 Hello, welcome to NPTEL NOC course on point set apology part 2, module 17. As promised last time, we shall do some function space study today, Arjala Ascoli's Theorums. So, in this section, let us address ourselves to determine compact subspaces of the Banach space that we have introduced namely continuous functions from X to R and continuous function X to C. So, both of them we handled simultaneously no separate proofs or no separate techniques, where X is a compact metric space ok. And then once you have a compact metric space, continuous functions are given supremum norm that is how it becomes a Banach space. As before we shall use the notation is K to denote either R or C. I will make the historical comments later ok. So, we need a new and important notion here of equicontinuity. Let Xt be any metric space, a family of functions f from X to K ok. It is what we starting data for us. We say A, this A is equicontinuous if for every epsilon positive, there exist a delta positive such that distance between X and Y is less than delta implies distance between f X and f Y is less than epsilon for all f inside A. So, this is an epsilon delta definition. So, why it is called equicontinuous you can see suppose it is only one function f. Then this is like uniform continuity, but epsilon deltas are the same for all f inside A. So, that is why it is called equicontinuous family. So, this is a new name for a new concept ok. So, every member of an equicontinuous family is automatically uniformly continuous. If you just read this one for f this uniform continuity right. Also it is easy to see that a finite family of uniformly continuous functions is equicontinuous because for each f you have delta 1 say f 1 delta 1 f 2 delta 2. Then you take the minimum of delta 1 delta 2 delta n that will work for all f's. So, finite families of uniform continuous functions is automatically equicontinuous. So, this kind of notion is important only when A is an infinite family ok. So, here is the famous theorem of Arzela Ascoli. Let xt be a compact matrix space. Then a closed subset A of the C remember C continuous functions from x to k. A is compact if and only if it is bounded and equicontinuous. Since we are working in a house door space you know this C H way is a Banach space. So, if you want compactness and so on you have to assume A is closed. So, that standing assumption that A is a closed family of continuous functions from x to k. It will be compact in the supremum norm of apology if and only if it is bounded and equicontinuous ok. So, this is the statement. The proof is one part is very easy. The second part is a little more you know new techniques illuminating. So, that is what we have to start with the compact A. Then it is clearly bounded because in any matrix space compact subsets are bounded. In fact, we know that it is totally bounded. To see if we continuity given epsilon positive let f1 f2 fn contained inside A be an epsilon 3 net. Because I am now using the total boundedness of this one ok. For each i ring to 1 to n whether f1 or fn right. Choose delta i such that modulus of x minus y is less than delta i. Imply f y of x minus f y of y is less than epsilon minus 3 by uniform continuity since x is compact. Indeed here I should put just distance between x and y ok. Modulus will make sense only when x is real or complex number for Rn or Cn. Let now delta be the minimum of delta 1 delta 2 delta n ok. You see I have already used that is only finitely many of them. So, as before indicated by my remark I am taking the minimum of delta i. Check this delta i as the property distance between x and y is less than delta i implies f x minus f y less than 4 all the f's. First it is true for all f1 f2 fn. Now you have to use the fact that this f1 f2 finished an epsilon by 3 net. So, you have to do little more triangle inequality business here ok. So, that much I am leaving it to you as an exercise. For all f inside A this will be true. So, from finite set to the whole of A the link is that this finite set is a epsilon 3 net. So, that comes from total boundedness alright. The converse part is much more involved ok. If these conditions are satisfied then you have to show that this is compact ok that A is compact ok. Starting with a bounded and epicontinuous family we should show that A sequentially compact. Then by our earlier theorem compactness of A will follow ok. Being a closed subspace of a complete metric space A is a complete metric space by itself right. Therefore, from the previous theorem A will be compact. A complete metric space all these things are equivalent etcetera we have seen earlier. Since x d is a compact metric space it is second countable also and hence separable ok. Therefore, x has a countable dense subset let us fix one such anyone any countable set which is