 Hi, so we started the LP spaces, and so I just recall you the young inequality, okay? What we proved last time, which is the following. You have A and B, two non-negative real numbers, and lambda, a constant in R, which is between 0 and 1. So what we proved last time is that the following equality holds. You have that A lambda times B, 1 minus lambda is less or equal than lambda a plus 1 minus lambda b. And we also saw that we have that the equality holds when A is equal to B. Okay, so now we will prove another somehow another version of this inequality, okay? So call it second version of the young inequality. So we have this time P and Q in R. So we have P larger than 1 and Q larger than 1, such that you have that 1P plus 1 over Q is equal to 1. They are called conjugated exponents. Then we have for any AB in R, and again we have AB non-negative. We have that A to the power AB, sorry, is less or equal than 1 over P, times A to the power P plus 1 over Q times B to the power Q. And these times we have that we have the equality here, if AP is equal to BQ. Okay, so let's see the proof. So the idea of course is to use the first version of the young inequality. So choosing for some appropriate choice of AB. Okay, so first of all we define lambda. That lambda has 1 over P, okay? And then we get that this, you get that 1 over Q is equal to 1 minus lambda. And then we define with capital A as A rises to 1 over lambda, and B is small b rises to the power 1 over 1 minus lambda. Okay, then we want to apply this first version of the young inequality for big A and big B. Okay, this capital A and capital B. Okay, so you have that A lambda times B, 1 minus lambda is less or equal than lambda A plus 1 minus lambda B. And then we plug in this inequality, our choice of A and B. So by definition we have that AB is less or equal than lambda A lambda plus 1 minus lambda B, 1 minus lambda. And so finally what you get is that AB is less or equal than 1 over P times A to the power P plus 1 over QB to the power Q. Okay, so we are done. Okay, now we will somehow start to introduce some function that then will turn out to be a norm in these LP spaces. And so we need to introduce the following definition. So you have F, a function F defined on a measurable set E, the values on the extended real line. Okay, so measurable of course. And we define the essential supremum which is somehow a generalization of the notion of supremum of F is the following. So I will denote it with S of F over E is defined as the infimum over T in the full extended real line. Such that we have F of X is less than T, almost everywhere in E. And analogously we define the essential infimum of F. So we will denote it with S of F over E. This would be the supremum over T belonging to the extended real line such that F of X is larger than T, almost everywhere in E. Okay, and then as I told you we try to introduce some norm in these LP spaces that we are going to define. So for the moment there will be just some other definition. So we start again by a function with the same property defined on a measurable set E and with the values in the extended real line measurable. And so we distinguish two cases. Okay, in the first cases we pick a P which will be an exponent which is in between 1 and infinity, but it can take the values 1 but it cannot be equal to infinity. And we define what will prove that it is a norm, but for the moment that just used this notation, is the integral over E of the modulus of F to the power P rises to the exponent 1 over P. While when P is equal to infinity, norm will be the essential supremum of the modulus of F over E. Okay, so just now keep in mind this definition. Okay, so we want to prove step by step that this will be norm in some appropriate space. And to do this we need, which is the so-called Euler inequality, and which will be a consequence of the of the Young inequality. Okay, so you consider two exponents P and Q in R. They are both larger or equal than 1. And okay, they must be conjugate exponents. So it means that 1 over P plus 1 over Q is equal to 1. Then we consider two measurable functions, defined on a measurable set E with values in R, measurable functions, okay. And such that we assume that this that will turn out to be the Lp in R for the moment to consider like a quantity is finite for F and the same is finite for G. Then we have that we can say something about the product. So we have that F times G, the norm of the product with exponent P equal to 1 is finite. And moreover we can control this in terms of these two norm. Okay, this is just this, okay, by definition. This will be less or equal than, okay. And moreover, again, we also consider the case when the quality holds here. Okay, so we have that the quality holds. When we have that F to the power, the modulus of