 Kdaj pokazim, da pojelim snigs duže delle historije načejne opot老師je. Načejne historije prihlede kabalje imejte so tudi skupovati teorim. Štešnje odgleda je, da je tudi tukale vsapak, zelo, da je vnitost in večost. In idem, čez pa jazба je benden Del. K evaluate. hind. Jujte pa kahoga, lebo vse...] risimo za odίνaj. Manja... ...runja del je to o BLB ...testo. A begina da sva spusten. pa kita da lahko vedel. Vte, BF, ki je tukaj razredno, in tez ešte zelo, če vorne bomo da tukaj nač backwards pristali, če mu sem w tukaj bolj, L to je minus 1, to je invertible, je kontinuos. Kaj je tukaj, da L to je minus 1 kontinuos? To je L, to je L minus 1. Tako je, da je... Zato, kaj sem tukaj, nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj nekaj, ko je je zelo dojevaj kaj je zelo dojevaj. I tako, če je nekaj konstant. Vse je poslednja z vse, da bo, bo, bo, je tukaj v l, v, v, o, je tako zelo na svoj smizitvi, najskega. Zato. there exists. So if I fix the ball here, B, E1, this says that, namely, L to the minus one, Bf of radius c is contained in B E. So given a neighborhood of the origin here... Or in this case, just the ball of radius one, Zato, da bomo zelo kratil, da bi imeli njih boljev, počeš sem počeš všeč, da bi mi dobro odbrali landa. Tudi možem hvorezno vse zelo je boljev avon, in da imamo mali boljev na vse, tudi na njih, da so dobro vse. In da je boljev, da bi se dela boljev, shrejuje to s njenim po blockedo, ki jih všimi, nuno so nekako reprevečiti. Is it OK? Is it clear the statement? The notation is the ball of radius C in the space f. So you see why it is called open mapping? Well because if you take the unit ball, which is open, you take the image of the unit ball, a potem je to vsebo, ker je vsebo. Kaj je to vsebo? Tukaj, na vseboj kaj je vsebo, ker je vsebo. Vseboj je vseboj, ker je vseboj, ker je vseboj. Vseboj je vseboj. If A is symmetric, the interior of A is symmetric. So, I read it. So, let X be a Banach space. Let lambda be a non-zero number, real number. Then for any subset of the Banach space, the closure of lambda A is lambda the closure of A. The interior of lambda A is lambda times the interior of A. If A is symmetric, what does it mean? It means that A is equal to minus A. Then A bar is symmetric. And if A is symmetric, then the interior of A is also symmetric. This is the first lemma, which will be useful in the proof. So, the proof of this lemma is, so, proof. Point one and two are a consequence of the fact that the map, if lambda is different from zero, then the map is a homomorphism, which implies one and two. Assume now that A is symmetric. Let us prove three. So, if A is symmetric, this means that this is equal to this. Therefore, we have this. But now it is sufficient to apply point one with lambda equal to minus one. So, with lambda equal to minus one, this immediately gives that also A bar is symmetric. And similarly you can check the same argument using two, that also the interior of A is symmetric with the same argument. So, this is lemma one. Now let me write down lemma two in the order. So, lemma two is the following. If A, now let me change letter. If K is convex, then K bar is convex. And the interior of K is convex. So, let us show point one here and point two. So, if K is convex, take two points X and Y, and take two numbers, lambda and nu in zero one, with sum up to one. We have to show that lambda X plus mu Y belongs to K bar. So, what do we know? We know that X and Y belongs to K bar. Therefore, given any ball, we have that lambda, we have that X plus say any ball belongs to K, intersection K is empty, and Y plus mu, any ball centered at the origin. Intersection K is non-empty. This means that X and Y belongs to the closure of K. Given any ball sent U. I have denoted it by U. Given any ball, we have this. What does it mean proving this? It means that we have to show that lambda X plus mu Y plus U, intersection of K is non-empty. Is it OK? So, I repeat, given any ball U, neighborhood of the origin particular, we know that X and Y are in the closure. Therefore, if I take any ball centered at X, which is exactly as this expression, then this ball does intersect K. And the same for this. We have to show that this is true. Therefore, we have to show that given any ball centered at this point, namely equivalently this, plus this plus the ball intersect K. So, from this we have that there exists U1 mu such that X plus U1 belongs to K. And from the second formula this, it follows that there exists U2 in U such that Y plus U2 belongs to K. Now, it is enough to consider lambda X plus U1. Now, since K is convex, this lambda times this plus mu times this is an element of K, because K is convex. Sorry? U is a subset of X. So, maybe I should write who is U. So, U is an arbitrary U, is an arbitrary ball of X centered at zero. OK? I use the notation U. OK. So, by convexity, a convex combination of these two points must belong to K, capital K. Therefore, convex combination must belong to capital K mu. Y, excuse me, you are right. Sorry. Thank you. Then, therefore, let me write it like this, lambda X plus mu Y plus lambda U1 plus mu U2 belongs to K. But now, you