 Good morning, welcome back everybody. So yesterday I finished the lecture with some elementary comparisons between spherical excesses in different points. You have noticed that the epsilon regularity theorems are actually stated in terms of something which have to do with the cylinder rather than a sphere, right, so this is for a graph more natural and so let me actually introduce the cylindrical excess. So in what follows, we will use the following notation for a cylinder CRP of the following form, or let's say for a cylinder CRX of the following form, so we have a disk in Rm of radius r and centered at x and then we multiply by r to the power n over here. So this is a subset of Rm. Okay, and then we have seen the formula that if we take the m-dimensional volume of the graph of a function u over this disk of radius r and centered at x and if we subtract the m-dimensional volume of the base, then we find that actually this one is equal to the integral, so to 1 half the integral over the graph of u intersected the cylinder of the distance between the tangent space or the m vector orienting the tangent space at any point p to the graph of u minus the m vector orienting the base space to the power two and this is in the house of measure in the point p, okay? So it looks natural actually to introduce therefore this notion over here as a notion of cylindrical excess and as we have done for the case of the sphere, we actually introduce a notion of cylindrical excess for a general vector m vector pi, so for a general plane pi and this is going to be 1 half the integral over the cylinder of the distance between the tangent space to the graph of u and your m vector pi squared, okay? And contrary to what we do with the sphere, if we omit pi, then we understand that the excess is actually computed with respect to the horizontal plane. So with the sphere, we were actually minimizing because we don't have any preferred choice. In the case of the cylinder, we are just using the horizontal plane. Notice also that this excess is not normalized, right? So for the sphere, we were dividing by omega m and then r to the power m. So here we are not dividing, what is most important is that we are not dividing by r to the power m. So if we were dividing by r to the power m, then this would be more similar to the spherical excess. So of course there is an obvious comparison. If I take the m plus n dimensional ball of reduce r centered at x u of x, this is contained in the cylinder. So it is therefore obvious that modulo the normalization constant, I can actually compare the excess with respect to a certain plane pi in the ball to the excess with respect to a certain plane to the same plane pi in the cylinder, simply because I'm actually integrating on a larger set, right? So of course then this inequality is just obvious. And then here I would have just the same thing. Okay, it's less obvious that I can actually have a reverse inequality. So while of course I am with a sphere of reduce r inside the cylinder of reduce r, if I am taking the sphere of reduce r, I'm missing actually something from the cylinder, right? But you could perfectly imagine that, so here say is the point x u of x. Okay, so you could perfectly imagine that if your surface is going through this point and the surface is sufficiently flat, okay? There's not much that I'm missing actually from the sphere to the cylinder. So let me make the following consideration. So fix a number eta bigger than zero and for scaling reasons in order to fix ideas, let us assume that actually r is equal to one. And let me assume that the excess on the ball of reduce one centered at x u of x, which is going to be my point p. This is the excess of the graph of u. Okay, so that this one is less than some epsilon one, some constant. Okay, so now this is the excess with respect to a certain optimal plane pi. So let pi be such that you're optimizing. So let's call this e, okay? And now without loss of generality, let us assume that you make a rotation and you actually put this plane pi to be just the horizontal plane, okay? Rotate and assume pi is actually the horizontal plane pi zero. Okay, so now since you have in our discussions, we are having, so this is p. So since we are considering a Lipschitz function with constant say less or equal than two, right? So it's actually a very easy exercise to see that if I take the cylinder of radius one half, so let's say actually even that the point p, just for the sake of argument is the origin. So if I take the cylinder of radius one half centered at zero, okay? Then the graph over this cylinder of the function that I'm looking at is certainly contained inside the ball, right? So since I have a Lipschitz bound and just a method of playing with the constant, it's not possible that the function sort of escapes towards the vertical part of the cylinder, okay? So now I just want to understand what is the maximal radius, rho of a cylinder for which the graph of the function over this cylinder is still contained inside my ball, okay? So here I define rho to be the maximum among all sigma such that if I take