 So, what were we doing last time? We assumed that the Fourier series of every continuous function to with 2 pi periodic converges point wise in order to obtain a contradiction. If the under that assumption we concluded by the Banach-Steinhaus's theorem that there must be a constant m such that inequality 7.3 in the display holds namely mod integral minus pi to pi f of t d n t d t less than or equal to m for all continuous 2 pi periodic functions. And we are going to assume that the function lies between minus 1 and 1 otherwise there will be a norm f on the right hand side. To get to the norm f on the right hand side I am going to assume that the function lies between minus 1 and 1. And we are going to restrict ourselves to real valued function. And we also are going to prove later that integral minus pi to pi mod d n t d t is approximately c log n where c is some positive constant. So, these integrals on the left hand side of 7.4 go to infinity as n tends to infinity. Now, how to arrive at a contradiction using 7.3 and 7.4? I indicated to you last time that the idea is to take f of t to be the signum function namely the sign of d n t. If it on an interval d n t is positive then on that interval you declare f of t to be 1. If for an interval d n t is negative on that interval declare f to be minus 1 then the left hand side of 7.3 exactly gives you the left hand side of 7.4. The right hand side of 7.3 says that it is bounded by capital M whereas the right hand side of 7.4 says that it is c log n and that will give you the contradiction. But the problem the technical issue here is that the signum function is not continuous. And so, we have to get round this difficulty of the signum function being discontinuous. So, let us see how to do that. First of all observe that d n t the Dirichlet kernel to get the formula for the Dirichlet kernel you must go back to chapter 1 to see the exact formula of the Dirichlet kernel. It alternates its sign in intervals of length pi by n. So, in successive intervals of length pi by n the sign will keep alternating. So, the signum function will also have the signs alternating between minus 1 and plus 1 on successive intervals of length pi by n. But the signum function certainly lies in l 1 of minus pi pi. Now, we can use Luzin's theorem. What is Luzin's theorem say that I can find a sequence of continuous functions f j such that f j converges to the signum function in l 1 norm. Further I can arrange it in such a way that the these functions f j vanish at the two ends minus pi pi. So, that we can take the two pi periodic extensions and the convergence in l 1 norm. So, let us work with these continuous approximations. So, what do we have integral minus pi to pi mod f j t minus the signum function goes to 0 as j goes to infinity convergence in l 1 norm. And so, integral minus pi to pi mod d n t d t which is exactly equal to integral minus pi to pi d n t times the signum function. Now, I add and subtract an f j and now I will take the absolute value. So, I will use triangle inequality. And so, now I will say that this is less than or equal to mod d n t mod sigma t minus f j t d t plus integral minus pi to pi d n t f j t d t the whole thing mod. But remember that f j s are continuous. So, I can certainly use the estimate that we obtained earlier 7.3 can certainly be used here. And so, we get that this piece is less than m. And now remember if you look at the Dirichlet kernel the formula for the Dirichlet kernel looking at the formula for the Dirichlet kernel I would like you to prove that the mod of the Dirichlet kernel is going to be less than or equal to capital N. And so, we are left with integral minus pi to pi mod sigma t minus f j t d t plus m. Now, the trick is to fix n and allow the j to go to infinity. So, once you fix n and allow the j to go to infinity we immediately get that this thing goes to 0. And so, this thing drops out and we get the result namely integral minus pi to pi mod d n t d t is less than or equal to m. So, our difficulty of using the signum function to obtain this estimate has now been circumvented. And so, now that we have this we will prove in immediately in the next few slides and the left hand side integral minus pi to pi mod d n t d t is actually c times log n and that gives you a contradiction. So, these integrals of mod the Dirichlet kernel, let us call them ln and these are called Lebesgue constant. And a great deal is known about the Lebesgue constants these days and the literature is quite vast we shall not need all those things and we shall not concern ourselves with these fine points of Fourier analysis. There are papers written on this Lebesgue constants. So, now let us look at the behavior the Lebesgue constants. So, since the Dirichlet kernel is an even function it is enough to look at the interval 0 to pi. So, what is the Dirichlet kernel sin n plus half t upon sin t by 2 and where does this change sin it changes