see, the ball is convex. So, U1 and U2 are elements of the ball, but the ball has the special property to be convex. Therefore, this convex combination is still an element of the ball. Belongs to U, because U is convex. And this exactly is equivalent to say that this... What does it mean, this? That there is a point here, such that this plus this belongs to K. And this point is this convex combination. OK? Which implies... OK. So, it remains to show the second part. Maybe I could leave this as a homework, maybe. So, you should prove. What is the second part? You should prove that the interior is convex. So, given two points in the interior and given two numbers, which sum up to one, a negative number, we have to show that this is in the interior, is in the interior. So, what does it mean that it is in the interior? There exists now. Now, there exists a ball. Now, there exists U ball, such that X plus U is contained in K. And Y plus U is contained in K, in the convex set. So, now the claim is that lambda X plus mu Y, now you should show that lambda X plus mu Y, then plus mu Y plus U is contained in K. Well, this is essentially trivial, because, again, what does it mean, this? This means that for any U you have that X plus small U belongs to K, and Y plus small U belongs to K. Then, now you take lambda X plus mu Y plus U This is Y. Then you take the convex combination of these two. This is an element of K, because K is convex, and it can also be written as follows, lambda X plus mu Y plus X plus U plus lambda U plus mu U plus U, because the sum is equal to 1. Therefore, we have shown that for any U in U, this belongs to K. And this is equivalent to say that this. So, now let us go to the proof. So, once we have these two lemmas, let us go to the proof. In the proof we will use the completeness of both E and F. Let us start with using the completeness of F. Step one, use completeness of F. Ok, and we will show that L of B has nonempt interior. We will first show that L of B has nonempt interior. Namely, there exists a positive constant that I denote by rho. Positive such that the ball is contained. Ok, so the first step consists in showing that this object here has nonempt interior. Remember, maybe I have I think that, yeah, B E. You are right. I don't remember my notation. This is E. Ok. Of radius one. Let me simply write it for simplicity. Let me use this simple notation. So, this is B. Ok. So, the claim is that this has nonempt interior or equivalently it contains a ball with positive radius. Other questions? Yeah. Yes, in the proof of step one we will use the completeness of F. I will show. Ah, we will show, sorry. Show. We will show that. And now you will see how we will use completeness of F. Maybe you can already infer that once you see the interior of the closure, something like the bear theorem is behind. Anyway. So, first we recall that the map is by assumption surjective in particular. Therefore, I can write F as the union of all L and B. Ok. This is surjectivity and also obviously it is the union of something which seems to be larger but it is not. Which is this. Ok. Now, if by contradiction of B is empty then then default then we have using the lemmas Yes. I want to show that LB bar has non-empt interior. So, assume by contradiction it has empty interior. I want to show non-empt. So, assume instead it has empty interior. So, the interior of this is empty. Sorry. So, it was the interior Ok. Then we have L or NB closure is equal to what? So, it is equal to N RB closure by lemma 1. No, sorry. This is simply by the linearity but by lemma 1 this is equal to this. Lemma 1. So, this is lemma 1. This is simply linearity. Ok. Then this is closed and therefore now the interior of L NB lemma N interior of LB. Therefore this implies that if this is empty then also this is empty for any N, N positive. Interior is empty for any N, larger or equal than 1. Which is a contradiction of using bare theorem because we are writing the complete metric space F as a countable union of closed sets. Therefore at least one of these closed sets has non-empt interior. Contradiction with bare in F. Is it ok for up to now? So, what we have shown is therefore that there exist so this has non-empt interior therefore it contains a ball. Now look at the statement. The statement is this. And this is not the statement. Why? Because there is a closure. Other problems? No? Is it ok? Because these are closed. Of course this is the reason why I add so I have this equality but in order to have a union of closed set it is better to close it. Non-empt interior. This is not like to say it is non-empty. One of those has a ball inside. Not only an event but a ball. Open ball. Ok. So we have we have not yet unfortunately we have found a constant row. Sorry, row and C probably will be the same. But we have not proven that this ball is contained here but this ball is contained in something larger the closure. But now if you want to eliminate say this closure there is another argument and so let me show you the second step. So this was step one. So now let us go to step two. So let me write down what we have proven up to now. There