 So today we start with the Banach spaces. So we know already what is a Banach space. Is a normal vector space. So B is a Banach, B is normed vector space, complete. See, so this is a generalization of Hilbert. So clearly any Hilbert space, the converse is not true. We have an infinite dimensional exam Remember, we have shown that L1 is not Hilbert. We have shown this using the fact that the parallelogram identity was characterizing the scalar product. So this has a norm which does not come from a scalar product. We used, remember, two functions with disjoint supports. The characteristic function. So, but is Banach. Actually, the class of Banach space is very large, which is infinite dimension. Infinite dimension, what do you have to think? For instance, in R2 the analogy is to have on R2 an ellipse. OK, this is a scalar product. But if on R2 you have any other convex body, something like this, or this, or any convex body, which is not an ellipse, not a polynomial of degree 2, say, or whatever you can imagine. These are norms. These are the unit balls of the norm. But they are not an ellipse. And so they do not come from a scalar product. Concern in differential geometry, the difference between Banach and Hilbert is the same that we have from a Thinsler manifold and a Riemannian manifold. Riemannian manifold has a scalar product on the tangent space, which changes with the tangent space. A Thinsler manifold on each tangent space has a convex body, changing from the point. Here we don't do geometry, so we have just only one body. So examples, very, very, very useful examples. The examples that we have, well, LP for P larger or equal than 1. So these are spaces of sequences. Simply generalizing the L2. So you can imagine, which is the norm. So is the sum. This is the norm on a smaller P. So these consist of sequences. Any point here is a sequence, x, such that this is finite. For any P. Of course, we also have small L infinity, which consists of sequences, which are bounded. So we have, is the set of real sequences, such that the L infinity norm, which is defined as simply the supremum of the absolute value, is finite. Is finite. All these, one can show that are complete. So one can prove that these are, of course, probably you are more used to define capital LP and capital L infinity. But these are also very useful spaces. In view also of our experience of small L2, we have understood a lot of things of infinite dimensional functional analysis looking only at small L2. So and small L2 is important because it is a standard model of separable in the best space. And for the same reason, somehow these are also quite important. Other examples. So if k is compact, say compact, then the continuous functions on k with the L infinity, with the sup norm. So f on k to r, continuous. And the norm is the supremum, which is a maximum, actually, of f, supremum of f of x, such that, which is a maximum. This is another example of Banach space. OK. Other examples maybe is, another example maybe is this, is continuous functions on small c0. Small c0 is the sequences such that, so infinitesimal sequences, which is maybe also this. So this small 0 here, so in general, in general, say, if you have a set, say topological set, let me call it capital X, then you can define this as the set of all continuous functions going to 0 at infinity, which means that maybe something like continuous, such that maybe for any epsilon positive there exist k compact, such that f of x is less than epsilon for any x, x minus k, something like this. So all continuous functions, such that they are very small out of compact sets. Maybe it's something like to say that they are 0 at infinity. Whatever we can mean by infinity here. But it doesn't matter. This is not very important. I mean, don't be, so effects of topological space, this is not very important. I mean, just this. So this is all continuous functions from n to r. So any sequence. However, with the property that it is infinitesimal. So if you don't like this. Other examples, so c0, l infinity with this notation. So with this notation, continuous functions from n to r. So continuous functions from n to r are just sequences, just a sequence, space of sequence. Because on n we have the discrete topology, so everything is continuous. Any sequence is continuous. So this means, so that l infinity actually, if you want, is at the bounded continuous functions from n into r, if you like. This is just notation. But notation that, however, do not confuse infinitesimal sequences with another notation, which is this. This is a different object that today I will not discuss. What can you imagine? What is this? The set of sequences. What does it mean with compact support? It's finite. So definitely 0. So this is another object, such that xk equals 0, definitely. But you