 Hello, so in the last capsule we were talking about weak convergence and norm convergence. We defined weak convergence. What is weak convergence? A sequence Vn converges to V, weakly means the difference Vn minus V dot product with W goes to 0 for every W. So, weak convergence is really weak because you can take our favorite Hilbert space L2 of minus pi pi and the principles of the very first chapter the Riemann-Lebesgue-Lehmann tells us that sin nx converges to 0 weakly, but sin nx is far from norm convergent. It is not norm convergent, it is not point wise convergence. So, weak convergence is very weak. What we proved in the last lecture is a very important theorem. In fact, what we have proved is a very special case or the important Banach-Lauglou theorem. What we did is norm boundedness implies there is a subsequence that converges weakly. Now, you might wonder is this some kind of a compactness argument. If you have been having these kinds of questions creeping up that means you are on the right track. Yes indeed we have proved a compactness theorem actually. What is happening is that you take the Hilbert space H infinite dimensional of course and you take the unit ball in the Hilbert space the closed unit ball is not going to be compact. So, you simply do the following you throw away the norm topology and replace it by a weaker topology. This weaker topology has the property that every sequence will have a convergent subsequence. So, what we have shown is that the unit ball in a infinite dimensional Hilbert space is sequentially compact with respect to the weak star topology. And of course, we are assumed that the Hilbert space is separable and this is been done only for Hilbert spaces the general theorem is the Banach-Lauglou theorem. I would not even state what the Banach-Lauglou theorem is you can read it in Goffman-Pedrick's book which I referred to last time. Now, how is this going to be important? It is going to be a very important ingredient in the basic step in the proof of the spectral theorem. Before we get to that let me just make one small comment we have seen that there is no hope of getting norm convergence out of weakly convergent sequences sin nx sin nx converges to 0 weakly there is no way we are going to be getting any kind of norm convergence out of this. So, norm convergence implies weak convergence the converse is not true but the converse will become true under additional hypothesis there is a peculiar additional condition from which we can actually recover norm convergence out of weak convergence. And that result is remarkable and it can be used at several places and it pops up at unexpected places. So, for records I have stated it very clearly in the slide theorem 91 suppose h is a Hilbert space and vn is a sequence that converges weakly to v and further assume that norm vn converges to norm v this condition note is a very special condition then we can conclude that vn actually converges to v in norm. Go back to the example of sin nx sin nx converges to 0 but norm sin nx does not converge to norm 0 remember. So, this condition is very special it is very important condition we can prove this result using the parallelogram law for Hilbert spaces we shall not do so because we will not use this theorem. Now let us get the spectral theorem we are now ready to prove the spectral theorem for compact self fed joint operators on a Hilbert space we are going to assume that the Hilbert space is separable there is no harm in doing that in this capsule and in the next one will be used in the proof of this important theorem. Now a few preliminaries which we should dispose of before we get to the spectral theorem theorem 92 suppose t from h to h is a self fed joint operator then inner product T v v is real for all v in h Eigen values of t if they exist are real we are not shown Eigen values exist that is going to be proved later, but right now we have got to put this disclaimer that Eigen values of t if they exist are real the Eigen vectors corresponding to distinct Eigen values are perpendicular to each other the first two results are of course significant only when the Hilbert space is complex the Hilbert space is real then one is a moot point, but in our spectral theorem we shall produce directly a real Eigen value. So, we really do not have to prove this result, but for completeness let us give an elementary argument to prove theorem 92 even in the complex case. So, first of all t v v because of self fed joint the t can be put on the right hand side and once I put the t on the right hand side I switch the factors and I put a complex conjugate and it immediately implies that t v v is real. Now the second one suppose lambda is an Eigen value of t with Eigen vector v then what happens what is t v v on the one hand it is going to be lambda v the lambda comes out it is simply in the product of v with itself it is lambda norm v squared. On the other hand because of self fed jointness t v v I put a t on the other side now t v is lambda v, but this time when the lambda comes out it comes out of the bar on top of it and there will be lambda bar times norm v squared now compare these two expressions we got and remember that v is a non zero vector. So, norm v squared is not going to be 0. So, we conclude that lambda must be equal to lambda bar and therefore, lambda must be real. So, Eigen values if they exist are real finally suppose v and w are Eigen vectors corresponding to two distinct Eigen values lambda and mu that means what that means that t v w which will be equal to v t w what does this translate to this translates to saying