sin at these points k pi by n plus half k equal to 0 1 2 3 and this points divide the interval 0 pi into finitely many sub intervals each of length pi by n plus half. So, we got so many sub intervals on each of these intervals we have the estimate 1 upon sin t by 2 greater than or equal to 2 upon t. Remember when t is between 0 and pi by 2 is between 0 and pi by 2 what are the inequality of the sin function. Remember that sin t is less than or equal to t on the interval 0 to pi by 2 that is all that I have used sin theta less than or equal to theta for theta between 0 and pi by 2. So, 1 upon sin t by 2 is greater than or equal to 2 upon t. So, what do we have using this we have that the more the Dirichlet kernel is greater or equal to 2 upon t mod sin t into n plus half the denominator in the Dirichlet kernel has been taken care of. Now, on the sub intervals of length pi by 2 times n plus half let us look at i k. So, you see that the Dirichlet kernel changes signs in certain sub intervals. In these sub intervals we will slightly cut these sub intervals a little bit. So, that let us look at t between pi times k plus 1 fourth upon n plus half and pi times k plus 3 fourths upon n plus half. So, I cut out these intervals on which the d and t maintains sign either positive or negative throughout I cut it ever so slightly. Once I cut it ever so slightly and I take the mod this mod will be bigger than or equal to 1 upon root 2. You see it is only the end points of these intervals that sin t times n plus half becomes 0. The Dirichlet kernel becomes 0 at those end points and I have taken a little inside. So, that mod of the sin function is greater than or equal to 1 half. So, we get mod d and t which is greater than or equal to 1 upon t because this 2 is an innocent constant and I had taken this to be 2 upon pi into 2n plus 1 upon 4k plus 3. So, on each of these sub intervals we have this lower estimate on the mod of the Dirichlet kernel. So, we get integrating over i k we get integral over i k mod d and t d t will be greater than or equal to the length of the interval times the constant 2 upon pi times. And so, if you do the algebra correctly if I multiply this by the length of the interval I get the lower estimate 2 upon 4k plus 3. Now, your sum over k equal to 1 2 3 up to n and then the integral over 0 to pi is certainly going to be greater than or equal to integral over the union of i 1 union i 2 union i n. And that is greater than or equal to this summation of 2 by 4k plus 3 k running from 1 to n and that sum is lower bounded by 1 plus half plus delta plus 1 upon n. And this 1 plus half plus delta plus 1 upon n is approximately log n for large values of n. You get this further lower estimate c tilde log n where c tilde is some positive constant. So, we have finished the complete demonstration about the fact that the Lebesgue constants go to infinity and how the Banach-Steinhaus's theorem gives you lots of continuous functions whose Fourier series will diverge at the origin. In fact, if you examine the Bayer category theorem what we established is that the set of continuous functions which are 2 pi periodic whose Fourier series diverges at the origin is a dense G delta set. So, a majority of continuous functions do exhibit the errant behavior that we are seeing. So, now let us take up the next issue in this chapter on functional analytic preliminaries for study of Fourier series and generalized Fourier series and that is Hilbert space basics. We already used some elementary facts about Hilbert spaces such as the least square approximation and so on so forth. Here we shall go a little deeper into the study of Hilbert spaces. So, recall that a Hilbert space is a vector space h endowed with an inner product v triangular bracket vw such that the norm is given by the inner product vv. So, take a vector v and take the inner product vv and take its square root and you get the norm and once you have the norm you got the distance function. And if this distance function is a complete metric then you say that it is a Hilbert space and the norm has a set of properties we already discussed those things before. So, I shall not repeat those things. So, if v and w are vectors in a Hilbert space these vectors are said to be perpendicular or orthogonal denoted as v perp w if the inner product between v and w is 0. And now let us take a Hilbert space h and let us take v to be a vector subspace v is a vector subspace and v perp the orthocomplement of v is defined as v perp equal to set of all vectors x which are orthogonal to everything in v. The first exercise is to show that v perp is a vector subspace of h it is practically self-evident and v intersect v perp is 0 because if a vector v is in v intersect v perp then it is orthogonal to itself. So, norm v squared is 0 therefore, v must be 0. The orthocomplement of 0 is a whole Hilbert space h and the orthocomplement of h is the 0 space. The