So last time we saw that how to write the geodesic in the space of probability measure. So we had the space of measure given a pretty nice space. Now I am interested in the case in which X is an open set. We were working actually with X in geodesic space. Jodezic space means that, so as d, be a jodezic space, which means that between every two points there is a unique, no, not a unique, just there is a minimizing jodezic, which has no reason to be unique, and the length of this jodezic is the distance between these two points. So x can be a mermanian manifold, can be Rn, can be a convex subset of Rn, but if you take, for instance, the Euclidean distance, and the general open subset of Rn, this is not a jodezic space, because if you take something like this, with d equal the Euclidean jodezic, this is not a jodezic space, because the distance between these two points is not given by the length of the jodezic. I mean, I want the jodezic to stay into the space, so I cannot do this. OK, so xd is, so this is not a jodezic. So when xd is a jodezic space, and I look to the space of probability measure on x with finite second moment, and I, for some x dot in x, I can say some or every x dot is the same, obviously, and you can endow this space with a metric, which is, so the distance in between two, let's say the distance square between two probability measure is just the infimum, actually the minimum given by the Kantorovich problem, so mongol plants, which are connecting mu and nu of, so now we have a metric space, which is p2 with the vastest time distance, and we have seen that mu t is a jodezic, if and only if you can find a measure, nu, which is a measure concentrated, a probability measure on jodezic cars, such that mu t is given by the push forward through the valuation map of eta, OK? So you've seen something which you can somehow summarize in the following way, a jodezic on the space of measure is a measure on the space of jodezic, OK? And so let me look to some, so say you are on RD or on Eremanian manifold and we have seen that if you start with a measure which is absolutely continuous with respect to the volume measure, OK? Or in case of RD, something like this, so the volume measure is just the back measure, when you start by a measure done like this, then the jodezic, so and then you take another measure, the jodezic between mu 0 and mu 1 is simply given by mu 0, where t t is just, so we have seen that between two measure, if the starting measure is an, the starting measure is absolutely continuous with respect to the volume measure, McCann's theorem gives you a map and this, and this, and the map is just given here with t equal one, phi is a second cave, so it's d squared over two concave map and you see this, this, the jodezics can be written explicitly through this map, OK? So in case you are in RD, in Rm, the map t t, you can write in this way, so this is one minus t times the identity plus t times gradient of u and u is convex and is the convex map such that grad u sharp rho 0 dx is d1, d is mu 1, OK, and this is the map given to you by the Brignetti theorem, OK, this u. You have seen that in RD, this expression takes this form, OK? So we have this nice representation formula for jodezic and what I would like to do now is to show you how this, I mean how these jodezics, I mean this formula can give you some nice information on the geometry of the space, which is on the space xd. So to do this, it's interesting to look at functional, define on the space of measure, which are convex along jodezic. So you have probably seen in the classes of Ambrosio that for instance, a lot of information can be derived by saying that the entropy is convex along jodezic, OK? And I'm going so to introduce, I mean to talk a little bit, because I'm going to give you a proof just in particular case, which is the one from which we can deduce some nice, I mean it's not the only one, it's the one just giving an example. So if you have a functional, say f, which is define the space of probability measure to r, say that this functional is displacement convex, this was introduced by McCann, so this functional is displacement convex simply if the map, if, OK, now I give it, this is not the best definition you can give, but so if for every jodezic mu t, the map, which has to say t to f of mu t is convex. So this is a map from 01 to r, so it's convex in the sense of calculus one, it's just a convex map of one variable. So let me give you some example of interesting functionals on the space of probability measure. So a nice functional, so there are essentially three kind of functional, which are of some interest in the space of probability measure, and these are like the first one, is the integral of v d mu for some function v, and these you can think to be like a potential energy. So you see if mu is the density of some particles, and on these particles, when there is a potential, as some force is given by a potential, this is essentially the total potential energy of these particles. Okay, and another nice functional you can look at is f t of mu to be mu y. These you can think to be like an interaction potential, so interaction energy. You see, right?  So the gradient of this, so this is the total interaction energy. And finally, which is the one I'm going to use, is something like this. So it's euro, say devol, this in case you can write mu as rho devol, and plus infinity otherwise. And this can be seen like an internal energy. And these are, I mean these are just three kind of functional, which have some interest. And then you would like to know when this kind of functional are displacement convex, and they're not going to give you the general case. So just let me mention, for this two is pretty simple, maybe you can prove it by exercise following the proof now I'm going to give for a special case of this functional. So for instance in Rn, f1 is displacement convex if and only if v is convex. In the same here, gain in Rn, f2 is displacement convex if and only if v is convex. And here the condition is slightly more complicated. I can give it to you the condition then I make the proof in a particular case and then you see that once you understand that proof you are going to do all this proof. I mean just as an exercise in this case, again in Rn, you can prove that u, so f3 is displacement convex if and only if the map R to, what is this, Rn, R to the n, u of R to the minus n is convex and decreasing. Again like positive for me decreasing me in what usually you call not increasing. But too much negation early in the morning it's a mess, so it's decreasing. So I will just show a particular case of this statement. So look to this functional. Un of rho to be minus the integral of rho to the 1 minus 1 over n devol, this in case, so I'm on the remaining manifold, OK, I'm giving slightly more general proof. So in case your measure is absolutely continuous, so let's write this way. So this is a function which is a priori is defined on the space of measure, but I'm just going to put it to be plus infinity in case mu is not absolutely continuous with this. And here there is a minus. So we have dysfunctional, which is the kind of internal energy, and I would like to know if this functional is convex on the space of geodesic. So I'm just now I'm working on a Riemannian manifold and I will try to show you that knowing that this functional is convex along the space of, along geodesic and vastestane space gives you some geometric information on the manifold, OK? So we start by a measure, which is, so our starting