AI Assistant
Transcript
00:29:760Paolo Guiotto: Okay, let's start with the sum of the… Exercises 8, 3, 4… V is, C1,
00:47:320Paolo Guiotto: And, we have… On this space, we… consider…
00:58:740Paolo Guiotto: The infinity Norma.
01:02:690Paolo Guiotto: a second null, which is called the VINORMA, Oh, where if so?
01:09:980Paolo Guiotto: which is defined in this way. We take the infinity norm of F,
01:15:790Paolo Guiotto: plus the L1 norm of F'.
01:21:420Paolo Guiotto: So this means, the infinity norm of F,
01:26:720Paolo Guiotto: plus integral from 0 to 1 of modus f prime x.
01:32:290Paolo Guiotto: Yes.
01:36:370Paolo Guiotto: And finally, there is a third norm, let's call it star norm.
01:42:00Paolo Guiotto: Which is, by definition… well, this would be the natural standard C1 norm, so the norm for this space, which is made of the infinity norm of F plus the infinity norm of FY.
01:59:530Paolo Guiotto: So we accept that all these are norms. In fact, the infinity norm is a norm. We know the V-norm is the sum of two objects that are norms, so easily you get everything, and also the star norms. So we accept that all these are norms.
02:16:480Paolo Guiotto: So there are these questions. Number one.
02:20:470Paolo Guiotto: Should that there exist a constant, C, capital C, such that infinity norm of F is controlled by C times the V normal of F, which is controlled by the star norm.
02:35:700Paolo Guiotto: of effort.
02:37:600Paolo Guiotto: Of course, this for every F in Z.
02:42:230Paolo Guiotto: So, in practice, the V-norm is stronger than infinity norm, the star norm is stronger than V nom, which is stronger than we should norm.
02:50:630Paolo Guiotto: Number 2… using, using… by using…
03:00:380Paolo Guiotto: these type of functions, FK of X, Equala CK sine… KPX, and the GK…
03:13:980Paolo Guiotto: of X, equal, CKX to power K.
03:19:570Paolo Guiotto: Well, basically, you have to show that they are not equivalent, okay?
03:24:410Paolo Guiotto: Using this type of functions show that,
03:31:720Paolo Guiotto: There cannot be two constants, little m and capital M, such that, for example.
03:38:650Paolo Guiotto: The V-norm is controlled by the infinity norm.
03:44:760Paolo Guiotto: And, the star norm… is controlled by… TV Norma.
03:52:880Paolo Guiotto: Okay, stop.
03:56:20Paolo Guiotto: Mmm… that's cheap.
04:00:200Paolo Guiotto: solution.
04:02:380Paolo Guiotto: So let's start with question one.
04:05:150Paolo Guiotto: This is a bound that we have to do with the generic AFA.
04:10:620Paolo Guiotto: Okay, so…
04:12:990Paolo Guiotto: the infinity norm of F, we know what is it. The V norm of F is, by definition, what? The infinity norm of F plus the L1 norm of F prime.
04:24:770Paolo Guiotto: So it is evident that since this is the infinity norm of F plus the one norm of F', this is positive, so it is greater
04:34:790Paolo Guiotto: well then, infill the norm of F, and we are done.
04:38:560Paolo Guiotto: Okay?
04:39:750Paolo Guiotto: For the second bound, Duh?
04:43:370Paolo Guiotto: the… we basically have to control the V norm by the star normal, okay? So the V normal, we have here, the V-norma is, well, the star normal, let's write the star normal.
04:57:700Paolo Guiotto: That was the infinity norm of F plus the infinity norm of F prime.
05:05:700Paolo Guiotto: Now, we want now to control the V-norm through this quantity. You see that one piece is the same?
05:14:670Paolo Guiotto: Huh? Now, so the point is how we could control this by this.
05:20:350Paolo Guiotto: But we have already seen that… this kind of bound, you know? So we have that, the,
05:27:20Paolo Guiotto: One norm of F' is the integral 01 modulus f prime x.
05:33:370Paolo Guiotto: Now, since these quantities inside are all controlled by the infinity norm of F',
05:41:530Paolo Guiotto: If we plug that into the integral, we get that this is less or equal than integral of infinorm of f prime.
05:49:600Paolo Guiotto: And, so this is a constant integral of 1 is 1, between 0, 1, and this is
05:55:670Paolo Guiotto: infinity norm of F'.
05:59:290Paolo Guiotto: And so from this, the conclusion is evident, because the…
06:03:640Paolo Guiotto: V norm, which is the infinity norm of F plus the one norm of F', will be less than the infinity norm of F plus the infinity norm of F',
06:17:520Paolo Guiotto: And this is the star known of F.
06:22:570Paolo Guiotto: Okay, question two, we have to show that these are not equivalent.
06:27:920Paolo Guiotto: It is suggested what kind of functions to use. However, even if we don't have this kind of suggestion.
06:35:830Paolo Guiotto: So, let's give a look to this control, no?
06:41:90Paolo Guiotto: If we have to prove that there is no M, Such that, de vi norma…
06:49:760Paolo Guiotto: which is the infinity norm of F, plus
06:53:840Paolo Guiotto: the one normal L prime, is controlled as F prime is controlled by M times the infinity norm of F.
07:07:690Paolo Guiotto: You see that, of course, the infinity norm controls itself, so the problem is that it shouldn't be possible to control the one norm of F' through the infinity norm of F.
07:20:170Paolo Guiotto: So, I need to build functions which are
07:24:620Paolo Guiotto: for example, bounded by a constant in infinity norm, but huge with F'.
07:35:570Paolo Guiotto: So the derivative with L must… must have a big, area.
07:42:500Paolo Guiotto: So… What kind of functions among these could do the job?
07:50:530Paolo Guiotto: Because if you do the derivative for this type of functions, you notice that, forget the constant, the constant probably is irrelevant.
08:00:130Paolo Guiotto: Also because this kind of bound is homogeneous, so if you have a factor that multiplies F, you simplify, so that factor is just a fake,
08:11:630Paolo Guiotto: a fake ID, you know? When you do the derivatives of these…
08:17:60Paolo Guiotto: you see that you get a constant K in front.
08:21:150Paolo Guiotto: No?
08:22:120Paolo Guiotto: So…
08:24:780Paolo Guiotto: I think it… let's see what happens if we use the FK, and I don't see why there is pi, I think that's also pi. That's because we want to have the integral between them on a period, perhaps. So, let's take, FK of X equal sine…
08:42:590Paolo Guiotto: K by N, so… So, F prime…
08:47:770Paolo Guiotto: is equal to k pi cosine k pi x.
08:54:810Paolo Guiotto: So D1, the infinity norm of FK, it is clear, this is a maximum for X between 0 and 1.
09:06:410Paolo Guiotto: of modulus of FK, so modulus of sine k pi X. Now, when X is between 0 and 1, this argument is between 0 and k pi, so k times a period.
09:25:410Paolo Guiotto: So you are looking at a function sine on K periods, and taking the absolute value of the modules. This quantity would be 1, not because you swipe in tier periods, so definitely the distance equal to 1.
09:42:50Paolo Guiotto: About the, one norm of FK prime.
09:48:680Paolo Guiotto: This is the integral 01 of modulus, the derivatives, so we get K pi modulus cosine k pi x.
10:00:160Paolo Guiotto: DX,
10:03:450Paolo Guiotto: Well, maybe we could just rescale the variable. So if we do a change of variable, this is y equal K pi X. This becomes the integral. When x is between 0, 1, we go from 0 to k pi.
10:21:120Paolo Guiotto: of… you see that this factor enters here, because DY is k pi.
10:27:830Paolo Guiotto: DX, so it is absorbed by the change of variable. And then we have modulus cos y DY.
10:37:250Paolo Guiotto: Now, remind that the goal is that we should get this quantity big when K becomes big.
10:44:20Paolo Guiotto: In fact, this is big not because the function is big, but because the area is big, you know? We are computing area of modulus cosine on K… on k half periods.
10:58:220Paolo Guiotto: No? So, you know, the integral from 0… 2… Well, that's a…
11:05:540Paolo Guiotto: from 0 to pi is this. For cosine, absolute value is this. The, this black…
11:12:870Paolo Guiotto: quantity is the integral 0 to pi of modules cos y. It's a number, it's a constant. We can compute it, probably is equal to 2, something like this.
11:26:560Paolo Guiotto: However, whatever it is, doesn't matter, it's a positive constant, K.
11:32:930Paolo Guiotto: positive.
11:34:750Paolo Guiotto: And, here, if you split the integration, as the integration from 0 to pi, then you have that one from pi to…
11:43:60Paolo Guiotto: 2 pi. Then, from 2 pi to 3 pi, and so on, until you reach k minus 1 pi to K pi.
11:53:920Paolo Guiotto: When there are apps, it is better if we call different fancy this constant, let's say, L.
12:03:500Paolo Guiotto: Now, each of these has value equal to L. Now, this has value L, this has value L, this has value L, this one has value L, and they are K of them. So, you see that the one norm of
12:17:640Paolo Guiotto: AK prime is proportional to that positive value.
12:26:510Paolo Guiotto: If you want, we can compute, it should be something like 2, I think. In any case, it doesn't matter, it's possible.
12:35:850Paolo Guiotto: positive, because it's, the integral, it's constant, not the integral on, half period of, modulus cosine. Actually, it's a period for modulus of cosine. In any case.
