 Hello, welcome to module 24 of NPTEL NOC on Introduction to Point Setupology Part 2. So, we shall continue the study of Stone-Weierstrass theorem, today we will take the complex case and then we will also study the extended version namely for locally compact spaces. So, theorem 5.36, the complex Stone-Weierstrass, let X be a compact rar-stores space, A be a closed subalgebra of C X C. So, this complex varied function contains x is locally compact rar-stores, which separates points this should be always there, contains a non-zero constant this is an option always, but this time we are taking only this hypothesis, there is one more closed under conjugation is very important. Now, we are taking the complex case you have to put an extra condition you have closed under conjugation, conjugation of what conjugation coming from the complex conjugation C, a function taking values in C, you know what is your conjugation of that, because you can follow it by conjugation C to C, so that is F point F bar. So, if F is there F bar should be also there that is the meaning of closed under conjugation, then this A is the whole of X C. If you remove this closed under conjugation and take C equal to r, this is the Stone-Weierstrass theorem. Corresponding to the theorem of Gaddy, there is a version here that I will leave to you as an exercise, so we will only concentrate on the main version here. Consider CXR as a sub algebra of CXC over the real, this is like a real vector subspace of CXR, which is a real Banach subspace. Let us put A underscore r here as A intersection CXR, CXR is a real sub algebra, A is a sub algebra complex sub algebra, but it is also real sub algebra, so when you take the intersection this will be a closed sub algebra of CXR, it is an r algebra, r vector space, real vector space. So, this is a sub algebra, it is a closed sub algebra, now something nice happens because A is closed under conjugation, so that is what we have to use. If I write an element of CXR as F equal to U plus IV, any complex value to function not element of CXR, CXT as a real part and imaginary part, you can always write U plus IV, where U and V are now inside CXR, they are continuous also. Since A is closed under conjugation, U minus IV will be also inside A, therefore their sum will be inside A, divide by 2 will be inside A, that is nothing but U. Similarly, I times this will be also inside A, when you add the 2, you get I times V divided by 2 will be just divided by 2I, that will be also V. So, you see that F belongs to A implies real part of F is inside A, r right, it is inside A as well as inside CXR, it is a real now, it is a r. Similarly, the imaginary part also inside A, because once A is there, I times A is there, F is there, I times F is there, minus I times F is there, you apply that one to this condition, the real part of minus I times F is the imaginary part of F, there are different ways of looking at it. So, in conclusion, closed under conjugation implies that for the real part of that, A intersection CXR, this has both real part and imaginary part of every element inside A will be inside here. Just you have shown that F belongs to A implies both real and imaginary parts are inside, okay, inside A, r. Now, given X not equal to Y, let F belong to CXC be such that FX is not equal to FY. As usual, you can write F as U plus IV, then FX not equal to FY implies either the real parts are different or the imaginary part are distinct. When two compressed numbers are different, they can be that the real parts are different or imaginary parts are different, okay. Now, U and V are inside A, r, right. So, A r also separates points, either I can take the imaginary part or I can take the real part or imaginary part. So, A separates points implies A r also separates points, thanks to closed under conjugation, without that you could not have concluded this one. For similar reasons, if C is a non-zero constant inside A, a non-zero constant could be what? Could be a complex number, one either real part or imaginary part must be non-zero. So, the same hypothesis will be true for A r also. A r also contains the non-zero constant. This time non-zero constant has to be real. So, you have to take real part or the imaginary part, whichever is non-zero, okay. They will be there inside A. So, all the hypotheses are satisfied by A r, right. Therefore, by our real part theorem, okay, we conclude that A r must be equal to the whole of C X r. But then, I times C of X r will be also inside A because after all A r is above this curly A and curly A is a complex vector space. So, I times this will be also inside A. Therefore, C X C, which is nothing but real plus imaginary, right. C X r plus I times C X r. Both of them are inside A. So, some is inside A. Therefore, this whole thing is A, okay. It cannot be bigger than, anything cannot be bigger than