 Hi. So, in today's capsule we are going to now introduce the Fourier transform. We are going to extend the definition of Fourier transform to all tempered distributions. Remember the class of tempered distributions is quite huge any LP function is a tempered distribution. Functions which grow like a polynomial they are tempered distributions. Things which are not functions Dirac delta derivatives of Dirac delta they are also tempered distributions. So, the class of entities for which the Fourier transform is defined has suddenly become pretty huge and this is possible because of this duality results. So, before we take up the definition of Fourier transform of a tempered distribution we proved a little lemma last time. Let us recall what this lemma says look at the slide theorem 112. If f and g are in the Schwarz space S of r then f hat g pairing is the same as f g hat pairing that is a straight forward application of Fibonni's theorem the proof simply writes itself out. Now f and g were in the Schwarz space. Now look at equation 10.8 now what we are going to do is that we are going to take the right hand side as the definition. Now what if f is not in the Schwarz space? What if f happens to be in S prime of r? What if f happens to be a tempered distribution? The right hand side of 10.8 would make perfect sense why because f is a tempered distribution and g was in the Schwarz class. So, g hat is also in the Schwarz class. So, u triangular bracket u with g hat makes perfect sense. So, let us now get to the formal definition of the Fourier transform of a tempered distribution. Suppose u is a tempered distribution then the map g going to triangular bracket u g hat is a tempered distribution where g varies over the Schwarz class. The tempered distribution given by 10.9 is denoted by u hat and is called the Fourier transform of u. So, because we are denoting it by u hat what does 10.9 read u hat paired with g is the same as u paired with g hat. Left hand side is object that we are trying to define right hand side is the definition. We have to check that 10.9 is a tempered distribution that means that we have to check that the map 10.9 is continuous and this will follow from the previous result that the Fourier transform is continuous as an operator from S r to S r ok. So, g n is a sequence in S r converging to g then g n hat will converges to g hat g n converges to g in a very strong sense S of r. So, g n hat will converge to g hat in a very strong sense again in S of r, but u our original tempered distribution was continuous from S r to the scalars. So, u paired with g n hat will converge to u paired with g hat by continuity. So, we conclude from this that if g n converges to g the pairing u g n hat converges to u g hat proving the sequential continuity of 10.9. So, 10.9 does define a tempered distribution. So, let us calculate some tempered distribution. The most basic example is you take a function which is in l 1 of r. So, because u is in l 1 of r in chapter 4 we define the Fourier transform of u directly as an integral and we got a function which is in l infinity. Now, will that function represent the same distribution as the one that we have defined now. u is in l 1 so u is a distribution. So, we have the Fourier transform of u as a distribution and we have the Fourier transform of u as an l infinity function. Do these two things match? Is there a consistency in the definition? There had better be a consistency in the definition in order for the theory to go through. So, what are the distributional meaning of u hat? The distributional meaning of u hat is that u hat is a distribution g maps to the pairing u g hat. But u is in l 1 remember u is in l 1. So, what is this pairing u g hat? It is simply the integral u y g hat y d y put the definition of g hat because g hat is a nice function. So, the Fourier transform of g hat is a simply given by integral g x e to the power minus i x y dx. Now, a simple application of Fubini will enable you to switch the order of integration. So, this right hand side now simplifies to integral over r g x dx integral over r u y e to the power minus i x y d y. This inside integral is been called capital U. This inside integral is being called capital U. Capital U is Fourier transform of u in the classical sense. Capital U is the Fourier transform of u in the classical sense. The Fourier transform of u in the classical sense agrees with the Fourier transform of u in the sense of distribution. What is the RHS again? U paired with g hat is the same as u hat paired with g. So, u hat paired with g is the same as capital U paired with g. So, u hat and capital U, they describe the same distribution, the same generalized function and so the definition is certainly consistent. So, we establish the consistency of the two definitions. Now, let us take up the Dirac delta. The Dirac delta is a favorite distribution. What is its Fourier transform? What is the definition of Fourier transform? Delta hat paired with g is by definition delta paired with g hat. But what is delta paired with g hat? It is g hat evaluated at the origin. But