AI Assistant
Transcript
00:04:480Paolo Guiotto: Okay, good morning.
00:11:380Paolo Guiotto: So, yesterday we introduced this operation, which is called the convolution of two functions, F and G.
00:21:480Paolo Guiotto: Which is defined through this integral.
00:25:330Paolo Guiotto: The idea was that, this operation…
00:34:700Paolo Guiotto: It's finally meeting.
00:37:540Paolo Guiotto: We wish to re-molarize a function f.
00:42:210Paolo Guiotto: So the goal is, actually, there is a…
00:47:200Paolo Guiotto: There are two goals. One is regularize the function, the other one is to approximate, because of course, F is not regular in general, if it is just integral.
00:58:820Paolo Guiotto: And the idea was to regularize by doing the convolution with a function
01:06:750Paolo Guiotto: because we had this delta epsilon, that, for this particular case, has these features. It's a function that is concentrated around the origin, somehow.
01:21:70Paolo Guiotto: So in this case, it's a function which is 0 outside the interval minus epsilon epsilon.
01:26:770Paolo Guiotto: And, it is equal to 1 over 2 epsilons, so when epsilon is small, it's very big, it is.
01:33:610Paolo Guiotto: And when epsilon goes to zero, this peak shrinks to one single infinite value at zero, at least intuitively.
01:44:400Paolo Guiotto: Now, we want to give a definition to this.
01:50:950Paolo Guiotto: So, let's introduce the concept of approximate unit.
01:57:400Paolo Guiotto: So, definition…
02:07:190Paolo Guiotto: We take a function, delta.
02:10:780Paolo Guiotto: which is a function defined. I do this definition, these constructions in the case of one-dimensional functions, so functions of real variable, but they can be, done in, in,
02:24:640Paolo Guiotto: any RDKs. So let delta be a function R to R, such that… Number one…
02:33:820Paolo Guiotto: This delta is positive.
02:36:570Paolo Guiotto: And number 2, we just need that the integral on r of this delta is equal to 1.
02:44:760Paolo Guiotto: We define, huh?
02:50:130Paolo Guiotto: Function delta epsilon.
02:52:840Paolo Guiotto: in the following way, we want that these delta epsilon have the same properties of delta, so they are positive, and with the integral equal 1. That's why you see a scaling factor here, which is 1 over epsilon, times delta evaluated at X over epsilon.
03:13:120Paolo Guiotto: So the idea is, in fact, that… well, let's finish the finish. And we call…
03:24:40Paolo Guiotto: The family of these functions, delta epsilon with epsilon positive, and approximate…
03:36:120Paolo Guiotto: unit.
03:40:460Paolo Guiotto: For example, the family we introduced, the bova.
03:45:740Paolo Guiotto: You remind that they were delta epsilon of X, they were defined as 1 over 2 epsilon, the indicator of the interval minus epsilon epsilon at point X.
03:58:720Paolo Guiotto: Mom?
03:59:770Paolo Guiotto: Now, we can see this, exactly in this form, because,
04:05:70Paolo Guiotto: This condition, this indicator is 1 when X is between minus epsilon and epsilon. That is, if you divide by epsilon, here epsilon is positive, when X over epsilon is between minus 1 and 1. So, this is the same of 1 over 2 epsilon.
04:25:340Paolo Guiotto: the, delta, say, of X over… well, say, the indicator, sorry, indicators…
04:35:240Paolo Guiotto: of minus epsilon, epsilon, evaluated… sorry.
04:41:630Paolo Guiotto: minus 1, 1.
04:44:390Paolo Guiotto: Evaluated at X over epsilon.
04:48:550Paolo Guiotto: Because it is 1 if and only if X is between minus epsilon and epsilon, or X over epsilon is between minus 1 and 1. So, as you can see, this is 1 over epsilon delta.
05:02:900Paolo Guiotto: of X over epsilon, with the delta of X equal 1 half the indicator of the interval minus 1, 1.
05:13:380Paolo Guiotto: And exactly you have that this delta is clearly positive, and the integral of delta, delta is this function.
05:23:830Paolo Guiotto: is the function defined on the interval minus 1, 1, equal to 1 half, in such a way that you have the total area is 1, no? Because it's a rectangle base 2, it's 1 half, the total area is 1, and 0 outside. That's the delta, the function delta.
05:43:720Paolo Guiotto: So this is a rescaling that makes, shrinks the function and rises the peak when we have distal types in.
05:54:920Paolo Guiotto: Other important examples of these functions are the following. This is the so-called
06:03:810Paolo Guiotto: perhaps is the most important. The Goshen… approximate unit.
06:16:700Paolo Guiotto: So we start from delta, which is a standard of the Gaussian, so let's take delta X defined by 1 over root of 2 pi.
06:28:230Paolo Guiotto: exponential minus X squared divided 2.
06:32:10Paolo Guiotto: So the shape of this function is well known.
06:42:680Paolo Guiotto: This is delta X. It is positive, and the integral of this is 1.
06:49:110Paolo Guiotto: It is the Gaussian integral.
06:51:370Paolo Guiotto: And you define delta epsilon, in this case, is equal 1 over epsilon delta of X over epsilon.
07:01:790Paolo Guiotto: If you do the calculation, you arrive to see that this is 1 over root of 2 pi epsilon squared. We write the epsilon inside the router, so it becomes square.
07:13:360Paolo Guiotto: E minus X squared divided to epsilon squared.
07:18:860Paolo Guiotto: Which is, what is a Gaussian distribution with this parameter, epsilon squared, which is the variance of the distribution.
07:27:200Paolo Guiotto: And when epsilon is very small, this delta epsilon is still…
07:33:460Paolo Guiotto: are still a shape similar to delta X, but it shrinks around the edge.
07:40:230Paolo Guiotto: We can actually, build other types of units, because sometimes, doing the convolution, we need to have much stronger properties. For example, we can do another construction is the following.
07:58:490Paolo Guiotto: is a… To take a function that is made like this.
08:03:560Paolo Guiotto: I first draw the pictures.
08:06:320Paolo Guiotto: You see that, the delta, the… this rectangle unit, has the,
08:15:340Paolo Guiotto: feature that is not continuous. There are these jumps.
08:19:10Paolo Guiotto: So, while this is continuous, it's very regular, actually you can compute any number of derivatives, but it has not a compact support. It is always different from zero in the interior line.
08:36:539Paolo Guiotto: Now, it is possible to do a sort of hybrid, so something which is zero of some interval.
08:45:750Paolo Guiotto: And it is extremely smooth, so something like C infinity, which is like this. This is not easy, to do, but it can be done with this delta of X.
08:59:730Paolo Guiotto: Which is equal to 0 when modus of X is greater than 1.
09:05:160Paolo Guiotto: Now, we take, an exponential.
09:08:670Paolo Guiotto: made in this way, 1 over 1 minus X squared.
09:13:930Paolo Guiotto: For models of X, less than… well, let's put the list greater than equal than 1, and here less than 1.
09:20:650Paolo Guiotto: Now, what happens when,
09:23:10Paolo Guiotto: Both is not going to be the definition. The unique product is, what? This is equal 1 is a really nice function, it's a smooth function.
09:33:780Paolo Guiotto: When X goes to lambda, from inside the interval, X squared goes to lambda with a smaller band. So this… then the interval goes to zero positive energy.
09:44:350Paolo Guiotto: No? So this fraction goes to infinity D, plus infinity with a minus minus infinity. The exponential goes to zero. So this makes sure that you have a continuous function.
09:54:260Paolo Guiotto: One of the derivatives, for example, here, This delta can be proved.
10:00:680Paolo Guiotto: That delta is… also a C infinity.
10:05:700Paolo Guiotto: Function in the real line, huh?
10:08:900Paolo Guiotto: It is, of course, positive.
10:11:440Paolo Guiotto: And, actually, we could, we should rescale, put the coefficient C here in such a way that the integral D1, where C
10:21:580Paolo Guiotto: Is such that the integral of delta on the real line is equal to 1, to make this the base function for an approximate unit.
10:33:770Paolo Guiotto: It's just a scaling factor, no?
10:36:90Paolo Guiotto: this one.
10:38:430Paolo Guiotto: This… the fact that delta is C infinity is not easy, but let's just have an idea, if you complete the derivative of delta, of course, this derivative is 0 for bottles of X greater than 1. For X less than 1,
10:57:240Paolo Guiotto: is the exponential, E minus 1 over 1 minus X squared, then there is the derivative of the exponent. Now, when you differentiate the exponent, in this case, it is a little bit easier, because it's an exponent of type 1 over.
11:10:850Paolo Guiotto: So it is, the formula is the square of denominator.
11:19:230Paolo Guiotto: And then we have, on the numerator, the derivative of the denominator. So with the minus, so it is, with plus, should be… however, it doesn't matter, change the sign here. The key point is that when X goes to 1,
11:37:890Paolo Guiotto: what happens? When you send X to 1 from the inside, so model of X goes
11:44:390Paolo Guiotto: to 1 with the smallest values, so it means X goes to minus 1 from the right, or plus 1 from the left. You see that this quantity goes where? Again, the exponential goes to 0, and it goes to zero, let's say, exponentially.
12:00:100Paolo Guiotto: The fraction goes to infinity, because you see that this quantity here goes to zero, so here's 0, and here you go to plus minus 1 divided 0. So, in principle, you should go to infinity plus minus infinity.
