 So, hello. So, in this fifth part of this Fourier analysis course, we are returning back to some of the classical aspects of Fourier analysis. In chapter 8, we looked at an application to celestial mechanics and again we used some of the ideas in the early chapters in this course namely chapter 1. Now, we look at again classical aspects and I am calling this chapter odds and ends. Some of those tiny little things that got left out because we had to cover many other things and so we take up these things and consolidate them in this chapter and what are the topics that we discussed in this chapter? Uniform convergence of Fourier series. We give some sufficient condition when will the Fourier series of a 2 pi periodic function will converge uniformly to the given function and will prove Dirichlet's theorem on point wise convergence at points where the function has one sided derivatives. This was mentioned in the very first part of this course, but I said that we will prove this towards the end in a later part. So, here it comes and then we will discuss an important topic called Riemann's localization principle and we will discuss the case of monotone functions, functions that might be discontinuous and yet monotone. For this we will need what is known as the Bonnet's mean value theorem. We will give a complete proof of Bonnet's mean value theorem because it is usually not found easily in textbooks. It used to be available in all the classical books but modern books on analysis somehow have left out these kinds of things and this is very essential proof of Dirichlet's theorem. The first of these four items is very easy namely the case of uniform convergence of Fourier series and we will take up this first. The theorem says suppose f from R to R is a 2 pi periodic continuous function and it is piecewise smooth. What do I mean by piecewise smooth? That means that I have a partition of the interval minus pi pi into finitely many non-overlapping intervals. On each piece the function is differentiable and at the interface at those junction points the points of the partition the function has one sided derivatives. These one sided derivatives may be different but they are both finite. So that is piecewise smooth function that the Fourier series of f of x converges absolutely and uniformly to f of x. Examples of such continuous 2 pi periodic piecewise smooth functions here is an example. For instance you can look at the function which is a triangular wave. Take mod x and extend it as a 2 pi periodic function. Remember that uniform convergence has not touched upon so far. We only talked about point wise convergence in the very first chapter. Uniform convergence requires a little more techniques but they are very easy. It is a nice application of the Bessel's inequality that we established in the second chapter. So let a and b and b the Fourier coefficients of f of x and a and prime b and prime be the Fourier coefficients of f prime of x. Because of our hypothesis of piecewise smoothness the derivative f prime is continuous except possibly at finitely many points and at these finitely many points the derivative has a finite jump discontinuity. In particular the derivative is an L2 function for instance. What is the definition of the coefficients a n prime equal to 1 upon pi integral minus pi to pi f prime of x cos nx dx. We can integrate by parts. We can integrate by parts and the boundary terms will cancel out and we can check that a not prime is 0 a n prime is n b n and b n prime is minus n a n. Since the function f prime is in L2 we can use Bessel's inequality and we can apply the Bessel's inequality to the Fourier coefficients of f prime of x. a not prime was 0 so there is no constant term summation n from 1 to infinity n squared mod a n squared plus mod b n squared less than or equal to 1 upon pi integral minus pi to pi mod f prime x the whole squared dx. Now we are in a position to establish the absolute and uniform convergence of the Fourier series. We begin with a very simple observation mod a n cos nx plus mod b n sin nx clearly less than or equal to square root of mod a n squared plus mod b n squared. So, if we show that the series summation n from 1 to infinity root mod a n squared plus mod b n squared converges then we can use Weierstrass's m test to conclude that summation a n cos nx plus b n sin x converges absolutely and uniformly on the interval minus pi pi. So, our task is to prove that 9.1 the series 9.1 converges and that is pretty easy. So, let us write this summation n from 1 to capital N root mod a n squared plus mod b n squared we are looking at the nth partial sum you introduce a 1 upon n in the business. So, you multiply and divided by n and use Cauchy-Schwarz inequality in the very first line. So, when you use Cauchy-Schwarz inequality you get summation n from 1 to capital N 1 upon n squared and the second factor will be summation n from 1 to capital N n squared mod a n squared plus mod b n squared. It is a simple application of Cauchy-Schwarz inequality and the summation n from 1 to capital N is clearly less than or equal to summation n from 1 to infinity 1 upon n squared everything here is positive