 Zdaj zelo smo izgledati te dve izglede, a potem zelo smo počkati. Znamenj, da jasnjih smo zvedali konvex set in počkati počkati. V Hilberti počkati, da je počkati počkati počkati počkati. Ok. Now, this is the convex set. Could be half space, could be convex set, whatever, but the important fact is that it is convex. Now, what happens if your C is instead a subspace? Remember that yesterday we found the Euler Lagrange equation using unilateral variation. Namely, we compare, given any direction from the inside, we compare this length with all possible lengths along this direction. And convexity means that since you are just only from one side, you can do variation only from one side. You can compare the length with this only with objects from one side, and therefore you get just only one inequality. You cannot get an equality. Now, on the other hand, now assume that you have now a subspace, and then there is really equality. So, the exercise was let H be Hilbert, and C a closed subspace. Then, so this is the point H. In the Euclidean world, this would be what we called X, the unique point X that maybe sometimes it is denoted by PCH. So, we have a map sending H into X projection uniquely defined. So, we have a map sending this into this. So, a map sending H into C. And this X is sometimes denoted by projection of H on C. Orthogonal projection. By orthogonal in the sense of the scalar product that you see from your unit bone. So, this is projection on the closed convex set C of the vector H. Now, yesterday we found a set of inequalities, but now we have that this must be equal to zero for any C in C. So, H minus X, any element of C. And we understand from the picture that there should be orthogonality. This is orthogonal to this. This is simply the meaning. Ok, in the proof. So, maybe this is a theorem. No, corollary, maybe. Corollary. So, yesterday we proved that H minus X, C minus X, is always less than or equal than zero for any C in C. Why is this so? Well, because capital C is in particular closed and convex. So, we can apply our Euler Lagrange inequality. It is clear, by the way, that any subspec is convex, right? So, we can apply. So, this says that H minus X, C is less than or equal than something which is independent of C on the right hand side. So, when I write a comma and then here I mean for any C. Just a way not to write the quantifier. Ok. Ok, now we have, so this is just from convexity. But we can switch now the design of C so that we can use the fact that capital C is a subspace. So, if capital C is a subspace, then we also have what we didn't have yesterday. So, we can change the, it is not. So, if this is a point, then also minus is, if this is the origin, say, then I have a point C, then also minus C is on the set. Therefore, we also have here with minus C. And therefore, this must be also larger than or equal to. Not only less than. I put minus C here. I put the minus outside. And so, I mean taking simply variation of this sort, this implies that you have a quantity necessarily equal to zero. Namely, essentially you can, instead of taking C, equal to zero. So, instead of taking C, you can multiply, say, any number Tc with T in R. So, if you use these new competitors, then you easily end up with a quantity. OK, so, this is the first, but the most interesting exercise of yesterday was the projection on polynomials. So, the projection on polynomials was the following. The homework exercise. Maybe we can complete also. So, the map here Pc is linear. Maybe this also check at home that this map, this projection map is linear, OK, for the subspace C. Now, projection on polynomials. So, we had T cubed minus the projection of, so, C is the space of, subspace of all polynomials of degree two. Remember? So, now I take the unique projection on polynomial of degree two. So, let me call this difference T, maybe tau of T. And so, we know that tau must be orthogonal to any polynomial of degree two. And this is, what does it mean? This, in the setting of the exercise, means that the integral on zero one of tau T C T dt, OK, is this notation clear? This is the scalar product between an element of L2 and another element of L2. And this is the scalar product integral, OK. So, the projection, the projected point tau T is orthogonal to Ct, OK. So, what we know is the following. So, we have to find this. And what we know is the following, is that, so, the integral from zero to one to tau T multiplied by alpha T squared. So, for any alpha, beta, and gamma, this must be equal to zero, OK. And so, now, well, from this you find the system, but maybe it is not in the case to make all computations. So, from this you find the system taking say alpha equal beta equal zero and then gamma equal one or alpha equal gamma equal zero and beta equal one, or put this equal one and the other equal to zero. This gives system, the system I write it to you is the following, OK. So, I don't do it here, just write the solution, but maybe plus equal to three. This is obtained with alpha equal beta equal zero and gamma equal one. Then the other equation is 15 alpha plus 20 beta plus 13 gamma equal 11. 