12:15:170Paolo Guiotto: But the point is that this exponential is staying up, no? Because if you look at the exponent, you see that it is this exponential side having P2 minus T times, over T when P goes to E3. It's the exponential that is the exponent. So, in fact, this goes to zero.
12:35:150Paolo Guiotto: So, the derivative is horizontal, and it glues perfectly with the derivative outside, which is 0, no? So, actually, this means that delta prime… delta prime of X goes to 0 when x goes to plus minus 1,
12:53:350Paolo Guiotto: And so this makes possible to extend this definition also at X equal 1 and make it a continuous function. You can iterate this argument and show in general that this is
13:05:430Paolo Guiotto: irregular function, but with the difference with respect to the Gaussian, that now it is… it has a support, so the set of values where it is non-zero, which is just limited interval, okay? This is interesting, I will just tell you in words, something later.
13:25:910Paolo Guiotto: Also, in this case, you can construct detail types, you know, which are…
13:31:400Paolo Guiotto: 1 over epsilon, this delta of X over epsilon.
13:36:260Paolo Guiotto: So this is just a shrinking of the axis and rescaling, because you want to keep the integral equal to 1.
13:43:510Paolo Guiotto: Now, what these, units, do, exactly what, intuitively, we actually
13:50:950Paolo Guiotto: have given, no, perhaps you have not given a justification, but what this, this, unit does when epsilon goes to zero, there is an important technical theorem that says the following. If you take a function F, which is in LP,
14:08:850Paolo Guiotto: RDR… And you take any approximate unit, doesn't matter which one, so delta epsilon with epsilon positive, B,
14:22:70Paolo Guiotto: and approximate.
14:26:660Paolo Guiotto: unit.
14:28:320Paolo Guiotto: Notice that this is… delta epsilon is positive, and integral of delta epsilon is finite. So these are L1 functions automatically, no? So this is a subfamily of L1. So you are allowed to do the convolution between an L1 and LP.
14:47:480Paolo Guiotto: So what happens when you do the convolution between F and this family? Then… F star, delta epsilon.
14:57:970Paolo Guiotto: This is now a function of XM.
15:01:490Paolo Guiotto: And, when epsilon goes to zero, the idea is that that epsilon concentrates, around the value zero, and that you derail that composition
15:13:110Paolo Guiotto: in some sense, should be the value of the function, so… and actually, what happens is that this go… this sequence goes to F in LP.
15:26:120Paolo Guiotto: So…
15:27:310Paolo Guiotto: the intuition is that, of course, this is not a proof, is that when you do this convolution, you are doing integral on r of f of y times delta epsilon of X minus y.
15:43:670Paolo Guiotto: DY. No, this is the definition of the convolution.
15:48:840Paolo Guiotto: Which is, wang.
15:53:260Paolo Guiotto: over epsilon integral on r of f of y, and this is delta of X minus y over epsilon.
16:04:580Paolo Guiotto: DY.
16:09:250Paolo Guiotto: Oh, changing variables, so calling this,
16:14:790Paolo Guiotto: Z. This becomes 1 over epsilon. Now, the integration variable is Y, so if we say Z equals X minus Y, it means that Y is,
16:27:850Paolo Guiotto: is X minus Z, so DY becomes minus DZ, algebraically.
16:36:210Paolo Guiotto: then when you do the change of variable, you know that you have to put the absolute value of this, so the integral remains an integral on r of f of y, which is,
16:46:520Paolo Guiotto: X, minus Z delta Z over epsilon DZ.
16:57:50Paolo Guiotto: Now, the idea is that, since you are avid production very small allowance, Alright, heck.
17:07:89Paolo Guiotto: You see… Visa
17:09:819Paolo Guiotto: bigger, and this is big. This integral saves only the value of f of x minus Z meet approximately equal to 0. So that's why this thing is approximately equal to F of X.
17:26:980Paolo Guiotto: this is the rough inclusion. Of course, you have to do a proof, and the proof is quite technical. There is a sort of,
17:36:890Paolo Guiotto: Idea of the proof in the notes, if you are curious.
17:43:710Paolo Guiotto: But that's the point, so…
17:46:430Paolo Guiotto: This way provides a way to approximate an effort with these functions. It does not say on the model that this function… these functions are better, but these are convolution, so they are healthy functions.
18:02:150Paolo Guiotto: Now, the key point is that if you assume that delta epsilon is a good function, regular, for example, the motion kernel, or the other, or the other unity I've shown here.
18:14:370Paolo Guiotto: with the compact supporter, it turns out that that convolution is also regular, is also differentiable. It can be made also to infinity, and this is, in fact, a second different
18:27:250Paolo Guiotto: PRM?
18:28:420Paolo Guiotto: That says… If… For example, Delta.
18:41:530Paolo Guiotto: Ease.
18:42:860Paolo Guiotto: The… Goshen.
18:46:770Paolo Guiotto: Congol… kernel, sorry, the… negotiate… Approxy it.
18:56:270Paolo Guiotto: unit.
18:58:770Paolo Guiotto: So, in particular, it is C infinity. These functions, F star delta epsilon, turns out to be C infinity functions on R.
19:10:560Paolo Guiotto: For every epsilon possible.
19:13:190Paolo Guiotto: So basically, combining these two facts, and this is because the intuition is the following. If you have to compute the derivative with respect to X of this thing, no, you have to differentiate with respect to X this, the integral of F of Y
19:34:20Paolo Guiotto: delta epsilon X minus y dy. But remind that that delta epsilon, for example, here is the Gaussian, the Gaussian distribution, so it is integral on R, f of y.
19:47:940Paolo Guiotto: For example, here it is E2 minus X minus y squared divided 2 epsilon squared over root of 2 pi epsilon squared for this example.
20:01:200Paolo Guiotto: Now…
20:02:560Paolo Guiotto: you see that we want to differentiate this with respect to this X, which is for the integral, not the integration by one, but a parameter. So here we are, we are in case of this usual power, differentiating an integral depending on the parameter.
20:21:00Paolo Guiotto: Now, the key point is that… If you differentiate formally, This is,
20:28:910Paolo Guiotto: equal to the integral of F of Y, you carry the derivative side, you differentiate with respect to X, you get basically the exponential kernel.
20:42:250Paolo Guiotto: this function here, and then you have the derivative of the exponent with respect to X that yields a factor like there is a 2 that comes down that simplifies the 2 downstairs, so we have something like an X minus y this.
20:58:970Paolo Guiotto: So basically, you are still doing a key convolution with the annual function, which is this one, which is a good function, so this makes sense, and so this means that you actually can come inside and deliver it, because this exponential keeps any power.
21:16:460Paolo Guiotto: Okay? So, this basically is the reason why you can do… you are allowed to do this derivative, and actually you can repeat no matter how many times you like, because anytime you will have a factor of this type that comes down from exponential, so imagine I want to do another derivative, however I get this to power 2, which is still controlled by the exponential.
21:41:30Paolo Guiotto: feels any power out of the feeling.
21:43:120Paolo Guiotto: Okay? So, that's why we can… we can say that if we do the convolution with an approximate unit, like the Gaussian unit, these functions are also similarly in function.
21:59:290Paolo Guiotto: So combining the two, this says that we can always approximate any LP function, which is normally a patch function, without any continuity property, with nice functions.
22:16:390Paolo Guiotto: in the natural metric of the space, in the RP normal.
22:21:410Paolo Guiotto: So this means that if we need… we can always say that we can approximate with the system we like, any empty function with the infinity function.
22:35:10Paolo Guiotto: This, in other ways…
22:38:200Paolo Guiotto: that the set of CP functions into LP is like the set of rational. They are tense, they are everywhere.
22:48:190Paolo Guiotto: Normally, we are used to think that LP function is maxed out a certain sim ability to see continuous function, huh? They're much more.
22:57:860Paolo Guiotto: But the fact, he says that we can approximate in the metric of that space, so we have people who are just normal, every function in the regular function. This is used a lot in differential equations.
23:15:580Paolo Guiotto: Okay, so, since we are a little bit late in the course, I don't insist too much on this part, because we're not going to use, particularly… even if we need somewhere to use this later.
23:31:240Paolo Guiotto: So, let's close here for the moment. We will return on the convolution.
23:36:600Paolo Guiotto: Let's close here this part.
23:41:00Paolo Guiotto: And, let's open the, the new part, which is… Schilbert, Space is…
23:54:730Paolo Guiotto: So, what are in-bet spaces?
23:57:570Paolo Guiotto: Basically, Hilbert spaces are normed spaces where the norm is induced by another geometrical structure, which is called discolor or inner product, okay? So, sometimes.
24:18:170Paolo Guiotto: V… Norma.
24:21:990Paolo Guiotto: is induced
24:27:630Paolo Guiotto: by… Inner product.
24:39:630Paolo Guiotto: What is this? Let's introduce the definition. Now, here, for this part, the real and the complex case are different, slightly different format.
24:50:820Paolo Guiotto: So, let's… V, B, a vector space.
25:00:720Paolo Guiotto: on R, on reals.
25:03:890Paolo Guiotto: So, a function.
25:09:460Paolo Guiotto: That will be denoted by this notation, with the bracket. It's actually a function of two variables, two vectors, defined on V
25:19:980Paolo Guiotto: times V, so on pairs of vectors, with values in R, for this case, is called…
25:30:230Paolo Guiotto: Well, to distinguish the two cases, we will use this terminology. It is called a scalar.
