 Hello. So, we continue with our discussion of Fourier series, but this time we are moving into the fourth part of this course where we bring in techniques of functional analysis into play. Specifically, we will use theorems like the Banach-Steinhaus's theorem, we will use ideas from Hilbert spaces. And so, these functional analytic techniques will be used to throw light on both issues in the first, second and third parts of the course. Namely, we will use the Banach-Steinhaus's theorem to show that there is a plethora of continuous functions whose Fourier series diverges. We shall use Hilbert space techniques to discuss how complete orthogonal systems arise. We will prove the general spectral theorem for compact self-adjoint operators on a Hilbert space. We shall discuss some ideas of Green's functions for the Laplacian. And we shall discuss the completeness of the Hermite functions in L2 of the real line and so on. So, these are some of the things that we shall be discussing in this chapter. This chapter is slightly lengthy, but that is the nature of things. So, let us recall a few facts from Banach spaces and Hilbert spaces. We shall not be proving all of them. We shall be giving proofs of a few selected results. So, first a norm linear space is a vector space v which is endowed with a norm function. Norm map takes the vector space to r. What are the properties of the norm? They are displayed in the slide. So, the first item is norm v must be non-negative. The second is norm v is 0 if and only if v is 0. The last one is a homogeneity property of the norm. Norm Tv must be mod t times norm v for v in the vector space and for scalars t. The scalars could be real numbers or complex numbers, but the norm only takes non-negative real values. At the triangle inequality norm v plus w less than or equal to norm v plus norm w for all v w in v. So, given a norm on a vector space we can define a metric on the vector space. How do we define a metric on the vector space? The distance between v and w is simply the norm or the difference v minus w. And if this metric is complete then we will say that the norm linear space is a Banach space. So, these are things that you have no doubt seen in your courses on functional analysis which is usually taught at the advanced undergraduate level or at the masters level across the country. So, the most basic example is the Banach space c of a, b. It is a set of all continuous functions in the closed interval a, b endowed with the sup norm. Norm f is supremum mod fx where x varies over a, b. We call this the sup norm on c of a, b. The next thing is that a closed subset of a complete metric space is complete and so we can look at various subspaces of c of a, b. For example, I might be interested in looking at those continuous functions which vanish at a and b and that subspace is a closed subspace and that will also be a Banach space. Another important and significant example is you take a equal to minus pi and you take b equal to pi and you look at 2 pi periodic functions on the real line and restrict them to the interval minus pi pi, which basically means that the function must take the same value at pi and at minus pi and that again is a Banach space. Since this is so important, we will give it a special name period minus pi pi. So, these are some of the examples of Banach spaces and there are lot more examples on Banach spaces that we shall use and some of them we already encountered before such as L2 of minus 1, 1. When we discussed the Legendre polynomials that was a Banach space. In fact, it is a Hilbert space. When we studied the boundary value problems, we came across again L2 specifically for the vibrating string y double prime plus lambda rho xy equal to 0 with y of 0 equal to 0 and y of 1 equal to 0 the Dirichlet boundary conditions. The relevant Hilbert space is L2 of the closed interval 0, 1 with respect to the measure rho t dt. That is another important examples. Generally, you have got the LP spaces LP on the closed interval 0, 1 for example, where P runs between 1 to infinity both 1 and infinity are included. These are the classical Banach spaces. The space L2 is one of the most important among all the LP spaces and L2 of r is important in the study of Fourier transforms. If x is any compact metric space, the closed interval 0, 1 is just one example of a compact metric space. Set of all continuous functions on x real valued or complex valued with respect to the sup norm, with respect to the sup norm that is a Banach space. Now, you take the set of all continuous functions of the closed unit disk mod z less than or equal to 1. The closed unit disk is compact and so when you end out with the sup norm, you get a Banach space. Now, you look at a vector subspace, look at the space of functions which are continuous on the closed unit disk mod z less than or equal to 1 and in the interior mod z strictly less than 1, the functions are required to be holomorphic. Again, that is going to be a closed subspace. Why is it closed? How do you show that it is closed? Let me give you a sketch. Though it is not there