measure is this one. Say we end up on a measure, which is, so again in sense, if I know that the measure is plus infinity somewhere, it's not so interesting. So just look the case in which we have absolutely continuous measure, then we know by McCann theorem that the geodesic is given by 0 dx. So this is our geodesic and again t t is just given by x p x minus t grad phi, and phi is second k. If you do the change of variable formula, as we did in the case of the quadratic cost, with the same computation, you can show that you can write down an expression, so this will be like rho t devol, yeah? So you would like to know which is rho t, OK? You can prove this measure to be absolutely continuous, but you would like to know which is its density, which is the relation between the map, the starting density and the density of this measure, and what you get here is that rho t is just given by rho 0 over j t, composition t t to the minus 1, where j t is just the determinant of gradient t t. I'm not interested in this point in regularity issues, OK? You actually don't need any, not too much regularity on the map to do this computation, so let me do the computation without caring about regularity issues. So this is being done, OK? So we have this, so we look to the functional u of rho t. This is just minus the integral of rho t to the 1 minus 1 over n in devol. Now I just use this formula to say that this is minus rho 0 over j t, composition t t to the minus 1, the 1 minus 1 over n devol. Then I make the change of variable y equal t t to the minus 1 over x, and this is just a computation that this gives you the integral of rho 0 to the 1 minus 1 over n times j t to the 1 over n dy. Yes? Rho t. J t is the determinant of gradient t t. Is t, is t, is t? This is also t, yes. It's the geodesics of time t. OK, so we have this, we have seen that this functional can be written this way, and so you see which is the, the advantage is that the dependence on t is just here. So if, so let me state our dream statement. If the map t into j t to the 1 over n is concave, then the map t to u rho is convex. OK, this is the reason for the minus here. OK, so we have seen that studying the convexity to this functional reduce to understand the concavity of this map. OK, so when is this map concave? OK, so the map, which has to say it to, so you take for any seconcave function phi, you can construct this map, you can compute say this gradient, then compute the determinant of this gradient, then you rise it to the 1 over n and you ask when is this map concave. So the first simple answer is in Rd. So that t t can be equally written as 1 over t identity plus t grad u for some u, which is convex. OK, so grad t t is simply, 1 minus t identity plus t action of u for some convex function. So we are trying to understand just the concavity of this map minus t identity plus t action of u rise to the 1 over n. OK, and then this is a simple lem that if you take a matrix a, which is symmetric, positive, then the map, which a associates the determinant of a to the 1 over n is concave. And this is a simple computation you can do. So you see that both these metrics are positive since u is convex. So you see that in Rn you have that t t is actually true that in Rn this functional is displacement convex. And what happens on a Riemannian manifold? So on a Riemannian manifold, so let me raise here. So the goal is to understand if this map is concave. Jt is the determinant of gradient t t. And t t is just given by x minus t grad phi, OK? So we want to understand when this map is concave. So in some sense we have to understand what is the structure of this map. So there will be a very nice computation. There is a very nice computation in Riemannian geometry to do this, which I cannot do with all the details, fortunately. So you see, what is this? This is a geodesic. So you take a point x, you have this minus grad phi of x, and you are shooting this geodesic in this direction, right? Then to compute the gradient of this map, I mean, you just go to the point x plus dx, right? You have this new gradient, which is minus grad phi x plus dx, and you shoot this other geodesic. And in some sense you have to divide by dx and send dx to 0, OK? This is the way you compute a gradient. So what are you looking at? So you have a field of parallel geodesics. So dx is very small, right? So you have a lot of geodesics, and you are taking a derivative of these geodesics with respect to the parameter. And you are looking to the fields, which you get in this way. So this is, say, you have a family of geodesic t parameterized by a parameter, in these cases, like dx. And then you are computing this derivative when dx equals 0, OK? So what is this? This is a Jacobi field, right? You know from Germanian geometry that if you have a family of geodesic, which depends on a parameter, and you differentiate with respect to this parameter, so you get a vector field. This vector field is going to solve the, this vector is going to solve the Jacobi equation. So, OK? So as I said, I cannot give you full details of these proofs and of the Germanian geometry, which is behind, although it's very nice. So a good reference for the computation that I'm going to sketch is chapter 14 of the Lenin book, whose name is something like Jacobi, no, there is an appendix, which is named Jacobi field forever, so you understand that. It's kind of related. OK, but what, and just give you some formulas. So you can write down this matrix in this way. So let me call this matrix Jt, which is J, which is J different from the other one. And so Jt, what is it? Jt ij is just given by grad tt ei dot ej, where these are vector field, which are time dependent, and that just a frame, you are parallel moving along your geodesics, OK? So eit is an orthonormal frame at the point gamma t. So gamma t is this geodesic. This is an orthonormal frame at the point gamma t, which is parallel transported along gamma t. You know that when you have a metric, you have the Riemannian, you have the Levy-Civitagon connection, so you can define what is parallel transport. So you start with an orthonormal frame at gamma of 0, and then you parallel transport this orthonormal frame along your geodesics, OK?  And in this frame, with that notation, the equation becomes very nice. So in this frame you get the following question for Jt, which is a metric, so Jt plus RT is equal to 0, where RT, well, the only, I can give you the right expression. And this is the Riemannian tensor applied to gamma dot t. So you know that the Riemannian tensor is something which you put down, put inside two vector, and you get a vector. Scholar product is with gamma t. So no, sorry, you put three vectors and you get a vector, right? So this is a vector. So then I take, so this is a number. Yeah, this is a number. So when I make i and j varying, I get a matrix, OK? But the only thing you need to know from this expression is that the trace of RT, which is a matrix, is just the Ricci tensor computes along gamma t, OK? So you can think to the Ricci tensor is a quadratic form. So when you compute, so this is actually, if you want to think it as a quadratic form, is something like this. You applied it to gamma t twice.  