 Welcome to NPTEL NOCA on Point Setupology Part O. So, today we will take module 22 Stone Weistras Theorem. We begin with the classical result due to Weistras on approximating continuous functions defined on a closed interval by polynomial functions and then go on to study some sweeping generalization of it popularly known as Stone Weistras Theorem. Throughout this section x will denote a compact host of space and later on locally compact host of space that I will tell you again C X R or C X C will denote respectively binoc algebra of all real or respectively complex valid functions continuous functions and we take the supremum norm because x is in a way compact. The problem of approximating elements of C X R or C of X A is formulated into determining when a particular sub-algebra is dense. So, density of certain thing just implies that you know when you take points in the closure they are the approximated functions from the set which is dense. So, that is the whole terminology. We begin with the classical result namely due to Weistras. Let f from A B to R be any continuous function then there exists a sequence of polynomials which uniformly convert to f on the interval A B. See each term here is very important. A B must be a closed interval and you start with a continuous function. You can approximate it by polynomial functions that is the way to remember but what is exactly the meaning of approximating here that this sequence of polynomials uniformly converges to the function f on the closed interval A B. For simplicity of writing down the proof I will assume that the interval is 0 comma 1 but this is no loss of generality because what you can do is you can make a change of variable linear change of variable actually in the domain itself namely by taking t going to t minus a divided by b minus a. When t equal to a this will be 0 when t equal to b this b minus a divided by b minus a equal to 1. So, interval A B will go to 0 1. So, this way you can change the coordinate. If you have a polynomial in t if you substitute t minus a divided by b minus a instead of t that will be again a polynomial in t. So, there is no loss of generality. So, from now onwards we will look at the closed interval 0 1. So, start with a continuous function defined on 0 1. So, this function I am assuming this notation is again the real variable function here. So, we may assume that the second assumption f 0 is 0 and f 1 is also 0. So, how do you do that? By considering now this time I am going to change the inner domain by taking g t equal to f t minus f 0 minus t times the constant f 1 minus f 0. Look at this one g of 0 is f 0 minus g 0 g of 0 is t is 0. So, this is 0 this is f 0 minus f 0 is 0 g of 1 is f 1 minus f 0 minus this is 0. So, this is f 1 minus f 0 this is f 1 minus f 0 this is f n 0. So, this new function has this property and if I approximated f by a polynomial approximated g by a polynomial I can go back by this. So, the corresponding thing is again linear change of coordinate. If you have polynomial polynomial plus some lambda times some constant t times some constant is again a polynomial. So, two such assumptions right in the beginning we are going to do namely the domain is 0 1 and the function is taking value 0 at both the end points. The advantage of this one is immediately we can extend f to the whole of r by defining it to be 0 outside of the interval 0 1. For each positive integer n put q n of t equal to 1 minus t square raised to n divided by alpha n for all mod t less than equal to 1. So, minus 1 plus 1 I am taking this function and I am putting 0 outside this one. So, look at this one when t equal to 1 or minus 1 this term is 0. Therefore, q n t is a continuous function on the whole of r. This alpha n is some constant here what is that constant it is just the integral of the numerator 1 minus t square raised to n dt minus 1 to plus 1. So, this is some kind of normalizing factor here. So, it is reflected in this property 20 namely this q n t is first of all non-negative ok. This is a non-negative function this is non-negative and here it is just 0 anyway. For all t q of minus t is q t because what appears here is t square ok. So, q of minus t is q n t. Finally, if you integrate this q q n from minus 1 to plus 1 ok this alpha n is a constant that will come out, but what is the integral of the numerator it is just alpha n itself. So, so the total integral here will be just 1. So, this is the normal why I have made it developed alpha n for this purpose one. The integral minus 1 to plus n q n of s d s is 1 ok. So, this is an auxiliary function which is going to help us in approximating the function with polynomials. Let us see how. Now immediately I am going to define the sequence that we are interested in namely p n t is defined to be 0 to 1 f s f is the given function ok and then q n of t minus s d s. So, what I have done is f is a continuous function these auxiliary