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00:17:620Paolo Guiotto: Good morning!
00:20:970Paolo Guiotto: Okay, so let's start with some of, some exercise… I do the exercise 9… 4, 5…
00:34:800Paolo Guiotto: This is, in LP Spaces chapter.
00:39:930Paolo Guiotto: So we have a measure space, X, F, mu, genetic.
00:49:250Paolo Guiotto: So, we have to show first that,
00:53:260Paolo Guiotto: If the measure of the space is finite, Then, the two norm… is stronger.
01:07:930Paolo Guiotto: Bye.
01:10:220Paolo Guiotto: L1 normal.
01:13:460Paolo Guiotto: And there is, a hint used for Schwartz inequality.
01:22:590Paolo Guiotto: So let's do, let's write the answer to each question.
01:28:790Paolo Guiotto: So, we have to prove that, huh?
01:41:420Paolo Guiotto: So, the genome is stronger if, basically, it controls the one norm. So, there exists a constant, C, positive, such that the one norm of F is controlled by
01:54:270Paolo Guiotto: the true norm of F through.
01:56:730Paolo Guiotto: constant.
01:58:570Paolo Guiotto: Now, the one norm of F is we start… since here we have to prove this. It means that for a generic F, we have to prove this. It is the integral on X of the absolute value of F in mu.
02:13:420Paolo Guiotto: Now, it says use the Posh-Sports inequality. Let's remind what this inequality says. Ipoche-Sports inequality says that the absolute value of the integral of F times G, d mu.
02:27:80Paolo Guiotto: is less or equal than norm of F… 2, sorry, times norm of G2.
02:36:710Paolo Guiotto: This is what the Kashivot says. Now, we have somehow to apply here.
02:43:80Paolo Guiotto: And how? Well, this is a standard trick. We just write modulus of F times 1, so we look at this as a product, where one of the two functions is 1. Since the two functions are positive, I can have the absolute value outside.
02:59:00Paolo Guiotto: It is the same. And now I use the Cauchy-Sports inequality. It says that it is less or equal than the two norm of this absolute value of F, so it is the integral on X of the absolute value of F to power 2,
03:15:00Paolo Guiotto: The mu to 1.5.
03:18:540Paolo Guiotto: times the two norm of G. This is G, so it is just 1, 1 squared d mu.
03:26:280Paolo Guiotto: to one and a half.
03:28:230Paolo Guiotto: So this is the L2 norm of F, times about this one, the integral of 1 is the measure of the set. So we have mu of X to power 1 half. Of course, this can be taken as constant C,
03:45:290Paolo Guiotto: Provided the measure of X is final, which is the case here.
03:49:520Paolo Guiotto: So here we have the deal.
03:52:60Paolo Guiotto: control.
03:53:400Paolo Guiotto: Question 2 says that, however, if measure… of X is infinite.
04:03:570Paolo Guiotto: in general, the previous conclusion is false.
04:08:760Paolo Guiotto: So, in general, the, true norm… is not.
04:18:880Paolo Guiotto: Stronger.
04:22:770Paolo Guiotto: one or normal.
04:26:230Paolo Guiotto: So, and it says, take X equals 0 plus infinity with the rebell measure.
04:32:320Paolo Guiotto: To build an example. Here, we have to show that the two norm is not stronger than one norm. So, stronger means it is written here, there is a constant such that the one norm is bounded by constant, the two norm.
04:46:690Paolo Guiotto: This false means that we cannot find a constant C, or equivalently, from this inequality, we have that
04:56:660Paolo Guiotto: this is not verified when we can have two norms, for example, bounded by a constant, and one norm unbounded. Of course, not for one single function, we need a sequence of functions. So, we need to find…
05:13:720Paolo Guiotto: Sweet.
05:15:100Paolo Guiotto: need… to find… FN, such that… the two norm of FN.
05:27:260Paolo Guiotto: is… bonded Okay? While… It bounded in N.
05:38:670Paolo Guiotto: While the one norm of the FN is unbelievable.
05:48:500Paolo Guiotto: In that.
05:51:550Paolo Guiotto: So…
05:52:860Paolo Guiotto: We need to find something for which the one normal, the integral of modulus of Fn, is big, while the two norm is not big.
06:05:410Paolo Guiotto: Okay? While this one, which is the integral of modulus Fn squared, well, let's take to the power 2, it is the same, it is not big, so I have to think something like this. So, a function for which the integral of modulus of F is big, but the integral of F squared is not big, is bounded.
06:24:510Paolo Guiotto: So we could take a function for which one of the two integrals is finite and the other is infinite.
06:31:620Paolo Guiotto: Now, we… here we can take, on 0 plus infinity, we can take, simple functions for which we know integrability.
06:41:880Paolo Guiotto: So, what kind of function we could consider? For example, normal functions are like powers, 1 over X to some exponent. So, if I take 1 over X to some exponent alpha.
06:54:890Paolo Guiotto: Well, here I can have, problems if I consider 0 plus infinity. If you want, we can eliminate 0, because
07:04:520Paolo Guiotto: a function of this type is never integral, okay? Because those alpha for which it is integral at zero are not integral at infinity, and vice versa.
07:15:10Paolo Guiotto: So, either I eliminate the 0, so it's the same, I take interval 1 plus infinity, or I eliminate the 0, for example. Here, I put the plus 1.
07:25:750Paolo Guiotto: No? You see? So in each of the two cases, I eliminate one of the two problems. So, let's say that we stay on 0 plus infinity, and we take something like this. Now, we know that for alpha equal… equal… we want something that
07:40:630Paolo Guiotto: the two norm is fine, but the one norm is not, no? So, ideally, I would take 1 over 1 plus X. Now, if I take this function, this function is not in L1, but it is in L2, no? This is…
08:00:580Paolo Guiotto: not in L1, because if you take the integral from 0 to plus infinity of the absolute value of the function.
08:09:330Paolo Guiotto: you get, since the function is positive, you get the integral from 0 to plus integer of 1 over 1 plus X.
08:16:250Paolo Guiotto: which is the log of 1 plus X to be evaluated from 0 to plus infinity, and you get plus infinity, no? So this function is not in L1, but it is in L2.
08:31:110Paolo Guiotto: Because in this case, when you take the integral 0 plus infinity of the square of the modulus, in this case, we have integral 0 plus infinity, now 1 over 1 plus X squared.
08:45:490Paolo Guiotto: And this integral is the evaluation of minus 1 over 1 plus X between 0 and plus infinity. At plus infinity, you get 0, at 0 you get minus 1 with the minus comes plus 1, no?
09:00:720Paolo Guiotto: So this function would be not in L1, but in L2. So, this is not 100% useful, because, you see, it is true that I have a function for which the L2 norm is bounded, and the L1 norm is not bounded. But in fact, that function is not even in L1, so I shouldn't be allowed to write the L1 norm. Now, I adjusted
09:25:230Paolo Guiotto: this by introducing functions FN.
09:28:270Paolo Guiotto: What I do is, I take that function, 1 over 1 plus X, and I cut off out of a finite interval, so, for example, 0N.
09:38:890Paolo Guiotto: Now, in this way, I made this as an L1 and L2 function now.
09:43:450Paolo Guiotto: This FN…
09:45:840Paolo Guiotto: So it's a function which is basically like that until point n, and then it is equal to 0.
09:54:690Paolo Guiotto: After. Now, this FN is contained in both in L1 and in L2, so the true norm makes sense. If you compute the one norm of Fn, now you get the integral 0 to plus infinity modules of Fn.
10:14:430Paolo Guiotto: Of course, the reference measure is the Rebec measure. Since Fn is 0, out of the interval 0N, this becomes the integral from 0 to n of 1 over 1 plus X absolute value, it is positive, so I leave like that.
10:29:90Paolo Guiotto: Now, this is the log of 1 plus X to be evaluated between 0 and N. At N, I get a log of 1 plus n.
10:41:170Paolo Guiotto: minus that 0 gets logged 1, which is 0. So the value of this norm is finite, and it is big, as we desired.
10:49:540Paolo Guiotto: When we compute the two norm.
10:53:390Paolo Guiotto: Well, I take the square of this.
10:56:590Paolo Guiotto: to do not carry around the root in the integral, so this is the integral 0 plus infinity modulus Fn squared.
11:05:180Paolo Guiotto: Now, it's similar. Again, function fn is 0 out of 0N, so the integral restricts to the interval 0 to n of 1 over 1 plus X squared, dx.
11:18:790Paolo Guiotto: So this is, again, the evaluation of minus 1 over 1 plus X.
11:24:830Paolo Guiotto: from x equals 0 to X equals 1, sorry, to X equals n. So, at X equals n, I get minus 1 over 1 plus n. At X equals 0, I get minus 1. I have to do the difference, so the second is plus 1.
11:42:630Paolo Guiotto: So this is the value. As you can see, the two norm is 1 minus something, 1 over n plus 1, which is small for n big. In any case, these values are less or equal than 1.
11:56:600Paolo Guiotto: Now we've got the contradiction. So…
12:00:270Paolo Guiotto: If, there exists a constancy such that the,
12:06:670Paolo Guiotto: What is the, we have to prove that.
