 Hello. So, in the last capsule we started the proof of the spectral theorem for a compact self adjoint operator on a Hilbert space. We could complete the proof I said that this proof will spill over into two capsules. We finished the difficult part of it namely we proved that there exists eigenvalues and there exists a eigenvector. So, existence of eigenvalues and eigenvectors were established and for that we used the Banachala-Glue theorem for Hilbert spaces which we took great care and we proved that. The Banachala-Glue theorem is a compactness result and the unit ball in the Hilbert space is not compact with respect to the norm, but it is compact in a very weak sense namely every norm bounded sequence has a weakly convergent subsequence and we use this critically to prove the existence of eigenvalues and eigenvectors. Now that the difficult part is done let us finish the proof and again so let us begin with a setup. H is a Hilbert space, T from H to H is a self adjoint bounded linear operator on H. Right now we do not need compactness we say that Y is a T invariant subspace this is a purely algebraic notion so this can be defined in the context of linear algebra. So, T invariant subspace means whenever V belongs to Y, T V also belongs to Y. So, T of capital Y is contained in capital Y if you like. Now let us prove that if T is bounded self adjoint and if Y is T invariant then Y perp is also T invariant so this is where self adjointness of T is going to come to play. So, let us take a Z in Y perp we have to show that TZ is also in Y perp namely TZ in a product with Y should be 0 for every Y in Y, but what is TZ in a product with Y it is Z in a product with TY because T is self adjoint I can simply put the T from the left factor to the right factor. But where is TY? TY is again in Y because Y is T invariant and where is Z? Z is in Y perp. So, Z is in Y perp and TY is in Y so Z in a product with TY is going to be 0 and that is all we want to show. So, we proved that Z belongs to Y perp implies that Z in a product with TY is 0 for all Y which means that TZ in a product with Y is 0 for all Y which means that TZ is in Y perp and the proof is complete. We are shown that if you have a self adjoint operator and if Y is T invariant Y perp is also T invariant. Now what we would do is that because Y perp is T invariant I can restrict T to Y perp and the restriction will be a linear transformation from Y perp to Y perp. Again it is obviously linear transformation and restriction of a continuous function is continuous. So, again the restriction is a bounded linear transformation and Y perp is a closed subspace. Remember that regardless of Y, Y perp will always be closed and the closed subspace of a Hilbert space is again a Hilbert space. So, T is a bounded linear operator from a Hilbert space to itself and the equation TX dot product with Y equal to X dot product with TY will hold for X and Y in Y perp. In other words this restriction T from Y perp to Y perp is again a bounded self adjoint operator and if T is a compact operator this restriction of a compact operator is also going to be a compact operator. So, this is an important reduction. Now you might have begin to guess that some kind of an inductive argument is going to be given and this is a kind of a preparation for that in inductive argument if that is the kind of thoughts that are creeping into your mind then certainly you are in the right track. So, now let us look at theorem 94 the completion of the proof of the spectral theorem. Suppose T from H to H is a compact self adjoint operator on a Hilbert space then there is an orthonormal basis of eigenvectors of H. Each nonzero eigenvalue has finite multiplicity. So, if you take lambda not equal to 0 and if lambda is a eigenvalue then there are only finitely many linearly independent eigenvectors associated with eigenvalue lambda that this is obviously the geometric multiplicity that we are talking about. There is no concept of algebraic multiplicity because we are in an infinite dimensional setup. We have already established the existence of an eigenvector and now we are going to apply the Zohn's lemma or the Hausdorff's maximality theorem to show that there is a maximal linearly independent orthonormal set consisting of eigenvectors. What is the Hausdorff's maximality theorem? Hausdorff's maximality theorem says that if you have a partially ordered set then maximal chains exist. So, what is this partially ordered set that we are talking about? We have a Hilbert space H it is an inner product space first and foremost. So, we can talk about orthonormal subsets of H. So, we are talking about subsets of H both finite and infinite subsets which are orthonormal and we want to further put the restriction that these are orthonormal sets of eigenvectors. We are talking about orthonormal sets of eigenvectors. That is family of orthonormal sets of eigenvectors is obviously a partially ordered set partially ordered by set theoretic inclusion. Now, the Hausdorff's maximality theorem maximal chains exist. How do I know that this is a non-empty family? To show that this family itself is non-empty remember we proved last time there exists a eigenvector and eigenvectors are by definition non-zero. So, normalize it. So, a singleton consisting of an eigenvector is obviously a orthonormal set. So, this partially ordered set is certainly non-empty and since it is non-empty I can apply the Hausdorff's maximality theorem and I can take a maximal linearly independent orthonormal set B in this Hilbert space H consisting of eigenvectors and this B is non-empty. Now, we want to claim that this B is a basis and that is exactly the first clause in the theorem. So, we have to show that B is a basis. What does it mean to say that B is a basis? It means the linear span of B must be dense in H or the closure of the linear span must be the whole of H. So, take the closure of the linear span of B. Take the closure of the linear span of B. It is a closed subspace of a Hilbert space. If it is not the whole of the Hilbert space H then it is a proper closed subspace. So, let us call this y and if y is a proper closed subspace