AI Assistant
Transcript
00:05:820Paolo Guiotto: So yesterday, we… Introduce the definition of former
00:14:870Paolo Guiotto: We computed the Fourier transform for a couple of examples.
00:21:910Paolo Guiotto: So let's do another important example, which is the Fourier
00:26:660Paolo Guiotto: Transform of what is called the rectangle.
00:33:270Paolo Guiotto: Rect A, say. Now, Rect A…
00:38:850Paolo Guiotto: It's a function, it's just an indicator of the interval well, here it depends.
00:46:650Paolo Guiotto: what we want for the scaling. So let's say it should be AF, AF. This is just to have the formula… the formula with A, otherwise we have with 2A.
00:58:180Paolo Guiotto: So it's a function which is a…
01:01:560Paolo Guiotto: equal to 1 on the interval minus A, half A half. A is supposed to be a positive number.
01:11:130Paolo Guiotto: NCO.
01:12:650Paolo Guiotto: I swear.
01:15:580Paolo Guiotto: Of course, it is an L1 function.
01:19:60Paolo Guiotto: And, let's compute the Fourier transform of, this, rector, A.
01:26:560Paolo Guiotto: This is, by definition, the integral
01:30:850Paolo Guiotto: on the real liner of the function, so indicator of interval minus a half.
01:38:70Paolo Guiotto: A half foot?
01:40:420Paolo Guiotto: Except… times the character E minus iCX dx.
01:48:440Paolo Guiotto: So, since the indicator is 0 out of the interval minus A half AF, the integral reduces to the integral on the interval minus A half AF of the exponential.
02:01:830Paolo Guiotto: E minus ICXA.
02:06:380Paolo Guiotto: Yeah, so… So here, basically, we have something E to lambda X.
02:13:910Paolo Guiotto: lambda can be 0 if X is 0, so we have these two cases. If X is equal to 0, we are integrating into 0, so 1 integral from minus A half to a half of 1X,
02:31:110Paolo Guiotto: And this is the length of the interval, which is just A.
02:34:660Paolo Guiotto: If C is different from 0, we can say that this is E minus ICX divided minus iC.
02:45:280Paolo Guiotto: to be evaluated between minus a half.
02:49:660Paolo Guiotto: 2 plus a half.
02:53:420Paolo Guiotto: Now, doing this calculation, we have 1 over minus iX. We can put the minus i at numerator, it becomes I.
03:06:320Paolo Guiotto: So… Down here, Xi, E2 minus I.
03:12:940Paolo Guiotto: Well, let's see, CA half minus E2 minus, or plus… I see a puffer.
03:25:440Paolo Guiotto: Now, perhaps it is better to keep the eye down here with the minus, because here we recognize that if we change sign.
03:35:330Paolo Guiotto: E2 plus IXCA half minus E2 minus IXCA half divided So let's put Dixie here.
03:46:990Paolo Guiotto: I, 2I, let's put the 2 here.
03:51:740Paolo Guiotto: That's, what is this?
03:55:270Paolo Guiotto: sine, right? So, sine. So, 2 over C…
04:06:560Paolo Guiotto: Sign… of,
04:11:50Paolo Guiotto: A C over 2. So perhaps it would have been better to use A over 2 as parameter.
04:19:950Paolo Guiotto: So we get that, maybe it's better if we say D…
04:25:520Paolo Guiotto: However, the Fourier transform of the indicator minus AA is equal to… If we use this scale…
04:35:200Paolo Guiotto: So, I'm replacing AF with A, so it's like if I multiply by 2A. So, it is 2A for X equals 0.
04:48:50Paolo Guiotto: And the 2 over XC sine HC.
04:54:180Paolo Guiotto: Oxy different from zero.
04:57:870Paolo Guiotto: Now, we can,
04:59:900Paolo Guiotto: Let me write this in this form. Let's give X here, and put an A also here.
05:06:770Paolo Guiotto: So, two-way… And this is, so 2AE…
05:13:640Paolo Guiotto: 2A, both sides, here's time sine C.
05:18:100Paolo Guiotto: HC divided 8C.
05:22:990Paolo Guiotto: This box C equals 0, box C different from 0. Now, you see that there is this function, sine t over t, that when T is 0, is not defined, but since the limit is 1, we can accept that conventionally, we give value 1, and we get a unit formula, 2A sine…
05:41:850Paolo Guiotto: A C divided 8C.
05:45:820Paolo Guiotto: with the agreement that sine 0 over 0 is by definition equal to 1, which is the reasonable value because of the fundamental limit. And this is the Fourier transform for this indicator of the interval minus 8A.
06:05:180Paolo Guiotto: Okay.
06:08:490Paolo Guiotto: So, of course, you may understand that there are not so many returns that can be completed explicitly.
06:15:630Paolo Guiotto: But there is a certain class of transforms that can be computed by using several methods. For example,
06:24:410Paolo Guiotto: There are methods coming from integration, with a complex variable that enlarge the class of the transform we can compute, but we do not, here, we do not use these methods for this,
06:41:960Paolo Guiotto: For our purposes, let's say.
06:45:430Paolo Guiotto: Okay, so now, yesterday, the Fourier transform arose from this,
06:53:680Paolo Guiotto: identity that we call the inversion formula. It says that the function f is a part of this factor 1 over 2 pi.
07:03:310Paolo Guiotto: the double Fourier transform of F with the variable, changed from X to minus X.
07:12:810Paolo Guiotto: Now, the key point is to understand when this formula is valid.
07:21:120Paolo Guiotto: Now, this formula, if you look at the formula, so you have the right-hand side, of course, there is the more, the most of the interest, because you have this double hat, no?
07:34:370Paolo Guiotto: why this is interesting, because if you have that your function is in L1, you can compute the first head, the first Boole transformer.
07:43:980Paolo Guiotto: But then, to compute the second hat, you would need to know if the function f hat is still in L1.
07:51:820Paolo Guiotto: So now we are going to discuss a bit this question. What are the properties of the Fourier transform as a function of its variable C? And in particular, when can we say that this is an L1 function in such a way that this formula makes sense?
08:10:230Paolo Guiotto: So, we are going to start talking about properties of Fourier transform.
08:28:670Paolo Guiotto: A first fact that,
08:32:919Paolo Guiotto: there are some immediate properties, are the following. Let's say that we have a basic bound that says, let's say, proposition.
08:43:730Paolo Guiotto: the Fourier transform, of F is, always.
08:52:710Paolo Guiotto: continuos…
08:59:90Paolo Guiotto: function.
09:08:120Paolo Guiotto: bounded.
09:09:890Paolo Guiotto: founded…
09:11:940Paolo Guiotto: continuous function, and this bounds also, modulus F, F of C, is always bounded by the L1 norm of F for every C in
09:28:10Paolo Guiotto: So, this is saying that the Fourier transform is a bounded continuous function, so…
09:36:50Paolo Guiotto: it doesn't look to be an L1 function, necessarily. First of all, because it is continuous, and we know that L1 functions are normally irregular.
09:45:670Paolo Guiotto: And the second one, bounded, yes, bounded, but we will see that, in general, the Fourier transform of a function is not L1, and the example is just the example we have above here.
09:59:10Paolo Guiotto: We will, stress this in a second. So, let's see the little check.
10:06:250Paolo Guiotto: This is, so let's say the boundedness…
10:14:230Paolo Guiotto: is easy, because it's just a straight bond. We take F at Xi.
10:19:510Paolo Guiotto: absolute value, this is the absolute value of the integral of FX E minus ITCX, EXA,
10:29:420Paolo Guiotto: Now, you carry, as usual, the modulus inside, so this is less frequent than integral modulus of F, and then when you have the modulus of the exponential, this is a unitary number, so you get 1.
10:44:780Paolo Guiotto: And this is the one norm of F, and this is this bound that you have here.
10:52:370Paolo Guiotto: Continuity?
10:59:470Paolo Guiotto: This follows from the continuity of integral depending on parameters, because, in fact, if you look at the F at C, this is an integral of FXE minus iCX, DX,
11:17:380Paolo Guiotto: Under the integral, you have a certain function, let's say G of XC.
11:24:430Paolo Guiotto: Where C is the variable of the outside function f at.
