 So, we continue from where we left off recall last time we discussed the different modes in which the Fourier series can converge to a function namely point wise convergence, mean convergence and Cesaro convergence. So, that is point wise convergence, mean convergence and Cesaro convergence. Now, let us continue with the discussion. So, let us recall from elementary calculus the problem of expressing a function f as a power series. So, f of x equal to a naught plus a 1 x plus a 2 x square plus etcetera equation 1.2. So, how do you write a function as a Taylor series? Now, if you take a C infinity function a C infinity of r means the class of infinitely differentiable functions. Of course, I can take a function which is infinitely differentiable and you must have seen in the calculus courses that the coefficients a naught a 1 a 2 etcetera are given by this equation 1.3. The coefficients are given by this and when you put a n equal to 1 upon n factorial the nth derivative of f at the origin the series 1.2 is called the Taylor series for the function. You know also examples of functions which are C infinity, but whose Taylor series does not converge to the given function. We do have a distinguished subclass of the class of C infinity functions that do have a representation of the form 1.2 which is valid on some open interval minus r r and for which this f of x is indeed the sum of its Taylor expansion where the a n's are given by 1.3. Now, let us ask what is the analog of 1.2, 1.3 in the context of trigonometric series. So, by way of analogy I try to bring out an analogy between expressing the function f of x as a Fourier series and expressing the function f of x as a power series or as a Taylor series. So, let us see what happens. So, now let us start with the function f of x and we want to write f of x as a series 1.1. What is the series 1.1? Let me recapitulate a naught plus a 1 cos x plus b 1 sin x plus a 2 cos 2 x plus b 2 sin 2 x. So, the 1.1 was basically a naught plus summation n from 1 to infinity a n cos n x plus b n sin n x. What I could do is I could do a term by term integration of 1.1. I could do a term by term integration of 1.1 and integrate both sides of 1.1 from minus pi to pi. Now, when you integrate a cosine n x and a sin n x from minus pi to pi that will disappear and you are simply left with integral f x dx from minus pi to pi equal to 2 pi times a naught. We immediately get the formula for a naught namely a naught equal to 1 upon 2 pi integral minus pi to pi f x dx that is equation 1.4. Similarly, I could multiply the equation 1.1 by cos k x and later sin k x. So, multiply equation 1.1 by cos k x and sin k x, f of x cos k x equal to a naught cos k x plus summation a n cos n x cos k x sin n x sin k x. Now, remember from trigonometry 2 cos a cos b is cos of a plus b plus cos of a minus b. Use this de-facturization formula and every term will disappear except the n equal to k term. When n equal to k, you get cos squared k x. Cos squared k x you will have to use the half angle formula or the double angle formula depending on how you look at it and you can write down immediately that a k will be 1 upon pi integral minus pi to pi f of x cos k x dx. Similarly, b k will be 1 upon pi integral minus pi to pi f of x sin k x dx. So, you got this equation 1.4 and 1.5. I will leave out the details because these details are just routine stuff. The point is you are multiplying the equation by cos k x and performing a term by term integration and all terms disappear except one of them and the one that survives will be a k and divide by 1 upon pi because the integral of cos squared k x dx you have to compute the integral from minus pi to pi and then you get a coefficient. You divide by the coefficient and that is how you get this equation 1.5 a k equal to 1 upon pi integral minus pi to pi f of x cos k x dx. Likewise b k will be 1 upon pi integral minus pi to pi f of x sin k x dx. So, that gives us the first definition formal definition of this course. Suppose f of x is an integrable function on minus pi to pi when I say integrable I mean Lebesgue integrable. So, Lebesgue integrable function on minus pi to pi we say that the trigonometric series 1.1 is a Fourier series. We say that the trigonometric series 1.1 is a Fourier series if the a n's and the b n's and the a naught are given explicitly in terms of 1.4 and 1.5. So, when the coefficients are determined by 1.4 and 1.5 it is a trigonometric series of Fourier series. The Fourier series when I say a trigonometric series 1.1 is a Fourier series it has a very specific meaning namely the coefficients have to be given by these two formulas 1.4 and 1.5 respectively. Now note that if the series 1.1 is not sufficiently well behaved as regards convergence usually the series 1.1 will only converge conditionally it will very rarely converge absolutely. And the validity of the deduction 1.4 1.5 is quite suspect we multiplied it by cos kx and did a term by term integration. Now that the validity of the deduction is suspect even if 1.1 converges point wise everywhere to f of x the coefficients may not be given by 1.4 1.5. A classic example was given by Fatou. Fatou wrote down this series 1.6 summation n from 2 to infinity sin nx by log n look at this infinite series summation n from 2 to infinity sin nx by log n. You can prove that the series 1.6 converges point wise everywhere. It converges point wise everywhere and it even converges uniformly on the interval delta to 2 pi minus delta. So, cut off a piece of size delta say from 0 to delta and cut off a piece from 2 pi minus delta to 2 pi. And on this sub interval closed interval delta to closed interval 