AI Assistant
Transcript
01:04:780Paolo Guiotto: Okay, good morning.
01:08:590Paolo Guiotto: So, last time we introduced the definition of norm.
01:15:40Paolo Guiotto: And, we have seen that on the same vector space, which, we can have several norms defined.
01:23:850Paolo Guiotto: We introduced this definition that, expresses a relation between two norms. We say that the star… no, they say the norm without the star is
01:40:640Paolo Guiotto: Stronger.
01:45:100Paolo Guiotto: then… This star, then, the star norm.
01:51:640Paolo Guiotto: If there exists a constant, C, That will be positive.
01:57:80Paolo Guiotto: such that…
01:58:520Paolo Guiotto: The norm of F in star norm is controlled above by modular factor C by the norm of F.
02:07:740Paolo Guiotto: in the other node. And this must be for every app.
02:11:100Paolo Guiotto: of the space.
02:14:550Paolo Guiotto: Now, the idea behind this definition is that
02:18:550Paolo Guiotto: Say that when you are small, when the vector is small in norm without the star, it is small also in this star norm.
02:28:800Paolo Guiotto: So, since a small vector will mean a small distance, basically.
02:33:720Paolo Guiotto: Whenever you are close to someone in the norm without the star, you are close to that same point, also in the other norm. And this will have as consequence that if a sequence is convergent in one norm, it will be convergent off in the other.
02:50:850Paolo Guiotto: Okay, now, we say that the two norms are equivalent, Sweet.
02:58:800Paolo Guiotto: Also… say… That's… Is to know.
03:10:760Paolo Guiotto: And… The star known.
03:15:900Paolo Guiotto: are equivalent.
03:19:390Paolo Guiotto: If each one is stronger than the other.
03:41:500Paolo Guiotto: And, just want to have,
03:44:150Paolo Guiotto: a unique condition, since you write two times that property, it's changing the order of the two, it means that there exists two constants, C
03:54:970Paolo Guiotto: and capital C positive, such that, let's say, the star norm of F is bounded above by capital C, the norm of F, and below by liter c
04:07:960Paolo Guiotto: the norm of F, and this happens for every F in vector space B.
04:16:200Paolo Guiotto: Now,
04:18:870Paolo Guiotto: It is not difficult, it is an interesting fact. It is not difficult to prove that in finite dimensions, all norms are equivalent.
04:31:870Paolo Guiotto: So, whatever is the norm you take, it's equivalent to any other one. So, in particular, if you are in RD, all norms are equivalent to the Euclidean norm. So, there is basically, let's say, one unique norm.
04:46:580Paolo Guiotto: respect, we, we define and define limits, so there is this theorem.
04:56:190Paolo Guiotto: if…
04:58:170Paolo Guiotto: the dimension of V is finite. Here, I… I mean the algebraic dimension, also the space is the vector space generated by a finite basis, no? So, V is the set of linear combinations, say, X, J, EJ,
05:18:470Paolo Guiotto: where VXJ are real or complex, this depends on which is the case. And, EJ is, say, from, say, 1 to
05:30:290Paolo Guiotto: The set of DJ is a set of vectors of the space. So, every vector can be expressed by a linear combination of a finite number of vectors. We also denote this by the span
05:45:200Paolo Guiotto: of, vectors E1, EV.
05:52:690Paolo Guiotto: So this is what… what does this mean, that the space is finally dimensional. Then… all norms.
06:05:790Paolo Guiotto: are equivalent.
06:11:90Paolo Guiotto: it is not, particularly interesting for us, because we normally deal within finer dimensional space, but it's something that is interesting to be known. So, I don't know if I do the proof in the full details, but let's see.
06:30:450Paolo Guiotto: Now, let… Let's take… And generic norm B.
06:43:820Paolo Guiotto: And in… Norma.
06:48:660Paolo Guiotto: And we noticed this.
06:50:960Paolo Guiotto: that if we take a generic vector F of the space, the vector will be written as a linear combination. Let's use the letters F, J, EJ,
07:03:440Paolo Guiotto: Then, by the triangular inequality, Of the… this norm.
07:11:240Paolo Guiotto: This will be less or equal than sum of… I write in this way, norm of EJ modus FJ,
07:22:930Paolo Guiotto: they are numbers, so I can… I can write in the order I prefer. Now, this quantity that you see here…
07:30:770Paolo Guiotto: is a quantity that I give a name. It looks like the one norm.
07:36:20Paolo Guiotto: And in fact, if we are in RD, and this is, for example, the Euclidean norm, the vectors say EJ can be the unitary vectors of the canonical basis, the norm is 1. But it is not important. This norm will be some positive numbers, and we give a name to this quantity.
07:58:90Paolo Guiotto: Let's give a… Let's call it the one norm of F.
08:04:230Paolo Guiotto: Now, I claim that, number one, this is a normal.
08:08:590Paolo Guiotto: the one norm, and as you can see from this simple equality, it comes that the one norm is stronger than the generic norm I'm taking on F.
08:22:380Paolo Guiotto: The second part will be to prove that also vice versa also, so that the normal
08:29:130Paolo Guiotto: the generic norm is stronger than this one, okay? So, the steps are step one.
08:38:950Paolo Guiotto: Well, actually, here there is apparently a problem with the definition, so that the step one is that we prove that, is a well
08:51:150Paolo Guiotto: defined… Norma.
08:55:980Paolo Guiotto: on… V. I don't do the details, but let's see what, what, what we should verify.
09:06:490Paolo Guiotto: Now, if you look at the definition, this one norm is some norms J modus FJ.
09:15:530Paolo Guiotto: And, where… The vector app can be written in this way, okay?
09:24:900Paolo Guiotto: Now, the problem is, what if I have two different representations of the same vector? Well, this is not possible if we assume that the vector CJ are a basis, so they are linearly independent. In that case, there is a unique possible representation of… there is a unique possible,
09:46:360Paolo Guiotto: set of coefficients, FJ, such that this factor is alpha, okay? So, first… This is well-defined.
10:07:840Paolo Guiotto: So, indeed, if, EJ, J from 1 to D, is a basis.
10:22:390Paolo Guiotto: Or… the made-off linearly independent vectors. We can always do that, no?
10:30:150Paolo Guiotto: If some of them is linearly dependent on the others, we can just throw away, because it won't add anything to the space, okay? Made.
10:42:500Paolo Guiotto: off.
10:43:620Paolo Guiotto: Linearly, this is the important requirement.
10:47:590Paolo Guiotto: linearly.
10:49:760Paolo Guiotto: independent.
10:53:200Paolo Guiotto: vectors…
10:57:50Paolo Guiotto: So this means that if the same vector F in V has two different writings, so we can write as F, J, J,
11:07:480Paolo Guiotto: and at the same time, I don't know, GJ, EJ, then you would have that, subtracting the two, sum F, J,
11:16:660Paolo Guiotto: minus GJ EJ, so this particular linear combination of vectors EJ would be the zero of the vector space.
11:30:110Paolo Guiotto: But since they are linearly independent, this linear combination can be 0 if and only if each of the coefficients is 0. So we would have FJ minus GJ equals zero for every J, and this would mean that the coefficients are the same, FJ equals GJ,
11:48:840Paolo Guiotto: Alright, VJ. So, you see that this definition
11:53:360Paolo Guiotto: does not depend on a particular representation, because in fact, there is a unique representation of the vector F in the basis EJ, okay? So, the quantity…
12:05:550Paolo Guiotto: that we call the one norm of F, defined as a sum, a norm EJ modulus FJ,
12:14:800Paolo Guiotto: is, well… defined.
12:20:820Paolo Guiotto: So, it does… It's not.
12:27:630Paolo Guiotto: Dependency.
12:29:850Paolo Guiotto: On average.
12:32:940Paolo Guiotto: And on that, on the deep… Representation.
12:38:980Paolo Guiotto: of F in.
12:42:560Paolo Guiotto: basis.
12:44:700Paolo Guiotto: EJ, because there is a unique representation. So, this means that we have a well-defined quantity. Now we should check that it is a norm. So you should check vanishing positivity, vanishing homogeneity, and triangular inequality.
12:58:550Paolo Guiotto: But you can see that this is basically an L1 norm, no? A one norm of RD, no? The unique difference is that we have some coefficients in front of the FJ. Notice that these coefficients, the norms of the EJ, are strictly positive.
13:17:530Paolo Guiotto: Because we are assuming that the back of CJ are linearly independent. In particular, they cannot be zero, okay? Otherwise, they would be linearly independent. So, we know that, these are positive, so easily.
13:33:610Paolo Guiotto: This one norm verifies The… characteristic… properties Oh.
13:49:730Paolo Guiotto: Norm.
13:52:00Paolo Guiotto: So, we can say that.
13:54:140Paolo Guiotto: I would suggest you to do the… Do… The… Check.
14:04:220Paolo Guiotto: Okay. Now, we know that this is a normal, and by definition, and by this inequality, by definition.
14:15:610Paolo Guiotto: We have that the generic norm is less or equal than the one norm.
14:21:20Paolo Guiotto: for every year, indeed. And this means that the one norm This one norm is stronger Then…
14:35:600Paolo Guiotto: The genetic score.
