 So we have to prove the Hambanak theorem in geometric form. We have seen the Hambanak theorem for functions. And now we say we see the Hambanak theorem for level sets, geometric form. We will split the proof into two sub theorems. So maybe theorem 1. OK, so let E be a normed vector space. And let C be a subset convex, open, open, and non-empty. Then we take a point outside that we denoted by x0, maybe. Let x0 be a point outside. So now we separate x0 and C. So there exists. Exists that we denoted by L, a linear continuous functional on E, so that the picture is the following. Essentially, we have a convex set and a point outside. This is open. This is x0. So there is a closed hyperplane separating so that C lies on one side of the hyperplane and x0 on the other side. And this is closed. So L is in a star. And so that say L of x is less than L of x0 for any point x in C. So C is from one side, x0 is on the other side. And therefore, in particular, the set where L is equal to L, in particular, L of x0 separates C and x0. So possibly by translation, possibly translating, we can assume we can suppose that 0 is in C. Remember that C is open, so it coincides with its interior. And therefore, 0 is an interior point. 0 is here. Now we want to apply the Hanbanak in the form of function. So we have to find the function g, a subspace capital G, and a sublinear function p. So capital G, so let capital G be equal to the span of x0. So this is capital G. Then we define g, linear function as follows. So g of lambda x0 is equal by definition lambda. So notice that it is important to remark, this is obvious, but let me write it, g of x0 is equal to 1, obviously. And g is continuous and linear. This is always with the induced norm. OK, why I stress this fact? Because then I want to show, so I want to find the function so that C becomes the one sub-level set. So C will be the set of point where some function p is less than 1, while this is, of course, outside and is exactly equal than 1. OK, so the function p, maybe I already wrote, yes, so this is g. Now we need p in order to apply Hanbanak. To apply Hanbanak, remember Hanbanak was the theorem, the proof of which was divided into two. There was one part in which you extend the g at one point outside, and then the other part which was abstract and based on Zon's lemma. So this I want to define it. And for any x in E, we set p of x alpha. OK, what does it mean? It means the following. So I have a point x. Any point x outside or inside doesn't matter. So I start by homotetizing C, since away. At some moment, there will be alpha large enough so that this is alpha C now. This is C. Alpha is very large, and this alpha times C. And therefore, now this point x that I have fixed before lies inside some alpha C. Then I take the smallest alpha. So now I reduce alpha in order to take the smallest one. And the definition of p at this point is this smallest alpha, roughly. It is an infimum. So this is the definition of p. Now we have to prove various facts about the connection between p and C. So let me state it more precisely. So in the order lemma, lemma 1, say, p satisfies the properties of ambana, in particular, so p. So p of lambda x is equal to lambda p of x for any lambda, non-negative index in the north space. So this is point 1. Point 2 is p of x plus y is less than maybe there is first. Well, OK, doesn't matter. I will say it during the proof. So we have this. And then we have there exists a constant. Let me denote it by m, such that OK. And then C. So this is the standard way to pass from a convex set to a function which is positively one homogeneous and sublinear, so that the original convex set is the unit ball of p. This is the crucial remark. So we have to show this lemma. And so let me erase something here. The function? Yes, the function p is defined on the wall of v. And it is finite. And we will see this now in a moment. Which is the question? Maybe? Yes, indeed. This was what I was thinking just before writing this. You're right. This could be empty in principle. The question is, maybe there is no alpha such that given x, there is no alpha such that x is an alpha c. Actually, this is not the case. Actually, this is finite. And we will see during the proof. So this set, you're right. Correct comment. This set is non-empty given x. This was the question. Now we will see during the proof. And this is a consequence of the fact that c is open. Because we will see it because c is open. Because, of course, if there is no alpha, then p is plus infinity. And that makes problem in the application of Antanak. Now let's see why all these kind of things. So point one. So remember the definition. Let me write it as follows, maybe. So p of lambda x, take lambda positive. Then p of lambda x, by definition, is the infimum of alpha positive such that alpha, the minus 1, lambda x belongs to c. Now let me multiply and divide by alpha here. The infimum is over all alpha such that this holds. Now lambda is positive. We change name, lambda is positive and given. So let me write this. So this is equal to lambda inf mu such that mu minus 1x is in c, p of x. So for any lambda positive, we