 Welcome to the new module on this course on differential equations. We will also be concerned in this course to do some theoretical aspects of differential equations and we will state and prove the existence and uniqueness theorem. For that purpose, we need to get familiarized with some of the concepts, basic concepts of real function theory. In this lecture, we will discuss about convergence of sequence of functions, two types of convergence we deal with namely the point wise convergence and uniform convergence, some concepts of real function theory. So, first we deal with point wise convergence of sequence of functions and also we will do uniform convergence of sequence of functions. Before we start the point wise convergence and uniform convergence of sequence of functions, let us start with the convergence of sequence of real numbers. So, definition, say I call it definition 1, a sequence of real numbers. Let us denote this by x n, we set to converge to the limit, call it x 0. If given epsilon greater than 0, a positive number, there exist a positive number denoted by n, such that absolute value of x n minus x 0 is less than epsilon for all n greater than this number capital N. So, a sequence of real numbers x n is to converge to a limit x 0, if this happens that for all n, small n greater than capital N, the difference between x n and x 0 that is made less than epsilon and we denote this by limit x n as n goes to infinity is equal to x 0. So, let us consider an example, we will consider an example of sequence of real numbers. So, example 1, let us consider a sequence of real numbers given by x n is equal to n by n plus 1, where n goes from 1, 2, 3, etcetera. And then we will see that limit of this sequence, limit n goes to infinity x n, which is by definition limit n goes to infinity n by n plus 1 is equal to 1. So, that is x n is given by n by n plus 1 and x 0 the limit is 1. So, we will see how this happens. So, 4, so given epsilon greater than 0, so what we are looking for? We are looking for x n minus x 0, which in our case n by n plus 1 minus x 0 is 1. We are looking for the situation that this difference is less than epsilon and we know that for this less than epsilon for a large capital N. So, we need to find a capital N such that for all n greater than that n, this difference is less than epsilon. Let us consider this n by n plus 1 minus 1, this quantity just by simplifying it is since n is positive this is 1 by n plus 1. So, we want this to be less than epsilon and a simple manipulation shows that 1 by epsilon is less than n plus 1 or 1 by epsilon minus 1 less than n. So, therefore, this happens 1 by n plus 1 is less than epsilon if n is greater than strictly greater than 1 by epsilon minus 1. So, what is the moral of the story is that given epsilon greater than 0, there exist an n capital N, which in our case we could compute this as 1 by epsilon minus 1 such that x n minus x 0. So, which in our case it is 1 by n plus 1 is less than epsilon for all strictly greater than n, this capital N is 1 by epsilon minus 1. So, hence the conclusion is the sequence x n is equal to n by n plus 1 converges to a limit 1. So, this implies that x n is equal to n upon n plus 1 converges to 1 as n goes to infinity. So, this is concerning a sequence of real numbers. Now, we will consider a sequence of functions. So, now convergence of sequence of functions. So, sequence of functions consider a sequence of functions denoted by f n such that each f n is a function from an interval a, b to r. So, revalued function defined on the interval a, b and we will now discuss about the convergence of this sequence of functions. When do we say that this sequence of functions converges to some limit function. Now, let x be any fixed real number in the interval a, b. Now, consider the sequence of real numbers comes to the sequence which is value of the functions f n evaluated at x. So, now this is a sequence of real numbers. Now, consider the sequence f n x of real numbers. Now, this sequence of real numbers for every fixed x is to converge to a limit call it f x. If we can find a capital for every epsilon greater than 0, we can find a capital N such that the difference between f n x and f x is less than epsilon for all n greater than capital N. So, now this f n x the sequence of real numbers is said to converge to a limit say f x for a fixed x in the interval a, b. If for every epsilon greater than 0, there exists a number n such that this difference f n x minus f x the absolute value of f n x minus f x is less than epsilon for all n greater than n. So, this sequence of functions when it is evaluated at a point x it turns out to be a sequence of real numbers. So, therefore, the convergence of sequence of functions at a point turns out to be the convergence of a sequence of real numbers. So, this sequence is said to be said to converge to a point a limit f x if this happens and if this happens for every x in the interval a, b. So, if this convergence happens at every x in the interval a, b then we say that the sequence of functions f n converges to the function f x point wise on a, b. So, this defines a point wise convergence of a sequence of functions defined on some interval a, b. Now, we give an example of a sequence of functions converging to a function