then y perp is not the zero space and y perp is then a Hilbert space in its own right and the operator is self adjoint. So, I restrict t to y perp as an operator from y perp to y perp. Now, that is going to be a self adjoint compact operator on y perp and therefore, it must have an eigenvector v0 and by normalizing it I may assume that this v0 is a unit vector. And so, now I can look at B union this singleton v0. Obviously, this v0 is not in B right because B is sitting in y and v0 is sitting in y perp and v0 is a unit vector. So, this particular set is a strictly larger orthonormal system of eigenvectors and that contradicts the maximality of B and the claim is proved and the first part of the theorem is not established. Now, we turn to the second part namely the statement concerning the multiplicities of the eigenvalues, the geometric multiplicity if you like. Each nonzero eigenvalue has finite multiplicity. Take a nonzero eigenvalue lambda and let v lambda be the corresponding eigen space and we have to show that this has finite dimension. That is another way of saying that the eigenvalue has finite multiplicity namely the eigen subspace must have finite dimension. How do you show that a subspace has finite dimension? One way to do that would be to show the unit ball is compact. Remember that in a Banach space a unit ball is going to be compact if and only if the Banach space is finite dimensional. So, take u0 to be the closed unit ball in v lambda. If we show that u0 is compact we are finished. So, take a sequence of vectors in u0 and remember that these v lambda is an eigen space. Remember so, T vn will be equal to lambda vn, but T is compact and vn's are coming from a bounded set it is coming from the unit ball. So, T vn is pre-compact. So, T vn has a convergent subsequence converging in norm and lambda is nonzero. So, I can divide by lambda and it follows that vn's itself must have a convergent subsequence which shows that every sequence that I take in u0 has a convergent subsequence namely u0 is pre-compact. Since I take in the closed unit ball it is compact. Now the same argument also reveals another important result and if I fix a number c positive there are only finitely many eigen values lambda with mod lambda bigger than c. When I say finitely many eigen values I mean counting multiplicities and so taking into account the geometric multiplicities there are only finitely many eigen values which exceed c in absolute value. Suppose the result is false then we can select an infinite sequence of unit eigen vectors because the result has been assumed false and which are pairwise distinct eigen values because each eigen value has finite multiplicity. So, there are infinitely many eigen values with mod lambda bigger than c that means that there must be infinitely many distinct eigen values in the first place and so I can select these sequence of unit eigen vectors with pairwise distinct eigen values forming a monotone sequence. Remember that we have proved that t is self adjoint so the eigen values are real numbers and every real sequence has a monotone subsequence. So, passing to a subsequence if necessary I might as well assume that the eigen values that I got from a monotone sequence. So, you got a monotone sequence either it was converged or it must go to plus infinity or minus infinity there are only three choices in each of these three choices will get a contradiction. The eigen vectors are mutually perpendicular remember I am taking distinct eigen values and so the distance between any two of them is root 2. So, the eigen vectors definitely are not going to have a convergent subsequence. Image vectors t, v, n will be lambda n, v, n and the image vectors must have a convergent subsequence because t is compact and the v n's are unit vectors. So, it means passing to the subsequence v, n, k we get the equation v, n, k equal to 1 upon lambda n, k t, v, n, k t, v, n, k's are converging lambda n, k's are a monotone sequence. So, either it must converge or 1 upon lambda n, k must go to 0 in any case v, n, k's must converge in norm that is a contradiction because v, n, k's cannot converge because the distance between any two of them is square root of 2 that completes the proof of the theorem. So, far we are established that for a compact self-adjoint operator on a Hilbert space there is an orthonormal basis of eigen vectors. Each non-zero eigen value is finite multiplicity the eigen values from a countable set and they can only have one limit point if at all and that is 0 and the eigen values can also be generated successively by maximizing the Rayleigh quotients analogous to the case of the real symmetric matrix. So, all these things we have established and now we should apply it to the study of two-point boundary value problems because these eigen functions are going to give us the generalized Fourier series. A general solution is going to be written as a superposition of these eigen bases and that is generalized Fourier series. So, rho from closed interval 0, 1 to r is a continuous function whose 0's form a set of measures 0. It is not the case that rho is identically 0 on some small sub interval or something like that. So, we are excluding those cases by saying that the 0's of rho can at most be a set of measures 0. Of course, rho can have isolated 0's for example. Now, L2 subscript rho denotes a set of all real valued functions such that integral 0 to 1 mod f t squared rho t dt is less than infinity. In other words we are taking the measure to be rho t dt with respect to this measure I am taking the usual L2 space with respect to the weighted measure. Now, this is a vector space over the real numbers and the inner product of two of the functions f and g is simply integral 0 to 1 f t dt rho t dt. The usual inner product I am not putting the bar here because I am looking at real valued functions only. So, with this prescription L2 rho 0, 1 is a Hilbert space. The proof is exactly the same as the usual L2 space on 0, 1 with cosmetic changes that I leave it to you to figure out or if you have done a general measure theory course this is L2 of a measured space where the measure is the