And I'm going to write this in this way, just to write less. OK. Now you want, actually, we are not in the full evolution of gradient t t, but just in the evolution of the determinant. And in some sense you know how to differentiate the determinant, then you play a little bit with this equation, and you end up with the following question. And just keeping some details, there are some nice computation. And you end up with the following, where I just introduce dt to be jt, the one over n. Which is, by the way, what we are looking, what we are looking to be concave, OK? So we are interested in understanding when this map is concave, and we see that this map satisfies the following differential inequality. So if the Ritchie curvature is positive in the sense of tensors, then the map t in jt to the one over n is concave, which means that the function, which is associated to rho t, minus the integral of rho t to the one over n, is convex. And what is really nice is that you can actually invert this implication. So if you know that this map is convex along every vast extent geodesics, then you can prove that the Ritchie tensor of your manifold is positive. So having positive Ritchie tensor is equivalent to have the convexity of this map along vastest geodesics. And you see which is the power, which was the starting point of what Ambrosio was telling you. You see what is the power here, because now you have given a completely, in sensency, you have able to give a formulation of positive Ritchie tensor without using derivatives.  But in sensence, I mean, you need a metric, which is at least C2, and maybe even more, because then you want to know, you want to take some derivative of the tensor, or you want to understand if geodesic equation is nice, and so on. So you need a lot of derivative for doing this. While to do this, you don't even need the metric, I mean, a tensor. You just need a distance to define vastest and geodesics. So this is a synthetic formulation, an integral and synthetic formulation of positivity of the Ritchie tensor. And this is, I mean, it's very strong as an implication. It allows you, first, to give sense in asking that general metric space have positive Ritchie tensor. And second, it's much more stable. I mean, it's easier to see that this fact is stable through very weak kind of convergence of metric spaces. It's the same example, you think, what is a convex function? A convex function is a function, whose second derivative is positive. But if this is your definition of convex function, it's very difficult to show that a sequence of convex function point-wise converging to a function, if you have this sequence of function point-wise converging to a function, then this function is convex. It's very hard with this definition. While if you take the definition that the function is convex, it slides below the second, then it's very easy to pass to the limit this definition. So in some sense, here is the same you are doing. You've passed from saying that a second derivative, so the Ritchie tensor is kind of a second derivative of the metric. I mean, the Riemannian is. Part of the second derivative of a metric is positive. You have translated this in a synthetic statement. And this is the power of this. And just using this fact, I can show you, I'm going to show you some nice property of the manifold, which are classical theorem in Riemannian geometry, but using this information. I mean, they are equivalent. So obviously, actually, it's pretty easy to prove this. You just need to test this inequality along enough geodetic. But so I'm going to prove nothing new, but I'm going to give you proofs, which are new of this geometric property. And so in this way, these are proofs, which can be extended in a more general framework. Because this, requiring this, does not need Riemannian tensor behind. So let me start Rd, again. We have the so-called Bruminkowski inequality, which I hope is the right way to write. Tells you that if you take two sets, E and F, to subset of Rn, then you define gt to b, this is just to write less. 1 minus ta plus tf, which is just the set, though too much fantasy. OK, so it's like a convex combination of these two sets. Then you know that the measure of gt to the 1 over n is greater or equal than the measure, than 1 minus t, the measure of e to the 1 over n, plus t, the measure of F, the 1 over n. OK, so this is the Bruminkowski inequality. OK, and actually this inequality implies, it's pretty easy to see that it implies the isoperimetric inequality. But how can you prove with this information the Bruminkowski inequality? What you have seen in the proof of the isoperimetric inequality is that we were just taking the set, a probability measure, I mean, just the normalized Lebesgue measure restricted to that set, and we are mapping this map to the other. OK, so what happens? So we take mu zero, in this case, to be just the Lebesgue measure restricted to our set e, normalized to be a probability measure, mu 1, the Lebesgue set, the Lebesgue measure restricted to F, normalized to be a probability measure, and we look to the optimal map, u. So, let's see. We look to the vases that geodesic, mu t, between mu zero and mu 1, and we know also that mu t can be written as rho t devol, dhe Lebesgue, sorry. And we know that our functional was convex along these geodesics. OK, so what we know was that minus the integral of rho t to the 1 minus 1 over n is less or equal than minus 1 minus t, the integral of rho zero to the 1 minus 1 over n minus t, the integral of rho 1 to the 1 minus 1 over n. OK. And now you see that there are too much minus. So this becomes a plus, plus. OK, so now it's just computation. So the first observation is that the support of rho t is contained in the set 1 minus t e plus t F. And the reason is simple, because you know that rho t is just, as we have seen, rho t, the Lebesgue is just 1 minus t, the identity plus t grad u, sharp rho zero dx. And grad u is pushing forward rho zero to rho 1. So what you have here, and rho zero is the uniform distribution here, rho 1 is the uniform distribution here. You know that these maps are just, so every point is sent somewhere. No, e and f, no one of that is convex. Brooming-Koskin equality rules for every set. And the word space is convex, so I can do this geodesic stuff.  But I'm not going to need that. Apart from some measurability issue, if you want. I mean, I think you're not sure that, not probably. I don't remember. There was some measurability issue behind Brooming-Koskin equality. I don't remember which one. But OK, we don't care about measurability issues. OK, so you see, right? You have just making a linear interpolation among points, a convex combination of points in e with points in f. At time t, so rho t will be something here, living here. So the support of rho t is just contained here. But now, we look to this expression. So we have the first integral. You can just localize. So this is the set we call gt. So you can just localize the first integral over gt. Outsider is nothing. Then you compute. You know what is this, right? You know these two measures. You just compute what are these functionals, which is trivial, right? Because rho 0 is just, you just integrate over e. And you have 1 over e to the 1 minus 1 over n. And then you have to multiply, you are integrating. So you have to multiply by the measure of e. So what you get on the right-hand side is just 1 minus t measure of e to the 1 over n plus t the measure of f to the 1 over n, OK? This is just a computation of this right-hand side. But now, you do, this is a concave function, OK? So you can just say that this is the same thing, obviously. Just multiply and divide by the measure of gt. Now, I do Jansen inequality. So since this is a concave function, I go in the other direction. So this is less or equal than gt, the integral over rho t, the 1 minus 