12:13:330Paolo Guiotto: the two norm is not stronger. Now, suppose that it is stronger, so the one norm is controlled by constant time, the true norm of F,
12:24:900Paolo Guiotto: Then, applying that to the FN, I would add that the one norm of Fn would be less or equal than C, the two norm of Fn. But the two norm of Fn, as we say here, the square of this is less than 1, so this is less than 1.
12:43:390Paolo Guiotto: less or equal C, while the one norm is log of 1 plus n, which is
12:49:500Paolo Guiotto: unbounded when n goes to infinity, no? So you get that we would have log of 1 plus n, or n plus 1, less or equal C, that's for every n natural, and that's impossible.
13:09:140Paolo Guiotto: Okay?
13:11:50Paolo Guiotto: Question number 3.
13:17:910Paolo Guiotto: Even, if, even.
13:23:150Paolo Guiotto: If the measure of X is finite.
13:26:340Paolo Guiotto: So, if the measure of X is valid, we proved in the first question that
13:32:850Paolo Guiotto: the two norms is stronger than one norm. That's all the ways. Now it says, even if the measure of X is finals, in general, the two norms are not equivalent.
13:43:390Paolo Guiotto: So what do you say?
13:48:40Paolo Guiotto: In general, B, one normal, and… The 2-normal.
13:58:210Paolo Guiotto: are not equivalent. I have to…
14:02:480Paolo Guiotto: prove this, no? So, since, we know.
14:10:310Paolo Guiotto: that.
14:12:520Paolo Guiotto: That, that was one. Now the first question of this, number one.
14:17:680Paolo Guiotto: Of this exhaust size.
14:21:20Paolo Guiotto: the, 2, 1, 1 mu of X.
14:28:440Paolo Guiotto: mu of X is finite.
14:32:830Paolo Guiotto: the true norm.
14:34:960Paolo Guiotto: is stronger.
14:37:980Paolo Guiotto: Bang.
14:39:380Paolo Guiotto: the one normal.
14:42:520Paolo Guiotto: So, to show that they are not equivalent, I have to prove that the opposite does not work, okay?
14:49:440Paolo Guiotto: We, to show… Yeah, too.
14:59:230Paolo Guiotto: In general, the one norm, is not stronger.
15:10:140Paolo Guiotto: Then… too north.
15:13:750Paolo Guiotto: And again, the game is the same. I want to find functions fn for which, the one norm is bounded, and the two norm is unbounded, okay? So we look…
15:31:20Paolo Guiotto: for FN, such that D.
15:36:150Paolo Guiotto: One norm is bounded.
15:39:990Paolo Guiotto: And the true norm… is unbounded.
15:48:200Paolo Guiotto: Now, how can we do that? So, the idea is similar. So, let's try to see… one norm means integral of modal of Fn.
15:58:860Paolo Guiotto: So this quantity must be bounded, and the quantity integral modulus Fn squared must be unbounded. So we need a function
16:07:270Paolo Guiotto: that when I square, It is not integral, basically, no?
16:13:130Paolo Guiotto: So again, I look here, I take as x the interval 0, 1, and I look at simple functions like power. And the right function is this one, f of x equals 1 over root of X.
16:28:240Paolo Guiotto: Because if you look at this, and you see what happens to the 1 and the two norms, so the one norm of this F…
16:36:520Paolo Guiotto: is the integral from 0 to 1, of course, here we are on 0, 1, of models of F. So, F is positive, is the integral 0 to 1 of 1 over root of X.
16:50:570Paolo Guiotto: Which is the evaluation of 2 root of X.
16:53:910Paolo Guiotto: between 0 and 1. So at the end, you get 2. So it finds its value. While, when you do the same for the 2 normal, so let's do it power 2, it is the integral of modulus F squared. In this case, we get integral of 1 over x
17:12:89Paolo Guiotto: Which is the evaluation of log of X.
17:15:300Paolo Guiotto: Between 0 and 1, and we get plus infinity.
17:20:720Paolo Guiotto: Now, the idea is to modify these functions to make them,
17:24:869Paolo Guiotto: the functions with the desired characteristics. So all we do, well, you see that this F is not in L2, because the integral is infinite, and this is because we have a problem at zero, no? So we eliminate 0 by restricting the function, so let Fn, no?
17:45:160Paolo Guiotto: of x be 1 over root of X.
17:48:630Paolo Guiotto: But we prevent of taking value 0, so we say indicator of, for example, 1 over n1.
17:57:820Paolo Guiotto: So we cut off the function when we are close to zero.
18:02:600Paolo Guiotto: The function is going, like, 1 over the router, and then we set the value equals 0.
18:10:190Paolo Guiotto: 4x between 0 and 1 over n.
18:14:550Paolo Guiotto: If we repeat the calculation, here we have that. The one norm of this is the integral 01 of modulus F, and since this f is 0, between 0 and 1 over n, the integral
18:28:560Paolo Guiotto: is the integral between 1 over n and 1 of what? 1 over root of X.
18:35:940Paolo Guiotto: Which is, the evaluation of 2 root of X.
18:40:880Paolo Guiotto: between X equals 1 over N and X equals 1, so it is 2 times 1 minus the root of 1 over n.
18:52:140Paolo Guiotto: Which is, in any case, less than 1292. So, bounded.
18:56:790Paolo Guiotto: And when we take the two norma.
19:01:180Paolo Guiotto: We take the square of this, it is equal to integral 02 modules FN squared.
19:07:710Paolo Guiotto: Now, Fn, again, is 0 between 0 and 1 over n, so the integral becomes the integral from 1 over n to 1, and when we do Fn squared on that interval, we get 1 over X.
19:20:240Paolo Guiotto: which is the log of X.
19:23:910Paolo Guiotto: to be evaluated between 1 over n and 1.
19:29:140Paolo Guiotto: At X equals 1, we get log 1, which is 0, minus at X equals 1 over n, we get log of 1 over n.
19:37:860Paolo Guiotto: Which, because of the properties of logarithm, it is log of n.
19:43:820Paolo Guiotto: So now you can conclude this.
19:51:910Paolo Guiotto: Well, there is the last question that we already mentioned. In certain cases, the one normal and two normal can be equivalent. For example, if the set X is finite.
20:02:20Paolo Guiotto: So… Number 4… In… some.
20:09:520Paolo Guiotto: K's… the one norm, and… Too normal.
20:16:990Paolo Guiotto: are equivalent.
20:22:230Paolo Guiotto: For example, take X equal, find it.
20:29:700Paolo Guiotto: sector.
20:32:110Paolo Guiotto: Well, and then continue. Here, we take also as measure mu, the counting measure.
20:46:40Paolo Guiotto: So let's say that the set X is made of,
20:49:770Paolo Guiotto: Final number of elements, we can always call them 1, 2, 3, etc, until the number of elements say D.
20:57:920Paolo Guiotto: No? So, a function f defined on x real-valued.
21:03:340Paolo Guiotto: is what? It's an object that, to each number from 1 to D, is a number. So it is identified once you know the values F of 1, f of 2, F of 3, and so on, F of D.
21:19:270Paolo Guiotto: So… but this is exactly an array of RV.
21:25:150Paolo Guiotto: And if you look at the one normal, for example, the one normal of F, it is the integral on X of modulus f d mu, no?
21:36:510Paolo Guiotto: But since the set X is just a final set, we can decompose as some on the subset made by the single element J of models of F immune.
21:53:60Paolo Guiotto: Now, since this set is made of one single element, the function on that set is constant, because it is just one single point, and the value will be modulus of F of J.
22:04:960Paolo Guiotto: So this is a constant that comes out of the integral, so I will have sum modulus f of j times the measure of that singleton. If the measure is the counting measure, each singleton has measure 1, because the measure points count
22:22:510Paolo Guiotto: It's the number of points in the set, so this is equal to 1.
22:27:330Paolo Guiotto: At the end, this is the sum modulus FJ, which is what we call the Manhattan norm for the vector, no?
22:38:180Paolo Guiotto: But as you can see, it's a norm for this vector. If you take the Euclidean norma.
22:44:680Paolo Guiotto: the two norms, sorry, the L2 norm squared. Doing the same calculation, the only thing that changes is that you have modulus F squared in mu. So everything is replaced by modulus F squared. So at the end, you will have sum of a J of modulus FJ squared.
23:02:520Paolo Guiotto: And so the altern is the root of this.
23:06:730Paolo Guiotto: That's the Euclidean null.
23:09:630Paolo Guiotto: And we know that on a fin-dimensional space, all norms are equivalent.
23:14:470Paolo Guiotto: So, since these two quantities, they are equivalent, So, equivalent… Because… on… are D… all norms… are equivalent.
23:37:130Paolo Guiotto: You get that also, as a consequence, also these two guys are equivalent.
23:46:550Paolo Guiotto: Okay, so in general, there is no relation, this exercise tells. There is no relation, general relation between the R1 norm and the R2 norm, okay? But if the one… if the measure of the set is finite, you have that 2 norm is stronger.