F to the power P is equal to some constant times the modulus of G to the power Q. Okay, so we consider, start by considering the case when P is strictly larger than 1, okay. And okay, by this relation we know that this implies that Q must be finite, okay. Okay, then we express the modulus of the product of F times G in this way. We consider, we divide and multiply for the same T positive. This is true for any T positive, of course. And okay, we, as before, we introduce some auxiliary function A tilde. We find that T times the modulus of F and B tilde would be 1 over T times the modulus of G. And we want to apply the Young inequality, the second version of the Young inequality to those new functions, A tilde and B tilde. Okay, so tilde P plus 1 over Q would be tilde Q. Okay, so we have to be reminded that these are equal to the modulus of this product. And then we substitute this choice here. We have 1 over P times T to the power P times F, modulus of F to the power P plus 1 over Q, T to the power minus Q times G to the power of Q. Okay, so we have that for any X in E, any T positive, we have that, okay, let's put it actually. You have that for any T positive, you have the following. You pass to the integral. Okay, and this is less or equal than 1 over P times T to the power P, minus P norm of F to the power P plus 1 over Q, T to the minus Q, G, Q, Q. Where if we define this A and B this way as T, Q. And now we choose, we make a choice for T here, and we choose T so that we have that AP is equal to BQ. Okay, so in this case we have that here we have this is equal to A times B for what we prove about the, you know, the Young inequality. And so we have that this, for this choice we have that T P F to the power P, P is equal to T minus Q to the power Q. And then we have that T P to the power P plus Q is equal to G Q, Q divided by P. And so for this choice of our star we have that is less or equal than A times B which is equal to T F P times T minus 1. And so we are done, okay. Okay, now when the case P is equal to 1 is somehow easier. Because when P is equal to 1, so Q is equal to infinity, plus infinity. So we have, immediately have that F times G. So point wise is less or equal than modulus of F times the essential supremum of G over E. And this is almost everywhere in E. And then if we pass to the limit, we get what we want, okay. Ten of the limits, we pass to the integral. Okay, this is of course. Okay, now we have still to clarify when the equality holds for the elderly inequality. Okay, so we know that the integral of this modulus is less or equal than T P over P. Just prove it, P plus, we know this by the young inequality, okay. So here we have the equality, since from these two, when T P plus Q is equal to G to the power Q, energy Q to the power Q divided by F to the power P divided by P. So after this, we observe that if the equality holds, so it must be, FG must be equal to T P over P times FP. So we have this would be T minus Q over Q times GQ. So call it star. Okay, now we look at the function, the quality in a point wise way somehow. So we have that by the young inequality, we have that the modulus of this product is less or equal than T P P times F of P plus T minus Q. Q times GQ, this is by the young inequality. And so basically we can say something about the sign of this quantity minus this, of course. So we have that this implies that you have that T P over P times F plus T minus Q to the power minus Q over Q of G. Q minus the modulus of this product is larger or equal than zero. So we know this and this. So somehow this is, this is a non-negative function. All this here, non-negative function, which has zero integral by this. So this means that it must be zero almost everywhere in E. So for almost every X in E, we must have that this function must be zero. So in particular you have that this equality must also GQ. And okay, let me also show that this is, okay, now let's put it that way. Star, this, it's a Q, yes, it's a Q. Okay, so we have that by the young inequality in the second version for these choices of little a T equal to T times modulus of F and B equal to T minus 1 and modulus of G. We have that T P F of P, so we know when it holds the equality. And so from this what we get finally is that the modulus of G must be equal to a constant times F to the power P minus Q. Okay, or also you can express this since these two are conjugate exponents as constant times the modulus of F times P minus 1. Okay, and this concludes the proof. And now we start to introduce these LP spaces. So we define the following set LP. Let me denote it now with this italic symbol, with the italic L as the set of function, okay, of measurable function. Now, measurable such that you have that this integral is finite. This is for P, which belong to this range