1 over n, right? But you know rho t is a probability measure, so its integral is 1. And this, so this is giving you, again, gt. So this is a very nice proof of the Brooming-Kolskij inequality, which is not the best proof you can do with optimal transport. But there's the nice advantage that the same proof works line by line in the case you have a Riemannian manifold with positive reach. Because we have only used the concavity of this function, or the convexity of the function with the minus, OK? So we only have to say what is gt on a Riemannian manifold. But what is, I mean, what is the substitute for this set on a Riemannian manifold? But you see, it has just to be the place where is leaving the measure, which is supported this measure rho t. So, actually, what I'm saying is the following, that with the same proof, with exactly the same proof, you have the following. You have that same proof. If mg is a Riemannian manifold with positive reach, the measure, you have this, every two sets, where gt is just the set of x, which can be written as gamma of t for gamma geodesic, which is the first, the starting point into e, and the ending point in the f, which is exactly what we get there. So you take all the geodesics from e to o f, from points in e to points in f, and you just take a picture at time t. This is your set gt. And the same proof gives you this inequality. So this is the Bruminkowski inequality on a Riemannian manifold with positive reach. And it has a very nice consequence, which is the following. So we take as e to be just a point, and f to be the geodesic ball of radius r as center x dot. So you see. You have a point, you are moving it. So if you think you are moving it on this geodesic ball, now what is gt? Well, I don't know. Or maybe I don't care. gt, for sure, is contained in the geodesic ball of radius rt at center x dot, right? Simply because, I mean, the length of this geodesic is. So up to the normalization is easy, because at time t, this geodesic is a length, which is t times r. But now we plug this information in the Bruminkowski inequality. Obviously, eR measure of a set e is just the volume measure of this set. We plug this information in the Bruminkowski inequality to get that the measure of the geodesic ball with radius rt of x dot to the 1 over n is greater or equal. In eR, you get 0 of t, the measure of the ball of radius r. Then you just call this s. Let me call this variable s. And then you get that Br Bs x dot to the n is greater or equal than s over r to the n, the measure of Br of x dot, or, which is the same, that the map from s to the measure of the ball of radius s center x dot over s to the n is decreasing. And this is the Bishop-Gromov inequality. So we have a proof of the Bishop-Gromov inequality through optimal transportation. So remember, we are working with positive reach, otherwise this is false. So this concludes the part of optimal transportation and geometry, which I would like. I mean, it's not conclude the argument. Just conclude the part that I would like to show you. Let me just mention that the same trick, I mean, using not so clear proof, you can work with reach bounded by k. And what I'm telling here is a little bit more, because I'm always taking care of what is the dimension. So in some sense, the right space is this CD0 in space, which you probably saw with Ambrosio, which are space whose reach is bounded below from 0 and whose dimension is bounded from above by n. For this kind of space, you can redo the whole. So we take, like, five-minute breaks, and then we change argument. OK. So this is the last part of the course, in which I will try to give you an idea about what is the regularity problem behind optimal transportation. So I'm giving you, I mean, it's not completely correct to say that I'm giving you an idea, because I'm going to work in a very special case, which is the case of the quadratic cost over Rn, where there is this nice link. I mean, more or less, this is always a link, but in this case, it's very clear with monjampere equations. So if that x and y are open subsets over n, my probability measure will be just, so at some point I need to put mu 0, it's just f dx mu 1 g dy. And I'm going to assume that f and g are in between two constants. OK. So this is on their support, which are x and y. Just taken as a cost function x minus y square. So branean theorem gives us convex function u, such that the gradient of u pushes forward f dx over g dy. And we have seen also in the branean theorem that u satisfies the following monjampere equation, position gradient, so which makes sense, because they are both bounded on their support. These functions are bounded from above. I mean, so what I need to write this is that this function is never 0. And this equation is satisfied in x. And if you want, there is a kind of boundary condition, which is the following, which is that the gradient of x is contained in y. And this is just related to this. And I'm going to call a map set like this a branean solution on the monjampere equation. It's called branean solution monjampere equation. So by the way, I think that the monjampere equation is not really, I mean, it has the name of monj, but monj didn't wrote this down in relation with the optimal transportation problem. Because it's optimal transportation problem was not with the distance squared, but with the distance. And you don't get the monjampere equation when you work with the distance. So to understand the regularity issue, so this is like an IPD. And you see, since u is convex, this PD is somehow elliptic, elliptic in the sense that if you linearize this PD, you get an elliptic operator. But in the case of optimal transport, there is this problem. So what is this object? OK, u is a convex map. I'm not claiming anything about the differentiability of u. And even less about the differentiability of the gradient of u. So what is this point? So there is a very nice theorem, which is due to Aleksandrov. So which is the equation of Rademacher theorem for convex function. Aleksandrov theorem tells you that if u is convex, then for ln almost every x, you know that u of y is just equal to u of x plus grad u of x, which is well-defined for almost every x, since every convex function is locally elliptic. So you have a Rademacher theorem. So after you don't need a Rademacher theorem to define this, but in case of convex function, plus a, 1 half, a, y.  So for almost every x, your function is a second-order Taylor expansion. And obviously, if a matrix like this exists, this matrix is unique. So I can just say that graduation of u is a. So this is kind of my definition. And there is another nice thing. So you have a convex function. Think this in R. You have a convex function. It's second derivative, it's positive. So you can take this convex function, take it's second derivative in the sense of distribution, and it has to be positive. But the positive distribution is a measure. And the same is true in every dimension. So, moreover, if you look to the eschan of u in the sense of distribution, so this is the distributional action, then this is a measure. It's a measure with value in the space of matrix. And you can write this measure as an absolutely continuous part with respect to the big measure plus a singular part. And the density of the absolutely continuous part is exactly the eschan, I mean the matrix which is appearing here. And this matrix is the matrix which is appearing here. So I have an information on the determinant of the absolutely continuous part on the eschan. And this is likely to not be enough to prove any regularity. So let me give you an example in 1D. For instance, you can take u of x to be 1 alpha x square plus modulus of x. Then the second derivative, so let me write this way. So the second derivative in the sense of distribution of this map is just 1 times the big measure plus twice a delta in 0. But the absolutely continuous part of this map, I mean the point wise eschan, in some sense, the one which is appearing there is just 1. So you see that knowing that this is 1 does not allow me to distinguish this map from this map, which is pretty obvious. I mean a point wise derivative cannot give too much. And this is actually what happens in the optimal transportation problem. So you cannot expect, even if your density is as smooth as you want and your domains are as nice as you want, you cannot expect in general to have any regularity.  