 We have two exercises for yesterday. One, remember was we take L1, 0 1, and we take the function that I have denoted by L of x. We know that the norm of L is equal 1, so this is a linear bounded operator. Frej龍. We take an element X in E, and we assume the norm of X in 1, of course, is equal to 1, Frow. Je sredu, to je različ um vzpo雯. We want to show that it is not possible that L of X This is not a spac, to dentist. Dлиbe se kaj. takko však ima, da je tudi drugi. Čak je tudi drugi, da je tudi drugi, da je tudi drugi? Tako da je tudi drugi, da je tudi drugi. To začneko, bo hmm... so, ki mali da je tudi drugi, tudi drugi, da je tudi drugi, tako, da je tudi drugi, ki je tudi drugi, ki se počena z deltavna z 0 do delta del X je nekaj pozitivne vrte. Zatim, da je to... nekaj nekaj, da je alfa... Alfa... Ok. Zato početno, da je bilo zelo počen. na drugi... nekaj, zelo se je... Noto, da je zelo, da je zelo, da je ni... je nekaj negrej z 0, da si je zelo... t. verg. t. t.Ézolo. The integral from zero to delta, tx of t, plus the integral from the delta to one, tx of t. So we split the integral in between zero and one, between zero delta and delta one, and then here t is less than delta, so this is less than delta, delta times the integral from zero to delta. So this way we want, this is the trick to forget about the multiplication by this factor here. So in this way there is no pre-factor here, and also here t is less than one, so also in this way we are now without the multiplication by t. So this is very similar to del one norm, so this is equal to delta, integral from zero to delta, x of t, t, plus integral from zero to one, minus integral from zero to delta. So in this way we now have the absolute value of x, but more importantly there is no more, there is no more multiplication by t here. So one is equal to that, so this is equal to the integral from zero to one, x of t, t, plus, plus delta minus one, delta minus one times, thank you, times alpha, so one is equal to this, and therefore finally the l one norm of x cannot be equal to one, cannot be equal to one, because if it is one then, I mean, one plus one minus delta alpha. So this says that if I have an element x, such that l of x is equal to one, then it is impossible to have that this is equal to one. So the proof runs as follows, assume that l of x is equal to one, so we have this, and so necessarily the norm of x is larger than one, because delta is less than one, you see, and alpha is positive. So this cannot be true. So x cannot be one, so we have shown that l of x equal to one implies, so this shows that this is not a max, as we said yesterday. So the idea essentially is to find a way to remove the multiplication by t. You see, there is no more t here, so that is very similar to the l one norm, actually it is the l one norm plus something, so this is, okay. Then the other exercise was, I think it was this one, yes, so I have the span of e one, en, et cetera. So remember, any element here is a finite sum, finite combination, final linear combination of this. We will consider this with the, and do it with the l one norm. So we look, we see this inside l one with the induced norm. So the, we want, now we take the g, g was span of e one r e one, and we have the functional g of lambda e one equal lambda. I think that this is where my notation, I hope. So this is a linear function on a subspace, so this is subspace of e. This is a linear, linear, linear, and also it is bounded because we can compute the norm of g, and so the norm of g is clearly one, okay. Is this clear? The norm of g is equal one. Is the supremum of this among lambda less than one? This is supremum, just one minute. So this is just one, this is just, let me call this maybe g zero, g zero, or g one, g one, okay. So this is just a linear map on a subspace, a linear functional on a subspace of e. Now we want to show that it can be extended in several ways. No, I mean, the norm is, these are always zero, zero, zero, one, zero, zero, zero. Sorry, these are the usual notation that we have used in the course. I mean, this is one, zero, zero, zero, zero, one, zero, zero, et cetera, et cetera, et cetera. So the notation is the following, e i equal zero, one, zero, et cetera. This is in the ith position. So now I was saying that this is the supremum of g lambda e one such that lambda is less than one, and therefore this is equal to one. So now we, so fix j in between one, two, in n, say, without zero, j bigger equal than one, and define, so for instance j equal to two if you define the following, the following, my notation is this, I think, yes, define the following object. So this is equal one if i is in n and i not zero, for instance take f2, just two. So f2 with j equal to two, so f2 of a1 is equal one because i is in one, because i is equal one, f2 of e2 is equal one, because j is equal to, now, as an example, and f2 of e i is equal to zero if i is bigger than two. This is just an example, is it clear? But for any j we have fj, for instance f2. Now we have to make it linear, so we extend this fj, so let me denote it yet by fj, so fj is the same symbol for the linear extension of fj on e, I am using the same symbol, still denote by fj, the linear extension of fj, this means that if I take a finite combination of e1, e2, et cetera, et cetera, then I use the obvious way to extend it. For instance, say f2 of lambda e1 plus mu2 is just, by definition of the linear extension, just an example, lambda f2 e1 plus mu f2 e2, and by that example, this is simply lambda plus mu as an example, so this is an example, for instance, remark, any fj extends g, for any j we have that fj is