 Tako, tako, znači, izgledajte, da zelo smo prišli, da je to prišlo. Tako, da sem češljala notacije, da bi me zelo. Tako, da sem zelo, da sem zelo, da sem zelo, da sem zelo. Beznice po пуšku морal je..če tako ne, nemožno.. In neki pojoma niščil neowo nešto. Niše kilerskim prihpijemo, odrda vič numar, tudi mi taj nekaz je vse cont Qurullo, da bo veseli spook behind. Profesal je in odm capitula as the expression for any x in c naught. So it is clear that, so maybe we will redo this now, last time maybe we didn't complete the proof, it is clear that conversely, so there is remark. So if l is in l1, Then it is clear that if I define l of x... ...as the right-hand side, then l is an element of c0, namely it is linear on c0 and it is also continuous bounded. Thank you. Tako. So this is easy. The statement of the existence is exactly the commers, so that any element of C0 star can be represented as follows. So in some sense there is a map and moreover maybe there is also something else here, moreover, moreover here. Zelo se ne zelo. Srečen norm, zelo vzal, so je to zelo, ne bo, da je neko spes, bo srečen za kontekst. Srečen je vsak z vzal. Tama je zelo vzal v c0 star, In to je norma v l1. Srečno, imamo to, da je map z c0 starov. Zelo, da je tko... Zelo, da imamo map z c0 starov v l1. Que grease which is a isometry also in isometry between C0 star. So I mean this is a way is not easy to understand what is the dual of an infinite dimensional vector space this result say that there is a way to realize the dual if you want, if it is necessary that the dual of C zero can be seen as elements of as elements of L1. And the duality is just the integral. So we try to prove this once more. So we take the elements E1, E2, EI equals 0, 0, I, 0, 1, sorry, 0, 0, et cetera, where 1 is the IT element position. So this is EI is an element of L1 with norm 1, of course. And then we introduce a notation. I think that was XM, was XM. XM was just X1, XM, 0, 0, et cetera, for any X in C0. So I just take the first M components of the vector X in C0. This was the notation. And the convergence in C0 implies that the limit as M goes to plus X minus XM is equal to 0. This is simply because on C0, we have the sup norm. And therefore, this is nothing else, the sup, because this is clearly the supremum, bigger than M, of X minus XK. Sorry, it is simply a supremum. So this is just since C0 consists of sequences which are infinitesimal, this obviously goes to 0. And therefore, this says that XM converges to X in the right topology. And therefore, now L is continuous. And therefore, L of XM converges to L of X. So we know this. So notation, I think it was L, it was, so notation LK is just L of EK. This was the notation, I think, I hope. So continuity of L implies this. Now this is equal to L sum from K to 1 to M XK. Now we use linearity. And therefore, this is equal, now L of EK was denoted by LK. So this is equal to LK. So this, we have that this object converges to this. And therefore, there exists the limit of this finite sum. So there exists the limit that we usually, of course, denote by this. And this is equal to L of X. OK, so what remains to show is that, so we have shown just this, but we have not yet the fact that this is an element in this space. So we have to show, of course, that L is bounded, but also that small L is in L1. OK, then L of X is surely less than the supremum of X, OK. So L of X actually is clearly less than or equal than the infinity norm of X times sum of K. OK, so this is equal to, OK. We still don't know, OK, this is maybe the second part. We still don't know that this is finite. So let us first show that this is finite by choosing properly the, so this is true for any X. This is true for any X. And so now I can choose a special X. And as we saw last time, one can take special X, choose X in the following way. So one take just the sign, so one wants to make this an absolute value of Lk. So just take Xk equal, say, the sign of Lk, so if one less than K, less than n, and zero else. OK, so one gets in this way, maybe, I don't remember if I use this symbol. Maybe, now I see that here there is another symbol. I don't remember. Why? Maybe? I don't know. Why? Sorry. Sorry. So last time was Yn, so probably so was capital N, maybe. Ynk, this. So sorry. OK, this is less than n. And so this is just the symbols that I used last time. OK, so for any capital N. So choose X equal to Yn in this expression. And so what do I find? I find that L of Yn. So this is just the sum up to capital N. To capital N, then I have the absolute value of Lk. Then this is less than or equal to, by definition of the norm of L. And then I have the norm of Yn. But Yn is just plus one, if this is positive, minus one, if this is negative, and therefore this is zero norm. The L infinity norm of Yn is one. OK, so this is equal to. And so this shows that for any capital N, this finite sum is always bounded by a number. This number is, of course, finite. So for any capital N, this sum is bounded by the same number dependent of N. And therefore the whole infinite sum. And therefore we deduce that this is less than