 Hello, so we continue with our discussion on the classical aspects of Fourier series. So these are the odds and ends that we talked about some of these we disposed of in the last capsule now we will take up the remaining parts and we come to monotone functions. Monotone functions can be pretty bad in fact, but they also have some pleasant features some redeeming features a monotone function has only countably many discontinuities. The discontinuities are simple jump discontinuities a monotone function is Riemann integrable. Given any countable density there is a monotone function which is discontinuous precisely at the points of D. So for example, you can have the set of rationals and I can construct of monotone function which is discontinuous at every rational and continuous at every irrational. A monotone function is differentiable almost everywhere the last property is highly non-trivial you can look at the third edition of Royden's real analysis. I only studied the third edition there is a fourth edition I understand, but Royden's real analysis is one of the finest books written for the sheer joy of reading mathematics in his style and here in chapter 5 he proves this last property using Vitale covering theorem and I urge you to look at the proof there we will discuss the other properties briefly. Item number 4 says that a monotone functions can behave pretty badly because it can be discontinuous at lots of places distributed densely. Nevertheless, we shall see that the Fourier series of a piecewise continuous monotone function is quite well behaved as regard this convergence this is the theorem of deletion that we shall establish presently. So a few properties of monotone functions would take a function f from AB to R which is monotone increasing and let us take a point C in the interval AB and let us look at the interval which is half open open from the C end and closed at the B end you can actually ignore what happens near B you are going to concentrate on what happens near C only and it is a monotone increasing function. So I am taking the infimum of f of t as t varies from c to b c excluded and this infimum let us call it l and let us show that the function f of t tends to l as t tends to c plus and that will prove that a monotone function has a right hand limit at each point. So let epsilon greater than 0 be arbitrary and invoke the definition of infimum there exists a t naught between c and b c excluded such that f of t naught is less than l plus epsilon. Now by monotone city we know that f of t is greater than or equal to l and it is less than l plus epsilon for c less than t less than t naught because already f of t naught is less than l plus epsilon. So if t is less than t naught f of t will be less than or equal to f of t naught. So f of t will also be less than l plus epsilon but l is a infimum so f of t will be greater than or equal to l. So the requisite delta is t naught minus c. So using the epsilon delta definition we already established that the monotone function has a right hand limit the case of the left hand limit is very similar and the proof can also reveals that the left hand limit is less than or equal to a right hand limit. What will you do for the left hand limit you will need a supremum of f of t for t less than c then you will establish that f of c minus is less than or equal to f of c plus that I am leaving it as an exercise. Now suppose I take 2 points c1 and c2 c1 strictly less than c2. Now let us look at the open interval c1 c2 and let us take the infimum of f of t on the open interval c1 c2 and the supremum of f of t on the open interval c1 c2. When I take the infimum of f of t on the open interval c1 c2 what do I get? I get the right hand limit of f at c1. The supremum of f of t on the open interval c1 c2 is the left hand limit of f at c2. So we get f of c1 plus less than or equal to f of c2 minus. And by the previous slide f of c2 minus is less than or equal to f of c2 plus and f of c1 minus is less than or equal to f of c1 plus. So I consolidated all these inequalities into one inequality. So now if c1 and c2 are two distinct points of discontinuity that means that this inequality is strict and this inequality is strict then the open intervals i c1 and i c2 that is open interval f of c1 minus comma f of c1 plus the open interval i of c2 f of c2 minus comma f of c2 plus these are disjoint non-empty open interval. They are non-empty because the function is discontinuous and they are disjoint because the previous inequality. So for each discontinuity c I can pick a rational number qc in this interval f of c minus comma f of c plus. In the function c going to qc is a injective function because of the disjoint nature of these intervals this assignment c going to qc will become injective and it perhaps the set of discontinuities into the rationals because I have chosen this qc to be a rational number. So I have an injective map from e into rationals so the set of discontinuities is at most countable because the set of discontinuities is at most countable the Riemann integrability immediately follows. Recall when is a function Riemann integrable a function is Riemann integrable on a close bounded interval if and only if the set of discontinuities has Lebesgue measure 0 a countable set has Lebesgue measure 0 and the job is done. So now suppose that e is any countable dense set of ab take any countable set in fact whether it is dense or not is irrelevant let us call this countable set e as r1 r2 r3 some enumeration of this set I have taken. Now you choose a convergent series of positive terms summation 1 upon 2 to the power n a geometric series what we do is that we prescribe f of x to be summation 1 upon 2 to the power j where you pick those j's for which rj is less than x that is a well-defined function. Now if x is less than y remember that there are more j's for which rj is less than y and there are fewer j's for which rj is less than x because x is less than y. So that immediately gives you that if x is less than y then f of x will be less than or equal to f of y and if this is dense then f of x will be strictly less than f of y we get if it is dense we get something better we get a strictly monotone function it is not difficult to verify that the function f of x that we are constructed is discontinuous at every point e and it is continuous on the complement of e. So this establishes some of the properties of monotone functions that we talked about early in the first couple of slides. Now we come to the Dirichlet's theorem which goes back to 1829 is the earliest rigorous result this is the general convergence theorem for piecewise continuous monotone functions. Suppose f from minus pi pi to r is a piecewise continuous monotone function the Fourier series of f converges to f of x at points