 Čutih, počutim, zelo naredamo vse teorema, z počutim, na konvek sez, v Hilberti spasih. Tegorem je to, da se vse nekaj, da se nekaj, da se nekaj, da se nekaj, da se nekaj, da se nekaj. Zato, kaj da bom razližila, da s比pusti preč, bilem blog P-k Open Space. Zda nekakvo PS, nekako H, nekakvo B space, in da, možem da, v sen vene, bilim po neko V, nekako P-komen ljudi v početnih s evosti gled endorse, Resakč, that closed here means sequenšal底 closed with respect to the strong norm so with respect to the norm induced by the distance. For when I don't write anything I mean, this is close with respect to the distance in h. for the long or Nas Durham dish. Pranje in š �čenje There exists a unique ... Nud Organization z štočef... ... Devil, that we denote by x, though we satisfy ... That distance between H and X Tudi, je zelo vsega odgovoriljna vsega vzgovijaj. Zelo da sem zelo komentato, če me zelo, kaj me ne patila. V sezhu odgovorile vsega h in zezir, je vsega h minus, zelo je, da je vršen začinjen, če je vršen, bo je nekozvršen, bo je to vršen. Zato bo je, da je vršen začinjen, če je vršen. Kaj je tudi? Kaj je vršen iz pomeče? Zelo je, da je vršen iz pomeče in vršen sub pomeče, Weekly space while a set sub set in the space is just the smallest possible, smallest in the sense the infimum of the distance. Which is nothing else, of course. Kaj je to teorem, zelo je naša izgleda, da se je vse konveč in tudi je in za vse, da je vse za vse, da je vse za vse, da je vse za vse, da je za vse za vse, da je za vse za vse. Zato je počak h, da h je vsega in vsega. Vsreč je, da h je vsega. Zato počak h. Zato počak h je vsega. Zato počak h je vsega. T wouldn't say that there is just only one, probably in this picture, is this X, which minimizes all possible distances. So the length of this segment is less than the length of all other segments. The idea is that if we could talk about tangent but we cannot. For the moment we cannot talk about tangent but the idea is that. Ok, infinite dimension this is 90 degree and this is the point which minimizes the distance. Ok, we cannot do this in infinite dimension. taj počet, pa ideja je počet. To je počet. Kaj je ideja nekaj? To je ideja. Počet način jaz počet. To je... To je... Nekaj ideja je počet. Taj počet. Počet način jaz počet počet način jaz počet način jaz počet počet način jaz počet taj počet način. To je... To je interseksja. Taj počet način. In taj počet način posez. There is just one intersection Why? Morality, to je nepečit. Just the idea, this is not the proof. Morality, why there is only one intersection? When there are no... When things can go wrong. Yes, of course, you have more. Well, the first is the point of minimal distance, no? The idea is, so you have two convex sets. One is the ball, the other is the sea. So you enlarge the ball until you touch the first time. And the claim is that you touch the first time ki je nekaj počet, nekaj počet, nekaj počet, in nekaj počet in infinitivno počet. Kaj je, da je počet, da je v finaj dimenzij, pa v istvu. Zelo, zelo je konveč, pa v istvu je to. To je konveč, je počet, je počet, je počet. Zelo, ki jaz sem v spasih, kaj je to unit, Kam bom se ona da je problem. Spodobno je izgovorila, katero se bolj jem je zelo po napravljenj snah, njen zelo po napravljenj, katero je ide dvler na bolj, kaj drugi vzlušal, ki so očilo imelje in refleti. N disappoint posledajte, da bolj je�� nesvetnim konvečnjem. but the definition of uniformity convex. And therefore the idea, okay... This ball is uniformity convex, is more or less like this, and therefore, even if s 나도 uniformity convex, but the ball is... So this is something behind, okay, so and you see this picture is very quickly done. Zelo, da je bilo izgledaj na bolj, kaj je, da je tudi na spračenost. To je neuklidian. To je nekaj počist, tako tudi je nekaj 90 gradov, začne, da je to nekaj počist, je počist. Počist je to, da je nekaj počist. Zobčite razvorega. Zdaj, se, ki H je v C, ne zelo včasne, naredi je zelo včasne včasne. Včasnje zelo včasne je mene delvega z konstitutnih zeljic, in je zelo alesem zelo včasne. Kako je zelo naredi, zelo včasne. je lahko, da gre zelo, ki zelo ta je tle. B Cinh se zato, da je tle. Bath. Rastče se je to, da vólj na zelo vzelo na tle. Bath. Bath. Čae je nrno. Kaj je prišel? Vač je tle. Ča je? we have zero equal to zero. Not only this, but H is unique. X is unique because if the distance is zero we know from the property of the distance, then there is just only one point having zero distance, exactly that point. So this is the unique solution of this minimum problem. So by the way, notice that you have to solve a minimum problem in infinite