 Hello, so last time we discussed compact operators and we looked at one example of a compact operator the Volterra operator. The example can easily be generalized and we are going to discuss a whole range of operators called Hilbert Schmidt operator. This is not the definition of Hilbert Schmidt but these are examples of Hilbert Schmidt operators. So the most basic example of compact operators are these classes of operators. The study of compact operators is a very vast chapter in functional analysis but we are going to only look at those aspects which are relevant for generalized Fourier series and Fourier expansions. So that is the example number 1 that you see on the slide. Number 1 the Hilbert Schmidt operator T from L2 of 01 to L2 of 01 given by Tfx equal to integral 0 to 1 kxt f of t dt. Now I had taken a compact interval 01 for convenience and for illustration purposes one could take more general measure spaces and one can look at L2 of those. I am also going to restrict myself to begin with continuous functions kxt. This kxt will be called a kernel function. So there is an integral operator with a kernel. So this is going to be a compact operator. First of all observe that since k is continuous on 01 cross 01 it is uniformly continuous. This uniform continuity of k is going to immediately give us the equicontinuity or the family Tfx. So let us look at this output Tfx input is f of x and output is Tfx. So now by uniform continuity given any epsilon greater than 0 there is a delta greater than 0 such that mod x minus y less than delta implies mod kxt minus ky t less than epsilon. Here I have taken the second variable t to be the same and it is the first variable x that is changing. So k of x t minus k of y t that modulus is less than epsilon. So now let us estimate it. So let us look at what is Tfx minus Tfy. Tfx minus Tfy will be I am applying the triangle inequality integral 0 to 1 the mod goes inside the integral kxt minus ky t into mod f of t dt. Now remember that by appealing to uniform continuity this is less than epsilon. The epsilon comes out and I am left out with mod f t dt and this integral is less than or equal to norm of f by Cauchy Schwarz inequality and I am going to assume that norm f is less than or equal to 1 the L2 norm mind you and so what we get is that this family Tfx is equicontinuous. To establish uniform boundedness we simply take the supremum m of kxt dt and proceed along similar lines mod Tfx is less than or equal to integral 0 to 1 mod kxt into mod f of t dt mod kxt has been dominated by m and again apply Cauchy Schwarz inequality you will get this L2 norm of f here and that is less than or equal to m. So all the functions Tf's are uniformly bounded by m so the family is equicontinuous and uniformly bounded so as soon as norm f is less than 1 the corresponding family of functions Tfx will be a pre-compact with respect to which norm with respect to sup-norm remember that the Oscoli Arzela theorem concerns sup-norm so we got that the family Tfx is pre-compact with respect to the sup-norm which means that given any sequence as a subsequence converging uniformly but uniform convergence will imply L2 convergence we are on a compact interval 0 1 there is no problem and so the family Tfx as f varies over the unit ball in L2 will be pre-compact in L2 norm also and thereby we have finished showing that this operator is a compact operator. Now how do these kinds of Hilbert Schmidt operators arise in the context of differential equations in the context of boundary value problems of ODE's with a 2 point boundary value problem that is equation 7.23 we are looking at y double prime plus lambda rho xy equal to f of x y of 0 equal to 0 and y of 1 equal to 0 we are looking at Dirichlet boundary conditions. So now let us look at this equation 7.23 and we are going to prove that when you try to solve 7.23 we can convert this into an integral operator and the solution y of this equation 7.23 can be obtained as an integral of f of x with k of xt dt where k is a certain kernel and we want to derive this kernel look at its properties and so on that is the job for the next few slides and for the rest of the capsule. So let us first assume that lambda is not an eigenvalue of the homogeneous problem. So when I put 0 on the right hand side lambda is not an eigenvalue so the problem with 0 r h s has only the trivial solution which means that the inhomogeneous problem 7.23 if at all it has a solution can only have one solution. Let us compare this with the elementary situation that we encounter in linear algebra courses namely solving systems of linear equations A x equal to b. So whenever system of linear equations A x equal to b if A x equal to 0 has only the trivial solution that means if the kernel of A is trivial then the inhomogeneous equation A x equal to b has at most one solution either it has no solution or it has a unique solution that we know from basic linear algebra and the situation here is very similar. So we are assuming that lambda is not an eigenvalue so the associated homogeneous equation has only a trivial solutions the inhomogeneous