 Hello, so today will be the last lecture or the denouement of this course on Fourier analysis and applications, last time we discussed the continuity or the differentiation operator with respect to the weak star topology on this piece of Schwarz distribution here continuity refers to sequential continuity. Today we begin by proving the corresponding result for the Fourier transform, the Fourier transform maps s prime r to s prime r, this Fourier transform as an operator will be sequentially continuous with respect to the weak convergence of distribution or the weak star topology as it were. So, theorem 123 that you see displayed in the slide, suppose u n is a sequence of tempered distributions converging weakly to u, then the corresponding sequence u n hat converges weakly to u hat. So, take an element g which is rapidly decreasing that is the Schwarz class and let us compute u n hat paired with g, for the definition of Fourier transform of a tempered distribution simply put the hat on the other factor, so the u n paired with g hat, but u n converges to u weakly, so u n paired with g hat converges to u paired with g hat that is what you see a red arrow and then put the hat back to where it belongs namely u hat paired with g and that completes the proof of the theorem. The proof is exactly similar to the earlier proof of continuity of the differentiation operator, a simple exercise find the weak limit of root pi by root epsilon e to the power minus x squared by 4 epsilon, either you can use a definition of weak limit or you can recognize this as a Fourier transform of e to the power minus epsilon x squared e to the power minus epsilon x squared converges weakly to 1, so by the continuity of the Fourier transform by theorem 123 the corresponding Fourier transforms will converge to the Fourier transform of 1 that is 2 pi times the Dirac delta, so I have given you the exercise and I have given you 2 different ways to proceed. Some more exercises consider the distribution u equal to exponential of i a x squared where a is a real number, now because a is a real number u is actually a L infinity function, so it is a tempered distribution. Now what I do is that I change it, I change a to a plus i epsilon, I change a to a plus i epsilon prove that u epsilon converges to u in the weak sense of distribution. Now take epsilon bigger than 0 and you just proceed in the usual way, so if you replace a by a plus i epsilon what happens to the exponential, the exponential becomes e to the power i a x squared and there is a minus epsilon x squared because the presence of the minus epsilon x squared suddenly the function has become from L infinity it has suddenly landed up in the Schwarz space S of r it is rapidly decreasing so we can compute the Fourier transform by using the integral formula. Proceed along the usual lines how you compute the Fourier transform in chapter 4 and put the definition complete the square and you land up in the integral that is displayed in the slide integral over r there is a factor in front of the integral exponential of minus lambda times x plus i chi by 2 lambda the whole square where lambda simply denotes epsilon plus i a just for simplicity I use an notation lambda. Now we want to reduce this integral to integral of e to the power minus t squared dt how will you transform this to an integral e to the power minus t squared dt, you must go back to chapter 1 see how we computed the Fourier transform of the Gaussian by using Cauchy's theorem right we shifted the contour of integration using Cauchy's theorem you imitate that idea and you complete this problem for a slightly different approach you can look at the book of stretch arts page 47 and that completes this exercise. There are many other exercise and here I am going to take a couple from stretch arts book and this is a very nice exercise that you see on page 50 before we take it up we must know what are radial functions and radial distributions in chapter 4 towards the very end we talked about functions which are radial and we proved that the Fourier transform is also radial but this was the context of the schwarz space now we want to do the same thing for a distribution when will a distribution be radial okay before that what are the radial function a function phi is said to be a radial function if phi composed with r is phi where r is a rotation matrix for example e to the power minus mod x squared is a radial function so anything which depends only on mod x is a radial function there is a way to describe which tempered distributions are radial there is a way to compose a distribution with a diffeomorphism we do not need this general result and that is discussed in chapter 6 of hormone does book the chapter says composition with smooth maps but the rule that we will apply it will be to distributions which are actually functions so we are going to be looking at functions which are locally integrable etc and those are going to depend only on the mod x