11:29:570Paolo Guiotto: So, we apply…
11:36:820Paolo Guiotto: B… Continuity.
11:40:860Paolo Guiotto: off.
11:42:180Paolo Guiotto: integrals, Depending on… parameters.
11:54:510Paolo Guiotto: We mostly applied the differentiability, but the philosophy is similar. What is the philosophy? If the function that you have in the integral, so in this case the G, is a continuous function of parameter.
12:08:910Paolo Guiotto: almost every X integral variable, so this is GX, let's say, sharper.
12:16:970Paolo Guiotto: This is a continuous function on R.
12:22:90Paolo Guiotto: almost every X, in R.
12:27:230Paolo Guiotto: In this case, the domain of parameter and the domain of integration coincide, but it's a coincidence. Yes, because this is F of X times exponential E minus iCX, and you see that dependence on C
12:42:340Paolo Guiotto: is here, no? It's just in this part. So that F of X is just a factor, no? So as a function of C, this is a continuous, because it's a combination of sine and cosine, no?
12:55:390Paolo Guiotto: And the second condition, you need a uniform controller
13:01:160Paolo Guiotto: in the parameter through an integral function of X. But when you take the absolute value of this G, what you get is exactly the absolute value of F.
13:13:330Paolo Guiotto: And this function is L1, because F is supposed to be L1.
13:19:420Paolo Guiotto: No? This also for every value of parameter X in the set of parameters, and almost every X in for the integration variable.
13:29:920Paolo Guiotto: And therefore, the theorem applies, and we concluded that the act is continuous on the real line.
13:41:390Paolo Guiotto: There is a more sophisticated result that we won't prove in such a generality, which is called the Riemann…
13:52:690Paolo Guiotto: fled back.
13:57:130Paolo Guiotto: Lemma.
13:59:470Paolo Guiotto: That says that, the Fourier transformer is not only bounded, but it goes to zero at infinity.
14:07:870Paolo Guiotto: So, if F is in L1, R.
14:12:360Paolo Guiotto: than that.
14:13:810Paolo Guiotto: the Fourier transform goes to zero.
14:18:50Paolo Guiotto: when C goes to plus-minus infinity.
14:22:720Paolo Guiotto: This is a little bit, tricky,
14:27:930Paolo Guiotto: It's not essential for our purposes, but it's something that it is useful to know, because sometimes,
14:34:410Paolo Guiotto: it is… it is helpful to recognize that something cannot be a Fourier transform. If it doesn't go to zero at infinity, it cannot be a Fourier transform. Notice that this is the case in all the examples, because if you look at the examples we have done here, this is the
14:52:120Paolo Guiotto: in the rectangle, look at as a function of T. So, if you look at this as a function of C, it's basically a sine T over T function.
15:03:190Paolo Guiotto: So, at X equals 0, the value is 2A, and then we have a sort of obscillations like that. They are dumped obscillations, so at infinity, you go to 0. If you look at the example of the exponential function, look at this. When X goes to infinity, you decay to 0.
15:22:990Paolo Guiotto: You see? And the same in the case of the Gaussian, even
15:28:410Paolo Guiotto: Here, the Gaussian… the Fourier transform of the Gaussian is still something like a Gaussian distribution, so in particular, it's quadratic, is exponential of a negative quadratic polynomial, so it goes to zero.
15:45:710Paolo Guiotto: Even, very fast.
15:48:80Paolo Guiotto: So, it's a general feature of this.
15:51:560Paolo Guiotto: Now, remarks, warning.
15:57:330Paolo Guiotto: In general, In general.
16:05:440Paolo Guiotto: the Fourier transform of F is not in L1.
16:12:80Paolo Guiotto: So, the Riemann-Lebach says it goes to zero, but going to zero at infinity is not a sufficient condition to be integral, no? 1 over X goes to 0 at infinity, but it is not integral at infinity.
16:24:610Paolo Guiotto: So be careful on that. In general, the free transform of F is not in L1, and the example is just what we have seen above. So let's take with A equal 1, for example.
16:40:980Paolo Guiotto: So if we do the Fourier transform of this, this is the rectangle with the A equals 1,
16:46:390Paolo Guiotto: So, from this formula, we have a 2 sine C over XC.
16:51:620Paolo Guiotto: to sine C over X, which is a known example of a function which is not in L1 in the real line.
17:03:140Paolo Guiotto: Because we have the trouble at infinity here.
17:09:300Paolo Guiotto: Okay, so what can be said is that weird transform goes to zero, but clearly, we may roughly say, not fast enough to be integral.
17:20:930Paolo Guiotto: So now the question is, how can I do to make the Fourier transform going fast?
17:26:390Paolo Guiotto: fast enough at zero, at infinity, in such a way that we have an integral function. And this is where we have the same property of the Fourier coefficient.
17:41:560Paolo Guiotto: You're reminded that yesterday we have seen that this is connected somehow with the regularity of the function. The more is the regularity, the faster is the decay at infinity.
17:56:50Paolo Guiotto: So, now we are going to see a very similar property to this one.
18:01:450Paolo Guiotto: that will involve the Fourier transform of the derivative, okay? So this is the next important factor, from which we will have, as a consequence, the Rieman-Lebeck lemma with the stronger assumption, basically. So, proposition.
18:21:240Paolo Guiotto: Well, let's say that the big team is, Pourier Transformer, and derivative.
18:29:200Paolo Guiotto: Because this is a very important,
18:32:660Paolo Guiotto: is a very important, connection between these two operations. So we have this proposition.
18:42:140Paolo Guiotto: So let's assume that… F being Aluana.
18:48:770Paolo Guiotto: B.
18:49:850Paolo Guiotto: Now, here we have to be a little bit careful when we write. So, I could write, say, a strong version of this theorem, which is, let's assume that F is a very regular function, so C1.
19:06:470Paolo Guiotto: with a continuous derivative, and such that the function and derivative are both integral. But in fact, what we need is that the function and its derivative are in L1. Now, the point is, what exactly means this?
19:22:260Paolo Guiotto: What do we mean with the derivative in L1, 4 and L1 function?
19:27:930Paolo Guiotto: So now we mean that there are different definitions. One is that the derivative, the traditional derivative, is well-defined at almost every point. So you could compute the derivative at almost every point, and what you get is a function, the derivative, which is in L1.
19:46:90Paolo Guiotto: Or here, what we need is just to know that the,
19:51:930Paolo Guiotto: the function f, can be written as… so the fundamental theorem of integral calculus falls, so F of X is equal to F of 0 plus the integral from 0 to X of f prime.
20:07:60Paolo Guiotto: of YDY. So, in fact, saying that F has a derivative in L1 means that there exists a function, the derivative, for which this identity holds, okay? Because this is basically what we need to write.
20:26:760Paolo Guiotto: But if you want to, to have for your convenience, that function f is in, in, L1, so in C1, it's perfectly fine.
20:41:880Paolo Guiotto: Now, if this happens, Dan, no?
20:46:910Paolo Guiotto: Of course, since the derivative is in L1, there is a Tuier transform for the derivative, and the formula that holds is exactly
20:55:340Paolo Guiotto: more or less the same of the Fourier coefficients. So here, it is simpler, because we have noted 2 pi scaling factor and the B minus A, so it's iC, the Fourier transform of F.
21:10:110Paolo Guiotto: whatever you see?
21:12:20Paolo Guiotto: Ria.
21:13:820Paolo Guiotto: So let's see first this factor, then we derive some consequences.
21:20:290Paolo Guiotto: So proof.
21:24:860Paolo Guiotto: Now, what we want to do is to compute the derivative of the Fourier transform, so we start from this…
21:33:60Paolo Guiotto: And this is the integral on r of f prime x e minus iCX DX.
21:44:310Paolo Guiotto: Okay, now, since you have a derivative, the natural thing to do is to integrate by parts, so by parts…
21:54:840Paolo Guiotto: We have that this is, and this is a little bit delicate, the evaluation f of x times C minus iCXM.
22:04:660Paolo Guiotto: from minus infinity to plus infinity, minus integrals on the real line. Now, derivative moves on the exponential, so we have f of x.
