 OK, so v zelo vseče. U spesu deličke, bilim? Zelo vseče. Vselo za spesu deličke in vseče zato njega u spesu deličke. Zelo R, G v spesu deličke. Zelo vseče. Kratič. In, ok, značujemo norma. Zato je to suprimum vsega, da je vsega funkcija f in lpe, ki ne bo 0. Značujemo vsega resulta. Vsega izgleda. Ok, kako lpe je vsega 1 in infinity, Zame si če endinga zaščelo, da je gazlost v ljube. Kaj je ineštavno vsega eneštavna eneštavna eneštave. Ko jel bi zgeljeno, da proroduje, je ovo, nespečno, dataj je kaj je vsegerzavno.If we do longer be in this way you can define f of f as the integral over e of the product between the two. Moreover we also prove that we can say something about the normal big f. Ok. So, beside this, we prove also lemna. So, we consider e a measurable set with finite measure. And we assume that there is a constant m positive, ...sat that we have that the absolute values of f of the integral over e o f times g is less or equal than m times the lp norm of f. And this must be true for any f bounded and measurable. Then we show that we can infer that g is an hq. And also we can control the hq norm of g by means of m. Ok. So, what remains to prove? And this would be the last theorem of the course. And then we will see some exercise in the preparation of the test. The so-called Ritz representation theorem. So, this is known under the name of Ritz representation theorem. Ok. Which test is the following? So, you have let f be a bounded linear functional on lp. And you can somehow represent this big f by means of a function g in lq. Ok. Then there is a unique function we will call g in lq. Where as usual with p and q are conjugate exponents. Ok. Such that f of f. We can represent big f in this way as the integral over e of f. Which is the test function times this function g. And we have also that there is a relation also between the norm of this big f and the norm of g. So, the norm of f of capital f as functional is equal to the norm of g as element in lq. Ok. Ok. So, for simplicity I will prove this when e is an interval. So, in the case when e for instance is not an interval but it is bounded. You can think that e has a set contained in an interval e. If e is unbounded you can split it into intervals. Ok. Ok. Just keep in mind that theorem holds for measurable set e. Ok. So, now we will see the proof under this hypothesis. Ok. So, we consider a sub-interval of e of the type. For instance, I don't know, we can think of it as 0,1. Just to fix the idea. So, we consider a sub-interval 0s where s is less or equal to 1. And we denote with as usual key of s the characteristic function of this interval. Ok. So, basically we want to proceed by step. Ok. We first, we try to test this capital f on this very elementary function key of s. Then we improve, we refine our argument. Ok. Let's define f of s by definition f of s. Ok. Ok. So, phi is a function between 0,1 and r. And now we claim that this phi is an AC. Ok. Phi is an AC. Ok. As usual, when you want to prove that the function is AC, you find a collection, a finite collection of non-overlapping intervals. So, for instance, we can denote this finite collection in this way. Take s i s i prime. We use this notation with the prime usually. So, be any finite collection of non-overlapping intervals, which have length of 0,1, which has length less than delta. Ok. Now we define a function f by means of this combining this notion and the intervals that we have just introduced. So, we define f as the sum over e of the characteristic function of s i prime minus the characteristic function of k s i time the sine of phi phi s i prime minus phi. Ok. So, we have that. Ok. So, let's compute f of f. Use this function f as this function. You have that f you use. We want to simplify this expression. So, we have that. This is by definition. Ok. f is linear. So, we can. Ok. So, basically what you can do all that. So, this by definition is phi. So, at the end you end up with this. So, we have that finally. Ok. So, now we recall our choice of this finite collection on over mapping integral. And then we have that the Lp norm as to the p is less than delta. And so, we have that. We have that sum is equal to f of f, which is less or equal than the norm of f as a functional times of p, which is less than delta 1 over p. So, basically we have that total variation of phi is less than this. So, if we take phi is in AC and taking 1 is to be less than epsilon, taking delta equal 2. Ok. So, we know that phi is in AC, we can use a previous theorem. Ok. Ok. So, by a previous result. Ok. We know that we can represent this phi in a special way. We know that there is an integrable function g. Ok. So, we have that in particular f of k of s, which is precisely phi of s, can be expressed as the integral between 0 and s of g t, where g is an integral function. So, here you can express this time characteristic function. Ok. From this we can immediately somehow generalize this. So, we know that any step function is a linear combination of characteristic functions. So, we can extend this relation to somehow more general functions. So, we know that every is a, ok, is a suitable linear combination of the type c i k o s i. And so, we have that psi is equal to the integral between 0 and 1. This is true for any, this is for any step function. Ok. Now, we want to generalize a little bit more and take abounded measurable function. Ok. So, let be abounded. Ok. If you remember, we showed that we can maybe, when we dealt with the measure theory, we showed that we can approximate this kind of function with step function. Ok. So, we saw that there is a sequence such that psi n converts to f almost everywhere in 0,1. And