dimension. Se je probrnje vnitrat vsdo varješnji. Navelj cambiamo, da se dokazujemo vso mirany. Moj dr overlač s včeljih. V neko danas, da se te minimuač naredilo, tako klasik это vnitrat vsdo varješnji. Na početku z radinii, pros Charlie je x ak to h, zelo da je zaštej. sreči, da h je nekaj v zivu. Zazaj da načaj... Zame načaj... Zazam, da je tudi standard metod i način kakoliv nič, da njemo vse, da tudi načaj vse. Zelo je. A načaj vse, da je tudi način? Tudi je to tudi, da je tudi vse. Tudi je to tudi. Tukaj, to je infimum, če je infimum z vsem H minus C, kaj je v C. Ok. Then we can pick a minimizing sequence, which is the following, sequence of points, let me denote it by C... C n. Let me denote it by C n, sequence of points of c, such that the limit as n goes to plus infinity of the C n h C n, n is equal to the infimum. This is by definition of the infimum simply. This is an infimum of this function. You can take a sequence of point here converging to the infimum, such that this h minus cn converges to the infimum. This you can always do. OK? Sorry for the notation. You have to be careful. Capital C, this is capital C. And this is small c. I will try to write symbols. However, these are points in c. So maybe I am using the letter c for this reason. So I will try to write cn very small. OK? So this always exists. There is a minimizing sequence. So now let me give you two proofs. One proof, first proof, we use the parallelogram identity. So from now on, therefore, this proof is adapted to the Hilbert setting, right, because we know that this characterizes Hilbert norms. So this proof cannot be, if we want to do the proof using the parallelogram identity, this proof cannot work in Banach. So we use the parallelogram identity. So remember, so our parallelogram identity is the following. So we have a, b, and therefore r's. One half a squared plus b squared must be equal. This plus this is just this plus this. So this is our parallelogram identity. And we try this with the following choice. h minus, say, cn, and b equals, say, h minus cm, with two different indices. The idea is that, which is the idea. We want to show that this sequence is Cauchy. Therefore, if you want to show that this sequence is Cauchy, we want to control the difference cn minus cm. And so, to control the difference, you see, if I take this minus this, I have the difference. And then there is something else. So I try this. So because we want, because we want to control cn minus cn. Hope, the hope is that cn is Cauchy. This is the idea. So let me apply that, so let me write it exactly like it is. So we have, say, I prefer maybe to multiply everything by one half, so that this is one four. And so we have one half. Now, h minus cn square plus h minus cm square. Equal, one four. And then we have the sum, which is 2h minus cn minus cm square. Plus, now the difference is what we want to control. So cn minus cn square. So let me write the right hand side as follows. This is equal to h minus cn plus cm over 2 square plus one four cn minus cm square. Because I put the four inside the norm, but the norm is squared, so this is actually a two. Two cancels with these two, and the two remains just only here. And therefore, so we have, so let me write this as follows, one four. What I want control, meaning control, meaning that I would like that this goes to zero. So this is equal to one half h minus cn minus one half h minus cm square plus h minus cn minus cm square. The opposite, yes, sorry, plus, plus and minus. Now, this is a minimizing sequence. So this converges by definition to one half the distance, because it is a minimizing sequence. So h minus cn, as m goes to infinity converges to the one half the distance between h and c. This is as n goes to infinity. And this for the same reason for as m goes to infinity converges to one half the distance c. Now, we observe that c is convex. And therefore, this cn plus cm divided by two is just a combination, convex combination of two points of c. So cn, cm both belongs to c, which is convex and this belongs to c. This belongs to c. This implies that this is the distance between h and one point of c. So this is h minus cn plus cm over two is the distance between one point and one point of c. So this is surely larger than the distance. It is clear this, is it clear? Therefore, when I put the minus, I find that this object here, so let me go here. So this is surely less than or equal than one half h minus cn square plus one half h minus cm square minus distance square. Well, then this goes to this. This goes to this. Therefore, when n and m is sufficiently large, this goes to zero. And