 Češče, da smo dobro izgledati, da smo način v teori. Prvom, da je nekaj remač. Zelo smo videli, da je izometri, da je zelo, da je zelo, da je izometri, da je zelo, da je izometri. in to je in izometri. V sveče in inilber spes, da bomo, da bomo, identifujemo to v to, da bomo. In tukaj, prišli do starov, da bomo, da bomo, identifujemo h s duvala, h starov. In včetno, da imamo tudi banak spes, tudi je tudi natučnja mapa, včetno je tudi in tudi b-dual. Tudi je tudi definicija tudi v dual. Kaj je tudi definicija tudi v dual? If I have a point x, then j of x is an element of e star star. Therefore, by definition, is a linear map sending star. Proste, da je tudi, kaj je tudi J of X, kaj je tudi, kaj je element istar star, je tudi, kaj je tudi, kaj je tudi, kaj je element istar star. Zato vzvečenje, da je vzvečenje, je to. Vzvečenje, zato vzvečenje, je to vzvečenje kajte dvej. Vzvečenje, je to vzvečenje, je to vzvečenje, kajte dvej. Kaj je to? Zelo, da v zelo, da jih je vse obježen, vzelo, da jih je vse obježen. Zelo, že jih je vse obježen, zelo, da je vse obježen. Tako to vidi to. A čekaj, izgledaj se vse element stripped star. To je vse element strip, to je vse element strip, to je vse element obi izgledaj. Parti celo je duvalitvanie v drzostih da je duvalitvanie v komod za glasboke. me vse imamo the same symbol for a different duality, is the duality between the dual and the original space. So, to be more precise, one should distinguish maybe using different symbols, and therefore one should write something like if you want something like this to remember that it's duality between different spaces. Zato vidiš, da v svoj nekaj vse je jazem jazem vse. Vse značimo, da jazem jazem vse. Vse je jazem vse star star. Vse je, da je vse suprimum, jazem vse f, kaj je tukaj nekako ni rukaj, ni nekako ni 1. Bejdevaj, jah se ki je sate z k ε, odunje se je k samo roasti a je x 1 plus x 2. Tukaj je tukaj f comma x 1 plus x 2, z kaj sem dreven. Now, what is this? By definition, the definition of this map is this, so this is fx, such that. But we have seen yesterday one of the corollaries of ambanak was this, that this is exactly equal to ex, if you remember. This is corollary of ambanak. What does this mean? Well, it means that the map J preserves the norms. This means that this is an isometry. J, what does it mean? Partial isometry. Isometry to some subspace for e double star star? Yes, double star star. So this way you can look, so this is a way to, so we can identify e with J of e, if you want. And this is an element of e star star. So at least we know that e, if we want, can be embedded into e star star. The point is that in general, this is not surjective. So in general, this is a proper subset of e star star. And not necessarily, so you cannot, in general, identify e with e star star. But just you can embed e inside e star star isometrically. In Hilbert is different, because in Hilbert we can, even more, because you can also identify with h star. So there are some spaces, so in general this map is not surjective. We have already almost seen this. When we studied space of sequences, we saw that if we have C0, then we pass to L1, and then from L1 to L infinity. Remember? I mean the dual of C0 was realized as L1, and the dual of L1 was realized as L infinity. And therefore the b dual of C0 was realized sort of L infinity, which is not C0. It's larger. So in general, the t is not surjective. This map J is not surjective. So in general, surjective. When it is, the space is called reflexive. When it is surjective, of course this kind of phenomena you don't see in finite dimension. In finite dimension you can always identify a finite dimensional vector space with its dual, and then with its b dual, and so on. So this fact that this is not surjective is a fact that you see in infinite dimensions for suitable Banach spaces. There are examples, as we already remarked, that of non-reflexive Banach spaces. But there are examples also of Banach and not Hilbert reflexive spaces. Do you have some examples in mind of non-Hilbert Banach reflexive? I don't think so. I don't think so. Even if at the moment I don't know exactly who is the dual of the measures, because you pass to the space of measures, for instance, continuous with compact support, you pass to the random measures. If they have, now if you take continuous with zero in infinity, it depends. It's a space of measure, but then taking the dual is much more difficult. I think it's more simple. Lp with p is strictly larger than 1, as strictly smaller than infinity. You know that when you take Lp, then the dual can be identified with Lp prime, p prime the conjugate exponent, and then you read dual once more and then you take the conjugate and the conjugate and then go back. Another exercise. No, this is not an exercise. This is far from being simple, but it is interesting to have this map J. Now an exercise, say a Banach space, and take two points, different points, then there exists L, which separates these points. The duality between L and X. Remember that in my notation, I'm switching notation for convenience, sometimes I use this, sometimes