 Let us look at how we will be using this or how one can use it. Let us look at some examples. Let us look at fn of x to be equal to x by n, n bigger than or equal to 1. fn is a sequence of functions defined on x to r. Let us specialize this for the particular thing. Let us take x is equal to r. Let us take so that we can choose special things. Let us take r to r. x is taken as the real line. Functions are defined on the real line. For every x fixed, fn of x is convergent. If x is fixed, it is like 1 over n that converges to f of x, which is equal to 0. So, fn converges to f, which is identically 0 point wise. Let us consider, let us take n equal to nk equal to k and x also equal to k, nk the numbers equal to k and the point x in the real line. So, what is f of nk xk? So, that is equal to 1 for every k. It is x by n. So, it is equal to 1 for every n. That does not converge to f of xk, which is identically 0. f is the function, which is identically 0. So, f at each xk is 0. So, that means what? It implies f of nk xk does not converge to f of xk. So, what we have done? We have said for each, there exists a sequence nk. So, I have found nk equal to k, xk equal to k. So, either this quantity is always 1 and this quantity, we know it is 0. So, what is the difference? It is equal to 1. So, there is epsilon equal to 1. So, that implies mod of f of nk xk minus f of xk equal to 1. Hence, the sequence fn x is x by n does not converge uniformly. So, the criteria I gave for not uniform convergence, that means we are able to find a sequence nk of natural numbers and a sequence xk in the domain, say that f of nk xk does not converge to f of xk or the difference always remains bigger. So that is what difference remains bigger than or equal to 1. So, this sequence, so that means, so I have applied this criteria to this sequence. So, this sequence converges point wise but not uniformly. So, let us look at some more examples. Let us look at consider fn of x to be equal to x to the power n for x belonging to minus 1 to 1. So, question is does fn x converge point wise? Does fn x converge point wise to anything? X is between minus 1 and 1. So, mod x is less than or equal to 1. If x is equal to 1, this is the constant function 1. So, it converges to the value 1. If it is between minus 1 and 1 in the open interval, then this number x having mod strictly less than 1, raised to power n, so goes on decreasing. So, converges to, we have seen that converges to 0. So, fn converges, yes, fn x converges to f and what is f? f of x is equal to 0. If mod x is less than 1, it is equal to 1 if x is equal to 1. So, it converges point wise. Now, x to the power n, I can suitably choose x, suitably choose n, so that this value becomes a, I can evaluate that easily. So, what do you think I should choose? So, let us choose that n k and x k. I have to choose these two points, so that if I, let us choose the n k to be equal to k itself. So, it is x to the power k and x, I can choose whatever I like. So, let us choose 1 over 2 to the power k. I have to choose a number between minus 1 and 1. Keep in mind the domain is between minus 1 and 1. So, let me choose this. So, let this be so, this implies what is fn k x k, sorry, not x to the power k, x sequence x k. So, what is this value? That does not help. That is k by, so let me choose slightly differently. So, this is minus, that does not a very good choice. So, let me choose it, I want it to be constant. So, let me choose it to be 2 to the power 1 by k. Will that help or 1 by 2 to the power 1? That was, this point goes out. So, I should not be choosing this point, because this goes out. So, let us choose 1 over 2 raise to the power 1 by k. That is, point is still between minus 1 and 1. Why I am manipulating all this? Because then this comes out to be equal to 1 by 2 for every k. Then this value comes out to be 1 by 2 for every k. And this does not converge to f. This value is not equal to 1. So, f of 1 by 2 to the power 1 by k, which is equal to 0. f is 0. The point wise limit is 0 everywhere, except at the point 1. The value is 1. This is in between. So, implies, so once again that implies, so hence f n does not converge to f uniformly. They are converging point wise, but they are not converging uniformly. So, the question comes, this is one way of testing when something is not happening. Can we give a criteria for something when something happens? So, we want to know, can we give a necessary and sufficient condition, saying that f n converges to f uniformly. So, let us look at that. For functions f n belonging to, so let us look