 Hello. So, we continue from where we left off. So, last time we saw that mere continuity is insufficient for us to prove that the partial sums of the Fourier series converges point wise to f of x. The problem is when you try to write the difference between the partial sums S n f x minus f of x and estimate its absolute value. You take the absolute value inside the integral, you immediately run into difficulty because the Dirichlet kernel is highly oscillatory kernel, it is not a positive kernel and the integral of mod d n t d t from minus pi to pi behaves like C log n as you see in the slide in the display in the display. Now, let us try to understand how to get around this difficulty, why is it that instead of continuity if we have holder continuity slightly better than just continuity then the problem is much tractable. So, before we take it up let us expand the Dirichlet kernel, what is the Dirichlet kernel again? Let us see it is basically 1 upon 2 pi, 1 upon 2 pi is an innocent constant factor sin n theta plus theta by 2 and divided by sin theta by 2. The numerator we use the addition formulas of sin, sin of a plus b is sin a cos b plus cos a sin b. When you use this formula, you will get one term sin n theta cos theta by 2, the other term that you will get will be cos n theta sin theta by 2, the sin theta by 2 will cancel out in the second term. So, that is exactly what we are trying to do right, what we do now is we expand the numerator in the Dirichlet kernel and we get at sum of two terms d n theta equal to 1 upon 2 pi cos n theta plus sin n theta into cot theta by 2. Likewise the integral 1.14 splits into a sum of two integrals, one of the integrals is cos n t times the difference f of x minus t minus f of x, f of x minus t minus f of x has been abbreviated to delta x t. It is very cumbersome to be writing f of x minus t minus f of x everywhere. So, I just abbreviated to capital delta x comma t. So, the integral splits into two integrals, integral minus pi to pi cos n t capital delta tx dt, the second integral integral minus pi to pi sin n t cot t by 2 delta tx dx, the second integral that you see in 1.5 that gives you the trouble. All right, let us move a little ahead. The first thing that we should do is to prove what is called as a Riemann-Lebesgue Lemma. So, to handle the first piece integral minus pi to pi cos n t times delta tx dt. Again let us assume f is continuous. So, that delta tx is a continuous function of two variables. So, now first piece, this first integral is an innocent piece. Why is it innocent? Why would it straight away go to 0 as n tends to infinity? Because of the next lemma called the Riemann-Lebesgue Lemma. What does Riemann-Lebesgue Lemma say? If you have a function g in l 1 of a b, if you have a function g in l 1 of a b, theorem 3, then integral from a to b cosine n t into g t, that goes to 0 as n goes to infinity. Likewise integral a to b sin n t g t dt that also goes to 0 as n goes to infinity. So, this is 1.16 in the display. So, we shall prove the Riemann-Lebesgue Lemma later and let us see how the Riemann-Lebesgue Lemma helps us. It helps us to cope with the first term in 1.15 namely integral minus pi to pi cos n t capital delta t x dt that goes to 0 as n goes to infinity. Now, to complete the proof that Sn f x converges to f of x as n tends to infinity, we need to secure that the second of the two integrals in 1.15, the second one minus pi to pi sin n t cot t by 2 delta t x must go to 0, this integral must go to 0. How do you do that? First of all, let us do a simple exercise. Let us define a function capital F of t equal to cot t by 2 minus 2 by t if t is not equal to 0 and f of 0 is 0. How does cot t by 2 behave near the origin? Well, what is cotangent? 1 upon tangent. How does tan theta behave? Tan theta is approximately theta. So, what is cot theta? Cot theta is approximately 1 upon theta. So, cot t by 2 behaves like 2 upon t. So, when I subtract 2 upon t, it is going to become nice because the ugly part in cot is the 2 by t part. So, when I subtract 2 by t from cot t by 2, the function becomes nice. So, this new function capital F of t which is defined to be cot t by 2 minus 2 upon t when t is not equal to 0 and at the origin it is defined to be 0. That is a nice continuous function. And so, the Riemann-Lebesgue Lemma says that integral from minus pi to pi f of t sin n t dt will be 0. So, what we do is look at 1.17, you modify the integrand, add and subtract a minus 2 by t. So, make it cot t by 2 minus 2 by t plus 2 by t. And now when you have cot t by 2 minus 2 by t, that is when you have capital F, use the next line here. So, what we need to show is the 2 is an innocent factor. What we need to show is limit as n tends to infinity minus pi to pi integral 1 upon t sin n t times this difference capital delta x t that should go to 0 as n tends to infinity. This is