have positive 1 homogeneity. And it is clear that if lambda is 0 also. So the same holds when lambda is equal to 0. And therefore, formula 1 is true for any lambda larger than or equal than 0. Is it OK? Now maybe the comment, the first initial comment would be actually his comment. So first thing to prove should be that this is not empty. So let us make the following remark, side remark, say. C is absorbing. What does it mean? It means that, namely, given any x in E, there exists a constant. Let me call it, I don't know, a constant r, such that x belongs to r times c. Let us see why. This is a consequence of the fact that. So take x, any x. So x belongs to some, since c is open, there is a ball, there is a ball, there is a positive radius, small r, such that the open ball is contained in c. So first remark, take small r. Since c is open and 0 is an interior point, we can pick there exists r positive such that br0 is contained in c. And then we take this small r. Then we take a point x in E, and we have that, of course, v. Now, this is not true, because the ball is open. So it is not true. I mean, if I have point x, I take the distance. It is not true that it belongs to the ball of this radius. Adjust just a little bit. This becomes true for any small epsilon. Other questions? If there are questions, please stop me so that. Is it OK? OK, now I want to put here r instead. So what happens if I put here r? Then I think something like this. 1 over r x plus epsilon. This is equal to this. So we have, is it OK? But br is contained in c. So this is contained in 1 over r x plus epsilon times c. And this shows that, hence, there is an alpha such that, hence, the set of alpha positive such that x belongs to alpha c is not empty, because one of those alpha is exactly this. So maybe this should be maybe the starting point of the proof. The starting point of the proof should be this. This set is non-empty, so that this is real valued. Of course, it is non-negative, but real valued. Then once we know this, we can prove 0.1. OK, so now 0.1 is proven. Let us show 0.2. So take now two points, x and y in E. Now we know that c is absorbing. So there are two numbers, say, lambda 1 and lambda 2, such that this holds. So let me denote it. Yes, so there exists lambda 1 and lambda 2, such that x1 belongs to, sorry, x belongs to lambda 1c and y belongs to lambda 2c. This means that there is some small c1, such that x is equal to lambda 1c1. So hence there exists some small c1 in c and c2 in c, such that x is equal to lambda 1c1 and y is equal to lambda 2c2. OK, this is x plus y. Therefore, is equal to lambda 1c1 plus lambda 2c2. And we want to use convexity of c. So in order to make a convex combination, it is natural here to divide by lambda 1 plus lambda 2. So we multiply and divide c1, c2. Now these are non-negative numbers, actually positive numbers, that sum up to 1, sum of this plus this is equal to 1. Therefore, this is a convex combination of two points, c1 and c2 of the set c. Therefore, this is an element of c. So this shows, therefore, that this belongs to lambda 1 plus lambda 2 of capital C by convexity. And now you remember the definition of p. But p is the smallest number such that this inclusion holds. And therefore, p is necessarily less than or equal than this number, because you remember the definition. It's the smallest number alpha such that x is in alpha c. So by definition, this implies that this is less than or equal than lambda 1 plus lambda 2. This holds up taking the infimum with respect to a lambda 1 and lambda 2 gives conclusion 2. So we have used convexity here. In point 1, we have not only convexity, but also absorbing, property absorbing, which is consequence of the fact that this set is supposed to be open. So convexity and openness give us two relevant points so that we are already in the setting of Hanbanak. Now we want this. Point 3, essentially, is again a consequence of the fact that c is open. Because we have already, essentially, first of all, p is non-negative by definition. So this inequality is immediate. Now let us show the second inequality, which we have already shown, because we know that it is absorbing. So we know that x belongs to, I think, something like 1 over r x plus epsilon c. This was the previous. I think it was this, no? Please check. It was this. So this again, remember the definition of p? p is the smallest number that this inclusion holds. Therefore, p is surely less than or equal than this factor in front. So by definition, this implies that p is less than or equal than 1 over r x plus epsilon. But this is true for any positive epsilon. Therefore, sending epsilon to 0 plus, this implies sending epsilon to 0, this implies p of x less than or equal than 1 over r x. So 1 over r is the constant that I have indicated here by m, capital M. Well, we cannot hope to have a sort of inequality like small m here, norm of x. This cannot be. We cannot hope. But at least we can hope this. We cannot hope this. For instance, take in R2 the following convex set. Now I make a finite dimensional observation, two dimensions. So take, for instance, the following convex set in R2, origin 