equal to g on g1, on the domain, any fj is equal to g, any fj is linear by definition, I have found for the moment infinitely many linear functionals extending g, for the moment, is it okay for the moment? Now what we have to show that, what about the norm now, so I claim that not only this, I will show that each fj is continuous and norm preserving, so I claim that actually this is equal to the norm of g, which is equal to 1. So this means that if this is true, of course, any fj is finite norm, so it is bounded, not only bounded, but yeah, sorry, for any j, g1, g1, this is, sorry, it was an notation, just the domain g, g, okay, sorry, g or g1, I don't remember, because at some moment I have written g1, yesterday was g, sorry, yesterday was g, so also today is g, not g1, sorry, g. So if this is true, this means that there is an infinitely, there are infinitely many linear continuous maps keeping the norm, preserving the norm of g and extending g. So this, if it is true, means, of course, that in the Hanbanak version of the extension theorem, we cannot expect uniqueness. Okay, what do we have to do? We have to compute the supremum of, so we have to compute the supremum, so this is, by definition, is the supremum of fj, let me call it x, divided by the norm of x, such that x is in the span, the span was e, maybe, I don't remember, e, okay, we have to compute this. So what is x? So take a finite, so take a finite set of indices, set of indices, and so we have to compute fj of the sum i, divided by the norm of the sum i, okay, so take finite set of indices, take numbers lambda i, not all equal to zero, so that this element, this is x, so that this x is non-zero. So we have to compute this, so this is, I mean, this is f of x, divided by x, for where x is, so this is x. The important point here is that what is a linear span, it is finite linear combinations, so this is finite sum, not the infinite sum, it's a finite sum, finite linear combination, so now we have to compute this denominator and the numerator, so what is the denominator? It's the sum of absolute values of lambda i, because this is the L1 norm, so we are using the small L1 norm on E, so remember that we are looking at E as contained in L1, so the small L1 norm is just the sum of the absolute value of the components, okay, do you remember? So if you have say, lambda 1 E1 plus lambda 2 E2 say plus lambda 3 E3, for the I, for the EI, this is, no this is lambda 1, lambda 2, lambda 3, 0, 0, 0, okay, in components, in coordinates, do you agree? Because this is one at the first position, in the second position, this is one only in the third position, so lambda 1 in the first, lambda 2 in the second, lambda 3 in the third, okay, it's okay, and L1 norm is just the sum of the absolute value of the components, which is written here, okay? Now we have to see what happens here, so we have to look at J, now if this can be various things, depending on the relations between J and capital E, okay, if capital E contains, contains J, if capital E contains J and 1, I would say, then this is just what? Absolute value of what? Absolute value of lambda 1 plus lambda J, do you agree? Because by definition, this is the linear span, and here however, I have this lambda 1, absolute value plus lambda J, absolute value, so this is surely less than or equal to 1, and then also the other cases, it is the same, only one of these two belongs to this and the other does not belong, then still this is less than 1, less or equal than 1, and if none of these belongs to this, still this is always less than or equal than 1, so in any case, we always have that this quotient is less than or equal than 1, and therefore this is less than or equal than 1, but it is also equal 1, because it extends the functional with norm equal 1, therefore its norm is surely larger than the norm of g, and the norm of g is 1, so it is at the same time less than or equal than 1, and surely larger or equal than 1, and therefore it is 1. I think that we have to make maybe the proof of Anbanak, so no uniqueness of the extension, this is clear however, you really don't need infinite dimensions to do this. There is another corollary, sorry, before doing the proof there is another corollary of Anbanak, so the third corollary, corollary, so assume that e is a norm vector space and take x in e, then the norm of x is defined as a supremum of f, x, such that, such that, which is also a max. So, remember, if we want to write the norm of a linear operator, then the norm of a linear operator is defined as a supremum. So, if I want to write the norm of a linear operator, then I take the supremum of all possible scalar product between the, in inverse setting. In inverse setting, remember, in inverse setting, if I have a linear operator, then there is the Ritz mapping, so that we have in Hilbert, we have this representation. And therefore the norm of the linear operator is the maximum of all possible scalar product between the representing vector h0 and h, as h runs in the boundary in the unit bowl. So, this means that essentially I am taking, so if I have found the representing vector h0, then I multiply it against any h, infinite dimension in the boundary of the unit bowl, and then I take the h and the norm of the scalar product, essentially h would be, say, parallel to h0. Now, here the sort of converse. If I want to find the norm of a vector now, this norm of a vector can be computed as a supremum, and now the supremum, however, is against all covectors, say, against all equal to 1. So, it's exactly sort of dual version of the norm. So, if you want to compute the norm of a linear operator, then you have