this shows that this goes to zero. Sorry, there is a square here. This square, sorry, here there is a square. So one half, one half minus one, this goes to zero. Is it clear? Is it okay? So cn is Cauchy. So we have used convexity of c here. So we have for the moment used that c is non-empty and convex. Still we have to use that it is closed. Now, cn is Cauchy, h is Hilbert. Therefore, cn converges to some point x in h. Now we are done. Because cn converges to x, the distance is continuous with respect to the strong convergence. So this limit is also h minus. This limit is then also equal to the distance. And therefore the distance from h to one point is equal to the infimum. So we have found a solution to our minimum problem. Is it okay? Uniqueness not yet, but for the moment, okay? Well, now, so these are equal, and this shows existence, okay? Because x is in c, right? C is closed, x is in c, and we have a solution, uniqueness for the moment, okay? So we have used closure, so non-empty. Convexity when we want to show that convex combination is an element of c and closure to know that this strong limit belongs to c, because c is strongly closed, okay? So that we have a solution. Now, uniqueness, so maybe there is a remark here. This minimum problem, no. The minimum problem dhx equal dhc is equivalent to this, to this. This is maybe home. The point is that this square from the outside of the infimum can go inside the infimum. Namely, a solution to this is a solution to this and converse, okay? Now, this function here has one property, it is strictly convex. Strictly convex on the convex set c, therefore it has a unique minimizer. We will prove uniqueness also in other ways. But this maybe is one. By strict convexity, I mean that, well, strict convexity, this is in a convex set. What is strict convexity? This is a scalar value function. So it is strictly convex. It is known what does it mean. Okay. Now, let me give you, so strictly convex. Let me give you now another proof of existence, which is not really completely self-contained. So another proof, so another proof is the following. So take again cn, let me still denote it by cn, minimizing sequence. This proof is probably more instructive than the previous one. The previous one is just magic because of the parallelogram identity. But as soon as you don't remember that there is the parallelogram identity, or as soon as you don't want to use the parallelogram identity, then you don't know what to do. I mean, that proof is perfect, but adapted to the Hilbert case. Now, minimizing sequence. And then we know that minimizing sequence means that the limit of h minus cn is equal to the infimum of h minus c in c, which is finite. Okay. This is what we have for the moment. Now, since this is finite, this means that we can assume that for any n, for any n there is a k such that for any n. Okay. So this is sequence of bounded numbers. This we can assume. Now, we also have that cn is less than or equal than h minus cn plus h by the triangular property of the distance. Okay. And therefore, also this is uniformly bounded. Because this is uniformly bounded and this is just a given number, independent of n. Okay. Therefore, there is another constant such that this is true. Now, this says that our sequence, our minimizing sequence from this, from this bound here is a bounded sequence. Okay. So the sequence cn is bounded, strongly bounded, of course. Now, you remember something that we have proven here last time, which says that bounded sets are weakly sequentially compact. Okay. In Hilbert. But this property is true in higher generality than this. But at least in Hilbert we know we have proven in L2. Okay, sorry. We have proven this in L2. So now assume that h is equal to L2, but this proof is much more general. So since this compact, weak compactness property we have proven just only in L2, we assume that h is equal to 2. So g is bounded therefore. So there exists a subsequence of cn subsequence. Sorry. There exists a subsequence in the sequence so that cnk converges. And there exists a point. And there exists a point. Let me call it x into h equal to L2, unfortunately, because of our such that cnk converges weakly to x. Now, the problem is that now there are two problems. First of all, one problem is the following. What about h minus x? So I claim that h minus x is less than or equal than the limit for sk of. So this is the following statement, more general statement. You have a bounded sequence, weakly converging to something. Then the norm is lower semi-continuous, but not continuous, just lower semi-continuous. Because your convergence is not strong, is