 Vzaj sem zelo vizio na zelo sej tudi, ki je vse transformočne. To je zelo veliko zelo, da smo začali pačenje in začali. To je veliko zelo, da smo vse naprej počantili, da smo vse naprej počenje. To je izgleda, da so vsi vse načinj, da je zelo. Počekaj smo prišli všim definitivnih. Počekaj, da vse je vse funkciju v L1, vse je vse vse vse, in izgledaj, da je vse funkcije kompleksne valide. Tako, težemo, da je bilo kompleks, da je bilo kompleks. Zdaj smo zelo. U hat xi. Na definicijstvo je integral n, xi x ovex dx. Tako, za vse ksaj in tvoj kopi na n, vse vse skala produt in tvoj ukledija, vse ksaj in tvoj x. Nelj je tvoj kompleks eksponentij, tvoj minus e ksaj x, tako vse malo tvoj minus. In tvoj vse vse vse vse u. ... ... ... ... ... ... ... V normu 1, zato je norma nekaj, da je u of x, kaj je v zemljenju v l1. Zato je vzelo, da je vzelo, da je vzelo, da je vzelo, da je vzelo. skovalo. To je vzgledajto. Tukaj, in na dimension 1 to je hat x je počas x u of x vx ker je odpravila z vsej vsega formula, minus i integral r, psi x, u of x, px, ok. To je zelo tako... Zelo, da u jezmi... Vzelo, da u jezmi zelo, kaj je vse, če je odd vs nekaj, če je even, če je zelo. Vse imamo ksi u hat, ki je vse. Kaj je u odd, nekaj zelo. Vse je tukaj, če je tukaj. In, če si u je del, tudi je to več v Leah'u z zelo in nočne infine. Ne? Skroj. We want to integrate on the positive half line then just multiply this by two. Exercise in one dimension, take u, which is the characteristic function of the interval minus one one. The it is clearly an nl1 function and we want to compute u of xi, u hat of xi. And well, it is not difficult because u in that case is even. A, ultimately, we have that this is just the integral over r of cosine xi x u of x dx. Now, if x i is equal to 0... Well, if x i is equal to 0, the computation is immediate. This is just only 2. This is … we know is equal to, say twice. zelo, da si je početnja, danes je tudi početnja v poživu ljenju, ali poživu ljenju mu je početnja v poživu ljenju 1 in vzeloj z 1 in vzeloj z 0. Tak je početnja vzeloj v 1 vzeloj z 0. Oko vzeloj s cos sign. Neste, da se je zeloj s 0, z 2 sin xi divided by xi. To je 2 sin xi. If xi is different from 0. Observe that always. Remark maybe in general. If I compute u hat at 0 this is always the integral of u. Which is of course in accordance with this. So at the end u hat has this expression. And so we can make from this simple calculation. We can make already some remark. U is of course with compact support. But u hat has not compact support in this case. So first remark is that u has compact support. But u hat has not compact support. So from the point of view of support there is a striking difference between u and u hat in this case. Another remark that we can do is that u hat is continuous. It is clear. We know that the limit of this as xi was 2 0 converges to 2. So this is a continuous function. While u hat, while u was discontinuous. So again there is a big difference in the regularity between u and u hat in this case. Then another fact is that it is continuous but it is infinitesimal at infinity. Limit of this xi goes to 0 is 0. Like xi goes to infinity is 0. U hat is continuous infinitesimal at infinity. The next exercise that I would like to leave you as homework slightly more general than this exercise number one. So this is home in one dimension. Take u, take two real numbers, one less than the other. And take u, the characteristic function of the interval a, b. So the previous case corresponds to a to equal minus one and b equal one. Now we want to slightly generalize this result. Then you have to compute this. Well, still this is surely true. And then we have something like one. One minus a, b. Something like this, a and b. So try to do this at home. So it is very important in Fourier transform make several exercises. So we will do a lot of this. U equal to other exercise. So the heavy side function e to the minus x. Remember that the heavy side is 1, 0. So u psi is what? Is the integral over r e to the minus x. By the way, maybe in several books there is a coefficient in front of this integral. In this case we have taken the coefficient equal to one. It is just a matter of normalization. So it may happen that in some texts you find something like 2 pi to some power depending on the dimension. Doesn't matter. We take the constant equal one. H of x between 0 and plus infinity. And this is, we are in one dimension. So this is simply psi times x minus x. So this is actually nothing but e to the minus x times 1 plus e psi. Which is equal to 1 divided by 1 plus x psi. So as you can see, as we can see from all these examples, the output is a complex valued function. So in general maybe it is better to think of u of the function u as say complex valued. Now try to do at home compute the Fourier transform of, sorry this is u of x. So v of x equal e to the xh of minus x. E to the xh of minus x. And then using, and you will see that this, the result will be, the result will be 1 divided minus i xc for an xi in R. And once we have done this exercise, compute this Fourier transform of this. Compute w hat of xi. W hat of xi. And which is how to compute this. Well