 Rečno, da smo za njega vidim, in puno this will be a relatively speculative talk, ... lighter if this is was how it was intended by organisers, ... but probably would be more conjectures than new results in particular. in na obstavne prvne analizie v bojvom moženjem, ne obstavne. Oga nekaj je však drugi počut. V moženjem prejv, da v se zelo pravno začeli, oče, tako, džegaj, no, počutujem, ko smo bojvom na obstavne prvne prvne prvne prvne. Stajem s prvne prvne prvne. Vesme se, da potrebeno. Uklidnje konceler, zelo obstavne domenje in plorisa parmonic function that vanishes on the boundary, when whose monjam pair is given by some density, I mean, essentially equivalent, you can think of u at least for now as smooth. So, the trivial estimate is this. So this is immediately from comparison principle, just comparing this u with the modulus of z square. Now, in fact, this is quite related to the energy density in the control of the yaw. So this is one of the proofs, gives you first by, you know, quite elementary way a localization to the Euclidean setting. And then you have to get a certain estimate. And what you need really is almost something like that with a difference that you need the constant, I mean, you need the constant to be better, slightly. So, somehow, to ensure that it goes to zero, as volume goes to zero, that's kind of crucial, you cannot just rely on the diameter. And I guess this, when I know this, I haven't been very active as you know for the last few years, but I noticed that actually this kind of, this estimate has been used in various situations. Now, this that you can have volume going, I mean that you have the dependence on volume, follows in particular from the definitely non-trivial, like the first one was trivial, but this is non-trivial local estimate of this kind. So, now you have the dependence on the diameter still, but if you use held or continue or whatever, as you obviously see that you can get in the first estimate, this dependence on the volume, provided that diameter is under control. So, now I have my first conjecture, is that you should have something like that without any control, I mean without caring about diameter. Well, it has to be, if u is zero on the boundary, it has to be like hyperconvex, but so equivalent to u, you can say strongly superconvex bounded. It is not different because by approximation, so if u can get 2, 4, f, n is 1. Oh, exactly, very good question, but because, you see, you actually believe that u came, it was a very good question. But, you know, in fact, and this is not so completely trivial and I have realized recently that people, even when they do n equa 1, often miss the following result, which reduces this thing to a radius method. So, this is, I think it is quite often overlooked. So, this result by Valenti from the 70s, well, generally for Laplacian, which says that if you, well, you have this Dirichard problem and you symmetrize the right-hand side as a Schwarz symmetrization, right? And then the solve corresponding problem in the symmetrize domain. Symmetrize domain, meaning ball, with the same volume, centered at the origin, and functions having, you know, whose sub-level, super-level set in this case are positive, they have the same volume. So, then this v is regularly symmetric, obviously, and the result is that it has to be below symmetrization of u star. So, in fact, not only this original estimate with L infinity norm, but if you want to have any estimate for, you know, LP norm of F, since symmetrization preserves LP norms and, in fact, other norms, to other, really, so it could be even useful for the openness conjecture, well, okay, but let's not talk about that. This result reduces all these kind of estimates to radially symmetric functions, and then it's very simple. You can get, I mean, for L infinity norms, this specific estimate, and, in fact, for, right, so, and then, okay, to get this precise, so, now domain has no restriction, it doesn't have to be, because talenti reduces this in the case of radially symmetric functions because the L infinity norm of u is bounded by the L infinity norm of v, and then for radially symmetric functions, well, yeah, you just compare it with, you know, modulo z square over r square in the bowl, r is the bowl of omega star, and you have to remember that the volume is v r square, so you get an estimate in terms of r, so therefore, there's nothing about f. Nothing about f? Why? There is an f. The f's, the L infinity norm of f star is the same as the L infinity norm of f. In, in fact, the Lp norm of f is the same. So, okay, talenti, okay, maybe I can write it down here. The main point with talenti is that the p norm of f star is always the same as this is just the property of asymmetrization, whereas the infinity norm of u is actually my talenti that's the the other way around. No, that's the way. Yeah, this way. So, so, so you do it for v, so you do, you know, the estimate for v, but then you have equality on the right hand side, so that's