weak. Therefore, you cannot pretend to have strong convergence. You cannot pretend to have strong convergence of the norms of the difference of the strong convergence. But at least the norm is weakly lower semi-continuous. So how to show this? Well, you take the scalar product h minus x, h minus cnk, h minus s, h minus cnk. And then you know that this is less than or equal than h minus x, by the Cauchschwarz inequality cnk. Then you know this. However, cnk is weakly converging to x. So by the weak convergence implies that this converges to h minus x, h minus x. So this converges, since the weak convergence is enough to say that this converges to h minus x, h minus x. Because this against this, so this is an element of your space, and this converges against all elements of your space. Remember, the definition of weak convergence says not that the components, but more, not only all components, but all projections on any vector of the space. Therefore, this converges to h minus x square. So the limit of this, therefore the limit of the four with what we find, we find that the limit of this is equal to this, so h minus x square less than or equal than, here there is the limit, because we don't know also the limit. So h minus x limit as k goes to plus infinity h minus cnk. Before, now h and x are different, and this implies that this implies this. So we have observed also that this converges so we have observed also that there is weak lower semi-continue of the norm, and so for the moment we are in a good shape. Now the point is unfortunately that we would like to say that x is in c, because this is just a weak limit, x is a weak limit. C is strongly closed, so it's not clear, because if you take a sequence converges strongly, then the limit sequence of elements of c converges strongly, then the limit is in c, but the sequence converges weakly is not clear. Well, this is a result that we will prove, so we have already a self-contained proof, the parallelogram proof, so that is a proof. Now I say that this is true. So general fact that for the moment we don't prove c convex in Hilbert, then c weakly, sequentially closed, weakly closed, if and only if c strongly closed. It is clear that if it is weakly closed, then it is strongly closed, because the sequence converges strongly converges also weakly. But the converse is not trivial. So this is easy, the more difficult, much more difficult, but this is a result. So if you accept this result, then you have also that c is in c, x is in c. So this proof is not self-contained because of this point here. Now, how we can conclude? Well, now we have that h minus x by just the lower semi-continuity and not the continuity is less than or equal than the limit over k of h minus c and k. But this is a subsequence of cn and the whole cn, so this is also the limit along the whole sequence, x minus cn, right? Because this is just a subsequence, this was already converging. So this was converges to the h capital c. And this is enough because this shows that we have shown the following, is this clear? So this proof is maybe more general, but for the moment it is not self-contained because of the result of strongly and weakly close. Okay. So now let us go back to our geometric intuition. So remember, so the proof is clear to everybody. The proof is okay, it's not okay, but the first proof is okay. The first proof is okay. So remember, we are in the following situation, this is h and this is our point of minimal distance. Why this? Because I am taking a strange unit ball. Maybe let me take this isotropic unit ball. But you have to think about a strange projection point. I mean you don't know how is the ball. So let us do a completely Euclidean picture, which is not the case, but just to understand. So this is x, this is h. Now there is the following result. We would like to say essentially that is reasonable to say that now, let me take a point out inside c, let me call it small c. Now I can consider the scalar product between h minus x and c minus x. So I can consider the scalar product between this vector and this vector. So h minus x and c minus x. Well, in R2, with this picture it is clear that we have a sign, there is a sign that this scalar product has a sign. Well, this is true also in infinite dimensions. So the theorem says under the previous assumptions, if x is a solution, so under the previous assumptions, the solution x of the minimum problem satisfies the following inequality. So this is, so it is inequality, why there is an inequality? Well, there is an inequality because this object c is not flat, it is not a hyperplane, it is just convex. So even in