 Hello everybody. So, in the last capsule you recall that we have discussed Fayer's theorem. We gave a complete proof of Fayer's theorem. Recall what is the statement of Fayer's theorem? Suppose you have a 2 pi periodic continuous function. You got a 2 pi periodic continuous function. We know that the Fourier series of F may not converge point wise. In other words, if Sn Fx is the nth partial sum of the Fourier series, then we know that Sn Fx need not converge. But it does converge in the sense of Cesaro. Sn Fx may not converge point wise, but it would converge in the sense of Cesaro. Indeed, it converges uniformly in the sense of Cesaro. That is the content of Fayer's theorem. So, we have continuity. We do not have anything better than continuity. We do not have point wise convergence of the Fourier series, but we have Cesaro convergence of the partial sums of the Fourier series. Not only that, we have an additional bonus that this convergence of the arithmetic means is uniform. So, now we see this in the slide on theorem 31 that you see displayed is the statement of the Fayer's theorem. The arithmetic means of S0, S1, S2, etc. converges to F of X uniformly. So, today we continue from here and we discuss a few applications of Fayer's theorem. An immediate corollary of Fayer's theorem is that the set of trigonometric polynomials is dense in the space of 2 pi periodic continuous functions in the following sense. If you take a 2 pi periodic continuous function for every epsilon greater than 0, there is a trigonometric polynomial PNX such that supremum F of X minus PNX the supremum is taken over mod X less than or equal to pi that is less than epsilon. I have put epsilon by root 2 pi for convenience. We will see why the root 2 pi comes in very soon. What is the trigonometric polynomial? It is simply a finite linear combination of 1 cos X sin X cos 2 X sin 2 X dot dot dot cos n X sin n X dot dot dot. So, you take the system of functions 1 sin X cos X sin 2 X cos 2 X and take finite linear combinations of these. Those are the trigonometric polynomials and you take a continuous function F of X which is 2 pi periodic. Let epsilon greater than 0 be arbitrary. What does Fayer's theorem say? Fayer's theorem says that there exists an n naught such that for n bigger than n naught the arithmetic mean 1 upon n into s naught plus s 1 plus s 2 plus dot dot dot plus s n. That difference F X minus this arithmetic mean is less than epsilon uniformly. But what is this arithmetic mean 1 upon n times s naught plus s 1 plus s 2 plus dot dot dot plus s n? What is sj? sj is the jth partial sum of the Fourier series. So, sj itself is a trigonometric polynomial. So, s naught plus s 1 plus s 2 plus dot dot dot plus s n is a trigonometric polynomial divided by 1 upon n you get a trigonometric polynomial. These arithmetic means that you see in the statement of Fayer's theorem the left hand side of 3.1 1 upon n times s naught plus s 1 plus s 2 plus dot dot dot plus s n that is a trigonometric polynomial. The sequence of trigonometric polynomials converges uniformly to F of X. So, Fayer's theorem immediately gives us that the trigonometric polynomials are dense in the space of 2 pi periodic continuous functions with respect to the soup norm. So, that was a very important corollary and now why is it important? Now the next exercise asks you to show that if F is in L 2 of minus pi pi, if F is in L 2 of minus pi pi then there exists a trigonometric polynomial P n X such that the L 2 norm of F X minus P n X is less than epsilon. Is this clear? If F of X is continuous, if F of X is continuous the previous theorem, the previous corollary says that the supremum mod F X minus P n X is less than epsilon by root 2 pi. So, let us square it mod F X minus P n X squared less than epsilon squared by 2 pi. Integrate from minus pi to pi, what do you get? Integral minus pi to pi mod F X minus P n X squared dx less than epsilon squared. Take the square root, what do you get? The L 2 norm, you get the fact that the L 2 norm of F X minus P n X is less than epsilon. So, we get that the trigonometric polynomials are dense in the space of 2 pi periodic continuous functions with respect to L 2 norm. So, from approximation in sup norm, we get an approximation in L 2 norm. Now from continuous functions, we must pass on to all L 2 functions. What we have shown so far? If F of X is a 2 pi periodic continuous function, then