point wise. So, example say called example 2. So, consider a sequence of functions f n sequence of function f n defined y. So, we consider a sequence of functions f n defined by f n x is equal to n x divided by n x plus 1 and for all x between 0 and 1 and here n goes from 1, 2, 3, etc. This is a sequence of functions infinite sequence of functions defined by n x by n x plus 1. Here a, b is interval 0, 1. So, a, b in our definition is 0, 1. Now, if look at the few terms of the sequence say for example, the first three terms when n is equal to 1, 2 and 3 say f 1 x is x upon x plus 1 and f 2 x is 2 x upon 2 x plus 1 and f 3 x is 3 x by 3 x plus 1 and so on. If you evaluate this sequence of functions at x is equal to 0 say let x is equal to 0 which is a point inside our interval, then f 1 at 0, f 2 at 0, f 3 at 0 and so on. We get the sequence f n at 0. So, which is a sequence of 0's. So, obviously at x is equal to 0, this is values the 0 sequence that converges to 0. So, limit n goes to infinity f n at 0, this goes to 0, 0 is a limit when x is equal to 0. Now, we consider for the case when x is a non-zero, but value in the interval. Therefore, every x which is strictly greater than 0 and less than or equal to 1, the sequence is the same f n x is n x by n x plus 1. If you look at the graph of the sequence, so this is x and this is y is equal to f x, the function y is equal to f n x and if you look at the graph of the sequence say this point is 1 and you will see that the graph may be f 1 x will have a form of this type is say f 5 x and if look at say f 50 x. So, this will be f 50 x and we can show it that limit n goes to infinity for x between strictly greater than 0 and less than or equal to 1, n goes to infinity n x by n x plus 1, this is equal to 1. So, this goes to 1 and 1 is the constant function given here. That can be seen just by simplifying this n x by n x plus 1, which if I take n common n of x plus 1 by n and as n goes to infinity 1 by n goes to 0 and n n n can be cancelled and you get this is equal to 1 as n goes to infinity. Therefore, for the values of x strictly greater than 0 and less than or equal to 1, the sequence of function f n x that converges to 1 as the constant function 1. So, in short we write the sequence of function f n. So, this sequence of function converges to a function f x. So, which is 0 when x is equal to 0 and which is 1 when x is strictly greater than 0 and less than or equal to 1. So, this sequence of functions in the example converges point wise to a function f x and the limit function. Look at the limit function, it is a discontinuous function. Also, if you look at each function each f n x in the sequence, each f n x is a continuous function on the interval 0 1 and this sequence of function converges to a function f x. The limiting function is not a continuous function which has a discontinuity at 0. So, one thing we observe that even if we have a sequence of continuous functions which converges to a function f x point wise is not necessary that the limit function f x will be continuous on the given interval. But in the existence uniqueness theorem, we want to have a situation where the limit function also must be continuous. So, what is a guarantee or what makes it sure or ensure that if you have a sequence of continuous functions and that converges to a function f x, what is a guarantee that the limit function is also continuous. That is a stronger requirement that is ensured if we assume with some different kind of convergence rather than point wise convergence. So, point wise convergence does not guarantee that the limit function is continuous even if each individual function in the sequence itself is continuous. So, we to guarantee that the limit function is also continuous. We need a stronger notion of convergence which is known as a uniform convergence. So, we now introduce what we call as uniform convergence of sequence of functions. So, uniform convergence of sequence of functions. So, let us recast the point wise convergence once again in a different way of putting. So, definition as we say definition 2 bar which is another way of defining the point wise convergence point wise convergence we have already defined. Let f n be a sequence of real functions. Let f n be a sequence of real functions each of which is defined on an interval called it a b. Now, the sequence of function f n is said to converge point wise to a function f on a b given epsilon greater than 0 for each x in the interval a b. There exist a positive number called it n. Now, this n is going to depend on epsilon and also on x. For each x and for a given epsilon if there exist a positive number n that depends upon both epsilon and x such that the absolute value of f n x minus f x is less than epsilon for all n greater than this n. This n is basically that depends upon epsilon and x. So, observe that in general the number n depends not only on epsilon it also depends on the point x. So, for a given epsilon and for the numbers n epsilon will be different for different points x and if you can find a single n which is available for all x in the interval a b then we say that the convergence is uniform convergence. So, uniform convergence is a case where for a given epsilon greater than 0 there exist a positive number n that number