in the slide, the proof is not there in the slide, I will give you a sketch of the proof. You take a sequence of continuous functions fn, all of them defined on the closed unit disk mod z less than or equal to 1 and these are all holomorphic in the open disk mod z less than 1 and let us assume that fn converge uniformly to f in mod z less than or equal to 1. You know that a uniform limit of a sequence of continuous functions is continuous. So, the limit function is certainly continuous. But what we want is we want the limit function to be in the subspace. Namely, the limit function f must be holomorphic on mod z less than 1. How do you show that it is holomorphic in the open disk mod z less than 1? Use Morera's theorem. What is Morera's theorem in complex analysis? You have a continuous function on a domain omega and integral over every closed triangle is 0 and the function must be holomorphic. And so, using Morera's theorem, you can complete the task of showing that this space, what is it? Space of all continuous functions of a closed disk mod z less than or equal to 1 which is holomorphic in the interior mod z less than 1. That is a Banach space. In fact, it is what is known as a Banach algebra. It is what is called as a disk algebra. It plays a very important role in analysis. The next example, again it comes from the first chapter on point wise convergence. Set of all 2 pi periodic continuous functions. So, what is this? Item number 5. Set of all 2 pi periodic continuous functions on the real line which are actually holder continuous with exponent alpha. So, first of all, my functions are 2 pi periodic and they are continuous. Moreover, they are holder continuous with exponent alpha. What is the definition of holder continuity? Mod fx minus fy less than or equal to some constant times mod x minus y to the power alpha. That is the definition of holder continuity. Recall from the first chapter and alpha is strictly between 0 and 1. 1 is included. When alpha equal to 1, you get what are called as Lipschitz function. If alpha is bigger than 1, only the functions that satisfy this property are constants and it is not interesting. So, alpha will be strictly between 0 and 1, 0 exclude 1 included. What are the norm? Norm is displayed here in the slide. Norm f equal to mod f of 0 plus take the difference mod of fx minus fy upon x minus y to the power alpha to the supremum of all these things when x is not equal to y. This supremum is finite from the very definition of holder continuity. And this thing is a norm and with respect to this norm, the space becomes a Banach space. Very, very important example of a Banach space. And the exercises prove the last two statements. For item number 4, I have already given you the argument. Item number 5, you can try to do it yourself. So, let us look at still more example. One example that I would like to work out in proper detail is the Bergman space. This Bergman space is extremely important in connection with various problems and theorems in analysis such as for example, Riemann Wabbing theorem. That is one place where the Bergman space makes its appearance. It appears in differential geometry. It appears in harmonic analysis in a variety of places. So, it is a very important example of a Hilbert space actually. In particular, it is a Banach space. So, let omega be a connected domain in the complex plane. Connectedness is for convenience and I am working with one variable. You can define this in several variables also, but we would not do anything with several variables. So, I shall restrict myself to one variable. So, A omega, what is A omega? A omega is a set of all holomorphic functions on omega. And L2 of omega is of course, well known to everybody and L2 of omega is a Hilbert space. In particular, it is a Banach space. Now I want to look at A omega intersect L2 omega. In other words, those holomorphic functions omega which are square integrable. Now in other words, f from omega to c is holomorphic and integral mod f squared over omega is finite. The integral is with respect to the Lebesgue measure on the plane dx dy integral mod f of x plus Iy the whole squared dx dy that should be finite. So, now this is the space A2 of omega, the A superscript 2 of omega. As the very notation indicates, for example, instead of L2, if I make it Lp, then I will call it Ap omega. All these are Bergman spaces. All these Bergman spaces are extremely important, but we shall look at A2 of omega just for convenience. So, the space A2 of omega is a closed subspace of L2 of omega, closed subspace of a complete metric space is complete. And so A2 of omega is also a Banach space. It is a Hilbert space. In fact, this space is known as the Bergman space. So, let us prove the theorem. So, to prove the theorem, we take a sequence fn of functions in A2 of omega, which converges in the ambient space L2 of omega. So, fn converges to f in L2 norm. What we need to show is that the limit f is actually in A2 of omega. In other words, we have to show that the limit lies in A omega because the limit is already in L2. In other words, f is