 Vse je bilo načine, da se načine, da je vse načine, da je vse načine. L1, kaj je vse začne, ovo je vse načine. To je vse načine, OK? Zato imeš nas načine, da je vse načine. Widel, da je vse načine, da to je ne in in in in in, da bi sem... Tako izproutim, da je to zelo izplavno karakterizacije. S njem da smo da mojimo števenje v vsemenje z 2, skorke z Vljaga, skorke z Vljaga, v bl إنlci na izpronvenje začnjala produkt izgleda parallelograma identična, tako zelo smo tukaj izgledali. Tukaj karakterizuje vse norma, izgleda z skala produktu. Zato vse možemo izgledati, da je to karakterizacija, izgleda z skala produktu, da je to ne Hilbert, L1 normo ne zte, z kvaližba. Počutiti, to je čestna, da da vse istreje, že L1 normo neеленi iz kvaližba. Je vsožasilo, da izprotojno, že Pay-dela vse za delom njič sem stvari. Tukaj nekaj koment nekaj izgleda na zelo, zelo na l1 norma. Zato, če je l1 norma v r2? Zelo na r2. Ne v l1, v r2. Če je l1 norma v r2? Tukaj. Vektor. Vektor v r2, če je l1 norma. To je modulos v r1 plus modulos v r2. Kaj je l1 norma? Zato. Zato. Zato. Vse. To je l1 norma v r2. Zato. Zato. Zato. Zato. Zato. Zato. To je l2 na vrlozna dimensiona ? To je l1, to je l2 in v l3 ? ? V ambulance ? ? Zato. Zato. Zato. tudi je del 2, del infinity. V vseh vseh je vseh konvex objev, kaj je lp. Vseh je vseh. Vse otvore, da se to zelo, da se je izgleda, da je izgleda, da se je izgleda, in da se je izgleda, da je izgleda, da se je izgleda. Zelo smo v finetih dimenzijaju. Vsa močer, kaj je to, kaj pa je zelo, vzelo rombiče obzir, in izvenoВ pa razredno. Vzelo rombiče je zelo v razredno. Zelo se pomoče, kaj je je ljudi, kaj je ir vsega, vsak je jelj nekaj, Actually, one is inside the other say something about the two norms, saying that one norm is larger than the other. So the question is, if this set is contained in this circle, which is the larger? Yes. Well you have to think about this picture. Over the rhomb, you have a linear one homogeneous function, which is the norm. What is the norm? It's something that if you cut at different heights, you always find the same convex rescaled set. So you have rhomb, smaller rhomb, and then bigger rhomb. And then you are linear. So this is the graph of your L1 norm. Clear? Because the norm is one homogeneous and is identified by the unit ball. Once you know the unit ball, you know the norm by the homogeneity. So you have this convex set, then you rescale it, or enlarge in a linear way. So what you see is when you cut, you always see the same convex set, but with different scaling. Not only this, but the function is linear. Now you have another object, which is larger. So this means that the function is smaller. Because when you cut, you have a larger set. You have to think about two cones, one inside the other. You have one cone, another cone. You cut, well, this slice is larger than this slice, but this was higher function than this. This cone, of course, is a standard cone with sections which are circles. This is not so standard because the sections are squares rotated. Is it clear? This is just to have some intuition about the relation between the unit ball and the norm. Unfortunately, here we are in infinite dimension, so we cannot really do. Of course, we can suspect that L1 and L infinity will have some strange behavior. Just because these two norms in finite dimension are non-strictly convex, have flat portions in the unit ball, this could be a source of difficulties in the infinite dimensional case. And indeed, there are problems of reflexivity and so on for L infinity, L1, etc. But this we will see. We have just to prove that the parallelogram identity is not true. And this will be sufficient to show that this is not true, but because we have a characterization. Remember that the characterization was proven showing that if this is true, then there is this function a of x and y, which is a scalar product. That proof was not so easy. Remember? We have just to exhibit two elements, x and y in L1. Let me call them f equal the characteristic function of the interval 0, 1,5. So this is the function, which is, the graph of f is this, 1,5, 1. This is the graph of f, namely it is 1, so the characteristic function of a set is 1 if x belongs to the set and 0 if x does not belong to the set. This is called characteristic function. Maybe sometimes it's called indicator function. But these words, I mean, we have always to be careful because sometimes the indicator function is something which is plus infinity if x is 0. So now let us call this characteristic function. So we agree that with the word characteristic we mean a function with two values only, 1 and 0 in the set and outside. It is characteristic function, characteristic function of a. So now take also g equal the characteristic function of 1, 1. So let us compute now the norm of f, the norm square of f. Sorry? We want to prove that this is not true. Well, but we have to show just, if you put here the l1 norm, then you want to show that this is false. This is the standard form of the parallelogram identity. OK, so let us try to show this. So the l1 norm of f, well, it is, I mean, 1 over 4. Now, what is g? G is this, and it has the same norm. Of course, this is the l1 norm. This is the l1 norm. Well, the one norm