 So, let us start looking at point wise convergence of sequences of functions. So, let us take a sequence fn, a sequence of functions. Each fn is defined on x2. So, we say that fn converges point wise to a function fx to r, if fnx converges to f of x for every x belonging to x. So, at every point, you look at the value of the function fn that gives a sequence of real numbers and that converges to f of x. So, that is point wise. So, that is same as saying in terms of definition of convergence of sequences that is for every epsilon bigger than 0. There exists some stage n which will depend on epsilon as well as on x such that fnx minus f of x is less than epsilon for every n bigger than this number n. So, this stage may depend upon epsilon and it may depend on the point. So, for different points the stage may be different. So, we had looked at some examples last time. So, we looked at one example was I gave you the graph of the function 0 to 1. So, here is 1 by n and you look at the graph to be say you can start at 0 itself does not matter where you start. So, you can go up to 1. So, this is a point 1. If you like, you can go to minus 1 does not matter. So, this is the graph of the function. So, if that is your fn then this fn converges to f and what is that f? f of x is equal to 1 if x is bigger than or equal to. So, if you are including 1 over n, then bigger than or equal to 0 and is equal to 0 if x is less than 0. So, that meant that fn converges to fx for every x. Each fn is continuous, but f is not continuous. We looked at some other examples also. So, second example we looked at was I think fnx equal to sin of nx divided by n, n bigger than or equal to 1. Then fnx converges to f of x which is equal to 0, identically 0 for every x. So, converges point wise to the function because sin is always bounded and third, but this example, this was an example particularly saying that if I look at fn dash of x exists for every n, but fn dash of x is not convergent. I think I should need to change that for example. If you want not convergent, let us put it square root of n here. Then the derivative will be square root of n into cos nx which will not be convergent. I am just repeating the examples that we had done last time. So, that essentially says that fn differentiable need not imply f differentiable. So, let me give you one more example of… Let us look at… I gave one example last time. Let me give another one here. Let us look at… Let us look at rationals between 0 and 1. Let… What shall I write? So, let us A be the set of x belonging to 0, 1, x rational. So, look at all the rational points between 0 and 1. A is a countable set. So, A is a countable set. So, let us write A to be equal to some R 1, R 2 and so on. So, give a enumeration of the rationals. It is a countable set. So, call it R 1, R 2, R n. We are not saying in the increasing or decreasing or anything. Just enumeration of rationals. So, define fn to be equal to the indicator function of the set R 1, R 2, R n. So, what does that mean? What is the meaning of the indicator function? If you recall, that meant that fn of x is equal to 1, if x is equal to R 1, R 2, R up to R n and is 0, otherwise. Of course, x is between. So, we are looking at this in the interval as a function 0, 1, to R. So, look at… This is a subset of 0, 1. So, given a subset A, which is a countable set, rationals in 0, 1 define fn to be the indicator function of the rationals R 1 up to R n in that enumeration. Then each fn is integrable, because each fn is a function which has only discontinuities at the points R 1 up to R n. Everywhere else is a constant function, 0. Only at these points the value is 1. So, it has only jump discontinuities at this finite number of points. So, it is a finite number of discontinuities. So, f is integrable. Where does fn x converge to? What is f of x? It is nothing but the… It will be 1 if x is any one of the rationals R 1 up to R n, otherwise it will be 0. So, it is the indicator function of… So, let me write it as 0. It is 1 if x is equal to R n and 0 otherwise. So, it is the indicator function of… So, it is the indicator function of the set A. A is the set of all rationals in 0 and 1. So, what does the indicator function of rational mean? It is 1 at all rationals in 0, 1 and 0 at all irrational points. It is discontinuous everywhere. If you look at the upper sums or lower sums in respect to any partition, the upper sum will be equal to 1. The lower sum will be equal to 0. So, this function is not f. Each fn is integrable. f is not integrable. So, this very simple way of looking at convergence of sequence is namely at every point, look at the sequence fn x and look at the limit if it exists that function. So, then you say fn converges point wise. So, what these examples illustrate that point wise convergence does not preserve continuity, does not preserve differentiability and does not preserve integrability. So, it is a reasonably bad way of looking at convergence of sequences. So, we like to define a notion