while when you consider P equal to infinity, this would be the space of F such that you have that the essential supremum of F over E is finite. So what we can observe is that LP is a vector space. So given to function F and G in LP V and some lambda, we have that for instance F plus G is well defined and lambda times F also. And so it is well defined in the sense that since you know that this norm are finite. So if you have F which belongs to LP and G which belongs to LP, then we have that, okay, by definition, these are finite. Okay, and so we know that if they are integral, we know that the measure where they are equal to infinity is zero. We saw this for integral function. It's true. And so we have that the sum is well defined. And okay, moreover, you have that inside this, you have that lambda F P would be equal to lambda P F of P, which would be finite as well. And then so what about the sum of F plus P? Okay, to prove that the sum of F plus G, sorry, is in LP, we will use this inequality, which maybe you can prove by yourself. And if you have A plus B to a non-negative number, rise to the power P, this is less or equal than to the power P times A P plus B P. So if you can apply this to with the choice, so you have F plus G to the power P is less or equal if you want to F plus G to the power P, which is less or equal than 2 to the power P by this inequality. And so you have F power P plus G to the power P. And so this means that you can bound the integral LP norm of the sum in terms of 2P of this sum here. And that we know by hypothesis that these two are finite, so also you have that F plus G would belongs to LP. Okay, so it's clear how we endow these spaces. So we will endow these spaces with the quantity that I defined before that then we will prove that are norms. Okay, so to prove that these are norm, we still need to prove that the triangle inequality holds and this will be provided by the Minkowski inequality. Okay, which tell us the following. If you have 1P, an exponent P between 1 and infinity, then the LP norm of F plus G is less or equal than the sum of the norm LP norm of F plus the LP norm of G. Okay, we start by somehow the easy case, which are from P equal to 1 and P equal to infinity. So for we observe that by the somehow point-wise triangle inequality, we can infer that F plus G, the modulus of F plus G is less than F, the modulus of F times modulus of G. And so for P equal to 1, just take the integral on both sides and so we are done. Okay, and the same is basically for P equal to infinity. So we have that essential supremum of F plus G in E is less or equal than the essential supremum of F in E plus the essential supremum of G in E. Okay, now we treat the case when P is strictly in between 1 and plus infinity. Okay, so we start to compute this norm P to the power P, so this is nothing as that F plus G modulus to the power P. Okay, we can write this as F plus G P minus 1 times F plus G. And then this is less or equal than plus G P minus 1 times F plus F plus G P minus 1 times G. Okay, now we want to apply the alder inequality to this first term. Okay, so we have this here. So you have that F plus G Q times F P, where of course you have that 1 over P plus 1 over Q is equal to 1. And in particular you have that Q is equal to P P minus 1, and so we can continue here. Okay, I just substitute the values of Q here, okay, times the LP norm of F. And so this is what? This is F plus G LP norm rise to the power P minus 1 times the LP norm of F. Okay, and then we plug this here. What we get is that F plus G is less or equal than another P of plus G times F plus, you have F plus G here. This is P P minus 1 times G P. Okay, you can do the same there. And so you collect F plus G P minus 1, and then we are done. Okay, of course if this sum is 0, then the inequality is 3, but it's nothing to prove. Okay, just to complete, just write quickly the fact that if F plus G is equal to 0, then the inequality is trivial. And then if F plus G is equal to plus infinity, then we use the inequality that I mentioned before. So it's that A plus B, the power P is less or equal than 2 power P times A P plus B P. And so we have that plus infinity must be equal to F plus G to the power P is less or equal than C F P G P. So means that at least one of these must be equal to plus infinity, okay? So at least A is equal to infinity, okay? So this really concludes the proof. Okay, so there is one, still one remark. So we want to prove that what we call LP norm, this is a norm, okay? So there is still something to fix in the sense that to fulfill all the requirement to be a norm, we still miss one of them in the sense that to be a norm, we should have that if the norm is 0, then the function must be 0, okay? But you can, this is true almost everywhere, okay? This is what