25:37:740Paolo Guiotto: Product.
25:42:570Paolo Guiotto: If…
25:45:880Paolo Guiotto: The following properties also. They smell something like properties for the norm, as you will understand. In fact, there is a… they are tightly related to those properties. If, number one, this property is called the positivity.
26:05:540Paolo Guiotto: When we do the scalar product between F and F, we have always a positive quantity for every F in V.
26:14:660Paolo Guiotto: Number 2 is vanishing.
26:20:710Paolo Guiotto: which says that when F scalar F, is 0.
26:26:240Paolo Guiotto: This is possible if and only if F is the zero of the space.
26:31:620Paolo Guiotto: Number 3 is, So-called homogeneity.
26:39:740Paolo Guiotto: Well, let's say that, Let's call linearity.
26:49:470Paolo Guiotto: This function is linear in the first argument.
26:54:600Paolo Guiotto: So, let's say this… is linear.
27:01:380Paolo Guiotto: in the… First.
27:06:70Paolo Guiotto: argument.
27:10:670Paolo Guiotto: So this means that if we do a linear combination like alpha F plus beta G, and we do the scala product with H, this is the same of alpha F scala H plus beta
27:26:250Paolo Guiotto: G, Scala H, and this is for every alpha, beta scala, so real here, and for every F and G
27:35:550Paolo Guiotto: And nature… Baktos Indy.
27:40:20Paolo Guiotto: And last one is so-called symmetry.
27:47:100Paolo Guiotto: that automatically makes a linear, also the scalar product also with respect to the second variable. So, the scalar product between F and G is the same of the scalar product between G and F.
28:01:70Paolo Guiotto: for every F, G, in B.
28:08:90Paolo Guiotto: So these are the… the characteristic properties of a scara product.
28:16:20Paolo Guiotto: So, remark… combining it.
28:27:250Paolo Guiotto: symmetry.
28:31:60Paolo Guiotto: And what… Linearity.
28:36:280Paolo Guiotto: We… have… that.
28:42:50Paolo Guiotto: This color product.
28:43:930Paolo Guiotto: It's also linear.
28:48:140Paolo Guiotto: linear… in the… Second… variable.
28:57:470Paolo Guiotto: So it is what we call a bilinear function.
29:01:370Paolo Guiotto: It is…
29:04:550Paolo Guiotto: billionaire.
29:08:330Paolo Guiotto: function.
29:11:340Paolo Guiotto: This is not the case for the complex product.
29:16:510Paolo Guiotto: Because you can easily see that if you do F scar alpha G plus beta H,
29:25:600Paolo Guiotto: You can flip the order by symmetry.
29:31:280Paolo Guiotto: So we have alpha G plus beta H, color F.
29:37:770Paolo Guiotto: use linearity, this comes alpha G, scalar F.
29:42:920Paolo Guiotto: Las pita.
29:45:210Paolo Guiotto: H, scalar F, and again, back symmetry.
29:52:30Paolo Guiotto: Alpha F, scholar, G, plus beta F, Scala H.
30:00:780Paolo Guiotto: Okay, examples.
30:07:260Paolo Guiotto: There are not so many examples. In fact, there is a deep reason why there are not so many examples, because all these bases can be identified, basically, in a unique main example. So we have…
30:20:800Paolo Guiotto: The well-known example is that, you know, from linear algebra, RD. So in RD, a vector is made by an array of D numbers, F1, FD.
30:32:250Paolo Guiotto: So G, another vector, is made of components G1, GDE, and the natural definition of scalar product for RB is this one, is you do the sum J1 to D of FJGGJ.
30:53:930Paolo Guiotto: The second and the most important example here is… you can easily verify that the properties for this. The second important example is the case of space L2X.
31:08:380Paolo Guiotto: So, where there is an underlying measure space, XF mu.
31:13:20Paolo Guiotto: In this case, We define them.
31:16:740Paolo Guiotto: for F and G in L2,
31:20:690Paolo Guiotto: We define the scalar product between F and G as the integral on X of F times G, d mu.
31:29:490Paolo Guiotto: Which literally is the extension of this formula, no? To replace the sum with the integral, the arrays are functions, so you may think that this is a sum on top of the extension field of exercise, no?
31:44:170Paolo Guiotto: Actually, here.
31:46:540Paolo Guiotto: to say that this is a well-defined scalar product, we need to justify something, because it is not evident that if F, G are in L2, their product, is integral. Okay, and moreover, there is another important fact. So, this…
32:06:580Paolo Guiotto: We put a 2… yeah, to…
32:09:170Paolo Guiotto: identify this with this case, is a well-defined Scarlet.
32:20:10Paolo Guiotto: product.
32:22:70Paolo Guiotto: But, with the vanishing property, that holds in a weak form. So, with… vanishing.
32:38:750Paolo Guiotto: In the form that if you have F scalar F equal to 0, then you know that we have this trouble with these functions that, if we consider as a vector space.
32:53:700Paolo Guiotto: that… We cannot say that when norm is zero, the vector is zero.
33:00:790Paolo Guiotto: So, it is more convenient to consider as a vector space where there are many zeros, and when none is zero, it means just F equals zero almost everywhere. And the same happens here. F equals zero, mu almost everywhere. So, almost everywhere, respect to the measurement.
33:17:570Paolo Guiotto: Let's, check these, huh?
33:22:170Paolo Guiotto: Well, first of all, it's well-defined, no?
33:26:300Paolo Guiotto: why I have to, to do this, to discuss this, is well defined.
33:35:720Paolo Guiotto: Because… If… F and G are in L2,
33:45:30Paolo Guiotto: So this means that, integral of F squared mu
33:50:860Paolo Guiotto: And the integral of G squared d mu is finite.
33:57:750Paolo Guiotto: I can say that also F times G is integral, Then, F.
34:04:780Paolo Guiotto: times G.
34:06:330Paolo Guiotto: is integral, this means belongs in L1,
34:09:850Paolo Guiotto: If you want, because we know that modulus of F times G is less or equal, so if we write as modulus of F times modules of G, there is this elementary inequality, AB less or equal than 1 half A squared plus B squared.
34:29:230Paolo Guiotto: So this is less than 1 half modulus F squared plus modulus G squared. So we are sure that integrating this integral of modulus FG in d mu
34:41:949Paolo Guiotto: This will be less or equal than 1 half integral on X modulus F squared delu.
34:49:620Paolo Guiotto: plus integral on X modulus G squared emu.
34:54:420Paolo Guiotto: And since they are both finite, that integral is finite, so this is finite.
34:59:410Paolo Guiotto: And this implies that this integral here
35:02:740Paolo Guiotto: is finite, and this is F times G belongs into L1, so the integral makes sense.
35:12:430Paolo Guiotto: it is easy to check that the properties of the Scarlet product are verified.
35:20:140Paolo Guiotto: Let's check…
35:25:930Paolo Guiotto: The… characteristic.
35:35:60Paolo Guiotto: Properties.
35:38:170Paolo Guiotto: So, positivity is straightforward, because evident, right? When you do the scalar product F times F, you have the integral of F times F, so the integral of F squared, the mu, which is clearly greater than 0 for every F in the space L2.
35:57:680Paolo Guiotto: And here we come to the vanishing. If F scalar F in the two product is 0, this means, as we see above, that integral of F squared d mu, is equal to 0.
36:12:30Paolo Guiotto: Now, F squared is a positive, measurable function, so when we have this, we know that this happens if and only if F squared is equal to 0 almost everywhere.
36:22:180Paolo Guiotto: This is the prospective measure mu. So, F is equal to 0 almost everywhere, and we have the vanish.
36:29:500Paolo Guiotto: The linearity and symmetry are… the Nietzi is straightforward.
36:36:370Paolo Guiotto: No, it… Brilliant.
36:41:230Paolo Guiotto: forward.
36:42:680Paolo Guiotto: And, this symmetry is evident now, because if you do F scalar G,
36:48:900Paolo Guiotto: This is the integral of F times G, d mu, but you can also write this as integral of G times F d mu, which is the product of F, G with F in energy.
37:04:700Paolo Guiotto: Okay.
37:05:870Paolo Guiotto: So this is the definition of a scalar product. What is now the Hermitian product, which is the analogous concept, but when the space is based on complex scalars? We have a different definition.
37:24:560Paolo Guiotto: If, V… is vector.
37:29:480Paolo Guiotto: space on C, we have…
37:38:610Paolo Guiotto: a different…
37:45:140Paolo Guiotto: definition.
37:48:230Paolo Guiotto: Well, this is because,
37:51:400Paolo Guiotto: We need. We, we need,
37:57:750Paolo Guiotto: a different definition, because imagine that we can do the same definition, we consider the same definition, but with scalars that are complex numbers.
38:07:720Paolo Guiotto: Notice what happens. You take a vector F, and according to that definition, this quantity should be positive, right?
38:17:870Paolo Guiotto: Now, imagine that you take IF.
38:21:750Paolo Guiotto: So again, this should be positive, because you have the same vector multiplied by itself.
38:27:260Paolo Guiotto: Now, if linearity holds, you can carry outside this coefficient, i, from here and here. And this would be I square scalar product of F with F.
38:40:200Paolo Guiotto: Which is exactly minus F, Scala F.
38:44:160Paolo Guiotto: And this would be negative.