dense ok. No need to be any specific here because after all x t is an arbitrary space we do not know anything more than that ok. Fix one such separable subspace subset of x that is all fine countable separable. Start with a sequence say f 1 f 2 f n. So, I am calling it as f naught inside A every sequence must have a sub sequence which is convergent is our aim to show that this f has a sub sequence to convergent. So, what we are going to do is we are going to produce a sequence of sub sequences ok. By that is almost like improving f each step we shall produce a sub sequence which is convergent finally. So, the first sequence is look at values of f 1 ok f 1 x 1 f 1 f 2 x 1 and all of this entire sequence evaluated at x 1 at one single point. So, now what we have what is a sequence inside k which is bounded because entire of A is bounded. So, f 1 f f n is also bounded by one single say m or something. So, each sequence here is bounded right a bounded sequence inside r or c has a sub sequence which is convergent. So, the property of the codomain is essential here ok it is happening in the codomain point wise. So, we are using that here. So, it has a convergent subsequence. So, we shall use this to denote. So, this convergent subsequence f i 1 x 1 it is like you know f 1 x 1 f 2 x 1 f 3 x 1 and so on. So, a sub sequence we have got ok f i x 1 we shall now work with the sequence f 1 which is f i 1. Forget about x 1 f i 1 ok this is subsequence of f 1 f 2 f n ok or anything. So, that subsequence we are calling it as f 1 now apply the same technique to this f 1, but with the point x 2 evaluate it at x 2 ok take a subsequence of this which is convergent like this inductively choose sequence is f k which is a subsequence of f you know f k minus 1 and convergent what is convergent f i k x k is convergent ok each time f i k x k is convergent f i k is a subsequence of f i k minus 1 ok. So, we have got a sequence of subsequences from starting with one sequence ok. So, here is a picture you can say f 1 x 1 f 2 x 1 and so on ok I have picked up f i 1 x 1 ok then there is there are many more terms here I have not bothered about that I am looking at the first term here ok. So, if I you only take that subsequence convergent is what I know similarly the second part I have a subsequence of this f i ok I am not bothered about this one. So, I am taking something the subsequence this may be equal to this one does not matter it is important that I am going forward. So, I have put it somewhere here that is subsequence of that ok next takes a subsequence of that one. So, ultimately what I want is these term are increasing i 2 i 3 etcetera ok. So, this is x 2 here. So, f i 1 x 1 etcetera. So, this is what I am looking at. So, what I am going to do is just s j equal to f j j for all j ok then what happens is this s j is a subsequence of the original f because indexing is going increasing only that is that is all we need for subsequence, but what is the property of this sequence that is what we want to know. So, in order to show that this sequence is convergent it is enough to show that is Cauchy sequence ok for each point ok if it is Cauchy what happens inside k k is just c or r all right that will be a conversion, but here what we are showing is the Cauchy sequence in the metric of what of c which is continuous assumptions from x to r or x to k ok which is uniform which is just the supremum norm. So, what you get is uniform Cauchy the Cauchy sequence just means that in each point if you take those sequences is uniform Cauchy therefore you will get uniform convergence automatically. So, let us do not bother about that much all that I am going to show is that inside the metric space that we are working namely inside c this sequence is a Cauchy sequence then since c is already complete that we have already a will be closed sorry a is also closed what happens the limit which exists because c is complete the limit will be inside a. So, we have formed a subsequence it is convergent inside a starting with any subsequence as convergent subsequence inside a will prove that a is complete ok. So, what is immediate is the fact that for each i the sequence s j of x i is convergent ok for each fix i at each i s j of x i will be convergent because s j of x i after certain stage is a subsequence of f i of x i right all that I have to do is i j is bigger than i. So, that once it is subsequence of that that is a sequence is convergent subsequence will be convergent. So, each s j of x i x j of x 1 s j of x 2 s. So, what we have achieved is a sequence which is convergent at this countable subset is not arbitrary countable subset it is a dense subset from the density we want to conclude that this s j is convergent at all the points. So, what we will do is we will just show that it is Cauchy sequence that is all