Okay, so I'm going to start my second lecture on stability result for geometric inequalities. So let me recall the last thing we did yesterday was about Fuglede result. So we are in the following situation. We have a set E, which has the same measure as the unit bowl. And we assume that E can be parameterized with respect to the sphere. So we assume that the boundary of E is of the form omega 1 plus U of omega, where U is a function on the sphere, and U is small as well as its gradient. So U in an infinity plus grad U in an infinity is less than some constant epsilon, which we'll see is universal. And what we want to prove is the following. The perimeter of E minus the perimeter of the unit bowl is greater or equal than a constant, which depends on the dimension, times the integral of U squared on the sphere. And actually, we'll see that this is not exactly what we are going to prove, because actually this result is false. And a simple way to see that this is false is you just take E. So if this was the unit bowl, and we take E just to be a small translation, this is a bowl, E was just a translation of the bowl, then of course the perimeter of E would be equal to the perimeter of the bowl. So this is not really true. So that's why you must be careful what I'm saying. It's not exactly what we are going to show. So if E was a bowl, we are not going to parametrize E with respect to a different bowl. We have to readjust the body center. So there will be a moment where I have to readjust the body center to prove this estimate, because otherwise this would be true. So this is true, a translation of E. So if E were a bowl, U would be zero. OK. And so let's start this proof. I really sketched yesterday very briefly the argument. So the first point is to write down the perimeter in terms of U. Maybe let me just very briefly explain to you where the formula that I wrote yesterday comes from. So the first fact is that the perimeter of E is equal to the integral of square root of 1 plus grad U square over 1 plus U square 1 plus U to the m minus 1, where the gradient means the tangential gradient. U is defined on the sphere. So the gradient you are just differentiating tangentially in the direction of the tangent to the sphere. So where does this formula come from? There is the following. I mean, let's say that I have the sphere here and then I have a path. This is boundary of E.  I'm parameterizing this surface on the sphere. So what you have to compute is the Jacobian of the transformation which maps a piece on the sphere onto the piece of the surface. So what is appearing here is a Jacobian determinant. So you have the map which to omega associates omega plus U of omega times omega. So let's say that I take E1, En minus 1, a basis on the tangent space. So here you have tangent vectors EI. And then you differentiate in the tangential directions. So if you take the derivative of this map of omega, omega is like that entity, omega into itself. It's like x.   At the point of omega times omega. So this is kind of the partial derivative of this map in the direction DEI. So this means that if you take the tangent vector EI here and you look where this vector is mapped by the differential of the map, it's going to be mapped in a vector VI, which is this one. So VI is just the image by the differential of the map of this vector. And then how do you compute the Jacobian? The easiest way, the more convenient in general, is just to take a wedge product of the tangent vectors. It's usually the easiest way to compute this kind of stuff. So the Jacobian of the map omega into omega plus U of omega times omega is equal to do the following. You take the image of VI, so you take the image of the different vectors on the basis, you look where they go, so you have these vectors VI, and you take just the wedge product and the minus one times, and you compute the modulus of this vector. What happens when you take wedge products of stuff like this? So EI are tangent vectors. So omega, you see, is a vector on the sphere. So EI and omega, this is an orthonormal basis. So observe that E1, EI minus one and omega is an orthonormal basis in arena. So you start to take wedges, so you have let me do maybe with just one doesn't change much, right? So this is what? The first one would be one plus U of omega E1 plus V1U omega. This is the first vector. Wedge, wedge, one plus U, EI minus one, plus DM minus one U, one plus U plus U, okay, yeah, the parenthesis is here. So what happens when you take wedge products? So in some sense, the wedge is multi-linear, like the determinant. So the only way you get something on trivial is that in every step you have to take different vectors. Like if I take omega two times, you get zero because of anti-symmetry. So this vector, the only way you get something on trivial for instance is, so this is going to be either you take E1, E2, E3 up to EI minus one, so you get whenever you take all the EI in the product, this is multi-linear, right? So I have to take one of these two pieces here, then one of these two pieces here, so on, one of these two. So the first term I take is the first one with all the first ones. So I get one plus U, M minus one times, and I get E1 wedge EI minus one. Then there is the case where I take one of them, I take an omega. So I can take, for instance, let's say E1, E1, E1, E1, and the last one I take omega instead of EI minus one. So I get one plus U, one times, two times, three times, M minus two times, and then on the last step I don't have it. So I get one plus U to the M minus two, then on the last time I get D, M minus one, E, M minus one of U, and then I get the wedge product E1, E, M minus two, omega. And then I keep going like this. So I take, for instance, E1, I miss one of the EI and then put omega in place. So you get plus something like one plus U, M minus two, D EI of U, and then you're going to have E1 wedge. At some moment you're going to miss the EI, so this was the I spot, and then EI minus one. And the sign could be either plus or minus, depends on the bimultinarity, one of the two. It doesn't matter because we're going to compute the models. And this is actually everything you have because you see that you can never take twice omega. So you can only miss omega one time. If you take two omegas, you get zero. So in fact, you have only this, this, and all the other terms are zero. So you can try on your own like to do a simple case with three vectors and you will see that this is the only thing you have. So now you have a modulus, and all these vectors are, these are M minus one vectors are orthogonal to each other. So you just have to take, use Pythagorean rule. So what is the norm of this? It's just square root. This is square root of this M minus one square plus one plus U M minus two square times sum over i of the EIU square. And what is this? You take this common factor out. You get one plus U M minus one square root one plus grad U square. This is grad U square one plus U, which is exactly what I have here. So this is the jacob. Again, in case you are not familiar with veg product, a simple way, I mean, you can think in three dimension, you can try to do this computation in three dimension and then this is just vector product in three dimension, just A vector B and you get that vector, right? And in that case, you can convince yourself that what I'm doing is correct. Okay, so this is just to justify this formula. Now what do we want to do? We want to start to go down, right? We want to estimate this from below. So the first thing I observe is that square root of one plus grad U square over one plus U square. So remember that U and grad U are small, so you can start to do a kind of Taylor expansion. This is going to be something like the first order one plus grad U square one plus U square with a factor one-half, so I put a two. And this is up to an error epsilon, which is going to be greater. So U is small as well, so this is really up to one. So this is really one plus one minus O of epsilon grad U square, yeah, that's all. So really this expression is greater or equal than this, over two. Just use Taylor expansion to see that this is true, using that this is small and this is small, bounded by epsilon. So this is the first bound. And then what about this?   And now you start to compute products. So you have a bit of time. You have the one one. So this is when you have the integral of one, then you have this term. So plus one minus O of epsilon integral of grad U square over two. Then you have this with one.  