2 pi minus delta you even have uniform convergence. But the sum function f of x is not Lebesgue integrable on minus pi pi. In 1875 Paul Dubois Raymond established that if the sum of a trigonometric series is an integrable function in the sense of Riemann then it is a Fourier series that means the coefficients must be given in terms of 1.4 1.5. The result was extended in 1912 to Lebesgue integrable functions. See the book of G. Bachmann, L. Naricci and E. Beckenstein Fourier and wavelet analysis Springer-Werlach 2000. This example has been taken from pages 290 and 220. A complete discussion is here in this book. So, not all trigonometric series are Fourier series. So, there is a distinction between Fourier series and trigonometric series. This distinction should be kept in mind namely if the coefficients a0, a n and b n are given by 1.4 1.5 then the trigonometric series is a Fourier series. Now, we come to the point wise convergence theorem. We shall now work only with Fourier series of a function f of x which is Lebesgue integrable on minus pi pi namely the coefficients are given by 1.4 1.5. We will prove a simple lemma. Lemma 2 1 plus 2 cos theta plus 2 cos 2 theta plus dot plus 2 cos n theta is given by sin of n theta plus theta by 2 divided by sin of theta by 2. That is equation 1.7 on the display. Let us prove this formula 1.7 by using simple trigonometry. You need a trigonometric identity 2 cos a sin b. You want a de-factorization formula for 2 cos a sin b that you need to recall from elementary trigonometry. So, we get 2 cos j theta sin theta by 2 equals sin j plus half theta minus sin j minus half theta. You got this de-factorization formula. Now, you simply put j equal to 1 2 3 and add and you get a telescoping sum on the right hand side. The sum telescopes and you are left with summation j from 1 to n with a 2 in front cos j theta sin theta by 2 equals sin n plus half theta minus sin theta by 2. Of course, you have to bring the sin theta by 2 on the left hand side. You are going to get exactly 1 plus 2 cos theta plus 2 cos 2 theta plus dot dot dot plus 2 cos n theta times sin theta by 2 is sin n theta plus theta by 2 as advertised in equation 1.7. And so, we have proved this lemma completely. Now, as a preparation for the point wise convergence theorem, we will obtain a integral expression for the finite sum. First, let us write the partial sum s n f x a naught plus summation j from 1 to n a j cos j x plus b j sin j x equation 1.8. We use the integrals representations for the coefficients a naught a j and b j. We use the integrals 1.4 and 1.5 of the coefficients and we put in the values of a j and b j what do we get? We get s n f x equal to 1 upon 2 pi integral minus pi 2 pi f of t into 1 plus 2 cosine x minus t plus 2 cosine 2 times x minus t plus dot dot dot plus 2 cosine n times x minus t dt. Now, we know exactly how to sum this terms within the parenthesis 1 plus 2 cos theta plus 2 cos 2 theta plus dot dot dot plus 2 cos n theta. We are just finished finding the sum of that. Remember, we got that that is basically 1.7. So, we use this your theta is what? Theta is nothing but x minus t. So, that expression that we have on the right hand side sin n theta plus theta by 2 this right hand side expression that you have in 1.7. Instead of theta we have over here what x minus t we have x minus t and that expression that we have over there the right hand side expression has been denoted by d n theta it has been denoted by d n theta. So, we write this thing as s n f x equal to 1 upon this 1 upon 2 pi has been subsumed in the formula for d n theta. So, integral minus pi to pi f of t times this expression divided by 2 pi using formula 1.7 we get d n of x minus t dt. So, now we will transform 1.9 slightly and we will do it as follows. We will make two simple observations. So, here in this slide you see that this expression sin n theta plus theta by 2 divided by sin theta by 2 and then 1 upon 2 pi this complicated expression has been given the name d n theta. The d n comes because it is called the Dirichlet kernel this is called the Dirichlet kernel and we make two simple observations. Suppose if p and q are two periodic functions on the real line and the period is 2 c, c is a positive real number and it is a periodic function of period 2 c then integral from minus c to c p t q x minus t dt is integral from minus c to c p of x minus t q t dt. So, 1.11 equation 1.11 is the simple transformation how do you prove equation 1.11 it is very easy and easy change of variables put x minus t equal to s in the left hand side integral and we break this integral the resulting integral will be broken into 3 integrals. Of course, now when I put x minus t equal to s and when t equal to c your x your s will be equal to x minus c and so there will be an x coming in the limits of the integral. So, you get a resulting integral that resulting integral you must break it up into 3 integrals from minus c to x minus c minus c to c and c to x plus c and what you will notice is that because of the periodicity because of the periodicity you will the two of the integrals will cancel out two of the integrals will cancel out and the middle integral will survive and you will get the right hand side 1.11 why would two of the integrals cancel out because of the periodicity it is here that we use the 2 c periodicity of the functions p and q and this equation 1.11 has been established why would you need equation 1.11 we will see in a moment before we take up before we use equation 1.11 let us make one more simple observation let us do the following let us take this equation 1.7 let us take this equation 1.7 and integrate both sides