finite dimension you cannot pretend that points of c have zero scalar product. Even in finite dimension you cannot pretend to have a quality here. You are just saying that this has scalar product non-positive. You cannot say more than this. You could say it if c was a half space, but this is just convex. So you cannot pretend so you cannot have more than the inequality. Right? This is the so-called Euler Lagrange equation of the minimum problem. So when you have a minimum problem, you solve it, you find the solution and then the minimizer usually satisfies, not d e, p d e, something like that. This is the analog, usually in physics called Euler Lagrange equation. If you have studied mechanics, you know that if you have a Lagrangian, you minimize it, then you find a system of these Lagrange equations. This is the analog. So we have conversely also, maybe. Conversely, if x in c satisfies this, let me call star, this family of inequalities, then x is a minimizer. So here there is something more, because usually in calculus of variations, what do you do is the following. You have a functional, you minimize it, you find the solution and then if you are lucky, the solution satisfies a p d, or not d, which is the Euler Lagrange. That is, a minimizer is a stationary point, in other words. This is obvious. If everything goes well, it is obvious, even in one dimension, right? A minimizer is a stationary point. Namely, a minimizer, here you have zero gradient. In final dimension, ne? Here it is written something else, it is written. If you have a stationary point, then this is a minimizer. If x satisfies this, then it is a minimizer. This is completely different stuff. It is not true in this case. So why one can hope that this statement is true? Because of the convexity. Actually, you don't have this picture, but you have a picture like this. If you have a stationary point, strictly convex, stationary point means minimizer, unique. This is the principle behind this theorem and these kind of things. Convexity here is really a very strong assumption. If you now remember, if you have seen at least once how you pass from being a minimizer to writing the Euler Lagrange equation, then you make a variation. You make the so-called first variation. You choose a direction in which you do a variation. And then we have to choose the direction to vary our. So we take a number that we denoted by t, t a number between 0 and 1. And then we take, so h is given. We have h in h, outside, say. And then we have x minimizer. And then we consider the following competitor. Let me denote it by v, 1 minus t h, 1 minus t, 1 minus t times x, x plus t c. So if you want, if you are confused by this, maybe it is convenient to think, since everything actually is invariant for translations, for the moment you can think that x is the origin, if you want. Just think that x is the origin. So you have to show that the scalar, so if x is the origin, then you have to show that the scalar product of h against c is less than or equal to 0. Thinking of x as the origin. Now, what do you know? Now you are here, I mean you have, now x is the origin. And then you take a variation in the sense that you compare, you compare the value of your functional with the value that you have if you move along this direction. But you are convex. So you can move in this direction only with t positive and not with t negative. Because if you take t negative, then you go outside and you cannot compare. So I have denoted this by v, right? So what does it mean comparing now? We know that h minus x is the minimum possible distances of the form between h minus c, right? So, ok, this is an element of c, right? Because I am taking a convex combination of two elements of c. Actually, so this is v as a competitor, say? And therefore now I know that this is the minimizer. v is a competitor, is a good competitor, so I know this. Competitor belonging to c. This minimizes the distance between h and all points of c. This is one of the points of c, therefore this is less than or equal to. Ok? Maybe it is better if I square it, because then I... Let me square it, ok? Ok, so this is equal to h minus 1 minus tx minus tc, square. H minus x plus t. So h minus x plus tx minus... Correct? H minus x plus tx minus c. So this is less than or equal than this. Ok? Now I expand this square. I have the scalar product I can do. So this is h minus x square plus t square x minus c square. And then finally there is what I want to show to be non-positive. So plus 2 t scalar product h minus x. The sign is correct. Yes. X minus c. Ok? So now I can... So this is less than or equal than this. Now I