so, we are reduced the task to proving 1.18. So, now comes the, now we bring in the hypothesis of Holder continuity. A function f from r to r is said to be Holder continuous of class alpha if mod f of x minus f of y is less than or equal to l times mod x minus y to the power alpha for all x y in r. Here l is a constant, alpha is a positive constant, l is also positive. Now, if alpha happens to be bigger than 1, then f is a constant, that is an exercise. So, the only interesting case is when alpha is between 0 and 1, 1 included, 0 excluded. Alpha should be strictly positive and alpha should be less than or equal to 1. When alpha equal to 1, then the functions which satisfy this inequality are called Lipchitz functions. So, Lipchitz functions are a special case of Holder continuous functions. So, now we are ready to prove the basic convergence theorem. So, theorem 5, suppose f from r to r is a 2 pi periodic Holder continuous function with Holder exponent alpha positive in the Fourier series of f of x converges point wise everywhere to f. This is theorem 5 in the, in the displayed slide. So, as I explained before, what is needed is to prove limit as n tends to infinity integral minus pi to pi 1 upon t f of x minus t minus f of x, that is your capital delta x t times sin n t dt should be 0. Well, now it is very clear how to proceed. We should split the integral from minus pi to pi into 2 parts and integral from minus delta to delta and the integral of the complementary part. So, mod t less than delta mod t greater than or equal to delta. So, that is how I split the integral into a sum of 2 integrals. So, in the first piece mod t less than delta, in the first piece mod t less than delta we use Holder continuity. What happens is that mod of x minus t minus fx mod less than or equal to l times mod t to the power alpha. Remember the definition of Holder continuity. So, we must use Holder continuity on this difference, but this difference is denoted by capital delta. So, over here capital delta of tx in absolute value is going to be less than l times t to the power alpha. So, we got the 1 upon t here this capital delta tx is less than t to the power alpha and I am going to take the absolute value. So, this integral is less than or equal to l times integral mod t less than delta dt by mod t to the power 1 minus alpha because mod sin is less than 1 anyway not a problem. Then the other integral integral mod t between delta and pi 1 upon t capital delta tx sin n t dt you compute this integral it will be 2 l upon alpha delta to the power alpha where the 2 come from is an even function. Then the second piece I have not done anything to the second piece yet. Now, let epsilon greater than 0 be arbitrary select the delta so small that 2 l upon alpha delta to the power alpha is less than epsilon by 2. So, the first piece here is already less than epsilon by 2 here. Now, we need to worry about the second piece. Remember that when mod t is between delta and pi when mod t is between delta and pi this business capital delta tx is a continuous function and now I can apply the Riemann-Lebesgue lemma because 1 upon t is also continuous because 0 is far away 0 is far away we are between delta and pi. So, I can apply the Riemann-Lebesgue lemma so this whole second piece goes to 0 as n goes to infinity. So, I can select an n naught so large that this particular piece is less than epsilon by 2 and epsilon by 2 plus epsilon by 2 is epsilon. So, for n greater than or equal to n naught it is this whole thing is less than epsilon and the proof is thereby complete. So, that completes the proof that when you have a holder continuous 2 pi periodic function the Fourier series converges point wise to the given function f of x that was a little bit of work but we will very soon see that we are going to derive some number of corollaries. In particular it is also true for Lipschitz functions because after all Lipschitz functions are a special case of holder continuous functions and our first example that we shall see presently is the case of a Lipschitz function well we have to prove still the Riemann-Lebesgue lemma we have not done that yet. So, to prove Riemann-Lebesgue lemma we shall make use of 2 results one is the Weierstrass's approximation theorem and the other is a theorem due to n in losing by the way there are several proofs of the Riemann-Lebesgue lemma and I have chosen this particular one. I will indicate some other proofs also okay the first is a classic theorem called the Weierstrass's approximation theorem I am pretty certain that most of you have seen the proof of the Weierstrass's approximation theorem in your courses. So, let us recall the Weierstrass's approximation theorem a continuous functions on a closed bounded interval a b can be uniformly approximated by polynomials. There is an informal way of saying it a little more formal way of