1 minus 1. This is C. This is C. This is open, convex, non-empty. And actually, it is the unit ball of a function p. So which function p? So p of x, y equal to what? Absolute value of y. Absolute value of y, which cannot be larger or equal than some constant square root of x square plus y square for some positive m. So we cannot hope to have this. And this is infinite dimension due to the fact that this open convex set can be unbounded. Because infinite dimension, if this would be bounded, convex and open, then whatever. I mean, finite dimension, if we are in this situation, then we can find small m. Because we can put a small ball inside, which gives us this inequality. But then a larger ball, a big ball outside of the norm, which would give us the converse inequality. Is this clear? So we cannot hope to improve this in this generality, this inequality. But openness gives us at least this. It remains to show property 4. OK, so take x in C. C is open. It means that there exists small epsilon positive such that 1 plus epsilon x belongs to C. This means that, equivalently, that x belongs to 1 plus epsilon to the minus 1 C. And this, remembering the definition of p of x, this implies that p of x is surely less than this pre-factor because it's the smallest one. So this is less than 1 plus epsilon for any epsilon. And this is less than 1. So we have shown that this inclusion. So take any point in C. Then, necessarily, p of x is less than 1. So this means this inclusion. Now we have to show the converse, the opposite inclusion. So this one. So take a point, take a point x such that p of x is less than 1. So under exist alpha, alpha less than 1, such that x, x belongs to alpha C, by definition. And alpha is less than 1. Now I take simply a combination of the origin and this point here. So I just write x equal to this factor here. So alpha x multiplied by alpha minus 1 plus 1 minus alpha alpha, the origin, OK? So this is, like to say, alpha minus 1x. So I have a point in C plus 1 minus alpha maybe. So I have this point in C, alpha minus 1x is in C. Then I have the origin, which is in C. Then I take a convex combination of the two points, because alpha is in between 0 and 1, OK? So alpha times this point plus 1 minus alpha times this other point is a convex combination of two points of C. Therefore, C is convex and this belongs to C. And so we have shown that given any point here is a point of C. So lemma 1 is proven. And now we can erase it. And we can finally use it to prove the theorem. So therefore, remember, we have the function G from G to R, such that G is linear, G is continuous. And G of x naught is equal to 1, OK? We have also this gauge, this Minkowski functional. And what about P of x naught? Well, P of x naught. So G of x naught is 1 here, but x naught is outside C. But C is the set of all point where P is less than 1. And this is outside. Therefore, P of x naught is larger than or equal to 1, OK? Because it does not belong to C by assumption. And C is equal by the lemma, OK? So at x naught, we have the inequality that, at least at x naught, we have the inequality that we want in order to apply ambanac. Hence, we have this, OK? Now, having a linear functional, less than or equal than a sublinear functional, at this point it's sufficient to show that this implies that G of x is less than or equal to P of x on the domain of G. Why is this so? Well, take, for instance, a lambda positive, and for a point of capital G, take the following point of capital G. And capital G is just a line. So I have to distinguish lambda positive and lambda negative. So this, but G is linear. So this is lambda G of x naught, which is actually lambda less than or equal than lambda P of x naught, because lambda is positive, and I have this inequality. But this is also homogeneous, and lambda is positive. So this is equal to P of lambda x naught, OK? So on the positive half line, say, half line, I have that G is less than or equal than P. Now I have to check what happens for lambda negative. So lambda negative still have this. This is negative. And therefore, since P is non-negative, we trivially have this inequality. What does it mean? It means that we have shown this inequality. Now we can apply, finally, ambanak. We have all assumptions to apply ambanak theorem. So there exists an extension, L. I don't know the symbol. In the statement, I wrote L. So L from E to R extending G and less than. I don't remember the symbol that I use in the statement. Was it L? Capital. So this is ambanak. Extending G means that, in particular, at the point x naught, L of x naught is equal to G of x naught, OK? Is equal to 1. So this corollary of ambanak, this is ambanak, maybe in the version of the norm of the space. Remember, we made the theorem, which was set theory. No continuity. A corollary on norm space, maybe the question you made the last time, which gives us continuity. So we know that this is continuous, OK? So actually, I'm applying the corollary. I don't remember if I wrote corollary or theorem, but it was a corollary or topological version of ambanak. OK. So we have this, OK? Therefore, on C, on C, this is true everywhere. At the