can cancel this with this. And therefore I find that 2 t h minus x x minus c is less than or equal than t square x minus c square. Now t is positive. And therefore I can divide by t. Sorry. And there is plus. That is not equal than 0. Larger than or equal than 0. Larger than or equal than 0, ok? Now t is positive. And therefore I can divide everything by t. Ok? And then I find that t x minus... So I find that t x minus c. Larger than or equal than 2 h minus x c minus x. Now, because this goes on the other side of the equation. Now t is positive. It is sufficient. This is true for any t positive between 0 and 1. Let t go to 0. Let t go to 0. And you end up that this necessarily must be less than or equal than 0. Ok? So sending t to 0 plus. This implies that c minus x is less than 0. And this is our Euler Lagrange inequality. So you have to be very careful here because what you have done is to compare this distance h minus x with the distance of... You take any point in this half line, which is not a line, it's just half line because t is positive. T is in between 0 and 1. But if you fix any c, then you move a little bit inside. So between 0 and c, essentially, if x is the origin. And then you compare the distance... You compare the distance... This distance, you know that it is smaller than the distance between h and 1 of this point. But your variation is unilateral. It's only from this side and not from this side. And therefore, you cannot hope to have an equality but just an inequality. If this would be linear, then the situation is different. But then we will go to this. So there is still something missing here because there is a claim also that if a solution of the inequality is a minimizer. This is more difficult. So to take a solution... So let x be a solution of that. So let x solve star. We have to show the following that this is a minimizer. So we take the difference here. We have to show this. Actually, it's more convenient to show again with the squares. It is more convenient. So let us do this. And we expand this. The idea is we have to show this. We expand this at some moment. If you are lucky, we will use also this after the expansion in the scalar products. So this is what? h2 plus x2 minus 2 hx minus h2 minus c2 plus 2 hc. Therefore, we can delete h2. So we end up with x2 minus c2 minus 2 hx plus 2 hc. Now we want that maybe we can write a equal x2 minus c2 minus 2 hc minus x2 hc minus x. Now we add and subtract here. Agree? x minus c. Yes, yes, yes, x minus c. My prefer c minus x, so I put the plus here. Now we have to use this. This is not present here, so we add and subtract. x here. So this is equal to x square minus c square. Then what? Plus 2 h minus x c minus x plus 2. So minus x plus x2 x c minus x. Hoping, is it correct? x square. So let me, let me expand this now. So, okay, yes, yes, probably is correct, yes, because then I expand this. So this is just equal to minus 2 x square plus 2 x c. C, yes, is correct, because now this everything becomes this minus 2. It's this and so it remains minus x square. Then minus c square. Then there is the object that I want to estimate. C minus x, and then there is this plus 2 x c. Okay, okay. So our assumption is that this is less than or equal than zero. So I hope that the computations are correct. I go here, it is more safe. So at the end we find that the different that we want to. So this difference is actually equal to what we have. This is by assumption is known. This is less than or equal than zero by assumption. And then there is a double product. Actually, I mean, there is a perfect square because this is a minus x minus c square, which is also less than or equal than zero. And therefore everything is less than or equal than zero. Okay. So we have shown that if a solution of the Euler Lagrangian, solution of the Euler Lagrangian equation is actually minimizer, due to complexity. Now we can give another proof of uniqueness of minimizers. If you don't like the proof that we have made before, now we can do another proof of uniqueness of minimizers. So let me erase everything, so I can erase. Therefore, now the initial problem, the original minimum problem, so x is unique, minimum problem has a unique minimizer. Unique minimizer is unique solution. So, well, assume that you have, assume that let me call x1, x2, maybe, yes, x1, x2, 2, so 2, minimum, 2 solutions, 2 minimizers, 2 solutions. Then they must solve the Euler Lagrangian equality. The statement, this, the original minimum problem, remember we have written that the statement was c non-empty closed complex, closed, there exist unique x in c, such that x minus h is equal