or equal to N. OK. We have therefore the inequality that, so we have that small, this element here is in L1. So this inequality says that this is in L1. Not only this, but it also implies that clearly the L1 norm of L is less than or equal to N. Now this inequality say exactly the opposite. OK. This implies this inequality. Now this is finite. This inequality says exactly the opposite. So it says that, because if you remember the definition of this norm, it's the supremum of this number divided by this norm. OK. So this is divided by this is always less than for any x. So we have exactly the converse inequality. And this concludes the proof of this exercise. OK. So this was the exercise. And they were quite similar. And maybe the same proof, actually. So maybe I don't want to repeat it. But let me write down the result, maybe. And also I hope that still I am using the same symbol of last lecture. So maybe it was this, g. I don't know. So assume that g is an element, is continuous, is linear and continuous on L1. Then there exists a unique d, d, maybe, d. Yes. Infinity. This was the letters. These were the letters. I don't know. So, such that g of y is equal yk dk for any y in L1. And conversely, of course, if you have an element in L infinity, so if you have given d and you define capital G as follows, then it is clear that capital G is continuous and linear on L1. And this statement says also the converse, the difficult part that any element of any linear continuous function on L1 can be represented as follows. OK. So there is always an easy remark that is the converse of this is easy. And the difficult part is that what is written here. Hoping that I am using the same symbols. OK. And also we have the isometry, namely that the norm of g is equal to the norm of d. So this is exactly the same kind of statement. And so observe that, so we have found the map from C0 star into L1. And we have found the map now from L1 star into L infinity. And so we have a map actually from C0 star star, no, sorry, C0 star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star star o daj je tukaj, da je po tebe vse zač. Vsake, blizk je tukaj, da je tukaj prav. In da malo so zelo, začnega je tebe zelo pomiliti na zelo. Sino je s tukaj 1. Učnega je tukaj 1. Tukaj ne se očč naja, tebe je tukaj l, ne očč tukaj, So jasno gdaj jamo v teromi tudi, da je sve zmizu pojelje je kar hv 맛있u tukaj M plus 1 infinity yk. Na srešive kajovu mi je hvalo jazno v l1 ruk. Tukaj mjelje spremstva vse. seasoned tukaj je zdaj pojelje za 0. Zdaj to je 0. Tko to je vsečen. Ako tvoje je kakva, tvoje je tvoje. Nisem, da je bilo vseč nekaj skat, da so dajte tvoje vseč, etc. Kajščno vsečen bi se učinila, da je vsečen vsečen. In ga je potravil prv, vsečen je vsečen. Dobro. Vespešno. So, we have some experience now on Banak spaces. And what I would like to tell you today is start the theory of Han-Banak theorems. So, I don't know if we will do all the proofs. If you want to check the details, you can go through the reference of brziz book on functional analysis. Brziz book is I think functional analysis and partial differential equations, maybe all sobole spaces and partial differential equations. Is a big book of spring. No, the new version, much bigger than the previous one. Okay, yeah. There are other books of breziz, but this is a book, the new version is maybe three years ago, zelo. OK. So let me, let me give you now the statement. So we start from an important chapter of function analysis, which are the theorems of ambanak, which have a lot of applications in the world, in the world theory. But please remember that we want, at some moment we have to go through the Fourier transform and other things. And so we cannot really dedicate too many time to this kind of very, very important results. So maybe I will skip some something. OK. So let me use the following notation. So I give you first a statement, which is a sort of statement where topology takes no role. So there is a statement essentially in set theory. And so I give you a statement without topology and I use the following symbols maybe. Assume that this is vector space, vector space. And then take a subspace. So let the theorem, let e be a vector space. And then let g be a subspace. Then assume that we have, assume that we use the symbol g. And that g from g into r satisfies the following, satisfies the following. So g is linear, and g is less than or equal p, where p from e to r has the following properties. The following properties are p of x plus y less than or equal to p of x plus p of y for any y, x and y into e. So this is 0.1. And 0.2 is p lambda x equal lambda p of x for any x in e. Maybe this book is something less than this. So let me make something weaker. Do this. OK, and then there exists f from e to r linear f equal to g on g. And f less than or equal to p on e. So let me try to explain first this statement. So in this statement, first of all, there is no topology. Just linear maps, or something less maybe than linear on vector spaces. So there is no continuity assertion here. I am not saying that this is continuous, then g is