38:46:990Paolo Guiotto: So you would have that ayah scholarly. Ayash should be followed by the same name.
38:53:240Paolo Guiotto: This cannot be unless it is zero. So the product should be basically cleaner.
38:58:430Paolo Guiotto: Okay? So you cannot work with that definition.
39:02:50Paolo Guiotto: That's why we need something that it is not linear.
39:06:140Paolo Guiotto: But actually, it turns out it is anti-linear. It means that it is linear in the first argument, it is anti-linear in the second, in the sense that you can carry outside the factor, but you have to take the conduit. This is the correction that has to be made. So, definition.
39:26:860Paolo Guiotto: If, V is a factor.
39:32:950Paolo Guiotto: Space.
39:34:320Paolo Guiotto: on. He… a function.
39:40:390Paolo Guiotto: with a function of two arguments, defined on V times V. This time, this function will be C-valued, ease…
39:52:490Paolo Guiotto: Cold.
39:55:600Paolo Guiotto: And here we use a different name, permitian.
40:03:80Paolo Guiotto: product.
40:07:180Paolo Guiotto: If… Well, the four properties are similar. Number one, positivity.
40:17:200Paolo Guiotto: if the… Following… Properties.
40:27:180Paolo Guiotto: So, the first one is the same, positivity.
40:31:410Paolo Guiotto: I do not rewrite exactly the same. The second one, again, is the same, vanishing.
40:40:970Paolo Guiotto: The third one is the same, linearity in the first argument.
40:46:300Paolo Guiotto: linearity.
40:48:80Paolo Guiotto: in… first.
40:51:880Paolo Guiotto: argument.
40:53:940Paolo Guiotto: But the last one is different. We say that it is anti… symmetric.
41:11:160Paolo Guiotto: So, yeah, we have that if we do F scalar G, this is not equal to G scalar F.
41:18:920Paolo Guiotto: But it is equal, rather, to the conjugate of this.
41:23:300Paolo Guiotto: Okay?
41:26:230Paolo Guiotto: So, here, Well, let's do a couple of remarks before we…
41:35:720Paolo Guiotto: Because this factor of the anti-symmetry changed a bit. So, the point is, is it linear or not with respect to the second argument? If you write F scalar alpha G plus beta F h.
41:54:580Paolo Guiotto: Now, as before, you can say that this is the conjugate, you can invert the order of the product between alpha G plus beta H and F.
42:05:580Paolo Guiotto: Now, we said that with respect to the first argument, it is linear, so we can say conjugate of alpha, G, scalar H,
42:16:820Paolo Guiotto: plus beta, sorry, scalar F.
42:23:570Paolo Guiotto: And here, H, F.
42:27:330Paolo Guiotto: Now, you know that the conjugate of the sum is the sum of the conjugates. The conjugate of the product is the product of the conjugate, so what happens here? You get alpha conjugate times G scalar F conjugate.
42:41:10Paolo Guiotto: Plus, beta conjugate.
42:43:680Paolo Guiotto: H scalar F conjugate.
42:47:180Paolo Guiotto: And again, by symmetry, or anti-symmetry better, this is back to F scar G plus beta, conjugate F scar H.
43:00:00Paolo Guiotto: So, you see that, it's somehow linear, huh?
43:07:130Paolo Guiotto: Put the time to get out of my essentials.
43:09:650Paolo Guiotto: So it means that if the coefficients are real, for example, if alpha and eta are real, so you do have scala 3G minus 4H, this remains because the contributions are the same.
43:22:100Paolo Guiotto: If they are imaginary, you have to change the sign, no? So, for example… When you have…
43:29:410Paolo Guiotto: F, Scala, I, G.
43:33:730Paolo Guiotto: I can carry out this I, but it will come out as I conjugate, so minus i
43:41:150Paolo Guiotto: F, scalar G. It works like this.
43:45:740Paolo Guiotto: Okay.
43:54:670Paolo Guiotto: So, example.
43:57:750Paolo Guiotto: The example number one is the fin-dimensional version that probably you have already encountered in linear algebra, no? If the space is CD,
44:11:970Paolo Guiotto: So, a vector here is an array of complex numbers, F1, FB,
44:19:00Paolo Guiotto: So, FJ are complex numbers, J equals 1 to DE.
44:26:30Paolo Guiotto: And the same for the second vector, G.
44:31:490Paolo Guiotto: Now, the definition of a scalar product for this case, for C,
44:37:670Paolo Guiotto: is similar to the Scarlet product we have for R, but
44:42:440Paolo Guiotto: we take, instead of the product FJGJ, we take the product FJGJ conjugate.
44:49:560Paolo Guiotto: J going from 1 to D.
44:53:110Paolo Guiotto: And the important case for us is the case of the space L2C.
44:59:880Paolo Guiotto: X, so this means the space of functions F, defined on X, C-valued.
45:08:350Paolo Guiotto: Such that the integral on X of modus f squared d mu is finite.
45:18:290Paolo Guiotto: In this case, we define the products between F and G that we still denote with the sub 2.
45:25:390Paolo Guiotto: as integral on X of F times G conjugate dimini.
45:34:540Paolo Guiotto: It is easy to verify, as we have done above, that this product is well-defined.
45:48:140Paolo Guiotto: on L2C… And the vanishing is weak, no? So, with vanishing.
46:03:200Paolo Guiotto: In the form that if you have F, scalar F,
46:08:150Paolo Guiotto: equals 0. So, as you can see, with G equal F, it becomes C integral of F times F conjugate.
46:15:780Paolo Guiotto: F times F conjugate is the absolute value in the complex
46:20:880Paolo Guiotto: numbers, the modulus of F squared.
46:25:340Paolo Guiotto: So, again, when you put the frequencies, you get a…
46:29:330Paolo Guiotto: if and only if integral on X of F, F conjugate d mu equals 0, but that's equal to modules of F squared.
46:40:750Paolo Guiotto: And again, we get the same conclusion, if and only if F equals 0 almost everywhere.
46:51:410Paolo Guiotto: Okay, so these are the two important,
46:55:380Paolo Guiotto: cases. Now, looking at these examples, in particular to the examples of the L2 spaces, In… D… Examples…
47:13:390Paolo Guiotto: of L2.
47:16:240Paolo Guiotto: It doesn't matter, I do not write R or C, okay? It's the same remark.
47:22:990Paolo Guiotto: we noticed… That, when we take,
47:29:800Paolo Guiotto: F, scalar F, in the L2 product, this comes equal, in both cases, integral of modules F squared.
47:40:40Paolo Guiotto: Which is exactly the square of the L2 norm.
47:46:720Paolo Guiotto: Or, in other words,
47:48:870Paolo Guiotto: The altonorm of F is the root of this color twilot between F and S.
47:57:530Paolo Guiotto: So there is this relation between the Scala product and the norm.
48:02:770Paolo Guiotto: Now, turns out that whatever is the scalar product, this formula defines always a norm on the space. So we have this proposition.
48:14:940Paolo Guiotto: So, let, B, B, vector space, on. R.
48:25:610Paolo Guiotto: or SUE.
48:28:410Paolo Guiotto: And, this be a scholar.
48:35:930Paolo Guiotto: If we are in the real case, or intermedian.
48:40:70Paolo Guiotto: If we are in the complex case, Products.
48:46:790Paolo Guiotto: Then… If we define this quantity, secting.
48:58:920Paolo Guiotto: I know… you notice that I have not said that V is a known space. V is just a bacterial space, where we are a Scala or NFL.
49:08:960Paolo Guiotto: If we define norm of F as B, by definition.
49:14:740Paolo Guiotto: the square root of F, scalar F, for F in B.
49:22:70Paolo Guiotto: Well, it turns out that this is a norm.
49:27:360Paolo Guiotto: Jeez.
49:28:710Paolo Guiotto: Well… defined… Norma.
49:35:660Paolo Guiotto: on V.
49:39:50Paolo Guiotto: So this is also called the canonical norm of the space.
49:44:470Paolo Guiotto: cold, but…
49:48:750Paolo Guiotto: I'm on the top.
49:54:820Paolo Guiotto: normal.
49:57:220Paolo Guiotto: False.
49:58:380Paolo Guiotto: B.
50:02:30Paolo Guiotto: Well, actually, this is a nice,
50:05:50Paolo Guiotto: Result that, we, we look at the proof, because,
50:11:930Paolo Guiotto: We just learned how to use these definitions, basically.
50:19:30Paolo Guiotto: So, first of all, we have to verify that this quantity is verifiable.
50:24:760Paolo Guiotto: But, of course, F-scar F is well-defined.
50:29:170Paolo Guiotto: There is a router, no? And the router requires that the applicant be greater confidential, which is the case, because we are the possibility.
50:38:290Paolo Guiotto: So, from positivity, that holds for both cases, scholar and Dmitian.
50:44:750Paolo Guiotto: Positivity implies that
50:48:00Paolo Guiotto: F scalar F is always greater or equal than zero for every F in V, and therefore, the root of F scalar F
50:59:210Paolo Guiotto: Makes sense, huh?
51:05:240Paolo Guiotto: So… norm of Earth.
51:08:530Paolo Guiotto: is well.
51:11:90Paolo Guiotto: define.
51:13:20Paolo Guiotto: Of course, here we mean… when we… in analysis, normally when we talk about the root, the square root, we mean the positive root of the positive number, no? So it's not plus, minus, it's just plus.
