 w squared is 1 plus dw squared divided by 2. So when I integrate 1, I actually get the size of the domain. And then I get the integral of dw squared over the ball of radius 1 half. And then I get higher order terms of the type dw to the power 4. But the size of dw is e to the omega. And we know, actually, that we know already that the integral of dw to the power 2 is actually almost e. It's controlled by the excess. So here, I then have something, which is of the order e. I mean, I have an error. And somehow, if I want the bound on the error, so then I know that the error is less or equal than a constant, and then I have e to the power 2 omega. And I have the integral of dw squared. So it's a perturbation of higher order term. So the biggest term is actually this one. Now, I assume, actually, that the set l, I mean, the complement of the set that I'm missing, is actually very large. So another thing that I could do is I could actually test the area minimality property by kind of smoothing w if it has too large a region where the gradients are too steep. And one way of smoothing w, of course, is just to convolve it by a smooth kernel. So convolve w with a smooth kernel. So let us say the smooth kernel is phi. And let me just, for the moment, put l as the size of the kernel. So then I define z to be equal to w star phi l. And now let me compute the Dirichlet energy. So of course, w star phi l will have a Lipschitz constant which is less or equal than the Lipschitz constant of w, up to constants. And so if for w, the area is essentially the Dirichlet energy, of course, for z, the area of the graph would also be the size of the domain plus the Dirichlet energy. So let me now compute, or let me get an estimate, on how big is this energy. And let me do it in the following way. So I want to actually estimate the integral of the z squared over the ball of reduced one-half. I'm cheating a little bit, actually, because if I am actually making a convolution, then I have to go to a ball which is slightly smaller. But let me ignore that. So then here I have a key estimate. It's over here. So this is, of course, the integral over the ball of reduced one-half of the derivative of the convolution. And now let me split the derivative of the convolution in the following way. So let me compute it as less or equal than, well, the convolution of the modulus of the gradient. And then let me split this dw into two parts. So I take here the indicator function of the set L. So it's the set where the function w coincides with my function v. And I convolve it. And then I have dw and then the indicator function of the complement of L. And I convolve it. And now I make a kind of brutal estimate. So when I'm computing the Dirichlet energy, I have to square this. I have a term which is the square of this quantity. And if I look at that term, this is actually here, if I look at that term, that term is obviously less or equal than the Dirichlet energy of w on the set L. This is what is coming from here, OK? Then I have the Dirichlet energy of this part. And I have the product of the, I mean, I have something which I can estimate by Cauchy-Schwarz inequality with the Dirichlet energy of this part to the power 1 half and the Dirichlet energy of this part to the power 1 half. Now I know that the Dirichlet energy of dw squared on the whole domain is less or equal than E. I will actually show you that the Dirichlet energy of this part is even less than E. So this one will actually be smaller. So let me just estimate everything by a constant times E to the power 1 half. And then I will have the Dirichlet energy of that to the power 1 half. OK, so the term which is left, which is the square of this, is actually higher order. So I will just estimate this part. OK, and now let me be sloppy and let me just say, OK, I have to estimate the L2 norm of this quantity, right? Well, the L2 norm of this quantity, first of all, remember that the modulus of the derivative is less than E to the power omega. So let me just estimate everything with the L infinity norm of this and just get the convolution of the indicator function of the complement of L star phi L. OK, so this is less or equal than E to the omega times a constant. And then I just have the L2 norm of the indicator function star phi L. OK, and now I want to estimate that L2 norm, and I use the usual fact that if I'm estimating the indicator function of L complement, convoluted with something, if I'm estimating the L2 norm, I can estimate this with the L1 norm of this guy, and then here I have the L2 norm of phi L. OK, and so what is the L1 norm of the complement of L? Well, that I have the L1 estimate, the weak L1 estimate, which I used before. So this is less or equal than a constant E to the power 1 minus 2 omega. And now I have the L2 norm of this phi L, and of course, this gives me just L to the power minus m divided by 2. So now I want to use this convolution of W with phi L as a competitor against W. So I certainly want my competitor to be sufficiently close to W. So