exist a constant row such that we have this. And my notation was this simple. Ok, so we have this. Now in step two step two something like b, rho, h f is contained in L to b. Ok, so and of course once we have this we have concluded by taking C equal to rho over 2. Ok, this is the conclusion. Right? So we have to enlarge at least of the factor 2 in this. And in this step we will use for the moment we have used completeness of the target now we will use also completeness of the source. We will use completeness of a in this step. Ok, so you see the proof is involved because we have really to to use the deep completeness assumption in order to have the result. Now it follows that L of b is dense so this implies this assertion here. This L of b is dense in b rho. Ok. So now we can so what does it mean? Is dense in b rho f? It means that if we take if we take a point here y star so let y star be any point of this ball we can find a point of b into b at distance as small as I want say x1 therefore let me write it as follows such that y minus L of x1 is less say than rho over ok, so this is density even y star there exist x1 ok. Now observe however that from this we also have the following. If I multiply by lambda positive and I take the lambda rho f then this is then we have still this by one of dilemmas so if we multiply everything by lambda this inclusion remains but the lambda goes inside by one of dilemmas ok. And therefore we have also something more that L of b is dense in b rho f but L of lambda b actually if lambda is positive f of lambda b is dense in bf lambda rho by this conclusion and therefore we can repeat this argument y star so we can repeat the argument and now so we have a point in the bowl of radius y star centered r over 2 so we can find x2 so x1 is in b this and there exist an x2 such that x2 is less of 1 over 2 and y star minus L of x1 minus L of x2 is less than rho over 4 ok. Exactly. Therefore we can construct by induction a sequence of points such that xn is less than 1 over over 2 over n minus 1 because you see here we have 2 and here we have power 1 so 2dn minus 1 and such that y star minus L of x1 minus minus L of xn is less than or equal than rho over 2 to the n plus 2 to the n 2 to the n. Let me check. Inductively. Is it ok for the moment? So let me repeat. How can I find this x2? Well I apply this with lambda equal 1 over 2. Ok. So with lambda equal 1 over 2 this is dense in this. But this is an element of that ball. So this belongs to b f rho over 2. So this is dense b over 2 is dense into this. Therefore we can find another element x2 in the ball b of radius 1 over 2 such that this new element minus this is less than say rho over 4. Is this ok? Therefore using induction I can find such a kind of and let me write it this as obvious list. Now l is linear so this is equal plus xn rho to the n. Now by construction we have that this converges by construction because it is less than 2. So now there is a criterion in banak spaces that says essentially that if this converges then also xn converges. Do you know it? It is completeness criterion. The converse is also true. If you have this assertion then the space is necessarily complete. Ok. Therefore xn use here completeness of e in this sense criterion of converging of absolute convergence of series. Say. And therefore xn converges to in e converges to some x star by the completeness of e by the completeness of e. Now we look at this and what about x star here and x star is less than or equal than 2. The sum of xn converges to x star and x star is less than or equal than 2. Ok. Therefore x star belongs to l belongs to 2b. Say. Maybe we also have less than or equal than 2b. Is this strict inequality? Maybe we have the strict inequality. Because we have the strict inequality here. Ok. So strict inequality here. So x star belongs to 2b. Now we look, we pass to the limit into this inequality. X, this is converging to x star and l is continuous. The norm is also continuous. Therefore we can pass to the limit here and find that y star minus l of x star is necessary 0. Y star minus l of x star is equal to 0 Equivalent in since this is a norm is equivalent to say this and so this concludes because we have taken any point in the ball and we have found a point x into 2b such that this holds. And therefore this is equivalent to say this green conclusion. Ok. So you see eliminating the closure here is not so easy. And so this concludes the proof of the theorem of open mapping theorem which is one of the basic theorems in function analysis. Let us see some corollaries of this show you some corollary we didn't choose we have used subjectivity only and completeness of F. In step 2 we have used completeness of E moment, yes, maybe not maybe you are right, strange however. No, well, well, it's not strange if I want to write l to the minus 1 in the statement. If I want to invert it so if I want in the statement write the symbol l minus 1 then I need it I need the injectivity so that this is a one-valued operator associating the point to a point. Ok. So the corollary is one corollary is the following. So corollary this is very interesting Assume