you are right. Thank you. We are also, because here this is, what is this? I mean this is 0 for positive x and minus 1 for positive, for negative x. So this is in l1 and then compute this. Yes. Remember that you must be in l1 always for the moment. OK. Other examples which are maybe, maybe which are, so I was saying how to compute this with the minus. Well you can use the previous results, because you can split this as the sum of this. E to the minus xh of x plus e to the xh of minus x. OK. And so I mean it is now, now you can see that if you take w hat of xi, by definition you have that this is the sum of the w hat of this plus, is the sum of the hat of this plus the hat of this. And so you have simply to sum up this at the end with this. Value of w hat, w hat, this e to the minus x. What is the question? The value of a function in l1 is not, it is irrelevant, u is in l1. I mean it is irrelevant, the field definition of u at one point. Yeah, yeah, whatever. OK. So there is however one much more interesting computation in the exercise maybe we could try to do together. Or at least I would like to leave you as homework, but this is much, much more difficult than the previous exercises. So let us make some space. So the exercise is the following compute, the Fourier transform of 1 divided by 1 plus x squared. So u of x in one dimension. So u of x is equal to 1, n is equal to 1, and u of x is 1 divided by, which is of course in l1 of r. So now we have the following problem. So we have the following problem, which is compute the following integral e to the minus x i x divided by 1 plus x squared dx. Do you know how to compute such a kind of integrals? So everybody knows? Is it so? So let me give you just for those that maybe do not remember completely the details, let me give you some hint in order to make this computation. OK, first of all, well, this is even. OK, so u is even. And therefore remember that u hat implies that u hat is also even. Do you agree with this? If you remember the expression in one dimension of u hat, which was the integral over r of u x cosine of psi x, this is clearly even with respect to psi. OK, so u is even. So u hat is even. So we can suppose either psi positive or psi negative. Take for instance psi negative. This is the limit as r go to plus infinity of the integral over the interval minus r r of e to the minus psi x divided x squared dx. OK, so our integral is now the limit between minus r and r over that integral. So now the idea, so observe that, now the idea is to pass exactly to, as he was saying, to pass to complex numbers. And so to compute this limit passing to the complex plane. So now there is a remark here that if I pass to complex plane, so now if I consider for instance the function 1 divided by 1 plus z squared here, which is the extension of the function u to complex numbers, here there is a singularity, there are two actually. Now in particular I am interested, so this is, so I have two poles, two poles of my function at minus i and i. So now the idea is to do this, say now I want to compute the limit as this interval invades the whole line. But now let me do this, take say for instance this half circle and for fixed r, so let me call this part, this part let me call the c r plus, plus in the sense that it is in the positive part of the plane like this. So the idea is now to add and subtract to this, so write this as the sum, now I am using sort of standard notation, and then I explain now e to the minus, minus the integral, the same quantity. So I add and subtract the integral of my function considered as a complex with z in place of x. Of course my path here does not touch the pole, and with this notation sum of this interval plus this half circle I simply mean that I am considering the line integrals in the complex. I mean this is a closed curve, so I simply integrate this on this loop oriented this way. And then I have to subtract what I have added before. For any positive r I can do this, there is no limit here in this procedure, r is fixed, everything is converging for r fixed, this path does not touch the singular point, so this can be done. The denominator is non-zero and it's okay for the moment. Now the problem is that I want to pass to the limit this sum and there are two points here. First show that this goes to zero as r goes to infinity, and I have taken xi negative, so minus xi is positive in this case, with my choice. So one point is to show that this goes to zero as r goes to infinity, r goes to plus infinity, and then the other point is to compute what? I mean there is a singular, I mean I have a closed loop, I am integrating a function with one pole, just one pole inside the loop. So I know that this is the so-called residues, so there is a computation of residues. The pole has multiplicity one, so it's not so difficult to compute the residues. So there are two theorems in the residues theorem, which show you that if you have a function which is not holomorphic inside this, it has a singularity, then you can compute, in this case there is a pole of multiplicity one, then you can compute this in some standard way. The other problem is to show that this goes