51:27:100Paolo Guiotto: Okay, so… We.
51:30:600Paolo Guiotto: have… Check… D… characteristic, properties…
51:45:390Paolo Guiotto: So, number one, they have similar names for some cases, positivity.
51:52:950Paolo Guiotto: Well, norm of F is the root of F, scalar F, so since we said this is the positive root.
52:03:420Paolo Guiotto: positive root of
52:07:770Paolo Guiotto: F, scalar F, greater or equal than zero, so it is positive. There is nothing to prove.
52:15:210Paolo Guiotto: Second, the vanishing.
52:19:620Paolo Guiotto: This, of course, comes from the vanishing of the product, because norm of F is equal to zero if and only if the root of F scalar F is equal to zero, but when the root is zero, the root is 0 only when the argument is zero. So if and only if F scalar F is 0,
52:39:250Paolo Guiotto: But this is zero because of vanishing. This is the vanishing of the scalar product, or the medium product. They both have the same property. This is possible if an elith is zero.
52:53:340Paolo Guiotto: Homogeneity, Well, this is likely different for the two cases, because for the real case, our case.
53:06:00Paolo Guiotto: you have that, you take norm of lambda, or better, let's use alpha as a scholar, alpha, F.
53:15:130Paolo Guiotto: According to the definition, it is… this is the root of alpha F alpha F.
53:22:280Paolo Guiotto: Now, here you see there is a little difference, because if you are in the RI case.
53:26:730Paolo Guiotto: You can say that the product is linear in both factors, so you can carry out that alpha, and you put in front of the product as an alpha squared.
53:36:650Paolo Guiotto: F scalar F. Now, the root of the product, product of the root, if the two are positive, which is the case, so we have root of alpha squared times root of F
53:46:900Paolo Guiotto: times F, and this is modulus of alpha times norm of F.
53:53:990Paolo Guiotto: for the CKs, There is a little difference here.
54:01:750Paolo Guiotto: Of course, you start writing, it is root of alpha F, star alpha F, but now it is non-linear in the both.
54:09:370Paolo Guiotto: It's linear in the first, anti-linear in the second, so you have, basically, that it comes out an alpha from this, and the alpha conjugate from the second, then you have still F times F.
54:22:390Paolo Guiotto: And now, again, this product, alpha conjugate, is models of alpha squared, and you get the same conclusion.
54:31:310Paolo Guiotto: And number 4 is the triangular inequality.
54:39:200Paolo Guiotto: This is, highly more complex here.
54:44:350Paolo Guiotto: And for the two cases, the C case, it's a bit more complicated, so we will limit to the real case.
54:53:510Paolo Guiotto: So, however, let's start computing, no? So, we have to show that the norm of F plus G is less or equal than norm of F plus norm of G, right?
55:06:170Paolo Guiotto: Now, since each norm here is defined through a router.
55:11:30Paolo Guiotto: It is convenient, as what we do with the L2 norm. We always compute the square to avoid carrying out… carrying around the roots. So we take the norm of F plus G squared.
55:26:50Paolo Guiotto: So, according to the definition, this is the product F plus G, F plus G.
55:34:470Paolo Guiotto: Okay, here there is no difference for the two cases, because it is true that this… the emitium product is not linear in the second variable, but it is additive. If you go back here.
55:48:960Paolo Guiotto: Look, suppose that you put alpha and beta equal…
55:54:420Paolo Guiotto: We say that this being a 55 meter, so particular, if I do F scalar T plus H2, this becomes F scalar G plus F scalar H, so it works. So we can say that, applying, two times.
56:12:580Paolo Guiotto: linearity, ad activity. This is F, F, plus F, G, plus GF… plus GG.
56:27:990Paolo Guiotto: Now, we recognize that the first
56:31:120Paolo Guiotto: And the last term are the squares of the norm of F and the norm of G. So here we have a norm of F.
56:40:440Paolo Guiotto: square.
56:41:830Paolo Guiotto: plus norm of G squared.
56:45:220Paolo Guiotto: Then, here, the story changes, depending on which case we are considering. Because in the real case, we know that G scalar F is the same than F scalar G. So, for the real case, I would write 2FG,
57:01:150Paolo Guiotto: While for the CKs, I will just stop here for the CKs, for the CKs.
57:10:310Paolo Guiotto: We know that when I birth the order, this becomes the conduit.
57:17:270Paolo Guiotto: So I would have extra G plus, extra G quantum.
57:22:150Paolo Guiotto: A number plus its conjugate.
57:25:660Paolo Guiotto: No? Z plus Z conjugated is 2 times the real factor. So, here we would have a slight difference.
57:35:910Paolo Guiotto: Two, real part.
57:38:480Paolo Guiotto: of F. Scalarji.
57:41:720Paolo Guiotto: So, it's a minimal difference, let's say.
57:46:590Paolo Guiotto: Okay, now…
57:48:360Paolo Guiotto: Do you mind of the proof that the L2 norm was a true norm, no? Because, basically, we did the same kind of start, no? We started computing the L2 norm of the sum, F plus G, doing the square.
58:05:120Paolo Guiotto: And we computed, and we had this identity written with integrals, because the first one was integral of F squared, then integral of G squared, then 2 times the integral of F times G.
58:16:180Paolo Guiotto: Now, the key point in that case What is it?
58:22:490Paolo Guiotto: This is… Maybe, yeah.
58:27:390Paolo Guiotto: Yeah.
58:30:550Paolo Guiotto: Seems to be… yeah.
58:33:730Paolo Guiotto: Yeah. This is… and in fact, we could write today F scalar G, right?
58:39:760Paolo Guiotto: No? Now, the key remark was the Cauchy-Sports inequality.
58:45:730Paolo Guiotto: No? That is, modulus of that integral less or equal than the product of the two knobs. This is a story that is related to integrals, but this, with the new interpretation, could be…
58:59:340Paolo Guiotto: which is more general, because for the at-to-face, that's a scalar model, but the scale model makes sense, whatever is the structure. So this could be the scalar between F and G plus the number of F, none of G. Now, it turns out that this inequality holds
59:17:470Paolo Guiotto: I'm sorry, ulcer in, in general.
59:23:710Paolo Guiotto: And this is a lemma.
59:28:610Paolo Guiotto: That we may call abstract… Cauchy-Schwartz inequality.
59:42:320Paolo Guiotto: So this says that the absolute value of this color product between F and G
59:48:490Paolo Guiotto: I hope I am not confusing you. When we talk about the scanner product into a vector space, we are not thinking 2L2.
59:57:10Paolo Guiotto: At the end of this story, it is possible to prove that L2 is basically the unique possible example of in-bet space, but this is another story. Let's say at this point, V could be any set made of any type of vectors, not necessarily a space made of functions that are measurable with respect to a measure.
00:16:840Paolo Guiotto: Okay, so it can be whatever. So, here I'm not thinking, but of course, this is an extension of the other one, so this is less or equal than norm of F,
00:27:810Paolo Guiotto: times norm of G. The fact that we are thinking to a general result is visible here, because I'm not saying to norm of F, okay? This is the norm defined by the scalar product for every F and G in V.
00:45:160Paolo Guiotto: Since it takes time to do the proof of this, it's not particularly complicated, but…
00:51:200Paolo Guiotto: It's an algebraic proof, I will skip, we accept this, and we finish. So, we can continue here.
01:01:260Paolo Guiotto: saying that, since this color product between F and G in absolute value is controlled by the product of the two norms, norm of f and norm of g, in particular, the scar product will be less or equal, no? So I can say that this is less or equal than norm of F,
01:20:40Paolo Guiotto: square plus norm of G.
01:22:710Paolo Guiotto: square plus 2 times norm of F times norm of G. And now I recognize that this is exactly the square of norm of F plus norm of G.
01:37:630Paolo Guiotto: So, at the end, you see that we started from the square of the norm F plus G,
01:44:660Paolo Guiotto: we obtain that it is less or equal than the square of norm of F plus norm of G.
01:52:600Paolo Guiotto: Since everything is positive, we get the conclusion that norm of F plus G
01:59:160Paolo Guiotto: is less or equal than normal BF.
02:02:410Paolo Guiotto: Last norm of G.
02:07:340Paolo Guiotto: What kind of thing?
02:09:230Paolo Guiotto: Mouth.
02:11:610Paolo Guiotto: Before we enter in, some example. It's nice to…
02:20:590Paolo Guiotto: Give a better look to this particular identity we have here, no?
02:26:700Paolo Guiotto: So, remark.
02:29:380Paolo Guiotto: The algebra suggested that norm of F plus G Square equals norm of F.
02:37:960Paolo Guiotto: square plus norm of G.
02:41:20Paolo Guiotto: square plus 2 times F scalar G. This is the real case, otherwise you would have two real F scalar G.
02:53:130Paolo Guiotto: Now, this, which is a formula that comes from the properties of, of,
03:03:390Paolo Guiotto: of the scalar product is a fact that we all studied in high school, basically. This is called the cosine
03:16:330Paolo Guiotto: stealing.
03:18:920Paolo Guiotto: You're in mine!
03:21:360Paolo Guiotto: trigonometry.
03:23:590Paolo Guiotto: No?
03:24:600Paolo Guiotto: So you have a triangle with two sides, no? Let's say that we call them 1F and the other G.