So just take a ball, take f equal 1 on a ball, then take 2 alpha ball and g equal 1 on the 2 alpha ball. So the optimal transport here is t is just x plus the sine of x1 times u1. So you are just sending alpha ball in this direction and alpha ball in this direction. And this is the gradient of the convex function plus modus of x1. And you see that this map is not continuous, but it cannot be continuous, because there is no continuous map from a connect set to a disconnected one.  There is topological reason why you cannot get a continuous map. So take y to be connected. But then, Kaffarelli show you that if you take this set, take a small perturbation of this example. So you just connect these two set by a very small width. Well, also in this case, I cannot write down the map, but again show that also in this case, the optimal map is discontinuous. If this trip is small enough, this map is discontinuous. So it seems there is no hope. But actually, there is a hope. And the reason is that you would like to make a link to the theory, to a well-established theory of Monjampere type equations. So we have seen that in the Brenier solution, optimal transport solution of the Monjampere equation, you have information just on the absolutely continuous part on the issue, in some sense. So you would like to have something which gives you information on the wall issue. But you cannot take the determinant of a measure, because it makes no sense to make the product of measure. So what is the right definition? So you note that when you have a convex function u, you have well-defined, I mean, you can construct this sub-differential, which is just the set of slopes of supporting plane, mx, so you have your convex function, you are just taking the slopes of supporting planes. In some sense, what is the determinant of the s? And I mean, if everything is smooth, just the measure of the image of the gradient. But so don't look at the gradient, just look at the sub-differential. So if u is convex, you define its Monjampere measure. So let's call it Ma of u. This is a measure, which is defining the following way. This is the Lebesgue measure of the image through the sub-differential of u of a. So what does it mean? This is just the measure of this set, union x in a of du of x.  So it's not completely trivial that this is a measure. I mean, you're like pushing forward the measure, pulling back the measure, not pushing forward, sorry. This is like a measure, yeah. You're doing what you cannot do with a measure, right? But since u is convex, you can prove this is a measure. So this is a measure. Yeah, it's like a pullback of a measure, right? Because this is the back measure you have to take on the target, and you're making a measure on the domain. So this is a bad operation to do with measures in general. The key point is that this map is kind of monotone, the sub-differential. OK, so this is a measure. And you say that you have an Aleksandrov solution. So u is an Aleksandrov solution to a monjamper type equation of the following form. Let's say, if the monjamper measure of u is equal to f over g grad u dln, OK? So in some sense, I'm asking also that there is no singular part in that measure. While in principle, it could be, right? Again, if you take, for instance, the map of that example before, so x squared plus modulus of x1, so if you have this map u, then if you compute the monjamper measure of u bar, this is given by 1 times the back measure plus, I guess twice, the hn minus 1 measure restricted to xn equals 0. Because this map, as when x1 is over 0. When x1 is 0, this map has a corner. And you see this corner in the monjamper measure, OK? This map was the map which was sending this ball in these two balls.  So this line is sent in a set of positive measure through the sub-differential. So this is the reason why you see this, right? The sub-differential here is going to feel the whole whole. So for Alexander of Solution, there is a nice regularity theory, which was due to Kaffarelli. So the key point here is to understand when a Brainiere solution is an Alexander of Solution. And you see that this is not an Alexander of Solution because the points, I mean, there are points where there is a non-trivial sub-differential. But the image of this sub-differential is outside the support of g, right? Because the support of g are these two balls. And this is outside. So in some sense, knowing that g is positive does not give, on its support does not give any information of what happens outside. So you want that if a point, I mean, you want to be sure that whatever happens here, the image of what is here through the sub-differential of u is inside the target. But this set is a convex set. So it's pretty natural to assume that the target is convex. So the theorem is the following. If you have x and y, y is convex, then you have your density at the x, gdy. So f and g are all bounded away from 0 at infinity. And then any Brignan solution and Alexander of Solution. And the sketch of the proof is the following. So you know that if a is a subset of x, then Alexander of Solution in x. If a is a subset of x, then since, I mean, when you have a gradient, the sub-differential is univalent, and it's a gradient. And it's just given by that singleton. So this is a trivial inclusion, where here I'm just saying nothing where the gradient does not exist. So since you have these, and you know by the change of variable formula, that this measure is just given by the integral over a of the point-wise session of u, this is less or equal than the monjamper measure by 2a. So to show that these two, so in particular, we know by the optimal transport that this is f over g composition grad u integrated over i over a. So what we get from these two, from these is that the monjamper measure of u is just greater or equal than f over g composition grad u, dila bag measure. I mean, this is true for every set. This is inequality for measure. So now to prove that they are equal, I just need to prove that they have the same mass on the whole set. If you have two measures which are point-wise bounded, but they have the same integral, they have to be the same. So to conclude, I only have to prove, it is enough to show u over x is equal to the integral over x of f over g composition grad u. Once I show this, I've proved that this, so what does it mean to be an Alexander solution, means that the monjamper measure of u is just f over g. So this will be my claim. So to conclude, I just have to show this. But since for every point the sub-differential is a convex set, d u of x is contained in the convex envelope of y. I know that this is true, but this is just y, since y