12:46:990Paolo Guiotto: So, the V norm of FK is equal to the infinity norm. We said that the infinity norm is 1,
12:59:70Paolo Guiotto: So, infinity norm of FK plus
13:02:240Paolo Guiotto: one norm of FK prime. This is equal to 1. This is equal to KL,
13:08:790Paolo Guiotto: And this should be less or equal than if there exists a constant little m times the infinity norm of FK, if this is true, which is 1. So we would have that 1 plus KL should be less or equal than m for every KL.
13:26:680Paolo Guiotto: natural. And that's impossible.
13:33:640Paolo Guiotto: Okay.
13:35:280Paolo Guiotto: So this is what happens with these two. And now for the other two, probably the functions, to be considered are these ones, no? X to K. So for the second bound.
13:49:20Paolo Guiotto: 4.
13:51:360Paolo Guiotto: Bound.
13:54:780Paolo Guiotto: which is false. At the end, a star controlled by constant DV norm.
14:02:250Paolo Guiotto: We have that. FK… well, let's take FGK of X equal X to K. Let's see what happens.
14:11:380Paolo Guiotto: Now, this dark norm of GKE
14:17:120Paolo Guiotto: is the infinity norm of GKE,
14:20:980Paolo Guiotto: plus the infinity norm of GK prime.
14:26:00Paolo Guiotto: The infinity norm of GK is, the, infinity you know, of GKE, He's, the maximum…
14:37:630Paolo Guiotto: on 01.
14:40:500Paolo Guiotto: of modulus of GK, so modulus of X to power K.
14:44:820Paolo Guiotto: And that quantity is 1, no?
14:48:130Paolo Guiotto: It is modulus of X to power K, X is prime if it's 0, 1. There's not even the modulus here. X to K, when x is between 0, 1, that's 1.
14:57:790Paolo Guiotto: for GK prime, the infinity norm of GK prime, that's equal to the infinity norm, the maximum.
15:06:10Paolo Guiotto: of the derivative is kx to k minus 1, still with x between 0, 1.
15:13:250Paolo Guiotto: Now, that K is a constant.
15:15:760Paolo Guiotto: Then you have still the maximum of X2K minus 1. For k greater or equal than 1, this is 1, so this is equal to K.
15:25:410Paolo Guiotto: So we conclude that this is equal to 1 plus K.
15:31:710Paolo Guiotto: for the V-norma, the GK,
15:35:500Paolo Guiotto: in V norm is equal to the infinity norm of GK, we already computed, it is equal to 1, plus the 1 norm of GK prime.
15:46:910Paolo Guiotto: So this is one.
15:49:60Paolo Guiotto: This is 1 plus integral 01 of the absolute value of the derivative, which is kx to k minus 1. The absolute value is not necessarily because function is positive. Now, we can integrate this, because this is the derivative of X to K.
16:09:320Paolo Guiotto: So this is the evaluation of X to K between 0 and 1. So for x equal 1, you get 1. For X equals 0, you get 0. So at the end, this is constantly equal to 1. So everything is equal to 2.
16:25:840Paolo Guiotto: So, as you can see, So… It cannot… B…
16:34:690Paolo Guiotto: that the norm of GKE, star norm, is controlled by a constant, the
16:45:130Paolo Guiotto: Well, if you want. It cannot be that the star norm is controlled by the V norm for every F in V,
16:53:880Paolo Guiotto: Otherwise…
16:58:520Paolo Guiotto: If we plug… to this, the GKE?
17:04:760Paolo Guiotto: Norm of GK.
17:07:400Paolo Guiotto: which is a YSC1 function, should be less or equal than the V-norm of GK, but this is 1 plus K, should be less or equal than M times 2 for every K, and this is impossible.
17:24:00Paolo Guiotto: And this ends the exercise.
17:30:00Paolo Guiotto: I told you to do the number 6 and number 9, I do the number 9.
17:35:620Paolo Guiotto: 8, 3, 9. Then I will, today I will add some solutions to the exercise file, so I hope you try to do the exercises,
17:46:650Paolo Guiotto: left it.
17:48:370Paolo Guiotto: Until this point. So, here we have V is C01.
17:57:530Paolo Guiotto: This is, similar to some of the exam exercises.
18:03:810Paolo Guiotto: And we define this quantity.
18:06:780Paolo Guiotto: Le Vinorma… as the integral from 0 to 1, modulus f of x, divided by root of X.
18:17:480Paolo Guiotto: Notice that, in principle, this is a generalized integral because of the singularity in zero. So…
18:26:440Paolo Guiotto: Number one, check that… check that, the V-norma is well defined.
18:37:480Paolo Guiotto: And, it is, normal.
18:44:310Paolo Guiotto: Norma on V.
18:47:440Paolo Guiotto: Number 2.
18:51:370Paolo Guiotto: Infinity Norm.
18:53:540Paolo Guiotto: is stronger.
18:58:390Paolo Guiotto: Then, Vina Orma.
19:01:270Paolo Guiotto: Number 3…
19:06:720Paolo Guiotto: Weird.
19:07:800Paolo Guiotto: We have these functions, fn of x, defined in this way.
19:13:50Paolo Guiotto: They are cubic root of N,
19:22:860Paolo Guiotto: for X between 0 and 1 over n.
19:28:190Paolo Guiotto: Then, 1 over cubic root of X,
19:32:960Paolo Guiotto: for X between one of the N and 1.
19:38:210Paolo Guiotto: Now, it asks, is the sequence FN contained in D?
19:45:640Paolo Guiotto: then compute…
19:48:950Paolo Guiotto: the norm of these FN according to the V-norm, and then what can you do about the relation between infinity norm and V norm?
20:00:190Paolo Guiotto: What about?
20:05:770Paolo Guiotto: infinity, and… We… Nope.
20:11:910Paolo Guiotto: Of course, in the question to part of this answer, Is already asked.
20:19:80Paolo Guiotto: So… We are first to check that this quantity is well-defined.
20:25:640Paolo Guiotto: I told you before, this is generalized indica because of this denominator that vanishes at zero. So…
20:33:670Paolo Guiotto: some care should be taken here to discuss the existence of that integral, okay? So, the V norm of F is, by definition, this integral.
20:50:440Paolo Guiotto: which is…
20:56:190Paolo Guiotto: generalized integral.
20:59:510Paolo Guiotto: at… X equals 0. There is no other problem for X positive, because for… X positive.
21:10:500Paolo Guiotto: Remind that V means function in continuous function on the interval 0, 1, so…
21:18:710Paolo Guiotto: the absolute value of F divided the root is a well-defined continuous function on 0 excluded until 1. So the only problem is the behavior at x equals 0.
21:36:310Paolo Guiotto: What can be said at X equals 0?
21:39:80Paolo Guiotto: Well, at x equals 0, the function, we don't know anything less than the function f is continuous, so f of x will go to the value of F0,
21:51:80Paolo Guiotto: No?
21:54:260Paolo Guiotto: So, the first, let's say, the temptation would be, well, then modulus f of x divided by root of x would be asymptotic when x goes to 0 to modulus f of 0, which is whatever constant divided by root of x.
22:13:800Paolo Guiotto: Now, there is a little detail here that this is not 100% true. This is if f of 0 is different from 0, because if f of 0 is equal to 0, you are saying that something is asymptotic to 0. Asymptotic means the ratio goes to 1.
22:30:870Paolo Guiotto: So if you divide by 0, you cannot say that the ratio goes to 1. Or if you do 0 divided the other one, the ratio is 0, it doesn't go to 1. So be careful. So this is if F of 0 is different from 0. So it doesn't work for all the possibilities.
22:47:800Paolo Guiotto: However, another more simple argument here is that the function is continuous, so it is
23:02:610Paolo Guiotto: Let's add that f of 0 equals 0. Well, actually, it sounds… should be easier, because the function is even helping, you know?
23:12:970Paolo Guiotto: But, now, let's stay on this framework before to change it. Is this function integral in 0?
23:22:710Paolo Guiotto: 1 over the root of X, that's a constant. So, constant over root of X. Is integral on 0, 1, yes or no?
23:29:870Paolo Guiotto: Yes, no? So that's the good news, okay? So I suspect that F equals 0, it should be easier than this one, because it's… it's even better. I cannot say it is asymptotic, but let's say that to make this a complete argument, I should discuss F of 0 equals 0.
23:48:360Paolo Guiotto: But since I realized that this is integral, I could notice I could… But,
23:57:930Paolo Guiotto: I could do better. And notice that if I take my function, f of x, divided root of x.
24:06:70Paolo Guiotto: to show integrability, I just… Can say that.
24:17:220Paolo Guiotto: What do we know about the function f is?
24:21:550Paolo Guiotto: Continuous, so it will be…
24:27:220Paolo Guiotto: bounded, right? And that's by a bias theorem, no? So this will be less frequent than a constant M divided the root of X. F is continuous, so modulus of F is continuous as well, and so modulus of F is bounded
24:48:380Paolo Guiotto: Bye.
24:49:580Paolo Guiotto: via the steering, okay?
24:52:780Paolo Guiotto: So, I can say that, then, my integral 01, modus f of x, blah blah blah, over root of x.
25:02:180Paolo Guiotto: is bounded by this integral, M over root of X, which is finite, I don't need to compute, and this tells that the norm of F, the V norm of F, is well-defined, is a finite quantity, whatever is F in the set B. Okay.