22:15:890Paolo Guiotto: Derivative is referring to the integration variable here, no? So it's respect to X, so this is the derivative with respect to X of F. So when you differentiate the exponential, you get the exponential, the minus x CX,
22:31:660Paolo Guiotto: times the derivative of the exponent with respect to X, and this is minus IC.
22:38:610Paolo Guiotto: DX.
22:40:720Paolo Guiotto: Now, it's like to delegate passages to understand what happens to this evaluation.
22:48:890Paolo Guiotto: Because when X goes to infinity plus infinity, for example, you have that this is just sine XCX minus… cosine minus sino, so this is cos minus CX plus CX is the same minus i sine
23:06:70Paolo Guiotto: CX. So, it is clear that when X goes to plus infinity, this thing, has not… the limit.
23:18:100Paolo Guiotto: Okay.
23:19:580Paolo Guiotto: So, it won't be helpful to compute anything. So, the unique hope is about this quantity.
23:28:140Paolo Guiotto: So now, it comes out that in general, if you have an L1 function, be very careful. So let's do a remark. We have never said this, but it's important, because there are certain beliefs that are wrong.
23:46:990Paolo Guiotto: Okay? In general.
23:53:100Paolo Guiotto: if f of x
23:56:20Paolo Guiotto: is an L1 function in the real line, and it is not necessarily true that there exists the limit, and the limit is equal to zero.
24:05:310Paolo Guiotto: Okay?
24:06:410Paolo Guiotto: In general, it is… nuts.
24:12:680Paolo Guiotto: granted that… That, huh?
24:18:380Paolo Guiotto: there exists a limit when x goes to plus or minus infinity of f of x.
24:26:450Paolo Guiotto: So, let's start with an example. Well, you can just do, with a function which is zero everywhere, except points where is non-zero, and maybe be given
24:41:60Paolo Guiotto: but in such a way that they are not relevant for the integrability. So what I'm thinking is like a function… is a function like this. Now, imagine that this is 0. I take a function which is 0, then at 1, I give a positive value, but for a small interval, in such a way that the integral is still found, okay? Then I… maybe I give a value which is bigger, and bigger, and bigger.
25:05:770Paolo Guiotto: bigger, but on smaller and smaller regions in such a way that I keep the function
25:12:110Paolo Guiotto: integrable, hmm? So, how can I do that? Well.
25:17:160Paolo Guiotto: I can say that this could be an indicator from, say, in general, from N to N, for example.
25:27:830Paolo Guiotto: plus something that we fix in a moment. Let's say that the value of the function is N here. So this tower is tall n. I want that area
25:38:930Paolo Guiotto: be not only finite, that will be true, because it's a rectangle, but I want that when I sum up all these areas, this is the integral of F, I get a finite value. So, for example, I can do that with putting here, I think the 1 over n cubed.
25:56:400Paolo Guiotto: Because if you… so this is formally the function sum of n times indicator of interval n plus 1 over n cubed, or any other number smaller than this one.
26:12:930Paolo Guiotto: Now, this function is well-defined. For every X, there is only one indicator which is on, all the other are off, so it's mostly equal to 0, and when it is not equal to 0, it's only one of the indicators which is equal to 1, and the value is equal to 1.
26:30:520Paolo Guiotto: Now, it is clear that, for example, the limit when x goes to plus infinity of this function f of x cannot exist.
26:41:950Paolo Guiotto: Because if you evaluate F, on points n, this is equal to N, this goes to plus infinity.
26:50:890Paolo Guiotto: But if you look at values, f of x, when x is whatever between n plus 1 over n cubed, n plus 1,
27:02:790Paolo Guiotto: This is constant equal to zero, so it will go to zero, no? Imagine that you evaluate along a sequence of points like this, and you see zeros everywhere, so the limiter cannot exist in this case, no?
27:18:590Paolo Guiotto: Second, this function is in L1, because if you look at the integral 0 plus infinity of modulus F, F is positive, so integral 0 plus infinity of F, you actually have the sum of these areas. So these areas are the length of the base, which is 1 over n cubed.
27:38:40Paolo Guiotto: times D8, which is N. That's why I took 1 over n cubed, because I get sum of 1 over n squares, which is final.
27:48:30Paolo Guiotto: This ensures that we have an integral function.
27:53:740Paolo Guiotto: This function is even unbounded, no? You can even see that the infinity norm of this is infinity.
28:04:280Paolo Guiotto: the essential supremum, no? Because this function cannot be bounded by anything. So, it's, it's, it's something that… it's not intuitive, but it happens, no?
28:17:310Paolo Guiotto: So we cannot think that, in general, the limit exists. But we can say that if the limit exists, then you may expect that cannot be anything else than zero.
28:28:960Paolo Guiotto: But… if…
28:34:310Paolo Guiotto: there exists a limit for some reason, if there exists a limit when x goes to infinity.
28:41:140Paolo Guiotto: of f of x equal to L, then necessarily that L will be equal to 0. Because if the limit is different from 0, it's a constant, for example, positively, you have an infinite integral, definitely, so there is no problem here, no?
28:58:400Paolo Guiotto: Now, if,
29:01:710Paolo Guiotto: as in the case of this statement, the function f is in L1 with derivative in L1, so this formula applies. So now, if…
29:13:600Paolo Guiotto: F and F prime are in L1, huh?
29:18:100Paolo Guiotto: We claim… that, the limit when x goes to plus or minus infinity of F exists.
29:31:720Paolo Guiotto: So, we are in the case that necessarily that limit is zero.
29:37:430Paolo Guiotto: So… L would be equal to 0.
29:42:950Paolo Guiotto: And why this?
29:44:690Paolo Guiotto: This, because…
29:49:890Paolo Guiotto: F of X is equal to F of 0,
29:54:890Paolo Guiotto: plus the integral from 0 to X of f prime.
29:58:350Paolo Guiotto: Now, what happens when you send that, for example, X2 plus infinity?
30:04:70Paolo Guiotto: Now, here, there is a limit. Basically, this is… to do a precise justification, you need the dominated convergence. I will explain you in a second. But intuitively, when you send X to infinity, you go to F of 0 plus integral 0 to plus infinity of f prime.
30:23:810Paolo Guiotto: Right? No? When you send X to plus infinity.
30:28:150Paolo Guiotto: And this is because you look at this as integral from 0 to plus infinity, f prime times the indicator of 0x.
30:38:650Paolo Guiotto: 0, X, well, let's say DY.
30:41:970Paolo Guiotto: This is the variable of this integral, let's call it Y. And now, imagine that you send X to… So, you see that? It's like you are passing the data inside the integral, and sending back this X to plus infinity.
30:56:370Paolo Guiotto: So point Y, trying to fix LY, you would decide when X goes to plus infinity, this indicator goes to an indicator of 0 plus infinity, so it becomes 1. And so this becomes F1, pointwise.
31:10:390Paolo Guiotto: And the second one, you have the domination, uniform in X, because if you put the absolute value here, you can draw only indicator, because it is, most people to 1. So models of this is bounded by model setback. So that's why, formally, to prove that the limiting system, when you send X to infinity, X system, you should use the dominator.
31:33:480Paolo Guiotto: books.
31:35:40Paolo Guiotto: Okay, so, however, since F has a limit, and this limit is 5, because the value of that is 0 plus this integral, F prime is the integral is 5, so this is a finite value.
31:55:220Paolo Guiotto: Because F prime is in L1.
31:58:300Paolo Guiotto: then this value, let's baptize L, must be equal to 0. So, at the end, we have that under our assumption.
32:09:60Paolo Guiotto: We know that the behavior at infinity is clear. The function must go to zero.
32:15:960Paolo Guiotto: So… when we have F, F' in L1, necessarily F of X,
32:25:420Paolo Guiotto: goes to 0 when x goes to plus minus infinity. That, I repeat, this is not true in general, with only a final one, okay?
32:35:450Paolo Guiotto: So, we need the distance.
32:37:910Paolo Guiotto: point. So, returning back to the proof, we know now that this quantity goes to zero, this is bounded.
32:48:220Paolo Guiotto: Because it's a modulus 1 number. So when you go to plus and minus infinity, that quantity will go to 0 as well.
32:55:960Paolo Guiotto: And therefore, we have this formula.