03:32:30Paolo Guiotto: So F plus G is this diagonal, because you can translate, and that's the diagonal. So, let's represent this F plus G. So, this says that the length of the square of this lambda, which is represented by norm of F plus G squared, is what?
03:51:390Paolo Guiotto: The square of this plus the square of this.
03:56:550Paolo Guiotto: Plus, well, as you remind, the story was two times the product of the lens times the cosine of the angle between the two vectors. So this would be, something like F…
04:11:60Paolo Guiotto: let's say… F… it would say that F plus G square equal. F squared…
04:24:90Paolo Guiotto: G squared plus G squared plus 2 times the product of the two lengths, FG, and cosine theta, where theta is this. It's the angle made of the two vectors, right? The two vectors. We don't see this because we see 2S color G, but…
04:42:680Paolo Guiotto: By Cauchy-Swartz's inequality, we know that the modulus of F
04:49:570Paolo Guiotto: G is less than normal F, Normal gene, right?
04:56:260Paolo Guiotto: Now, excluding the generated cases like F or G, or both equal 0, so assuming that they are non-zero, so that the two norms at the right-hand side are different from zero, so if…
05:09:600Paolo Guiotto: F and G are not zero. This is equivalent to modulus of FG
05:17:40Paolo Guiotto: Scarology divided by the product of the two norms, Less or equal than 1.
05:24:310Paolo Guiotto: Well, since this is less frequent than 1, we can always think that it is the cosine of something.
05:32:390Paolo Guiotto: Now, actually, this something is generally interpreted as the angle between two vectors F and G.
05:40:470Paolo Guiotto: This is the case if F and G are, for example, plain vectors, and that's the standard scalar product. You can prove that this quantity is exactly the cosine of the angle. If you want to do that, you can do, for example.
05:55:210Paolo Guiotto: imagine that one of the vectors is along the x-axis, and the other is a generic vector. If you do this ratio with all quantities, you will see that it is exactly the cosine of the angle between these two vectors. This, if the scalar product is the scalar product of R2, and these are vectors of R2,
06:14:10Paolo Guiotto: And this is, of course, the Euclidean norm, which is the norm associated with that.
06:20:480Paolo Guiotto: But we can take this as a definition, so we may think that even if F and J are two functions, for example, of the space L2, so we do not pay to arrows, no.
06:30:300Paolo Guiotto: No? But they are two functions, two vectors of this base. We can detect what is an angle between them. And this angle is defined as, again, your data such that cosine of data is equal to this squad.
06:43:220Paolo Guiotto: Okay, if… provided we are in the real case otherwise, that would be a complex number, so we should take it to a real part of, etc. So, but if we do that, you see that our…
06:56:490Paolo Guiotto: We'll… we'll say that…
06:59:980Paolo Guiotto: We say that the scalar product between F and G is exactly norm of F, normal G times cos theta. Now, so if cos theta is equal to scalar product between F and G divided by norm of F,
07:16:180Paolo Guiotto: norm of G, it is like if we say that the scalar product between F and G
07:21:780Paolo Guiotto: is equal to, norm of F, norm of G,
07:26:740Paolo Guiotto: times cosine theta. So, as you can see, we can make this identity
07:33:900Paolo Guiotto: this identity here, which comes from the algebraic properties of the Scarlet product, exactly as this one, which is a geometrical identity that expresses this remarkable fact called the cosine theorem, no?
07:50:510Paolo Guiotto: And this also gives the idea to introduce a new concept. A particular case is when that angle is 90 degrees, so pi half. In that case, cosine of theta is 0. So we introduce this definition.
08:07:520Paolo Guiotto: It's very important and efficient. We say, that.
08:14:430Paolo Guiotto: Two vectors, F and G, R.
08:17:720Paolo Guiotto: orthogonal.
08:24:90Paolo Guiotto: If…
08:26:359Paolo Guiotto: 10 scar product, F scalar G, is equal to 0, because F scalar G equals 0, this means cosine theta equals 0, and cosine theta is zero, all for theta equals 90 degrees, or,
08:44:479Paolo Guiotto: It's minus 17 degrees, and it's on, okay? So they are perpendicular.
08:50:20Paolo Guiotto: In this case, we may remark another interesting fact.
08:54:520Paolo Guiotto: In this case.
09:02:330Paolo Guiotto: seen above. Now, what happens to that geometry? It becomes like this. You have F,
09:09:729Paolo Guiotto: you have an orthogonal vector, G, so this angle is the 90 degree angle.
09:16:630Paolo Guiotto: And this is the vector F plus G.
09:20:279Paolo Guiotto: Now, this triangle is rectangle.
09:23:689Paolo Guiotto: And, the identity we have seen above.
09:27:920Paolo Guiotto: if F squared G is 0, becomes just…
09:33:760Paolo Guiotto: norm of F plus G squared is equal to norm of F squared plus norm of G squared
09:44:170Paolo Guiotto: Which is what… something that we know since, the, more or less, primary school.
09:50:420Paolo Guiotto: You see, what is it?
09:54:260Paolo Guiotto: Exactly, it's the Pythagorent, yeah.
10:02:990Paolo Guiotto: So we call it Pythagorean theorem, even if these are vectors, so they can be whatever. They are not necessarily… we are not talking about the triangle in the plane, you see? But it has the same kind of…
10:19:440Paolo Guiotto: interpretation.
10:21:60Paolo Guiotto: Another,
10:23:370Paolo Guiotto: nice, there are… it's plenty of this kind of geometric… you… you start smelling that there is some geometric flavor behind this structure, okay?
10:34:670Paolo Guiotto: Because now we are talking about angles, orthogonal, things like this. Another remarkable fact is this,
10:44:170Paolo Guiotto: Nice identity, which is called the parallelogram.
10:49:730Paolo Guiotto: Para le grand.
10:54:770Paolo Guiotto: IED.
10:56:170Paolo Guiotto: This says that, Of course, lead… V… B… Effective base… They keep the… weed.
11:18:180Paolo Guiotto: And, well, let's say in a product, it doesn't matter if it is real or complex in a product.
11:28:630Paolo Guiotto: De Norma, B, D.
11:33:110Paolo Guiotto: canonical.
11:37:580Paolo Guiotto: Nor.
11:39:450Paolo Guiotto: So the root of, norm of F is root of F scalar F.
11:44:610Paolo Guiotto: Then, the following identity… happens, which is, norm of F plus G.
11:53:260Paolo Guiotto: square plus norm of F minus G.
11:56:880Paolo Guiotto: square is equal to 2 norm of F.
12:01:180Paolo Guiotto: square plus norm of G.
12:04:410Paolo Guiotto: square. And this solves for every F and G in B. This identity is called the parallelogram identity because its geometrical interpretation is the following. Imagine you have F and G, like that.
12:20:220Paolo Guiotto: Now, F plus G is this vector.
12:26:150Paolo Guiotto: You're doing the sum, no, with the…
12:31:620Paolo Guiotto: physical roots, no? There's the sum of forces, so this is F plus G.
12:37:830Paolo Guiotto: While F minus G is the sum of F plus minus G, no? Now, if you do minus G, it is like that.
12:46:840Paolo Guiotto: It's the opposite. When you do the sum, you get something like this.
12:51:510Paolo Guiotto: Which is actually this one.
12:55:240Paolo Guiotto: So this factor here is F minus G. Geometrically, it's the same, it's just a parallel
13:01:750Paolo Guiotto: So, you see that here there is a parallelogram, which is this one, so where you have seen F and G here.
13:08:620Paolo Guiotto: And it says that the norm of F plus G squared plus the norm of F minus G squared to the sum of the squares of these two diagonals.
13:18:270Paolo Guiotto: It's equal to 2 times, or better, is equal to the parameter, no?
13:23:120Paolo Guiotto: 2 times the sum of the squares of the two sides, no? F, F, and G.
13:32:750Paolo Guiotto: And actually, this is a simple calculation that we can do.
13:38:660Paolo Guiotto: Because if we do the norm of F plus G,
13:43:150Paolo Guiotto: square, I can use the formula that says this is norm of F squared plus norm of G squared, the cosine theorem, let's call it, plus 2F scalar G. I do in the real case, if you want, you can do in the complex case, it's similar.
14:02:60Paolo Guiotto: And if you do with the F minus G,
14:05:50Paolo Guiotto: is the same, because the norm of F squared remains, then we would have the norm of minus G squared, but that's a norm, so this is a norm of G squared. And finally, we have 2Fs color minus G.
14:22:40Paolo Guiotto: So, as you can see, it doesn't matter if we are in a real or complex case, because that minus is a minus 1, and even the medium product is linear when the coefficients are real. So this is minus 2F scalar G.
14:41:10Paolo Guiotto: Now, you sum these two lines.
14:44:430Paolo Guiotto: And you get the conclusion, because you see, at left, you have the sum of the two norms, or the two… the squares of the norms, or the two diagonals, and at right, you have two times the sum of the squares. So, you have the conclusion.
15:04:20Paolo Guiotto: It's an important identity that,
15:08:740Paolo Guiotto: We need to keep in mind.
15:14:350Paolo Guiotto: Another useful remark is the following.
15:19:160Paolo Guiotto: Fact that we will use.
15:22:270Paolo Guiotto: Proposition.
15:24:560Paolo Guiotto: Now, E, inner product.
15:29:00Paolo Guiotto: Scala or Hermitian is a function on V cross B real or complex value, no?
15:37:70Paolo Guiotto: On this space, now we know that there is a natural norm.
