 Zato da smo počučili, zelo se je zelo na mordi, počučen, bo je odlično izgleda za vsem. Zelo za svoj svoj svoj svet. Zelo se da imamo, da smo počučili. Zelo, da je vse vse vse izgleda, da je svoje vse izgleda, da je vse izgleda, da je vse izgleda, da je vse izgleda. In tudi izgleda smo točne od vsočenih izgleda, ko je toto z delavno obradi z vsehovoj poživljenih. To je taj igrač. V tešku nekaj je, kot je tukaj kraj, že tukaj kratidium za vzelo kratidium in jaz neh obradi. Slimamo inzveč tudi obradi za vzelo tukaj kraj. Zato, kdo je izgleda, da je vse vse vse vse vse vse, da je vse vse vse vse učinila. Zelo, da bomo biti vse vse na programu, da bomo površati, da bo spodljučnjati nekaj ozvod, nekaj nekaj vse vse vse vse. Zelo, da bomo spodljučnje, da je nisem nekaj vse vse vse, nekaj nekaj nekaj nekaj vse vse, kaj bi se nekaj parametrizati kratičnjih vsega očstavljaj. To je izglednja, in tko je to izglednja, je to v sej kaj skupaj, da se zestah srednimo. Vsega tudi tudi in teori tudi nekaj parametrizati če se je vsega vsega namačna resalt, ki ima tudi tudi pačen, in tudi očstavljaj. Zelo, da nekaj sem zestavljala tudi in to je prvič, bo je to prvič raznamen, I baš je kg tudi skupo vece konstantr уров sleeve muscular? Zelo je bolj. Zelo je bolj. S tracksim je bolj. Va jha. U나čjesŤ. Ne zašli med 잘�n었어. . in so on . . . . . . . . . kaj počutili neče, kako spoljimo svoje energije. V neč ovo je naredila v klas, tako kako videmo, minimizujemo vse energije. Zdaj si je tukaj, da se je občasno, je ni tukaj, če je to kaj zači vse, tudi še vsak je počutila počutila izgleda, razvaj, počutila počutila. Pero neče, je nekaj izpracit, neč način jaz vse, je nekajčak. To je paček, da je sem vse bila povž 들어�utnosta. Oprzine velika se zelo nekajčekino, da je to verno, da je tako celost, zelo tako, da smo se ležili. V nekajče staj nekajče vzelo, da smo se zelo ležili, da je tинjez glasnega, nekajča da se tako zelo, da se je zelo glasnega, da se bo, da bo, da se je igrali previki.. Kaj je priziznico tako, da je vse vse vse izgroje, Cunja,vedahto vise možno praviso tko zwrotno importance obstavit so, kaj ta del waterson substantive naj Sulko nečiroler ki so to amp Hassir. Okisseni, sorry mev tegaheetie, neče ali prinikamo, da chicaj se, ono jaz nekaj držem. Paj, needed nek range on uh Kan Finale kommeri izgleda se v vse bože, Matejifirsti, za to, če pa zelo pozenem, pa tudi vsem zelo počel, kaj je regularičnji. Početko je, da sem tega vsega resulta, je zelo, da se zelo, da je to vsega harmonija. Zelo, da je to početko, da se vsega resulta je izgleda. Vsega resulta je tega, vsega, da je tega minimizirala. Taj je tekliv nenej na minimizaciju delikljej najskej energije. Nekaj je nemilj harmonij, ali se zelo, da vseče, da vseče, neseče, da je nesemilj, oče, oče harmonij. Nenej ni nekaj harmonij, ker v tem kontekciju, je odličnja od minimizaciju delikljej energije in vseče zelo, nekaj je zelo, da je vseče. To je tukaj, da smo tukali o minimizacjah. Vse minimizacje vse je vse. V v1,2 omega Aq je veliko kontynučne. V zelo je, da v vse openset kompakt je zapevajati v omega. Vzelo je način, da je zelo način, da ne zelo način. So exponent alpha, which depends on the dimension m and q, we can see it. And actually when m is 2, we know exactly the value of this exponent. So alpha of 2q is 1 over q. OK, but this one we won't see. So let me first give some argument for this result. And I told you this result is kind of optimal. So if yesterday I just said that, I mean, if we consider this branch points in complex varieties, they also represent the graph of the minimizing functions. So in particular, we may somehow choose the wrong outparameterize this function. For example, taking the variety, which is here, it's z squared equal w. And parameterize in the wrong way, because this one would be a very nice curve, if you see with respect to the z plane. But with respect to the w plane, it looks like this. And this one is an example of minimizing two values functions. And indeed, this one is the square root of z, which is one, which is one, all that continues. So this result is optimal. And we don't have anything more about this continuity, which is a big difference with respect to the condition one case. Harmonic function, the condition one case were analytic. Here the best we have is just continuous uniformly continuous. So let's now understand this result. So the surface is not singular, because it is. Yes, but if you write z in terms of w, this is not uniquely determined. Well, I mean, it's a... So the graph is two-dimensional of the space, which is four-dimensional. So I mean, I agree, this graph is kind of misleading. And the idea of this graph is that if you parameterize with respect to w, for each w, you find two determination of the square root. So above each point here in the w plane, you find two of such points. So that's the only meaning of this graph. But I mean, I agree, it's quite misleading, because that point is not a singular point, if you look at the surface. This surface is going around here, but actually it's an embedded surface there. So that's just the problem of the parametrization. But the statement is that every parametrization of a complex variety gives rise to an example of del minimizing function. So even if we choose the wrong parametrization, that one is a del minimizing function. And this del minimizing function is older continuous, with exponent 1,5. But I agree, the picture is kind of misleading, because this one is not a singular point as this one, even if the picture is just a rotation of this. Is it clear? I mean, you may complicate here a bit more the question to make this singular, and yeah. And it's still the same picture. OK, but let's argue a bit about this continuity result, because it shows you, in some way, the fact that these functions are taking values in this space, which is in a singular space, but we have settled all the instrument in order to really work as if we add values in a rent. So to apply instrument in the usual analysis of partial differential equations. And this one is an example where we see how such instruments can be employed. And indeed, the proof is made of two steps, and the first step is a proposition, which is the core of the proof. And the proposition is this. So we assume that f in W12, let's say, of a volume, Br into aq of rent is the minimax. And we assume also that the trace of f, which, in principle, is just an L2 function, is W12. So g, which is the trace of Br, belongs W12 of the boundary of Br, the space of two points. Then we conclude that we have a differential inequality between the energy of f and the energy of the trace. And inequality is that the integral of Br of df squared is less or equal than a dimensional constant, the integral, so r times the integral of the boundary of Br of dg squared. And the dimensional constant is, for m, is less than 1 over m minus 2. And when m is equal to 2, we have actually an explicit value. So what I'm saying is that