25:20:810Paolo Guiotto: Now we discussed the root position, and we have to verify the characteristic properties for this quantity, okay? So let's now…
25:31:140Paolo Guiotto: Let's check.
25:33:220Paolo Guiotto: that this… Is, normal, with the usual all, properties. Positivity.
25:44:640Paolo Guiotto: Well, this is evident, because it is the integral of a positive quantitative, no? So, norm of F…
25:53:300Paolo Guiotto: is integral 01 of modulus f of x divided the root of x.
25:59:850Paolo Guiotto: This is greater or equal than zero, so that everything will be greater or equal than zero, no matter who is F.
26:07:670Paolo Guiotto: Number two, vanishing.
26:12:100Paolo Guiotto: Norm of F equals 0.
26:15:760Paolo Guiotto: If and not if this integral 01 modulus of fx root of X, DX,
26:24:740Paolo Guiotto: is equal to 0. Now, this is not the Riemann case, but… so here you have two possibilities. Either you accept that probably it will work similarly, because it is not a Riemann integral, but it's a generalized integral.
26:40:440Paolo Guiotto: still of a continuous function, so what should be expected? That the function is zero everywhere. Whose function? This one, so accept at most X equals zero. Or…
26:52:970Paolo Guiotto: You can use the LeBague version, which is weaker. It says modulus F of X
26:59:320Paolo Guiotto: Divided the root of X equals 0, but this time almost everywhere.
27:04:680Paolo Guiotto: Oops.
27:05:600Paolo Guiotto: X in 01, which is… It's not yet sufficient to conclude. Well, first of all, I can say
27:14:410Paolo Guiotto: Of course, the denominator cannot make this zero, so modulus f of x equals 0 almost everywhere, and therefore f of x equals 0 almost everywhere. Now, if I want to catch the everywhere.
27:31:540Paolo Guiotto: I need to put something more on the table.
27:35:00Paolo Guiotto: What do we know about this, F?
27:40:270Paolo Guiotto: is continuous. So, what do you think? That the function which is continuous can be only equal to zero almost everywhere, and not everywhere?
27:49:710Paolo Guiotto: We have seen in one of the exercises, I told you exactly that it's an exercise, but it is useful to know this. Now, this F must be continual… must be constant equal to 0 for every X in 01, so it isn't everywhere, this, because…
28:07:990Paolo Guiotto: F is continuous.
28:10:960Paolo Guiotto: No? Just imagine, no, you have this function, which is from 0 to 1. Imagine there is one single point X where the function is non-zero.
28:20:220Paolo Guiotto: F of X is positive. Because it is continuous, it will be positive somewhere around there. So there will be a set down here, an interval, a range of X, with positive measure where f is positive. So you cannot say that F is equal to…
28:37:30Paolo Guiotto: zero almost everywhere, okay? That's the argument.
28:42:30Paolo Guiotto: Okay, so we have the vanishing gun.
28:45:110Paolo Guiotto: homogeneity…
28:48:200Paolo Guiotto: Well, if here you say homogeneity and triangular inequality are straightforward, because really, they are, it is not wrong.
28:56:620Paolo Guiotto: Okay, you see, now if you put an alpha, you can easily carry out this alpha homogeneity, and
29:03:960Paolo Guiotto: triangular… inequality.
29:12:10Paolo Guiotto: evident.
29:16:220Paolo Guiotto: So let's say that we stop here for this part.
29:19:560Paolo Guiotto: Now we have the second question. The second question, is… so this is the answer to question one. The infinorm is stronger than V norm, okay?
29:33:660Paolo Guiotto: So… we have… to show… So since this is a direct, statement, so that direct… sorry, there exists
29:49:160Paolo Guiotto: a constant C, such that the V norm is controlled by this infinity norm of F for every F in V. So I have to find a way to bound this by this, essentially.
30:06:50Paolo Guiotto: So the V norm of F is, by definition, that integral 01 modules FX
30:13:960Paolo Guiotto: divided the root of X, DX.
30:17:210Paolo Guiotto: So, what could we do here?
30:35:500Paolo Guiotto: So remember, the guy here, the infinity norm of F,
30:39:790Paolo Guiotto: is the maximum, maximum because F is continuous, of modulus FX when x is in 0, 1. So…
30:50:260Paolo Guiotto: In Canada, we would say that…
30:54:20Paolo Guiotto: Less or equal… Yeah, this quantity is bounded less or equal than infinity number of f.
31:01:430Paolo Guiotto: Density is multiplied by root of x, or divided, it is positive, so the function increases, so that integral is less than integral of infinity norm over f divided by root of X.
31:14:20Paolo Guiotto: Now, this is a constant, so it comes out, and it remains the infinity norm of F times integral 01 of 1 over root of X, that if you want to compute, otherwise, you know, it is finite, and you call it C.
31:29:380Paolo Guiotto: Okay.
31:31:860Paolo Guiotto: The important fact is that it does not depend on F. No, that's… that's why it's a constant for F.
31:39:250Paolo Guiotto: Okay, question 3.
31:41:730Paolo Guiotto: Now, it has to consider these functions, FN, defined in this way.
31:48:620Paolo Guiotto: So we have a constant function between 0 and 1 of n. You can have also an idea on how these functions are made, so why not?
32:00:530Paolo Guiotto: So we have… the interval is 0, 1.
32:03:390Paolo Guiotto: We have that from 0 to 1 over n, and the idea is that this n can be also big, so I have to imagine that probably this is a small interval.
32:12:990Paolo Guiotto: The function is equal to cubic root of n. So, when n is big, this is big.
32:18:970Paolo Guiotto: Well, I can imagine that this number would be about here.
32:23:180Paolo Guiotto: And the function is flat.
32:25:180Paolo Guiotto: Until that point.
32:27:410Paolo Guiotto: And then it follows another function, which is 1 over the cubic root of X.
32:32:660Paolo Guiotto: That function is like 1 over X, it's a decreasing function, no?
32:37:550Paolo Guiotto: The only question, and this will be important because we have to establish if these functions are in V, so if they are continuous, is what happens at 1 over n. In fact, the definition is a little bit dirty, because you see the 1 over n is in both lines.
32:54:90Paolo Guiotto: No?
32:54:970Paolo Guiotto: But there is no problem here, because if you look at the second line, when you put X equals 1 over N, you get 1 over the root of 1 over N, and that is root of n, no?
33:07:960Paolo Guiotto: So, we can say that there is no problem with the definition, and since the root is a continuous function for X positive here, we can say that they glue perfectly. So, what I see, here, it would be something like this.
33:25:580Paolo Guiotto: Of course, the plot is not in scale, because this quarter is the value a taxable one, which is 1.
33:35:240Paolo Guiotto: So DFN is something like this. So if I freeze N, I can say that Fn is definitely a continuous function on 0, 1, so this means that our sequence belongs to the set.
33:53:70Paolo Guiotto: Okay?
33:54:860Paolo Guiotto: This was the first question. Now we have to compute the V-norm of this.
33:59:940Paolo Guiotto: The V-norm of this is, by definition, the integral from 0 to 1 of modulus FNX
34:08:719Paolo Guiotto: divided the root of X, the X.
34:13:80Paolo Guiotto: Now, this integral is, clearly different if I integrate from 0 to 1 over n, and then from 1 over n to 1, so let's split.
34:23:50Paolo Guiotto: from 0 to 1 over n, the function is there constantly equal to the cubic root of n.
34:29:400Paolo Guiotto: So, the absolute value is this, divided by root of X.
34:33:719Paolo Guiotto: And then I have the integral from 1 over n to 1. Here, the function is 1 over the cubic root of X,
34:41:540Paolo Guiotto: So, with the absolute value, 1 over… well, let's put in powers, maybe it's better. It is X to 1 3rd.
34:51:659Paolo Guiotto: And downstairs, I have X to 1 half.
34:56:350Paolo Guiotto: the X.
34:57:980Paolo Guiotto: So let's do the calculations, because here it is required to compute explicitly this quantity. So we have n to 1 3rd, we put it outside, then we have integral of 1 over root of X,
35:11:110Paolo Guiotto: This is, if I put a 2 here, this is the derivative root of X, so it will be a valuation of root of x between x equals 0 and x equals 1 over n, plus
35:25:460Paolo Guiotto: This one is the integral from 1 over n to 1, and there we have…
35:31:580Paolo Guiotto: 1 over, or better if you write in a unique power at this point. X2, well, we have a negative exponent, 1 third plus 1H, what is this? 1.
35:46:440Paolo Guiotto: 1, 2… 1 over 6th… What is 5 sixths?
35:54:540Paolo Guiotto: So this is the exponent.
35:57:870Paolo Guiotto: Okay, so here we have n to 1 third times, when we evaluate, divided 1 over n to 1 half minus the evaluation is 0, 0.
36:09:780Paolo Guiotto: Plus, we can compute the primitive. This comes from X to this plus 1, so 1 minus 5 sixths is 1 sixth, right?
36:21:110Paolo Guiotto: X to 1 sixth divided 1 sixth to be evaluated between 1 over n and 1.
36:29:360Paolo Guiotto: So, at the end,
36:32:110Paolo Guiotto: Yeah, 1 over n to 1 half minus 1 third, this is 1 sixth, right?
36:39:420Paolo Guiotto: I hope.
36:42:940Paolo Guiotto: plus, that 1 sixth is 6 times X to 1 sixth, so 1 minus 1 over…
36:51:760Paolo Guiotto: N to 1 sixth.