to zero. This is a strategy, one strategy to two. And maybe it is better that I state you a lemma for such kind of integrals. Yes, maybe at least once it is better that I state a lemma. So let r be positive, let omega be positive, and I want to integrate, I mean to bound, and let, let me call it f, we say continuous on CR plus. And then I can e to the omega z, f of z. Now this integral is performed on CR, plus z is z. So now there is a plus here, f is continuous, r is positive, omega is positive. Then there exists a constant, sorry, CR plus, so let me just parameterize it, for instance, more generally than this. So CR plus is the image of t into say rho of t e to the i t, where rho is piecewise c1, just to allow piecewise c1, instead of c1 is just to allow just some two or three or four corners. If I would like to take another path, just to, so this piecewise c1, but it is important that rho does not vanish. So the inf of rho 0 pi is positive. So I can parameterize, maybe I should change symbol. This is for convenience, say just half circle. Here is any curve also like this, something different from the circle. So t is in between 0 and pi, rho is strictly positive, r is big, but positive and big and positive. This never vanishes, so that CR plus does not touch the origin. Then there exists a constant, let me denote it by c star. C star is independent of r, depending only on rho, but independent of r. This is less than or equal, then I have some c star here, then I have maybe omega, and then the supremum of f, over the image of the curve. So the idea is to apply this lem. As you see, I have taken omega positive, so I can exactly apply that kind of lem with minus psi in place of omega. Yes. But the interesting fact is that there is some homogeneity, when you compute this, there is a cancellation between r at the denominator and r at the denominator, and at the end this c star is just something which depends on the infimum of rho, on the maximum of the derivative of rho prime, and so on, but does not depend on r. So I was saying that you could try in doing, so one of the difficulties, of course this is a difficult exercise, it is clear. If you apply now, taking omega equal minus psi, so this is e to the e omega z, like this, f is one over one plus z square, and then you could try to see this. There is one possibility, one possibility to try to use such kind of lemmas. Maybe this can be done also in another way, maybe this can be estimated separately, you don't need this, just to recall you that there is some sort of tool in order to reach such kind of integrals, maybe you can do this directly. You have to show this and then try to compute this. This is one way, maybe. So if you have already done the exercise, it is very good. This lemma is just to recall you that there is some tool to estimate integrals similar to this, similar to this if you want. It is clearly important that if you restrict this one over one plus z square to the big circle, then this goes to zero. You have to use this, clearly. Now, let me introduce just one notation. Notation. If u1 and u2 are two functions in l1, then let me introduce this u1x, u2y for almost everywhere x and almost every y in rn. And then I can try to compute the Fourier transform of this. So the exercise is to show this and show that this is just the product of the two Fourier transforms. And this more generally holds for n copies u1, u2, u3, and so on. You can generalize it easily. Now, corollary. Corollary of what we have seen exercises on corollary of u hat of 1ab hat and the general case of this and star. Check that if u is the characteristic function of a plurirectangle, that is the product of characteristic functions of segments of intervals, just the product. Then your hat is infinitesimal at infinity because we know that if u is the characteristic function of one interval, then we have the explicit expression of u hat, which is infinitesimal. Remember, it's 1 divided i xi times e to the minus i xi a minus e minus i xi b. That is infinitesimal. And therefore, since when I have a product of characteristic functions of intervals, then I have the Fourier transform of the product splits into the product of the Fourier transforms. Each of the element of the product is infinitesimal and therefore this is also infinitesimal. So, now we are in a position to prove the following theorem. So, let u be in l1, then u hat is in l infinity. Moreover, we have that u hat in l infinity is less than or equal than u in l1, hence the operator linear, operator f takes l1 in l infinity and is continuous. So, you see, up to now we have never said which was the target space of f. Now we see that if this is true, surely the target space, the image of f stay inside this. So, I can look at this operator for taking u here and u hat here. And since I have that l infinity norm of the image is less than a constant 1, the norm of the source space by definition of linear operator this becomes a bounded linear operator. Remember our theory on linear operator. Moreover, u hat is continuous and infinitesimal infinity. So, the phenomenon that u hat is continuous is actually more general than the examples that we