from minus pi to pi when you integrate 2 cos theta 2 cos 2 theta that are 2 cos n theta from minus pi to pi you get 0 the only term that will survive will integral 1 from minus pi to pi that will be 2 pi and you divide by 2 pi after after you have finished integration and what you get you get this formula 1.12 you get that the Dirichlet kernel when you integrate from minus pi to pi you get 1. Now what we do is the following how do we use these two simple observations we use the simple of the first simple observation we got Sn of fx equal to f of t into dn of x minus t I write it as f of x minus t times dn t using the first observation the second observation that I want to use is that you got this last equation 1.12 multiply this by f of x so you get f of x equal to integral minus pi to pi fx times dn t so you got 2 integrals Sn of fx equal to integral f of x minus t dn t dt the second thing is f of x equal to integral minus pi to pi fx dn t dt subtract when you subtract you get Sn fx minus f of x minus pi to pi integral f of x minus t minus f of x times dn t dt you got this very important result 1.4 this is going to be the crucial point in the proof of the point wise convergence theorem of Fourier series. So, pay good attention to this equation 1.14 this is going to be crucially used and so now to prove point wise convergence we need to show that Sn fx converges to f of x everywhere that means that this integral appearing in the right hand side of 1.14 must go to 0 as n goes to infinity. Now we will examine how this integral will go to 0 whether it will go to 0 in the first place and if not why would it not go to 0. So, let us try let us try an obvious attempt that is not going to work a naive idea that is going to fail let us try it out so that we get some experience on what not to do. So, one thing is that is very tempting to do the following I will take the absolute value of both sides of 1.14 and take the absolute value inside the integral integral of remember integral of h t dt some function is given to you you are integrating it and taking the absolute value less than or equal to the integral of the absolute value there is a triangle inequality for integrals. So, I will take the absolute value of both sides of 1.14 right hand side I will take the absolute value inside the integral sign I will get an inequality what inequality do I get I get this inequality 1.14 a. So, mod Sn fx minus fx less than or equal to integral from minus pi to pi mod f of x minus t minus f of x mod d and t dt. The natural thing to do would be to try and show that the right hand side of 1.14 a goes to 0. Now, the first thing that will come to your mind is uniform continuity of the function. So, you appeal to the uniform continuity of a function in a neighborhood of the origin remember x minus t must be close to x. So, t must be small t must be small means when t is in minus delta delta when t varies over minus delta delta this p is mod of f of x minus t minus f of x should be less than epsilon. And so, we so, we replace this by epsilon well what about when mod t is bigger than delta when t is far away from the origin or when these 2 points x minus t and x are far apart then we then this is not small, but we still have the fact that function is continuous and therefore, it is bounded. And so, we can still say that mod of f of x minus t minus fx is less than or equal to m. So, secure that the right hand side goes to 0 what would we need we broke this integral 2 pieces from minus delta to delta and there we use the fact that this absolute value is less than epsilon. So, epsilon comes out and we get integral minus pi to pi mod d n t dt. Let us see what do we get we get 2 things we get that we need epsilon times this that was the first piece epsilon times this that must go to 0 right in both the pieces we get in both the pieces we get the integral of d n t in the first piece it is epsilon times integral mod d n t in the second piece it is m times integral mod d n t. So, what we need is that we would need the fact that this integral from minus pi to pi mod d n t dt should decay to 0 as n goes to infinity, but here our luck has forsaken us because that is false integral from minus pi to pi mod d n t dt not only will fail to decay to 0 it is actually go to infinity integral minus pi to pi mod d n t dt behaves like a constant multiple of log n as n tends to infinity. So, this is the problem. So, there is no way to salvage the argument in fact, the problem is the moment you took the absolute value inside the integral moment you wrote mod s n f x minus f x is less than or equal to integral from minus pi to pi mod of this into mod of this all has let loose we do not have any control whatsoever on the behavior of this integral it is not even not even bounded it is actually going to infinity as n goes to infinity. So, this idea is going to fail it is not going to give us point wise convergence in fact, we know that point wise convergence must fail when the function is merely continuous just continuity and nothing else will not do we need a little more than that we need a little more than continuity in the next capsule we shall exactly see that when we have holder continuity even with positive holder exponent they can salvage the argument, but before that we will need a couple more preparations. So, this is the reason why it will that this simple looking argument does not work ok. So, I think it is a good time to stop this over here this capsule over here and will continue to see will continue the discussion we will see how to prove the convergence when you have holder continuity of some positive exponent alpha. Thank you very much.