 Hello. So, we continue our discussion on Fourier transforms of tempered distributions and we studied the one variable case in detail. Now, we go on to the several variables. Now, the Dirac delta, the Dirac delta is a as a distribution in R n. Now, the first question is show that the Fourier transform of the Dirac delta is a constant function 1. It will immediately follow from the definition exactly as the one variable case. So, exercise 16 that I showed you in the first slide we will skip over. This next question that we want to discuss is the Fourier transforms of the derivatives of the Dirac delta. When I said derivatives, what kind of derivatives are you talking about? Suppose we are talking about R 2, am I talking about del 2 by del x del y or am I talking about del 2 by del x squared or am I talking about del 2 by del y squared. So, take a multi index alpha alpha 1 alpha 2 alpha n and you take del del x 1 alpha 1 times del del x 2 alpha 2 times delta del del x n alpha n times. Compute the Fourier transform of this just apply the definition put the hat on the G and then transfer the derivatives out there and you can see that the Fourier transforms of the derivatives of the Dirac delta are basically monomers. So, if I take del del x 1 to the power alpha 1 del del x 2 to the power alpha 2 etcetera it will be chi 1 to the power alpha 1 chi 2 to the power alpha 2 to the power alpha n to the power alpha n. There will be a constant coming in because the Fourier transform exchanges differentiation and multiplication by the coordinate function, but there is an I factor floating around. So, there will be some I to the power mod alpha or something like that I leave it to you to figure out the details. So, that is easy find the Fourier transform of the derivative of the Dirac delta which what is the answer the expected answer is going to be a monomial. What about the Fourier transform of monomial suppose I take a monomial x to the power alpha what will be the Fourier transform no prices for guessing it will be the corresponding derivative of the Dirac delta, but there will be some constant because of the 1 upon I factor. The Fourier transform of the monomial x to the power alpha is going to be a derivative of the Dirac delta the corresponding derivative of the Dirac delta and the Fourier transform of the derivatives of the Dirac delta are the monomials. So, 16, 17 and 18 are through let us take the next question question number 19 I take the unit disk in R 2 x squared plus y squared less than or equal to 1 and I am looking at the characteristic function of D that is a function which is identically 1 on x squared plus y squared less than or equal to 1 and 0 outside the disk and that is a nice L 1 function if you like and so you can compute the Fourier transform using the directly the formula, but I do not want to compute the Fourier transform I want to compute the derivatives of this temper distribution because in L 1 function it is a temper distribution. The characteristic function is not differentiable in the classical sense, but in the classical sense you can still compute the derivative on x squared plus y squared less than 1 in the interior of the unit disk you can compute the derivative it is 0 in the exterior of the derivative x squared plus y squared greater than 1 it is 0. So, the derivative of the characteristic function of the disk D that is going to be 0 inside the disk and it is going to be 0 outside the disk. So, the support of this is going to be the circle x squared plus y squared equal to 1. So, it is going to be a distribution the derivatives of this distribution is going to be a distribution which has support along the unit circle x squared plus y squared equal to 1. What is this distribution? Let us find out. Let us compute the x partial derivative del u by del x. So, what is the formula del u by del x paired with g triangular bracket which is what throw the derivative on g factor put a minus sign it is u paired with del g by del x, but u is a function u is a nice function it is a l 1 function. So, the pairing is basically an integral over d del g by del x dx dy that is what you see in the slide here. Now, use a Gauss divergence theorem to simplify it. So, you got del g by del x and you can apply the Gauss divergence theorem and you can get minus integral over the boundary n 1 g ds. This equality how you apply the Gauss divergence theorem I am going to leave it to you as a vector calculus exercise. Gauss divergence theorem and you will ask me there should be a vector field in the background. What is that vector field? I can give you that vector field. g i plus 0 g that i j are the standard unit vectors in r 2 or an r 3 i j k are the standard unit vectors if you want to use the undergraduate vector calculus notations then i g plus 0 j plus 0 k what are the divergence of this del g by del x and so the double integral that you see is the divergence and so it is going to be what it is going to be n dot the vector field what is the vector field again g i what is the outer normal n n 1 i plus n 2 j. So, n 1 i plus n 2 j dot product with g i plus 