continuous, f is continuous, and so on. Then we have a small g, which is less than or equal to something. And this something is defined everywhere on the ambient space. And it is almost a semi norm. So remember, why this is not exactly a semi norm, it's slightly less. Essentially because as I, so if I want a semi norm, the definition remember was sublinearity, homogeneity, but with the absolute value here for any lambda in r. And then also that p was taking non-negative values, which actually was a consequence. You remember, I left you an exercise maybe, that once you have a stronger form of 2, then automatically p takes non-negative values. Do you remember? So with this kind of statement for the moment, there is no reason for which p takes non-negative values. So it's slightly less than, OK. Then the assertion says that there exists an extension of small g. What does it mean? It means that there is a new function defined in the ambient space, so also outside capital G. So defined everywhere. Keeping linearity, OK. So of course here, so linearity is important. We are doing linear function analysis. G is a linear space. Small g is a linear map on the linear space g and f is a linear map on the linear space e. So f is equal to g on the small space, so that it is an extension, is a linear extension. And the difficult part is to prove that it keeps the bound. So it's not any linear extension, but if initially in the small space, so this inequality is valid only in small g. I have not written it because it is clear, because g here is defined only on capital G. So this is clear that it is valid only on g. But then this becomes valid on the whole space. So it is an extension keeping some constraint. So the remarks maybe are. There is for the moment no continuity assertions here, just linearity, but there is a bound. And maybe the difficult part is. And also there is no claim on uniqueness. Actually there is no uniqueness. This just says that there exists one. But in general there are very many possible extensions. So maybe before proving this, before proving this I would like to make some remarks. So this is actually a very abstract statement, first of all, very, very abstract. Indeed it is related to set theory. It is implied by the zorn lemma. Maybe I don't know exactly maybe. So it is implied, I know it is implied by zorn lemma. I don't think it implies zorn lemma. But anyway, I don't know. So what happens if I put, if I make this really a semi norm. So if p is a semi norm, then notice that I have that minus g of x is equal to g is linear. So g of minus x. So if I have this, this is less than or equal than p of minus x, which is equal to minus 1 p of x, which is equal to p of x. So if g is a semi norm, which means that this is written in the stronger form by putting the absolute value here and by removing this constraint. Then I also have not only, from this I also have that. And so I find that g is less than or equal than p. So now let me do some, maybe some first consequences of this before doing the proof in the order. So corollari. Ok, so assume that e is normed. So I assume something more. Now I insert the topology in the statement. So I assume that e is normed. And then as you take, now g as before, but not only this assume that g is continuous. Linear, of course linear and continuous. So linear, continuous. This is a subspace, ok. Then there exists f from e to r, which is lightly more, because it's linear and continuous. F, linear and continuous. Then f is equal to g, f to g on g. And then it keeps bound. Ok, so this is an extension. So there is continuity here inside. So actually therefore f, f is an element of star. So it's a linear and continuous map. And also I don't increase the norm. That is the element, the norm of f is the norm of g. So which is the remark. This is another statement. Now we have continuity. So g in particular is supposed to be continuous. Then there is an extension, which is continuous and therefore is an element of the dual of e. And we have norm preserving extension. So this is norm preserving extension. I don't increase the norm. Norm preserving. So now let us prove the corollary as a consequence of the hambannak that I wrote before. Sorry, this statement is not present. So this is the statement. Ok. I have to find p in order to apply so the proof. Look for p so that I can apply hambannak. So that we can apply hambannak. And the p of x is not difficult because I define p of x and I have g. It's norm so I have g star and then I take p of x. This is my definition of p of x. Now I have to check that this p of x satisfies the assumption of hambannak. So p is actually more than a semi norm. So p surely satisfies 1, 2 in the stronger form also. Satisfies 1 and 2. Not only but we also have in the stronger form because this is a norm and not only a semi norm. And not only actually less than a semi norm. It's more than a semi norm. And also we have that we have to check that g is less than or equal to p on the domain of small g and this is clear, right? Because we always know that g of x is less than this product. This is immediate by the