the supremum, and here you divide by x, and here you have x is varying. Now, you want to find, on the other hand, the norm of a vector, then your supremum now is over f in the unit bowl. F is varying. So, just dual, exactly dual, and this is a maximum also in this case. So, there is some gain here. So, in general, we have seen that if you want to compute the norm of a linear operator, this is a soup and not necessarily a max. But now if we do this step, in this step this soup becomes magically a max. So, it's not completely symmetric, the situation, right? It is infinite dimension, but in infinite dimension it is not. So, we gain something here. This is attained. The previous exercise shows that in general for a linear operator, this is not attained. But for indiscreet, okay? Well, fx is always less than by definition of normal a definition of normal linear operator. This is always less than or equal than x, then f times x. So, and therefore if f is non-zero, then I have that this is always less than x. So, this shows passing to the supremum that at least we have that this supremum is always less than or equal than this. Is it okay? No. So, let me repeat it. I repeat it. For any f zero, we know that we have this. Is this okay? Is this okay? This is simply by the definition of norm of f. If you look at the norm, this is a supremum of this and therefore this is immediate. So, once you have this, if this is non-zero, you can divide. Okay? Hence passing, now this is true for any f. The right hand side does not depend on f. So, taking the supremum this immediately implies that the supremum and therefore as I said we have we have not shown this equality but just for the moment this. Now, so now give an x one of the corollaries maybe cornell corollary two. Yes, let me write down corollary two here was that that was this. Don't remember the notation, please help me. So, given x0 into e, there exists not necessarily unique but there exists such that I don't remember the notation but it was such that something like this. So, now we take advantage of this, this was corollary two of ambana. So, now we have given x we apply corollary two so there is some f0 such that such that x is equal to f0 into a star such that and then f0x is equal to the product x ok. Now the point is that I want to use I want to use this f0 in order to show this inequality I want to show this inequality now so let me so I want to take the maximizer so in the maximizer here is exactly essentially is f0 not quite because it does not satisfy the constraint so let me write it for instance this we know another way to write it is fx such that norm of f is less than 1 ok. Now I want to insert here f0 can I I'm not sure because I don't know I do not control the norm this f0 is the I want to use f0 to show that this supreme move is larger than f0 scalar product with x but f0 does not satisfy this so now I force it to satisfy so I take another parallel linear parallel to f0 with satisfying the correct constraint so I just scale f0 so I take say f1 just a multiple of f0 I go parallel to f0 so this is just f0 I want it to have norm equal to 1 and therefore I divide it by its norm simply so remember x is given and so I simply divide it by this so this is parallel to f0 because it is a number times f0 and the norm of f1 is by this is 1 is clear? now f1 finally is in the constraint so now f1 is inside the unit bowl ok so let me compute the f1x now f1x is equal to f0 f0x divided by the norm of x and f0x is is so this is simply equal to x right? because f0x is x2 is x2 divided by x so now f1 satisfies the constant against x makes this and therefore this supremum finally this supremum is certainly larger than what I obtain f1x divided by x which is equal to x ok? fine now I think that sooner or later we have to go to the proof of Van Banak so which is the situation of Van Banak I don't right here now the statement is a statement of set theory but we have the notation however so I think that we have some g subspace of a vector space we have a linear g and there is also something globally defined which is p ok? so this is linear this is almost a semi norm but slightly less than a semi norm so capital G is subspace capital E is a vector space small g is linear this is sub linear and positively one homogenous just positively one homogenous and they are related by this so g is dominated by p and we want to find an extension of g in the wall space a linear extension but still the difficult part is that you could produce several linear extensions but the point is that produce one which is below p everywhere this is the point now let us start by trying to so obvious remark if g coincides with e there is nothing to do because we have already globally the linear g so we can assume that this is not true so we can assume this and therefore there is a point outside so let me denote it by I don't know x naught take x naught in e so x naught so the first the first point is now to extend in some way your linear functional to the span so you have g and then you have x naught you take the linear span of g union x naught so the first point of the proof extend the linear span of g union x naught so for the moment we do not pretend to extend keeping linearly keeping the constraint ok so what do we do we define f so any point in the linear span of g union x naught can be uniquely written if x is in the linear span of g union x naught then there is a unique way to write it as lambda x naught plus some element that I have denoted by I don't know doesn't matter any element here can be written as x plus lambda x naught uniquely for some x in g and lambda real ok any