orthocomplement of a vector space is always a closed subspace even the original vector subspace w is not closed the orthocomplement will by default be closed. And if I take a vector subspace v, v perp will be v closure and then the usual results from elementary linear algebra if I take a set of mutually orthogonal non-zero vectors then this set will be linearly independent. And of course, we have got the Gram-Schmidt's ortho normalization process which I would like you to review from elementary linear algebra courses. So, all these things are completely elementary. So, I shall not devote any time on this I will leave you to standard books or maybe your class notes if you have taken your functional analysis courses which I am sure most of you have. Now comes a very important notion the notion of basis and Hamel basis. Here we are only going to be talking about Hilbert spaces. When I use the word basis I am going to talk about Hilbert spaces only. There is a notion of basis namely shorter basis which is defined for Banach spaces. But that notion is slightly non-trivial and shorter basis may not even exist. But we are going to confine ourselves to Hilbert spaces only. So, now when we have a Hilbert space we know what is a complete orthonormal system. A complete orthonormal system is called a basis for a Hilbert space. For example, if I take L2 of minus 11 then P0x P1x P2x the Legendre polynomials forms a complete orthogonal system and so it will form a basis. So, we got an example. Another example is L2 of minus pi pi. From the very second chapter in this course we saw that 1, cos x sin x cos 2x sin 2x dot dot dot is a complete orthogonal system. So, it forms a basis for the Hilbert space L2 of minus pi pi. So, now these bases are obviously countable. Now the Hilbert space is first and foremost a vector space. And right away in linear algebra we know that every vector space has a basis. Now that notion of basis is different from the notion of basis that we have been talking about. What does it mean to say that 1 cos x sin x cos 2x sin 2x dot dot is a complete orthonormal system? It means that if I take finite linear combinations and I take their closure then I get the whole vector space. So, when I use the word basis in the context of Hilbert space there is a closure operation which is an operation in analysis. You take finite linear combinations the linear span and you take the closure of the linear span. In other words the linear span must be dense in H. If I take the Legendre polynomials p0, xp1, xp2, xdelta I get all the polynomials because it is I am taking finite linear combinations. And now polynomials are dense in L2. So, if I take the closure I will get the whole of L2 of minus 1, 1. But in elementary linear algebra courses you do not take the closure remember. So, what is the basis you simply demand that the linear span must be the whole space. It must be linearly independent of course a linearly independent set is called a basis if it is linear span is the whole of the vector space that is your elementary linear algebra definition. So, now we must be careful to separate out these two notions. So, let me straighten out the records once and for all and you see all the information in this slide the notion of basis in the sense of elementary linear algebra if we ever need it we will call it a Hamel basis a Hamel basis. The word basis without the word Hamel here in this course will always mean a complete orthogonal system namely it should be linearly independent and the closure of the linear span should be the whole of H. I do not mention the orthogonality here I can always subject B to the Gram-Schmidt's process and if B is not orthogonal already I can make it orthogonal by applying the Gram-Schmidt's process. So, there is no harm in putting the orthogonality clause in the definition that is perfectly fine. So, there is a two different notions of basis basis and Hamel basis. Hamel basis are basis in the sense of linear algebra. Now, these two notions are very different why are they very different let me explain to you a Hilbert space which is infinite dimensional cannot have a countable Hamel basis. In fact, if you take a Banach space which is infinite dimensional you got lots of Banach spaces LP of the closed interval 0 1, C of AB with soup norm the Bergman space lots of Banach spaces which are infinite dimensional a Banach space which is infinite dimensional cannot have a countable Hamel basis. This follows from the Bayer category theorem and I leave it as an exercise for you to do this using the Bayer category theorem. Now, metric space is said to be separable if it has a countable dense subset. So, for example, the real numbers is a separable metric space because the rationals form a countable dense set and the same definition applies to a Hilbert space after all a Hilbert space is a metric space to begin with. So, a separable Hilbert space is a Hilbert space which is separable