 Hello, so we continue discussing those exercises that I left out last time, so there are three exercises that we discussed. First one was exercise 1 that you see in the slides, we explained how the parsimall formula could be used to calculate this integral for sine squared chi d chi upon chi squared over the real line. For the second one I given you a hint, I said if you take this Cauchy distribution little f of s, then this Cauchy distribution is a Fourier transform of some known function capital f of s. Now by now you may have guessed or may have gone and referred to the nodes and you would have found out what this f of s is. Let me give it to you, it is e to the power minus s mod x, s is positive we are going to take s positive e to the power minus s mod x. If you take the Fourier transform of that you are going to get little f of s. On the other hand if you take the Fourier transform of little f of s you are going to get back e to the power minus s mod x because e to the power minus s mod x is an even function and f s of chi is also an even function of chi. So, if you have two even functions f and g then if g is f hat then f will be g hat. So, you take the convolution of f s and f of t and take its Fourier transform. So, f s star f t hat, but the convolution theorem is f s hat into f t hat, but f s hat is going to be e to the power minus mod x, f t hat is going to be e to the power minus t mod x and so the product is going to be e to the power minus s plus t mod x and then we need to take the inverse Fourier transform. That is what you will give you the convolution f s star f t efficiently. Trying to directly use the definition of convolution to compute the integral is going to be messy, but you could try it and you could see whether you get the same result. Then we discuss the Fourier transform of this particular function 1 upon cos a x. I already told you which contour to integrate. You must integrate it over a rectangle in the contour, put down the definition of the Fourier transform. The problem amounts to computing this integral i chi and then you must integrate it over a rectangle using Cauchy's theorem which holomorphic function are equal to choose f of z equal to cos chi z upon Cauchy's a z. a is positive real number. You can take chi also to be positive real and cos chi z upon Cauchy's a z. You take a rectangle as I indicated here and you must choose the height of the rectangle in such a way that the integral over the top is a multiple of the integral over the base. So, z along the top part of the rectangle will be x plus i t. So, you get Cauchy's of a x plus a i t. So, the t happens to be pi by a. If I choose the t to be pi by a, then I get Cauchy's of a x plus i pi which is minus Cauchy's x. So, the new integral that you will get will be similar. Of course, in the top part you will have to use the addition formula for cosine and you will get two terms and the algebra is fairly routine and you can carry it out. Of course, there is one small thing there is a simple pole of the function f of z inside the rectangular contour and there is an isolated pole i pi by 2 a and you have to find the residue of the function at a simple pole. So, how do you find the residue of the function f of z at a simple pole p? You simply have to multiply by z minus p and take the limit as z tends to p. You can use Lopital's rule for the purpose and so that is a routine Fourier transform calculation using complex analysis. Let us take the next problem f and g are in L1 show that the convolution f star g is again in L1 further the L1 norm of f star g is less than or equal to the product of the L1 norms of f and g. It is a simple application of Fubini's theorem. So, what is f star g? f star g of x is integral f y g of x minus y but now we have to integrate again with respect to x. So, it is a repeated integral and you take the modulus of everything and you reduce it to a double integral or switch the order of integrations. The Fubini's theorem has to be used and you will easily get that it is less than or equal to norm f times norm g. Now, if f and g are non-negative then you can say slightly more and I will leave that to you. This inequality is a very particular case of a more general inequality called the Young's convolution theorem. Let me explain to you what the Young's convolution theorem is. If f is in Lp and g is in Lq and I try to take the convolution of f f and g, if I try to find what is f star g, write it as an integral. Integral f of y g of x minus y dy will this integral be finite and if so in which Lp class will it lie? That is answered by the Young's convolution theorem namely it will be the convolution of f and g will be in Lr where 1 upon p plus 1 upon q equal to 1 plus 1 upon r that is the Young's convolution theorem. So, if p is 1 and q is 1 then r equal to 1 will work in fact r equal to 1 is the only r that will work and so