functions q n which I have defined here I am convoluting it this is a convolution all the way goes back to here ok. So, Weistras also has used it. Now let now now what I have to observe is that this formula I will tell you that p n t is actually a function of t first of all because the variable s is getting integrated here it is actually a polynomial function. Why? Let us look at suppose this polynomial is a constant then this is the whole constant if suppose there is a t term here that t would have come out it will be t times integral 0 to 1 f s d s. If there is a t power n the t power will come out and then there is a function of s left out here. So, when you expand t q n t minus s in terms of t and s it will be some t power n q power n and s power n and so on, but t power n shall come out and then what is left out is some coefficient right. So, p n is actually a polynomial function for each n what we claim is we claim that this sequence p n t converges uniformly to f on 0 1 ok. So, statement is very easy statement is very clear. So, only thing you have to there is some q n t what was that you may not remember. So, you may have to remember this one. In fact, there are many other functions also which will do this job there is no uniqueness here ok. So, this is my personal choice you may say not exactly mean by many other people also use it and at this time I will tell you that there are many proofs of vice truss theorem none of them go beyond the ideas of vice truss, but in computational simplicity or something else some other things they have achieved some different things. For example, what I like is one proof which is there in Rudin's book on principles of mathematical analysis. So, you can have a look at that also ok. So, now we have to prove this p n converges to f uniformly ok. So, first observation is for all t inside 0 1 we have this interval is contained inside t minus 1 comma t plus 1 right. If t is within 0 1 for example, the t is 0 then this is minus 1 to plus 1. So, it contains 0 1 and if t is 1 it will be 0 to 2. So, thereof it contains. So, it is always this is of length how much 2 which is only the half 1. So, it is always contains itself and f s is 0 outside 0 1 ok. Therefore, when you take the integral minus t plus 1 to t minus 1 to t plus 1 f s times something since f s is 0 outside this interval it is as if we are taking the integration from 0 to 1 ok. So, that is p and t if you take 0 to 1 it is p and t. So, I can rewrite it as integral from t minus 1 to t plus 1 the same function I can write like this. Now, you will see the advantage of writing like this namely substitute t minus s equal to u typical thing done when you are doing a convolution ok interchanging the variables here ok. So, substitute t minus s equal to u what happens to s s becomes t minus u ok and t become t minus s becomes u. So, what happens here p and t is p and t I am substituting this one it will have a new form namely it will be minus 1 to plus 1 f of t minus u q n of u d u ok q n of t minus s this one of course d s is minus of d u. So, you have to interchange the up end lower end upper integral. So, this t minus 1 becomes 1 and t plus 1 becomes minus 1. So, that is why you get plus minus 1 to plus 1 along with the sign here d s is minus d u ok you get fine now what is the why I am doing all this the point is now there is symmetry q n remember was a symmetric function. So, this property of q n q n is positive q n minus then equal to minus of you know q n of minus t equal to q n t in this interval ok. So, also the integral of minus 1 plus 1 of q n s is 1. So, these properties can be used for useful now ok in this form all right. So, you will see each of these statements will be useful now. So, that is all computational now, but it is interesting and quite entertaining. The simplest thing is Bernoulli inequality for modulus of t less than equal to 1 we always have 1 minus t square raise to n is bigger than 1 minus n time when you binomial expansion the first term is 1 the first two terms are 1 minus n minus t square you can ignore the rest of them if you put inequality like this ok. This is easy to prove in there are several ways of proving this this is just elementary calculator. And hence alpha n which is integral minus 1 to plus 1 1 minus t square raise to n d t is nothing, but by symmetry it is 0 to 1 twice of that 1 minus t square raise to n d t the same function, but that is now greater than equal to this two is as it is 0 to 1 by square root of n only I am taking 1 minus n t square d t this function is smaller than this function they are all both of them positive in this interval provided you take only up till here there are other terms which you can ignore because you are taking only this is bigger than equal to this one, but now you integrate what you get is this is bigger than equal to 4 by 3 times square root of n that itself is bigger than 1 by square root of n. There