then we can, ok, we can use either the bounded convergence theorem or the Lebesgue convergence theorem, but it is the same. We know that since we have that f n, sorry, f minus psi n p is uniformly bounded and tends to 0 almost everywhere in e, then we can apply the bounded convergence theorem. Ok. And we can infer that they converge in lp. Ok. We have that f minus psi n tends to 0 in lp norm. Ok. Moreover, we can say more. So, we have that the functional f is bounded. Ok. We can estimate this difference. Ok. If you want, we can just proceed by step. So, we use the linearity. Ok. And then we use the definition of supremum. And we know that this goes to 0. Call it, so call this star. And moreover, we also know g times the constant c, because psi n are uniformly bounded by the same bound of f for f. Then, ok. So, by the normality convergence theorem, we have that f g is equal to the limit. So, by, you combine star and we have that, we finally have that f of f. And this is true for any bounded measurable function f. So, we need not to extend for any function f in p. Ok. We can also, from this we can also deduce a bit more. We can deduce that since f is less or equal than my previous result, the one that I stated at the beginning of the lesson, we have that this plays the role of m, if you want. We have that g is in the queue. And moreover, we can estimate the norm of g by the norm of f. Ok. So, what remains to show? So, it remains to show that f of f is equal to, for any f. Ok. Here we can, we have that f in p, can be approximated. So, we have f in p, then the random epsilon positive. There is a bounded function, bounded measurable function. Ok. Call it psi again, such that is less than epsilon. Ok. So, we know that the relation is true for this particular psi, because it is bounded. So, we know that f evaluated on psi is indeed equal to psi. And now we want to estimate this difference f of f minus this. Ok. We want to prove that this is arbitrarily small. Ok. So, as usual we split, we add and subtract of f psi plus psi g minus f times g. This is less than equal than f. We can bound this by the norm of f as function minus epsilon. Ok. This is by the other inequality. For instance, we can collect here the same quantity. We have this. Finally, what we prove is that epsilon is arbitrary. So, we prove that indeed f of f is equal to, for any f in p. And equality, this follows from a previous lem, ok, by the previous proposition. Ok. The fact that g is unique, you can do it by yourself. You assume that there are two different, and you will get a contradiction. Ok. Ok. Show you exercise, which was one of the ones in the homeworks. Ok. So, you had to prove that you have a sequence fn, which converts to a function s in measure. And you know that there is an integrable function g, such that you have any fn is bounded by g. And then you had to prove that they converge in L1, basically. Ok. So, there are many ways, actually. So, now I propose one, but two of you, so Delia and... Propose other approach, which were correct, as well. Ok. Yeah, so Delia and Andris, they propose other approach, which were correct. Ok. So, somehow here Delia were to split this domain e, which, in principle, could be unbounded, in a bounded domain, and in an unbounded part, and to deal separately in those two parts. So, you fix some epsilon. Ok. And we know that fn minus f, somehow you cut. You use a cutoff within this interval here. And you see that this converts to fn minus f, almost everywhere, as k tends to plus infinity. And then here, this point, you can use the Lebesgue convergence theorem, or the monotone convergence theorem. So, by the monotone convergence theorem. Ok. Here n is kept fixed. Ok. What is moving is this k. So, by definition of limit, we know that there exists some n such that outside this interval we have that fn minus f was less than, for instance, epsilon over 3, just 2. Eh? This is n. No. What do you expect here? What? n. And, ok, it's related with k. I call it n, but it's related with the index k. Here n is kept fixed. Ok. Ok. Now we look at the complement of this domain. And we have that. We recall that by the absolute continuity of the integral. Ok. We have that for any delta. Ok. For each n given some epsilon positive resist delta, such that for any set, which is small, set A, such that the measure of A is less than delta, we have that this integral is small as well. Ok. We can put over 3, just because then we will gather many times together. Ok. So, we call it this. Ok. So, we may assume with no loss of generality, that delta is less than epsilon over this number. And we have that. Since fn converts to f in measure, there exist sum and prime, such that the measure of the set where fn minus f is larger or equal than delta is less than delta. And this is true for any n larger delta. Ok. So, we define as A, so the set, which will play this role, as the set of fn, this set here, ok, where fn, fnvex minus fvex is larger or equal than delta. Ok. And then we combine all these three parts somehow. So, we have that. So, we want to estimate this, ok. So, we split it in three parts. Part that we encounter at the beginning, so this unbounded domain, plus intersected minus n and n, so, which is bounded set, plus the complement of. And then we have this part was less or equal than epsilon over 3 by what we call star, plus this part, ok, we are here. So, this is less than epsilon over 3 because of this, plus we have this is less than, this is AC intersected minus n, n, and this is can be bound by delta because we are outside this set. And so, you put all these things together, so, you have that epsilon over 3, plus epsilon over 3, plus