36:55:380Paolo Guiotto: So, at the end of the day, this is 6 minus 5 of N to 1 sixth.
37:05:800Paolo Guiotto: This is the calculation.
37:08:180Paolo Guiotto: If, people are not doing it.
37:11:300Paolo Guiotto: computational errors.
37:14:380Paolo Guiotto: This is… this seems to be the V norm of this. Now, this is the calculation, and now what can you draw about the relation between the infinity norm and the V norm? So, of course, let's see what is the V-norm of this, sorry, the infinity norm of this, FN.
37:31:660Paolo Guiotto: And, well, you don't need the…
37:35:360Paolo Guiotto: You just need to look at this figure, and the Infinidian armies.
37:44:530Paolo Guiotto: Yeah, the… root of n.
37:51:600Paolo Guiotto: Okay, so what do we have here? The infinite army is big.
37:55:650Paolo Guiotto: the, vino is bounded by 6, basically, you know? This is 6 minus something is less so equal than 6.
38:03:730Paolo Guiotto: So, it means that the Venom is bounded, the infinitum is bigger, you cannot control the inflatome with the V-norm.
38:10:510Paolo Guiotto: But we did use…
38:15:230Paolo Guiotto: that there cannot be a constant C such that the infinity norm can be controlled to that constant by the V norm of F for every F in V.
38:27:710Paolo Guiotto: Otherwise.
38:33:80Paolo Guiotto: If we plug into this relation, DFN, we would have that infinity norm of Fn, that is cubic root of n, should be less or equal than C times 6 minus 5 over n to whatever.
38:50:810Paolo Guiotto: Which is less than 6C.
38:53:930Paolo Guiotto: And this should be for every, let's say, C…
38:57:600Paolo Guiotto: the V norm of Fn, which is equal to this.
39:02:660Paolo Guiotto: And this should be for every N natural.
39:05:930Paolo Guiotto: Of course, this cannot be possible.
39:12:830Paolo Guiotto: So, in particular.
39:17:140Paolo Guiotto: We proved that the infinity norm is stronger, now we can say that they are not equivalent.
39:23:80Paolo Guiotto: Infinity Norm, and… V-norma… are not.
39:31:370Paolo Guiotto: equivalent.
39:34:840Paolo Guiotto: And with this, the exercise bleuishes.
39:37:860Paolo Guiotto: Now, let me, one second check on the… Ondi, where is it?
39:46:910Paolo Guiotto: On the exam… File… We have… okay.
39:56:850Paolo Guiotto: Because there is a number of exercises that you can start doing here.
40:10:870Paolo Guiotto: Well, maybe before to give you… or whatever. For example, you can do… do… Except size.
40:22:760Paolo Guiotto: 2… Let's give me a few of them.
40:36:80Paolo Guiotto: 12.
40:48:320Paolo Guiotto: 17,
40:50:440Paolo Guiotto: from… exam.
40:55:590Paolo Guiotto: sense.
40:57:300Paolo Guiotto: I will, put the solutions,
41:01:390Paolo Guiotto: for these exercises, maybe by the end of the week, okay? So, take some time, we also have the…
41:08:830Paolo Guiotto: beyond work to do. Maybe next time, next week.
41:13:630Paolo Guiotto: At the end of the week, I will give you a second of work on this part, but we have still to do something here.
41:20:890Paolo Guiotto: Okay, so,
41:29:30Paolo Guiotto: We continue, right?
41:31:720Paolo Guiotto: Okay, so let's, return now on the discussion about the L…
41:40:890Paolo Guiotto: the discussion we started yesterday about the LP spaces. We were proving that L2 is,
41:50:740Paolo Guiotto: is a norm spacer, no? So this was a particular case of the general statement here. We proved that L2 is a vector space, so basically this means that if we sum two L2 functions, we have still an L2 function.
42:08:320Paolo Guiotto: Which is not an immediate consequence of the fact that the integral is linear, okay?
42:13:660Paolo Guiotto: And now we want to see that this quantity that is naturally defined, the 2 normal, the P norma, is a norm on L2, okay? So…
42:29:240Paolo Guiotto: let's say, proof…
42:38:190Paolo Guiotto: off.
42:40:220Paolo Guiotto: L. P.X.
42:43:790Paolo Guiotto: normal.
42:47:140Paolo Guiotto: Space… Gaze.
42:53:90Paolo Guiotto: P equal to continuation.
42:59:150Paolo Guiotto: So now we have to prove that the two norm is a norm, okay? So the goal is,
43:07:120Paolo Guiotto: The goal is, the two norm.
43:11:440Paolo Guiotto: is a norm.
43:15:900Paolo Guiotto: We don't… we cook.
43:20:650Paolo Guiotto: vanishing.
43:24:970Paolo Guiotto: So, we remind that the two norm is defined in this way, is the root of the integral of the square of F in mu.
43:37:750Paolo Guiotto: Now, to check that this is a norm, we have to do the usual checks. So, first, positivity.
43:47:320Paolo Guiotto: But we would say it is evident, it is the root of a positive number, okay? So we say evident. Remind that… here, when we talk about the root, any root, we mean the positive root of the number, if the number is positive, okay?
44:04:630Paolo Guiotto: The vanishing, and this is the point, the critical point of this.
44:09:840Paolo Guiotto: We say that, however.
44:12:600Paolo Guiotto: With tools we have, this is easily handled. So, suppose that the two norm is zero. This means that it can be possible if and only if the integral of modulus F squared mu is 0.
44:27:370Paolo Guiotto: But since this function here, inside the integral, is positive, the integral is zero, if and all if the function f squared
44:36:880Paolo Guiotto: F is 0 almost everywhere, with respect to the measure mu, of course, whatever is the meaning of measure 0 in mu. And now this means that F is equal to 0 almost everywhere, so we get a bunch, and nothing more than this.
44:52:50Paolo Guiotto: Homogeneity is evident.
45:00:200Paolo Guiotto: And what remains is the triangular inequality, which is a little bit… the more… the more… the more difficult part of this, triangular
45:12:740Paolo Guiotto: inequality.
45:17:470Paolo Guiotto: So we have to check the norm of F plus G is less or equal than the norm of F plus norm of G. So let's start with the norm of F plus G.
45:26:260Paolo Guiotto: the 2 norm. Now, since the 2 norm arises with the router, let's take the square first to eliminate that exponent 1 half. So this is the integral on x of modulus F plus G squared.
45:41:660Paolo Guiotto: Now, the proof here, it's a bit diff…
45:44:410Paolo Guiotto: different if we are in the real case or in the complex case, because if we are in the real case, this is the square of F plus G.
45:52:870Paolo Guiotto: In the complex case, we cannot say that. It's a bit more complicated, because it should be F plus G times the conjugate of F plus G, okay? So, to make the proof easier, because that's a technical point, it's not a…
46:05:280Paolo Guiotto: Particularly important point, we take the real case, okay? So we assume that FG are real valued.
46:15:220Paolo Guiotto: the proof is slightly more complex if functions are complex values. So this is F plus G squared.
46:24:540Paolo Guiotto: So, we develop, of course, the square. We get integral of F squared plus G squared plus the double product F.
46:32:710Paolo Guiotto: times GD mu. It's similar to what we have seen yesterday, to verify that if two functions are in L2, the sum belongs to L2.
46:43:320Paolo Guiotto: Now, as you recognize here, when you decompose this integral into integral of F squared plus integral of G squared.
46:51:900Paolo Guiotto: d mu plus 2 times integral of F times G d mu.
46:57:770Paolo Guiotto: Now, the first two terms are known quantities. This is the norm of F, the L2 norm of F squared, and this is the L2 norm of G squared. So we have the sum of these two guys.
47:13:40Paolo Guiotto: Remind that the goal is norm of F plus G less or equal than norm of F plus norm of G, so we need something based on norm of F and norm of g, and this looks to be good. But then we have this 2 times the integral FG immune.
47:30:670Paolo Guiotto: Now, it is at this point that we have a very important and remarkable inequality, let's say, lemma, which is called the Cauchy-Swartz inequality.
47:44:180Paolo Guiotto: That's perfect.
47:53:310Paolo Guiotto: This inequality says that if you take the integral of F times G in mu, and you take the absolute value of this, this is less or equal than
48:07:170Paolo Guiotto: The product of the 2L2 norm.
48:15:670Paolo Guiotto: Now, let's suspend the Koshiv-Waltz inequality for a second, and let's see what happens if we apply the inequality to the calculation we are doing, okay?
48:27:370Paolo Guiotto: So… From… let's just copy.
48:33:620Paolo Guiotto: We started from F plus G, the L2 norm squared, and we obtained that this was the L2 norm of F squared, the L2 norm of G squared, plus 2 times the integral F times G, d mu.
48:50:660Paolo Guiotto: Now, this guy here.
48:53:300Paolo Guiotto: According… because of this inequality, in particular, since the absolute value is less than, it means that the integral is between
49:03:620Paolo Guiotto: above the product of the two norms below minus the product of the two norms. We just need the last side of this story. This is less or equal than norm of F2 times norm of G2.
49:18:320Paolo Guiotto: So this all becomes less or equal than the L2 norm of F.
49:23:720Paolo Guiotto: plus the L2 norm of G, the 2 squared, plus 2 times the norm of F2 plus norm of G2.
