Lecture 54

Welcome back. In the last lecture, we were discussing about how to approach to prove if F is in Schwarz space of R n, then F hat is also will be in Schwarz space of R n. So, towards that end, let us first look at we what is the derivative if I take and then the Fourier transform. So, by definition, this is going to be d alpha 1, we are taking alpha is a multi index and d x 1 alpha 1 up to F, this Fourier transform that means, this is R, this is n copies, n terms and e to the power minus 2 pi i x 1 xi 1 e to the power minus 2 pi i x n xi n and then this is d x 1 to d x n. Now, F is in the Schwarz space. So, first what we can do is that we will take out all the partial derivative up to alpha n minus of 1 and then all these are independent of x n. So, then this one we can get that this is the derivative up to x n minus of 1 and then this is what one will get is that the 2 pi i by as we notice that if I am dealing with only del alpha n by del x n F of x e to the power minus 2 pi i x n xi n d x n, then by integration of parts and F is a Schwarz class function. So, every time when I take the integration, this is going to vanish. So, this one is going to be minus of 1 to the power n with the repeated minus 2 pi i xi n and then the integral over R, this is F of x d x n is going to remain over there. If I am applying n minus 1 by delta x n minus of 1 alpha n minus of 1, if I am applying it over there, then again I will get a factor of this is alpha n, then I am going to get a factor of minus 2 pi i and then minus 1. So, this by combining this simple observation what we are going to get this is going to be 2 pi i xi to the power alpha and then this is F hat of xi. With the similar kind of thing what we can show is that 2 pi i x I take to the power alpha and then I take F x, then I take the Fourier transform of this xi, then this is going to be d alpha of F hat of xi. So, these two fundamental relation between the multiplication with the function with a polynomial and take the Fourier transform and then you take the derivative and take the Fourier transform, they are consistent what we have seen it in R. So, with this observation one now we can we can convince ourselves that if I am looking at xi alpha, then d beta of F hat at xi, then I know exactly what this is d beta of F hat xi, this is not this is going to be the Fourier transform of some Schwarz class function which is essentially this. So, this is xi alpha minus 2 pi i x beta F, then this hat at xi. Now, this is nothing, but I one can get that this is some constant factor d alpha of minus 2 pi i x beta F, this entire Fourier transform of xi and what. So, this is going to obviously, this is a Schwarz class function. So, we can write it down and you can take the derivative inside to get that this we can control it over there. So, just as in the case of R this is what we had done. So, this will put it that if F is in Schwarz space of R n, then F hat is also in the Schwarz space of R n. So, once we have this, then natural we must try to aim to get the Fourier inversion theorem as well as the Planck's rule theorem for the Schwarz class function. And what we have seen is that in R as well as in the circle also in the finite abelian group what we need is a powerful tool of convolution. So, here if we define let. So, convolution let F and G they both belongs to the Schwarz space, then define as usual F convolution of G at x this is equal to integral over R n F of x minus of y G of y d y which by the change of variable we can write this as F of y G of x minus of y which is again G convolution of F we can call it as if you wish. So, now as we can see that one easily can see that d alpha of F convolution of G this is equal to d alpha F convolution of G which is going to be shifted and there is nothing wholly about here we are taking both F and G are in Schwarz space. So, we can write down these two and thus this F convolution G like just like in the case of R one can prove that F convolution of G this belongs to the Schwarz space of and therefore, F convolution of G hat at xi this is integral of R n e to the power minus 2 pi i x dot xi d x and then this by opening up the definition of F convolution of G and then making the change of order of the integral exactly as in the case of R we will get that this is F hat of xi into G hat of xi ok. So, once we have the convolution nicely exactly the same way defined as in the case of R. So, we are now in a position to talk about good kernel because they play a very important role in the Fourier analysis of R and as well as they have helped us to get the inversion and Planck's rule theorem. So, now, what is a good kernel? The definition is same. So, we can take k delta to be greater or equal to 0 integral over R n of k delta of x d x is equal to 1. Now this is a family remember. So, for eta greater than 0 for any eta mod of x greater than eta k delta of x d x this goes to 0 as delta goes to 0. This means it is going to be concentrated near the origin and the tail of this is very small we can control. So, the all the relevant portion is going to come from the neighborhood of the origin ok. So, this is what we have seen how useful they are in our analysis of R. So, in fact, if we if we look at this condition I can always replace this condition by this condition when that is an easy exercise for you to do. If I say that k delta of x d x is equal to 1 and this condition I am replacing it by R n mod of k delta of x d x which is lesser equal to m for all delta. That is uniformly bounded instead of the positive thing what we can take and then the third is this goes to 0. So, this can also yield us the same result. So, now let us see that if let F belongs to Schwarz space then F convolution of k delta of x converges to F of x as delta goes to 0 uniformly. So, this again the standard trick what you do is that integral over R n F convolution k delta is F of x minus of y k delta of y d y and minus F x which one will replace by F x into 1 by F x k delta y d y. Now if you take the modulus this argument is same F of x minus of y minus of F of x and then this is mod of k delta of y d y. If we are assuming k is non-negative then the mod can be replaced otherwise you can keep the mod to be. And so, then what we do is that we break we know that F is uniformly continuous on a compact set I mean. So, in