24:04:680Paolo Guiotto: Otherwise, and this is basically the unique general relation we have, no? If the measure of the set is infinite, we have examples for which we can say that the two norm is not more stronger than one norm.
24:22:740Paolo Guiotto: And if the measure is finite, we cannot even say that the two norms are equivalent, because of the other example. The only case when they are equivalent is when the space is finite dimensional, and this is the discussion. This ends this XSR.
24:41:280Paolo Guiotto: Okay, I want to do now the, 946.
24:53:320Paolo Guiotto: It says, F is in L2. R,
24:59:670Paolo Guiotto: Question 1, which we basically already discussed there.
25:04:140Paolo Guiotto: Is it true… Is. It.
25:07:910Paolo Guiotto: True.
25:09:770Paolo Guiotto: that… F is also in L1.
25:16:420Paolo Guiotto: So it says, prove, in general, if true, disprove with a counterexample, if false.
25:22:260Paolo Guiotto: We already met something like this, because, here, this was this, the question too, no?
25:30:290Paolo Guiotto: We had an example of a function
25:36:140Paolo Guiotto: Which is in L2, exactly, and is in the next statement, but not in L1.
25:42:240Paolo Guiotto: So we may work out with this example. For example, we can take the same example, and which works on 0 plus infinity, we cut to zero, on the negative part, take f of x equal indicator of 0 plus infinity.
25:59:240Paolo Guiotto: X times 1 over 1 plus X. And then you have the example. So this is an L2 function, but not L1 function.
26:14:670Paolo Guiotto: 2, it says… but now, assume that you know something more, that, only this, so some additional assumptions. Suppose that you know that also X times F
26:28:770Paolo Guiotto: is in L2.
26:32:60Paolo Guiotto: So X times F means what? Means that the function X, F of X, as a function of x is in L2, so you know that integral on R of XFX.
26:45:180Paolo Guiotto: Absolute value squared, EX, this is finite.
26:50:510Paolo Guiotto: Then prove that F belongs in L1 in this case, and we have also a binder.
26:57:730Paolo Guiotto: That says the one norm is bounded by square root of 2,
27:03:480Paolo Guiotto: L to norm of F plus L to norm of X times F.
27:15:570Paolo Guiotto: Okay, so let's see how to do this. So let's start trying to prove that F is in L1, so we have to consider the one norm, which is the integral on R of the absolute value of F.
27:33:670Paolo Guiotto: Okay.
27:35:460Paolo Guiotto: So… There is no heat here, okay. However, we know that this integral is finite.
27:46:360Paolo Guiotto: So this integral has F squared, which I do not have here.
27:51:800Paolo Guiotto: So I can create this square. One way is, the… the trick of the Cauchy-Sports inequality, no?
28:00:880Paolo Guiotto: You have seen here now, this is F to power 1, has been controlled by F to power 2.
28:07:800Paolo Guiotto: But the problem is that I cannot apply the Cauchy force inequality on an unbounded set, because if the measure of x is infinite, this coefficient, mu of X, would be infinity. So I cannot just say that, okay, let's do this.
28:23:490Paolo Guiotto: This is, by Schiffords, this is true, integral on R of modulus F squared to 1 half, which is good, because it is the L2 norm, this finite. But then I have also integral on R, because the other factor is 1, of 1 squared dx to 1 half.
28:42:430Paolo Guiotto: But that's equal to plus infinity, so…
28:46:240Paolo Guiotto: I cannot control in that way.
28:48:850Paolo Guiotto: Okay.
28:50:290Paolo Guiotto: But we know that we can do that on a finite measure domain, okay? Not on the interior line, but on a limited part, we could do. And then… then, let's… it says that we know also that this is finite.
29:07:840Paolo Guiotto: So, what should I do to recreate here the integral on R of XF squared?
29:15:940Paolo Guiotto: to one half after applying the Cauchy-Fort's inequality.
29:21:230Paolo Guiotto: What should I put here?
29:31:900Paolo Guiotto: It is similar, because it is like, if you say, you take the integral of modules of F,
29:37:170Paolo Guiotto: and you multiply and divide by X. So you write 1 over, you want modules of X, modules of X, F of X,
29:47:950Paolo Guiotto: Then I use Cauchy Schwarz, and this becomes 1 over X modules of X squared.
29:55:440Paolo Guiotto: Which is not good, again, because this integral is good at infinity, but not good at zero.
30:04:860Paolo Guiotto: You see?
30:06:210Paolo Guiotto: So this is also infinity, so it's not good. Now, the point is that you should try to combine these two things.
30:14:860Paolo Guiotto: So the first bound would be 9,
30:18:300Paolo Guiotto: Instead of having as integration thing, do the fine.
30:24:420Paolo Guiotto: The second one would be good if we could do that away from sphere.
30:35:590Paolo Guiotto: So that could be now an idea, no? So, we have one bound, the first one, which is good on any final interval, the second bound, which is good away from zero.
30:48:40Paolo Guiotto: So I could say, let's divide the real line in three parts. On this part, the middle part, which is bounded, I will use the first bound somehow, and in the other two parts, which are away from zero, I will use the second bound.
31:05:410Paolo Guiotto: And so the choice I can do, for example, is let's put minus 1, and let's compute the one norm of f in this way, integral from minus infinity plus infinity.
31:16:920Paolo Guiotto: modulus of F, I split into minus infinity to minus 1 modulus F, plus minus 1 to 1 modulus F, plus 1 to plus infinity modulus F.
31:30:410Paolo Guiotto: Okay, now, let's look at the middle term. This one will be controlled by the direct Cauchy force inequality, as we have done above, because this integral minus 1 to 1 modules of F
31:44:350Paolo Guiotto: Because of Cauchy-Schworth. I do not write DX, but it is always a dx. This is… these are all the bag integrals here. So this will be less or equal integral minus 11 of F squared.
31:57:20Paolo Guiotto: to 1 half, then I have an integral minus 1 1 of 1 square, to one half.
32:03:820Paolo Guiotto: Notice that this is the integral of 1 between minus 1 and 1, so it is the length of the interval, which is 2, no?
32:11:710Paolo Guiotto: So this number is 2, and you get root of 2 times
32:18:840Paolo Guiotto: this integral, which is not the anterior L2 normal, because it is the L2 normal between minus 1 and 1, but that will be less or equal than the full integral, so I can say less or equal than integral on R of modulus F squared to 1 half
32:35:910Paolo Guiotto: And that's now the L2 norm, so I get…
32:38:890Paolo Guiotto: This is less than or equal than root of 2L2 norm of F.
32:45:390Paolo Guiotto: So now you understand that, why there is this root of 2 here, where it comes from, no?
32:53:620Paolo Guiotto: It comes from this. So, I have this first bound, integral minus 1, 1 modules F is bounded by root of 2 norm of F, L2.
33:04:890Paolo Guiotto: Let's take, aside this, and now let's work on the other two integrals, which are similar. So let's take, for example, this one, the integral from 1 to plus infinity.
33:16:700Paolo Guiotto: Now, on this interval, we cannot use the Cauchy-Swartz bound.
33:22:350Paolo Guiotto: Because that bound is with a constant equal to plus infinity. But we use three cohesive words, but before, we recreate the factor X in front of F. So I multiply and divide by X, positive here.
33:37:990Paolo Guiotto: So 1 over X, X times F, if you want to put your modulus wherever.
33:46:90Paolo Guiotto: Now I use Cauchy-Force with this product, no? This is 1, and this is 2. So I get less or equal, again, Cauchy-Force, integral from 1 to plus infinity of XF squared
34:00:860Paolo Guiotto: to power 1 half times integral from 1 to plus infinity of 1 over X squared to 1 half.
34:11:290Paolo Guiotto: Now, you see that that integral of 1 over X squared is no more singular, because that's integral on integral 1 plus infinity. We can compute the integral, this is minus 1 over x, and that's the primitive, to be evaluated between 1 and plus infinity.
34:28:440Paolo Guiotto: At plus infinity, you get 0. Minus.
34:32:449Paolo Guiotto: At 1, you get minus 1, so the value is plus 1.
34:37:469Paolo Guiotto: So all this is plus 1. To expound 1 half, this is 1.
34:44:159Paolo Guiotto: So at the end, I have this, that the integral from 1 to plus infinity of modulus of F is controlled by 1 times this quantity, which is DL
34:57:680Paolo Guiotto: too Norma.
34:59:310Paolo Guiotto: But on 1 plus infinity, here, if I want to get the final bound, I don't have to, to… to take all the integral from minus infinity plus infinity. So let's take the alginorm of F between, or let's say, on…
35:19:630Paolo Guiotto: on, 1 to plus infinity. They are to norm on 1 to plus infinity.
35:26:890Paolo Guiotto: So we get this… The, L2 norma on… 1 to plus infinity.
35:35:830Paolo Guiotto: Because we have the other integral, that one from minus infinity to minus 1, which is
35:43:960Paolo Guiotto: Analogous to this one, repeating the same calculation, we get, again, the L to norm of F, but this time on the interval from minus infinity to minus 1.