And then all the rest, which can be bounded by minus, let's say O epsilon integral of U square. So this is the kind of estimate we have. Now, this first term is just the perimeter of the ball. It's the area of the sphere. So we are saying that the perimeter of E is greater or equal than the perimeter of the ball plus this term. There are these two other terms and then a smaller rest. And what we want to do is to show that this quantity, not only is non-negative, but actually controls U square from below, from above. So this is greater than constant times U square. OK, so we want U square. So first of all, what about integral of U? So integral of U, we can use the following fact. So what is the volume? So now we have to start to use the volume constraints. So the volume of E in polar coordinates is the integral over the sphere, the integral from zero to one plus U of omega. So this is in the omega. And this is rho to the n minus 1 d rho, right? Polar coordinates. So this is nothing else than the integral of one plus U to the n divided by n over the sphere. Integral of rho to the n minus 1 from zero to one plus U is one plus U to the n over n. But this volume must be equal to the volume of B1. And the volume of B1 is the integral on the sphere in the case U equals zero. So integral on SM minus 1 of one over n because it corresponds to the case U equals zero. So this means that this integral, the integral of one is U to the power n over n is equal to the integral of one over n. So we are going to expand this term and use that these two terms are equal. So I discover that the one over n cancels. So I get that zero is equal. So one over n cancels with the first term. So I get the second one, which is U, plus n n minus 1 over 2, and then I divide by n.  Sometimes maybe I go just fast on this computation and you can check them. OK, so the good thing here, and now that I have a relation between U and U squared, which means that here I can substitute and at the place of integral of U I can put minus n minus 1 over 2 U squared. So this implies that the perimeter, I go back to my formula, perimeter of E minus perimeter of B1 is greater or equal than 1 plus over epsilon, plus or minus, that's it. Integral of grad U squared over 2. Now you substitute here this term and you just put these two terms together.  OK, so this just comes using this equation and substituting here. OK, so this is where we are. Now, as I mentioned yesterday, once you are here, what's the idea? You would like to use some kind of Poincare inequality, something which tells you that the gradient in L2 controls the function in L2. Here is the Poincare inequality on the sphere. So let me state it inequality on SM minus 1. So you want to know if the gradient of U squared controls with some constant, I'm going to write it in a second, the integral of U squared. Again, remember what I already mentioned yesterday, in general inequality like this cannot be true if you plug U to be a constant, this is 0 but this is not 0. So you have to keep constant. So if you assume that the integral of U is 0, so if you consider just U, which has 0 average, then you have this inequality with a constant n minus 1. Wow, it's the same, right? Yeah, I can never, I mean, if the average is 0 then the average is 0. In addition, you know also when you have equality. So equality if and only if U is equal to xi, one of the coordinates, i from 1 to n. So you just take coordinates in a random. So these are what I call the first eigenfunction. I mean, after the constant, the constant U equal 1 is the first eigenfunction and these are the second eigenfunction on the sphere for the Laplacian. Now, so this is the first one. So what are we saying here? So you have the case U equal 1, so you have constants. Now once you kill constants, you have this inequality and the quality cases are given by these special functions, the coordinate functions. Okay, convex combination of them also, of course. These and then the convex combination.  So it's like, so I'm going to discuss this in a second using the spherical harmonics, but the point is this, it's like when you do, I mean, just Fourier series, right? So you have the constant, the first, the Fourier expansion, the first term is the sinus, but once you're orthogonal to the sinus, what do you mean, sine and cosine, once you're orthogonal to them, then in some sense you get to sine 2x and cosine 2x and you get a better inequality. The constant improves. So actually what happens is that if you assume here that U not only has zero average, but the integral of Uxi is also zero for every i, so you take a U which is L2, orthogonal to both constants and to the function which saturates this inequality, then you end up with a better constant, 2n. So the constant improves every time. In some sense, every time you kill, you're kind of, you have to think, you are in an Hilbert space, L2, and then you have the space generated by constants, which is a line. You say, okay, if I remove that line and I go to the orthogonal space in L2, I'm orthogonal to the constant function, then I get an inequality. Now you have, you say, I know the function which satisfies this inequality, I go orthogonal also to them, and then I end up with the next eigenvalues. In some sense, you have a matrix, an infinite matrix, and you are killing one by one or the eigenvalues, and you're going down. So I'm going to explain this actually a bit better in a second. But anyhow, these are two informations on the sphere. That you have these and you have these. Now if you look at the first one, we see that the first one is actually not enough because here, if you forget epsilon and you see that I have a two, well let me put the two inside, you see that m minus one here is exactly m minus one there. So the inequality on the sphere, even if u was to be zero average, this quantity cannot completely absorb this one because here I have an error also. I have no epsilon which goes against me. So the first inequality is not enough for me. So if I want to have a hope, I kind of have to go to this inequality. I want to use this one. And in addition, in my case, I don't have this. In some sense, the only thing I know is this. That interview is like interview square. OK, so let me cancel a second this. Hopefully you've wrote it down and it's not too difficult to remember. Let me see what happens in our case. So in our case, we have two things. The first thing is that the integral of u is equal to minus, so let's say, is bounded by a constant integral of u squared, which is what I wrote there. But now I can use that u as mole. So this is bounded by constant times epsilon integral of modulus of u. And now on the sphere, the sphere is compact, del one norm is controlled by del two norm. So this is bounded by constant epsilon del two norm of u. So the average of u is a small fraction of del two norm. Now, so this is like a way to say that the average, OK, it's not zero, but it's almost zero. It's very, very small. As I show you in a second ago, the average condition is not enough for me because I will get a constant m minus one. I have to go to the next one. So what's the trick now? So the trick on observation is the following. So e, the boundary of e, is parametrized over s m minus one. But I can, let's say, adjust, move e so that some kind of barycenter is the same. e is equal, is the origin. So there is a small argument here. In some sense, if you have a set which is very close to the sphere, up to some small translation, I can do it in such a way that it matches exactly the origin. And this is not going to change the perimeter. So up to translation, I'm going to assume that this is a k. So without loss of generality, we can assume that the integral over e of the function x is zero. I like to do this computation as an exercise. It implies that the integral of u times xi is bounded by constant epsilon, no more of u than two. So the fact that the barycenter of e is zero and the boundary of e is parametrized by u, you can do polar coordinates and you will see that the scalar problem between u and xi is very small for every i. It's similar to the argument that I did before for the volume and then just writing polar coordinates and compute the integral. OK, so we have these two informations in our case, that both this guy and this guy are small. OK, I will come back to this. Let me cancel this for a second. In some sense, our point now is to estimate from below this quantity. But if you have a function on S1 on the circle, so a periodic function, you can expand it in Fourier modes. You have the analogous of this in higher dimension. If you have a function on the sphere, you can expand it in an orthonormal basis in L2, which is given by spherical harmonics. Spherical harmonics are just orthonormal basis in L2 which are given by a function on the Laplacian. On SN minus 1, so this is just a fact in function analysis, if you want, every function in L2 can be written as u equals sum over j and k. Let me just use the same notation, k, of some coefficient, ajk, yjk, where yjk are functions on the sphere and they satisfy minus Laplacian of yjk equal, a constant which I think is k, k plus m minus 2, yjk