49:34:960Paolo Guiotto: And now at right, you recognize there is a square.
49:39:100Paolo Guiotto: That's exactly the square of norm of F to plus norm of G2.
49:46:770Paolo Guiotto: So if you do this square, you get the sum of these squares plus the table product.
49:51:890Paolo Guiotto: So now.
49:53:370Paolo Guiotto: Look at the beginning. We have the square of the L2 norm of F plus G is less or equal than the square of the sum of the two norms. So, now we take the roots.
50:06:790Paolo Guiotto: And since everything is positive, we concluded that
50:11:110Paolo Guiotto: the norm of F plus G2 is less or equal than norm of F.
50:18:70Paolo Guiotto: 2 plus norm of G.
50:21:10Paolo Guiotto: 2. And that's the end of the proof.
50:25:130Paolo Guiotto: Okay? The triangular inequality.
50:28:380Paolo Guiotto: Of course, it remains to prove the Cauchy-Swartz inequality.
50:33:870Paolo Guiotto: Okay, so let's do now the proof.
50:38:430Paolo Guiotto: of Nico Shish words, inequality.
50:46:810Paolo Guiotto: Now, there are different proofs, more or less elegant. I will do a proof which is basically based on the original Hoshia idea.
50:58:250Paolo Guiotto: Which is not the most elegant proof. The most elegant is an algebraic proof.
51:03:560Paolo Guiotto: That we will see later, in a few weeks, when we will treat the Hilbert spaces.
51:10:50Paolo Guiotto: Now, this proofs… this proof is based on this first remark, that if one of the two quantities at right is zero, so if one of the two norms of F or G is 0, the inequality is trivial.
51:27:230Paolo Guiotto: So, we noticed that if, let's call it, for Shishwartz, let's give this name CS.
51:34:970Paolo Guiotto: if, norm of F.
51:39:90Paolo Guiotto: equals zero, or norm of G.
51:43:680Paolo Guiotto: equals 0.
51:45:370Paolo Guiotto: What happens? Well, because of the vanishing, if when the norm is zero, it means that, for example, F is equal to zero almost everywhere, right?
51:57:40Paolo Guiotto: And therefore, when you compute the integral from, on X of F times G, one of the two factors is zero almost everywhere.
52:08:730Paolo Guiotto: You remind that we accepted the rule that zero times whatever is zero, okay? So it means that this is zero almost everywhere, and the integral of something which is almost everywhere equal to zero, it is equal to zero.
52:23:690Paolo Guiotto: So the left-hand side would be 0, and therefore, it would be less or equal than the right-hand side, which is 0 as well, no?
52:31:410Paolo Guiotto: norm of F times norm of G.
52:35:350Paolo Guiotto: Okay, same if, G, norm of G is 0.
52:40:00Paolo Guiotto: So, in this case, the Cauchy-Swartz inequality is true.
52:47:310Paolo Guiotto: But of course, this is a sort of degenerate case. Now, let's assume that both the two norms are different from zero.
52:56:570Paolo Guiotto: So we can assume,
53:03:880Paolo Guiotto: that the two norm of F
53:07:110Paolo Guiotto: And the norm of G are different from 0.
53:11:440Paolo Guiotto: In this way, we can now divide by the right-hand side, carry into the left-hand side, and translate into what? So the quotes becomes equivalent to proving that modulus… so you see that when you take this, you divide by this number.
53:31:620Paolo Guiotto: You carry here in the left-hand side, so you have a factor of 1 over.
53:36:290Paolo Guiotto: This factor is positive, you carry inside the modules, so you have absolute value of 1 over the integral.
53:42:740Paolo Guiotto: But that factor is a constant you carry inside the integral, so at the end, you have this absolute value of integral of F times G divided by the two norm of F, the true norm of G,
53:56:830Paolo Guiotto: This thing in absolute value must be less or equal than 1.
54:03:300Paolo Guiotto: So we can transform the inequality in this way.
54:08:340Paolo Guiotto: Now…
54:10:310Paolo Guiotto: Let's take the absolute value. First of all, let's carry the absolute value inside the integral. So, absolute value of the integral is less or equal, because of the triangular inequality of the integral, it is less or equal than the integral of the absolute value.
54:26:820Paolo Guiotto: which is integral on X of modulus of F.
54:31:290Paolo Guiotto: Let me write it this way, divided by the 2 norm of F times modulus of G divided the 2 norm of G.
54:43:10Paolo Guiotto: Now, my goal is to show that this is less or equal than 1, okay? So I want that this be less or equal 1.
54:52:390Paolo Guiotto: Okay?
54:54:500Paolo Guiotto: How do we do this? And that's what is nice of this proof, is that we need that trivial inequality. You remind that
55:04:10Paolo Guiotto: We have seen other times that 2AB is always less or equal than A.
55:09:380Paolo Guiotto: B squared plus B squared.
55:11:960Paolo Guiotto: None?
55:13:780Paolo Guiotto: And this comes from the fact that A minus B squared is always greater or equal than 0.
55:20:200Paolo Guiotto: So, this is my A, and this is my B. So I can say that this guy here
55:27:550Paolo Guiotto: is less or equal, because I'm going to increase the integrand. So, let's put the two…
55:36:360Paolo Guiotto: As denominator here.
55:39:20Paolo Guiotto: So, 1 half the sum of the squares. So, I increase the integral, I have an integral of 1 half the square of A, which is models of F squared, divided, the two norms.
55:52:460Paolo Guiotto: of F squared, last modulus of G squared divided the 2-norm of G squared.
56:02:50Paolo Guiotto: in the meal.
56:04:630Paolo Guiotto: Now, the proof is basically finished, because we see what is written at the right-hand side, and we recognize that it is 1. Why? Take the photo one half and carry outside. So this is one half.
56:20:980Paolo Guiotto: Then, split the integral. Integral of the sum is sum of the integrals. You have integral of X of modulus F squared divided the L2 norm of F squared.
56:33:40Paolo Guiotto: D mu plus the same 4Gs, modulus G squared divided the square of the L2 norm of G.
56:43:770Paolo Guiotto: You see that they are basically the same type of integral, no?
56:47:940Paolo Guiotto: Now, look at this.
56:51:580Paolo Guiotto: This is…
56:53:420Paolo Guiotto: Now, this denominator is a constant, no? It's a number. I can carry outside of the integral. I have 1 over the L2 norm of F squared times the integral of modus f squared d mu, but what is this?
57:12:700Paolo Guiotto: This quantity here is… you know, I'm gonna…
57:18:990Paolo Guiotto: is the 2 normal of F squared.
57:22:910Paolo Guiotto: Divided by himself, so this is equal to 1.
57:28:50Paolo Guiotto: And the same happens here, of course. So basically, there we have 1 half times 1 plus 1, 2 divided to 1.
57:39:100Paolo Guiotto: So this would finish the proof of the Cauchy-Swartz inequality.
57:44:520Paolo Guiotto: What is nice with this argument is that it does something also.
57:48:860Paolo Guiotto: It gives something, more than this.
57:54:850Paolo Guiotto: In particular, it says, when do we have equality?
57:59:690Paolo Guiotto: It's a curious factor.
58:02:20Paolo Guiotto: No? When… this less or equal that you have here in Kuchish force becomes an equal.
58:10:850Paolo Guiotto: Well, to be an equal, everything must be equal, no? Otherwise, if you have a less or equal that becomes strictly less, it will never be equal.
58:20:190Paolo Guiotto: So let's go back from the end, and let's see where… you see there is a unit passage where there is a lesser equal, which is this step here. So that's the unit passage, because
58:31:320Paolo Guiotto: From the end, you see that everything is an equality, and the unique step when you have a lesser equal is at this stage, when we pass it from this integral here to this integral here.
58:45:810Paolo Guiotto: Okay?
58:46:830Paolo Guiotto: Now, that step is based on the fact that A times B is less or equal than A squared plus B squared over 2, but when it is equal.
58:58:20Paolo Guiotto: when this is equal. Well, when 2AB is equal to 2A squared plus B squared, and this is when A minus B
59:09:240Paolo Guiotto: Square is zero.
59:11:310Paolo Guiotto: That is when A is equal to B.
59:16:300Paolo Guiotto: Okay, so now, to have an equal… It must be that,
59:23:130Paolo Guiotto: This one would be here, be equal to this one, and that's for some X,
59:31:300Paolo Guiotto: That these are the conscious, not actually conscious. So these actions are there not.
59:36:790Paolo Guiotto: then X by X, this be true.
59:40:640Paolo Guiotto: At least for almost every act, because if there is a set, a positive measure, when you have a strict sign, that won't be never less, no? If the equality is, strict on a positive measure set, the integral will increase strictly.
00:00:490Paolo Guiotto: So the possibility is that that product must be grade equal, sorry, grade equal to this one, almost everywhere. And this is where the two quantities, so that's the A equal to the B, almost everywhere. So…
00:17:620Paolo Guiotto: This argument says that, moreover.
00:24:140Paolo Guiotto: Moreover… The absolute value of integral FGD mu
00:32:660Paolo Guiotto: Equal to the product of the two.
00:38:260Paolo Guiotto: If and only if…
00:40:350Paolo Guiotto: We said that the A must be equal to B almost everywhere. Our A is modulus of F divided norm of F, so modulus of F divided norm of F squared must be to modulus of G divided
00:57:600Paolo Guiotto: norm of G squared.