32:59:890Paolo Guiotto: So, the Fourier transform of the derivative
33:03:900Paolo Guiotto: is equal to… we said that the evaluation at plus minus infinity vanishes, so it remains the integral. You have this factor here, minus xi, that comes out, it does not depend on X, and so we get plus xi.
33:22:200Paolo Guiotto: plus IXC, then integral on R of F of X P minus iX dx, and that's the Fourier transform of F, F hat X.
33:37:850Paolo Guiotto: As advertised.
33:42:210Paolo Guiotto: So, from this formula, We have a number of, consequences, so corollary.
33:52:170Paolo Guiotto: So, the first consequence is that
33:55:490Paolo Guiotto: we can improve the bound for F at infinity.
34:01:740Paolo Guiotto: So, in fact, if we take the absolute value from this equation, So… from…
34:11:110Paolo Guiotto: So let's assume that F and F prime be both in L1.
34:16:650Paolo Guiotto: from the identity, the Fourier transform of F equal IC
34:22:760Paolo Guiotto: DF prime. They create a form of F,
34:26:60Paolo Guiotto: So we get that the Fourier transform of F is 1 over iC, the Fourier transform of F'.
34:35:639Paolo Guiotto: At least for every C different from zero.
34:39:730Paolo Guiotto: So now we have that modulus of… So let's say, modulus… of F at C,
34:50:380Paolo Guiotto: Is, well, is less or equal than
34:54:550Paolo Guiotto: 1 over xZ, when I take the absolute value, i as modulus 1, becomes 1 over modulus of Xi.
35:02:670Paolo Guiotto: And about this one, I use the simple bound, no? The bound that we have seen above.
35:11:580Paolo Guiotto: the general weak bound, that any fluid transform is bounded, is controlled uniformly, pointwise, by the L1 norm of the function. In this case, it will be the L1 norm of F',
35:27:200Paolo Guiotto: So here I have the L1 norm of F prime.
35:32:20Paolo Guiotto: So, in particular, you see that if we have this extra assumption, F prime is in L1, we re-obtain the Riemann-Lebeg result, the Fourier transform goes to 0.
35:46:950Paolo Guiotto: Okay? Goes to zero at least as 1 over C.
35:53:550Paolo Guiotto: This is not sufficient for integrity, because we know that C is notable, but iterating this mechanism, so if we now assume that F, F prime and F second is also in L1, then we can apply the formula.
36:11:980Paolo Guiotto: not only to F', but also to F second. So we have that the Fourier transform of F second is I see, the Fourier transform of F'.
36:24:380Paolo Guiotto: which is IkeC, so in total, IkeC squared, the Fourier transform of F.
36:32:840Paolo Guiotto: So, in general, what we get is that the written form of the second derivative is the second power
36:41:860Paolo Guiotto: of xC multiplied by… effect.
36:46:700Paolo Guiotto: So, in some sense, Fourier transform transforms derivatives into multiplication by powers.
36:57:150Paolo Guiotto: these powers I exceed.
36:59:770Paolo Guiotto: This is an important fact for differential equations, for example.
37:04:890Paolo Guiotto: I will show something later on this. And more in general, if we have F, F prime, F second, and so on, until the antiderivative are all in L1,
37:21:380Paolo Guiotto: then we will have this formula. The antiderivative of F has for its form, which is ix to power n.
37:30:940Paolo Guiotto: F at C.
37:36:250Paolo Guiotto: And from the bound for the second derivative, we get also this, that modulus f hat C
37:44:650Paolo Guiotto: is now controlled by… I divide by IkeC squared, so voxC different from zero. I put the models, I get one of the models C squared, and then I will have the L1 norm of the second derivative.
38:02:00Paolo Guiotto: So this is sufficient to conclude that in this case, F hat belongs to L1.
38:10:600Paolo Guiotto: So we have this, interesting test. F, F prime, F second, in L1,
38:20:800Paolo Guiotto: we have that F hat is in L1.
38:27:900Paolo Guiotto: So this provides a sufficient condition to have that the Fourier transform of F is in L1.
38:38:150Paolo Guiotto: So the message is, however, the higher is the regularity of F, the faster is the decay of f hat at infinity.
38:49:80Paolo Guiotto: Actually, there is a sort of, vice versa of this,
38:56:520Paolo Guiotto: Which is, the,
39:00:630Paolo Guiotto: the following factor that says the higher is the decay of F,
39:07:810Paolo Guiotto: And the more regular is the Fourier transform. So now we are going to talk about the derivative of the Fourier transform.
39:17:190Paolo Guiotto: So, proposition…
39:22:20Paolo Guiotto: Now, assume that our F is in L1,
39:26:640Paolo Guiotto: Here, to… to gain regularity for the Fourier transform, I need to have dec at infinity for F.
39:36:650Paolo Guiotto: How do we say this with this condition? Suppose that when you multiply F by X, the variable.
39:44:200Paolo Guiotto: So, XF of X, okay? So, here you are multiplying your function by a factor that is not bounded at infinity. It's big, no?
39:56:590Paolo Guiotto: Now, suppose that these two are in L1,
40:00:110Paolo Guiotto: So it means that you have a function which is integral, we cannot say that it goes to 0 and infinity, because it supports this one. But, let's say that in the intuition that decay somehow infinity, but this says that mass decay of solving also X.
40:18:640Paolo Guiotto: So you cannot become that, for example, you have X equal to 1 over X squared, which is not multiple at zero, but forget, the problem is infinity, no? You want to take 1 over 1 plus X squared.
40:30:150Paolo Guiotto: Okay? So now you have a function everywhere, you find it, it is integral, no? No problem. But even when you divide by X, you get X divided 1 plus X squared.
40:41:410Paolo Guiotto: So that's, like, 1 over X at plus minus infinity, and that's not even.
40:46:250Paolo Guiotto: So it means that 1 over X squared does not arrive fast enough to have this, no?
40:53:860Paolo Guiotto: So, if you have these two, then it turns out that the Fourier transform is differentiable, so there exists the derivative with respect to its variable.
41:06:40Paolo Guiotto: let's say Xi of the Fourier transformer.
41:10:280Paolo Guiotto: And it turns out that this is the Fourier transform, Offer.
41:17:130Paolo Guiotto: Now, it's exactly the dual, no, operation that you have here. Here you say that the Fourier transform of the derivative is the multiplication by x.
41:28:760Paolo Guiotto: of the FDA score.
41:30:670Paolo Guiotto: Now, sort of dual operation happens. The derivative of the Fourier transform is the Fourier transform of the multiplication.
41:38:330Paolo Guiotto: The unique difference is a sine, so here you have minus I. Since this is in the Fourier transform operation, I do not like to write X, otherwise it's a mess, you do not… you do not understand what is the variable. So let's say it's I sharp, F sharp.
41:58:180Paolo Guiotto: evaluated at point X, so this means the function X goes to XF of X, okay?
42:07:120Paolo Guiotto: That function, that we are assuming it is integral, so we know that, that the cat makes sense. And this happens for every scene.
42:18:260Paolo Guiotto: react.
42:20:660Paolo Guiotto: So, also, this one is,
42:25:180Paolo Guiotto: a relatively straightforward, operation, because, we know that, By definition, effort.
42:38:80Paolo Guiotto: C is, by definition, integral on R, f of x e minus iCX.
42:47:70Paolo Guiotto: So, if we want to compute the derivative.
42:50:780Paolo Guiotto: Naturally, we would say that the derivative with respect to X of the Fourier transform, we need to differentiate that integral.
43:02:150Paolo Guiotto: So, why don't we carry the derivative inside? So, can we say that this is the integral of the derivative respect to C of whatever you have inside?
43:14:580Paolo Guiotto: Notice that if this is possible, what is the derivative of the inside? Of course, this is independent of X, so it's a constant, and it remains F of X. Then you have to differentiate this exponential that now with respect to C.
43:33:300Paolo Guiotto: So, it is the exponential.
43:35:670Paolo Guiotto: E minus iCX times the derivative of the exponent with respect to C. You get minus iX that you put in front of F.