is convex. So the measure of d u of x, so actually, to conclude, it's enough to show this, because the other inequality follows from that. So d u of x is less or equal than the measure of y, but the measure of y is just integrating 1 over y, which is just the integral of g over g in dy over y. Now I use the optimal transport to write this as the integral of f over g grad u over x. So here, I'm using that grad u sharp. And this concludes, right? So we have shown that when the target is convex, any Bernier solution is an Alexander solution. So now the task is to understand, which is the regularity of Alexander solution under the assumption I wrote there. And this is, again, a theorem due to Kaffarelli is an Alexander solution, let me write this way, determinant of the action of u equal. So it's actually, it's enough to have something like this. To be coherent, I don't remember. Well, lambda was bigger or less than 1? Landa is here. OK, if you have an Alexander solution of this, and u is 0 on the boundary of omega, then u is c1 alpha. So this almost immediately gives you that solution of optimal transport with convex target are the potential is c1 alpha, so the gradient is just c alpha. This almost give you what we want for the optimal transport, except that here I'm doing this theorem for solution, which are 0 on the boundary of omega. So here there is a subtle thing that you can always reduce to this case when you know that your function is strictly convex, but then you have also to prove that the reason why you can reduce to this case when the function is strictly convex is just that you, since a function is strictly convex, if every tangent plane touch just in one point. So if I take this plane, I move it a little bit up, and then I just subtract a linear function, which is something I can do, this is a technique we are going to see, but if I take this plane, I move it up, and then I go in this situation just subtracting a linear function, so I'm looking u minus l. So this is l, then I look to u minus l plus t, this is this one, and then you see that since subtracting linear function does not, obviously does not influence the determinant of the action, then you see that you are reduced to this case, but to do this, you need to know that your function is strictly convex. Otherwise, if your function has a line, then when you move up this plane, you are going to get two big sets. So modulo proving that in the case of optimal transport, you actually have that the solution is strictly convex. This is something which needs to be proved. This is not trivial, but I'm going to show you a proof, which is almost the same technique, so maybe. I'm not claiming that you can do it by exercise, but probably you can understand the proof once you see. The regularity of Alexander's solution then gives you regularity of Brine's solution. So I'm going just to sketch the proof of this theorem, and this gives, as a corollary, Brine's solution with y convex rc1 alpha, which gives that the map t, which is the gradient of u, is c alpha, OK? So, and OK, to get from this, this, you just need to prove strict convexity, but just. So I would like to prove that theorem, more or less, at least to give you an idea of the theorem. And to prove that theorem, I need. I mean, the proof is pretty by end. There are a lot of nice ideas behind, but almost no technique, no difficult technique. So I have to say this is one of my favorite proof in PDE. And it also, I'm trying to show you how it use a very nice idea, which is very common in PDE, is that if you have qualitative information plus compactness, you get quantitative information. It is something that in PDE you use a lot of time. OK, so let's start with this lemma, which is due to Alexandrov. Alexandrov tells you that if you have two convex function, no, let me write this way. So if u is convex, u is convex, u is zero on the boundary of omega, then, so if you have a convex function, which is zero on the boundary set, the function is going to be negative. So minus u of x is less or equal than a constant, which depends just on the dimension, times the distance from x to the boundary of omega. Then you have something like the diameter of omega to the n minus 1. And then you have the measure of the sub differential of u of omega. OK? So it tells you that if this measure is big, then your function has to go down a lot. And if you think you're saying that, I mean, the determinant of the s and you can think more or less to be the curvature of the graph. And it's comparable, at least. So if you have a lot of curvature, you have to go down a lot. And the proof is based first, and this is something I'm going to use again. If you have two convex functions, u less than v, with u equal v on the boundary of omega, then, what is u less than v? The sub differential of v in omega is going to be included in the sub differential of u in omega. OK? And the proof is the following. So you have u, you have v. And I want to show that any tangent plane, any slope of a tangent plane to v is also tangent plane to u, right? But what I do is just take a tangent plane to v. So this is a plane with a slope p, which is in the sub differential of v of omega, right? Then I move this plane down, and I move it up. So in parallel move it down. Then I move it up until it touch u at some point. And in that point, it's a supporting plane of u, OK? So this is the proof. This picture is the proof, OK? Well, I'm also explained. No, it takes the case in which v is 0 and u is this one. So this plan cannot do. The point is that you are going to touch at the boundary. In this way, you don't touch at the boundary. OK. So you have this first observation, which is to be useful again. OK. So this is the first. The second thing is just OK. We have our function u as in the theorem. We have a point, we call it x bar. Then we look to the cone, which is generated by the boundary and this point x bar. And let me call this cone c of x bar of x, OK? So we know that c by convexity, c of x bar of x is going to be greater or equal than u, right? And it's equal to u on the boundary. So part one tells you that the sub differential of c of x bar of x of omega is going to be less or equal than the sub differential of u in the wall omega, right? But now c is a cone. So if you have something, which is a supporting plane to a cone in some point, it's going also to be a supporting plane in the vertex. So this is nothing else that vc x bar of x bar, OK? Which makes sense. This is not a trivial set, OK? Because you have a lot of supporting plane. So it's giving you something with measure. So we need to estimate from below the measure of this set. And this is just simple, right? Because you have a cone, which is x bar.  And which is that slope? So this slope is just the diag, what is this? This is, so recall that this was minus u of x bar as length. So this slope is going to be, what is this? So this slope, so what, the tangent of this angle is minus u of x bar over 2 diameter of omega, right? Something like that. So you know that the ball of radius minus u of x bar over diameter of omega is included here. And this is not enough. Sorry, there was a mistake in this formula. Here there is the power n, OK? But then you have a special plane. So let me draw. If you take this plane, you see that this plane here is going to be supporting, right? And which is the slope of this plane? The slope of this plane is going to be, what is? So the slope of this plane, which is this angle here, so the tangent of this angle is going just to be, again, minus u of x bar over the distance of x bar from the boundary of omega, right? So you have that this is included, but also the point minus u of x bar over distance of x of omega is included over there, right? But now we know that this is, so this is x bar. Since this is the sub differential at the point, is a convex set. So you have a convex set, which contains, so dc x bar at x bar is convex set, convex. And it contains a ball of that radius and that point. So it contains the convex envelope, right? Of this set. So OK, you have a ball and the point, which is different. And you know that all this cone is going to be contained in the C sub differential of that cone.  