if I make a convolution with a kernel L, which is some power of E, say E to the power theta, then when I plug in in this estimate, so when I plug actually everything back in in this estimate, so you see what I get. So over there I have the integral over b 1 half of dz squared. So I have the integral over L of dw squared, and then I have a constant times E to the power 1 half. Then I have plus omega minus 2 omega, so I have minus omega. Then I have plus 1, which is coming from here. And then I have L to the power, and then I have E to the power minus m divided by 2 times theta. I'm not understanding phi. Phi is a smooth kernel. Phi is a smooth kernel compactly supported on the ball of reduced 1, which I'm actually descaling for making the convolution. I mean phi L is the usual kernel that you use for modifying functions. I mean phi L of z is equal 1 over L to the power of m phi of z divided by L. Usual smoothing. OK, so now what you see over here is that you have 3 half minus omega, and then you have this theta that you can actually choose. And what you discover is that if theta is sufficiently small, then you actually have that this is less or equal than the integral over L of w z squared plus a constant, and then you have E to the power 1 plus something, say kappa. So now I want to use this convolution of w as a competitor with v. So I have to show you by some computations that I can patch or glue the function w and the function z. Well, actually the function z because it's the one that I'm using as a competitor with v. So let me give you just a brief sketch while the patching, so the patching costs in energy again a power of E, which is 1 plus something. So the reason is the following. So for instance, let's say that you want to patch v and w. So what is a typical way of patching v and w? Well, a typical way would be to fix some radius. So fix a radius r, or maybe we should call it sigma. Let's say this is a sigma between, I don't know, 1 over 4 and 1 1⁄2. And one way of patching v and w would be to make a linear interpolation along the radii between v and w. OK, so for instance, if you introduce radial coordinates, one way of doing this would be to define a new function. So the new function, say, the new function v tilde is equal to v outside the ball of radius sigma. So v tilde of x is a scaled version of w. So it's like something like w of sigma divided by sigma minus a layer, and say the layer is e to the power beta times x. OK, and here you want to interpolate between v and w on this annulus. So here is the interpolation. And the way I do interpolation is something like this. I do a linear interpolation on the radii. So sorry. Why don't you retell the function vw? Inside the ball of radius sigma minus e to the power beta. So it's a small, slightly omotetic version of the function w. And this interpolation is in the annulus between v sigma and v sigma minus e to the power beta. And what I would like to do, I mean, if I were in one dimension, what I would like to do is I would like to define v tilde of, say, the point x. So here you want to interpolate between a point, say, sigma minus e to the power beta times some, say, let us use xi, and sigma times xi, right, where xi is on the sphere. And here, I mean, for any thing which is in here, so it's something that you can write it as lambda sigma minus e to the power beta times xi plus 1 minus lambda sigma times xi. And here you would like to have a linear interpolation. And the linear interpolation is then going to be something like lambda v of xi plus 1 minus lambda w of xi, OK? So if you carry on some tedious computations, once you actually do this interpolation, you want to know how much it costs in terms of the Dirichlet energy in this layer. Yeah, probably. Then the integral over the annulus of dv tilde squared will actually look like this. So you will be able to estimate it with a constant times e to the power 2 beta, right? And you will have the integral over db sigma of the sum of the two Dirichlet energies, OK? And then you will have actually something to pay over here. Here you will have a constant. And you actually have to divide by e to the power 2 beta. And then you will have here the integral of the difference of the L2 norms between w and v, OK? But now remember, w and v, so you can imagine if I make, I mean, if I choose my sphere correctly, the Dirichlet energy over the sphere is comparable to the Dirichlet energy in total. And the Dirichlet energy, total Dirichlet energy of both of them is around e. So you can easily imagine that this one over here behaves like e to the power 1 minus 2 beta. If I choose my radius sigma via a Fubini type argument correctly. And now what about this dangerous term over here? Because I am divided by e to the power 2 beta. Well, the term actually is, after all, not that dangerous at all. OK, actually there's an integral because I'm integrating over an annulus. So there's actually an e to the power beta over here. So that term is actually not dangerous at all. Because remember, w and