for every j from 1 to some function capital N of k. What am I saying here? You can look at the Laplacian on the sphere and this is a symmetric operator. Then you can look at this as an operator on L2, this is an inverse space, and you can look at these eigenfunctions. What I'm saying is that for each eigenvalue, k is like an index on the, so these are my eigenvalues. For every eigenvalue, so k is parametrizing my eigenvalues, so this is my eigenvalue, lambda k. For every eigenvalue, I have a set of eigenfunctions which I'm going to call y of jk. I have a second index because I have more than one. These are finite, this is kind of, this followed because the operator is compact. There are finitely many eigenfunctions and the set of all the eigenfunctions of the Laplacian on the sphere forms a complete orthonormal basis in L2. All this family, so here you have in some sense yjk which varies over two indices, so k goes from one, I mean, is infinite, over the national number and j goes from one to some capital N of k, which is the dimension of the eigenspace associated to the eigenfunction lambda k. And every function u can be written in this way and this is an orthonormal basis, so orthonormal basis. And so if you have an orthonormal basis in L2, you can write the L2 norm of u in terms of the Fourier coefficients. What we have, the first thing is that the L2 norm of u squared is the sum of ajk, sum over j and k, sorry, ajk squared, the square of the Fourier coefficients. And if you use that this eigenfunction of the Laplacian, it's easy to see that the gradient squared exercise is the sum j and k of k, k minus, I think it was the plus N minus 2, ajk squared. Just, as you see this, just compute here the Laplacian, just use that the integral of gradient squared is equal to minus, you integrate by parts, so this is minus the integral of the Laplacian of u times u. And now u, this is the sum ajk, okay let me write down, so this is minus the integral, so you have the first term is Laplacian of u, but Laplacian is linear, so Laplacian is the same, these are numbers, these are functions, Laplacian entering inside, so this is the sum of j and k of ajk Laplacian of yjk times sum over l and m, a, l, m, y, lm. And now you use that the Laplacian of yjk is equal to a multiple of yjk, so just use the formula for the Laplacian, and then use that the R of the normal, so the only term which survives are where j and k is equal to lm, and you get this, so I like to finish the computation, but just use the fact that these are the functions. So you have this, so you have these two formulas, is the analogous on S1 where you have the Fourier coefficients and here you have sum over k, just the coefficients, so it's k in dimension one, sorry, k squared. So where are we now? We are here, and let's go back to my information. So we know, so integral of u, by assumption in my case is bounded by constant epsilon, normal of u l2, but what is this integral of u? The integral of u, so the key point is that the first eigenfunction, so what is the first eigenvalue? The first eigenvalue is k equal to zero, so Laplacian zero, and the functions which have Laplacian zero in the sphere are constants. So y, actually y10 is one. So this is the first eigenfunction. Well, you have to divide by the measure of the sphere, but let's say it's one. So what I'm saying here, this is nothing else that a10 is the first coefficient in my Fourier expansion, because this exactly corresponds to the first eigenfunction of the sphere, which is at the constant. And the second ones, one has to show this, but the second ones are exactly the function that I wrote before. So the integral of xi u being small in l2, well, actually this yxi are exactly, is nothing else than yi1. So the coordinate functions are exactly the eigenfunction on the sphere for the next eigenvalue, that these are the first ones. So here, this term is nothing else than the second term in my Fourier expansion. So these are ai1. So here in this Fourier expansion, what do I have? I have a10 squared plus sum of a11 squared plus the sum for j and k, k greater or equal than 2, of ajk squared. What am I saying?  It's just a fraction epsilon of the total norm. So we know a10 squared plus sum over i a11 squared is over epsilon norm of u l2 squared. So this term is just a very small fraction of this. So in particular, this implies that if I look only at this term, so the normal view in l2 squared is bounded by 1 plus constant times epsilon and the sum starting from 2, jk, k greater or equal than 2, ajk squared. In some sense, the only terms which counts in the l2 norm are the coefficients in the Fourier expansion starting from the second one because the 0 and the first are very small. So these are all the two norm this year. So this is a fraction of the total norm but this is almost everything. What about the gradient? The good thing about the gradient is that there are these coefficients in front. So this coefficient is small. So for k equals 0, this is 0. So it's not very useful. For k equals 1, you get m minus 1. But for k greater than 2, this term is getting bigger and bigger. So in particular, so you observe that k, k plus m minus 2 is greater or equal than 2n if k greater or equal than 2. And so here, the gradient of u, l2 squared, I say, well, this is greater if I just consider the sum starting from 2, k greater or equal than 2 of these, k, k plus m minus 2 ajk squared. See, I forget the first two terms. I don't care. But this is greater than 2n times the sum j and k, k greater or equal than 2 of ajk squared. And this is almost all the two norm. You see? Just this sum is almost all controlled to normal, not to an error epsilon. So this is 2n, 1 plus constant epsilon, norm of u, l2 squared. Like this. So in a sense, this explains also what I said before, why the Poincare inequality on the sphere gets better every time you start to kill again functions. If u is orthogonal to the constant as 0 average, it means that this term is 0. If u is orthogonal also to the xi, it means that all these terms are 0. And so in the Fourier's function, you are that the l2 norm is exactly equal only to this term. And in the gradient, you can just use that this is greater than 2n. And this is why you get a better constant in the Poincare inequality. So the l2 norm of the gradient controls the function with a better constant. And every time you keep your orthogonal to, so in a sense, if you knew that your function was orthogonal also to the next eigenfunctions or to the next, let's say, to the spherical harmonic with k equal to, you would say, OK, let me get rid of the second term, I start from 3, and you get an even better inequality. OK, so the more you are orthogonal to some to the eigenfunction, the better gets the inequality every step. So in our case, here is what we get. And now you see that we are more or less done because what was the inequality we had before?  And now we can just apply this inequality to say that this is greater than 2n, then, OK, 1 plus c epsilon denominator is the same as 1 minus c epsilon denominator integral of u squared over 2 minus m minus 1 integral of u squared over 2 minus over epsilon integral of u squared. I see 2n is greater than m minus 1, so this is, I can say, so 2n minus m minus 1 is m plus 1 and they have a small error, so it's going to be greater than n integral of u squared, n over 2 integral of u squared if epsilon is sufficiently small. So you see, I just did, the argument at the end is not that complicated. I mean, I just expanded everything. This is just simple terror expansion. At some moment, I used the volume constraint to write everything in terms of u squared. I didn't want the terms integral of u. And the key point from this step to this step was to use the fact that by being volume equal to the sphere, so we average almost 0, and by the center constraint, so the two information, integral of u almost 0, and integral of xi u almost 0, gave us this inequality with a very good constant, which was better than this. So it is normal that we had to do something. Remember, because until I didn't fix the barycenter, my set could have been just a translation of a ball. And in the case of a translation of a ball, there is no way I can gain something. So it is normal that I had to do something to adjust the barycenter at some moment because otherwise I would not be able to distinguish...  