points of capital C, this is also true. But P is less than 1. So you see, the set C is on one side with respect to L equal 1. And so this concludes the proof. Is it OK? Is it clear? So this is the first part. This is the difficult part, because now we have to separate two convex sets. And the notation, I don't remember, maybe was a and b. OK. Theorem, normed vector space, and a and b subsets of e, both non-empty, then a and b convex. I don't remember last time, but assume that a is open and assume that they are disjoint. OK. So the first one is non-empty convex and open. The second one is just non-empty convex. Then there exists, let me denote it by, exists alpha here, such that L equal to alpha separates a and b. So now the situation is more or less like before. We have, now this is a. Then we have b. They are convex, et cetera. I am putting some closed hyperplane in between, so that a is just from one side and b on the other side. OK. So the proof is based on the previous theorem, proof. So proof. It is enough. So let us consider the following c. Let me denote it by c, I think. Yes. This is this. So what is this set? So we can add and subtract vectors. So this is, do not confuse this with the difference of sets. Difference of sets is another symbol, this one, usually. Do not confuse this. So paying attention to this notation, this is just the union by definition of all possible differences such that, which is actually also equal to the union of a minus b such that b. Now what about c? Well, homework maybe. I leave you to check that c is convex. It's not difficult, so remarks. So c is convex, non-empty. As an exercise, you could, well, no, it doesn't matter. So c is convex, non-empty. Open. Open why? Because each of this given b, small b, this is a minus b, is open. And so this is the union of open sets. So this is open. And also the origin is outside. So 0 is not in c. Why 0 is not in c? Because if 0 were in c, this means that 0 can be written as small a minus small b. Therefore, the a small a is equal to small b. And therefore this contradicts this. Is it clear? Because so we can apply the previous theorem. So now we apply the previous theorem with the choice c is c and 0 is x0. So apply theorem 1 with 0 equals x0. x0 equals 0. And so we find l such that, and alpha also, l such that l of a minus b is less than l of 0, which is 0, for any a in a and b in b. OK. This, but l is linear. Therefore, this is like to say this for any a in a and b in b. And therefore it is enough to choose now alpha, such that alpha is in between the supremum of this and the infimum of this. We know that the supremum over a of this is less than or equal to the infimum over b of this. So there is at least one alpha in between. And therefore l equal to alpha is the hyperplane. The closed hyperplane is the closed we are looking for. So this gives us theorem 2. Now I leave you to look at the book of Brezzis, for instance, for the separation theorem in the form that we have already written. So assume now that a and b, a convex, non-empty, closed, b convex, non-empty, and compact. Then a and b are strictly separated by a closed hyperplane. So look at the book of Brezzis. We have already written the statement on last time. OK, I would like to make now an exercise. And the exercise is the following. Maybe I leave you partially as homework. So look at this in the Brezzis book, strict separation. And exercise, consider the set a, the sum over j from 1 to n, lambda j j, such that lambda j is in r, and then for any 4j, 1n, and lambda n is positive. So take, and n is also varying, such that n is n, lambda j is in r, but the last one is positive. And then b, define b as minus a. Define b as minus a. This is subset over 2. And ej are the usual 0, 0, 0, 1, 0, et cetera, et cetera. Now try to show that 1, a and b are convex, non-empty. And 2, they are disjoint. Then 3 for any l in l2 star. So can a be separated by closed hyperplane? So you have two disjoint convex sets. And however, it is almost clear that if point 3 is true, then they cannot be separated. Because if the image is the wall of r, it is impossible to put a from one side with respect to a number and b from the other side. So it is clear that if 3 is true, they cannot be separated. So the point is try to prove that 3 is true. We have already seen a lot of consequences of Hanbanak. So now we pass to another big theorem, which is the Banach Steinhaus. This is another big theorem. So we have to prove it carefully. So we have now, let me denote it by e, Banach space, and Banach space, normed vector space. And then we have a family of, let me denote it by Li, let Li be a family of linear continuous functionals from maps from e to n, where capital I is a set of indices, not necessarily countable, not necessarily countable. So we have a big family of indices, and then you have a family indecised by capital I of linear continuous map from a Banach space into n. And assume that, suppose that for any x, the soup over I of x is finite for any x. Then the statement seems impossible. This seems impossible. Soup over I Banach Steinhaus. So what is the meaning of this result? Starting from a