35:59:860Paolo Guiotto: Okay, so now let's try to put together everything. So, what we have is then…
36:06:580Paolo Guiotto: the one norm of F, that we split into these three integrals. You see that the first
36:15:850Paolo Guiotto: And the last…
36:17:260Paolo Guiotto: are bounded by the same kind of quantity. So let's write first the middle term that is less or equal root of 2
36:27:480Paolo Guiotto: the two norm of F, plus we have the 2norm of XF
36:34:360Paolo Guiotto: On minus infinity to minus 1.
36:40:170Paolo Guiotto: and the 2-norm of XF, On 1 to plus infinity.
36:50:210Paolo Guiotto: Okay.
36:52:170Paolo Guiotto: Now, if you look at this, what we obtain, it starts looking similar to what we have to obtain, because we have root of 2
37:01:720Paolo Guiotto: norm of F to norm of F, and that's already there.
37:06:440Paolo Guiotto: And now, we need to show that the remaining part is bounded by root of 2D norm of XF, okay? So the conclusion holds once I prove if… that this is less or equal than root of 2,
37:23:460Paolo Guiotto: the L2 norm of XF, the L2 norm on the real line.
37:29:400Paolo Guiotto: the full L2 nore. Now, the point is, is that, true, no?
37:35:150Paolo Guiotto: or false? Let's see. Now, let's focus only on this part. To say if this is true, I would say, let's take all this…
37:45:280Paolo Guiotto: No? Remind that the L2 norms are roots, so it is better if we square everything, and we say that this is, if and only if, the square of
37:57:760Paolo Guiotto: XFL2 norm on minus infinity to minus 1.
38:05:110Paolo Guiotto: Plus.
38:06:310Paolo Guiotto: the L2 norm of XF,
38:09:250Paolo Guiotto: On 1 plus infinity, the square of all this is less or equal than the square of the right-hand side.
38:17:940Paolo Guiotto: which is 2, the altenome of XF on the interior line, squared.
38:27:420Paolo Guiotto: Oh, this… this is,
38:30:710Paolo Guiotto: This is better, because… let's now restore integrals here. So, this is equivalent to… when you do the square of the left-hand side, you do the square of the first factor, right?
38:44:550Paolo Guiotto: plus the square of the second, plus the double product. But the square of the first is the integral from minus infinity to minus 1 of modulus XF squared. That's the square of the first.
38:57:110Paolo Guiotto: Last, the square of the second is the same, integral 1 to plus infinity of XF squared.
39:04:790Paolo Guiotto: And then we have the double product, so 2, norm of XF…
39:12:600Paolo Guiotto: L2 on… I do not write,
39:17:420Paolo Guiotto: I don't need to write here, as you will see.
39:21:310Paolo Guiotto: Normals of XFL2,
39:24:730Paolo Guiotto: 1 plus infinity. Now, all this, the point is, must be less or equal than to… that, L2 norm is the integral from minus infinity plus infinity of XF squared.
39:44:780Paolo Guiotto: Now, as you can see, these two together
39:51:70Paolo Guiotto: They are the integrals from minus infinity to minus 1 plus the integral from 1 to plus infinity, when they together are less than the integrals from minus infinity to plus infinity.
40:02:600Paolo Guiotto: So, what they need to prove is that these two, this factor here in the dot, this tail here, is less than this.
40:12:710Paolo Guiotto: You see?
40:14:190Paolo Guiotto: So, I can say that this, in the circle in red, is alone less or equal than the full integral from minus infinity plus infinity of XF squared, right?
40:25:960Paolo Guiotto: So basically, this is true.
40:28:930Paolo Guiotto: Provided to that product, XFL2, on minus infinity, to minus 1.
40:40:130Paolo Guiotto: excess.
40:42:00Paolo Guiotto: L2, 1 to plus infinity. This is less or equal than the integral from minus infinity to plus infinity of XF squared.
40:54:800Paolo Guiotto: So it remains this, huh?
41:00:930Paolo Guiotto: Oh, now, reminder, these are…
41:03:360Paolo Guiotto: 1 half integral from minus infinity to minus 1 XF squared, and this is integral from 1 to plus infinity XF squared to 1 half.
41:20:820Paolo Guiotto: You see?
41:22:250Paolo Guiotto: And now, there is an elementary inequality that makes this… you're reminded 2AB is less than A square plus B squared.
41:34:150Paolo Guiotto: So use this, too. A, B.
41:37:890Paolo Guiotto: This guy is less or equal than the square of A, which is integral from minus infinity to minus 1 of XF squared.
41:48:450Paolo Guiotto: then the 1 half squared disappear. Then I have the square of B, which is integral from 1 to plus infinity of XF squared, and that's less or equal than this one. So you have the chain. Now it is completely closed.
42:05:720Paolo Guiotto: So, we have the conclusion.
42:14:960Paolo Guiotto: Okay.
42:25:630Paolo Guiotto: Okay, now today I will, publish
42:31:380Paolo Guiotto: some of the solutions of these chapters, nine… 10, and
42:40:140Paolo Guiotto: Not 11, because we have to do today, but…
42:44:690Paolo Guiotto: 12. We already done that, so…
42:47:640Paolo Guiotto: on completeness. In particular, I recommend you to do these exercises.
42:53:530Paolo Guiotto: So, let's say that for Chapter 9,
42:58:290Paolo Guiotto: Since you will have the solutions.
43:01:700Paolo Guiotto: This afternoon, basically.
43:05:260Paolo Guiotto: So, do… 9… 4… 7… Well…
43:15:440Paolo Guiotto: 8 and 9 are more difficult, okay? So, 3-star exercises. So, it is not important if you succeed, but this one is important, okay?
43:30:940Paolo Guiotto: It's not particularly difficult, but it uses what we have seen. So, at least this one.
43:38:610Paolo Guiotto: about Chapter 10… So… Do, 10… The Exercise 10, 3, 1…
43:51:330Paolo Guiotto: The two, we have done in class, the number three, the number 5…
44:01:820Paolo Guiotto: Well, here you can do all the exercises.
44:05:200Paolo Guiotto: 5, 6, 7… But… 8… I already left you to do these exercises.
44:16:860Paolo Guiotto: from Chapter 12, which is,
44:20:430Paolo Guiotto: A chapter where there is completeness.
44:24:150Paolo Guiotto: do exercise. We don't do because we are a little bit late in time, so I want to do the application of Cauchier sequences to the Banach fixed point theorem, which is pretty important for solving equations.
44:39:120Paolo Guiotto: But I'm pretty sure you will see in any… in other courses. So do exercises 12, 4…
44:47:620Paolo Guiotto: Let's see, one… 2… The three we have done in class,
45:00:190Paolo Guiotto: Well, I would, I would tell you to do the numbers.
45:09:190Paolo Guiotto: But not easier, for… Five, huh?
45:23:110Paolo Guiotto: 6, and 7. The last exercises are on fixed point theorem.
45:32:700Paolo Guiotto: Okay, so we have still to… to do one…
45:39:210Paolo Guiotto: One topic here, before we leave this part.
45:44:670Paolo Guiotto: So we continue without any break, right?
45:48:620Paolo Guiotto: So this part is, this operation is called convolution.
46:00:610Paolo Guiotto: So now, the convolution with this strange name is an operation that arises for a main motivation, which is the following.
46:12:630Paolo Guiotto: In general, And now… P function.
46:20:350Paolo Guiotto: Well, let's, let's here talk about LP, RD.
46:26:710Paolo Guiotto: with the LeBague measure, so MD lambda D,
46:32:470Paolo Guiotto: So this is, the LeBague measure.
46:39:360Paolo Guiotto: So a measurable integral function, with respect to the Libert measure is not a regular function, no? It's…
46:51:520Paolo Guiotto: not, regular. Regular means, you can put the objective you like, continuous, differentiable.
47:02:260Paolo Guiotto: And so on.
47:03:910Paolo Guiotto: With a certain number of derivatives, etc.
47:07:580Paolo Guiotto: It's, in fact, a mess. A measurable function is a real mess. Now, is there a way to regularize a generic function of this type, obtaining a nice function? So, for example, continuous? The question is.
47:31:860Paolo Guiotto: Is there… Wait… to… regularize.
47:46:540Paolo Guiotto: And F in the space LP.
47:52:250Paolo Guiotto: R. T.
47:55:40Paolo Guiotto: Now, the idea is that, for example, it's a very practical problem. For example, this is a very… is the base of the signal theory. So, you have signals can be modeled through functions, no?
48:12:220Paolo Guiotto: For example, an audio signal is a function of one variable, no? An image, for example, a plain image, like the picture, no?
48:24:320Paolo Guiotto: can be, defined through a function of two variables. At every point, you tell what is the color of that pixel, okay?
48:33:770Paolo Guiotto: And, well, now… These kind of images are subject to noises.
48:41:20Paolo Guiotto: So normally, when you take a picture as well, when you… when you transmit a voice, you know, to a signal, there are lots of noises that affect the signal, and at the end, you don't have the original signal.