C X C. So, that is the end of the proof for complex case, all right. Now, let us go to the extensions of this one, the locally compact case. Once again, Alexander of one point compactification plays the role and our job will be quite simple when you pass on to the Alexander's compactification. Then, we can apply these theorems. So, I have told you, I repeat, you may state and prove a theorem similar to this 5.35 due to Gadi in the complex case, okay, which we shall leave you as an exercise. We shall now discuss the case when X is locally compact and R star. As usual, we will now combine both the cases. This K denotes either R or C, okay. So, take a function F from X to K. We say F vanishes at infinity. This is a phrase I am defining. I am not defining infinity here. So, there is no infinity as such. But we are defining what is the meaning of F vanishes at infinity. Understand? What is that? If for every epsilon positive, there exists compact substrate C of X such that modulus of X is less than epsilon for all X inside X minus C in the complement of away from a compact set, I should be able to control the value of Fx, modulus of Fx is arbitrary. Whatever epsilon I have chosen, no, C can be chosen. C will depend upon epsilon as well as F, of course, okay. Such that away from C, modulus of Fx will be less than epsilon. That is the meaning of vanishing at infinity. Just let me give you some examples. If X is compact, of course, every function from X to R vanishes at infinity, vacuously. Why? Because I can take C equal to the whole of X and X minus C is empty. So, there is no condition. That is all. On R, okay, look at this function G, Gx equal to e power minus X square, okay. At X equal to 0, this is 1, alright. And then it just tapers down and as X goes to infinity, this tends to 0, okay. So, here the model is there. Why we are using this word? As X goes to infinity, the limit, this function tends to 0. So, we have converted that into an arbitrary topological space by this definition here, vanishing at infinity. Outside a compact set, it will become less than epsilon. Given any epsilon, you can choose your compact set such that that condition is whole. That is the meaning of this term, okay. So, this example will motivate the definition here. A constant function on a non-compact space vanishes at infinity if and only if it is a 0 function. Take any constant function, take your epsilon to be such that it is less than modulus of that constant, okay. That is possible if the constant is not 0. Then that condition will not be satisfied at all, right. So, it will not vanish at infinity. So, it vanishes at only if it is a function, 0 function, okay. This is a strong conclusion here, you see. Because we are interested in subspace is having, must be having non-zero constant. If non-zero constant, then this condition will not be satisfied now. So, these two conditions are somewhat, you know, opposite of each other, all right. So, let X be locally compact host of space. By the way, this definition is there for all spaces. Now, we are only interested in locally compact host of space. Let X star denote its one point compactification. I want to specifically say that it is one point compactification when I mean the. So, I am referring to Alexander of compact space. A continuous function f from X to K is the restriction of a continuous function f hat from X star to K such that f star of f hat of f, the star is 0 if filled only if f vanishes at infinity. So, the definition of vanishing at infinity has been given a different meaning here. Take a function which vanishes at infinity in a locally compact host of space. Then you can extend it to a continuous function on the one point compactification by sending the infinity or the star to 0 and conversely. If you can do that and you can always define f hat of star equal to 0, but it may not be continuous. After defining this, if it is continuous then this f must be vanishing at infinity. So, this is just a consequence of the definition of X star. Namely, what are the opens of sets of X star which contain the point 0, point infinity. What are the neighborhoods of infinity? They are nothing but the complement must be compact and the host of course, so compact and close. Compliment must be compact, so that is the hyper. So, that will automatically give you this one. So, I am not going to explain anything more than that here. The proof is easy. So, now comes this extra notation we have here. Let C0 of Xk denote the space of all continuous functions on X which vanishes at infinity. Then C0 of Xk is a closed sub-algebra of the Penang algebra of all bounded continuous functions on X. When X is locally compact host of space, it can also be identified with the maximal ideal M star. This notation is now familiar to you. M X was what? All those which vanishes at X. So, the same thing M