point-wise estimate, so for any x, given x, this number is finite, but depends on x. The statement says not only this, but then, if I then take the supremum over x, this is still finite. So this is a uniform estimate starting from a point-wise estimate. Before doing the proof, again, we could see... I would like to leave you some exercise before doing the proof. Again, home, let and ln on c0 to R. So c00, remember, c00 is this. What supported continuous functions from n to R? This means simply sequences that are 0 at some point and then definitely 0. Sequences which are definitely 0. C00 show that over n ln of x is finite. Sorry, ln over x is equal to n xn. n xn. Find this follow. So show this for any x. So we have assumption 1. So 1 is satisfied, but equal over n equal plus infinity. So this is a failure of Banach-Steinhaus. It's an example where Banach-Steinhaus doesn't work. I mean 1 is true, but 2 is false. Is there an explanation for this? So 2 is false. Evidently, if Banach-Steinhaus is true, there is some assumption missing here, which could be the missing assumption? Yes. C00 is not complete. This is, of course, with an infinity norm, as usual, is not complete. Remember, c0 is complete with 1, 0 only. This is complete. This is not. Actually, this is the completion of this. The relation between these two is that if you complete c00 with infinity norm, then you get c0. This is always a difficulty when you have to distinguish, in general, continuous function with compact support or continuous function which are 0 at infinity. So this is usually the completion of this. So this is not complete, but this is complete. So please do not, when you look at distribution theory or measure theory and so on, pay attention if the symbol is 0 or c. Because one is not complete and the other is complete. This space, we'll see something, I think, about this space is very, very common in measure theory. Continuous function with compact support in general. So the reason for which this is not true, point 2 is not true, is that this is not a Banach space. While in the assumption here is Banach. Completeness is important because the proof of this theorem is based, one of the possible proofs, proofs is based on Bayer theorem, which happens to be true in complete metric spaces. So please, at home, try to check these two inequality, these two assumptions. Another consequence that I would like to do before proving Banach's denouche is the following corollary. So let now Ln be a sequence of elements of this. And assume, suppose, that for any x, there exists a limit and let denote this limit by L. Then, three points. Well, the first is just an immediate consequence of a Banach. And then, OK, deliminf. This is also maybe a good exercise. It's easy. It's just a consequence of, of course, if you don't want to do it by yourself for some reason, you'll find this in the Brezis book. So check this corollary by yourself. It's very easy. The first one, just a direct consequence of a Banach. Finally, another interesting consequence of a Banach is something which says that if you have a weakly convert in L2, but more generally, I'm saying L2 because weak convergence, we have just stated only in L2, small L2, but this is true in much more generality. It concerns the fact that in general, weakly converging sequences are bounded. They're not necessarily strongly converging, but at least bounded. So we have the following corollary again. So let in the same, under the same assumptions of a Banach, OK, the same assumptions. So corollary, so let E be a Banach space and let B contained in E be a subset with the following property. For any L in a star, L of B is bounded. Then B itself is bounded. This says the following. If you have a subset of your Banach space and you know what is L of B and you know that if you take essentially the components because you know at least in L2, possible linear continuous functionals in small L2 are just the projections on the given coordinate that gives you that projection. So this says that all projections in particular, not only projections, but in particular, all projections are bounded. So if you know that your set is such that all of its projections is bounded, then the set itself is bounded, more or less. If you think about small L2, OK, the proof of the corollary is not an exercise. It's an application of Banach-Steinhaus with the following position. We apply Banach-Steinhaus with the following choice. Maybe there is a conflict of notation. May I change notation here? Sorry. There is a small conflict of notation. Let me call this capital G. Capital G and capital G. I'm Banach. I've already erased. OK. So we know that if this is, OK, we know that with this choice, surely E is a Banach space. Well, this is obvious. F was N. Thank you. N was F was N. Sorry, I now understand the question. This was N. Sorry. OK. And this is a set of indices now. And of course, in general, it's not countable. OK. Now, for any b, we can define the duality between, for any b, we want equal g star. So the duality between F and b. OK. So we know that this is bounded by assumptions. So define this, OK? Assumption, by assumption, Lb, L