n will work for all x in the interval a b then the point wise convergence stand out to be uniform convergence. So, we will define write as a definition. So, call it definition 3. Let f n be a sequence of real functions each of which is defined on a given interval a b then the sequence f n is said to be or is said to converge uniformly to a function f on a b given epsilon greater than 0 there exist a positive number n and now in this case n depends only on epsilon such that the absolute value of f n x minus f x is less than epsilon for all n greater than this capital N and for every x in the interval a b. So, here the capital N that positive number capital N depends only on epsilon not on x n depends only on epsilon not on x. So, that is true for all x in the interval a b. So, geometrically what does it mean? So, this means that so given epsilon greater than 0 the graphs of y is equal to f n x for n greater than capital N the graphs of y is equal to f n x for n greater than n lies between the graphs between the graphs of y is equal to f x plus epsilon and y is equal to f x minus epsilon. So, that is if you have the interval for this is the limit function f x say this is f x plus epsilon and this is f x minus epsilon then uniform convergence ensures that for all small n greater than the capital N the function f n lies f n x lies between f x plus epsilon and f x minus epsilon. So, n greater than capital N. So, we will see an example of uniform convergence. So, an example so this is a third example. So, consider the sequence of functions f n defined by. So, we consider a sequence of functions defined by f n x is equal to n x square divided by n x plus 1 where x is between 0 and 1 and n is equal to 1, 2, 3 etcetera. So, obviously, for x is equal to 0 when I fix x is equal to 0 f n 0 is a 0 sequence and that converges to 0 n goes to infinity f n 0 is 0. So, at x is equal to 0 the given sequence of function converges to 0 point wise and also for x strictly greater than 0 and less than or equal to 1 f n x which is given by n x square by n x plus 1 and we can show it easily by a manipulation n x square by n x plus 1 which is n x square by n we take common x plus 1 by n and this is equal to. So, n and n get cancelled and as n goes to infinity denominator becomes x and the numerator is x square. So, x square by x that is x. So, this sequence this converges to a function f x which is equal to x on x if it is true including 0. So, therefore, it converges point wise to the function f x is equal to x. So, graphically we can see that the graph of these functions for this is 1 is y is 1 then f x is equal to x is a limit function and you will see that for f 1 looks like a continuous function f and f may be f 5 and as n go is becoming larger and larger it is clustering towards the function f x is equal to x. So, what is a conclusion the conclusion is the sequence f n x which is n x square by n x plus 1 this converges to a function f x which is equal to x on the interval 0 1 and one thing we can observe that this time the limit function is also continuous the limit function f x which is equal to x on 0 1 is continuous and observe that each of the function in the sequence f n x is continuous on 0 1. So, each function in the sequence is continuous and this sequence converges to a function f x which is also continuous. So, this time we got a better result that the limit of the sequence of continuous functions is also continuous and this happened because the convergence in this situation is uniform convergence we will show that the convergence is uniform. So, how do we show so for a given epsilon greater than 0 we should be able to find an n which is independent of x such that f n the difference between f n x and x is less than epsilon for all small n greater than that capital n which is independent of x. So, we will show for this example. So, f n x minus f x by definition of the sequence this is n x square by n x plus 1 minus the limit function is x just doing the algebraic manipulation you see that this is nothing but x upon n x plus 1 and what we want is we want. So, given epsilon greater than 0 we are looking for x upon n x plus 1 is less than epsilon for all n greater than a given capital n which is independent of x. So, we are looking for this one yes simple manipulation if we apply here this one this can also be written as x upon epsilon is less than n x plus 1 and if you divide throughout by x then 1 by epsilon is less than n plus 1 by x. So, in other words 1 by epsilon minus 1 by x is less than n or if n is greater than 1 by epsilon minus 1 by x and 1 by x that there you can have a bound that n is greater than 1 by epsilon. So, x varies in the interval 0 1. So, the greatest value it can assume is 1. So, 1 by epsilon minus 1. So, therefore, it is obvious that this difference f n x minus f x which is x by n x plus 1 and which is less than epsilon for all n strictly greater than n that n is 1 by epsilon minus 1. So, this n is independent of the point x. So, that is true this n will work for all x. So, hence the convergence is uniform and we have a uniform convergence of sequence of functions and it converges to f and also we got an additional property that the limit function is also continuous when each individual function the sequence is continuous. So, we