at the space. So, let us consider all functions f x to r f bounded. So, we are going to look at bounded functions. So, what does it mean? So, that is, if I look at mod of f x and look at supremum over x, that is finite. That is what a bounded function means. And we will call this as the bound for that function. I did not think I gave a name for this. Probably I gave. So, let us write bounded x to r. So, space of all or the set of all by space, it is not a space, set of all bounded functions x to r. For every f belonging to b x r, we can define, we had already done it, the L-infinity norm, that is the supremum of mod f x, x belonging to x. This exists. So, call that as, so this has those properties. I am listing again, it is bigger than or equal to 0, equal to 0, if only if f is identically 0. Because if the supremum is mod of f x is 0, that means f x must be 0. So, if and only if, second alpha f is equal to, I am just going to write this repeating the infinity thing that we had done. So, because supremum of mod alpha, mod alpha comes out and the third one, that f plus g is less than plus mod d. That is also obvious, triangle inequality, because the supremum of mod f plus g will be less than or equal to mod f plus d is less than mod f plus mod g. So, supremum will be less than or equal to supremum of those things. So, this gives a metric on b x r. So, norm of f minus g, so that is, that is a metric d infinity f g. So, why all this being done? So, let us assume, suppose f n, f belong to b x r, a root script b x r, a root script b x r, a root script b x r. f n converges to f uniformly. So, assume these are bounded functions and they are converging uniformly. So, that will mean what? So, that is, for every epsilon bigger than 0, there is a n naught depends on epsilon such that mod f n x minus f of x, such that for every x belonging to x, this is less than epsilon for every n bigger than n. Now, if this is happening for every x, then I can take the supremum over x, which exists. So, implies supremum x belonging to x of mod f n x minus f of x is also less than epsilon. For every n bigger than n epsilon, because for every x, something is happening. So, for the supremum, that will also happen. So, what implies is norm of f n minus f goes to 0 as n goes to infinity, saying that what is this quantity? This is same as this, supremum of mod f n minus f, that is norm. So, norm is less than epsilon after some says that means this goes to 0. So, what we have shown is f n converges uniformly. So, hence f n converges to f uniformly implies norm of f n minus f. All for bounded functions goes to 0 as n goes to infinity. So, let us look at, can I say that the converse also holds? So, conversely. So, what we will suppose norm of f n minus f goes to 0 as n goes to infinity. Suppose that happens. So, go back in the, let us way we write epsilon delta if you like, the reason matter actually. So, implies for every epsilon bigger than 0, there is a stage n epsilon such that norm f n minus f is less than epsilon for every n bigger than n epsilon. That is the meaning of saying something goes to 0. What is this quantity? This is nothing but supremum x belonging to x. That obviously, implies for every x belonging to x. If supremum is less than epsilon, then for every term it should be also less than epsilon. So, implies mod of f n x minus f of x is less than epsilon for every n bigger than or equal to n epsilon. So, what is the meaning of that? That means the stage is not depending upon x at all. So, that means, so that is f n converges to f uniformly. So, basically what is saying? If for every x something is small, then the supremum over x also should be small. Consequently, saying if supremum is small, then every term must be small. So, that is the thing used if and only if that is all in both the things. We are not saying anything great. We are saying if supremum over some things is less than epsilon, then each term must be less than epsilon. Conversely, if each term is less than epsilon, then the supremum must be less than epsilon. That is all. Nothing more. We are not saying anything. But what we are saying is interpretation in terms of the metric. So, f n, so the theorem says, so let us write the theorem, f n and f belong to bounded functions x to r. So, let f n converges to f uniformly if and only if norm of f n minus f goes to 0. So, essentially we are looking at sequences in the metric space BXR under the L infinity norm. We are looking at sequences. What is the meaning of convergence of a sequence in the metric space BXR? That is given the name as uniform