 Hello, we continue from where we stopped last time, recall that last time we discussed Kronecker's theorem, namely if alpha is an irrational number and you take multiples of alpha, alpha 2, alpha 3, alpha etc., and take their fractional parts, these fractional parts are dense in the interval 0, 1. We proved this theorem, but the Weyl's theorem considerably sharpens classical theorem of Kronecker. Let me explain this to you. You take this list 3.6, the fractional parts of multiples of alpha and you look at the number of elements in 3.7, fractional part of alpha, fractional part of 2 alpha, etc., fractional part of n alpha. How many of these lie in a specified closed interval a, b? So, take a closed interval a, b and think of each time you take k alpha and it is take its fractional part. Think of it as an experiment and in the first n it rates of this experiment. How many times does this fractional part k alpha hit the interval a, b? So, it is like a game of dots where it is a linear dot board and the bull's eye is the closed interval a, b and in the first n it rates namely equation 3.7. k n is the number of hits in the interval a, b, number of points in 3.7 that lie in the given interval a, b. Then k n by n is the frequency with which you hit the bull's eye and you take the limit of k n by n as n tends to infinity. It is a asymptotic relative frequency with which these numbers enter the specified interval a, b. And 3.8 tells you that this asymptotic relative frequency is precisely the length of the interval. So, suppose for example, if the length of the interval is one fourth then the last equation 3.8 says that approximately one fourth of the numbers in 3.6 they lie in the closed interval a, b in the long term. Of course, the first few several iterates anything can happen, but eventually in an asymptotic sense roughly one fourth of the elements in 3.6 will appear in the interval a, b if the length of the interval is one fourth. So, now we see that Wiles theorem is a quantification of Kronecker's result. Kronecker's result is a corollary of Wiles theorem and it is a considerable sharpening. We want to prove Wiles theorem following the treatment in this book by Andrew Browder, Mathematical Analysis and Introduction, Springer-Walagh 1996. Now, Kai will denote the characteristic function of the closed interval a, b. Now you take the fractional part of j alpha and whether this fractional part of j alpha lies in the closed interval a, b or not that is decided by the value of the characteristic function Kai evaluated at brace bracket j alpha. If this value is one then the fractional part j alpha has landed in the interval a, b if it is 0 if it landed outside the interval a, b. So, now how do I count the number of times the elements in 3.7 land in the closed interval a, b simply add the values of the characteristic function at these points and that will exactly be K n. K n by n is the arithmetic mean of Kai fractional part of alpha plus Kai fractional part of 2 alpha plus dot dot dot Kai fractional part of n alpha 3.9. So, this suggests that we should look more generally for any given integrable function f of x we should replace the characteristic function Kai by f and we should look instead of 3.9 we should be looking at 3.10 we should be looking at the arithmetic mean of f of fractional part of alpha plus f of fractional part 2 alpha plus dot dot plus f of fractional part n alpha that let us call that expression ln of f. Let us look at theorem 3.5. So, let alpha be an irrational number and let f of x be a bounded Riemann integrable function on 0 1. The averages 3.10 converges point wise to the integral 0 to 1 fx dx. In particular I could take the function f of x to be the characteristic function of the interval a, b then I get exactly what I want then I will get that limit as n tends to infinity 1 upon n the characteristic function evaluated at these fractional parts brace bracket alpha the average is exactly b minus a the integral of the characteristic function of an interval is the length of the interval. So, we get the displayed equation 3.11 that is exactly what we wanted to prove we wanted to prove 3.11 instead of proving 3.11 we will prove this more generally we will replace the characteristic function Kai by a general Riemann integrable function f and we will prove that the right hand side is integral 0 to 1 f of x dx. It may be a little puzzling to you why is it that instead of proving a specific result we simply make life more complicated by replacing the characteristic function by general function we will see that this generalization is easier to prove rather than the specific example. So, let us make two simple observations about this expression ln f 3.10. Obviously ln f is linear ln of c 1 f 1 plus c 2 f 2 is c 1 ln of f 1 plus c 2 ln of f 2 the second important observation is monotonicity f and g are real valued and f less than or equal to g implies ln f less than or equal to ln g these are two important properties that will be heavily used in the proof of theorem 35. Let us as a special case do a simple calculation let us verify the correctness of theorem 3.5 for some innocent looking function f of x what function am I going to take exponential of 2 pi i k x where k is a integer let us do it for this special case f is a one periodic function f s period 1. Now, why have I chosen this particular f x if you think about it if you take the function f of x and evaluate it at these values the brace bracket g alpha what are the brace bracket g alpha it is simply j alpha minus square bracket g alpha square bracket g alpha as an integer and exponential of 2 pi i k times an integer is 1. So, the expression is very simple f evaluated brace bracket g alpha is simply exponential of 2 pi i k j alpha the sum that we are trying to compute is simply 3.12 it is 1 upon n f of brace bracket alpha plus f of brace bracket 2 alpha plus dot dot dot plus f of brace bracket n alpha is simply a finite geometric series you simply get you simply get it from here you see that when you put this value over here you will get a finite geometric series and the finite geometric series can immediately be