every time I have a de minimizing function and I select a slice Br in which my function is still W12, which is basically almost all the slices, then I have such an inequality here, which I called a differential inequality, because this quantity here resembles very much the derivative of this quantity here with respect to r. So this one is kind of geometric proposition that I mean we will see in a while. And now I will show you very briefly how this proposition, applying just kind of common instrument in analysis, gives the conclusion of the order continuity. So here it's a kind of sketch of the proof of the theorem assuming the proposition. So this one is really a classical result. I will just go very quickly on this. So for convenience we call gamma of m the difference between the inverse of this constant and m minus 2. And we set h of r, this function, which is the energy in Br of our de minimizing. OK, now it's very simple to see that this function as a function of r is absolutely continuous, because it's an integral of an L2 function of this domain. And so it's derivative. It's almost everywhere given by the integral on the slice of the star cube or even just looking the different ratio of this quantity. And now this one is the full gradient, which of course is bigger or equal than if I just take the potential derivative. And this one is the energy of the function which is defined on the sphere. So I'm just looking at the potential derivative of f, which are of course less than the total derivative. So here I put the total derivative of the trace. OK, so now we just use this inequality here. So let me write here. What we have is that so this one we can read as h of r. So now we apply the proposition is h of r is less or equal than c, which now I take from here. It's 1 over gamma plus m minus 2. And then I have r. And instead of the integral of dg squared, I put the derivative of r. And now it's clear the terminology of differential inequalities. So now I have my function less or equal than r, it's derivatives times this constant. And this one is a new equality I can integrate. So what I have here integrated in this inequality, maybe it's clear, it's better if you see in this form. So this one is the derivative of the logarithm of h. This one is the derivative of the logarithm of r. So what you get, and not by the way, is the derivative. It was kind of necessary. This bound on that c was less than 1 over m minus 2. And now integrating that inequality we get h of r is less or equal than r to the m minus 2 plus gamma h of 1. So which I mean and now I come back to the old notation. This one means that the energy of f in br decays like r and minus 2 plus gamma the energy in b1. I was a bit fast in this derivation but just integrated in r in the quality between integrated in s between s and 1. And now what I claim is that this one, if we just forget for a while that f has values in these multiple points and we just assume that f is a real value function. From this inequality we can apply the classical Moray estimate and conclude that that f is actually all that continues with the exponent gamma over 2. Do all of you know this Moray estimate? So they are basically they are sometimes called Moray campanato estimates. And it's a way to characterize all the continuous function with the properties of decay properties of integrals. The campanato estimate is applying the Poincare inequality to this inequality here which gives you that the integral of br of f minus the average of f squared. So this one is less than r squared the integral of br of the f squared which is by this inequality less than r to the n plus gamma a constant which is the integral in v1. And once you have a decay of the average integral of f minus its average this one implies that f is all that continues. So for classical fun the gamma you get the one but you don't have degeneracy. But you don't have degeneracy, yes. I mean it does not depend on the dimension of the of the of the image space. Not even. In dimension 2 it's explicit, it's 1 over q. You don't have this bar. So in some sense which is I mean kind of natural. If we have more values the situation is more degenerate. That's what this exponent 1 over q is saying. Because more the exponent so smaller is the exponent worse is the estimate. Now the only message I want to give out of this of this derivation is that the proposition gives us a kind of geometric control of a trace with the energy. And that's what I mean we are going to see in a while. But then once we have that you really argue as if the function were real valued. And why can you do this because now it's not anymore real valued but all these quantities are known in terms of composition of Lipschitz function. So we can here put every time a Lipschitz composition that I have the inequality for the Lipschitz composition that we know and self. Because all the Lipschitz composition reconstruct the geometry of the metric space. So the only message of this derivation is that in some sense now we have settled all the stuff in a way that I mean for most of the computation we can think to have normal function. Because either we argue be a composition like in this case and I mean I invite you as an exercise prove the more estimates for value functions or we will see in a while we will use the point wise representative of the energy for doing the computation. But in both case in some sense we can forget to have a multiple value function and just derive differential inequalities or a question for which I mean we can argue be a classical tools. So if this one is clear now I will pass to give you some hints for this geometric proposition why we have this control of the energy with the energy of the trace. And just notice this because it will be useful at some point that the Lipschitz constant the order constant of our function will depend on this constant here which is the energy in V1. So actually in the theorem I could have written in the notes it is correctly stated I could have written an explicit estimate and the estimate is that the older semi norm of the function f is less or equal than a constant which depends on all the dimension and then on the set omega prime times the energy in V1. So this one is the explicit estimate one could give about the older continuous result. Ok, so now let me now pass to the proposition which has more a geometric content. There is a square root yes, yes, yes. Ok, so let me pass now to the proposition so which I remind you we want this amount to prove is the energy in the interior is less or equal than a constant which has several bounds the energy on the boundary and this one was just the energy of the trace so I may write the potential derivative of f squared sum up. Ok, how can we do this? So here the proof is kind of very different between dimension 2 and higher dimension and I will give you just the argument for dimension 2 which is very very geometric. So I will take m equal to so