I use this. But this is a notation. So there are sufficiently enough linear continuous functionals on a Banach space, at least to separate points. So do you have some suggestions for doing this? Yes, you take X minus Y, you take X minus Y, which is different from zero, then you take the linear span, the linear span, let me call it maybe G, span over R, over X minus Y, and then you can define some this, say G of lambda X minus Y equal to, for instance, lambda times norm of X minus Y for any lambda into the real. Also probably it works, also it's OK. We can take also equal to lambda. OK. So this is linear continuous functional on small g, sorry, on capital G, and therefore by Han Banach, this can be extended into F, so there exists a continuous linear functional on the whole ambient space extending G. So let me call, sorry, let me call this capital L, and not F, let me call it capital L. Change notation a little bit. So you see what happens to, and then one X that then L of X minus Y is non-zero, because L of X minus Y coincides with G of X minus Y, which is this norm, which is non-zero, and therefore L of X is different from L of Y, because this is linear. There are several linear functions. Other exercises before going to other consequences of the Han Banach. So what is hyperplane in a normal space actually? So in normal space. Well, we can give the definition in two ways of hyperplane. Let me denote it by M is a hyperplane. One would like to say that it has co-dimension. It is a vector space of co-dimension one. One should make this rigorous. Co-dimension one linear space. So there are two possibilities. One is the following, is that N can be written as M times R, let me denote it X naught. So there exists one point out of M such that this. Or, well, it is an exercise to show that equivalently, in equivalent ways, is to say that there exists linear continuous, so assume we are in Banach just for simplicity. Just for simplicity. And there exists such that M, there exists L linear such that there exists a linear function N such that M is equal to, and there exists alpha in R such that is the set where L is equal to alpha. Hyperplane. Maybe not the vector space, but also in a fine hyperplane. So translation of the vector space far from the origin still is an hyperplane and therefore it is a level set of a linear function. So this is the definition of hyperplane and L star is non-empty. The star is non-empty. Let me see. Well, this is true in normal space, but also in normal space. I think that I am just using, well, no, I am using the continuity. So I am using the topological version of Banach space. So the question is that can we say that this is true in more general. So the proof of the topological version of Banach was do we use completeness or not? Maybe not. No. So this is okay also in normal space. I wrote Banach for simplicity. So, well, this is a hyperplane, sort of hyperplane like in finite dimensional space. It is something which is affine, could be also a subspace that is not so important. I mean if it pass through the origin or not. The important point is to say that it has co-dimension one in the sense that the outside of m, say, is one dimension somehow. This explains the word hyper in front of plane. Now, of course, this hyperplane is closed. Another exercise, which, however, we have already done essentially, is that this hyperplane is closed if and only if L is bounded. Do you agree? We have already seen this. Not exactly. We have seen that the kernel of a linear functional is closed if and only if the linear map is continuous or bounded. This is a little bit more. It says not only L equal to zero closed, but for any alpha L equal to alpha closed. Well, they are homomorphic. L equal to alpha is homomorphic to L equal to zero. So, exercise, for any alpha L equal to alpha closed if and only if L is continuous. So, you add the word here bounded. If you add here bounded, then you find that L equal to alpha is closed. And vice versa. Now, examples of, so these are, for what concerns, hyperplanes and example, other exercises maybe. Now, let me go to a subspace, general subspace, assume that S is contained into some Banach, if you want. And S, finite dimensional, then you should prove that S is closed. So, this is just to have some intuition of what are subsets of infinite dimensional spaces. We have seen something. Remember, we made an exercise on the interior part of a vector subspace. The interior is empty. So, like in finite dimensional space, if you take a line in R2 or a plane in R3, they have empty interior. A plane in R2 have co-dimension 1, so it's a hyperplane. And if a finite dimensional vector space, subspace is closed, this is an exercise. And then the closure of S is a closed subspace. So, this is item 1, item 2. Of course, here there is no assumption on the dimension of S. Take any subspace, maybe could not be closed. Why? We don't know, could not be closed. If it is not closed, we close it. We take the closure. And then the exercise, they says that this is still a subspace, say Banak, simplicity. This is an exercise. Now, it is