43:48:120Paolo Guiotto: And that's, you see, that this is the Fourier transform of minus i, the variable f
43:58:170Paolo Guiotto: evaluated at C. So if you can differentiate under integral sign, you get exactly the formula announced above. So the question is, can we do this
44:22:930Paolo Guiotto: So, of course, yes, if we verify that the assumptions of the differentiation and the integral sign are verified.
44:33:500Paolo Guiotto: Yes, if… differentiation.
44:39:830Paolo Guiotto: ERM.
44:42:310Paolo Guiotto: Also, huh?
44:43:750Paolo Guiotto: And what are the assumptions? Number one, we need to verify that there exists the derivative with respect to X of the integrand, which is F of X
44:54:680Paolo Guiotto: E minus ICX, and that's evident because we are differentiating in X, so F of X, as we said, is a constant, and this is E minus iCX times minus iX.
45:10:440Paolo Guiotto: This holds what? Well, for X, you have F, that forces almost every X, because F is just L1, so almost every X, but that's exactly what is required in the integration domain. And for C, there is no restriction. Whatever is C,
45:29:200Paolo Guiotto: in Ria.
45:31:130Paolo Guiotto: And second requirement is that we need to bound this derivative, so models of DXC, of that stuff.
45:39:940Paolo Guiotto: must be bounded by something that is independent of X, and it is integral. But what is the bound of this? When we do the modulus, we have exactly modulus f of x.
45:55:360Paolo Guiotto: Well, let's say minus IX and the exponential.
46:00:920Paolo Guiotto: The exponential is, as usually, a module 1 number, and so what remains is the absolute value of, well, even the minus can be simplified, XF of X.
46:14:880Paolo Guiotto: And this is L1 by hypothesis.
46:21:610Paolo Guiotto: So, we are now, in,
46:26:170Paolo Guiotto: In all circumstances, to apply the differentiation theorem, and we get the conclusion.
46:32:880Paolo Guiotto: Yeah.
46:37:720Paolo Guiotto: So, you see that,
46:40:290Paolo Guiotto: Fourier transform converts derivations into multiplication by powers, and vice versa, differentiation of the Fourier transform is Fourier transform of,
46:53:240Paolo Guiotto: of multiplication by power. So it's like if these two operations, Fourier transform, sorry, differentiation and multiplication by power, they sort of commute under the Fourier transform. There is some algebraic flavor under this.
47:14:130Paolo Guiotto: For example, an interesting consequence of this is an important subset of this space of functions, which is the Schwartz space.
47:31:80Paolo Guiotto: Also, there should be an H somewhere.
47:41:940Paolo Guiotto: I do not remind… I was reading this.
47:45:190Paolo Guiotto: Oh, well.
47:46:530Paolo Guiotto: sports… space.
47:51:980Paolo Guiotto: The Schwartz space is a very particular subspace of function, particularly important for the Fourier transformer. It's roughly, space
48:05:620Paolo Guiotto: of symphony functions, so… The highest possible regularity, basically.
48:16:340Paolo Guiotto: That decay faster.
48:20:630Paolo Guiotto: at infinity, and practically decaying faster than any polynomial. Not only the functions, but all the derivatives.
48:29:990Paolo Guiotto: Such that, Day.
48:37:270Paolo Guiotto: they… decay.
48:43:750Paolo Guiotto: faster.
48:46:690Paolo Guiotto: Then… Annie.
48:50:930Paolo Guiotto: polynomial.
48:57:140Paolo Guiotto: Together.
49:01:550Paolo Guiotto: we…
49:07:400Paolo Guiotto: Well, let's say. That's… well, it's not correct to say this. So, let's say, such that… They decay…
49:19:790Paolo Guiotto: Faster than putting on a date.
49:24:650Paolo Guiotto: Good.
49:27:840Paolo Guiotto: Sweet.
49:29:980Paolo Guiotto: there.
49:32:210Paolo Guiotto: derivatives.
49:34:370Paolo Guiotto: It's longer to write in words than to write in CMOST, faster than any polynomial.
49:41:220Paolo Guiotto: Now, you may wonder, is there any function like that? Well, the main prototype is this function, is the Gaussian. E minus X squared is a function that belongs to C infinity, and you understand that
49:55:850Paolo Guiotto: the function is clearly decaying faster than any polynomial, because it's decaying faster than any power. But also, derivatives, when you do derivatives, the derivatives are still E minus X squared times a polynomial. So, E minus X squared is strong enough to kill any kind of polynomial.
50:15:790Paolo Guiotto: Now, a formal definition is the following. The Schwartz space is the set of functions F in C infinity, realign.
50:26:460Paolo Guiotto: So, how do we say that they decry faster than any polynomial? Well, there are two equivalent ways to define this. I don't have here the issue of notes, so… Well, we can say that when you take the function f, and you compute a certain derivative, so let's say the nth derivative of F,
50:49:730Paolo Guiotto: and you multiply by any power, X to K, this quantity goes to 0 when x goes to plus-minus infinity. And this happens whatever is the exponent k of the power, and whatever is the order N of the derivative.
51:09:260Paolo Guiotto: So this is the formal, the formal definition. Somewhere, somewhere you may find an equivalent, basically, definition that says that these functions are bounded, because it's the same. If this is bounded, it is clear that the function and the derivative must kill any polynomial.
51:28:980Paolo Guiotto: Now, a remarkable feature of the Fourier transform is that
51:35:90Paolo Guiotto: A corollary of, of previous,
51:39:420Paolo Guiotto: properties is the following, that Fourier transformer maps Wie Schwarzspace.
51:51:380Paolo Guiotto: Well, actually, it should be checked that a Schwarz space is contained in L1, but you may mention they decry so fast that they will be also integral. Well, let's say in a moment.
52:03:400Paolo Guiotto: In any case, when you take a function of the Schwartz space, the Fourier transform maps into a function of the same set. So, into…
52:13:560Paolo Guiotto: itself.
52:17:910Paolo Guiotto: Well, I don't want to, basically, technical, but let's… let's do some check. So, sketch.
52:36:440Paolo Guiotto: So, faster… Schwarz space is a subset of L1R.
52:49:150Paolo Guiotto: That's clear, because, for example, you take X squared times function f of x, you know that this goes to 0 when x goes to plus-minus infinity.
53:03:800Paolo Guiotto: So it means that, in particular, if you go to 0, you must be bounded. So there exists a constant C such that x squared f of x absolute value of this will be bounded by a constant when x goes to plus, minus infinity.
53:21:960Paolo Guiotto: But this means that modules f of x is controlled above by C over X squared when x goes to plus minus infinity.
53:33:880Paolo Guiotto: And since the function is continuous, it's regular, you have no problems with integrability at any finite interval. The only question, maybe, is the integrability at plus-minus infinity, but if you decay faster than this function, you are integral at plus-minus infinity. So F integral at
53:53:870Paolo Guiotto: plus minus infinity, plus F continuous means that F is integral on the real line, so F is in L1.
54:09:100Paolo Guiotto: And question two, we should prove that if F is a Schwartz function, if F is a Schwarz function.
54:20:170Paolo Guiotto: then the Fourier transform of F is still a Schwartz function. That is what it means the Fourier transform maps the Schwarz space into itself.
54:33:780Paolo Guiotto: Now, what should be checked? I should check, number one, so, let's say, first thing to check, is the Fourier transform Sinfinity function?
54:44:770Paolo Guiotto: Question 2, is the fluid form decaying faster at infinity together with its derivative? So, is it true, now the variable for the fluid form is C. Is it true that C to K, the,
54:59:620Paolo Guiotto: K… the nth derivative with respect to X of the Fourier transform is decaying fast at infinity when C goes to plus infinity. So these are the two questions, no?
55:13:340Paolo Guiotto: So, for example, 1, let's prove that F at is C1.
55:19:850Paolo Guiotto: Okay?
55:20:980Paolo Guiotto: Then, similarly, you prove that it is C infinity. So, how do I prove that f at is C1? I need the derivative of F at. Who gives me the derivative? It's this
55:32:580Paolo Guiotto: second statement. This statement says that the Fourier transform is differentiable, so I'm going to use this. Now, to use this, I need to know that F and XF are in L1. Who says that? The fact that F is Schwartz's function. If it is Schwartz's function, F is in L1, we already proved it. But also, XF is in L1.