delta. We assume that delta is less than epsilon over times 2. So, this comes. This is. Ah, ok. Ok, so this is by the absolute convergence of the integral, this second term here we are within A, and A is a bounded domain, is a domain which has small measure, and these are integral function. And here we are, ok, here we are outside A, but at least we know that the function here, this difference is small, ok. And then we are done. If you start from a bounded domain, probably you don't need the first part, you just need these two, ok. This is the second. You don't need to do the first step, ok. Ok, now I'd like to show you an example of a function, which is quite of a sequence, which converts in measure, but not almost everywhere. Ok, probably you already know this. Ok, for instance, we can work in this closed interval, 0, 1. And we consider, somehow we split n, integer n, with a compose n has power of 2, plus some rest. R is in between 2k. Ok, we have that. If we define this sequence of function here, it follows the characteristic function of R2 to the minus k, R plus 1 over x. Somehow you can look at it as, this is 0, 1. Has a train, which passes through 0, 1 infinitely many time, and each time you reduce the size of the development, ok. The size of the train. And so what happens is that for any x, 0, 1, there are infinitely many n, such that fn of x is equal to 1, and also infinitely many n, such that fn of x is equal to 0. So there is no, there is no point wise limit, ok. Ok, but what about the convergence in measure? Ok, this is one fact, but we have that, the fn converts in measure to 0. Ok, so we want to estimate the set that is epsilon. Ok, for instance, you can, in this case you can fix epsilon between 0 and 1. But please. Hmm? There is infinity, infinitely many values, but what if this, the whole set of infinity many values of 1. Ok, you can, you can find in them, ok. Yeah, fn of x converges to 0. In measure, ok. There is no point wise convergence, no point wise limit, ok. fn of x converges to 1. Equal to 1. Equal to 1. Equal to 1. There are infinitely many n for that fn of x is equal to 1. And 1x? Yeah, fix x and you move n. R is in between 0 and 2k, ok. So each time you decompose n in this way. Ok, you fix n and you decompose n in this way and you define fn of x in this, ok. And then you see, I mean, you can think at it, I mean, as, I mean, these are the characteristic functions, I think like this, ok. Like something that goes to pass through 0, 1 and go to 1 and then come back and then come back. And each time it reduces the size of the internet. So this is just to have the idea that you can formalize. But this is the way, ok. It passes to 0, 1 infinitely many times and this means that each time it reduces the size of this interval. So you cannot have point-wise convergence. But since it reduces this size, you have point-wise convergence in measure, ok. So this is what I'm going to formalize. So you have that fn minus 0. Ok, this is equal to R2 minus k. Ok, this is this and plus 1 to minus k. Ok, sorry. Ah, yes, it's larger than the time. We want to prove that it converges in measure, ok. So we want to prove that this set has a small measure. So now we have that measure of this set. It is equal, it is less equal than R2 minus k. And this, as k goes to less infinity because as m goes to infinity as k goes to less infinity. So we have that. So we have that, ok. And I want to show you another another exercise. May I erase this part? So here you have a sequence of function fn. Ok. We are within a set of finite measures, ok. We have a sequence of function fn in lp. Ok, such that we have that fn converge to f, some function f, almost everywhere in e. And we know that f also belongs to lp. Ok, we also enforce this extra hypothesis. Assume that there is a constant m positive such that this sequence of function is uniformly bounded in lp norm. This is for any n. Ok, then what we want to prove is that for any g in lq. So again we have this relation between the two. We have that a g, yeah. Ok, so this is what we would like to prove. Ok, so first of all maybe we should first answer to this question. So do you think that under this hypothesis fn converge to f in lp? In vlp norm, or not? Because in that case would be very easy, no? So is it true that under this hypothesis or not? Of course the answer is no, because otherwise would be too easy. And you can, for instance you can also fix this by an example. So take p equal to 2 and we can define e equal to 0, 1 and define fn of x as a square root of n times the characteristic function of the interval 0, 1 over n. Ok. And take f equal to 0. So we have that indeed. fn of x, fn converge to f which is equal to 0 almost everywhere in 0, 1 because you fix some x and if n is arbitrary large it step over the part where it is equal to square root of n. So this is true. Then. So what about the norm of fn is this bounded? The l2 norm, this would be sorry this is 0, 1 and ok, actually we are within this set n, 1 over 2 and this is equal to 1, right? Ok, but we don't have convergence ok, because of this now because we have that this is 0, 1 this is just to be minus 0 so this is what, this is just ok, of 1 over 2 this is just 1 over n and to the power 2 here this is n and this is equal to 1 so it doesn't go it is always equal to 1 so it cannot go to 0 ok, so we have to prove this using other arguments Is it true and you don't prove it? No, we don't prove it we prove it that this is not true by a counter example I mean, what we want to prove is that is this, ok? If under this hypothesis we have this if this would be true, this would be trivial, but this is not true because we showed it