c sigma of x and intersect it, okay, so if I take the graph of u over the ball of radius rho of the ball of radius sigma centered at zero, so because I'm actually assuming now that x is equal to zero, this is still contained inside the ball of radius one centered at zero, okay? And I can actually make first of all the following observation. So when I look at the distance between the tangent plane at some point q, let's say it's the point y, u of y and the horizontal plane, okay? So it's not difficult to imagine that the distance between these two planes is actually comparable to the modules of the derivative of the function, okay? So this is just a very simple geometric exercise. So this is controlled by a constant times the modules of the u and it controls from below a constant to the minus one times the modules of the u. And here of course, I'm using actually crucially the fact that the Lipschitz constant of u is less or equal than two. So my constant is depending on these two, all right? If I take a Lipschitz constant to be very, very high and I have very steep gradients, then this inequality is not true anymore, okay? But now once I know this inequality, what I can actually discover is that in this b sigma, if I integrate over b sigma du squared, right? This is going to be less or equal than the excess and since b sigma is actually contained inside my sphere, it's going to be less or equal than the excess on the whole sphere with a constant that I have to pay, right? So I have to pay a constant for two reasons because I'm actually using this instead of the modules of du and because I'm actually integrating over the graph instead of integrating over the base, okay? Actually, that makes things better because when I'm integrating over the graph, I'm actually integrating an area factor which is bigger or equal than one, okay? So here I can then have this very simple inequality and I know that this is less than epsilon one by my assumption, okay? But then on the other hand, I also know that the derivative of u is less or equal than an absolute constant because I have the Lipschitz constant over there. So it's less or equal than two. I can actually interpolate and I can say that the l to m norm, for instance, in the ball of radius sigma, right? It's less or equal than a constant and then I can interpolate between this l two bound and the l infinity bound and if I made the computation correctly, I get epsilon one to the power one over two m, okay? So now why did I do this? Well, I did this because if I control the derivative of u in l two m, since my ball is actually m dimensional, I have more as inequality which tells me that I'm held there and so it also tells me that I have an l infinity bound on or I have a uniform bound on the function u, right? So the oscillation of the function u is then going to be bounded on the ball of radius sigma is then going to be bounded by a constant then times epsilon one to the power one over two m. This is just more is a madding but then I am assuming that the origin is in the graph of the function. So this means actually that so the l infinity norm of u, it's in fact bounded by a constant times epsilon to the power one over two m, okay? Now notice the following. So if I'm taking the maximum radius rho over which I have this inclusion for very simple continuity reasons, it means that for this particular radius rho, my graph must be hitting the sphere of radius one, right? Otherwise, if it's not hitting the sphere of radius one, then I can just make rho slightly larger and that's not the maximum one. Okay, then by a very simple consideration, so here there's a point say z. So the point z as modules of z equal to rho, okay? And what actually happens is that this point is u of z, z in coordinates. So modules of z squared plus modules of u of z squared must be equal to one. And now you get a bound on your function rho, right? On your rho. So because you actually get that one minus rho squared which is less or equal than modules of u of z squared, this is actually less or equal than a constant times epsilon one to the power of one over m, okay? So this means actually rho has to be pretty close to one. In particular, if you remember how I fixed this eta and what do they want to do with this eta? If eta is small, I can then choose epsilon one, very small, so that the cylinder of radius one minus eta centered at zero intersected the graph, is still contained in my ball, okay? And then of course once again now I can compare the excess in the cylinder with the excess on the ball. And so module of choosing the excess, very small, what is the philosophy of this? The philosophy of this is that whether I'm using the excess on the sphere or the excess on the cylinder, actually the two will be essentially comparable, right? So I just have to wiggle my radius a little bit. So now when I control, so when I have the integral over the cylinder, c one minus eta intersected the graph of u of my usual whatever, tp graph of u minus pi zero squared, I will be able to control this with the integral over the ball of radius one. And okay, so here I just