is in R2, finite dimension. L1 norm is this. This is an infinite dimensional space. This is a norm on R2. I mean, of course, this is a sort of generalization of this, but there is a big difference between the two. The difference between the finite dimension and the infinite dimension. Now, what is the l1 norm square of this? Do you agree? It is 1, because the sum of f plus g is 1, and I have to integrate 1 between 0 and 1, which is 1. Now, which is the, OK, which is this, see, also 1, because when we take the absolute value, now, f minus g is something like that, but when we take the absolute value, we don't see this problem of sign, so it becomes 1. So, this implies that 1 4 plus 1 4, which is 1f, cannot be equal to 1. So, this is actually is equal to 1f, and this is equal 1f, 1 plus 1, which is 1. So, this is the first exercise. Maybe another exercise, sorry. Another exercise, well, I don't find it, but it's very simple. I mean, maybe this is home c0 continuous functions with the l2 norm. So, if I put the l2 norm on this space, this is not Hilbert, is not complete. Therefore, this is an example of space with an inner product, which is not complete, OK, in infinite dimensions. It's not difficult. You can find a sequence of... Well, the hint is that if you take a continuous function, say, which is minus 1, if you take this function here, where here, say, this is 1 over n, this is 1. So, you have a continuous, piecewise linear, continuous functions like this. This is an element here. And this converges in l2 to a function, which is discontinuous, so it is not in the space. Because it converges to function with jumps. So, it is outside the space. Well, this is a hint. Try to prove by yourself, OK. Another remark related to the discussion of yesterday is the following. So, yesterday we have constructed, say, so consider the ball of radius 0 in l2, small l2, and we have seen that we have constructed, I think, k, I don't remember the symbols, OK. Zero, if k is different from n, r over 2, if k is equal to n, I think we have constructed something like this. And what we have seen is that the distance between yn and ym was always equal to r over square root of 2. So, we have, say, an element here. Of course, this is equal to another element here and so on for infinite directions. So, first direction, second direction, third direction, et cetera, et cetera. There are infinitely many. But the distance between l2, between these two, so this is a sequence, this is another sequence. But the distance between these is always constant. This is what we, everything is inside a big ball. Everything happens inside a big ball. So, every big ball, one direction r over 2. Another axis r over 2. Another axis r over 2. But we have infinitely many axes. Everything is inside the ball. Now, the claim is that now consider ball centered at one of those, say b, centered at one of those, say, generic, of radius r over 2 square root of 2. So, now I would like to know, first of all, whether this is really contained in br. So, take a point, so let me call it z, a point into a generic one of those balls and then I want to measure the distance in the center. So, the center was x. Yesterday, maybe yesterday, sorry, yesterday, well, it is easier to do for x equal to the origin, but maybe yesterday was x. Well, okay, this was x, this was xk, and this was xk plus, so, just a translation of the previous one. xn, xn, this was yesterday, okay? Now, the distance from a point z, I want to measure the distance from z to the center, now the center is x. So, the distance, now we have the triangular property of the distance and this is the center, yn. So, this is less than or equal to the distance of z from yn plus the distance from yn to x, which is r over 2. Okay, and this is, by definition of the radius, is less than, so this is the distance, sorry, and maybe jumping some, less than z, yn, yn, x, so, which is less than or equal to the distance z, yn, plus r over 2, and then this is less than or equal to the distance of, from, between z and yn, is, by definition, less than, is less than this, which is less than r, okay? So, that we have really this picture, now the point is, is it true that these balls are disjoint one, each other? Okay, so the problem is, is it true that b, r, 2 square root of 2, yn, intersection b, r, 2 square root of 2, ym is empty, provided this, is it true? Now, so, let me, let me draw, assume we are in this condition, so take a point in, assume that we are in this condition, take a point in the intersection, okay, let me call it, let me call it z, z. So, b, r, square root of 2, yn, intersection b, r, 2 square root of 2, yn. Let me compute the distance between yn and yn. So, I have a point z, here. Now I have ym and yn. So, now what can I do? Again the triangular property of the distance. So, the distance between this and this is less than the distance between this and this. So, the distance between yn, z, plus the distance between, okay. And then, we know that this is strictly less. So, r over 2, square root of 2, plus r over 2, square root of 2, okay. And this is r over square root of 2, which contradicts, since there is a strict inequality, the balls are open, this is a strict inequality, it is a contradiction with