in r3 our examples are going to be in r3 and the in fact that examples are going to be specific we are going to be looking at the distributions mod x to the power minus 1 and mod x to the power minus 2 they are locally integrable functions and they decay as x goes to infinity so they both define tempered distributions in r3 and what radial means is very clear here and so we will not elaborate on this concept anymore just that these are radial functions and the Fourier transform of one of them is going to be the other one with a constant factor thrown in so let us get to the example itself so that is theorem 124 the locally integrable functions 1 upon mod x and 4 pi by mod x squared both define tempered distributions on r3 how do you make them as tempered distributions you are going to tell me what does it do to a function g in the schwarz class this g comes from the schwarz class g maps to integral gx dx upon mod x to the power k remember I will write dx in polar coordinates and so k is either 1 or 2 and I will get r squared sin phi dr d theta d phi remember that r squared will cancel with the mod x to the power k where k is 1 or 2 and then you will be left with simply integral of gx times either r or nothing and gx is a rapidly decreasing function so there is no problem as far as the convergence of this integral is concerned and you can see that 10.21 does make 1 upon mod x and 4 pi by mod x squared into tempered distributions if you want you can directly prove that they are continuous if gn converges to g in the schwarz class the corresponding integrals converge first part is very clear now we turn to the second part of the theorem how to find the Fourier transform of 1 upon mod x the Fourier transform computation is a very different matter call this distribution 1 upon mod x is u and we proceed as usual so what are the definition of Fourier transform you see the last display u hat paired with g is simply u paired with g hat the hat goes on the other side and so now we use the definition of Fourier transform of a rapidly decreasing function as an integral so right hand side basically becomes integral g hat dx upon mod x we use the definition 10.21 and the definitions of Fourier transform both and the left hand side u hat paired with g is simply this okay now what we are going to do is that we are going to put the definition of g hat and that is one more integration e to the power minus ix dot y gy dy of course the next step is pretty obvious we are going to switch the order of integrations we are going to switch we are going to integrate with respect to x first let us try to do that and let us see what happens the inner integral is e to the power minus ix dot y dx upon mod x I am going to put x equal to az and a will be thrown to the other factor it will become a transpose y and I am going to select the rotation matrix a in such a way that a transpose y is simply aligned in the e3 direction again you must go back to chapter 4 when we proved that a Fourier transform of a radial function is another radial function and you see the same technique was used there and so I will leave that for you to verify and I can choose this a in such a way that we will get mod y times e3 and this becomes z and z can be written in terms of polar coordinates and you get an integral e to the power minus rho mod y cos phi remember there is a rho squared sine phi one rho cancelled because the denominator we had a rho remember mod x will be the same as mod z because a is a orthogonal matrix and mod z is rho and so one rho get cancels and you left with the rho d rho and the radial coordinate goes from 0 to infinity and there are two angular coordinates theta and phi theta goes from 0 to 2 pi and that is why picked up a 2 pi here and the phi variable goes from 0 to pi e to the power minus i rho mod y cos phi sine phi d phi how convenient put cos phi equal to t so sine phi d phi will be minus dt and the limits of integration will become minus 1 and 1 and you can actually perform this integral e to the power minus rho mod y t perform that integration and so one more integral has been dealt with only the integral with respect to rho remains and what you see you see trouble in front of you you see integral 0 to infinity sine mod y rho d rho that is an oscillatory integral and you need to deal with oscillatory integral by now you got sufficient experience in dealing with oscillatory integrals whenever you deal with an oscillatory integrals what comes to your mind is x of minus epsilon mod x squared trick of course you will try to throw in e to the power minus epsilon mod x squared and take the limit as epsilon goes to 0 we have seen this several times in the past so no need to dwell on this but if you put in this e to the power minus epsilon mod x squared you will run into a messy integral you may be able to come out of it but it will give you some little bit of trouble but there is no need to choose e to the power minus epsilon mod x squared some