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Transcript
00:10:770Paolo Guiotto: Okay, good morning.
00:34:280Paolo Guiotto: So… The topic of today is,
00:40:340Paolo Guiotto: A few basic concepts of, Norman spaces.
00:46:530Paolo Guiotto: Normally, when we, when we do… when we study a problem, Typical mathematics solves for…
00:59:230Paolo Guiotto: equations and optimization problems and things like that, at least analysis, let's say.
01:06:840Paolo Guiotto: In advanced applications, we work with functions, so equations where the unknown is a function, like different… partial differential equations, for example, differential equation as well.
01:20:380Paolo Guiotto: So, the natural environment, the objects on which we work are functions.
01:28:140Paolo Guiotto: of some variable, VI complex, doesn't matter.
01:34:570Paolo Guiotto: So we need, to be able to discuss operations like limits.
01:43:40Paolo Guiotto: Where we have, for example, a sequence of functions, okay?
01:48:350Paolo Guiotto: Now, the idea is to set up the natural environment, is to think to a set where the objects are functions, the points are functions.
01:57:720Paolo Guiotto: And, this type of sets normally have some algebraic structure.
02:04:430Paolo Guiotto: And this structure, I just remind you, because you should know very well, is the structure of vector space.
02:17:500Paolo Guiotto: A vector space is a set.
02:24:550Paolo Guiotto: set, say, V.
02:28:630Paolo Guiotto: on Weecher.
02:34:20Paolo Guiotto: to… Operations.
02:42:420Paolo Guiotto: Algebraic operations are defined.
02:49:350Paolo Guiotto: So, one is the sum.
02:54:360Paolo Guiotto: So we can give a meaning to, given two objects, F and G, in the set. We can define the object F plus G, and this object must be in the set as well.
03:09:980Paolo Guiotto: And the other one is the product.
03:16:520Paolo Guiotto: But in general.
03:18:30Paolo Guiotto: In certain cases, probability between objects of the set is defined, but in general is not defined.
03:25:120Paolo Guiotto: What we mean here with product is product by scholars.
03:32:830Paolo Guiotto: What we call the scholars.
03:35:520Paolo Guiotto: So this means, for us here.
03:39:440Paolo Guiotto: Scholars can be elements in a more general structure than this one, but for us, scholars would be real numbers or complex numbers.
03:49:380Paolo Guiotto: So it means that if you take an element of the space V,
03:54:990Paolo Guiotto: And the number, let's say alpha, That can be real, or… If we need the complexer.
04:03:860Paolo Guiotto: Then, it is well defined, this product, alpha F, and it is an element of the set G.
04:11:360Paolo Guiotto: Such that a certain number of properties are verified.
04:16:200Paolo Guiotto: So, for example, such that, the, number 1, say.
04:22:240Paolo Guiotto: Number one, the sum is commutative, so F plus G is equal to G plus F, whatever, F and G in the set V.
04:34:210Paolo Guiotto: And number two, the sum is associative, so if you do F plus G plus H, this is the same alpha F plus G, and then
04:43:960Paolo Guiotto: with some H for every F, G, and H. Then there exists a vector, which is called the zero of the space.
04:54:230Paolo Guiotto: There exists a vector of 0. Let's call for a second V, then we will use always the same letter, the same symbol, 0.
05:01:970Paolo Guiotto: there is just a vector in… in the space, in the set VE,
05:07:650Paolo Guiotto: Such that, F plus 0
05:11:330Paolo Guiotto: is equal to F for every F. And number 4, there exists the opposite to each element. So, for every element of the set, there exists another element in the set.
05:25:900Paolo Guiotto: Such that F plus G is equal to 0.
05:30:550Paolo Guiotto: This element, G, is denoted by definition as minus F, so we call minus F is, by definition, this G.
05:41:870Paolo Guiotto: Now, it can be proved that, once you have these properties, the,
05:48:760Paolo Guiotto: Provided that there exists a unique zero, let's say.
05:52:800Paolo Guiotto: provide the zero is unique, that the opposite is unique as well, so there is just one opposite.
06:00:130Paolo Guiotto: Then we have, a few properties in relation with the product by Scalas.
06:07:390Paolo Guiotto: For example, you notice that
06:10:200Paolo Guiotto: the product is not an operation between two objects of the same nature, because one is a number, and the other is an element of this set V that can be whatever. In our mind, there will be always an example. V will be a set of functions, so these F… that's why I'm also using these letters, because
06:30:120Paolo Guiotto: In our applications.
06:32:500Paolo Guiotto: at least in this course, but as you will see in most of the other courses, where you meet this, this, type of structure, we always have in mind application to spaces of functions, so the elements F, G, etc. are functions.
06:52:230Paolo Guiotto: So, as you can see, the operation is not just the algebraic product of numbers. So, what can be said about commutative? It doesn't make sense, we do not write, F times alpha. Normally, we put the scalar in front. There is an associativity, that means that if you do alpha.
07:14:340Paolo Guiotto: do not use the word Scala product, because that's another operation, okay? This is the product by Scala.
07:23:120Paolo Guiotto: So, alpha, let's use still times beta F, this is the same of alpha-beta, the scalar alpha-beta time vector F, for every values of scalars alpha-beta, real or complex, depending on the case, and for every F.
07:41:660Paolo Guiotto: In the exact view.
07:44:260Paolo Guiotto: Number 6, when we multiply by 1, by the number, 1F, we get F, so 1 is the unity.
07:57:950Paolo Guiotto: For every F in V. And then, I don't know if I'm forgetting anything here, but, however, these are very natural algebraic things. The distributivity, so if you have something like alpha plus beta.
08:15:580Paolo Guiotto: times F, this is the same of alpha F plus beta F, and similarly, alpha times F plus G,
08:24:50Paolo Guiotto: it is equal to alpha F plus alpha G. So this is for every alpha beta, scala's for every F, in D.
08:33:679Paolo Guiotto: For every alpha, Scala, for every F, and G, these are in B.
08:40:299Paolo Guiotto: So, the set V with all these properties, a sum and a product by scalars that fulfill this property, is called a vector space, okay? The elements of the space are called vectors, and
08:57:300Paolo Guiotto: And the numbers are called scalas. So you already have seen plenty of examples of this, because you have done linear algebra course.
09:07:970Paolo Guiotto: But let's say that, examples…
09:17:100Paolo Guiotto: Well, some of these examples are well known, and it is a good idea to have them in mind, because they can provide some…
09:27:760Paolo Guiotto: reference to simple examples. So, for example, if I take the set we made by real numbers.
09:34:970Paolo Guiotto: It's a bit, you can say, exaggerated to think this as a vector space, but it is a vector space, where the product alpha F is, in this case, the algebraic product alpha times F, okay? Here are alpha and F,
09:52:880Paolo Guiotto: are both the same kind of objects, are both numbers. Notice that this can be made a space, a vector space, only when alpha is real, because if you multiply a complex number by a real number, you get a complex number.
10:08:370Paolo Guiotto: And if the space is made of real numbers, you are out of that space. So here, alpha real F, real.
10:17:90Paolo Guiotto: So this is B, and this is D, referral to scholars.
10:22:860Paolo Guiotto: If I take as B equals C,
10:27:700Paolo Guiotto: also the same operation, alpha F is alpha dot F. This can be a vector space with the scalars, real, of course, but also, in this case, complex.
10:40:720Paolo Guiotto: No? Because if alpha is real, and F is an element of the space, which is in this case a complex number, alpha F is a complex number.
10:50:220Paolo Guiotto: So you are still in that space.
10:52:610Paolo Guiotto: Of course, you are in the space also if you use scalars which are complex numbers. Then we have the classical finite dimensional spaces, like what we call RT, which is the set of arrays with D coordinates, say F1,
11:11:520Paolo Guiotto: FD… with these components, FJ, that are real numbers, J equals 1 to D.
11:22:350Paolo Guiotto: In this case, you know that the two operations, F1, if you have two vectors, F1, FD, this is the vector F.
11:33:830Paolo Guiotto: plus, G1 GDE,
11:37:960Paolo Guiotto: which is the vector G. A natural definition of sum is that you sum component by component, so it is the classical F1 plus G1,
11:50:10Paolo Guiotto: then F2 plus G2, then FD plus GD.
11:55:750Paolo Guiotto: And the product by Scala time vector.
12:01:390Paolo Guiotto: alpha times the vector of F1FD is, by definition, we multiply components, each of the components, by alpha, so…
12:09:990Paolo Guiotto: alpha F1, alpha F2, etc, alpha FD. Of course, if in this case, the spaces are D, so the components must be real, alpha cannot be taken as a complex number. So in this case, alpha will be real.
12:26:720Paolo Guiotto: But we can do similar… the same kind of structure with C, D. So, the set made of arrays.
12:37:340Paolo Guiotto: F1… FD, where these components, FJ, are now complex numbers.
12:44:630Paolo Guiotto: J from 1 to B.
12:47:890Paolo Guiotto: same… same, definitions, for…
12:55:860Paolo Guiotto: sum and, product by Scalar.
13:00:700Paolo Guiotto: alpha.
13:04:990Paolo Guiotto: can be.
13:06:720Paolo Guiotto: both real, Or… complex. It depends on what we want, okay?