this, okay. So, at the end, the problem that we find here, which is really a basic problem, is that inside a fixed ball centered at the point, we can choose a countable, countable many disjoint balls of the same radius inside. That is the problem. So, geometrically we have, now I cannot draw, but inside the ball of fixed radius there are countably many disjoint balls of the same, of a constant, positive radius. Is it clear this statement? So, br x contains a countable set of of pairwise disjoint balls of radius root of 2, okay. And we have already observed that the centers of this ball are a sequence, which has no converging subsequences. So, you have this many balls, the centers, and this has no converging subsequences. And this is the problem of the strong topology in L2. But this remark is curious, because now I am saying something which is not related to the, strictly related to the to the program, but it is remark that maybe can be of interest, is the following. Now assume that you have a sort of a Lebesgue measure on small L2. I don't say what is a Lebesgue measure now, but assume that I have a set function positive, no negative set function which measures at least the Borel set. So, take a set function, which measures the Borel set of L2. Assume that reasonably that mu of a ball of any ball is finite. So, this measure like the Lebesgue measure if you have a bounded set this is a finite volume reasonable. So, first of all it gives a volume to any Borel set, no negative volume. It gives a finite volume to a bounded set, this is reasonable. What is also reasonable is that it is countably additive. What does it mean? If I take a countable union of these joint Borel sets I measure the volume the volume is the sum is the infinite sum of the volumes of all sets if they are disjoint. Well, and assume also that mu is translation invariant. So, mu translation invariant means that if a volume of a set you translate by a vector then the volume of a translated set is the same as the previous one. So, for any Borel set A, for any vector X, assume this now, so these are properties that are all satisfied by the Lebesgue measure at least on the Borel sets in a random infinite dimension. I don't require to have such those properties on all sets but if I require just to have only on Borel sets, this is possible infinite dimension. Do you know this? Ok. Now the point is that we want to give some structure on a 2. We are trying to understand the geometry of the bowl difficult, the geometry of this infinite dimension space is difficult Is it possible to make integration? One of the structures most useful in analysis is to make construct measures so that you can integrate you can make probability you can do whatever. And now this implies that mu is identically zero. Can you see this? Can you see this? Of course, if this is true then there is an interesting point here. There cannot be a measure in the sense that we are usually use it integration we integrate, we split the integral all properties of the measure that we use comes from but if this is true then we cannot so the point is if you want to make integration in L2 and so essentially stochastic analysis probability and so on we have to weaken some we have to decide which of these assumptions is too strong. So this is maybe not related to our discussion in function and discourse but maybe something of interest for having a more global picture in infinite dimension so why this is true? It's very easy because you take the measure of the bowl the measure of the bowl is finite the big bowl br but the bowl contains a countable number of this joint bowl of the same radine now the measure is increasing I mean I have not written here but the measure is increasing if the sets increase the volume increases so now the measure of the bowl the measure of the bowl is certainly larger than the measure of the union of these small bowls this joint so bigger or equal than the measure of all b r over 2 square root of 2 yn is it okay for you? No problems? So I have the big bowl the measure of the big bowl the big bowl contains a countable bowl number of bowls of these joints so the measure of the big bowl is surely larger than the measure of the union of the bowls because they are contained but then they are this joint also and therefore since now they are this joint it is a property that the measure of the union is the sum of the measures okay br 2 square root of 2 yn clear now these bowls are one translated of the other they are constructed by translation as you can see therefore by this property they have always the same mu because they are one translated before this number is a constant well is a constant but you see it diverges unless this constant is 0 so if you want to give a finite so either the measure of all these small bowls is 0 and therefore the measure is 0 or the measure of the big bowl is plus infinity which is a contradiction of this now I mean this is a rough statement it says that I am assigning 0 to the measure of any bowl essentially because the claim is the hypothesis that the measure of a bowl is finite but then if it is positive say also if it is positive and finite then necessary actually is 0 so we have this assumption then okay any bowl so change