definition of g star norm. So this is immediate. Then we can apply hambannak. So hambannak says that there exists hence we can apply hambannak which gives us f from e to r linear satisfying f of x less than or equal than norm of g g star x for any x in e. So because we keep the bounds. So f on the whole set e on the whole vector space is less than p. This is written here. F less than p everywhere. So it is linear and not only this but also this, right? So not only this but also with the absolute value. This is really a norm. There is the absolute value. So this says that f is linear and it also says that the norm of f is less than or equal than the norm of g. Because f of x divided by x takes the supremum with respect to x and you end up with this. And this is always less than this number. Is it clear? So the proof is simply look for the correct p in order to apply hambannak. So p is this. We can apply hambannak and we have this. So p is not only linear but it is also bounded and this is finite by assumption because g was bounded on g. So this was finite in particular. Also this is finite and therefore f is bounded. Is linear and bounded. And well, so the norm is this but the opposite inequality is immediate. So it is immediate on the other hand. So this is the difficult part but showing this is immediate. Is this clear? So linear extending g. So there exist f linear continuous extending g and with the same norm. So this is maybe the form in which we usually find the hambannak theorem. So it is less abstract in some sense. So this is the first corollary. Another corollary. So corollary. Still in the topological in the continuous version in the topological version so let x not be in E. Then there exist such that f not x not remember the notation and this is simply by notation means this. This is actually equal to f not x not. Or if you want and hence also equal f not square and also equal x not square. F not x not. Ah, excuse me. Sorry. Sorry Mark. So in this way so we have found if this is true Assuming for instance to normalize things sorry Mark, if this is true Assume to normalize things so that x not is in the unit bowl then f not is also in the unit bowl and we have found a map from the unit bowl so the map taking x not into f not so there is a map from the unit bowl of E into the unit bowl of the dual sort of duality map say between this and this ok now this map is not, so I mean there is no claim that f not is unique here so given x not there can be several f not in general so this is a multi-valued map take a point here into several possibly many points ah and also there is no claim that this map capital T is linear here there is, it is not written that capital T is a linear map ah so it is an, it associate to a point in the unit bowl another point in the other unit bowl ah and, but this map is not necessarily linear and not necessary it can be multiple-valued in some cases it becomes single-valued and in some cases it becomes linear also it is true ok, this is a remark which is not so not so easy what happens to this corollary in Hilbert space in R2 in R2 with the with the Hilbert norm in that case this, but also in Hilbert you are right, it becomes a linear map and also ah so in that case somehow you have your unit bowl your point X naught and then somehow you associate to this point X naught in the unit bowl in R2, say you associate this ah tangent line which is up to a factor it somehow identifies a linear map as a kernel ah and then in R2 if you identify this with its unit normal one of the two, say unit normal then then I mean you have you can say that you can put this point X naught to another point here which is this direction so I mean you have a point on the unit bowl so take the tangent this is normalized this is the unit bowl then you take the tangent here and this is this is a linear is an affine space but if you translate it at the origin it is a linear space then this is the kernel of a linear map and if you normalize probably this becomes a linear map with norm 1 now you can identify this plane its unit normal in the Euclidean sense so if you do this identification at the end you have a map taking and then you normalize properly this normal the direction is the unit Euclidean normal then you normalize it properly so at the end you have a map taking a point here into another point somewhere here and so at the end you have a map taking this convex set into this convex set not exactly because the Gauss map is the map from a surface into s2 into s1 I mean just the normal so it's more general in some sense because this is not any surface this is just the unit bowl convex object so the Gauss map take a point on the surface on the embedded surface and are associated to eat the unit normal this is different in the sense that this is just a way to put in correspondence the two unit bowls the bowl of E and the bowl of the dual somehow however there is this identification hyper plane unit normal which is always hidden somehow at least in finite dimension because it is always difficult to think what it is a linear map how can I draw it so maybe if I