 Hello, so let us recall that last time we defined the Fourier transform of a function in L 1 of r and we also discussed several computations of Fourier transforms. We did the Fourier transform of the characteristic function of minus 1 1 it was 2 sin chi by chi. We discussed other examples such as the Fourier transform of 1 upon x square plus a squared which is related to the Cauchy distribution in probability theory and we discussed the Fourier transform of e to the power minus a x squared. The most important example is that of the Gaussian. The Fourier transform of the Gaussian was another Gaussian. This is something that we saw right in the first chapter, but we again looked at it from a different perspective. We did this calculation using complex analysis. Now, we shall introduce the Schwarz space of rapidly decreasing functions. This is a very convenient function space to work with. This Schwarz space is a subset of L 1 of r. So, all the elements in the Schwarz space will have a Fourier transform. So, this is a convenient class of functions introduced by Laurent Schwarz in his influential work theory, the distributions Hermann-Paris as I indicated in the slides. This function space has the advantage that it is a vector space and is closed under differentiation as well as multiplication by polynomials. If you want a function space to be closed under differentiation, that means, if I differentiate the function, it should again be in my vector space. So, it must contain. So, it must consist of c infinity functions only. It cannot have anything else. Further, we want the functions to decay and it should decay even after multiplication by polynomials. So, these are some very stringent requirements and we can also show later that functions which are in L 1 can be approximated by this Schwarz class S. In fact, this space S of rapidly decreasing functions is dense in L p of r for every p between 1 and infinity, 1 included infinity excluded. And one first proves the results on Fourier transforms in S and resorts to approximation techniques. The situation is reminiscent of the proof of the Parseval formula where in the context of Parseval formula, we first proved it for a convenient class of functions and we approximated it. Let us get to the definition. The space S consists of infinitely differentiable functions f of t such that for any natural numbers m and n, I differentiate the function n times and I multiply it by t to the power m. And the result should again be bounded. So, it should be bounded. So, for example, taking n equal to 0 and m equal to 0, the function f of t itself must be bounded. Taking m equal to 0 and n equal to 1, 2, 3, etc., the derivatives must all be bounded. And then by taking n and m arbitrary, not only are the derivatives bounded, after multiplying the derivatives by powers of t, it must still be bounded. So, there is a lot of demands. This class of functions is a very convenient class because of these decay properties. If this is true, then this is going to remain true when I multiply this by c and take linear combinations. And so, it means that if I take a polynomial p of t, then p of t times any derivative should be bounded. The function e to the power minus a t squared obviously has these properties. e to the power minus a t squared is certainly in the Schwarz class because it is rapidly decreasing. You differentiate it any number of times and you multiply by any polynomial, it will remain in the Schwarz class. For the same reason, p t times e to the power minus a t squared, that will also be in the Schwarz class. So, we already have quite a good supply of elements in the Schwarz class. Are there any others? There are plenty of others. First, let us take the hyperbolic cosine cosh a t where I am going to take a not equal to 0. In fact, I am going to take a positive because cosh is an even function. So, I am going to take cosh a t. Cosh a t grows exponentially fast. Cosh a t grows exponentially fast. So, 1 upon cosh a t will decay to 0 very rapidly. Of course, one should be sure that cosh a t does not vanish because if it were to vanish, then I cannot take its reciprocal. But it is an easy exercise for you. Directly from the definition, you can check that cosh a t does not vanish for any reality. Remember a is already real. And so, 1 upon cosh a t is definitely a smooth function. And it decays very rapidly as t goes to infinity. You differentiate it any number of times and you multiply it by any power of t, it will decay rapidly. And so, these functions cosh a t inverse, they are in the Schwarz class. What about sinh a t? 1 upon sinh a t also has the same behavior as t goes to infinity. There is only one small problem. One tiny little problem we have to deal with. Where does sinh a t vanish? A is real positive and t is real. Sinh a t vanishes exactly at one value namely t equal to 0. So, 1 upon sinh a t is not going to be smooth. It is not