The translations are my enemies because they are the only transformation which do not increase the perimeter. I just take a set translating, it's the same. So at some moment, I had to kill that. And that was the moment where I fixed the barycenter that I killed kind of the degeneracy of my operator. The perimeter did not decrease on that transformation. Okay, once you are here, let me do a simple fact. L2 norm controls L1 norm. So up to constant, I mean, maybe you will get some measure of the sphere. You get this. And what is this? So here is my sphere. And my set E is some perturbation. Like this. So you see this height is u of omega. So the L1 norm of u is controlling exactly these volumes. Right? It is exactly this area. And this area is exactly the symmetry that they defined yesterday. The L1 difference between the set and the ball. So this is constant. The difference between e and the ball. And you have a square outside. So we got the deficit delta controls e delta b1 square, which was exactly what I wanted. And you see that this argument is kind of, I mean, I think this argument already shows that kind of the deepness of the result because it's already very non-trivial in this setting. I mean, getting this power 2 is already sharp in this very special setting. So the fact that one can get power 1 half all the time is already non-obvious in this setting, so you can imagine that for a general set it's not obvious either. Any questions up to now? The second derivative isn't. Yeah, which is behind that one kind of inequality. Yeah, in some sense, philosophically, you can think about that, this power 2 in the following way. I'll just kind of repeat what comment on that. You can think in the following way. An infinite dimensional space, the space of sets, and I have a function which is delta v. And here I have the space of balls. So b1 and translations. So my function delta is zero at this point, the deficit. The deficit banish on the ball. And then it's non-negative. So delta loop is something like this. It's the graph of delta. And here, so what I want to show is that delta v is greater than a constant and then kind of a distance. So this is a metric between e and the space of balls up to translation. That's the kind of estimate we are trying to prove, right? What I'm saying, but you see that usually, what does it mean for a function? If you were in a ren, a function in a ren. I'm saying that the second derivative, I mean, if you think in a neighborhood, what I'm saying is that the second derivative of my function is positive definite. I'm saying that f of x greater or equal than constant x squared. So my function is uniformly convex near the minimum. So the minimum is non-degenerate. So this is kind of a saying that the h of the delta function of the deficit is strictly positive. It's greater than a constant. That's very philosophical. It doesn't mean anything. But so essentially what we are trying to prove by proving that the constant is, the exponent is one-half, one-half here or two here, is to say that the h of our function at the minimum is uniformly positive definite. So we have a non-degenerate minimum point. The first integral time in the Taylor expansion is the second one. It's a uniform positive. It's not like four-power or so on. So it's not flat. OK. Maybe... Yeah, let's take a break now and then I will start again in 10 minutes with the next part. OK, so let me go on with the stability issue. So what we saw is this argument of Fuglede which proves the sharp stability in equality, power one-half, in the case of perturbation of the sphere. So we have a W1 infinity perturbation of the sphere, a Lipschitz perturbation. I'm going to do... I want to show you kind of a different approach to the stability question which is kind of general. It's going to treat automatically every set, not just perturbation. And then probably either today or at the end of the lecture, most likely next time, I will actually show you a different approach which combined with Fuglede argument allows to prove the stability in equality for every set. That will be more based on a compactness argument. But in the hour coming, I want to show you just a complete proof of the stability in equality for the super-metric... stability result for a super-metric inequality. So my argument is going to be based on optimal transport. Actually, you saw a proof last week by Guido. I'll show you how to prove the super-metric inequality using optimal transportation. And actually, since it doesn't change much, I want to show you that this proof is very flexible and can be used not only for the super-metric inequality but for what is called the unisotropic super-metric inequality. It's just going to be a matter of two notations more. But this inequality is very interesting for applications because it has been introduced by Wolf, a crystallographer, so he's used the study of crystals. But mathematics also has several interests. So there is the following. Up to now, we have the super-metric inequality with this perimeter. And the minimizer was a ball. Why a ball? Because the ball has the most symmetries. It's the most symmetric set you can think of. Now we're going to change the geometry of our space and we're going to say, okay, I want to change my ball. My ball up to now was round. I want to take as ball an arbitrary convex set. So I define my new ball in my space. It's like when you take a Banach space. You put a norm. The ball is not any more round. If you put a different norm, the ball can be different. Okay, so unisotropic, isoperemetric inequality. So I'm going to take let K be a bounded open convex set which contains the origin. So I have the origin and take K. And now I'm going to say K is my new ball. How do I do this? I want to define kind of a norm for which this is the ball. So to do this, I'm going to define first a dual norm. So I'm going to say that the dual norm of a vector nu, by definition, is the supremum of an x in K of x dot nu. This is usually how you define a dual norm in a Banach space. You take the unit ball and then you define the sup over the unit ball. How do you define your norm as the dual of the dual? We are defining dimension. So the norm of your x is the sup norm of nu star less than 1 of x dot nu. And if you take this definition, I'll let you check that K corresponds to the set of points x which have norm strictly less than 1 with this definition. This is not properly speaking a norm because norm of x and norm of minus x may be different. But apart from that, let me call it a norm. It has triangle inequality. It's one homogeneous. It's positively one homogeneous. Now what is the perimeter of a set? So geometrically, so the perimeter of a set, you can think of this, the classical perimeter. So for the smooth case, you can take a set E and then you can take the epsilon neighborhood. So you take E epsilon to be the epsilon neighborhood of E, the two points which are at most epsilon. Roughly speaking, the perimeter is going to be, so if you take the measure of E epsilon, what is this measure? It's going to be the measure of E. Then you have a first order term in epsilon, which is the volume of this tube. And this tube at the first order is really proportional to epsilon times the length of this. So this is epsilon times the perimeter of E. And then you have lower order terms. So the perimeter of E, you can think of it as the limit. So perimeter of E is the limit as epsilon goes to zero of E epsilon minus E over epsilon. But what is E epsilon? E epsilon, a way to define it is through what is called the Minkowski sum. So I'm going to say that this is equal to E plus epsilon times B1. So it's nothing of the set of points of the form X plus Z, where X is in E, and Z is in, let me write, epsilon B1. By epsilon B1, I just mean epsilon times B1, which is B epsilon. But more convenient to write in this way because I'm going to write epsilon K in a second. So this is B epsilon. So I like to check this. So this is the Minkowski sum by definition if you have two sets. And you do a sum, which exactly means that you start to take all possible sums of points in one and the other. And I like to check that the epsilon neighbor is exactly the same as this. Because what you're doing is saying for every point X, you're taking a ball of centered X of radius epsilon, and then you do the union of all these balls. That's what I'm really doing. And this way you get the epsilon neighborhood of the set. So this means the perimeter of epsilon is, for a smooth set at least, is the limit as epsilon goes to zero of E plus epsilon B1 minus E over epsilon. So once you see this definition, you see that to define the perimeter really, you don't need many things. There is just a bad measure and there is the unit ball. So if I say that now, in my case, this is my new unit ball, I can just define a perimeter replacing B1 with K. So what I'm going to do here, we define. So here I'm not going to be as rigorous as I was yesterday with the official perimeter of the soup of the versions, blah, blah, blah, because if I want to do the same here, I should start to use theorem about the structure of sets of finite perimeter. But if you have a bit of the reduced boundary, blah, blah, blah, so a bunch of technical stuff. So just think that my sets are all smooth, because