 OK, so, I'm gonna start my class on stability result for functional inequalities. First of all, I would like to thank the organizer for inviting me to this beautiful place. It's nice to wake up and see the sea. So the purpose of my class is to investigate some geometric and functional inequalities, some classical results like, for instance, the superimetric inequality. And first of all, I'll show you how you can prove such inequalities, and then understand how to understand whether these inequalities are stable and how to prove stability. So, at the beginning I want to start from the very beginning. So, what I'm gonna do is to present you the superimetric inequality now, and then I'm gonna try to give you the classical proof of the superimetric inequality, so back to the 50s. So, for some people who are experts in this, first part can be a bit boring, but I think it's interesting to see how to do this stuff. So, the superimetric inequality is the following. Let's say that I have some domain E in a ren. Let's ask for a moment that the set is bounded and smooth. Let's say open. And what the superimetric inequality says is that we can control the volume of the set E through the surface measure. So, the perimeter of E, so this is the perimeter, controls up to a constant and the optimal constant is the following n times the measure of the unit bolt to the power 1 over n measure of E, m minus 1 over n. So, the perimeter controls the volume up to a constant and actually this inequality is sharp. You have equality whenever actually if and only if E is a bull. So, if E is a smooth set you can define the perimeter in tons of ways you can think this is just you parameterize locally the boundary and you just compute the surface area and then you integrate and you find the total perimeter. So, for a smooth set the notion of perimeter is kind of not difficult but I want to give a more general definition of perimeter of course is scaling invariant it must be otherwise it could have been true which means the following that so if you change the set E by lambda E then what happens is that the perimeter scale like, so the perimeter change with a factor lambda to the m minus 1 while the volume change with a factor lambda to the n so if you take a set you delete the volume is multiplied by lambda to the m minus 1 the volume is multiplied by lambda to the n but then because of this power you will see that the factor lambda cancel each other which is natural. Ok, so now the first question is how can I prove such an inequality? So we want to prove this and so the goal is to prove the superimetric inequality so I know that last week Guido gave a proof of this inequality using a mass transport I'm gonna come back to that proof most likely in the next lecture which I think is tomorrow but at the moment I want to just stick to the classical so I will give a more geometric proof of this inequality so first of all in so generality I can always assume that the volume of v is 1 just take the set and delete it make it volume 1 and so what is our goal so to prove since now this number is fixed this one what you want to prove is that the perimeter of a set of measure 1 is always greater than this constant so our goal is prove is always greater than a dimensional constant which is this one how can we prove this so the strategy is kind of a classical problem in the calculus of variations so we want to minimize the perimeter of e among all sets e such that measure of equal 1 so it's just a minimization problem you say I take almost the sets of volume 1 and I want to minimize the perimeter so up to now I didn't even define what the perimeter is when the set is kind of a general let's say so for the moment since I didn't say let's say e is still smooth in some sense I could say I just want to minimize among the smooth sets so what's now what's the strategy when you have a problem like this there is what is called the direct method of the calculus of variation and that method consists just in the following first of all you say let's take a minimizing sequence so I want to understand this problem so I'm going to take a sequence e k volume k equal 1 such that the perimeter of e k converge to the infimum of the perimeter of e among all sets so you consider this number the minimum possible is an infimum because we don't know that there exists a minimizer you just take a minimizing sequence and then you would like to prove so up to sub sequences you want to prove that e k converges in some sense to some set infinity I use this abbreviation sub sequences without loss of generality let me move them so up to that sequences we get that they converge to some set infinity and then what you want to prove is that prove two things first of all that the volume of e infinity is still one oh sorry so that this is still an abysmal set and then that the perimeter of this set infinity is less so equal than the limum over k of the perimeter of e k and since this limum was equal to the infimum equal one I'm worried that if I write two down some people in the back are not going to see so I'm going to avoid to write two down I apologize for this time so you see that if I can find you can prove that the volume is still one and the perimeter is lower semi continuous than I found my minimizer because my perimeter is less so equal than the infimum must be equal to the infimum so this is the strategy now there are several issues in this argument because at the moment I just have smooth sets and it's very difficult if you take for instance a minimizing sequence first of all show that during a sub sequence you need some compactness and then here you have to prove slower semi continuity for the perimeter and still I didn't define exactly you can just think again the smooth surface area but then it's not obvious how to prove that so what do you do when you have such a problem the first thing you want to do sometimes is to a larger class of a missable sets so maybe let me cancel here so before giving this definition let me just make an analogy analogy most of you already saw when you want to minimize the Dirichlet integral so when you minimize the integral let's say of grad square let's say we use infinity so you would like to minimize the integral of the gradient of a function square among those smooth functions which have some bounded e equal g on the boundary of omega in general you are not going to be able to find a minimizer in infinity functions if you try it is difficult to find prove that there exists a minimizer in infinity because you don't have enough compactness in infinity so what do you do you say I don't know how to do this I move to sobolev classes so I minimize the integral of grad square not amongst infinity function but among functions which are just w12 u equal g on the boundary of omega you prove that a minimizer exists in that class so that if u bar is a minimizer then u bar is infinity and this is like regularity for harmonic function you prove that the minimizer is harmonic and you prove that it is infinity so now it doesn't really matter this part I don't want to discuss I'm just saying maybe some of you already saw this kind of procedure where you relax the problem you find a minimizer in a bigger class in that case w12 is infinity so we're gonna do something analogous we're gonna enlarge the class so not only smooth sets even if we know that the minimizer is a smooth set it's a ball and and then find the minimizer there so definition of perimeter how can we define the perimeter of a set so definition of perimeter definition of perimeter let's try to do aneuristic so let's say that let e be smooth by smooth let's say boundary of e is infinity then what I can say so the perimeter of e is the integral on the boundary of the function 1 with respect to some surface measure by definition is the integral 1 so it's the total mass of the surface measure now 1 this grid is equal if I consider the soup over all vector field so all x from rn into rn x of class c1 model of x less or equal to 1 and then here I place 1 by the scalar product between x and the normal ds so it's clear that this quantity is less or equal to this one just because x is less or equal to 1 the