interesting, one could wonder, are there non-closed subspaces? Because this suggests that there are, right? So, are there non-trivial, non-closed subspaces of a Banak space? So, let us try to see if we can do an example. So, this is our homework. Now, exercise. So, take the Banak space smaller than 1. And then take the vector subspace defined as follows. V is the linear span over R of the following vectors. So, E, N. What is this? So, as usual, we use the standard notation. So, EI is 0, 0, 1, 0, etc. At the height position. Then we have a vector space. Because this is the span. So, by definition is a vector space. Finite linear combinations of elements of this sort. Finite linear combination. With real coefficients. So, this is a vector subspace. Vector space contained in a 1, for instance. The point is that it is disclosed. Is this closed? It is not closed. Because there is an N. Is not. Do you agree? I don't want to say what is this. But I just want to say that there is one element here. Which is outside here. This is enough. Sorry, sorry. No, okay. Sorry, sorry. Is V closed? Sorry. You have to erase this. Sorry. And so, it is not. So, I just show that there is one element in the closure. Which is outside V. It is enough. And yes, exactly. E1. E1 is not in V. Because it is not in V. On the other hand, it is easy to see that E1 minus this difference obviously goes to zero as N goes to infinity. So, there is a sequence of elements of V. Elements of V converging to E1. And so, E1 is in the closure. So, E1 is not in V, but is in the closure of V. And therefore, as a consequence V is a subspace, which is not closed. Is a proper subspace, which is not closed. I think that this phenomenon we don't see in finite dimension. We don't see it. Any subspace in finite dimension is closed. Okay. So, maybe it is useful in functional analysis to isolate what kind of statements are really due to the infinite dimensions and what they are still true in finite dimension. Maybe just to start, right? So, this we don't see in finite dimension. It is written here. And so, let's see if there is something else. Yes, maybe this one. Exercise. Let V be contained in E. Hyperplane. If V is not closed, then V is dense. So, we know that there are non-closed subspaces. Now, here there is a little bit more. Is a hyperplane. Well, it is not a subspace because maybe I find. Here is a hyperplane. Assume it is not closed. Then you close it. The closure is still a subspace, right? But the hyperplane is called dimension one. So, if you close it, you have all the space. This means it is dense. So, try to make this rigorous. But the proof is exactly this one. So, if it is not closed, you take a point outside. And then you close it. This is still a subspace. Remember the definition of hyperplane. And then you cover all the space. So, this is a home. Let me see if there are other exercises. So, now we go back to the ambanak. Let me see other consequences of the ambanak. OK, so maybe I can state the theorem. So, assume that E is a banaq space. And let A be a subset of V. And B is a subset of V with the following properties. So, assume that A is open convex. B is convex. Assume that the intersection between A and B is empty. Then, so you have two convex sets. Maybe this is a convex set. And then you have another convex set like this. So, this is A. This is B. Then, the theorem says that there exists a closed hyperplane with the following properties. Notice it is closed. This means that we have to look for linear continuous functional. But the point is that there is a closed in the picture, something like this, say, which separates the two convex sets. It says that one convex set is in a half space and the other convex set is in the other half space. How can I say this? OK, so this is L equal to alpha, maybe. L equal to alpha, the green line, closed hyperplane separating A and B. What does it mean separating? It means that, namely, there exists, that is, edesta, exists a linear continuous linear bounded, that, well, you have to think of, you have this, the graph of, you have the graph of your hyperplane and this means that this is in sub-level alpha and this is in the super-level alpha. So, there exists linear bounded such that say, for instance, A is contained in L less than or equal to alpha and B is contained in this super-level, alpha super-level. Or, in other words, for any A in L of A is less than or equal to alpha and for any B in B, L of B is larger or equal to alpha. This is equivalent to this. So, you have to think about, yes, simply the word separating is OK. The point is that here there is this closed. So, you have to look for something which is continuous. So, it's not trivial. And so, this is one version and then there is, before doing the proof as yesterday, I think that we will state the theorem, then we make some corollaries, comments and try to prove it. So, this is, so now, crucial assumptions here are this and convexity and of course, they must be disjoint, otherwise, in general, you cannot. OK, so let me write, so this is theorem one. Now, let me state theorem two. So, in Banak, A and B, contained