know what I mean by definitely. So after some index, all components are 0. So that maybe, such that xk different from 0, only for a finite number of indices, set of indices, set. So only for a finite set of indices k, xk is non-zero. And then it is 0. So these are the continuous functions with compact support that should not be confused with the continuous function infinitesimal on n. So they are different. Actually, one could show that this is the closure of this. It is clear. I mean, if you take all possible limits, say, of sequences with compact support, then you don't get a sequence with compact support, but at least you get something which goes to 0. This is, I mean, natural. But do not confuse this with this. And indeed, this is another symbol. So it is 2c00, different symbol. So I don't want to talk about this. So we will say something maybe about this, but not about it. So we have seen that L2, so the first theorem that we, one of the first remarks on L2 was that L2 was separable, separable. What does it mean? It means that it meets a countable dense subset. Actually, separability is, so I mean, and L2 is separable and is the model of any separable infinite dimensional Hilbert space. There is this, maybe, remark. So L infinity is not separable. So this is a Banach space, which is non separable. And so this is already one new phenomenon. So that makes things more difficult. Can you imagine why it is not? So the idea is the following. It's not difficult, the idea. I mean, the idea is, supposite is, I'm sure that you have seen the other statement, capital L infinity of omega is non separable, right? But for me, infinite dimensional vector space are so difficult that I prefer to start with spaces of sequences. Can you imagine why this is so? Let us try to understand why. So concerning L2, the idea was to take, essentially, sequence, which are 0, definitely, but with entries, which are rational. Can you remember this? And so using this, since the tails go to 0, it was very easy to show separability. Can you remember this argument? So the idea was, well, you have to control the tails. But the tails go to 0, because the series, the infinite sum is converging. So actually, I more or less forget about the tails. But then I have just a finite number of entries, and then I have rational numbers, which approximate all these finite entries. This was the idea of separability of L2. And this argument was working exactly because of the norm. In L2, it was crucial that the infinite sum was converging, so that the tails of the sum was going to 0. Now, however, the norm is something different. The L infinity norm is something much, much different. Do you remember, for instance, in R2, finite dimension of this L infinity norm? Just to have some intuition, we have already said this in two dimensions, which is the unit ball. Please, think of unit balls instead of algebraic formulas. What is the unit ball? Just the square, unit square. I mean, the square centered at the origin with sides 2, minus 1, 1, minus 1, OK? The only thing that we can say in two dimensions is just that, why this is bad? Well, it is not uniformly convex. It is not strictly convex. Not only that, but it is also no smooth. Because we can imagine no strictly convex smooth sets. We can easily imagine. But this is worse. It is not only, it is piecewise linear, but it is also, it is no smooth because it has convex. But this is the only thing that we can say. It is finite dimension, nothing more. But this is infinite dimension. And so the idea is, assume we have, so let me denote it by qn. This is a subset of l infinity. And assume it is countable. It is countable because n is in n, and is dense. Now the idea is that, by the way, I can find a point x at a given distance positive from any of this. So let me denote it by maybe xn, this. So this is equal to qnn. This is equal to 0, or maybe 2, or 0. If qnn is, say, less or equal than 1, and qnn if is larger than 1, for any n. And so now I have, what do I have? I have x, which is a point of l infinity. It is clear that this is in l infinity because the n's are just 2 and 0. So this is clear an element of l infinity. So assume that I have a countable dense subset qn. Any qn is a sequence. So the index of the sequence is lower, as usual. Or less is our convention. So this is our xn. Now I take the distance between x and any element in l infinity. What can I say about the distance? Well, I have to take the supremum between the difference of the components. Well, if this component, the n component is less than 1 in absolute value, then this is 2. If it is larger than 1, then this is 0. And therefore, this is always larger than or equal than 1 for any capital n. Xn equal to qn plus 1. Is it plus 1? Ah, it's the same. It doesn't mean the same. It's completely the same. This