saying it is in the second line in other words given a continuous function f on the closed interval a b and an epsilon greater than 0 there exists a polynomial p such that the sup norm of f minus p is less than epsilon. In other words the supremum of mod fx minus px over a less than or equal to x less than or equal to b is less than epsilon. The second theorem that we need is a theorem of losing and the theorem of losing says that if a function f is in l1 and epsilon greater than 0 then there is a continuous function g such that the l1 norm of f minus g is less than epsilon. These are the two theorems that we shall use to complete the proof of the Riemann-Lebesgue lemma. So, let us look at the Riemann-Lebesgue lemma the Riemann-Lebesgue lemma says that if you have a l1 function g integral gt sin nt dt goes to 0 as n tends to infinity. So, let us do the following let us take a simple example take g of t to be 1 take the constant function 1 integrate the constant function 1 times sin nt from a to b you can integrate sin nt from a to b you get minus cos nt upon n put the limits a to b and the denominator you are picked up a n and that goes to 0 as n tends to infinity. So, we have already checked it when g is a constant function you can check it for when g of t is t you can prove it for g of t equal to t square because you can actually compute these integrals right when g of t equal to t you are going to be integrating t times cosine nt dt and you have to integrate by parts. But when you integrate by parts you are going to pick up an n in the denominator again it will go to 0 in particular for all k equal to 1 2 3 etcetera you can check the Riemann-Lebesgue lemma is valid when g of t equal to t to the power k k runs from 0 to infinity over the integers. So, Riemann-Lebesgue lemma has been verified. Now the Riemann-Lebesgue lemma is true for g 1 and it is true for g 2 it is certainly going to be true for c 1 g 1 plus c 2 g 2. So, it is going to be true for polynomials. So, the Riemann-Lebesgue lemma is true for polynomials g. Now we want to prove it for a general g take a general l 1 function take any l 1 function by Luzin's theorem we can choose a continuous function h such that integral from a to b the l 1 norm remember integral from a to b mod g t minus h t dt less than epsilon by 3 for example. Now your h is a continuous function since h is a continuous function appeal to Varstas approximation theorem appeal to Varstas approximation theorem and you got a polynomial p of t such that the supremum of h t minus p of t is less than epsilon by 3 times b minus a. If the supremum is less than epsilon by 3 times b minus a integral of g t minus p of t what is integral of g t minus p of t absolute is going to be g t minus h t absolute plus integral of h t minus p t absolute. The first one is less than epsilon by 3 the second one will also be less than epsilon by 3 these two things will become less than 2 epsilon by 3. So, using these two very simple approximations we immediately get that integral from a to b mod g t minus p of t dt is less than 2 epsilon by 3 ok. So, now we are ready integral a to b g t sin n t dt the cosine is similar the cosine integral is similar is integral a to b add and subtract a p t g t minus p t into sin n t dt absolute plus integral from a to b p of t into sin n t dt the whole thing absolute take the absolute value inside the first integral not the second one. Second one keep the absolute value outside in the first integral take the absolute value inside integral a to b mod g t minus p t dt mod sin n t is less than 1 no problem. The first piece integral a to b mod g t minus p t dt is less than 2 epsilon by 3 we just proved that the second piece integral a to b p of t sin n t dt but we are checked that the Riemann Lebesgue Lemma is true for polynomials. So, the second piece integral a to b p of t sin n t dt goes to 0. So, there exists an n naught such that for n greater than or equal to n naught the second piece is also less than epsilon by 3. So, all in all they get integral a to b g of t sin n t dt the whole thing in absolute is less than epsilon for n greater than or equal to n naught and the proof of the Riemann Lebesgue Lemma is thereby completed ok. So, that is a very nice proof. Let me suggest a second proof let me suggest a second proof take for example, an L 1 function take an L 1 function on the interval a to b g g belongs to L 1 of a to b. Now, suppose if g would be a step function what is a step function? Step function means take the interval a to b chop it up into finitely many pieces call a a equal to t naught t naught t 1 t 2 t 3 t k t k equal to b. So, just take finitely many points in the interval a to b or each piece t naught to t 1 it is constant t 1 to t 2 it is another constant t 2 to t 3 some other constant. So, it is discontinuous, but it is a step function. Now, for a step function