So there is, so is a constant, a1 over cn. So what you get minus u to the power of x to the n over the diameter of omega to the power, I think it should be omega. So this is the measure of this cone. This is going to be less or equal than the measure of the C sub differential, which is going to be less or equal than the measure of the differential of the cone. And this is exactly what I have here. So we have this nice thing. How can we make this lemma useful? To make this lemma useful, in some sense, you see here there are a lot of bad things. Like diameter of omega, distance of, OK. Why bad things? The point is this one, monjamper equation as an invariance, which usual elliptic equation does not have. So monjamper equation is invariant through composition with affine maps with determinant 1. So if you take your function u, you compose this function u with an affine map, which has determinant 1. This is going to give you a solution of the same equation. So in some sense, since a stupid map with determinant 1 is like epsilon x, y over epsilon, so you can just stretch one component in one direction a lot and compress the other one. So you see that in some sense, you can have solution of the monjamper equation, which are as stretched as you want. So you have in some sense to get rid of this fact. So to get rid of this fact, use the so-called John lemma. John lemma is a nice lemma in convex geometry, which tells you that if omega is convex, then there exists an ellipsoid e, such that e is contained in omega, and omega is contained in n times e, where the dilation is done, for instance, with respect to the center of mass of the ellipsoids. Not for instance, with respect to the center of mass of ellipsoids. OK. Then when you have John lemma, you can define a convex set omega. It's normalized. If the ball of radius 1 is contained in omega, omega is contained in n times the ball of radius 1. So if this ellipsoid is the ball. So the consequence of these two is that for every convex set omega, there exists a fine transformation. Let me call it a fine transformation. Omega is? Yeah, yeah, yeah. Yeah, sorry. For every one, is there a fine transformation such that A of omega is normalized. They just send these ellipsoids on the ball. And in this case, I will write something like omega is comparable to the ball of radius 1, just as a notation. OK, but as a consequence of Alexandrov tells you that if is an Alexandrov solution on omega, and omega is normalized, so it's comparable to the ball of radius 1, so the Alexandrov solution of, let me write this way, determinant of the shton of u in between two constants, then you see that the inf of u minus the inf of u to the minus inf of u to the n is less or equal than a constant, which depends on n. Then you have the distance of the infimum, that's called xm, from the boundary of omega, then you have the diameter of omega. But omega is normalized, so its diameter is something which is a constant, depending just on n. The diameter is normalized, and moreover, then you have the measure of the sub-differential. In the full estimate, there was this diameter of omega, like to the n minus 1, measure of u of omega. But now you see, this can be bounded by a constant, which depends just on n, because omega is normalized. This can be bounded by lambda times the measure of omega, and the measure of omega is something, which again is a constant, which depends only on n. So cn lambda. So you see that the distance in a normalized domain, and the distance on the minimum point, bounds, which is the minimum point. So now notice that I have used just this inequality. But what I would like to say is also this quantity is bounded from below by something which is universal, where by universal I just mean it depends on n and on lambda. And how can do this? So say, so what does it mean that that's, so you see we are using here that your function is not carving too much, but it's not carving even too little, right? So this means that this infimum has to go down a little bit, otherwise you cannot have enough grade, enough curvature, right? So the lambda is something like if omega is normalized, then minus the inf of u is bounded by a constant, which depends just n and on lambda from below. So I want to say that my function goes down at least something. And how can I prove this? So say this is my u. I know that omega is normalized. So I know that here inside there is the ball of radius 1. And then I take v, which is something like x square minus 1 over 2 times some small universal constant epsilon. So this is a parabola like this. And I am claiming that if I take epsilon universally small, my function is below that parabola. So I get this estimate with epsilon here. So I just have to take epsilon, which depends just on n and on lambda. But say this is not the case. If u is not below, this means that u crosses this parabola somewhere here. Because on the boundary is below, for sure. Now if it crosses, and you go to compute, so you can apply Alexander of Lemma and say that the sub-differential of u in the crossing region is in measure less or equal than the sub-differential of the image of the parabola. But then you can compute this measure. So if you compute the one for the parabola, this is epsilon to the n times the measure of the crossing region. Maybe there is some constant somewhere. So this is the sub-differential of the parabola. The measure of the sub-differential of the parabola. So the determinant of this parabola is epsilon to the n. But this is greater or equal than 1 over lambda with my notation, which are changing line by line time the crossing region. So you see that if you take epsilon small enough, depending only on lambda, this is impossible. So you get that this infimum is bounded from below. But if the infimum is bounded from below, the conclusion of all this thing is that if the infimum, so the conclusion is that if omega is normalized, then the distance of the point where you realize the infimum from the boundary of omega has to be greater or equal than 1. I mean, not 1, but just is comparable to 1. It's a constant, which depends just on n and on lambda. So in a situation where you are in a normalized domain, you know that the point where you take the infimum has to be far away from the boundary. And this is all you need somehow. Yeah, zero boundary condition. Yeah, but I'm not going to do it. I mean, if you are strict convex, you can always localize and go. I mean, there are solutions which are not strictly convex and which are not regular. So if you are strict convex, you are regular, otherwise no. So it's enough to prove the. So we know this. Now the other lemma is the following. If u is a solution with zero boundary condition and so on, then the set, so for every linear function l, the set where u is equal to l, is just given by one point. So here I'm taking fuzor solution. Let me write