in E, assumption one, two and three. So, one, A is non-empty and is convex. Two, B is non-empty and is convex. They are disjoint. So, here, you see, I have not written open. Now, I add closed here. And compact here. So, what could happen in this picture? In this picture, concerning here, I have to know, I will state that this is, but in this picture it could happen the following. I could imagine this open convex set, little bit larger, but open, still open. OK, this is A, which is open. So, the boundary in finite dimension, the boundary is not in A. And then, I can have another convex set, like this, just going to this point. OK. Still, we can separate. Since we have here the large inequalities, we can still see that this is still reasonable, even in this condition. When this point is exact, this goes here, say, this point belongs, say, to B, but does not belong to A, because A is open, so that they are still disjoint. And still, we can imagine the green line. Now, here the situation is slightly different, because now you are assuming, say, that A is closed. So, A is closed. Assume, you are in finite dimension. So, A is closed, now you add also the boundary. But B is more, now in finite dimension, B is even compact. And they are disjoint. So, now this situation is not allowed anymore, because this endpoint would belong both to B and to A. So, one imagines something like, OK, this is closed, this is closed, OK, maybe it's infinite, convex. I don't know, but closed. And then this is compact, however. Now, the idea is that, well, under this condition there is a little bit space in between, and so we can gain some epsilon. What does it mean? It means that there exist positive epsilon, and there exist linear bounded. Such that now is strictly separating. I mean, one is in the alpha minus epsilon sub level, and the other is inside the alpha plus epsilon super level. So, there is some space, such that A is contained in the, is contained, something like this, OK. So, this is the statement of theorem one. Of course, here compactness will be crucial in some sense, to get to have this small space. You see, infinite dimension, what could you say infinite dimension? You could say, well, I have a closed set, a compact set, they are disjoint, then they are at positive distance. Well, here we are not infinite dimension, we are in infinite dimension, but at least in finite dimension there is some distance in between this and the other one, OK. So, we gain a small space, and still here one gains, in this sense one can have this small space, so in some sense I could enlarge a little bit one of them, and fine, I could enlarge it, take points of distance, say epsilon, and then still have this statement for the enlarge set. Now, however, we will see the proof, but before then, before doing the proof, I need some, I would like to show you some corollary of theorem one and theorem two in the order, so, ah, well, this is the first corollary, corollary density, I will tell you now, in which sense this is called density criterion. OK, so assume that f is a subspace of E, subspace of E, and assume that this is not, this is strictly contained in E. Then there exists, let me call it L in a star, this is a linear continuous function in star, such that L of x is equal to zero for any x in f, and L is not identically zero. This is the density criterion. Now, in which sense it is a density criterion, then we will prove it a density criterion. Density criterion, this is usually as follows, assume that you want to show that some subspace is dense, assume that we want to show, this is not the proof of the density of the corollary, it's just consequence of the corollary, then I will prove the corollary, OK? How I use this corollary? Assume that we want to show that some subspace, some subspace f of E is dense. This means this. Then it is enough to show, then it is sufficient to show that given L in star, L x equals zero on f for any x in f implies that L is identically zero, OK? This is very useful. So, in general, assume that you have a subspace, you want to show it is dense, while you take any linear continuous function vanishing on f, and then you prove that this, if you prove that this vanishes everywhere, then necessarily f was dense. Because if it is not dense, then you contradict this statement. Is it clear? OK? So, this is usually used as follows, this corollary. OK. Now, how can we prove this corollary? So, let us go through the proof of the corollary now. So, we have proof of the corollary. Take a point, so if f is a subspace, but we know that by assumption this is strictly contained in e, so we can take a point outside. We can take a point outside, and now we apply theorem 2 with the following choice. I don't remember which was the compact and which was the closed set. B was compact. B is equal to x naught and A is equal to f bar. So, f is a subspace by in homework f bar is a subspace. OK? In particular, f bar is convex. So, f bar is convex, closed, non-empty. Well, non-empty subspace. Add non-empty here, OK? So, f bar is in the assumptions of theorem 2. B is a compact convex non-empty