 OK. So today we start with the part of functional analysis. So we leave for some moment the PDE equations. And we start on apparently different subject, which is functional analysis. OK, so let us take vector space on R. So we confine ourselves considering always vector space on R and not on C. So the theory on C is sometimes more complicated. So we prefer to consider real vector space. So probably everybody knows what is a vector, a scalar product on V. Everybody knows what is a scalar product. So it is maybe it is not necessary that I write it. It is just linear. It is a function of two variables, which is linear. So linear separately in both in the linear in. So let me write it. So alpha x plus beta y comma z is equal to alpha xz plus beta yz for any x, y, z and for any alpha and beta in R. So we have linearity. We have a similarity linear in the other variable. So say x alpha beta plus alpha y plus beta z is equal to alpha xy plus beta xz. So we have linearity in both the two variables in the two entries. And we have also xx bigger or equal than 0. And then we have x0, xy say equal yx, x0 equal 0. And xx equals 0 implies x equals 0. So it is linear symmetric on the same vector x is non-negative. And it is non-degenerate, namely, xx equal to 0 if and only if x equal to 0. So this is a scalar product. It may happen that in general one should relax these assumptions. So for instance, this can be relaxed by assuming only this. So in principle, there could be vectors which are orthogonal, non-zero vectors which are orthogonal to themselves. This could happen. But for us, in this course, it will not happen. So this, in principle, can be relaxed. But for us, we keep this property. Also, it may happen that we have a scalar product which is sometimes negative. And also, this could happen, but we don't care. So this is our definition for the course. So we take this. OK, now given a scalar product, we can consider following definition, x equal maybe x square equal to xx. Now this is a norm. This is a norm. So what is a norm in general? So this is a definition. It's a scalar product. And then we have also a norm. So this is a norm. If, essentially, we have alpha x equal to alpha x for any alpha in r for any x in v. And then it is, so this is one homogeneous, one homogeneity. Then we have sublinearity, which means this for any x and y in v. And then we have, OK, from this it follows that the 0 vector has 0 norm. But say we add x equal to 0. So this is a norm. So this is a scalar product. This is a norm. So the exercise homework is if this is a scalar product, then that is a norm. So home, if this is a scalar product, scalar also called inner product. Sometimes it's called inner product. Inner same is the same. Inner product then is a norm. So this is a norm. So this is a very, very easy exercise. OK, so given a scalar product, we can always construct a norm. And the converse is not always true. So one of the questions is when. So we can go from this to this. And the question is when we can go from a norm to a scalar product, OK? So let me just make some remark, maybe some examples. Examples, well, in our n, all we know that x, y equal by this is a scalar product. Not only this, but if we have, say, a matrix, capital A, which is symmetric and positive definite, a matrix n times n matrix, symmetric and positive definite, then this is also a scalar product, OK? Yj, sorry, Yj. Associated to this, we have, by this exercise, this is the standard Euclidean scalar inner product. And then associated to this, we have the standard Euclidean norm. All of us know that this is the standard Euclidean norm. This is a slightly less standard. And it is just another norm, Riemannian norm. This is called the Riemannian norm in our n, where the metric, Riemannian metric, is identified by these numbers, OK? It is useful to remark that given a norm, we can consider the unit set of all points x in v, such that x in norm is less than or equal than 1. Essentially, once we know this, now this set has various properties. We will go to the property of this set. But essentially, a norm is identified by its unit ball. So if you know what is this, then you know the norm. And conversely, if you know the norm, you know what is this. We will see this. In this case, of course, the unit ball in our n is just a sphere. And in this case, it is an ellipsoid. It is just an ellipsoid. OK. Other examples, maybe? Yes, examples can be made on spaces of matrices. Say, if you have matrices, n times n matrices, m1, mn, then you can take the trace of mn, for instance. These are matrices, square matrices. So you can consider this object here as a scalar product between two matrices. Or maybe it is more correct to take the transposed matrix here, just