normal is less or equal to 1 so the scalar product here is less or equal to 1 but actually you have a quality because you can just take x to be equal to nu on the boundary of e so you have your set which is a smooth set you have your new v you have your normal and locally you can find as smooth let's say c1 vector field which is equal to the normal on the boundary so let's say and you take the normal you extend it in a small two-baral neighborhood and then with a cutoff function you put it to zero and in this way you get that this quantity is equal to this quantity but now once you look at this expression you can recognize that this is just the flux of the vector field on the boundary and this is the soup over x c1 so capital X will always be a vector field for me model of x less or equal to 1 the integral over e of the divergence of x and now since I put the derivative on x you see that this is actually a definition which makes sense for every set e and actually just to make sure that the integral still converges let me gain one centimeter more so this is the same if I do by density the soup over x which is c1 and compactly supported so here by little c I mean compactly supported and now this the space of x which are compactly c1 compactly supported is dense in the other space and then you just put the divergence so all the soups are the same so you see that this is e so you see that this last formula soup this makes sense for every Borel set because this is a continuous function and they are just integrating over as Borel set continuous function and then they take a soup in one so now this justifies the definition to use this last formula as a definition of perimeter and this is what George did more than around 60 years ago so definition this is the one adopted by George we are in the 50s let e be a Borel set we define p of e to be by definition the supremum x c1 and compactly supported models of x less or equal than 1 of the inter over e of the divergence of x p as finite perimeter if this number p of e is finite so this is a supremum it could be plus infinity it's gonna be plus infinity for very rough sets but whenever it's finite we say that our set is finite perimeter so now we have a definition of perimeter and the good thing is that this definition gives a lot of nice properties so what I want to show now is that with this definition you have lower semi continuity of the perimeter the perimeter along sequence can only go down and you have compactness of sets which are the two properties I needed to prove existence of minimizers so first thing lower semi continuity on this first part I will try to give some proofs also for people who are not used to this to this setting and then as we move on maybe skip some argument and just give sketches but at least for this first two properties I will give the proofs two or three properties so lower semi continuity so the lemma is the following if ek converges in l1 lock to a set e so I will use this notation to say that the characteristic function of ek converges in l1 lock to the characteristic function of e so the characteristic function is just the function which is one inside key means one and zero outside the characteristic function is zero infinity so I just wanted to fix the notation so if ek converges to in l1 lock then the perimeter of e is less equal than the lemma of the perimeter of e maybe everything is plus infinity but you still have lower semi continuity and the proof is one line is just the following that x c1 vector field mouse of x less equal than 1 we have that the integral over e of the divergence of x is equal to c1 and compactly supported is equal to the limit over k of the integral over ek of the divergence of x so this follows by that one lock convergence so this integral converge and now this is less or equal so remember the definition of perimeter for every ek the perimeter is the supremum of this quantity is among all set pk so this is less than the perimeter so this is less than pk each of these quantity is less than the perimeter of ek and then I can just put lemma because this is less than this point as a number so we have that this lemma is greater equal than this quantity and now x is arbitrary so don't just take the soup over x you have this lemma is greater equal than this you take the soup over x so this proves the lower semi continuity if you have questions please interrupt lemma here well here I don't know that there exists a limit so this is just so this I know that there is a limit and this is equal to this because they set ek are converging in a 1, 2 so this is a limit now I am just saying for every fixed k this quantity is less than this quantity so it means that the lemma of this which always exists is greater equal than the lemma of this but this is a limit because I already know that this is a limit so I could write here lemma if you want just to in general I don't know if this is a limit so this is why I have to take a lemma I could take a lemma soup but it will get worse information so it is stupid to take a lemma soup ok so this is the first property and now actually here so for people who knows this is probably a bit boring but let me do also very briefly the regularization argument so I want to show that if you give me a set of finite perimeter I can always approximate it with smooth functions the argument is very elementary so lemma if e is bounded I am just gonna consider the bounded case anyhow then I can find a sequence U epsilon infinity actually compactly supported in a ran such that U epsilon converges in L1 to the characteristic of V and integral of the gradient of U epsilon over a ran converges to the perimeter of V so I am saying that I can always approximate sets with smooth functions and you see by this that in some sense the perimeter is morally speaking the integral of the gradient the L1 norm of the gradient of the function so the idea is this so if you take a set E smooth set how can you approximate with the smooth functions instead of having an interface 1 0 you must do a smooth interface here in an epsilon neighborhood so you do a smooth transition between 1 and 0 and if you do a construction like this so that is just the geometric idea if you do an interface between E1 and 0 in a smooth way and you compute what is the integral of the gradient of a function like this roughly speaking the gradient is gonna be all in this direction so the gradient roughly speaking is proportional to the area because you get in some sense the picture from the side here would be something like this over a width epsilon you have 1 0 the derivative 1 over epsilon here the slope is 1 over epsilon so when you integrate the gradient you have the integral over this tube the gradient is over the 1 over epsilon and the area of this is perimeter of E times epsilon so this is kind of a picture illustrating this but let me give you the proof which is actually completely formal it means pure computation so how can you regularize a set just convolve so take row epsilon same thing in compactly supported convolution kernels so smooth function which converge to a delta and just define u epsilon to be characteristic of a convolution pro epsilon so if you want the easiest thing you can come up with works that's good and then with this I claim that this satisfies all the properties and the argument is also very simple so first of all u epsilon converge to the characteristic of E or this is a classical fact you can prove as an exercise if you take a characteristic of a function you convolve with smooth when epsilon goes to 0 this converge to kd now there are two inequalities one is the same as before so I'm gonna take now take x to be c1 in compactly supported x less than 1 so in one case I have the integral over e divergence of x is equal to the limit over epsilon of u epsilon divergence of x as before this converge to this now here you can integrate by parts these are smooth compactly supported functions so this is the limit epsilon goes to 0 of minus the integral of gradual epsilon scalar product with x here we are on a random and now x as modulus less than 1 so this quantity is for sure less than the gradient of the integral of the