equation if it has a solution that solution will have to be unique. So now let us examine whether 7.23 has a solution at all. So if it has a solution then that solution will be unique and so that will define for me an operator the input is f of x and that unique solution will be the output and thereby you get a mapping from L2 of 0 1 to L2 of 0 1 and that will be linear. Suppose you solve 7.23 with f of x and you solve 7.23 with g of x again then I can call the solutions y and z then y plus z will be the solution of the corresponding problem with f plus g on the right hand side. So the solution operator is going to be linear with respect to this input f of x and this linear transformation will map L2 of 0 1 to L2 of 0 1 and that is exactly what we want to check. So let us assume that y 1 and y 2 are two solutions of the homogeneous problem y double prime plus lambda rho x y equal to 0 and let us take initial conditions for y 1 and y 2. What are the initial conditions I am taking? y 1 of 0 is 1, y 1 prime of 0 is 0, y 2 of 0 is 0, y 2 prime 0 is 1. So have I taken these initial conditions? What is the Ronskian at the origin? The Ronskian at the origin is an identity matrix and so the value of the Ronskian is 1 and so the solutions y 1 and y 2 are linearly independent. Once you have got two linearly independent solutions of the homogeneous equation, how do I solve the inhomogeneous equation with f of x on the right hand side? The method of variation of parameters that you teach your undergraduate students at the sophomore level which we need to recall. So according to the method of variation of parameters, the solution of the inhomogeneous problem which is known as the particular integral or the particular solution is obtained how? It is obtained in the form 7.26 namely v 1 x y 1 x plus v 2 x y 2 x. This technique goes back to Lagrange and Lagrange was motivated by this in his researches in celestial mechanics, specifically the orbit of comets. I have written a paper on the geometrical interpretation or the method of variation of parameters. If your curiosity has been aroused, you can consult this paper. So now let us get on with the solution. So v 1 and v 2 have to be determined. How do I determine v 1 and v 2? There are two equations determining v 1 and v 2. The first one is the osculatory condition v 1 prime y 1 plus v 2 prime y 2 equal to 0. And the second condition is v 1 prime y 1 prime plus v 2 prime y 2 prime is f. Now remember that these are two equations for v 1 prime and v 2 prime. And what are the determinant of these two equations? y 1, y 2, y 1 prime, y 2 prime. They are all skin of y 1 and y 2 and the round skin of y 1 and y 2 is never going to be 0. It is going to be constant and it is going to be 1 throughout. But the Abel-Lewel formula, the round skin is going to be constant because there is no y prime term in the differential equation and the value of the round skin at the origin is 1. So when you solve these two equations using the Kramer's rule, the denominator is going to be 1. There is no problem. So let us solve them by Kramer's rule. You can immediately solve them and you will get v 1 x. In fact, you are going to get v 1 prime x equal to minus y 2 x f of x and v 2 prime x is going to be y 1 x f of x. So you need to perform one integration and we integrate from 0 so that you make it a definite integral. So v 1 x is minus integral 0 to x y 2 t f of t dt, v 2 x is integral 0 to x y 1 t f of t dt. So we have found v 1 and v 2 and we need to substitute over here in 7.26 and write down the particular integral. So we get the particular solution is integral combining these two things 0 to chi y 1 t y 2 chi minus y 1 chi y 2 t f of t dt and let us write this as K of chi t f of t dt where the kernel K chi t is basically K chi t equal to y 1 t y 2 chi minus y 1 chi y 2 t. For t less than or equal to chi and K of chi t is 0 when t is greater than or equal to chi. The kernel is continuous on 0 1 cross 0. Why is the kernel continuous on the rectangle 0 1 cross 0 1? Because what is the value of the kernel when chi and t are equal? When chi and t are equal what is the situation? It is 0 y 1 t y 2 t minus y 1 t y 2 t. So for t less than or equal to chi along the boundary t equal to chi it is 0 when t is greater than or equal to chi anyway it is 0. So, the function is defined in two different ways in the two triangles t less than or equal to chi and t greater than or equal to chi correct and along the interface the two prescriptions match. So, by the basic gluing lemma from elementary matrix space theory the combined thing is continuous. So, K is a continuous function the kernel is continuous and so we are going to get a Hilbert Schmidt operator, but we are not quite done. We have a small problem namely we cannot simply take y of x equal to integral 0 to x K of x t f of t dt that will not do. We need to make sure that the boundary conditions are satisfied. We need to make sure that the boundary conditions are satisfied and in order to do that we have to modify the solution by adding c 1 y 1 x plus c 2 y 2 x where c 1 and c 2 are arbitrary constants and we need to