13:17:540Paolo Guiotto: Okay, so these are the, let's say, the standard examples. Now, let's see something…
13:25:430Paolo Guiotto: you can see also these as functions, huh? Notice that it… Notice… That an array, of this type, F1.
13:39:310Paolo Guiotto: FBE.
13:41:20Paolo Guiotto: Well, it is determined, of course, by these D numbers, F1, FD.
13:46:540Paolo Guiotto: Right.
13:47:620Paolo Guiotto: that we can see as… it is equivalent… it's a little bit of an exaggeration to do this, okay? To a function defined on the set 1D,
14:00:860Paolo Guiotto: With values in reals.
14:03:520Paolo Guiotto: Because if you have a function defined on a finite set, that function, F, is determined…
14:11:610Paolo Guiotto: by the values, it takes on the domain, which is made of these D numbers. So by F at 1,
14:21:40Paolo Guiotto: F at 2, etc, F at D.
14:25:570Paolo Guiotto: No?
14:26:570Paolo Guiotto: You can say that any function is determined by the values it takes, okay?
14:32:810Paolo Guiotto: So, in this case, a function is determined by these D numbers that we can relabel as F1, F2, F3. So, to a function, there is an array, to an array, there is a function. This correspondence is a projection.
14:50:160Paolo Guiotto: So we can see, that these examples can be seen as, particularly simple cases of sets of functions, so functions defined on finite sets.
15:04:650Paolo Guiotto: So this means that we could take something like,
15:08:980Paolo Guiotto: an extension of this, we could say…
15:12:40Paolo Guiotto: are infinity, what could be our infinity? Imagine that if we take these components, but they are infinitely many, no?
15:20:270Paolo Guiotto: So in this case, R infinity would be a set made by an array where we never stop with the components. We have components D, but then we continue.
15:30:740Paolo Guiotto: Now, what is this kind of object?
15:33:510Paolo Guiotto: An infinite list of numbers is a sequence, is what in mathematics we call a sequence, and we normally use this notation. Well, instead of using letter D, let's use letter N, where N is natural.
15:47:320Paolo Guiotto: So, this is the set made of sequences of real numbers, for example, or complex numbers, where these are real or complex, depending on what we want to do, we then naturally.
16:02:710Paolo Guiotto: And you can do the same operations, no? You take a sequence, FN, that is a vector for the space, so if you take F and G in the space V, this means that F is an anterior sequence FN, G is an anterior sequence Gn.
16:20:640Paolo Guiotto: No? What you can do is F plus G will be the sequence made of Fn plus GN.
16:28:910Paolo Guiotto: Okay, and what will be lambda F? It will be the sequence lambda FN.
16:35:270Paolo Guiotto: By definition. As you can see, these are two well-defined operations, because they give a value. So this one, no? This is a well-defined operation of some. Of course, you should verify that
16:52:380Paolo Guiotto: This sum… you should check that this sum verifies all these properties, but, I mean, it's a very straightforward check.
17:02:460Paolo Guiotto: For example, here, the zero of the vector space will be the sequence constantly equal to zero. So 0, 0, 0 is an array made of infinitely many zeros in all the components, no? This is the zero of the space.
17:20:280Paolo Guiotto: So, again, we can look at one of these,
17:26:250Paolo Guiotto: arrays, and in fact, this is the original definition of what is a sequence. What is a sequence? Formally, it is a function defined on naturals. We call F sub n the value of that function at integer n, no?
17:40:420Paolo Guiotto: So, a sequence can be identified with a function f defined on natural, real, or complex valued.
17:51:880Paolo Guiotto: Well, the identification is FN, F sub n, say, is the value of this function f at integer n, no?
18:02:10Paolo Guiotto: So, also, these ones are functions, in fact, no?
18:05:420Paolo Guiotto: So the next step will be that let's take a more general domain, no? So what if we replace this domain by a set, no? So we have generic functions defined on some domain.
18:18:60Paolo Guiotto: So we can say, you know, for example, this could be
18:22:870Paolo Guiotto: So if we have a set X, Genetic.
18:30:190Paolo Guiotto: Sette.
18:31:730Paolo Guiotto: We can say, let's take… let's define this way, then we introduce a symbol. Let's define this as the set of functions defined on X, real, valued.
18:43:780Paolo Guiotto: Or complex valued, if you want to see version.
18:47:910Paolo Guiotto: So, well, actually, there is a notation for this. It is this one. It is R2X.
18:57:520Paolo Guiotto: This is the notation to say all the functions real-valued defined on X, or C, to the accident.
19:07:160Paolo Guiotto: Is this the notation? And you can define a sum. What is the natural sum of two functions?
19:15:190Paolo Guiotto: Of course, you could do this operation in other ways, but if you look at these definitions, the natural definition of sum would be it is a function.
19:27:230Paolo Guiotto: Whose value at X is the sum of the values of F and G at point X. So this is the function that at point x is, by definition, F of X plus
19:39:770Paolo Guiotto: G of X. And this is the definition for every X in capital X. As you can see, this gives you, which is still a function defined on set capital X, real RC values. And the multiplication by scala
19:57:810Paolo Guiotto: will be a function, again, that at point x is lambda times f of x.
20:05:650Paolo Guiotto: still for every X in X. These are the natural definitions, so we will never repeat anymore. It is clear that when we deal with the spaces of functions, we have these natural operations of sum and product by Scala.
20:23:300Paolo Guiotto: Okay, so this is the algebraic structure.
20:27:430Paolo Guiotto: Now…
20:29:130Paolo Guiotto: Of course, we don't do any analysis with the algebraic structure, because the, let's say, the core business of mathematical analysis is the operation of limit, no? So the typical problem is, here.
20:44:290Paolo Guiotto: Well, now I, I know you will, you will,
20:49:570Paolo Guiotto: You will complain about this, but now, a typical problem
21:00:30Paolo Guiotto: problem is…
21:02:790Paolo Guiotto: Now, you will complain, I'm sorry, but let F be a vector's… no vector, yeah, let V be…
21:11:760Paolo Guiotto: F vector space.
21:16:280Paolo Guiotto: on ions, it's the same, huh? And, take a sequence of vectors in the
21:23:210Paolo Guiotto: Now, I'm going to confuse you, because I take a sequence of Fn, and you may think it is this object here.
21:30:630Paolo Guiotto: It is not.
21:31:890Paolo Guiotto: Okay, so let FN be a sequence
21:40:650Paolo Guiotto: off.
21:41:720Paolo Guiotto: vectors, of V, no? So, f of n is a vector for every n natural.
21:55:700Paolo Guiotto: There is nothing to visualize. Typically, when we imagine a vector space, we imagine something like R2, no, in our intuition. So this means we have a sequence of vectors, F1, F2, F3, going around in the space.
22:14:60Paolo Guiotto: And we want to define that this sequence goes somewhere.
22:19:160Paolo Guiotto: No? So, we want… Sweet.
22:24:570Paolo Guiotto: want.
22:26:280Paolo Guiotto: to give.
22:29:110Paolo Guiotto: I mean, England.
22:33:520Paolo Guiotto: 2… FN goes to sum F of the space V when
22:41:500Paolo Guiotto: we send n to plus infinity.
22:46:510Paolo Guiotto: Now, to do this, we need to say the idea behind limit is that the distance between Fn, and F gets small, provided that is big enough, no?
22:57:990Paolo Guiotto: Intuitively.
23:07:640Paolo Guiotto: FN goes to F, huh?
23:10:840Paolo Guiotto: If and only if the… let's write distance.
23:16:620Paolo Guiotto: between Fn, and F goes to 0.
23:22:730Paolo Guiotto: Now, the problem is, of course, what is the distance between vectors? What do we need?
23:29:230Paolo Guiotto: Now, if we go back to R, so on V equal R, we say that Fn, in this case, the sequence of vectors is just a sequence of numbers.
23:44:80Paolo Guiotto: goes to number F of the space, so this is a number, if and only if the distance is measured by the absolute value of the difference between the two.
23:55:550Paolo Guiotto: Huh?
23:56:430Paolo Guiotto: modulus of Fn minus F goes to 0.
24:01:370Paolo Guiotto: Okay? Now, if we have vectors, FN minus F makes sense, the difference between two vectors.
24:09:790Paolo Guiotto: So formula 1 plus the opposite of the other, that's what we would say. Deliverance means we do the difference of these two factors. And the problem is, what is this absolute value? We need something that, has, is a quantifier that should be…
24:25:950Paolo Guiotto: computed for every vector, because that time is a vector, should provide as the absolute value of a positive value.
24:34:930Paolo Guiotto: Actually, there are properties that often we have seen them in the properties of the norms. This concept is the concept of norm. So we have this important definition.
24:49:00Paolo Guiotto: So, let V be a vector space.
24:59:370Paolo Guiotto: a function.
25:01:470Paolo Guiotto: Because at the end, this is a function.
25:05:80Paolo Guiotto: A function, which is denoted by this symbol, these two bars.
25:11:350Paolo Guiotto: It is defined on V, so this means that we can calculate this function on any vector of the space, and it provides positive numbers, positive and defined
25:23:570Paolo Guiotto: Numbers.
25:25:550Paolo Guiotto: is called… Norma?
25:33:680Paolo Guiotto: on V… Well, let's say… And, the… Fair.