we can infer. So somehow we don't have this, we don't have that if the norm is 0 then the function is 0 because if you change the values of the function in the set of measure 0, you obtain a different function, okay? So to avoid this problem, somehow it's very easy. You identify functions which are equal outside the set of measure 0. So somehow you consider, instead of considering this space here, you take the quotient of this set over function which are equal outside of set of measure 0, okay? So basically we introduce an equivalent relation, so we introduce that FG, so we say that F is in relation with G, if and only if F is equal to G almost everywhere, okay? So in this way we identify, we are considering class of equivalence actually instead of function, but we identify functions that are equal up to a set of measure 0, okay? So just to be formal, first I introduce this space with this capital L. Now I will define in this way with this other L the space, the quotient space, okay? So this will be the space whose elements are indeed class of equivalence by this construct with equivalence relation, okay? So of course we want that F is, there must be of course measurable function and we want that the LP norm of F is finite, okay? This is a norm at linear space. So finally what we have is that if we endow this new space with this norm, it is a linear, is a norm at new space, okay? So we have that if now I use this somehow heavy notation, but then we will get rid of this bracket, okay? If B is 0, that means that F is equal to 0 almost everywhere, so we have also the common E. So we have that for any lambda positive. We have that lambda F B is equal to B, okay? Still with the brackets. And we have that triangle inequality all because of the Minkowski inequality, okay? So the triangle inequality all because of the Minkowski inequality. Okay, yeah, yeah, yeah, yes. Okay, okay now this, we see a little somehow exercise, okay? Okay, we see that if we are within a space, within a set of finite measure, if the domain of this function, the domain E has finite measure, then we have this somehow syntotic behavior of the norm. So we have that you can compute the L infinity norm of F as the limit as p, exponent p tends to plus infinity of the Lp norm of F, okay? So we start by considering the definition of essential supremum. We have that, this is the smallest number, such that we have that F is inequality all, almost everywhere in E, okay? So we have M is equal to this. And then we consider a number M prime less than M. If we have that M prime is less than M, okay? Then the set, sorry, then the set A of X in E, where F of X is larger than M prime must have a positive measure because otherwise we would find that M prime is the essential supremum, okay? Okay, and in particular, since we know that the domain E is finite, also the measure of A of course will be finite. So we have, we won't, if we estimate the Lp norm of F, this is larger or equal than the integral over A of F of p, the power 1 over p, which is larger or equal because we are within A of M prime times the measure of A, which is finite, to the power 1 over p. So since we notice that the measure of A to the power 1 over p tends to 1, as soon as p goes to infinity, what we get is that we obtain a bound from below of the limit of this quantity is larger or equal than M prime, and this is for any M prime less than M. So you have that many qualities preserved up to M. Okay, on the other side we have that, this is the easy part somehow. If you compute the norm, this is, you have that this is, by definition is, okay, it's F, p1 to the power p, this would be less or equal than M power p1 over p, and this is equal to the measure of E times 1 over p. So for the same reason we have that the lim soup, this time of Fp, as p goes to plus infinity, is less or equal than M, and so we are done, okay, we combine these two, star and two star, and we prove, and we prove the thesis. Okay, so now the question is, here we require that the measure of E is finite. So do you think that we can remove this or there are some counter examples for which, for instance, if we are within a set of infinite measures, this property doesn't hold anymore. So things are very easy functions. A very easy function which is in Nell infinity, so it's bounded, but does not belong, but whose integral in, for instance, in the whole real line is not bound. The easiest function of all, the easiest function of, that you can think, is a constant function, with a constant different than zero. You have that in, so things have, if you have f of x, a constant function with c positive, for instance. So what you have, you have that this belongs to L infinity of R, okay, because the norm will be c, it will be the constant, but of course if you, if you do the