15:42:990Paolo Guiotto: which is induced by the Scala product.
15:45:690Paolo Guiotto: It was outed that, with respect to that norm, this color product is continuous, in which sense? So, any inner product
16:02:370Paolo Guiotto: is continuous.
16:06:260Paolo Guiotto: in each… Awful.
16:11:890Paolo Guiotto: it's… components.
16:18:240Paolo Guiotto: So we call this property continuity. What does it mean? It means that if we have a FN that goes to F,
16:25:440Paolo Guiotto: according to the norm, when, unless it's differently stated, when we talk about norm, and there is an inner product, we mean the canonical norm associated to that product. So imagine that Fn goes to F,
16:40:540Paolo Guiotto: And you consider the product between FN and some G,
16:45:70Paolo Guiotto: And you want to know what happens when you send N to infinity.
16:48:870Paolo Guiotto: Now, you may expect that if FN goes to F, this quantity will go to F, scalar G,
16:56:390Paolo Guiotto: And similarly, if you do G, Scarlet, F, N, this will go to…
17:02:900Paolo Guiotto: G scale F, so it will be continuous in both components. And even more, you can easily see that it is also continuous even if both components vary. So, if also GN goes to G in normal.
17:20:200Paolo Guiotto: and you do the scalar product of Fn. Gn…
17:24:300Paolo Guiotto: this goes to the product of F times G.
17:27:780Paolo Guiotto: So this is important because it says, if you want to pass to the limiter, you can always do that in the scaffold, you have to do in one of the two variables, provided the sequences you have,
17:44:970Paolo Guiotto: They are both a limit in the normal, in the canonical norma, induced by the scalar product.
17:54:440Paolo Guiotto: let's see the little proofer. We… I do the first one. If you want, you can try to do the third one, which is a little bit more tricky.
18:06:940Paolo Guiotto: Just write it. I'm basically saying that it's… the third one, it's like if the limit of the product is the product of the limit.
18:15:190Paolo Guiotto: You see? So it sounds like this, kind of problem.
18:19:670Paolo Guiotto: Now, to show the first one, I take the difference between FNG and the candidate limit, F, scalar G, and do the absolute value of this. And the goal is to show that this quantity,
18:36:00Paolo Guiotto: goes to zero, okay? Now, I can use additivity, linearity. This is the modulus of Fn minus F star RG.
18:47:620Paolo Guiotto: And now we use Koshish water.
18:50:890Paolo Guiotto: That says that the absolute value of this color quota is less than the product of the norms. So norm of Fn minus F times the norm of G.
19:00:310Paolo Guiotto: And now we have the conclusion, because this goes to zero, by hypothesis.
19:10:20Paolo Guiotto: No? Because we know that Fn goes to F in normal, and therefore this will go to 0.
19:19:650Paolo Guiotto: As a consequence, also the absolute value at the left goes to zero. So this is another important fact to know.
19:29:10Paolo Guiotto: Okay, so what is a nilbot space?
19:32:310Paolo Guiotto: definition.
19:35:70Paolo Guiotto: Es… evacu space…
19:41:730Paolo Guiotto: equipte.
19:46:340Paolo Guiotto: we… And in, product.
19:55:560Paolo Guiotto: is cold.
19:59:240Paolo Guiotto: Kilba Space.
20:06:790Paolo Guiotto: if it is complete with respect to the norm induced by the Scala product, or if it is a Banach space with respect to the norm, if
20:19:930Paolo Guiotto: Eat. These.
20:21:860Paolo Guiotto: at Bana.
20:25:840Paolo Guiotto: space, huh?
20:27:820Paolo Guiotto: With respect to the norm, induced the…
20:35:460Paolo Guiotto: Well, let's say, with respect to the canonical norm.
20:42:410Paolo Guiotto: Canonical… Can't be called.
20:47:900Paolo Guiotto: Norma.
20:50:700Paolo Guiotto: norm of F, which is the root of F. Scholar F.
20:57:210Paolo Guiotto: So, example, fundamental example, we already know that LP spaces are complete, we have not proved, we accept, but the fundamental example is just that the L2 space with the real scalas, or L2 space with complex scalas, are
21:17:90Paolo Guiotto: both, huh?
21:19:40Paolo Guiotto: Hilbert.
21:22:960Paolo Guiotto: spaces.
21:29:870Paolo Guiotto: Okay.
21:32:690Paolo Guiotto: So, here there is a…
21:37:850Paolo Guiotto: Some exercise at the end of, the chapter… We have,
21:49:140Paolo Guiotto: 10 minutes, so let's do an exercise.
21:55:110Paolo Guiotto: For example, exercise,
21:59:450Paolo Guiotto: There is not… there are not a large number of exercises here for the moment, because we are basically given the definition of what is a scalar product, a medium product, few properties.
22:13:170Paolo Guiotto: And, introduce this definition of feedback space. But as you can see, the fundamental examples, are basically these two.
22:23:410Paolo Guiotto: And at the end, it's an L2 space.
22:26:740Paolo Guiotto: So it's not like when I have a norm space, I can be the number of norms. I can… I can put on L2 different norms than the canonical L2 norm, but those norms won't come from a Scala product.
22:42:20Paolo Guiotto: Okay?
22:43:190Paolo Guiotto: So, that's something more, because with this kind of product comes the idea of, angle, perpendicularity, so it comes a sort of geometry behind, and normally a norm
22:57:190Paolo Guiotto: contains up to a certain point, because you have distances, but you have not angles. Normally, a norm does not give you the idea of perpendicular. How would you say that two vectors are perpendicular? That's why you need the idea of angle. So here you have some more structure, and this…
23:17:70Paolo Guiotto: create a very, very strong restriction on the kind of spaces you can consider. So there are not…
23:24:220Paolo Guiotto: large number of examples of full bed spaces. At least at this stage. Of course, one can consider complicated spaces, but they are basically based on L2.
23:34:770Paolo Guiotto: So this example says we take V equal the set of functions in C1,
23:41:290Paolo Guiotto: 0, 1, so it's an attempt to take a space of continuous function, basically.
23:49:300Paolo Guiotto: with F of 0 equals 0.
23:53:880Paolo Guiotto: We define this product, F, scalar G, real.
23:59:880Paolo Guiotto: product is the integral between 01, but not of F times G, it is F prime X times G prime of X.
24:10:830Paolo Guiotto: I don't know, but I'm sure that you have already heard about subolem spaces, right?
24:17:550Paolo Guiotto: Okay, so basically this is the construction of the Sobel space, no? This would be a natural product on a space which is made of L2 functions.
24:28:440Paolo Guiotto: that you have a derivative in a weak sense, same almost everywhere, and this derivative is itself an L2 function.
24:37:420Paolo Guiotto: Since we have not yet introduced this concept here, I have to cheat a bit, so I take the space of C1 functions, which is a lot more than what I'm saying, because C1 functions is they are continuous, so they are L2.
24:50:610Paolo Guiotto: and derivative continuous, it is L2. So, in particular, these functions are such that F and F prime are L2, okay?
24:59:970Paolo Guiotto: But as you see, this is more, because the derivatives are continuous as well as the function, so it's a bit more than just F, F prime in L2.
25:09:430Paolo Guiotto: The space where FF prime is in L2 is the smaller space, but you need to understand what does it mean F'. I will return on this later, but now let's do this cheap version.
25:23:320Paolo Guiotto: So what we have here is one check… Yeah, this is, scalar product.
25:38:220Paolo Guiotto: I have not said, but the scholars here are real. And, it asks, determine if V is an alert pace. Is V…
25:47:920Paolo Guiotto: Killbert.
25:50:910Paolo Guiotto: space.
25:53:210Paolo Guiotto: Well, let's see quickly a solution.
25:57:370Paolo Guiotto: So, number one, we have to check that this is well-defined, and it verifies the
26:02:420Paolo Guiotto: characteristics of a Scala product. First, it is well-defined. So, F, Scala G is… well… defined…
26:14:90Paolo Guiotto: If this is not clear yet, well-defined is because whenever you define something, unless it is trivially well-defined, you have to justify. Here, there is an operation which is non-trivial, the integral with derivatives, because
26:31:650Paolo Guiotto: what we know is that FG are C1, so in particular, the derivatives F prime and G prime, are continuous, so that product, F prime G prime, is a continuous function on 0, 1, and therefore, the integral is well-defined.
26:54:280Paolo Guiotto: I know, it's a formal passage, it's nothing particular, but sometimes it might be non-trivial.
27:04:210Paolo Guiotto: Let's check…
27:09:950Paolo Guiotto: The… characteristic.
27:13:680Paolo Guiotto: Drop of this.
27:16:930Paolo Guiotto: So, number one, we have positivity.
27:20:320Paolo Guiotto: If I do F scalar F, this is integral 01 of f prime times f prime, so f prime squared.
27:29:30Paolo Guiotto: And this is clearly greater or equal than zero.
27:32:770Paolo Guiotto: Vanishing.
27:35:290Paolo Guiotto: If F, scholar F, is 0,
27:39:750Paolo Guiotto: This means that integral 0 1 of f prime squared is 0.
27:45:350Paolo Guiotto: Here, we have the usual factor that integral of a positive function equals zero.
27:52:480Paolo Guiotto: In this case, we can use the strong version, because the function here is the square of F', F prime is continuous, the square will be continuous as well. So this is a continuous function on 0, 1, so we can say that this function, F' square, is constantly equal to 0.