a measurement. Now, a more interesting example for us is another scalar product on this space. What is this space? So it's the set of all x into r omega. Now, countable product. So the notation, by this, we mean countable product of vi, where vi is r. So it's a countable product of r. Maybe we can also use the notation r to the n. OK. So this is the set of all, if you want real sequences, this is a sequence, right? Because this is nothing else identifiable by all maps from n to r, right? Therefore, this is a sequence, real sequence. So this consists of all x, which is a real sequence, such that, however, this is finite. This is maybe the most important space that we will consider in these lectures. It's very complicated. It's very difficult to understand what is this. But there is a scalar product on it. So x and y, xk, yk, prove that it is a scalar product. An inner product is an inner product. Maybe, as homework, try to construct something which is not an inner product. Home, for instance, check that maybe this is not an inner product, for instance. Try to understand why this is not. OK. Maybe to understand why this is an inner product is also useful to make the following remark, which maybe, again, I can leave you as homework because it's very simple also, but very useful. So of course, this is the first example that we see of vector space with infinite dimension. So now, what is the dimension must be defined. But for the moment, it is rather clear that this is an infinite dimension, not vector space. And all the difficulties in functional analysis come exactly because of the infinite dimension. So maybe another homework. OK. Home x, y square less than. Now, I use the notation. Maybe, let me write it for n x, y, n, d. This is called the Cauchy-Schwarz inequality. I think this is useful so that if two elements are in this space, so this is finite, this infinite sum. So if this infinite sum is finite, if this infinite sum is finite, then also this infinite sum is finite. And we have that in the quantities. Well, maybe a hint, if you want. Consider the following polynomial, quadratic polynomial, that I denote by p of lambda. x plus lambda y, x plus lambda y, where lambda is real. So consider this quadratic polynomial, which is x, x plus, let me write this lambda square y, y plus 2 lambda x, y plus x, x. And then go on by yourself in the sense that, OK, this is a quadratic polynomial. It is non-negative because of the properties of our scalar products. Not only this, but since we have linearity and we have symmetry, this is also this. So linearity plus symmetry gives us this. And then, OK, this is non-negative. Therefore, we can say something about the determinant of this quadratic polynomial. And from this, something we have to deduce. You have to deduce this, OK? Or maybe also when there is equality, when there is equality, when there is equality here. Assume that, so of course there is equality if x and y, either x or y are 0. But assume for which x different from 0 and y different from 0, is there equality here? Yeah, they must be parallel. But check by yourself. They must be parallel, OK? Parallel in the sense that one is the multiple of the other. So let me go on quickly with this, which is very simple. So let me write down some, so keep in mind, so this space, space of sequences, which are integral such that this sum is finite. Of course, one could consider a more general space. But since now for the moment we have scalar products, let us confine ourselves. But this is really the generalization of our n. So this generalized generalizes our n, OK? So this is, I mean, when we work usually in analysis, we think about our n, right? In functional analysis, maybe it is better as a model in mind is to have this. So infinite dimension, but maybe the simplest object with infinite dimensions, somehow. Where we have a sort of Euclidean structure, Euclidean norm, somehow, like this, OK? If you consider this, if you consider this is not maybe the most natural object, at least generalizing this. Because for instance, this object here, inside this, there are several points with infinite this. If you want to put a distance here, and you want to consider this as a natural distance norm, then you immediately see that this has a lot of points with infinite distance one each other. So I mean, if you take just a sequence here, such that this is infinite, then this means that the distance of that sequence from the origin is infinite. So if you want to put on this a distance of this sort, then a lot of points have distance infinite one each other, which is maybe something that we want to avoid. So maybe it is better to have in mind this smaller space instead of this. Now, it is interesting to remark