even going to be continuous. Multiply it by t. t by sinh a t is going to be a smooth function because limit as t tends to 0, t by sinh a t exists. It is 1 upon a. So, assign the value 1 upon a at the origin. And with this assignment, t by sinh a t is an element of the Schwarz class. These are easy exercises for you to do. Some other examples of Schwarz class, some more exotic example, here is one. Take the familiar gamma function. I hope you all remember the gamma function from elementary calculus courses. So, take the gamma function gamma z. What is gamma z? Integral 0 to infinity e to the power minus t, t to the power z minus 1. This function is holomorphic in the right half plane. This function gamma z is holomorphic in the right half plane. So, in the right half plane, I take a vertical line, vertical line a plus i t, where a is real positive and t is an arbitrary real number. In other words, I restrict the gamma function to a vertical line in the right half plane, take its absolute value and square it. Mod gamma a plus i t, the whole squared. That is an element in the Schwarz space s and I leave it as an optional exercise. You probably want to use the Stirling's approximation formula or some other technique that you may think about. Some are the properties of s. If you take two functions f and g in s, then obviously f plus g is also in s and take a function f in s, c times f is also in s. This class s is a vector space. It is very evident that this class is a vector space, but something more is true. I already indicated to you that if a function is in s, its derivative is also in s. If a function is in s and I take a polynomial p t, then the product polynomial p t times f of t is also in s. More generally, if you take two functions f and g which are in s, then the product f into g is also in s. Here of course, polynomials are not in s. They are special. They are special cases where the product is in s. Instead of a polynomial, if I replace it by e to the power t, answer is false. It is not true that if f of t is in s and I take the exponential function e to the power t times f of t is in s. You have to find a counter example. For every f of t in s, you take the absolute value of f of t and integrate it over r and that is finite. And so elements of s are absolutely summable. They are in l 1 of r and the Fourier transform is defined for every function f in the schwarz space s. Now, we are going to show that if you have a function f of t in s and the Fourier transform also lies in s. This is one of the most pleasant features about the schwarz space that it is invariant under Fourier transforms. Now, you might ask what happens with l 1? Why cannot we take a function in l 1 and look at its Fourier transform? Let us take a simple function in l 1, the characteristic function of minus 1 1, the characteristic function of minus 1 1. What is a Fourier transform? Is a Fourier transform again in l 1? No. So, I have given you a simple example of a function in l 1 whose Fourier transform is not in l 1. So, this is the problem of working with l 1. l 1 is not very convenient to deal with when it comes to Fourier transform. Whereas, s is very convenient. This schwarz class is very convenient. It is invariant under the Fourier transform. Also, let us look at the list of examples of Fourier transforms that you have computed. Try to find some eigenvalues of the Fourier transform. Try to find some of the eigenvalues of the Fourier transform looking at the examples. The Fourier transform maps s to s. The Fourier transform is a linear transformation from s to s. And so, it is of interest to know what are its eigenvalues. I can give you another exercise. We know that Cauchy-A.T. inverse is in s. Try to find the Fourier transform of Cauchy-A.T. inverse using complex analysis, using Cauchy integral theorem. Try to find the Fourier transform of this. And we can see whether there are more eigenvalues and more eigenfunctions and we will get all of them very soon. The first theorem, if f is in s, we know that the Fourier transform is also in s. But we have not proved it. We just stated it, but we need to look at the proof of this. First thing is, let us write the definition of the Fourier transform. f hat of chi equal to integral minus infinity to infinity e to the power minus i T chi f of T dt. Let us differentiate under the integral sign with respect to chi. Differentiating under the integral sign is permissible because when I differentiate under the integral sign with respect to chi, each time I differentiate, I am going to pick up power of T and a constant. Constant, of course, comes out of the integral and these powers of T will get clubbed with this f of T. But f of T is the Schwarz class. So, T to the power n times f of T is also in the Schwarz class. So, T to the power n f of T is in l 1. So, this differentiation under the integral sign is permissible. And now, let us write the integrand