14:38:110Paolo Guiotto: Now, the difficult part of the proof is to prove the vice versa.
14:42:810Paolo Guiotto: So, step two… Step 2 is,
14:48:320Paolo Guiotto: That this… the generic, norm is stronger then… T1 law.
15:04:150Paolo Guiotto: Now, let's see what should be done to get this conclusion.
15:08:570Paolo Guiotto: By definition, a norm is stronger than another, if and or if there exists a constant, such that, positive constant, such that you can control the weaker, let's say, the one norm.
15:20:700Paolo Guiotto: So, one norm of F is less or equal than constant E.
15:25:850Paolo Guiotto: No, I'm all there, but…
15:29:950Paolo Guiotto: the generic girls, normal, but whatever, yeah, in V.
15:35:680Paolo Guiotto: Now, if F is 0, Both quantities are zero, so this inequality is always verified.
15:43:750Paolo Guiotto: If f is different from 0, both quantities are different from 0, and so we can say that this is equivalent for F different from 0,
15:55:170Paolo Guiotto: For efficacy of there is no interest, it's true.
15:58:480Paolo Guiotto: Of saying what? I divide by this quantity.
16:04:240Paolo Guiotto: Here, I carry this on the other side. So, it means that the norm of F divided in one norm of F
16:12:790Paolo Guiotto: And they carry the constant on the other side, to say that this should be greater or equal than 1 over C, or if you want to call this quantity K, no? That should be positive.
16:26:130Paolo Guiotto: for every F.
16:28:150Paolo Guiotto: in V. So, it means that the ratio between the two norms should be bounded below… it is always positive, because it is the ratio of two positive numbers.
16:39:280Paolo Guiotto: So it is bounded below by something, zero. But what we want, it is bounded below by a positive constant, that's the difficult point. And that quantity cannot be close to zero, arbitrarily close to zero.
16:55:650Paolo Guiotto: So, what is the idea now here? How can I get this? So…
16:59:720Paolo Guiotto: We should prove that, huh? We.
17:02:460Paolo Guiotto: Shoot.
17:05:250Paolo Guiotto: Rouve.
17:08:260Paolo Guiotto: that this is basically the thesis. There exists a constant K positive, such that norm of F
17:16:180Paolo Guiotto: divided by 1 null of F is greater or equal than constant for every
17:23:109Paolo Guiotto: for every error in V, different from 0. For 0, we cannot say anything, because it becomes 0 over 0, but for vector equals 0, the conclusion is always true, okay, whatever is the con.
17:38:290Paolo Guiotto: So, this property looks like if I'm saying that this quantity is bounded below by a constant, by a positive constant.
17:49:460Paolo Guiotto: I should think that this quantity that we have at left as a function of vector F,
17:55:260Paolo Guiotto: has a minimum, is bounded below, so if I prove If we prove That.
18:06:940Paolo Guiotto: There exists, this is the key idea, there exists the minimum
18:12:930Paolo Guiotto: I just say minimum, because I can always say there exists infamum, which is the moral minimum, but if it is minimum, it is achieved, so it becomes a strictly positive constant, okay? If I prove that there exists a minimum for this quantity.
18:32:850Paolo Guiotto: when F is in D different from 0,
18:38:160Paolo Guiotto: I'm done, because this quantity has a minimum for F. It means that there exists a specific factor of F, F minimum.
18:50:40Paolo Guiotto: Such that these ratios are always greater or equal than the ratio with F meaning. And since that particular ratio is the ratio between two positive numbers, it would be positive, and it will be provided for the constant A. So, you see, if I prove this.
19:08:740Paolo Guiotto: This will imply that So there exists a vector, let's say, F-min.
19:16:800Paolo Guiotto: different from zero, such that the norm of F divided to one norm of F,
19:23:980Paolo Guiotto: is greater or equal than the value this quantity takes at that vector. So, F min divided by one norm of that F min.
19:35:780Paolo Guiotto: which is, I don't know what is it, but it doesn't matter. Definitely, it is a positive quantity, because it is the ratio between two norms of the same vector different from zero, so it cannot be zero.
19:48:110Paolo Guiotto: Because the vanishing holds, no? No, miss zero only if the vector is zero. So this will provide the value for the constant K that makes this
19:58:70Paolo Guiotto: 2.
19:59:50Paolo Guiotto: Okay, so this solves for every F different from 0.
20:04:40Paolo Guiotto: So now, the point is, we should prove that that function has a minimum.
20:11:90Paolo Guiotto: And, now how can we take in this? How can we attack this problem?
20:16:320Paolo Guiotto: Now, minimum of a function…
20:19:470Paolo Guiotto: What kind of function is this? So let's define the function, first of all. Let…
20:25:290Paolo Guiotto: Let's call it T over F.
20:28:650Paolo Guiotto: is, by definition, this function. Norm of f divided one norm or that.
20:36:460Paolo Guiotto: Now, what can ensure a minimum?
20:43:240Paolo Guiotto: Yes, plus what? The continuity plus another property, because you are thinking to buy a stress. And what says bias stress? It says that if you have a continuous function.
20:57:480Paolo Guiotto: on… On a pump back.
21:00:650Paolo Guiotto: And here, there is a little problem, because the full space minus the zero is not compact, you know? V is RV, I take out the zero, it's not compact.
21:10:840Paolo Guiotto: So, we need the first to make this thing compact.
21:14:670Paolo Guiotto: Okay, so what we do is, let's take a subset, which is definitely a compact set, that can be, for example, the set of points where the one norm is equal to 1, okay? So, and…
21:32:50Paolo Guiotto: and… set.
21:35:400Paolo Guiotto: S.
21:36:860Paolo Guiotto: as the set of vectors in V, where D1 norm Music, but for example, 2-1.
21:46:850Paolo Guiotto: This is a subset of vectors which are non-zero, because none cannot be 1, okay? What is it this, by the way? Well, intuitively, since the one norm is the set of vectors which are linear combinations, F, J, EJ,
22:03:840Paolo Guiotto: Such that that quantity is sum of these numbers.
22:08:540Paolo Guiotto: normie j modulus FJ is equal to 1. So if you want to have an idea, think about that if
22:19:430Paolo Guiotto: If,
22:21:500Paolo Guiotto: if the norms of these IJ, which are numbers, but assume that, for simplicity, they are all equal to 1,
22:29:220Paolo Guiotto: like, what happens with the canonical vectors for the canonical basis of RD, and the norm is the Euclidean norm, for example, or one of the norms we have seen they are called nomical 1. So if this is equal to 1, that set is the set of vectors, some
22:49:100Paolo Guiotto: FJ EJ… Such that the sum of the moduluses of DFJ is 1. What is this thing?
22:57:210Paolo Guiotto: Wow.
22:58:140Paolo Guiotto: If it is a sum of J squared equal 1, it would be a sphere.
23:02:530Paolo Guiotto: It is not a sphere, because it is modulus of FJ equal 1, but if you want to see… suppose that we have a dimension 2 to make simple, no? In dimension 2, if D is equal to 2, it would mean that this is the set of vectors, F1E1 plus F2E2,
23:22:530Paolo Guiotto: Where modulus of F1 plus modulus of F2 is equal to 1. Do you know what is this?
23:30:550Paolo Guiotto: In the plane, So, modulus of X plus modulus of Y equals 1 in plain XY is what?
23:41:560Paolo Guiotto: Is a box like that.
23:46:150Paolo Guiotto: So this is the set where modulus of X plus modulus of Y is equal to 1.
23:53:680Paolo Guiotto: And, so it's a sort of, the boundary of a square.
23:59:760Paolo Guiotto: Now, in a higher dimensions, it's more complicated, but you understand what kind of object it is. In any case, in any case.
24:10:200Paolo Guiotto: This set is closed and bounded.
24:14:400Paolo Guiotto: And, so it is a compact. So, S is… Compacta?
24:21:390Paolo Guiotto: Well, here there should be a little detail. Come back with respect to what?
24:26:220Paolo Guiotto: Because, our norm is, on this space is the one norm. Okay, so it is compact in,
24:35:290Paolo Guiotto: One Norma.
24:37:00Paolo Guiotto: It doesn't change, because this is basically the Euclidean norm, it's equivalent.
24:42:760Paolo Guiotto: And what about the function t?
24:45:790Paolo Guiotto: So, the function t, when we restrict on S,
24:51:40Paolo Guiotto: So if we take vectors in S, I made this also to simplify a little bit the function t, because when I am in S, the denominator is 1, it is just norm of t.
25:04:60Paolo Guiotto: Now…
25:05:640Paolo Guiotto: We have to be careful, because this function is doing what? This t is going from S to positive values, 0 to plus infinity.
25:18:340Paolo Guiotto: It is taking a vector F,
25:21:920Paolo Guiotto: And it, it's a normal bath.
25:25:810Paolo Guiotto: So, when I say I want to use the Weisser's theorem here, saying that that minimum exists because I have a continuous function on a compact set, no? My compact set would be that S, okay? And this function must be continuous
25:46:860Paolo Guiotto: Here, there is, A little but important detail, continues with respect to what?