25:44:280Paolo Guiotto: Spades, sat, say.
25:46:900Paolo Guiotto: It keeps the width a norm, huh?
25:50:200Paolo Guiotto: is called… normed space.
26:04:230Paolo Guiotto: If… now, this function that will extend the concept of absolute value modulus must verify these characteristic properties. Number one, the first property is called positivity.
26:22:120Paolo Guiotto: Well, it is tautological in the definition, but it is better to refresh.
26:27:80Paolo Guiotto: Positivity.
26:30:720Paolo Guiotto: That norm of F is always greater or equal than zero for every F in B. It's redundant, because I… I wrote here that the norm takes values in 0 plus infinity, okay? But when you have to check if something is zero, you have never to forget that it must be positive.
26:49:300Paolo Guiotto: Okay, cannot be a negative quantity.
26:52:300Paolo Guiotto: Second, it is called vanishing.
26:58:600Paolo Guiotto: It says that norm of f is 0,
27:02:890Paolo Guiotto: So, norm geometrically should mean the length of the vector, or the distance to the origin, zero, is 0 if and only if the vector F is the vector zero, is the zero of the space.
27:18:630Paolo Guiotto: Number 3 is called the homogeneity.
27:26:40Paolo Guiotto: It says that we take the norm of alpha F, where alpha is a scalar.
27:31:870Paolo Guiotto: So, let's say, real or complex.
27:35:230Paolo Guiotto: And F is a vector.
27:37:880Paolo Guiotto: And this is, well, modulus of, in R, on R, modulus of the product, it is the product of the moduluses, okay?
27:48:520Paolo Guiotto: So here, we cannot say models of alpha times models of F, because F is a vector.
27:54:870Paolo Guiotto: We cannot say norm of alpha times norm of F, because alpha is a scalar, so that is not norm. So, it's a bit mixed, so we have models of alpha times norm of F.
28:08:260Paolo Guiotto: But this is the exact extension of the property that the models of the product is the product of the models.
28:15:540Paolo Guiotto: And number four, Perhaps the most important is the triangular inequality.
28:26:750Paolo Guiotto: As you know, the modulus of the sum is not the sum of the moduluses, no, in general. You can have the sum is equal to zero, so modulus is 0, and the two numbers are not 0, so the sum of the moduluses is definitely possible.
28:41:990Paolo Guiotto: So you cannot say that norm of F plus G will be equal to norm of F plus norm of g, but in general, it will be less or equal than norm of F plus norm of G.
28:54:520Paolo Guiotto: for every F and G, indeed.
29:00:830Paolo Guiotto: Okay, so these are the, this is the concept, its main,
29:07:900Paolo Guiotto: characteristics, sir. So, let's see examples.
29:17:00Paolo Guiotto: I suppose you… it's not the first time you've heard about this, no? At least, maybe you have not seen, in a general, definition with the vector space, perhaps you have seen, in, Rn, no, the Euclidean norm, etc.
29:35:780Paolo Guiotto: So, let's say that I give for known that, for example, if V is RD,
29:45:690Paolo Guiotto: this quantity, norm of F,
29:49:720Paolo Guiotto: This will be called… I will put a 2 here, you will understand later why we use these indexes.
29:58:790Paolo Guiotto: Because this is a special case of a large family of quantities. This is called the Euclidean Norma.
30:10:660Paolo Guiotto: Well, this is the norm that measures distances according to the Pythagorean theorem.
30:18:440Paolo Guiotto: Okay, so it is defined, so if the vector is F1FD, so I write the components here.
30:28:720Paolo Guiotto: This is, by definition, the root of F1 square plus F2 square, etc, plus FE squared.
30:39:930Paolo Guiotto: So, in a little bit shorter formula.
30:43:620Paolo Guiotto: I write the sum for J going from 1 to D of FJ square, and then instead of writing the root, I write 2 exponent 1 half.
30:54:370Paolo Guiotto: The why there is this 2 is because there is the 2 here.
31:00:340Paolo Guiotto: Okay, and to make that a norm, you need to put that exponent to 1 half.
31:06:130Paolo Guiotto: It is not straightforward to verify that this is normal. The difficulty is… part is the triangular inequality. However, since I will… we will prove this on a more general class, for the moment I don't do the proof. I suppose that perhaps you have even seen the proof of this, okay?
31:26:160Paolo Guiotto: Okay. We noticed that, and this will be a very important factor, that on RD, we have different norms available. There is not just one norm.
31:40:810Paolo Guiotto: So, still on RD, let's introduce this one.
31:45:280Paolo Guiotto: I don't know if you… if you have, never seen. This is the one norm.
31:51:860Paolo Guiotto: And this is usually called in Manhattan.
31:57:990Paolo Guiotto: Norma.
32:00:540Paolo Guiotto: Because it is an alternative way to measure distances.
32:04:600Paolo Guiotto: I will explain why it is called the Manhattan Norm. Now, this is the finder, so if F is the vector with components F1, FD, this is the finder.
32:18:230Paolo Guiotto: as the sum, J1 to D, of modulus FJ.
32:26:540Paolo Guiotto: Okay, so now you have… you see that there is a sort of similarity between these two guys, because the one here is the exponential one here, and if you imagine…
32:39:730Paolo Guiotto: Outside, you have a 1 over 1, which is 1. Here, you have the 2. So they actually belong to a large class of family, which is the family of the so-called PIN norms, which are defined this way. The PINORM is the sum
32:56:830Paolo Guiotto: J going from 1 to D, modulus FJ to exponent P,
33:03:500Paolo Guiotto: And this must be raised,
33:06:170Paolo Guiotto: to have the homogeneity, 2 exponent 1 over P. This is for P, greater or equal than 1, less than plus infinity.
33:18:10Paolo Guiotto: This is just called the PINORMA.
33:25:10Paolo Guiotto: Now, this one is much, in general, for a generic P, is much more complicated. It requires a very non-trivial inequality to prove, especially… well, actually, the first three properties are easy.
33:39:190Paolo Guiotto: The last one is difficult, okay? However, since these are particular cases of something we will see later, which is the LP normal.
33:49:890Paolo Guiotto: You remember that L1 is integral functions, there will be P integral functions soon, so we will return on this later. These are examples just for your
34:00:950Paolo Guiotto: knowledge. This one is called the Manhattan Norm. The idea is, because if you are on the Manhattan island, and you want to compute the distances between two points.
34:13:389Paolo Guiotto: in town, you will never use the Euclidean distance, because if you have to go from one point to another, the right distance for you is the sum of these lengths.
34:26:889Paolo Guiotto: And if you look at the… at how this quantum can be obtained, imagine that this is a point, this is another point. As you will see, these are subtle by the sum of the absolute values, of the difference between the coordinates. So, this comes from the…
34:43:389Paolo Guiotto: Another important norm we have on our deal
34:47:850Paolo Guiotto: is the infinity Norma. There is also this one.
34:53:120Paolo Guiotto: This one is defined as the maximum Of the modulus, of the…
35:03:400Paolo Guiotto: Of the coordinates.
35:07:800Paolo Guiotto: just for an example, just to do once, we will do several times this type of check. Let's see that one of these, one which is not too complicated, is a norm. We will do for the Manhattan norm.
35:26:280Paolo Guiotto: And similarly, you can do for the infinity norm. These two are easy. The P-norm, in general, is complicated, okay, including the case P equal to, which is the Euclidean norm.
35:36:840Paolo Guiotto: So, let's tack…
35:42:530Paolo Guiotto: that, this one.
35:47:710Paolo Guiotto: is a norm.
35:51:990Paolo Guiotto: So, the one norm is defined by
35:58:10Paolo Guiotto: sum J going from 1 to D. We remind that F is a vector of RD. It will have D components that we call F1FT.
36:10:450Paolo Guiotto: And this is defined as the sum of the absolute values of the components.
36:17:170Paolo Guiotto: Now, there is something that whenever you define a quantity, you should check that that quantity is well-defined. This is not the case here, because this is what is a definite sum of quantities, so it is here that it is well-defined.
36:32:280Paolo Guiotto: Later, we will have to deal with some quantities which are defined through multiplic derivatives and things like that, and if the concept is not similar to those quantities? Okay, and we will talk on this later. So this quantity…
36:49:890Paolo Guiotto: So, let's say that the one norm of F is well… defined that.
36:59:770Paolo Guiotto: for every F in RD, there is nothing to say. So, let's check the properties. Number one, positivity.
37:08:580Paolo Guiotto: Well, this is evident, no? Because the norm of Earth, the one norm of Earth.
37:15:840Paolo Guiotto: is the sum of modulus of FJ of positive quantities, so it will be positive. There is nothing to say.
37:25:710Paolo Guiotto: Number 2, vanishing.
37:33:150Paolo Guiotto: Norm of F equals 0 if and only if the sum of J, of the modulus of FJ is 0.
37:43:460Paolo Guiotto: Now, this is simple, because this is a sum of positive numbers, right?
37:50:710Paolo Guiotto: So, this sum can be 0 if and all if.
38:00:310Paolo Guiotto: There's… Yes,
38:05:300Paolo Guiotto: All of them must be zero, because if just one of them were positive, the sum could be positive.
38:12:780Paolo Guiotto: Okay, this is because of the apple bottle.