think as a linear map as its kernel and proper normalization then at least I have a hyper plane which is the kernel now the kernel can be identified since it is co-dimension 1 in finite dimension it can always be identified with the normal so you have a normal you have a tangent tangent you have a normal you can identify so at the end after this identification I can write a map between these two and if these are ellipses actually becomes a linear map of course the problem is when your convex set is not an ellipse maybe a square or something else and always also multivalent so but I don't want to say much just to have some idea take this statement if you want and so let us try to prove the corollary so we want to apply in Banach so there is for the moment there is no vector subspace and we define it we choose a vector subspace just one dimensional so we find we have only X naught we want to apply in Banach so we need the capital G because in Banach had the capital G inside the big ambient space so I have X naught so I just take the span of X naught this is the span ok and then I just so I have defined this ok and then I have small g to define small g well if X naught is 0 then F naught is 0 so there is nothing to say I mean this is true everything is true so it is so we can assume right so assume that if X naught is 0 and it is ok so assume that X naught is non-zero in this case we can take the linear span of X naught which is a line in the vector space E maybe infinite dimensional vector space E ok now I just have to define g from g to R so any element of capital G can be written as some number x naught and therefore I simply define equal lambda and in this way in this way I also have that so g must be linear g must be linear of course g is a linear space g must be linear so let me compute the norm of g so the norm now we have to understand why so this is equal to the supremum of g lambda X naught such that such that lambda X naught is less than or equal than 1 ok so I can put equality here X naught equal to 1 over lambda right because in the same I can put X naught so this means X naught equal 1 over lambda and therefore if X naught is equal 1 over lambda this is equal to X naught so this explains the reason for which we put X naught square for the moment so the supremum you take this is equal to lambda absolute value is equal to 1 over X naught so we have this so this gives us the norm of g ok now we can apply the corollary that I have the first corollary, this is corollary 2 so I have raised corollary 1 so let us apply now corollary 1 so that from corollary 1 of ambanak from corollary 1 we find that there exists f to a star which keeps the norm and extend g extending g keeping the norm but the norm of g is X naught and so the norm of f naught is equal to X naught so I think that this is enough because at X naught at X naught f of X naught is equal to g of X naught because this is an extension and so this means lambda equal 1 so this is X naught square so sorry let me call f equal to f naught so f and f naught are the same so this is f naught let me call it f naught ok f naught of X naught is equal to g of X naught because f extends g and on capital G they are the same small g of X naught is X naught square so we have this equal to this and moreover these are obvious because we have this relation now I would like to make an exercise now before continuing we still have to prove the theorem of Anbanak but still I think that we need an exercise so I raised this so let us do together this exercise take so let us now we have the Banak space small l1 it is clear that capital L is linear let us check that capital L is continuous let us compute the norm of capital L and so the exercise is the following 1, L is continuous 2, compute the norm of L 3 so the norm of L remember is the supremum supremum of L of X such that X is less than or equal to 1 and then the point is that to show that is not attained the supremum is not attained the supremum in 2 is not attained what does it mean it means that it means that there exists no X in the unit ball such that well so there exists no X in the unit ball such that L of X which means that this is not a maximum this is the meaning so is this continuous ok, so let us check L of X is less than or equal than the integral the X which is less than or equal than since T is less than 1 ok this implies if you agree that this is less than or equal than 1 ok, so L is bounded now we want to find the norm of L that the definition is this so we know that the norm is less than or equal than 1 is the norm equal to 1 or not, repeat please such kind of sequence in L1 0 in L1 0 it is defined around 1 ok, so let us take XL equal to 0 if say 6000 and say now I want it to be norm 1 so I take 1 minus 1 ok, so we are taking a function so this is 1, this is 1 minus 1 over N this is N so what up, so this this is so the L1 norm of this map this function is just 1 ok because this is height N and this is 1 over N so the product is 1 so it is in the unit ball so we can use it to test this supremum and see whether this is a maximizing sequence what does it