12 with now beta equal gamma equal zero and alpha equal one. And the other one, no, sorry, this, no, this is alpha equal gamma and this is beta equal one and then beta equal gamma equal zero and alpha equal one gives 10. This is the system it all obtains. Well, apparently it turns out that b c t cube is equal to, I hope that this is correct, no, sorry, minus four five t plus one over 20. But you don't agree with this four, right? Maybe it is three. Well, OK. So, now the main point of today is to prove, at least to start to prove the Ritz representation theorem. The Ritz representation theorem, I don't, now I just comment on it then we slowly go into the, so what is the meaning of the Ritz representation theorem? So, first of all we give a definition. So, let me denote by, so h is Hilbert, let me denote by this the set of all l from h into R say l linear and continuous. So, this is a set of functionals. So, it is a function, it is a linear, sorry, linear continuous functional on the Hilbert space. Now, given any l, so maybe a remark, so remark proposition say, so proposition, the following are equivalent, assertions are equivalent, l, so let maybe, let l from h to R be continuous, be linear, then the following assertions are equivalent, l is continuous at zero, l is say continuous, l is bounded, now I will define it and l is continuous at any point, l is continuous at some point, at some point h. So, given a linear operator, the following facts are equivalent, l is continuous and l is bounded, essentially. And continuity can be checked just only at one point or at some point, for instance at the origin, which is easier as usual. Now, what does it mean that l is bounded? So, definition, l is bounded, l linear is bounded, if there exists a constant, positive, such that l over h, less than or equal for any h in h. So, why the word bounded? Well, bounded because it means that if I take a ball, l over ball is bounded, this is clear, right? It takes bounded sets into bounded sets. So, it takes bounded sets of your illber space into bounded sets of your image space as the target space, which is just, for the moment, just the reals. Is it clear, the meaning of bounded? I mean, if you take a ball which is bounded, then it is contained in a ball. Therefore, this is less than some capital R. And therefore, this absolute value is less than c times r, so it is inside some big interval. Is it clear? So, this means bounded. So, the proposition says that once you have a linear map, it is equivalent to check continuity of your boundedness. It's equivalent. And you can check just only continuity at zero. So, maybe home you could prove. So, here we prove that one, here we prove this. Then some r rows are immediate. The two implies one. The two implies four, et cetera. So, it is enough that you show, say, for instance, that one implies two at home. But you just use the linear structure of your vector space. So, let us prove that one and three are equivalent. So, assume that L is continuous at zero. So, now, if this is true, this means that L of H is equivalent to say linear map from your real best space with the scalar values, bounded here is equivalent. So, you can, if you want, if you prefer, this can be replaced once I have proven that r rows, this can be replaced to bounded. Is equivalent. So, now, let us prove this. So, assume that L is continuous at zero. So, then, so, if you take, so, there exists, let me call it there exists delta. There exists a positive delta such that if you take the counter image, say, of the unity interval, a name for the, ah, maybe. Then, by continuity, we know that this is open. So, it is contained in some ball of radius delta. Centret at zero. Centret at zero. Because this is an open set, therefore, the map takes open set into open set and therefore we can find some delta. I mean, this is open in R. If by assumption L is continuous, assumption one, I am proving this. If L is continuous, then this is open counter in the preimage of an open set is an open set and therefore it contains some ball. Centret at zero. What does it mean? If any h in h with h norm of h less than delta, then we have that h belongs to this preimage. OK? So, this means that L of h belongs to this. And therefore, the absolute value of L of h is less than one. So, what I've shown is that there exists delta such that if this is less than delta, then L of h. So, let me write it here. So, I've shown that there exists delta such that h less than delta implies L of h less than one. And actually, this is enough because of the homogeneity of the fact that L is linear. So, I would like to prove, you see, I would like to prove this without any restriction. So, now take any h. Take now any h in h. And so, we have an inequality just only for those h satisfying this. But this does not. And now I force it to satisfy by dividing by the norm and multiplying by delta. So, now maybe I change letter. Take h like this. So, now define this as h. So, assume that this is non-zero. So, divided by its norm multiplied by delta. In this way we are almost in a position to apply this implication but not quite because this is equal to delta and not less than delta. So, I cannot really apply this implication to this choice of h, right? If I would have less than or equal, then I could. But I have the strict inequality for the moment. So, just now it is also clear that I can suppose that this is non-zero, right? Because if it is zero, I am already here. Now, it is sufficient now to take small epsilon and not less than delta. Well, it is not so... The problem is not so serious because I can take now a small epsilon positive and define instead the new h which