or at least non-negatives. So, we get these inequalities are very easy to establish what is summation n from 1 to infinity 1 upon n squared right from chapter 1 we got it as pi squared by 6 but we really do not need the exact value all we need is some universal constant c and then there is this integral minus pi to pi mod f prime x the whole squared dx and so the partial sums are bounded and hence summation n from 1 to infinity root mod a n squared plus mod b n squared converges and so summation a n cos nx plus b n sin nx converges absolutely and uniformly and that is exactly what we wanted to establish and we have proved our first result on uniform convergence of Fourier series. Now what is the advantage of this large number of examples fall under the scheme of things. So, now having discussed uniform convergence we will take up the next item in the agenda namely point wise convergence of the Fourier series at places where the function has one sided derivatives. So, large number of examples will fall under the scheme the main point is that with regard to convergence of Fourier series at a specific point x naught the behavior of the function far away from x naught does not matter what we of course need is that the function must be in l 1 the Fourier coefficients must make sense otherwise we cannot get started. So, to get started we assume that the function is in l 1 of minus pi pi and at a specified point x naught in the vicinity of x naught we are going to assume that the function is reasonably well behaved namely we are going to assume the existence of one sided derivatives at x naught away from x naught say outside a delta neighborhood of x naught it really does not matter the function can be very badly behaved. So, we take up this as the next theorem, theorem 98 suppose f is a 2 pi periodic function on R which is continuous at x naught and piecewise smooth what does piecewise smooth mean in the earlier case we took continuous and piecewise smooth here we are going to assume continuity only at x naught and here we are going to use piecewise smooth everywhere which basically means that I can break the interval into say a which is equal to t naught less than t 1 less than t 2 delta less than t n and t n is b taken a partition of the interval a b and on each sub interval t j t j plus 1 the function is smooth at the end point the function has one sided derivatives and at the junction points t 1 t 2 t 3 etcetera the one sided derivatives exist and the function may even be discontinuous at these junction points t 1 t 2 t 3 for example, I can take this function which is a signum function f of x equal to 1 on the interval 0 to pi and minus 1 on the interval minus pi to 0 the function is evidently differentiable on the closed interval 0 to pi it is also differentiable at the closed interval minus pi to 0 at 0 the value does not really matter we can take whichever you want. So, you got this kind of a prescription now so for such a function so it is a piecewise smooth 2 pi periodic function and the one sided derivative exists at x naught the Fourier series of f converges to f of x naught at the point x naught we must employ the notations of chapter 1 and recall that d n t is the Dirichlet kernel which is the most important thing as regards point wise convergence and s n f of x naught is basically the sum of the first 2 n plus 1 terms a naught plus a 1 cos x plus b 1 sin x plus dot dot plus a n cos n x plus b n sin n x the first 2 n plus 1 terms in the Fourier series that is the nth partial sum which we just call it nth partial sum for convenience s n f comma x naught and what exactly is happening is that we must look at the difference s n of f comma x naught minus f of x naught s n of f comma x naught we computed in the very first chapter it is a convolution with the Dirichlet kernel it is integral from minus pi to pi f of x naught minus t d n t d t and remember that the integral d n t d t from minus pi to pi is 1 so f of x naught can be written as integral f of x naught d n t so that this difference s n f x naught minus f of x naught can be combined into 1 integral minus pi to pi f of x naught minus t minus f of x naught d n t and we need to understand the behavior of this integral as n tends to infinity so now what we do is that we split the integral into 2 integrals an integral from 0 to pi and an integral from minus pi to 0 and the discussion is parallel for the 2 integrals so we will just discuss the first one the other one is left for you to check so let us go from 0 to pi f of x naught minus t minus f of x naught d n t d t we have to show this goes to 0 as n tends to infinity now we also need to recall the expression for the Dirichlet kernel the Dirichlet kernel has an innocent factor of 2 pi which I am putting on the left hand side and you have to use the addition formula for the sin and you are going to get a cos n t term and a sin t by 2 term in one of the terms the sin t by 2 cancels out you simply left with cos n t the other term you got a sin n t and you got cos t by 2 upon sin t but there is cot t by 2 it is a cot t by 2 that is causing all the problems and let us do the following as a multiply and divide by