is just to have, this is even easier because this has just two values, 0 and 2. Just so it is in mid. So this, the point x in l infinity consists of either 0 or 2, 0, 2, 2, 0, 2, 0, 2 and so on. Maybe it's easier even, so just two values. So is this clear? And what does it mean, however? It means, well, this is accountable, cannot be dense because I have a point in the space with distance at least 1 from the set. And this contradicts the density. So this shows a remarkable property of l infinity. OK. Now, I read a bit more quickly. We can go through, we can go to the definition. So assume that e and f are Banach spaces. Then we can define this. It's the set of all l from e to f, l continuous, l linear continuous. Our notation yesterday was when a is Hilbert, we have considered, we have used the following notation. And f was r. Our notation was, I think that we agree maybe to use the star. This was the notation of yesterday. So this is a much more general situation for two reasons. One is that e is not only Hilbert, but is Banach. And the second is that the target is not the field of numbers, but is another Banach space. So it's much, much more general object. OK? Fine. So, well, maybe home. So definition, maybe? Definition. So we know what does it mean that it is continuous. OK. This is clear. Because these are normed space, so that there is topology, there is, we know. But definition, what does it mean that l from e to f linear, what does it mean that it is bounded? Well, bounded means, as yesterday, is bounded if it takes bounded sets in e into bounded sets in f. So let me write it in a little bit roughly like this, but it is clear. I take a bounded set in the source, and then must be bounded in the target. OK? Or equivalently, so if it takes a ball, so. Equivalently, there exists a constant c such that l of x less than or equal than c of x for any x in e, which is the difference with respect to yesterday. Well, the definition is exactly the same. The difference is that yesterday here we just have absolute value of l of x. Because yesterday l of x was a number. Today is. So if I don't now specify which kind of norm I'm using, it is clear from the context. This is the norm of e, and this is the norm of f. Even if I'm using the same symbol, sorry, but it is clear. Is it, do you agree? So we don't need to make heavy notation in this case. It is clear. So please check that this is the equivalent way to say that the map l takes bounded sets into bounded sets. And check that again, like yesterday, we have equivalence between continuity and boundedness. So l is bound l linear. Then l is bounded if and only if l is continuous. The same proof, exactly the same proof of yesterday works. So really you do the same proof of yesterday. There is nothing new here. Now let me, ah, yes, maybe, again, I can like exactly, like yesterday, I can give a norm to l. So check that what it is interesting now is to take the smallest possible c for which this holds for any x. So, again, this is, and again, it is like yesterday, the statement that we had yesterday, OK? So now, let us do some exercise, because in infinite dimension everything is not so intuitive. So I think it's good to have some intuition. So we have already observed that we know a lot about a linear map if we know something about the kernel. So maybe an exercise here. So let l into f linear, then l is continuous. So there is, now, so we can recognize, given a linear map, whether it is bounded, looking just only at the kernel. Now this is not, I mean, OK, let us try to, if you want to help me, I mean, that there is another which is immediate. Everybody knows. Well, this is immediate because 0 is closed. So, OK, so now let us try to show this. What can we do? So we can assume that l is not identically 0, because, so assume that l is not identically 0. Yes, so, and then, so, OK, so take x out of the kernel. So we can do this, because now, so now we are assuming that l is not identically 0, therefore this is not empty, and so we can take a point outside. So what he is saying now, since this is point outside, we know that it is non-zero, say, so there exists a ball, so now we know that this is open. So we can take a ball, there exists a ball, say, br of x, center the text of radius r, such that it does not intersect the kernel. So we can now take this, take this. So we can assume, therefore, that, say, either positive or negative, we can assume it is positive. And therefore we have that l of y is positive for any y in br. And now, what do we do? Yes, for the beginning, yes. There is a special case. Sorry, this is the assumptions, f is equal to r. So up to now is OK? OK, now, so l of y is bigger than 0 for any y. So if you, now I take a point z in the unit ball, and then I translate it for the, there, so I have that x plus r of z