point an element of can be written uniquely as this for some x for 1x in g and lambda in r ok so so we have to define f on such kind of points and of course we want it to be linear ok so this we define as f of x plus some number alpha f of x naught lambda sorry f of x naught lambda alpha where now f is g so at the end our definition will be this and we have to find alpha where alpha equal to f of x naught of course must be chosen properly chosen so I have capital G x naught outside this is the origin now I want to extend small g on the linear span of g union x naught but the point is that is not clear that I can do this because of course but the point is to compute alpha so that this is still true is it clear for the moment the strategy so this is not we are far from the end of the proof but the strategy is first to extend at one point outside keeping the constraint so now claim so we have to choose alpha we have to choose alpha so properly so that so that f is less than p ok this is the point so now the claim is the following assume that we are able to assume that we prove the following two kind of inequalities so g of x plus alpha less than or equal than f of x plus x naught p of x plus x naught and g of x minus alpha larger or equal p ah sorry sorry, I change in the letters so this I call x but this I call in another way so g y so assume that we prove we are able to show to find alpha such that for any x and g and y in g so we find alpha we are able to find alpha such that for any x and y in g that inequality what we want to prove we want to show that this is less than p ok, our aim we want to show we want to show that g of x plus alpha is less than or equal than p of x plus x naught we want to show this I claim that if we show this then this follows is it clear so I mean we want to find alpha so that this holds equivalently find alpha so that this holds ok claim if I prove that for any x in g I have this then this holds so claim assume then this for any x in g and therefore then this and therefore we have our extension ok so now I concentrate on this and try to show that this implies this argument ok so there is no anymore lambda here this is essentially for lambda equal 1 and minus 1 is enough essentially the idea is that I have to show this but essentially if I prove it for lambda equal 1 or minus 1 then this is enough this is the claim lambda since our p is just positively one homogeneous and not one homogeneous so remember our p satisfies this lambda positive et cetera and there is not the absolute value here and lambda real it's just this for lambda positive and no absolute value so I have to distinguish lambda positive so take for instance lambda positive and I want to show that did the first one of this implies this for lambda positive ok then I will show that for lambda negative the second one implies this for lambda negative so first take lambda positive and so we have that so our assumption p of x plus x not so what do I do I think I multiply I consider x over lambda here so I take one over lambda so yes x over lambda so I have g of x over lambda plus alpha this is less than or equal than p x over lambda plus x not do you agree because I have this so if I have this g is a subspace so I can consider in place of x just x over lambda which is still an element of capital G so I insert here in place of x x over lambda and this is still true here one over lambda this is linear now I look at this it's very similar not exactly the same so let me multiply I want to show this almost the same not exactly so I just multiply by lambda which is positive the inequality remains so I just have g of x plus lambda alpha less than or equal than lambda p of x over lambda plus x not ok now simply remember that our p is positively one homogeneous and lambda is positive so now by the property of positively one homogeneous of p I just put inside and it immediately check that this is equal to p of x plus lambda x not so what I have shown is that the first of this the first inequality implies what I want to show only for lambda positive now I would prefer maybe to leave you as an exercise that the second one home work the second one implies this for lambda negative so home is almost the same trick of course now lambda is negative so when you multiply by one over lambda you have to reverse inequalities so there is some ok so home work assume for any y in g this is larger than y minus x not for any y in g then g of x plus lambda alpha is less than or equal than p of x plus lambda x not for any lambda negative ok this is home work so using the homogeneity we are reduced to the case lambda equal one and minus one ok we are reduced to find alpha that this is true now the point is is there some alpha satisfied so that these two inequalities families of inequalities are true at the same time so let us try to rewrite this so alpha must be at the same time from this alpha must be less than or equal than this minus this p of x plus x not minus g of x but at the same time alpha must be alpha must be is it correct what I have written yes sorry because this is wrong ok sorry so please correct this family of inequalities with less than or equal to ok so at the same time alpha must be less than this minus this but also larger than this minus this so at the same time g of y minus p of y minus x not so if we want to prove the claim we have to find alpha such that this is true for any x and y in g we need this for any x and y ok this is equivalent to this so there should be some separate number separating these two so let us rewrite it in another way so observe now that g of x plus g of y is equal to g of x plus y which