have seen, it is always true, at least when u is in l1. And also the phenomenon that it is infinitesimal is always true. Now, the difficult part of this theorem there are some easy parts and the difficult part, the difficult part I believe is this one. In any case, let us see that the first part is immediate and remember u high of xi is by definition u hat of xi minus xi scalar product u of x dx and therefore this is less than or equal than this, which is the l1 node for any xi. So, for any xi u hat of xi the norm of this vector is less than this and therefore we immediately have since this is true for any xi we immediately have the estimate. So, the first part is just immediate from the definition. Of course, it is crucial here that x is real. It is crucial that x and xi are real. This is, while u could be complex but x and xi are real. Now, to show that u hat is continuous also we can take a sequence xi n, xi k, a sequence of points converging to xi and then we have to show that to show that the integral sorry, here is an n. We have to show that the integral of e to the minus i xi kx u of x dx converges to the integral over n e to the minus i xi x u of x dx and this is just an application of the dominated convergence theorem because you have pointwise convergence of the integrants and each integrand is bounded because this has norm one each integrand is bounded by a function which is integrable, which is u. So, this is an immediate application of the beg theory, dominated convergence theorem. So, continuity of u hat is always true and therefore u hat is defined as all point xi and is continuous at all point xi. Now, what remains is the infinitesimal part and this one way to prove it is maybe to use density argument and what we have seen up to now namely characteristics of plural rectangle. So, fix epsilon, we know maybe remark. So, if u is the characteristic function of a plural rectangle then the statement is true. Therefore, u hat is infinitesimal. Not only this, but if u is a finite sum up to some index of coefficients and characteristic functions of r i finite sum, then still the statement then u hat is infinitesimal. Now, the idea is to to use the fact that this class of function is dense in l1 and to use this assertion to prove that actually this is true for any u. So, let epsilon positive fix epsilon positive and take w of this form. So, we know that the set of function of this form is dense in l1. So, we can take w of the form 1 of the form. How do you call this function? Step functions. Finite linear combination of characteristic function of plural rectangles. Step in one dimension of the form 1 which is close to u in l1 less than epsilon. So, this is density of this class of function and then there is also the fact that we have to use that now w hat of xi is of course u hat of xi minus w hat of xi plus w hat of xi. Therefore, u hat of xi is surely less than or equal than u hat of xi for any xi. W hat of xi plus w hat of xi. Now, this is surely less or equal less than or equal to the infinity norm of this difference. Less than. This is bounded by by the l1 norm. The previous estimate. But this is small by by assumption. By this assumption, by density assumption. So, this is less than epsilon plus w hat of xi. Well, now it is sufficient to take r so that w hat of xi is less than epsilon for any xi bigger than r. So that for r sufficiently large if xi has norm larger than r we have that this u hat is less than epsilon plus epsilon. Which gives you the final part. So, the difficult part of this result actually is the fact that the u hat is infinitesimal infinity. And for this, our strategy was to reason by density and to look at the Fourier transform of the characteristic of intervals. Fine. Now, I leave you to check the following list of results very useful for computations. We have no time to prove them, but they are easy. So, I write it as a theorem. Yes, but they are actually they are not difficult. They are based maybe on some change of variables. Sort of change of variables. OK, so let me write down just a statement. I am sorry, I don't prove this, but I repeat it's easy. So, let u in l1 and y given in rn and bnn times n invertible, real invertible matrix. Then, the following statements hold one v of x equal the translation by y of u imply e to the minus i xi y u hat xi. So, there is a sort of translations at the level of the original function u transform at the level of the Fourier transforms as multiplication by this kind of exponentials. Plus, yes, you are right. No, no, no, no, no. So, v of x now sort of dual statement e xy u of x then v hat xi, think it is this this sort of transformation within translation and multiplication of complex exponentials. Then three, v of x equal to u absolute value of the determinant of a u hat transposed of a times xi. For v of x equal complex conjugate v hat of xi equal u hat of minus i conjugate other results these are immediate but maybe this is not theorem. So, let u in C1 nu a unit vector, u is in L1 and the partial derivative with respect to nu partial derivative is also in L1 and I add this in order to be sure as x goes to plus infinity then this in particular if n