38:16:250Paolo Guiotto: It's not true that if you're the sum of zero necessarily returns back to zero. 1 plus minus 1 equal 0. But since they are positive, this is true.
38:29:170Paolo Guiotto: So this means that the modules FJ equals 0. That's for every J equals 1D, so FJ is 0.
38:39:680Paolo Guiotto: for the same Js. But this means that the vector F
38:44:590Paolo Guiotto: Is the vector made by all zeros.
38:48:450Paolo Guiotto: in all the companies, and this is the zero of our space RD.
38:54:810Paolo Guiotto: Okay.
38:56:370Paolo Guiotto: So, vanishing works. Homogeneity…
39:01:590Paolo Guiotto: This is straightforward, we take alpha F, we compute the norm, this is sum of J, of modulus. You know that the product works component by component.
39:14:750Paolo Guiotto: So the jth component of the product is alpha times the jth component of F. So I write here alpha F j.
39:24:440Paolo Guiotto: Then, we know that for the modulus, there is homogeneity, that means that the modulus of the product is the product of the modulus. So, we have modulus of alpha times models of FJ.
39:37:620Paolo Guiotto: Now, these models of alpha is a constant factor that we can factorize. This is models of alpha times the sum of j of the modules of EFJ.
39:49:840Paolo Guiotto: And this is nothing but our one norm of S. So at the end, we see modulus of alpha times
39:57:580Paolo Guiotto: one norm of F for every alpha, real.
40:04:40Paolo Guiotto: for every, F in RD.
40:09:820Paolo Guiotto: Number 4 is the triangular inequality.
40:15:820Paolo Guiotto: Also, this one is pretty straightforward, because we have norm of F plus G, 1…
40:25:30Paolo Guiotto: It will be sum over J. What is? This is the sum of the absolute values of the components of F plus G. But you remind that F plus G has component F1 plus G1, F2 plus G2, etc. The J component will be FJ plus GJ.
40:42:550Paolo Guiotto: Now, here, of course, you will use the triangular inequality for the modulus.
40:47:600Paolo Guiotto: That says that the modules of the sum is less or equal than the sum of the models is
40:54:110Paolo Guiotto: So you get this, so you are now summing these quantities. So if you replace these modules by these quantities, the sum will increase, because they are all bigger. So you have less or equal than sum of j of modulus FJ plus modulus GJ.
41:14:80Paolo Guiotto: And now you split the sum into two sums, sum of DFJ first, plus sum of the GJ,
41:23:770Paolo Guiotto: Second one. This one is the norm of F,
41:27:290Paolo Guiotto: 1, this is the norm of G.
41:30:550Paolo Guiotto: Wow. And since you have at certain step a less or equal.
41:36:10Paolo Guiotto: This means that norm of epa G1 is less frequent than the sum of the two norms.
41:42:640Paolo Guiotto: So I… I would say you do, check, so exercise, Check.
41:52:110Paolo Guiotto: that.
41:54:420Paolo Guiotto: also the, infinity norm is S null.
42:01:660Paolo Guiotto: on.
42:04:690Paolo Guiotto: Basically, the same norms can be also defined on CD with the same definition, so I don't, I don't insist on this because it's quite boring.
42:15:450Paolo Guiotto: Now, let's come to, Some cases where we have, let's say, two functions.
42:24:260Paolo Guiotto: We said before that also vectors of RD, arrays of RD, can be seen as functions defined on finite sets, okay, very special cases.
42:36:770Paolo Guiotto: Now, let's introduce some class of spaces, and, well, let's define another example, which is the so-called uniform
42:51:500Paolo Guiotto: It's normal.
42:54:670Paolo Guiotto: So we've… we introduced first space, LET.
42:59:240Paolo Guiotto: V, B, by definition, we call this B of X. That stands for bounded functions.
43:13:330Paolo Guiotto: on X. X is here a genetic setter.
43:17:790Paolo Guiotto: So, it is, by definition, the set of functions F, defined on X, real, or C-valued. At this stage, for most of what we say, there is no difference between real and C cases. So, I will do all calculations with reals.
43:37:570Paolo Guiotto: But the same calculation works on the complex case. Where there is a difference between real and complex is where we will… it will be at the next step, so which is a specification of node space.
43:53:210Paolo Guiotto: And these are spaces where the norm is generated by a Scala product. In that case, there is a difference between the two.
44:00:50Paolo Guiotto: the two situations. But here, there is no difference, so let's stay on the real case.
44:05:780Paolo Guiotto: So these are functions which are banded. What does it mean is there exists a constant M, such that
44:15:150Paolo Guiotto: A modulus F of X is bounded by M for every X in this set, capital X.
44:25:80Paolo Guiotto: So this is the set of bounded functions.
44:28:770Paolo Guiotto: Now, first of all, let's verify that with usual operations of sum and product by scalars, this is a vector space. V is a vector.
44:44:150Paolo Guiotto: space, huh?
44:46:730Paolo Guiotto: on… if we are in the real… in the case of real-valued function, we can only consider real scalars. If we have C-valued functions, we can consider both real or complex, depending on what we need to do.
45:02:230Paolo Guiotto: Let's check.
45:08:980Paolo Guiotto: this, huh?
45:10:980Paolo Guiotto: I've not yet introduced any norm. Okay, I'm just checking that this is a vector space. What should I check? Well, I should check that the sum
45:21:170Paolo Guiotto: takes two elements of that set and yields an element of the same set. That's what I need to check. Letter F and V be two vectors of V, that means two bounded functions on X.
45:37:710Paolo Guiotto: No?
45:38:740Paolo Guiotto: So, the claim is… Well, that's what we call.
45:47:180Paolo Guiotto: Claimer.
45:49:460Paolo Guiotto: I know it's, these are not particularly… important…
45:57:780Paolo Guiotto: checks, but this is to point out that whenever you have a space, there will be defined operations, quantities, you should always check, in principle, unless
46:13:570Paolo Guiotto: you accept, but you should always check that the structure you have is really a structure of the type. For example, the claim here is I have two vectors, I check that F plus G belongs to V.
46:27:860Paolo Guiotto: You see, if, I, I, I am in the case, I said above.
46:35:110Paolo Guiotto: The case of, this case, Pancho suave all day.
46:41:590Paolo Guiotto: It is clear that the sum of two functions real value is a function of real value, okay?
46:49:500Paolo Guiotto: But if an ally to sanctify that convolution, that…
46:53:710Paolo Guiotto: bounded functions, so not only do we have bound, but you have bounded. And it is true that when I sum two bounded functions, I get a bounded function, otherwise I don't get out of that set.
47:06:440Paolo Guiotto: or put the property you want, you have continuous functions, not space made of continuous functions.
47:15:670Paolo Guiotto: You should verify that when you sum two continuous functions, you get the continuous functions. This is a simple process, because you know that you can sum continuous functions, we get continuous functions, because continuity is preserved by these other drag operations. Like, it's no treatment.
47:32:390Paolo Guiotto: Okay? Of differential, no? You have two different functions, that sum is differential because of a theory that says the derivative of the sum of these derivatives providing the two functions are differentiable. Otherwise.
47:45:590Paolo Guiotto: you should verify this, you see. So, whenever you have a property, a characteristic property of a space, you should always check that that property is preserved by the algebraic structure, otherwise you have not a vector space.
48:02:640Paolo Guiotto: Well, this is, basically obvious, because, since F,
48:08:560Paolo Guiotto: belongs to V, this means that there exists a constant M, such that modulus F of X is less or equal than M for every X in capital X.
48:19:170Paolo Guiotto: Now, be careful, because now it is very easy, but soon it will become a little bit more complicated, because we will consider these are not interesting spaces, but there are more interesting spaces, similar to this one, that are constructed on measurability, that these checks are required.
48:37:620Paolo Guiotto: So, same for G. Of course, the bound, this constant M, is not a universal constant that holds for all the functions. It means, for a function, you have an M. For another function, you have another M. This is not… because they are all bounded, so I can say there exist
48:55:830Paolo Guiotto: I don't know, another M, let's say M tilde, such that modules G of X,
49:01:290Paolo Guiotto: is less or equal than this other constant for every X in X.
49:06:220Paolo Guiotto: So, from this, of course, it is clear that when I take modulus of X of X plus G of X, why am I computing this? Because this is, you know, by definition, it is exactly the value of F plus G at point x, no?
49:22:610Paolo Guiotto: We have seen above, now, what are the natural definitions of sum and product
49:28:600Paolo Guiotto: by Scala in the case of functions. These are pointwise sum and pointwise multiplication by Scala. Now, this is, according to the triangular inequality for the absolute value that we know since first year. This is less or equal than modulus f of x plus modulus G of X.
49:47:410Paolo Guiotto: And now, by this, I did use that since the first one is less than M, the second one is less or equal than M tilde, their sum will be less or equal than this new constant, M plus M tilde, and this solves for every X in capital X.
50:03:330Paolo Guiotto: This means that now I can conclude that also the function f plus g is bounded.
50:10:630Paolo Guiotto: And therefore, it is an element of my space, B of X.
50:17:60Paolo Guiotto: So this sum keeps, functions in that, in that space. The same, should be for the product, no? Claim…
50:28:670Paolo Guiotto: Alpha F belongs to bounded function for every alpha scalar wheel.