as e to the power minus i T chi T to the power n f of T 1 plus T square into 1 plus T square inverse. The bracketed term is bounded in absolute value. And so, we multiply by minus i chi to the power k. Multiply by minus i chi to the power k and we put 1 i below. We put 1 i below and we say that is equal to minus 1 to the power n integral minus infinity to infinity d dt of e to the power minus i T chi T to the power n f of T dt. Integration by parts and using the fact that the derivatives of T to the power n f of T also lie in S. We conclude that chi to the power k dn f hat of chi is also bounded for every n k. So, this so, what we are done is that we are differentiated the Fourier transform n times and you are multiplied by chi to the power k. These i's that float around are innocent constants that have been introduced for our own convenience. So, that the expression of the left hand side can be conveniently written like this. And we can check that this particular object over here is bounded. So, we conclude that f hat is in the Schwarz class. Now that we know the S is invariant under the Fourier transform that is if a function is in the Schwarz class its Fourier transform is also in the Schwarz class. We know this. So, now we can do other things we can prove that if f is in the Schwarz class. In fact, differentiate the function and then take the Fourier transform that is the same as first taking the Fourier transform and multiplying by chi except for the existence of this factor i here. So, 1 upon i d dt. So, Fourier transform of 1 upon i d dt f is chi times f hat. On the other hand if I first multiply by the independent variable and I take t f t and then take the Fourier transform it is minus 1 upon i d d chi f hat of chi. So, these two formulas go hand in hand. The formulas essentially tell you that the Fourier transform exchanges 1 upon i d dt and multiplication by chi except that in one of the formulas there is a minus sign involved. To prove the first part simply integrate by parts and you will get the first equation. So, let us do the second one I leave the first one for you as an exercise with a hint to do the second part you differentiate the integral with respect to chi. So, first write the definition of Fourier transform f hat of chi equal to integral minus infinity to infinity e to the power minus i t chi f of t dt differentiate with respect to chi and take the derivative under the integral sign it is minus i t e to the power minus i t chi f of t dt you divide by minus i and you get the second formula straight. So, these two formulae have been established and recorded as theorem 42. Let us now come to the Hermits differential equation and Hermits polynomials. The Hermits differential equation is this ODE y double prime minus 2 x y prime plus 2 lambda y equal to 0. So, this is the Hermits ODE this ODE appears in quantum mechanics. You can for example, refer to any books on quantum mechanics for example, you can look at Robert Resnick's quantum physics or you can look at quantum mechanics by Pauling and Wilson Linus Pauling and Wilson that is published by Dover. So, this differential equation has a series solution. So, you have probably derived the series solutions of this differential equation in your elementary differential equations course, but let us do something else. Let us make the substitution y e to the power minus x squared by 2 equal to u. So, let us may or y equal to u e to the power x squared by 2. When you make the substitution, the differential equation for u assumes a simple form u double prime minus x squared u plus 2 lambda plus 1 u equal to 0. The beauty of the transformed equation is that there is no middle term minus 2 x y prime. Now, assume that u is a solution of the transformed ODE then if I formally take the Fourier transform then you will realize that u hat also is a solution of the same differential equation. The reason is because when you take the Fourier transform of u double prime, I am going to get minus x squared Fourier transform of u. When I take the Fourier transform of minus x squared u, I am going to get the second derivative of u hat. You are going to use theorem 42 twice and the differential equation remains invariant under Fourier transform thanks to theorem 42 in the previous slides. So, the next problem is to show that the equation y double prime minus 2 x y prime plus 2 lambda y equal to 0 when the parameter lambda is a natural number including 0 has a polynomial solution of degree exactly n. That is the Hermits differential equation that you see in the previous exercise when lambda takes a positive integer value has a polynomial solution. Now, you must prove that there is only one polynomial solution up to scalar multiples. Any one of these polynomial solutions of degree n is denoted by h n of x. Unlike the case of the Legendre polynomials, there is no universally accepted normalization for these Hermit polynomials. We