belongs to br of x. And therefore, I have that l of x plus r of z is bigger than 0. So again, I am using the structure of the vector space. So this is clear, right? And this is because of this. Now, l of z is what? Is l of x is l of x plus, is 1 over r, l of x plus rz minus, no, I prefer to change sign. Sorry, I prefer to add and subtract l of x. So l of x minus l of x plus rz divided by 1 over r. So it is 1 over r l of x minus l of x plus rz. Now, what do we know? Well, minus, there is a minus. You are right, there is a minus here. Thank you. There is a minus. It's OK. And now, what do I know is that this is positive, therefore this is negative. And so this is less than 1 over r l of x. And with the plus l of z is now equal to 1 over r l of, I think, is enough. Because now, I take also minus z. And therefore, I also have that l of z is less than or equal than 1 over r l of x. Because if this is true for z in b1, also for minus z. So once I have for z, I have this inequality, then I take minus z, which is still here. This does not change. And therefore, putting minus z here, I have this. But this and this gives me that the absolute value of l of z is less than or equal than 1 over r l of x. OK? So taking minus z gives this. Now, this together give me that the absolute value of l of z is less than or equal than 1 over r. And this is enough. This is enough to show that it is bounded. Because it takes the unit ball into a bounded set. And once you take it unit ball in a bounded set, so this means that l of b1, 0 is bounded. And this is enough to show boundedness. So this is the first remark on linear maps. Linear functional. Actually, I use the word functional because this is scalar valued. Otherwise, maybe I can use linear operator or linear map. Some other words. So the kernel is a subspace. It is very important to understand whether it is closed or not. In any case, it is important to understand what is a subspace. So let me try to make another exercise, which says the following. Assume that s is contained in e. And assume that s is a strict, is proper subset. Is not a wall of e. And assume that s is a subspace. OK? Is a subspace. Then the interior of s is empty. OK, so assume. So this says something. I mean, difficult to imagine. But you have a proper subspace. It can be infinite dimensional. Could have say co-dimension 1, in some sense. But still, it is empty interior. OK, so assume it is not. So assume that this is not true. And so there is a point, let me denote it by x not into s. And then there is a radius such that the ball, the open ball, the open ball centered at x not of radius r is all contained in s. The idea is then, the idea is quite simple now. I mean, you translate this ball at the origin. So you have a ball in your subspace. OK. Then you translate the ball at the origin, remaining inside your subspace. Then you have a ball centered at the origin in your subspace. But then any point of the space can be reached as a homotetized of a point of the ball. And then the wall space is equal to s. Because the homotetized point still belongs to s. Because s is a vector space, subspace. This is the idea. So first translate at the origin. So you have an open ball centered at the origin inside s. However, once you have this, any point of the ambient comes from some point in this ball. But then this is a vector space. So also that point must belong to s. And therefore, s is equal to e, which is a contradiction with this. So this is the idea that we write it. So br of 0 is equal to br of x not minus x not, which is an element of s. Because s is a vector space. So differences of elements of s still lie in x. So this is the set of all differences. This is a subset of s. So this is an element. Is this clear? Because s is a vector space. So actually, the assumptions here really do not need any completeness. This proof now works for any normed vector space. Just normed, it seems to me. Because s is a vector space. So completeness, I think that completeness here is not the point. So you have this. And now take any point x into e. And then you scale it back, dividing. So you just take x. So assume it is not 0. Then you just scale it to the boundary, say, of the unit bowl. I cannot use the word boundary here. But it is clear what is this. And then I multiply by r. So that I'm going inside the boundary of this. And then I multiply, say, by r over 2, just to be sure. So this. And then I multiply by lambda. So this is just inside r0. Because it has distance r over 2 from the origin. And lambda is just the inverse of this. No, lambda is a number. I mean, it's just given x, lambda is a number. So lambda is just the inverse of, OK. So I'm simply writing x as follows. What I'm doing is simply this. It's very clear. This is br0. I have a point x. And then what I do is this. I isolate