50:35:460Paolo Guiotto: And for every F, bounded.
50:39:460Paolo Guiotto: But basically, it is the same argument.
50:42:280Paolo Guiotto: Similar.
50:46:100Paolo Guiotto: See me now. Check.
50:49:350Paolo Guiotto: Okay, now we know that this B of X is a vector space, and we keep this with a norm.
50:57:520Paolo Guiotto: Of course.
51:00:00Paolo Guiotto: one of the first messages that come, comes from this example of RD is that on the same space, we may have different norms. There is not a unique norm.
51:15:850Paolo Guiotto: Okay?
51:17:160Paolo Guiotto: However, let's say that in certain space, there are the most natural norms. In this case, what is the most natural norm? The natural norm is what is called the uniform norm.
51:34:920Paolo Guiotto: The uniform norm is a quantity which is defined. Let's def… let's now define.
51:47:850Paolo Guiotto: We will call this infinity normal. You remind that we already introduced this symbol, no? That was for RD. And you will see that there is a similarity.
51:57:770Paolo Guiotto: We call this, by definition, the,
52:02:850Paolo Guiotto: Now, these functions are bounded, this means that in absolute value, they are bounded above by something.
52:09:530Paolo Guiotto: So we take the best of these upper bounds. So we take the supremum.
52:15:100Paolo Guiotto: We would like to take the maximum, but as you know, when we deal with infinite sets, this is…
52:21:120Paolo Guiotto: not granted.
52:22:970Paolo Guiotto: So we take the supremum, that, however, in some sense, plays the same role. If the maximum exists, they are the same, they coincide.
52:31:520Paolo Guiotto: Take the supremum of modules F of X when x is in capital X, no? So you take the set made of the values of modules of X, these are called numbers, and take the biggest possible of these numbers, more or less.
52:48:320Paolo Guiotto: Okay, they're the best upper bound for the reset.
52:52:30Paolo Guiotto: Now, you call this, infinity normal,
52:57:540Paolo Guiotto: And, this is called the uniform.
53:02:200Paolo Guiotto: Norma.
53:04:210Paolo Guiotto: We check that this is really a norm.
53:09:110Paolo Guiotto: The infinity Norma?
53:11:430Paolo Guiotto: So, we normally prefer to write this with the dot, because we say this is not the norm of a vector. It's the norm as a function, okay?
53:23:110Paolo Guiotto: So norm of F is a number, okay? It's a certain number. Norm of F equals 57, no? For that vector, the norm is 57. And the norm is a function.
53:38:60Paolo Guiotto: The function that takes a vector and gives you the number of that vector. So when I talk about the norm, I get to the function, not the norm of a specific vector. So that's why I would normally use a dot or a sharp, things like this.
53:55:320Paolo Guiotto: This is, norm.
53:59:850Paolo Guiotto: on… B of X.
54:03:980Paolo Guiotto: Oh, I suppose that, we go… Non-stop, until 10.50.
54:10:900Paolo Guiotto: Okay, before we tack this, huh?
54:13:740Paolo Guiotto: Look, we use this symbol that… it's confusing because we use this notation for a completely different thing, no? If you go back here, we defined the infinity norm of F, that here is a vector of RD, as the maximum of the modulus of the FJ.
54:32:870Paolo Guiotto: This is actually the same thing, because if you think about it, you remind that we can identify
54:39:290Paolo Guiotto: arrays of numbers… with functions, no? So this is a F of J,
54:47:890Paolo Guiotto: So we are doing the maximum of the values of modulus F of J. When taking rate, you certainly would dominate the areas 1 to B.
54:57:850Paolo Guiotto: So that maximum exists because it is the maximum of a higher number of numbers.
55:04:370Paolo Guiotto: So the maximum, they can cease.
55:07:900Paolo Guiotto: And therefore, I would write also here, supreme of modus F of J, when J is from one of D and I have the other definition. So, in other words, this one is special case of that one.
55:21:480Paolo Guiotto: Guerra?
55:23:120Paolo Guiotto: You take as capital X, the set 1, 2, 3, D, okay?
55:29:70Paolo Guiotto: Is that clear? Do you see? So… It's just a little remark.
55:38:160Paolo Guiotto: remark.
55:41:450Paolo Guiotto: If X.
55:43:450Paolo Guiotto: is the set, finite set 1, 2, 3, D.
55:48:980Paolo Guiotto: So, as you can see, functions,
55:52:470Paolo Guiotto: functions on X can be identified with vectors. Real valued… well, of course, they are bounded, because a finite number of numbers is bounded always. So, you can see that dB of X is just RB,
56:10:720Paolo Guiotto: is identified, can be identified. Of course, formally, they are not the same objects, because B of X are functions, are they arrays, but you can identify, you know, the two, and the infinity norm…
56:25:260Paolo Guiotto: the infinity norms.
56:27:410Paolo Guiotto: of F, in this case, it becomes the maximum of the modulus of F, J,
56:35:720Paolo Guiotto: When J is 1, 2, 3,
56:39:640Paolo Guiotto: D. And of course, I talk about the maximum, because this is the supremum of a finite set, so there is a maximum, and the two coincide, because when maximum exists, supremum and maximum are the same.
56:53:790Paolo Guiotto: So, as you can see, this is exactly the quantity we introduced about. So, it's a special case, that one.
57:00:820Paolo Guiotto: So if you want, you don't need to find out the solution, but I suggest you write directly the check, the exercise I left you here. Do the direct check on that definition without passing through this kind of
57:15:100Paolo Guiotto: of proof. Okay, so now, let's stay on the general case here.
57:22:940Paolo Guiotto: We have to verify that this quantity is well-defined, supremum.
57:30:730Paolo Guiotto: Does always exist as supremum of a set.
57:38:500Paolo Guiotto: Nope.
57:39:670Paolo Guiotto: Generally.
57:42:290Paolo Guiotto: In which case, the supremum does not exist.
57:46:950Paolo Guiotto: Yeah, in that case, we set supremum equal plus infinity. Okay.
57:55:620Paolo Guiotto: So, let's say that one of the features… that's just an agreement to say that in R, no, on R, each set has an infimum and a supremum. Okay, so let's say that supremum will always exist, but
58:10:650Paolo Guiotto: Here, it's a bit more, because you remind that that maybe it's a detail. In the definition of norm, a norm must be a phyt.
58:18:210Paolo Guiotto: valued function. Cannot be… take the value equal plus infinity.
58:22:600Paolo Guiotto: Norm equal plus infinity does not make any sense, okay? So, the point is still there, because is that quantity well-defined? So, is that quantity finite or not? Is that supremum finite or not? This is the question. You see?
58:43:740Paolo Guiotto: I know these are formal details, okay?
58:47:280Paolo Guiotto: But you are going to work on details in your professional future.
58:52:880Paolo Guiotto: So it doesn't matter what you are doing, what you are modeling, there will be just a little, a little comma in your machinery that doesn't work. And you have to… you have to be able to find out that.
59:08:240Paolo Guiotto: critical point, okay? So even details are, matters. Okay, so… Test that?
59:17:750Paolo Guiotto: I don't know what… what's going on with dispensing that… First, this quantity is well.
59:27:640Paolo Guiotto: defined.
59:32:510Paolo Guiotto: Why?
59:34:810Paolo Guiotto: I'm, that is, huh?
59:40:720Paolo Guiotto: That is, norm of F, infinity norm, is finite for every F in B of X.
59:49:210Paolo Guiotto: Of course, there is a reason for that, and the reason must be solved in the definition of P . B of X is made of bounded functions.
59:59:80Paolo Guiotto: So what does it mean? It means that if F, indeed.
00:07:90Paolo Guiotto: F in B of X means that, you know, the definition says that there is just a constant M, such that modulus F of X is less or equal to that constant for every X in capital X. I don't know where these numbers are. They are positive.
00:25:370Paolo Guiotto: So this is the rear liner. We have here.
00:29:220Paolo Guiotto: We have somewhere here these numbers, modulus f of x.
00:33:870Paolo Guiotto: I change X, I change the number, so maybe for another X, I have here modulus FX prime here. But what I know is that there exists a bound, M, that cannot be crossed.
00:47:610Paolo Guiotto: So, all these numbers are between 0 and M.
00:52:10Paolo Guiotto: So, in particular, this means that…
00:56:900Paolo Guiotto: The set of numbers, modulus f of x, with X in capital X.
01:04:30Paolo Guiotto: is… What would you say?
01:09:670Paolo Guiotto: Ease?
01:22:210Paolo Guiotto: This is set,
01:24:120Paolo Guiotto: This is the set of… made of all this number. So, these are the numbers modules F of X when X range in the capital X space, okay?
01:36:740Paolo Guiotto: So all these numbers are there, between 0 and a certain number, capital N.
01:41:560Paolo Guiotto: So that set is… Yeah, it's contained, and a more appropriate word is bounded… On which side?
01:58:950Paolo Guiotto: quite below, by zero. But here, since we are talking about supremum, the interest in the upper bound is upper…
02:09:330Paolo Guiotto: bounded.
02:11:850Paolo Guiotto: So this means that…
02:13:990Paolo Guiotto: This is the axioms of realism, no? If you have an upper bounded set, the supremum of that set exists, and it is fined, okay? Supremum…
02:27:160Paolo Guiotto: of this sector.