why it reduces omega star. But omega star is yeah, omega star the thing is that the volume of omega star is p r squared. So, no, they are in both defined in the whole. So, in fact, yeah, the, I've been thinking a lot about the symmetrization problem for the complex mojampere which now, in fact, I've resulted in writing down that in such a generality it cannot fold for the complex mojampere. There are results that I will mention for the real mojampere by talenti, again, in dimension two and so in higher dimensions. They are not completely satisfactory to my purposes but I will show you. Yeah, but the general, I have a, again, general conjecture. So, say, in Kowoji's local LP estimate the constant should be really determined by radially symmetric function. So, it's kind of a simple to give a Kowoji estimate for radially symmetric function. It's maybe harder to write down those constants in general conjecture despite the fact that there cannot be like a similar analogous result for the symmetrization. It's still, I believe, that the constant should be somehow determined by radially symmetric functions. Well, OK. So, that's dimension one and now here and that's essentially the Alexander of Backelman-Pucci estimate which boils down to the similar problem for convex functions. So, bounded convex domain zero on the boundary convex and then so I have the really the L1 norm on the right-hand side because, OK, there you don't have to care about higher norm because it's already true for a one norm which it is not for in the complex case. So, you have this general estimate and it's known that it depends on the diameter. It's just the comparison with the ball. Right, so if you that's also how to improve this dependence on the domain. Now, there is this symmetrization result for the Mohsen pair but in so but the problem is that the right-hand side is symmetrized, the F is still symmetrized in respect to the leg measure but the function in respect to the different measure namely precisely the square mass integral w and minus one which is which is multi-volume of this w and minus one of omega is multi-volume of you take omega and minus one times unit of ones. But, you know is a parametric inequality this is actually bigger than in general than the volume. So, it's not yet like optimal. What is the denominator? What is the denominator? Omega n is the volume of the unit of one. Okay. So, all right. I will have omega n later on as the future documentation. Okay. So, well, the results I can prove that in nevertheless despite this and I also this kind of counter example for symmetrization is really symmetrized exactly the same way for the real one. But despite that you still have the result for for which is alexander kerman Pucci estimat so maybe let me quickly prove it because it turns out it has a relationship Okay. So, what I do okay. I can assume the origin is in belongs to the domain of u is a plane and is equal to minus one and here zero on the boundary I have this minimum of course it doesn't have to be symmetric and what I do I take this function which agrees here on the minimum and it's defined along line segments connecting to the boundary and zero on the boundary. Right. So, this is this v that is like that and then in fact this is the domination principle essentially the total mass of this still below has to be bigger than the total mass of the v which means really the estimated reduce to only this cone functions right? That's the origin this is the origin it's inside this is where the minimum is attained this is the origin and its value is minus one but zero is value zero on the boundary Right. So, the key observation is that the monjam pair of this this is the volume of the gradient image is really the volume the gradient image is the dual of omega that's the that's the thing and now yeah so that's the optimal constant in this ABM really is one over the infimum of all dual with respect to points inside the domain and you know so if you want to estimate this from above means you also estimate this from below which is you know the Burgan Milman in equality and as Vlasis was talking about but I needed for for non-necessarily symmetric domain it's not an issue really and so Maastron Tomis Rubinstein proved recently that you can take proved that you can have this inequality for this particular constant although I should also mention that that you know if you don't care I mean if you don't really care about dimension it's maybe there is a you know simpler simpler argument so by by Jones theorem really you can get some constant much worse but still depending on the on the on the on the on the dimension ok and then one more remark so there is this the the what is estimate for true was proved earlier by Cheng and Yao well I should that was really yeah in the written version that appeared in the Bedford survey paper and there was also kind of worked out by SEGRA so that this really I didn't mention so alexander bakerman putchi also also that's that's the usefulness or