holomorphic on omega. That is exactly what we need to show. First we shall establish that the limit f is actually continuous. Since continuity is a local property to check that something is continuous, it is enough if we show that it is continuous on each closed disk sitting in omega. So, let us work with a closed disk of radius 2 R sitting in omega. It is a closed disk mind you, it is compact. We need to recall the mean value property for a holomorphic function. You know the mean value property also is true for harmonic functions and we already discussed this in the earlier chapters in the first part of the course. So, what is the mean value property? Assume G is holomorphic on omega and you got this disk dr, which is sitting inside omega. Then you calculate the mean value of the function on a circle. So, you take a point p, fix the center of the disk is fixed, you calculate G of p plus rho e to the power i t dt. You evaluated the function on points on a circle of radius rho centered at p and you are integrating it over the circle with respect to the arc length and you are divided by the total perimeter of the circle. So, this is the average value of the function G on the circle and this average value is exactly going to be the value of the center. This is the mean value property which you are no doubt studying in your complex analysis courses. What we do is that, let us multiply this equation by rho and integrate with respect to rho. So, left hand side what is it going to be? It is going to be integrated with integral rho d rho and that is going to be rho squared by 2 and you are integrating it over 0 to r. It is going to be r squared by 2 g p. g p is constant. Right hand side is going to become a double integral 1 upon 2 pi integral 0 to 2 pi integral 0 to r G of p plus rho e to the power i t rho d rho dt. What is rho d rho dt? It is the planar Lebesgue measure dx dy and p plus rho e to the power i t can be written as z where z is an arbitrary point in the closed disc of radius r centered at p. So, this repeated integral is basically this double integral and I divide by r squared and cancel out 2 I get 1 upon pi r squared. 1 upon pi r squared is the area of the disc. So, you get the solid mean value theorem. In other words, the integral of the function g on the disc of radius r centered at p gives you the value of the function at the center. Of course, you have to divide by the area of the disc. Now, as an exercise I want you to imitate the same argument and prove the following theorem. Prove the solid mean value property for harmonic functions assuming the mean value property over spheres. You know that if you have a harmonic function in R n, then you proved the mean value theorem over spheres and I want you to use the mean value theorem over spheres to prove the mean value theorem over the solid ball. So, now let us continue with our discussion of the Bergman space. So, take 2 points z and w in the disc of radius r centered at p. Remember that every start we took a disc of radius 2 r, we took a disc of radius 2 r and we took the concentric disc of radius r and we take 2 points z and w in this smaller disc. And now if I take with z as center, if I draw a disc of radius r with w as center, I draw a disc of radius r. Both these discs are going to comfortably sit inside d 2 r p. What I do is I write the solid mean value property over d r p and d r w. So, what does it say? It says the g z, the center, the value of the center of the disc will be equal to 1 upon pi r squared times integral of the function g over this disc. Again, the value of the function g at w, the center will be 1 upon pi r squared the integral over this whole thing and a subtract. Of course, the both the discs have the same area 1 upon pi r squared and I am integrating and the integrals over the overlap d r z intersection d r w over the intersection the integrals will cancel out. And so, I am going to get the integral of g over what over the part of the union. That is you take d r z union d r w and remove the lens shaped intersection. So, when you take 2 discs and when you remove the lens shaped intersection, what is left over? You get 2 crescent shaped things. The union of these 2 crescents is a symmetric difference. In set theory, what is the symmetric difference of 2 sets? A delta B, what is A delta B? A union B minus A intersection B. So, from A union B, you remove A intersection B and what is left over is the symmetric difference of the 2 sets. So, integrating g over the symmetric difference. Now, let us apply the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality will tell you that take the absolute value, take the absolute value, take the absolute value inside, it becomes less than or equal to and apply the Cauchy-Schwarz inequality. The integral of mod g squared square root but integral of mod g squared square root will be less than or equal to the integral of mod g squared over the whole domain omega and that will be the L2 norm of g on the whole domain omega. And what are the other factors that you will get when you apply the Cauchy-Schwarz inequality? Double