 So, now we start the third chapter in this course, this is concerns the Cesaro Summability. Remember I talked about several different modes in which the Fourier series converge, point wise convergence, mean convergence, Cesaro convergence, the very first chapter concern point wise convergence, where we needed the function to be better than merely continuous, we wanted to hold the continuity. In the second chapter, when we asked for mean convergence, it was enough for the function to be in L2, the partial sums will converge in L2. Now, we come to Fayer's Theorem or Cesaro Summability and mainly we shall be concerned with continuous functions. One can also talk about Fayer's Theorem in the context of other norms, where the arithmetic means converge in say L1 or L2. We will not discuss that right now, maybe later if time permits. We shall see how the course unfolds. Let us at least do the case for continuous functions. So, before we begin, let us recall some ideas from real analysis, simple theorems from real analysis that you are already familiar with, the Cauchy's First Limit Theorem. Theorem 27. Suppose Bn is a sequence of real or complex numbers converging to L, then the sequence of arithmetic means 1 upon n, B1 plus B2 plus dot dot dot plus Bn also converges to L. This theorem is really very simple if you think about it. So, let us assume that you have got a class, a mathematics class of say 25 students and you compute the class average of the examination. Say the class average is say 62 marks. Now, I enlarge this class from 25, I put in a thousand more students in the class, but these remaining additional students have marks close to 62, then the averages will remain close to 62. So, what this means is that you got a sequence of numbers B1, B2, B3, B4 etcetera. As we go down this sequence, the numbers Bn's are all coming close to L. So, it is like the first 25 terms of the sequence may not be close to L, it may toggle up and down. But as we go down the sequence, more and more terms are coming closer and closer to L. The farther you go, the closer it is to L and there are infinitely many terms that are close to L. In fact, all but finitely many terms are close to L. So, as you take averages of larger and larger strings, you are going to get numbers which are closer and closer to L. So, intuitively this theorem is pretty self-evident, but let us do a formal proof. So, let us do some preliminary algebra. So, 1 upon n times B1 plus B2 plus Bn minus L, I just pull the 1 upon n outside, write it as mod of B1 minus L plus B2 minus L plus delta plus Bn minus L. So, split the right hand side into 2 pieces, 1 upon n times mod B1 minus L plus delta mod Bn1 minus L plus 1 upon n times mod Bn1 plus 1 minus L plus delta plus mod Bn minus L. Now, this is true for any n1 and any n bigger than n1, but we shall specify the n1 presently. However, before we do that, so this is the Bn's were assumed to be convergent, they must be bounded. So, there is an m bigger than 0 such that mod Bj minus L must be less than or equal to m for all n. So, what we do is the first piece, we replace B1 minus L delta Bn1 minus L, all of them we replace by capital M. So, we get n1 upon n capital M. This is what I meant by saying that the first few terms of the sequence may be wild, it may deviate from L a lot, but in any case these deviations are bounded and so this bound is taken to be m. So, we got this piece n1 m upon n. Then the other thing is 1 upon n times mod Bn1 plus 1 minus L plus delta plus mod Bn minus L. Now, we bring in the epsilons, let epsilon greater than 0 be arbitrary. There is an n1 such that mod Bn minus L less than epsilon by 2 beyond n1 and it is this n1 that we use over here. And the second piece, we replace all these things by epsilon by 2 epsilon by 2 L. So, mod Bn1 plus 1 minus L less than epsilon by 2 plus delta mod Bn minus L less than epsilon by 2. How many are there n minus n1 by n that many epsilon by 2, but n minus n1 upon n anyway is less than 1. So, I just replace it by epsilon by 2. And we are left with the first few terms where the deviation was probably large, but never mind there is an n in the denominator and this capital M is a constant and I can arrange the this n to be so large that n1 m upon n is also less than epsilon by 2. And all in all I will get epsilon. What do I get? So, we select the n0 so large. First of all it should be larger than n1. So, that whatever we are done so far is valid and also this n0 should be so large that we ensure that this piece n1 capital M upon n is less than epsilon by 2 when n is bigger than n0. For that to happen we must choose the n0 to be larger than 2 n1 capital M upon epsilon. So, that this n1 m upon n that will be less than epsilon