convergence, which is uniformly. Obviously, the corollary of this f n converges to f uniformly implies f n converges to f point wise, because saying f n converges to f uniformly means the norm of f n minus f goes to 0. And saying point wise is f n x minus f of x absolute value that goes to 0. So, that is one of the terms where supremum is being taken. So, in converse, we have already seen converse we not told. There are point wise, not converse, not true. We have already seen many examples of sequences, which are converging point wise, but not uniformly. Here is another way of interpreting this theorem. So, see we are saying that f n converges to f uniformly means in the metric space BXR under the sup metric, under the L infinity metric, the sequence f n converges to f. And we have already seen that if a sequence converges in a metric space, then it is always Cauchy. Every convergent sequence is Cauchy. Can we say that in this metric space Cauchy also implies convergent in the metric BXR? We want to know whether Cauchy is equivalent to saying a sequence being Cauchy is equivalent to saying it is being convergent. So, let us write that. So, what we start with note f n converges to f uniformly implies the sequence f n is Cauchy in BXR with L infinity metric with metric. Because, the sequence f n converges to f because this is equivalent to saying that norm of f n minus f goes to 0. So, convergence implies Cauchy in any metric space. What is convergence? a n converges to a. If a n is converging to a, a n must come closer to a after some stage. So, if I take any two terms after that stage, they should be close to each other anyway. We have proved that every convergent sequence is Cauchy. In the real line, every Cauchy was also convergent. So, we are using the fact here that every convergent sequence in a metric space is also Cauchy. If you like, you can write down the proof. Because, if it is or if you, let me write the proof here once again so that you feel a bit. So, implies Cauchy. So, here is the proof. So, let us say f n converges to f uniformly. So, that is same as saying for every epsilon bigger than 0, there is a stage n such there is a stage n naught such that norm of f n minus f is less than epsilon for every n bigger than n naught. So, that implies for every n and m bigger than or equal to n naught. Let us look at norm of f n minus f m. For Cauchy, we have to say that two are close, but this is less than or equal to norm of by triangle inequality proof of the norm. So, this is less than f n minus f plus f m minus f. So, that is less than 0. So, if you want to be very nice, you can make it epsilon by 2 and equal to epsilon. So, this is the proof. Every convergent sequence is Cauchy. I am repeating the proof. That is all nothing more than that. Let us prove conversely every. So, converse is also true. So, conversely let f n be a Cauchy sequence in B X R. We want to say it is convergent. Then there exists a function f belonging to B X R such that f n converges to f uniformly. That is same as saying norm of f n minus f converges to 0. So, every Cauchy sequence in B X R is convergent. So, let us prove that. If I want to prove every Cauchy sequence is convergent uniformly, if it is going to be uniform convergence, I know uniform convergence implies point wise. So, first of all, I should be able to say that f n converges to f point wise. Here, the problem is given something is Cauchy, I do not know what is f. What is going to be the limit? I have to locate a function and make a guess, make a conjecture. That also is the limit in the L infinity norm. So, how do I get hold of that? So, the clue is f n converges to f uniformly. If I am able to find such an f, then uniform convergence implies point wise convergence. So, f n should converge to f point wise also, whatever that f may be. So, that gives me the clue that I should try to show that f n is point wise Cauchy. Once it is point wise Cauchy, by the property of real line being complete, it will converge somewhere and that I will call as the function f and then prove f n converges to f uniformly. So, to make a guess, we will look at the known properties. So, the first thing is note for every x belonging to x fix, the sequence f n of x is a Cauchy sequence. So, this happens. That is the Cauchy sequence. For x fix, look at the values. Why? Obviously, because if you like f n of x minus f m of x is less than or equal to norm of f n minus f m. For every x, that is true, because the right hand side is a supremum over all x, left hand side is some x is fix. And if this is going to 0, then