AI Assistant
Transcript
00:00:00Paolo Guiotto: Okay, good morning.
00:14:760Paolo Guiotto: That's not rantic about it.
00:21:990Paolo Guiotto: Thank you.
00:25:40Paolo Guiotto: So we are a little bit,
00:28:160Paolo Guiotto: late, so I have, to, to, to…
00:33:00Paolo Guiotto: to cut something from the program. So, what we're going to do today is, we close this path on inbound spaces, and in particular.
00:43:890Paolo Guiotto: we discuss, an important example of orthonormal basis, the orthonormal basis for S space L2, big L2.
00:54:90Paolo Guiotto: And then from that, topic, we will, move to Fourier Transform, as you will see.
01:05:120Paolo Guiotto: That will be the final part of this course.
01:10:530Paolo Guiotto: Just a quick communication, I've been informed that all of you who are enrolled are automatically, you can enroll to the Uniweb list, doesn't matter if you are
01:27:110Paolo Guiotto: officially enrolled in the degree or not, okay? So, please enroll to the partial exam list that will be next week on Thursday, right?
01:42:10Paolo Guiotto: Okay, so let's talk about this, ortho… Normal… basis.
01:55:410Paolo Guiotto: of L to, say, AB.
02:00:600Paolo Guiotto: Now, for…
02:02:300Paolo Guiotto: a little bit of simplification, so I will write everything in general, but then it will do the check in a case A equals 0, just to understand, okay?
02:12:760Paolo Guiotto: So, actually, it is easier if we consider the complex version. So, letter… L2.
02:25:90Paolo Guiotto: C.
02:27:00Paolo Guiotto: AB, B, the space of… measurable functions.
02:33:540Paolo Guiotto: on AB, C-value de… Such that the integral from A to B modulus F squared DX is 5.
02:46:850Paolo Guiotto: Equipped with a slight modification of the usual Scala product, or back to Hermitian product here.
02:55:360Paolo Guiotto: This is to make the vectors, we are going to write a unit vector automatically. Keep the…
03:06:620Paolo Guiotto: with… this product. We take a product of FG,
03:12:600Paolo Guiotto: in this L2 space as 1 over… we divide by the length of the interval, so the difference between this product and the classical product is just a factor.
03:23:300Paolo Guiotto: Integral from A to B of FG conjugate DX.
03:30:450Paolo Guiotto: Now, since this is basically the path for a factor, the old Scala Hermitian product, we have that this one is an intermediate product as well.
03:42:430Paolo Guiotto: Now, it turns out that, there is this, TRM?
03:51:110Paolo Guiotto: D… Family.
03:57:790Paolo Guiotto: EN. Well, here, for convenience, we will take the index n in integers, positive and negative, so in Z.
04:07:630Paolo Guiotto: And the family is defined as EN of X equal exponential of I2 pi divided B minus ANX.
04:21:860Paolo Guiotto: So, X is the variable, is in the interval AB,
04:26:580Paolo Guiotto: N is an integer, positive, negative, in Z.
04:32:460Paolo Guiotto: Now, these functions, by the way, are a combination of sine and cosine. They are precisely cosine of 2 pi over B minus A and X.
04:45:40Paolo Guiotto: Plus, I, sign.
04:48:520Paolo Guiotto: 2 pi over B minus A and X. And these two functions, so the system made of all sides of this, cosines of this, or these functions.
05:04:170Paolo Guiotto: This is the base for the case real badger. I will tell you something later. But it's much better to consider this one, because the computations are easier and better, instead of
05:19:340Paolo Guiotto: Carrying around the sine and cosine. I'll let you notice that these functions
05:25:450Paolo Guiotto: because of this scaling, are just periodic functions of period, the length of V minus A. Because if you take X and you add B minus A, this variable, so you have, these times B minus A, that is 2 pi
05:43:230Paolo Guiotto: Tupayana for justice.
05:49:470Paolo Guiotto: Equal to zero.
05:53:330Paolo Guiotto: Sope.
05:54:400Paolo Guiotto: We have that, we have a dysfunction, The family Ian, And in Z… Ease N.
06:07:230Paolo Guiotto: orthonormal.
06:13:350Paolo Guiotto: basis.
06:16:310Paolo Guiotto: off.
06:18:430Paolo Guiotto: L2C.
06:21:320Paolo Guiotto: Baby.
06:24:50Paolo Guiotto: Of course, there are many autonomal bases, it depends on what kind of application you want to consider. This one is important because the abstract for the S-series
06:37:260Paolo Guiotto: with respect to these functions is the… what is called the classical Fourier series. So, in particular.
06:48:860Paolo Guiotto: for every F, in L2.
06:52:710Paolo Guiotto: C.
06:54:200Paolo Guiotto: AB… the function F, Can be represented.
07:00:860Paolo Guiotto: as an infinite sum, some… well, here n is not a natural, it's an integer, so it will be sum for n going from minus infinity to plus infinity of
07:13:840Paolo Guiotto: F scala EN times EN. If you want… if you wonder what is this sum, well, this is the limit
07:24:60Paolo Guiotto: when big n goes to plus infinity of the sum for N going from minus capital N to plus capital N of the same things. F, E, N scalar EN times EN in that scalar product that we introduced above.
07:45:550Paolo Guiotto: So, this means that every function f can be represented as an infinite linear combination of these functions EN, which are combinations of sine and cosine, so which are combinations of functions periodic.
08:03:120Paolo Guiotto: with period exactly the interval AB. So these functions, these functions are… Periodic.
08:16:980Paolo Guiotto: And if you want to, to say that period is B minus A, so we say B minus A periodic. So also this one is B minus A periodic.
08:32:360Paolo Guiotto: This was the origin, in fact, of all these,
08:36:130Paolo Guiotto: story, because, initially, Fourier discussed just
08:40:880Paolo Guiotto: everything starts from this problem that was studied by Fourier. Fourier tried to represent every periodic function as a superposition, a combination of simple elementary periodic functions as cosine and sine. So the problem Fourier studied was, is it possible to
09:04:500Paolo Guiotto: represent any periodic function of period, say, B minus A on the interval AB, as a combination, suitable combination of sine and cosine.
09:15:480Paolo Guiotto: with the same period. Now, so this means that you want to write F as a sum of sine and cosine with suitable coefficients.
09:27:630Paolo Guiotto: It can be proved now, I don't… I don't have the time to do all this, that this is equivalent to show that the function f can be represented as an infinite combination of these functions EN that you see here. These functions are called characters.
09:53:990Paolo Guiotto: Now, we will not prove that, this is completely this theorem. We will just limit to verify that this system is an orthonormal system.
10:04:100Paolo Guiotto: to show that it is an orthonormal basis, we should also prove that for every function F, you can say that F is the sum of its Fourier series, okay? So, we just limit,
10:17:360Paolo Guiotto: we… Limit… Check, okay?
10:25:80Paolo Guiotto: that the family, Ian, is, or to normal.
10:39:430Paolo Guiotto: system.
10:43:530Paolo Guiotto: Of factos.
10:45:610Paolo Guiotto: The proof, that, The Yari base is complicated, it's not,
10:51:470Paolo Guiotto: trivial, so it's not like the little l2 that you have immediately, the fact that the canonical basis of little l2 is a basis. Here, it's more complicated, because the problem is that you have to show that
11:06:160Paolo Guiotto: each function can be… I mean, so you have a concrete object that is now a complex object, a function, that can be
11:13:770Paolo Guiotto: exactly represented as the sum of what you see at the right-hand side. You may notice that at the right-hand side, I will return this after this little check, you have an explicit expression that can be computed by F. So the problem is, you can write always the right-hand side, given an F. The problem is add the two the same. This is the
11:37:420Paolo Guiotto: key problem. So that… that proof is complex, and we… we are not interested now in this.
11:44:310Paolo Guiotto: Let's check that these are orthonormal vectors. So let's do the scalar product between NEM and NEM in that product. This is 1 over B minus A, the scaling factor, integral from A to B
12:00:960Paolo Guiotto: E2, so EN is, E2, I2 pi divided by B minus A and X. Then the other one is E2, I2 pi B minus AN.
12:18:640Paolo Guiotto: X conjugated.
12:20:980Paolo Guiotto: the X.
12:22:660Paolo Guiotto: Now, you see, the advantage of doing calculations with exponentials is that when you do the conjugate, this comes a minus in the exponent E minus i, 2 pi over B minus AMX, and then when you multiply two exponentials, you get an exponential. Well, if you have 200 sine and cosine.