summed and this is the sum of this finite geometric series and I have the 1 upon n factor 3 1 n and it is very clear that if k is not equal to 0 it is immediate that this goes to 0. So, what we have done is that we have calculated the left hand side of 3.11 instead of the characteristic function I use this exponential function the left hand side happens to be 0. Now we must calculate the right hand side the right hand side is integral 0 to 1 x of 2 pi i k x let us assume that k is not 0 and I immediately get that this is 0. So, the theorem 35 has been verified for the special case when the function f of x is the exponential function e to the power 2 pi i k x we assume k is not equal to 0 what happens when k is 0 when k is 0 the function f is a constant function function f is a constant function these l n's are all 1 and the integral is also 1 and theorem 35 is trivially true when you take k equal to 0. So, we have proved 3.13 and f is the exponential function e to the power 2 pi i k x. So, the theorem has been established for the special case and so by linearity it will be true when I replace f of x by a linear combination of these exponentials x of 2 pi i k x when I take linear combinations of these things what do I get I get trigonometric polynomials. So, the result has been proved for trigonometric polynomials trigonometric polynomials of period 1. So, far we have been working with trigonometric polynomials of period 2 pi how do I pass on from a trigonometric polynomial of period 2 pi to a trigonometric polynomial of period 1. For example, sin x is a 2 pi periodic function, but to make it a 1 periodic function I rescale the variable x. So, instead of looking at sin x I will look at sin 2 pi x instead of looking at cos x I will look at cos 2 pi x. So, passing from a 2 pi periodic function to a 1 periodic function simply means scaling the variable x to 2 pi i x. So, now when I take linear combinations of these I get trigonometric polynomials of period 1. Now, remember that as a corollary to failure's theorem we have established that when f of x is a 2 pi periodic function on the real line and epsilon is greater than 0 then I can find a trigonometric polynomial of I can find a trigonometric polynomial g x of period 2 pi such as supremum mod f x minus g x is less than epsilon. So, I told you now simply replacing x by 2 pi x will convert it into a corresponding theorem for 1 periodic functions. So, now we use this simple idea. Then now we take epsilon greater than 0 b arbitrary then there is a 1 periodic function g x such that sup 0 less than or equal to x less than or equal to 1 f x minus g x less than epsilon by 3 that is 3.14. Now, let us use linearity of ln ln f minus integral 0 to 1 f x dx. I am going to add and subtract a g. So, I am going to write this as ln of f minus g plus ln g minus integral 0 to 1 g x dx plus integral 0 to 1 f of x minus g of x and they are taken the absolute value inside the integral and it becomes less than or equal to. Now, f x minus g x is less than epsilon by 3. So, when I integrate it over 0 to 1 this last piece is clearly going to be less than epsilon by 3. Also, f x minus g x is less than epsilon by 3 means when I evaluate f and g at a bunch of points and take the arithmetic mean that will also be less than epsilon by 3. So, I said because of 3.14 first and the last summands are both less than epsilon by 3. So, now we need to look at the middle summand what are the middle summand mod ln g minus integral 0 to 1 g x dx. Let us focus on this. The added advantage of this is g is one periodic trigonometric polynomial and the result has been proved for one periodic trigonometric polynomial. So, this expression goes to 0 as n tends to infinity. So, there exists an n naught such that this expression is less than epsilon by 3 for n bigger than n naught. And so, all three pieces are less than epsilon by 3 for n bigger than n naught. So, we have proved the theorem. So, this completes the proof for one periodic continuous functions. So, theorem 35 has been established for one periodic continuous functions. But now we wanted for all Riemann integrable functions. So, from continuous functions we must pass on to Riemann integrable functions and that is going to be very easy the moment you do one little exercise. And what is that exercise? That exercise has been described as theorem 36. Suppose f from 0 1 to R is a bounded Riemann integrable function. Then given any epsilon greater than 0 I can find two continuous functions g and h from 0 1 to R such that f of x is sandwiched between g x and h x on the interval 0 1. Second g 0 will be equal to g 1 h of 0 will be equal to h of 1. Third condition integral 0 to 1 the top minus the bottom one integral 0 to 1 h x minus g x will be less than epsilon. So, let us prove this theorem. Let us see how to prove this theorem. It is a exercise, but I will give you some hints as to how to go about doing this exercise. Of course, we are not going to stop with this we are going to extend g and h as one periodic functions that is exactly the reason why we ensure condition 2. Because g 0 equal to g 1 and h 0 equal to h 1 the one periodic extensions of g and h will be continuous. Therefore, I can use the result that we have proved for g and h. So, first let us select a partition of the closed interval 0 1. So, 0 equal to t naught less than t 1 less than t 2 less than t n equal to 1 and let us select this partition. So, that the upper Riemann sum minus the lower Riemann sum is less than epsilon by 3. Capital U f and capital L f are the upper and lower Riemann sums of f with respect to this partition. And as usual when you study Riemann integration we introduce these terms upper Riemann sum lower Riemann sum and we talk about this measure the partition or the norm of the partition and capital M j and little m j are the supremum and infimum of the function on the j th sub interval closed interval t j minus 1 to t j for instance and capital M and little m are the supremum and infimum of the function f on the entire interval 0 to 1. So, that we get this chain of inequalities little m less than or equal to little m j less than or equal