what is the Fourier transform evaluated at the origin? It is simply integral g x dx. But what is integral g x dx? It is one paired with g. The distribution 1, the l infinity function 1, when regarded as a distribution is exactly this. And the l infinity function 1 regarded as a distribution is simply one paired with g. So, delta naught hat paired with g is one paired with g. So, delta naught hat equal to 1. That is an important conclusion that the Fourier transform of the Dirac delta is the constant function 1. Now, what are the Fourier inversion theorems say? If f is the Schwarz class, then f hat hat is simply 2 pi times f of minus x. There is a reflection. Remember, when you take the Fourier transform twice, the Fourier transform is not a period 2 map, it is a period 4 map. You must remember that. So, when you take the Fourier transform twice, there is going to be an inversion in sign. So, if the inversion theorem were to hold for tempered distributions also, then I will apply one more hat to this equation. And I will get one hat equal to delta naught hat hat. And delta naught hat hat is supposed to be 2 pi times delta naught at minus x. But since the Dirac delta is at the origin, the inversion in sign does not matter. So, the inversion theorem were to hold, then we will get that the Fourier transform of 1 is 2 pi times delta naught. But we are not proved the inversion theorem yet. We are going to do it later. So, this equation 10.10 should now be established from first principles. Namely, we must use a definition of Fourier transform. So, one hat applied to g is by definition one applying to g hat. One applying to g hat is simply integral g hat y dy over the real numbers. How do I evaluate this integral? You have to use the e to the power minus epsilon y squared trick. So, write this as limit as epsilon goes to 0 outside integral g hat y e to the power minus epsilon y squared. First step, second step, put the definition of g hat as an integral. We will get two integrals, switch the order of integration and finish the job. You have to use the dominated convergence theorem and you will land up with integral of e to the power minus z squared dz over the real line which is root pi. That calculation will come in and you will get the final answer to be 2 pi times g 0 and you would have verified 10.10. That is a second example on Fourier transform calculation. More examples on Fourier transform calculation. What is the Fourier transform of e to the power i x? What is the Fourier transform of cos x? What is the Fourier transform of sin x? After all cos x is e to the power i x plus e to the power minus i x upon 2. If you calculate the Fourier transform of e to the power minus i x a where a is real, when a is real that is a L infinity function. So, it is a tempered distribution. So, if you can calculate calculate the Fourier transform of this then you can calculate the Fourier transform of sin x and cos x. How do you calculate the Fourier transform of this? Apply the definition call this u call this object u u hat paired with g is the same as u paired with g hat u paired with g hat means integral u of x g hat x dx u of x is e to the power minus i x a and now you have to put the definition of g hat as an integral and you switch the order of integration. But before you do that make sure to throw in the e to the power minus epsilon y squared use the e to the power minus epsilon y squared trick to do the job and you will get the Fourier transform. Use the e to the power minus epsilon y squared trick to compute the Fourier transform of the signum function what is the signum function this x upon mod x if you like. So, compute the Fourier transform of the signum function the signum function is h x minus h of minus x where h is a heavy side function. So, you can use the heavy side function to create signum function and I want you to calculate the Fourier transform of the signum function. So, denote the signum function by u namely h x minus h of minus x you call it u and use a definition of the Fourier transform namely u hat paired with g is simply u paired with g hat u paired with g hat means you will it is a integral it is an integral over 0 infinity and an integral over minus infinity 0 the two integrals will combine to give you one integral after change of variables and. So, u hat is given by the prescription g maps to integral 0 to infinity g hat chi minus g hat of minus chi d chi. Now, you need to evaluate this integral again throw in the e to the power minus epsilon chi squared and then perform the switching of the integration and you finish the job. It is a extremely important calculation and please do this as it is going to be very useful it is going to be is going to teach many things. Now, in chapter 4 we have seen that the Fourier transform exchanges differentiation and multiplication by the coordinate variable or to use a physicist's lingo the momentum and the position operators are dual operators with respect to the Fourier transform it converts the momentum operator into the position operator and vice versa. Of course, there is a minus sign floating