this approximation happens in L 2 norm. But now we must leave the realm of continuous functions and we must prove this L 2 estimate, this L 2 approximation for all F in L 2 of minus pi pi. Let us see how to do that. Trinometric polynomials are dense in L 2 of minus pi pi. So, the function that we are going to take is not continuous, it is simply in L 2. So, suppose if F is in L 2 of minus pi pi, then given any epsilon greater than 0, there is a trigonometric polynomial P n X such that the L 2 norm of F X minus P n X is less than epsilon. Let us look at the proof of this. The proof proceeds in four easy steps. First, let epsilon greater than 0 be arbitrary. Now by Luzin's theorem, by Luzin's theorem, there is a continuous function G from minus pi pi such that integral from minus pi to pi mod F X minus G X the whole squared DX less than epsilon squared by 8. Let us step 1. Now let us look at step 2. Let us take M to be the supremum of G X. Now we will work with G X norm. Let M be the supremum of G X. Then there is a delta greater than 0 such that the integral of mod F X squared DX for mod X bigger than or equal to pi minus delta and put a throw in the factor 2. And then integral M squared DX constant mod X bigger than or equal to pi minus delta throw in the factor 2 less than epsilon squared by 8. Of course, I could have put the 2 on the right hand side, I could have written it as epsilon squared by 16. In other words, the contribution of mod F X squared and M squared from the ends of the interval, you got the interval from minus pi to pi. So, cut out a delta piece here and cut out a delta piece here. Look at the piece from pi minus delta to pi and look at this piece from minus pi to minus pi plus delta. From the 2 ends, you take a small piece of length epsilon on either ends, you chop it or the contribution from these small pieces, these tiny pieces or these little tuqras as it were. That contribution is less than epsilon squared by 16. That is very easy to see because F is in L2, mod F squared is in L1. So, when you have a function which is L1, then if the interval over which you are integrating is terribly small, then the contribution is going to be made terribly small. So, this is a very easy step, but it is a useful step. How do we proceed from there? Now, let us take step 3. Now, what we will do is that we will take a continuous function G, then we will take a continuous function capital G X such that capital G X equal to small G X on the major part of the interval, mod X less than or equal to pi minus delta. Remember, we cut out those 2 little pieces on the 2 ends on the remaining large piece minus pi plus delta to pi minus delta. On the large piece, this capital G X equals little G X. Now, you define capital G to be 0 at plus minus pi. At plus minus pi, you define the function to be 0. Now, what does Tidges' extension theorem tell you from general topology? What is Tidges' extension theorem? You have got a metric space. You have got a metric space capital X. You have got a closed subset A. You have got a closed subset A and you have got a continuous function on the closed subset A into the real numbers. Then, this continuous function will extend continuously to whole of X. And further, the extension will also have the same bounds as the original function. This is the important Tidges' extension theorem from general topology. So, here in this particular context, what is the metric space? The closed interval minus pi pi. What are the closed subspace that we are talking about? It is mod X less than or equal to pi minus delta and throw in these 2 end points plus pi and minus pi. So, on mod X less than or equal to pi minus delta, I had defined the function capital G X. At the end points, I had defined this function capital G X. This capital G X is continuous on this closed subset that I am talking about. But Tidges' extension theorem, this will extend continuously to the closed interval minus pi pi. It will extend continuously to capital X, which is minus pi pi. Further, the bound on capital G is the same as the bound on little g, which is capital M. And now, this capital G is a continuous function from minus pi to pi and it vanishes at the 2 end points. So, take its 2 pi periodic extension. Extend it as a 2 pi periodic function. This 2 pi periodic function will be continuous because it is 0 at both ends. So, now, let us look at integral minus pi to pi mod f x minus