that even in infinite dimensions, we still have something. So if I have in the plane, if I have two points, x and y, and then I consider this and this. So I have a parallelogram. And then I have x plus y and x minus y. And there is a relation between x plus y square and x minus y square. So 1 half of this is equal to x square plus y square. So if I take the length of this square, and then I sum with the length of this square, I have that this is 1 half the length of this square plus the length of this square, parallelogram identity. Well, this is true also in infinite dimensions. It is true in any, if you have a scalar product, so if then this is still true. This identity holds. Well, it is immediate. So check this by yourself. It is immediate proof. So home. So in any vector space with an inner product, this identity holds. This is the homework. Is it clear? A more curious exercise. So home. Suppose that p from v in R is not exactly a norm, but satisfies the following. p of lambda, alpha, lambda, alpha. When I wrote the norm, I used alpha or the lambda, alpha. Alpha p of x, p of x plus y less than or equal to p of x plus p of y. Suppose just this, then. So from this, it is clear that p of 0 is equal to 0, because I just take alpha equal to 0 here. So this is 0, and this is a finite number, because we have real values. So since this is finite, and this is 0, and the product is 0, so this is clear. Less clear is that p of x is over less than or equal to 0 for any x. Try to see whether this is true or not. Can you prove this? So this is not exactly a norm. Maybe this is called just a semi-norm. This is just a semi-norm. Maybe this is a semi-norm. Which is the different with the norm. We don't know that if this is 0, then x is 0. This we don't know. This we cannot say. So there is the word semi, not exactly a norm. So we cannot deduce that, I say once again, we cannot deduce that if p of x is 0, then x is 0. This we cannot deduce. And indeed, analogously, there is the notion of semi-inner product, in which you cannot say that if x, x is 0, then x is 0. This you cannot say. So this you cannot say. So try to see whether it is true. Another homework, if is a norm, then has this following Lipschitz property. In particular, it is continuous. So we have seen, so remark, it is a scalar product. The parallelogram identity holds. Now we would like to see when a norm comes from a scalar product, and there is a theorem here, theorem. Let this be a norm such that the parallelogram identity holds. Define, so if we have the parallelogram identity, look at this. Define this object here, 1, 4, x plus y square minus x minus y square. So if the norm, what is this object is not so clear, right? However, if the norm comes from a scalar product, what is this? Is this a scalar product? Because x square cancels with this x square. y square cancels with this y square. 2xy sum with this 2xy, so we have 4xy. 4 divided by 4 gives you x, y, x comma y, right? If, is it clear? Yes? But the problem is that we don't know that this comes from a scalar product. So the theorem says that if the parallelogram identity holds, then this is a scalar product. And this is the answer to one of the previous questions when a norm comes from a scalar product. If and only if the parallelogram identity holds, OK? Now the proof of this theorem is not easy. We will try to do it now. But the important fact is that you have a general picture in mind. So we have scalar products, which gives us norms. And in general, a norm is not coming from a scalar product in view of this, essentially, because not all norms satisfies the parallelogram identity. We will see that not all norms satisfy the parallelogram identity. Are the questions on this? Is this picture clear? OK. So now we have to prove this theorem. So I erased one Lipschitz property of the norm. I have to erase this. So let me try to prove this. And this must be so proof. So what do we see from this definition, OK? So first of all, we see that a of x and 0 is 0. a of 0y is 0. OK, this we see from the definition. Then what else we see? We see that a of xy is also equal to ayx. So symmetry. Symmetry it is clear. a of xx also is bigger or equal than 0, because this is 0. Because of the property of a norm, OK? No, not the semi-norm norm, OK? So this we see. And also we see maybe that a minus xy is equal to minus axy. And also ax minus y equal minus axy, OK? These properties are clear from the definition. Now the more difficult point is linearity and homogeneity. So for linearity, so let us start with a of x1. So let me call this x1, yes, y, plus ax to y. So I start from this, and I would like to prove that this is equal to ax1 plus x to y. This I would like to prove. It's not so clear from the definition. It's not so clear because we cannot expand. I mean, for the moment, it does not come from a scalar product. So I cannot expand it as a scalar product with x plus y with itself. So what can I do? So OK, so we keep the definition. 