mean it means that to look whether L of XN converges to the supremum 1 in this case so we have to compute an integral I think so L of XN is the integral from 0 to 1 of T XN of T T now this is the characteristic function so this integral concentrates it is always 0 in between 0 and 1 over N so the integral is just in between 1 minus 1 over N and 1 T XN of T but now we recall that XN is equal to N therefore this is a constant that goes outside the integral in this integral ok so this is equal to N integral 1 minus 1 over N and 1 T DT so this is what N and then we have T square over 2 1 1 minus 1 over N namely it is N 1 minus 1 minus 1 over N square which goes to 1 as N goes to infinity you compute you have 1 minus 1 plus 2 over N minus 1 over N square so this is multiplied by N over 2 so you find this is equal to 1 minus 1 over 2 N which converges to 1 is it ok? so, what does it mean? it means that so L of XN tends to 1 but we also have that 1 is larger than L and so hence N is equal to 1 ok so we have found the norm of the operator of the functional so the functional is bounded and its norm is equal 1 ok, 1 is it ok? now the difficult part is that is it true or not that there is one at least one maximizing function so, is it possible to say so the point is so we have to answer question 3 so question 3 says is it possible that there exists or not a function X a function X with norm 1 such that this is 1 do you agree that this is the question is it clear statement statement of the question is it clear so, I mean you have to solve a maximum, a supremum problem to find the norm, right? the norm is the supremum of such and such modulus of ah, yes, yes, sorry this is ok this is the question ok, so I was saying you want to find the norm of an operator of a functional in this case scalar valued and to find the norm you have to solve a maximum problem supremum of something itself the point is there exists a maximizer such that the supremum is a maximum that is the question so there is a what do you think seems difficult because you see assume that now that X is positive so for some reason you can somehow do this yes, but just for simplicity just to argue assume that you look for a positive X then it seems very difficult but this is true, right? because you want that so if assume for simplicity that X is positive then you have that this should imply this but you are multiplying by a factor so if this is one it's an integral but you are multiplying by a factor which is always less than one so it seems impossible, right? that if this is one then this is one then this must be smaller than one not equal one this is more or less the so this seems to be the answer seems to be no and indeed the answer should be that there exists no such an X maybe I can leave you this as a homework for tomorrow, okay? and we will do as a first time of tomorrow we will do in class this part okay? so home for tomorrow morning is 0.3 that sort of this inequality it is a con now we will see exactly in which sense, yes so continue maybe another remark before continuing the list of corollaries maybe another exercise actually so exercise I said that in general the extension, the ambanak gives you an extension but not uniqueness in general uniqueness is not true is not true even in the strongest form when you have the continuity ambanak the topological say version of ambanak theory so you don't have uniqueness so let us consider the following the following the linear span of let me call this yes, the linear span in 1 n, etc what does it mean? this means these are finite linear combinations of elements finite linear combination the word finite is important it is not infinite sum it is just a finite linear combination of these vectors so any of these vectors say is in L1 for instance and so also the linear span is still in L1 and then take g, for instance the span of the first one so r e1 g of lambda e1 say is equal to lambda then this is linear this is continuous and well this is a linear continuous map functional on g now home compute the norm show that so now we want to extend so we know that there is an extension let us try to find several extensions so fj of ei equal to 1 if e1 j 0 is different from 1 and j so let us consider this and extended it by linearity extended by linearity on the span of e1 so these are linear of course these extend of g so there are infinitely many extensions of g if this is true for any j this is an extension of g so these are linear extensions but we have to show that they are continuous so show the most difficult part is that fj star is equal to actually so this is equal to 1 and this is also equal so what does it mean so I have a vector space capital E I have a linear subspace capital G and the linear continuous map on capital G with norm 1 what I am saying is that there are infinitely many extensions keeping the norm linear extension keeping the norm so if this is true it is clear that ambanak cannot give you uniqueness just one extension so we will do this exercise tomorrow again and then we will continue with the corollaries of ambanak and then the proof of ambanak