25:57:60Paolo Guiotto: Okay, continuous with respect to the normal that makes the set closing boundless, so the one normal. So notice that we can say that the T is continuous
26:09:510Paolo Guiotto: With respect to the one norm, and that's because of the first part of the argument, because…
26:19:180Paolo Guiotto: If you compute the evaluation between two different points, you take T of F, minus T of G,
26:31:150Paolo Guiotto: And you want to show that when F is close to G, that this quantity is close to 0. This is continuity. So we are going to try to estimate this variation in terms of the distance between F and G, but which distance? One distance. Now, what I have here, this is, by definition, norm of F,
26:50:240Paolo Guiotto: my norm of G, the generic norm.
26:54:890Paolo Guiotto: Now, there is the triangular inequality for this norm. This triangular inequality says… triangular inequality…
27:04:120Paolo Guiotto: says that norm of F… sorry.
27:09:50Paolo Guiotto: Normal VAF, Minus G.
27:20:190Paolo Guiotto: No, let's… let's see in this way. Norm of F equals… I can write as norm of F minus G plus G,
27:28:800Paolo Guiotto: Then I use the triangular inequality, this says it is less than norm of F minus G plus norm of G.
27:36:490Paolo Guiotto: So we carry now this at this side, and we have norm of F minus norm of G is less or equal than norm of F minus G. This is a way to write also the triangular inequality. Now, if you flip G with F,
27:55:150Paolo Guiotto: it happens that norm of G minus norm of F, it is also less or equal than norm of G minus F, which is the same of norm of F
28:07:180Paolo Guiotto: Minus G, right?
28:09:830Paolo Guiotto: So, as you can see, if you look at this quantity here, Norm of F minus…
28:15:980Paolo Guiotto: norm of G, it is bounded above by the norm of F minus G.
28:22:430Paolo Guiotto: And this comes from the first of these two inequalities. But if you look at the second, multiplying by minus 1, everything.
28:31:450Paolo Guiotto: Here, you got that norm of F.
28:36:910Paolo Guiotto: minus norm of G becomes greater here, so you see, if I multiply by minus everything, this becomes greater or equal than minus this.
28:48:220Paolo Guiotto: You see?
28:49:790Paolo Guiotto: Now, what happens when I put the minus in front of norm of G minus norm of F? It becomes norm of F minus norm of G. So I can say that it is upper bounded by norm of F minus G, and lower bounded by minus norm of F minus G.
29:03:830Paolo Guiotto: And gluing these two properties into a unique one, this means that the absolute value of norm of F
29:10:860Paolo Guiotto: minus norm of G.
29:14:30Paolo Guiotto: It is less or equal since that quantity. If between minus this and plus the second one, it means that in absolute value, this is less than this.
29:25:140Paolo Guiotto: So we get that the absolute value of the difference is less or equal than the norm of F minus G.
29:32:290Paolo Guiotto: Now, this is the consequence of the triangular inequality. Going back to this calculation, we can say that, so…
29:40:880Paolo Guiotto: Modulusoft TF.
29:43:750Paolo Guiotto: minus TGE.
29:46:490Paolo Guiotto: Which is modules of normal wear.
29:49:800Paolo Guiotto: minus norm of G. Because of this inequality becomes less or equal than norm of F minus G.
29:57:980Paolo Guiotto: Which is less or equal, and this is the first step.
30:06:80Paolo Guiotto: or if you want, from this, that was, however, the first step, the norm is controlled by the one norm, and this is controlled by the one norm of that factor. So we read here one norm of F minus G.
30:22:790Paolo Guiotto: So this says that when F goes to G in one norma, so in the space, in the starting space, so ER,
30:33:250Paolo Guiotto: T of F goes to T of G in reals, so that function is continuous.
30:39:910Paolo Guiotto: So this means that T… is continuous.
30:45:70Paolo Guiotto: as a function, strong.
30:51:870Paolo Guiotto: S.
30:52:870Paolo Guiotto: with the one norma, to…
31:01:110Paolo Guiotto: Now, we have all the ingredients. We have a continuous function, we have a compact set.
31:08:540Paolo Guiotto: So this function has a minimum. So we can say, now, visa still applies.
31:20:620Paolo Guiotto: There exists the minimum
31:23:670Paolo Guiotto: of this function T of F when F belonged to that set S, which is a sort of unitary sphere.
31:32:980Paolo Guiotto: Okay, now I want to go back, because I wanted the minimum for every T.
31:39:680Paolo Guiotto: But this can be, realized in this way. Now, if F…
31:46:690Paolo Guiotto: is in V, and it is different. Let's give a name to this name. Let's call it capital K.
31:54:860Paolo Guiotto: If F is a generic vector different from 0,
31:59:80Paolo Guiotto: Now, what can be said about the ratio norm of F divided one norm of that.
32:08:210Paolo Guiotto: Well, it's a little trick. You just write as 1 as norm… 1 over norm… one norm of F, times…
32:16:600Paolo Guiotto: Norm of F1.
32:18:560Paolo Guiotto: Now, this is a Scala, That you can carry inside the normal by using geomogeneity.
32:27:500Paolo Guiotto: Because it is a positive scalar, no? So if I put the absolute value here, I do not change anything, no? So it is something like the modulus of lambda
32:38:130Paolo Guiotto: times norm of F. This is norm of lambda F.
32:43:400Paolo Guiotto: So this becomes the norm of F divided, the one norm.
32:50:20Paolo Guiotto: of F.
32:52:940Paolo Guiotto: I can do that, because norm, without anything, is a norm, okay? So you can carry in and out the factors with the absolute value. And if you look at this, and you compute the one norm, what happens to this guy?
33:09:270Paolo Guiotto: And if you now take this F divided by the one norm of F, and you compute the one norm of this vector.
33:19:940Paolo Guiotto: you can do the same game. You can carry outside the factor, one of a norm 1, because this… also, one norm is a norm.
33:30:670Paolo Guiotto: times the one norm of F. This is 1. So this means that this guy here belongs to the set S.
33:39:50Paolo Guiotto: So this is in the set S.
33:43:100Paolo Guiotto: It is in the set where the one norm is equal to 1. And therefore, as you can see here, I have that my quantity, norm of F, divided one norm of F,
33:59:350Paolo Guiotto: is equal to the one… to the norm, sorry, of F divided the one norm of F.
34:06:540Paolo Guiotto: which is a vector of S, and this quantity, the norm of the vector, when the vector is in the set S,
34:15:469Paolo Guiotto: As a reminder, when we look at the function t restricted to vectors of s, this is the norm of the vector.
34:23:260Paolo Guiotto: So this is T, what we call the T.
34:27:159Paolo Guiotto: of that vector, F divided by one norm.
34:31:530Paolo Guiotto: of F, but that is bounded below, it has a minimum equal to K, so that quantity will be greater or equal than that number K.
34:41:830Paolo Guiotto: And this solves for every vector, for every vector F.
34:46:850Paolo Guiotto: So at the end, we have that not only the norm of F is bounded below by a constant, but in general, the ratio, norm of F divided the one norm
35:02:360Paolo Guiotto: of F is bounded below by a constant for every F in the space P, F different from 0. And that's the conclusion.
35:13:830Paolo Guiotto: So it's not an easy theorem.
35:17:940Paolo Guiotto: It has, some interesting, good, idea.
35:25:460Paolo Guiotto: However, it says that on finite dimensional spaces, all norms are equivalent.
35:31:660Paolo Guiotto: This fact is false when we switch to infinite dimensional spaces. So, cases where we have spaces of functions.
35:42:640Paolo Guiotto: So, this factor…
35:48:240Paolo Guiotto: is false.
35:53:630Paolo Guiotto: when… the dimensions… of the Eden Bi.
36:01:950Paolo Guiotto: So we cannot say that all norms are equivalent.
36:05:640Paolo Guiotto: And basically, we have examples.
36:08:590Paolo Guiotto: With what we have seen last time. So let's see an example.
36:17:690Paolo Guiotto: What we have to learn here is how to check if a norm is stronger than another, and how to disprove this.
36:29:160Paolo Guiotto: What have you to do to prove that they are not… one is not stronger than the other?
36:34:200Paolo Guiotto: Now, we will take, as set V, a simple set, simple, relatively simple, the set of continuous functions on the interval 0, 1.
36:44:630Paolo Guiotto: And on this space, we take two norms. The first one is what we call the natural norm for this space, which is the infinity norm. So, the infinity norm that I remind you, in this case, it is defined as the maximum of the modulus of f of x.
37:03:470Paolo Guiotto: when x is in 0, 1, no? In general, the infinity is the supremum, but since here we have continuous functions, we can also write maximum.
37:13:970Paolo Guiotto: And the second one, we already checked, it is a norm, so we don't have to check anything again here, is the integral from 0 to 1 of the modulus f of x.
37:25:330Paolo Guiotto: the X.
37:26:750Paolo Guiotto: Now, we will see that the infinity norm is stronger than one norm, but they are not equivalent.
37:33:730Paolo Guiotto: So, we have… That, huh?
37:41:720Paolo Guiotto: Infinity Norm.
37:43:840Paolo Guiotto: is stronger.
37:48:610Paolo Guiotto: Then… One normal.