12:42:280Paolo Guiotto: you have a mixed product, product of two sines, product of two cosines, then you need to use, you know, those formulas like sine of X times sine of y can be expressed by sine of X plus Y, sine of x minus Y, and cosine, and whatever, so it's a little bit more involved, the calculation, but it's basically the same. In this case, we have a direct
13:06:460Paolo Guiotto: calculation, so E to i, there is a factor of 2 pi over B minus A, then you see that the unique difference is the presence here of N and of N with minus, so N
13:21:220Paolo Guiotto: minus M, say, times… X.
13:26:860Paolo Guiotto: the X. Now, this calculation is, in fact, just something like E2 integral from A to B, E to lambda X, you see? So, it's something that can be easily done, because if lambda is different from zero.
13:45:110Paolo Guiotto: we can say that this is the derivative of E lambda X divided by lambda, so we have the evaluation of E lambda X divided lambda between A and B. So…
13:58:110Paolo Guiotto: This will give, let's write here, 1 over lambda times e to lambda B,
14:06:280Paolo Guiotto: minus E to lambda A, where, of course, lambda here is all this box here.
14:14:720Paolo Guiotto: This is the lambda.
14:16:740Paolo Guiotto: Well, lambda is zero.
14:19:150Paolo Guiotto: And this can be 2 pi over B minus A is not 0, i is not 0. It can be only when the factor N minus m is 0, so when n is equal to M. Here, you cannot say that the calculation is the first line, because it does not make sense, but if lambda is 0, you are integrating e to 0. So 1, so this is the integral from A to B,
14:43:990Paolo Guiotto: of E to 0, constant equal to 1.
14:47:840Paolo Guiotto: BX, so it is exactly B minus A.
14:51:350Paolo Guiotto: Now, this is the formula for the integral. Let's continue here. So, we have 1 over B minus A times
15:00:710Paolo Guiotto: that integral. Now, we have two cases, according… that box is different from 0 or equal 0. It is different from 0 for N, different from M, so N different from M, and the other case would be N equal M.
15:19:740Paolo Guiotto: Let's do the second one, which is easier. So we have 1 over B minus A times B minus A times this.
15:28:180Paolo Guiotto: So, 1 over B minus A times B minus A is 1.
15:32:740Paolo Guiotto: So this means that the second line means that when I do the product EN is color EN, in these two products, I get one.
15:41:970Paolo Guiotto: So, in particular, this says that the norm of these vectors is equal to 1. The L2 norm… we still use the L2 even if it is not properly the L2 norm because of the scaling factor, but you understand it.
15:56:260Paolo Guiotto: It's something which is directly proportional to the… to the L2 node.
16:01:570Paolo Guiotto: For N different from N,
16:05:100Paolo Guiotto: The integral yields 1 over lambda, so I can write 1 over lambda is i, 2 pi divided B minus a times n minus m.
16:17:300Paolo Guiotto: And then I have an exponential E to lambda, which is that box, B minus E to lambda A, so E.
16:27:80Paolo Guiotto: I2 pi over B minus A times… well, let's give a name to this number.
16:36:230Paolo Guiotto: It's now just a natural… an integer, k, k. So we will have B minus e to i2 pi over B minus a, say, K,
16:50:870Paolo Guiotto: A.
16:56:390Paolo Guiotto: Okay, now it should be that this quantity is equal to zero.
17:01:780Paolo Guiotto: Let's see if this is the case. So let's take just this parenthesis.
17:08:380Paolo Guiotto: stuck here.
17:10:780Paolo Guiotto: So I can write this as, e to i 2 pi over B minus A.
17:18:609Paolo Guiotto: KB… well, let's write B minus A plus A, let's see what happens, minus E2, this one we leave E2 pi over B minus a KA.
17:32:470Paolo Guiotto: Now, when we split this exponential, it becomes…
17:37:70Paolo Guiotto: E to i, 2 pi divided B minus a times KB minus a.
17:44:150Paolo Guiotto: times the exponential EI2 pi over B minus A K.A.
17:51:200Paolo Guiotto: But this, once you simplify the B minus A here, it is e to i multiple of 2 pi, which is equal to 1.
18:03:890Paolo Guiotto: So this coefficient here is just 1 times that exponential minus itself that you have here.
18:12:300Paolo Guiotto: And so, at the end, we get… Everything is equal to zero.
18:18:310Paolo Guiotto: So, this means that all this calculation yields zero. So, from this line, we see that this color product between EN and DM in the L2 product is zero when N is different from N.
18:33:910Paolo Guiotto: So, they are perpendicular when the index is different.
18:37:790Paolo Guiotto: and they have norm equal 1, so they are unit factors. Of course, I repeat.
18:45:800Paolo Guiotto: This does not prove that it is an orthonormal basis. This just proves that this is a system of perpendicular vectors with length equal 1, which is the precondition for an autonomal basis. Then we should prove that this is an orthonormal basis.
19:02:400Paolo Guiotto: Okay.
19:03:890Paolo Guiotto: I want to return a second on this series.
19:07:700Paolo Guiotto: Let's give a look through the series. D… Abstract.
19:17:730Paolo Guiotto: 48. Serious.
19:19:950Paolo Guiotto: for… this… basis.
19:26:360Paolo Guiotto: is… so, F equals sum
19:30:750Paolo Guiotto: Well, we can just write quickly N in Z of F scala EN times EN,
19:40:340Paolo Guiotto: that we can explicitly write as sum N in Z. Now, this coefficient, the Fourier coefficient, is 1 over B minus A integral from A to B,
19:56:240Paolo Guiotto: F of X times scalar with the conjugate of EN, so it is E2 minus
20:04:160Paolo Guiotto: I2 pi over B minus A
20:08:420Paolo Guiotto: and X, you understand that if the length of the interval AB is 2 pi, this formula simplifies a bit. But we will keep in this general form, because later, when we do the introduction, the passage to the Fourier transform, we somehow need to take this interval going to the infinity, so going to the inferior line.
20:30:860Paolo Guiotto: So this is the Fourier coefficient. You see an X here, but this X is integrated, so it's just a constant, it's not quantity depending on X. So if you prefer, maybe, let's call it the integration variable in a different name, something like Y, F of Y in Y, here.
20:51:200Paolo Guiotto: And this is multiplied, so this is a coefficient which is multiplied by the character, and the character is E2I, it is EN, e to i 2 pi over B minus A and X. So the variable X is here, is only here.
21:09:720Paolo Guiotto: These are coefficients. Normally, the notation used, you will see it's the same notation we used for the Fourier transform, is this one. We call these numbers F hat N.
21:23:260Paolo Guiotto: And this is a sort of discrete version of the Fourier transform. So, our series takes this shape, sum, N instead of F hat n, PI2 pi over B minus a NX.
21:43:760Paolo Guiotto: Now, this is the… what we call classical
21:51:60Paolo Guiotto: Fourier series, because it's in a specific base, in a specific environment, and this is, let's say, the original Fourier series of Fourier.
22:03:440Paolo Guiotto: Now, an important fact to be noticed here is that the sense of this identity
22:10:70Paolo Guiotto: is that the right-hand side is what? We have to go back to the meaning of this series. This is a series of vector in the Hilbert space you are working, so in the L2 space.
22:23:970Paolo Guiotto: So, it means that this series is convergent according to the norm of that space, which is the L2 norm.
22:32:890Paolo Guiotto: So that's why here we specify, it's, let's say, unnecessary, but it is better to understand that this is an identity in L2, in the sense that this series
22:53:70Paolo Guiotto: converges.
22:57:790Paolo Guiotto: according…
23:02:580Paolo Guiotto: to…
23:04:170Paolo Guiotto: L2, let's say, C, this is not important, it's because we are considering complex valued function, but if your function is the L-valued, you can use this. AB,
23:15:100Paolo Guiotto: So, basically, in L2. Why this is important? Because we know that L2 convergence is a little bit strange. For example, it does not imply point-wise convergence.
23:28:80Paolo Guiotto: And, in fact, it has been discovered through a little bit complicated examples that you may have exactly that phenomena that we have seen, no, with convergence in LP. You may remind of an example where we had a sequence of simple functions, that unit functions, basically. They are converging in L1, for example, to 0, but they are never 0.1.
23:52:880Paolo Guiotto: converted.
23:54:10Paolo Guiotto: Exactly this phenomenon happens here. It can be built. An example where this series is never pointwise convergent, but it is still L2 convergent. This may happen, it's not something that we can avoid.
24:10:380Paolo Guiotto: And, of course, since normally you use this kind of decompositions.
