 Hello, so recall that we started discussing Dirichlet's theorem on the convergence of the Fourier series for a monotone function and this concerns point-wise convergence. So what's the hypothesis? The function is piecewise continuous monotone that means that I can break the interval minus pi pi into finitely many pieces of non-overlapping intervals such that on each interval the function is monotone and at these points of overlap of two adjacent intervals. Two adjacent intervals have just one point in common and on that particular point the function has jump discontinuities and at points of continuity the Fourier series will converge to f of x and at points of discontinuity it will converge to the arithmetic mean of the right hand limit and the left hand limit at that particular point. The theorem is clearly stated in the slide as theorem 101 and I'm just recalling to you what we have discussed last time that is theorem 1.1. We have to prove this theorem and we started preparing for the proof of theorem 1.1 and we proved a couple of partial results and we used a result of Bonnet. Bonnet's mean value theorem for integrals this was discovered by Ossian Bonnet to simplify Dirichlet's theorem. You see the theorem of Bonnet in the displayed slide. We are going to prove Bonnet's theorem today. We proved this auxiliary result theorem 103 integral 0 to infinity sin omega t by t dt is signum omega times pi by 2 and if you take a monotone function then for any s bigger than 0 integral 0 to s f of u sin omega u by u du limit as omega tends to infinity is pi by 2 f of 0 plus. This also we established last time. Now let us straight away get on to the proof of Bonnet's mean value theorem. To prove Bonnet's mean value theorem we need a result called Abel's summation by parts and that is theorem 1.4 that you see in the displayed slide. What is the Abel's summation by parts? Summation j from 1 to n a j times b j minus b j minus 1 equal to minus summation j from 1 to n a j minus a j minus 1 b j minus 1 plus a n b n with understanding that a naught and b naught are both 0. So, if you look at this formula you see a striking resemblance to the rule for integration by parts and the proof of course is quite clear a simple rearrangement will give you this identity. After all both sides are finite sums and you can use this summation by parts formula in many results in analysis particularly in delicate tests for convergence such as the tests of Abel and Dirichlet you can consult Rudin's principles of mathematical analysis for other applications ok. So, now let us get to the proof of Bonnet's mean value theorem. Assume that f is monotone decreasing. Remember that monotone functions are Riemann integrable and you have taken a function g and g is assumed to be continuous and f g is Riemann integrable and the idea is to approximate the integral of f g by its Riemann sums and these Riemann sums are finite sums. How do these finite sums look like? For example, you see the display in the slide they look like these and one should apply the Abel's summation by parts formula to these things and we will continue from there and the Riemann sums will involve a choice of a point c j in each of the partitioning intervals. The points of the partition are t 1, t 2, t n, t naught is the left end point. How do I choose these c j's? They will be described in the next display and these Riemann sums we have and to this we are going to apply the Abel's summation by parts formula ok. So, g is assumed to be continuous and so, but the usual mean value theorem integral of g u d u from t j minus 1 to t j is t j minus t j minus 1 the length of the interval times the function evaluated at the intermediate point. Now this is pretty easy this follows immediately from the Lagrange's mean value theorem. So, now this is the choice of c j for cooking up our Riemann sums that you see here alright. So, what do we get? The Riemann sum is summation j from 1 to n f of c j times integral t j minus 1 t j g u d u that is exactly this particular part g of c j t j minus t j minus 1 has been replaced by this ok. So, this integral from t j minus 1 to t j can be written as a difference. Now you can see that we are ready to apply Abel's formula for integration by parts. In Abel's formula for integration by parts summation by parts summation is basically a special case of an integration after all. There is a minus sign that we pick up summation j from 1 to n f of c j minus f of c j minus 1 integral a to t j minus 1 g u d u plus the last term that you see and that is in blue and the one term that has been in red. And as the measure the partition goes to 0 this last term will go to f of b minus into integral a to b g u d u as a partition becomes finer and finer. Now we deal with the term in red which is j equals minus summation j from 1 to n f of c j minus f of c j minus 1 integral a to t j minus 1 g u d u. Let us first introduce a new function phi of x which is minus integral a to x g u d u which is obviously a continuous function and it will have a supremum and an infimum capital M and little m and so this object is going to be sandwiched between capital M and little m. Multiply both sides by f of c j minus f of c j minus 1. Remember that we are assuming that f is monotone so the inequality is going to be preserved. I am going to get m times f of c n minus f of c 1 less than or equal to j less than or equal to capital M times f of c n minus f of c 1. So, I have summed over j and now I am going to allow the partition to become finer and finer the measure the partition goes to 0 and in the limit we get m times f of b minus minus f of a plus less than or equal to limit of j less than or equal to capital M times f of b minus minus f of a plus or the ratio j upon f of b minus minus f of a plus lies between little m and capital M and so since the function phi was continuous j upon f of b minus minus f of a plus is going to be the value of the function a integral a to c g u d u and then we get in the limit integral a to b f of t g t equal to minus f of b minus minus f of a plus integral a to c g u plus f of b minus integral a to b g u d u that is the term in red was this and the term in blue was the last one whereby we get integral a to b f of t g t equal to f of b minus time integral b to c g u d u I combined two of the integrals and I got I simplified the previous expression plus f of a plus integral a to c g u d u that completes the first part of Bonnet's theorem and now we need to deal with the other two parts of Bonnet's theorem. So, what we have proved is the first part namely the first display that you see here now we subtract from both sides f of a plus integral a to c g u d u from both sides and we will get integral