1, 4 x1 plus y plus x2 plus y squared minus x1 minus y squared minus x2 minus y squared. And then we make the following position. So we apply the parallelogram identity with the choice x1 plus y and x2 plus y. So let me write here the parallelogram identity once more. So we have 1f x plus y squared plus x minus y squared equal to x squared plus y squared. So let me apply it with the choice x1 plus y in place of x and x2 plus y in place of y. So if this is x and this is y, then I have that x1 plus y squared plus x2 plus y squared is equal to 1f x1 plus x2 plus 2y squared plus x1 plus y minus x2 minus y squared. So now what I am trying to do is to compute this. And I see that from the parallelogram identity, this is equal to this. Now I want to compute separately this. And I apply, again, the parallelogram identity now with the choice x1 minus y and x2 minus y. And so the second addendum can be written as follows. So x1 minus y squared plus x2 minus y squared is actually equal to 1f. And then I have this plus this. So x1 plus x2 minus 2y squared plus. And then I have the difference. So this minus this. And therefore now I can substitute this with this and this with this. And I obtain that a of x1 y plus a of x2 y. Now I substitute and there is a cancellation here. So I have 1 4 multiplied by this 1f this x1 plus x2 plus 2y squared minus x1 plus x2 minus 2y squared. Sorry, the 1 half is also here. Thank you. Now what is this? Well, this is 1f a of what? x1 plus x2 2y. It is this. So let us keep in mind that for now if you agree with this, now I cancel everything. I just write the result. Is it OK? So let me now forget this computation. And let me just write what I found here. So this is just as a consequence of the hypothesis, parallelogram identity. I wait a little bit. Maybe I can already write here a of x1 plus x2. So now if I take x1 equal to x and x2 equal to 0, since a of 0y is 0, I obtain that a of xy is equal to 1 half a of x2y. And therefore, we don't have yet the homogeneity. But in the second variable, at least, we can put out outside and inside the two for the moment. OK. Now from this, it follows that a of x1 plus x2y. If I apply this with a choice of x like x1 plus x2, then I have 1 half a of x1 plus x2y. And so this is equal. And therefore, we see that there is linearity in the first variable, additivity in the first variable. Now by induction, since this is true at this level, then by induction for any x, since we have linearity in the additivity in the first variable, then we have that this is true for any m in n by induction. Not only this, but remember that I can change sign here so that this is true also for any m in z. OK, now let me go here. We have a of mx over 2y. Now is what? Here is a, now I use the symmetry, mx over 2. This is equal to m over 2 by this property, a of yx. And therefore, this is equal to m over 2a of xy. By induction, once more, from this we deduce that a of this is equal to this for any m integer, for any n natural. Now we also know that the set of the addit numbers is dense in r. So the set of all numbers of that form is dense in r. And we also know that a is continuous. y, a is continuous. y, a is continuous. Because norm is Lipschitz. And therefore, it's continuous. And this is the difference of two continuous functions. OK, a is continuous. That is dense. This implies passing to the limit. That we have one homogeneity in, so we have a of lambda xy alpha, sorry, equal to alpha a of xy for any alpha. I think that this is all, because then we have symmetry. We can exchange x and y. So this concludes. So you see, the proof is not trivial at all. Even if the tools that you can use are just two or three objects, two or three tools, it's only that. But if you mix them in a proper way, then you end up with this non-trivial result. So you need actually all the properties of the norm, continuity also. So this is a scalar product. It's linear, homogeneous. It's one homogeneous. Symmetric, positive, definite, non-negative. So maybe, OK, definition. Let the norm be. So we say that this uniformity convex for any x in v such that x is less than or equal than 1. For any y in v such that y is less than or equal than 1, for any epsilon positive, such that x minus y is bigger than 1 minus epsilon, there exists delta positive such that x plus y over 2 is less than or equal than 1 minus delta. This is uniformity convexity. So this says that if I have a point here and I have a point here, and the distance between x and y is not 