 Okay, so last time we proved the Euler inequality, do you remember? So basically it tells you that if you start by two functions, one of which is in LPOV, the other one G, take it in LQV with where P and Q are conjugate exponents, which means that then you have that the product of the two functions belongs to L1 and especially you can bound this L1 norm in terms of the product of the LP norm of F and the LQ norm of G. Okay, and beside this we prove the Minkowski inequality, which somehow we observe that it provides a tool to prove the triangle inequality for the LP norm. Okay, so we go on with our analysis. Okay, so just with the following theorem, which is known under the name of Chebyshev inequality. Okay, so you start by P less than infinity and take F of function which belongs to a P and we define Mt as the measure of the point where the modulus of F is larger than T. So we have that you can bound this Mt in this way. Okay, of course this is a consequence of another version of the Chebyshev inequality that we already proved, okay. Okay, so it's easy to follow that by the former version, by the somehow former version of the Chebyshev inequality. So we apply our former version to the modulus of F raised to the power P and to T instead of T, T to the power P, okay. So we have the measure of this of this set is less or equal than 1 over T to the power P to the times the integral of V over F P and this is precisely 1 over T to the power P times the LP norm of F to the power P. So basically what we get is that the measure, okay, is what we want, okay. T being this two quantity the same, this is, sorry, less or equal. Okay, so an immediate consequence of this Chebyshev inequality is that somehow that the convergence in the LP norm is stronger than the convergence in LP and so on. And moreover, if you have that sequence Fn converts to a function F in LP then you can extract a subsequence of the original sequence which converges pointwise everywhere, almost everywhere to the function F, sorry. And consider a sequence of function Fn in LP such that we have that goes to zero, okay, as it tends to plus infinity. This is the convergence. This is what I mean when I say this means that Fn converts to F in LP. Okay, then what we want to prove is that Fn converts to F, so we want to use this, okay. So it's clear how you use this, you fix an epsilon. Okay, of course instead of F you consider the difference between Fn and F, okay. So you have that for any epsilon positive. We have that measure of the set where Fn minus F is less or equal than epsilon is less or equal than one of epsilon to the power p times Fn minus F. Okay, so this goes to zero to plus infinity so you can make this measure arbitrarily small. So and another corollary, corollary of this proposition somehow is that if you have that, if you have a sequence Fn which converts to F in LP, there exists a subsequence actually you can claim this just for a subsequence such that subsequence called it Fnk for instance, such that Fnk converts to the same F almost everywhere, okay. So this is a corollary of this proposition in the sense that this proposition tells you that the convergence in LP is stronger than the convergence in measure and we know that if a sequence converts to a function F in measure then we can extract a subsequence which converts almost everywhere, okay. Measure this is for previous proposition. Even from this we deduce that there exists, so we prove this, no? There exists a subsequence such that Fnk converts to F almost everywhere, okay. Okay, so what we proved last time, we proved a lot of things about the norm in LP. We finally proved that it is indeed in norm. What it remains to prove is that the space LP E is complete, okay. Because of course for application only complete spaces are interesting, okay. So the theorem which is known under the name of Ritz Fischer, I'm sure that indeed LP is a complete space. Okay, so we distinguish two cases. The cases in which P is in between 1 and infinity but doesn't take the values infinity. And then we will consider the case when P is equal to infinity. Okay, so we want to prove that the space is complete. So we start by a QG sequence in LP. Okay, so that sequence in LP. And then by this previous proposition, okay, we have that what we somehow we already observed. We have that we can estimate the measure of of this set to work in this way. But epsilon of course is an arbitrary small number. I'm sorry, now since we are starting by a QG sequence, let me and this goes to 0 as infinity. So in particular we have that the sequence fn is a QG sequence in measure. Okay, then for what we already proved, we proved that there exists measurable function f and a sub sequence and sub sequence f and k call it sub c. Okay, such that we have that f and k plus 2f almost everywhere. Okay, and of course it's clear that f would be the candidate for the convergence in LP. Okay, now we use the fact that our fn is a sequence in LP. Okay, so since the QG sequence in LP, we have that for any epsilon positive that exists an integer and such that for any index n and then larger than n, we have fn minus fn p is less than epsilon. Okay, we can take it. Okay, now if you remember when we construct in theorem where we prove that a sequence, a QG sequence in measure admits a sub sequence which converts point-wise almost everywhere to a function f, we notice that nk is increasing with respect to k. I mean, we prove that with no loss of generality, we can assume that this index nk is increasing with respect to k. So we still need this and so we have that we