int, okay, you can remove c, it's positive, p, 1 over p. This cannot be, this is unbounded, okay? So you cannot have this. So actually this requirement is essential, okay? Okay, now we prove a related result, which is the monotonicity of the alpino. Okay, so again we consider, we consider a set e with finite measure, so then we have the following. We have, if we have two exponents, for instance, b less than R, then we have that 1 over the measure of e1 over p, times fp is less than the same quantity with p replaced by R. So basically if you start by a function, if you are within a set with finite measure, and we start, you start by a function which belongs to LR, and you automatically know that f belongs to Lp. And okay, and if, and let, to exponent p less than R, okay? Okay, so the proof is just a matter of tuning the exponent in a proper way and to apply the order inequality. Okay, so we start by the Lp norm of f to the power p. Okay, this is nothing else that you can see it as f of p times 1, for instance. Okay, now we use the order inequality. Okay, we use the order inequality with some exponent that would be fixed later on, 1 alpha prime equal to 1, so you have that this is less or equal than e fp times alpha over alpha times the measure of v times 1 over alpha prime. Okay, so somehow we want that this p times alpha be equal to R because we have to put this in relation with this part, so we have that if we choose, this is times 1, this is, okay. So if we choose alpha equal to R divided by p, this is an admissible choice because we know that this is larger than 1. And so if you do some computation, you have that alpha prime the conjugate exponent is equal to R divided by R minus p. Okay, so we substitute this choice here and so you have p over R, right, p over R times measure of e, yeah, it's larger than, alpha is R divided by p, we choose it like this, and it's larger than 1. And so you have that alpha prime, the conjugate exponent with respect to alpha prime, yeah, it's this. Okay, you have, this is R minus p divided by R, and so we say that we are done, okay, because then we rise to everything to the power 1 over p and p, 1 over p. You just observe that R minus p divided by R is 1 over p minus 1 over R, and so we are done, okay, because this is the measure of e minus 1 over p as we recall then. Okay, so also in this case this hypothesis cannot be omitted, okay. So do you, do you see why? So can I cancel here? R here. R, yeah, this is, you make this choice, okay. First you say that just generally you apply the other inequality with some exponent then you fix the exponent in that way, you tune them in that way. Here. It's R. R, ah, yeah, because from time, I changed the way, I, yeah, you are right, yeah. Okay, here. You have, maybe it's, ah, p, ah, sorry, it was p, ah, maybe there are also, yeah, yeah, yeah, yeah, because, no, no, it's like this now, because here you take 2 1 over p. Okay, so, ah, okay, so why you need that the measure of e must be less than infinity? Sorry? Yeah, yeah, yeah, now what I mean, okay, now I understand what you would say. So what I mean is that this inequality means that if f, if you know that f belongs to L R with R larger than p, then you can say that f belongs to L p, okay. So this inequality tells you that if the measure of e is finite, then it's enough to know that if, no, this is what I mean. So if we are not anymore within, within a set of finite measure, can we still infer this implication? This is my question. So the answer is no, of course, because otherwise, and so you can consider this example. So for instance, we show, I think that we show this fact, but you know, of course, that, okay, here we are within a set of infinite measure 1 and 2 e, this time is disunbounded offline. So we have this is, this integral is finite if alpha is larger than 1, while it's equal to plus infinity if alpha is in between 0 and 1. So actually the requirement that the measure of e is finite to have this kind of implication is necessary, okay. Okay, and again, this is, we already proved this that any constant function with the constant difference from 0 belongs to L infinity of, but does not belong to, does not belong to LP if we are within, within a set of infinite measure, okay. Okay, so just another remark and then we can stop. Maybe this is, you can think at it, so a function may belong to LP1 for any, for any P1 less than another exponent P2, but must not and not belong to LP2. Okay, if P2 you can, maybe you can formalize this, if P2 is less than infinity, then you can choose this function x minus P2 in LP1, 0, 1. So for any P1 less than P2 and if P2 is equal to plus infinity, then you can use the logarithm of x which belongs to LP1 for any P1 less than infinity, but not to infinity. Okay, so for today I think we can stop here.