is delta h, script h, divided by script h plus epsilon. In this way surely now we have that this, this now is less than delta. Okay? Hence, I can put this h here and therefore I know that this is less than one. Okay? So, l of h is less than one. But l of h is... But l of h is what? So, l of delta script h, divided by script h plus epsilon and I know that this must be less than one. Now, I use the... the homogeneity. I mean the fact that capital L is one homogenous so that this is what? This is this object here actually is equal to what? Delta is positive. One over script h plus epsilon l of h less than one. Okay? Bilinearity. Hmm? This says that l of h is less than h plus epsilon divided by... Now, this is true for any epsilon and therefore we deduce that l of h is less than or equal than h over delta. And therefore constant c that we require is just one over delta. Okay? So, the constant c exists and it is equal to one over delta. This is true for any h. Hmm? Do we agree? And c is one over delta. Okay? So, we have shown that one implies three. Namely continuity implies boundedness. Now, we show this is that boundedness implies continuity. Hmm? So, we know that there exists a constant. So, fix epsilon positive then and take say delta equal to epsilon over c then if h is less than delta it follows that l of h is less than epsilon. Okay? Because we have less than or equal than c times epsilon c. Merci. Yes, yes. It is true. It is liptious. Namely l of h1 minus h2 it is less than or equal which is the liptious continuity of h. For linear maps. So, please at home fix the missing points in the proof. Okay? Of the proposition. Okay, now so there is this interesting characterization once you have linearity then continuity is equivalent to taking bounded sets into bounded sets. This is very convenient. And now so we have to study a little bit the linear continuous maps functionals maybe. Just one comment here if l is bounded we know that there exists by definition a finite number such that that is true. So, this means essentially that l of h divided by h once h is non-zero is less than or equal to h non-zero. Now it is clear that which is the interesting constant here. Well, the interesting constant is the smallest c for which this is true. Because of course if it is true for c equal 100 then it is true for c equal 200. And so on. So, what it is interesting to find the smallest c for which this is true. The liptious constant. So, it is interesting to consider the smallest positive number such that l of h is less than or equal c h for any h. Well, this is denoted by this symbol here. Is this symbol. So, at home we should try to show the following. So, this is called the normal. This is called the normal. So, we have that so you could try to prove one of this equivalent way of defining the norm of a linear function. Ok. It is up to you. What you prefer, what you understand better. Any of these these expressions give you the length of the linear function and the norm of the linear function. Now, this is a norm. You can check it is a norm. Ok. And it is a norm and the space of I mean this is a norm and actually l of h is a linear space is a vector space. And this is a norm on that vector space. And this is a norm. Now, let me just remember what we know in finite dimension. Because now again because of the infinite dimension the notion of dual space is really something nontrivial. Let me remember, recall you just only what it is known just only what it is known in finite dimension. Ok. Now, I have not defined what is the dimension of the Hilbert space. But I can write without ambiguity this. This simply means that you have a finite dimensional vector space with the Hilbert norm. Ok. With the scalar product. With the linear product. So what it is known about lh. Well, lh is just what is lh. lh is just a copy of rn for some n endowed with the scalar product of Riemannian type. Right? So what is h here? h is just somewhere isomorphic to somewhere n on which we put an ellipsoid. A fixed ellipsoid. So we have a special Riemannian flat. Riemannian manifold. The ellipsoid does not change with the point. So it is very special Riemannian because the scalar product does not move with the tangent space. It is just always the same. You have always the same ellipsoid on the tangent space. So very special, easy Riemannian. So together with an inner product or equivalently an ellipsoid the unit ball of the norm. Ok. So you already know what is Riemannian manifold at least in the embedded case. Well you have a surface on a surface you have your tangent spaces since it is embedded your tangent spaces you can see them are inside the space and then on each tangent space you have an ellipsoid and the ellipsoid changes if you change the tangent space because the ellipsoid is the scalar product and the scalar product may depend on the point. In this case it does not depend just one fixed ellipsoid. Is it clear? In the connection. I mean you are working just one chart just one chart this is geometry in one chart without any I mean just one ellipsoid. Ok. So what is this? Is it the vector space? Yes. Which is the dimension? N. So this is isomorphic to Rn and therefore these two are isomorphic. They can be identified. So once you have finite dimension by the way of course L of H you don't even need to write continuous or bounded because any linear map is automatically