48:56:320Paolo Guiotto: So, in fact, if you could see the original signal, you would see a very irregular and noised signal, no? So, the point is, suppose that you see something that is very extremely irregular like that, that you imagine has a certain
49:14:410Paolo Guiotto: shape like this. Now, how can you get the shape cleaning up this from the noses? Now.
49:21:470Paolo Guiotto: So that's the scope of regularization.
49:24:890Paolo Guiotto: And it is really made, when you think to the typical image format, like JPG. JPG is a standard that makes exactly this, no? It takes an image, and it regularizes the image together.
49:39:160Paolo Guiotto: Actually, there, it's for compression regional, not to simplify the image, but not too badly.
49:44:920Paolo Guiotto: To get a less heavy image, less heavy file.
49:50:70Paolo Guiotto: Now, there is a natural idea on how to do this, because we know that if you have very big irregularities.
50:00:660Paolo Guiotto: The natural way to regularize, no, is to do an average of values, no? Imagine we have a class of people with different hates, for example.
50:12:710Paolo Guiotto: And so this is a very irregular shape, and we want something smooth. We do the average, so the average maybe is not the age of any one of us, but it will be an average, so it will be a constant that works for everyone, no?
50:29:300Paolo Guiotto: So, averages are the way to do this. So, how do we do this with the function?
50:36:420Paolo Guiotto: Now, the idea is the following. Take a function at some point X,
50:41:370Paolo Guiotto: And you want to replace the value of the function f at point X by an average of values of F taken around the point X. So you may decide
50:53:360Paolo Guiotto: to take… to fix the size of the interval around which you want to do the average. So let's say we are here talking about a function of one real variable. So we take an interval like X minus epsilon X plus epsilon. In the idea, this epsilon is a small number, no?
51:11:900Paolo Guiotto: And, how do we compute the average of F around that polymer?
51:16:570Paolo Guiotto: Of course, X is varying on the real, so it's not a discrete set, but we have a way to do continuum sums, and that's provided by an integral. So we take the integral from X minus epsilon to X plus epsilon of the values of F, let's call them f of y, in the variable of y, and we divide by
51:40:100Paolo Guiotto: The length of the interval, which is, in this case, 1 over 2 epsilon.
51:44:860Paolo Guiotto: So we call this quantity F epsilon over X.
51:50:630Paolo Guiotto: And if we do this X by X, we define a function, no?
51:56:150Paolo Guiotto: Which is, again, epsilon is fixed in this.
51:59:980Paolo Guiotto: the picture, and this function replaced the value path at point X with the average of values around the point X with this range of minimal equals this size is 2x, not the length of the
52:17:140Paolo Guiotto: Now, this integral that we have written here can be reworked and rewritten in another form, so we write 1 over 2 epsilon.
52:29:840Paolo Guiotto: We start putting this as an integral on the interior line, f of y, and to make the restriction, we put an indicator here of interval x minus epsilon X plus epsilon.
52:44:440Paolo Guiotto: of Y. So that indicator is equal to 1 only when Y is in the interval x minus epsilon x plus epsilon.
52:54:120Paolo Guiotto: Now, this means that this indicator is 1 if and only if
53:01:870Paolo Guiotto: Y is between X minus epsilon and X plus epsilon. That is, carrying X on the other side, this is Y minus X between
53:17:260Paolo Guiotto: Below minus epsilon, above epsilon.
53:21:150Paolo Guiotto: So that indicator is 1 if and only if Y is there, which is equivalent of saying that Y minus X must be in minus epsilon epsilon, which is equivalent of saying that the indicator of the interval minus epsilon epsilon
53:37:530Paolo Guiotto: Evaluated at Y.
53:40:320Paolo Guiotto: minus X, or X minus Y. I prefer to write this to have the standard notation. It is the same.
53:48:10Paolo Guiotto: Belongs between minus epsilon and epsilon.
53:51:370Paolo Guiotto: So, this means that we can rewrite our integral as 1 over 2 epsilon, integral on real line of FY
54:02:830Paolo Guiotto: Indicator minus epsilon. Epsilon evaluated at X minus Y, DY,
54:11:480Paolo Guiotto: Now, if I give this factor 1 over 2 epsilon to the indicator.
54:17:410Paolo Guiotto: This is the integral from, on r of f of y
54:22:360Paolo Guiotto: Times a function, which is this, 1 over 2 epsilon, indicator of minus epsilon epsilon.
54:32:700Paolo Guiotto: evaluated at point X minus YDY.
54:45:30Paolo Guiotto: Okay, now, this, this, if we call this function delta epsilon.
54:54:440Paolo Guiotto: Now, this function is, is like what?
54:57:890Paolo Guiotto: So if you look at the function delta epsilon of t.
55:02:10Paolo Guiotto: equal 1 over 2 epsilon indicator of interval minus epsilon epsilon evaluated at T. As a function of t. This function is zero everywhere, except on that small in our
55:17:600Paolo Guiotto: intuition, interval minus epsilon epsilon. So this is the interval minus epsilon epsilon. D function, delta epsilon is zero outside that interval. And in the interval, delta epsilon, the function, is equal to 1 over 2 epsilon. 1 epsilon is big.
55:36:270Paolo Guiotto: No? It's a very big number, 1 over 2 epsilon, and so the function is like that.
55:43:930Paolo Guiotto: So now you understand what is the shape of this function.
55:49:320Paolo Guiotto: No? And when you restrict epsilon, so you make epsilon smaller, what happens? That you get a function that,
55:57:600Paolo Guiotto: Paza.
55:59:220Paolo Guiotto: is 0 out of a smaller interval, and on that interval is bigger, you see?
56:14:680Paolo Guiotto: Now, you may imagine that when epsilon is 0, I'm not averaging F, no? In the idea, no? Because it just means that you are doing the average on an interval of length 0. So, there is no average, it is F. So, I should expect that when epsilon goes to zero, this quantity
56:33:980Paolo Guiotto: this function that I called here F epsilon should go to the function f. So the main question is, is it true that this f epsilon of X goes, in some sense, to f of x when epsilon goes to 0?
56:54:50Paolo Guiotto: And of course, in which sense?
57:03:530Paolo Guiotto: For example, we may notice that if I take this effort, with F integral, let's say.
57:13:320Paolo Guiotto: for simplicity, L1, but it actually can be LP, whatever is B. That function is automatically a continuous function.
57:22:670Paolo Guiotto: So DF epsilon is more regular than F, because if F is integral, it does not mean that it is continuous, okay?
57:33:460Paolo Guiotto: For example, we could prove…
57:43:90Paolo Guiotto: That this proposition holds.
57:47:610Paolo Guiotto: Suppose that you have a function F in L1R,
57:53:590Paolo Guiotto: Then, the function f epsilon is a continuous function in the real line for every epsilon positive. So, you remind that all this problem arises from the problem of building a regularization of F.
58:11:30Paolo Guiotto: So, this says you built a regularization. Of F is, if we are able to prove that when epsilon goes to zero, we go to F of X. Now, just to have an idea on how this works.
58:29:140Paolo Guiotto: I don't want to do the proof here, because it is not this most important factor. Let's say a sketch.
58:40:890Paolo Guiotto: Off.
58:42:720Paolo Guiotto: Proof.
58:47:180Paolo Guiotto: The point is that f epsilon of X is, by definition, one of a 2 epsilon integrals from x minus epsilon to X plus epsilon of F.
58:59:10Paolo Guiotto: Now, 1 over 2 epsilon here is fixed, because epsilon positive is fixed at this point. We are not yet taking the limit, no? We are just saying that for epsilon positive, this… this family of functions, this F epsilon.
59:14:210Paolo Guiotto: are all continuous. So, let's focus on the… how this function depends on X. Where is X? You see that X is here, in the endpoints of this integral.
59:24:680Paolo Guiotto: Now, there is a general factor
59:27:640Paolo Guiotto: That follows from dominated convergence, general Fact.
59:34:910Paolo Guiotto: If you have an integral function, L1R,
59:39:740Paolo Guiotto: The integral function, the integral from A to X of F,
59:45:740Paolo Guiotto: So I'll call this function GA of X.
59:52:210Paolo Guiotto: This function is always continuous.
00:00:970Paolo Guiotto: This follows…
00:08:520Paolo Guiotto: from… dominated the… convergence.
00:14:680Paolo Guiotto: let's see very roughly why, you know? If you send an XN to some X star.
00:23:390Paolo Guiotto: You want to show that?
00:28:370Paolo Guiotto: G of XN, GA of XN goes to GA of X star. You want to show this.
00:37:590Paolo Guiotto: No? Because that's the continuity. Now, what is GA of Xn? According to that definition, GA of XN is the integral from A to Xn of F.
00:50:160Paolo Guiotto: That doesn't look a problem of,
00:54:570Paolo Guiotto: alleviate the problem with the integer of a sequence at N. You see? N is in this index that changes to the endpoint of the sequence.
01:07:410Paolo Guiotto: However, you can easily transform into that kind of problem, because that's call the integration variable T, or Y, let's say Y.
01:17:270Paolo Guiotto: We are using what?