modulus and so I get the limit so that's less than limit over epsilon going to 0 of gradual epsilon so in this way I get and then you take the sup over x so for every x I have that this is less than the limit sup over x give 1 inequality so it tells me that the limit of these is greater than pd so by this we get p less so equally month integral of gradual epsilon how to get the other one that's just by using the property of the convolution convolution decreases norms and in this case I will just do by putting the convolution kernel on the x so let's go back to this formula here so let me do it moving from there probably yeah ok so I'm going to start now from here so minus the integral so I want to prove the opposite inequality minus the integral of gradual epsilon dot x again this is the integral of U epsilon divergence x and what is U epsilon this is the integral of key convolution of rho epsilon divergence x now the convolution when you have three functions you can always move the convolution from one to the other if you take an even kernel if it's not even you have to change inside let's say that this is even so I can move the convolution from here to here so this is the same as characteristic of V times divergence of x convolution rho epsilon and now the divergence in the convolution commute so this is the same as integral key of E divergence of x convolution rho epsilon and now this is nothing else than the integral of E of the divergence of x convolution rho epsilon but this now is c1 and compactly supported and the modulus of x convolution rho epsilon is still less than 1 because if you take a vector field which has modulus less than 1 you convolve it the modulus remains less than 1 so this is less than the perimeter of E because the perimeter of E is the supremum so perimeter of E is greater or equal than this for every x and so by taking the sup over x we get that the integral of grad U epsilon is less or equal than P over V for every epsilon but this proves the opposite inequality because I have that this limoff is greater or equal but each of them is less or equal than the other one so they converge so for every epsilon I have this inequality ok so questions? ok why did I do this because now I want to prove compactness and for the compactness proof I'm going to use a Poincare inequality and in some sense every time you have an inequality for functions by the fact that you can approximate sets of final perimeter by smooth functions you can pass the inequality to the sets and is what I'm going to do so compactness is the following now I know that sometimes I write a bit fast so if you feel that you need bit more time sometimes to copy let me know I know this from my students in Austin it's not useful if you tell me at the end of the week that I was writing to pass so I consider a family ek in a ren a sequence of sets and I'm going to assume that the perimeter of ek is bounded by a constant capital M uniformly, so for every k and I'm going to assume that my sets are uniformly bounded so let's say ek are contained in some ball b of r for some r greater than zero so a family of sets which are bounded and perimeter is uniformly bounded there exists a sub sequence ekj converging in l1 to some set e and e of course is still contained in b r so what I'm saying is that if you give me a set a family of sets of final perimeter which are uniformly bounded and whose perimeters are uniformly bounded then this family is pre-compact in l1 up to such sequences they converge in l1 to some set e of course this set e is going to contain in b r and then by the lower semi-continuity this property will also be true but this is just the compactness so I've been thinking for a while whether I wanted to do this but I'm going to give the proof exactly because I think it's nice if you have never seen sets of final perimeter I think this is one of the main reasons why sets of final perimeter are useful I mean all these definitions of perimeter are useful because you have compactness and then once you have compactness you can prove existence of minimizers and once you have existence of minimizer then you can work with the minimizer so this proof is not like the previous ones it's not just one line of computation with the right limit so so the first thing the sets must always be bounded the reason why you want to be bounded because otherwise you have a set of final perimeter which escape to infinity just take a ball and start to translate it then this ball just escape at infinity and the limit you don't get anything you can always have compactness in l1 lock but if your sets are not bounded you cannot ensure that the mass is conserved exactly because just take balls which convert to infinity in the limit you get the empty set so the mass could escape that's why you need a boundedness so first step is the Poincarian equality so this one I'm gonna give for granted and there are I can give you a reference for the proof it's really two lines by Fubini for every unit cube let's say q1 so I take a cube in a ren I have that the integral over q1 of a function u minus its average over the cube is bounded by dimensional constant cn integral over q1 of the gradient of u is true for every u in c1 of the cube so you give me a c1 function on the cube the integral of the gradient over q1 controls the l1 norm of u but you see that you have to subtract an average because if you didn't have this you could just take the function u to be a constant and then here you would get zero while this would not be zero if you didn't subtract the average so the reason why you subtract the average is exactly to kill the constants once you kill the constants then this inequality is true so the gradient is controlling the function in l1 so this inequality is in q1 now I want to put it in a cube qr as a small cube if I want to go to a small cube so now given now let v for instance be a function in qr where qr is a cube of side r so you have now a small cube qr and you want to write analogues in equality in a small cube how do you do? you take v, you rescale v to q1 apply the inequality and go back so let v be in q1 qr be a cube define v of rx in this way u is c1 sorry, there is no compati support u is now c1 in q1 and then you apply to u and then go back to v let you do this as an exercise so you just define u in this way apply the inequality and then just do the change of variable and then go back to the cube qr and what you get ok or I can just keep going with this there is gonna be another popping out go back to v and we get that the average sorry, the interval of the qr of v minus the average of v is less than cn it's in constant and now we have a r in front the interval of qr of the gradient of v so when you go from a cubo scale 1 to a cubo scale r you get an r in front this is by dimensional reason here you have gradient which is like one over length and you need a length in front to match the dimensions so this is the inequality and now I want to apply this inequality not to c1 function but to sets of finite perimeter so now 2 let u first be a function which is c1 c in a ren take a grid of side r in a ren so you take a ren and you divide it in cubes of side and then on each of these cubes we're gonna apply this inequality so what they have is that the integral over a ren of the gradient of u while this is the sum over h of the integral of qh of the gradient of u so I'm gonna call qh the family of cubes these are cubes of side r and now on each qr I apply this inequality so I get so this I put it here as a denominator so 1 over cn times r sum over h and now I have the integral over qh of u minus the average over qh of u so this inequality is true now for functions which are c1 I put compactly support to ensure that the integral converges and now by approximation I stand it to set of an perimeter in the following way so now apply this bound this estimate to the function u equal characteristic so just take the where e is just a bounded set of an perimeter and then so you apply the inequality for these functions and then you let epsilon goes to zero and we know that this left hand side converges to the perimeter so apply this to u and then you let epsilon 10 to zero