otherwise we should go through technical results in geometric measure theory. But in some sense everything can be made rigorous. There is no issue there. I want to define pK of E, so the perimeter of E depending on K, to be the limit as epsilon goes to zero of E plus epsilon K minus E over epsilon. So in some sense what you do, you do an epsilon neighborhood, but what the epsilon neighborhood is not any more measured with the classical equivalent distance, but with the norm associated to K. And actually there is a formula for this, which is not real to prove, is in books of complex geometry, that tells you that this is the same as the integral over the boundary of E of the dual norm of the exterior norm. So here you have the set and here you have nu of E, the unit outer normal, unit equivalent sense, just classical normal, and then you weight it, you see, the classical perimeter would just need to say put one here. The classical perimeter you integrate one. Here instead of you, you integrate some function of the normal, where this function of the normal is this dual norm. So this means that although this is a unit vector at every point, depending on the direction, because this norm is not the classical norm, depending on the direction this weight is going to be different, this number is going to be different depending on the point, who is going to depend on every point of the direction of the normal. So this is my new perimeter. We have an isoperimetric, so this perimeter has been introduced by Wolf, 1905, Wolf was a crystallographer, and actually he kind of proved, but as a crystallographer, so there was not real mathematical proof, but he said that what is called the Wolf inequality, so the perimeter of a set is greater or equal, so he didn't really prove it, he's kind of claimed it, than this. And here you can recognize the superimetric inequality of yesterday, where I'm just replacing b1 by k. So that's also why yesterday I wrote down equality in that way with measure of b1 inside, because you just replace b1 by k at every step and you get the anisotropic isoperimetric inequality. So this is called Wolf inequality, and you have equality if and only if e is equal to k up to translation and dilations. So e is equal to x plus rk, so you take a k, you can dilate it, and translate it. OK, so let me define a deficit associated to this. So this inequality is, of course, a scaling variant, you can always dilate e, but actually you can also dilate k. If you dilate k, this is going to change, but this is going to change in the right way, otherwise the inequality will be false. So without lots of generality, just to make things scale one, I'm going to assume that measure of e equal measure of k equal one. Just to avoid the degeneracy. Now I'm going to define delta k of e to be the error. So it would be pk of e minus n. I mean, I copy it, but I could put one. I understand what you're saying. So delta k of e is this quantity, which is non-negative. And we're going to define an asymmetry, which is the infimum of this immediate difference between a and translation of k, where x is in a random. So the symmetry is del one norm. So here I'm not going to compare any more a with. Oops. So if this is k and this is e, I want to measure del one distance between a and k, e and k. And so the question that I want to ask, that I want to address is, do there exist some constant c and k and the exponent alpha and k positive such that, so a k of e is less or equal to c and k delta k of e alpha and k. So it is exactly the same as yesterday and it's written k is everywhere. But the question is the same as yesterday. So this question has been first investigated in this anisotropic context by Esposito and O4, where they proved something like alpha and k goes like one over n squared. There is an explicit formula that now I don't remember, but they proved that result for some exponent which goes roughly in the dimension as one over n squared. And then that's exactly the generalization I mentioned we used yesterday, what we did with Francesco and Aldo, so Maggi and Pratelli. In 2010 what we proved is that alpha and k is always one alpha and actually c and k can be bounded by something like a number which is less than 200, I don't remember, and to the power something around 11, maybe 9, not different from 9, something like that. So there is, this point is always one half, no matter what is k, and there is a constant which doesn't depend on k, depends only on dimension. The fact that you get a constant depending on k of constant based geometry, and I think what is most important is the exponent, that we will always get one half. And the starting point to study this stability test, to prove this estimate that I want to show you, it's the proof of the superimetic inequality which was given to you last week by Guido, that I'm going to repeat right now to show you that it works exactly with no changes in this setting. So last week he gave you a proof of the superimetic inequality using optimal transport. So let me prove the superimetic inequality. So I want to prove this unisotropic superimetic inequality. So proof of the superimetic inequality. So what Guido did was the following. He took a set E and he took the unit ball and he transported E onto the unit ball. So in this case the ball is not the ball anymore, it's k. So I have my set k, E. How do you transport a set to another? You associate to E a probability measure. So I'm going to define mu to be the characteristic function of E over the Lebesgue measure of E to be x. So it's a constant density inside E. And I'm going to define nu to be characteristic function of k over volume of k dy. So by optimal transport theory, what we know is the following. If you give me two absolutely continuous probability density, there exists a transport map to a map T, which is the gradient of phi. So there exists phi converts such that T, the map T, which is defined as the gradient of phi, sends mu onto nu. So this means the push forward of mu to the map T is equal to nu. And in addition, this condition gives you a Jacobian identity for T. It tells you that the determinant of the gradient of T is equal, so I don't know exactly what you saw last week, but maybe if you have a question, let me know. Let me assume that you know this, that you get, I guess he did it, you get this number. So this comes from the fact, so why this is true, that the Jacobian terminal T is this constant. The point is that you're sending a uniform density, so this is a constant density, onto a constant density. So if you send constant density to constant density, the Jacobian must be constant at every point. And the only constant which matches the volume, so you have the volume of this guy that must be equal, I mean here you have renormalization. In order to, since these volumes may not be different, it's the same, the only way to get constant density here is that if the determinant is this number, you have a volume, right? If the volume were the same, you see that you get 1. If volume of Vm, volume k are the same, you get determinant 1. But if you're sending a ball of radius 2 onto a ball of radius 1, the Jacobian will be 2 to n. So that's, there is a constant number and the only possible number is that. So that's the Jacobian determinant. So I'm going to use this result. So let me show you now two properties, actually three properties of the transport map. So property A, norm of T of x less than or equal to 1 for every x in A. Why this is true? Because by definition, k is the unit ball for my norm. So since a point here is sent here, this point, so this is because k is the set of points which have norm less than 1 and my map is sending here onto k. Property B, I'm just going to copy it, is that one, determinant of grad T equal k over E. So just copy that, just to side. Then property C. So property C, I guess you already saw this as well, property on the divergence of T. So what is the divergence of T? The divergence of T, T is the gradient of phi, so this is the Laplacian of phi. Let me, I'm going to go a bit fast because I assume you already saw this proof. So this is the, if lambda i are the gain values of the action of phi, this is the sum of the lambda i. The Laplacian is the sum of the gain values and these are non-negative because phi is convex. And so the sum, this is, I can write as n 1 over n sum of the lambda i and this is an arithmetic mean of non-negative numbers, this 1 over n. So this is greater than n times the geometric mean of the gain values and the product of the gain values is the determinant. So this is n, the determinant of the action of phi which is the same as the gradient of T. Please stop me if you have questions, but you should have seen already this. And so now I can prove the superimetric inequality because it's actually going to be just two lines. So PKOV, we say it is the integral on the boundary of the dual norm of the normal. Now I use property A, so T less than or equal to 1 to say that this is greater if I put T. Now here you have a vector and another vector, norm and dual norm. Norm times normal norm is greater than scalar product because the dual norm are defined as 2. So it's greater than the integral of the boundary of T dot nu E. Now by Stokes, this is the introvery of the divergence and now you see greater or equal n integral over E of the determinant of grad T, 1 over n, and then I use B to say that this determinant is K1 over n, E1 over n, and then I have volume of E coming from the integral and so you get exactly this, which is exactly the inequality. So you see the proof is exactly the same. There is no extra difficulty in this anisotropic case. Okay, so here we are. We have a proof, two lines proof. The argument as it is is due to Gromov. 