that lambda v w equal to v mu w. Now both mu and lambda are real. So, when they come out of this inner product there is no question of putting any bar on top of it. So, we will get that lambda minus mu times inner product v w is 0 and lambda and mu are distinct. So, this is not 0. So, the only other option is that the inner product of v and w is 0 and that completes the third part also. So, now we come to the existence of Eigen values and let us recall what we did in chapter 6. In chapter 6 we did the variational characterization of Eigen values. We translated the problem of the Dirichlet problem for a 2 point boundary value problem y w prime plus lambda rho x y equal to 0 with Dirichlet boundary condition into a variational problem. Before doing so we reexamined the situation in elementary linear algebra and we discuss the spectral theorem in that particular context. So, let us recall that briefly because we are going to use those ideas here in the context of Hilbert spaces. So, take a real symmetric matrix A and you take v transpose A v and that is a quadratic form you take this quadratic form you restrict it to the unit sphere norm v equal to 1 and the unit sphere in R n is compact it is finite dimensional here is where the finite dimensionality becomes crucial and the compactness of the closed unit ball in R n is crucial the supremum that you see. So, in a product A v v is attained at some vector v 1 let us call it and this v 1 is going to be the eigen vector and the supremum is going to be the eigen value. How do you continue? You take the orthocomplement of v 1 and you slice the unit sphere by this orthocomplement and you get a sphere of smaller dimension and you keep continuing you take the supremum on the sphere of smaller dimension and that will be the next eigen value lambda 2 and the supremum will be attained at v 2 that will be the next eigen vector v 2 and by construction v 1 and v 2 were orthogonal and so on and so forth and infinitely many steps we get a complete sequence of eigen vectors and we proved the spectral theorem that way. Now unfortunately we are in a infinite dimensional setting instead of a matrix we have a self adjoint compact linear transformation and the Hilbert space is infinite dimensional we could start out with T v v and we could take the supremum but unfortunately we are no longer in the situation where the unit ball is compacted compactness of the unit ball is not available for us what we have is a weak substitute namely the weakly convergent subsequence that we have it is this weak substitute that we need to employ in order to prove the theorem. So, this is the crux of the proof of the spectral theorem once we are through with this part the rest of the matter is routine and the same idea of maximizing the Rayleigh quotient instead of taking a v v we simply take T v v take the supremum of that we know that this is real and we can use the weak compactness of the unit ball which is why the Banachala Gullu theorem was so important we have proved the Banachala Gullu theorem in the Hilbert space setting before carrying out this project let us make a couple of simple observations. Now, suppose if T v v is 0 for all v then T must be 0 why is that T v v is 0 means instead of v replaced by v plus w instead of v replaced by v minus w we get these two equations and subtract and we get T v w plus T w v equal to 0 and from this we will get that T v w equal to 0 for all v w in H and that will mean that if I take w equal to T v we will conclude that T is identically 0. So, that is how we prove the theorem. So, clearly we may assume that T is not the 0 operator and so the Rayleigh quotient cannot be identically 0 T v v is called the Rayleigh quotients replacing T by minus v we know that T v v or minus T v v one of them must be real number which is positive. So, the Rayleigh quotient certainly assumes some strictly positive values. So, the supremum of T v v will always be positive. So, now we continue. So, theorem 93 suppose T from H to H is a compact self fed joint operator on a Hilbert space then you take the Rayleigh quotient T v v until the supremum over the unit sphere this supremum will be an eigenvalue and this will be attained at an eigenvector. So, although the unit ball is not compact in the norm topology the weak compactness suffices for our purpose call the supremum lambda and the supremum exists and why does the supremum exist simply take the absolute value and apply Cauchy Schwarz inequality and it is positive by the remarks made at the beginning. So, take a sequence of unit vectors v n such that T v n v n converges to lambda this is called as a maximizing sequence. Since T is a compact operator this T v n will have a norm convergent subsequence here we are strongly using the fact that T is a compact operator this is very essential. So, I can work with this subsequence I can throw away the original sequence and replace it with a subsequence if you like and I am going to continue calling it v n itself. So, I am going to assume without loss of generality the T v n itself converges to some y let us say further the v n's are norm bounded because they are unit vectors and so, they have a weakly convergent subsequence the version of the Banakala Gullu that we have proved for Hilbert spaces and so, they have a weakly convergent subsequence and I am going to pass on to the subsequence and replace v n by the subsequence I am going to work with the subsequence only and I am going to rename it as v n itself. So, without loss of generality we may assume that v n minus v naught inner product with w goes to 0 for every