is less than or equal than p of x plus y which is less than or equal than p of x plus x not plus p of y minus x not do we agree so let us repeat we need to show that there is some alpha so that these two inequalities are true for any pair x and y in g ok remark g is linear so take any two points x and y in g g is linear and its domain is a linear space so this is an element of the domain and I can write this this is clear I also know by assumption that g is less than p so we have this now add them subtract x not use the sub additive of p so you have this and therefore so you have separately this but this is this inequality namely for any x and y you surely know this and therefore you can put something in between by the properties of the real numbers is this clear or not its ok proof so there is alpha there is at least one alpha which makes the job so what we have done up to now is that we have found the way to extend just one point outside just in the linear span just using one point outside g as you see it from the proof the difficult part is to keep the constraint the linear extension is easy but keeping the constraint means to cleverly choose alpha otherwise you don't have the constraint so this is just the first part of the proof which is somehow constructive and now there is the not constructive part to conclude the proof so now you consider so now this is ok let me consider now the family of course I have problems with the notation p now consider following set p p is the set of all h they are domain it's domain linear subspace so now I consider an abstract object the set of all linear maps functionals defined on subspace subspace subspace containing g containing extending such that h is equal to g on g and keeping the constraint and h less than or equal than p so what is this this is a set of linear functionals abstract what we have proven up to now what we have proven that this is non-empty all the effort that we have done up to now is just to show this p is non-empty why? because f that we have constructed is an element of p because f is an element of p now the idea is to take the largest possible h what does it mean the largest possible h? h is a linear functional so now the candidate of our of our theorem is to take the maximal extension so the largest possible elements of p so this is an abstract problem of set theory unfortunately this is given by Zorn Lem but Zorn Lem says that there exist such a maximal element upper bound maybe provided that p satisfies special property so take an element c as a subset c contained in p which has the following property which is totally ordered which is the order of p I have to end of p with so when I say that h1 denotation I think h1 is less than or equal or contained let me just check denotation yes less than let me just let me just use this obvious notation this is not bad because h1 and h2 are scalar valued so maybe not bad to say that this is less than or equal than this if well if the domain of h1 is contained in the domain of h2 and h1 extend h2 h1 what does it mean h1 extends h2 significantly they coincide it means that they coincide in the smallest h1 equal h2 on the h1 it is an extension so in this way I am comparing to elements of p so inside p is a big abstract set and then saying what does it mean that one is less than the other of course not all elements of p can be compared to each other we are not as in the real numbers that any real number can be compared with any other real number not the case you have an element here in principle one is not less than or equal than the other so I cannot compare all elements one each other but I can compare some maybe one some other so I am considering a big abstract set p and I am undoing it with an order relation partial order relation in the sense that I cannot compare in principle all elements one with each other so I cannot say that I am higher than you but maybe I can say that I am higher than you I mean it is a sort of not higher maybe it is not the best way to say but well if you have say points with arrows you can order a finite set of points for instance with arrows connecting with arrows and you can say that one is larger of the other if there is an arrow in between the two with the proper orientation so for instance I can compare say this is less than this for instance but I cannot compare this and this so there is an order relation which say okay this is larger than this okay there is an arrow going from here to here with this orientation but then there is no arrow between this and this so they are not related by the order so we want so the claim is that Zorn lemma should give us the largest possible extension but to apply Zorn lemma there is an assumption the assumption is that if you take C totally ordered ordered then there is a maximal element of C so you take a sub chain totally order of P then there is a maximal element okay and then it is very easy to show that there is a maximal element because if you have HI and C consists of HI for some strange set of indices capital I so this is a family of linear maps in P E set of indices I don't know how large it is and HI is in P and then how is naturally defined of C I take simply the union so I simply take the following H so the domain of H first of all first I isolate the domain of H the domain of H is the union of all possible domains union of the domain of HI okay now I have not only to define the domain of H by define H of X how do I define it well I take a point X inside here this means that X is inside sum of these at least one and therefore I define it as X belongs to some HI some D HI so X in the union so P is non-empty