37:53:330Paolo Guiotto: But… They are not equivalent.
38:04:130Paolo Guiotto: So, as we will see soon, these are parts of a general family of norms, and they are extension of exactly the one norm of RD and the infinite norm of RD. So, on RD, these two quantities would be equivalent.
38:21:860Paolo Guiotto: This would be the case when the domain is not the interval 0, 1, but it is a finite set.
38:27:890Paolo Guiotto: Okay, they are equivalent, but when the domain is infinite, they are no more equivalent, and we can see with this example. To understand, if possible.
38:39:300Paolo Guiotto: here, then we will see how to prove, disprove these… these things. But, let's see what do they measure, these two quantities.
38:49:950Paolo Guiotto: And,
38:51:30Paolo Guiotto: what could be the relations between these two. And remind always that the meaning of being stronger, if you want, intuitively, is if you are smaller, if you are small for this quantity, if this quantity is smaller, it will be small also this one. It doesn't matter if the constant is huge, it's 1 million. The idea is that you can take the right-hand side, the norm of F,
39:15:200Paolo Guiotto: small as you like, and this will make the left-hand side small, okay? So, when… when you have a relation like this one.
39:24:430Paolo Guiotto: norm is stronger than star norm, it means that when norm is small, star norm must be small as well. Now, let's look here.
39:33:320Paolo Guiotto: To see why one should be stronger than the other, and why they shouldn't be equivalent.
39:38:810Paolo Guiotto: Because here, we take the interval 0, 1,
39:43:640Paolo Guiotto: We imagine for simplicity that we deal with positive function. It is not important, because they both, you see, take the absolute value, okay?
39:52:910Paolo Guiotto: So, if the function is negative, you plot the absolute value, it is positive. So, let's take, for simplicity, the directly positive functions. So, let's take a function like this.
40:06:130Paolo Guiotto: No? A continuous function between 0 and 1.
40:09:500Paolo Guiotto: So, what the infinite norm does, it takes the maximum of the absolute value. Here, the absolute value is F, because F is positive. So, it takes the maximum point, the maximum, sorry, value of F. So, the value of the norm, we may say, is the length of this interval.
40:29:330Paolo Guiotto: So this length is the infinity norm of F.
40:35:170Paolo Guiotto: Now, the second one is the one norm takes the integral.
40:39:750Paolo Guiotto: So, the area.
40:41:210Paolo Guiotto: the area between function and the x-axis. So the second quantity, rather, computes this green value, the area of this. So the green quantity is the one norm of F.
40:57:330Paolo Guiotto: Now, you understand that.
40:59:270Paolo Guiotto: If the random thing is more.
41:02:610Paolo Guiotto: means that the function is small, because that's the maximum. If the function is positive, that's the highest point of the function. If you say that that is more, it means that the function is close to the ground. So you will expect if this right thing is very smaller, the function will be smaller.
41:20:890Paolo Guiotto: As well, and therefore the area, the green area, will be small. So, intuitively, this suggests that including norm should be stronger than one norm.
41:31:480Paolo Guiotto: But, on the same ambulance, we may say that, what if the green area is small? Can we say that the meantime must be small?
41:42:810Paolo Guiotto: Here, it's more complicated, because…
41:45:870Paolo Guiotto: You may imagine that the function has a very high peak, but with a very small area.
41:52:290Paolo Guiotto: So if you take a function like this.
41:56:570Paolo Guiotto: Imagine that our function is like this.
42:00:50Paolo Guiotto: Is it around zero, and then you have a huge peak like this, huh?
42:05:30Paolo Guiotto: So this will have… of course, it's… this is not a function, but it doesn't matter. This is a big…
42:14:470Paolo Guiotto: Infinity normal.
42:17:640Paolo Guiotto: But… small.
42:21:40Paolo Guiotto: One more.
42:26:350Paolo Guiotto: So, you understand that? You cannot…
42:29:730Paolo Guiotto: small segment here and big green area. If the red segment is small, the green area will be small.
42:38:160Paolo Guiotto: But you can add a small area and bigger treadmill.
42:44:440Paolo Guiotto: So it means that you can have it this small, but this big, so you do not control this one by this one.
42:51:520Paolo Guiotto: So these seconds, the intuition suggests that this is not stronger than this one.
42:59:440Paolo Guiotto: Okay? Now, let's see what can be proved formally, and how we can justify. And as you will see, the example will be very close to what we are drawing here. So, let's say, number one.
43:17:150Paolo Guiotto: The infinity norm, so since we are talking about the norm, let's say, as a function, let's put the dot, is stronger
43:29:130Paolo Guiotto: Ben.
43:31:660Paolo Guiotto: D1 norm.
43:34:330Paolo Guiotto: So, what does it mean? By definition, it means that… so there exists a constant. I have to prove to exhibit this constant for some reason, justify the existence of this constant, such that I can control the infini… the one norm of F,
43:50:770Paolo Guiotto: with the infinity norm of F, and this happens for every F in the space B. So in this case, for every F continues on 01.
44:00:960Paolo Guiotto: Well, an important thing here is that the constant is independent of S.
44:07:750Paolo Guiotto: Otherwise, that would be always true, no? Okay? It seems stupid, but there are people who believe that this is dependent on F, and they prove something, but this time, the constant's still depending on F.
44:22:950Paolo Guiotto: That's not the proof of this. You must obtain something that is independent of
44:29:920Paolo Guiotto: Second, how can we prove this? We have to prove this for a generic app, so you don't have any specific information on this app, apart from the fact that the function is continuous.
44:40:230Paolo Guiotto: Okay, so this argument works in this way. I take the quantity I want to bound the one norm, so the one norm of f, which is by definition integral 0, 1, modus f of x, dx.
44:56:970Paolo Guiotto: And now I have to try to bound up over, so say this is less or equal constant infinity norm. So, where can I put the infinity norm here? That's the point.
45:10:340Paolo Guiotto: The infinity norm is the maximum of the moduluses of F.
45:15:170Paolo Guiotto: So, what can be said here? If I look at inside the integral, there is modulus f of x. Definitely, modulus f of x, it is less or equal than the maximum of modulus F, let's use another letter, because
45:29:150Paolo Guiotto: Here, I'm… I'm taking an X fixed, no? For an X fixed, but generic, I can say that modulus F of X will be less frequent than the maximum of all possible modules F of Y for all Ys between 0, 1,
45:45:250Paolo Guiotto: Among which there is also X, okay? Because our X is 4.
45:51:650Paolo Guiotto: X in 01.
45:54:320Paolo Guiotto: where I have to integrate models of axes less than this, which is the infinite norm.
45:59:850Paolo Guiotto: By the way, it is the infinity norm, and it is a constant.
46:04:860Paolo Guiotto: So since you have that this quantity is bounded above by this one, by the properties of integrals, if you increase the integral, you increase also the integral. So you can say that this is less or equal integral 01 of infinity norm of f in the X. But now this is a constant.
46:25:50Paolo Guiotto: So you can carry outside, and it remains integral from 0 to 1 of 1, which is 1.
46:32:70Paolo Guiotto: So we get here, this is equal to infinity norm of F. So the conclusion is that
46:39:820Paolo Guiotto: one norm of F is less or equal than infinity norm of F. As you can see, this does not depend on any specific F, so it works for every F.
46:52:760Paolo Guiotto: Continuous until 1.
46:56:600Paolo Guiotto: And so you see that the constant we have here is just 1. So that's the value of C.
47:04:50Paolo Guiotto: Of course, if you prove with 2, with 3, with 1 million, it's… it's okay, you know?
47:09:920Paolo Guiotto: Okay, so now we have the first factor proved. Now, let's go to the second one. So the second one is that the one norm
47:21:880Paolo Guiotto: is not.
47:24:120Paolo Guiotto: Stronger.
47:27:840Paolo Guiotto: Stan… infinity norm.
47:32:970Paolo Guiotto: Now, how do we prove this?
47:36:360Paolo Guiotto: So this is sort of, it's a logical negation of the property. So you have to prove that there is not a constant, formally.
47:46:700Paolo Guiotto: This means that there is not a constant, let's still call it C,
47:52:800Paolo Guiotto: such that norm of F, infinite norm, is controlled above by C, the one norm of F, and this for every F, continuous
48:04:760Paolo Guiotto: on Zoom.
48:08:330Paolo Guiotto: Now, how do we, show this?
48:12:810Paolo Guiotto: Well, a possibility is the following. Basically, it is equivalent.
48:17:920Paolo Guiotto: if I, am able to build a family of F,
48:26:320Paolo Guiotto: For which, for example, the one norm is bounded by a constant.
48:32:890Paolo Guiotto: But the infinity norm of F is not bounded by any constant. This property cannot be true.
48:40:470Paolo Guiotto: Okay? And this is basically equivalent. So, if we prove bet.
48:51:40Paolo Guiotto: There exists a family.
48:53:970Paolo Guiotto: F, indexed by N, N natural.
49:01:210Paolo Guiotto: Such that… So let's say the one norm of Fn is controlled by a constancy.
49:12:420Paolo Guiotto: But…
49:13:910Paolo Guiotto: the infinity norm of Fn cannot be controlled by a constant, for example, because it goes to plus infinity.