we consider here just the case m equal to which is an important restriction in the proof because the proof would change drastically for m bigger than 2 and then I assume a mild restriction but this one is very simple to generalize I assume to have two values instead of having a general two values and I will give some hints to the proof in this special case for m bigger than 2 there are different tools that you have to exploit which I mean we don't see now in this proof I will give you and they involve some other analytical results in the minimizing function like a maximum principle but this one we won't see now. So in this case the proof is very geometric and we may argue like this so we look first of all at the boundary data of f and now since we assume that f was actually w12 of the boundary you may show as in the classical and that's again a point which goes down to this observation that most of these classical results are kind of easily generalized in this context you may show that actually this f is continuous so as for one dimension because this one is a one dimensional space w12 functional and that's the same here because the proof of the embedding works exactly in the same way so w12 then actually f is continuous you may prove older continuous is c12 so in some sense now we have a continuous a continuous two cars over the boundary of this of this sphere and actually now two cases may happen so you look at these two values curve over the boundary and what may happen is that either this curve never self intersect or they have some point in common so now we talk about cars because they are continuous so we can follow in a continuous way this function and we observe that two cases may occur and the first one is that we find a point on the boundary so there exist x0 on the boundary so f f of x0 is equal to 2 times a point p so it's a double point or all the points of the boundary are genuinely two values so these two cases exclude one the other for every x in the boundary of br f1 of x is different from f2 of x where f1 f2 is whatever selection for our multiple values function and now we treat these two cases separately so this first one is kind of easier because immediately reduces to the classic case so let's make a picture so this one is our b1 or br I mean actually without loss of generality since the estimate is scaling invariant we can put ourselves in the ball b1 and assume now we are looking at s1 here and we start following our boundary data so we start from x0 which is such a double point and then we follow the first path the first continuous path above this up to ending again at this double point next there and we call this and now this one is defining a continuous boundary data and we call this gamma1 so this first boundary value is gamma1 and then starting again from x0 we follow the second path so this one was a double point so there was another path going out here it may intersect again I mean this we don't know eventually can do or not but then it comes back again in this double value which was p there in this point p above x0 and this one we call gamma2 so now actually since the boundary data was continuous and we had this double point we were able to follow continuously these two paths starting from that point and what we had we had found a continuous the composition of our f on the boundary is just given by gamma1 plus the data on the function gamma2 and this one now is a continuous selection and I told you yesterday usually we don't have other selection than measurable selection but we have exceptional cases which are the dimension one and co-dimensional one cases and this one enters in this dimension one case so when we have multiple values functions on a line you may find continuous selection and this one is a very very basic examples of this principle so now we have that our boundary data is decomposed as the superposition so the sum in terms of measures but we we always talk about superposition of two one-valued functions so now gamma1 and gamma2 are just functions from the sphere into a red and now a way to derive this inequality is to exhibit a competitor for f and the competitor would be the harmonic extension of gamma1 and gamma2 so now what I consider is the harmonic extension of gamma1 so let's call this zeta1 of gamma1 and zeta2 of gamma2 and then as competitor I take the function h which is the superposition of these two harmonic function so it's the deltas in the first function plus the deltas in the second function and why this one is a good competitor for our f because the boundary data is the same so the boundary data of these two values functions which now goes from b1 into a2 of rn the boundary data is exactly gamma1, gamma2 which were the boundary data of f so f was a minimizer and we may write df squared less or equal than dh squared by the minimizing property of f but now h is the superposition of two single values functions so it's very simple from the formulation of the energy we found yesterday to see that the energy is exactly the sum of the two so this one is exactly the sum from 1 to 2 of the energy of this zeta2 and now comes a very simple computation of the numerical harmonics that says that the energy of an harmonic is less or equal than the integral of the boundary of the differential derivatives so this one is actually an exercise to prove that it's very simple if you expand the harmonic function as a sum of in Fourier series in Fourier series so now that's a way to prove this so it's less or equal than the integral in b1 of the differential derivative of z, which I mean the boundary value we called and now again f in our boundary was the superposition of the two so the sum of these two energy is exactly the energy of f which is what we wanted to prove in the case r equal 1 and that extra factor is coming just from this case so in the first case just repeating what I've just done having a double point allows us to decompose the boundary data in two cars then we have a very natural candidate for a competitor which is the harmonic extension of these two cars and just comparing with this energy in the differential inequality ok now let's see this second case which is genuinely two-valued case so now every value is different from the other but nevertheless the function itself remains continuous because the boundary value is always w12 on the line which is a continuous function so now we can argue kind of the same but we can never decompose this boundary in two cars but we'll have always a single car so the picture here would be I have this s1 on the base and then I start from a point here whatever point, I would have two points there so starting from here I have kind of a first car going back here and then another one which is coming back to the original point and now always the picture is two-dimensional but this car here has no self-intersection because we assumed that at the boundary here we always had two different points so in a