02:33:70Paolo Guiotto: Exists, and it is finite.
02:36:250Paolo Guiotto: But this is not for free.
02:38:290Paolo Guiotto: It is called completeness Axiom of Real Numbers. So, it's the trademark of real numbers. It doesn't work on rationals, okay? It's something which is a milestone that makes the difference, that makes everything possible with reals. So, completeness…
03:01:630Paolo Guiotto: axiom.
03:04:360Paolo Guiotto: However, now we know that this means that our quantity, infinity norm of F, exists, it is finite, it is well defined.
03:14:270Paolo Guiotto: This is the good position that we have to check in.
03:18:450Paolo Guiotto: Second, let's check…
03:24:730Paolo Guiotto: Let's check.
03:29:110Paolo Guiotto: Verify.
03:31:890Paolo Guiotto: the… characteristic.
03:39:800Paolo Guiotto: properties.
03:46:410Paolo Guiotto: So, what are the characteristic properties? Number one, positivity.
03:51:950Paolo Guiotto: And this is, basically evident, because, infinity norm of F,
03:58:300Paolo Guiotto: Since it is the supremum of a set of positive numbers, as we said, all these numbers are between 0 and certain bound M. In particular, they are bounded below, so the supremum cannot be here.
04:12:990Paolo Guiotto: negative. Otherwise, all the numbers would be below that supremum, so they should be down here.
04:21:460Paolo Guiotto: Okay, so the supremum must be positive. Is greater or equal than zero because…
04:30:320Paolo Guiotto: modulus f of x is positive for every x, and therefore, the supremum of modulus of f of x
04:42:70Paolo Guiotto: when x is in capital X, will be greater or equal than 0.
04:48:70Paolo Guiotto: vanishingly.
04:53:510Paolo Guiotto: What if normal of f infinity norm is zero?
04:58:370Paolo Guiotto: This means that, again, we go back to the definition, the supremum of all these quantities, models f of x.
05:07:890Paolo Guiotto: when X is in capital X, Is equal to zero.
05:14:70Paolo Guiotto: Now, what… what is the goal? It is the vanishing. We want to prove that F is zero. Zero in which sense? Is the zero of the vector space? What is the zero of this vector space?
05:26:800Paolo Guiotto: Remind that, the goal is…
05:32:990Paolo Guiotto: to prove that F is the zero of the vector space, which is the space of bounded functions. What is the zero here?
05:43:110Paolo Guiotto: the zero function, no? This is, this is function… Constantly.
05:56:570Paolo Guiotto: equal to zero. Okay, that's the zero of this function.
06:01:610Paolo Guiotto: No? Is the function f of x equals 0 for every X? So that's the goal, no? So the goal is to prove that f of x must be equal to 0 for every x in capital X.
06:15:830Paolo Guiotto: Okay, that's the goal. Can I deduce fees from this factor?
06:21:760Paolo Guiotto: I would say, what if this is false? What if the computer is not verifiable?
06:28:800Paolo Guiotto: If at least one of these values is different from the other.
06:36:930Paolo Guiotto: If…
06:41:500Paolo Guiotto: If there exists, say, an X star, such that F of X star
06:47:410Paolo Guiotto: is different from zero, so you would have that the absolute value of FX star would be
06:55:640Paolo Guiotto: strictly greater than zero. And then what about the supremum? So the infinity norm of F, which is the supremum.
07:07:370Paolo Guiotto: Of all these numbers.
07:14:800Paolo Guiotto: Question.
07:16:30Paolo Guiotto: No, it is not. But… It is…
07:22:100Paolo Guiotto: Greater, it is greater. No, because this is the maximum.
07:25:860Paolo Guiotto: It doesn't matter if you are solid on what precisely is supremum, but supremum is a moral maximum, so it's a maximum. If one of the numbers is greater than 1, the maximum must be greater or equal than 1.
07:41:700Paolo Guiotto: That's what we need, because this would say that it will be greater or equal than
07:45:950Paolo Guiotto: models FX star, which is definitely positive, and this would say that the infinity norm of F
07:52:550Paolo Guiotto: must be strictly positive, and that's in contradiction with the initial Assumption, no?
08:01:460Paolo Guiotto: Okay, so, since there cannot be an X where F is different from 0, it means that F of X must be equal to 0 for every X in capital X. And that's the conclusion, because this says F is the zero of the space.
08:20:490Paolo Guiotto: Number 3, the homogeneity.
08:24:800Paolo Guiotto: This is easier, because we have alpha F.
08:30:160Paolo Guiotto: infinity norm. We have to take the supremum of numbers, alpha f of x.
08:39:930Paolo Guiotto: Because the function alpha f evaluated at point x is alpha times f of x when x is in capital X.
08:49:10Paolo Guiotto: Of course, now, this number here is modulus of alpha times modules f of x.
08:56:760Paolo Guiotto: So we are doing the supremum of a set of numbers, where these numbers are all multiplied by
09:05:470Paolo Guiotto: The same constant factor alpha.
09:10:890Paolo Guiotto: So, say, by 2. All numbers are multiplied by 2. What would you expect about the Supremo?
09:17:729Paolo Guiotto: There is a non-trivial fact that should be checked, but we accept, no? What will be? That this factor can be carried in front of everything, so this will be modulus of alpha times supremum modulus F of X.
09:33:300Paolo Guiotto: However, if we want to be precise, we should verify that this is correct, okay? However, it's intuitive. Now, all numbers are multiplied by the same number. Take the maximum of this number, you multiply by these models of alpha, you have the maximum of the models of alpha times these numbers, okay?
09:53:60Paolo Guiotto: And this is exactly models of alpha times the infinity norm of L.
09:58:960Paolo Guiotto: Number 4, the, triangular inequality.
10:04:750Paolo Guiotto: So here, we have the infinity norm of F plus G.
10:10:870Paolo Guiotto: Which is the supremum of Madrus.
10:15:280Paolo Guiotto: F of X plus G of X.
10:19:80Paolo Guiotto: Such that X belongs to capital letters.
10:25:40Paolo Guiotto: Okay, so let's take this quantity as, above with the model sulf of X.
10:31:500Paolo Guiotto: What can be said about this?
10:33:710Paolo Guiotto: modulus of F of X plus Q of X. We can, of course, use the triangular inequality for modulus, and say that this is…
10:44:410Paolo Guiotto: less or equal than modular F of X.
10:47:430Paolo Guiotto: plus modulus G of X.
10:51:890Paolo Guiotto: So, these numbers…
10:55:160Paolo Guiotto: Each of these numbers is bounded above by X. So if I replace inside theta, I'm taking numbers which are X by X bigger, so the supervum will become bigger.
11:08:810Paolo Guiotto: So… but, instead of focusing in that way, we can do a, not literally, intermediate work, because…
11:19:70Paolo Guiotto: You may not wish also that this one to be here alone, huh?
11:23:630Paolo Guiotto: is one of those that you use to compute the norm of F.
11:28:300Paolo Guiotto: The norm of F is the supremum of the modules of f of x. So if you take a modulus f of x, that's one of the numbers you use to compute the maximum numbers, pretty much. And therefore, these numbers will be
11:44:230Paolo Guiotto: What, with respect to this one?
11:47:590Paolo Guiotto: If this is the maximum of this, feces…
11:57:570Paolo Guiotto: How will you finish?
12:04:730Paolo Guiotto: So, I'm safe.
12:09:240Paolo Guiotto: If it is norm of F,
12:14:400Paolo Guiotto: Is equal to the supremum, or if you prefer the maximum, okay?
12:19:00Paolo Guiotto: of the modulus F of X.
12:21:660Paolo Guiotto: when X is in X. Now, the question is, what is, what can I write here between these two quantities?
12:31:80Paolo Guiotto: What relation can I write?
12:34:450Paolo Guiotto: Absolutely.
12:37:390Paolo Guiotto: Order, or… Exactly.
12:40:450Paolo Guiotto: So this will be… the right. And this also for every X in capital X.
12:50:460Paolo Guiotto: Okay? So, since these numbers that you see here Each of them
12:57:670Paolo Guiotto: is less or equal than the infinity norm of F,
13:03:530Paolo Guiotto: If I further replace that number, the infinity norm of F, this number increases again, you know? And also this one, for the same reason, will be less or equal than the infinity norm of G.
13:16:910Paolo Guiotto: And this holds for every X in capital X.
13:22:130Paolo Guiotto: So what I'm saying is that…
13:24:650Paolo Guiotto: All of these numbers that you see inside the Supreme Home.
13:29:430Paolo Guiotto: Are bounded by the constant, which is the sum of these two numbers.
13:34:990Paolo Guiotto: So now, what about the supreme one? Let's write this down here, better. So, modulus f of x glass.
13:45:10Paolo Guiotto: G of X.
13:47:80Paolo Guiotto: is less or equal infinity of F,
13:50:710Paolo Guiotto: plus the infinity norm of G,
13:53:970Paolo Guiotto: for every X in capital X. Notice that this quantity here is a constant
14:00:490Paolo Guiotto: is a constant in X, of course. So, at the left-hand side, you have something that depends on X. At the right-hand side, you have a constant. So now, what would you deduce from this respect to the supremum of the quantities modulus F of X plus G of X?