g x the whole square. Now, this integral, I will break it into a bunch of integrals. The first piece is over mod X less than or equal to pi minus delta. But when mod X is less than or equal to pi minus delta, what is capital G X? Capital G X is the same as little g x. So, that is what we get the first piece. And then the second piece is integral over mod X greater than or equal to pi minus delta. When mod X is greater than or equal to pi minus delta, what is this mod f x minus g x the whole square? Expand mod f x squared plus mod g x squared minus twice f x g x. And I will get 2 times integral from mod X bigger or equal to pi minus delta mod f x squared plus mod g x squared, right? And now, this piece mod f x minus g x the whole squared is less than epsilon squared by 8. That is how we chose, that is how we chose the G, remember? We got this epsilon squared by 8 and I get from there, it is certainly this first piece is definitely less than epsilon squared by 8. Now, we need to manipulate this. Mod G has been replaced by m because my bound is preserved for the extension. So, I get mod f x squared plus m squared and I know from step 2 that this is going to be less than epsilon squared by this is going to be less than epsilon squared by 8. So, put that also over here and I get epsilon squared by 4. This total contribution is less than epsilon squared by 4. So, the L2 norm of f minus capital G is less than epsilon by 2. Step 4, I have already indicated that we will be extending this capital G as a two-periodic continuous function, which is perfectly fine because the capital G vanishes at both the ends. Now, by Feier's theorem applied to capital G, I can apply Feier's theorem to capital G. There is a trigonometric polynomial px such that mod capital G minus capital P is less than epsilon by 2. Now, finally, we let us apply the triangle inequality. The L2 norm of f minus p is less than or equal to the L2 norm of f minus capital G plus the L2 norm of capital G minus capital P, correct? The L2 norm of f minus capital G is less than epsilon by 2 from the previous step and by choice of this p, the L2 norm of capital G minus capital P is less than epsilon. So, all in all, we have proved the theorem. We have proved the theorem which is stated as a title of this slide. Trigonometric polynomials are dense in L2 of minus pi pi, namely given any epsilon greater than 0 and given in a L2 function f is a Trigonometric polynomial p such that the L2 norm of f minus p is less than epsilon. That concludes this very important corollary of Feier's theorem. We are going to use this very important corollary of Feier's theorem to complete the proof of Parseval's formula of chapter 2 that was left out. So, let us proceed to that. Now, let us recall Parseval's formula from the last chapter. Parseval's formula says that if f is in L2 of minus pi pi, then 1 upon 2 pi integral minus pi 2 pi mod fx squared dx. The left-hand side of equation 3.4 that is displayed in a slide, that is the energy of the signal. The right-hand side of 3.4 tells you how to calculate the energy of the signal using the Fourier components mod a0 squared plus 1 half summation j from 1 to infinity mod aj squared plus mod bj squared. We have to prove this theorem. We have not proved it completely. We only got the inequality, the Bessel's inequality we got. Remember? It is useful to recall at this point that if Pn is a trigonometric polynomial of degree capital N, then its nth partial sum will agree with Pn for n greater than or equal to capital N. You must think about this. It is very easy to prove. You take a trigonometric polynomial Pnx, alpha naught plus alpha 1 cos x plus beta 1 sin x plus dot dot dot plus alpha capital N cos capital Nx plus beta n sin capital Nx. I just stop it with that and calculate the Fourier series of Pnx. What are the Fourier series of Pnx? Pnx itself. Pnx itself is a Fourier series of Pnx. So the nth partial sum Sn of Pnx is going to be exactly Pnx when little n exceeds capital N. So the first comment here in this slide is actually a completely trivial comment, but it is going to be explicitly used very soon. So I thought it is best to put it up here. You must think about it. If it is not clear to you, a few moments of reflection will convince you of this. Now let us proceed. Let us take a function f in l2 of minus pi pi. Then by Pythagoras' theorem, we proved this already that f minus Sn is