around and you must do the bookkeeping carefully. Now, the question is will that formula carry over to tempered distributions remember if I have a tempered distribution I can differentiate. So, 1 upon id dx can be applied to a tempered distribution if I have a tempered distribution I can multiplied by a polynomial in particular I can multiplied by a coordinate variable. So, what is the story it is summarized as theorem 114 suppose you use a tempered distribution then as distributions we have the following equalities the you take the distribution you differentiate throw in the I and then take the Fourier transform that is same as first taking the Fourier transform and multiplying by the coordinate variable that is the first identity the second identity says that take the tempered distribution multiply by the polynomial x and then take the Fourier transform that is same as taking the Fourier transform first and then applying the 1 upon id dx except that here you will find a minus sign. So, you have to prove these two results the proofs are very routine and they follow immediately from the definition of Fourier transform of a tempered distribution and of course we will have to use this corresponding formulas for the Schwartz class that we established in chapter 4. So, we have to use those formulas in chapter 4 to derive these formulas let us do the first one and I leave the second one as an exercise to the audience. So, pick an element in the Schwartz class S of r it is a rapidly decreasing function and I want to calculate 1 upon id dx u hat applied to g by definition it is just put the hat on the other side and take 1 upon id dx u applied to g hat and now push the derivative of the other side how do I differentiate it distribution you simply throw the derivative of the other side and put a minus sign. So, it is u paired with minus 1 upon id dx g hat, but use the formula in chapter 4 g is the Schwartz class is rapidly decreasing. So, minus 1 upon id dx applied to g hat is the same as multiplying g by chi and then taking the Fourier transform. So, so far we have come here, but now again apply the definition of the Fourier transform of a tempered distribution. So, put the hat on the other factor u hat. So, it is u hat paired with chi g, but how do I pair u hat with chi g it is the same as chi u hat paired with g remember the rule for multiplying a distribution of the polynomial simply you can put the polynomial over here on the other side and you can take it from the other side on the left hand side whichever way. So, here we first put the hat on u by the definition of the Fourier transform and put the polynomial chi on the left hand side and you get this. So, we see that 1 upon id dx u hat paired with g is the same as chi u hat paired with g in other words 1 upon id dx u hat is the same as chi times u hat and the other equation 10.12 is proved in exactly the same way there is no change only changes are cosmetic. Now we have ready to prove the inversion theorem. So, let us now state and prove the inversion theorem precisely suppose I take a tempered distribution u and I take a Schwarz function g I want to understand what is u hat hat what is this tempered distribution u hat and u is a tempered distribution u hat is a tempered distribution again take its Fourier transform and I can ask what is u hat hat u hat hat is the same as u except for the switch in the sign. So, what do I mean by switching the sign in a distribution what exactly is the definition of that and that is clarified by theorem 115 suppose u is a distribution and g is a Schwarz function then u hat hat there is a tempered distribution how does it act on g the way it acts on g is 2 pi times u acting on the reflected function g of minus x. So, this is the avatar of the Fourier inversion theorem for distributions. So, the same thing carries over for tempered distributions also. So, let us look at the left hand side the left hand side is u hat hat paired with g that will be first put the hat on the other side. So, it will be u hat paired with g hat put the hat again on the right hand side will be u paired with g hat hat, but g is the Schwarz class. So, we are back to chapter 4. So, g hat hat as per chapter 4 is 2 pi times g of minus x the 2 pi being a scalar comes out innocently. So, it will be u paired with g of minus x and that is the right hand side and we have finished proving the theorem as an immediate corollary we get a very important result. Suppose I give you an L1 function suppose I give you an L1 function you know that the Fourier transform is in L infinity the Fourier transform may not be in L1. Example take the characteristic function of minus 1 1 take the characteristic function of minus 1 1 that is certainly in L1 take its Fourier transform what is the Fourier transform 2 times sin chi by chi sin chi by chi is not in L1. So, the Fourier transform of an L1 function may not be in L1, but now I am giving you more hypothesis I am telling you that u is in L1 and u hat is also in L1 under this additional hypothesis I can take u hat hat this u hat hat is going to be 2 pi times u