a to b f of t minus f of a g t d t equal to f of b minus minus f of a plus integral c to b g u d u. So, now we observe that f x minus f of a plus is decreasing and non negative and then we get the second part of Bonnet's theorem right capital f of t equal to little f of t minus f of a plus the proof of the Bonnet's theorem is complete. Now, we come to the proof of Dirichlet's theorem recall that the nth partial sum for the Fourier series is S n f x that is the notation that we have used and that is integral from minus pi to pi d n t f of x minus t d t that is a convolution or the function with the Dirichlet kernel and in view of the fact that the functions and d n are both 2 pi periodic we can write this integral as integral from x minus pi to x plus pi d n t f of x minus t d t. Again writing out the expression for the Dirichlet kernel write out the expression for the Dirichlet kernel there is going to be a nice term a cos n t term and there is going to be a bad term sin n t times cot t by 2 term. The cos term is an innocent term by Riemann Lebesgue Lemma that goes to 0 as n tends to infinity the term that we need to worry about is the cot t by 2 term and that is a displayed here is 1 upon 2 pi integral from x minus pi to x plus pi sin n t cot t by 2 f of x minus t d t. Now add and subtract a 2 by t because cot t by 2 behaves like 2 by t. So, add and subtract a 2 by t. So, cot t by 2 minus 2 upon t plus 2 upon t cot t by 2 minus 2 by t that is continuous assign its value at 0 appropriately it becomes continuous. And so again Riemann Lebesgue Lemma and again the term will go to 0. So, we just have to deal with sin n t upon t that is the only term that we need to deal with. So, what we need to deal with is that as n tends to infinity the limit of s n f x is the same as the limit of this expression 1 upon pi integral x minus pi to x plus pi sin n t by t f of x minus t d t. So, far we have simplified our life. Now we need to look at the intervals on which the function is monotone and remember that I break the interval minus pi pi into finitely many intervals which are non overlapping to adjacent intervals have only the end one end point in common. So, these intervals are I 1 I 2 I 3 dot dot dot I k. So, minus pi to pi is I 1 union I 2 union the data I k and the I 1 I 2 I k are non overlapping intervals and f restricted to each I j is continuous and monotone. And of course, these intervals have to be translated by x the integration is not from minus pi to pi the integration from x minus pi to x plus pi. So, we have to look at x plus I 1 x plus I 2 etcetera that is the interval that we need to be looking at. So, if x is not in I j if x is not in I j then I have to look at x minus I j will have intervals x minus c and x minus d and x is not in this closed interval. So, both these numbers x minus c and x minus d will be non zero and they will have the same sign. And so, in this integral over x minus I j let us make the substitution n t equal to u. Basically this interval is c to d. So, this integral goes from x minus d to x minus c. So, I am going to put n t equal to u and I put n t equal to u what am I going to get the limits of integration are going to change to n times x minus d n times x minus c f of x minus t is f of x minus u by n and I am going to get a plus or minus 1 upon pi that depends on the sign of this x minus c and x both of them have the same sign and this sign is plus is going to be plus if that sign is minus is going to be minus. Now, I am going to allow the n to go to infinity I am going to allow the n to go to infinity and then I will realize that this particular interval goes off to infinity both n times x minus d and n times x minus c both go to plus infinity or both go to minus infinity and so this integral collapses to 0 as n goes to infinity. So, this is an easy case. Now, let us consider the case where x belongs to one of those intervals I j that means that x minus b and x minus a are of opposite signs and let us also assume that x lies the interior of a b. So, both are non-zero and then we split the integral on x minus i j into a sum of two integrals x minus b to 0 and 0 to x minus a and so we have to tackle these integrals separately. Again, I am going to make the same substitution n t equal to u here this will be an integral from minus infinity to 0 and other will become an integral from 0 to infinity. So, as usual I will get if x minus a or x minus b is 0 then one of these integrals will collapse that will happen when x lies in the end point of i j and to deal with the limit as n tends to infinity we use the lemma that we proved that when f is monotone remember limit as omega tends to 0 integral 0 to s f of u sin omega u upon u equal to pi by 2 times f of 0 plus use this result and you deal with these integrals and we will get the value pi by 2 times f of x minus plus f of x plus if x happens to be one of the end points. So, if x is not an end point then f is continuous and these two values will be equal and you will simply get 2 times f of x. So, and there is a 1 upon pi outside and that will cancel out and you have done the job. So, as I explained if x is one of the end points then it will line 2 adjacent integrals there will be a contribution coming from both of them and that is how you get this arithmetic mean in the limit n tends to infinity s n f x and that is the expression that is there in the statement of the Dirichlet's theorem. So, the proof of Dirichlet's theorem is thereby completed with all the auxiliary propositions thrown in. It was a technical result using many different ideas such as Abel's summation by parts formula the mean value theorem of Aussie and Bonnet and so on so forth, but it is well worth going through this because a large number of situations that you encounter they fall in this category. For example, you can take a function which is piecewise continuous say represented by line segments or parabolas or whatever and you can write down the Fourier expansions and we will get various useful formulas that are of importance in classical analysis. So, I think with this we have completed this chapter on odds and ends those parts of classical Fourier analysis that got left out in chapters 1, 2, 3 they have now been incorporated in chapter 9 because these are for lesser importance and we had other important things that had to be done in the middle. So, with this I will close this chapter and we will proceed to the next chapter which is again returning back to Fourier transforms and we will develop more technology in the next chapter and we will return back to chapter 4 and also some parts of chapter 1. So, we shall take a relook at classical analysis from a modern perspective. Thank you very much.