 So, in the last module, we discussed Hermite polynomials, Hermite functions as Eigen functions or the Fourier transform as an operator from S to S. Now, we will continue with the general discussion of the Fourier transform as an operator from S to S. We shall first prove the inversion theorem that you see on the slide. Suppose f of t is a function in S and the function can be recovered from its Fourier transform via the inversion theorem equation 4.10 that you see displayed, f of t equal to 1 upon 2 pi integral minus infinity to infinity f hat of chi e to the power i t chi d can notice that the inversion theorem the inversion formula the right hand side is very similar to the definition of Fourier transform except for the appearance of this factor 1 upon 2 pi and the change in the sign here. In the definition of Fourier transform, there was an e to the power minus i t chi in the inversion theorem there is a plus sign ok. The proof of the inversion theorem involves one trick, but before we bring that in let us see what happens when we try to prove this theorem directly and naively. Let us try to prove that RHS equal to LHS start with the RHS of 4.10. What do you see in the RHS of 4.10 1 upon 2 pi integral minus infinity to infinity e to the power i t chi f hat of chi. In place of f hat of chi you simply put the definition of the Fourier transform the t variable already been taken up. So, write f hat of chi as integral from minus infinity to infinity f of x e to the power minus i x chi dx and you switch the order of integration switch the order of integration and the integral with respect to chi assumes the form integral minus infinity to infinity e to the power i t minus x chi d chi. So, what have I done again? I put in the definition of the Fourier transform as an integral try to switch the order of integration, but I ran into this meaningless integral integral minus infinity to infinity e to the power i t minus x chi d chi. I say this is a hitherto meaningless because there is a way to assign a meaning to this integral and that is what we shall try to do now. We shall try to make sense out of this formal computation make it rigorous and to make it rigorous we resort to what is known as the x of minus epsilon x square trick. So, what we do is the following we start with this right hand side of 4.10 what is what is in the right hand side of 4.10 there is an innocent 1 upon 2 by a factor which has suppressed over here because integral from minus infinity to infinity f hat of chi e to the power i t chi d chi. What I do is I introduce an e to the power minus epsilon chi squared factor in inside and take the limit as epsilon goes to 0 and that limit can be taken outside the integral and we can do this because this f hat of chi is rapidly decreasing function and so the limit can be taken in and out of the integral after I introduce this e to the power minus epsilon chi squared. Now that I taken the limit outside the integral now I can put the definition of f hat of chi. What is the definition of f hat of chi integral minus infinity to infinity f of x e to the power minus i chi x dx and then I try to switch the order of integrations. I first perform the integration with respect to the chi variable. So, instead of the earlier integral that we got integral minus infinity to infinity e to the power i chi t minus x d chi that is what we obtained earlier now we have the additional minus epsilon chi squared in the exponent. It is this additional minus epsilon chi squared that makes things go through. So, let us do that the inner integral we recognize as a Fourier transform of the Gaussian. So, that is exactly what I have written in the last line in the slide and we have already computed the Fourier transform of the Gaussian and that is another Gaussian and we know the formula for the Fourier transform of the Gaussian namely root pi by root epsilon e to the power minus x minus t the whole squared upon 4 epsilon. Remember there is a epsilon here and the Fourier transform the epsilon will come in the denominator of the exponent and there will be a root pi by root epsilon coming in. So, all in all we get these things. Now, I need to make a simple change of variables x minus t equal to root 4 epsilon s simply put x minus t equal to root 4 epsilon s this expression simplifies considerably it is simply e to the power minus s squared and f of x becomes f of t plus root 4 epsilon s. Again, because e to the power minus s squared is a very rapidly decreasing function the limit which was outside the integral can be taken back inside the integral. But when I take it back inside the integral I simply get f of t whereas, I am integrating with respect to s. So, f of t can be taken outside the integral integral of e to the power minus s squared is remember root pi there is a root pi outside. And so, we get this extra factor of 2 pi and I divide by 2 pi and I get the left hand side of the theorem. So, that completes the proof of the important Fourier inversion theorem. In the next slide I will give you a bunch of exercises that are based on this e to the power minus epsilon t squared trick. The first exercise is an important exercise it says that when you integrate the Fourier transform remember if f is in the Schwarz class f hat is also in the Schwarz class and so, f hat is an L 1 function you can integrate it. So, integral minus infinity to infinity f hat chi d chi equals 2 pi times f of 0. Again, if you try to put in the definition of the Fourier transform if you try to put in f hat of chi equal to integral minus infinity to infinity f x e to the power minus i chi x dx directly and switch the order of integrals we encounter the same difficulty. Again the inner integral becomes integral minus infinity