this point at distance r over 2. And then I re-multiply by this. So this point is in the bowl. And each line, the line pass into the point. So this point is in the bowl. And therefore is in s. But this is a vector space. So all these lines belong to s. And therefore, x belong to s. So x belongs to s. And therefore, this is true for any x. And therefore, s is equal to e. So what I'm using here is just to have a distance. So the norm is just to have the vector space structure. I don't use any completeness in this kind of line. So you see, still difficult to imagine. There are infinite dimensional subspace. Doesn't matter. They at least have always empty interior. OK. So now theorem. So theorem. Now it is clear that lef. Now let us come back to lef. It is clear that this is a vector space. We can sum two linear maps. We can multiply by a number. And it is also clear that this is a norm. The supremum of this is a norm on lef. And so the theorem says that this is actually complete. Actually, this is complete. So is complete. And it is interesting to remark that so this is actually lebanac space. But it is interesting to remark that the proof that I will do now, the assumptions are just e normed. So you don't need to have that the space of all maps, linear continuous maps from e to f, to have completeness of this. You don't need that the source is complete, but just need that the target is complete. OK. If what I am saying is true, in particular, so in particular, there is something which is improving when I pass to the dual. I mean, assume that I have a normed space. Well, this is a normed space. I can define star, the dual space. Well, then star is banac. Even if e were not banac. So there is some improvement in doing this star operation. Is it clear? This is because this is and r is complete. So we improve in doing this star. So it is maybe doing star and star and star several times. Actually, two times is enough. Doing two times star sometimes. So now, so we have to take a Cauchy sequence of element. Let me denote it by ln Cauchy sequence. Cauchy sequence of in f. I have to show that this converges with respect to the norm we are considering. Converges to an element of lef. So what does it mean? It is Cauchy. So it means that ln minus lm. That's denor equal. So maybe in inequality that I've never remarked. But this is immediate, however. But it is important maybe for the first time to say, is that if you have l from e in f l, if you have an element of script l, then it is always true that l of x, this is immediate. Is it OK? Because this is just the supremum of all these quotients. So this is the supremum of all l of x divided by norm of x. So this is large. So this is something that I will use even without addressing it. Now, the ln is a Cauchy sequence. And we have, therefore, this inequality. For any nm, we have Cauchy sequence. Moreover, for any x in e, we have this inequality as a direct consequence of this. Therefore, this means that for any x, hence for any x in e, the sequence of element of f is Cauchy in f. Do you agree? Yes, it's clear, right? Simply because of the Cauchy assumption here and this inequality. So if this is small, definitely in nm, then for any x, fixed x, this is small, definitely in nm, which is exactly to say that for a given x, this is Cauchy. But now, as I said, I have that f is complete. So f in complete means that for any x, now f is complete by the completeness of f, so we can define the following. We can give the following the definition. We can define l of x. This is a definition for any x in e. We can define l of x as the limit. This defines a functional, a map. So I have a map x in l of x. Do you agree? This I can do. Simply because this is Cauchy and therefore it converges and therefore given x, I take the limit and I call this limit l of x. So of course, what do I have to show? I have to prove is linear. L is continuous. And also that ln converges to l. And also this. Is it OK? Do you agree? So I have defined a map, a sort of abstract map, because it's a limit. But still it is a map. And so I want to show that this is linearity. Well, linearity is almost immediate. Because ln of x, we know that for any n, for any x, for any y, we have this. So simply in passing to the limit in this relation implies that l is linear. I mean, you can multiply by a number alpha and still you can pass to the limit and so on. So it is clear that l is linear. And now, remember that we have, for any epsilon, that exist such that we have this, for any x. For any epsilon, for any x, we have this. Then we can pass to the limit here. I want to fix one of these indices. So remember that this is also larger or equal than, so remember, this inequality may be a minus b. We have this inequality, so I can use it here to get this, which says that, so this is actually