weighted Lebesgue measure either approach you can take. This is a space which is the correct setup for studying the Sturm-Lewel problem y double prime plus lambda rho y equal to 0, y of 0 equal to 0, y of 1 equal to 0. Let us see why that is so. We make the observation that the operator t from L2 0, 1 to L2 0, 1 given by t f x equal to integral 0 to 1 k x t f of t rho t dt is a compact operator when the kernel is continuous and is self adjoint precisely when the kernel is symmetric. We approve these things on the measure is the usual Lebesgue measure that is when rho t is identically 1 but the same proof simply goes through with cosmetic changes and so I leave the proofs to the audience because essential ideas are all being discussed. It is just a matter of writing down. Now, we transform the Sturm-Lewel problem into an Eigen value problem for a compact self adjoint operator of the form 7.39. So, that we can apply the spectral theorem for the Hilbert spaces and we can re-derive the results of chapter 6. So, let y of x be an Eigen function namely equation 7.40 y double prime plus lambda rho y equal to 0 with Dirichlet boundary conditions and I am taking the Eigen value lambda we know 0 is not an Eigen value. So, might as well assume that lambda is not equal to 0 we integrate 7.40 twice in succession and when you integrate it once we get y prime of x equal to y prime of 0 minus integral lambda y t rho t dt integrate again and use the fact that y of 0 is 0. The second time when you integrate you are going to get a y of 0 which is which drops out and we are left with this. We will get a repeated integral we switch the order of repeated integrals and that you know how to do integral 0 to x y of t rho of t dt integral d s and this can be integrated immediately and you get the solution in the form y of x equal to x y prime 0 minus lambda integral 0 to x x minus t y of t rho of t dt. So far so good. Now the other boundary condition has to be put in when you put in the other boundary condition we will get y of 1 equal to 0 of y prime of 0 equal to lambda integral 0 to 1 1 minus t y of t rho t dt. So, you substitute this expression for y prime of 0 above and we combine the integrals and we get y of x equal to lambda times integral 0 to 1 x into 1 minus t y of t rho of t dt minus lambda times integral 0 to x x minus t y of t rho of t dt and let me combine these two integrals as one integral lambda integral 0 to 1 g of x t y t rho t dt everywhere there is a rho t dt and so the other terms have been combined as g of x t. So, what is this prescription g x of course its prescription will differ from 0 to x and from x to 1 and you have to write them separately. When you look at this integral and write the g of x t from 0 to x I have to take both of these and from x to 1 I will have to take only this and this will have to be dropped out. But what do you get at the end g of x t equal to t into 1 minus x if t less than or equal to x g of x t equal to x into 1 minus t x is less than or equal to t. Lo and behold the Green's function is symmetric and lambda inverse is an eigen value the compact self-adjoint operator given by equation 7.4 to the Hilbert Schmidt operator given by g of x t and we know that the eigen values of 7.42 form a countable sequence with 0 as the only possible accumulation point. So, the eigen values of the original stem level problem forms an infinite sequence going off to infinity because of the presence of the inverse here. Conversely if y of x is an eigen function for this Hilbert Schmidt operator 7.42 with eigen value lambda inverse namely this equation holds y of x equal to lambda times integral 0 to 1 g of x t y t rho t dt then we have to show that y of x is an eigen function for the boundary value problem y double prime plus lambda rho y equal to 0 with this Dirichlet boundary conditions. Let us check that what is the starting equation ty equal to lambda inverse y differentiate this equation we get y prime equal to put the lambda on the other side lambda times integral 0 to 1 g prime x t y t rho t dt we have an expression for g prime remember here it is. So, you can calculate the derivative with respect to x from here and what do we get we get two integrals you just calculate the derivative and put them together and then you differentiate again y double prime equal to lambda times minus x y x rho x minus from here you will get this 1 minus x y of x rho of x which is exactly lambda rho x y of x the minus sign which exactly shows that it is a solution of the boundary value problem. So, we have transformed the solution of a boundary value problem for a ODE with Dirichlet boundary condition into a eigen value problem for a Hilbert-Schmidt operator and we re-derive the theorems of chapter 6. So, we have re-examined the contents of chapter 6 from a abstract perspective. Let us just summarize all these things in another theorem, theorem 95 the Sturm level problem y double prime plus lambda rho y equal to 0 y of 0 equal to 0 y of 1 equal to 0 has a countable set of eigen values tending to infinity each eigen value is simple and the eigen functions form a countable orthonormal basis for l 2 of 0 1 with respect to the weighted measure rho. In other words every function f in l 2 of 0 1 with measure rho t dt can be written as a Fourier series f equal to summation n from 1 to infinity c and yn where the y n's are eigen functions of the Sturm level problem. But this gives you the l 2 theory of course this will not give you the point wise convergence or anything like that that is of course is a very very different ball game. I think this will be a very good place to stop this capsule and also stop this lecture and in this lecture we have revisited many classical parts of Fourier series in the modern functional analysis setup. And moving forward in the next chapter we are going to look at an application to celestial mechanics. We are going back to the very first chapter and we will see how those results which we derived in the very first chapter answers a very specific question in celestial mechanics. It is a very nice short chapter consisting of two capsules we will take up next. Thank you very much.