24:17:190Paolo Guiotto: For example, this is the basis of, digital, digital, sound processing, no? So you, you, instead of using… you have your signal, which is the function F,
24:31:370Paolo Guiotto: To know F, this says, basically, you need to know these numbers, F at N. These numbers are normally infinitely many numbers, no? For each natural n, either one of them is sufficient.
24:44:530Paolo Guiotto: But as you may expect, since this is a series, probably these numbers are going to be small for and large.
24:52:940Paolo Guiotto: And moreover, since these functions think in terms of sine and cosine, when N is big.
25:02:640Paolo Guiotto: These functions have a high frequency.
25:06:210Paolo Guiotto: So, let's say that the contribution of these terms with N pig is the contribution produced by noise, this kind of phenomenon with very regular material.
25:21:720Paolo Guiotto: So we may think that we could cancel noise, cleanup signal, and simplify by using a finite array of these
25:31:810Paolo Guiotto: F and H. This is the base of signal theory. So, we do approximate this, we define some in the best possible way, no? And the digital signal processing is based on this. So, it's… it's important
25:51:420Paolo Guiotto: to know, how this thing approximates. If you know that this thing approximates, but never point-wise, you may wonder, so what is the approximation of the right-hand side.
26:04:440Paolo Guiotto: Now, the next thing is that if we… we can easily see, we can have a cheap result that shows that if we have some regularity on F, so if F is not only L2, but it is maybe differential, continuous differentiable, it has a certain number of derivatives.
26:25:420Paolo Guiotto: This convergence
26:27:130Paolo Guiotto: can be made pointwise, and even more… even stronger than this. This is… is because of a phenomenon, which is the following.
26:36:930Paolo Guiotto: Let's, do this as a proposition.
26:41:630Paolo Guiotto: Suppose that our F is a C1 function on interval AB,
26:51:900Paolo Guiotto: with, say, periodic functions, so with the values at end points, the same value, with F of A equal F of B.
27:05:120Paolo Guiotto: So, it's a… there are, let's say, two… two strong requirements, irregularity, C1, and the fact that the function must be periodic. For an L2 function, saying that the final value is equal to the initial value has no meaning, because the value at two points
27:23:960Paolo Guiotto: No? Cannot be… cannot decide the function. You know, the function is undetermined apart for a measure zero set of points, so if I say F is in L2 with F of A is equal to F of B, the condition F of A equals F of B does not tell.
27:40:100Paolo Guiotto: anything, in fact, okay? Because if I change the function on a measure zero set of points, in particular these two points, I have the same function for the measure, let's say, but it does not verify this condition. For C1 makes sense.
27:54:990Paolo Guiotto: Now, in this case, we can have this relation. Now, the function f prime is continuous, so it will be in L2, and we have this formula, the Fourier coefficient of the derivative.
28:09:330Paolo Guiotto: is equal to, well, let's suspend, because it should be minus i to pi.
28:17:960Paolo Guiotto: the Fourier coefficient of F.
28:21:530Paolo Guiotto: sorry, n minus psi2 pi n, the Fourier coefficient of F, at integer n for every n in Z.
28:32:750Paolo Guiotto: Well, this is a simple application of integration by parts, because we have that the Fourier coefficient for the derivative is… the formula for the Fourier coefficient is here. Now, what you want is the scarabaddle between F'
28:51:790Paolo Guiotto: and EN. So, it is 1 over B minus A integral from A to B of F prime y, e2 minus i, 2 pi over B minus a, and
29:08:900Paolo Guiotto: X. Probably I forgot the scaling factor, we're gonna say, but let's see what happens. Now, what we do is we integrate by parts, switching the derivative on the other parts. So, by parts…
29:24:880Paolo Guiotto: we get that this is 1 over B minus A. Now, we have the evaluation, F of YE minus i2 pi over B minus A and X between A and B, minus
29:41:350Paolo Guiotto: then derivative moves on the other factor, so I have F of Y,
29:46:210Paolo Guiotto: And then I have… sorry, there is a Y here, sorry.
29:52:330Paolo Guiotto: The derivative with respect to the variable, which is just Y here, it is all the NTF factor
30:01:220Paolo Guiotto: minus i2 pi over B minus AN, so the inferior formula here is wrong, so let's correct in a moment. And then we have the exponential E minus i2 pi
30:18:330Paolo Guiotto: of the B minus A NYDY.
30:22:460Paolo Guiotto: Now, we already said that this path here is periodic. The value at B is equal to the value at A, so you have that when… since this,
30:36:640Paolo Guiotto: Let's say the value at B for the exponential is equal to the value at A, is what we have,
30:43:860Paolo Guiotto: done here. If you see, this was, this first was the exponential at point B. With this little transformation, we have the exponential at point A. So, this function is periodic of period
31:01:220Paolo Guiotto: B minus A, as we said. So the value in B is equal to the value in A, and since for F, we are assuming that also F is B minus A periodic, so this contribution will be equal just to 0.
31:18:330Paolo Guiotto: So we have here…
31:23:640Paolo Guiotto: So…
31:27:00Paolo Guiotto: Okay, so we have minus, minus, plus. Let's take out this de facto, I2 pi divided B minus AN
31:36:660Paolo Guiotto: And then, what you have is one of the B minus a integral from A to B of F of Y, E minus i to pi over b minus a
31:48:280Paolo Guiotto: and YDY, but this is the Fourier coefficient of F. This is the F, hat… N.
31:56:660Paolo Guiotto: So the formula is not the one that doesn't matter because that spelling is irrelevant. So let me correct the formula.
32:07:130Paolo Guiotto: It is, I… 2 pi over B minus A, And this is the formula.
32:16:550Paolo Guiotto: Because you see the factor here.
32:19:120Paolo Guiotto: Now, in particular, this means that…
32:22:730Paolo Guiotto: you can divide, by n, this formula, and it says that for n different from 0,
32:34:410Paolo Guiotto: you have this, that the nth Fourier coefficient is 1 over n, then we have the factor i2 pi divided B,
32:45:530Paolo Guiotto: of an A… No, sorry, the contrary, probably.
32:51:330Paolo Guiotto: Yes, I'm dividing, so… well, there should be 1 over… 1 over I, which is minus i, so… it doesn't matter, this is just a constant, B minus A divided 2 pi.
33:03:770Paolo Guiotto: And then we have the Fourier coefficient for the derivative. We have these four graphs.
33:09:980Paolo Guiotto: Now, this formula
33:12:880Paolo Guiotto: is important because it says that our… what you have to see here is that our FN hat is 1 over n, which is something which is going to 0, times a constant here.
33:28:770Paolo Guiotto: Times this 1.
33:30:820Paolo Guiotto: That we don't know if it goes to zero, but if we do, in general, let's say, for a G, this bound modulus of G, of the n Fourier coefficient of G, which is 1 over B minus A, the integral from A to B, G times, let's write just ENY,
33:52:180Paolo Guiotto: be wise.
33:54:970Paolo Guiotto: So, this is a trivial bound that we can do. We carry inside the absolute value. This is less or equal than 1 over B minus A integral from A to B modulus G of Y.
34:08:770Paolo Guiotto: Then, when I do modulus y NY, the character is a unitary complex number, because it is E to i theta, no? This is something like e to i theta. So the modulus of this will be equal to 1,
34:24:830Paolo Guiotto: So we get that this is just bounded by 1 over B minus A integral from A to B modulus G of Y.
34:37:530Paolo Guiotto: which is a finite quantity, no, here. If you want, we can say this is the one norm of G.
34:46:810Paolo Guiotto: Or if you want, since we are on a finite measure set, you can easily see that this is less than the 2 norm of G.
34:55:659Paolo Guiotto: Which is the reference norm for this space. So, in particular, it is bounded by a constant. You see, this quantity depends on N,
35:05:590Paolo Guiotto: But this is a constant. This is a constant in N.
35:12:640Paolo Guiotto: So this says that… this formula says that our F at N, in fact, goes to 0 at least as 1 over n.
35:24:810Paolo Guiotto: Now, if it goes to 0 as 1 over n, and you go back to the Fourier series here.
35:30:690Paolo Guiotto: It means, like, imagine that moving base here of 1 over N. These are numbers with modules, but 2, 1. That series is not converted, because it's basically after 1 over N times 1 over n, it's not one series.
35:46:420Paolo Guiotto: But if we iterate this formula, so we apply again to… but now to F prime, we can say that
35:56:30Paolo Guiotto: We have a corollary.
36:00:100Paolo Guiotto: Let's say that if you now ask that F is not only C1, but C2, so there are two derivatives.
36:10:890Paolo Guiotto: And both F… and F prime are… B minus A periodic.
36:25:180Paolo Guiotto: So, this means that, in particular, F prime verifies the assumptions of this.