shall use this Hermit polynomials very soon. The next problem is to rewrite the Hermits ODE in self adjoint form. What is self adjoint form? Take this differential equation that you see in exercise 14 y double prime minus 2 x y prime plus 2 lambda y equal to 0 multiplied by e to the power minus x squared. When you multiply by e to the power minus x squared, the first two terms will combine to give you d dx of e to the power minus x squared y prime. So, that is how you get the self adjoint form of the Hermits differential equation. Now, use this particular form to show that if m and n are distinct natural numbers, then e to the power minus x squared by 2 times the mth Hermit polynomial is orthogonal to e to the power minus x squared by 2 times the nth Hermit polynomial. That is, these two functions that you see displayed in the last line are orthogonal in the Hilbert space L 2 of R. At most one of the solutions of the transformed ODE lies in S. Now, the tricky question is the following. How do you know that the Hermits equation has a solution which is lying in the Schwarz class? What we have proved is we have proved that if differential equation in U has a solution which is in the Schwarz class and the Fourier transform is also a solution of the same differential equation. Now, I am telling you something else. Now, I am telling you that this differential equation in U cannot have two linearly independent solutions which are in the Schwarz class. Only one at most can be in the Schwarz class up to multiples of course. So, now this will tell you that if it has a solution in the Schwarz class, then the Fourier transform which is also a solution of the same equation must be a multiple of U. So, that solution U must be an eigen function of the Fourier transform operator. Now, if lambda belongs to n, then the h n e to the power minus x squared by 2 where h n is a nth Hermite polynomial. Now, this polynomial times e to the power minus x squared by 2 lies in S and this particular object h n x minus x squared by 2 is a solution of the U equation. So, what exactly are we saying? We have to remember that in the first exercise, you have the Hermites differential equation y double prime minus 2 x y prime plus 2 lambda y equal to 0. You must solve the Hermites equation using power series and you will realize that when lambda is an integer, this Hermites differential equation has a polynomial solution n and that polynomial will have degree n and that degree n polynomial is called h n x. That is a nth Hermite polynomial. Thanks to this first exercise that you are done, it follows that if h n is the nth Hermite polynomial multiplying by e to the power minus x squared by 2 furnishes as a U which is a solution of this differential equation. So, this differential equation is invariant under the Fourier transform and it has a unique solution in the Schwarz class. Therefore, h n e to the power minus x squared by 2 and its Fourier transform both satisfies the same differential equation. So, h n e to the power minus x squared by 2 Fourier transform must be a multiple of h n e to the power minus x squared by 2. In other words, we have an eigen function of the Fourier transform operator. Okay. Now, what about solving exercise 16? How do you show that there is at most one solution of this U equation in the Schwarz class? You use the Abel-Lewel formula. You use the Abel-Lewel formula. What are the Abel-Lewel formula in elementary differential equation? y double prime plus p x y prime plus q x y equal to 0. Second order OD, take the Ronskian W. Take the Ronskian W. Then you got a differential equation for W, right? You get a first order OD for the Ronskian and this first order OD for the Ronskian can be integrated and you can determine the Ronskian. Now, if p is 0, if the differential equation has no middle term, y prime term, then this Ronskian is going to be a constant. So, that is the case with the U equation, right? There is no U prime term. So, if I take two solutions, U1 and U2 of the U equation, the Ronskian is going to be constant and if I take two linearly independent solutions, the Ronskian should be a non-zero constant. Now, what are the Ronskian U1, U2, U1 prime, U2 prime? If U1 and U2 are both in the Schwarz class, what happens is the first row U1, U2, U1 prime, U2 prime. What happens to this as you go to infinity? It goes to 0, but the Ronskian is constant. So, it must be identically 0. So, U1 and U2 must be linearly dependent. So, if I take two solutions of this differential equation which are in the Schwarz class, then the Ronskian must be identically 0 and so, those two solutions must be linearly dependent. So, one of them must be a multiple of the other. So, this way you can determine a large number of eigenvalues and eigenfunctions for the Fourier transform operator. I think it is a good time to stop this lecture here. We will continue this next time. Thank you very much.