the thesis any bowl has measure 0 okay which is say something which is strange from the point of view of the volume and so now you have to decide what you want to do which if you want to construct so from this observation there is starting point of the famous Gaussian measure so actually it is possible to construct a measure in infinite dimension leading to a lot of properties in probability and so on and what do you what do you weaken is this assumption here so people prefer to renounce not to assume that you have translation in variance okay so I don't want to say anything more than this just to let you know that one cancels essentially this assumption once you don't have this assumption then you cannot you cannot say that this is a constant if it is not a constant then you can make it converging okay now so now we have understood that we have two small number two small number of compact sets with respect to the strong topology now I will use a new word strong topology is the topology induced by the distance so from now on the strong topology is the topology is induced by D on L2 the natural one but then this is too strong because there are the number of compact sets is too small because any compact set has an interior so now we weaken a little bit so definition let xn be a sequence of L2 sequence in L2 we say that and let x in the point of L2 we say that xn that xn converges converges weekly to x and we write in the books sometimes one finds the following way of writing just only half a row just only half or maybe this is another notation the symbol W here stands for weekly on the other hand the symbol stands for strong this is in the sense of the distance strong and weak just notation we say that xn converges weekly to x if xn against y a scalar product with y converges as a number as a real number to xy for any y in L2 so this is another notion of convergence ok remark if xn converges weekly to x then it converges for the product topology for any k and these actually are numbers so pay attention to the notation this xn is a sequence so you may use this symbol this is a sequence of sequences any xn is a sequence is a vector with infinite many entries ok so pay attention to this because with the small symbol there is something complicated behind ok by the way is this clear why is this clear I don't want to talk about orthonormal basis because the existence orthonormal basis in nontrivial in infinite dimension is true in a smaller 2 but when you use we have to be careful the word basis in infinite dimension is a complicated object there are Hilbert basis Schauder basis Hamel basis etc etc so it is better for the moment skip the point of the basis as well as skip the definition of dimension for the moment but I mean it is just enough to take this element which is by the way an element over 2 0 0 0 1 at the height position 0 0 0 0 0 0 1 0 0 0 and then I test xn against this EI and this gives me exactly the k component is it ok as a y you take each ek so indeed take y equal ek equal 0 0 1 0 this is k position with this choice of test y vector y we obtain this remark maybe another remark that we can do before so exercise if xn converges strongly to x then xn where this converges not only this but also it converges weekly and xn converges weekly to x can we prove this how should you prove for instance that if I have strong convergence I have convergence of the norms strong convergence is convergence between strong convergence is the convergence distance of the space this is strong convergence ok because we have this inequality remember we have xn minus x because the distance is is continuous so remember the distance within two element in a Hilbert space is the norm of the difference we are in a vector space so that it is true and then so if the right hand side converges to zero the left hand side converges to zero is it clear ok fine so this is immediate now how can I prove this that actually strong convergence implies weak convergence well so I have to show what do we have to show this so what do I do I take the difference between this and this ok and so I take xn minus y minus x y which by the linearity of the scalar product with respect to the first variable is just this do you agree now I have proven the schwarzhalder inequality schwarz sorry the schwarz inequality no kosci schwarz inequality and therefore we have xn minus x product y now this y is an element of the space therefore is finite norm so this is a finite number and this is going to zero product is going to zero is it ok so at least we have a convergence say which is weaker than the strong convergence now we would like to know as we will see we cannot hope that so in general xn converges weakly to x does not imply so weak does not imply this and this we will see because we will prove actually that the sequence of centers of the balls that we have considered before we know it has not converging has not strongly converging subsequence right but it will have weakly converging subsequence however if we reach the assumptions so this we will prove it is not in general too so weak convergence this is a big problem I mean in functional analysis it is always very useful if you will continue in the study of PDE and functional analysis you