36:32:780Paolo Guiotto: Now, if you think to F', if F is C2, F' is C1,
36:38:60Paolo Guiotto: And F prime is periodic, we are saying. So, for applying to F prime, we have that F second, which is the derivative of F',
36:48:980Paolo Guiotto: at 10, this is, we said, equal to I2 phi over B minus A,
36:57:140Paolo Guiotto: F second is the derivative of F', so using the same formula, you would say that this is the relation between the Fourier coefficient for the second derivative and the Fourier coefficient for the first.
37:11:780Paolo Guiotto: But this one was itself i2 pi over B minus A, the Fourier coefficient of F.
37:19:730Paolo Guiotto: So now we get… sorry, there is an N here, so here there is another N. So at the end, we have II is minus… then we have 2 pi divided B minus A squared.
37:33:500Paolo Guiotto: N squared, F at N. So we have this relation.
37:39:480Paolo Guiotto: from Weecher, for… And different from zero.
37:45:880Paolo Guiotto: We have that, F at N, we can, isolate this.
37:53:400Paolo Guiotto: is what? Is 1 over n squared
37:57:520Paolo Guiotto: Then, let's say that when we divide by this, we write just a constant, so say C over N squared, the Fourier coefficient of F second at N.
38:10:400Paolo Guiotto: that now we can bound exactly as above, so we can say that modules of F at N
38:17:820Paolo Guiotto: is C over N squared modulus of F second hat N,
38:26:110Paolo Guiotto: But as we said, the modulus of a Fourier coefficient is basically the one norm is bounded by the one norm of the function. So we can say less or equal C over N squared, the one norm of F second.
38:44:790Paolo Guiotto: Which is finite, because the function is 2, so the second derivative is continuous, that's a constant.
38:51:540Paolo Guiotto: Now, this is saying that under this further assumption, the N Fourier coefficient now goes to 0 faster than 1 over n squared.
39:02:440Paolo Guiotto: And now, this is sufficient to conclude that we have a strong convergence, because… So… if we… consider.
39:17:470Paolo Guiotto: the… Korea is serious.
39:20:680Paolo Guiotto: some.
39:21:960Paolo Guiotto: over N of F hat N, the character NX,
39:29:290Paolo Guiotto: Now, the point is that, under the assumption we have seen, this series is uniformly convergent, which is much more than point-wise convergent. So, under…
39:46:180Paolo Guiotto: the… assumption that F is C2.
39:51:910Paolo Guiotto: on the interval AB.
39:55:330Paolo Guiotto: with F and F prime, B minus A periodic.
40:03:270Paolo Guiotto: So definitely not for a generic L2 function, no? But at least we can say that the flavor is, if F is regular enough, then this series
40:18:720Paolo Guiotto: The Fourier series.
40:21:150Paolo Guiotto: is uniformly
40:27:900Paolo Guiotto: convergent.
40:29:750Paolo Guiotto: Now, you're reminded, uniformly, convergent refers to the uniform norm, to the infinity norm.
40:36:740Paolo Guiotto: That is… Converges.
40:43:320Paolo Guiotto: It's… converges in… Infinity.
40:49:940Paolo Guiotto: norm.
40:53:810Paolo Guiotto: So, in particular, it will be pointwise convergence, because uniform convergence is stronger.
40:59:780Paolo Guiotto: And you can see this, this follows by… This, follows.
41:10:30Paolo Guiotto: by… via stress.
41:17:470Paolo Guiotto: Thanks.
41:19:690Paolo Guiotto: This is something we have seen last time.
41:24:340Paolo Guiotto: So, let me remind you. We introduced this discussing the problem of convergence of a series of vectors.
41:36:760Paolo Guiotto: And we have seen that if we are in a balanced space, so we are in a space with the norm that makes the space complete, Cauchy sequences are convergent.
41:49:340Paolo Guiotto: If we have
41:51:110Paolo Guiotto: Convergence for the series of norms of vectors, which is a numerical series, then we have convergence for series of vectors.
42:01:40Paolo Guiotto: So, device test consists in checking that the series of norms in the norm you are considering is convergent.
42:09:670Paolo Guiotto: We do this in which setup? So, we take a space V, well, since here.
42:17:960Paolo Guiotto: We are doing sums of objects, the character are combination of sine and cosine, they are continuous functions. So the space will be the space of continuous function on AB. D norm will be the uniform norm on AB.
42:36:390Paolo Guiotto: So, when I apply the uniform norm to this series, to the Fourier series, so to that series, this is a coefficient, and then I have the character, which is the function.
42:51:30Paolo Guiotto: Now, what happens here? This is a coefficient, it's a number, it's a… in this case, it's a complex number, and this is the function.
43:00:250Paolo Guiotto: So, we treat that as a normal, so we carry outside this color, becomes modus F and
43:07:730Paolo Guiotto: At times the infinity norm of the character.
43:12:650Paolo Guiotto: But what is the infinite norm of the character? This is the maximum for X in AB of modulus ENX.
43:25:770Paolo Guiotto: But you remind that the character is a unitary exponential, it's something like EI something, right? But this something is a real number. So when you take the models of this, you get one.
43:39:250Paolo Guiotto: So you are doing the maximum of once, and you get 1. So the norm is just constantly equal to 1.
43:47:460Paolo Guiotto: So this sum reduces to the sum of modulus FN hat.
43:54:350Paolo Guiotto: And since now, we have,
43:56:820Paolo Guiotto: The good bound, the modular FN at, is controlled by… this can be engloved into a constant, divided by N square, so I have that this is controlled by K over N squared.
44:13:130Paolo Guiotto: So, this will be bounded by the harmonic series sine k over n squared, which is convergent.
44:21:640Paolo Guiotto: So, the Weissler's test applies, because we proved that the series of norms is convergent. So, according to the Weissler's test, we can say that the series of vectors
44:35:460Paolo Guiotto: Which is this one.
44:37:190Paolo Guiotto: Yen.
44:38:470Paolo Guiotto: is… is uniformly convergent.
44:46:290Paolo Guiotto: Now, this is a subtle, point.
44:50:250Paolo Guiotto: Yes, but uniformly convergent to what?
44:54:650Paolo Guiotto: Of course, we may expect that it will be… we hope that it will be convergent to F,
45:00:980Paolo Guiotto: Right? Now, the point is,
45:06:960Paolo Guiotto: Let's say that this defines a function G for the moment.
45:11:730Paolo Guiotto: The question is, is G equal to F?
45:19:230Paolo Guiotto: This is yes, because F is the function that is the sum in L2 of the Fourier series.
45:30:160Paolo Guiotto: As we know, this convergence is not a point-wise convergence, okay?
45:36:360Paolo Guiotto: So, let's say that this is a limit of partial sums, Sn, where Sn is the sum for N going from minus capital N to capital N of these functions, FN at EN.
45:53:310Paolo Guiotto: Now, this is a limit in L2, so what we are saying here is that this SN goes in L2 to F, but we cannot say that it goes pointwise, because we know that, in general.
46:08:500Paolo Guiotto: for NLP convergence, we do not have 0. But, at the same time, remind that we just proved that this convergence in L infinity, uniform normal to some G.
46:21:190Paolo Guiotto: Now, we want to prove that they are the same. How can I get the same?
46:25:300Paolo Guiotto: Well, I use a fact that we know from the LP conversion. We don't know that if the sequence's point was convergent, but certainly there exists a subsequence, S and, say, K,
46:38:490Paolo Guiotto: Such that this subsequence is pointwise convergent to F, at least almost everywhere.
46:48:220Paolo Guiotto: But the same subsequence, since the full sequence SN converges to G, will still converge to G in uniform norm, and uniform is stronger than point-wise, so in pointwise.
47:04:310Paolo Guiotto: And this will be every X. So the conclusion is that now I can say that the two functions, G and F, must be the same at least almost everywhere.
47:17:870Paolo Guiotto: And finally, we have the conclusion. Why? Because what we know about this F, F is a good function here, is the C2, so in particular is continuous.
47:28:870Paolo Guiotto: G is continuous, because it's uniform limit of continuous functions.
47:35:270Paolo Guiotto: And when two continuous functions are the same almost everywhere, they must coincide everywhere, no? You take the difference, the difference is zero almost everywhere. There cannot be even a single point where it is different from zero, no?
47:49:260Paolo Guiotto: So these forces G identically equal to F for every X in AB.
47:57:890Paolo Guiotto: So, this argument shows what?
48:02:270Paolo Guiotto: Let's take the principal messages here, because we are going to, to…
48:11:930Paolo Guiotto: find these things, again, in affiliate transform.