99% of something interesting in geometry and deeply is done by Gromov. So let's see. We want to, what we want to do now is to make this argument quantitative. So what is delta? Delta K, remember, we want to prove that the deficit controls the symmetry. So what is the deficit? The deficit is this guy minus this guy. So we want to say that if the difference between these two guys is small, then the sets must be close. But if this is small, it means that every time here I add smallness. So I didn't lose much in these inequalities and I want to use this information. Every time you want to prove a stability in equality through a proof, you always want to do one check. The quality case. If you have a proof where you're not able to use it to prove the quality case, you're not going to be able to use it to prove a stability result which is stronger. So you want from this proof to deduce the quality case, at least. This is like step zero. Otherwise, don't start because there is no hope you can prove something stronger if you're not able to do it, at least that. So let's do the step zero first. At least formally. Step zero, a quality case. This is kind of a check for yourself. You want to see if you have hope to do more. A quality case. So if delta K of E is zero, it means that from the first to the last step I didn't lose anything. So this implies that the integral of E, the divergence T, is equal to N integral of E of determinant of grad T, one over N. So this was the sum of the game values and this is the product of the game values one over N. So this means that when you apply arithmetic geometric inequality, you don't lose. So the only way you don't lose in arithmetic geometric inequality is if all the numbers are the same. So this is a function of X and this depends on the points. So let me put an X to say that these are the eigenvalues of the point X. At every point you have different eigenvalues. So this implies that you have equality in AMGM for every X, which means that lambda one of X is equal to lambda N of X for every X. So all the eigenvalues are equal. But in addition, the product of the lambda I of X in my eigenvalues is the determinant and this is K over E. So this is a constant independent of the point. So they are all equal, but the product is constant. So this implies that lambda I of X equals some number alpha positive for every X. So all my eigenvalues are all equal and all equal to the same constant. But what are these? These are the eigenvalues of the action of phi at the point X. So you have a symmetric matrix which has all the eigenvalues equal for constant. It's a multiple of the identity. So this implies that the action of phi at every point it's alpha times the identity matrix. But the action of phi, this is the same as the gradient of the transport map. So the gradient of the transport map is the action of phi, and it's a multiple of the identity. Alpha, which is a number times the identity. So I'm saying that my gradient is a multiple of the identity at every point. But it is constant. So this implies, model of smoothness, whatever, that T of X, so which maps have gradient alpha times the identity are, is the map, alpha of X plus up to a translation. So T of X must be alpha of X plus X dot for some X dot. I want to keep this argument, so that's why I'm trying to remain there. OK? OK, but what we have, we know that T of E is equal to K. Because T is sending A on to K. And T of X is equal to alpha of X plus X dot. So this implies that A is K minus X dot over alpha. T is mapping E on to A, E on to K, T is translation plus direction, so the sets were the same. In particular, if you remember that I assumed, I mean, here I'm not using it, but if you renormalize so that everything is one, alpha is one, and you don't have even the constant alpha. Here I'm trying just to, I mean, I'm keeping alpha just to make the argument a bit general here and try to understand what's going on. OK? So this is good. This is a good starting point. It means that at least the quality case we can get it. So that's the shooting. We have some hopes. Is it clear the quality case? At least formally. I never worried about smoothness, but besides that, I hope it's clear what's going on, because now we have to make quantitative what I just made. Kind of, I did in the quality case. So let's go back. OK, we like this strategy, and in particular we discovered something from the quality case that one in a quality that we want to use is this one. It's the fact that from a quality here we had the gain values equal, gave us that the matrix was a multiple of the identity here. So now we want to make this quantitative, like, OK, now things are not going to be equal to zero, but maybe we can say that are almost zero in some sense, very small. So let's go back now to our problem. Now delta K is not zero anymore, but it's only small from the proof. So I want to, so quantitative result from the proof there are two information that I get that I'm going to write down so that then I can also cancel the proof. So one, which is the obvious one to use, that delta K there or between these controls the difference between these two terms. This minus this. So this, we want to use it. We discovered it that is useful. There is a second one which is, you still don't know that I want to use it, but I know it because I know the proof, which is this one. So there are two things. So property one that I want to use is that delta K, so this minus this, controls this minus this. Integral over the boundary of E, nu E star, one minus norm of T. Second property, delta K of E is greater or equal than the integral over E. So that minus that is divergence of T. Let me put over N here minus determinant of grad T over N. We are not going to worry about constants. So these are two information that I get from the proof. Let's start with the second one to work with the second one. So here, what is this? This is the sum at every point is the sum of the lambda i one over N minus the product of the lambda i one over N. So you have arithmetic mean minus geometric mean. OK, maybe, and I know, so this is, OK, here I don't want to have half a popping out. So let me maybe write here in the box E equal K equal one, which is going to imply that the determinant of grad T is one now. Let me, so that I don't have a strange number around. So now the determinant is one because I normalize. So in this quantitative estimate, it is useful that I normalize stuff. OK, so you have an arithmetic mean. Well, this number is actually one. And you know that if you have a quality, all the numbers are equal to one. You would like to say some stability result, some quantitative result. Now this becomes a pure algebraic fact. I mean, it's an exercise if you want. It's not trivial, but you do it. You can do it.  