are heavy inequalities here no no economy you may be able to prove this alpha n is bigger than equal to 1 by square root of n in many other ways I do not care I want one proof. So, alpha n is bigger than equal to 1 by square root of n this all I wanted ok. Next thing is for every delta between 0 and 1 we have q n of t is less than equal to square root of n into 1 minus delta square t square remember what was q n t q n t was 1 minus t square raise to n divided by alpha n this alpha n we have estimated is bigger than 1 by square root of n. Therefore, I can ignore this alpha n now ok I can simplify it becomes it comes in the numerator here q n t less than equal to square root of n time 1 minus delta square whole square provided t itself mod t itself is bigger than equal to delta. So, in this interval delta to 1 this inequality holds ok now recall geometric expansion 1 by 1 minus t square square of that is nothing but 0 to infinity n plus 1 there is a square from here that is why n plus 1 into t raise to 2 n it is a polynomial which is a power series in t square. So, you get t square t to the 4 and so on. So, t raise to 2 n this is valid for mod t less than 1 this is geometric series ok. This implies that if you take n plus 1 term here n plus r n n term whatever it must tend to 0 right limit of n tends to infinity n plus 1, but t power 2 n is 0 ok whenever mod t is less than equal to 1 this in turn implies that of course, you can n to n plus 1 n plus 1 you can restructure n also n into t power 2 n is also 0 ok sandwich theorem upon taking square root and putting t equal to 1 minus delta square we get limit of square root of n in 1 minus delta square raise to n is 0. So, only for this kind of purpose you have selected this function the polynomial function ok. If you can choose something simpler you will get a simpler problem ok. So, what I have what is that you see this right hand side here tends to 0 and tends to infinity provided this is always true now if we use delta less than equal to t less than equal to then we will we can pass on to q n here ok. So, let us see what happens now by uniform continuity of f f is continuous and we are restricted only in the closed interval 0 1 right. So, therefore, it is in the only continuous given epsilon positive there will be delta 0 less than delta less than 1 I can always choose this delta to be less than 1 such that f t minus f s modulus less than epsilon to whenever t minus s is less than delta both t and s are inside r I have written inside 0 1 of course, but because we are thinking of f s function defined on the whole of r outside 0 1 we have extended it to be 0 remember that. So, this is true whatever uniform continuity first we apply for closed interval 0 1, but rest of them is 0 ok. So, there is no problem for that. So, this is valid now delta has been chosen this statement was true for any delta between 0 and 1 ok. So, now we will use this data and combine all these various properties 20 what is 20 let us just recall 20 I told you these are three properties here symmetry of q n non negativity and integral q n c is normal normalness integral minus 1 plus 1 is 1 and 21 is a formula for p n t 22 is the is the changed formula after change of variables ok. Then 23 is q n t is dominated by this term ok which converges to 0 ok that is 24. So, if you combine all these things what you get is p n t minus f t modulus we have to estimate this one right we have to show that this is less than epsilon ok. Irrespective of what t is provided modulus of that various thing whatever ok that is what we have to show this converges. So, irrespective of what t I must be able to choose n. So, that is uniform convergence. So, this is equal to modulus of I am just writing down the formula for p n t and using the fact that q n s t s integral is 1. So, I can multiply by f t which is a constant as far as integration is concerned. So, f of t minus s q n t is p n this f t into q n s integral is just f t because integral of q n s is 1. So, that has been used here. So, there is nothing else here, but now modulus of this one is taken in the inside the integral ok. Modulus of the integral is less than or equal to integral of the modulus is a elementary property of Riemann integration functions of real value function ok. When you take the modulus inside what you get is f of t minus s minus f s modulus q n s is non-negative. So, it comes out. So, is q n s d s ok this less than or equal to I will put because I cannot put equality here because this is only inequality. Now, this minus 1 to plus 1 I am breaking it into three parts. One is minus 1 to minus delta either is minus delta to plus delta in the last one is delta to 1 of the same function. So, write down these three things. Now, in the first interval minus 1 to minus delta ok what is happening this will be less than or equal to m f t is less than or equal to m has been chosen. So, this is a general