is equal one in one dimension in one dimension I see that if I want to transform I need C1 just to be to define point wise this if and I need this in L1 so that this is well defined so you see that there is this transformation I pass from the derivative to multiplication by polynomial of degree 1 multiplied by E then there is another theorem these are useful to make computations so let u in L1 nu unit vector define v of x as the scalar product x nu times u of x if v is in L1 then there is a sort of dual statement again f of v d over d nu v is in L1 u hat minus i v hat so in particular in one dimension in one dimension the Fourier transform of the product of x times 1 because nu is one say u of x u of x and the Fourier transform of x times u multiplied by evaluated at xi is equal to minus 1 over i minus 1 over i u hat prime of xi that is it i u prime easy to understand to remember this kind of formulas in one dimension so let me just point out so for instance from this formula here you see the following if u is in L1 and this assumption is satisfied and u prime is in L1 then the product of xi and u hat is infinitesimal at infinity so the most regular is your function u say for instance it has u in L1 and the first derivative in L1 say in the sobolev space w11 then not only u hat is infinitesimal u hat multiplied by a linear polynomial is infinitesimal and so on similarly in a sort of dual way if u is in L1 and u multiplied by a linear polynomial is in L1 then you see u hat is differentiable with this kind of statement so there is a sort of relation between differentiable properties of the functions and the fact that the transform is infinitesimal at infinity there is this sort of now exercise that maybe the last one maybe I can give you let me see if I can give you I give you this exercise as homework let n be equal to 1 and u of x equal to e to the minus x2 write the od e satisfied by u what does it mean this more precisely if you differentiate once you find that u prime is equal to minus 2x u do you agree in this sense check that all assumptions in one of those theorems are satisfied so that you can Fourier transform check that you can Fourier transform that you can Fourier transform so what I mean is this minus 2x u let me do this more generally fix a positive number t so let me do a slightly more general exercise fix a positive number t and consider these functions now t is a parameter and then you have this differential equation here ok, check that you can Fourier transform all members you can Fourier transform the left hand side Fourier transform the right hand side Fourier transform all members of this p od e find the solution of the transformed equation now I have to tell you something it is not immediate to transform well how can we transform this Fourier transform we can start from the transformation of e to the minus x square because if not we can make a change of variable say x over 2 square root of t t is positive and all the theorems that have now erased in particular those with the matrix A matrix A was just multiplication in this case by by one over matrix A is in one dimension but more general is just this so using some kind diagonal matrix so you can compute determinant and so on you can use that formula to be able to to make the transform of this once you know the transform of this so you can use this so you can reduce to e to the minus x square so in transforming this means that you have to integrate over r e to the minus i xcx in one dimension e to the minus x square dx and maybe here there is some sort of trick so I have to compute e to the minus i xcx minus x square dx so let me look at this quantity as it was sort of square perfect square it is not so let me add them subtract so this is what I am missing here is square maybe add and subtract 4 minus xi x xi x minus x square so now this becomes maybe here the solution of this you had yes there is a square root of pi I think that well the solution I think is like in one dimension it is a constant maybe square root of pi e to the minus xi square over 4 I think xi square without I mean we are real now so xi square over 4 maybe yes so I leave you as an exercise to proceed to do such a kind of integrals so from this formula you see that almost this function is almost a fixed point of f I mean up to this number here this is e to the minus x square and this is almost e to the minus this divided by 4 so this exponential here is almost sent into itself by the map f almost just up to some constant so maybe I leave you to continue this exercise you should look at this kind of formulas and so I over 2 square minus maybe minus x square minus like xi so this is equal to minus xi square over 4 minus this so you can write this integral integral over R e to the minus sorry e to the minus x plus xi over 2 square this is another of those integrals that can be done using complex analysis now I have no time to do this but try to do by yourself to end up with this kind of exercise kind of exercise well the story is very long this is just the beginning I'm sorry I cannot continue but let you know that this is extremely useful for partial differential equations in any case