0 j what is going to happen it is going to be n 1 g and that you are going to integrate over the boundary of the domain del d. So, that is how we get this. So, we get that the distribution del u by del x is obtained by restricting this function g on the circle x square plus y square d equal to 1 and integrating n 1 g d s. Similarly, you can calculate del u by del y again it is going to be minus integral over the boundary n 2 g d s. Now, I am going to ask you what is x del u by del x plus y del u by del y what is y del u by del x minus x del u by del y please compute these two and one case you are going to get 0 in the other case you are going to get something interesting. So, I leave it to you to complete the problem. So, now let us prove the theorem of Liouville that I have promised last capsule the last capsule we proved that if you take the Laplace's operator Laplacian of u if u is a tempered distribution Laplacian of u is also a tempered distribution what is its Fourier transform use those formulas or the Fourier transform exchange in the derivatives with multiplication by the coordinate function remember there is going to be an I floating around in one of the formulas there is going to be a minus sign floating around, but you are applying the formula twice because it is Laplacian. So, Fourier transform of Laplacian of u is minus mod chi squared u hat that is the function you need to keep in mind ok. So, now we are going to prove Liouville's theorem what is Liouville's theorem in complex analysis it says a bounded entire function is constant but an entire function satisfies the Laplace's equation right if f is entire then Laplacian of u is 0 Laplacian of v is 0 f is u plus i v Laplacian of f is 0. So, an entire function is a harmonic function it satisfies the Laplace's equation and I am saying that the entire function is bounded which means it is a tempered distribution it is a tempered distribution, but why talk about R2 why talk about complex analysis we will work in Rn we will generalize Liouville's theorem and say that if you have a tempered distribution in Rn which satisfies the Laplace's equation then the distribution is actually the polynomial further if you have a tempered distribution which satisfies the Laplace's equation and if that distribution happens to be in LP then u must be 0 if p is between 1 and infinity infinity excluded and if p is infinity that is if the function is bounded harmonic function then it must be constant. So, theorem 118 is a profound generalization of the classical Liouville's theorem from one complex analysis that you teach in undergraduate complex analysis courses and the proof is going to be very simple because we are developed all the machinery that we need to establish such results. So, let us take the Fourier transform of Laplacian of u equal to 0 as I said you are going to get minus mod chi squared u hat is 0 since 0 is in the right hand side I do not have to put the minus sign which means that the tempered distribution u hat vanishes away from the 0 because if chi is not equal to 0 I can divide by mod chi squared and u hat will be 0. So, u hat will be 0 away from the origin which means the support of u hat u hat is a distribution which has point support namely the origin and last time I stated the theorem without proof that a distribution with point support is a finite linear combination of Dirac delta and its derivatives. So, it is going to be summation mod alpha less than or equal to n c alpha delta not to the power alpha that is you have to take alpha derivative the Dirac delta and so what is u? u is going to be obtained by applying hat again and doing the reflection business. So, when I apply hat again I am going to get u of minus x when I take the Fourier transform of the derivative the Dirac delta remember the very first thing we did in the today's capsule that is going to be a monomial x to the power alpha and there is going to be some minus signs and powers of I floating around. So, in anyway those constants are going to be clubbed with this c alpha. So, u is going to be a polynomial in the variables x 1, x 2, x n. So, u is a polynomial and the only polynomials which are in LP are 0 if p is not infinity and the only polynomial that is in L infinity is the constant polynomial and that completes the proof of Liouville's theorem an interesting theorem in partial differential equation that we picked up since we are done a lot of work ok. Now, we talk about a concept called restriction or localization this concept is very familiar from elementary calculus you take a function say sin x on the real line and you want to talk about restricting the function to an open subset. So, you want to restrict sin x to the open interval minus pi by 3 to pi by 3 for example. So, there is a restriction operation and you will denote the restriction by f vertical stroke g we want to discuss the corresponding notion for distributions how do you localize a distribution restricting a function to an open set means localizing the function to the open set just taking the restriction the way to do that would be to take those