 Hello, so let us recapitulate that in the last capsule we looked at examples of compact operators arising from two point boundary value problems in ordinary differential equations. In fact, we specifically took y double prime plus lambda rho xy equal to 0 with y of 0 is 0 and y of 1 equal to 0 and we saw that the solution operator can be written as integral 0 to 1 g of xt f of t dt where lambda is not an eigenvalue and this kernel g of xt called the Green's function was symmetric. We also discussed what happens when you take other types of boundary value problems such as the Laplace's equation in higher dimensions for a ball with Dirichlet boundary conditions. We do not necessarily get a kernel which is in L2 of the product domain. Now, the symmetry of the Green's function is related to self adjointness or the operator on a Hilbert space. Let us recall now self adjoint operators on a Hilbert space and let us prove the spectral theorem for compact self adjoint operators on a Hilbert space. So, recall first that if a is an n cross n real symmetric matrix and the inner product of axny is the same as the inner product of xny for xny in rn and the triangular bracket of xy denotes a usual dot product in rn. The definition of self adjoint operators on a Hilbert space is modeled along this. Let us take a bounded operator. A bounded operator t from h to h on a Hilbert space is said to be self adjoint if the inner product tx,y is the same as the inner product x,ty for all xy in h. Again, I want to emphasize that this definition is for bounded operators on a Hilbert space. In fact, an important component of operator theory on a Hilbert space concerns unbounded operators and this is not the definition of self adjoint if the operator is not bounded. This would be the definition of a symmetric operator. Self adjointness of unbounded operators is a tricky business and we are not going to discuss those here. So, I want to emphasize and I want to flag this word bounded operator. A bounded operator is self adjoint if the inner product tx,y is same as the inner product of x,ty. Let us examine this when we have a Hilbert Schmidt operator. When is a Hilbert Schmidt operator with continuous kernel self adjoint? We are going to be dealing only with ordinary differential equations. Regular stimuli will problems and so my kernels will be continuous. So, it is not a problem. If you are studying partial differential equation like the Laplace zene etc., you will encounter kernels which become infinite and there is a different story altogether. We want to keep the notions simple. Now, if the kernel is complex valued, then one has to deal with complex conjugates. Here, we are only going to look at real kernels because these problems are going to arise from a two-point boundary value problem with a real density row x. Remember, the row x was the density of a string or a membrane or something like that and these densities are real valued functions and so for the most part, we are going to be dealing only with real valued functions and so there is no need to put the complex conjugates. Hilbert Schmidt operator that we get will be self adjoint and if and only if the kernel is a symmetric kernels. That is what we are going to check. So, that is the next theorem. Suppose K x t is a continuous function on the closed rectangle 0 1 cross 0 1, then the Hilbert Schmidt operator t from l 2 0 1 to l 2 0 1 given by t f x equal to integral 0 to 1 K x t f of t dt self adjoint if and only if the kernel is symmetric. The Volterra operator is not self adjoint because the kernel is not going to be symmetric for the Volterra operator. Directly look at what it is and you can check it directly. So, in particular since the Green's function for a regular-sterm level problem with Dirichlet boundary conditions is symmetric, those Hilbert Schmidt operators arising as solution operators are automatically compact and self adjoint. So, we are going to be working with the real Hilbert space as I explained to you. Just for simplicity, you will have to make minor modifications with the Hilbert space is complex. So, f and g are in l 2 of 0 1. So, t f in a product with g is integral 0 to 1 integral 0 to 1 K x t f of t g of x dx dt. On the other hand, what will be f in a product with t g integral 0 to 1 integral 0 to 1 K t x f of t g of x dx dt. Now, the condition that t f g equal f t g translates to the condition double integral K x t minus K t x f of t g of x dx dt equal to 0 for all l 2 functions f and g. Now, this implies using for example, the Weier-Strauss approximation theorem if you like that K x t equal to K t x. Well, one way to do do this would be take f to be a polynomial and take g to be a polynomial in one variable and more specifically take f to be a monomial t to the power i take g to be a monomial x to the power j. And so, t to the power i x to the power j this equation holds and take the linear combinations. And so, this holds for polynomials in two variables t and x and polynomials are dense. And so, I can replace this f of t into g of x by any continuous function psi of t comma x. And since continuous functions are dense in l 2 it will be true for all l 2 functions of two variables. And so, that will mean that K x t minus K t x will be identically 0 because I can take that l 2 function of two variables to be this itself. And I would get integral 0 to 1, integral 0 to 1 mod K x t minus K t x the whole square d x d 2 to be 0. That is how you prove this result. So, this completes the argument. Eigen values recall that in linear algebra you take a linear transformation t from r n to r n. The following are the equivalents t is injective and t is surjective. This follows immediately from the rank nullity theorem. Remember that the domain and the co-domain have the same dimension n. So, what does the rank nullity theorem tell you? The rank nullity theorem tells you that n will be equal to rank of t plus nullity of t. So, if nullity of t is 0 then rank of t will be n. If rank of t equal to n then nullity of t will be 0. From that it immediately follows that injectivity