55:54:170Paolo Guiotto: So we notice that F is in L1,
55:58:610Paolo Guiotto: Okay, we already proved it. But also, XF is in L1.
56:03:170Paolo Guiotto: And this is because you adopt the same argument we have seen here. Now, I wanted to show that XF decays at least as 1 over X squared, so it means that X cubed F must go to 0. So X cubed F goes to zero.
56:22:220Paolo Guiotto: So, in particular, modulus X cubed F is bounded by a constant at infinity, and so in particular, modulus XF
56:35:380Paolo Guiotto: will be less or equal than constant x squared.
56:39:400Paolo Guiotto: at infinity. So this says that XF decays sufficiently fast to be integral, so this says XF is in L1.
56:50:880Paolo Guiotto: So we have these two conditions, F in L1, XF in L1 are verified, so we have that there exists the derivative with respect to C of the Fourier transform of F.
57:07:130Paolo Guiotto: Right? And what is this?
57:09:430Paolo Guiotto: is a Fourier transform. It's the Fourier transform of something, that something is minus i, the variable F.
57:17:560Paolo Guiotto: C.
57:18:910Paolo Guiotto: And we know that, in general, any Fourier transform of an L1 function, we proved, is a continuous function. So this is a continuous function.
57:31:840Paolo Guiotto: Because…
57:35:740Paolo Guiotto: This is the general fact we proved above. If we have a G which is in L1, the Fourier transform of G must necessarily be a continuous function.
57:46:280Paolo Guiotto: So we gain that the derivative of F hat not only exists, but it is continuous. So this means that F hat is a C1 function.
57:58:390Paolo Guiotto: It is clear that now you should repeat this iterate in general for any kind of derivative, and with a little bit of work, you will get the general fact. So this should prove that F hat is infinity. Then we should prove that multiplying by a power
58:16:640Paolo Guiotto: A derivative, this goes to zero.
58:19:360Paolo Guiotto: But…
58:20:400Paolo Guiotto: Suppose that I take C, just the first power and the first derivative, to make easier, no? C, D, the C…
58:33:680Paolo Guiotto: F at.
58:36:910Paolo Guiotto: Actually, let's do easier. There is also the zero derivative. The zero derivative is the Fourier transform. So, the problem is, let's do with the zero derivative. So, C times F at C. How can I say that this decayed fast enough,
58:55:910Paolo Guiotto: in such way that you go to 0 when X goes to infinity.
59:02:450Paolo Guiotto: Now,
59:08:500Paolo Guiotto: How can I see that?
59:12:360Paolo Guiotto: So this is the…
59:31:470Paolo Guiotto: It's not the…
59:41:500Paolo Guiotto: Well, if I put the IC this, no?
59:45:770Paolo Guiotto: This is what? This is the formula of,
59:50:370Paolo Guiotto: the Fourier transform of the derivative, right?
59:53:500Paolo Guiotto: No? So, I have that this is, this. I can do that because, of course, being F in the Schwarz space, also F' is in the Schwarz space, so in particular, it is in L1.
00:09:50Paolo Guiotto: So this formula makes sense. Now, and this now, the conclusion is automatic, because we know that any Fourier transform
00:17:550Paolo Guiotto: goes to zero at infinity. Now, this is the Rieman-Lebacc lemma. This says that this goes to 0 by Riemann-Lebacc lemma when C goes to plus minus infinity. Because I see F hat is F free transform, it's the free transform of F'.
00:36:500Paolo Guiotto: So, this means that this box goes to 0, i is irrelevant, when X goes to infinity. So, you can iterate this argument with any derivative, any power, and you basically get the conclusion. So…
00:53:930Paolo Guiotto: If we want to have a new picture, we could say that there is a one
00:59:670Paolo Guiotto: There is the Schwartz space,
01:04:520Paolo Guiotto: Actually, you can, you can even prove that the Schwarz space, by… by doing the convolution.
01:12:530Paolo Guiotto: the… with the approximate units, for example, with the Gaussian units, you can show that, actually, the Schwarz space is dense in L1, so it's…
01:26:60Paolo Guiotto: It's not like the figure you see here. You can approximate any L1 function with the Schwarz functions. And the same happens with the L2. So L2 is not contained into L1 and does not contain L1, but the Schwarz space is still inside also L2.
01:44:690Paolo Guiotto: So it's a common basement for the two spaces. And what the Fourier transform does, the Fourier transform is defined on a one.
01:55:20Paolo Guiotto: We don't know exactly, and it's actually complicated. It's extremely complicated to characterize
02:02:120Paolo Guiotto: how is… what are the characteristic properties of a Fourier transform? So is there a test that says, this is the Fourier transform of something if and all if these properties are verified? There is not such test. We know that if we do the Fourier transform of an L1 function, we get a bounded function, it goes to zero at infinity, but there is not an if and only if.
02:27:00Paolo Guiotto: You can have functions going to zero at infinity and bounded, like the… that… that are not fully transformed of anything. This is a complicated example, but let's say, more or less like the sign, Xi over Xi.
02:45:720Paolo Guiotto: Okay, so, let's see some user… of these, properties…
03:08:930Paolo Guiotto: Well, the exercises are all on the convolution.
03:23:430Paolo Guiotto: Yeah, we have still, okay, let me show this example.
03:36:430Paolo Guiotto: Also, we have this function, FA of X, which is defined this way, 0 when x is less than minus 2A, or X greater than 2A. Then we have X plus 2A,
03:52:600Paolo Guiotto: for X between minus 2A And zero.
03:59:290Paolo Guiotto: and minus X plus 2A.
04:03:140Paolo Guiotto: when X is greater than 0, less than 2A.
04:08:60Paolo Guiotto: So this function is this.
04:12:920Paolo Guiotto: A is a positive parameter, so let's say that we have minus 2A here, plus 2A here.
04:21:10Paolo Guiotto: And, the function is 0.
04:23:850Paolo Guiotto: Out of this interval.
04:26:260Paolo Guiotto: Then it grows linearly, like this.
04:30:570Paolo Guiotto: This is value 2A at x equals 0, and it decreases linearly in this way. So this is the function FU.
04:39:20Paolo Guiotto: Now, it says that, compute… DX of FA.
04:50:540Paolo Guiotto: And… Did you reserve.
04:55:920Paolo Guiotto: F-A-S.
05:00:40Paolo Guiotto: But let's see what this problem wants to do. So, we have first to compute the derivative. Of course, we could compute directly with the definition.
05:11:210Paolo Guiotto: It's not a complicated calculation, but we can do in this alternative way. So let's start computing the derivative with respect to X of this FA. Now, it is clear that the derivative is 0 when x is less than minus 2A and greater than 2A.
05:30:880Paolo Guiotto: At minus 2A, there is no derivative.
05:34:650Paolo Guiotto: From, minus 2A and, 0, derivative is 1.
05:42:990Paolo Guiotto: And from, 2A… from zero, 2A derivative is minus 1.
05:51:960Paolo Guiotto: So…
05:56:270Paolo Guiotto: We can say that derivative exists apart for these three points, minus 2A, 0, and 2A, so particular almost everywhere.
06:06:20Paolo Guiotto: And… So…
06:12:960Paolo Guiotto: probably we have to find some connection between the free transform of this and the free transform of FA. In any case, this is an L1 function, so we can compute the Fourier transform.
06:25:540Paolo Guiotto: the Pourettance form of, FAE
06:30:40Paolo Guiotto: Now, this is the Fourier transform of the derivative. So, it is, I see the Fourier transform of FA.
06:44:500Paolo Guiotto: And what about the Fourier transform of F… so we can say that, in particular, the Fourier transform of FA, which is the goal, is one of IC,
06:57:980Paolo Guiotto: at least FOC C different from zero, then we will see what happens for X equals 0.
07:03:460Paolo Guiotto: The Fourier transform of the derivative of FA.
07:09:190Paolo Guiotto: Now, the Pourier transform of the derivative, since the derivative is what? The derivative looks like a sort of indicators. It is zero here.
07:21:500Paolo Guiotto: This is minus 2A, it is 1 from this to 0, then it is minus 1 from 0 to minus 2 plus 2A, and then again 0.