as a metric space the cardinality of a Hamel basis is uncountable in any infinite dimensional Hilbert space. Now, to see this we proceed as follows suppose you have a Hamel basis which is actually countable then how do we arrive at a contradiction we arrive at a contradiction by making a couple of simple observations. First of all a finite dimensional subspace of a Hilbert space is always closed in fact it is to any Banach space you take any Banach space and you take a finite dimensional subspace it is always going to be closed subspace and because the original space is infinite dimensional a finite dimensional subspace will have empty interior because if you take an interior point in a vector subspace then I can blow it up I can first of all translate it and assume that 0 is an interior point and I can blow up the neighborhood by scalar multiplication and I will get the whole space. So, a proper closed subspace will always have empty interior a finite dimensional subspace has empty interior because my ambient space is infinite dimensional and it is closed that is a first important observation the second observation is suppose I have a countable Hamel basis call it v1 v2 v3 dot dot dot take the first n vectors v1 comma v2 comma dot dot dot vn take the linear span call it en then en is a finite dimensional subspace so it is closed en is a proper subspace it has empty interior so en is a closed subspaces with empty interior. Now since the Hamel basis generates the whole space look at the union of en's n from 1 to infinity that will be my whole Hilbert space and so the whole Hilbert space has been written as a countable union of closed nowhere dense sets and that evidently contradicts the Bayer category theorem. So, the Bayer category theorem will immediately tell you one of the most beautiful applications of Bayer category theorem is that a infinite dimensional Banach space cannot have a countable Hamel basis. Examples of basis again example when I say basis in a Hilbert space I mean it is a basis in the sense of analysis that is it is a linearly independent set whose linear span is dense and there is no harm in assuming it is orthogonal. So, L2 of a comma v if a is minus pi and b is pi then we have this sequence 1 cos x sin x cos 2 x sin 2 x etcetera that is going to be a basis from a Hilbert space L2 of minus pi another example is the Legendre polynomials. So, that is also a basis in L2 of minus 1. Now if you take 1 x x squared etcetera it is linearly independent subset of L2 of minus pi pi subjected to the Gram-Schmidt's process I am going to get an orthogonal sequence and use the fundamental orthogonality lemma that we have established earlier to prove that you are going to get the Legendre polynomials we already discussed this in the last chapter. The Hermite functions we already encountered these Hermite functions before but let us recapitulate it in the light of Hilbert spaces. The Hermite polynomials are solutions of Hermite's differential equation y double prime minus 2 x y prime plus 2 lambda y equal to 0. Let us look for a power series solution for this differential equation 7.6 y equal to summation n from 0 to infinity a n x to the power n. We proceed exactly as we did for the Legendres differential equation you get a recursion formula you got to differentiate able to compute y double prime you got to compute minus 2 x y prime etcetera you got to put those things in the differential equation and what will you get the last display in the slide summation n from 0 to infinity x to the power n a n plus 2 n plus 2 into n plus 1 minus 2 n a n plus 2 lambda a n equal to 0 equating the coefficients to 0 we get the recurrence relation a n plus 2 equal to 2 times lambda minus n upon n plus 1 into n plus 2 a n. Now you must put n equal to 0 1 2 3 etcetera and you are going to determine the successive coefficients a naught a 1 a 2 etcetera and then the solution y of x will be a naught times all the even powers of x and a 1 times all the odd powers of x and you see a remarkable fact that if lambda is an integer then one of these two infinite series will terminate and you are going to get a polynomial solutions so for example if lambda is an even integer then I could take a naught to be 1 a 1 to be 0 and I get a polynomial solution if a lambda is say 3 for example and I could take a 1 to be 1 and a naught to be 0 and I get the second parenthesis in the display and I will get a polynomial of degree 3. So if lambda is an integer n positive integer n then the Hermits differential equation admits a polynomial solution which is a polynomial of degree exactly n and these are the sequence of Hermit polynomials. We are going to use these Hermit polynomials and we are going to cook up a set of functions called the Hermit functions and that will give you a basis for L2 of the real line. This will be a good place to stop this capsule and we shall continue this next time. Thank you very much.