01:02:240Paolo Guiotto: Yeah, almost at the end, because this means that models of F sorry, modulus of F.
01:12:580Paolo Guiotto: Must be… now, you carry this on the other side, for example, the ratio between the two norms becomes just a constant, so k times mole…
01:23:840Paolo Guiotto: Lulus of G.
01:25:530Paolo Guiotto: So this means what?
01:27:520Paolo Guiotto: That the two functions must be proportional
01:31:650Paolo Guiotto: Because this means that F must be plus or minus KG almost everywhere, and this means that the two functions must be proportional almost everywhere.
01:43:240Paolo Guiotto: So the only possibility when the two… when that inequality becomes an equality is when the two functions are proportional.
01:53:150Paolo Guiotto: So… say that Cauchy Schwartz, systems.
02:01:240Paolo Guiotto: Into.
02:03:120Paolo Guiotto: And… Equality.
02:09:150Paolo Guiotto: If and only if, F is proportional.
02:14:440Paolo Guiotto: to G, or vice versa.
02:18:590Paolo Guiotto: you may say, but it's the same. Yeah, but it could be that, if I say one is zero, the other must be zero as well, no. Otherwise, the constant could be one is zero, and the other is non-zero.
02:33:60Paolo Guiotto: However, this was that sort of remark, so let's put it here as a remark.
02:41:820Paolo Guiotto: the proof of the push-ish port ends here, and this is just a note.
02:47:110Paolo Guiotto: Okay, so now we have the, we have seen the complete,
02:53:540Paolo Guiotto: proof, showing that, DL2 is,
03:00:880Paolo Guiotto: is a norm space, we keep with that norm. The same happens with the LP norm. The construction is more complex. It is given…
03:11:860Paolo Guiotto: I mean, it's not impossible, but it's pretty hard. It's given in a set of exercises, 942, 943, and 944 for the general PKs. So, if you want to…
03:26:760Paolo Guiotto: To try to test yourself, but that's a very hard work, so it's not important, we don't do.
03:37:710Paolo Guiotto: Now, we introduce another important space, which is the space called the Linfinity X space.
03:48:730Paolo Guiotto: Of course, there is a reason why we would call this an infinity. There are two reasons, basically. One is…
03:55:300Paolo Guiotto: is because we can start this discussion by thinking to the concept of bounded function when the function is measurable, and that's how we tackle this. The other is that, in fact, it turns out that this is the limit when P goes to infinity of the LP space.
04:12:160Paolo Guiotto: However, I don't enter into these details, but this is just to motivate why there is this notation, okay? Now, let's start the question with the following apparent question.
04:26:490Paolo Guiotto: So, what is, Bounded… measurable… function.
04:39:480Paolo Guiotto: Well, apparently there is no problem. I have a measurable function, I say it is bounded if a modulus of f of x is less or equal than a constant for every X.
04:49:770Paolo Guiotto: Yeah, but there is a problem here, because we may say.
04:59:290Paolo Guiotto: F.
05:01:330Paolo Guiotto: measurable on X.
05:03:910Paolo Guiotto: is bounded.
05:07:70Paolo Guiotto: If and all if there exists a constant m such that modulus f of x is less or equal than m for every x in capital X.
05:18:110Paolo Guiotto: Well, this…
05:19:430Paolo Guiotto: This definition, which is actually the application of the concept of bounded function to this F, has basically two problems. Number one.
05:31:30Paolo Guiotto: Normally, or better, not always, measurable functions are defined everywhere. Now, imagine if you have functions with singularities, and you do not define the function, but the set of singularities is negligible for the measure, you don't care.
05:48:590Paolo Guiotto: So you don't need to have the function defined exactly everywhere. It is sufficient always to consider the function defined apart for a measure zero set of points.
05:59:30Paolo Guiotto: So, that condition could be better recasted as, or exists a constant.
06:08:940Paolo Guiotto: M, such that modulus F of X is less or equal than m almost every x into capital X. And this sounds better, no? But if you look at this second definition, this function might not be bounded.
06:24:710Paolo Guiotto: At all.
06:26:170Paolo Guiotto: Picosa… Let's say, let's call it a 2 and 1 is, huh?
06:33:220Paolo Guiotto: Definition 2… Implies… That, huh?
06:44:720Paolo Guiotto: F… might… not B.
06:51:250Paolo Guiotto: Bonded.
06:55:620Paolo Guiotto: at… All.
06:58:430Paolo Guiotto: For example, take this stupid example, we are on X is R, the real line, with the bag measure, mu is LeBague measure.
07:12:930Paolo Guiotto: Take the function f of x.
07:16:500Paolo Guiotto: defined this way. It is zero for every point X real, except Those that belongs to naturals.
07:25:940Paolo Guiotto: And it is equal to N,
07:28:120Paolo Guiotto: when X is equal to N natural.
07:31:520Paolo Guiotto: So this function is a function which is, Completely equal to zero.
07:40:310Paolo Guiotto: So if you would plot this function, you would see 0 here, 0 here, 0. Then when you arrive at 1, it is value 1. When you arrive at 2, the value is 2. When you arrive at 3, the value is 3, and so on.
07:54:80Paolo Guiotto: It is clear that this function is not bounded, because you can take a values figure as we want.
08:01:470Paolo Guiotto: So this is not, strictly speaking, bounded function, but in the second of definition, 2E is bound.
08:10:300Paolo Guiotto: Now, which one of these two definitions should be better for a measurable function? The one, which is the real bounded function, or the two where we might not have bounded functions? Well, if you think to the measurable functions, it seems more reasonable than the second definition.
08:27:350Paolo Guiotto: to be the definition, because also, you remind that, for example, we modify a function on a measure zero set of points, we consider the same function into the vanishing, no? The vanishing says that when norm of the difference between two functions is zero, the two functions are the same almost everywhere.
08:47:359Paolo Guiotto: So they can be different on a measure of zero set of points, but we should consider as if they are the same function.
08:53:640Paolo Guiotto: So there would be no difference between this function, which is unbounded, and the function constant equal to 0, which is constant equal to zero, bounded in particular. But from the point of view of measure, we should look at these two as the same thing. So if one is bounded, it should be also the other one.
09:11:920Paolo Guiotto: Okay? So, the point is that for measurable functions, we need to introduce this new concept, a function which is not necessarily bounded, but it is bounded almost everywhere. We call it essentially bounded function, okay?
09:30:850Paolo Guiotto: So, let's just formalize this definition.
09:39:399Paolo Guiotto: So, F, measurable function on X. So, we understand that there is an underline XF mu measure.
09:51:490Paolo Guiotto: space.
09:53:520Paolo Guiotto: is… Side.
09:59:430Paolo Guiotto: Essentially.
10:02:250Paolo Guiotto: This sent… shall we?
10:06:250Paolo Guiotto: Bounded.
10:11:780Paolo Guiotto: If… There exists a constant, M,
10:18:600Paolo Guiotto: Such that, in modulus of f of x.
10:22:520Paolo Guiotto: is less or equal than M, not for every X, but for almost every X in capital X.
10:32:530Paolo Guiotto: So this is going to be what is the set of bounded functions for measurable functions.
10:40:730Paolo Guiotto: Okay, and now what we want to do is to introduce a way to measure the length of these functions. So if you take bounded functions, we already met this as one of the first examples, probably here.
10:56:130Paolo Guiotto: Yes, we consider it just one of the first, the same.
11:01:280Paolo Guiotto: More or less.
11:02:460Paolo Guiotto: stupid examples of, spaces of functions. It was the set of bounded functions. And we defined as, as,
11:14:580Paolo Guiotto: Well, actually, it's a vector space, and we defined as normal, natural norm, what we call the infinity norm. That was the supremum of the values, absolute values of F . Now, what we want to do is to extend this operation.
11:33:10Paolo Guiotto: It depends on the type of functions we have considered. For example, we cannot make this supremo, because the supremo can be infinite.
11:41:320Paolo Guiotto: Okay? We need something else that replaced this Supremum, but that plays more or less the same role. Okay, and that's what we are going to do. But first, we notice that
11:52:860Paolo Guiotto: So, okay, let's… we call, we… call.
11:58:840Paolo Guiotto: L infinity X.
12:02:290Paolo Guiotto: the set… False.
12:06:60Paolo Guiotto: All essentially bounded.
12:11:200Paolo Guiotto: functions.
12:14:830Paolo Guiotto: on X.
12:16:960Paolo Guiotto: Well, it is a simple check. I leave to you to do, proposition.
12:22:820Paolo Guiotto: This is a vector space, L infinity, X, is a backdoor space.
12:32:700Paolo Guiotto: we don't… usual.
12:37:300Paolo Guiotto: sum.
12:38:700Paolo Guiotto: And… Products.
12:44:10Paolo Guiotto: by SCARAS.
12:48:680Paolo Guiotto: So the point is, you take two essentially bounded functions, you sum, and you get an essentially bounded function.
12:56:620Paolo Guiotto: Should be easy to prove. But, let's see, do the proof, write the proof, to see what… how should we prove this, if it is so trivial, or if it demands some argument, okay?
13:08:960Paolo Guiotto: Existize it.
13:11:690Paolo Guiotto: Right.
13:14:360Paolo Guiotto: The… Proof.
13:18:800Paolo Guiotto: Okay, now… We want to introduce the infinity Norma.