so the left hand side will converge to the perimeter of e so this is I am going to put maybe this back so I get that cn times r times the perimeter of e is greater or equal than the sum over h of the integral over qh and here now I have characteristic of e minus the average over qh of the characteristic of e so I get this information just took this inequality and applied to qa and let epsilon goes to zero ok so let now compute what is this expression so this is sum over h and now so when I integrate over qh I have two possibilities well maybe let me first observe what is this so this is the average so this is one over the measure of qh and then you have the integral of the characteristic of inside qh which is the measure of e intersection qh because this is one when a point is in e and qh so you get this so this is just a number which is equal to this now what is this so characteristic of e is going to be either zero or one depending whether the point is e or not so I have two possibilities here either I am inside so I can integrate if you want this in qh inside e so when I am in this set this function is one so I get one minus e intersection qh over qh and ok maybe let me let me rewrite this here because otherwise I don't have space so this was so I have either this or I can integrate in qh minus e in this case this function is one sorry it is zero and then I get only this so I get e qh over qh but now this the one you can rewrite it as measure of qh over measure of qh common denominator common factor so you see immediately that this is actually equal qh minus e over qh and so you have qh minus e minus e here you have qh intersection e divided by qh and here you have qh minus e qh intersection e over qh these two integrals are exactly the same and you get a factor two so I go maybe it is very faster you can recheck by yourself that this is correct you get this qh minus e qh intersection e divided by qh so this is the formula ok so we discovered that the perimeter times are very small constant if I take r very small this I can make this as well as I want which at the moment is not very peeling it is not very clear what to do but here comes the key point so maybe let me draw a picture with just a big grid just to try to understand very few ques so I have my set e and I want to approximate e so the idea is the following I want to show that e can be very well approximated with unions of ques so a set which is just either one or zero inside a cube so which ques I want to keep to approximate e I am gonna say ok if e gives at least half of the measure to the cube I take it it is less than half I leave it so I am gonna take this maybe this, this is too small too small this, this is enough maybe this as well this, probably this these are the cubes that I want to keep so the cubes that I want are the one where I have at least half of the measure and the one I don't care so up to reorder the cubes I am gonna assume the following that e intersection qh is greater or equal than qh over 2 for every h from one to some number capital N and e intersection qh is strictly less than qh over 2 for every h greater or equal to n plus 1 so I just reorder the cube so I change names and I am gonna take just the ones where e gives at least half of the measure this is a finite set because it is bounded so you cannot have this for infinitely many cubes so otherwise it would have infinite mass so let's do in this way so I am just gonna take these cubes and now the nice thing is the following that I can keep going in the so what I proved up to here is that this quantity controls that sum but now in this sum let me sum first up to capital N and then so I split in two sums from one to capital N to capital N plus one to infinity so I get two i's then I have the sum from h from one to capital N of that quantity but what happens here so if I am less than capital N you see that this is at least one half of this so this ratio is greater than one half so I can say that this is greater than one half so I get one half qh minus e for the first capital N and then for the other ones I use the opposite inequality that if this is less than one half it means that this is greater than one half so this ratio is greater than one half so I get plus the sum from h, from N plus one to infinity of one half qh intersection e and now if you define the set t to be the union of the cubes of the first capital N cubes so this set the yellow one so just the union of the first N cubes then you can see so here there are some two and one half which simplifies and this sum is nothing else that t minus e plus here you get e minus t exercise and so what we proved is that t so if this is bound that this orbital is small controls the L1 distance so this is nothing else the characteristic of t minus characteristic of e so that quantity controls the L1 distance between my set e which is also an arbitrary set of finite perimeter and t which is a finite union of cubes so let me pause to finish the proof then we make a break so let me stop let me keep this so step 3, last step consider the following metric space x rm to be the familial sets e such that e is containing the r and perimeter of e is less or equal to m so this is and I am going to put a metric which is the L1 metric so I consider this the familial sets which are bounded in mr perimeter bounded by m endowed with L1 topology and I want to prove that this set my goal is to prove that this set is compact how can you prove that the set is compact you can suffice to prove that is closed and totally bounded so closed x rm is closed by lower semi continuity of perimeter if you have a familial sets e k which converges to set e if you have a familial sets e k here in this set which converges to set e while the limiting set will be in mr and by lower perimeter the perimeter will still be less than m so this set is closed totally bounded what does it mean? it means that I can find so totally bounded means that for every epsilon there exists a family finite family f1 f capital N in x rm such that norm of q f minus norm of q fi it is at the minimum i from 1 to n this norm is less than epsilon so it means that for every epsilon I can find a finite set such that each element sorry, this is true for every f in x rm so each element f in x rm is at most a distance epsilon from one of the sets so I have an epsilon net inside my space finite epsilon net so this is what I need to prove which is very smart offer you remember this part so how do we prove it? so to prove it indeed so why this is true choose r sufficiently small such that cn times r times m is less than epsilon cn is the constant of Poincare r is the number you are choosing and capital M is the bound of the perimeter and then by what we proved three seconds ago then for every e contained in x rm there exists a set t finite union of cubes side r such that characteristic of t in l1 is less than cn rm which is less than epsilon and just observe just note that the number of such t since all my sets are in br I have my grid inside br and now how many sets t can I construct which are union of cubes while I only have finite in many possibilities so I prove that there exists a finite set in this space actually it doesn't even matter it doesn't have necessary to be here but this is true for a different time this doesn't even matter if there exists a finite set anyhow then you can do by triangle in equality if there exists a finite set t such that every element e inside x rm is at most a distance epsilon from one of these so I found a finite cn set which proved that the space is bounded and then close to the bounded implies compactness ok, so maybe let's take a small break ten minutes and then in the next hour I will show you the isoperimetric inequality ok, so what we saw up to now is that with this definition of perimeter which are lowest in continuity and compactness so once you have these two properties kind of make the machine work and use the argument that I described at the beginning this is kind of direct method of the calculus of variation take a minimizing sequence this will converge so remember what was our goal so the first goal e minimizer for the problem perimeter of e