other modification of this trick will also be sufficient for our purpose so instead of using a e to the power minus epsilon mod x squared trick you use e to the power minus epsilon mod x instead and the same logic will go through but this time you will get an integral that you can quickly evaluate namely the inner integral becomes e to the power minus epsilon rho sine mod y rho d rho now we must go back and consult tables of Laplace transforms that you taught your undergraduate students the Laplace transform of sine and the Laplace transform of cosine so what you see the inner integral is the Laplace transform of sine or go ahead and integrate it using integration by parts twice whichever method suits you but I am going to make my life easy simply by recalling the formula for Laplace transform of sine you are going to get a mod y by mod y squared plus epsilon squared and one mod y cancelled off you simply left with integral over r3 g y dy upon mod y squared plus epsilon squared well now we need to take the limit inside the integral the dominated convergence theorem will immediately come to your rescue remember that we are in r3 when you write polar coordinates in r3 it is r squared dr d theta d phi sine phi and there is a r squared below and r squared above so the fact that there is a mod y squared below is going to give you no difficulty whatsoever you can use a DCT and we will get the result it is basically this epsilon will go away and you get integral g y dy by mod y squared which is exactly 1 upon mod y squared paired with g there is an 4 pi factor of course which always remains and we are established the theorem and so I explained to you how to prove the last equality it is an application of the dominated convergence theorem now let us discuss Fourier series again from the point of view of distribution theory so in the very first chapter we talked about point wise convergence in the second chapter we talked about L2 convergence and third chapter we talked about Cesaro convergence in the final chapter we talk about distributional convergence of Fourier series so four different modes of convergence we are discussing and the weakest one and the most useful one is here so let us go back to one of those examples from chapter one the Fourier expansion of mod x and that is displayed here in the slide now because the presence of 2k minus 1 the whole square to the denominator the series converges uniformly to mod x the 2 pi periodic extension of mod x and uniform convergence of a periodic function will imply weak convergence in the sense of distributions and in other words the partial sums s and fx converges uniformly to mod x and so it will also converge in the sense of distributions so why are we weakening it why are we doing this why are we getting rid of uniform convergence and harping on weak star convergence because differentiation is continuous if I know that if a sequence of distributions converges weakly the derivatives will also converge weakly so I can differentiate sn once twice etc so let me differentiate sn once the sequence of derivatives will converge weakly when I differentiate what happens the cosine becomes a sine and the minus sine goes away 1 2k minus 1 goes away and you get the next one so from the right hand side I get 4 times summation k from 1 to infinity sine 2k minus 1x upon 2k minus so the left hand side when you differentiate mod x you get the signum function remember that we are looking not at the mod x on the entire real line if you do the mod x on the entire real line the derivative is a signum function but you are looking at the mod x which has been periodically extended beyond minus pi pi so you are going to get a periodized version of the signum function what is it this signum function of x is 1 on the interval 0 to pi it is minus 1 on the interval minus pi to 0 extend this as a 2 pi periodic function and you get this result this falls under the purview of Dirichlet's theorem that we discussed in the last chapter namely this converges pointwise except at the discontinuities and at points of discontinuity it converges to 0 we have seen this in the last lecture but now I want to differentiate it once again when I differentiate the signum function I get the Dirac delta in fact I get the twice the Dirac delta and I will get a twice the Dirac delta placed at 0 pi minus pi etc so I am going to differentiate this term by term now once I differentiate it term by term I go outside the realm of classical analysis and I must resort to distributions the series of derivatives will no longer converge pointwise it will converge in the sense of distributions so what happens a sequence of derivatives the one of the 2k minus 1 which was the denominator that goes away you simply got summation k from 1 to infinity cosine 2k minus 1x and when I differentiate the signum function I get the Dirac delta but here I get a minus 1 to the power n and I want you to tell me why is this minus 1 to the power n appearing