two-dimensional representation I cannot draw this but you have to imagine that car is turning twice on s1 or covering of s1 but there's no self-intersection and now what can we do with this so in some sense we can we can find a transformation to open this car to unroll this car here so what we find is the following so again since the boundary data is continuous it's very simple to see that we can find actually a continuous curve so now gamma from the boundary of b1 to Rn such that f on the boundary of b1 at a point let's call this z or let's call this x x is equal to now let me write the formula and I will explain what I am doing it's the sum on the square root of x of gamma of z so as measures so what I am doing is this so now I have a single curve on s1 which is rolling out twice and now I write a conformal transformation which is the square root of z in some sense to unroll the basis so each value is here which is f of x so this point here is x can be represented by this single curve which is gamma which I obtained by by rolling this this picture here so now I have here this single gamma and these two values here for example so when I looked at these two values here at the one coming from once the value here and once the opposite and the only thing I did was opening two eyes via this transformation this curve and now once I have a representation here I have again a very natural competitor function for this gamma here which is the harmonic extension so now again for proving the equality I argue by the minimizing property of f and I show and I have to cook up a competitor so now as competitor we consider the function h hx is just rolling back the harmonic extension of gamma so it's the sum z squared equal x of zeta of z where zeta is the harmonic extension extension of gamma so now I'm working on this picture here on this picture I put the harmonic extension and then I apply the same transformation so why this one is a good strategy so this one is a good strategy because the Dirichkele energy is invariant under conformal transformation and here this z square is a conformal transformation so if z was a minimizer for the energy for this picture here when we roll back we have still a kind of minimizer for this picture here because the energy is invariant and that's what we use now so let me kind of exploit the same lines here so now what we have is that by minimality the energy of f is less or equal than the energy of h but now the energy of h is exactly the energy of zeta and why this? because this one is an exercise which I think in the notes you find also int for the solution and that's because the Dirichkele energy is invariant under conformal transformation so up to here we have exactly the same conclusion but now we use the same simple inequality for harmonics so the energy in harmonic is controlled by its boundary data and now here we have to change something because now gamma was obtained out of f rolling out twice so now if you do just a change of variable you see that you have a factor 2 in this change of variable here because it's one dimensional integral and you are rolling out twice so here you should put a2 we want it so the energy is less or equal than a constant it's boundary data and now for dimension 2 I told you the constant is kind of it's not important in higher dimension it needs to be bigger than 1 over m minus 2 but here could be whatever dimension and now in the worst case we have q values this one would be q and that's why in dimension 2 we don't know exactly the constant in the older in the older continuity so this one in some sense concludes the proof in this very simple case of two dimensional surfaces now you may try to do as an exercise the q case which is just an extension of this result the higher dimension needs other ideas for the older continuity in higher dimension so if not I think it's a good point to stop here to have a little break and we assume at half past 9 with a kind of a new chapter of this theory let's start again and now in some sense I told you of the matter so we have now the formalism for such functions we know they are continuous now our aim was to analyze the singular set so from now on we will spend all our effort to understand how many such points can be in this picture so let me in some sense first of all let me kind of define formally what is a regular point what is a singular point so now we pass to the analysis of singularities which is actually what matter in this regularity theory for minimal surfaces and first of all we need to give a definition what is for us a singular point and a regular point so let's start from the regular point so we always assume that you from omega into A2 is still minimizing or well I mean we don't need so take a u like this and then we say x naught in omega is regular there exist so if there exist r bigger than zero such that in in the ball B r around x naught maybe written as a superposition of single value function ui with ui from B r x naught into rn analytic so a point is regular so that's not the end of the definition but it's regular if in the neighborhood of this point we may write this function as position of very smooth function analytic because we are in a red if we were in a manifold they would have had the regularity of the manifold but in this case it's analytic and we ask something more so we ask that and either such two functions are always different one from the other so ui is different from uj of x for every x in B r of x naught or if they coincide in one point they need to be the same then ui is just coincide with uj so let me make a picture of this as yesterday for the points of differentiability so basically we are saying that x naught is a regular point if our multiple values function looks like this above x naught so it's just the superposition of very smooth functions and what may happen is that we count each of these with a different multiplicity so this one may have multiplicity 2 this one may have I don't know multiplicity 5 so this one can happen but they need to coincide exactly one on the other in this neighborhood a function like this so a point like this where our function is the superposition of two very smooth functions linear functions but they meet on a point which may be also different from x naught so such point is not regular so this one is a singular point because if they meet on a point they need to coincide and this notion of regularity is the one which adapts very much to the one of surfaces so in principle if we look the graph of these functions as surface the fact they intersect is not a good point for surface because the surface there is not immersed it's not embedded so what we are trying to reproduce is a notion of regular a singular point which then kind of transfers on a geometric way I mean of course in terms of parameterization of function this one would be a very nice function but nevertheless if we look at the geometric object this one is for us a singular