functions which are compactly supported in g. So, we take a tempered distribution on the real line again we shall discuss it only for the one variable case the multivariable generalization is routine. So, take a tempered distribution in one variable then the restriction of this tempered distribution to an open set g simply means you pair you with those g's which have compactly supported in capital G that is equation 10.17 that you see in the slide. So, this is the business that we are talking about. So, we use the same notation you restricted to g to denote this map 10.17 the support of you is the complement to the largest open set. So, take the largest open set g such that the localization of u to this g is 0 and this g is an open set and the complement of this open set g is the support of the distribution. So, the support of the distribution can be reformulated in this language of localization. Again you know from elementary analysis supposing a two functions f and g and you know that f restricted to g equal to g restricted to g and this happens for all the open sets in f and g are equal. So, localizing is a way to compare two functions in small neighborhoods and see. For example, if you want to discuss whether the function is continuous you want to discuss whether the function is holomorphic you localize the function and you check continuity holomorphy these are local properties to. So, for example, if you if a function f is defined on an open set in a complex plane and to understand whether this function is holomorphic you take a small open ball or an open disc restrict the function to the open disc and check whether it is holomorphic in that in that open disc. So, restricting is a very important idea localizing is very important today. Now suppose if you have two distributions u and v such that u localize to g is the same as v localize to g that means that u minus v when localize to g is 0 that means that the support of the distribution u minus v is outside of this g that is its intersection with g is empty support is contained in g complement. So, or we say that u and v agree on g u and v agree on g in particular if two distributions u and v agree on r minus 0 then what is the support of the difference u minus v the support of the difference u minus v simply the singleton 0 which means that u minus v is a finite linear combination of Dirac deltas and its derivatives it means that u and v differ additively by and Dirac delta and finite linear derivatives p v 1 over x when localize to r minus 0 is simply the smooth function 1 upon x. Now this raises the following question let us turn this around the question that can turn around is suppose I give you a smooth function on r minus 0 1 upon x is an example another example could be 1 upon sinh x hyperbolic sin x 1 upon hyperbolic sin x is a perfectly nice function away from the origin right what is hyperbolic sin x 1 half of e to the power x minus e to the power minus x the derivative is always positive is a strictly increasing function and it vanishes at the origin and it does not vanish anywhere else. So, 1 upon hyperbolic sin is smooth on r minus 0 is there a tempered distribution on r whose restriction to r minus 0 is 1 upon hyperbolic sin. So, given a smooth function f definition 120 given a smooth function f on an open set g in r can you extend this smooth function to a distribution on the real line such that the distribution localized to g is the given smooth function. In other words it is an extension problem you are given a smooth function f on an open set can you extend it as a distribution 1 upon x is smooth on r minus 0 of course 1 upon x is not going to extend smoothly on the real line it won't even extend continuously on the real line forget about smoothness but I am not asking for a continuous extension I am not asking for a smooth extension I am asking for an extension which is a distribution which is a tempered distribution tempered distribution can be ugly no problem it can be like Dirac delta and its derivatives pv1 over x and also all sorts of things. So, the question is there a distributional extension. So, for example that is the next question is does the smooth function 1 upon sinh x the hyperbolic sin does it extend as a tempered distribution on the real line and is this extension unique. Obviously the extension cannot be unique right because if you have one extension add to it a Dirac delta or its derivatives I am going to get another extension. So, the extension is certainly not unique. Let us write 1 upon sinh x is 1 upon x into x upon sinh x minus 1 plus 1 upon x as a simple routine algebra look at the first piece here 1 upon x into x upon sinh x minus 1 remember x and sinh checks both have a simple 0 at the origin they are both holomorphic functions having a simple 0 at the origin. So, x upon sinh checks does not have a 0 at the origin the origin is a removable singularity and x upon sinh checks is a perfectly nice function near the origin and its value at the origin is 1 and a subtracted of 1 look at the Taylor series for x upon sinh checks what the Taylor series for x upon