continuous in finite dimension. So this is just the space of linear maps. Automatically they are continuous all bounded is equivalent. And also this is isomorphic to H itself and therefore if I do it twice L of L of H again I find something which is isomorphic. Is it clear? This is the picture. So you don't need to require continuity and this is nothing else another copy of Rn. The norm is not the same. If you are starting if you are starting from ok, if you start from around sphere Euclidean exactly then that is exactly the Euclidean. But if you start from an ellipsoid then this is the norm here is not the same ellipsoid. It's just modification. But anyway still we can also write the basis if you have a basis here then we also write explicitly another basis here. You know this is the analogy that we have between if you want vectors and covectors in differential geometry. One vectors and one covectors just the same. Ok, so in finite dimension at the end when we perform this operation of taking linear continuous functions we don't find any new space. We always have the same copy of the same n dimensional vector space. Well the point is that in infinite dimension this is not true anymore unfortunately and this probably you already know. I don't know if you know. But standard examples are not in Hilbert setting but in Banach setting it is well known that if you start from a Banach space in infinite dimension then you take the dual and you take the dual of the dual and you don't come back to your initial space. Unfortunately L1, L infinity dual of L infinity is not L1. We will come slowly to this point. Just to let you know that now we are just generalizing the notion of dual space. Sometimes denoted also by well maybe I am sometimes this but also sometimes also this. Now I don't exactly remember I don't know exactly. I don't know to remember but the look of Brezzis maybe is h star. We can adopt maybe h star also h prime is also very common as a notation. So remember that this is all why this distinction? Well because in principle this is called the topological dual. Topological dual meaning linear continuous functions. But if you just require linear functions then you have another notion of dual much larger which is called algebraic dual. And that could be denoted with another symbol. So anyway we will never consider algebraic dual here. We are interested in topological dual so we require continuity. Ok. Ok. Now there is theory. Example so it is interesting to remark that there are in infinite dimensions linear discontinuous maps there are. It is not true that any linear map is continuous. Ok. We will see examples but let us see first examples of linear continuous map. So take fix once for all an element, non-zero element of your Hilbert. And then you can consider the following map for any h. Check at home Ok. And the Ritz theorem is concerned with the converse statement. The interesting point of the Ritz theorem that now I will try to explain is exactly the converse. It says that all linear maps are of that form. This is much more surprising. Theorem now I state the theorem. So the example says these are all linear maps for any h not this is a linear map linear continuous map. So it is an element of the topological dual. Now theorem so let h be a Hilbert space and let l be linear continuous functional on h then there exists a uniquely let me denote it by h not such that l of h or over this is called the Ritz theorem so it says exactly the converse statement this statement says that these lh not are bounded linear function this is the opposite take any bounded linear function then it is of the form it is of the form l of h not for some h not actually for just one h not this is much more much more unclear in some sense right this is the content of the Ritz theorem so the consequence of this one consequence of this so one consequence of this is that we have a map from this we can find a map taking l to h not so it takes any point in the dual and associate uniquely another point in the Hilbert space not only this so let not only this but this is linear this map taking this into this is linear and also this is an isometry because it preserves the norm if this has some norm in the sorts then the image has the same norm so the isometry between two matrix is an isometry between two matrix spaces one h star and the other is h so this is a consequence of this now now let me let me make some comment in finite dimension a picture of what does this mean in finite dimension at least we try to see what does this mean finite dimension again this is well defined so what is a linear functional and on zero say on trivial linear function on Rn how can I identify it how can I think about a linear function ok this is some what do you think when I tell you ok this is a linear function from Rn to R what do you think yeah I identified with one vector so one thing would be ok and L so L from h to R is just linear in particular continuous well if I know the kernel of L so this is a linear map from R2 to R so if I know the graph is just a plane is a plane because it is linear ok so now if I know the kernel I don't really know exactly the linear map but almost in some sense because if I know the kernel so the kernel is a subspace of of h say of codimension 1 so I have in the source space in Rn I have a subspace of codimension 1 say for instance a hyperplane this does not