48:15:800Paolo Guiotto: So this says that, the family of characters is an orthonormal basis for this L2 space.
48:25:780Paolo Guiotto: We used, just for convenience, we could do every calculation in the real case by using sine and cosine, all sines and cosines together, so it's a family made of two families, let's say, of sine and cosine. In that case, for example, we could use an integer, only n naturals.
48:49:670Paolo Guiotto: Because if you take minus an, what happens? So sine is even, so it doesn't change.
48:54:620Paolo Guiotto: So, if you put n equal minus 3, it's the same of having n equal 3 for the cosine. For this sign, if you take n equal minus 3,
49:04:990Paolo Guiotto: Sine is odd, so you take minus outside, so it's still the same function, it's just multiply by minus, but the linear combination is not a new vector of the basis.
49:15:410Paolo Guiotto: So, for the ELKs, the basis is made by cosine for N from 0 infinity.
49:23:420Paolo Guiotto: For n equal 0, the sine is 0, so you eliminate, because 0 is not a vector for a base. For n equal 0, sine is 1, this is a vector, so the first element of the base is just a constant.
49:37:170Paolo Guiotto: And this is the basis for the yield basin, okay?
49:42:310Paolo Guiotto: But for, let's say, computational simplicity, it is better to consider the complex version, so we take the space L2, we see variet functions.
49:54:510Paolo Guiotto: we give to this space a particular scalar problem, which is the standard L2 product, rescaled. This is, otherwise we should put the factor in front of the exponential. If we use the ordinary scalar problem, we are going to scale that at those exponentials, because they want that normal equal to 1.
50:13:260Paolo Guiotto: So, with this setup, which is the standard setup, however, we have that this is a normal basis. So, in particular, we can take stress.
50:22:50Paolo Guiotto: every function has an abstract Fourier series respect to that base. This series is now a combination of sine and cosine, and it is what is called the
50:33:510Paolo Guiotto: classical Fourier series.
50:36:170Paolo Guiotto: A remark is that this is, in general.
50:41:390Paolo Guiotto: is convergent in N2, so this may give some problem, because, perhaps for a fixed X, this quantity is not convergent.
50:51:460Paolo Guiotto: Now, what we learn here is that if we ask some regularity for F, we gain some behavior for the Fourier coefficient to go faster. In fact, you have seen that.
51:05:720Paolo Guiotto: If the function is C1, so the first derivative continuous and periodic, the behavior of FN is like 1 over N.
51:17:40Paolo Guiotto: If you ask more, so, say, FC2,
51:20:850Paolo Guiotto: you get that the behavior of F and F is like C over N squared.
51:27:920Paolo Guiotto: So, you may expect the higher is the regularity, the faster is the Fourier coefficient in going to zero.
51:34:870Paolo Guiotto: So this makes, now, if it is fast enough, makes that series to be uniformly convergent.
51:43:130Paolo Guiotto: So, it makes, in particular, pointwise convergent. It's much more, so you may expect that there will be a much relaxed condition. This is a bit strong, but in any case, that condition is always of the same types. You require… you need some regularity to have pointwise convergence.
52:03:60Paolo Guiotto: And under this condition, the series is uniformly convergent, so it's also pointwise convergent, and therefore, it makes sense also to use the Fourier series as a pointwise approximation of the function.
52:18:100Paolo Guiotto: Okay. Now, let's,
52:21:860Paolo Guiotto: do a transition to the full return. If we stop here, we do not have time to explore more than this.
52:30:260Paolo Guiotto: So, transition…
52:37:360Paolo Guiotto: to… Fourier transforms for.
52:39:970Paolo Guiotto: Now, basically, this says that you can express an L2
52:47:610Paolo Guiotto: periodic function as a combination of sine and cosine. This is roughly the message.
52:54:870Paolo Guiotto: Now, we apply this, result to an interval that we take symmetrically back to the origin. So, since we have… we're going to, let's call the period the length of the interval capital T, so this will be the interval minus T half, P half.
53:14:850Paolo Guiotto: So, on… L2C… interval minus T half… behafe.
53:25:330Paolo Guiotto: We have…
53:29:160Paolo Guiotto: We can rewrite the fact that a function f is equal, in L2 sense to the sum in Z of the Fourier coefficients F at N times the characters E and X.
53:49:50Paolo Guiotto: Where now…
53:53:810Paolo Guiotto: The Fourier coefficient FNF at N is 1 over… the length of the interval is now capital T.
54:04:650Paolo Guiotto: We integrate from minus T half to T half.
54:08:980Paolo Guiotto: the function f of y minus i 2 pi over TNXY,
54:30:320Paolo Guiotto: Okay, let's keep like that.
54:32:730Paolo Guiotto: DY.
54:36:240Paolo Guiotto: And the character ENX is equal to exponential I, still 2 pi over capital T, and X.
54:50:10Paolo Guiotto: Now, the idea is that now we informally will send this capital T to infinity.
54:57:190Paolo Guiotto: Because what we want to do is to consider a function which is simply L2 from minus infinity to plus infinity, not periodic.
55:06:280Paolo Guiotto: And what we will have is a sort of extension of this formula. Let's see, informally, what happens. So let's plug this. We have that, so F…
55:18:600Paolo Guiotto: is equal to the sum, and in Z, of, these,
55:25:890Paolo Guiotto: coefficients, so 1 over T integral minus T half, T half, f of y, E minus I to… Bye.
55:39:500Paolo Guiotto: over TNY, DY, then there is E to i to pi over TNX.
55:55:100Paolo Guiotto: Now, what happens when we,
55:58:230Paolo Guiotto: when we, let t to infinity.
56:03:880Paolo Guiotto: What?
56:05:170Paolo Guiotto: it cannot be done so easily, because if you send t to infinity, you have that this goes to zero. So, for T going to plus infinity.
56:15:30Paolo Guiotto: This goes to zero. This integral could go reasonably from minus infinity to plus infinity, but of what? Of f of y e to 0, because t goes to infinity.
56:29:490Paolo Guiotto: And then we have another e to 0 here, so you see that you cannot do just t equal plus infinity, and that… and that's it. No?
56:39:630Paolo Guiotto: But here, there is something that we can do to rework this expression in a way that we can read something.
56:50:550Paolo Guiotto: For example, we write this as sum and in Z.
56:56:640Paolo Guiotto: The scaling factor 1 over T will be written at the end of this. So let's start with the integral. So we have integral minus T half T half.
57:07:310Paolo Guiotto: F of Y, Then, the key point is to emphasize this object here.
57:15:60Paolo Guiotto: to pi over TN, that you see also here, to pi over TN.
57:20:990Paolo Guiotto: Because, before we write anything, we can say this, Now…
57:28:660Paolo Guiotto: Let's start from n equals 0. What is 2 pi over TN? 2… it is equal to 0.
57:36:950Paolo Guiotto: So when n is equal to 1, we have 2 pi over t. Imagine that t is going to be big, so this quantity is going to be small, okay? So it's a little number, let's say here, 2 pi over T. Then for n equal 2, we have 2 pi over T times 2,
57:56:760Paolo Guiotto: Then we have a 2 pi over T times 3, and so on. And for N negative, we do the same on the left. So, minus 2 pi over T, minus…
58:08:820Paolo Guiotto: 2 to pi over T, and so on.
58:12:580Paolo Guiotto: So, these points are like a subdivision of the real line.
58:18:890Paolo Guiotto: And the subdivision that when T is big, is going to zero.
58:23:520Paolo Guiotto: And also notice this fact that this scaling factor you have in front, that 1 over T, is not the length of this interval.
58:33:890Paolo Guiotto: But it is more or less the length. In fact, the length of that interval is always the quantity 2 pi over T, no? These intervals are all of lengths.
58:47:450Paolo Guiotto: So, if, for a moment, we call these points of subdivision.
58:52:820Paolo Guiotto: Let me use a sweetable letter that can be… what?
59:02:270Paolo Guiotto: let's call them… so, let's call this C0, C1, C2, C3…
59:11:300Paolo Guiotto: This will be C minus 1, C minus 2, and so on. So, let's introduce CN equal to pi over T.
59:21:890Paolo Guiotto: N with an EZ.
59:24:850Paolo Guiotto: You may notice that the delta between two consecutive of these points, C and plus 1, minus CN…
59:35:190Paolo Guiotto: is exactly 2 pi over t. So we can see this if we put
59:42:950Paolo Guiotto: the missing factor 2 pi here, since this is just a constant factor, I multiply
59:49:550Paolo Guiotto: by 1 over 2 pi outside. I look at this as a delta between two of these points. So let's write all this formula with this CK. What happens? Look at the exponential inside here. We have E2 minus i.