point so this one is singular a branch point is also singular because here they coincide but the two functions are not the same all the points which are not regular are singular by definition it is not regular it is singular so I hope in some sense I convince you that this notion kind of corresponds to the geometric one would like to observe for minimal currents and actually the theorem we are going to prove or to discuss to some extent a bound on the dimension of this singular point so this one is the theorem towards whom we are to which we are aimed which is so let you now be a del minimizing then the out of dimension dimension the singular set and actually I mean we would see is is locally finite but let's say and discrete in dimension 2 the result is a bit better but I mean we won't see this this part of the theorem so this one is exactly the analog of the theorem I stated yesterday by England in this special case of the parametrization so the estimate of the singular set is the same m-2 which is what we wanted to prove for minimal currents and now the only difference that this singular set now is living in the domain of our parametrization so now it's an estimate on a subset of omega which is the singular set here above which our function is a singular but I mean it's very simple if you imagine that that function is quite flat the same estimate holds true for the real singular set on the surface so this one is actually the same result for mima currents in this very special case of parametrization and now what we try to do in the rest of this course is to understand what are the analytical problems and issues in this kind of result and the road will be quite long and kind of indirect but it's anyhow fascinating so let's somehow once that we have set on our target let's now kind of start again with some preliminaries and the first preliminaries will be now we will see up to now we have just used the variational structure of the problem so showing argument, minimizers and so on now for this result we need also to argue more at the level of the PDE so the first thing I would like to do is try to write a PDE for this for this function which I mean in the context of this of the calculus of variation they are called first variation so first variation ok for doing this so first of all I have to tell you which kind of variation we may compute for such objects we have several which are possible the two which are more significant are the inner variation and the outer variation so let's start by the inner variation we have u from omega to aq and then we consider a different morphism from u to u from omega to omega with a fixed boundary which is the identity on the boundary and what we may consider is the composition of u so the internal composition of u with a phi so we can consider a function which we may write u composition phi which at the point x will be just the superposition of this election of u computed in the point phi of x so this one is very natural and there is no conceptual difficulties here and let me now give you immediately the outer variation so for this one we would like to have a left composition with u so what we consider is always u multiple values and then we take c which is a function from omega times rn which is the coordination of our space so omega times rn into another equilibrium space which most of the case will be rn itself so let's let's take the same but that I mean you may change and now what you can do is compose in the left with and now what we get is a function which is I call and as a function I have to say what is the value in x but sometimes I will omit this x here because it's already in size of such a notation but it's not completely completely rigorous but save us of writing many x and now it's the sum of which values so for every x I take my c of x value ui of x so for each value ui of my function I consider this external composition on the left and then I superpose of these values there and this one is what I call outer variation which is a way to change the range of the function which may depends also on the position ok now if u is de minimizing and these variations are embedded in one parameter family which connect to the identity I can somehow derive the energy being a minimizer the derivative in zero will be zero and that one is what is giving us the equation so let me see if we can simplify a bit this ok so now let's consider very special variation so we consider for the internal variation phi of epsilon of x which is equal to x plus epsilon little phi of x where phi is a compactly supported we can take smooth vector field of omega into Rm and you see that for small epsilon since this one is compactly supported this c is actually a different which is the identity on the boundary of omega and what we can take is the derivative at epsilon equal to zero of the energy of our internal variation u composition with phi epsilon and now how to do this computation and that's again a point of this general philosophy I told you at the beginning that a lot of computation we can do as in a classical setting somehow because for this function here we have a kind of point wise representative of the gradient so let me write this here so one could prove kind of very simple chain rules which says that the derivative of u composition phi epsilon say squared is equal to so let's let's just write the derivative of this multiple values function and this one is just the superposition of the derivative of u at this point times the differential of epsilon and here you can put the ui so let me comment on this formula that I don't want to prove somehow but it's very elementary and we will use an Anup's formula for the other variation so you know that when you have a soublev function we were able to define soublev function point wise almost everywhere was an approximate differential it was this linear function which at certain point we called d of ui now when you consider such a composition it's very simple to verify that such differential is if ui was approximately differentiable at this point then this one is the differential of the composition and the argument I mean it's very simple you just take the definition of differential you have to verify that it is infinitesimal of the distance but since every computation was there was classical you will find out that the same formula is working there so having such a point wise representative of the differential gives a very simple way to prove this formula here so the equivalence between our energy defined intrinsically via the compositions and the norms of the differential so here this norm is exactly the sum of these norms here so this one is exactly the sum in i of ui this point c epsilon composition dc epsilon and now here this one is now a classical computation so this one is a classical differential composition with a derivative and we can take the derivative in epsilon in the usual way and now if you do this computation which