star is all those functions which take 0 value at the point star which is the infinity. You can write it as infinity or you can write it as star. So, this is also straightforward now. So, it is like remark I put it as a lemma so that I can refer to this one again and again. That is all. Now, we can state the extended stono extra theorem. Start with any locally compact host of space. Let A contained inside C0 of Xk be a closed sub-algebra which separates points of X and is a closed under conjugation. If Kc. So, this closed under conjugation is not necessary or even if you put it is R-less if K is R. That is the only case I have to mention these two separately. Otherwise, the proofs are all the same. You know together I can handle the case R or C together. As soon as Kc you should assume this extra condition. That is all. Then the conclusion is that either A is C0 of Xk the entire thing or you have one extra point X0 inside X now such that this A is Mx0 I am putting because you should not confuse it with the Mx0 has a different meaning in Cxr, Cxk. So, this is vanishing at 0 at x0 and also it is vanishing at infinity. I can have put star here if we notation but Mx0 I have put fine. So, these are the 0s of course these are 0s at star and they are 0 at one extra point. So, the sub-algebra can be this one or it should be the whole of it. So, this is the conclusion of extended stone-wise truss theorem. So, we are not going to study Cxr in particular here the whole space. We are only going to study those which vanishes at infinity. So, that is the key for us so that we can use the compactification of X. A little bit caution is necessary here before we apply the result for compact case. I told you that we want to convert the problem into studying the continuous functions on compact by going to one point compactification. But you have to be bit cautious here namely consider the case when X itself is compact then C0 of Xk is nothing but C of Xk which we have seen that every point every function now vanishes at infinity that we have seen. And we can directly apply the result in the compact case. So, we do not have to prove this one but we have to consider this one because a locally compact torsional space is also a compact torsional space is also a locally compact torsional space. That is why we have to check that the statement is correct in that is all we are not proving that. So, that part is taken care. Therefore, we can now come to the case when X is locally compact. So, I repeat consider the case when X is compact then Cx0 C0 of Xk is nothing but C of Xk. Hence we can directly apply this in a compact case. Now come to the case when X is non-compact. As observed before the only constant function which vanishes at infinity is 0. Therefore, C0 of Xk does not have any non-zero constant. In particular, the sub-vajibras of C0 of Xk also have no non-zero constant. Therefore, the regular Stone-Vaistra theorem cannot be applied here. So, you have to take the Caddy's version the other part. Now as a sub-vajibra of Cx0 X star of K, A may not separate points of X star. So, that is also a problem. We have assumed that A separates points of X. But when you pass to X star and think of this as sub-vajibra this may fail to separate points also. Because there is an extra point right. So, we have to be careful. So, let us study these things case by case. So, assume first that A separates points of X star. Remember, A is a sub-vajibra of C0 of Xk. Therefore, you can think for each F inside here you can put F hat and think of that as a function from X star. It is a unique F hat. What is F hat? F hat of star is going to 0. That is the only way you can extend. So, you can think of A as a sub-vajibra of Cx star K. We can then apply the standard theorem 535 Caddy's version to conclude that A is Cx star K or there is a unique point X in X star such that A equal to Mx. So, this is the 2 parts are there. But A is already in C0 of Xk which is not the whole of C0 of X star of K, C of X star of K. The first possibility is ruled out. Therefore, it is a second possibility. But this X cannot be inside X because then A does not separate X and star contradicting the assumption. So, we say A separates points of X star. If there is another point X and star, both of them will be 0. So, that is not possible. So, this X must be equal to star. Therefore, A is nothing but C0 of X. Wonderful. So, this is the same point. Suppose, I assume A separates points of X star. Now comes the little more complicated case. Suppose, A does not separate X star. Then what I am going to do? In passing, we note that since A separates points of X, the extra assumption that A separates points of X star is the same thing as equal to assuming that for every point X in X, there must be a function F belonging to this is that Fx is not equal to 0. For each point, this function