by 2 if n is bigger than n0. And that completes the proof. Now, the converse is obviously false. The original sequence may fail to converge, but the sequence of arithmetic means may jolly well converge. For example, take the sequence to be 1 0 1 0 1 0 1 0 alternately. The sequence does not converge, but take the arithmetic means. Suppose I take the arithmetic mean of say the first n terms. So, 1 plus 0 plus 1 plus 0 plus 1 plus 0 n terms divided by n. What am I going to get? I am going to get a limit of half. So, the average is still converge the original sequence does not converge. So, the convergence of a sequence of arithmetic means may be regarded as a generalization of the notion of ordinary convergence. Namely, if a sequence Bn converges to L then the sequence of arithmetic means also converge to L. On the other hand, if a sequence does not converge at all the averages may still converge. So, let us say that a sequence Bn converges in a generalized sense if 1 upon n times B1 plus B2 plus Bn converges to L. What the theorem that we have proved now says that ordinary convergence implies generalized convergence. So, it is indeed a widening of the notion of convergence. So, this generalized convergence has a name and it is called Cesaro convergence. We say the sequence Bn is said to converge in the Cesaro sense to L. The sequence of arithmetic means converges to L namely limit as n tends to infinity 1 upon n times B1 plus B2 plus zeta Bn equal to L. So, what Cauchy's first limit theorem says is that ordinary convergence implies a Cesaro convergence. Converse is not true as I said 1 0 1 0 1 0 the sequence which alternates between 0s and 1s is convergence in the sense of Cesaro, but it is not convergent in the ordinary sense. Now, we have seen that if f of x is a 2 pi periodic continuous function then the sequence of partial sums of the Fourier series Snfx the nth partial sum of the Fourier series of f need not converge point wise. This was the example due to Paul Dubois Raymond. The theorem of failure says that although the sequence Snfx may not converge point wise that is ordinary convergence may fail for the sequence Snfx, but Cesaro convergence holds. But what are the hypothesis? The function f of x is continuous. So, for continuous functions the sequence of partial sums of the Fourier series may not converge, but it does converge in the sense of Cesaro. But something better actually happens not only does Snfx converges in the sense of Cesaro to f the convergence is actually going to be uniform. The sequence of arithmetic means actually converges uniformly to f. This assertion is called Feier's theorem and we must now prove Feier's theorem that is the immediate goal. Let us recall the definition of uniform convergence. A sequence of functions fnx all defined on a common domain e is said to converge uniformly to f of x if supremum mod fnx minus fx over e goes to 0 as n goes to infinity. We already used this notion earlier to prove Riemann-Lebesgue Lemma. So, some exercises on Cesaro convergence. Suppose Bn is a monotone increasing sequence of real numbers and the sequence of arithmetic means is also monotone increasing. If a sequence is decreasing sequence of real numbers then the sequence of arithmetic means is also decreasing. Suppose Bn is a sequence of positive real numbers converging to L then the sequence of geometric means gn equal to the product B1, B2, Bn raised to the power 1 by n also converges to L. Remember since I am taking the product none of the Bn's must be 0. It should be strictly positive real numbers. Further if L is positive if L is positive then we can look at the sequence of harmonic means. What is the harmonic mean of n numbers? It is a reciprocal of the arithmetic means of reciprocals. So, take the n numbers to be B1, B2, Bn take the reciprocals 1 upon B1, 1 upon B2, 1 upon Bn take the arithmetic mean of the reciprocals 1 upon n times 1 upon B1 plus 1 upon B2 plus 1 upon Bn and take its reciprocal. So, it is the reciprocal of the arithmetic mean of the reciprocals. That is the harmonic mean. The harmonic mean also converges to L the expression for the harmonic mean you see on the slide. So, those are three exercises on Cesaro convergence or Cauchy's first limit theorem and you can do them. Now deduce that if Bn is a sequence of positive real numbers such that the ratio Bn plus 1 by Bn converges to L and the nth root of Bn also converges to L. This important limit is called Cauchy's second limit theorem. You use the previous result on this geometric sequence. Use the second result to show Cauchy's second limit theorem that Bn plus 1 by Bn converges to L implies the nth root of Bn also converges to L. In particular, use a special choice of these