where measure of e is 1 and then the second goal would be to prove that this minimizer is a ball so what you would like to do is to adopt the method of the calculus of variation to prove that existence of minimizer there is only one problem here is that in the compactness theorem I had an assumption which was the boundedness which is not given here is here minimizing but maybe my sets are not bounded now there are two approaches here either you prove in your argument that you can reduce yourself to bounded sets and it's not too complicated but it's not pretty trivial either so you should prove that if you take a minimizing sequence it can be bounded of course you could say I can always take balls which escape at infinity but what you can prove is that up to take your sets and translate them back at the origin up to a translation your sets are bounded and then you make the machine works but actually there is a simpler approach which is the following instead of minimizing consider this problem what I'm gonna do is that consider another problem this one and I'm gonna put an artificial constraint E contained in the R for some R large so I'm gonna impose in my problem that my sets are bounded inside a large ball why can I do it? well philosophically we know that the minimizer must be a ball so the fact that they put or not this constraint shouldn't make much of a difference because some moment my minimizer at the end will be a ball so the constraint is kind of just artificial and technical so I'm gonna consider this problem I'm gonna prove that the minimizer for this problem is a ball and once you know that the minimizer for this problem is a ball for every R then sometimes you just let R go to infinity and you obtain so there is a slightly more argument that I'm not gonna do but in some sense once you know that this is true for every R that the minimizer is a ball and if you get also the minimizer without this constraint it's a ball this does not gonna prove that the unique minimizer is a ball but it's gonna prove that one minimizer is a ball which is already what you need so I'm not gonna do the quality case I'm just saying that one of the minimizer is a ball so we are considering this problem now with a constraint so by compactness lower semi-continuity there exists E R so this is just what I described at the beginning you can try to do it as an exercise so how do you prove that this is a minimizer take a minimizing sequence the perimeter is gonna be bounded the sets are bounded so that you have compactness and so I have to assume the convergenial one to something this limiting guy will have the right volume so we continue this is gonna be a minimizer so compactness lower semi-continuity now gives us the existence of a minimizer so goal for us now the goal is that E R is a ball and all that it is now but I will explain to you the strategy how to prove that the minimizer is a ball this is based on the so called Steiner symmetrization which is very much used in the geometry measure theory so idea so the first technique is Steiner Steiner symmetrization so what is Steiner symmetrization so this is the following so let's take some direction here let's say on the sphere and now I'm gonna split my R n into the space orthogonal to V and this is the direction of V so I just take assistant coordinates let's say that my set is somewhere here what I'm gonna do is the following so for every point Y which is in the hyper plane orthogonal to V I'm gonna consider the line passing through Y in the direction of V I'm going to measure this intersection now I'm just gonna copy the intersection I got this is a segment I'm just gonna copy this segment here in a symmetric fashion so just take this length and I do this for every direction so I look at this length and I copy it here and in this way I'm gonna define a new set symmetric by construction with respect to the hyper plane and such that so as I said for every line here the length here is the same my picture is not here but I think as a metric set where this line is equal to this line even if you cannot see from the picture so this new set is gonna be called E SV S by symmetrized and V because it depends on the direction V so for every V you can do a symmetrization by Fubini you have that the volume of V is equal to the volume of ESV because the volume of V is just by Fubini you can compute it and for every Y you look at the length of this segment in your integral with respect to Y and since you preserve the length by construction by Fubini theorem this volume is equal to this volume and the key property is that now the perimeter has decreased so this is the key property of Steiner so claim perimeter of ESV is less or equal in the perimeter of V so im gonna show you very roughly the argument I don't want to be taken on this point because if you want to do a rigorous proof you have to be allowed with almost everywhere and stuff like that but let's forget about these technical points so let's think that there is no almost everywhere and let's think that our sets are nice so this is another problem argument rigos V you have to deal of how nice is the set if you want to use some property of sets of any perimeter let's think that the set is nice but the general idea is the following it just becomes more technical to do the full argument so let me draw so I want to prove the claim let me draw a piece of E so E is this and let's try to understand let's say I just look at the perimeter in a strip here so I want to I'm gonna show something stronger actually that in every strip the perimeter is decreased so here I have my set E and let's say that the perimeter can be parameterized by functions so I'm gonna call let's say FI NGI the different parts F1G1 F2G2 what is ES ES it's gonna be some set like this I'm gonna take the same strip here and I have this part of the bound so this is symmetric with respect to this line and what is this function here well so how much is this length of a segment so the length of a segment is the difference between F1 minus G1 plus S2 minus G2 and so on so the length is the sum of FI minus GI so this length is the sum of FI minus GI but then I have alpha on the top, alpha on the bottom so this card is some if you go from except for very rare points every time I go I enter I have to exit so I enter so I have a G I accept an F the only case he is here but this is a set of measure 0 this is a very except one set it's a very unlikely set in this case there are 3 points so it doesn't come for the big so that's what I say that there are few technical points but in some sense the main idea is there so this boundary is the sum of FI minus GI over 2 this is the function that I have minus now what is the perimeter the perimeter of a graph by the area formula is so if you give me a few and you want to compute the area the area is so the perimeter is 1 plus gradio square I'm gonna use this so in our case what is the perimeter on the left the perimeter on the right so the perimeter of E I mean I write perimeter of E but really I just compute the perimeter in this strip is the sum over I of the integral 1 plus grad FI square this counts all these top parts and then for the bottom I have square root of 1 plus grad GI while here so perimeter of ESV is the integral of the square root of 1 plus the gradient gradient is sum of sum of grad FI minus grad GI square so this is the top part and then the bottom is the same so I put 2 in front so this is the perimeter of the other one and now all is behind I just want to show that this is greater than this and this is just kind of a Jensen inequality it's based on convexity maybe let me just very briefly do it so behind I'm gonna use that X so let's say T 1 plus T square, square root of 1 plus T square this function is convex what is it behind so this is a sum let's say up to capital N there are capital N leaves capital N functions so let's see so what I'm gonna use is that the sum let me just write it here in between 1 plus grad FI square plus 1 plus grad GI square up to N this is greater or equal then 2N 1 over N so I'm gonna go a bit fast and then you can recheck everything later 1 plus grad FI minus grad GI now you put this is an