identify L but almost I mean it is enough to know the value of L just in one point outside so because it is clear that the same kernel can be shared by many linear functions but once I know the value at one point outside since this is codimension 1 I know this is dimension 1 they are outside and they know L everything so so this is the graph of my linear map I know I know the kernel and I know the value in one point and then I know the whole linear map so if you want now let me so this is the graph now assume I am in h and this is the kernel this is h this is the origin and now I have also the unit ball some strange unit ball this is the unit ball of h of the Hilbert space then what I can do is to consider the translation to this the picture is not perfect let me do it in another way so this is my kernel this is my kernel this is my unit ball then what can I do what I can do is this now that I have an infinite dimension I have just one intersection here this is the origin so this is the kernel just translated so this is the unit ball this is the kernel just translated of the proper quantity and then I mean what I can do is to identify to associate with this kernel and then there is the problem of normalization this point here you see to this linear subspace hyperplane I have the unit ball and then I can associate this point here this is close to the 2h0 this is all essentially so given your L this is up to a constant factor essentially it is your so the this point here is of course a point of h so what I am saying is that I am associating to this vector subspace, the kernel of L I am associating this vector essentially it is somehow in the Euclidean case it is clear that they are orthogonal just now I have to say indeed one point is to say h0 but geometrically one of the idea is that you fix a hyperplane and then you can associate to this hyperplane the intersection of a parallel of it or minus it also with the boundary of the unit ball with that parallel translation and this unique intersection essentially is your h0 so this is a way maybe to associate to a one co-dimensional space subspace a vector like this of course this is I mean there are various things to understand one we are in infinite dimension so it is not clear what is the boundary of the unit ball we don't know we have never defined the tangent space to the boundary of the unit ball very difficult we cannot not only that but again I think that it is true that in infinite dimension the kernel is one co-dimensional so infinite infinity minus one but let us go into the proof of the the Ritz representation so we define m to be the kernel of n this is a definition so proof define the kernel of n then there are two cases well if m is h if it covers all your space then l is identically zero everything is in the kernel so l is identically zero and therefore we can simply take h0 equals zero and it satisfies all assumptions because this is zero this is zero equal zero is okay and you is the unique object okay so we can assume that m is properly contained in h okay hence we can take a point outside let me denote it by take a point outside g0 so g0 is outside of m yes thank you because I take the projection soon yes so maybe I remark it here is enough okay remark now we will use that m must be closed why? m is a closed subspace h well it is a subspace because it is a kernel it is closed because l is by assumption is continuous okay so the kernel m is the counter image of a closed point therefore it is closed and continuous therefore m is closed okay and this is exactly here the point where I use it because now I have a point outside I can project it uniquely on m and let me denote it by pm so I have a unique projection so take so this is pm g0 so this point here is uniquely defined by what we have seen at the beginning because this is a closed linear subspace therefore we can consider and actually also we know that of course p m g0 is different from g0 because g0 is outside the kernel and this is on the kernel because this is closed therefore we can take now the difference g0 minus pm of g0 and we can divide it by the norm and let me denote it this object here as f0 this is well defined because the denominator is non-zero and this is in the unit ball of course so what I am doing is the following I take this difference and I normalize it so now if I imagine that this is the origin for the moment then I have my unit ball ok so let me so this is too large so this is let me denote it just by x ok this is also x and this is also x then I have my unit ball and therefore so this is f0 so I am assuming that this is just the origin for simplicity ok so that is f0 now the claim is that h0 we are looking for h0 unfortunately I have raised the black board but we look so give an m we have an m this is m the kernel we are looking for some h0 and we will take at the end h0 parallel to this f0 as I was trying to explain 10 minutes ago ok so now let me try to show you so at the end h0 we look for h0 parallel to f0 so for the moment we have never, we have not used so we have maybe understood which is the direction of of h0 but still we don't know how long it must be the norm of it because for the moment we have just used on capital L we have just used the information on the kernel but capital L is not identified only by the kernel it is identified by kernel and something outside so for the moment we have