0, so no, sorry, let me take this closer, x and y. And the distance between x and y is not 0. It's bigger than epsilon. So if the distance is bigger than epsilon, if this here and this difference is bigger than epsilon, then the sum divided by 2 is not on the boundaries. So if I take now the sum divided by 2, I lie strictly inside the unit ball. Even if I start from, say, one point in the boundary of the unit ball and another point in the boundary of the ball and I know the distance is a little bit positive, then when I take the difference and divide it by 2, sum and divide it by 2, then I do not fall on the boundary of the ball, but I am inside the ball of a small quantity delta depending on epsilon just. So what is an example for which this is not true? Assume that x is here and y is here. Then what I see? Assume that x minus y is not 0. I mean, it's just this epsilon. Then I take the sum divided by 2, and this is not less than or equal than 1 minus delta because it falls exactly on the boundary of the ball. So this is a way to say that somehow the ball is sort of, the ball is sort of more than convex. I don't want to say strictly convex because strictly convex could be a different concept. It's more or less like to say that it's strict convex, but not exactly. It is this. And of course, this definition, I don't differentiate. I don't say that the curvature is positive or something. I cannot say it because I don't know what is the curvature. I mean infinite dimension. I cannot say anything. This object is just a convex set. The unit ball of the normal just convex. Nothing else. Infinite dimensions. So the result is, theorem, if this comes from a scalar product, then this is uniformly convex. Therefore, this says something about the geometry of the space. So this informs us about the geometry of the unit ball. So in a Hilbert space, infinite dimension, the unit ball is always an ellipsoid, finite dimension. Do you agree? It's always an ellipsoid. And of course, the ellipsoid is uniformly convex. It's smooth. It's strictly convex. It's everything. This says that also in infinite dimension, we cannot speak about so easy about ellipsoids and so on. But at least we can say that still there is some uniform convexity hidden in the geometry. So now the proof. So let me consider x plus y divided by 2. I want to work on this. So let me consider this. I want to say also what is this? OK, let me write down. So what do we know? We know that the parallelogram identity holds. So x plus y squared plus x minus y squared equal x squared plus y squared. This we know. So in particular, we know. Now let me divide it by 2 once more. So if I put a 2 here, and if I put a 2 here, then I have 1f, right? Then I have 1f, OK? And therefore, from this, I can find that this is equal to 1f. There is a mistake? 1 over 4. Why 1 over 4? This is 1 over 4 outside. Yeah, but the parallelogram identity has 1 over 4. OK, so let me just write this minus x minus y over 2 squared so that I have almost everything that I need because this is my thesis. And this is something related to my hypothesis, OK? So what do I know by assumption? I know that this is less than or equal than 1. This is less than or equal than 1. And I know that this divided by 4, so this divided by 2, is larger than y over 2. Therefore, I know that this square is less than or equal than epsilon squared over 4, right? So this is equal to 1 minus epsilon squared over 4. 1 minus epsilon squared over 4, OK? And this is almost what we need because now we just only have 2 to take the square root. So we have x plus y divided by 2 now is less than or equal than the square root. The square root, 1 minus epsilon squared over 4 over 8. There is an 8 here. And this square root is less than or equal to 1 minus epsilon squared. Yes, but now I don't, you mean this, yes, yes. But what I can do also is just do this trick and define this equal minus delta. So the delta is 1 minus square root of 1 minus epsilon squared over 4 without doing any Taylor expansion or anything, OK? Which is positive, delta is positive, OK? So in particular, the unit ball of L2, we know something, is uniformly convex because we have a scalar and inner product on the 2. Fine. So this has been done essentially without any, just only working on the algebraic inequalities, linearities, sublinearity, subadditivity, and so on. Now we need on V some more structure. On V, we had, well, we have used continuity on, so our vector space actually, when we talk about continuity of A, maybe it would be better to say that, OK. But now we put some more assumptions and we will get it more clear what is in Hilbert's space and so on. So definition, everybody knows what is a Hilbert's space. So let me denote it by h, maybe. So