by a counter example that this is not true ok, so somehow the convergence in LP is much stronger than this ok, so we want to prove this we have to use to combine somehow the other inequality and the absolute continuity of the integral ok, so I recall you that so, bye now we have that for any epsilon positive there is this some delta positive such that for any so for any set a contained in e a small measure such that the measure of A is less than delta then the integral in particular we have that the absolute values of G raised to the power q is an integrable function so we can apply the absolute continuity of the integral to this function is less than epsilon ok, just for convenience let me take one over q here ok, now comes the theorem that maybe we we forgot the theorem of Egorov-Severini ok, we have that for any delta positive there is a subset such that measure of A is less than delta we have that fn converts to f uniformly outside this set of small measure ok, so now we can we can estimate this difference ok, so take this difference then G minus fG is less than ok, now we use this the composition here A and E minus A ok, now ok, here we use the yield inequality on both integrals ok, so how can we bound this part? so what we know is that by hypothesis we know that we know this, right? so how can we infer a similar bound for f which theorem can we use which covariance theorem which somehow so what we know is that this the fatulema ok, fatu is enough so fn converts to f almost everywhere are positive this is less or equal so you just need a bound this is less or equal then the means so by fatulema and this is uniformly bounded by this so we can continue here and say that this is so equal then 2 times M times epsilon because we saw that by the absolute continuity of the integral this is less than epsilon plus ok, so we have fn minus f in L infinity g you can bound this by the 2 norm of g on the whirl hill that's not important times the measure of a but here ok, we are within a set of finite measures so this is ok, this is bounded this is bounded because g is in the queue and this is small why? why this is small? ok ok ok ok ok, put p here ok ok, because we are interested ok here you have fn minus f in L infinity here ok, here you can bound this by 2 times so you have fn minus f in Lp is less equal than the sum of the 2 Lp norm ok, maybe it's too small is it ok? over? over A, yeah yeah ok, but if you want here what is important here you can also bound again this by the integral over E here it's enough a rough bound A is important for g ok, to make this small, this you can also bound in a kind of rough way ok, but this by Igor of Severini we know that this is small because we know that outside this set of small measures we have uniform convergence, ok so this is if you take an sufficiently large, this is would be 2m times epsilon plus epsilon ok, so you have this is fixed the measure of V minus A times 1 over B, this is finite because it's finite, it's a finite measure so if you choose epsilon sufficiently small you can make this quantity, this sum as small as you want ok, so maybe you can do just another exercise so very about absolute continuity just to conclude ok, so we have E and interval and we consider a function which is defined over I with values in R B monoton ok, assume increasing ok, but you can just to fix the idea but it's the same if u is the crazy, ok and let I is equal to V or rather maybe contained in I just to be sure ok, then we have this equivalence, then u is in A C of A B if and only if we have that the integral of the absolute values of the of the first derivative of u prime is equal to mu P minus u A ok, so since we assume that is monoton increasing we can get rid of these absolute values ok, of course interesting part is that to prove that if it is true then u is in A C because the other way it comes directly from fundamental theorem of calculus for A C function, ok so this side so if you start if u is in A C then it's straight, ok we know that the function in A C can be expressed as the indefinity integral of the integral prime ok, this is 3 ok, so let's prove the other side ok, here the idea is somehow something that we already know so we want to write u as the sum of two function in A C ok, to do this we define this function wx as the integral between A and x of u prime t in dt ok so we know that by the Lebesgue theorem that u prime is integrable so w is in A C ok, now we want to to use something more about this function so take for instance y larger than x and we want to evaluate wy minus wx this is what by definition this is the integral between x and y of u prime dt in dt and ok, now we can about u we just know that it is a monoton increasing but we can use the Lebesgue theorem and tell us that this is less or equal than u y minus ux and so we have that what we get is that ux minus wx this is a particular function is less than uy minus wy so it's increasing but the same time we have that if we evaluate on the end point a and b so at some point we have to use this of course so just evaluate this function at the end points this is by hypothesis this is equal to ub minus ua so at the end what we get is that wb minus ub is equal to wa minus ua so it's increasing but it's at the same point has the same value so it must be constant ok, so we are done of course a constant function we may see function is that ux and u has wx plus ux minus wx this is in ac because it was just defined as the definite integral of the integral function and this is in ac actually this is more than ac in this constant but in particular it is in ac so the full function is easy so we are done ok, so we can stop here