 So, we continue with our discussion of Hilbert spaces from the Fourier analysis perspective and also from the differential equations perspective. So, you see a synergy here between Fourier analysis, functional analysis and differential equations. So, our next question is about a brief discussion about how non separable Hilbert spaces are important in certain parts of analysis and that concerns as I said in the previous capsule, the almost periodic function. So, let us look at these things briefly and maybe if time permits later, we could come back to almost periodic functions and look at it from the Fourier analysis perspective. I will certainly give you a couple of references for these things. In case we do not return to this subject. So, do non separable Hilbert spaces exist? If so, are they useful? Of course, one can always define something and we can always bring into existence a creature from a purely mathematical perspective, but it is very significant if that entity arises from other considerations such as Fourier analysis or physics or geometry. In fact, non separable Hilbert spaces do arise naturally and what are almost periodic functions? They play a very important role in modern dynamical systems, celestial mechanics for example. So, the definition goes back to Harald Bohr who is the brother of the physicist Niels Bohr. So, a continuous function f from the real line to the complex plane is said to be almost periodic if for every epsilon greater than 0, there is a positive number L epsilon such that every interval of length L epsilon, I emphasize the use of the word every interval of length L epsilon that every has been marked contains a point C such that mod f of x plus C minus f of x less than epsilon for all x in R. In particular, for every epsilon greater than 0, there is an epsilon approximate period. You see if f of x plus C minus f of x were 0 for all x, then C will be a period and f will be a periodic function with period C, but we do not have f of x plus C equal to f of x. What we need is mod f of x plus C minus f of x less than epsilon for all x. In other words, the function is not periodic, but it is epsilon approximately periodic and and that epsilon approximate period is C. Unfortunately, unlike periodic functions and epsilon approximate periods will not form a subgroup. You remember, right in the first part of the course, if you have a function f which is periodic with period C, then it will have periods C, 2 C, 3 C, etc. The periods will form a subgroup of the real line and it will form a cyclic subgroup unless the function is 0 or a constant. Here if you have just an epsilon approximate period C, then 2 C will be a 2 epsilon approximate period, not an epsilon approximate period. This is an important fact. It is very tempting to say, okay in this inequality simply replace x by x plus C, then you will get mod f of x plus 2 C minus f of x plus C less than epsilon. But if you try to add the 2 inequalities and appeal to the triangle inequality, what you are going to get is mod f of x plus 2 C minus f of x is less than 2 epsilon. So 2 C will be a 2 epsilon approximate period that is not good enough. So unlike the periodic case, the epsilon approximate periods will not form an additive subgroup of R. So the clause in the definition that every interval of length epsilon, where I flagged the term every, that is significant because this phrase that every interval of length epsilon contains an epsilon approximate period is a strong condition and this condition is a uniformity condition that no matter where this interval of length epsilon is located in that interval of length epsilon, you will find an epsilon approximate period. So it is somewhat of a uniformity condition and you should compare this clause with the corresponding definitions of uniform continuity in elementary analysis. It is exactly an analog of that. So it is a uniformity condition. It is completely trivial to observe that a periodic function is certainly almost periodic. Converse is not true as we shall see several examples. An almost periodic function is uniformly continuous. To prove this uniform continuity, you will use the fact that the interval of length epsilon can be located anywhere in the real line and every interval of length epsilon should contain an epsilon approximate period C. The set of almost periodic functions will be denoted by AP and AP will be a vector space. It will also be an algebra because if f and g are almost periodic, f into g will be almost periodic and also if you take a sequence of almost periodic functions converging uniformly, the limit function will also be almost periodic. So this algebra of almost periodic functions is closed under uniform limits. So it is a very pleasant space to work with. Now we want to make this almost periodic functions into an inner product space. So in order to make it an inner product space, we need to define an inner product and in order to define the inner product, we need to bring in a notion called mean value of an almost periodic function. What is the mean value of an almost periodic function? Suppose f is almost periodic. Then we look at the limit t goes to infinity 1 upon 2t integral from minus t to t f of x plus t dt. As such, if you stare at this limit, the limit is supposed to depend on x. If it is independent of x, then that is a result that you need to prove. If the function is almost periodic, then this limit is independent of x and the limit exists uniformly with respect to x. Remember, when the definition of a limit given in epsilon greater than 0, there is a t0 such that the mod of integral from minus t to t f of x plus t dt minus the limit L less than epsilon for capital T greater than or equal to t0. This t0 will certainly depend on epsilon. This t0 will be independent of x, that is uniformity with respect to x and the limit is called the mean value of the almost periodic function f. So almost periodic functions have a mean value. Now what happens? You take two almost periodic functions f and g and I know that the product is