state an important theorem in this regard when we have uniform convergence. So, theorem 1 let f n be a sequence of let f n be a sequence of functions which converges uniformly to a function f x on an interval a b and suppose that each function f n x is continuous on a b. So, this sequence converges uniformly to a function f x on a b and we assume that each function f n is continuous on a b then the conclusion is then the limit function then the limit function f x. So, limit function f is continuous. So, this is an interesting result that guarantees that if a sequence of function converges to a function and each of the function the sequence is continuous then the limit function is also continuous provided the convergence is uniform. So, therefore, a sequence of continuous functions converging uniformly to a function f then the function limit function f itself is continuous. In the existence and uniqueness theorem we will also use the idea of interchange of limit and integration of sequences of functions. So, we now see a result that deals with interchange of limit and integration. So, interchange limit integration of sequence of functions. So, we state a theorem without proof. Proof can be seen in any one of the standard text on real analysis or calculus. So, theorem called it theorem 2. So, let f n be a sequence of functions f n b a sequence of functions defined on some interval a b. So, assume that f n converges uniformly to a function f on a b. So, f n is a sequence of functions on a b and that converges uniformly to a function f on a b. Also, suppose that each function f n is continuous on a b each function is assumed to be continuous on a b then the conclusion is. So, then for every alpha and beta such that a is less than or equal to alpha strictly less than beta less than or equal to b limit n goes to infinity integral alpha to beta f n x d x is equal to limit integral alpha to beta can take the limit inside n goes to infinity f n x d x which is nothing but integral alpha to beta f x d x. So, what does it say if you have a sequence of functions that converges uniformly and each of the function in the sequence is continuous then you can interchange the limit and the integration. So, you can take the limit inside the integral and which is same as you know taking the limit of the integral or integrating the limit of the function. So, we will see one example in this regard. So, example 4. So, consider a sequence. So, functions which we have already considered. So, f n x which is equal to n x square by n x plus 1 the interval x is less than equal to 1 and less than equal to 0 n is equal to 1 2 3 etcetera. So, we have already seen that this sequence of functions converges uniformly to a function limit function f x is equal to x. So, limit n goes to infinity f n x is equal to f x which is x we have seen from the previous example for x between 0 and 1 and this convergence is uniform convergence. So, f n converges to f uniformly and also f n is continuous for n is equal to 1 2 on the interval 0 1. Now, let us see what will be the value of the limit. So, limit n goes to infinity integral 0 to 1 f n x d x which is equal to theorem says that this can be the limit can be taken inside. So, limit integral 0 to 1 limit f n x d x which is integral 0 to 1 the limit is f x d x that limit is f x is x. So, which is 1 by 2 now if you take the LHS. So, if you take the left hand side and treat it separately. So, LHS take the limit n goes to infinity integral 0 to 1 f n x d x which is limit n goes to infinity integral 0 to 1 n x square by n x plus 1 d x which is equal to limit n goes to infinity integral 0 to 1. If you do simple algebraic operation if you divide this polynomial by the denominator this can be shown that this is nothing but x minus 1 by n plus 1 by n into n x plus 1 d x. So, this is equal to if you do the integration. So, limit n goes to infinity. So, if you integrate it you get x square by 2 minus integral of 1 by n is x by n and integral of 1 by n into n x plus 1 is natural logarithm of n x plus 1 divided by n square and if you evaluate the integral at 0 and 1 you get this is equal to this is equal to limit n goes to infinity 1 by 2 minus 1 by n into ln of n plus 1 divided by n square is plus. And now if you take the limit that limit happens to be so this is equal to 1 by 2 and show that this is equal to 1 by 2. So, this limit is equal to half. So, the limit the LHS is 1 by 2 minus 1 by n plus ln of n plus 1 by n square which is half. So, the RHS is equal to LHS is equal to RHS. So, therefore, this verifies that as limit n goes to infinity integral 0 to 1 f n x d x is equal to integral 0 to 1 limit n goes to infinity f n x d x which is of course, is equal to integral 0 to 1 f x d x. So, conclusion of the theorem the conclusion is that the theorem is verified. So, in this lecture what we have seen is one say sequence of functions are given there are two types of convergence as we discussed. One is a point wise convergence and other one is the uniform convergence. The advantage of uniform convergence is that it preserves the continuity of the terms of the sequence to the limit to the limit function. If each of f n is continuous and f n converges uniformly to f then the limit function f itself is continuous on the given interval a b.