of minus x how do I prove this. So, from this theorem 115 we immediately conclude that if u is in L1 and u hat is in L1 then u hat hat is simply 2 pi times u of the reflected variable and that completes the theorem on Fourier version. Now, so far we have been talking about tempered distributions of one variable what happens about tempered distributions in several variables go back to chapter 4. In chapter 4 we develop the Schwarz class S of r in one variable for notational simplicity and at the end we simply indicated how to discuss Schwarz class in several variables and we even talked about radial functions and therefore, a transform is also radial and all those things we proved in chapter 4. Here again we have been confining ourselves to the one variable case for notational simplicity how do I discuss the tempered distributions in several variables well we should start with a Schwarz space in several variables S of r in and I have to put a topology in a Schwarz space exactly as we did earlier. Only thing is that now instead of K and L we will have to take multi indices I will have to take a multi index alpha which is alpha 1 alpha 2 alpha n and I want to differentiate with respect to x 1 x 2 x n correspond into this multi index. So, del del x 1 alpha 1 times del del x 2 alpha 2 times that are del del x n alpha n times so that is the kind of a mixed partial derivative that we need to take. So, when does a sequence f nu converge to f in the Schwarz space you take the difference and you take the alpha derivative where alpha is a specified multi index and you multiplied by any polynomial whatsoever q x you can take a monomial if you like x to the power beta but does not matter you can take a polynomial as an equivalent definition. So, supremum over r in and that must go to 0 for every choice of multi index alpha and for every choice of polynomial q x if this happens then we say that f nu converges to f in the Schwarz class as it happens the one variable case the several variables also S of r n is a complete metric space and this notion of convergence arises from a complete metric as in the one variable case this is not going to be a normal space as in the one variable case it is going to be a locally convex topological vector space and hence the dual space which is a rich space S prime of r n is a space of tempered distributions in several variables. In other words the tempered distribution in several variables is simply a continuous linear map from S of r n to a scalars. So, it is a linear transformation with a target space a scalars and this linear transformation should be continuous. So, there are no new ideas there are just cosmetic changes in passing from one variable to several variables as far as the definitions are concerned the same thing is true for Fourier transform to define the Fourier transform you simply use the same prescription as before use a tempered distribution and u hat is its Fourier transform I will tell you what u hat does to g. So, the pairing of u hat with g is simply u paired with g hat this prescription g maps to u paired with g hat is a continuous map from S of r n to the scalars. So, this also defines a tempered distribution the support of a tempered distribution of several variables is defined exactly as in the one variable case it is the smallest closed set outside which the distribution is 0 and I leave it to you to formulate precisely the definition of a support just the same thing simply carries forward verbatim. So, now the theorem on point supports theorem 117 a tempered distribution whose support is a single point origin what kind of distribution is it of course the Dirac delta will be supported the origin the derivatives of the Dirac delta now the derivatives are multidimensional derivatives del del x 1 to the power alpha 1 del del x 2 to the power alpha 2 del del x n to the power alpha and that kind of differentiation you have to take on the Dirac delta they those will also have point support namely the origin and their finite linear combinations will have point support the converse is true and that is theorem 117 suppose if u is a tempered distribution whose support is a single point namely the origin then this distribution is a linear combination of finitely many derivatives of Dirac delta. So, namely u equals summation c alpha delta naught to the power alpha mod alpha less than or equal to 1 there are finitely many derivatives of Dirac delta and its linear combination. So, this theorem is extremely important we are not proved this for one variable case we are not we are not going to prove it for several variables also in the next capsule we are going to give a nice application of theorem 117 to prove Lieuweil's theorem which is a generalization of the Lieuweil's theorem for complex analysis in elementary complex analysis you proved a Lieuweil's theorem we are going to generalize it to harmonic functions but we are going to use distribution theory to dispose of the proof quickly and efficiently. I think this will be a very good place to stop this capsule we will continue this next time. Thank you very much.