have the same thing. Okay, so these are all kind of small lemmas which we will use at a certain point in both the proof of Andren's theorem and the Georgian's theorem. But so now let us actually come to the most important tool that we will use in both theorems. So let us now for the moment fix ideas and say that you are on the cylinder of radius r centered at some point x. Okay, and assume that the normalized excess, so assume that one over two, okay, so assume r to the power minus m times the cylindrical excess, which is the thing which scales correctly. So centered on CRX, assume that this is somehow very small, so small than some constant epsilon two. Okay, so you are in such a situation now, apply scaling and translation. So scale the function back to radius one and the point x translated to zero, then you have the assumption that the excess of the graph U on the cylinder C one zero compared to the horizontal plane pi zero, because when I'm actually leaving the plane away, then that is horizontal. This is somehow very small. This is epsilon two. And then as if we have observed, this means that the integral of a gradient of U squared is actually less or equal than epsilon two times the constant. So let us actually call this E as we have done other times. So in fact, as an intermediate step here, I will have a constant times E. And then here I have a constant times epsilon two. Okay, so now let's look at the how the elitist would maximal function of the derivative of U at some point y so let's say this is the supremum over r bigger than zero of the average of the U squared. It's actually the hardly readable multiple function of the U squared on the ball of radius r centered at y. And let me do the supremum overall radii between one half and zero. And let me assume actually that the point y is in the ball of radius one half centered at zero. Okay, so then you have the classical weak L one estimate for this maximal function. So you actually know that if I want to truncate so if I want to take as a following set so say the set L where the maximal function of the U squared in y is less or equal than some say E to the power two alpha, right? So the hardly readable maximal theorem will tell me or the weak L one estimate will tell me that the sides of the complement of K on the ball of radius one half K is L, sorry, yes. The complement of L, then this is less or equal than the integral of the U squared. Here I have a constant and then I have to divide by E to the power two alpha, right? So here it gives me actually a constant times E to the power one minus two alpha, okay? So I have an efficient Lipschitz approximation now because it's a classical exercise in sub-space theorem in sub-space theory to actually prove that if I restrict U to the set L, the Lipschitz constant of the restriction is actually E to the power alpha. Well, a constant times E to the power alpha. Okay, but so far I'm actually having this observation only for a Lipschitz function. This is always valid. It's actually valid for a general Sobolev function. Now I want to use the fact that the graph is area minimizing to actually improve the estimate on this L, in a sense. So in fact, I have a Lipschitz approximation theorem which tells you that I can, I mean, if you are area minimizing, this truncation with the maximal function is actually much more efficient as an approximation than for a general Lipschitz function. Okay, and this we will call improved Lipschitz approximation and let me give you the exact proposition in its full generality. And then, I mean, here in the lecture notes you can actually go through the proof in every details. It will be impossible to give you all the details of the proof, but I want to give you somehow the kind of core idea, the core estimate why this improvement is going to happen. Okay, so here is the proposition. So the proposition is saying, so there are constant epsilon bar bigger than zero and gamma bigger than zero. So these are only dimensional constant such that the following holds. So take a function V, which is going from the Bologgius R centered at X into Rn. So let this be a Lipschitz map and assume the Lipschitz constant is less or equal than two. Okay. So assume the graph of U is area minimizing, V is area minimizing and assume that this normalized excess. So this is going to be R to the power minus M and then you have the excess of the graph of V on the cylinder of reduced R centered at X is less than epsilon bar. Okay, so if we set roll, okay, so here there are actually constant epsilon bar, gamma and also constant C, V here than zero. So if we set roll to be R times one minus a constant times this E to the power gamma, okay, I actually find a set K inside the Bologgius roll centered at X such that the following two estimates hold. So first of all, what the set K misses from the ball is actually a set of measure bounded by a constant times E to the power one plus gamma. And then the other thing is that the Lipschitz constant of V restricted to K is less or equal than E to the power gamma. Actually, I think in the, so let me see, I think actually in the