set. By this inclusion, they are disjoint. So, the intersection of B and A is empty. So, we define the assumptions of theorem 2. Therefore, we can find the linear continuous, there exist alpha, let's say there exist l linear and continuous, bounded, linear and bounded, and then there exist alpha, let's say l of such that. I am strictly separating. So, I even don't take epsilon, but I have the strict inequalities here. So, l of A is less than alpha, less than l x naught. And this for any A in A. I even don't need epsilon here, but I just need the strict inequalities, OK? I have not used one assumption, well, not completely at least, but now I remember that A is a subspace, because it's the closure of a subspace. A itself is a subspace. So, this inequality is not possible, unless l is identically equal to zero, because this is less than a number, but then I can change sign to A and I can multiply it by any real number, 1000 and minus 1000. This inequality is not possible, you see? This is a subspace, I can change sign. I can multiply for any big number here, positive and negative. OK. And therefore the conclusion is that, unless this is vanish identically on F, this is the only possibility. This equal to zero. If it is not zero, I can multiply by big positive or negative number. So, the only conclusion that it is possible, so this is impossible, unless this is equal to zero, which is the conclusion. So, this is one assumption, sorry, one corollary, and then let me show you another interesting corollary. So, this is the density criterion. So, assume that star is separable, then e is separable, and the converse is not true. Remark. E is separable. It does not imply in general that the dual is separable. Sorry. Why is this so? L1 is separable, L infinity. Sorry. L1 is separable. We know that we can realize the dual of L1 as small L infinity. This is an exercise that we have made. But L1 star is not. On the other hand, we can use corollary. So, this is corollary 1. This is corollary 2. The claim is that we can use corollary 1 to show corollary 2. And therefore have this interesting conclusion on separability of corollary 2. So, if it is separable, star separable, we can choose, implies that we can choose a countable dense subset of the star. So, there exists, let me call this fn contained in star countable dense subset. Now, for any n, now fix epsilon positive, take for any n, first take xn, take for any n, take for any n, xn into e with the following properties divided by xn larger than or equal to 1.5. So, I take for any n in correspondence on this, the definition of norm of fn, which is the supremum of this divided by this, you can always choose this. Remember in the definition. Remember definition of, do you agree with this? Do you agree? It is the supremum of this divided by this. And therefore I can. So, now I have a family, countable family of xn, and then I can take the span who can consider the linear span with rational coefficients of xn. So, we can consider the span of xn. It is, of course, yet countable because I'm countable, countable contained in e. And it is also dense and dense in the span over the real. So, let me call this v. So, if I show that v is dense, I am done. We agree? Because then this is dense in this, this is in dense in e, and therefore the span over q is dense. Is it okay? No? Which is the point? Are there questions on this? This is okay? So, I have chosen for any n one point of your space. Okay. Satisfing this inequality. Then I would like to show that, now I'm looking for a countable dense subset of e, right? Because I want to show that it is separable. So, the idea is, okay, I take all linear finite combinations of this with rational coefficients. And this is this, and this is yet countable. Since q is dense in r and these are finite sums, this object is dense in this. And I call this v. Now, I try to prove that v is dense. Try to prove, let us try to show, let us show that v is dense. If we show that v is dense, then this is dense also. Because this is dense in this, this is dense in v, and v is dense in e. And therefore also this is dense in e. Okay? Fine. So, we have to show this now. Let us try to show this. And now we apply the corollary one. The density criterion. We want to show that some subspace, because this is a subspace, v is a subspace. The closure of a subspace is a subspace. We want to show that this is dense, okay. And therefore we apply the density criterion. I mean the consequence of the density criterion was, I want to show that some subspace is dense, so I take any linear function on v and I show that this is zero. Vanishing on v, then it is identically zero. Okay, so let, let me denote it by L. From e into R, L linear and bounded. And assume that I know that L, Lv is equal to zero for any v in v. And then we have to show, okay, so now fix epsilon positive. So now we know that fn is dense. Therefore, there is fn minus L is less than epsilon. Okay. So there exist n, depending on epsilon, such that fn minus L is less than epsilon. Okay, this is by density because fn is dense in vista. Now, no, I mean this is just to say