07:32:640Paolo Guiotto: So, we could somehow use indicators. How? Well, for example, we could split this into the difference between two indicators. One could be this one, the green, which is,
07:49:800Paolo Guiotto: 0 here, except the interval minus 2A to 0. So the green one is the indicator of the interval minus 2A
08:01:800Paolo Guiotto: Zero.
08:03:260Paolo Guiotto: And let's say the blue one…
08:11:110Paolo Guiotto: Well, this would be minus…
08:14:10Paolo Guiotto: indicator of 0 to 2A. So we can say that the Fourier transform of the derivative of FA is the Fourier transform of indicator minus 2A0,
08:32:830Paolo Guiotto: Minus indicator of 0 to 2A.
08:37:450Paolo Guiotto: So, by linearity, we could split this calculation into the Fourier transform of minus 2A0,
08:45:260Paolo Guiotto: minus the free transform of indicator 0 to 2A.
08:51:180Paolo Guiotto: Now, even these ones can be computed directly.
08:55:740Paolo Guiotto: But we can also reduce this to known quantities. In particular, I'm now trying to restore the indicator of the rectangle. Now, direct…
09:09:649Paolo Guiotto: is an asymmetric indicator. That's the feature of the rack. Direct A is this, you remind, minus A, A, and 1 here.
09:22:210Paolo Guiotto: This is the rect…
09:25:20Paolo Guiotto: A, what we computed if we had transformed. So, now, what can be said about the indicator of minus 2A and 0?
09:37:680Paolo Guiotto: Well, this is… now, I'm always in trouble with these things, so it is 1 if and only if X is between minus 2A and 0, right?
09:49:410Paolo Guiotto: So, if I add XA, both sides, X plus A between minus A and A. So, it is the indicator of minus A, A, evaluated at X plus A.
10:07:520Paolo Guiotto: So when I do the Fourier transform of indicator minus 2A0,
10:15:430Paolo Guiotto: evaluated at point C, I am doing the indicator of minus AA, but translated with plus A here.
10:28:760Paolo Guiotto: Now, do you remind the formula?
10:31:540Paolo Guiotto: Translation, is transformed into multiplication by character.
10:38:480Paolo Guiotto: Maybe the plus or minus… I reminded, well, we have seen yesterday this formula, so let's quickly see what is the relation.
10:49:400Paolo Guiotto: We computed somewhere here. For the transform of X minus M, is multiplied by E minus Ixm, so since we have plus, it will be plus.
11:02:370Paolo Guiotto: So, we have here…
11:04:840Paolo Guiotto: and E to ICA times the indicator of now of the, the free transform of indicator minus AA.
11:18:750Paolo Guiotto: But that is, the rectangle, so we have T2IXCA, times… Is, 2…
11:30:810Paolo Guiotto: A sine HC over XC, if I'm not wrong.
11:38:990Paolo Guiotto: And similarly, for the other rectangle, for the other indicator, from 0 to 2A,
11:46:630Paolo Guiotto: It will be complementary, now it will be, like, if we transform off the indicator minus AA,
11:54:970Paolo Guiotto: Evaluated at this minus age, So this will be E minus ICAE.
12:05:270Paolo Guiotto: times the Fourier transform of the rectangle, which is 2A, Sign CAE.
12:13:350Paolo Guiotto: over CA.
12:16:10Paolo Guiotto: Okay, now we can put together these things.
12:21:460Paolo Guiotto: So we have that.
12:24:530Paolo Guiotto: This is the Fourier transform of the derivative of FA. So, the Fourier transform of DXFA is the difference between these two Fourier transforms.
12:39:30Paolo Guiotto: So I take the first minus the second. You see that there is a common 2A sine HC.
12:47:90Paolo Guiotto: over 8C, that multiplies… E to ICA.
12:53:790Paolo Guiotto: minus E to minus ICA.
12:58:600Paolo Guiotto: And this, if I divide it by 2i and multiply by 2i.
13:06:780Paolo Guiotto: This is, again, sine of Xia, right?
13:13:100Paolo Guiotto: This is the Euler formula. This is sine of CA.
13:20:440Paolo Guiotto: So, at the end, we have… 2… 4… Ayy.
13:29:230Paolo Guiotto: I… Sine… square.
13:35:430Paolo Guiotto: AXC divided by 8C.
13:40:80Paolo Guiotto: This is the Fourier transform of the derivative of FA.
13:45:950Paolo Guiotto: Now, to go back to the original problem, we had that this, the Fourier transform of FA is 1 over IC, what we computed. So I have seemed to divide by IXC.
13:59:580Paolo Guiotto: So the Fourier transform of FA at point C is 1 over ITC.
14:06:290Paolo Guiotto: the Fourier transform of the derivative of FA So, 1 over IC.
14:14:680Paolo Guiotto: of this thing. So we have, 4AI… sine square… HC… divided by XC.
14:27:950Paolo Guiotto: I can give the X to this, perhaps I put a square here, so I add the square here, I cancel the I, so we have 4.
14:38:960Paolo Guiotto: Well, that's exactly the square.
14:42:490Paolo Guiotto: the square of 2A sine HC over 8C.
14:53:830Paolo Guiotto: Now, this formula, if you remind, is Foxy different from 0, because we divided it by xi. But Foxy equal to 0
15:08:330Paolo Guiotto: Well…
15:09:450Paolo Guiotto: We don't need to do any calculation, we can use what we know. F A is… FA hat is a continuous function, because it's the Fourier transform of an L1 stuff.
15:22:110Paolo Guiotto: So, what I can say… I can say that F at A at 0 is just the limit when C goes to 0 of F
15:31:760Paolo Guiotto: A hat at C.
15:34:970Paolo Guiotto: And again, I am reduced to the limit sine t over t, so, it's, one…
15:44:440Paolo Guiotto: squared, so it gives a 4A squared. So, again, we can take sine T over T equal to 0, and use that formula also for t equals 0.
16:02:650Paolo Guiotto: Okay… Let's see another nice property of the Fourier transformer.
16:12:40Paolo Guiotto: which is Fourier transform, and… convolution.
16:23:130Paolo Guiotto: So, you're reminded of this operation we introduced there. It happens the following.
16:30:170Paolo Guiotto: Now, suppose that we have two L1 functions.
16:34:720Paolo Guiotto: If we have two L1 functions, the convolution product is in L1, and it is also an L1 function, no? F star G is in L1,
16:48:810Paolo Guiotto: So, the Fourier transform of this thing is well-defined.
16:52:650Paolo Guiotto: What is remarkable is that the Fourier transform of the convolution product is the algebraic product of the Fourier transforms.
17:04:530Paolo Guiotto: Also, this has basically a simple and straightforward proof. It's just a matter of writing the symbol, so let's start with the Fourier transform of the convolution.
17:20:760Paolo Guiotto: So, first of all, it creates a transform. This is integral on R of F star G, evaluated
17:34:410Paolo Guiotto: Now, let's replace into this formula what is the convolution. This is a double integral. It's an integral.
17:44:800Paolo Guiotto: on R of F of Y
17:48:890Paolo Guiotto: G of X minus Y DY.
17:53:400Paolo Guiotto: So we plug now this integral into that formula.
17:57:200Paolo Guiotto: And then we will see what happens by switching the integration.
18:02:600Paolo Guiotto: order. So we have integral on… we can say, let's split directly on R2, according to Fubini theorem.
18:12:970Paolo Guiotto: This is, F of Y.
18:16:600Paolo Guiotto: G of X minus Y.
18:20:80Paolo Guiotto: E minus IX, and then we have a double integration, DXDY.
18:27:20Paolo Guiotto: Now, the point is that we recreate here the argument, so by adding and subtracting Y, and then we split this into E minus i times C times X minus Y, and E minus iCY.
18:46:230Paolo Guiotto: We now apply back the Fubini theorem.
18:50:150Paolo Guiotto: But switching the order of the two integrations.
18:53:950Paolo Guiotto: So, doing a first integration in,
18:57:900Paolo Guiotto: No, first do in the integration. So here, the first integration was in Y and then in X, okay? X is the last integration, because it comes from the words of one. Now, we do the opposite. So we first integrate in X and then in Y.