will several times have the problem that you would like to have a sequence of solutions of something sequence of something strongly convergence and you always have weakly convergence so how to improve the convergence from weak to strong this is really common problem people working in functional analysis because once you have strong convergence you are able to pass to the limit in several expressions but if you have just only weak convergence you cannot really pass to the limit so remember that weak convergence very often is not enough in these arguments however if we know something more so weak and also something intermediate can we conclude that xn converges strongly this y in ilber space in l2 how we conclude this can we conclude this what do we do so let us take the difference we have to show that this converges to 0 we have to show this that this converges to 0 however this is a scalar product so we can split xn2 plus x2 minus 2 xnx xnx so you see the exercise is organized exactly so that this will converges to 0 because this by assumptions converges to x2 this is x2 then this is true for any y in l2 this is true for any y in particular for y equal to x x is an element of the space therefore this converges to xx so this is equal to x2 and this converges to xx which is twice the square norm and therefore this is equal to 0 so remember at the end conclusion strong implies weak weak does not imply strong strong implies convergence of the norms and convergence of the norms plus weak implies strong this is for the moment the situation actually this situation is very general it is not only true for l2 if you look at the Brezis book of functional analysis you will find much more general statements on this so if everything is clear we can continue so well what it is not easy maybe is to imagine what is weak convergence can you somebody have some geometric hints on this no strong is easy convergence in the distance geometric space I think that everybody know of us can understand what is strong convergence even in infinite dimensions the distance between the two points goes to 0 the distance has the triangular property the ball is convex not so bad but what does it mean geometrical is not easy for me at least for me how can you imagine this well it is more easy to imagine but this convergence of coordinates what does it mean it means you have a sequence you project on a fixed coordinate and then the projections converges xk xk and convergence but this is true for any k for all projections on the axis if there are axis I mean but this is something more we are not only projecting on on the coordinates you are projecting even on on any not only on the coordinate axis but you are projecting on any direction of the infinite dimensional space which is not so easy at least if you have some intuition more than this we should not lack such a convergence well one should show first of all now I don't know if we will show maybe yes well it's rather easy well once you have a notion of convergence it would be very nice to have the uniqueness of the limit usually it is possible to show we will show maybe that if the sequence weakly converges to x then x is unique there is only one weak limit not two or three so uniqueness of weak limit is something that one should prove first remark second remark is that well do you know in general in topology what is related to uniqueness of the limit in general if you have a topology say then you take sequence converging in that topology when you can ensure that when the topology is housed so when two points have this joint neighborhoods in this case you are sure that that limit is unique and this is true also for the weak topology so it is possible to show that this concept actually comes from a topology the neighborhoods are not so easy but the topology is housed of so the topology inducing weak convergence is housed of topology and you will find these statements in the breziz book for instance if you want so that you can ensure uniqueness of the weak limit the convergence is very weak but at least still we have uniqueness of course a two weak convergence could have more than one limit but this is weak but not too weak so that still fortunately there is just one weak limit this is the then another remark maybe that is useful so consider given y in L2 we can consider the following the following linear the following linear map following linear functional so given y you fix a projection direction see y and then you consider the following map the map that at any x into L2 it associates this number scalar product between x and y what is weak convergence weak convergence makes this map continuous by definition namely if xn weakly converges to x then lyxn converges to lyx by definition is this clear so we have with the weak convergence topology we are making continuous and infinite the remaining number of linear maps a joint is a word that for the moment is mysterious why you say a joint there is no duality I have never I have never I have carefully avoided the word dual for the moment I will use it of course my effort is to make things as simple as possible avoiding then of course at some moment we will have the dual I am just only saying that this convergence implies that the family