what is an Hilbert's space? Is a vector space endowed by with an inner product and such that, you know, what is this? Therefore, there is a norm. So it becomes, say, a metric space. So the distance between two points is x minus y. This is a distance. This becomes a metric space. OK, everybody knows, I think, what is a distance. So check that this is a distance. So triangular property, non-negativity, symmetry, et cetera. And then this is called Hilbert when it is, when what happens? It's complete. It's complete. So h is complete, OK. So and Banach space, which is the difference. Well, the difference is that Banach space, you don't have the inner product, but you just have the norm. Therefore, again, you can measure the distance. And when it is complete, it is called the Banach space, OK? Everybody knows. Completed, h is called Hilbert space. OK, so now you know what is a separable space. Separable space, no? Definition, h is called separable. If it admits a subset, which is dense and countable, a countable dense subset separable. So the theorem, now we want to understand a little bit what is this L2, OK? So let me state the theorem L2 is a separable Hilbert space, OK? Therefore, this says that it is separable and it is complete. Other questions? So now the idea of today is that we want from now on to understand a little bit something about this space. This theorem is important because at the end, essentially, all separable Hilbert spaces, all are isomorphic to small L2. Yes, in any infinite dimensional separable Hilbert space is isomorphic to L2. There is a way to write an isomorphism between this Hilbert separable infinite dimensional space and L2. So this says that this is a model like Rn. I mean, if you know that V is a vector space of finite dimension, then V is a copy of Rn for some n, OK? And therefore, you always think about Rn. Now this theorem, so the theorem that I mentioned is important because it says that if you have a separable infinite dimensional Hilbert space, then we can think of it as a smaller two. That's why it's so important to understand what is this. Of course, there are Hilbert spaces which are non-separable, unfortunately. There are. And not so easy to find examples, but there are. And this does not cover all Hilbert spaces. But I mean, usually, the Hilbert space that one has to work with is separable. So this is OK. So now we have to exhibit a countable dense subset of L2. So let me call it D is the set of all point Q in L2 such that Q has only finite number of non-zero coordinates. Q has only finite number of non-zero. I mean, Q is a sequence, right? Q is a sequence. But I'm saying that, definitely, this sequence is 0. So it's only finite number of non-zero elements. By element, I mean an element of the sequence. And Q and the non-zero element are all rational. And the non-zero element, OK? So we know what is an element of D. Just the finite, the 0 from just an index to infinity and before, rational numbers, OK? So D is countable, OK? Now we want to show that this D is dense, OK? So to do this, let us fix epsilon. So let fix a point x in L2 and epsilon positive. So since x is in L2, we know that the sum from 1 to infinity of x k square is finite, OK? This is finite. So therefore, we can choose N bar in N such that the sum from N bar plus 1 infinity x k square is less than, say, epsilon over 2, I think is enough. Yes. And then, however, for any index k from 1, 2, N bar, choose a rational number, Q k, so that Q k minus x k epsilon over 2 N bar, OK? Epsilon over 2 epsilon over N bar. 2 square root over N bar. You need the square root. OK, maybe you are right. And I will adjust the square root at the end, OK? So now I define Q as Q1, Q n bar, 0, 0, OK? This is an element of D, OK? So Q is an element of D because it is just only finite number of entries and each entry is rational. Now I have to see that Q is close to x in the distance of L2 by epsilon, so that then is dense. OK, and so I have just to compute the distance in L2 between Q and x, which is nothing else x minus Q, which is nothing else by the square root. Now there is the famous square root. The sum of x k minus Q k square plus the sum x k square, OK? Which is less than or equal to what? So less than or equal than epsilon square. Yeah, OK, less than or equal than. Here we have maybe epsilon square over 2, over 4, over 2, OK? And here what do we have? Epsilon square over 2. Seems to be, you are right because there is a square, which is epsilon. Hence, for any x, for any epsilon that resists Q in D such that this is less than epsilon, this means that D is separable and that L2 is separable, OK? Nice, then we have to show that that is complete. So hence L2 is separable. Now we have to show it is complete. So