in femon h minus x, maybe. This was the claim, given a point h there is a unique projection point nearest to c. This means that this is a minimum, of course. This was the claim, we have shown existence, we have said that it is unique by strict convexity. Now we give another proof of uniqueness of this x. Is it clear the statement? Problem, original problem, sorry. This is the original problem, this is the original problem, minimum problem. So, this infimum has a solution that is this inf is a mean. We have shown that this if is a mean and now we show that x is unique. So, now assume that you have 2 minimizers, two solutions of these two points in c. Namely, we have that x1 is in c and it satisfies h minus x1 scalar product c minus x1 less than or equal than 0 for any c. So, since it is a minimizer, we know it satisfies the Euler Lagrangian equality. Now x2 also is a minimizer by the contradictory assumption and therefore now we have also this for any c in c. So, we have this, is it okay? So, we have at our disposal now two inequalities and we want to find that at the end these are equal. So, what do we do? You see we can test, we have the freedom of testing this for any c but we know that x2 is in c. So, we can put x2 here and similarly x1 here because x1 is in c. So, this we can do. Okay? This must be true for any. So, in particular true for one of them. So, and therefore in particular we have h minus x1 x2 minus x1 less than or equal than 0 and also h minus x2 x1 minus x2 less than or equal than 0. Now, we add the two inequalities so that we find what we are adding the two inequalities we have. So, x2 minus x1, scalar product this actually minus this, right? So, h minus x1 minus x2 minus x1. Is it okay? Do you agree? No, you don't agree? Okay, so this is just the norm square and therefore this is necessarily equal to 0 but we know that this is a norm and therefore x1 is equal to x2. This implies so this is so you have now a complete proof in various ways of the result. Now, this projection on convex set is very, very useful in general in functional analysis. So, please remember and also is the base to show the Ritz isometry. But we will go through this but first let me say something on this. Maybe this is a coroll. Yes, coroll. Assume something more on c. Now assume that so let h be an inverse space and let m be something more be the subspace which is in particular convex but much more. So, let m be subspace. However, we need to the not all subspaces of an infinite dimensional not all subspaces of an infinite dimensional hyperspace are closed. Therefore, you have to add that this is closed if you want to closed. So, in particular we know that m is convex and closed non-empty subspace convex and closed. So, in particular we know that we have our family of inequalities but this is more now. Can you imagine what is it? I mean we are in this situation. Now we have a subspace. So, here there is our h. If the picture is Euclidean, really complete Euclidean then what happens? It happens that h is essentially orthogonal to m. Sorry, if you take this point here then this h minus x is orthogonal to this. Right? This is what happens in finite dimension. Or remember that this picture is Euclidean. If your ball is strange then this is orthogonal to this. Apart from this remark so this is m. So, given h there is a unique x by the previous result. We know that h minus x, c minus x is non-positive. Ok? But actually you can say much more here which is exactly the following. Then h minus x c let me use the symbol c now is equal to zero for any c. Sorry for the symbol c. Maybe we could use small m. I don't know what to use. So, this is, I would like to leave you for homework and we will do this tomorrow. I prefer to leave you this as home. This will be the first thing that we will do tomorrow. And then I would like also to leave you another exercise for tomorrow that we will do together. And so exercise, homework. So take h equal to l2 of zero, one. So we are slowly going outside the small l2 and going outside any ill-be space. Consider the closed subspace b of h. Consisting of all polynomials polynomials of degree 2. Compute the point of v m. Well, m and v are the same in this sense. Compute the point of v. Let me call it m. Consisting of all polynomials of degree 2. Compute the point of m of minimal distance from t-cube. So, you have, this is your h. This is a point in l2. Well, you want to find, so now your closed subspace consists of polynomials of degree 2. So you want to find the point the polynomial of degree 2 to this polynomial of degree 3 with respect to the l2 norm. This is the problem, homework.