this is going to 0. So, that prove that because f n is given to be Cauchy in L infinity norm, that implies point wise Cauchy and hence limit n going to infinity f n x equal to f x. So, I define exists for every x. f n x for every x fix is a Cauchy sequence. So, it must converge. That limit, I call it as f of x. So, for every x that is convergent. So, it has a limit. So, limit is given a name. It depends on x. Limit will depend on x. So, it is a function of x. So, let us call it as f of x for exists for every x. So, now I claim f n converges to this f uniformly. This converges to f uniformly. Now, if I want to prove f n converges to f uniformly, all the f n's are given to be in the space B x r. We do not know where is f, whether f is bounded or not. So, let us look at note. We prove f belongs to B x r, that it is bounded. So, to do that, I have to estimate mod of f of x. I know that f n converges to f of x. So, for that, let us look at f n Cauchy in B x r. We have already seen for every epsilon bigger than 0, there is a stage n naught such that norm of f n minus f m is less than epsilon for every n and m bigger than n naught. So, this is another thing I should recall. We also proved a theorem for sequences that if sequence of real numbers, if it is quotient Cauchy, it must be bounded. So, recall Cauchy sequences are also bounded. So, implies f n is Cauchy. So, it must be bounded. It implies mod of f n supremum over n is equal to f n equal to some number say m is finite. What is the metric? What is the metric? L infinity metric. So, it is a Cauchy sequence. So, it must be bounded. So, this is bounded. Now, that is okay. Now, I can prove. Now, mod of f n x, now, mod minus f m of x is less than or equal to norm of f n minus f m. That is okay. Yeah, good. Note for every x belonging to x, this is true and this I can make it less than or equal to, if it is m, it is 2 times m. If this is true, supremum over n, so, we lesser mod f, norm of f n minus plus norm of f m. So, 2 times. So, this is true for every n and m bigger than something. So, let m go to infinity. Then, you get implying f n of x minus f, because it is converging point wise. f m x converges to f of x point wise. So, this is less than or equal to 2 m for every x belonging to x. Is that okay? That implies that f n minus f belongs to b x r. You can take the supremum. So, implies supremum over x of this quantity is less than 2 m. So, that means f n minus f belongs to x and that implies f also belongs to b x r. So, that is good enough. We can conclude x also belongs. It has to belong actually, because f n is Cauchy and we are trying to prove it is convergent. So, is that okay? Because I was trying to work it out. So, basically this being a Cauchy sequence, it is bounded. So, that means all the norm of f n must be less than some number. So, that is this quantity and that using the fact that f n x minus f m of x is less than norm, because that is the supremum that is remains bounded. So, for every x this is bounded. So, you can let m go to infinity. So, that means f n x this goes to f of x point wise convergence is already there. So, this is less than or equal to 2 m and this is happening for every x. So, I can take the supremum now. So, supremum over x of this quantity means this is finite less than 2 m. So, supremum of this and that means f n minus f belongs to b x r and is a vector space. B x r is a vector space. So, f also belongs to. So, finally, f n converges to f in b x r. Actually, that is already there. So, let us prove of that for every epsilon bigger than 0. There is a stage and not say that f n x is less than or equal to 2 m. So, this same idea repeated again norm of f n minus f m less than epsilon for every n bigger than n naught. So, basically what we want to do is we want to show it is in this. So, let us write for every x belonging to x mod f n x mod minus f m x is less than epsilon for every n bigger than n naught. Now, let m go to infinity for every n and m bigger than n and m coshiness. Let m go to infinity implying for every x mod f n x minus f of x is less than epsilon for every n bigger than n naught. This happening for every x implies norm of f n minus f is less than epsilon for every n bigger than n naught. So, basically in coshiness n and m, let one of the things go to infinity, so that f comes into the picture. At every stage, we are doing that. So, implies f n converges to f in b x r. So, what we are saying is if we are looking at the space of bounded functions, proving something converges uniformly is equivalent to proving that it is coshy in the supremum norm.