Fx, so this just means that it is not contained in any maximal ideal Mx. That was the assumption. If this is not true means equivalently what happens when A does not separate points. Since A separates points of X, but does not separate points of X star, it just means that now you have to have analyzed this incorrectly. This can happen only if there exists a unique point X0 X such that A is inside Mx0 intersection Cx0 which is Mx0 0. As soon as this condition is satisfied for each point X, Fx is not equal to 0. There is one F. It will separate points of X and star. In addition to separating points within X, it will separate points of X star the whole thing. So under the assumption that it does not separate X star, we have concluded that A is contained inside in this ideal. You have to show that it is equal to this ideal. So, star is not yet over. That is the only choice now. It cannot be the whole thing, it has to be this ideal. Now consider subspace y equal to X minus X0. Throw away X0. When you throw away X0, you have got an open subset that is also locally compact of star. So that type of this is not a chain. Now I have to make some more cases here. Suppose X0 is not an isolated point. Let us assume that X0 is not an isolated point. That is easier case. Perhaps the other one is too easy. That is why I am considering it later. Inclusion map I from Y to X induces an algebra homomorphism of algebra which I write as I star from CXK to CYK. Y is a subspace. So what is this? I take Y to X inclusion map and then take the function F from X to K. So that is I star. So it is like restriction map actually. But restriction map is injective now. This is always there. This restriction algebra homomorphism injective map algebra homomorphism always there. It is injective because X0 is not an isolated point. It just means that the subspace Y is dense in X. Dense subset if two functions agree, they agree on the whole space. So this we are using it again because Y is X is host of space. Therefore this I star of now. So we have got an injective mapping. Now look at the image of MX0, intersection CX0 of XK. This map I am restricting it to C0 now. C0 is what? Those things which vanishes at infinity. Vanishes at X star. A star is not here but vanishing at infinity makes sense. Intersect with MX0, that means those which vanishes at X0 also. So I star of that is nothing but C0 of YK. What is being of C0 of YK? Again those which vanishes at infinity. Now look at this one. That this X add an infinity that star but this Y not, this Y has two of them. This X0 which you have thrown away that is also part of infinity now. So around the neighborhood of that point, you can cut off and then whatever is left out, you cut off from the other side also you get a compact. So you can think of this also being now part of infinity now. This point which you have thrown out. So this is important to see. This is nothing but now C0 of YK. Without referring to anything, any extraneous points and so on, what is this one? This is just all continuous functions which vanishes at infinity in our definition. Now consider the one point complication Y star of X, Y star of Y star of Y star. Since X0 is the unique point such that A is contained inside MX0, it follows that A separates points of Y star. Uniqueness was important there. Now you see A did not separate points of this X star. That was the starting assumption but after throwing away X0 and bringing that in the sense to along with a star so that the bad point has been brought back together with infinity. That is the whole idea here. Now what happens is X0 was the unique point in inside MX0. It follows that this A now separates points of Y star. Therefore we can conclude from case one that A must be C0 of YK which is by we have defined is image of this one. I am just writing MX0 intersection this itself because I star is injective mapping which is MX0, X0 0. So that is the end of one case. What is the sub case? Sub case is when X0 is an isolated, not an isolated point of X. Now we come to the case wherein X0 is an isolated point of X. Here you do not need much topology just algebra now. When X0 is an isolated point how does CXK look like? At the point, the isolated point you are free to define the function whichever way you like. So that is a free point. So it can be assigned free real or complex number any number in K. Therefore what happens is CXR, CXK is nothing but CYK where Y is the complement of this point X0 is thrown out. Cross with K. This second factor corresponds to arbitrary values taken at the point X0. That is what CXK is. With what is this product structure I want to tell you namely take the inclusion map I from Y to X which we have taken earlier with I star being the first projection. Now the restriction map from here to here becomes the first projection. The second projection is