v agree on a set whose complement has measured less or equal than e to the power 1 minus 2 omega. But I can actually gain on the following fact. So since v and w are both lip sheets right on the set where they don't agree, they're actually pretty close. OK, so if you carry on all the computations that are actually done carefully in the notes, you discover that here you can actually gain something which is likely super linear. And even though you are dividing by something which is a power of e, if you set your exponent beta correctly, then you actually get that this one is also gaining something which is less or equal than e to the 1 plus something. OK, now let me come to the conclusion. So after I've done the patching, I have a new competitor. And what I discover is that the area of the new competitor, so after the patching, in the ball of radius one half, it will actually be in the ball of, I mean in some other ball, but say, for simplicity, let's fix ideas and say that everything is done in the ball of radius one half. So in b one half, I have a competitor. And the volume of this competitor, which I will call v prime, so the volume of this competitor is going to be the size of the ball of radius one half. Then there's going to be the Dirichlet energy, or dv squared, but intersect it, the set L, where everything is nice. And then there's going to be an error, which is of the type e to the power 1 plus kappa. OK, and why actually now all of a sudden in the set L I'm actually putting dv squared instead of dw squared? The reason is because w was the extension of the restriction of v to L. So v and w actually they coincide on L, and therefore they have the same gradient. OK, now this competitor must have volume of the graph, which is bigger or equal than the volume of the graph of v. So the m dimensional volume of the graph of v on the ball of radius one half must be therefore bigger or equal than b one half. And then there is one half the integral of over b one half intersect L of dv squared. OK, and then there is this error over here. So I have here minus a constant times e to the power 1 plus this kappa. OK, so let me now look at that volume over there. So what is that? Sorry? Yeah, it's the other way around. Yes, sorry. Right. Yes. OK, so now let me actually expand that volume over there. So if I subtract the ball of radius one half on these other side, I get the excess. OK, so what I actually can then compute is that I have the integral. OK, maybe let me put it in the following way. So here I take the integral over b one half intersect L, and I take the area functional. OK, and then let me subtract the corresponding volume over here. So that's one term. And then I have what is left over here. So this is the b one half minus L. And here I have the square root of 1 plus dv squared plus all the jets. OK, minus b one half minus L. OK? And this I discover is less or equal than one half integral over b one half intersect L dv squared plus a constant e to the power 1 plus kappa. OK, now make a tailored expansion of this portion. So in this portion is actually where I know that the maximal function of dv squared is small. So in particular, modules of dv is less or equal than a constant times this e to the power, whatever we fixed at the beginning, like omega. So you can actually tailor expand that. And when you tailor expand that, you find exactly the Dirichlet energy over that portion plus higher order terms. So you find exactly this plus higher order terms that you can stick into this e to the power 1 plus kappa. OK, and then what you find over here is actually this difference. And so the Dirichlet energy now cancels with the tailored expansion of this guy. And you get this. OK, but now this quantity over here, although, so this is actually on the complement of L, so this over here, which is on the complement of L, controls a constant times dv squared. So the final inequality that you have is this one. So now remember what the complement of L was. The complement of L was the place where the maximal function is large. So what you actually have discovered is that this integral over where the maximal function is bigger or equal than e to the power 2 omega of dv squared is less or equal than a constant e to the power 1 plus kappa. Now although, so this omega has to be sufficiently small, but this kappa actually that you have gained over here, if you make all the computations carefully, OK, can heat up the 1 over e to the power something that we had before, right? So this kappa can actually be made independent of omega. So now when you make the Wickel 1 estimate, you're making this Wickel 1 estimate with this integral over this portion, which is actually better because it's super linear with respect to e. And then if you choose your truncation level possibly with a much smaller omega, you actually don't heat up what you have gained here with kappa and then you are left with the estimate that you wanted. OK, so that's it. Sorry, even this time I went extra time, but not that much as yesterday.