integral 1 dx dy square root but integral of 1 dx dy will be the area of the symmetric difference and it will be square. Of course, the 1 upon pi r squared is always there. So, this is what we get 7.1. We get a beautiful estimate for the difference between gz and gw. Now, we must understand how to find the area of this symmetric difference. We can exactly compute the area of the symmetric difference but I am not interested in knowing exact values. All I want to show is g is continuous. All I need to show is that this right hand side of 7.1 converges to 0 when w converges to z. That is all that I really need. For that any rough estimate on this area will do. So, here is the exercise. Draw pictures and show that the area of drw, symmetric difference with drz. The area of the symmetric difference is less than or equal to 2 pi r mod z-w plus some stuff which goes to 0 faster than mod z-w. So, it will be big O of mod z-w, the whole squared, the equation written in red and numbered is 7.2 in the slide. How do you do this exercise? A hint has been given to you. Now, let me explain to you pictorially what this hint means. So, you have got two congruent disks, two disks of radius r. They are overlapping disks and they intersect along a lens shaped region. So, take the two centers of the two disks z and w and draw a straight line. And then look at this lens. And remember that z and w are going to be very close. So, this lens is going to be pretty thick. It is going to be a very thick lens. And you draw the line passing through the centers. That is the bisector of the lens. Now, what you do is you inscribe with this, see where this horizontal line intersects the two faces of the lens. With that as a diameter draw a disk. So, what is this capital D? Capital D is this largest possible disk that you can fit inside this lens. What I will find out? I will find out the diameter of this disk and the radius of this disk. So, just draw the picture and try to find out what is the radius of this largest possible disk that you can inscribe in this lens. And you know the area of this disk D. So, now what is the symmetric difference? The symmetric difference is the union of those two crescents. What is the area of the left crescent? Area of the crescent will be less than or equal to the area of drz minus area of D. The same thing for the crescent on the right hand side. These two crescents will have equal areas. 7.2 has been established. So, when you take the square root, what do you get? You get that mod gz minus gw absolute value less than or equal to the L2 norm of g some innocent constant pi r squared root of mod z minus w. Basically, that is what you get root of mod z minus w times other things which are innocent and they are bounded things. So, what do we get? We have got this estimate mod gz minus gw is less than or equal to some cr times root of mod z minus w. Let us apply this estimate to my fn. My g that I am interested in is those terms of the sequence fn's. So, we get mod fnz minus fnw less than or equal to some innocent constant cr which depends only on the radius r, the L2 norm of fn square root of mod z minus w where z and w are in Dp. Remember that was the starting point. We took 2 points z and w in Dp. Now, the fn's are converging to f. So, the L2 norms of fn's are all bounded. So, all these things are bounded. CR depends only on r and you got the square root of mod z minus w. And so, it means that the sequence fn is equicontinuous on the disc Dp and it is also uniformly bounded in the disc Dp because r is fixed. These L2 norms are bounded and it is going to give you uniform boundedness as well. So, by Oskoli-Arzela theorem, the limit function f must be continuous. The limit function must be continuous. So, continuity has been established. For uniform boundedness, you will have to use again the solid mean value property for the disc and you can finish it off. Or you can fix w to be p and then you can show that the fn's converge at p at the center of the disc. You will again have to use the mean value property but you cannot escape the solid mean value property one way or the other. Now, we do the following. Now, we can use dominated convergence theorem. We can use dominated convergence theorem to prove that the integral of f over every triangle is 0. So, by Morira's theorem, the limit function will be holomorphic. So, how to do that? You take a triangle which is lying inside the disc Dp. fn's are holomorphic. So, integral of fn zeta D zeta over the triangle is 0 by Koshy's theorem. And I want to take the limit of this and I want to take the limit under the integral sign. Use dominated convergence theorem. We have all these wonderful estimates we have. And that completes the argument that the limit function is holomorphic. And we have proved that the Bergman space is a closed subspace of L2 of omega. L2 of omega is a Hilbert space and so the Bergman space is also a Hilbert space. One of the most important examples in analysis and we have completed that. I think it is a good place to stop this capsule and we will continue this in the next capsule.