evaluation of the function at the point X0. That is the second projection. So if I know what are two projection maps here I know the product structure. In other words every function here can be written as I star of the function comma f evaluated at X0. That expression will be unique. Therefore C0 of XK is nothing but C0 of YK cross K. K has to be free. See vanishing at infinity is not affected with this K at all here. Because whenever you take compact set you can include the single point inside that compact set. Away from that compact set vanishing is same thing vanishing at this part. So C0 of XK is C0 of YK cross K. And MX0 the ideal is nothing but CY of K because this K factor will go away. Things which vanish then K factor will go away. Therefore we are in the situation wherein A is contained inside C0 of YK and separates points of Y star namely Ks1. So sub case B is reduced to Ks1. Therefore we conclude that A is nothing but C0 of YK which is MX0 X0. This was case 1. So this completes the proof that the theorem that for locally compact spaces how does a closed subalgebra look like. Locally compact top-top spaces take a closed subalgebra of the set of the algebra of all functions which vanishes at infinity. For that we have this. This is one extension. I do not say this is the only extension. There can be many other possibilities here. This is one of the popular things. You can also have a look at Siemens book for such versions. So thank you. Maybe I will let you know a few of the exercises here before closing up. They are not directly related to this one but since we are studying function spaces I think these are relevant. You prove that continuous functions on A, B, this binoc algebra is separable. Separable binoc algebras are more and rarer and they are very important. F belongs to C01. Instead of A, B I put 0, 1 just for writing down. It is actually applicable to CAB also. Put Mn of F as fx multiplied by xn and integrate. So this is a weight function. So call that as Mn for n equal to 0, 1, 2, 3, up to n. With one function you have put several these functions or Mn of F are these constants here. Put Mf equal to the sequence M0f, M1f, M2f and so on. So this is the definition. So I have taken f and then I have produced a sequence here into whatever. So this r plus r plus r show that this map M is injective on C01. What does it mean? One of these if f and g are different at least one of them Mn for some n will be different. That is what we have to prove. Now this exercise is directly related to the Stone Weistras theorem. Through the following n-variable version of Weistras approximation theorem. Weistras approximation theorem was only for in the interval, closed interval to r. Now we will also prove it for n-variable. What is that? I have given the version also here. Any continuous real-world function on a closed rectangular box in that rn can be approximated by a polynomial in n-variables. They deduce from this that any continuous real-world function on a closed and bounded subset of rn instead of the box can be uniformly approximated by a polynomial function. Okay. Here also you can put the uniformly approximated. Show that any real-value or complex-valued function can be uniformly approximated by a polynomial in a complex variable z and z bar. A polynomial in z and z bar is different from just polynomial z and taking some conjugation, okay. For example, x can be written as z plus z bar by 2, right. So that is the whole idea here. Contrast is the fact that from complex analysis the following fact is there. If you do not know that, you may learn it somewhere. Namely, look at the simple function f z equal to 1 by z. This function cannot be approximated uniformly on the unit circle by a polynomial in z. Unit circle is closed and compact. Only unit circle you take, okay. Do not take the whole disk. The whole disk does not make sense because 1 by z is not defined on the whole disk. At 0 it is not defined, okay. Try to find a sequence of uniformly, a sequence of polynomials is converged. It is not possible, okay. But why can't we, why it should be true? I mean because why can't we apply Stone-Weisstrass theorem for complex case to this function? After all this is a nice function defined on c minus 0, right. And then I am taking a compact subset there, the circle. So what is wrong? So you have to explain that, okay. Don't make this mistake that Weisstrass theorem can be applied here, okay. Not Weisstrass theorem, Stone-Weisstrass theorem. Why? Just tell me why this is one, there is one line, okay. That is this exercise. There are some more exercises. These exercises will be there in the, in the PDF file I am going to give you anyway, okay. So you do not have to depend upon the slide but I would like to show it in the slide. That is all, alright. Thanks a lot. So let us meet next time with a new topic.