even larger than without the absolute value. So ln x minus lm x, so without the absolute value. And so I have now put in together all these inequalities. I have ln of x less than or equal lm of x plus epsilon x. This is true. For any epsilon, there exist n bar, such that this. So in particular, I can fix one of the indices. So for any epsilon exist n bar, such that, for any n bigger requirement bar, for any x, we have this. Passing to the limit as n goes to infinity, gives me lx less than or equal, then ln bar x plus xx, which gives me the boundedness of l. OK, so we have that l is bounded. We have it is linear. Therefore, we have shown at least, so what remains to show is now that, so what remains to show is now that, ln converges to l in norm. So remember that we have, I have already erased that, for any epsilon, for any x, there exist n bar, such that for any n bigger requirement bar, for any x, we have ln of x minus lm of x less than or equal than epsilon of x, we have this, we have this. Now, before we have fixed just one index, now it is not sufficient, we have to pass to the limit as n bar, now it is not sufficient, we have to pass to the limit in this as m, for instance, or n, one of the two goes to infinity. So sending m to infinity, we obtain that for any epsilon, there exist n, such that for any n bigger or equal than n bar, x ln of x minus l of x is less than or equal than epsilon, this, and this is exactly to say, therefore the supremum of x, the say different from 0, ln of x minus l of x divided by x, so passing to the supremum with respect to x in this inequality is less than or equal than epsilon, and this is by definition, is the difference of the norm. So first pass to the limit in one index, use the definition of l, and then take the supremum with respect to x, and then use the definition of norm. So we have shown that for any epsilon positive, there exist n bar into n, such that for any n bigger than n bar, then this is less than epsilon, which is like to say that ln minus l goes to 0 in norm, do you agree? OK, so we have that this is a Banach space, so from now on we will always know that the dual of a Hilbert space is actually complete, is Hilbert, the dual of a Banach space is a Banach space, and more generally the set of all linear bounded maps from one norm into a Banach is Banach, a proposition. So now the point is that we would like to understand more or less what is the dual of some space of sequences. So let us now start from C0, so let be an element of C0 star. There exist x in l1, such that l of x, l of y. By the way, sorry, notation, sometimes l of x in the books is denoted by lx sometimes. Just a notation, notation. x equal xk, let me denote it by x, no, let me denote it by l, sorry, I change notation, it is better. So that l of x is equal to, now I have this lk, and then this xk from k1 infinity. Moreover, sorry for denotation, this l, so it is clear, this is a space of sequence, this is an element of this small l, say small l, small l, this is small l, this is l1, I hope the denotation is clear. So this is a sort, remember, this reminds a little bit the Ritz theorem. The Ritz theorem was saying, then if I have a linear continuous map on l2, then I can find h0 in l2, so that this was true and that was an isometry. So what remains here? Well, you see, this is c0 star, this is l1, this is Banach is not Hilbert, again this is Banach is not Hilbert. The duality between l of x is still represented by this, and again I have that the norm is preserved. So if I associate to l, this small l, I know that the norm is preserved. This existence and this is also, what we will see next is what happens now if I start from l1 star, I will get an element in l infinity. And so at the end, we will conclude something about the star of the star. But anyway, so let us try to prove this assertion here. So two assertions maybe, so there is another assertion, l1 star here, then there exist this here, then this is in l1, and this is l1 star, and this will be l infinity. So we will try to prove these two assertions. So let us for the moment start with the white one. So now concentrate on white. Ok, so we use the usual notation EI equals 0, 0, 1, 0, et cetera. This is in the height position. Ok, this is clearly an element of C0. Ok, and also of course the supremum is equal to 1. Ok, so now take any point x in C0 and we use also this notation. So we use xm, take a point x, take an index n, and then we take the first m elements, so the first relevant part, 0, 0, 0, et cetera. This is a notation, ok? Just a notation for any x notation. Ok, so what do we know? Xm is just also, I can write it also as the sum from 1 to m, xk, xk, from i, k, xk, k. So now is it true or not, the following crucial point is this, is this true or not as m goes to infinity? Why is this