to capital M j less than or equal to capital M. Now, we consider the step function. So, what is the upper Riemann sum? The upper Riemann sum is simply the Riemann integral of a step function. So, let us take the step function which assumes the value capital M j on this sub interval t j minus 1 t j. If you integrate the step function h naught you are going to get the upper Riemann sum capital U f. Similarly, integrating another step function I will get the lower Riemann sum. So, what you do is that you start with this step function h naught and we start modifying it appropriately to get the requisite continuous function h. How do we modify it? We are going to choose an eta bigger than 0 and the eta will be specified very soon. What we will do is we will cut we will take this interval t j minus 1 t j and we will cut a small piece from the left and a small piece from the right. So, we will look at t j minus 1 plus eta t j minus eta equation 3.15 and we will define h of 0 and h of 1 to be both M and on this sub interval t j minus 1 plus eta t j minus eta we define h of x to be equal to h naught of x. So, the function h x agrees with the step function over a very large part of the interval 0 1 namely on each sub interval it agrees with h naught except for a small tiny piece that I cut out from both the ends. Now, what we have is that we have a defined a continuous function h on the union of all these sub intervals 3.15 as well as the 2 end points 0 1. Now, how to proceed to proceed from there is very easy we can define the function h of x in whatever way we want on the remaining intervals t j minus eta to t j plus eta. So, I urge you to draw some pictures and see what is going on. So, the interface point is t j and I cut out a piece from the right and a piece from the left. So, on this open interval t j minus eta to t j plus eta the function h of x is not yet defined and I must define the h of x over here also. So, that it becomes continuous you must continuously join it that is all that there is to it and you can do it in whatever way you want as long as the integral is small. The contribution of the integrals from these pieces must be terribly small. So, let us do the following. Let us take the point t j. Let us see the interface point t j and declare the value of the function to be 2 m at this interface point t j. And then now we have to define it on the pieces t j minus eta to t j and t j to t j plus eta simply join them by line segments simply join them by line segments and you will be through and this function h of x will be greater than or equal to f of x. Similarly, you can define the other function g x starting with the lower Riemann sum take the associated step function modify the step function as I explained instead of working with capital M work with little m instead of working with capital M j work with little m j you do the same thing. Now, what you will get is that what is the difference between h x and g x the difference between h x and g x on the interval 0 to 1. When you compute this integral on those pieces on those pieces 3.15 it is going to be capital M j minus little m j times the length of the interval and then what happens is that you get if you add over all the j's the contribution of this integrals from those intervals 3.15 will be in total less than u f minus l f what will be left over we have to now look at the contributions from these open intervals t j minus eta to t j plus eta h x minus g x and I am summing over j, but remember that this difference is going to be this h of x is going to be at most 2 m this g of x is going to be at least twice little m. So, this difference is going to be at most 2 times capital M minus little m and then you are going to add the lengths of these intervals and then you are going to take into account the length of these intervals will be 2 eta. So, all in all this contribution from this sum is less than 4 n eta times capital M minus little m. Now we could have chosen the eta so small at the very outset. So, this whole thing is less than epsilon by 3. So, if you do that then this thing is less than epsilon by 3 this integral from 0 to 1 h x minus g x d x is less than epsilon by 3 and we would have proved the result that we are looking for. So, theorem 36 has been established and this exercise has been worked out except for some details such as drawing some pictures which I urge you to do ok. So, now let us see how this helps in completing the proof of Herman Weyl's Equidistribution Theorem. So, let us take a bounded integrable function f of x and let epsilon greater than 0 be arbitrary. Select g x and h x as in the last theorem. Let us estimate mod ln f minus 0 to 1 f x d x less than or equal to mod ln f minus ln g plus mod ln g minus integral 0 to 1 f x d x. I will add and subtract a g x term. So, less than or equal to mod ln f minus ln g plus mod ln g minus integral 0 to 1 g x d x plus integral 0 to 1 g x minus f x d x. Now that will be less than or equal to mod ln f minus ln g plus mod ln g minus integral 0 to 1 g x d x plus this last piece is going to be less than epsilon because h x minus g x integral that itself is less than epsilon and f x minus g x is non-negative and less than h x minus g x the h minus g and the f minus g the f minus g is less than h minus g and f minus g is non-negative and so the larger piece itself is less than epsilon. So, this will be less than epsilon this last piece will be less than epsilon. And so now there is an n such that the middle term is less than epsilon for n bigger than n 1 why because g is one periodic continuous remember we have proved the theorem for one periodic continuous functions and so that. So, this middle piece will be less than epsilon for n bigger than n naught. So, we get the 2 epsilon plus ln of f minus g now you must use monotonicity of ln f minus g is less than epsilon and so ln of f minus g will also be less than epsilon and we get that the whole thing will be less than 4 epsilon or something and the proof of Weyl's theorem is completed. So, this concludes the proof of Weyl's theorem and I think this is a very good place to stop this capsule. Thank you very much.