19:16:590Paolo Guiotto: So when we integrate in X all these things, you see that there are factors like this, F of Y, and this, they do not depend on X.
19:27:510Paolo Guiotto: So when I integrate on X, these are constants that I can carry outside. I do this, writing maybe this.
19:36:980Paolo Guiotto: two factors here, F of Y, E minus IXY, DY, and what remains is G of X minus Y,
19:48:290Paolo Guiotto: E2 minus iCX minus Y.
19:52:810Paolo Guiotto: And now, of course, integration is on R in both cases, and here, in the innermost integral, we change variables. Set, for example, U equal X minus Y.
20:08:430Paolo Guiotto: So, since x is between minus infinity plus infinity, u will be between minus infinity plus infinity, so integral in r of g of u, e minus iCU, and the dx is just U.
20:25:350Paolo Guiotto: But this integral is now, which is inside, is no more depending on Y.
20:32:40Paolo Guiotto: So, it can be seen as a constant for this second integration, so you can carry all this block out of the other integral.
20:42:300Paolo Guiotto: and you get integral on R of GU E minus ICUDU.
20:51:880Paolo Guiotto: times… an integral on R of what remains, which is F of Y E minus I, C, Y,
21:02:270Paolo Guiotto: DY. And you see, now, there are the two Fourier transforms. This is G heter C,
21:10:260Paolo Guiotto: And this is F hat C.
21:14:390Paolo Guiotto: And you get the conclusion.
21:17:300Paolo Guiotto: Now, this formula has an acute consequence, which is… it's not… this is… which is important for this formula, but this explains why there cannot be an identity for the convolution.
21:32:700Paolo Guiotto: So now the convolution product is… has similar properties to the algebraic product. It is commutative, it is associative.
21:41:570Paolo Guiotto: It is distributive, okay, but there is not the unity. So, there is not a function, sorry, unity.
21:54:630Paolo Guiotto: for…
21:56:300Paolo Guiotto: the convolution product. What should be the unity? If F star, let's call it delta, would be equal to F for every F in L1,
22:12:620Paolo Guiotto: for… some.
22:15:790Paolo Guiotto: delta in L1. So, suppose that there is a delta that is like a unit in this product. Now, when you multiply in the convolution product with delta, you get the same function.
22:27:820Paolo Guiotto: Now, applying the Fourier transform to both sides, you would have that F star delta hat would be equal to F hat.
22:37:250Paolo Guiotto: So, this would mean that, because of the property, we come to see that F hat delta hat would be equal to F hat for every F in L1.
22:49:770Paolo Guiotto: So, for example, you take an F whose f hat is never zero, like in the case of the Gaussian or the exponential, those functions are never zero, no? So, this in particular will imply that delta hat must be identical equal to 1,
23:06:910Paolo Guiotto: But that's not possible, because of Riemann-Leberger Lemma.
23:12:230Paolo Guiotto: Impossible.
23:16:120Paolo Guiotto: Because… off.
23:20:310Paolo Guiotto: Riemann Lebeg Lemma.
23:24:840Paolo Guiotto: Riemanebach says that all Fourier transform decays at infinity, no?
23:31:290Paolo Guiotto: So delta hat of C should go to 0 when C goes to plus or minus infinity, and this is impossible if delta is 1.
23:43:630Paolo Guiotto: Okay, so, of course, this is not the most important factor.
23:51:630Paolo Guiotto: Okay, so let's do, let's give some exercises to do. Do exercises, 18, 5, 1…
24:12:750Paolo Guiotto: I would say, until number 6.
24:16:960Paolo Guiotto: Now… Mmm… let me… let me… Do…
24:26:620Paolo Guiotto: Okay, I will… tomorrow, I will do… Numbers 3…
24:39:00Paolo Guiotto: for… And maybe 6.
24:44:520Paolo Guiotto: Let's do the… we have about…
24:48:920Paolo Guiotto: 5 minutes. Let's see if we can, do the… Exercis, 18… 5… 2, which is compute…
25:02:620Paolo Guiotto: The Folliers transform of this thing.
25:05:650Paolo Guiotto: of XE minus X squared convolution with E minus X squared.
25:14:330Paolo Guiotto: So it's a calculation that uses what we have seen this morning, basically.
25:21:890Paolo Guiotto: Now, to do this calculation, we have, compute, of course, X is the variable, so let's… it's a little bit horrible, but sharp E minus a sharp square.
25:34:930Paolo Guiotto: convolution, E minus sharp square, we can use, immediately the property we have seen, that,
25:45:310Paolo Guiotto: So, in principle, you should do… let's take these two functions. Let's compute the convolution, then we do the free transform, no?
25:52:840Paolo Guiotto: This will take a certain time. Or, you could apply the Fourier transform directly, and use the fact that Fourier transform of the convolution product is the product of the Fourier transforms.
26:05:200Paolo Guiotto: And now the problem is to compute these two Fourier transforms. One is the transform of a Gaussian thing, so…
26:15:420Paolo Guiotto: It is something that we should know. Now, knowing exact formula is never easy. By the way, you will be a loader to carry with you,
26:29:940Paolo Guiotto: A paper with formulas you think, are… the only thing you cannot have during the exam is exercises.
26:39:220Paolo Guiotto: So, do not carry exercises, problems, solutions, but you can carry materials like formulas and so on. Even… you don't have time to, you have never seen a problem to learn at that moment, so it would be…
26:58:150Paolo Guiotto: Just a waste of time.
26:59:900Paolo Guiotto: Be prepared, carry with you some… some… a paper with the most important formulas that you think would be usable. For example, the formulas of Fourier transforms.
27:11:220Paolo Guiotto: I never remained, but this is, what is it?
27:19:730Paolo Guiotto: So, it is the Goshen, no?
27:22:770Paolo Guiotto: So it is with, is this one, with 2 sigma squared equals 1, basically?
27:31:870Paolo Guiotto: So imagine that 2 sigma squared is 1, no? You see?
27:36:480Paolo Guiotto: If this coefficient here is 1, it is E minus the square. So if we get root of pi, because there is 2 sigma squared, which is 1, exponential, then I multiply by 2, divide by 2, so I get C squared divided 4.
27:54:920Paolo Guiotto: So if I'm not wrong, should be, root of pi…
27:59:720Paolo Guiotto: E minus C squared divided by 4. This is the transform. About the second one, sharp E minus sharp square.
28:10:210Paolo Guiotto: This sounds, the… what? You see?
28:15:540Paolo Guiotto: transform of X times F, no? This looks like the formula we have seen here for the derivative of the Fourier transform.
28:27:40Paolo Guiotto: No? So I need a minus i, I put…
28:30:600Paolo Guiotto: this minus I, and I, of course, I have to multiply and subtract and divide, so I have minus i sharp E minus
28:42:200Paolo Guiotto: sharp square XC. Now, this is the derivative with respect to X of the Fourier transformer
28:50:240Paolo Guiotto: of the function E minus sharp squared that we already computed. So Dixie, this is root of pi E minus C squared divided 4.
29:02:650Paolo Guiotto: So this is root of pi E minus c squared divided 4, and when I do the derivative of the exponent, I get minus 2C divided 4 minus C.
29:14:890Paolo Guiotto: tough.
29:17:160Paolo Guiotto: So, at the end, well, this 1 over minus i is plus I.
29:24:70Paolo Guiotto: So minus, root of pi C half E minus C squared over 4.
29:33:860Paolo Guiotto: So these are the two Fourier transforms, and we now can go back to the original problem, the Four returns form of that convolution.
29:41:810Paolo Guiotto: So the Fourier transform of sharp
29:44:850Paolo Guiotto: E minus sharp square start. E minus sharp square at C is equal to this minus i.
29:55:590Paolo Guiotto: root of pi over 2C E minus C squared over 4 times the other one is root of pi E minus X squared over 4.
30:12:410Paolo Guiotto: So at the end, you see that these two cancels to minus pi, so minus i pi half C, exponential, and this is minus C squared over
30:27:660Paolo Guiotto: And that's the… We exhaust.
30:33:820Paolo Guiotto: Okay.
30:36:320Paolo Guiotto: Let's stop here.