to infinity e to the power minus i chi x d chi. Again this is a problematic integral it is an example of a oscillatory integral and so, to circumvent this difficulty we need to introduce the e to the power minus epsilon chi squared as before. So, let us restart integral minus infinity to infinity f hat chi d chi equal to limit epsilon goes to 0 integral minus infinity to infinity f hat of chi into e to the power minus epsilon chi squared d chi. Now, you put the definition of f hat of chi integral minus infinity to infinity f of x e to the power minus i x chi dx. Now, we can safely switch the order of integrations and you will get limit as epsilon goes to 0. The x integral later and the chi integral first integral minus infinity to infinity e to the power minus epsilon chi squared minus i x chi. The innermost integral is the Fourier transform or the Gaussian that we have computed that is root pi by root epsilon e to the power minus x squared upon 4 epsilon. Again, the same change of variables x equal to 2 y root epsilon and that gets rid of this root epsilon from the denominator. We get 2 root pi integral minus infinity to infinity f of 2 y root epsilon e to the power minus y squared dy. Now, we may pass to the limit as epsilon goes to 0 under the integral sign. Use the dominated convergence theorem you get 2 root pi integral minus infinity to infinity f of 0 e to the power minus y squared dy f of 0 is a constant it comes out integral from minus infinity to infinity e to the power minus y squared dy is root pi root pi root pi becomes pi I get 2 pi times f of 0. For those of you who may have seen a little bit of distribution theory the Dirac delta basically it says that the integral of f hat of chi d chi is 2 pi times the Dirac delta tested with f. So, this says that the Fourier transform of 1 is 2 pi times the Dirac delta. So, this last equation can be recast in the language of distributions as saying that if I take the distribution 1 and take its Fourier transform it is 2 pi times the Dirac delta. But to make sense of this last sentence you will have to develop the theory of distributions carefully we will do that if time permits later. Next exercise sign a t by t sign a t by t is not an l 1 function nevertheless I am asking you to compute its Fourier transform. So, technically this question is not completely well posed but what happens is that I would like you to write down the definition of Fourier transform formally integral sign a t by t into e to the power minus i t chi sign a t by t is an even function. So, it is sign a t cos t chi by t integral and you can use a de-factorization formula and try to use some complex analysis or something and try to get a expression formally. Of course, all these manipulations are formal but it will turn out that they are valid because although sign a t by t is not an l 1 function it is an l 2 function and very soon we will see how to define the Fourier transform of an l 2 function and the result will be another l 2 function. So, in anticipation of what is going to come you could try to do this calculation at a formal level by putting in the definition of the Fourier transform and try to compute the resulting conditionally convergent integral in closed form and see whether what you get is an l 2 function. The third problem is to compute the Fourier transform of 1 upon t squared plus a squared you see the appearance of the Cauchy distribution again is cos t chi upon t squared plus a squared. In the earlier capsule we computed this using complex analysis but now I want you to differentiate it twice under the integral sign and obtain a second order ODE for the integral exactly as we try to compute the Fourier transform of the Gaussian in the first part of the course. So, write down the integral the integral is a function of chi called i chi. So, i chi will be equal to integral you could go from 0 to infinity cos t chi because an even function divided by t squared plus a squared. So, try to differentiate with respect to chi twice and obtain a second order ODE but again you will run into an oscillatory integral you will have to throw in an e to the power minus epsilon t squared term. Use the x minus epsilon t squared trick to obtain a second order ODE for this particular function 2 upon pi integral 1 to infinity sin tx dt upon root t squared minus 1 for x bigger than 0. Can you recognize this as a familiar function also prove this by using Laplace transform methods. This exercise is from Hans Weber's partial differential equations of mathematical physics which was published in 1900. It is based on Riemann's lectures. The formula appears on page 175. This representation of the Bessel's function is due to Mahler and Sonnen. You can look at G N Watson's treatise on the theory of Bessel's functions. Now, let us move further with the general theory of Fourier transforms. We know that if you take an l 1 function then we have the formula for the Fourier transform f hat of chi equal to integral over the real numbers f of x e to the power minus i x chi dx. We can take the absolute value and take the absolute value inside the integral you get an inequality. So, mod f hat chi less than or equal to integral over R mod fx into mod e to the power minus i x chi but mod of e to the power minus i x chi is 1 and so that disappears and we simply have integral over R mod fx dx. In other words the Fourier transform is bounded and it is bounded by this number or stated differently the Fourier transform is an element of l infinity and the l infinity norm of the Fourier transform is less than or equal to the l 1 norm of the original function. In other words the Fourier transform is a continuous linear transformation from l 1 of R into l infinity of R it is not