can assume that nk is larger than k and if k is larger than n and n is larger than n, then k is larger than capital N and then we have, okay, that we estimate the norm and the norm of this, okay, which is, write it as a norm, is less than epsilon and this is for any nk larger than n. Then we also use, so we want somehow to combine the converges in p with the convergence, with the point-wise convergence. On the other hand, we also know that we have that fn minus fnk, this model is to the power p, converse, take n fix minus f, this functions almost everywhere in e, k tends to plus infinity and then here, so we use the Fatou lemma, so we have that limit fn minus nf, so the limit is less or equal than the limit as k tends to plus infinity minus power p and this is less or equal than epsilon. So what we prove is that fn minus f is tends to zero and tends to, so what remains to prove is that the limit f also belongs to p, okay, okay, so we want to estimate the norm of f, okay, we use the triangle inequality, so this is less or equal than epsilon for an sufficiently large and we have this sequence, the quesish sequence, so it's uniformly bounded by some hand, okay, and so this is finite, okay, this is for any n larger than, okay, now we it remains to prove the case when p is equal to plus infinity, okay, so we fix two index and then two positive integer and we have that, we recall how it is defined the l infinity norm is the essential supremum over e of fn minus fn, so by definition of essential supremum that there exists a set which depends on those two integers such that the set has zero measure, l infinity norm can be viewed as the norm of the supremum, the classical norm over e minus this set of of measure zero, okay, so the idea is that we want to remove this set of measure zero over all the possible integer n and m, we are doing anyway countable operation, okay, so you define a as the union over n and m of e and m, okay, so since we are doing countable operation we have that the measure of a is zero and, okay, so we define b as the set e, so our domain minus infinity set of measures zero, so we have that fn is, so now we focus just on the set b, so our original sequence of fn is a Cauchy sequence in the sense of the uniform convergence, okay, the Cauchy sequence in the sense, in the classical sense of the uniform, of the uniform convergence in b, we have that supremum of fn minus fn with b goes to zero for n and then that goes to zero, so hence we have that the limit exists for the uniform limit and so we can extend f outside of b, so in a for instance you can put it equal to zero, so yeah, yeah, yes, so when p is equal to plus infinity we define the lp norm in that way, okay, this is just a definition, so if you remember the essential supremum, if you look at the definition is, yes, it's the infimum and it's defined up to a set of measures zero, which are this one, okay, so this would depend on the index, so we have that this set here, meaning this definition is just the usual, the classical definition of supremum outside set of measures zero, this set here, okay, we have just to look at the definition of essential supreme, okay, and then the idea is that we want to use what we already know about the classical uniform convergence, so we remove this set of measures zero, we work with outside this set, we find that f resists and then we say, okay, we can extend f in a, for instance putting a equal to zero is not important because a is a set of measures zero, okay, so you can find it as you wish somehow, so f resists to zero, and in this way we have that it converts to f, and of course the limit that belongs to infinity because it's, okay, now we introduce, so somehow within this juice space, we can define bounded linear function, okay, so we consider p in this set, p of e, and we want to define the dual space of this, of this Banach space, okay, so we will, okay, we will denote with Lp, this is the dual space of Lp, so how it is defined, so this is a space of functional, so we have g, functional g between, takes values in Lp, so now takes value in Lp, the elements are in Lp, and takes values in R, okay, g is linear and continuous, okay, this is a vector space, particularly you have that, if you take a linear combination of f1 plus, this is trivial, this is a linear function, and they're bounded, so somehow our final aim would be to, to characterize this, this functional g, okay, and of course we will proceed by step, so we need to prove this proposition, so you consider p between 1 and infinity, and we have that, if we consider the conjugate q of p, we have that each function, each function g, which belongs to q of v, define a bounded, sorry, ah g, sorry, thank you, yes, I, yes, yes, thank you, bounded linear functional by this definition, you take the product of the tool, okay, and moreover you can also say something about the norm of this big f, which is precisely equal to the norm of g, the q norm of g, okay, okay, the fact that f is linear is clear because the integral is linear, okay, so the function is clearly, okay, since the integral is linear, so, okay, we prove the fact that it is bound, that it's somehow, it's easily followed by, by the other inequality, okay, okay, so we have, so by definition of, so the norm of f is functional is equal to, okay, it's less or equal than, than the norm of gq, it's the supremum of this quotient, so, so we are done, so what remains to prove is that, is this equality, okay, okay, so just f in infinity, f different from zero, okay, okay, so we want to prove is equal to the iq norm of