orthogonal to Sn. When we looked at the least square approximation in the previous chapter, we must go back to the previous chapter and you must check that we have proved 3.5. That Pythagoras' theorem says that norm f minus Sn squared plus norm Sn squared is norm f squared because Sn is orthogonal to f minus Sn. The thing written in blue is orthogonal to the thing written in red in this equation 3.5. Now we will see that 3.4 will now follow from 3.5 if we can show that this red thing goes to 0. If we can prove that this red thing goes to 0, then allowing n to go to infinity over here will give you 2 pi times this left hand side of 3.4. If I multiply the left hand side of 3.4 by 2 pi, I will get 3.5 and I have to just check that this blue thing that is written here norm Sn fx squared. What will be norm of Sn fx squared? The thing written in blue here norm Sn fx squared, if you compute it, it is a finite sum, you are going to get exactly 2 pi times mod a0 squared plus pi times summation j from 1 to n mod aj squared plus mod bj squared. That is exactly going to be norm Sn fx squared. When you calculate norm Sn fx squared, I repeat, you are going to get just this part except that j will go from 1 to n. So, obviously when I allow n to go to infinity, when I allow the n to go to infinity, this item here norm Sn fx squared will converge to 2 pi times mod a0 squared plus pi times summation j from 1 to infinity mod aj squared plus mod bj squared. So, the result 3.4 does indeed follow if we show norm f minus Sn fx squared goes to 0 as n tends to infinity. The last line in this slide is what we need to establish right now and we proceed for that. Let us use 3.5 to f minus pn where pn is a trigonometric polynomial with n getting equal to n. Instead of f, we apply the previous thing to f minus pn. So, what does 3.5 read? f minus pn minus Sn of f minus pn norm squared plus Sn of f minus pn norm squared equal to f minus pn norm squared. Now what is Sn of f minus pn? It is Sn of f minus Sn of pn which is Sn of f minus pn. Remember that when little n is greater than or equal to capital n, I get Sn of pn is the same as pn. The first comment I made in the last slide. So, what do we get? Norm f minus Sn fx squared plus norm Sn of f minus pn squared equal to norm f minus pn squared. That is exactly what we get whereby we conclude that norm f minus Sn of fx is less than or equal to f minus pn norm. I just knock this term off this becomes an inequality. But now we know given any epsilon greater than 0, there exists a trigonometric polynomial pn such that this right hand side norm f minus p capital n is less than epsilon. And so we conclude that for little n bigger than capital n, we have that norm f minus Sn fx is less than epsilon. So, this concludes the proof of Parseval formula that was left out in chapter 2 and now we have completed it. We need not have waited for this chapter. We could have done this in the last chapter itself. One could instead of using Feyer's theorem, we could have used Weiersta's approximation theorem. What is Weiersta's approximation theorem? That a continuous function from minus pi to pi can be approximated by polynomials. Now from algebraic polynomials a0 plus a1t plus a2t squared plus dot dot dot plus ak t to the power k. You have to now pass from algebraic polynomials to trigonometric polynomials. There is a way to do that and one could directly obtain the Parseval's formula as a corollary of the Weiersta's approximation formula. That is not difficult, but rather I thought this would be a very nice application of Feyer's theorem, a very important application of Feyer's theorem. So, the next thing to look at in the next capsule will be how Feyer's theorem allows us to prove a very important result in number theory. So, we are going to see another application of Fourier analysis to problems in number theory. We already saw in the very first chapter how using Fourier analysis we get the beautiful Riemann's functional equation, which is extremely important in analytic number theory. Here we look at a different aspect of analytic number theory, uniform distribution modulo 1. The literature on uniform distribution modulo 1 is very vast and what we will prove is a remarkable sharpening of a classical result of Kronecker, namely Weier's Equidistribution theorem. I think this is a very good place to stop this capsule. We will continue this next time. Thank you very much.