average so I'm gonna put the average inside because of the convexity so this is greater or equal to N square root of 1 plus 1 over N sum over I grad FI minus grad GI over 2 square this you put N inside to simplify the N so this is greater or equal then twice square root of N plus now you use the capital N 1 say that this is greater or equal in some sense it's a pointwise inequality it's not an integral inequality just using really the integrants a pointwise one greater than the other so in particular then the total integral ok so once we have that we actually discover something interesting here so this was like all the chain of inequality that we're gonna use to say that this quantity is greater or equal to this quantity but we find something very interesting and this is why it was crucial for us to prove first that the minimizer exists because if a minimizer exists it means that whenever I take a minimizer and I apply Steiner the perimeter cannot go down because it's a minimizer the volume is the same so necessarily the perimeter can never go down and if the perimeter cannot go down I discover some information because it means that in that chain of inequality I have equality so what happens when I have equality so if equality holds the first thing to discover is that e-intersection y plus rv is a segment for every y in vpe so so in some sense here you see I had my set e and here I had the point y the direction was v and I take this intersection and what I claim is that this cannot happen if I have equality this cannot happen because you see that here if I have equality here I have capital N and here I have a 1 so 1 and capital N must be the same so it means that there was only one function so there was never this picture so this is the first property that my set is whenever I intersect with every possible line is always a segment and second property so here is where almost everywhere starts to come in and becomes decade and there is the second fact that you would like to say that if equality holds actually the two sets are the same up to translation which is not true in general so upset that this set so if I take this set and you do a Steiner simulation and you do a Lipset this is already symmetric so you see the Steiner is going to be this so in this case I just translate this and translate this so the perimeter is the same but the sets are different because there is not just a translation it is translated differently so it is not true that if I have equality my sets are the same but it is true if the set is convex so the second fact is that if equality holds you see here is a problem that the sets are disconnected then E is equal to E SV up to a translation so my set essentially the only thing I did if I had equality was to translate the set down there was nothing else this is based on Jensen there is a moment where you apply the inequalities from here to here and you show that you had equality in some sense the function on the top and the function on the bottom must be the same one the opposite of the other otherwise you gain something when you apply Jensen equality so anyhow these are the two crucial properties which come from the proof now we are going to apply them and we will see that this implies that is a ball our minimizer so be a minimizer no, I mean what I do and if the set was like this and I do a symmetrization I get something like this and it is not a rotation no, no, no, because you see the lengths are not even the same the only thing you can do is only a translation ok, so let ER be a minimizer then we know that let me call it E so let E equal ER be a minimizer and we go to the perimeter E bar E bar symmetrize SV for every V so if I have a minimizer I can do Steiner in any possible direction and the perimeter can never go down and so the first thing that this implies is that E bar is convex why? well, you see that if my E bar was not convex this would mean that I have two points X1 and X2, such that the segment is not containing between but then just take V pointing in this direction and you see that this contradicts the first property if I had two points it just contradicts this fact I could find something where there is not a segment so my set must be convex and so now I go back here this is convex and I have equality so I can apply the second property this two together tells me that E bar equal E bar SV plus TV for every V so for every direction my set is equal to itself sorry is equal to symmetrize now from this I want to deduce this is a ball it would be kind of easy if I didn't have the translation so I want to get rid of this the fact that I have to translate and that T depends on V there is some translation which depends on the direction let me kill the translation in the following way so the first thing I'm gonna do is that up to a translation can assume equal sorry E bar equal E bar E I for every I 1 to N where E I is the canonical basis why is this the case let me do intro dimension let me imagine that M my set let me do already a fantasy it was a ball so what I'm saying is that ok if I symmetrize with respect to the first V1 I know that this set is equal to itself up to a translation here so what I know is that the following E bar is equal to E bar S E I plus some vector T I E I so the T I depends on E I and so you just define so just apply the transformation E bar maps into E bar minus sum over I of T I E I so just kill all the translation and the point is that since the directions are orthogonal these translation don't mess up with each other it's not that the problem is that if you take random directions that first you do a translation to adjust one and then you do a translation to adjust the other but then the second one destroy what you did for the first but since you're doing only orthogonal translations then in some sense you don't mess up with each other you adjust all the I of them so after you subtract the set you kill exactly the first T I then you have that in this way now you have that E bar is equal to E bar S E I and so this means that your set is symmetric E bar is symmetric for X I maps into minus X I so if you change one coordinate the set remains the same but this means in particular the same is true if you go from X into minus X because you just flip one by one of the coordinates so this implies that E bar is equal to minus E bar the set is symmetric with respect to the origin but now if it is symmetric with respect to the origin it means that there are not any more translations so E bar equal minus E bar and then I know that E bar is equal to E bar as V plus some T V translation will defend some V if this is true you can just apply a minus everywhere and you see that the only possible T admissible is zero now you have a set which is symmetric with respect to the origin the only possibility here is that T equal to zero forever V which implies that now E bar is equal to E bar as V for every V and now I claim exercise E bar is a ball the idea is very easy in some times because if you add let's say if it's not a ball it means that you take a circle and they have a point here but maybe there is no point here so my set will be something maybe smaller than the maximal circle so this is like the maximal circle which is included but then so this point is so this is in E let's say X bar and Y bar this is in E this is not in E and then just do a symmetricization with respect to the half plane this point is mapped onto this one and vice versa so if you take V orthogonal to this plane so this angle is equal to this angle you see that this point is mapped to this and this point is mapped to this but this is in E so this must be in E so you discover that whenever there is a point there is all the circle in the inside and then there is a ball so small exercise here so the minimizer is a ball which implies that so what we proved since the minimizer is a ball the minimum of perimeter of E where E is one E is contained in BR where R is large we need R large just to make sure that there exists at least a set of volume one inside BR that's the only reason I mean R must be large enough so that there exists a set this is a number and this is the perimeter of the ball of volume one so