it follows from the Young's convolution theorem that if f and g are in L1 then f star g is again in L1. Here I am asking you to prove a very special case of Young's convolution theorem directly using Fubini theorem. On the other hand the other interesting case when p and q are both 2 if p is 2 and q is 2 then 1 upon r must be 0 so r must be infinity. So, the Young's convolution theorem says that if you take two functions f and g in L2 then the convolution should be in L infinity. The product will be in L1 of course, but the convolution will be L infinity that is what the Young's convolution theorem is trying to say. Can you prove that if f and g are in L2 then f star g is in L infinity? Let us start very modestly let us take f and g to be the Schwarz class because the Schwarz class S will be contained in every LP. So, certainly this S Schwarz class is dense in L2. So, let us begin with f and g in S we certainly know that the convolution integral makes sense. So, f star g of x is a well defined real number there is no problem. What we need to do is to get this estimate that f star g of x is less than or equal to the L2 norm of f into L2 norm of g. Can you use the Cauchy Schwarz inequality on the convolution integral f star g of x is integral f of y g of x minus y d y. So, it is a integral over the real line of a product f of y and g of x minus y. If we use the Cauchy Schwarz inequality then you will get this inequality here. Now the next thing that we need to do is to try to see whether this inequality will persist if I resort to some kind of an approximation argument. Suppose I fix x for the time being and I take a f in L2 and I take a g also in L2. Now since the Schwarz class S is dense in L2 I can take a sequence f n's converging to f in L2 and I can take a sequence g n's converging to g in L2 and I certainly know that f n star g n x will be less than or equal to f n L2 norm g n L2 norm. The question is whether these f n star g n x will they converge to some number if they do then that would be the right candidate. Of course then we have to answer a tricky question. Suppose I take two different sequences converging to f and two different sequences converging to g will the corresponding sequences converge to the same limit. All these questions should be thought through and I am leaving that as an exercise for you. The next problem on computing convolutions. So take the characteristic function of the interval minus 11. We have been using this already a few times. Its Fourier transform is what? Two times sine chi by chi. We just used it in exercise one. But now let us take f star f. I want you to explicitly compute f star f. Note that the characteristic function of minus 11 is not going to be continuous. It is not a continuous function. It is discontinuous at minus 11. But when you take f star f you are going to get a continuous function. So that is what you are supposed to check. The next problem is to show that if I take f star f star f that is I take f star f and again take its convolution with f. I am going to get a class c1 function. More generally if I take characteristic functions of i1, characteristic functions of i2, characteristic function of i3, etc. I take k intervals bounded intervals and I take their characteristic functions f1, f2, f3, f3, fk and take the convolution f1 star f2 star fk. Then this convolution k-fold convolution will be of class ck minus 2. You can try to check that. In fact the proof of this last claim is available in Rudin's functional analysis. Now let me now give some background behind this problem. Suppose I ask you to produce a c infinity function with compact support. I want a c infinity function whose support is the interval say 0, 1. How do you go about doing it? One way to do it is let us take for example the convolution of the characteristic function of 0, 1, 0, 1, 4, etc. I keep taking the convolutions of these intervals and these repeated convolutions I can expect that they would converge f1 star f2 star fk. I created a sequence of smoother and smoother functions as I take more and more factors and hopefully the limit will be a smooth function and the support will be the sum of the intervals i1 plus i2 plus ik. And you can use this idea to produce non-trivial elements in the space of smooth function with compact support. Again I refer you to Rudin's functional analysis for such an argument. The next problem sin x upon x is not in l1. Remember integral sin x upon x from minus infinity to infinity is a conditionally convergent improper Riemann integrable. The mod sin x upon x diverges. So sin x upon x is not an l1 function but it is an l2 function. Sin squared x upon x squared is integrable. So sin x upon x being in l2 has a Fourier transform. How do you find the Fourier transform of sin x upon x? Because we know by Plancherel's theorem that sin x upon x will have a Fourier transform because