t let us divide by t and call the ratio g t so what is g t f of x naught minus t minus f of x naught upon t now remember that we are assuming that f has a one sided derivative so as t approaches 0 from the right this 9.3 has a limit as t goes to 0 because we divided by t we end up multiplying by t so t cos n t that is quite innocent t cot t by 2 now this object has a limit as t goes to 0 and by assigning the value of this as 2 at t equal to 0 we prove that the thing actually becomes continuous by so by prescribing this value properly at the origin this actually becomes a continuous function you will call it a removable discontinuity what we need to do is that we need to appeal to the Riemann-Lebesgue lemma so what happens is you take a small piece 0 delta and the function g t has a limit as t goes to 0 so it is bounded on 0 delta ok so because of that this function g is certainly in l 1 because this function g is in l 1 t times g t is also in l 1 so t g t cos n t this thing goes to 0 by Riemann-Lebesgue lemma and what about the other term the other term is t cot t by 2 that also became continuous and g is certainly in l 1 so that into sin n t dt that also goes to 0 by Riemann-Lebesgue lemma so the Riemann-Lebesgue lemma comes to our rescue and the proof is completed the conditions on hypothesis of f can be considerably relaxed but we are not really interested in giving you the most general theorem in this direction we can for example consult the monumental 2 volume text of Antony Sigmund called trigonometric functions on page 52 you can see some generalizations a large class of functions that are of interest in engineering will be piecewise smooth so will well fall into the scope of our theorem alright so the same argument the same ideas can be used to prove one more result called the localization principle that is theorem 99 suppose f of x is a 2 pi periodic function which is integrable on minus pi pi that is f is in l 1 and the function is 0 on a certain open sub interval right i is a sub interval of minus pi pi and open sub interval this open sub interval could be terribly small then the Fourier series for the function f of x converges to 0 at all points of i in particular if you take 2 functions f and g so if you have 2 functions f and g which are both in l 1 and the 2 functions agree on a certain sub interval i then on this sub interval i their Fourier series will behave alike that is they will both converge or maybe they will both diverge throughout this open interval i now why is this result important or interesting let us contrast this thing to the case of a power series now suppose I give you a function f of x which is analytic and it is identically 0 on a certain open sub domain g now I will take a point p in this g and you what happens to the power series the function is 0 at p and it is 0 in a neighborhood of p so that first derivative is 0 the second derivative is 0 the third the power series itself is a 0 power series and so the 0 power series evidently converges to 0 and so this kind of result which is completely trivial in the context of analytic functions but look at it in the context of Fourier analysis now what is given to you f of x is 0 on a small open interval i so this open interval could have a very small length but away from this open interval the function f of x may be non 0 the function f of x is non 0 integral from minus pi to pi fx cos nx need not be 0 integral minus pi to pi f of x sin nx need not be 0 the Fourier series is not 0 all the coefficients in fact can be non 0 and it is a non trivial to see that the sum of the series must be 0 the fact that the sum of the series is 0 throughout i then becomes very significant because you could not have predicted this simply by staring at the series you need to do analysis in order to draw this conclusion so the localization principle of Riemann is a non trivial result but it is not trivial you use the same circle of idea the Dirichlet kernel take a point x0 in i and take a small delta neighborhood of x0 x0 minus delta x0 plus delta contained in i and then do the same thing that we did earlier the nth partial sum of the Fourier series sn fx0 is the convolution f of x0 minus t dnt dt and then f of x0 is f of x0 dnt dt integral from minus pi to pi the difference you get integral from minus pi to pi f of x0 minus t minus f of x0 dnt dt and split the integral split the integral over the piece mod t less than delta and mod t bigger than delta now let us look at the integral over mod t less than delta what happens when mod t is less than delta anyway f of x0 is anyway 0 if mod t is less than delta this is also 0 and so this integral itself is 0 what happens when mod t is bigger than delta over this interval we need to look at the Dirichlet kernel carefully 2 pi times dnt will be cos nt plus sin nt into cot t by 2 this cot t by 2 is actually continuous on the interval delta less than mod t less than pi and so Riemann Lebesgue Lemma can be used on the interval delta less than mod t less than pi so there is no problem and the other factor what are the other factor f of x0 minus t minus f of x0 that anyway is in L1 so there is no problem and so the