fact, it is uniformly continuous everywhere. So, for the continuity of F what we can get is that there will exist a delta some small neighborhood of x in which F of x minus of y minus of F x is very small. So, you choose that neighborhood let us say that is to be our eta mod of y less than eta plus mod of y greater or equal to eta. Now in the first integral I will use the continuity of F and in the second integral use the third property of good kernel. So, this will give you this is less than epsilon. So, that is what the proof we have seen this proof time and again many times. And so, the that is why it is called a good kernel and here what I can do here I need to take the modulus. So, for so good we get exactly this no new idea is needed to prove this result in the higher dimension we have just re-employed the idea of the one dimension here. Now do we have enough example of good kernels. So, as usual the generic example what one can take is that take k of x is equal to e to the power minus pi mod x square. And if we define k delta of x is equal to 1 over delta n k of x by delta then as you can see that this is certainly non-negative all this k delta. Now if we are integrating then we are going to get 1 and the third property is going to be true over here because if we are taking e to the power minus pi mod x square by delta square then this is 1 by delta to the power n then this is dx and mod of x greater than eta. If we are taking this then this by making a change of variable what we get that this is minus pi mod y square dy and then this is mod of y greater than eta by delta. And as delta goes to 0 then this eta by delta that goes to infinity and that becomes tail of the integrable function tail of the integration of the Gaussian. So, this is going to be 0 as delta goes to 0 and there is nothing wholly about this Gaussian. So, you can play the same game by take k any k in the Schwarz space such that integral over k delta or you can take k to be non-negative and k of x dx is equal to 1 then if you define k delta of x 1 by delta to the power n k of x by delta then this argument is going to work. So, like just in the case of R we have got enough good kernel with us to deal with. Now another important thing what we have seen which is very useful to derive the inversion formula as well as the Planck's rule is the multiplication formula. Now what is the multiplication formula? Let f and g belongs to the Schwarz space of R n then integral R n of f of x of g hat at x dx this is equal to integral over R n f hat of x g of x dx because if I write down the left hand side this is going to be and then this is R n this is g of y e to the minus 2 pi y dot x dy dx and then if we make the change of the order of the integration what we are going to get is in R n then this is g of y and then f of x e to the power minus 2 pi i y dot x dx dy as we have noticed earlier y dot x is equal to x dot y therefore, this becomes integral over R n g of y f hat of y dy okay. So, I think now we are all set to derive the inversion formula as well as the Planck's rule theorem for the higher dimension. So, what should be the statement of the inversion theorem? This is if f is in Schwarz space then f of x is equal to integral over R n f hat of xi e to the power 2 pi i xi dot x d xi that is what is going to be our inversion formula. So, because f is in Schwarz space. So, we can define the integration of f hat because f is also going to be in the Schwarz space. So, now the proof is idea is exactly the same as that of R let g of x g I define g delta of x this is equal to e to the power minus pi delta square mod x square. So, therefore, so what we can get is that the g delta hat at xi this is 1 by delta to the power n and e to the power minus pi mod xi square by delta square that is what we are going to get. Now from our multiplication formula what we know is that this says that the integral over R n of f of x g delta hat at x dx this is going to be R n f of f hat at x g delta of x dx. So, therefore, as you can see that if RHS goes to if delta goes to 0 e to the power minus pi delta square mod x square this goes to 1 therefore, the RHS is R n hat at x. And if I look at the LHS, LHS is nothing but f convolution of g delta hat at 0. Now what we have seen just before is that g delta hat at 0 is a good kernel therefore, this is going to converse to f at 0 because it converges uniformly this. So, hence we get f of 0 this is equal to integral over R n f hat of xi d xi. Now in order to get for f of x consider tau of minus of f at x. So, you consider this g of x this is equal to this at 0. So, therefore, what we get is that tau minus of x f at 0 this is equal to integral R n tau minus of x f hat at xi d xi by our previous this is nothing but f of x. And then what we have seen is that if I am translating the then taking the Fourier transform it is going to be heated by character. So, this is R n then this is f hat of xi e to the power 2 pi i x dot xi d xi because here I am taking the translation with minus of x that is what gives us the inversion formula then the Planck's rule. So, what is the statement is mod of f of x square d x this is equal to integral over R n mod of f hat of xi square d xi and the proof exactly as what we can guess to work define f of x f tilde x bar then what we have seen is that f convolution of f tilde at 0 this is going to be equal to by definition f hat at xi f tilde hat at xi d xi and f tilde hat at xi by simple calculation we can easily see that this is. So, this is equal to then R n mod of f hat of xi square d xi and this is nothing but f of x then f tilde of minus of x d x and which is equal to R n f of x f of x bar d x which is nothing but R n mod of f of x square d x and that is what is going to be the Planck's rule formula for R n. So, more or less till this point what we have seen is that the study of the Fourier analysis in the higher dimension is more or less in the same spirit as that of the Fourier transform in one dimension and we have got inversion and Planck's rule. Next we would be definitely be interested to see that how this higher dimensional Fourier transform is going to help us in other areas or rather we would like to see the applications of the higher dimension Fourier transform in some concrete problem in mathematics as well as in engineering and science. Thank you.