onto l infinity of R it is into. We will prove that the Fourier transform is injective. We will prove that the Fourier transform is injective. So, obviously from that argument it cannot be onto because if it were onto then it will be a isomorphism between l 1 of R and l infinity of R and that is impossible. I would like you to think why is that impossible. So, the Fourier transform is a continuous linear map from l 1 of R into l infinity of R I emphasize that it is into and not onto. They are going to prove later that the Fourier transform is injective from l 1 into l infinity it obviously cannot be surjective because if it were surjective then this will become an isomorphism and the Fourier transform will establish an isomorphism from l 1 of R onto l infinity of R that is plainly false. That is not correct at all. So, it is into please remember this in mind. Can we say anything more about the Fourier transform? In fact, we can not only does the Fourier transform map l 1 into l infinity it actually maps it into the class of continuous functions which decay to 0. So, the image is a subspace of C of R set of all continuous functions not only that it lands up in C 0 of R actually not only is it continues it decays to 0 as x goes to plus minus infinity. Now, we shall prove a version of the Riemann-Lebesgue Lemma for Fourier transform. Remember the Riemann-Lebesgue Lemma from Fourier series let us look at the corresponding result for Fourier transform. Before we state and prove the theorem for Fourier transform let us recall the Riemann-Lebesgue Lemma from Fourier series and give a second proof of the Riemann-Lebesgue Lemma which dates back to Riemann himself. So, let us recall if f is in l 1 of minus pi pi then the Fourier coefficients of f decay to 0 the a n's and the b n's in the Fourier expansion of f decay to 0 as n tends to infinity. Let us first prove it assuming that f is continuous let us assume that f is continuous and let us extend f to the entire real line by declaring that f of x equal to f of pi when x is bigger than pi and f of x equal to f of minus pi when x is less than or equal to minus pi. This extension is now bounded of course and it is uniformly continuous on the entire real line. Now, let us look at the integral i equal to integral minus pi to pi f of x sin nx dx. We want to show that this goes to 0 as n tends to infinity. So, let us put x equal to y plus pi by n in this integral over here. What happens the limits of course will change and you will get minus pi minus pi over n plus pi minus pi over n and dx will be equal to dy and then you will get f of y plus pi by n and then you will get sin of n y plus pi because it is sin of n y plus pi you pick a minus sign. The appearance of the minus sign here is critical. So, now let us add the two results. So, 2 i equal to f of x minus f of x plus pi by n into sin nx dx. What I have done is that this integral minus pi by n to pi minus pi by n I have written it as an integral from minus pi to pi plus some two other things. Those two other things will be i 1 plus i 2 where i 1 is this integral from minus pi minus pi by n to minus pi and i 2 is the integral pi minus pi by n to pi. So, I have written this as an integral from minus pi to pi plus i 1 plus i 2 and that integral from minus pi to pi I have clubbed with the previous integral and I get this. Now, let us look at the contributions from each of these three pieces. So, first let capital M be the supremum of mod fx on the real line then obviously, I can estimate mod of i 1 and similarly for the other integral and that goes to 0 as n goes to infinity. So, mod i 1 certainly goes to 0 mod i 2 certainly goes to 0 we need to worry about the third piece over here minus pi to pi fx minus f of x plus pi by n into sin nx dx let us take up that now. Now, I take the absolute value inside the integral and mod sin nx I ignore because it is bounded above a 1. Now, let us invoke uniform continuity let epsilon greater than 0 be arbitrary there exists a delta greater than 0 such that mod x minus y less than delta implies mod fx minus fy less than epsilon by 2 pi. Now, we can select an n naught which is so large that x minus x plus pi by n in absolute value is less than delta if n is greater than n naught. In that case this right hand side will be less than epsilon by pi and I am integrating over a interval of length 2 pi and so the 2 pi and the 2 pi will cancel out and I will simply get an epsilon whereby we conclude that this piece also goes to 0 and therefore, i goes to 0 as n goes to infinity. So, we have completed the proof of Riemann Lebesgue Lemma for a continuous function from minus pi to pi. How to pass from continuous functions to l 1 functions very simple Luzin's theorem. Luzin's theorem says that continuous functions are dense in l 1 and a simple density argument will now give you the Riemann Lebesgue Lemma for l 1 functions. This last part I am leaving it as an exercise for you because this kind of argument has appeared several times in the past. So, this was a beautiful proof of the Riemann Lebesgue Lemma it goes back to Riemann himself. Now, we shall try to give a similar argument for the Fourier transform. We will prove a Riemann Lebesgue Lemma for Fourier transform namely if you take a function in l 1 of the real line and take its Fourier transform the Fourier transform will decay to 0 as chi goes to plus minus infinity. Later we should also prove that the Fourier transform is actually continuous. So, that actually gives you a lot more information that if you take a function in l 1 not only does the Fourier transform maps into l infinity it really maps into a very small subspace of l infinity. I think with this we shall close this capsule here. Thank you very much.