g, so somehow it's better to divide the proof in three cases, we consider three cases, okay, the first one is when p is strictly in between one and infinity, uh, so, so we want to prove this, so we have to select somehow in a, in a proper way a function f which realize the maximum, to prove that the maximum coincide with, which realize the supremum, okay, so we have, we want to select a function p which realize the supremum, okay, p and q are conjugate exponents, yeah, from time to time I forget, usually we always use this notation p and q for conjugate exponent, okay, okay, so a way to define this way is the following, so you define f as the modulus of g rise to the power q over p times the sine of g, the sine function, it depends on the point, yeah, yeah, it's the sine function, this is of, I mean, f of x, g of x, sine of v, okay, so where g is positive is, is one or is negative is minus one, okay, so we have that f to the power p by definition is g is equal to, okay, let me, this is, call it star, this is, okay, to see this, this is, sorry, or just computation, but this is finite q over p times the sine of g, so this is f times g, this is the modulus of g, and so since these two are conjugate exponents, so you have that q over p is equal to q minus one, and so you have that, this is, sorry, q minus one, this is p and q, so we continue here, they add an arm of f as functional, I mean, this is, I mean, the definition of an arm of a function, okay, so you compute for any f in the domain, in the space in this case, you compute f in, you take the supremo, okay, this is a general definition, doesn't not, yes, because, so f is a functional from lp to, so g is fixed, you fix g in lq, and you define, so g is in lq, and this is a way to define the function, okay, but the functional, um, takes values in function in lp, where p is the conjugate exponent of q, okay, so they have this and, okay, from this, so, okay, since we know that our special choice of f leads to this, we have that, and we know that since g is in q by, by hypothesis, then this we have that f is in p, because we still did not check, okay, now we know this, so it's, we can, we can, we can test our functional capital f on, on this small f, okay, and see what happens, so, okay, so we have, by definition, this is f times g, and we use this, and this is g, q, and, so we are done, right, because, times, so we have that the supremo is achieved in this choice of, for this choice of f, and actually the norm of f is equal to the IQ norm of g, and now we have to consider the case when, when p is equal to infinity, and for instance q is equal to 1, the conjugate exponent, so in this case we have that our functional f goes from l infinity v leads values in r, and, okay, it's defined as we know, so we have that this time take f as precisely, it's just the sign of g, okay, of course f is in infinity because it takes finite values, and we have that is equal to, to this, so we are done, okay, so the last case is when t is equal to 1, and q is equal to infinity, goes from n1 to 11a, we want to prove, I just want to remind you that the supremum of the f in l1, which are not the zero element of, of this quotient, is equal to the n infinity norm of, of g, okay, so if g is, okay, we start by observing the trivial case when g is equal to zero, of course there's nothing to prove, okay, so consider a case when g is not identical to zero, so it's, it's different from zero on a set of positive measure, then we claim the following, and we, and we prove the following, but first we, we state the following claim, we have that for any integer k positive such that the l infinity norm of g is larger than k, then there exists a function f in l1 of e such that this function f in l1 is equal to 1, and such that this integral is larger than k, so in that way, so in that way we will have that, that we mean over k of this function f, which of course belongs, and depends on k, g is larger than, okay, so fk times g, the function, right, okay, so we focus on the set ek defined in this way is the set of the x in e such that g is larger than k, okay, so by definition of essential supreme we have that since we have that the norm, the l infinity norm of g is larger than k, this set ek is, is positive, is positive measure, okay, then we distinguish somehow within this part of the proof too far the case is, in the case when, so we know that ek is positive measure, so we assume that beyond being positive is finite, okay, if the other case will be treated using this one, okay, then we can, we set function f that we want to construct in this way, take 1 and divide by ek, the measure of ek, this makes sense, times the characteristic function of ek times, one more time the sine of, the sine function of g, okay, now we compute the function for this choice of f, so we just substitute and what we get is the following, we have a g because the sine of g times g is the model of g times ek, and okay this is, I'm actually doing the integral over ek, so this is larger than, this is precisely equal to k, okay, and so we are done, and now the case when, okay, moreover we also have that the l1 norm of f is equal to 1, okay, and the other case when the measure of ek is plus infinity can be treated somehow analogously to this, we just observe that if infinity, okay, then we can find a measurable set fk, call it fk, such that you have that fk is contained in, in ek, it has a finite measure, but finite measure but positive measure, because for instance you can observe that in general you can split this thing that we already saw that in k