it is in l2. Now one way to do it would be to simply proceed formally. Simply take sin x upon x multiplied by e to the power minus i x chi and integrate. Sin x upon x is an even function. So the integral will be integral from minus infinity to infinity sin x upon x into cos x chi. Sin x cos x chi. Try to use a de-factorization formula. If you try to use a de-factorization formula and try to perform the integration. Of course you will run into some difficulties. You could try to see whether you can use the e to the power minus epsilon x squared trick to get around the difficulty. Again I leave the amusing calculations for you to check and you may actually get the result. Now that result ought to be the characteristic function of minus 1 1. Because that is what the inversion theorem will predict. The inversion theorem will predict that the Fourier transform of 2 sin x upon x must be the characteristic function of minus 1 1. You can see whether this formal argument with e to the power minus epsilon x squared trick produces the characteristic function of minus 1 1 with a half factor thrown in. Next is the Titchmarsh convolution theorem. Suppose f and g are two functions in L1. And assume that f vanishes outside the interval i and g vanishes outside an interval j. Then you have to show that the convolution f star g vanishes outside the interval i plus j. Remember i and j are intervals in the real line. So what is the symbol i plus j mean? It means the set of all points x plus y where x is picked from i and y is picked from g. Set of all points x plus y such that x is in i and y is in j that is i plus j. And you have to show that the support of f star g is contained in support of f plus support of g. In other words f star g will vanish outside the set support of f plus support of j. This is a inequality. Now you will wonder will this inequality become inequality? Will the inclusion be strict? Answer is yes. The inclusion will in general be strict. But what is remarkable is that if I take the convex hull of support of f and convex hull of support of g and convex hull of support of f star g and work with the convex hulls then you get an equality. Convex hull of support of f star g equals convex hull of support of f plus convex hull of support of g. The next problem phi is in the Schwarz class and I cook up a new function summation n from minus infinity to infinity phi of x plus 2 pi n. You must go back to the first chapter and recall the derivation of the Jacobi theta function identity. So there we started with e to the power minus tau t squared and I took a summation n from minus infinity to infinity e to the power minus tau into t plus 2 pi n the whole square n goes from minus infinity to infinity. And we use that 2 pi periodic function. We found the Fourier series or the 2 pi periodic function and we use the basic convergence theorem and we equated the Fourier series in the original function and we got the beautiful identity which was the theta function identity. I want you to do the same thing instead of e to the power minus tau t squared. Take any general phi in the Schwarz class and imitate the same thing, imitate the same thing, find the Fourier coefficients of this capital phi x. Of course, as in the previous case of Jacobi theta function identity here also you will have the job of exchanging summations and integrals. First you have to show that when a sum little phi x plus 2 n pi n going from minus infinity to infinity the sum is a smooth function that also you have to check. And the Fourier transform has to be computed and the Fourier transform has to be evaluated at the integer points k and you are going to get the Fourier coefficients. And what you are going to get the original function capital phi equal to its Fourier series that equality is called the Poisson summation formula. It is a beautiful generalization of what you did in the first chapter. So, these were some of the exercises that I would like you to discuss and I would like you to work out in detail. I have given you lots of hints and I sketched the solutions of many of these exercises. Please carry out the details, please work out the details because that is the only way you learn the fundamental principles of Fourier transform. The next item is to look at Fourier transforms in several variables. So, we will talk about radial functions. We will prove that the Fourier transform of a radial function is another radial function. And we will see that the Fourier transform of a radial function in odd dimensions is a sine transform and in even dimensions it is given by a Bessel transform. First we have to discuss Fourier transforms in several variables. We will have to introduce a Schwarz space in several variables. Again, it is not difficult to do this. I think we will do this in the next lecture. I have just given you a preview of what is to come. Thank you very much.