choose these constants c 1 and c 2 in such a way that the boundary conditions are satisfied. So, let us put x equal to 0 y of 0 equal to 0 that straight away gives you c 1 equal to 0 because the left hand side will be 0. So, c 1 y 1 of 0 what is y 1 of 0 y 1 of 0 is 1 remember and y 2 of 0 is 0. So, this goes this becomes 1 and the integral becomes 0. So, straight away c 1 is 0 now you put x equal to 1 you want y of 1 to be 0. So, this one becomes 0 what happens is that we will get c 1 is anyway 0 you will get c 2 times y 2 of 1 plus integral 0 to 1 K of x t f of t dt but x is going to be 1. So, here you are going to put a 1 here and integral 0 to 1 K of 1 t f of t and the c 2 value is minus 1 upon y 2 of 1. Now, here I am dividing by y 2 of 1 how can I conveniently divide how can I be sure that y 2 of 1 is not 0 we are going to make sure of that because y 2 of 1 is 0 it would mean that y 2 of 0 is already 0 y 2 of 1 is already 0 and y 2 is a eigen function with eigen value lambda and that is a contradiction because we are expressly ruled out the lambda being an eigen value we are not allowed the lambda to be an eigen value. So, y 2 of 1 is non 0 and so, division by y 2 of 1 is legal and we obtain my c 2 as well. So, let us put the value of c 1 and c 2 and let us write the solution properly the solution y of x is now incorporating the c 2 1 upon y 2 of 1 integral 0 to 1 K x t y 2 of 1 minus K 1 t y 2 of x f of t dt that is exactly what we got and this entire expression you take the y 2 under the integral sign and combine all these things and call it g x t f of t dt. So, what is g of x t K x t y 2 of 1 minus K 1 t y 2 of x upon y 2 of 1 that complicated expression is called g of x t this is called the greens function for the boundary value problem remember when we are talking about a greens function you may have already encountered this word greens function in undergraduate differential equations courses. It is very popular to see a section on greens functions in all these textbooks, but we have to remember one thing that the you cannot talk about the greens function for a differential equation you have got to give me the differential equation and you have got to give me the problem. Are you talking about an initial value problem then you are going to tell me what the initial conditions are are you giving me a boundary value problem then you have to tell me what the boundary values are. So, the greens function is something that corresponds to a specific problem the differential equation together with the prescription of the side condition. For example, you can take the Laplace's equation or the Poisson's equation you can say solve Laplacian of u equal to f and this u is 0 on the boundary you are given with the Dirichlet boundary value problem and then you can ask for the solution and you can say that the solution u can be written as integral of g x t f of t dt and you will say that the g is a greens function it is not correct to say it is a greens function for the Laplacian you should say it is a greens function for the Laplacian with the boundary value problem for the boundary value problem. So, you have to specify the problem if you are talking about the Neumann problem then the greens function will be different. So, when you say greens function the question is greens function with respect to which kind of problem initial value problem boundary value problem other types of boundary value problem whatever it is. So, here we are talking about greens function for the boundary value problem for the differential equation y double prime plus lambda rho ty equal to f of t the solution has been expressed as a Hilbert Schmidt operator with an L2 kernel why is a kernel L2 it is actually continuous k is a continuous function why y 2 of 1 is a non-zero constant and so everything is continuous. So, it is a Hilbert Schmidt operator that is it is L2 with respect to the product measure. So, let us examine the kernel in more detail and so the when t is less than or equal to x the kernel takes this form when t is greater than or equal to x the kernel takes this form. So, let us compare the two suppose I switch the roles of x and t suppose I switch the roles of x and t and think of g t x then I get y 2 t y 2 x. So, this part remains the same look at the second part if I switch the roles of x and t what happens instead of y 2 t I get y 2 of x instead of y 1 of x I get y 1 of t. In other words the greens function is symmetric. So, we got this beautiful result that the greens function g of x t equal to g of t x what is the cause of this symmetry of the greens function could we have anticipated that this greens function would be symmetric. For example, if you look at elementary books on partial differential equations. For example, look at any book on partial differential equations and turn to the chapter on the Laplace's equation the Dirichlet problem for the Laplacian on a domain omega and there in that context again you will see that the symmetry of the greens function is proved as a theorem there. Why is it that in these problems the greens function is symmetric it could be an