w that is v n converges to v naught weakly. So, now we got v n converging to v naught weakly T v n converging to y in norm that is very strongly. So, we are going to use both these things in our work. So, now let us take the inner product of T v n and v n add and subtract a y add and subtract a y the first piece goes to 0 because T v n converges to y in norm and the second piece v n converges to v naught weakly. So, we get 7.36. So, T v n v n converges to y comma v naught but T v n v n converges to lambda remember v n was a maximizing sequence. So, we get the first result which is in red saying that the inner product of y and v naught must be lambda. Next, T v n v n is real and so the limit lambda is also real. Now, let us apply self adjointness of T T v n z through the T on the z part we get v n T z but v n converges to v naught weakly it is v naught T z put the T back on the v naught and we get v naught of z and at same time because T v n converges to y in norm T v n z will converge to y z. So, again compare the two results we get that T v naught z equal to y z inner product and z is arbitrary and that means that T v naught must be y correct. So, now, we get lambda which is inner product of y and v naught we also get that y is T v naught. So, we get that lambda is basically T v naught v naught and so we see that the supremum is attained at v naught and because lambda is positive this v naught cannot be 0 and so certainly the supremum is attained and it is a non-zero vector. Let us prove that it is actually a unit vector we have to show it is a unit vector. So, fix an m and v n converges weakly to v naught and m is fixed. So, inner product v n v m goes to v naught v m. So, we see that norm v naught v m will be less than or equal to 1 letting m tend to infinity we get that norm v naught is less than or equal to 1. Now, on the other hand T v naught by norm v naught v naught by norm v naught is less than or equal to lambda because lambda is a supremum of all these things T v v where v is a unit vector. Now, you clear the denominators and we see that lambda which is T v naught v naught is less than or equal to lambda norm v naught squared lambda is non-zero it is positive. So, norm v naught squared must be greater than or equal to 1. So, combining the two equations we get norm v naught is equal to 1. So, we have finally, shown that the supremum of this Rayleigh quotient what is the Rayleigh quotient inner product for T v with v where v is a unit vector that is the quadratic form T v inner product with v restricted to the unit sphere the supremum is attained and it attained at a unit vector v naught. The most difficult part of the spectral theorem is now over. Now, we have to show that this point v naught that we captured is an eigenvector corresponding to the eigenvalue lambda this part of the argument is exactly similar to what we have done in chapter 6. It is exactly for this reason that we took this spectral theorem carefully for a matrix in chapter 6. So, we did a dry run for a elementary and well known case. So, that now we are ready to attack the infinite dimensional setting. So, the argument is very similar perturbed v naught ever so slightly by adding a T h v naught plus T h is not a unit vector divided by the length it becomes a unit vector clear the denominators we get that the inner product T v naught plus T h v naught plus T h less than or equal to lambda times norm v naught plus T h squared expand the two sides the right hand side will be norm v naught squared that is 1 and then we got a v naught inner product with T h and then we are going to get a T h inner product with v naught and then the T will be coming out and there will be a T squared term and here we got T v naught v naught and that T v naught v naught will lambda the lambda will cancel out and so on. We will get the terms involving T the 2 T into T v naught h less than or equal to 2 T lambda v naught h there will be a T squared term which I have conveniently ignored because I am going to divide by mod T and I am going to allow the T to go to 0 so anyway it is going to disappear. So, now when I divide by mod T and let T go to 0 through positive values and through negative values we will get two inequalities these inequalities will combine to give a equality T v naught minus lambda v naught inner product with h should be 0 and h was arbitrary. So, we conclude that T v naught will be equal to lambda v naught and we approve that v naught is an eigenvector in particular eigenvalues and eigenvectors exist if the operator T is a compact self-adjoint operator on a Hilbert space. So, the most difficult part of the spectral theorem is now over and we must go back to chapter 6 and read the proof of the spectral theorem for a real symmetric matrix and compare that proof with this proof. The stumbling block was the non-compactness of the unit ball or the unit sphere in a Hilbert space with respect to norm topology, but you do not work with the norm topology you throw away the norm topology and you replace it by the weak topology and you invoke the Banach-Allagallou theorem and you get around the difficulty and this was a remarkable thing. I think this will be a very good place to stop this capsule and we will continue this in the next capsule we will complete the proof of the spectral theorem. The reason why I would like to bifurcate this into two parts is that the difficult part has been addressed here, the routine stuff which will be addressed in the next capsule. There is a very clear polarization in the proof and so, this will be a very good place to stop. Thank you very much.