01:20:700Paolo Guiotto: You can always transform this into, for example, integral form A to a B fixed, F of Y…
01:29:520Paolo Guiotto: But if you want even more, you can put the interior line.
01:33:710Paolo Guiotto: So, let's say that we put the integral on R of f of y, and then you have indicator of A
01:42:340Paolo Guiotto: Xn in the variable Y.
01:46:120Paolo Guiotto: Now, this problem is a problem of type, limit of integrals depending on parameters. You call these functions f and of y.
01:59:260Paolo Guiotto: Now, what happens when you send N to infinity? To apply the negative convergence, you have to check what happens to FN when you send N to infinity, Y fixed, but this is F of Y, which is a constant in N, times the indicator A2XN,
02:18:100Paolo Guiotto: of why?
02:20:420Paolo Guiotto: Since this XN goes to X star, you can easily understand that this will go to F of Y indicator of A, Y, sorry, A, X star.
02:37:620Paolo Guiotto: of why?
02:40:480Paolo Guiotto: Okay?
02:41:930Paolo Guiotto: This happens not for all Y, because for Y equal X star, it is not… it's dangerous, but let's say that if this is X star, and you take a Y here.
02:55:150Paolo Guiotto: Since XN goes to X star, it will be definitely close to X.
03:01:600Paolo Guiotto: So, for example, if you are below X star, and you evaluate the risk indicator, it means that this indicator is 1, in this case, and it will remain equal 2, because the point itself goes to X star.
03:16:50Paolo Guiotto: If there's a point here, so you are outside of the interval A, etc, the indicator would be equal to zero.
03:24:810Paolo Guiotto: And it will stay equal to 0, because point XN will not come here. It will go to X star. So, this indicator goes to this one for all Y except for Y equal and X star. So, let's say almost everywhere, except for one point. So, this means almost everywhere.
03:44:590Paolo Guiotto: And second, you have also the domination, because models of Fn, Y, which is models of F, Y, times the indicator.
03:55:550Paolo Guiotto: The indicator, however, is less or equal than 1. This is less or equal than modulus f of y. And we know that, by our assumption, that this function is integral.
04:06:170Paolo Guiotto: So, Le Bergh theorem applies, no, Le Berg-dominated convergence theorem applies, dominated.
04:15:780Paolo Guiotto: convergence.
04:17:610Paolo Guiotto: applies… And we concluded that the function GA of X is continuous.
04:27:180Paolo Guiotto: Because when you pass to the limit, the FN of Y, you see, goes to the F of Y indicator ax star, and this means that this guy goes to the integral from A to X star of F of Y, which is exactly the function we call the GA at point X star.
04:49:130Paolo Guiotto: So we get the continuity. Now, the message is that an integral function, so a function where the variable is one of the endpoints, is continuous in the endpoint.
05:01:570Paolo Guiotto: So, in particular, this one, which can be written, for example, splitting the integral. So…
05:13:400Paolo Guiotto: F epsilon of X, which is 1 over 2 epsilon.
05:17:660Paolo Guiotto: the integral from X minus epsilon to X plus epsilon
05:22:340Paolo Guiotto: of F, you can divide this into… you put an arbitrary point, for example, 0. This becomes integral from X epsil, x minus epsilon to 0 of F plus the integral from 0 to X plus epsilon of F.
05:38:710Paolo Guiotto: Now, as you can see, this is, well, absolute is a constant, this function of the time GA, no, is an integral function, but the value is in one of the endpoint of the integral. That's continuous function of X. And this one is similar, you revert the order, I change inside, yeah? So these two are like…
06:00:490Paolo Guiotto: G… this is a G0 of X plus epsilon, and this is minus G0 of X minus epsilon.
06:10:90Paolo Guiotto: The minus is because this integral is minus integral from 0 to X minus epsilon of L, no? Since you know that the continuous…
06:25:560Paolo Guiotto: So this gives F epsilon continuous.
06:32:760Paolo Guiotto: So this says that this operation actually works, no? You take a bed from the point of view of regularity function, L1, and you have at least a continuous function, okay?
06:49:150Paolo Guiotto: For example, we cannot expect to have a stronger regularization. For example, can I approximate with a smooth function, not only continuous, but also differentiable, with the tangent?
07:04:980Paolo Guiotto: I would need to compute the derivative, no? How would I compute the derivative of this F epsilon with respect to X? So, so…
07:19:590Paolo Guiotto: If…
07:21:260Paolo Guiotto: F is in L1, this function, f epsilon of X, which is the integral on R of fy times that function we call delta epsilon X minus y.
07:35:580Paolo Guiotto: DY with the delta epsilon.
07:39:10Paolo Guiotto: of, T equal 1 over 2 epsilon, indicated minus epsilon epsilon, T.
07:48:940Paolo Guiotto: This F epsilon is, we discovered it is continuous on R.
07:56:710Paolo Guiotto: So, can we do better? Can we get, can we say that this ephepsial is, for example, differentiable? It has also derivative, so it is not only continuous, it's smooth, no?
08:09:300Paolo Guiotto: Can we say… can… we say… that… F epsilon.
08:19:189Paolo Guiotto: is also… differentiable?
08:25:140Paolo Guiotto: Well, the idea is that if you look at this function, we should, we should,
08:32:100Paolo Guiotto: Differentiate, so we should compute what? The derivative of that F epsilon with respect to X.
08:42:310Paolo Guiotto: And as you can see, what kind of thing is this?
08:46:240Paolo Guiotto: It is an integral where X is the parameter, not the integration part.
08:51:800Paolo Guiotto: So that's the kind of objects we have seen,
08:56:770Paolo Guiotto: That can be treated by differentiating an integral side. So, roughly, I would say, okay, let's carry the derivative inside, this will become integral of f of y, then we have to differentiate with respect to Y, this delta epsilon of X minus y.
09:15:490Paolo Guiotto: Well… As you can see, this function, delta epsilon, is not very regular, no? Because of the jump.
09:25:130Paolo Guiotto: So, we might… we might have problems here with this, especially to control the derivative, you know? We need to apply the differentiation and the integral sign. I do not need only to differentiate the function, but to control the derivative. And that jump is not a good thing.
09:45:80Paolo Guiotto: However, We may think, why should we consider these…
09:53:420Paolo Guiotto: Yes, they are simple, no function. 0 out of an interval, and then constant in that interval. But they could be replaced, in principle, I might expect, by something like this.
10:07:290Paolo Guiotto: A more general object could be…
10:09:960Paolo Guiotto: a sort of function that is centered, around the origin, but it is smooth, no? Why… why this, maybe this is a good alternative. So…
10:26:530Paolo Guiotto: To do this.
10:33:80Paolo Guiotto: we need… Delta, epsilon.
10:37:590Paolo Guiotto: differentiable.
10:40:60Paolo Guiotto: And… which is not the case.
10:50:60Paolo Guiotto: So, with this delta epsilon, we cannot expect to get better than a continuous regularization of F.
10:59:90Paolo Guiotto: So, we did… These… Delta episodes, we… Cannot.
11:10:810Paolo Guiotto: Expect that… Anything…
11:19:10Paolo Guiotto: more… Then… then these F-psilon are just continuous.
11:26:570Paolo Guiotto: But what if we… What?
11:33:60Paolo Guiotto: if, we use… regular… Delta epsilon.
11:44:910Paolo Guiotto: Now, this kind of function, this kind of function is not actually a function, it's a Fermi of function. As we say, there is this parameter epsilon. If you ma- immediately, if you put epsilon equals zero, this function would be zero everywhere, except at zero value.
12:03:920Paolo Guiotto: If you look at these functions, they have also some other features. For example, these delta functions are positive, no? You see?
12:13:540Paolo Guiotto: And moreover, if you complete the integral of the diagonal, you discover that the total mass of this detection is 1 of this period of this rectangle, so base is 2 epsilonicating is 1 of the two epsilon period is 1.
12:32:30Paolo Guiotto: So, the features of these, functions went back so far, possible.
12:37:550Paolo Guiotto: semantic respect is already.
12:40:510Paolo Guiotto: And… people are equal to value.
12:46:850Paolo Guiotto: And when x goes to zero, you keep in the equivalent 1, but the function is going to be zero at all, except that one single point when it is equal to the same thing.
12:57:80Paolo Guiotto: But this function has a name. It is a well-known object in physics is stored by the VR.
13:03:260Paolo Guiotto: A function that should be zero, where until you get a single point within the correct. That function, of course, simply does not exist with these features, no? But what makes sense, and it's not a unique function with these features, but definitive functions, where when you send epsilon to zero, somehow you get that they concentrate around a single point.
13:27:330Paolo Guiotto: And this is what should make this property, its approximation.
13:32:780Paolo Guiotto: Okay? Now, what we want to do is to give a precise meaning to all of this.
13:39:470Paolo Guiotto: The first thing to do is to recognize that
13:47:150Paolo Guiotto: Here, this integral, Defines, in fact, a new operation.
13:55:940Paolo Guiotto: which is the version, made, from two functions, here, F and that direction, building a new function defined star. This operation is what is called the convolution of F in that direction. So let's start defining this.