28:12:110Paolo Guiotto: Not almost everywhere, only, but everywhere. So this means that F prime is constantly 0, and in particular, also here, there is behind a non-trivial factor. The fact that from derivative equals 0, F is constant, is a non-trivial stuff.
28:28:50Paolo Guiotto: It depends on the Lagrange theorem, so it's something that comes… even if it seems evident, no? But this is just because we are working on intervals. If the function is defined on two intervals, derivative equals zero.
28:40:590Paolo Guiotto: can come with the function known constant. You take two constants, one equal plus 1 for X positive, minus 1 for X negative. It's not constant globally.
28:50:890Paolo Guiotto: Okay, but derivative is 0. So, here, there is always something behind non-trivial, so this means that F is equal to constant on the interval 0, 1,
29:02:480Paolo Guiotto: And now, we have to remind that in the definition of V, there is this condition, f of 0 equals 0.
29:10:10Paolo Guiotto: And, since,
29:14:750Paolo Guiotto: F belongs to V, F of 0 is 0. This implies that this constant is 0.
29:21:650Paolo Guiotto: So, at the end, F is identically equal to 0, which is the zero for this space. Number 3 is evident, because… no, number 3 is the linearity.
29:33:780Paolo Guiotto: linearity, no? If you do alpha F plus beta G, scala H,
29:42:510Paolo Guiotto: This is integral 01 of derivative of alpha F plus beta, G times H prime.
29:51:890Paolo Guiotto: Derivative is linear, so derivative of a linear combination is a linear combination of derivatives.
29:59:50Paolo Guiotto: And you multiply all this by H prime, and clearly you see that it becomes alpha integral s prime H prime, which is…
30:07:90Paolo Guiotto: F scalar H here, plus theta integral of G prime h prime, which is G scalar H here.
30:15:590Paolo Guiotto: And number 4, symmetry.
30:22:500Paolo Guiotto: it's evident, because F-scar G…
30:25:840Paolo Guiotto: You see, these integrals are 1F prime, G prime, you can flip the order, G prime, F', and that's G squared H, F.
30:37:470Paolo Guiotto: So we are done, no? We did check.
30:40:880Paolo Guiotto: Now, let's come to question number two. Is V… is V an Hilda space?
30:57:800Paolo Guiotto: Now… It might be difficult if you are… if you have not yet a long experience with these things.
31:07:270Paolo Guiotto: But…
31:08:530Paolo Guiotto: what is the norm of this, that this guy induces? Well, norm of F, the norm induced by that product, root of F scalar F, so let's write the square of the norm.
31:23:70Paolo Guiotto: This is F scalar, F is the integral 01 of F prime square. So, this is the L2 norm of F'. That's what is this norm.
31:35:980Paolo Guiotto: So…
31:38:40Paolo Guiotto: when… whatever is the idea of a limit, this will be more or less a limit in L2, no? FN goes to F in this norm.
31:51:970Paolo Guiotto: If and only if the norm of Fn minus F, which is the L2 norm of Fn prime minus F prime.
32:02:180Paolo Guiotto: goes to zero.
32:04:80Paolo Guiotto: So you see that a limit, F, here, is basically an F such that the sequence of the limit of mass converges in L2.
32:13:630Paolo Guiotto: So, if I take the derivatives, it is basically a negative space. The problem derivatives, these negative space. Actually.
32:22:90Paolo Guiotto: They are not continuous, the realities are continuous. So it is like if I'm saying, okay, I have that the realities are continuous functions.
32:35:270Paolo Guiotto: And they should converge in H2, which is an integral naught.
32:40:720Paolo Guiotto: Does it remind anything to you?
32:45:520Paolo Guiotto: So this is a square.
32:47:360Paolo Guiotto: So we already met something like this. We have not seen particular example of completeness, but we have seen something…
32:58:430Paolo Guiotto: Let's see, where is it? So, that's the Cauchy property.
33:05:200Paolo Guiotto: Here we are. I just took a run. This is a space of continuous functions. Okay, it's not the space, we have C1. Yeah, but forget of C1. We have… we are dealing with F' that must converge in L1, L2 in that case.
33:21:260Paolo Guiotto: But let's look at this example. This was… you have a space of continuous function.
33:28:340Paolo Guiotto: And, you will keep this with the L1 norm.
33:32:130Paolo Guiotto: And we have seen that we can build a sequence, like the green functions.
33:37:350Paolo Guiotto: that is a sequence in that space, that is not convergent in L1 norm, but it is a Cauchy sequence. That's because, actually, this sequence is really convergent in L1.
33:49:470Paolo Guiotto: No? That sequence converges in L1. The problem is that the limit is not… not is not in L1. It is not in that space, it's not a continuous function. That's the problem. So we may build something similar. Or even, why don't we take the same example?
34:04:540Paolo Guiotto: But be careful not with these… these are not the FN, these must be the derivatives, okay? So, the idea is, let's build an example where this FN are the FN prime. So, FN are… the integrals of this, are the primitives.
34:21:640Paolo Guiotto: You don't even need to write the FN, because your norm is working only on the FN prime, so who cares? So, let…
34:34:240Paolo Guiotto: Well, here it is 0, 1, so we have to adapt the example. Let Fn prime
34:42:380Paolo Guiotto: Okay? Be the following. Let's do graphically. This is 01.
34:47:110Paolo Guiotto: So my Fn prime is this. It is, I don't know, minus 1 here, there is the midpoint one half somewhere, plus 1 here.
34:56:630Paolo Guiotto: And we go strike in that way.
34:59:260Paolo Guiotto: with the idea that this, this part will become steeper and steeper, no? So this is my FN prime.
35:09:170Paolo Guiotto: Now, this is… this function is the derivative of something you integrate, and you have to put the constant in such a way that you make the value at 0 plus 0. So you define FN
35:23:100Paolo Guiotto: of x equal, for example, integral from 0 to X of this FN prime, Y, DY,
35:33:190Paolo Guiotto: So now, this function, by definition, has derivative Fn prime, and you add the constant plus constant. Well, actually, this one, it's okay, because at 0 is 0. And you put X equals 0,
35:49:410Paolo Guiotto: you have value 0, as you want for this space, no? So for this FN, FN at 0 is equal to 0.
36:00:00Paolo Guiotto: We have that this FN is continuous, the derivative of Fn is the function that you see here, which is continuous, so that FN is C1.
36:13:950Paolo Guiotto: I could compute that is FN, because FN is a function, so if I want, I can write the formula, but it's not really useless, okay? I prefer to deliver the notation like that. Now, so I know that this sequence is a sequence contained in V.
36:32:760Paolo Guiotto: Now, if you look at the norm of FN minus FN,
36:39:910Paolo Guiotto: in the space. Now I'm trying to show that it is… convince you that it is a Cauchy sequence. This is the L2 norm, so that's square, the L2 norm of Fn prime minus Fn prime.
36:53:150Paolo Guiotto: too north.
36:56:20Paolo Guiotto: Okay? But these, these F, if these are the FN prime.
37:01:190Paolo Guiotto: Well, the bound for the two norm is the same, integral 01 modulus fn minus FM squared. You remind what was the idea? The two functions are the same, except on a small interval of length at most 1 over n.
37:18:560Paolo Guiotto: Where the difference is bounded by 1, so I can say that this will be less or equal than, integral form 1 half minus 1 over n, 1 half plus 1 over n, 1 squared, if you want, dx.
37:34:900Paolo Guiotto: But at the end, this is, again, 2 over N.
37:39:390Paolo Guiotto: Okay, and is 4M larger than N.
37:43:760Paolo Guiotto: So this means that that difference can be made smaller as you like, provided that they are big enough. So this means that this sequence is a crochet sequence.
37:55:680Paolo Guiotto: So this, FN… is a. Cause she.
38:04:590Paolo Guiotto: sequence.
38:06:620Paolo Guiotto: on V… But it is not convergent on V.
38:12:450Paolo Guiotto: But… FN cannot be convergent to F according to the norm, because this would be equivalent to say that
38:23:570Paolo Guiotto: with this element in V, that, this Fn prime
38:30:240Paolo Guiotto: minus F prime in altenome goes to 0.
38:35:670Paolo Guiotto: Now, what is the limit of DFN prime? It's the same of the limit we have seen in the counterexample. It's this function. It's a function which is minus 1 from 0 to 1 half, and plus 1 from 0 to… so let's call it G.
38:54:660Paolo Guiotto: So this happens with the GE.
38:58:30Paolo Guiotto: Which is, however, not an element of V. So you'll remind, with that example, we proved that.
39:05:780Paolo Guiotto: this FN… In this case, this FN con convert us to that G, but that G…
39:12:860Paolo Guiotto: is not the derivative of someone in V, because it's… it is not continuous. If it would be a derivative, it should be
39:20:470Paolo Guiotto: a continuous function.
39:22:660Paolo Guiotto: So, the conclusion is that, V is not.
39:28:580Paolo Guiotto: Kilbert space.
39:31:670Paolo Guiotto: Okay.
39:33:530Paolo Guiotto: That's…
39:37:490Paolo Guiotto: Well… Do some of the exercises.
39:43:200Paolo Guiotto: 13… 3… From 2 to… Well, in principle,
39:58:60Paolo Guiotto: 27.
40:01:210Paolo Guiotto: Okay.