of ly is a family of continuous linear continuous is a family of linear continuous maps maps so once I know that I am making continuous I enforce continuity by hands I make continuous by hands all these linear maps for any why a lot of maps and indeed actually the weak topology is exactly the topology, the natural topology which makes continuous that family so you can reverse the argument and define weak topology as the maybe the how do you call the coarsest topology which makes these families continuous people working in topology knows this kind of topologies very well so I have chosen the opposite root I mean I have defined weak convergence I am observing that with this convergence I am making continuous all these kind of linear maps this is immediate from the definition clear usually in more advanced books you find what is the weak topology is the coarsest topology which makes this family of linear map continuous and then at the end is this it's nothing else than this so this is the remark and then finally sorry for all this long introduction maybe useful for understanding so as you see all difficulties starts from the example of the bolts and therefore now we observe that this is the good, the right notion of convergence because it gives us compactness that we don't have for strong topology therefore now you can expect the following theorem so let xn be be bounded then it admits xn admits a weakly converging subsequence a weakly converging subsequence so to an element converging to an element 2x 2 now why this is a fundamental result because we know that what does it mean xn to be bounded bounded means bounded in the strong topology the topology of the distance it means that xn belongs to the ball of some radius r centered at some point z this means bounded object content in some ball but we know that this is not strongly sequentially compact relatively compact so it does not have strongly converging subsequence but it has a weakly converging subsequence so concretely concretely you always find the following assertion take a family of functions which is bounded in the strong norm then it admits a weakly converging subsequence not strongly but a weakly so you have some limit is it okay so let me remember once more recall once more I have a bounded sequence this in general strongly converging subsequence theorem of yesterday example of the center of the balls but it has always a weakly converging subsequence you will find such a kind so you will have this is a very general result in any Hilbert space well in a 2 for the moment for us now the proof what is our assumption is that there exists k positive such that this belongs to a big ball for any n so we have this bound this is our starting the only starting point of the proof what we know is just only this boundedness in some norm in this norm from this bound we have to be able to extract something weakly converging to something and now we start again what at least we can infer from this we can infer something on the components because this is the sum k to 1 infinity square root maybe it would be better to use k square x n k square so it is immediate therefore that for any k this number is less than or equal oh sorry k there are too many k's so let me use another ok too many k's this is capital K ok, so now for any i, for any component this is k fixing i we have that this is true for any n and therefore consider i equal to 1 so we use the standard as you will recognize in a moment the standard diagonal argument so so consider xn1 this sequence so this is a sequence of real numbers in a bounded set so it is a converging subsequence that I call n1 a sequence of indices this sequence of indices is such that like yesterday I hope the notation this converges to x1 with x1 less than k along that subsequence ok now I am considering the sequence x2 and therefore I have a new subsequence n2 contained in n1 subsequence of n1 such that x2 converges to sum x2 with x2 less than or equal to k now this can be done for any k so that my notation is xi n k k converges to xi for any i, for any k k for any i less than k so I do this k number of times finite number of times I extract k subsequence one from the other and so for all components less than or equal to k, I have convergence along that common subsequence using a diagonal argument I can do this countable number of times so that I have there exists a sequence of indices of indices that now for simplicity I still denote by n so I extract the countable number of subsequences sequence of indices still denoted by we already made yesterday this kind of argument such that xi n converges to xi for any i in n and for any i xi is less than or equal to capital k do you agree? by the diagonal argument argument remark we have by assumption that the sum of x i n from i1 to infinity square is less than or equal to k square this is our fundamental assumption in particular what is finite sum for any l before passing to the limit here I can erase the n because of a finite sum and so this implies passing to the limit as n along the sequence n that I have extracted along this new sequence I can pass to the limit in this finite sum as n goes to infinity and therefore I can put xi here so the sum for 1 to l xi square is