta estimates, to some c3 beta estimate. So from now on, the goal is actually Amgren's theorem, which will tell you that under the same assumptions, with the purely geometric proof, so purely quote unquote geometric proof of c3 beta regularity. And in fact, we will have a c3 beta regularity, which is kind of the same flavor as the one over there. So we will be able to estimate du in c2 beta in some smaller ball. At the beginning, maybe the ball is going to be much smaller, but then by a covering argument, you can actually get the ball of reduced 1 half. So say this is going to be some ball of reduced tau, and this is going to be less or equal than a constant, and then e to the power of 1 half. OK. So how am I going to do this? So this is going to be a much more sophisticated construction, and it's taking advantage of one key fact. So it's not taking advantage of the fact that the function du is c1 alpha, but it will take advantage of this kind of equivalent thing that we have the excess decay. OK. So the idea is the following. So here we have our cylinder. So take this base of the cylinder, this cross section. OK, so I write it over here. And then inscribe inside here a, OK, so maybe make it slightly smaller, so make it the ball of reduced 1 half or the disc of reduced 1 half. OK, then put inside a square or a cube and decompose the cube in smaller cubes in a grid which has 2 to the n 0 cubes. OK, so let me be more precise, with formulas. So take minus sigma sigma to the power m inside the ball of reduced 1 half. OK, and I believe that in order to do this, you need to choose sigma to be equal to 1 over 2 square root of m. OK, and then subdivide the cube minus sigma sigma to the power m in 2 to the n 0 times m cubes l in a regular grid. OK, so to fix some notation, if l is a cube of the grid, I'm going to use l of l for the side length. So first of all maybe, OK, let me be slightly more precise. So subdivide minus sigma sigma to the power m in 2 to the k m cubes. And k is an integer which is bigger or equal than n 0. OK, so this is a regular grid of fine, which is rather fine, as k somehow is rather large. And call ck the collection of sub cubes. So now for each l in ck, OK, let xl be the center of the cube. We let pl be equal to xl u of xl. So the point of the graph of the function u, which is lying on top of xl. And let also l of l be the side length of the cube, which should be 2 to the minus k times 2 sigma. So this is the side length of the cube. Now maybe in the notes, I take half the side length actually for l of l. Well, it doesn't matter somehow. This factor 2 is certainly not disturbing. OK, so now here I have my cube. So here's the point xl. So what I'm going to do is I'm going to go up, pick up the point pl, OK. And then I take a ball, which is centered on pl on the whole space, which I will call bl. So bl is going to be the ball of a certain radius. So there is a certain constant m0 times l of l. And then this is centered on the point pl. Maybe there is a square root of m over here. OK, m0 is just sufficiently large. It's a constant which we fix together with this n0. In fact, the purpose of n0 is to be sure that this ball is contained in the cylinder of radius 1. OK, so this m0 will be decided at a certain point. It's actually a fixed geometric constant. And this gives the ball which is kind of a certain factor larger than the side length of the cube. And then I will choose the fineness of the grid sufficiently large so that, although I'm enlarging the side length of the cube by certain factor, the corresponding ball is still on the cylinder of the radius 1 half, OK? OK, so now, before I get messed up with the notation, maybe it's time to take a look at the lecture notes. Maybe also drink my tea. Right, so very good. Now, I have d-the-Georgi excess decay, which is telling me very well this d-excess of the graph of u in my ball bl is actually decaying, OK? And it's decaying like a constant, then the side length of the cube, which is comparable to the radius of the ball, to the power 2 minus 2 delta, sum delta. So now, at a certain point, I will want this delta actually to be sufficiently small. And what I know by the d-Georgi excess decay is that the smaller I want delta, the smaller I have to choose my initial excess in the initial cylinder, OK? So this is related to this eta error that we were having in the excess decay. So the smaller we wanted eta, the more we had to take the epsilon small in the previous proposition, OK? So this delta is now a fixed