of continuity and at a point x0 of discontinuity it converges to the arithmetic mean of the right hand limit and the left hand limit at x0 at plus minus pi the Fourier series converges to one half of f of pi minus the left hand limited pi and the right hand limited minus pi. To simplify Dirichlet's argument Ossian Bonnet established the mean value theorem for integrals that bear his name. Bonnet's mean value theorem is stated below and this theorem 102 let g from a b to r b continuous and let f from a b to r b monotone. So we have got a continuous function and we have got a monotone function and you are integrating the product f t g t dt from a to b the product is evidently Riemann integrable because the discontinuities are countable. Then this is nothing but f of b minus integral from c to b g t dt plus f of a plus integral from a to c g t dt for some c in the closed interval a b. So Bonnet's theorem asserts that there exists a c with this property. Further we can see a little more if you know that the function is non-negative and monotone decreasing then we can reduce this to one piece integral a to b f t g t dt equal to f of a plus integral from a to c g t dt. If f is monotone increasing and non positive then there is a c such that integral a to b f t g t dt equal to f of b minus integral c to b g t dt. So you got these three pieces and it is not difficult to prove the Bonnet's mean value theorem but let us go further with a Dirichlet's theorem on Fourier series. We need a preliminary lemma. What do the lemmas say? Theorem 103 we have the following limit integral 0 to infinity sin omega t upon t dt is signum omega into pi by 2 that is signum omega is 1 if omega is positive it is minus 1 if omega is negative it is 0 if omega is 0. The second result is that you have taken a monotone function and you have fixed a s positive and you are integrating from 0 to s f of u sin omega u by u du and I am taking the limit as omega tends to infinity I am going to get pi by 2 times f of 0 plus. Here I have not stated whether it is monotone increasing or monotone decreasing it should not matter because if it is true for monotone increasing it will be true for monotone decreasing because I can multiply by minus 1. So I have just written it as monotone. Secondly I require in Bonnet's mean value theorem some positivity or negativity or whatever it is again that will not matter. Suppose for example I want the function to be non-positive so what I do is that once I establish this I can subtract from f a large constant and I can make it strictly negative or I can add to f a large positive constant and I can make it strictly positive. So then you have to understand what happens to the limit as omega tends to infinity integral 0 to s constant times sin omega u upon u du remember I had tampered the function by adding a constant. So what is the result of this tampering the result of this tampering is going to be simply integral 0 to s sin omega u by u du put omega u equal to v then I will get sin v by v dv 0 to infinity because omega is going to infinity and that limit is pi by 2 remember. So tampering the function by adding a constant or subtracting a constant is not going to disturb this equation. So I may as well assume that the function is monotone decreasing and non-positive or non-negative whatever I like so that we can apply the Bonnet's mean value theorem. The first part is standard it is clearly enough to prove it for omega positive because if omega is negative sin is an odd function the both sides will change sin that is all make the change of variables t omega equal to u and the integral simply becomes integral 0 to infinity sin u by u du and we just have to prove that sin u by u du integral from minus infinity to infinity is pi but this is elementary complex analysis apply Cauchy's theorem to this holomorphic function e to the power i z minus 1 upon z this is holomorphic on the punctured plane but applied to a semicircular contour indented at the origin. You have probably seen this kind of stuff in your baby complex analysis course so I will leave it to you to verify this and we straight away go to the proof of the second part which is more interesting. So as I explained before there is no harm in assuming that f is monotone decreasing and non-positive so that I can apply the Bonnet's mean value theorem and add and subtract f of 0 plus now once I add and subtract and f of 0 plus this integral that you see that is very clear what is happening to it as omega goes to infinity I just explained this will go to integral 0 to infinity sin v by v dv that is pi by 2. So we need to simply tackle the first term so let us look into the first term in detail. First of all let epsilon greater than 0 be arbitrary now we select an r bigger than 0 such that mod f of u minus f of 0 plus less than epsilon by a for 0 less than u less than or equal to r where the a will be some large constant that will be specified below. So now what happens is that let us look at this integral from r to s f of u minus f of 0 plus upon u sin omega u by u d u remember that my r is now fixed it depends upon epsilon but once I choose my epsilon the r is sealed and this denominator u should not cause any problem because r is strictly positive and fixed so as omega goes to infinity this will go to 0 by Riemann Lebesgue Lemma. So all we need to do is to deal with this integral from 0 to r f of u minus f of 0 plus upon u sin omega u by u d u but now we apply the Bonnet's mean value theorem here is where the monotonosity of the function is going to play a critical role and by Bonnet's theorem this is going to be equal to f of r minus f of 0 plus integral from c to r sin omega u by u d u. Now we are ready the integral term of course is bounded bounded in absolute value by say a and so we see that the integral from 0 to r f of u minus f of 0 plus upon u sin omega u d u is less than epsilon in r in absolute value the proof of the theorem is completed we shall now prove the theorem of Bonnet since the theorem is usually not given in modern texts as explained to you and a proof is available in George Gibson's advanced calculus page 277 for convenience we state the theorem here in blue and we will prove this theorem in the next capsule and we will need for this purpose Abel's summation by parts formula this proof of Bonnet's mean value theorem is a beautiful application of the formula for summation by parts by Abel and this formula is used frequently in analysis particularly when you want to discuss conditionally convergent series etc you can see for example Rudin's principles of mathematical analysis already in chapter 3 you will see this result stated there stated and proved and we will discuss these things in the next capsule and I think it is a good time to stop here thank you very much.