almost periodic. Needless to say, if g is almost periodic, the complex conjugate g bar is also almost periodic. That is an obvious remark. And so take the mean value of f g bar. If you take the mean value of f g bar, you define that to be the inner product. Of course, you have to check the properties of inner product that is completely trivial to check and this inner product gives us to a norm namely the triangular bracket ff square root and that is the norm of f. This inner product makes AP, the set of all almost periodic functions a pre-Hillbert space. What is a pre-Hillbert space? A pre-Hillbert space is just an inner product space. The problem is that this inner product may not give rise to a complete metric. The metric resulting from this norm need not be complete and in fact it is not going to be complete. But you know that every metric space has a completion and this completion will be a, in this case will be a Hilbert space. When you take the completion of a pre-Hillbert space, you get a Hilbert space. And this Hilbert space is the Hilbert space of almost periodic functions. A simple exercise, suppose if f and g are actually 2 pi periodic continuous functions, then what is this mfg bar? The mean value of mfg bar will be the usual L2 inner product of f and g that we have studied in the first chapter. Now let us take an example f of t equal to e to the power i lambda t and gx equal to e to the power i mu t where lambda and mu are real numbers. Take that they are almost periodic, they are actually periodic. e to the power i lambda t and e to the power i mu t are both periodic functions and so they are almost periodic. Try to calculate the mean value of ff bar, you will get 1 upon 2t integral from minus t to t e to the power i lambda t plus x e to the power minus i lambda t plus x because you are taking the complex conjugate. But now you are going to get simply 1, so norm of f is 1, norm of g is 1 and when you try to calculate the inner product of f and g, you are going to get 0 if lambda not equal to mu. That is a very easy integration and I leave it to you to check. The space of almost periodic functions contains this uncountable family e to the power i lambda t where the lambda varies over the real numbers. And if I take lambda and mu to be distinct then they are orthogonal and they are unit vectors. So this space AP contains an uncountable family of pairwise orthogonal unit vectors. If you have an uncountable collection of pairwise unit vectors, the distance between any two of them is going to be root 2 Pythagoras's theorem. And now if you take any of these and take a ball of radius root 2 by 3, take any element of this uncountable family, take a ball with that as a center and with radius root 2 by 3. I am going to get a collection of pairwise disjoint balls, uncountable family. So obviously this will not have a countable sub cover and remember for a metric space, second countability and separability are equivalent. And so this family is not second countable and so obviously it is not separable either. So the space of almost periodic functions is not separable as a metric space and so the Hilbert space completion which is a larger space will also not be separable. So we do have a very important naturally occurring example of a non separable Hilbert space. As I mentioned, I will give you references. I have already given you the reference of Ries-Norge functional analysis. There is an Indian reprint which appeared in 2007 and the relevant pages are 254 to 260 where it gives a certain theorems which are relevant to Fourier analysis. The other reference, Corinneau is a very classic work on almost periodic functions which appeared in 1989 as an English translation published by Chelsea publishers. There are other references to almost periodic functions. Another classic reference is Harold Bohr's book on almost periodic functions. So with that example, we conclude this general discussion of Hilbert spaces. Now we come to the spectral theorem and what avatar the spectral theorem takes in a Hilbert space. So this is a long chapter in functional analysis and to discuss spectral theorem in its full glory will take us too far of a field. We will need several more lectures, perhaps a new course on this. So there is one important special case that is relevant to us in the context of generalized Fourier expansion namely the regular-sterm level problems. The case of the vibrating string was an example of a regular-sterm level problem and that gives rise to a certain part of spectral theorem which is a spectral theorem for compact self-fit joint operators on a Hilbert space. So regular-sterm level problem given as to compact self-fit joint operator on a Hilbert space, it is this particular version of the spectral theorem, a very basic and almost elementary version of the spectral theorem that we are going to discuss and the proof proceeds along predictable lines as we shall see. And we have seen the classical approach in chapter 5 and 6 to the-sterm level problems. Now we will look at the same problem in a functional analytic setting. So we begin with the relevant terms and definitions and some basic examples. Recall that an operator T from H to H is seem to be a bounded operator or a continuous operator if norm Tx less than or equal to c times norm x for all x in H. We began this chapter with this discussion. In other words the image of the unit ball, suppose I take norm x to be less than or equal to 1, then this inequality reads norm Tx less than or equal to c. The image of the unit ball U in H is again bounded in H, norm bounded. Compact operators are stronger. We want the image of the unit ball to be pre-compact in H. So H is a Hilbert space and operator T from H to H is said to be compact if image of U under T is of the closed unit ball U is pre-compact in H. Recall what is the pre-compact set in a metric space, a set A in a metric space is said to be pre-compact if its closure is compact. Now this definition of course will carry over word for word for any Banach space. It does not have