Bn's to prove that the nth root of n factorial upon n goes to 1 upon E. This last result here is a very weak form of the celebrated Stirling's approximation formula. Stirling's approximation formula says that when n is very large, when n is much larger than 1, n factorial is approximately equal to n to the power n into e to the power minus n into square root of 2n pi. This last exercise is a very weak form of Stirling's formula. I would like you to think about why this is a weak form of Stirling's formula. So, all this was a review of some classical analysis that you have been studying in elementary courses and probably some of you will be teaching these things in your elementary courses. So, it is a good recapitulator. So, now let us state and prove Feijer's theorem. Suppose f of x is a 2 pi periodic continuous function on the real line, just continuity nothing more than that and Sn fx is the nth partial sum of the Fourier series of f of x. Then Sn fx Cesaro converges to f of x. In other words, the arithmetic mean limit as n tends to infinity 1 upon n S0 f of x plus S1 f of x plus dot dot dot plus Sn f of x is f of x equation 3.1 in the display and the convergence is uniform. This last word is important. So, let us begin with the formula for the partial sums. What is the formula for the partial sums? We already derived that formula for the partial sums. The jth partial sum Sj f of x is given by integral from minus pi to pi dj x minus t f of t dt equation 3.2 in the slide where dj is the Dirichlet kernel. What is the Dirichlet kernel? It is recalled for you in the last line in the slide. dj theta is 1 upon 2 pi sin j theta plus theta by 2 divided by sin theta by 2. So, now what we do is we take this expression for Sj and calculate the averages. Put j equal to 0, 1, 2 up to n add them up and divide by n and we see what we get. Let us try. So, that is what I said S0 of f of x plus delta plus Sn of f of x will be integral minus pi to pi. All the things have f of t remember. So, the f of t will remain separate f of t dt times the other things has been clubbed as Kn theta f of x minus t. But what is this Kn theta 1 upon 2 pi. In the Dirichlet kernel there is a sin theta by 2 in all the terms and then in the numerator I will be putting j equal to 0, 1, 2, 3, etc. When put j equal to 0 it is simply sin theta by 2 and j equal to 1 it is sin 3 by 2 theta, etc. What do we get sin theta by 2, sin 3 by 2 theta plus dot, dot, dot sin 2n plus 1 theta by 2. Multiply the numerator and denominator by 2 sin theta by 2. We get 1 upon 4 pi sin square theta by 2 numerator 2 sin square theta by 2 2 sin theta by 2 sin 3 theta by 2 plus dot, dot, dot 2 sin theta by 2 sin 2n plus 1 theta by 2. 2 sin a sin b is cos of a minus b minus cos of a plus b remember from the addition formula for cosine. So, use that you get 2 sin square theta by 2 is 1 minus cos theta. 2 sin theta by 2 sin 3 theta by 2 is cos theta minus cos 2 theta, etc. The last one is cos n theta minus cos n plus 1 theta. Series telescopes lot of things cancel out. You are simply left with 1 minus cos n plus 1 theta upon 4 pi sin square theta by 2. And that is a very nice thing because that is non negative. Let us not forget that we have to divide by n plus 1 because we are taking averages and we are taking averages of n plus 1 things not n things s naught s 1 s 2 s n sequence begins with a 0 th term. So, we get after dividing by n plus 1 we get k n theta equal to 1 minus cos n plus 1 theta upon 4 times n plus 1 pi sin square theta by 2 1 minus cos a is 2 sin square a by 2 remember 1 2 cancels out we get sin square n plus 1 theta by 2 divided by 2 times n plus 1 pi sin square theta by 2. This 3.3 the right hand side of 3.3 is called the fair kernel. Fair kernel like the Poisson kernel is also a positive kernel. This positivity of this kernel will be very significant. And just as you had it for the Dirichlet kernel and Poisson kernel the fair kernel also has the property that when you integrate it from minus pi to pi k n theta d theta you are going to get 1 equation 3.4. Why is this true to prove 3.4? Remember that the same thing is true for Dirichlet kernel integral minus pi to pi dj theta d theta is 1 remember for each j. Now put j equal to 0 1 2 3 up to n and add it and you can prove 3.4. So, with 3.4 at our disposal and with the previous formula that we obtained for this we shall continue to prove the failure's theorem. We have proved that 1 upon n times s naught of f of x plus s 1 of f of x plus zeta s n of f of x which is equal to integral minus pi to pi k n of x minus t f of t dt. And I can do the usual change of variables put x minus t equal to s and I can write as integral minus pi to pi k n t f of x minus t dt. And since the