k n epsilon positive by if we continuity of we have a delta positive such that d x y less than delta implies s j of x minus s j of y is less than epsilon 3 for all j ok. You can choose any anything here given epsilon I choose epsilon by 3 to do that one accordingly there is some delta. Since x i is a dense set it follows that if you take union of all open balls of any positive radius v delta x i I varying from 1 to infinity all these x i's are belonging to a dense subset this will be the whole space of it every point must belong to x 1 of it ok. Since x is compact we will get a finite subset x i 1 x i 2 x i k such that x is in the union of iron into 1 to k b delta of x i. So, we have got an epsilon net for x itself now ok right. So, epsilon net was coming from implicitly for the family a here now it is coming for x itself that is the role of continuity here ok. So, what is the net result now for each fixed r look at s j of x i r ok i r i 1 i 2 i k only k of them ok. All these are convergent we know that one right and hence a Cauchy sequence they are Cauchy sequences ok out of all of them just the iron 1 to k of them I am looking at they are Cauchy sequences therefore we can find a uniform n naught such that n and m are bigger than n naught s n of x i r minus s m of x i r is s n epsilon by 3. Each 1 2 3 up to k will give you n 1 n 2 n k you take the maximum as n naught then for that it will be true for all r r equal to k right. Finally, given x belong to x now let i r be such that x is in one of the balls because x is union of this then x will be at a distance delta less than delta from x i r therefore we can use this ok. What we get is for n and m bigger than n naught I have this inequality so I what I do s m minus s n as soon as n m bigger this one I break it into three parts ok add and subtract s m of x i r s m of x i r added and subtracted then s n of x i r finally s n of x. So, s m of x minus s m of x i r is less than equal to epsilon by 3 by this inequality ok. So, this one comes from this inequality, but x is inside this ball so similarly x i x is inside this ball which will give you the other two inequality. So, you have s m of 3 epsilon by 3 epsilon by 3 this is true for all x now the right hand side is independent of x right. Therefore, the norm of s n minus s m instead is less than equal to epsilon. So, this means that the sequence s n is uniformly Cauchy Cauchy in the norm that we have been we have been using the namely the Banach space norm there ok. So, that is the proof. So, you see how equicontinuity has the news there ok. So, here are a few remarks here the total boundedness is replaced by nearly boundedness here, but what what we have here is already under assumption is that x is compact you know compact matrix space that is how all these things are. However, the equicontinuity together with compactness is taking care of that ok. Remember in that general theorem we want a total boundedness and completeness then only it was compact, but we are not putting total boundedness on x ok it is on c we are working on c for that the equicontinuity helps us this type of results began with the work of Ascoli in 1884 or 1883 sometimes. Ascoli and Arzela are both Italian mathematicians almost contemporaries. Ascoli introduced the notion of equicontinuity and proved the if part of the theorem which was more complicated here ok. However, ten years later Arzela improved upon it by proving the only if part I say improving upon it means because he also clarified his paper is much more more readable also induction the notion of equicontinuity equicontinuity belongs to Ascoli, but the present day form has been you know its contributions many other authors are there like Frechier, Schwarz etcetera they were enriched it with many other versions which are quite often more general and so on right. The the original version of Ascoli is only for closer intervals inside R to R we immediately can realize that you know R can be replaced by C there is no problem, but there are many more other versions wherein the co-domain can be replaced by what are called as uniform spaces and so on ok. So, ascoli Arzela Ascoli theorems have been of great use in differential equations, complex analysis especially Montel's theorem, Pietro's theorem etcetera. For a more general result than the one discussed here you may see Kelly's book which we have been referring to all the time all right. So, here are some easy exercises for you show that a metric space, second countability, lidel-offness and separability they are all equivalent. The second exercise show that a totally bounded metric space is separable and hence second countable. So, you see how different concepts are related of course only when you are working with metric spaces. So, that is for today let us meet next time. Thank you.