you should draw the graph of the signum function and you should understand what happens when I go from pi to 2 pi and from 2 pi to 3 pi in which way the graph ascends and descends accordingly the derivative will pick up a plus sign or a minus sign and at that because of discontinuity you are going to pick up a Dirac delta so I leave the amusing details for you and we get 10.22 10.22 has to be understood in the sense of distributions summation n from minus infinity to infinity minus 1 to the power n time the Dirac mass placed at pi n the sum of Dirac deltas left hand side is obviously a distribution and the right hand side is an infinite series that does not converge in the classical sense but it converges weakly in the sense of distributions. Now I take the Fourier transform of both sides of 10.22 both sides are distributions and both sides converge weakly in the sense of distribution that is the partial sums of the two series on either side converge in the sense of distributions the Fourier transform is a continuous operator in the sense of distributions and so I can take term by term Fourier transform of both sides. So calculate the Fourier transform of delta pi n remember this is not the Dirac delta at the origin the Dirac delta at pi n so go back and put the definition of the Dirac delta and put the definition of the Fourier transform and evaluate the Fourier transform of both sides term by term and look at the equation that you get the resulting equation will be the Jacobi theta function identity in disguise. Let us now do another exercise instead of taking the Fourier transform of both sides let us directly pair both sides with a Schwarz function gx which Schwarz function gx am I going to choose no prizes for guessing I am going to choose my gx to be e to the power minus tx squared t real positive and so what happens to the left hand side it is gx paired with delta pi n what am I going to get I am going to get g of pi n what about the right hand side cos 2k minus 1x paired with gx is simply integral gx cos 2k minus 1x dx. So what do I get that is exactly what I got on the right hand side integral from minus infinity to infinity e to the power minus tx squared cos 2k minus 1x dx left hand side is summation minus infinity to infinity minus 1 to the power n e to the power minus t pi squared n squared the g evaluated at pi n. Now let us combine this let us simply for the left hand side is pi times 1 plus 2 times summation n from 1 to infinity minus 1 to the power n e to the power minus pi squared n squared t and the 2 factor was taken the right hand side so 2 cosine a is e to the power i a plus e to the power minus i a each of these 2 pieces is a Fourier transform of a Gaussian you know how to calculate the Fourier transform of a Gaussian it is another Gaussian and the result is displayed over here the last line let us go a step further. So I am going to take the last equation and I am going to replace t by 1 upon 4 pi squared tau in the previous equation and just do a simple algebra here 10.23 what you are going to get is when you replace t by 1 upon 4 pi squared t that is a routine algebra you are going to get equation 10.23 1 upon 2 root pi tau times 1 plus twice summation n from 1 to infinity minus 1 to the power n e to the power minus n squared upon 4 tau equals summation k from minus infinity to infinity e to the power minus pi squared 2k minus 1 the whole squared tau equation 10.23. Now go back to chapter 1 go back to chapter 1 and recall the formula for the Jacobi theta function identity that transformation formula that was equation 1.25 and I repeated it here for your convenience and I want to compare 10.23 and 1.25 before comparing I am going to put t equal to pi when I put t equal to pi cos nt is minus 1 to the power n and you see we get the left hand side of 10.23 on the right hand side I put when I put t equal to pi I get exactly the right hand side of 10.23 so you see that the equation that we got 10.22 is exactly the Jacobi theta function identity in disguise. So now we want to look at this in a slightly different modern perspective the equation 10.22 that we got the left hand side was a bunch of Dirac deltas right here I can split this sum into two sums those with odd n's and those with even n's. So let us do that let us separate out the two and we get each of those two pieces is a sum of the form 10.24 it is an infinite series of Dirac deltas placed at mc where m varies over the integer that is a it is an integer lattice the set of all mc's forms a cyclic subgroup of r and the series 10.24 converges weakly and the sum is a tempered distribution now we want to study this tempered distribution closely I am going to omit some of the routine computational details. So I am going to take a Fourier transform of both sides as before so u hat will be take the term by term Fourier transform that Fourier transform of Dirac delta is basically e to the power minus i y mc and I am going to combine the exponentials