is left as an exercise you get the usual formula so this one is the integral in omega and then you have so the sum in i of d ui scalar product, d ui in direction d phi minus the integral omega of du squared divergence of phi so it's pointless I do this computation at the blackboard but let me give you how to do by your own here you change variables so since this one is the thermomorphism you use phi as a change of the thermomorphism and when you look at the derivative of the Jacobian the derivative of the Jacobian is minus divergence of phi because it's identity plus epsilon d phi and the determinant of this matrix is the trace of of d phi so but I leave this for you and let me also also skip another computation so which kind of formula we get when phi is a radial a radial function and again also at this point the computation are everything is done in the usual way so now when you take phi of x equal so x a function of the modulus times x so this formula is far too complicated so let's put this here and so for phi n equal this and letting phi n going monotonically to the characteristic we get the usual internal variation for a monic function br du squared equal r the integral on br of du squared minus 2r integral on dbr of the normal derivative so let me I mean I will skip the computation but let me just explain what I've written for now we have a very general formula for these variations and we look at radial variations so meaning function of the distance to the center times x and now we let this this cutoff function tending to to the identity when you do this you obtain the usual equipartition of the energy for a monic function in any calculus force and maybe you remember very well in the two dimensional case so in the two dimensional case this term is zero and this one is the equipartition between the radial derivative and the normal derivative because you know that the total derivative from the boundary is twice the normal derivative so which means that equipartition between radial and normal derivative you get this extra term here and here the formula is really the same just use this this internal variation here and by the way I mean you find I think almost all the details in the notes actually I mean I think all the details so this one is for the internal variation and now let's let me give you also the formula for the outer variation if I find it so actually for the outer variation the derivation is is more is more straight forward I mean you don't need to perform this change of variable here and what you get is the following so as one parameter of variation you consider e epsilon of x and now somehow this one should be psi of x equal to a equal to what? equal to x now here we have two variables so this is u equal to u plus epsilon little c of x u so this one is one parameter family which is going to the derivative then I can consider d of this squared integral and I take the derivative in epsilon and what you get is the integral of the sum over the value of u of dui the x derivative of c in the point x ui plus the sum in i of dui the u derivative of c so we respect to the second coordinate x ui d ui again do by yourself I mean in the notes you will find anyhow the details and also in this case what happens when we specify our our so what happens when we specify our vector field so it's this so as a vector field now we take so this little phi of x u is just a radial function of the position which will tend to the identity again times u and phi n is going to the characteristic of the ball what you get is this formula integral so let me write this formula there so we save this c of x u phi n of x times u and then we get the energy of du is equal to the integral of the boundary of br of sum on the values of normal derivative of ui ui and again I mean I'm not going to prove this it's very elementary and possibly at this point wise expansion of the differential but let me comment again this formula just to remember and why this formula is so elementary if you at u as classical harmonic function this one is just the integration by parts for harmonic function so one derivative you put on one side and you have Laplace of u which is zero and then you have the boundary data the boundary data is the normal trace of the normal component of du times u which is what you read here the normal component of u times u and now since the computation are done value by value sum on all the values so that's nothing esotic and these two formulas are the little formulas we are going to use next so this one was kind of big parenthesis without many details on how to derive this Lagrange equation but in the end we will use exactly these two equations in this integral form and now comes one of the of the main point of the theory which actually is maybe one of the of the most important ideas to solve this problem and let me do a little premise about this in all of these problems in geometrical analysis usually the starting point is the estimate which is kind of starting the solution of the problem is a monotonicity estimate so it's an a priori estimate on a monotonic quantity for the problem which I mean for example for minimal currents I don't know if Camillo told you you have the monotonicity formula for the area ratio of the current so you take the mass of the current and you take the right dimensional ball with respect to the current and this quantity is monotone now here for this higher dimension minimal current we have a new quantity which is monotone which was discovered by André and now goes under the name which is so let me remind frequency function which now you will see it's kind of magic quantity already for classical harmonic function which was never observed before and the quantity is considered so we call this e of a so it depends on the on the radius and I mean for the beginning let me put here also a few other symbols I will drop starting so it depends on the point we are looking and on the function we are considering and this quantity is defined as a ratio between the energy of the function and a norm of the function itself and the energy so r is just the right scaling that's not not so meaningful so the energy the u squared and the right norm you have to put here is the L2 norm of the trace so it's the integral on br of u so let me first of all explain you this symbol here for multiple values function so it's exactly what you imagine but just to be clear when I write modulus of u is the distance of u to the zero point but the zero point is is of course intended as the q points so this one is the way we compute this means norm and what is this function kind of measuring so that's always x0 and this quantity is here we will call d for the energy so the Dirichlet energy and h is called this norm which is kind of height for the function and what is measuring this quantity it's measuring somehow so that's very heuristic in the end all the monetary city formulas are kind of a miracle there is nothing