to be a Hilbert space. The same definition will simply go through for any Banach space such as L1 of closed interval 0 1 or L infinity or C of AB with sup normal whatever it is. So that as far as the definition goes everything goes for word for word. Now if H is a finite dimensional Hilbert space that every operator is compact. In general compact operators will form a distinguished subclass of operators on a Hilbert space. We look at some more interesting examples of compact operators on a Hilbert space. We begin with the favorite example L2 of 0 1. The first example is a very fundamental example the Volterra operator T from L2 of 0 1 to L2 of 0 1 given by Tf of x equal to integral 0 to x ft dt. This operator will arise when we try to solve an initial value problem for a differential equation which initial value problem are we looking at y prime equal to f of x y of 0 equal to 0 nothing fancy just a first order ODE and even a variable separable ODE. So you simply integrate y prime equal to f of x. So what is y of x y of x is integral from 0 to x ft dt that is exactly the operator that is displayed Tf of x is integral 0 to x f of t dt. Now remember that since we started out with an f which is in L2 and L2 of the closed interval 0 1 is certainly sitting inside L1 of the closed interval 0 1 right that is basic Cauchy Schwarz inequality and when you integrate an L1 function from 0 to x you are going to produce a continuous function. So Tf of x is actually going to be continuous it is going to be better than L2. So actually what is happening is that the operator T is mapping L2 of the closed interval 0 1 into C 0 1. Now let us do something better instead of just observing that Tf of x is continuous let us prove that is uniformly continuous. So let us assume that x is less than y so Tf of x is what Tf of x is integral 0 to x ft dt what is Tf of y integral 0 to y ft dt take the difference and take the absolute value mod Tf x minus Tf y less than or equal to integral from x to y mod ft dt the mod goes inside the integral apply the Cauchy Schwarz inequality. You will get root mod x minus y times the L2 norm of f but now we are going to look at what happened to the unit ball remember. So norm of f is going to be less than or equal to 1 so mod Tf x minus Tf y less than or equal to square root of mod x minus y we are actually established that Tf of x is not only continuous it is actually holds and continuous of exponent half. So now we want to prove that the Volterra operator is a compact operator. So as I said we are going to restrict ourselves to the unit ball norm f less than or equal to 1. So what is the inequality we got mod Tf x minus Tf y less than or equal to root of mod x minus y Tf of 0 is 0. So let us put y equal to 0 we get what mod Tf x is less than or equal to 1. So what we are obtained is that this family of functions Tf that we get is actually a bounded above in sup norm. So in sup norm these Tf's are all bounded by 1 and all the Tf's are holder continuous to exponent half and with the same constant C in the middle namely 1. So this inequality mod Tf x minus Tf y less than or equal to square root mod x minus y shows that the family of Tf's where f varies over the unit ball is actually equicontinuous and uniformly bounded. Now we know that if you have a family of functions which is equicontinuous and uniformly bounded on the closed interval 0 1 then it is pre-compact with respect to sup norm. But pre-compactness with respect to sup norm will imply pre-compactness with respect to L2 norm because convergence in sup norm is stronger than L2 convergence. In other words every sequence has a subsequence which converges uniformly and hence it converges in L2 norm and how did I conclude all these things I used the classical Arzela-Oscoli theorem. What are the classical Arzela-Oscoli theorem? If you have a family of continuous functions on the closed interval 0 1 if it is equicontinuous and uniformly bounded then it is pre-compact with respect to the sup norm that means that its closure is going to be compact with respect to sup norm. Every sequence will have a subsequence converging in sup norm. So recall the definition of equicontinuity a family S of continuous functions on the closed interval 0 1 is said to be equicontinuous if given any epsilon greater than 0 there exists a delta greater than 0 such that for all the functions in S mod x minus y less than delta implies mod fx minus fy less than epsilon the delta does not depend on f the delta does not depend on f. Of course if this set consists of just one function then this is simply uniform continuous. The same delta works for all the functions of the given family. A family S of continuous functions is said to be uniformly bounded if there is a common m such that mod fx is less than or equal to m for all the functions f. So you got uniform boundedness and equicontinuity. The Askoli-Arzela theorem says that a family of continuous functions on the closed interval 0 1 is pre-compact if and only if it is equicontinuous and uniformly bounded. So in our case we took the Volterra operated t and we computed tf and the f varies over the unit ball of L2 the image consists of a family of continuous functions and they were uniformly bounded we took y equal to 0 and we got the uniform bound 1 and it was equicontinuous because it was holder continuous to the same exponent half and we simply got mod tfx minus tfy less than or equal to square root of mod x minus y that will immediately give you equicontinuity. So by Askoli-Arzela theorem it is pre-compact with respect to the sup norm and hence it will be pre-compact with respect to the L2 norm. The proof of Askoli-Arzela theorem is not difficult at all but we shall not stop to prove it here because the time is limited. A very nice proof is given in Rudin's principles of mathematical analysis the third edition 1976. So the Volterra operator has been shown to be a compact operator. I think with this we shall stop this capsule here and we will continue with this in the next capsule. Thank you very much.