result follows from the Riemann Lebesgue the Riemann's localization principle theorem 99 has been established okay let us discuss the behavior at a point of discontinuity a model case is when the function has just one discontinuous point the origin you would look at the signum function f of x equal to 1 if x is positive minus 1 if x is negative look at it in the interval minus pi pi take the 2 pi periodic extension of the signum function and draw the graph the graph looks like a square wave train sigma x this is called the signum function extended as a 2 pi periodic function what are the Fourier series for sigma x use the formulas a for a0 a n and b n it is an odd function so the cosines are all 0 and the a0 term is 0 only the sign series will remain and we can quickly calculate the coefficients it is 4 upon pi times sin x plus 1 third sin 3x plus 1 5th sin 5x plus delta now by theorem 99 the Riemann's localization principle which are the 2 functions I am going to take I am going to take the constant function 1 and I am taking the signum function I could look at this in the open interval 0 pi on the open interval 0 pi we know that the 2 series are going to behave alike so the series converges to the signum function at points except the origin and plus minus pi plus minus pi and origin are excluded and all other points the sum of this series is going to return you back the original function sigma x what happens at the origin at the origin the series converges to 0 all the terms are 0 simple and why does it converge to 0 because it is the arithmetic mean of the right hand limit and the left hand limit namely it is 1 half of sigma of 0 plus plus sigma of 0 minus so at the point of discontinuity the Fourier series converge to the arithmetic mean of the right hand limit and the left hand limit now we want to show that the same thing happens in far greater generality then but we shall use the case of the signum function to prove the more general theorem signum function is easy because we could directly see what the Fourier series is and from the by looking at the Fourier series we know that at the origin it converges to the arithmetic mean of the right hand limit and the left hand limit and away from the origin we were able to use the Riemann's localization principle to prove that the series converges to the actual function so now theorem 100 suppose f is piecewise smooth on minus pi pi and has one sided derivatives at x naught and it is discontinuous at x naught that is it has a jump discontinuity there. The Fourier series of f converges at x naught to the value one half of f of x naught plus plus f of x naught minus here again I would like to make a small clarification remember that the one sided derivative means that on the open interval here the meaning of the word one sided derivative is you take the open interval x naught to x naught plus delta the function is differentiable on this interval x naught to x naught plus delta and the derivative has a limit as you approach x naught from the right similarly from the left that is the meaning of saying that the function has one sided derivative at x naught and the one sided derivative of the jump discontinuity. We reduce the problem to the continuous case. So, let alpha be the jump ok. So, alpha is the right hand limit minus the left hand limit or the saltis of the function. Now, use the signum function and you correct the function f of x by subtracting of a multiple of the signum function which multiple alpha by 2. Why have I chosen alpha by 2? Because if I choose this alpha by 2 exactly alpha by 2 then this new function f of x becomes continuous at x naught. Calculate the right hand limit and calculate the left hand limit it is displayed here and take the difference. The difference will exactly be 0. So, capital F is continuous at 0 and it has the one sided derivatives also at 0. So, we have proved the theorem when the one sided derivatives exist and the function is continuous. For that we know that the Fourier series converges to f of x naught namely f of x naught plus or f of x naught minus the two values are equal and it equals the arithmetic mean of the little f of x naught plus plus little f of x naught minus upon 2. You can check this and I am leading this for you to check. What is the partial sum for little f? The partial sum for little f is the partial sum for capital F plus alpha by 2 times the partial sum for the signum function x minus x naught. Look at the definition of little f. Little f is capital F plus alpha by 2 signum function translated by x naught and the partial sums will simply be linear. So, now the second term goes to 0 because we already proved the theorem for the signum function. Remember that for the Fourier series for the signum function converges to the signum function at points of discontinuity it converges to the arithmetic mean of the right hand left hand limits. So, the second term goes to 0 at x naught and the first term converges to f of x naught plus plus f of x naught minus upon 2 and so the result is established. I think this will be a good place to stop this capsule and we will continue this in the next capsule. Thank you very much.