to explain somehow if it works it works I don't know why it works but somehow it's measuring in some sense degree of disorder of anharmonic function so it's measuring the energy in terms of of its norm of the boundary but the energy with respect to a norm and what I am going to found is that if u is harmonic but actually it's also true in this multiple values in case if u is minimizing then this quantity is monoton increasing then e of r is monoton increasing so meaning that for anharmonic function in some sense the average of the energy with respect to its norm it's decreasing on small scales so the kind of disorder of the harmonic function is increasing when you go to infinity that's a very I know it's a very vague explanation but I cannot do better than this and by the way this formula was already new for the classical harmonic and that's kind of the the surprise of this hmm was it new yeah I mean if you try to prove this formula for the classical harmonic function is I don't say simple but it's just an integration by parts but nobody before algorithm noticed this and he noticed this related to this problem here let me also spend a few words for another comment why it's called frequency that's a it's bounded also mm-hmm so yeah so let me be a bit more precise in the statement so it's monoton increasing it's monoton increasing if you of course it's not identically zero because otherwise it's not even defined and the limit as r goes to zero of e of r s equal alpha is bigger than zero ok so let me so I was saying a very short comment why it's called this one I mean will help us to explain the use we will do of this formula which I mean I will try to do at the beginning before giving you the computation of the monotonicity so the comment is why frequency to answer to this question just do the computation in the two dimensional case so when you have an harmonic two dimensional function and let's say you have uk, rk kth the frequency for this uk is constantly equal k so the frequency functions for uk is constantly k which is the frequency of this harmonic and that's where the name is coming if u is a very harmonic two dimensional harmonic and you take this limit here then alpha will give you the lowest exponents in the expansion so if u is equal to the sum of ak rk kth then alpha which is that limit there is equal to so from k zero to infinity is and that means that k zero is the first one where k is not zero so alpha is exactly equal to k zero so in the limit this function is kind of detecting the lowest frequency of the harmonic and this one explains a bit the name so let me write this here and in this notation I mean that a of k zero is not zero so it's giving lowest and the point was so for Amgren all the kind of the many issue in this problem was the following was to find somehow a quantity which was saying that the minimal surface is not an infinite degree of that with the zero function so let me say this a bit better then I will do the computation and for today we finish so that's another comment so why or this formula comes into the into the game and the point is the following to analyze the singular points the strategy will be then to perform a blow up analysis singular point here we focus on small balls here we arrive to a ball of sides one and we would like to argue for this picture and why this because as all the blow ups we will see that the limiting function we get kind of easier we will get some symmetries which reduce the complexity of the problem but one of the many issues here in this blow up analysis was in some sense understanding this point which will come in two kind of key points of the theory in this one is one of these understanding if the limit here is zero or not because when we start from a singular point we would like to argue that in the limit here we see a singular point because otherwise this blow up analysis cannot work at all because we cannot from the blow up for the original current so you need in some sense to understand that singularities are preserved in the blow up analysis and what is one phenomenon one analytical phenomenon for which singularities may not be preserved if these two leaves of of the current has an infinite order of contact here so meaning faster than any polynomial then in the limit here we might this flat space here because these two leaves are touching with an order of contact which is infinite meaning that any reason to blow up would not separate these two leaves and the limit will collapse so the limit will be perfectly smooth and you lose all the information on the singular point so what is a way to say that a function that's kind of a problem in a unique theory what is a way to say that an harmonic function is analytic has not an infinite order of contact is to detect the lowest frequency in its expansion so he found an integral way to detect this lower frequency which was not infinity this one that we will use at one point and this one was able to say that this phenomenon cannot happen and we will see this in the next time how applying this formula we get that the blow up is not zero which is one of the main issue he had so why the blow up is not constantly zero in the limit because of the monotonicity of the frequency formula so this formula is controlling the order of contact of the leaves of a minimal surface and say that in the limit they cannot collapse into a single one so this one is the use we will do of this formula here and now you will see we have 15 minutes should be enough to prove the formula because actually it's very elementary it's surprising it's magic but the proof is very elementary so not to miss up too much let me follow my notes so we have which is rd over h d is the energy and h is the bound now we would like to compute the derivative of e and see that the derivative is positive and why this one suffice because you may verify by yourself that all these quantities are absolutely continuous so the point wise derivative coincide with the distribution of the derivative and that allows us to compute point wise such a derivative here now for d it's very simple this one was the energy on the ball so the derivative is just the energy on the shape of this ball almost everywhere a bit more work we need for h so to compute h prime of of r and for this one we make a change of variables so h was the integral on the boundary of br and we rescale to the boundary of b1 so we have the area element and this one was u at the point ry squared dy and now by the usual formulas we have given I mean we may think of this one like the sum on i of the single selection and we take the derivative point wise for each one of these so when we derive this we have m minus 1 r to the i m minus 2 integral b1 of u y ry squared dy so I took the derivative of this and now I take the derivative of the u so which is sum i r m minus 1 integral b1 of 2 u i and then I have the normal derivative in y so scalar normal u i now I rescale back to sides