00:07:140Paolo Guiotto: Instead of writing 2B over TN, I write just CN.
00:14:110Paolo Guiotto: Y, so it's in black.
00:17:480Paolo Guiotto: DY. Then I have E2i, another X here.
00:29:710Paolo Guiotto: X, putting a 1 over 2 pi in front of everything. A CKCN plus 1, minus CN.
00:42:450Paolo Guiotto: Now, if, for a moment, we call all this stuff, let's say a function capital F,
00:50:660Paolo Guiotto: Since Y is not seen outside, it is in the integral, right? So you don't see any Y outside, there is only X, no? This is a function of CN and X. So here we have something like this. So 1 over 2 pi
01:08:320Paolo Guiotto: sum of N in Z of this function fcn EXA.
01:17:00Paolo Guiotto: And, we could call this, simply DXN.
01:23:830Paolo Guiotto: This is to say that this looks like what kind of object?
01:33:740Paolo Guiotto: What object does this kind of, shape. Huh?
01:39:290Paolo Guiotto: An integral, exactly, no? So we're imagining that it's like if we are dividing the real line into infinitely many points, between them, there is a space which is going to be small, because T goes to infinity.
01:55:600Paolo Guiotto: And so we can see this as an integral, so let's say that we force this to be 1 over 2 pi.
02:03:140Paolo Guiotto: That will be the integral foxy that varies from minus infinity to plus infinity.
02:08:410Paolo Guiotto: Of the function, what's going on?
02:12:880Paolo Guiotto: But there is… It's not charged.
02:22:810Paolo Guiotto: The opposite of this.
02:28:400Paolo Guiotto: It seems it is. Of, this function, F, stake C now, X, Dixie.
02:38:10Paolo Guiotto: Now, what is the function f?
02:40:150Paolo Guiotto: Let's go back here.
02:41:950Paolo Guiotto: So the function f as function of C, you just change letter X CN into letter XC. You see that there is still integral from minus T half plus t half, but since T is going to infinity, we may expect that this will go to the integral from minus infinity to plus infinity. Now, we have not generouses here, so we may expect that this will be the integral
03:06:890Paolo Guiotto: From minus infinity to plus infinity, of FY… E minus I.
03:16:240Paolo Guiotto: CY.
03:19:550Paolo Guiotto: DY.
03:21:930Paolo Guiotto: Then, you have a second integration, minus V plus infinity.
03:29:540Paolo Guiotto: Yeah, I forgot the exponential, where… where should I put here?
03:34:410Paolo Guiotto: E to this exponential, E to IC.
03:42:420Paolo Guiotto: X. This is the function we call the F.
03:47:140Paolo Guiotto: CXA.
03:49:190Paolo Guiotto: And this is integrated in C.
03:52:710Paolo Guiotto: So, the informal calculation we made is… So, let's say that informally.
04:06:470Paolo Guiotto: We… Proved.
04:10:640Paolo Guiotto: this.
04:11:760Paolo Guiotto: that when you send T to infinity, the space L2 minus TF, TF will become L2 from minus infinity plus infinity. So, if F is a function in L2, the real line.
04:28:400Paolo Guiotto: then this identity holds, because you remind that the starting point is F, So, F…
04:37:60Paolo Guiotto: Now, since this, even if it is not pointwise a convergence case, remind that this is function of the same variable that you see here. These are only coefficients, okay? So DX enters in the character. So I may say that F of X
04:55:750Paolo Guiotto: Is equal 1 over 2 pi, integral on R,
05:01:260Paolo Guiotto: integral on r of f of y
05:05:640Paolo Guiotto: E minus ICY, DY. Now, once you have done this integral, you get the function of C,
05:14:670Paolo Guiotto: and you integrate again versus this factor, which is the character EICX in the variable C, and what you get is F.
05:26:640Paolo Guiotto: Well, I mean, it seems nice, but we have to understand what is the interest in this, and if it is, of course, correct, true.
05:39:280Paolo Guiotto: What is nice is that this quantity here, this round parenthesis, which is, let's say, at least informally, the limit of this thing here.
05:51:950Paolo Guiotto: So this is what we call the Fourier coefficient with d hat. Now is what we call the Fourier transform of F.
06:00:750Paolo Guiotto: As you can see, once you have done the integral, this will be a function of this quantity, X. So this will be a function of, say, the variable x.
06:12:570Paolo Guiotto: And this… the formula takes this shape. f of x is 1 over 2 pi.
06:18:680Paolo Guiotto: the integral on r of half at C e to iCX. Now, if we look at this second formula, it looks like the first one.
06:31:580Paolo Guiotto: the formula for a different form, now of FX. There is a little difference, which is the minus that you have here, and that you don't have here.
06:43:120Paolo Guiotto: But we can create a minus, so we can put a minus here, and give a minus to this X. And so this will be 1 over 2 pi.
06:55:140Paolo Guiotto: the, by definition, the Fourier transform of the Fourier transform of F, evaluated at minus X.
07:04:860Paolo Guiotto: So we get these remarkable formula.
07:09:930Paolo Guiotto: Now, this formula is called the inversion formula.
07:22:60Paolo Guiotto: And the main problem we will see is when this formula holds, so under which assumptions, for example, is it true that for F equals NL2,
07:34:840Paolo Guiotto: this identity is verified. We will discover immediately that this does not seem the case, okay?
07:42:480Paolo Guiotto: So, now, we are ready to formalize the definition of Fourier transform, of the function, of a function f. So, for the moment, I don't put any condition. We will understand immediately what are the natural conditions under which the Fourier transform is well-defined.
08:00:110Paolo Guiotto: So, let F, for a moment, be just a measurable function on the real liner.
08:08:490Paolo Guiotto: we call… Pourier.
08:15:980Paolo Guiotto: Transformer.
08:20:450Paolo Guiotto: of F, D function, It's a function.
08:30:270Paolo Guiotto: differently from the Fourier coefficients, which are a function of a discrete variable, the integers. Yeah, it's a function of real variable. F at X, by definition, equal the integral in the real line of F of Y
08:47:370Paolo Guiotto: E to minus iCYDY.
08:53:580Paolo Guiotto: Or, since, we usually…
08:56:149Paolo Guiotto: Use letter X, let's use letter X here. It doesn't matter what is the letter, of course.
09:05:130Paolo Guiotto: This is the definition of the fluid transform, provided
09:13:240Paolo Guiotto: The… integral.
09:16:510Paolo Guiotto: mates.
09:19:10Paolo Guiotto: sense.
09:23:69Paolo Guiotto: Now…
09:24:250Paolo Guiotto: So this is the object we have into this formula, and this formula was derived by assuming, informally, that F is in L2.
09:34:40Paolo Guiotto: The point is that this quantity makes sense exactly if and only if F is in L1. So this is the first remark.
09:46:859Paolo Guiotto: F at C is, well, defined.
09:53:960Paolo Guiotto: If and only if F is in L1R.
09:59:100Paolo Guiotto: So this is first a big problem, because L1
10:02:870Paolo Guiotto: And L2, in the real line, they are not one included in the other. We remind that when the domain has finite measure, L2 is in L1. L1 is bigger. But when the domain is in finite measure, L2 and L1,
10:19:160Paolo Guiotto: They are not one in the other. I will return in a second on this. This because…
10:31:20Paolo Guiotto: So, F at C.
10:34:510Paolo Guiotto: Makes sense.
10:36:920Paolo Guiotto: Stop.
10:39:310Paolo Guiotto: If and only if the quantity you have inside this integral is integral, right? So, F of, let's say, sharp, the variable is X, C here is a parameter. This is an L1 function in the real line.
10:58:580Paolo Guiotto: Now, since the character, we still call character, this coefficient, this function, is a continuous function in the real line, and we assume that this is a measurable function, so in particular, they are both measurable, the product will be measurable, this thing is measurable, okay?
11:18:130Paolo Guiotto: So, this means that, F…
11:21:910Paolo Guiotto: sharp E minus IT. Sharp is a measurable function.
11:29:550Paolo Guiotto: And, to be integral, to be.
11:35:120Paolo Guiotto: L1, we need the… That's the integral.
11:41:600Paolo Guiotto: of the absolute value, the X, be fined.
11:47:380Paolo Guiotto: But what is the absolute value? Is the absolute value of f of x times the character E minus ICX?
11:56:220Paolo Guiotto: And this is exactly modulus f of x times modulus of the character.
12:03:360Paolo Guiotto: But the character is a unitary exponential, so the modulus of this number is just 1.
12:10:150Paolo Guiotto: So that integral is the same of the integral for modulus FX.