true? First part is 0. The element x is in C0. So this means that it is an infinitesimal sequence. So if I remove from the sequence the first element, then it remains the tails and the tails as a supremum which goes to 0. Therefore this as m goes to infinity goes to 0 because x is in C0. So this is very important because it says that we now have an element which is continuous on C0, right? L is continuous. And this says that xm is an approximation of x. Therefore we expect L of xm to be an approximation of L of x because of this property. So L of xm, well, xm, this is an element, of course it is, this is an element, so this is just bilinearity of some numbers, L of xk, e k. Maybe this, just a word here. This is maybe another way for who of them that know what is a shoulder basis, this is more or less says that this is a shoulder basis for C0. That doesn't matter, doesn't matter. OK, so we have this. So let me denote it by what I am looking for in the sequence. So this is, for me, is Lk by definition. OK, so this is k from 1 to m, small l. Lk, now L and E are very similar. OK, Lk, e k. Now, as I said, this is going to 0. L is continuous. Therefore, this converges to L of x, because L is continuous. So L of x is the limit of some, which we, so this is a limit, this is real, these are real numbers. OK, just the usual limit is real numbers. So this is xk, e k. There is something wrong here. Xm is that Lxk, which I didn't know. E k, yes. No, no, there is something wrong here. This is OK, this is OK, OK. I have simply reversed the role. Sorry, this is a number, this is a vector. That simple problem, sorry. As usual, when you do linear algebra, so this is Lk, this is, sorry, this is xk. Sorry, these are real numbers, real numbers. Real numbers and real numbers, not elements of C0, so xk. And then, sorry, this is xk. This is just convergence in real numbers, OK? Sorry, so this is the standard image through a linear map of an element of the basis. So let me call it Lk. The notation Lk is OK, but there was a mistake here. OK, so L of x exist, and this is already something in the statement of the theorem, right? Because the first statement of the theorem was saying, let L be a linear functional, then there exists Lk. So what do I have to prove now? Problem. Well, I have to show that L, so now let me indicate by L this, I have to show that this is in L1. Not only this, then once it is in L1, so problem this and this also. This is what remains to show. But at least we have this representation, realization, representation of L. So this is true for any x. OK, now the idea is to choose special x in order to make the series of the abstracts. So the point here is to show that this is finite. I have to show this, OK? And I have to show this, and I will choose special element x, that I call now y, yn in C0. OK, now yn in C0 is cooked exactly in order to make this product the absolute value of Lk. How can I do this? Well, this, OK. So yn, ynk is equal to the sign of Lk if 1 less than or equal to n0, else, OK? Well, so this is plus minus 1 or 0. Therefore, this is converging to 0, is infinitesimal, OK? And therefore, this is an element of C0. So I can put it here, and it gives me, so it is allowed to put a yn here. It is allowed. And so L of yn is exactly what I want, because it is the sum k1 to n. Now here I have the sign plus, if I have the sign plus Lk, positive Lk, if the sign of Lk is positive, it is negative, and so at the end I have just the absolute value, but summed only up to capital N, OK? So this gives me the absolute value of Lk. Not only this, but also maybe I should observe that yn in absolute value, no, not in absolute value, sorry, in C0 is equal to 1. These are unit vectors. And this is the sum, OK? OK, now L of yn is this. So L of yn is this, which is less than or equal than the normal by the usual inequality that we discussed before. This is always less than or equal than this, OK? But this is 1. Therefore, this is equal to L. And this shows that for any capital N, this is less than a number. Therefore, this infinite sum is converging, OK? So this implies that the sum from 1 to infinity of Lk is less than or equal than L. And in particular, this problem is OK, hence L is in L1. Not only is in L1, but also this is less than or equal than this. So, I leave you to continue the proof in the sense that you have to show now that you have to show this, OK? This we will do tomorrow in the next lecture. So please, for the next lecture, try to show this and try to show also the green part of the proposition. So, realization of the dual of small L1. And I can say that the proof is very, very similar. So it is doable. So, conclusion summarizing, you have to show this inequality at home and then to show, to prove the proposition concerning L1, small L1, the dual of small L1, OK? OK.