that given epsilon there is some element of fn which is epsilon close to L. So fix epsilon, in the ball centered at L of radius epsilon you take one fn is dense. Now we want to use this inequality and this assumption here. So we proceed as follows. So we have that epsilon is larger than fn minus f which is larger. This is a norm and so it is larger than fn minus f, fn minus f comma xn corresponding to this. So to any fn, so we have chosen this n epsilon, right? So we have fn epsilon, maybe. Let me use n epsilon and given n epsilon we have xn epsilon. So let me use this xn epsilon here. Maybe it is better to use n epsilon here. So xn epsilon, fn epsilon minus L. I am making some mistake in the notation. This is capital L, sorry. This is by definition of norm of course which is the supremum of this quotient. So what is this now? So we know that xn epsilon, remember xn epsilon is in v. Yes, of course. So xn epsilon is important to remark that xn epsilon belongs to v. So this is equal to what? We have fn epsilon comma xn epsilon minus L comma xn epsilon divided by this. And now we are going to use this. So if in place of v, we take xn epsilon, this is equal to zero. Because xn epsilon is in v. This means that this is equal to zero. But let me go slowly. This is zero, right? This is equal to zero because of star and star star. Where star is this? And star star is this. So this is simply this xn epsilon divided by xn epsilon. And now we use this assumption. So by this assumption, by this choice, we have that this is larger than 1 half the norm, as he was saying exactly this. So you see interestingly we have found that in the end of this inequalities fn epsilon for any epsilon positive exist an epsilon such that fn epsilon is less than 2 epsilon. So from one hand this says that the longest subsequence fn epsilon is going to zero and the longest subsequence it is also going to f because we have this. So take epsilon say equal 1 over k and 1 over n. Then you can find a sequence of indices such that along this subsequence fn converges to l. At the same time epsilon is 1 over n and so this converges to zero. So the limit f is equal to zero necessarily. l is equal, so sorry, I am confusing f with l. So sorry, I repeat. Along the subsequence this is going to l. This is going to zero. Therefore l is equal to zero. And this says so we have to show that l is equal to zero. We have shown that l is equal to zero. Therefore necessarily v was dense. As a consequence of corollary 1 of ambanak. So this is a quite interesting result of structure of this strange infinite dimensional spaces at the end. I think that I have said all consequences that I want to say on theorem 1 and 2. So now we have to show those theorems. So proof of theorem 1. So you remember the proof of theorem ambanak theorem of yesterday was made into two steps. The first step was how to define the extension at the point on the linear span of one point outside keeping the constraint and then there is the zorn abstract zorn machinery. Now we want to use the theorem also in this case but still we start with the computational part assuming that now I don't remember if a and b maybe a was open convex of a was so assume that b is just a point for simplicity. So assume for the moment now we start to do the proof in the simple case that we have just one point outside. So you want to separate an open convex set with a closed hyperplane separated the open object and the point outside which is a convex set. So here there is no time today to conclude the proof. Of course we have five minutes. What I can say just to make a picture to convince you which is the strategy. Ok. So let me try to draw some picture in finite dimension sorry R2. Ok. This is a convex a and this is a point outside x0. Now the first thing that we will do is first of all a is open so a is convex and open so by translation it is very convenient to put the origin of our vector space inside a. So we translate the origin so that maybe the origin was here. Ok. Now we translate everything so that now the origin is here. Ok. Somewhere I don't know here. Origin. So by translation we can do the following interesting assumption that 0 is in a. So remember that a is open so a coincide with its interior. So now we have a convex set with the origin in its interior. Convex set with the origin in its interior. Ok. This is the origin. Now we want to apply so there is a general principle when you have a convex set with the origin in its interior this corresponds essentially to a semi norm it is essentially the unit ball of a semi norm and conversely if you have a semi norm then take the unit ball where the unit ball is a convex set contain the origin in its interior so let me make just some comments here. Ok. So find a dimension. You have a convex body. Ok. Like this. The origin is here. Well, this is a convex set with non-empt interior and the origin is in the interior then it is not difficult to define a one homogeneous function thinking of R2. Now this is in R2. It is not difficult to find a norm essentially having this as a unit ball. Not quite a norm exactly because this