failure kernel integrates to 1 remember the previous line equation 3.4 multiply both sides by f of t and you get the equation f of x equal to integral minus pi to pi k n t f of x dt. So, you subtract you subtract and you get 1 upon n s naught of f of x plus zeta s n of f of x minus f of x will be integral minus pi to pi in both you see the same failure kernel k n t f of x minus t minus f of x. And then take the absolute value and call the absolute value of the difference delta n x. So, delta n x less than or equal to integral minus pi to pi k n t mod f of x minus t minus f of x dt. There is no need to put the mod on the failure kernel because the failure kernel is a non negative kernel. So, now we do the same trick as before break the integral from minus pi to pi into sum of 2 integrals mod t less than delta and mod t bigger than delta. In one of them you could use uniform continuity and so on. So, to do that efficiently we must bring in the epsilons and deltas in the argument let epsilon greater than 0 be arbitrary we must appeal to uniform continuity of f of x and there is a delta greater than 0 such that mod f of x minus t minus f of x less than epsilon by 2 for mod t less than delta. And the integral that we see in the last slide should be broken up into sum of 2 integrals integral mod t less than delta and integral from mod t greater than or equal to delta. In the first piece mod of f of x minus t minus f of x is less than epsilon by 2 and the other integral is integral k n t dt. Now when you integrate it with respect to mod t less than delta or whether integrated from minus pi to pi it really does not matter because anyway this integral is 1 so the whole thing is going to be less than epsilon by 2. So, you can just forget about this mod t less than delta and you can conveniently put epsilon by 2 it is this thing that we need to be careful about. Here we use the fact that the function is continuous and so it is bounded call m to be the supremum of the function and this thing will be less than or equal to 2 m. 2 m times integral mod t greater than or equal to delta k n t dt. Now we need to deal with the second integral and we have to show that that is less than epsilon by 2 for all n greater than or equal to n naught. So, let us do that mod t is between delta and pi mod t is between delta and pi. So, mod t by 2 is less than delta by 2 and pi by 2. So, now on this interval the sin function is an increasing function. So, sin square delta by 2 is less than sin square t by 2. So, in the phasor kernel the sin square t by 2 which appears in the denominator gets upper bound. So, 0 less than or equal to the phasor kernel in the denominator you see sin square t by 2 replace the denominator sin square t by 2 by sin square delta by 2. Numerator sin has been replaced by 1 1 upon 2 pi anyway is less than 1. So, this whole thing is less than 1 upon n plus 1 sin square delta by 2. Now the delta is now fixed it depends only on epsilon and we got this 1 upon n plus 1 very conveniently. So, what we get all in all 0 less than or equal to delta n x and the first piece was less than epsilon by 2. The second piece was 2 m times integral of the phasor kernel in the interval delta less than or equal to mod t less than or equal to pi and that is now less than or equal to 1 upon n plus 1 sin square delta by 2. The 2 m will now become less than or equal to 4 m pi in the numerator. Now, we need to select the n naught so large that this thing this piece is less than epsilon by 2 when n bigger than or equal to n naught. So, what is the requisite n naught? If you take n naught to be bigger than 8 m pi upon epsilon sin square delta by 2 then n bigger than n naught will imply this piece 4 m pi upon n plus 1 sin square delta by 2 that will be less than epsilon by 2. So, epsilon by 2 plus epsilon by 2 will be less than epsilon so 0 less than or equal to delta n less than epsilon. So, this difference delta and the averages minus f of x in absolute value is less than epsilon for all n bigger than or equal to n naught. The n naught depends on what? The n naught depends on epsilon and delta delta depends on epsilon and so n naught depends only on epsilon it is independent of x. So, the convergence is uniform and so we established the Feijer's theorem. Now, we are going to give applications of Feijer's theorem. One of the most important applications of Feijer's theorem is that the trigonometric polynomials are dense. Remember, we use this very crucially in the proof of Parseval formula and remember in chapter 2 we did not complete the proof of Parseval formula and we said that we will have to wait until we do Feijer's theorem. Here is the time to do it. We will take it up in the next capsule. Thank you very much.