in but but it reduces to the convex case so if you have a non-convex function you take the envelope the lower envelope of convex functions and you only have so you only have to compute the mojampare this set of contact to the initial one so that is way of reducing so you can also use it for parmony functions this way and then you use this point wise inequality between real and complex mojampare equations and they both easily give you the kowođe estimate for p which is probably the simplest way of proving this way the uniform estimate in the kalabialo theorem just that you at least and this also gives you the global kowođe estimate on compact-color manifold for any p bigger than 2 was this over them separate in yaw? that's it? yeah, that's it it's also the kowođe estimate for p is due to general yaw but they didn't invite no, they didn't invite for the global right because for the global you need a streak of localization okay so that's and now the next two slides will be kind of side remarks so first is and is using this method you can prove a certain isoparametric inequality so let and this uses this method that kind of typical for those in this symmetrization result so again now assume our domain is smooth strongly convex and v is as before so it's you know this cone function a final long line segments so we know that the volume of the dual is equal to the možan pair but you can integrate by parts so really now since this is smooth near the boundary this integral is really equal to the boundary integral and hn minus one is precisely the Gauss curvature of the boundary times the modules or the gradient of v to the n and now you do you apply helder estimate so you can by helder inequality you can have this lower bound in fact the numerator is constant depending only on the dimension for the context case and for the denominator there is this general formula for any v not just ours due to rightly how this behaves under differentiation so and now you get hn minus two so if you differentiate this you know this kind of evolution of hyper surfaces is the Gauss curvature and hn minus two is instantimetric in order of one I think so the question of the you have hn minus the Gauss curvature is the product hn minus two is the next one of degree n minus two I guess I will do some differential geometry that some curvature tensors I haven't really studied the order of v d is not behind smooth no, I am not no, I think it should happen I mean this at least without this was smooth once but maybe there are some general differences this v is smooth because this v because our domain now we assumed as a smooth boundary so away from the origin it is smooth right it is essentially after constant the Minkowski functional right but our v is very special because really it is the sub levels that are proportional to omega so this way to the right hand side really you can easily compute what it is so this way you really get this parametric inequality because there is equality just for both I was thinking oh I found something new in convex but of course I proved something that was known before and 75 of course different methods but ok, the result is actually old this hn-2 it wasn't clear what it meant if it was not on the boundary yeah so it's really a boundary motion in fact, yeah I can tell you yeah I tried a lot of these approaches in the complex case as you know in the levy curvature and so on and so on and so on so so so so so so so so so so and so on maybe it's actually an interesting thing to prove counterpart of Reilly formula for complex complex stuff so levy levy curvature and so on ok and the other the Census whose like upside remark oh poslust this is getting late so I try to be quickly Tako tudi na 20 ročih, vzgleda je zelo čo je tudi tako zeno, neτ ta monžan presama in tudi, nekaj tudi modelne čl ata vzgleda vzgleda vzgleda. Pravno, je niče in vzgleda je, da tudi početni提redno član ko tvoje boje pojeb in Poel. Tukaj drugič je didzve vzgleda nekaj nekaj simetir,除 toga z ovener. Danes.. Tukaj, da te začeliš, da je to zelo ampak konveks, da je ni objez, pa to je na zelo na diagonal, vzelačno, za nozelnje domene in vse. Tukaj, da se bo, in zelo, da kaj smo pričeli, da je to što so skupnili za volumne delje. Tako je, da je to, da je ampak konveks, minimu v tudi, da je tudi predstavljena. Potenšelj nekaj, da je to nekaj zelo. Zato sem vsega, da je kovajski Tessier in Aleksandr Aleksandro Fenšel. Kaj je tukaj? Tukaj. Ok. Ok. Proste, zelo sem zelo, vzelo, da je zelo, zelo, da je zelo, zelo, da je zelo, zelo, da je zelo, zelo, da je zelo, da je zelo, da je zelo, da je zelo, but this is because you can solve. And here it's not really obvious what kind of linear operator you have to use because it's probably by... Right. And let me, the last thing, let me mention that it's a funny wise that you can also have opposite inequalities. And there's this result by Urban Sagrell, global one, where you actually have the opposite inequality. Ok, but that's just the side. Ok, so that's...