and surjectivity are equivalent for a linear operator t from r n to r n. In infinite dimensional spaces the above equivalents fails. So, let us take a Hilbert space h which is infinite dimensional and let t from h to h be a continuous linear map. A complex number lambda is said to be the spectrum of t if it satisfies one of the following conditions t minus lambda i fails to be injective, t minus lambda i fails to be surjective. So, these are distinct notions in an infinite dimensional setup. In finite dimensional setup they are equivalent notions they are one and the same notions. So, the spectrum of a linear transformation can consist of two types of objects. Those lambdas for which t minus lambda i fails to be injective those lambda for which t minus lambda i fails to be surjective. We are mainly concerned with the failure of one. We say that lambda is an eigenvalue of t if t minus lambda i fails to be injective. In other words lambda is said to be an eigenvalue of t if there exists a nonzero v such that t v equal to lambda v. Anyway, so spectrum will consist of both these pieces in case of one we will call it an eigenvalue. Here there is no question of finding the characteristic polynomial and stuff like that we are not talking about an infinite dimensional Hilbert space. Let us recall the open mapping theorem from functional analysis. Suppose x and y are metric spaces and f from x to y is a mapping then f is continuous if and only if inverse image of an open set is open. So, f pulls back open sets to open sets then you say that the function is continuous. Of course, open can be replaced by closed equivalently if inverse image of a closed set is closed. Now, let us look at a bijective continuous map then we can say that f inverse from y to x is continuous if it pulls back open sets to open sets. In other words, if f inverse inverse of g is open whenever g is open, but f inverse inverse is f remember I am talking about bijective continuous maps. So, this leads to an important notion of an open mapping the requirement is that if o is an open set in x f inverse inverse of o should be open in y right, but f inverse inverse of o is the same as f of o. So, f of o must be open whenever o is open continuity of the inverse will require the direct map to have the property that f of o is open whenever o is open. Again, we could replace the word open by the word closed everywhere and again we will get the requirement for f inverse to be continuous. So, this is a useful definition of map f from x to y x and y are metric spaces again is said to be an open mapping if whenever o is open in x f of o is open in y. A mapping f from x to y is said to be a closed map if whenever o is closed in x the image f of o is closed in y. So, we have three notions continuity openness and closedness continuity means inverse image of an open set is open equivalently inverse image of a closed set is closed. Open mapping means direct image of an open set is open closed means direct image of a closed set is closed. A bijective continuous map from x to y is a homeomorphism if it is an open mapping or it is a closed mapping. To conclude that a bijective continuous map is a homeomorphism you need that it should be an open mapping or a closed mapping otherwise you cannot conclude. Of course, if this domain is compact and the core domain is housed off then it will follow because compactness is a very strong concept. How are you going to check whether a mapping is an open mapping? It is usually quite a task to show that a mapping is open or a mapping is closed it is not easy it requires some work. If x is a compact metric space and y is any metric space then a continuous map from x to y is automatically a closed map because if you take a closed subset of x that means you are looking at a closed subset of a compact metric space. Close subset of a compact space is compact and continuous image of a compact set is compact. So, f of c which is going to be compact and y is a housed off space and in a housed off space compact sets are closed so f of c will be closed. So, we get it for free a continuous map will be closed provided the domain is compact and the core domain is housed off. So, that is one case where we can easily figure out whether a mapping is closed or not. Another situation that is very frequently appearing is the following item number 2. G is an open subset of the complex plane and f from g to c is a non-constant holomorphic function and g is connected. The open mapping theorem in complex analysis says that f is an open mapping. The open mapping theorem in complex analysis is a non-trivial business by the way. It is a non-trivial fact that a holomorphic function from a connected domain to c if it is non-constant it will be an open mapping. A similar result but in multivariable calculus. So, let g be a connected open set in Rn and f from g to Rn is a smooth mapping. Once it is a smooth mapping then df what is df it is a Jacobian matrix and the Jacobian matrix is non-singular for each x. Then you say that this function f is a local difumorphism such a f is called a local difumorphism. A local difumorphism is an open mapping that is a part of the inverse function theorem. It is one of the clauses in the quite non-trivial inverse function theorem that you study in multivariable calculus. For example, look at Rudin's principles of mathematical analysis chapter 9. You will see the proof. Now, theorem 88. Suppose h is a Hilbert space and t from h to h is a surjective continuous linear map. So, we have continuity and it is surjective onto then t is an open mapping. So, directly you can get that t is an open mapping just from continuity and surjectivity. This is not trivial it is quite some work to prove this but it is not difficult either but I will not do this. So, let us come back to the Volterra operator and let us look at the kernel of the Volterra operator. What is the Volterra operator again? It is there on the slide tf of x equal to integral 0 to x f of t dt. First of all, note that since the image of the Volterra operator contains only continuous