14:16:300Paolo Guiotto: So, definition.
14:20:550Paolo Guiotto: Let… But for the moment, we do not put the hypothesis, so let F and G be two functions.
14:28:640Paolo Guiotto: on the real line, well, in general, on RD, but here I will stay on dimension 1, okay?
14:37:810Paolo Guiotto: Real valuable.
14:39:990Paolo Guiotto: We… Fold.
14:43:810Paolo Guiotto: convolution.
14:50:740Paolo Guiotto: Awful.
14:52:390Paolo Guiotto: F. Reads.
14:55:190Paolo Guiotto: G.
14:58:130Paolo Guiotto: the… function.
15:04:30Paolo Guiotto: F star G.
15:06:530Paolo Guiotto: at point X is, by definition, the integral on the real line of f of y
15:14:550Paolo Guiotto: G of X minus Y.
15:18:130Paolo Guiotto: DY… Provided that this integral makes sense.
15:25:10Paolo Guiotto: Provided.
15:29:210Paolo Guiotto: the integral.
15:32:280Paolo Guiotto: Maids.
15:35:20Paolo Guiotto: sense.
15:38:980Paolo Guiotto: Now, when this makes sense, we have a proposition, which is called, jung.
15:49:110Paolo Guiotto: TRM.
15:53:370Paolo Guiotto: That says, in general, you can do the convolution between an L1 function and an LP function.
16:03:200Paolo Guiotto: So, if F is in L1, and G is in LP.
16:12:470Paolo Guiotto: R… with P… Greater, equal than 1, less or equal, plus infinity.
16:21:370Paolo Guiotto: Ben… deconvolution, F star G, is well-defined.
16:29:10Paolo Guiotto: And not only is well-defined, it belongs to the LP space.
16:34:500Paolo Guiotto: LPR.
16:36:640Paolo Guiotto: And moreover, we have a bound for the norm, And…
16:43:670Paolo Guiotto: This is the so-called Jung inequality that says the LP norm of the convolution is less or equal than the algebraic product of the norm. So the one norm for F, and the P norm for G.
17:02:360Paolo Guiotto: This is the young… inequality.
17:11:560Paolo Guiotto: So… Let's see the proof. We do the proof in the case P equals 1, which is not particularly…
17:20:190Paolo Guiotto: difficult. It's actually a straightforward calculation.
17:25:200Paolo Guiotto: 4P different than one.
17:28:640Paolo Guiotto: It is more complicated, and it requires an application, a smart application of the Coxish words inequality for P equal to, and of older inequality for P generic.
17:39:720Paolo Guiotto: So let's stay on P equal 1. There is some exercise on this copy different than 1.
17:46:10Paolo Guiotto: So we just check that the function is in L1 and the inequality, so by doing integral on R of models of F star G at point x.
17:59:770Paolo Guiotto: What I do not check is the measurability, which is a little bit tricky here, okay? So, however, we accept, okay? So now, this is the integral on R, so that's, by definition, would be the L1 norm
18:17:420Paolo Guiotto: of the convolution of F with G, right? So, I'm applying this, I'm showing that inequality with P equal 1.
18:28:750Paolo Guiotto: So it is absolute value. Let's copy the definition. This is the integral in r of f of y g of x minus Y is DY.
18:42:130Paolo Guiotto: And then outside, you have it, the integration lengths.
18:45:740Paolo Guiotto: It's a straightforward argument, because first of all, we carry the modulus inside with the triangular inequality, so when we carry inside the modulus, the integral increases. So, since here it is integrated, it will increase as well. So, less or equal, integral, integral.
19:04:980Paolo Guiotto: modulus F of Y,
19:07:830Paolo Guiotto: times modulus of GX minus Y, I'm using a modulus of the product is equal to the product of the moduluses.
19:17:700Paolo Guiotto: And now we use here the Fubini theorem that says this, you see, is an iterated integral, no? You do first integration in Y, and then integration in X.
19:31:720Paolo Guiotto: Well, we can see this as a double integral, so it's like integral on R2 of the product, f of y.
19:39:470Paolo Guiotto: Modulus F of Y modulus GX minus Y.
19:43:970Paolo Guiotto: So, I'm using, Fubini, in reverse direction, no?
19:49:980Paolo Guiotto: normally you use too busy to say. You have to compute a double integral. How do you compute a double integral? You split into two nested integrations in one spot, huh? This one, you would say, okay, let's start computing first integration in Y, and then we do the integration in X, no? And then the opposite, no, this is important to do this. And now, again, I apply another time.
20:14:10Paolo Guiotto: That's the beauty.
20:15:450Paolo Guiotto: That basically inverts the order of these two connections. I don't need to hit the first an X, and then the bar.
20:23:490Paolo Guiotto: So… This is Fubini, and again, Fubini.
20:32:960Paolo Guiotto: So we have an integration in R, an integration in R, but now we integrate first in X. So the first integration is of modulus FY, the function is always the same.
20:44:640Paolo Guiotto: But now, I start integrating in X, and then I integrate in Y.
20:50:440Paolo Guiotto: What changed here is that now, for this second integration, the factor F of Y is a constant, so you can carry out of this integral, so it is equal integral on R of modules F of Y,
21:06:830Paolo Guiotto: Then, it remains the integral on R of modulus G of X minus y.
21:13:350Paolo Guiotto: in the answer, then you integrate in Y.
21:17:200Paolo Guiotto: Here, at this stage, you do a change of variable.
21:20:760Paolo Guiotto: No? If you call U equal X minus Y, it's just a translation, no? It is X equal U plus Y. So, when you do this translation.
21:34:740Paolo Guiotto: Since X varies between minus infinity plus infinity, U will vary between minus infinity plus infinity, no? Because it is just X minus a constant. So for the inner law function, right?
21:51:270Paolo Guiotto: So, here, if you look at this integral, you have to fix that. You change by law, U equals X minus Y. So, Y is a parameter, the biom is X.
22:02:490Paolo Guiotto: So this means that X is equal to U plus 5, so DX becomes just U.
22:11:750Paolo Guiotto: So this becomes the integral on R of modulus. We call the X minus y U g.
22:20:340Paolo Guiotto: And then we have… this is inside that big integral, so modulus f of y.
22:27:930Paolo Guiotto: DY here integrated on the real line.
22:32:250Paolo Guiotto: But now, as you can see, this innermost integral.
22:36:890Paolo Guiotto: is independent of Y, so it's a constant for Y, and therefore it comes out of this
22:44:450Paolo Guiotto: exterior integral. So what I have at the end is integral on R of modules G of U.
22:53:260Paolo Guiotto: DU times integral on R of modulus F of U, sorry, F of Y dY.
23:01:600Paolo Guiotto: And these are nothing but the L1 norm of F and G. So this is the one norm of G times 1 norm of F.
23:12:530Paolo Guiotto: So this says at once that if these two columns are fine.
23:20:540Paolo Guiotto: Well, this is fine, this is 5. Going back to the beginning.
23:26:770Paolo Guiotto: The initial integral is fine, because that's an inequality. You see that there is a certain point assigned like this one. So this integral here is less or equal step.
23:39:520Paolo Guiotto: That product, these two constants, so we proved them.
23:44:520Paolo Guiotto: So… integral on R of the absolute value of the convolution
23:53:220Paolo Guiotto: DX is less or equal than the product of one norm of F times one norm of G. This says that
24:03:410Paolo Guiotto: First of all, If the F and G are in L1, these two quantities are final.
24:12:490Paolo Guiotto: F and G are in L1. So this implies that also this one is fine, and this means F star G belongs to L1.
24:23:230Paolo Guiotto: And second, that you see there is also the L1 norm of the convolution.
24:31:70Paolo Guiotto: And so you get the Jung inequality, because it says the L1 norm of the convolution is less than the product of the L1 norms of F and G, and that finishes the proof.
24:44:630Paolo Guiotto: Okay, so this is, what's up?
24:51:420Paolo Guiotto: And I went with the battle.
24:55:70Paolo Guiotto: This is, the, the…
25:00:630Paolo Guiotto: Well, the definition of this operation.
25:03:520Paolo Guiotto: For example, let's show with a calculation. Well, actually, as you may understand.
25:11:350Paolo Guiotto: to compute this operation in practice is almost never possible. There will be a remarkable property of the Fourier transform that sometimes makes this calculation a bit easier. So, compute
25:29:390Paolo Guiotto: the convolution of these E minus A models of X convolution with the E minus B models of X, with A and B positive constant.
25:47:480Paolo Guiotto: Now, just to be clear, here we have F of X equal E minus a modulus of X.
25:56:890Paolo Guiotto: Now, this function is made like this. For X positive is E2 minus AX. A is positive, so it is a negative exponential. It is like that.
26:08:890Paolo Guiotto: So the value at 0 is 1. And for X negative, since there is the model is symmetric, it's like this.
26:16:240Paolo Guiotto: So, it is easy to see that this function f is in L1R.