49:22:580Paolo Guiotto: You cannot have that inequality votes, because you plug FN, this should be FN. You click on, that's the constant times the FN10. But this is bounded, right? And say this bound, and when you send N to infinity, this explodes.
49:39:990Paolo Guiotto: So the idea is to look for… it's enough to look for functions. I, I looked,
49:47:440Paolo Guiotto: We will need infinitely many functions, not just 1 or 10, or 1 million. Because if you take any finite number of functions.
49:55:610Paolo Guiotto: No? F1, F2, whatever, F1 meter.
50:00:220Paolo Guiotto: You take all your functions, you compute the relative norms, the infinity of each of N, and the one norm of each ep n, and you will have that there is a coefficient there, no? So there is a coefficient C1 for F1, C2 for F2, C3 for F3, and so on.
50:16:610Paolo Guiotto: So when you find the number of F, you find and find the number of C. We take the biggest possible, the maximum of those C, and that C will work for all these at N.
50:29:40Paolo Guiotto: So, for a finite number of vector, this equally always worked, okay?
50:35:520Paolo Guiotto: But that's when the number of F is infinity, this is equality as a problem, okay? So that's why we need to do that. And the intuition is to make a formal
50:48:30Paolo Guiotto: Construction of this.
50:50:630Paolo Guiotto: However, the idea is I want to take an F, which is very high.
50:56:620Paolo Guiotto: But with very small area. Now, we don't need to,
51:01:530Paolo Guiotto: To find strange things, because we can think to a function made like this.
51:08:450Paolo Guiotto: For example, we take this function 0,
51:12:190Paolo Guiotto: Okay, and then we give to this function a very large value here. For example, we want to reach quote N, this to make the pick of 8N, and I take just a straight function like this.
51:29:650Paolo Guiotto: Okay, so let's see what happens. Here, you will have to put a pointer.
51:36:520Paolo Guiotto: where the function changed definition. I don't need even to write the analytical formula for this event, okay?
51:45:90Paolo Guiotto: But the point is that I want… this is high, because as you can see, the infinity norm of this, whatever it is, this FN, the infinity norm of this FN
51:57:590Paolo Guiotto: It is equal to…
52:02:340Paolo Guiotto: N, because it is the maximum
52:05:820Paolo Guiotto: of the modulus of the Fn of x when x is in 0, 1,
52:13:100Paolo Guiotto: function is positive, so you can throw away the modulus, and then the maximum of F is N. The maximum value of F is N. So we have this condition.
52:24:960Paolo Guiotto: Now I want to fulfill also this one. I want that the one norm be bounded, not too big. The one norm is the area, and so it's the area of this triangle here.
52:37:680Paolo Guiotto: Because the other part will not give any contribution, it's zero the function.
52:42:110Paolo Guiotto: So now, if you take, for example, this is equal to 1 over n.
52:47:450Paolo Guiotto: This is a triangle base equal 1 over N, H equal N. So the area is 1 half N times 1 over n, so it is 1 half.
52:57:390Paolo Guiotto: So the… I don't need to compute anything, any integral, to write any function to say that the one norm of this will be equal to 1 half times 1 over n times n. So, it is constantly equal to 1 half.
53:14:960Paolo Guiotto: And this makes the second condition be verified.
53:18:800Paolo Guiotto: So now, you see that it is impossible that there exists a constant C, no?
53:23:760Paolo Guiotto: If… The one norm, where… Stronger.
53:34:270Paolo Guiotto: Dan.
53:37:220Paolo Guiotto: The, if it is normal.
53:40:480Paolo Guiotto: then there would be a constant C, such that… norm of math.
53:46:160Paolo Guiotto: infinity norm of F, less frequent constant, one norm of F.
53:50:890Paolo Guiotto: this for every F, but then when I plug this FN inside this inequality, I would have that the infinity norm of Fn should be less or equal constant, the infinity… the one norm of this FN.
54:05:770Paolo Guiotto: But at left, we have N. At right, we have C over 2. So, this should be verified for what? For every N natural. And that's, of course, impossible.
54:18:660Paolo Guiotto: and less or equal than constant for every n. That's not possible.
54:23:920Paolo Guiotto: So you get a contradiction, and this means that it is wrong to say that the one norm is stronger than Finn's norm. So they are not equivalent at the end.
54:34:390Paolo Guiotto: So, as you can see, there is a sort of well-defined strategy. You have to prove that one norm controls the other. There is no way. You have to bound one quantity by the other one, apart from multiplying factors.
54:49:230Paolo Guiotto: And this must be for generic vectors, or for generic function, etc. You want to disprove that one bounds the other, you just need to find a specific F for which the dominant is small, or bounded, and what is supposed to be smaller is huge, okay?
55:11:160Paolo Guiotto: So as usual, we continue until, Yeah.
55:16:370Paolo Guiotto: Okay, let's see an example, very, very standard problem. Example age… to seek some.
55:26:850Paolo Guiotto: So, here we have the space V is the set C1.
55:31:170Paolo Guiotto: 0, 1… So, you know, C1 means what? It means functions F, which are continuous on 0, 1,
55:41:390Paolo Guiotto: Such that… there is just the derivative.
55:46:330Paolo Guiotto: of F, and this function is itself a continuous function.
55:53:240Paolo Guiotto: the functions.
55:54:680Paolo Guiotto: Differentiable function whose derivative is continuous.
55:59:80Paolo Guiotto: On this set, we define… We define this norm.
56:09:570Paolo Guiotto: the V norm called, is by definition this modulus of F at 0, plus infinity norma.
56:19:600Paolo Guiotto: of F prime.
56:24:500Paolo Guiotto: Well, actually, the first, question is, check…
56:31:770Paolo Guiotto: that this V, equipped with this norma, is a norm.
56:41:20Paolo Guiotto: space.
56:42:630Paolo Guiotto: Well, let's, solve this.
56:46:860Paolo Guiotto: So, the answer? Well, first of all, V is a vector space. This is a…
56:53:400Paolo Guiotto: straightforward, because if you do the sum of two C1 functions, you get the C1 function. Because of all properties of the derivative, the derivative of the sum is the sum of the derivatives, and when you sum continuous functions, you get a continuous function, okay? So V is a vector
57:13:600Paolo Guiotto: Space, but let's say, easy.
57:18:420Paolo Guiotto: A little bit, more delegated, and not particularly complicated, is the check of that quantity is well-defined, it is a norm.
57:28:400Paolo Guiotto: So, this is a norm.
57:35:50Paolo Guiotto: Well, first of all, it is well-defined.
57:39:190Paolo Guiotto: It is wealth.
57:42:60Paolo Guiotto: defined…
57:44:200Paolo Guiotto: Because, well, F with a first F of 0, the value of f at 0. There is no problem. Then we do the infinity norm of f prime.
57:55:980Paolo Guiotto: And this requires a little bit of justification, because…
58:02:270Paolo Guiotto: F prime, we know that if F is in V, F' is continuous.
58:08:940Paolo Guiotto: Huh? E is continuous.
58:11:970Paolo Guiotto: When?
58:14:520Paolo Guiotto: F is in V, which is the set C1.
58:19:240Paolo Guiotto: So, the infinity norm of F' makes sense.
58:28:860Paolo Guiotto: Okay, now we know that it is well defined.
58:32:200Paolo Guiotto: Let's check the, characteristic property. Check… Off.
58:40:240Paolo Guiotto: characteristic, good.
58:42:490Paolo Guiotto: properties.
58:48:110Paolo Guiotto: Well, clearly, as you can see, the V-norm is a sum of two quantities, both positive, no? So, it is clearly positive. So, positivity…
59:00:940Paolo Guiotto: norm of F, the norm is modulus of F at 0, so this is the absolute value greater or equal than zero, plus the infinity norm of F prime, which is a positive quantity as well, so that the sum will be positive this for every F.
59:19:780Paolo Guiotto: In V.
59:21:660Paolo Guiotto: Second, vanishing, that here… is the sage.
59:28:50Paolo Guiotto: the… slightly less…
59:31:450Paolo Guiotto: Trivial, check. Norm of F, the norm equals 0 means what? Means modulus F0 plus infinity norm of F prime equals 0.
59:45:560Paolo Guiotto: Now, this is the sum of two positive quantities that is equal to zero. So to be zero, both terms must be equal to zero, so we have that.
59:55:630Paolo Guiotto: At the same time, modulus F0 is equal to 0, and infinity norm of f prime is equal to 0.
00:04:590Paolo Guiotto: But this means that from the first weak gain, f of 0 must be equal to 0,
00:10:780Paolo Guiotto: From the second, well, we know that infinity norm is a norm, no?
00:16:260Paolo Guiotto: on the space of continuous function. This guy is an element of that space.
00:22:440Paolo Guiotto: So, that's the vanishing for that normal. When this happens, if and only if F' is the zero of that space, the space of continuous function. So, it is identically equal to zero.
00:37:590Paolo Guiotto: Now, here we know that derivative is constantly 0, and this means that, since we are on an interval.
00:45:830Paolo Guiotto: on 01.
00:47:790Paolo Guiotto: Necessarily, this happens if and only if the function f is… constant.