13:25:00Paolo Guiotto: So, the infinity Amish should be…
13:27:120Paolo Guiotto: the supremum of modules f of x, but that we cannot take, because in this case, we have an essentially bounded function. This one is
13:37:00Paolo Guiotto: Essentially bounded.
13:41:290Paolo Guiotto: function.
13:43:500Paolo Guiotto: But it is an unbounded function, okay? And… and at the same time, unbounded.
13:53:480Paolo Guiotto: function.
13:55:380Paolo Guiotto: So in this case, if you take the soup of the values of F,
14:02:30Paolo Guiotto: When X is in R, in this case, this quantity is equal to plus infinity.
14:10:190Paolo Guiotto: Okay.
14:11:600Paolo Guiotto: So we cannot take this. And this is not interesting, because we are taking in this supremum points that are negligible for the module. So, how should we define this, these things?
14:24:860Paolo Guiotto: Now, to come to the definition, we know that we start from what does mean to be essentially bound. It means that there exists a bound, M, for which, modus f of x is less or equal than that bound to almost every X, no?
14:42:720Paolo Guiotto: Of course, if M is a bound, any number greater than M would be a bound as well, no? Because if F is bound by 1, let's say, almost anywhere, it would be bounded by 2, by 3, by 1 million.
14:59:10Paolo Guiotto: Okay? So the set of M for which this happens is the set of numbers
15:06:750Paolo Guiotto: with this characteristic, if there is an M, there are all N bigger than this one, okay?
15:14:800Paolo Guiotto: So… What should be the bound for N? Should be the best possible of this constant N.
15:23:100Paolo Guiotto: And the best possible should be which one? The minimum, right? We can apply the minimum, we can write by the minimum, that's lower bound, but that's the idea, okay?
15:35:500Paolo Guiotto: Okay, so… If F is in L infinity X, we know
15:47:160Paolo Guiotto: that there exists a bound m such that modulus of f of x is bounded by that constant M almost every x in X.
15:57:510Paolo Guiotto: Clearly, for every other, let's say, K greater than that M, it holds…
16:10:200Paolo Guiotto: that modulus F of X is less or equal than k almost every x, no?
16:19:690Paolo Guiotto: So, this sector…
16:24:890Paolo Guiotto: The set of, of, bounds, is. Then…
16:35:500Paolo Guiotto: the set of M, let's say, such that modulus F of X is less or equal than m almost every X in capital X. This is not a set of X.
16:48:640Paolo Guiotto: is a set of values that bound our function, F. Now, if you want to have a figure.
16:55:520Paolo Guiotto: You may have a function here, not that some point is big, you know, but at some other point is even bigger, but there is a constant.
17:05:80Paolo Guiotto: M, this M, is a bound such that your function is between plus m and minus M, no?
17:15:670Paolo Guiotto: This is a bound.
17:18:190Paolo Guiotto: Now, what we want to find is the best possible of this bounds. The best possible is not the biggest possible, but the contrary, the smallest possible, something that should bound as much as possible the function, no?
17:33:580Paolo Guiotto: Now, this quantity here is the quantity we are going to define to be the norm of F, the infinity norm of F.
17:45:20Paolo Guiotto: As we said, it should be the best of these M's, so the minimum.
17:50:590Paolo Guiotto: In general, since this is a set of real numbers, we… it's dangerous to write minimum, but we can always do by writing infinim. So, we put this as definition.
18:02:490Paolo Guiotto: Infinity norm of f is, by definition, the infamum of the constants M, such that modulus F of X
18:12:570Paolo Guiotto: is less or equal than M almost every X in capital X.
18:20:120Paolo Guiotto: Now, this, this quantity is well-defined. There is not… even if it is defined to a non-trivial operation. It is well-defined because we are doing the infamo of a set where there is at least one N, and actually there are infinity, because as we noticed.
18:38:180Paolo Guiotto: Once you have 1M, all numbers greater than him are bounds for the function. So if you have a bound, there are infinitely many bounds, okay? For an essentially bounded function, we have at least one bound, so we have automatically infinitely many bounds.
18:54:550Paolo Guiotto: That set is a set of positive numbers, of course, the M are positive.
18:58:860Paolo Guiotto: Kim cannot be negative, and this is so old.
19:01:680Paolo Guiotto: So it is a set of positive number non-emptive, it will have a moral meaning, okay? So that quantity is now well defined for, every, function F in this class of essential boundary functions.
19:19:880Paolo Guiotto: What we aim to prove is that this is a normal, It turns out that proposition…
19:30:510Paolo Guiotto: The infinity norm is a norm.
19:36:310Paolo Guiotto: on L infinity X, huh?
19:39:730Paolo Guiotto: And, as usual, when you deal with measurable functions, the vanishing will be in weak form.
19:47:370Paolo Guiotto: Weed.
19:49:620Paolo Guiotto: vanishing.
19:55:380Paolo Guiotto: in… weaker.
19:58:960Paolo Guiotto: Forma.
20:00:660Paolo Guiotto: That is, norm of F, infinite norm of F, equals 0 if and only if F equals 0 almost everywhere.
20:13:370Paolo Guiotto: Okay.
20:15:170Paolo Guiotto: Let's see why this is…
20:20:770Paolo Guiotto: So, we say that this quantity is well-defined, so we don't need to repeat this. So, let's check…
20:29:220Paolo Guiotto: Let's check… D… characteristic, properties.
20:40:970Paolo Guiotto: So, positivity… Well, this is evident because,
20:47:410Paolo Guiotto: the set of M, where modulus of F is less or equal than M, almost every X,
20:57:550Paolo Guiotto: This is a set of bounds.
20:59:960Paolo Guiotto: So it has nothing to do with the space X, where there is the variable. It's a set of numbers, and what kind of numbers? We say that if you are there, you are positive, you cannot be negative, so this is contained into 0 plus infinity, whatever it is, and therefore.
21:18:680Paolo Guiotto: The infamum of this sector will be greater or equal than zero, cannot be less than zero, huh?
21:26:830Paolo Guiotto: vanishing.
21:31:290Paolo Guiotto: This is, this is non-trivial, even… but you see, if you go back to the case of the infinity norm, traditional infinity norm, the proof that the infinity norm is a norm is sort of trivial stuff, no? It's a…
21:47:710Paolo Guiotto: an argument without any kind of problems. Now, vanishing is very straightforward. Homogeneity, triangular inequality, everything is simple.
21:58:40Paolo Guiotto: But with the case of the infinity Norma, it's… the infinity norma, it's much more complicated, because look at this. Well, it's a nice exercise on this concept, measure, etc.
22:13:600Paolo Guiotto: So, take this equal to 0. What does it mean? It means that the infamum of the constants M for which modulus of F
22:23:200Paolo Guiotto: less or equal than M almost everywhere. This thing, is equal… this increment is equal to 0.
22:31:470Paolo Guiotto: Now, I want to conclude that the goal, that's the goal of this proof, is F must be equal 0 almost everywhere, okay? That's the goal, the big goal.
22:44:440Paolo Guiotto: Okay, how can we withdraw this from the assumption?
22:50:510Paolo Guiotto: Now, let's think about this infamom is zero. So, you are on the real line, and you know that the infamom of the set is zero.
23:00:10Paolo Guiotto: Here, you have to remind what does it mean that infamom is. Now, infamum is the best lower bound. It's a lower bound, and it is the best one. It means that as soon as you go.
23:11:160Paolo Guiotto: on the positive side, this number here, so let's take… let's call it epsilon positive, is no more, is no more an upper bound… a lower bound for that. So it means that there exists the number of the set, an M, is smaller than that epsilon.
23:30:980Paolo Guiotto: that verify this condition, okay? So, it means that since 0 equal infamom.
23:38:830Paolo Guiotto: This means that for every epsilon positive, we can find an M such that M belongs to that set there, the set of upper bounds.
23:51:510Paolo Guiotto: And M is less than epsilon.
23:55:190Paolo Guiotto: So, in other words, there exists an M,
23:58:390Paolo Guiotto: less than epsilon, such that modulus F of X is less or equal than m, which is less or equal… less or equal irrelevant than epsilon, almost every X.
24:13:780Paolo Guiotto: in capital letters.
24:16:70Paolo Guiotto: So what he's saying is that no matter how we choose absolute positive, we can say…
24:25:290Paolo Guiotto: is basically aggravated, because the set value is not below epsilon is a measurable effect.
24:31:850Paolo Guiotto: So now this is not yet the conclusion. The conclusion is much stronger, it's F is zero, not on smooth. F is just nothing.
24:41:510Paolo Guiotto: Now, the question is, how can we…
24:43:970Paolo Guiotto: arrive from this, no? F is less than epsilon for every altitude. The set where F is greater than epsilon has measured 0, let's just say, no? Or equivalently, we can say that
24:59:360Paolo Guiotto: For every epsilon positive, the measure of the set where modulus of F is greater than epsilon is 0, because F is always
25:08:690Paolo Guiotto: at least… reacts is less than absentee.
25:14:870Paolo Guiotto: Now, I want to get that the set… the conclusion is F is equal to 0 almost every layer, so F non-zero must have relation zero.
25:25:370Paolo Guiotto: So the conclusion… you see, they look… they are very close, but not yet.
25:30:520Paolo Guiotto: This means that measure where f is different from 0,
25:35:760Paolo Guiotto: And to make this similar to this one, I will write like that. Modulus of F, Whoa.