26:22:200Paolo Guiotto: exactly because A is positive, if and only if A is positive. For A equals 0, you would get e to minus 0, so constant equal to 1, definitely not in L1. For A negative, this would be a positive exponential, no, because of that minus.
26:41:730Paolo Guiotto: And similarly, for G, which is the other exponential, e to minus B modulus of X, this is also in L1.
26:51:20Paolo Guiotto: for B, pause.
26:52:860Paolo Guiotto: So we have 2L1 function, the convolution makes sense.
26:57:80Paolo Guiotto: So, F star G… is well.
27:03:320Paolo Guiotto: defined.
27:05:460Paolo Guiotto: Now, we can compute applying the definition, no? The definition says that E minus A… I put this because this is the function, I do the convolution with this other function.
27:18:660Paolo Guiotto: And I evaluate at point X.
27:22:240Paolo Guiotto: Okay? Now, what is this? By definition, this is the integral in the real liner. The definition says F of X minus Y… F of Y
27:33:780Paolo Guiotto: So F is E minus A modus Y.
27:38:280Paolo Guiotto: times G of X minus Y, E minus B modus X minus y.
27:45:150Paolo Guiotto: in the wild. So now I have to compute this integral, which is not impossible. That's a bit, tricky, because we have to do, some calculation. So, we do the calculation, for, for X positive, so let x be positive.
28:05:100Paolo Guiotto: for X negative is similar.
28:08:710Paolo Guiotto: So, it is clear that this is the real line where there is Y integrated from minus infinity to plus infinity. I have at a certain point my X, which is assumed to be positive. So, since… because of that modulus.
28:28:360Paolo Guiotto: You see that modulus of Y, this changes when a Y is positive or negative, okay?
28:34:210Paolo Guiotto: And these models change when X is… when Y is greater than X or less than X.
28:41:100Paolo Guiotto: So, because of this.
28:43:170Paolo Guiotto: This suggests that I should split that integral from minus infinity to 0, then from 0 to X, then from x to plus infinity. So I divide my integral from minus infinity to 0, plus integral from 0 to X, plus integral from X to plus infinity.
29:02:840Paolo Guiotto: Let's see what we have to write here. When I am for Y, the integration is in Y, okay? So this is respect to variable Y. Here, X is the parameter, fixed, and assumed to be positive for this calculation.
29:17:880Paolo Guiotto: Okay, so for, Y negative, the E2 minus A modules of Y. Modulus of Y is minus Y, so this becomes E2AY.
29:29:530Paolo Guiotto: What about that modulus of X minus Y?
29:32:770Paolo Guiotto: since I am less than 0, and in this case, less than x, the modules of X minus Y is X minus Y. So I have E2 minus BX minus y.
29:45:120Paolo Guiotto: For the middle case.
29:47:520Paolo Guiotto: I am in the Y here between 0 and X, so now Y is positive, so the models of Y is Y, the first factor is E2 minus AY, but the second is still this one, because I am still below X, so E2 minus BX minus Y.
30:07:280Paolo Guiotto: Finally, the third… the third integral, Y is now here, from X to plus infinity, so it is positive, and therefore E to minus ay. And now, since it is Y minus X to be positive.
30:23:340Paolo Guiotto: the E2 minus B models X minus Y will be E2 minus B times Y minus X.
30:33:530Paolo Guiotto: Now I have to compute this integral, so be patient. Let's take out here E minus BX. Then I have integrals from minus infinity to zero. E, what is A plus B, right?
30:51:150Paolo Guiotto: A plus B times Y,
30:54:450Paolo Guiotto: DY. That's the first integral. Plus, second, again, I have E minus, here I have E minus BX. Then, in this case, integral 0 2X, e to b minus a times y, EY,
31:13:690Paolo Guiotto: Third integral, the factor is E to BX.
31:19:10Paolo Guiotto: Then I have integral from X to plus infinity.
31:23:280Paolo Guiotto: And here we have E to minus A plus BY,
31:28:120Paolo Guiotto: I hope to have done the algebraic steps properly. Now, this is E minus BX. These are simple, because this comes from E to A plus P.
31:44:390Paolo Guiotto: Y divided by A plus B to be evaluated from minus infinity to 0.
31:51:710Paolo Guiotto: plus E minus BX. Here, I have the evaluation E.
31:57:130Paolo Guiotto: to B minus AY divided B minus A, this times from 0 to X.
32:05:790Paolo Guiotto: And finally, the third one is E2BX.
32:09:650Paolo Guiotto: it is e to minus A plus BY divided minus A plus B.
32:18:80Paolo Guiotto: And now it is from X to Blessing Fiend.
32:22:280Paolo Guiotto: Okay, we have e to minus BX. Do the first evaluation. At 0, when you put Y equals 0 into 0, 1, so it is 1 over equals…
32:34:100Paolo Guiotto: at minus infinity. When Y goes to minus infinity is the mind that A and B are positive, that explanation, A plus BY goes to minus infinity, so explanation plus the field, this is the factor, and so we have just 1 over A plus B.
32:50:970Paolo Guiotto: The second, plus E minus BX. Now, when we evaluate at X, so let's put the factor 1 over B minus A in front. When we evaluate at X, we have E to B minus AX minus, when we evaluate at 0, e to 01.
33:12:440Paolo Guiotto: And third, we have plus E to BX.
33:17:590Paolo Guiotto: Let's take out the factor minus over A plus B,
33:23:590Paolo Guiotto: Now, when we evaluate at the plus.
33:27:960Paolo Guiotto: This one, Y equals infinity equals 50 exponent is negative infinity, so 0.
33:35:540Paolo Guiotto: Minus. Divided by the X, we get E2, another minus.
33:40:920Paolo Guiotto: So 0 minus E minus A plus B.
33:46:660Paolo Guiotto: X.
33:49:630Paolo Guiotto: Okay, so…
33:52:160Paolo Guiotto: Now, we have to simplify as much as possible. We have e to minus BX divided A plus B.
34:00:230Paolo Guiotto: Then we have plus E to minus B plus B, so E2 minus AX divided B minus A minus E minus BX divided B minus A.
34:16:880Paolo Guiotto: The last one is PLUS.
34:19:120Paolo Guiotto: E plus BX minus, so it remains E2 minus AX.
34:24:150Paolo Guiotto: divided A plus B. So you see that you get lots of simplifications.
34:29:550Paolo Guiotto: So, factorizing E minus AX… that multiplies 1 over B minus A, plus 1 over… B plus A.
34:43:680Paolo Guiotto: And then we have plus E to minus BX.
34:48:30Paolo Guiotto: This one is 1 over B plus A minus 1 over B minus A.
34:56:830Paolo Guiotto: So we get… E minus A.
35:01:140Paolo Guiotto: X, that's B squared minus A squared.
35:06:820Paolo Guiotto: What is the common… This B plus A plus B minus A, so it is 2B.
35:14:990Paolo Guiotto: plus E minus BX.
35:18:240Paolo Guiotto: Again, here, B squared minus A squared is B minus A minus B, so it is 2A, yeah.
35:27:490Paolo Guiotto: So… We have 2 over B squared minus A squared.
35:36:760Paolo Guiotto: Of course, I should have that A is different from B, because if A is equal to B, probably I should, I should do a different calculation
35:47:620Paolo Guiotto: Here, no? Okay, but let's say that we have done the calculation for A different from B, this is B, E to minus AX plus A, E2 minus…
36:02:110Paolo Guiotto: Yes.
36:03:590Paolo Guiotto: I hope that the calculation is correct, but… I cannot be 100% sure.
36:09:950Paolo Guiotto: Well, it seems to be correct.
36:13:110Paolo Guiotto: Okay.
36:14:440Paolo Guiotto: Now, perhaps there is a wrong sign…
36:17:660Paolo Guiotto: for the factor, of, of, A…
36:23:880Paolo Guiotto: Okay, however, there might be… I see that in the notes that probably this is a minus, but
36:30:470Paolo Guiotto: Okay, this is the calculation for X positive. What for X negative? Well, you can easily see that you change X with minus X, you get the same value. So, for generic X, you get this with modulus of X for
36:49:310Paolo Guiotto: generic, excellent.
36:55:760Paolo Guiotto: Okay.
36:57:190Paolo Guiotto: I think, we can…
37:01:480Paolo Guiotto: stop here. Let me see if there is something… well, if you want to waste some time.
37:08:30Paolo Guiotto: I do not particularly recommend these exercises, but…
37:12:220Paolo Guiotto: 1, 11 for 1, 2, these are exercises similar to this one, compute the convolution.
37:21:850Paolo Guiotto: Maybe you can do the number 3?
37:29:730Paolo Guiotto: Well, if you want, there is the number 6, which is the proof of the Jung inequality for P equals 2.
37:37:630Paolo Guiotto: It is guided, so you should be able to figure out how to do this.
37:44:640Paolo Guiotto: Okay, let's stop here, thank you.
37:49:810Paolo Guiotto: Oh, am I still… Why this appeal, right?
38:04:640Paolo Guiotto: Unlessfei, I'm recording.
38:06:900Paolo Guiotto: Seems, yes.