12:17:800Paolo Guiotto: So, as you can see, this is finite.
12:20:340Paolo Guiotto: If and only if F is in L1.
12:26:540Paolo Guiotto: So the Fourier transform with that formula is well-defined only for L1 functions. And now, let's just… we have already said this, but in general.
12:42:670Paolo Guiotto: So, let's say that in general, since here we have a specific space, L1R is not contained in L2R.
12:55:100Paolo Guiotto: And also, L2R is not contained in L1.
13:01:380Paolo Guiotto: So there are functions that are in both, no? Because if you take, for example, any function.
13:09:350Paolo Guiotto: which is zero of a finite measure set, you have that this function is certainly in L2, and therefore, if it is in L2, it is in L1.
13:20:520Paolo Guiotto: But in general, for example, if you take a function which is in L1 but not in L2,
13:28:420Paolo Guiotto: For example, this function here, 1 over root of X.
13:35:530Paolo Guiotto: Well, let's say that we take X positive, so we put an indicator for X positive.
13:42:900Paolo Guiotto: If I take this one…
13:45:540Paolo Guiotto: To be in L1, you need the integral of modulus F, which is positive, so F itself, find it.
13:52:410Paolo Guiotto: So the root of X is integral in 0, but not at infinity, so I have to help at infinity, and so I will put a factor here, 1 plus X, for example.
14:02:720Paolo Guiotto: this. Now, if you have an integral on r of modulus of f, first of all, this reduces to the integral 0 plus infinity of 1 over root of x times 1 plus X.
14:16:760Paolo Guiotto: You can easily verify, and even to compute the… you can compute this integral to see that this is fine.
14:23:910Paolo Guiotto: Because, this function, f of x.
14:27:740Paolo Guiotto: for X going to 0, when x goes to 0, you see that the quantity 1 plus X goes to 1, so it is asymptotic to 1 over root of X, which is integral at 0. While, when you go to plus infinity.
14:45:640Paolo Guiotto: an hour.
14:47:200Paolo Guiotto: It is 1 to be irrelevant, so 1 plus X is like X times another root of X is like X to 3 halves, which is integral at plus infinity. So it is integral at both sides, it is integral. But when you do the integral of modulus F squared.
15:06:460Paolo Guiotto: This is now integral 0 plus infinity. When you square, you get X times 1 plus X.
15:14:520Paolo Guiotto: square, which is good at infinity, but bad at zero, no? This modulus F squared is asymptotic at zero.
15:25:00Paolo Guiotto: To… 1 of X, which is not integral.
15:29:870Paolo Guiotto: at zero.
15:31:940Paolo Guiotto: So this is an example of a function, F, that belongs into L1, but not into L2.
15:40:860Paolo Guiotto: And similarly, you can build an example of a function which is in L2, but not in L1. For example, if you take this same function, so let's take positive numbers, first of all.
15:58:520Paolo Guiotto: Then, 1 over 1 plus X, yeah.
16:05:740Paolo Guiotto: should be L2, but not L1. If you take the integral on R of modulus of F, well, F is positive, so you can
16:15:430Paolo Guiotto: just write F. Then it is 0 and 0 minus infinity 0, so you can reduce to 0 plus infinity. This is the integral of 1 over 1 plus X.
16:25:690Paolo Guiotto: And this is clearly equal to plus infinity. It is the log, no?
16:30:340Paolo Guiotto: But, if you take the integral of modulus F squared, this is the integral 0 plus infinity of 1 over 1 plus X squared.
16:42:260Paolo Guiotto: You see?
16:43:730Paolo Guiotto: And now, this is, this is spotted easily.
16:49:690Paolo Guiotto: So this function f belongs to L2, but not in L1.
16:56:40Paolo Guiotto: So, in general, you can have this sort of figure about these two spaces. We have L1,
17:03:230Paolo Guiotto: which is, like, say, like that. We have L2, which is like that.
17:08:860Paolo Guiotto: So there is something which is in L2, but not in L1, there is something which is in L1, but not in L2. There are functions, it's plenty of functions which are involved.
17:18:470Paolo Guiotto: Okay?
17:19:970Paolo Guiotto: We will see a very important class later of these functions, which are regular functions. We can have that C-infinity functions are dense in both. We mentioned this fact, the regularization, no?
17:34:800Paolo Guiotto: We can approximate in any LP norm any LP function through C-infinity functions.
17:41:40Paolo Guiotto: So, it means that, in fact, it's not like that.
17:44:810Paolo Guiotto: They are different, but it's plenty of common functions everywhere, no?
17:50:450Paolo Guiotto: Okay, so this is just to say that, that definition
17:56:100Paolo Guiotto: the definition of the classical Fourier transform is an L1 definition, so this is what we call the L1 Fourier transform.
18:07:550Paolo Guiotto: L1 Fourier transform.
18:11:310Paolo Guiotto: Okay, so the first thing to do is to do some examples.
18:17:480Paolo Guiotto: We already computed air transformer. It was an import… very important case, so examples.
18:33:320Paolo Guiotto: So, the number one is the case of the Goshen.
18:38:420Paolo Guiotto: Which is also important for probability. So the Gaussian distribution will be returned better in probability parts, is a function f of this type.
18:49:820Paolo Guiotto: E minus X minus M squared divided to sigma squared. No, usually, usually, we consider this, we scaled by factor 2 sigma squared.
19:04:320Paolo Guiotto: This to think as a probability distribution, but if you want just to take the exponential, you have to adjust
19:11:640Paolo Guiotto: the, formula. Now, it turns out that the Fourier transform for this
19:21:900Paolo Guiotto: What's going on?
19:24:60Paolo Guiotto: So the Fourier transform of this… is, E2 minus,
19:32:590Paolo Guiotto: well, minus 1 half sigma square c square, plus I CM.
19:43:60Paolo Guiotto: Should be like that.
19:45:740Paolo Guiotto: I think we computed for M equals 0, More or less, no.
19:53:780Paolo Guiotto: Yeah.
19:55:160Paolo Guiotto: So… If you remind, we did this calculation as an application of differentiation.
20:00:800Paolo Guiotto: And we have done this calculation, if that is, we…
20:06:160Paolo Guiotto: with M equals zero and without the scaling. We have seen that the formula was this one. The transformer, that is root of 2 pi sigma squared times the exponential, no? So let's restart from this.
20:25:850Paolo Guiotto: We… have… Sin… That, huh?
20:34:920Paolo Guiotto: If we do the Fourier transform of E minus D squared.
20:42:420Paolo Guiotto: Divided 2 sigma squared, so we did this.
20:46:620Paolo Guiotto: At point C, this is root of 2 pi sigma squared E minus exactly this quadratic part, 1 half sigma squared Xi squared.
20:59:360Paolo Guiotto: Okay.
21:00:680Paolo Guiotto: Now, if you want to have the rescaling, a first important remark Is the following.
21:10:580Paolo Guiotto: It's, basically evident, but it is better to know. The Fourier transform is a linear… operation.
21:23:740Paolo Guiotto: So if you do the Fourier transform of a linear combination, alpha F plus beta G, this is the linear combination of the Fourier transforms. And that's evident because, basically, what the Fourier transform is, is an integral, no?
21:41:470Paolo Guiotto: Yes, it is an integral of F multiplied by the character, but the linear combination is immediately reflected into the integral.
21:50:500Paolo Guiotto: So this is important because it says that if you have a factor, we can multiply, divide, we can take in and out of the Fourier transformer. So, for example, we could carry this and write in the denominator here.
22:04:400Paolo Guiotto: So we have the, the case of…
22:08:00Paolo Guiotto: of the, let's say, the centered motion with mean equals zero. If you want to have decays we have there, well, that can be easily done directly, or we can do a little more general property, remark number two.
22:29:850Paolo Guiotto: So you see that respect to the centred case, which is down here, and this one, there is a translation in the argument.
22:39:470Paolo Guiotto: So the function you have above, instead of having X squared, has X minus n.
22:45:890Paolo Guiotto: So there is a general formula that says, what happens to the Fourier transformer of a function where I do a translation in argument? Something like this, no?
22:58:740Paolo Guiotto: That can be easily seen, because you just plug into the formula, and this is how normally we prove the most elementary properties. So this is integral of F of X minus M, E minus ICMX, DX.
23:16:700Paolo Guiotto: What will you do? Of course, we will change variable. We call y equal X minus m, so the integral remains an integral in R. F becomes F of Y, E minus IC, instead of having X,
23:32:850Paolo Guiotto: So here I have X equals X minus M plus M, so this is our Y, so Y plus M.