14:21:180Paolo Guiotto: for X in capital X, which is, by the way, the infinity norm of F plus G.
14:29:500Paolo Guiotto: So what could you deduce now?
14:31:830Paolo Guiotto: Exactly, no? These numbers are controlled by this fixed constant, so the maximum of that, those numbers cannot be bigger of the constant, no? Will be less or equal than
14:48:00Paolo Guiotto: norm of F, infinity norm plus infinity norm.
14:52:40Paolo Guiotto: of G. And that's, finally, the triangular inequality.
14:57:50Paolo Guiotto: Okay.
15:02:420Paolo Guiotto: Okay, so this is the uniform norm. The uniform norm is the natural norm for the set of bounded functions.
15:10:910Paolo Guiotto: And, of course, the set of bounded functions on a generic X is not particularly interesting.
15:19:10Paolo Guiotto: Let's say a subclass of this, which is a very natural space of functions in our, let's say, experience, is the following.
15:32:800Paolo Guiotto: Let's take as V the set of continuous functions, on…
15:41:650Paolo Guiotto: a closed unbounded domain. So, on a domain D,
15:45:910Paolo Guiotto: Where D is the domain of RV, closed.
15:52:810Paolo Guiotto: and bandwidth.
15:55:520Paolo Guiotto: So this is what we call a convey.
16:00:960Paolo Guiotto: So, for example, because this
16:04:270Paolo Guiotto: will be an example we use often. B equal set of continuous functions on interval AB. So here, we are in dimension 1, no?
16:19:140Paolo Guiotto: Now, V, first of all, is a vector space.
16:30:450Paolo Guiotto: Bye.
16:31:790Paolo Guiotto: I don't write the formal argument, but, you know, this is a…
16:37:280Paolo Guiotto: a vector space, because when you sum continuous functions, you get a continuous function. When you multiply by a constant, you get, again, a continuous function. So, you are… the linear operations are in this… in this class.
16:55:350Paolo Guiotto: Is a vector space, it's, let's say, natural.
17:05:450Paolo Guiotto: Norma?
17:08:370Paolo Guiotto: Pizza.
17:09:840Paolo Guiotto: Now, since the domain is closed and bounded.
17:14:190Paolo Guiotto: Because of the Weioster's theorem, these functions are bounded.
17:19:280Paolo Guiotto: Because they have minimum, maximum, okay? So in absolute value, they are bounded.
17:24:630Paolo Guiotto: And therefore, it means that this is, in fact, a subset of the previous B, no? Set of bounded functions is a subset of bounded functions on the
17:38:830Paolo Guiotto: by… via SaaS.
17:41:900Paolo Guiotto: EOM.
17:44:620Paolo Guiotto: So, its natural norm would be this infinity normal.
17:51:730Paolo Guiotto: which is the supremum of modulus F of X when x is in now D, because it's called the domain D.
18:05:560Paolo Guiotto: Now, since the function f is continuous, also absolute value of F is continuous.
18:11:990Paolo Guiotto: And therefore, this supremum is actually a maximum because of the Weiss test theorem. So, we can also write that this is a maximum
18:21:410Paolo Guiotto: of values modulus f of x.
18:25:740Paolo Guiotto: when X is in G.
18:28:730Paolo Guiotto: We don't have to repeat any check, because this is just because of biased trust theorem.
18:35:650Paolo Guiotto: It is Weissian that says that is actually a maximum, okay?
18:41:290Paolo Guiotto: So sometimes we write also maximum of modulus of F on D, or also maximum of modulus F of X when X belongs to D. So we use these notations.
18:59:500Paolo Guiotto: Okay, so this is the natural… that's what we will consider as the natural norm for this type of set.
19:09:880Paolo Guiotto: But exactly as for the case of RD, where we have seen on RD, there are different norms well-defined, okay?
19:21:390Paolo Guiotto: Also here, we have different norms. Let's see another norm.
19:27:560Paolo Guiotto: So, example…
19:33:380Paolo Guiotto: So I'm still doing examples of spaces with norms. So let's take as V, and I'm going more and more into some specific examples. Take V, the set of continuous functions on a closed and bounded interval AB,
19:50:420Paolo Guiotto: which is a vector space with usual operations of sum and product by scalas, and define what we call the one norm.
20:01:560Paolo Guiotto: which is the integral from A to B of modulus F of X dx.
20:10:990Paolo Guiotto: So, let's define this by definition. Now, also here, you may notice that I am using a notation we already met above. We checked that in RB,
20:25:380Paolo Guiotto: This, this quantity, defines an order, no?
20:31:410Paolo Guiotto: Imagine what is this. This is a sum that an integral is a sum, no? A sum of what… this is a sum of absolute value of FJ that can be seen at F of J.
20:44:40Paolo Guiotto: And this is… this last one is an integral of absolute value of models.
20:52:250Paolo Guiotto: of a absolute value of X, so it's the sum of models of X, model of X. Of course, you know that, that is not just the sum, because it's a more common thing, but in some… in some sense, this reminds the other one, okay?
21:08:870Paolo Guiotto: So, you may expect a similar check, even if here we have something a bit more complicated, because the type of vectors are different, no? Before, we had arrays of D numbers, here we have an entire continuous function. Now, it turns out that this
21:29:20Paolo Guiotto: equipped with this norm, so the set of continuous functions on interval AB equipped with this quantity is normed.
21:40:370Paolo Guiotto: space.
21:43:730Paolo Guiotto: Let's check.
21:51:530Paolo Guiotto: Okay, so we accept easily the fact that V is a vector space, we are inside them, no?
21:57:960Paolo Guiotto: The ease.
21:59:880Paolo Guiotto: vector.
22:01:860Paolo Guiotto: space.
22:04:70Paolo Guiotto: It's because of thumb, and… proud of.
22:11:240Paolo Guiotto: Aw.
22:12:280Paolo Guiotto: Continos.
22:14:130Paolo Guiotto: functions.
22:17:210Paolo Guiotto: R, Continos.
22:20:970Paolo Guiotto: functions.
22:24:680Paolo Guiotto: Okay. Now, we have to check that this… let's focus on the norm.
22:29:730Paolo Guiotto: Let's check.
22:35:550Paolo Guiotto: that.
22:38:200Paolo Guiotto: Well, here, since there is a non-trivial, calculation, the integral.
22:45:420Paolo Guiotto: Not every function is integral. This one is the Riemann integral, by the way, but we could also think to the LeBague integral.
22:53:400Paolo Guiotto: In this case, we will take to the Riemann integral. Let's check that. Number 1…
22:59:760Paolo Guiotto: The one norm is well-defined.
23:05:840Paolo Guiotto: Number two, that it… it verifies Verifies… D.
23:14:170Paolo Guiotto: characteristic, properties.
23:22:750Paolo Guiotto: Well, number one requires a minimum, a minimum justification, so that comes from what? Since F is continuous, since F in V means it is a continuous function from A to B,
23:39:610Paolo Guiotto: Then, also, the absolute value of F, it is composition of continuous functions, F, followed by the modulus.
23:48:240Paolo Guiotto: is a continuous function on AB.
23:53:240Paolo Guiotto: We know that continuous functions are Riemann integral, So, a modulus of F… is… Riemann, Integral.
24:06:670Paolo Guiotto: on AB, and this just means that the integral
24:12:30Paolo Guiotto: from A to B of modulus F of X, DX,
24:17:740Paolo Guiotto: is well-defined, and this is, by definition, our quantity, the one norm.
24:26:670Paolo Guiotto: So, in this case.
24:28:580Paolo Guiotto: the ex… yeah. No, I'm… I mean, if it is not needed, I shouldn't use it. Okay.
24:43:240Paolo Guiotto: It's a distribution to… We will check that the function is a continuous, continuous moving.
24:50:780Paolo Guiotto: Or we need the, the model, we should only support individual.
24:55:850Paolo Guiotto: No, for Rieval integral, you should check. The function is measurable, and the integral of the absolute value is fine.
25:03:690Paolo Guiotto: Here, we are using a factor that we mentioned that refer to the Lebec measure.
25:09:710Paolo Guiotto: That is, if it is Riemann integral, it is also the Beg integral when we are on closing bound at the intervals, and the two integrals coincide.
25:19:990Paolo Guiotto: So, automatically, since I'm here, I'm dealing with continuous functions, which are Riemann integral.
25:27:930Paolo Guiotto: they will be Lebague integral, and the two integral are the same, so we don't need anything specific from the LeBague integral, because we are working on a restricted class of objects, so the older fashion, the integral is sufficient.
25:45:760Paolo Guiotto: Okay, now, let's check the characteristic properties.
25:50:240Paolo Guiotto: However, I want to make you clear that the fact that continuous functions are Riemann integral is a difficult theorem.
25:58:480Paolo Guiotto: It's not a trivial fact. It does not cancel automatically. It demands a proof which is complicated.
26:05:540Paolo Guiotto: So, I don't know even if you have never seen this kind of proof. Probably you heard, because just to compute integrals, you have to start from some classes of functions that you all know for sure they are integral.
26:18:620Paolo Guiotto: And usually, in first-year calculus courses, you learn that continuous functions are integral, but they… probably you don't see the proof of this factor.
26:29:580Paolo Guiotto: You have to be sure that this is a complex fact, it's a non-trivial fact.