bound. So, I am using that. So, f t minus s minus f s there are two terms modulus of this difference is less than or equal to modulus of this plus modulus of that both of them are less than or equal to m. Then integral of q s d s as it is I do not know what it is first I have got this much. In the second term I am putting f f t this one is less than f s n by 2 ok between minus delta to plus delta. So, this is where I have used this one now. So, this f s f of t minus f of t this is less than or equal to epsilon by 2. So, that goes away q n of d s remain the third third part again I am estimating this part is 2 m times 0 to 1 q n of s d s ok. So, different estimates in three different this part and this part is same similar estimate. Now, what happens this is same thing as 4 m times square root of n into 1 minus delta square d s. So, this is where I have used q n is dominated by this term ok this term is also dominated by that and minus 1 to delta and delta to 1 these two intervals are integrals are the same ok. So, I can bring them together 4 m times this one this is where minus of q s q of sorry q n of minus s is q q n of s is just ok symmetry is just ok. So, this term is big is smaller than integral of minus 1 to plus 1 minus 1 to plus 1 is 1. So, this also less than equal to 1. So, it is just epsilon by 2. So, what we have shown is p n t minus f n t is less than equal to this one this square root of n has come here. If you choose n sufficiently large ok this can be made less than epsilon by 2. So, epsilon by 2 plus epsilon by 2 less than epsilon. So, no no t is involved here for all t n is sufficiently large what sufficiently large that will depend upon your epsilon only because this term goes to 0 as n goes to 0. So, this proves Weisstrass theorem ok. Let us take a one small step before closing up today towards more general results now. For generalizations the only thing that we use from above classical result is the following corollary which can be proved in different ways. You do not have to prove the full theorem Weisstrass theorem ok. So, what is the corollary corollary is one that a b a clothed subalgebra of C X R ok. Remember subalgebra etcetera we have we have defined in the part one cloth means this there is a topology here clothness with respect to topology this is algebra. So, it is a cloth subalgebra of C X R if f is inside a then f modulus of f is inside a. So, this is what we want to prove why here is a proof. So, this is directly by Weisstrass approximation given epsilon positive choose a polynomial p 1 and p 1 polynomial means real coefficients everything right so far p 1 instant R t such that mod t is approximated by p 1 namely norm of mod t minus p 1 is less than epsilon by 2. On the entire interval this time interval is minus 1 to plus 1 I am taking because mod t I am equal to this. Remember Weisstrass theorem was proved for all intervals all closed intervals ok there is a polynomial. Let that polynomial look like a naught plus a 1 t plus a n t power n ok put p t equal to p 1 t minus a naught if this constant term is disturbing me. So, I will throw away and let us look at the rest of the terms p t p 1 t minus a naught then modulus of a naught is less than equal to epsilon by 2 why because you put t equal to 0 here this is just modulus of a naught is less than epsilon by 2 and mod t minus p now is epsilon by 2 this a naught is missing that will less than epsilon now epsilon plus epsilon by 2 plus epsilon by 2. So, it is epsilon ok norm of this is less than epsilon that is why I have chosen here epsilon by 2. So, now what I have got is a function p polynomial p without a constant term just like t a right mod t is approximated by this function. Now given any a belong any f belonging to a any element of a first of all we may assume that f is naught 0 ok. So, what we want to prove is that mod f it belongs to a right, but if f is 0 function mod f is also 0 function there is no nothing to prove algebras have always zeros anyway sub algebras they have. So, we may assume f is naught 0 identically. So, that it is norm is also naught 0 now you divide by the norm take g equal to f by norm ok. Now look at g is also continuous function of minus 1 plus 1 ok earlier it could be any polynomial function is now minus 1 plus 1 is a continuous function is inside a why because I have just divided by this is just a scalar function. So, polynomial this sub algebras are vector spaces after all since a is an algebra it follows that even if I take some constant you know a 1 times this 1 a 2 times this 1 square and so on and then add them up each term will be inside a some some will be also inside. So, any for any polynomial is to in particular if you put p g see the constant term is missing here that is important I have no control of constant a 1 times g is there because g is there a 2 times g square is there. So, something total is there. So, p g itself is there ok moreover