25:45:800Paolo Guiotto: how can I write F…
25:48:120Paolo Guiotto: non-Z, so the thesis is F equals 0 almost everywhere. So, measure of F, if you want, let's do an intermediate step.
25:58:220Paolo Guiotto: Model saw a measure of F different from 0. This is…
26:05:30Paolo Guiotto: 0, no? This is the conclusion. But I want to put this into this form, so modulus of modulus F,
26:14:560Paolo Guiotto: Strictly positive, that's right, equals zero. It's the same, no? So the thesis is, I have to prove that the measure where model size is positive is
26:25:410Paolo Guiotto: Okay? Because this is equally from zero. In fact, it's equally from below the absolute value, otherwise that would be zero.
26:34:580Paolo Guiotto: Here I am almost, no? Because this is the measure of what was that greater than some epsilon, small as likely. So how will you get to the conclusion, you see?
26:47:260Paolo Guiotto: His wife still needs to be a bit of a…
26:53:160Paolo Guiotto: Let's, let's be… doesn't matter if we are rules for the moment? How do we get that from this? By…
27:06:320Paolo Guiotto: You're not going to communicate. Bye!
27:09:410Paolo Guiotto: Sending returns, absolute zero. So by doing that.
27:14:130Paolo Guiotto: Really? No? Now, so let's see what happens when I send epsilon to zero. Okay, now this should come, the idea, it's a continuous product here. Take epsilon equal 1 over n.
27:26:810Paolo Guiotto: you have measured modulus F greater than 1 over n.
27:31:860Paolo Guiotto: What happens to this set when you increase N?
27:35:760Paolo Guiotto: So this is the set EN.
27:40:370Paolo Guiotto: If you are greater than 1 over n.
27:45:510Paolo Guiotto: U will be greater than 1 over n plus 1.
27:49:160Paolo Guiotto: Definitely, no? Because 1 over n plus 1 is molar.
27:53:460Paolo Guiotto: So, Ian is… with the N plus 1 is… If you are there.
28:05:850Paolo Guiotto: It is contained, no? If you are greater than 1 over n, you are greater than 1 over n plus 1. So you have the increasing sequence of set, nice, because that's the case where the continuity works always, no? That is going up to what?
28:20:900Paolo Guiotto: The set where… If you… if you do that, so this will go by continuity.
28:27:700Paolo Guiotto: The measure of the set, the case of the continuity from below, the set is the union of these sets, no?
28:37:90Paolo Guiotto: of F greater than 1 over N. Now, the question is, is this, sorry, modulus of F?
28:44:90Paolo Guiotto: Is this set here modulus of F positive? If this is the case, we are done.
28:51:800Paolo Guiotto: Think about that. If F is a model set is positive, you pick an X, model set of X is positive.
29:01:350Paolo Guiotto: So, is it possible that we have to be using one of them?
29:12:660Paolo Guiotto: take X. Model s is positive. Whatever it is, I take an N big enough such that 1 over N is smaller. So, definitely, my X will be in one of them. So, if you are clearly in the unit.
29:27:790Paolo Guiotto: And vice versa, involving the union, you have, yeah, because in each of them, bottles of that is positive, so they are the same.
29:35:30Paolo Guiotto: So this set is equal to this one, and since we are doing the limit of zero, we get that this is measured. And that's the conclusion.
29:43:610Paolo Guiotto: So…
29:45:660Paolo Guiotto: As you can see, the vanishing is not, is not, so simple, no? But at the end, we obtain the vanishing. Now, homogeneity…
29:57:410Paolo Guiotto: It's, mmm…
30:00:960Paolo Guiotto: It's simple, because you have that infinity norm of alpha F is the what? Is the infamom of the constants M, such that minus alpha f x is less than or equal than M.
30:18:940Paolo Guiotto: Almost everywhere.
30:21:990Paolo Guiotto: Now, what I'm doing holds for alpha different from 0, for alpha equals 0 is trivial.
30:29:00Paolo Guiotto: Because if alpha is 0, 0 times f is zero. So the infinity number malfunction constantly equal to 0 will be zero, no?
30:38:50Paolo Guiotto: And so you get that this will be modulus of alpha 0 times norm of F, whatever it is, it's 0. So for alpha different from 0, we can say you carry outside this modulus of alpha, modulus of x.
30:53:710Paolo Guiotto: you put this on the right-hand side, this, so you have… this is the infamum of numbers m, such that modulus of FX is less or equal than m divided modulus alpha almost everywhere.
31:10:880Paolo Guiotto: But then, if you put a modulus alpha down here and modulus alpha here, they are the same.
31:18:70Paolo Guiotto: Now, reliable constants. So, call these numbers K. So, this is the infamum.
31:26:90Paolo Guiotto: off.
31:27:330Paolo Guiotto: modulus alpha times K, such that modulus f of x is less or equal than k almost everywhere.
31:37:270Paolo Guiotto: But you're doing here of a set of numbers which are of this form, a constant models of alpha timesca. So they are all multiplied by the same factor.
31:50:340Paolo Guiotto: So, imagine the infamous factor times… So you can carry this factor outside, so this becomes modulus alpha, infamum.
32:03:710Paolo Guiotto: of K, such that models FX is less or equal K, almost everywhere.
32:12:860Paolo Guiotto: And now, you recognize that this is nothing but the infinite norm of S.
32:19:590Paolo Guiotto: Okay, then in this one, it's not completely simple. And then we have the triangular inequality.
32:36:500Paolo Guiotto: Hmm.
32:38:650Paolo Guiotto: But for the triangular inequality, An important fact that is useful to know is the following.
32:48:470Paolo Guiotto: Well, the infinity normal is itself one of the number M. So, in fact, that infamum is really a minimum, and the minimum is the infinity normal.
33:00:450Paolo Guiotto: Why?
33:02:980Paolo Guiotto: Notice that,
33:11:340Paolo Guiotto: modulus F of X is less or equal than infinity norm of F almost every x.
33:20:310Paolo Guiotto: Let's see why.
33:23:330Paolo Guiotto: This because,
33:26:470Paolo Guiotto: We know that if this is the infinity norm, it's an argument similar to this one we have seen here.
33:34:960Paolo Guiotto: So I will try to… to sketch, maybe you can write… try to write the precise argument as exercise.
33:43:290Paolo Guiotto: So, take your norm. The norm is an infamous, right?
33:47:630Paolo Guiotto: is an infinum of constant M, so it means that as soon as you take a number which is bigger than the infamum, let's say infinity norm of F plus something, there exists an element of the set, so a bound M, which is below that constant. So, you can say that for every epsilon.
34:07:380Paolo Guiotto: There exists an M.
34:09:620Paolo Guiotto: That belongs to the set of, bonds.
34:16:250Paolo Guiotto: Such that this M is less or equal than infinum plus epsilon.
34:23:460Paolo Guiotto: So since that M is a bound, in particular, you get modulus f of x is less than m, which is less than the infamum
34:33:610Paolo Guiotto: The normal plus epsilon, almost everywhere.
34:39:570Paolo Guiotto: Okay, this can be said.
34:42:300Paolo Guiotto: So if you take a number which is slightly bigger than the infinity, the function is bounded almost everywhere, but that number, so that number, if you take the infinite number, the infinity number, and you add something, then that's a number.
34:59:230Paolo Guiotto: Traffic bought that.
35:01:510Paolo Guiotto: Now, what you have to prove, and that's similar to the previous argument, is that from this, you get
35:10:420Paolo Guiotto: this one.
35:19:470Paolo Guiotto: So, epsilon equals 1 over n, and then take the limit, and you will see that this, comes, okay?
35:28:570Paolo Guiotto: So, eps should equal 1 of N.
35:30:910Paolo Guiotto: And then, try to see what… so you have to understand the argument we have seen here, and try to repeat in this different… it's not important if you do not succeed, but try, okay? So if we have this, one second we finish…
35:49:130Paolo Guiotto: We can say that modulus f of x is less or equal infinity norm of F.
35:54:790Paolo Guiotto: Almost everywhere. And the same holds for G. So modulus G of X,
36:01:90Paolo Guiotto: is less or equal than infinity norm of G almost everywhere.
36:07:30Paolo Guiotto: So, if you take F plus G, modulus of F plus G,
36:13:880Paolo Guiotto: Now, this, because of the triangular inequalities, is less than modulus F plus modulus G, which are each less than their respective norm, so norm of F plus norm of G,
36:29:610Paolo Guiotto: Almost everywhere. But that's the conclusion, because it is saying that it's not there.
36:37:290Paolo Guiotto: is one of Vienna, for which models is less than almost everywhere.
36:44:660Paolo Guiotto: So that number is one of them that are in the sector.
36:51:170Paolo Guiotto: When you take the infamo of DM such that modulus FX plus GX is less or equal than M almost everywhere, in that set, there is also this one.
37:06:190Paolo Guiotto: So, what about the informum? The infamo is the minimum, it will be less. So, it will be less than norm of F,
37:14:160Paolo Guiotto: plus norm of G.
37:16:320Paolo Guiotto: But this is, by definition, the norm of F plus G, and there you have the triangular inequality.
37:27:380Paolo Guiotto: Okay.
37:32:610Paolo Guiotto: On this, so, try to do Exercise 945.
37:44:750Paolo Guiotto: Maybe 6?
37:53:130Paolo Guiotto: Well, let's do these two for the moment, Jose.
37:56:900Paolo Guiotto: Okay.