accident that this particular greens function is symmetric here and the greens function that you saw in the chapter on Laplacian in a pd textbook is also symmetric. There must be a deep reason for the symmetry of the greens function this is related to the fact that the two-point boundary value problem with Dirichlet boundary condition gives rise to a self adjoint operator. We are going to define the self adjoint operator very soon. So, and the self adjointness is reflected by the symmetry of the greens function. Comment since there is a parameter lambda in the differential equation the greens function g x t will obviously depend upon the parameter lambda. So, technically speaking we should be writing not g x t we should be writing it as g x t comma lambda for the greens function and we are going to expressly postulate that this lambda is not a eigenvalue. As I explained there is a corresponding greens function for the Poisson's equation with Dirichlet boundary condition and the solution can be expressed as a greens function and the greens function again is symmetric in that context. And I want you to consult books on PDEs one recommendation is Fritz John partial differential equations the fourth addition for instance chapter four is about the Laplace's equation you can consult that book find the expression for the greens function for a ball and verify that it is symmetric. Determine the greens function for y double prime plus lambda y equal to f on the closed interval 0 1 with Dirichlet boundary condition much simpler because this row is given to you. So, you can use the previously given expression and you can actually write down the formula for the greens function. The greens function for a ball in R n now when you look at this book by Fritz John that I talked about in fact you should read the text very carefully it is a very beautifully written book on PDEs it is a very classical book but even today if you want to look for a very good and authoritative account for the classical aspects of PDEs Fritz John is the right book to read actually it is well worth reading Fritz John. He was a student of Richard Courant and he already mentioned Richard Courant several times in these course. So, the basic fact is that the greens function in the context of the Laplacian is a correction to what is called as the fundamental solution of the Laplace's equation. The fundamental solution of the Laplace's equation is the Newtonian potential why is it called the fundamental solution let us not get into that it is called the Newtonian potential. So, you are the greens function turns out to be the Newtonian potential plus a correction term in fact the Newtonian potential will satisfy the Laplace's equation except when x equal to chi at chi this function has a singularity and this singularity is required and you want this greens function to vanish on the boundary and so you want the greens function to be such that the Dirichlet boundary conditions are satisfied in order to satisfy the Dirichlet boundary condition you will have to modify by adding the correction term this correction term is going to be a harmonic function in an open neighborhood of the closed ball B is a harmonic function so it is going to be very nice it is going to be smooth in an open set containing the closed ball so this correction term is going to be particularly nice so the difficulty is going to come from the singular term the Newtonian potential and so will this g of x chi will this be an L2 kernel will it be in L2 of B cross B that is a question will it be an operator of the type that we have encountered so far I leave it to you to examine whether this one this piece whether that is in L2 of B cross B it will not be in general in fact if n is greater than or equal to 4 this will no longer be in L2 of B cross B so the solution operator is again of this form u of x equal to integral g of x chi f of chi d chi though you got the solution as an integral the kernel is not going to be in L2 of the product domain there will be a problem when n is greater than or equal to 4 so it does not fit into the scheme of things that we have been looking at so far so far for the ODE y double prime plus lambda rho x y equal to f where rho is a continuous function the green's function was actually continuous on the closed interval 0 1 cross 0 1 and so it is in L2 here the situation is slightly more complicated when the number of space dimensions is larger than 4 it fails to be an L2 kernel with the on the product domain so the next item that we should be looking at in the next capsule are self-fed joint operators so now we are going to be discussing compact self-fed joint operators in a Hilbert space specifically the kind of operators that arise when we solve a two-point boundary value problem a regular stability problem with Dirichlet boundary conditions so as a preparation for that we will need to recall some of the notions from functional analysis such as self-fed jointness we are going to be looking only at bounded operators and compact operators in fact so there is no problem and we will continue this next time thank you very much