functions it is not surjective. So, straight away the image is not the whole of L2. We are looking at the Volterra operator as a mapping from L2 of 01 to L2 of 01 our Hilbert space is L2 of 01 and the image is quite a small subspace C of 01. So, it is not surjective. So, it is not an invertible mapping. So, 0 is already in the spectrum. In fact, if you take any compact operator 0 will be in the spectrum. There is a general fact 0 will always be in the spectrum of all compact operators. Is 0 an eigenvalue? What are all the eigenvalues of the Volterra operator? We shall see the Volterra operator has no eigenvalues whatsoever. We shall also see that the spectrum of the Volterra operator consists of the 0 elemental on 0 is not an eigenvalue and yet it is in the spectrum. t minus lambda i fails to be surjective. The question is that could there be nonzero elements in the spectrum? There are no eigenvalues. Just because there are no eigenvalues it does not mean that the spectrum has 0 alone. We have to show that if lambda is not equal to 0 then t minus lambda i surjective. We have to look into that aspect. It may happen that t minus lambda i is injective but t minus lambda i fails to be surjective. That could be the reason why lambda is in the spectrum. So, both cases have to be looked at. So, let us look at the question of eigenvalues. First let us show that 0 is not an eigenvalue. Suppose 0 is an eigenvalue. What does it mean? That means that there is a nonzero vector f such that t f is 0 that is integral 0 to x f of t dt is 0 for all x 0 function. Now, this means that integral 0 to y f of t dt is also 0 and so subtract integral from x to y f of t dt is 0. So, the integral of f over every interval is 0 and therefore using the standard facts from basic measure theory integral of f of t dt over every measurable set A is 0. But this immediately forces f to be identically 0 and that is a contradiction. The last part I am leaving it as an exercise integral of f of t dt over A is 0 means f is 0 almost everywhere. Remember that f is an L2 function. So, this is identically 0 means 0 almost everywhere. So, this last part is a simple exercise in measure theory and so 0 is not an eigenvalue. Let us take lambda to be nonzero, a nonzero complex number. Let us show that lambda cannot be an eigenvalue. Suppose lambda is an eigenvalue, this time I am taking lambda nonzero because I am going to divide by lambda. If it is an eigenvalue then there is must be an eigenvector f that is f is not identically 0 and tf equal to lambda f. What does that mean? What is tf integral 0 to x f of t dt must be lambda fx. Now, the left hand side when you take a function in L2 and you integrate it from 0 to x I am going to get a continuous function. So, left hand side is a continuous function and lambda is not 0. So, divide by lambda and I conclude that f of x was continuous to begin with. But now f of x is continuous to begin with and I am integrating a continuous function from 0 to x I am going to get a smooth function. So, left hand side is now a smooth function and I can write it as 1 upon lambda and that means that f itself was smooth. And by bootstrapping argument I keep proving that f is smooth. So, the integral from 0 to x is a twice differentiable function. So, f itself was twice differentiable and so on. So, by this way you will prove that f of x must be infinitely differentiable function. So, if at all lambda is an eigenvalue lambda not equal to 0 and the corresponding eigenfunction f must be infinitely differentiable. So, because of that I can differentiate 7.27 and I will get the differential equation f of x equal to lambda f prime x. And now immediately that will tell me that f of x must be c times e to the power x by lambda c cannot be 0 because f is not 0. And now if I take this c e to the power x by lambda and put it here in 7.27 I will realize that 7.27 does not hold because the right hand side will be e to the power x by lambda where the factor of c will cancel out and the left hand will be integral 0 to x e to the power t by lambda dt will be equal to e to the power x by lambda that will give you a contradiction check that. So, 7.27 fails and if lambda is not equal to 0 it cannot be an eigenvalue. The proof of 3 is more involved we need to show that if lambda is not equal to 0 then we show that t minus lambda i is surjective. We are just proved that if lambda is not equal to 0 then t minus lambda i is injective now we are going to show that it is also surjective. So, lambda not equal to 0 means that t minus lambda i is going to be invertible. We are used the open mapping theorem by the way remember that t minus lambda i is invertible and the inverse should be a bounded linear operator remember or the continuity of the inverse is because the direct function is open mapping by the open mapping theorem. So, surjectivity of t minus lambda i means what given any l 2 function g I need to solve this equation 7.28 integral 0 to x f t dt that is t f minus lambda f equal to g. Now it is very tempting to simply go ahead and differentiate 7.28 and obtain a differential equation, but that temptation has to be tempered because g is only l 2 and so nobody said that g can be differentiated. So, we have a problem right there. So, let us assume to begin with that g is a smooth function and then let us obtain a formula for this f. So, I am going to start out by assuming that g is the smooth function and I am going to assume that this equation has a solution which means that this particular object can be differentiated and I will obtain a differential equation I will solve that differential equation and then I will figure out how to remove this unnecessary hypothesis that g is smooth. So, all these things will involve some work and I think it is best to do it the next capsule where we will prove that if lambda is not 0 t minus lambda i fail is surjective and so if lambda not equal to 0 then lambda cannot be in the spectrum of the voltage operator. I think with this let us close this capsule and continue with this in the next capsule. Thank you very much.