00:54:470Paolo Guiotto: F is constant.
00:57:320Paolo Guiotto: And since we know now this.
01:00:160Paolo Guiotto: we added information that f of 0 is 0, the function must be constantly equal to 0, okay? At one point, it's 0. It is constant, it is constantly equal to 0. So we get, at the end, F identically equal to 0, and this means that that is the zero of the vector space.
01:18:110Paolo Guiotto: Now, for the homogeneity and triangular inequality, it is straightforward, because if you look at the definition, the norm of alpha F, the V norm, you have to do alpha F evaluated at zero, but this is alpha times f of 0.
01:35:590Paolo Guiotto: Plus…
01:36:790Paolo Guiotto: you have to do the derivative of alpha F and take the infinite norm. But the derivative of alpha F, since alpha is constant, this is alpha F prime. That's the linearity of the derivative.
01:50:160Paolo Guiotto: Then, you can split this absolute value into the product of modulus of alpha times modulus f of 0, and for the second one, you use the fact that you know that infinite norm is a norm. So you can carry out factors as modulus of the scalar.
02:06:420Paolo Guiotto: alpha times infinorm of f prime. And as you can see, this is exactly models of alpha, V norm of F.
02:16:900Paolo Guiotto: And, the triangular inequality is similar, no? So you have the V norm of F plus G,
02:27:50Paolo Guiotto: This is, by definition, 1. F plus G evaluated at 0, absolute value.
02:33:300Paolo Guiotto: But the value at 0 of the sum is the sum of the value, by definition of sum. So this is modulus F0 plus G…
02:45:120Paolo Guiotto: of the derivative of the sum, F plus G.
02:49:630Paolo Guiotto: But the derivative of the sum is the sum of the derivatives, so this is F prime plus G prime.
02:56:90Paolo Guiotto: Okay, now we use the triangular inequality for the models.
03:00:340Paolo Guiotto: So modulus F0 plus G0 is less or equal than F0.
03:05:100Paolo Guiotto: modulus of F0 plus modulus G0. Plus, we know also that infinity norm is a norm. So, norm of F' plus G prime is less or equal than norm of F prime of
03:18:820Paolo Guiotto: plus norm of G prime.
03:21:850Paolo Guiotto: Now you put together these two and these two, and you get norm of F plus norm of G. So this is V norm of F plus
03:32:90Paolo Guiotto: Venom of G.
03:34:180Paolo Guiotto: And that's it for the chat.
03:37:500Paolo Guiotto: Okay, now it says…
03:41:200Paolo Guiotto: Well, here I forgot to tell you a definition, but we will give this in a second. So, there is a second question. This was the number one. The number two says.
03:54:670Paolo Guiotto: Let now W be the set of continuous functions on 0, 1.
04:03:840Paolo Guiotto: equipped with the natural norms, so norm of F in W is the infinity norm of F.
04:11:630Paolo Guiotto: We don't need to check that this is enormous space, because we already know this, okay?
04:16:590Paolo Guiotto: Now, it, the question is, the following. Check that,
04:26:990Paolo Guiotto: There is this notation. The space V with the norm.
04:32:790Paolo Guiotto: There is this, hooked arrow.
04:37:220Paolo Guiotto: That means, is embedded. Now we… I'll give the definition, embedded.
04:45:330Paolo Guiotto: into… W with the W null.
04:51:320Paolo Guiotto: So, definition.
04:55:360Paolo Guiotto: In general, If we have two normal spaces, so V
05:01:650Paolo Guiotto: with a certain norm. And W… with a certain norm, W.
05:09:470Paolo Guiotto: we say, that.
05:15:300Paolo Guiotto: As in, in this notation. The space V With its normal.
05:24:690Paolo Guiotto: is embedded.
05:30:960Paolo Guiotto: into… W, with its normal.
05:37:10Paolo Guiotto: Beef.
05:40:370Paolo Guiotto: First of all, if V is contained, so it is a subset.
05:47:950Paolo Guiotto: So number one, V must be contented into W. And number two, we want that this inclusion is also an inclusion between norms, in some sense. And this is that the V norm is stronger than W norm.
06:06:10Paolo Guiotto: on what? Not necessarily on vectors of W, because W may be… on W, the V norm is not defined, the space is bigger, and this is an example, because you see the V norm here.
06:20:570Paolo Guiotto: is a norm defined on the space of C1, and it uses the derivative. You cannot compute this quantity for a function f, which is just a continuous function, because if there is no derivative, you cannot compute this quantity.
06:36:890Paolo Guiotto: So, what we want is that, the… VINORMA… is stronger.
06:48:580Paolo Guiotto: than… W normal.
06:54:120Paolo Guiotto: But on V, so on vectors for which both make sense. So we would say that the V norm, the W norm of F,
07:04:370Paolo Guiotto: It's less or equal than some constant, so there exists a constant.
07:09:320Paolo Guiotto: for which this is bounded by the V norm of F for every F in the set V.
07:19:990Paolo Guiotto: Okay, so this is the general definition. Now, the exercise asks to check that this inclusion holds. This is not a set… just a set inclusion. It is an inclusion, not only of elements, but also of… between… with the relation between norms.
07:38:580Paolo Guiotto: So, elsewhere?
07:43:310Paolo Guiotto: So, our V is, C101, And our W is C01.
07:54:110Paolo Guiotto: So it is clear that the inclusion holds. C1 functions are continuous functions.
08:01:500Paolo Guiotto: Now, let's see…
08:09:430Paolo Guiotto: if… we have that the V norm is stronger than W norm.
08:16:420Paolo Guiotto: the VINORMA… is stronger.
08:23:410Paolo Guiotto: land.
08:25:920Paolo Guiotto: W. Norma on… V.
08:30:680Paolo Guiotto: That is, let's translate this into a specific property that should be checked.
08:37:670Paolo Guiotto: That is, if there exists a constancy, Such that…
08:42:189Paolo Guiotto: The W norm of F… the W norm is… the W norm is the infinity norm. So I want to bond the infinity norm of F
08:52:979Paolo Guiotto: by a constant times the V norm of F.
08:57:689Paolo Guiotto: The V-norm is a modulus of F at 0, plus infinity norm of F'.
09:06:279Paolo Guiotto: And this must be for every F in C101.
09:14:870Paolo Guiotto: So, the goal is, to basically…
09:19:520Paolo Guiotto: forget the constant will come as a byproduct of this estimation. You want to estimate this with this.
09:28:529Paolo Guiotto: So, if you look at the left-hand side, here we have the maximum of the models of F.
09:34:920Paolo Guiotto: But the maximum of the values of it.
09:37:370Paolo Guiotto: At the right-hand side, there is one granular diaphor.
09:40:850Paolo Guiotto: and the maximum of the values of the derivative. So the problem is, how can I estimate the values of the function with values of the derivative?
09:52:910Paolo Guiotto: And now, what are the relation, what are the relations between the function and its derivative? That's the problem.
10:01:510Paolo Guiotto: So what is, what I did there?
10:04:870Paolo Guiotto: So, do you know formulas that… we know, of course, the definition of derivative, no? This is F prime equal limit, etc. This gives F prime a function of F. But what I want is the contrary. I want F
10:18:820Paolo Guiotto: In function of F prime. Who gives that?
10:26:610Paolo Guiotto: Yeah, of course, the opposite, operation. Now, we needed to put this on a more,
10:33:40Paolo Guiotto: exact form. So, D integral, No? F is the integral of what?
10:40:460Paolo Guiotto: Off.
10:43:700Paolo Guiotto: Integral is the opposite of…
10:46:650Paolo Guiotto: derivative. So, the relation is F is the integral of the derivative. What's that relation exactly?
10:55:770Paolo Guiotto: So… recall… That… There is a fundamental theorem of integral calculus, So, D… fundamental… TRM.
11:16:400Paolo Guiotto: of integral… Cariculus.
11:21:700Paolo Guiotto: Says.
11:25:440Paolo Guiotto: But under certain importance, let's write first what is the conclusion. You remind this formula, perhaps. F of B minus F of A equal integral from A to B of F prime x dx.
11:41:240Paolo Guiotto: No?
11:42:220Paolo Guiotto: Now, under what circumstances is this valid?
11:53:820Paolo Guiotto: good circumstances, so what is the requirement of… on F? F must be differentiable, otherwise you don't have the derivative, but then you have also an integral, and so you need some assumption of an F prime.
12:07:460Paolo Guiotto: Now, this formula holds when the function f is exactly the C1 function, so function continues with derivative continuous.
12:18:380Paolo Guiotto: Provided the…
12:23:190Paolo Guiotto: F is in C1.
12:26:470Paolo Guiotto: interval AD.
12:29:200Paolo Guiotto: then this formula applies. Okay, now I want to use this.
12:32:900Paolo Guiotto: tool.
12:34:240Paolo Guiotto: to estimate this, the infinity norm. Okay, so now the infinity norm of F is… the maximum…
12:43:630Paolo Guiotto: of the absolute value of FX when x is between 0, 1.
12:49:610Paolo Guiotto: Okay, so I now needed to, somehow, to use this formula into this.
12:55:220Paolo Guiotto: modulus f of x. Well, let's first take f of x, then we put the modulus.