23:44:50Paolo Guiotto: DY.
23:45:700Paolo Guiotto: So now this splits into E minus IXY times E… minus ITM.
23:56:850Paolo Guiotto: So this is a constant factor for the integration variable, which is Y here.
24:03:450Paolo Guiotto: So we can carry outside. We have E minus ICM.
24:09:440Paolo Guiotto: And then it remains the integral of FYE minus iCYDY.
24:17:820Paolo Guiotto: And this is the Fourier transform of F at point C.
24:24:350Paolo Guiotto: So we get that this is equal to…
24:27:910Paolo Guiotto: E minus ICM times F at C.
24:35:370Paolo Guiotto: So this is the formula, this is the so-called formula of the Fourier transform of the translated function. It is not the translation of the Fourier transform, but you see, translation becomes multiplication by a character.
24:52:950Paolo Guiotto: Okay?
24:55:110Paolo Guiotto: So, going back to our example of the Gaussian, of the… of the Gaussian with mean M, we have that, the hat of minus, sharp minus M.
25:10:300Paolo Guiotto: Square over 2 sigma square root of… 2 pi sigma square.
25:17:280Paolo Guiotto: at point C.
25:19:690Paolo Guiotto: So you see that here we have a translation in the argument.
25:24:340Paolo Guiotto: And therefore, this, becomes a factor, minus HCM, and then we have the transform of the, minus, sharp squared divided 2 sigma squared over root of 2 pi.
25:40:440Paolo Guiotto: sigma square C.
25:43:490Paolo Guiotto: This is what we computed above.
25:46:120Paolo Guiotto: So it is E minus iCM, E minus 1 half sigma square C squared.
25:55:400Paolo Guiotto: So, probably, I did something wrong. It's quite normal, it's impossible to memorize these things, so this is not a plus, but a mind.
26:07:900Paolo Guiotto: Okay.
26:11:810Paolo Guiotto: Good.
26:14:490Paolo Guiotto: So, let's see… Couple of other…
26:19:610Paolo Guiotto: Example. So this is, this so-called exponential.
26:31:610Paolo Guiotto: Now, what we mean with exponential is not the function e tox, because that function is not in L1.
26:39:220Paolo Guiotto: So, we take this function, FAX equal
26:47:690Paolo Guiotto: E minus a absolute value of X.
26:52:300Paolo Guiotto: Which is a function.
26:56:00Paolo Guiotto: is an even function. A is supposed to be positive to have the integrability. So this is a function that goes to 0 at plus minus infinity, and it is equal to 1 at 0. So this is the shape of this function.
27:14:890Paolo Guiotto: We can easily compute the Fourier transform for this function. Well, let's first of all remark that clearly.
27:25:950Paolo Guiotto: This FA belongs to L1, the rear line.
27:32:980Paolo Guiotto: No? I think there are no questions on this, no?
27:37:120Paolo Guiotto: Because you have to integrate E minus ax at plus infinity and DAX at minus infinity. They are both easily integral, we can even compute the value of the
27:46:950Paolo Guiotto: And, what about the Fourier transformer, FA hat C?
27:53:390Paolo Guiotto: This is, by definition, integral on the real line of that function, FA, so E minus A modulus of X, E minus ICXDX.
28:09:700Paolo Guiotto: Okay, so now, to compute this integral, we split the integration from minus infinity to 0, plus that one from 0 to plus infinity. When we go from minus infinity to zero, the modulus of X is minus X.
28:25:220Paolo Guiotto: So we have an E to AX here, E minus… well, let's put together, so minus ICX.
28:34:400Paolo Guiotto: DX, plus integral from 0 to plus infinity. This time, modulus of X is X.
28:41:420Paolo Guiotto: So we have E2 minus AX minus IXDX.
28:49:830Paolo Guiotto: Okay? So…
28:53:50Paolo Guiotto: what have we to do? You see that, basically, we have E2 something, A minus IC, that's a constant for the integration variable, which is X, times X DX. And similarly here.
29:08:590Paolo Guiotto: E2 minus A plus ICX EX.
29:15:740Paolo Guiotto: So in both cases, we have to do something like integration of e to lambda X.
29:22:610Paolo Guiotto: Where lambda… Is this number here.
29:26:580Paolo Guiotto: Which is non-zero. Why?
29:28:830Paolo Guiotto: Because to be zero, it's a complex number. Remind that A in this example is a positive parameter, real in particular. C is the variable of the Fourier transformer.
29:48:150Paolo Guiotto: I never said, but Xi is real. It's supposed to be real.
29:55:740Paolo Guiotto: I implicitly used always that factor, because I said, for example, that that was a unitary exponential, but see, the variable for the fleet transform is real.
30:06:430Paolo Guiotto: Okay.
30:07:620Paolo Guiotto: So, to be zero, that number must be both A and C equal to 0. C can be 0, but A cannot be 0 because A is supposed to be positive. And the same for the other part, so we can directly compute the integral, which is E to,
30:26:410Paolo Guiotto: A minus side C, X divided.
30:30:250Paolo Guiotto: my A minus IXC.
30:33:250Paolo Guiotto: To be evaluated from minus infinity to zero, plus
30:38:370Paolo Guiotto: E2 minus A plus ITX divided minus A glass, I, C.
30:48:110Paolo Guiotto: To be evaluated from 0 to plus infinity.
30:51:770Paolo Guiotto: Now, what happens to this evaluation? Let's give a look to the contribution at infinity.
30:57:340Paolo Guiotto: For example, the first one is E to A minus IC times X, when X goes to minus infinity.
31:07:840Paolo Guiotto: You see what happens? Let's take the absolute value, I will ensure that it goes to zero, this.
31:13:600Paolo Guiotto: So, when I do the absolute value, I have the absolute value of E to AX, that's a real stuff, okay? Then I have E minus ICX, and this is a unitary complex number. So, when you split the absolute value like that, this is 1,
31:31:290Paolo Guiotto: and this remains e to ax. When x goes to negative infinity, since A is positive, the argument of the exponential goes to minus infinity, the exponential goes to 0.
31:44:450Paolo Guiotto: And the same E2 minus A plus x, this one goes to 0 in the same way. So what remains is just the contribution at zero. That, be careful with signs.
31:59:120Paolo Guiotto: And in the first case, I take with sine plus, because it is valued at 0 minus value at minus infinity. When x is 0, the exponential is 1, so I get 1 over A minus IX.
32:12:460Paolo Guiotto: Then here, I have the value at plus infinity minus the value at 0, which is, again, 1 for the exponential over minus that, so with the minus becomes plus.
32:23:490Paolo Guiotto: So, plus 1 over A plus IC,
32:28:610Paolo Guiotto: And if we do the common denominator.
32:32:10Paolo Guiotto: this is A minus XC times A plus XC, this is Z times Z conjugated, so it is models of Z squared. So it is A squared plus C squared.
32:44:230Paolo Guiotto: And when you do the common denominator, dividing, you get, above here, the sum of these two terms, so A plus IX plus A minus IXC, so the two imaginary parts disappear.
33:01:20Paolo Guiotto: And that's what we finally get. The Fourier transform… well, let's write it in this way. The Fourier transform for E minus A absolute value
33:11:520Paolo Guiotto: Use the dot or the shot, what you like.
33:14:260Paolo Guiotto: Evaluated at C is equal to 2A over A squared plus C squared for C, real.
33:24:610Paolo Guiotto: And that's the formula for the Fourier transform of this particular function.
33:32:60Paolo Guiotto: Just a remark, we finish, with this tool.
33:38:400Paolo Guiotto: is that, you notice something here, maybe, well, in the first case, we can get, complex values when N is different functions.
33:51:850Paolo Guiotto: So, if n is 0, there is no term here, and the function is a real function. So, even if the integral is a complex value, it becomes a real function.
34:04:780Paolo Guiotto: And the same that happens here.
34:07:820Paolo Guiotto: And this is because these two functions, in particular case for Gaussian, when M is 0,
34:14:440Paolo Guiotto: the Gaussian, this… the general Gaussian would be centered at M, no? So if M is here, you would have a function like that.
34:26:700Paolo Guiotto: symmetric with respect to M. When M is 0, that function becomes even, and that's the property that makes a real-valued de Fourier transform. This is an exercise
34:38:500Paolo Guiotto: So, maybe you can try to do by tomorrow. So, do.
34:45:450Paolo Guiotto: We have still to do some calculation. We will restart tomorrow, but do… Exercises, sir.
34:56:190Paolo Guiotto: 72… 1…
35:04:180Paolo Guiotto: And 4… Before is that on this property?
35:09:500Paolo Guiotto: Okay.