It's a matter of inequality in numbers that if you have an arithmetic mean minus a geometric mean, this is greater up to a constant. So let me start to put this is up to a constant to this. The sum over i of lambda i minus one square divided by lambda N, where I ordered the game values by sides. So this is a point-wise inequality. So I'm claiming that if, what I'm claiming is that if you give me N numbers, so if you have one over N sum of ai minus one and product of the i equal one, so if you want this is greater or equal than sum of ai minus one square divided by the max of ai for every ai such that the product of the i, sorry, is one. I mean, I'm just writing in small what I just wrote here, actually. So I'm just saying this step from here to here is a pure fact about real numbers. You give me N numbers, so this product is one, and then the arithmetic mean minus one controls that quantity up to a constant. So here there is a constant, which depends on N, so the dimension, because we have N numbers. So this is a fact, and you can find a proof in the paper or I can give you a reference, but it is an exercise. So what we know is what then? That delta controls the integral of the sum of the game values. So these are all functions of x, remember. The game values depend on x, so at every point we have different game values. But what we would like to say is that the game values are close to one, so here we have this division by a lambda N, which is annoying. How do we get rid of it? Just by older. So this part, it's kind of easy. What we have is that this, so let me look at what it was. Sum over i of lambda i minus one. This is the same as the introvery of the sum of lambda i minus one divided by square root of lambda N multiplied by square root of lambda N. Why I'm doing this? Because here I want to get something like this. Because now you can say that a times b is less than or equal to epsilon i squared plus one over epsilon b squared for every a and b positive. It's just young inequality. So here I can say that this is less than epsilon times this. Sorry, no, one over epsilon of this and epsilon of that. So it's less than one over epsilon, the introvery of the square of this and the square of this is the same as the square of this. I mean, it's the same as this. There is a square outside, but the square outside and the inside are comparable up to constant. So the sum of lambda i minus one square over lambda N plus epsilon and then you have the square of this which is lambda N. OK, so now this term we know is controlled by delta. So this term we know to control. What can we say about this? Well, the good thing here is that I can do the Fourier. Just subtract one and I read it. So I just do a trivial inequality, triangle inequality. So I just subtract one and then I read the term. And now this for sure is less than this because here I have the sum of all of them. Here I just have the last. So this is less than the sum of lambda i minus one. So what I'm saying, I'm saying that the integral of the sum of lambda i minus one is less than or equal to one over epsilon square root delta, here delta, up to constant. So constant over epsilon delta koV plus epsilon times the same quantity plus epsilon measure of this one. OK, now if epsilon is small, I can just take this on the other side. I can absorb this with this term. And then I optimize in epsilon. So I just first take the other one. So I get one-half integral of e over e of sum of lambda i minus one is less or equal than constant. And then I have delta k over epsilon plus epsilon. And now this is equal to constant square root of delta koV if we choose epsilon equals square root of delta koV. So that's the best choice I can do. So that's my epsilon. So up to here it was free. This was true for every positive epsilon, but then I chose epsilon to be the square root of delta. So what I discovered is that square root of delta controls this quantity. So let me probably write that down. And what is this quantity? This is the distance between the gain values of my matrix and one. So this is exactly the L1 distance between the action of phi and one. So maybe let me cancel this and I'm going to write a new inequality which is going to be two prime. So equality two, information two was this. Now two prime, new information, constant square root of delta koV is greater or equal than the integral over E of the distance between grad T minus the identity. This is the action of phi. It's the same as that. Because I mean point y, this is a norm. I mean the sum of the gain values is controlling exactly how far is this symmetric matrix to this symmetric matrix. I can do this because these are symmetric matrix. So controlling the gain values controls all the matrix. If this is symmetric, I would have problems. So it is crucial that I have symmetry. So this is my new information. So you see in the previous case what we say is that if delta is zero, the gradient of T is the identity. Here we're saying if delta is small, the square root of delta controls the L1 norm of the gradient from the identity. Now what I did before, I say the gradient is the identity. So T is a translation. There was the alpha, but alpha is one now. So if gradient T is the identity, it was exactly the entity who would say T is a translation. Problem, so because this is, okay let me say, so this is what? This is the interval of gradient of the map T of x minus x. You can see this way. That's the same. So up to now, in contrary, the gradient of T minus x. And I would like to control T minus x. I would like to say that the transport map is very close to be the density map after translation. Okay, so what do I need? You know that some function has very small gradient on a set, and you would like to say that the function is small on that set. So this recalls, this can recall to us what we did also this morning, kind of a Poincare inequality. The gradient controls the function. So I'm going to write two inequalities here. The first one that comes to your, I mean at least they come to my mind, when you have these and you would like to control the transport, so possible inequalities, inequalities. I'm going to write two. So the first one is Poincare, which tells the following that if you have a function, let's say you have a set E, and you take the gradient of F. Okay, here you have some assumption on E now. So it's going to be greater than a constant, what I call CP of E, the constant will depend on E. The integral of F minus its average. So Poincare inequality tells you your function, the gradient of the function controls how far your function is from the average with a constant which depends on E. You take any book, it will tell you if you have a leap sheet's connected domain, you have a Poincare inequality. Second possible, which is another one. So here I have an information on T on the boundary of E. So if I want to put this information together with these, maybe I could try to use these to get some information on the boundary. How can I do this with what is called a trace inequality? Trace inequality tells you that the gradient of F inside controls, so I'm going to call it CP of E, how far the function is on the boundary from its average. So it's like a control of F on the boundary of the domain. So this is what is called a trace inequality. So you go to the boundary. In general, this inequality is stronger because you are getting information on a lower dimensional set. So here are two possible inequalities. And since here I have T on the boundary, I'm going to apply the trace one. So let's combine 2 prime plus trace to get the following. That there is a constant, and now here, so there is a constant, C square root of delta K of E is greater or equal. So I'm going to apply this to this function, to the T of X minus X. And then you get here C T of E integral over the boundary of T minus X. So here you should add a translation, but it really would be plus the average. So another vector. But let me write directly this up to a translation of E. What am I saying? I'm saying, okay, my transport map must be the identity plus a translation. Okay, either I keep the translation here or I can just take my set at the very beginning and translate it back. And then I kill the translation in the transport map. So up to a translation of... I mean it's irrelevant, but I don't want to enter having X dot everywhere because it just mess up things. So it's clear that if the transport map is a translation I can just kill it at the very beginning by putting my set already in another position because I always move my set around. So let's say that here we have this information. And let me move on. So now I say I have this that I'm going to call two seconds and I have one. Let's put one and two together. One and two seconds together. Plus two seconds. Just use triangle inequality. You get that delta K of E plus constant over CTE delta K of E with a square root is greater or equal than T minus X. Okay, here there is another thing. So here I have a priori of the clean norm but all the norms in a random equivalent. So instead of the clean norm I can use my norm up to constant. So here I put my norm. Here I use that these and these are equivalent. These will put constant which may depend on K. But at the moment I don't want to... These are universal constants anyhow. So I can put here the other norm and then I put this one and here I'm going to also use that... Okay, this is the dual norm on the unit normal or the other normal. So all these vectors are on the sphere. So the norm on the sphere is going to be bounded from above from below. So this is bounded between two constants, this number. So here I use that these so the new E star is bounded by some constant positive. So I can get rid of this and then I put one minus T. Actually I can put just the modules because T is less than or equal to one. Also here I can put the modules. This is a non-negative quantity. T is less than or equal to one. I should have put the modules from the beginning. Please interrupt me