is not symmetric with respect to the origin, right? The origin is here and this is not symmetric. Symmetric would mean that if you have a point here then you have also the opposite inside but you have a point here the opposite is outside. So this is not symmetric. If it were symmetric then it is also convex and bounded convex and bounded then it is easy to find, to look at this as the unit sub-level set of one homogeneous of a norm. And then I will tell you this is called the gauge that there is a way to define a function in a natural way such that this is to construct the norm starting from the unit ball. Conversely, it is much more easier. Assume that I have given you one homogeneous convex function cut at level 1 that object is convex that contains the origin in its interior. So cut at level 1 put it in the horizontal plane this is the one level subset containing the origin somewhere. So now which are the difficulties? Well first of all this point is not this object is not symmetric with respect to the oil. This is the reason for which in the Banach theorem of yesterday we have written lambda positive and not modulus of lambda. Remember? We have written p of lambda x equal lambda p of x but only for lambda positive and this was to include non-symmetric sets because if you put this for any real then essentially you are symmetric. So if you want to include the cases in which you have no symmetry you cannot ask for this strong assumption just ask for this for positive lambda. So first difficulty is this second difficulty in finite dimension just to understand that you can have a convex body but this say could be infinite. So this this creates some degeneracy in your norm it's not really a norm actually because when this is infinite then here there is a direction essentially where you pay nothing I mean it's degeneracy here in the norm. So for instance the norm coming from this convex set is not equivalent to the Euclidean norm. So you cannot put yes of course you can put a Euclidean ball but you cannot put a Euclidean ball outside. This you cannot and so you expect that if you want to have an inequality inequalities between this norm and the other you can have just only one inequality and not two inequalities. They cannot be controlled one each other because this is unbounded. So I mean asking convexity of A and open says that you can assume that zero is in the interior but you cannot assume that in finite dimension for instance that this is bounded we can do the proof in any case but keeping in mind this this now which is the definition the natural definition of passing from a convex object to a norm or semi norm or whatever I mean even less than a semi norm but inside the statement of unbounded. Ok so if you have C in general so in general you have say C convex with zero in the interior so with non empty interior in particular then there is a function defined in the ambient space which is called maybe the gauge function of C itself and is the what do you do the idea is the following you take the definition what is it it is exactly sorry ok I am changing the name for maybe I am wrong Minkowski functional Minkowski functional also gauge gauge in France maybe minkowski let us call this Minkowski so what we are doing here we are doing the following so we have a convex body this is the origin we want to define at a point x a function then you do this you take well first you should prove that this set is absorbing points it absorbs points in the sense that if you omotetize it sooner or later you will include any point x so first you have to show that this is obvious in finite dimension but this is true also in infinite dimension once this has the origin in its interior it is convex then it absorbs points namely given any point you can scale your C so large with an alpha so large so that alpha times C becomes so large that includes x like a ball then you take a point outside then you take one thousand the radius of a ball sooner or later you include the point x and then you take the smallest alpha such that you have included x so you start from outside there is a big alpha and at some moment the smallest alpha you can have then this is the value of B at x so we will show on the next lecture that this is exactly the function such that its one sub level set is C and and you see now the idea is the following well I have a point outside here above this I have a linear sort of linear function now I define also say g of x on the linear span equal lambda so that the g of x is one it is rather clear that this is outside remember the idea that if you have one convex state inside the other and they take the linear functions then the linear functions corresponding to the smallest set is above the linear function corresponding to the larger set and therefore here it is rather clear that this p should be above this g well if this is true then we are almost in the condition to apply hand banach because we have a seminorm so almost a seminorm p linear function g and so on this is more or less the idea of the proof we cannot do this in 5 minutes we will do this, this is just the idea