it is an even function right and it vanishes when I subtract of 1. So, x upon sinh checks is 1 so I subtract of 1 and it vanishes and so what are the next term in the Taylor series x squared and the x squared will cancel with the 1 upon x. So, this first piece out here is a perfectly nice function it's a smooth function and it's a tempered distribution because it it doesn't grow it grows in fact it's bounded and the second piece is 1 upon x and you know 1 upon x has an extension to the whole real line what is the extension it is p v 1 over x it extends as a tempered distribution. So, the answer is yes that it does extend and the extension is not unique what about 1 upon x squared remember that distributions can be differentiated as many times as you want take p v 1 over x and differentiate and see what you get you will get some other distribution try to restrict it or localize it to r minus 0 and see whether you get 1 upon x squared the next problem gamma i x modulus the whole squared on r minus 0 remember when x is not 0 i of x is purely imaginary x is not 0 and real number i of x is purely imaginary and non-zero where are the poles of the gamma function the poles of the gamma function are 0 minus 1 minus 2 minus 3 etcetera right. So, mod gamma i x the whole squared gamma function is a is a holomorphic function of the complex plane except for poles at 0 minus 1 minus 2 minus 3 etcetera and so what is this mod gamma i x the whole squared remember that what is gamma z bar it will be gamma of z bar gamma is real when x is real remember gamma restricted to the real axis is real of course it has got poles remove the poles of course so it means that gamma z bar is gamma of z bar. So, what is mod gamma i x the whole squared is gamma i x gamma i x bar in other words is gamma i x into gamma minus i x but gamma minus i x means multiply and divide by minus i x. So, gamma minus i x is going to be gamma 1 minus i x upon minus i x but what is gamma z into gamma 1 minus z the Euler's reflection formula pi by sin pi z. So, this object mod gamma i x the whole squared is basically pi upon minus i x sin i x sin i x is i sinh x and so the answer is this particular function simplifies to pi by x sinh pi x and now the question I am asking you is 1 upon x sinh pi x that is a smooth function on r minus 0 does it extend as a tempered distribution on the real line I think after doing question 20 and 21 it should become easy now the next question is a difficult question now you can say 1 upon x has been extended as a tempered distribution 1 upon x squared has been extended as a tempered distribution of the real line 1 upon x cube 1 upon x to the power 4 you will believe they will extend as tempered distribution. What about e to the power 1 upon x squared that is also smooth on r minus the origin will e to the power 1 upon x squared extend as a tempered distribution on the real line No, answer is a big no. How do you prove that there is no tempered distribution on the real line such that its localization to r minus 0 is exactly e to the power of 1 upon x square. That's a difficult question and that's why it's an optional exercise. So the problems of this kind of extending distributions from an open set to a whole space. For example, I could give you a smooth function like 1 upon 1 minus x square minus y square. Let us look at 1 upon 1 minus x square minus y square. This function is smooth everywhere except along the unit circle x square plus y square equal to 1. So, from the plane remove the circle x square plus y square equal to 1. On the complement 1 upon 1 minus x square minus y square is a perfectly nice smooth function. Is there a tempered distribution which when localized to the plane minus the circle gives you exactly 1 upon 1 minus x square minus y square. This kind of problems in other words, problems of an analogous nature in several variables become very interesting and they are highly non-trivial and call for sophisticated techniques from algebraic geometry. And these matters are closely related to another problem called the problem of division in distribution theory. This problem of division has been studied by a number of mathematicians, Malgrange, Aranpris, Hormander, Lausaceux and a whole lot of people have studied the problem of division and the literature on this problem on division is very vast. So, in the next capsule, we will take up an interesting discussion on distributional solutions of ODE's with polynomial coefficients. Now, we know that a distribution can be differentiated and a distribution can be multiplied by polynomials. We are going to talk only about tempered distributions in this course. And so, when I say distribution, I tacitly mean it is tempered and these differential equations that we are looking at have polynomial coefficients and it is an interesting question whether they have solutions which are other than the classical solutions. The next capsule will take up this very interesting question and we will see some new features arise when you cross a singularity and then this will be a very good place to stop this capsule. Thank you very much.