now look at mod g minus p g norm of this one see remember the polynomial all these Weisstrass theorem was only for continuous functions defined on a closed interval. Now, we have gone into arbitrary space ok, but now the image is minus 1 plus 1. So, everything is happening in the image. So, mod g where is taking value minus 1 plus 1. So, it is like g is a variability that is precisely what I am doing g is like a variability t equal to g x ok. So, that is the function the mod g minus mod p g minus p g the norm of that ok this is same thing as supremum of all x belong into x such that modulus of g x minus modulus of p g x take the difference take the modulus and take the supremum right that is the definition of the super norm. But this supremum norm is less than or equal to t varying between minus 1 plus 1 both this this this value is between minus 1 plus 1 whenever g x is inside x g x is inside minus 1 plus 1 ok. So, I take t between minus 1 plus 1 ok modulus of t minus p t you take the supremum of all these elements. So, this may be larger because all t inside 0 1 may not be attained. So, this is the larger set the supremum larger set is larger at the most so this is less than this is less than this one ok. So, that is equal to but what is this one this is the same thing as norm of mod t minus p see I am I am not if you do not like this symbol you should write you know something like lambda is my lambda of t equal to mod t. So, that function lambda minus p nor it is less than epsilon ok. So, mod g minus mod p is less than epsilon that is what we have got ok. So, this just means that given every epsilon there is a g like this that p g is an element of a which just means that this mod g is in the closure of a, but closure of a is a itself because we have assumed that is closed ok. Therefore, mod f which is multiple of norm if you multiply by this scalar that will be also inside once mod g is inside a norm of times mod g will be also inside a, but that is mod f ok. Next we consider an elementary algebraic result namely instead of studying the big algebra C X R we just study the algebra R cross R think of R is a ring R cross R is a ring structure right. So, what is that ring structure I am telling you is a 1 b 1 into a 2 b 2 is just coordinate was a 1 a 2 b 1 b 2 ok. This is not like you know it is not like complex number very complicated multiplication is there. So, this is the algebra which is a product of R and R ok. Look at this algebra any closed sub algebra of R 2 has to be 1 of the following 1 2 3 4 5 it can be 0 ok. It can be 0 cross R it can be R cross 0 it can be the diagonal delta R or it can be the whole of R 2. So, these are the only possibilities of sub algebra why of course 0 is there of course R 2 is there these are sub algebra that is very easy to verify right. But why are they are the only one that is easy because a sub algebra is first of all a vector subspace vector subspace of 2 dimensional vector vector space right R cross R is 2 dimensional vector space overall has to be either 0 dimensional 1 dimensional or 2 dimensional 0 and 2 are taken care 1 and this 5 1 2 3 2 3 4 these correspond to 1 dimensional subspaces. How do you show that a 1 dimensional space is spanned by some vector right some vector if that vector is of this form 0 comma some R then it will be this one if so this form it is this one. But other one is I want to say it is it is actually delta x itself why this is true. So, this is one thing which bothers us right. So, you have to do a little more algebra than vector spaces ok otherwise you are classifying all vector spaces all vector all the lines are there other lines do not come here at all is what you have to see right ok. So, we will see that that will be the end of this. So, take a b belonging to a which is non 0 element of this sub algebra. So, I am looking at this 1 dimensional k but it is sub algebra therefore, a b square must be also inside a but a b square is by definition a square comma b square a b a b is a square p square right this must be also inside this 1 dimensional space. So, it must be a multiple of a b that lambda is some real number a square b square is lambda times a b for some lambda inside a. The case when a is 0 or b is 0 correspond to 0 cross R or R cross 0. So, let us forget about that the other case is when a and b are non 0 a square equal to lambda a and a is non 0 implies a is equal to lambda. Similarly, b square equal to lambda b implies lambda equal to b. So, we have got a equal to lambda equal to b. So, therefore, this is some a a that just means that this curly a is nothing but a diagonal the positive diagonal a a. So, that is the theorem that is the lemma here. So, there are only 5 sub algebra you will see that the entire thing will be reduced to this 2 dimensional k f ok. That means and the proof of Stone-Weierstras theorem that we are going to study alright. So, that we will do next time. Thank you.