is bounded. Lb, L is bounded. Lb, the set of Lb is bounded, which says that the supremum over b of Lb. And the notation is not quite happy. Sorry, the notation is not so good. But anyway, sup and b, Lb, N is finite. OK. This is true for any L. Is it clear? I mean, by assumption, this set of numbers is finite. It's bounded. Sorry. The supremum of this set of numbers is finite. And this is true for any L. L is in E, which is g star. So from this point twice, now we have that for any L, this is bounded. This is a point twice estimate. We can pass to the uniform estimate, OK, by Hanbanak of Lb. So maybe more precisely, let me write it with a constant, this supremum, because it is useful. So there exists a constant, absolute constant, independent of, so that for any b in b and for any L in E, B, L, then we call them c times b times L. OK. This is exactly like to say, equivalently, that the supremum of, OK, is this. OK. By Hanbanak, by Banak-Steinhaus, for any L in g star, we have this. This is equal. By definition, it's the duality between L and b. OK. I want to show that b is bounded. So, well, now observe that the norm of b is what? Is the supremum by one of the, OK, is the supremum of what Lb divided by L, OK, sup over L. This is one of the corollaries that we have seen last time. So this means that for any b, the norm of b is this, which is always less than or equal than c. Hence, this is exactly like to say that B is bounded, because all elements of B have a norm which is bounded. There is a dual version of this. The proof is the same, changing the symbols. So maybe it is better that I write it to you, but try to prove it by yourself. So let now we change the, assume now we have that for any x in g, the set of all Lx such that L in b star is bounded. Then b star is bounded. OK. Try to prove by yourself this. The proof is similar to the previous one. However, you have to choose different set of indices, different passing to the dual. It's not so difficult after one. I've seen this. OK. So five minutes. Let us try to start at least the proof of Banach-Steinhaus. Not much time, unfortunately. So OK. For any n, we introduce the following set. The set of all points in E such that Li of x is less than or equal than n for any i. Remember the assumption. Let me write here the assumption. Sup Li x less than or equal to infinity for any x. This was the assumption. OK. Now, easy fact. Then em are closed. And also the union is equal to e. Well, it is immediate by definition that you have this conclusion, but it's easy. Try to prove it by yourself that this inclusion also is a direct consequence of this. So this follows. The inclusion follows from one. Therefore, we are writing now the complete metric space as the countable union of closed sets. And therefore, there is at least one. So there exists at least one with non-empty interior. OK. And therefore, we can pick a point x0 into e. We can take radius positive such that the ball of radius r is contained in e. OK. This is a consequence of bare theorem. And this is the fact that it is non-empty interior. OK. So this is contained in this. Now what is this? This is the set of all points for any i. Now let us write explicitly this. This inclusion. What does it mean? So it means that if I write a point of this as x0 plus r, let me denote it by z, maybe, where z is in the unit ball. So all points of the ball centered at x0 of radius r can be written as x0 plus rz, where z are points in the unit ball. OK. And so we have that li of, so any point here, as li less than and not. OK. So li x0 plus rz in norm is less than or equal to not for any i. Do you agree with this? Is it OK? Is this clear? It is simply, I'm saying that all points here, which are written as x0 plus rz, are in size this set. This means that the norm of that point, of li of that point is less than or equal to not. OK. So we have this for any i and for any z in the unit ball. We have this. So this is equal to li of x0 plus rlz. And now, so this is larger than or equal than rli of z minus li of x0. And I hope that this will give me what I want because now I have that this is less than or equal than and not. So let me write it here. So r norm of li of z is less than or equal to and not plus li of x0 for any i in i or any z in b10. Now, passing to the supremum with respect to z here, passing to the supremum with respect to z is less than and not plus li of x. I am taking simply the supremum with respect to z. Therefore, the supremum with respect to z less than or equal than 1. So I have this. But you see, this is finite because now x0 is fixed. And so x0 is fixed. The supremum over i of this is finite is a number by assumption. And therefore, so this is less than or equal than n0 plus some big number by assumption. If you look at the assumption, because x0 is fixed. Now I divide by r, which is positive. And therefore, I find that li is less than or equal than which is independent of i. So taking the supremum with respect to i gives the assumption and gives the conclusion. So sorry, today I have a little bit quick because there are several things to do yet. And I'm still at the beginning of factual analysis.