claim of the proposition I'm doing without this constant, okay, everywhere. I mean, you can easily imagine that if I have an estimate with the constant over here, everywhere, and if I just make my, with a certain power gamma, then I make my gamma slightly smaller than I can actually eat up the constant with the epsilon bar over here. So I can eat up any constant with this epsilon bar to the power something by taking epsilon bar eventually smaller. Okay, so you see the difference between the approximation that we got here where the Lipschitz constant is E to the power alpha but the set that we are missing is E to the power one minus two alpha. And you can see here that instead we are getting a super linear estimate. So there the estimate is sublinear in terms of the excess. Here the estimate is actually super linear in terms of the excess. And the reason why we are able actually to do this is because the graph is area minimizing. So let me give you a sketch of a proof. So first of all, sorry? Yeah, yeah, definitely. Yeah, otherwise it's pretty silly, right? Okay, so first of all, when you want actually to prove that theorem we will apply this scaling and translating invariance and assume that X is equal to zero and R is equal to one, right? So we are then reduced to this situation that we were discussing before. Okay, and then let me not discuss how you actually eat up only a small fraction of your radius. You can easily imagine that if I can prove the estimate between, I mean, going from the radius R to the radius R divided by two you can then adjust the technicalities so to actually get just a little bit in the inside, okay? So let me focus on how you actually improve your estimate going from R equal one to rho equal one half, okay? So then it's just a technical fact that you can in fact eat up only a little bit of your initial radius R over here. Okay, so the point is actually the following. So the point is that when you are estimating this thing over here, right? The full estimate of the weak L1 estimate, it's, I mean, the full power, it's actually giving you a domain of integration here which is rather small. So in fact, the weak L1 estimate actually tells you the following that if you're going, I mean, if you're looking at the points Y in the ball of radius one half where the maximal function of du squared is bigger or equal than sum E to the power two omega, okay? This is actually less or equal. This is actually less or equal. Then one over E to the power two omega but here there's an integral which actually takes place of where the maximal function is bigger than one over a constant E to the power two omega. So you're actually not integrating over the whole domain. You're actually integrating over a sub domain where you know that the gradients are very high. And now the idea is that we actually try to estimate this quantity over here by making an energy comparison argument. So we try actually to get a much better estimate for this term. So before we used that this term is less or equal than the constant times E and now we actually want to use that this term has a much better estimate. Okay, so how am I going to do that? So the idea is that first of all I make my truncation. So I make my truncation and I define a function w which is a Lipschitz extension of V over L. Okay, so this is the complement of L. So L is the set where the maximal function of the powers of the power two of the function is less or equal than E to the two omega. So we let W be a, say E to the two, E to the omega Lipschitz extension of V restricted to L where L is the set where the maximal function of du squared is less or equal than E to the two omega. Okay, so now this V is not too far from, I mean this W is not too far from V. We will see in a moment that you can imagine that although W, I mean W is not too far from V. What I would like to do is to use W as a competitor for the area minimizing property of the function V. Okay, but of course W and V don't agree everywhere. So if I fix some radius and I want to stick in W instead of V inside this radius, I have to kind of patch the function W which does not have the correct trace to the function V. Okay, so I will do this now in a moment. I will tell you why this is possible, but imagine just for a second that actually it's W the minimizer and not V. So imagine W is the minimizer. Now, why do I want to actually imagine that? Well, I want to imagine that because if W is the minimizer, since the Lipschitz constant of W is actually less so equal than e to the power omega, I can actually tailor expand my area functional and discover that the area of this graph is very similar to the Dirichlet energy of the function W. Okay, so remember the formula, the volume of the graph of W, say on the ball of radius one-half, okay? So this one is the integral and here you have the square root of one plus DW squared and then you have higher order terms, I mean, quartic terms actually in DW. Okay, so now make a Taylor expansion of this and you will have that this is equal to the volume of the disk because the Taylor expansion of square root of one plus DW.