P of H is this I take a point X in the H this implies that there exist an index such that X is in the HI and then I try the following definition I try the following definition and I check that this definition is well is well this is well defined because of the properties that this is totally ordered so one is inside the other and therefore there is no ambiguity in define this is well defined for a totally ordered subset C in P this is the this is a maximal element of C and therefore Zor's lemmas gives us the what we want so by Zor Zor implies that there is an upper bound that we denoted by F DF in R now what we have to prove well it remains to show it remains to show this the point is to show this that F the domain is everything well and this is again true why it is true if this is not true then this is contained in this and there is one point outside but if there is one point outside we know how to extend F outside by the previous step the concrete previous step and that is a proper extension of this because now the domain contains strictly the F the maximality of F so this proof is interesting because it consists of two parts there is one computational part in which you try to extend outside on the linear span of G and one point outside this is concrete you find alpha not terrible but not easy then there is this abstract part because it is terrible in the sense that it is not constructive there is this extension you don't know very much about it how to construct it there is but this extension of course is based on the previous step so this proof is divided into two as you can see no continuity arguments just set theory in the last part what would happen if we assume now just remark rough remark little bit rough in the sense that I don't want to go into too many details into rough remark assume that E is something more has a norm is not only a vector space now let me put it a norm and assume even that it is separable is it possible in this condition to make a rather a proof less abstract than this so why is this proof so difficult because we don't have structure we don't have norm we don't have separability we just have a vector space so when you have this you have set theory the idea would be why it is so difficult but assume for the moment that for is a smaller 2 separability is not always true in infinity is not true but very often it is true so why don't we assume and so under this condition what would be there is a natural way to do this proof one way would be take a point outside x0 outside and we know how to so let me denoted g1 the span of g union x0 and we know how to extend it call this small g1 maybe so separable means that we have x0, x1 etc etc countable number of points thenz in e countable thenz in e ok take one of these points outside and extend it on the span as we have done before but this extension we can also keep the norm because under this stronger condition we have seen that also the norm ok so the norm of g1 is equal to the norm of g ok this is one of the remarks that we made yesterday if we have a little bit more structure we will have this so g is below the norm g is below the norm now this is really a norm and so we can make the norm preserving extension as in corollary 1 on g1 ok then the idea is to do on g2 so why don't we extend now on g2 which is the span of g1 union x2 2 yeah assume also this g is continuous thank you g is continuous you have a bound using a norm the constant is given by a norm so small g is continuous and we extend it into g1 into a continuous g1 on this now I do the same and extend it in span of g1 union x2 so I end up with a continuous extension g2 and then I proceed by induction doing this on each gn plus 1 equal the span of gn union xn plus 1 so this way and I have the extension gn plus 1 now function on this span and then now the idea is now we have to globally define f so and the idea is now why don't we try to extend it by continuity extend it by continuity so take x into e take a sequence converging to x so let me indicate it by xn sequence converges to x sequence of element of the dense subspace subset so x is in the dense xn in the denotation maybe was d I don't know extended to d dense subspace subset so and define f of x equal the limit of of what do we have say gn xn something like this so this is sorry this is very rough I don't want to enter in the details just a way to say that in order to make somehow a more proof made by hand assuming normal space separability maybe we can extend on each point of the dense subs subset and then extend by continuity outside and this is also this also works I think works only under these assumptions so this is yes ah separable yes I think that this works in general it is not never also in that case you don't have uniqueness but I think that in a separable here where space you can at least I believe that this kind of more concrete proof can be done by extending outside of the union of so you have essentially you have defined your your f just only here on the union of the gn ah because you proceed inductively with a countable induction and so in the union of the gn you have defined your f out this union is dense and so at any point text you take sequence of element here extend and you simply extend it by continuity these linear functions it works I think yes but doesn't give you any uniqueness you don't I think that you don't have uniqueness even in finite dimension by the way there is no reason if you have a linear function defined on a line in R3 I think it's not yeah you have to preserve known the point is always to preserve known of course ok well maybe we continue the discussion ok