26:33:940Paolo Guiotto: So, we say we… we say that this is easy, but there is behind a block, which is not at all an easy fact.
26:43:840Paolo Guiotto: Now, and also in the check that this is a norm, as you will see something on TV and cams, because
26:50:730Paolo Guiotto: Now, we know that it is a norm, it is well-defined. Let's check the characteristic property. We start with positivity. Well, this is evident because of the one norm of F.
27:04:300Paolo Guiotto: is the integral from A to B of absolute value of F.
27:10:280Paolo Guiotto: Now, since this is greater or equal than zero, you will use the monotonicity property of the integral, even for the Riemann integral, this is greater or equal than integral from A to B of 0.
27:22:800Paolo Guiotto: which is equal to zero, no? So you get that the one norm is positive.
27:28:310Paolo Guiotto: be vanishing. In this case, this is the difficult part of the check, because it is here where you have to invoke something we have seen, which is, suppose that one norm is 0. This means that integral from A to B of modulus f of x
27:46:580Paolo Guiotto: DX is equal to 0.
27:49:940Paolo Guiotto: What do we want? We want to prove that F is 0, right? So that's the goal. So now we have to draw something from this.
28:00:610Paolo Guiotto: Well, this is what is… it is an integral of the positive quantity equal to zero.
28:06:690Paolo Guiotto: We have seen this factor two times, once for the attack integral, and another for the Rieman integral.
28:14:480Paolo Guiotto: For a continuous function, okay? So, if we take the Levang version, that's very weak.
28:20:860Paolo Guiotto: We may think this is a Lebeche integral, so here you see that the two are different, because…
28:29:920Paolo Guiotto: recall…
28:34:90Paolo Guiotto: This was a general fact. If F is measurable on… well, let's stay on the… well, F is…
28:43:740Paolo Guiotto: maybe we use a letter, different letter G, is measurable on… I adapt, okay, the statement for the case we are considering here.
28:54:440Paolo Guiotto: And it is positive.
28:58:580Paolo Guiotto: And the integral from A to B of G, DX, is 0.
29:05:220Paolo Guiotto: Then… For the LeBerg integral, what is the conclusion?
29:13:410Paolo Guiotto: G of X equals 0, Almost everywhere.
29:22:940Paolo Guiotto: Almost every X.
29:24:850Paolo Guiotto: And this does not mean for every X.
29:30:180Paolo Guiotto: Because almost every means except for a measure zero set of points that can be used,
29:37:990Paolo Guiotto: But even if it is, made of one single point, it says this is not constantly equal to zero. So it is not the zero of my space.
29:47:880Paolo Guiotto: But, we have also checked that if we know something more about this G, so G is now continuous.
29:56:330Paolo Guiotto: positive.
29:57:720Paolo Guiotto: and the integral on AB of G is equal to 0, then in this case, we have a stronger conclusion, which is G of X is now constantly equal to 0, identically for every X.
30:12:400Paolo Guiotto: It is this second statement that we are going to use, because that's much stronger than the first one. Of course, it's stronger because we know something more on G. If G is only measurable, you can deduce only G equals 0 almost everywhere. But now we are…
30:29:380Paolo Guiotto: we know that G is a bit more… a bit, much more, it is continuous, then you get a stronger conclusion. So we use this here.
30:41:370Paolo Guiotto: applied, of course, to absolute value of F, and we get that modulus
30:46:680Paolo Guiotto: F of X is equal to 0 for every X in interval AB.
30:54:310Paolo Guiotto: Okay, but from this, it follows that f of x is equal to 0 for every X.
31:00:750Paolo Guiotto: And this means that the function f is the zero of the space.
31:08:50Paolo Guiotto: And so we have the vanishing.
31:10:30Paolo Guiotto: The other properties are simple, because we have now the homogeneity.
31:19:620Paolo Guiotto: We have the one normal of alpha F,
31:24:110Paolo Guiotto: This is the integral from A to B of modulus alpha FX.
31:29:900Paolo Guiotto: the accent
31:31:570Paolo Guiotto: Of course, inside the integral, we have this is equal modulus of alpha times modulus f of x.
31:38:320Paolo Guiotto: The modulus of alpha is a constant factor we can carry outside bilinearity. This becomes models of alpha times integral from A to B of modules f of x.
31:51:630Paolo Guiotto: DX, and that's the modulus of alpha times the one norm of F.
31:58:230Paolo Guiotto: And we have derogenic.
32:00:490Paolo Guiotto: And for the triangular inequality.
32:06:950Paolo Guiotto: It's, pretty similar, because we have the norm of F plus G, one normal, is equal to integral from A to B of modulus, f of x
32:18:930Paolo Guiotto: plus G of X.
32:28:320Paolo Guiotto: Yes, huh?
32:30:960Paolo Guiotto: Now, what we do?
32:33:830Paolo Guiotto: We apply the triangular inequality, so we said that inside the integral, this is less or equal than modulus f of x plus modulus g of x.
32:45:340Paolo Guiotto: And this holds X by X.
32:48:750Paolo Guiotto: Okay, so for every X in AB.
32:52:650Paolo Guiotto: But this means that this function
32:55:560Paolo Guiotto: modulus of F plus G is bounded by the function modulus of F plus modulus of G. When we increase the function, we increase the integral, and therefore we have that this is less or equal than integral from A to B of modulus of FX
33:14:510Paolo Guiotto: plus modulus of GX.
33:18:80Paolo Guiotto: the X. Now we use the linearity of integral. The integral of the sum is the sum of the integrals, and we get the conclusion, because we get integral from A to B of modules F plus integral from A to B of models G, and the first one is the one norm of F.
33:35:560Paolo Guiotto: The second one is the one norm object.
33:40:820Paolo Guiotto: And we got the conclusion.
33:50:920Paolo Guiotto: Okay.
33:54:470Paolo Guiotto: Well, since we have just 2 minutes left, I would prefer to stop here.
34:02:440Paolo Guiotto: And so, next time we will restart from this… let's say that we now learned what is a normal.
34:11:540Paolo Guiotto: We have seen examples, and an important fact is that on the same space, we can have different norms.
34:20:380Paolo Guiotto: So, now, A need, a natural need arises.
34:26:400Paolo Guiotto: Is there some kind of relation between different norms? Why this is so important? Remind that norm will be used to define limit, so to define how close I am to the limit, okay, the distance between points.
34:44:930Paolo Guiotto: So, is it possible that I am close to a point in some distance and not in some other? So, distances are small?
34:53:400Paolo Guiotto: in a respect to a norm and not small respect to another. This is going to be a very important problem, because this will mean that the norm changes the meaning of convergence. So you can be convergent in one… in respect to one norm and not respect to another.
35:11:780Paolo Guiotto: So, what is a convergent sequence depends on what is the… how do you measure distances. So, we need a definition of comparison between norms. That's a natural definition. We will start from this point.
35:26:160Paolo Guiotto: next time. Maybe you can also do some of the exercises. You can try to start,
35:35:110Paolo Guiotto: At least the first two do.
35:38:90Paolo Guiotto: Exercise 8, 3, 1.
35:41:940Paolo Guiotto: And, 2. Because, from the exercise tree, you need the concept of equivalent. No, well, actually.
35:50:90Paolo Guiotto: The number 2 is the exercise I left, so basically there is only exercise one.
35:55:30Paolo Guiotto: And, last thing, I published them.
35:59:550Paolo Guiotto: on the exam session, I sent you a message, so I'm repeating what I sent you in… I wrote you the message. I published the file with some past exams. You can start looking at the exam. I will upload the solutions.
36:16:750Paolo Guiotto: Along this course. And I also published the first homework, where you have to do, basically, three exercises. Two are taken from the exam set, okay?
36:27:870Paolo Guiotto: And another one, I wrote just the assignment in the homework.
36:32:240Paolo Guiotto: So you have about a week until next Friday.
36:36:170Paolo Guiotto: more than a week, because next Friday evening, to upload your solution. Please, do not write your name on the solution.
36:44:590Paolo Guiotto: I can't see each of you, because when you upload, you upload with your ID, therefore I see the file under your name, so it's not possible that I see a mess.
36:56:600Paolo Guiotto: I need to do this because the next step, after Friday, next Friday.
37:01:890Paolo Guiotto: is that I will… I don't know yet how to do this, but the idea is I will give, to each of you the paper of another… of your colleague, and you have to mark his or her paper, okay? And this must be anonymous.
37:18:530Paolo Guiotto: Please, I know that ChartGPT is able to provide the solution to these problems.
37:26:150Paolo Guiotto: But try to do by your own. I don't look at how you solve, how much… even if you are not able to solve all the three problems, but 2 out of 3, or 1 out of 3, it doesn't matter.
37:38:950Paolo Guiotto: do your solution, okay? Because the scope here is not to verify anything about how you solve.
37:46:930Paolo Guiotto: But it will be rather to verify how do you grade your colleague.
37:52:250Paolo Guiotto: Okay, this means that you must understand… you must do a work in understanding what your colleague wrote, and how to assess. It's not easy, okay? So, let's see what happens. It's an experiment we opted in QB5.
38:11:90Paolo Guiotto: Okay.
38:12:240Paolo Guiotto: Have a nice weekend.