less than or equal to mk this is true for any l mk is independent of l and therefore I find also that the sum from i to 1 to infinity xi square is more than or equal to mk namely sorry k square, thank you sorry and therefore x has finite norm so so as yesterday I mean we find convergence for the product topology our bound on uniform bound on the norms of xn gives that x also is bounded in the norm and therefore the element is at least in l2 and what remains to show is that actually so we have a candidate but the convergence for the moment is too weak because it's just convergence of the components now we have to project on all transverse non coordinate directions y to have weak convergence so it remains to show it remains to show xn converges that is xn minus x y converges to 0 that is for any y in l2 converges to 0 this we have to show now for any z in l2 let me and then and for any m in n I introduce a symbol which is the tail the tail of z in the sense that tau, the tail of z is equal to 0 0 0 and then zm plus 1 zm plus 2 et cetera so I cut the first important part of z and I keep from m plus 1 I think my notation is this ok so now let us write xn minus xy which is the product sorry let me use the symbol i from 1 to infinity fix m from 1 to m so fix m and fix y and so now I have xn i minus xi yi plus sum from i from m plus 1 infinity maybe this is already simpler writing tau mx n minus tau mx y scalar product ok do you agree because this is 0 before so I have already written here all the first m terms of the sum the next m terms the terms after m are these so it is this because this is 0 before before m ok so now I can evaluate the absolute value and I have xn minus x absolute value less than or equal now I take well let us start for the moment writing just this in absolute value now I want to control details for the moment so why now concerning details what can I do is that I can write schwarzselder and do this so remember now why is fixed why does not move why is fixed and then I have this by schwarz schwarz well I can but maybe it is not I could write nothing changes but also with this y is ok because I am multiplying by 0 so I can use tau m of y I can use y is whatever you want if you want I can put tau m of y so that this becomes tau m of y ok now I have to say something about this saying that this is maybe finite or uniformly bounded and then something on this so what can I say why is fixed so why is fixed and therefore since why is fixed I can choose an index let me call it I don't know index no I don't so fix epsilon positive take m in n so that this is very small what does it mean very small it means that no before doing this maybe I bound again this with the sum of the of the norms so please and now I do this plus this is less than or equal than this tau m of xn plus tau m of x and then still multiply by the same object tau m of y ok I do this so less than or equal same sum plus and then I put the plus here ok plus of the norms now what can I say about tau m of x well this is sure less than k because x itself is less than k so this first now remember here it is written that the square norm of x is less than k square ok so this is less than k right but this is also less than k because we have the uniform bound ok this is less than k and this is less than k so I can continue this inequality with so I have fixing y and then also I have this xn minus xy less than or equal than the sum up to m of what of x, y and y, y norm then we control in the moment but this is equal to 2k the tail of y but y is fixed so I can choose now ok so I have this ok now I fix epsilon and take m fix epsilon take m so that essentially then epsilon over 2 ok so that say 2 so epsilon divided by 2k sorry 2k the sum from m plus 1 to infinity y k square 1 half is less than epsilon over 2 epsilon just for simplicity I can do this because as I said k is the big constant uniform bound that we have at the beginning ok then y is fixed is not moving, it does not depend on anything that's the most important part so this says that I can take for given y fixing epsilon, fixing y we can take large enough so that this is less than epsilon ok we take that m epsilon over 2 ok so this part of the tails are under control therefore that becomes less than or equal than y square sum up to m x i n minus x i square once more schwarzselder in koszischwarz plus epsilon now still n is free so now I can choose n large enough now this is a finite sum so I can choose n bar now we remember that x i n is converges to x i for any i and so now I adjust the finite sum epsilon is fixed m is fixed y is fixed now I can choose capital N choose capital N so choose n bar such that x i and this sum to the half x i n minus x i square is less than or equal than epsilon for any n bigger than capital N bar and therefore at the end what do we find we find that this that for any epsilon there is a capital N bar such that this is less than 2 epsilon is it clear sorry for all this for all n epsilon delta et cetera et cetera so let me repeat give an epsilon positive I found n bar such that this is less than 2 epsilon epsilon plus epsilon for any n larger than n bar and this concludes the proof of weak convergence of a subsequence obtained by a diagonal argument