that you have a vector space such that if you have such that it is banak with two different norms making e-banak and suppose that there exists a constant such that this is less than or equal than then there exists another constant namely the two norms are equivalent. Ok. In finite dimension all norms are equivalent but clearly one of the big problems in functional analysis that in general norms are not equivalent but the strong if you know that both the two norms make e-banak then you have the norms are equivalent hence the two norms are equivalent. Now the proof is just the direct application of of the previous big theorem because it's sufficient to consider the identity from this banak space into the other banak space the identity you see by assumption one by one the identity is linear is bijective of course but assumption one says that it is also continuous between these two banak spaces well and then by the previous theorem inverse of course it is injective subjective the inverse which is still the identity still the identity id the identity is also continuous which is exactly like to say continuous which is equivalent to say of the corollary ok so equivalence of norms maybe I can I can also leave you as an exercise home so let e equal direct sum so let e be a banak space x and y closed subspaces this is the direct sum it means that any element the assumptions e equal x means that any e in e can be uniquely written be uniquely written as then the projections try to prove by yourself this theorem let us see other consequences so this is home now let us see other consequences of this theorem closed graph so let e and f be banak and let be linear so there is another criterion to see whether or not l is continuous so remember that l is continuous if and only if it is bounded not only this there is another criterion then l is continuous this is interesting if and only if the graph of l is closed in in this so what is the graph so the graph is clear we know what is the graph the graph is the set of all x, l of x in e such that let's see in e now the norm that we put here this is another vector space and the norm that we put it is endowed with this norm endowed let us show theorem so one implication actually is true without using any kind of open mapping theorem or closed but the other one is more difficult so let us see the proof so assume that l is continuous take a sequence of points in the graph let me write it sequence of points in graph l and assume that they converges to something converges to x and y in here what does it mean that xn converges to x with respect to the norm of e and this converges to y with respect to the norm of f because we have chosen this norm ok well now but if xn converges to x and l is continuous then l of xn necessarily converges to l of x which is like to say this is because l is continuous but l is continuous then this and this is like to say that is in the graph so we have shown that if I have a sequence of points of graph of l converges to something then something belongs to graph of l and this is like to say that graph of l is closed ok no is it ok what is the missing point take a sequence of point in this set this is a linear subspace of the direct sum so take a sequence of points converging to something in the norm in the correct norm in this norm what does it mean that this is closed it means that the limit belongs to the set so assume that this converges to something in this norm but this means by definition of this norm that xn converges to x in E and l of xn converges to y in F ok so but since xn converges to x in E and l is continuous then l of xn converges to l of x ok and therefore y is equal to l of x but what does it mean that the pair x and y belongs to the graph no yes it's ok and so graph of l is closed so you see this is very easy I don't use any kind of deep theorem here just observe that I need this rather natural norm one possible natural norm is this assume that the graph of l is closed in E direct sum F so so now I want to apply the so the corollary the corollary was saying if I have ok let me define the following graph l is closed graph l is a subspace first of all is a subspace of E F it is closed and therefore is a banach space closed hence graph l is a banach space therefore I can consider the following map between two banach spaces in the map from graph l into E which is defined as follows ok, this is continuous well first of all it is linear it is continuous it is injective and it is surjective also so essentially our graph is something like this this is E, this is F this is our graph which is a linear subspace of the direct sum of the algebraic direct sum so it projects down on E it is clearly continuous because it is clear that inverse is less it is less it is clearly continuous since it is continuous it is invertible and so on this means that also the inverse is continuous therefore there exists a constant such that x plus l of x is less than or equal to x this is the opposite it is the corollary corollary which says that l of x is less than or equal to c minus 1 which is equivalent to say that l is bounded or continuous now