it is very easy to prove Riemann Lebesgue Lemma because what is a step function? A step function is a linear combination of characteristic functions of the intervals right. Suppose that is a characteristic function of an interval t j t j plus 1 then what happens to the Riemann Lebesgue Lemma integral t j t j plus 1 sin n t d t you have to show that goes to 0 that is clear you can integrate and check. So, it is true for characteristic functions of intervals therefore, it is true for step functions and step functions are dense in L 1 that is another way of proving the Riemann Lebesgue Lemma whichever one is to your taste take up that take up that particular proof. So, the Riemann Lebesgue Lemma is proved. So, now, let us take stock of the situation. So, what is the issue? The issue with continuous functions f of x is merely assumed to be continuous then we know that the point wise convergence fails, but it was believed by many that the point wise convergence holds including initially until Paul Dubois Riemann after several abortive attempts at proving it produced a counter example in 1875. Using ideas from point set topology specifically the Bayer category theorem you can show that the majority of the continuous functions display such an errant behavior. A simplified proof was given by Stephen Banach. In fact, you use the Banach Steinhaus's theorem to produce a counter example. The Banach Steinhaus's theorem not only produces a counter example it tells you that this kind of errant behavior is true for a majority of the continuous functions. We shall give this proof using the Banach Steinhaus's theorem later in the course, but now let us continue a little further. Having proved a nice theorem we should look at a few examples. Let us look at a simple case namely f of x equal to mod x when mod x is less than or equal to pi. Do a 2 pi periodic extension extended 2 pi as a 2 pi periodic function. So, the mod function is like a v the mod function is like a v and then you just extend it and you get like a sawtooth you get a infinite sawtooth profile. Now from the graph it is very clear that the function is continuous. In fact, it is Lipschitz continuous. So, it is even better than holder continuity. Good. Now we know that the function f of x equal to mod x is an even function. It is a nice even function and the extension is also an even function. So, when you take an even function what happens? What is the formula for BN? You see it right here displayed in the slide. Here it is BN equal to 1 upon pi integral minus pi to pi f of x sin n x t. Sin is an odd function. Our f of x is an even function. So, product of an even function and an odd function is an odd function. When you integrate an odd function from minus pi to pi you are going to get 0. So, all the BNs are 0. So, the Fourier series will not involve the BNs it will only involve the ANs. What is A naught? A naught is 1 upon 2 pi integral minus pi to pi mod x dx. But being an even function it will be twice integral the integral from 0 to pi and it is a 1 upon 2 pi. So, it will be 1 upon pi. So, what will it be? It will be 1 upon pi integral 0 to pi x dx and that is easy to integrate and you see the A naught is simply pi by 2 A naught is simply pi by 2. Now, you need to calculate AN. AN will be 1 upon pi integral minus pi to pi mod x cos n x mod x is even cos is even the product is an even function will be twice integral from 0 to pi. So, 2 upon pi integral 0 to pi x cos n x you will have to perform an integration by parts and you have to be careful when you do the calculation. Because you make a mistake you are going to get it all wrong you do the calculation carefully and diligently you will get the ANs put it in the equation for the Fourier expansion f of x equal to A naught plus summation n from 1 to infinity AN cos n x because the bn is 0 and this is exactly what you get. Our basic convergence theorem this must hold for all values of x. Since it is a Lipschitz function the Fourier series will converge point wise to the given function f of x everywhere. So, we get this beautiful representation of mod x as a Fourier series. Now, what do you get when x is 0 put x equal to 0 the cosine term becomes 1. So, you get when x is 0 you get pi by 2 equal to summation 4 upon pi denominator 2 k minus 1 the whole square. Do a little bit of rearrangement you are going to get pi squared by 8 you will get pi square by 8 equals 1 plus 1 upon 3 square plus 1 upon 5 square plus 1 upon 7 square plus dot dot dot. You get that the sum of the reciprocal of squares of the odd numbers is pi square by 8. As an exercise determine 1 plus 1 upon 2 square plus 1 upon 3 square plus 1 upon 4 square plus you are supposed to get pi square by 6 do this as an exercise now. I think this should be a very nice place to stop this capsule and we will continue this in the next capsule. Thank you very much.