of convergence which is reasonably well behaved. So, that is called uniform convergence. So, let fn x to r be a sequence of functions. We should say n bigger than or equal to 1. We say the sequence fn converges uniformly to a function f on x to r. If you want something stronger than point wise convergence, it says if for every epsilon bigger than 0, there exist some stage n which depends only on epsilon such that for every x belonging to x, fn x minus f of x is less than epsilon for every n bigger than n epsilon. So, what we are saying is convergence of fn x at any point brings it closer to f of x and that closeness is not affected by the point x. So, same stage works for all the points. So, for every x, this is true. Let us look at some examples. Actually, we have one example ready here namely that fn x which was defined as sin n x divided by square root of n, say x belonging to r. So, we want to claim that fn converges to f which is identically 0 uniformly. So, for that we only have to note that if I take fn x minus f of x, so what is that? f is identically 0. So, it is absolute value of sin n x divided by square root of n which is less than or equal to 1 over square root of n. So, irrespective where is the point x, the distance between fn x and f of x is always less than 1 over square root n. So, that implies given we can choose say a stage n epsilon such that fn x minus f of x is less than or equal to or is less than epsilon for every n bigger than n epsilon because we can choose. So, given epsilon n is such 1 over square root n is less than epsilon. That is all you have to choose n large enough so that 1 over square root of n is less than epsilon. So, then this will imply that this is true. So, fn converges to f uniformly. There are many examples of functions which converge uniformly. Let us look at, we will give more examples later on. Let us just note one thing. So, note what is the meaning of saying fn does not converge uniformly. All functions are x to r. So, fn is x to r and f is from x to r. Saying that fn does not converge uniformly to f means what? So once again I am trying to bring your attention to the point that if you want to understand something is true, you should also understand when something is not true. Equally important to understand both ways. So, let us write once again fn converges to f uniformly meant was equivalent to saying for every epsilon bigger than 0, for every x belonging to x, there exists a stage n epsilon such that fn x minus f of x is less than epsilon for every n bigger than n naught. So, what is fn not converging to f uniformly? So, that is equivalent to saying so for every we should change it to, so there exists epsilon bigger than 0, so that for every x the statement is not true. And what is statement that is not true? There is a stage after which something is small that should not happen. That means after every stage I am able to find a point where this thing goes back. So, that means there exists a sequence of point n k and there is a sequence n k of natural numbers and a sequence x k of points in x such that mod of fn k, the places where the things are going bad. So, that is same as n k x k minus f of x k is bigger than or equal to epsilon. For every x the same stage works. So, it does not work for every x the same stage does not work. That means there are points x k where this thing does not work. For every k there is a point where this does not work. That means there is a stage n k say that f of n k minus f of x k is not less than epsilon, it is bigger than or equal to epsilon. For every x it does not happen. So, there is a sequence x k for which things go bad. And what goes bad? This is going bad. That means for every n I am able to find some stage after which the things go bad. So, that stage is for every k n k. So, f of n k x k minus f of x k is not less than epsilon. So, two things are going bad. Something was happening for every epsilon that is taken care of for every x and for every f. When things go bad that means there should be points where things are going bad. So, that is the point x k. And for x k what is going bad? f n of x k minus f of x k is not less than epsilon. For what n? At least every given any stage I can find something after which. So, that is same as finding a sub sequence x n k n k. So, this is what it means saying that the sequence does not converge uniformly. So, this gives us a variable of testing when is something not converging uniformly. For example, we will take ordinary sequences of numbers saying that a sequence a n converges to a. That means what? a n comes closer to a after some stage. If a n is not converging to a that means what? That means there is a epsilon. Say that whatever stage you give me there is a something stage after which. So, there is a sub sequence a n k. So, a sequence not converging to a means there is at least one sub sequence it is not converge. But here this was happening for every point also. So, there is a sequence of points where things are also going bad.