let's say Birro such that measure of Birro equal one and so we proved that the mirror of that is that guy and then you have to let R goes to infinity to say that actually this minimizer is that guy as well and so the minimizer is the ball because the ball is gives the number so here there is a small argument letting R goes to infinity I mean it's not really the point okay any question? I did the classical part which is existence of a minimizer the fact that the minimizer for the supermetric inequalities of ball now we want to understand next step which is kind of the topic of the lecture is the stability so we have now the ball and we want to understand is in some sense the ball a stable minimizer so what do I mean by this the kind of question I mean interested in is the following so actually I'm going further than where I should be because I didn't prove that the ball is the unique minimizer but that's kind of shitty because I'm gonna now work directly on some more tissue but let's say that I proved that so the question I won't understand so now the stability question let's say that you believe me that the ball is unique minimizer so now stability issue so let me take again let e as volume 1 and let me define delta v to be the following quantity the perimeter of e minus n b11 over n let me write this just for this is 1 so this is the inequality what we proved is that this quantity is non-negative the question I claim also that if delta is 0 is a ball then the question is if delta v is very small does this imply that e is almost a ball and I'm gonna make precise this so this is a kind of stability question I'm talking about is it true that if my perimeter is slightly bigger than the perimeter of the optimal guy then I'm not far from being the optimal guy so let me start with an example which was in dimension greater than 3 so if n is greater than 3 I can do the following picture let's say if n is equal to 3 I can do the following picture so you take a ball and then to the ball you add a very long and tiny tentacle so what happens here so if you take a ball and then you add this very long tentacle what is the perimeter of the tentacle the perimeter of the tentacle this is like a cylinder so the perimeter is like the length times the basis so for every length no matter how long it is I can always take the basis very very very small in such a way that the perimeter is super small so this arbitrary no matter how long it is I can always stretch enough the basis so this works only up to the dimension 2 the perimeter is just the length it is from dimension 3 where you get to multiply by the basis so this means that I can have a set which is just given by a ball plus a very long tentacle and this is can have perimeter as close as I want to a ball so this guy so the perimeter of e can be as close as I want to the optimal guy so this means that I can hope to say that so delta is extremely small in this case but I cannot hope to say that my set is uniformly close to a ball because that's false on the other end you see that to do this picture the volume of this tentacle is extremely small so at least in an L1 sense I can still hope to prove something so let me define kind of the distance between sets the kind of quantity I want to control so let me define AOV which is the infimum of the symmetry difference between e and x plus b rho where b rho is the ball which is the volume 1 which is the volume of e infimum on go translations so what does it mean what is the difference of this let me explain with a picture so I have my set e e now I take a ball which can be centered in different points so this is x plus b rho this has the same volume as e then I measure this the symmetry difference is this is the area which is not in combo so this is the L1 norm so this is nothing else actually that maybe I lost some parts but this is the same as characteristic of e minus characteristic of x plus b rho maybe b rho sorry so what I am measuring here is the L1 distance between e and the ball and of course I don't know where my set e is in space so I take the infimum among all possible translations because I have to take my inequality will not tell me what is the set and so the question would be is it true that delta small implies a of v small and this question is actually not to differ there is again the problem of being badness or the badness but you can prove it by a contradiction argument using compactness lower semi continuity you say suppose that it is false it means that I have a sequence of sets such that this is going to zero and this is not going to zero you strapped as a sequence because the sets of finite perimeter are compact it converges to something and this guy is going to have this zero and this non zero but the only guy which have this zero is the ball contradiction so that is kind of besides the compactness that you need to be careful about boundedness is not too difficult to prove something like this what is more difficult to prove something quantitative so I don't want to prove just a qualitative statement this small implies this small I want to prove that this quantity controls this quantity in some explicit way and so the kind of estimate I want to investigate is the following so this is the first question answer by compactness by contradiction so this is not too difficult is in the same spirit of what we already did I mean compactness and so on so now new question is the following do there exist a constant c of n and an exponent alpha of n positive such that A of is lessor equal than c of n delta of to the power of n so can I find a constant depending only on the dimension and then exponent depending only on the dimension such that this A of is controlled by constant delta of we can tell you delta comes because of deficit that's how people call it and A because of a symmetry just for information this is like a symmetry so we want a quantity statement we don't want to say just one is smaller, the other is smaller here we are saying that when delta goes to zero A of e goes to zero in a very precise way so you look at this this inequality and you wonder can this inequality be true and usually when you want to prove such inequality you start with examples and you wonder actually what is the best thing you can prove so the first example you can come up with is so remark you can try the following example you take in two dimension the ball and then you take an ellipsoid here max is one plus epsilon and one over one plus epsilon this is your set e and then when epsilon goes to zero you can kind of expand as a function of epsilon both quantities the symmetry and the deficit so you just have to compute what is this volume and you have to compute the perimeter of an ellipsoid for epsilon small so you can expand and then you see that you can plug in the power of epsilon which come up in both terms these are both functions of epsilon you look at the dominant terms and then you see that a of n with this example when epsilon is very small you see that a of n necessary must be less than one half in sens that if you plug an inequality with a of n greater than one half in this example you get something false and you see that the exponent here the bigger it is the better it is because this is a quantity going to zero so if this is bigger this goes to zero faster so you have a stronger inequality and so here I got an upper bound how good can it be not better than one half with this example now some results what was proved so there was a result Fuglede that I am going to describe to you maybe now just slightly sketch it and then we will see it better next time so Fuglede in 89 proved alpha of n is equal to one half if e is closing a strong sense in we will see w1 infinity to b1 to the ball so in sens Fuglede consider the case where your set is as smooth perturbation of a ball and proved that the result is true with one half in every dimension then all in 91 and all 92 proved the result alpha n is equal to one fourth so they could prove that in any dimension for every set the inequality is true with one fourth and then in 2008 Fuglede proved the general result alpha n is equal to one half for every case and so here we are in 08 and then