you can see this as the union over n of ek intersected minus n, okay, this has finite measure and so on, and then once we do this we define the function f using this set fk, and we proceed as before, okay, as before, okay, so this concludes this proof, we see another lemma which will be helpful to, to characterize this, this functional, okay, now we consider the, the exponent p between 1 but doesn't take the values plus infinity, so let g be an integrable function over, over e, we want e to have finite measure, and we assume that there exists a constant m such that we have the following bound constant, so we have that the absolute values of f times g is less or equal than m times the lp norm of f, and this must be true for any f for all bounded and measurable f, okay, so I mean somehow it makes sense to consider the lp norm of f because we're saying that the function f must be of course measurable and bounded, and we are within a set of measure, of finite measures, so this norm is finite, okay, so in particular they are in lp, okay, then the conclusion is the following, then if we know this we can infer something about g, we know that g is in a q, okay, with as usual I mean with p and q conjugate exponent, right, and the lq norm g can be bounded by n, so also in this case we distinguish two cases, the first one when p is in between in the open interval 1 and plus infinity, and then we will consider the case when p is equal to 1, so we define starting by g a sequence of measurable functions in a usual way, measurable and bounded function, okay, so we define a sequence and call them g n, so in this way, so they are equal to g of x and the modulus of g of x is less or equal than n, or otherwise you just put it, you define it as zero, it's a cut off somehow, okay, and so starting from this function g n we define other auxiliary function fn such that they are g n or q over p, so this is a choice that we already made before the sign of g n, okay, so we have compute the lp norm of this function fn, and this is, okay, by definition is just the lq norm of the function g n to the power q over p, and okay, just let me observe that the modulus of g n to the power q is equal to fn times g n, which is equal to fn times, so what about this norm, the q norm of g n to the power q, okay, by definition it is the integral of e on the modulus of g n to the power q, then we use that fact, so here this is e fn times g, and this is, now I use the hypothesis, this is by hypothesis, oh, it's less or equal than m over p, which is equal to m times, I use that quality, okay, so now since we have that they are conjugate exponents, so q minus 1, so we have, we have that, okay, so let me write it this way, which is less or equal than m, so we have that, okay, so this means that this integral is less or equal than m to the power q, and moreover we have that g n to the power q converged point-wise to g to the power q, everywhere in e, so we are allowed to apply the tato's lemma, okay, these are bounded, hence we have that g is in that q, and it's normal, it's indeed bounded by this n, so this concludes the part concerning when p lies between one and infinity, okay, so for the other case, case when p is equal to one, okay, we argue in this way, okay, so we, somehow we, we define this, this set f, which is the set of the point in e, where the modulus of g is larger than m plus some, plus some epsilon, okay, so g of x is larger than equal to m plus epsilon, so somehow I mean our goal would be, would be to prove, would be to prove that the measure of f is, is zero, okay, again we perform a suitable choice of this kind of test function f, so let f be defined as the sine of g times the characteristic function of this set f, okay, so then we compute the, what we are interested, so the n one norm of f, we use the definition, so this is the integral of the modulus of f over e, we use this, this would be larger or equal than times g, okay, so basically we see that if the measure of f is positive, then we get a contradiction, sorry I did a, sorry I did a mistake, so I'm sorry, okay, this is the, the sine of g times q of f is the measure of f, okay, then we have times the measure of f is equal to times l one norm, now we use the hypothesis, actually we call that, now we get a contradiction, so we get that if the measure of f is indeed strictly positive, so we can cancel out and we would obtain that, that m is larger than the measure of m plus epsilon, which of course is a contradiction, okay, so it's impossible, so it's a contradiction, and so we have that indeed the measure of f is zero, so we have that the n infinity norm of g is less or equal to positive proof, okay, so on the next time we will can anticipate what we will see, so it's the following theorem and this will conclude somehow these parts of the course, which is the Ritz representation theorem, which tells you the following, so if you have f bounded linear functional in a p, p to r, then you have a way, a unique way to represent it in the form that we use in these lectures, then there is a unique function g, which is in my q, okay, where p and q are usual or conjugate exponents such that you can represent this function and this is what the n is representation, in somehow in a quite convenient form you have that f of f is defined as g, and moreover we have, you can say something about the norm of f as a linear function and just that the norm of f as a functional is equal to the lq norm of g, of this somehow representation function, okay, so but we will prove this next time, I think that for today we can stop.