let us take a sequence, a Cauchy sequence. Let me denote it by xn, a Cauchy sequence. Pay attention. Each xn is a sequence, OK? So this is not so easy. This is a sequence of sequences, Cauchy sequence. And I want to show that x con n converges to some element in L2, OK? So what does it mean that x con n is a Cauchy sequence? It means that for any epsilon positive exists, let me call it n bar, such that for any mn bigger than n bar, we have that the distance in L2 between these two elements is less than epsilon, OK? Less than epsilon, OK? Less than epsilon. So let me write more explicitly what is this. So this is the distance in L2. Therefore, this is the sum. Therefore, the sum from k to 1 to infinity of xnk minus xnk squared is less than epsilon squared. And this happens for any n. So for any epsilon, the positive exists n bar, such that this happens for any n and m bigger or equal than n bar, OK? OK, now I just take just one component of this sum, OK? And in particular, for any k, for any k, we have that xkn minus xkm is less than epsilon for any k, because I just take one element of this sum. So and this gives me an information about this sequence, xkn. Now, think about k to be fixed. Now, this is a sequence in n. See, if k is fixed, this is a sequence in n. And this is a Cauchy sequence in r. Now, is it clear? So this, if considered for k fixed, this considered as a sequence with respect to the index capital N, is now Cauchy, because fixed k for any epsilon exists n bar, such that this is less than epsilon for any n. So this is Cauchy in r now. And we know that r is complete. So it is very important for us to work with objects, with values, in a complete field, OK? For simplicity, we work in r, we could work in c, and so on. But now, what matters here is that this is complete. So this means that this sequence converges as n goes to infinity, to some point of r that I denote by xk. So xk n converges as n goes to infinity, to some point xk, denoted by xk, OK? Therefore, I have, so you see what's happening. I mean, if I have a Cauchy sequence in the distance of the space, then all projections converges, which is sort of saying something like, do you know what is the convergence of all projections? Which topology, I mean, on L2, now we are considering a topology which is induced by the distance. This is our L2, right? So we are working with topology induced by the distance. This is our interest. However, here we are finding a convergence of all components. So do you know what is the topology that implies convergence of all projections? Is the product topology. So we have shown that at least we don't have convergence for the moment. We don't have convergence in topology of L2, but we have at least convergence in the product topology. This is not enough. Indeed, the two topologies on L2 are different. So at least we have something. So now the point is, of course, we have a candidate for being the limit in L2, not in the product topology, right? This is the natural candidate for any K. So now we have to show that xn converges to x in L2, where x is equal to x1, x2, et cetera. And these are obtained as limits as before. So this is what remains to show. And we have to show you're right. We have to show that first you're right, x is in L2, where x is this, you're right. The first thing to do is to prove that this object is in L2, and then that the convergence takes place in L2. This is what remains to do. Time is over. We will do this tomorrow. Remember that we are here for the moment. We are here. Just another small exercise for you. So we continue tomorrow. We need 10 minutes more. And so just a homework. Consider the following object. Is it possible, does it contains an L2 ball? I mean you have this cube. This cube has a name. It's called Hilbert's cube. It's a Hilbert's cube. You see, the sum of 1k squared converges. So it is an Hilbert's and it's immediate to check that c is contained in L2, because the sum. So you have this infinite cube. And each time you increase the dimension, the sides of the cube goes to 0, smaller and smaller. And the point is, can I put it inside the ball of L2? Try to prove that this is not possible. Of course, this is the first hint that says that the problem is infinite dimensional. Because it is true. It is not possible. In finite dimension, it is always possible. Because all norms are equivalent. So you can always do this, infinite dimension. You can always do this. So all norms are equivalent. Now the point, can you put? Now, you cannot. I mean it is impossible. And try to see why. Try to prove that this is not possible tomorrow. So tomorrow we will continue the proof. Need some 10 minutes, 15 minutes. Then we will continue on the geometry of L2, smaller.