with opposite signs as 2 cosine mcy. Now what I want to do is that this summation here converges in the sense of distributions and this is the second derivative of something it is the second derivative of what look at 10.26 so look at the series on 10.2 so the series on the right hand side of 10.26 is a uniformly convergent series and the sum is a nice function f of y when you differentiate the series 10.26 twice in the sense of distributions mind you I get 10.25 so 10.25 has been written as 1 plus f double prime y it will be very nice if I can find a better formula for f of y if I can sum this series on the right hand side then it will be nice that is what we will do now what you do is you start with a sawtooth function the sawtooth function is s of x equal to x on 0 to 2 pi extend this function as a 2 pi periodic function and you get a sawtooth function and you can write its Fourier series take that Fourier series and do a term by term integration take the Fourier series and do a term by term integration and what you will get will be pi squared by 3 minus twice summation m from 1 to infinity cos mx upon m squared so in some sense the summation cos mx upon m squared I have an expression for that namely I just have to take the sawtooth function and integrate it and I need to replace x by cx so that I get cos mcx upon m squared and so I can calculate explicitly what this f of y is going to be and that f of y has been displayed over here minus pi squared upon 3c squared plus pi times mod y by c minus y squared upon 2 in this interval now the first term is a constant and and the last term is an even function and the and the middle term is also an even function so you go to extend it as an even function as a 2 pi periodic even function with period 2 pi by c the period is now 2 pi by c because I scaled the variables I won't take the second derivative remember but I am going to take the second derivative in the sense of distributions the constant term will disappear when I take the distributional derivative of minus y squared by 2 I am going to take a minus 1 what are the distributional derivative of mod y it is the Dirac delta and it will be weighted by 2 twice the Dirac delta so you get 2 pi by c into the Dirac delta at the origin if I get the Dirac delta at the origin I am going to get a Dirac delta at 2 pi by c 4 pi by c etc because it's a periodic function f of y is a periodic function with period 2 pi by c so I get the whole bunch of Dirac deltas equation 10.27 the f double prime has been computed put this value f double prime in the previous equation 10.2 further one goes away I cancel to this minus 1 and you are left with this beautiful result theorem 125 this theorem 125 is called the Poisson summation formula for c not equal to 0 you take a lattice of Dirac deltas concentrated at the points mc the Fourier transform is another lattice of Dirac deltas concentrated at 2 pi m by c with a factor of 2 pi by c thrown in front of it now let us evaluate both sides at a function of the Schwarz class so if I take a phi in the Schwarz class and apply both sides of 10.28 to phi left hand side becomes summation phi of mc right hand side becomes summation phi hat of 2 pi m by c with a factor of 2 pi by c in front of it in chapter one the relevant phi was the Gaussian and we got the Poisson summation formula for a special case in chapter one this is the general theorem the most general Poisson summation formula now what happens that this is in one variable what about several variables what happens if I take Rn and take a lattice of Dirac deltas in Rn the formula becomes even more interesting in several variables the Poisson summation formula assumes a much more interesting aspect in several variables the book of Strichards page 120 contains a beautiful discussion on lattices and dual lattices you think of a lattice of Dirac deltas as a crystal structure and the Fourier transform corresponds to the x-ray diffraction pattern that emerges out and experiments have shown that bright spots appear at points of the dual lattice so this physical experiment with crystals and x-ray diffraction patterns arising from crystals has a beautiful mathematical description namely the Poisson summation formula the student can scarcely do better than undertake a systematic study of Strichards book after the scores the best thing to do would be to take up Strichards book and look at those topics that we could not touch upon in the scores here is a list of topics the Pochner's theorem on positive definite functions Heisenberg's uncertainty principle in quantum mechanics the Paley Wiener theorems and a glimpse into a beautiful area of analysis called micro local analysis wavelets and the Haar systems and how that gives rise to Vietta's formulas beautiful formulas in classical analysis and a glimpse into pseudo differential operators and I wish you good luck with your studies of Strichards and beyond and this is the end of this course. Thank you very much