if there is something... I'm just using elemental effect now. If you were in the Korean case, I wouldn't even worry about this but really I'm just using that the norms are equivalent. This norm is bounded from above and below and stuff like that. And now you're there and you just say, okay, let's apply triangle inequality. So trivial triangle inequality tells me first of all that this is... So let me copy first this and then the modules of the difference is greater than the difference of the modules. So this is norm of T minus norm of X and then triangle inequality I can get rid of the T. So this is greater than the integral of the boundary of E of norm of X minus one. I just used twice triangle inequality. So what did I discover? Did I discover... Okay, this you can even forget because delta is small so square root is, of course. This can be absorbed in this. Square root of delta times a constant which I'm going to come back to controls this quantity. And this quantity, what is this? Let me draw a picture. So let's say that this is my set K and now I have my set E. So this is the origin, let's say. And I take a point on the boundary X which is there. I take this line. So K is the unit ball. So here on the boundary norm is equal to one. This point is norm one. And this... So what is this length? This is norm one. So gradient distance is comparable to the norm distance. They're all up to dimension constant. This is comparable to this. So norm of X minus one is roughly speaking this length. And so what am I doing here? I'm integrating on the boundary norm of X minus one. So for every direction I'm integrating this length as X varies on the boundary. And so it's not difficult to convince itself that this controls E delta K. Oops, sorry. Once you integrate here the problem of all these regions is convincing enough because X minus one is exactly this length and then you just move X around and when X is here you move, wow. You move there, wow. There is a Jacobian factor to be careful of but it works, believe me. So there is... I mean it's a bit technical because you have to compute exactly the Jacobian factor but geometrically it's very obvious that it has to work because X minus one is actually doing this length and then you're just integrating over the boundary. So what would you get? For every A I prove the following. E delta K up to a translation. It is a translation. Bound by constant which depends only on the dimension in K delta K of E, C of E. OK. So this is... could be a thing nice but actually this is not what I wanted because what I proved is that square root of delta control C delta K with a constant which depends on itself. I mean a constant depends on the set. And in particular I can give you tons of examples set where that constant just plus infinity. No, plus infinity. Yeah, the denominator is zero so that is plus infinity. Actually I'm going to show you right now what are our enemies. So here is the problem. The constant depends on E. I wanted a constant depending only on the dimension. OK, but what are our enemies? So I will five minutes and we should sorry that time should be over now. I'm going to take five more minutes. So what are our enemies? There are two different enemies. So let me go again to the inequality, CT. So trace. So you want to see you want to look at whether this inequality you want an inequality like this. Sorry, this is E. This is Boundary V of F minus the average of F for the boundary. OK, first enemy is connected. This is CT of E is zero. If you have two connected components you get zero. Why? Because you can always do in such you can put two constant so you take F equal alpha and F equal minus beta and you choose alpha and beta in such a way that the average is zero and then the average is zero so the gradient is going to be zero and then this interval is not zero. The interval of both of F is not zero. So I can always make the average zero this has gradient zero inside E but this is not zero. So this is our first enemy. When F is disconnected you don't have any trace constant so that inequality is completely empty. I mean this is just plus infinity I didn't prove anything. Second back case is this one. You take a smooth set even very close to the ball and then you assume that it has a kind of a cusp like this. In this case you can construct examples so maybe let me blow it up here let me do a bigger picture. So the boundary looks like this. You can take a function F which is like F equal zero F equal one here and then you do a smooth interface like it goes very smoothly from zero to one and then roughly speaking the gradient is going to be all concentrated in something like this and then here on the boundary term you will have an integral over this and you can check the point is that this length is much smaller than this and this is like the gradient term on the right hand side and so you can make if this is a cusp like older continuous you can make the ratio be in this length and this as small as you want and so you contra need inequality. So by taking a sequence of function F which goes from zero to one very fast on a very small piece you show that CT again is zero. So this is another bad case but we have a hope so hope is the following if you give me a set and you start to make kind of the set bad so you start to add cusps around so this will be your enemy every time you add a cusp you pay perimeter it costs you perimeter to draw a cusp so you think how many cusp can I have cusp could be infinite I can always make infinite cusp very small but every time there is a cusp this area cannot be that huge because if I have a very big cusp every time I pay a certain amount of perimeter so the idea is that ok let maybe I can cut them away and after I cut them away I have a nicer set so let me state the reduction lemma lemma there exists delta n depending on the dimension positive and actually this is specific such that if delta k o v is less or equal to delta n delta n then the following is true there exists a subset g I call it g because g is good it's a good set such that the three following properties are true so the measure of e minus g is very small constant delta k o v the deficit of g is controlled by the deficit of v and key property the trace constant of g is bound from below by dimensional constant cn so the set g is done in this way maybe you had also multiple connected components in e you just throw them away you cut all the cusps and you get a set g and what I'm claiming is that this set g is almost full measure you cut them very little fraction of e by doing this you control this by the deficit so the deficit tells you how much mass is going to be so g has a universal trace constant one minute and I finish so now what do you do you go back to the proof I just gave but instead of applying the proof to e, apply it to g so by the previous proof applied to g what we get is that g delta k is bounded by constant now by the trace constant of g square root of delta k of g but now the trace constant of g is bounded from below by a dimensional constant so this is bounded by universal constant and in addition delta k of g is bounded by delta k of e so square root of delta k of e so this is just from the previous proof and this is from the lemma the lemma that I just stated and now I use that now I just do triangle inequality e delta k is less or equal than e minus g plus g delta k e minus g is controlled by the deficit constant delta k of e and g delta k is controlled by constant square root of delta k of e and so the sum of the two is given in a square root because delta is small and the term is controlled by the square root and so this is the proof of inequality with a general constant so the only thing I didn't prove to you is this reduction lemma this lemma is kind of tricky but it's based on a kind of the very beginning when we first give a proof was based on kind of a Zorn type argument we said let's take the worst the largest set by e for which in some sense so in some sense we looked what it means not to satisfy so we take a subset of e and we think what does it mean that this subset does not satisfy the trade inequality and so these are the sets that we don't like and then we take the smallest possible one if you want or the complement if you want is the largest possible one in volume we at the beginning we used a Zorn type lemma then we removed it but we took a maximal one and then we say we prove that the maximal set actually has to satisfy this property and it's based on the kind of elementary argument but it's a bit long so I don't want to enter into this but so modulo this reduction lemma you see that the proof is at the end and modulo the fact that every technical point about regularity but this is the full stability result for the anisotropic case so I'm completely out of time so I stop here and apologize for the delay