this the right way. So determinant of the action, let me say, is comparable to one. And u is zero in the boundary of omega. So this is here. I'm saying that the solution of that equation is strictly convex. So it can just one is. So for every linear function, such that u is above l, then the set where it coincides is given just by a point. So it cannot coincide with a linear function on the whole segment or something like that. And the idea of the proof is the following. So the idea of the proof is the following. So say that that point is, so the cardinality of this set is greater or equal than 2. Well, actually it's infinity because it's a convex set for sure this. Say that it's not one. So what happens? There are two possibilities. Or the set where you coincide with this linear function crosses the domain or not. It cannot cross the domain. Because if it crosses the domain, you are zero on the boundary of the domain. You are above this linear function and you are zero on the wall. So you have a convex function, which is zero here. And it's linear on this line. So the only possibility is that it's zero everywhere. But it cannot be zero everywhere because we show that the infimum has to be something. So this cannot happen. So the other thing which can happen is that this set has an extremum point inside omega. But if it has an extremum point inside of omega, then you can subtract that linear function. Say you are in this situation. Then I can look to the function v, which is u minus l. So v is positive and v is zero on the set where u equals l. And so you have the situation of having a function, which is zero for a while and then it grows. And this function is my solution to the Monjomper equation. But then what I'm going to do here is just that I can take a very... So, no, this picture is not that good. I know this is just a proof by drawing. But if you want, this proof is written... Well, it's written in a lot of place. But I try to write it down on my PhD thesis in a way that it's something in between a very, very rigorous proof, which maybe can obscure the ideas and something which is these draws, which, boy, I like to tell this proof. So, I hope it could be more... I mean, there are details over there. So you are just telling me very rough ideas. So I have these. Then you see that if you take planes, which are very, very flat, right? If you take planes that are very, very flat, you see that this point is going to be...  So you have something like this, right? And you see that if you move the plane the right way, the minimum point is going to the boundary of this set, right? You have a function which is this. So if you cut with plane in the good way, you see that this point is going to... So, let me make... So, if you go to cut in this way, the distance from this point to the boundary of the set where u is less than this linear function is going to be less... I mean, it's like a little o of the global length of this section, of this piece, right? So, but this is impossible by the Aleksandrov estimate, because, OK, these sets are not normalized. But, I mean, if you look to the picture in this way, the minimum point is like this, then you have this diameter. So this is the diameter of the set where u is less, say, this plane, which is kind of epsilon L, say. OK, and you know that xn over the diameter is going to zero.  Well, I mean, not determinant one, in order to have the new function is still a solution of the Monjampere equation, you are going to get a normalized set with a solution with an infimum which is going to the boundary, because, although, I mean, neither the diameter, neither this distance are affining variance, the ratio it is. So if you make an affine transformation, this thing is going to be preserved in the limit. OK, so what you get here is that this point is going too close to the boundary of what you can allow by the equation. So this cannot happen again. OK, so your function is strictly convex. OK, and the same. And now there is just this. So this is a qualitative property. This is what I was saying, and then I conclude with the last lemma. So this is a qualitative property of the solution. But the class of solution on a normalized domain of the Monjampere equation is a compact class. And this is easy, because you know that the oscillation is bounded by one, and once the oscillation is bounded by one, you know that your function is locally Lipschitz in the right way, and you have a family of locally Lipschitz functions which is point-wise converging, the limit is convex, and you can easily pass to the limit in the equation. For the definition of the equation I gave. I mean, so it's the key of giving that definition. I mean, whether it's stable with respect to convergence. So the last lemma is again a picture, say that u is a normalized solution. So omega is almost a ball. Then take the cone which is going, so let me call this cone c1. So it's the cone which goes from zero to the infimum. Then I take c one-half, which is the cone, so this height is like one-half minus one-half the infimum of u. And I draw this cone here. So this is what happens. And what I claim is that there exists a universal delta such that positive, such that c one-half is less or equal than one minus delta c one as function. So they cannot be equal, because if they were equal, then the function would not be strictly convex, because outside here the function has to be above this cone, below this one, so if they are equal, then there is a point where the function coincide with the segment. But I'm claiming that no matter which is your solution, no matter which is your domain, if you have the domain is normalized and the solution is, I mean, and the determinant is bounded, then you have the universal delta. The compactness. So you say, say it is not true. Say you have a sequence of normalized domain omega k, solution u k and numbers delta equal one over k such that c one-half is greater or equal than one minus one over k c one. OK? And then you pass to the limit. You can pass to the limit to everything, because your class is a compact class, so up to a set sequence. So it is a solution, so it is a domain omega infinity which is still normalized. A solution u infinity and number zero such that c one and the point if you want x infinity so this is true for some point xk. This cannot be true everywhere. And the point c one x infinity where the two cones coincide. But then this gives you that u infinity cannot be strictly convex. So you get, from a qualitative property you get a quantitative property and now you iterate this quantitative property.  If you go to the level one-fourth you see that you have to be again a little bit below. And if you go to level one-eight you see that you have to these angles has to open a little a fraction and so on. But this is giving you c one-alpha at this point. Your Lipchitz constant is getting better and better and better. So you are c one-alpha at that point. C one-alpha? C one-alpha is like the cone which is like the cone which is generated by the level ok. I think to write it down and take me ten minutes. But yeah. But you see that at the minimum point your Lipchitz constant is getting better and better and better. So this gives you that you are c one-alpha in that point. But for convex function up to subtract a linear function every point is a minimum point. And since subtracting a linear function leaves you the equation invariant this gives you that any I mean your c one-alpha at every point. And this is because for a linear and this is thank you.