13:01:380Paolo Guiotto: So, F of X, What could be the idea?
13:15:670Paolo Guiotto: I don't have any idea. And remind that there is also the value of X at 0 somewhere, so we need also to add that value.
13:24:240Paolo Guiotto: 0 plus the integral… Exactly. F0 plus integral 0 to X, F prime.
13:32:640Paolo Guiotto: So now, since X has been used, I changed later, I call it Y.
13:38:250Paolo Guiotto: DY.
13:39:410Paolo Guiotto: Now, this is the key passage. After this, it is more or less straightforward, because now you take the absolute value of F of X,
13:48:460Paolo Guiotto: This will be the absolute value of what you read here.
13:52:00Paolo Guiotto: Can you use the triangular inequality, less or equal? Modulus F0, which is good, because it is part of what there will be at right. At right, I will have this quantum, you see?
14:04:890Paolo Guiotto: Okay, so that model step of 0 is inside that quantity, so I keep there, plus integral 0 to X, absolute value here, F prime y dY.
14:19:210Paolo Guiotto: Okay, now, as you may expect, from this integral, we come out at the infinity norm of f prime.
14:26:320Paolo Guiotto: Now, we already seen something like this now, integral with the function infinity norm. So what we do? We first carry the modulus inside, triangular inequality of integral. So this is less to equal integral 0 to X modulus f prime y.
14:44:30Paolo Guiotto: DY.
14:45:550Paolo Guiotto: And this quantity inside here, These are bounded by…
14:55:660Paolo Guiotto: Just by… the infinity normal of f prime.
15:01:830Paolo Guiotto: Okay, so if I plug this constant into the integral, first of all, the integral will increase, so my inequality will continue to be… to increase, so I have this plus integral 0x, infinity norm of f prime. Now, that's a constant.
15:19:100Paolo Guiotto: I carry outside, and the integral is integral of 1 between 0 and X. It is equal to X. So…
15:30:480Paolo Guiotto: Probably one second.
15:55:940Paolo Guiotto: Let me check if we're still recording.
15:59:810Paolo Guiotto: It seems it is okay.
16:04:750Paolo Guiotto: Okay, so, now we have… this is modulus F0,
16:10:140Paolo Guiotto: plus infinity norm of f prime, then there is the integral which is equal to X.
16:17:660Paolo Guiotto: Okay, so we, we, we got, we got this modulus F of X less or equal modulus F of 0,
16:29:170Paolo Guiotto: Plus infinorm of, F times X.
16:35:200Paolo Guiotto: Okay, we are almost at the conclusion, because what we want is the infinity norm of F.
16:41:920Paolo Guiotto: But that's the maximum of the modulus f of x, so I can say that when I take the maximum.
16:49:330Paolo Guiotto: of modulus f of x, which, for X in 01.
16:55:890Paolo Guiotto: Of course, it will be less or equal than the maximum.
16:59:250Paolo Guiotto: for X in 01 of this right-hand side. So this modulus F0 plus infinity norm of f prime times X.
17:12:280Paolo Guiotto: But, at right, we read infinity norm of F.
17:17:980Paolo Guiotto: Less or equal. The maximum here is attained when x is 1.
17:23:130Paolo Guiotto: So it is less or equal modulus F0 plus infinity norm of F', and that's exactly the V norm.
17:33:270Paolo Guiotto: of F. That was the W norm of F, and here we are with the conclusion.
17:41:180Paolo Guiotto: So, we proved that the W norm is controlled by the V norm, and this solves, of course, for every F for which we apply the theorem, so C1, and that's the space V, no? So, for every F in V.
17:58:650Paolo Guiotto: Which was exactly the God.
18:03:410Paolo Guiotto: Okay?
18:05:820Paolo Guiotto: So, I'd say that there are basically all the exercises at the end of the chapter to do.
18:13:670Paolo Guiotto: So, do… maybe we will do some tomorrow. 833… Mmm, 4…
18:23:640Paolo Guiotto: Like, you can do basically all of them, let's say, until… They're not. Okay.
18:29:970Paolo Guiotto: Maybe, I would say, for tomorrow.
18:34:100Paolo Guiotto: Mmm… I would say you do.
18:38:490Paolo Guiotto: particular numbers… for… the…
18:52:780Paolo Guiotto: Do you think so?
18:54:950Paolo Guiotto: and 9.
18:57:290Paolo Guiotto: at least do these three. Two of three, I will do the solution in class. Okay?
19:06:720Paolo Guiotto: Okay, now we are a bit familiar still.
19:12:870Paolo Guiotto: A little bit more than 10 minutes. So now we are familiar with the definition of vector space, normed space, norm. An important family of normed spaces is the class of LP spaces.
19:30:820Paolo Guiotto: Now, we've met… we already met the definition of L1, no? So, let's remind that, let… X… F…
19:41:690Paolo Guiotto: mule B.
19:43:910Paolo Guiotto: a measure.
19:46:660Paolo Guiotto: face, huh?
19:48:80Paolo Guiotto: We already introduced the SATS L1, We… defined… L1.
19:59:580Paolo Guiotto: We just say L1X, but sometimes we will write to get
20:04:960Paolo Guiotto: If we need to be extremely detailed, we will write everything except mu, or sometimes we will just write L1, if the space is clear.
20:19:800Paolo Guiotto: Now, this is the set of functions of F, which are measurable on X, and such that the integral on X of modulus of F is found.
20:32:740Paolo Guiotto: This is his old one.
20:34:750Paolo Guiotto: Now, a simple factor is the following, that L1X Is a vector space.
20:53:420Paolo Guiotto: And…
20:59:720Paolo Guiotto: the quantity that we will call one norm. As you will see, the same symbol is returning, but this is, in fact, the same quantity we considered a few minutes ago.
21:12:840Paolo Guiotto: So the one norm in general is defined as the integral of modulus F d mu.
21:20:80Paolo Guiotto: The only point here is that this is not 100% a norm, because we have a little difficulty with the vanishing, okay? In any case, we will consider as a norm, so we say that it is
21:34:480Paolo Guiotto: And this quantity is a norm.
21:39:970Paolo Guiotto: on L1.
21:42:150Paolo Guiotto: Excellent. Wid.
21:45:400Paolo Guiotto: vanishing.
21:51:20Paolo Guiotto: in… weak… form.
21:58:600Paolo Guiotto: But you will understand it is norm of F, one norm of F equals zero, does not mean that
22:06:190Paolo Guiotto: F is constantly equal to 0, but F is equal to 0 almost everywhere. So it means that here there is a little formal problem.
22:21:500Paolo Guiotto: Either we consider the space as a vector space, and the function constantly equal to zero is the 0, and this is saying that the vanishing is not necessarily verified, because you may have that the norm is zero, but that is not constantly equal to zero. So that would be a problem.
22:40:760Paolo Guiotto: Or, we change the definition of the space, which is a formal trick, to identify those functions that are equal to zero almost everywhere, as in they are a unique function.
22:54:690Paolo Guiotto: There is an algebraic operation to do this. It is explained in one of the notes. We will never use, so I do not spend time on this. If you are curious, you can read. To make this a true norm.
23:07:180Paolo Guiotto: on a true space. So we prefer, just to say, L1 is this, and we accept that when we have the vanishing, there is this little ambiguity that says it is not necessarily equal to zero, but it is equal to zero almost everywhere. So let's see the little proof here. It's not particularly complex.
23:29:740Paolo Guiotto: First of all, it is a vector space.
23:32:690Paolo Guiotto: Well, we already know that, we already know this, we… I really did.
23:43:150Paolo Guiotto: No.
23:44:440Paolo Guiotto: Because it is in the basic properties of the integral.
23:48:780Paolo Guiotto: that if you have two functions, F and G, in L1, then there's some… actually, any linear combination of these is in L1, so F plus G is in L1, and alpha F is in L1.
24:05:700Paolo Guiotto: So, the sum is well-defined.
24:08:940Paolo Guiotto: On that space. So let's just check that this thing is a norm.
24:16:800Paolo Guiotto: So let's… Check… So, this… the conclusion would be L1 is a vector…
24:28:60Paolo Guiotto: Well, of course, there should be the algebraic properties, but we never verify.
24:33:500Paolo Guiotto: Let's check the… characteristic… properties… of D1 normal.
24:46:220Paolo Guiotto: So, positivity is evident.
24:50:820Paolo Guiotto: The one norm is,
24:53:300Paolo Guiotto: clearly greater or equal than zero, because we are integrating a positive function, no? The modulus of F.
25:00:160Paolo Guiotto: This one is integral on X of modus F d mu.
25:06:210Paolo Guiotto: And this is greater or equal than zero, so this makes this quantity positive. This for every S in L1.
25:14:400Paolo Guiotto: And, let's come to the vanishing, which is just the,
25:19:120Paolo Guiotto: The different property here, so vanishing.
25:26:350Paolo Guiotto: So we have that… suppose that the one norm of F be equal to 0. This means that integral on x of absolute value of F, E mu, is equal to 0.
25:38:70Paolo Guiotto: You remind that we have seen this was a consequence of Shevyshev inequality that says that when this happens, modulus of F
25:48:190Paolo Guiotto: It's equal to zero almost everywhere.