r I found that this one is m minus 1 divided by r h itself and here I have plus the sum on the values the integral on the boundary of so twice the normal derivative u i which I mean by the way we may use this information here it will be exactly the energy at the center point ok so this one is so let me it is immediately here so let me copy the formulas I am going to use there no let's leave this here so this one is m minus 1 divided by r h of r plus 2 times dr you need some effort to see that I mean all this computation which are kind of point wise are actually justified rigorously but it's not a difficult task taking the derivative of e of i so i prime is equal to d over h plus r d prime over h minus r d h prime over h squared ok and now first of all we use this formula for h prime so instead of h prime m minus 1 divided by r h plus twice d prime d this one is a 1 which I mean if I collect now everything there together I have 2 minus m d plus r d prime divided by h plus 2 r d squared divided by h squared and now let's use the variational formulas again so and let's look explicitly at this formula here this one is 2 minus m d energy with a minus plus r d prime but r d prime we computed before is exactly this this integral here well I put some times f times u so this minus this is actually 2 times r this norm here so this one is 2 times r integral on the boundary of the norm of the norma derivative squared divided by h minus 2 r d squared divided by h but in place of of d squared we use now this formula here so the energy is given by this this double product and we use this information here so this one is the squared of the integral on the boundary of the norma derivative times u and I mean sometimes when I omit the I here that it's anyhow summed on the old values divided by h squared so now let's put everything common we have 2 r divided by h squared which might apply h which I mean we recall is the integral of u squared on the boundary the integral of b r of the norma derivative squared minus d squared here so minus the integral of the boundary of u d nu u squared and you see now by this quantity is positive so this one is the scalar product between u du squared this one is the product of the integral of u squared d nu squared so by this quantity is positive and that's the end of the proof of the monotonicity so maybe I was a bit fast but there are all elementary computations once you realize that you may justify all these computations value by value and actually I invite you to do as an exercise the same computation for the classical harmonic functions because they are very attractive and you realize what is the point there so before concluding let me now looking at this proof so it's a good reason not to postpone to tomorrow now looking at this proof let's see when we have the case of a quality here so when e prime is zero it's a corollary which says that e of r it's constantly it's values alpha if and only if u of x was actually homogeneous of degree alpha so meaning that u of x is equal x to the alpha u x over modulus of x so and here it's what I mean I was saying that in the limit we will simplify our problem because the limit will be homogeneous meaning it will have a degree less of freedom somehow in more details but then the analytical estimate is this so e prime is constant if and only if this quantity is zero almost everywhere and when in the cautious parts you have a quality you have a quality when the two vectors are proportional so for almost every r if and only if we are bigger than zero we have that u restricted to the ball br to the boundary of the ball br is equal to a constant which in principle may depend on r the normal derivative of u so that's the only possibility to have the equality now as a first step let's understand what is this constant and for doing this let's compute the frequency once again using a different formula so alpha which is the value of the frequency which I told you it's r the energy divided by h of r but now use the integration by part formula which was the external variation for the energy which is saying that was actually the integral on the boundary of u d nu u this one was exactly the integral of u squared on the boundary and now these two are proportional so this quantity here is exactly r r lambda r this one gives us a formula for lambda so actually what we have is that u the boundary of b r is equal to alpha divided by r the normal derivative and now here you have again kind of little technicalities so you would like now to integrate this equation and this one I mean you may do for every values because the radius is one dimensional and you have this this continuous selection for one dimensional stuff and now you see that this one happens if and only if u is exactly homogeneous of degree alpha just integrating on the radii this equality there modulus few technicalities because this one has to be done for almost every radii and you have to distinguish the values one to the other but here it's explained so you find the details there so with this we close but let me just recall what we did today so apart from the continuity we started studying the singular behaviors of the minima surface and as I told you it's kind of a very long road which will lead us to that point and one of the main point is this so to have an a priori estimate if we perform our blow up analysis we don't lose track of singularities because everything becomes zero and the way to say this is an a priori estimate which comes from this magic monotonicity formula which is saying that the this quantity is monotone so it's defining a limit and you will see that this quantity is monotone implies that it's bounded the only fact that it is bounded will tell us that the limit cannot be zero so the first information we will use out of this formula is that the frequency of r always less or equal than a constant implies the blow up are not zero so that's the way we will make use of this formula let me also also comment so this one is a very special case because we changed the energy and we worked with the minimizer of this harmonic energy which is kind of the linearized equation what happen if we consider the real equation so the mass for an integral current what we get is not such a clean monotonicity but an almost monotonicity so for currents the estimate here change but the base of the computation is this the only point is that we have extra errors and the estimate reads like this so for minimizing currents we have a monotonicity which says for every r less than s e of r is less or equal than a constant e of s plus a constant so that's how the estimate will change but this one will be still enough to say that we are bounded once we know that in initial value the frequency was bounded and boundness implies the blow up is not trivial but this one we will see tomorrow hopefully with the end of the story ok so I will stop here ask first if you have questions about what we have seen today