just understood maybe this is the direction where I have to find h0 now I have to find how long is this vector and now I have to use on capital L that I have not only its kernel ok so f0 is 1 this is written here because this is a unit bowl now and therefore ok well it is clear that the following are obvious I mean f0 is non-zero obvious and f0 does not belong to m because f0 is non-zero so f0 is non-zero so this point is outside this so f0 is non-zero and then what else well and f0 is let me say is orthogonal to m so we know that f0 let me introduce ok f0 against m is equal to 0 this is these are the properties that we know on f0 ok this is because remember f0 is a projection is along I mean is g0 minus its projection therefore it is orthogonal to m ok so now take any h and consider the following quantity h minus now lambda I have to find it let us consider the following quantity h minus lambda f0 and I want this to be in m so if I want this to be in m then this must be equal to 0 and therefore lambda now so I can take lambda as equal to l of h divided by l of f0 I can do this because l of f0 is non-zero so this I can do summarizing h define this quantity with lambda equal to this ok so if we define lambda equal to this then this is equal to 0 therefore h minus l of h divided by l of f0 belongs to m right so given any h I just there is something missing here sorry given any h I subtract I subtract from h is normal part I would say so I subtract from h it is normal it is orthogonal part to m because this is orthogonal so this is orthogonal to m I subtract it is projection on the orthogonal line so what remains is tangent I mean is in the kernel and therefore once we know that is in the kernel we know that this must be orthogonal this is an element m so must be orthogonal to f0 and so we write down now this so the idea is now we have identified a direction say take any h take the tangential component of h along this I mean subtract to h it is normal component so it remains a point in this capital M and therefore it is orthogonal to f0 ok so what does it mean it means that scalar product so given any h any h in h the scalar product of this object with f0 must be 0 that is h minus L of h over L of f0 times f0 comma f0 this must be equal to 0 is it ok up to now is it clear it's ok clear and hence I think that we are done more or less because we have that now so h comma f0 must be equal to by the way f0 has norm 1 so L of h divided by L of f0 do we agree because we are using that we have this normalization ok and now we are done because you see we have understood who is h0 h0 is just a multiple of f0 because it follows that L of h equal to h h0 where h0 is by definition do we agree so the idea is completely geometric apart from the fact that we have to find this quantity here which is due to the fact that we cannot know all the linear map knowing only the kernel but at least if we know the kernel we understand which is the direction of h0 and it is this orthogonal to this and this is the identification metric because if I change the unit ball I change h0 also infinite dimension if I want to do this infinite dimension even in R2 if I change my scalar product I mean I rotate a little bit the ellipse then there is another h0 apart from the factor the scaling factor L of f0 I am identifying this vector one codimensional vector space comma a value of L outside with this direction with the proper length so this is the duality map between the dual and h and this is the it is never easy to really manage with this duality mapping even infinite dimension so check that the norm so check by a term that L of h that the norm of this we know so the norm of h0 is L of f0 and check that this is also equal to norm of L now uniqueness this gives us existence so uniqueness of h0 well by construction some sense but assume that there is another one so h0 prime assume that we have two of them making so we know that L of h is equal to h0 but also h0 prime for any h in h so this means that h0 minus h0 prime is equal to 0 for any h therefore it is enough to test this with h equal to this difference this is true for any h right ok so if I take inside this I take h equal exactly this difference I find that the norm of this difference must be equal to 0 and therefore I find that h is equal to h0 sorry h0 sorry h0 minus h0 prime ok ok, so for today is enough finally from the next time we will start with Banach spaces I think that now you have sufficient experience on infinite dimensional vector spaces with an inner product therefore and this is a deep theorem by the way even if the proof is not difficult it is just a consequence of the projection theorem once you know how to project on a closed convex set then you have automatically this theorem so ritz I mean if you want to look from the outside now you see that ritz is just a consequence of the projection on convex set so really the projection is something important but this has a lot of consequences also from the conceptual point of view which are never easy to realize in any case we have the experience of some infinite dimensional Hilbert space so starting from tomorrow Banach we go quickly to Banach and maybe we will skip some proof that you will find in the Brezis book because I mean we have not so much time at our disposal and I would like to come also to Fourier analysis and prove everything that I will say tomorrow