25:51:520Paolo Guiotto: But not necessarily everywhere.
25:53:910Paolo Guiotto: In fact, if you take a function which is different from zero at a single point, and you know that measure of the point is equal to zero, as for the case of the Lebec measure, points have not.
26:05:650Paolo Guiotto: have no measure, measure zero set, single… single don't, then you have a function which is zero almost everywhere, so the… we are in this case, the norm would be zero. And this means F equals zero almost everywhere.
26:22:890Paolo Guiotto: The homogeneity and the triangular inequality, we can say they are straightforward.
26:40:280Paolo Guiotto: Okay, so this is the L1.
26:43:470Paolo Guiotto: space, huh?
26:44:970Paolo Guiotto: Now…
26:46:230Paolo Guiotto: moved by this definition, we extend this to a larger class of spaces, which are the LP spaces. Definition.
26:58:470Paolo Guiotto: So, we take an exponent P that will be between 1 included, that's the case of L1, and plus infinity.
27:08:160Paolo Guiotto: Actually, the definition could be done for P less than 1. The problem is that when we define the norm, the norm is no more a normal for P less than 1. That's why we consider only this, this exponent. I call it exponent because, as you will see, they are exponents. We define
27:27:730Paolo Guiotto: LP… Xer.
27:31:970Paolo Guiotto: as the set of functions which are measurable on X, and now the difference is that the integral of modulus F to the power p is paramount.
27:47:220Paolo Guiotto: Important example, very important example, is P equal 2.
27:52:110Paolo Guiotto: We define this is the space, and… The Pinorma, will be…
28:00:410Paolo Guiotto: By definition, that integral of models of F to power P
28:05:800Paolo Guiotto: Now, if I take this, this, will not be enough, because you can see easily, the opportunity fails, no? You put an alpha here, and you complete the alpha alpha, you have an alpha in the R.
28:18:790Paolo Guiotto: This would become modus of alpha to P times modus sub n.
28:23:660Paolo Guiotto: The modus of alpha to P comes outside, that is the P norm over. But outside, you would have modules to the P norm. So to make modules alpha, you have to make the P root of this. So, you have to rise all this quantity to the exponent 1 over P.
28:42:100Paolo Guiotto: Now, it turns out that this proposition
28:49:290Paolo Guiotto: Lpx.
28:52:160Paolo Guiotto: Equipped with that Pinorma.
28:56:850Paolo Guiotto: Is. A. Normed.
29:01:780Paolo Guiotto: space.
29:04:490Paolo Guiotto: Wita.
29:07:270Paolo Guiotto: vanishing.
29:12:550Paolo Guiotto: In WIC.
29:15:830Paolo Guiotto: form.
29:17:520Paolo Guiotto: So, as executive 4L1, the P norm of F equal to 0, If, and only if,
29:26:700Paolo Guiotto: F is equal to 0 almost everywhere.
29:33:150Paolo Guiotto: Okay, I will not do the proof for a generic P, I will do the proof only for P equal 2, which is a bit easier, but the main ideas can be extended.
29:46:950Paolo Guiotto: Now, we will start the proof.
29:51:830Paolo Guiotto: Because it is quite long. But this is because by doing the proof, we learn some important inequalities.
30:00:690Paolo Guiotto: Okay, so I will do, say, the details for P equal2, then I will tell you what happens for P different from 2, because sometimes you might need, to know
30:13:200Paolo Guiotto: how it works. So, the first thing, which is non-trivial, is that this is a vector space. So, L2 means, first of all, let's write, it is the set of functions F, measurable functions, such that
30:29:820Paolo Guiotto: the integral on X of modulus F squared is finite.
30:40:170Paolo Guiotto: By the way, we may notice that if you take L2,
30:45:630Paolo Guiotto: the set X is made of a finite number, let's say D numbers.
30:52:830Paolo Guiotto: And the measure, the sigma algebra F is part of this set.
31:02:800Paolo Guiotto: This is DX.
31:05:660Paolo Guiotto: And the measure mu is the counting measure.
31:10:110Paolo Guiotto: counting… measure.
31:15:190Paolo Guiotto: Well… If you… if you plug all this, you get that this is nothing but R. D…
31:24:220Paolo Guiotto: equipped with the Euclidean norm.
31:27:860Paolo Guiotto: You have to think that functions are, in this case, functions defined on a finite set. So a function is known as soon as you know the value it takes on 1D. So you know F of 1, f of 2, F of D, but this is an array of D numbers. So functions are arrays of D numbers.
31:46:820Paolo Guiotto: And when you compute the integral.
31:49:370Paolo Guiotto: since each number counts 1, this becomes the sum of the squares, and you take the root, and that's the Euclidean norm, okay? This… all this to say that this is the, with the, the…
32:03:230Paolo Guiotto: Euclidean.
32:05:10Paolo Guiotto: No.
32:07:270Paolo Guiotto: So, the Euclidean space, final dimensional Euclidean space, is a particular case of this one.
32:13:330Paolo Guiotto: Okay, in just one minute, let's see that this is a vector space.
32:19:620Paolo Guiotto: So, step one… L2, X, huh?
32:26:350Paolo Guiotto: is a vector.
32:29:220Paolo Guiotto: space, so…
32:32:150Paolo Guiotto: Here, there is something which is not automatic, and it does not follow what we know about the properties of integrals. So that is, I want to show that if FGR in L2
32:44:170Paolo Guiotto: Then, also, F plus G is in L2.
32:50:290Paolo Guiotto: Now, what does it mean, this?
32:52:710Paolo Guiotto: To say that two functions are in L2 means, apart from the fact that they are measurable, okay, it means that the integral of F squared d mu is finite. I write F squared, I should put the absolute value.
33:09:580Paolo Guiotto: Okay? Now, if f is real valued, I can write F of F squared is the absolute value of F squared. Of F squared is the same.
33:18:740Paolo Guiotto: If F is C-valued, you have to put the modulus, otherwise F squared is a complex number, okay? But however, let's… it's not the important thing, this one. So let's fix ideas and take the real case.
33:30:850Paolo Guiotto: So, this means that we know that these integrals are finite.
33:35:160Paolo Guiotto: And the conclusion here means that integral of F plus G squared is fine. So…
33:44:550Paolo Guiotto: The point is, how can I get My conclusion from the assumptions.
33:50:990Paolo Guiotto: Now, the natural idea would be, take this one.
33:54:860Paolo Guiotto: integral on X of F plus G squared, do the square, because you will recognize there is something that we know. In fact, this is integral of F squared plus G squared, then there is the double product, 2FG.
34:12:470Paolo Guiotto: DMU.
34:14:760Paolo Guiotto: Now, a part of this quantity is definitely fine, because you have the integral of F squared, which is supposed to be fine, you have the integral of G squared, which is supposed to be fine. Now, the problem is, is this last integral defined, the integral of F times G?
34:31:880Paolo Guiotto: So this can be decomposed into integral, of F squared, emu.
34:38:450Paolo Guiotto: plus integral of G squared immune, and both these are finite.
34:44:750Paolo Guiotto: So, real, let's say, positives. But this is to say they are fun. Plus.
34:49:720Paolo Guiotto: 2 integral on X, F, G, d mu. Now, how can I prove that this one is also 5?
34:57:620Paolo Guiotto: And this is thanks to an elementary inequality, which is the key of everything here.
35:04:520Paolo Guiotto: elementary inequality.
35:08:740Paolo Guiotto: Which is this. Whenever you have two numbers, A and B, the double product is always less or equal than the sum of the squares.
35:19:780Paolo Guiotto: And this comes from a trivial fact.
35:25:440Paolo Guiotto: comms.
35:27:700Paolo Guiotto: from… the fact that if you do A minus B squared, this is positive.
35:35:160Paolo Guiotto: If you do this, you get this square, that's B squared minus 280.
35:40:260Paolo Guiotto: Okay? This must be positive to count the 2AD on the other side, the 2AD last time. A squared plus B squared.
35:48:860Paolo Guiotto: So, if we apply this, we can say that then 2FG
35:54:680Paolo Guiotto: is, of course, less or equal than F squared plus G squared.
36:01:510Paolo Guiotto: Okay, so now you can integrate.
36:08:950Paolo Guiotto: And the right-hand side is fine, it's because of the assumption.
36:17:550Paolo Guiotto: So this explains why, and if you do this to the P power, it's a bit more complicated. For LP, this is a sort of list of exercises at the end of the chapter.
36:31:620Paolo Guiotto: There is exercise 942 and 943944 to prove this for the general case. Do exercise…
36:42:250Paolo Guiotto: 943. That says, if F and G are in LP, then also their sum, F plus G, is in LP.
36:54:50Paolo Guiotto: And this is made by something similar, but more complicated, because as you understand, when P is a generic exponent, you don't have the formula for squares. So, if P is… maybe if P is an integer, you could, in principle, do something. But what if P is a fraction, no? 3 halves…
37:13:990Paolo Guiotto: What do you do? Okay, so there is this exercise that drives you to this. Okay, let's stop here. For today, tomorrow, we continue this prover, showing that the norm is a true norm.