so this closed kind of this case and but then we will see with optimal transport proof that we did with Francesco and Aldo one can get again alpha n is equal to one half but in more general cases we will see this and then there is also proof by Cicalese and Leonardo that I will sketch because it's kind of short and very it's an interesting idea which kind of connects a bit also with what you saw instance in Camille and then Emmanuel lecture so it's connected to regularity of minimal surfaces so this is like 2012 and again alpha n is equal to one half it's a different proof it's kind of much shorter but it involves in some sense more results but then it's a much shorter proof in some sense you have to use regularity for minimal surface you want to use it, the proof is short so one half is the optimal exponent maybe in five minutes let me just kind of give you an idea of how you get for glad result and then next time so tomorrow we will we will see more details I just want to maybe to start so that you can also try on your own to see where you can go the proof is not too difficult but it's marked in some point but I can give you the strategy so for glad for glad I consider the following case so it's supposed that the boundary is parameterized over the sphere so I have my sphere let's say I'm going to consider unisphere so I'm actually messing up with the volume sometimes before I said that e has volume one here I'm saying he is close to the volume b1 this is not volume one here the volume of e is the volume of b1 sometimes I have to fix a number it looked one before because I wanted to be more easy but here it's easier to fix so what I'm using implicitly saying here is that the volume of e has to be fixed to be the volume of b1 I can always choose one number and it's more convenient in such a way that I don't have a radius rho fixed in such a way that the volume of the volume of radius b rho is one either way anyhow the required is came by just for convenience that I fix one volume so here I fix this so we assume that the boundary of e it's parameterized by function u over the sphere so for every, you have the origin you have directions omega and then you have this point which is u of omega sorry 1 plus u of omega times omega so if you is 0 this is just omega so this point is the sphere and if you is not 0 be some set like this so is the set of points of the form 1 plus u of omega times omega such that u in L infinity plus grad u in L infinity is less than some small epsilon so you have a function which is more in L infinity so this is the the set I want to consider and now the goal is to prove that the perimeter of e minus the perimeter of the buoy is greater or equal so of course the volume is fixed it is greater or equal than constant integral of u square over the sphere and then by an older inequality we can prove constant asymmetro v so this is kind of the strategy prove that this is greater than this and then show that this is greater than this so you want to prove that in a quite how do you do let me just maybe sketch the strategies just the points and then tomorrow we see the details so you have to write down what is the perimeter of e I give you the formula so if you have a function integral of 1 plus grad u square 1 plus u square 1 plus u to the m minus 1 integrated over the sphere so this is the kind of area formula over the sphere then you take this expression and you develop so you start to do an expansion in u so you get that this is you have the dominant term is 1 then you put 1 then you are going to get something like so tomorrow we are going to see something like this term give you something like integral of grad u square over 2 and then you have terms in u m minus 1 integral of u and then you get plus m minus 1 m minus 2 over 2 square plus lower order terms lower order terms which you can absorb with some epsilon around so this is kind of the perimeter and now you want to see that this quantity is coercive this quantity this error the deficit allows you to control u square it is not completely obvious first you see that this is quadratic and this is linear so how you get rid of the linear term is the idea there 2 more minutes so since the measure of v again is equal to the measure of v1 you discover that the integral over the sphere of 1 is equal to the interval of the sphere of 1 plus u to the n this comes from the equalizing the volumes and then you develop this in n so you get so you see the first order is 1 so the 1 simplifies then you get a term in u and then the strongest term is the first and the second and the sum must be 0 because the 1f is cancelled so you get 0 is more or less n integral of u plus n over 2 intro of u square and then you use this information to plug in here and get so you plug so here you use this to get a term in u square there and say you get the perimeter of v minus perimeter of v1 is something like up to errors grad u square over 2 minus n minus 1 integral of u square over 2 plus small errors something like this and now you look at this inequality and you say OK if this quantity controls this quantity I'm done so I'm trying to have I'm using a kind of a Poincare inequality on the sphere I want that the gradient controls the function the gradient in L2 must control the function in L2 OK usually when you have a Poincare inequality you have a mean zero condition because otherwise you put a constant this is zero, this is non-zero but the mean zero condition is implicit almost here because if you think this has a lower order term because this is u, u is very small so you see OK the interval of u is approximately zero u is some lower order term so in some sense u cannot be that big and in fact you can also see here the only constant for which this inequality is true is zero so we are in good shape because you cannot be a constant so you cannot be the first counter example to this inequality then you say OK what is the Poincare inequality on the sphere when I have functions which are not constant there is a Poincare inequality on the sphere and it has a constant which is exactly m-1 so the Poincare inequality on the sphere this quantity is non-negative nothing better that's the best you can see but in some sense you have this inequality which a priori you can say only non-negative and then you have errors which could go all in the wrong direction so in some sense here you are kind of stuck because this is a sharp inequality on the sphere tomorrow I will discuss better so this means that it looks that you are stuck that you cannot get anything out of this you remember one other thing that you never you have still one degree of freedom in this parameterization is to say OK my set is here but is there a sphere better than another with respect to whom I can parameterize myself maybe I was stupid my set is just exactly a ball and I just took a ball I have the unit ball and my set was just a translation of a ball and parameterizing myself with respect to another ball so in some sense I am stupid because you are zero in that case you see I should have parameterized myself with respect to myself not with respect to another ball because then you are zero but I am not seeing it that I am zero so you say OK so then I can choose a better ball and which one I can be the best OK let's say I take the same body center so you can say OK among all possible balls whom I can be parameterized let me choose the ball for which I have the same body center so you impose that the integral on the boundary of e for instance of x you can always impose that this is the same as the interval b1 of x the function x you fix the body center on the boundary this can always be done up to translation once you impose that you have an extra condition which we are going to see gives you a better one-carrying equality in the sphere and one because you are going to be orthogonal to the first eigenfunctions on the sphere the constant for this inequality is not anymore m minus 1 it is going to be 2n and here this becomes that this quantity is strictly positive and controls q squared so there is going to be an argument of parameterized with respect to a better one but OK for today I think it is enough so I stop here