just few comments few comments on so remember this is a criterion to check whether a functional in operator is continuous so you have not only to check in inequality improving boundedness but you could also alternatively look at the graph ok, now some final comments some final comments it happen very often very often the operators the operators are not defined as follows between two Banach space but very often actually they are defined but it happens that we are in the following situation what does it mean this ok, l is defined in a linear subspace usually linear subspace of e with the induced norm from this is called domain of l so it is not globally defined on the whole Banach space but just in a linear subspace the fact is that you have to write this inclusion because you have to remember what is the norm so on a linear subspace you have the norm of the ambient space so you write this example let me show you an example simplest example but very interesting so take e equal l1 of r f equal l1 of r domain of l say continuous maybe with compact support considered as a subset of e notice that is this subset closed ok this operator is so you see this operator is clearly well defined for functions which are differentiable but in particular for c1 functions with compact support so this is well defined of course it is not defined in the rule of l1 because we don't know what does it mean differentiating a function in l1 but we know what does it mean what it means differentiating point wise a smooth function so this is well defined and necessarily here and not here we look at c1c with the induced norm the image so l of f this is continuous with compact support so in particular is in l1 ok so this is an example so what is the graph what do you think what is your definition of graph actually this is much more general than before the previous case corresponds we have studied up to now just the case of equality but this is much much more general ok is this clear I mean this small apparently small modification is not a small modification ok what is the graph ok try to show that the graph of l is not closed well I can consider take for instance this function f here the derivative of this f is out of the points where it is not differentiable is just 1 minus 1 ok this is f now take a sequence fn in the domain the domain of your operator more or less are the following something like this so something like say so this is fn this is a prime n so take the sequence fn have compact support so fn are elements of your domain suppose that fn converges to the graph of fn so to some fg what do we know well fn converges to this in l1 and f prime n converges to this in l1 but then it is clear that f is not in the domain of l because it is just Lipschitz and not c1 for instance ok so the graph of l is not closed but we would like somehow to force to find the natural extension so that it is closed so let me give you the following definition maybe exercise home the closure the closure of the graph is the graph of a linear map if and only if the following try to show this so this is a criterion to check whether or not this is a graph of something so we can give the following definition that the operator definition l from dl is closable if the closure of the graph is the graph of a linear map is a graph of course again the graph of l is a subset of e direct sum f where we use the usual norm when we talk about closure we are considering that norm so and the theorem is is that that operator in the example is closable the operator is closable so the idea to show that this closable is to use the criterion this criterion here proof use so and the idea is the following you simply have to take a sequence of function so you have a point a point in the closure so what does it mean you have a sequence take a sequence fn into your domain c1c fn converges to 0 in l1 and then you have that f prime of n converges to some element let me call it instead of y let me call it just g and what do you have to show using this criterion is that g is equal to 0 and so what do we do so take a test function phi in c1c and consider fn prime phi then what we know this we can integrate by parts without boundary terms because phi is compact support and also fn is compact support everything is compact support so integral phi prime dx minus now what do we know we know that fn converges to 0 in l1 and this implies that this converges to to 0 on the other hand f prime n converges to g in l1 therefore this converges to g phi dx we conclude that the integral of g phi dx is equal to 0 for any phi c1c ok and so what we conclude from this that ok is almost that we were equal to 0 and therefore this shows that the operator is closable well from this point it starts the theory of generalized derivatives sobolev spaces and so on so we don't have time to do this now but this is just maybe one starting point of sobolev spaces ok well the next lectures we will dedicate I think the remaining time we have just three lectures only to distributions theory so tomorrow and the next lectures we will study with some care as far as possible time permitting distributions theory ok