 Welcome back to the basic concepts of real function theory. In my previous lecture, I have given the properties of convergence of sequence of functions. We dealt with two types of convergence, one is point wise convergence of sequence of functions and also uniform convergence of sequence of functions. And further in the proof of existence and uniqueness theorem, we deal with the series of real functions. Now, we will discuss the uniform convergence of series of real functions. So, consider the infinite series summation n goes from 1 to infinity u n x. So, an infinite series of real functions, real functions u n's, each of which is defined on some real interval say a b. Now, to talk about the convergence of the series, we first form a sequence of partial sums of the series. So, consider the sequence of partial sum. So, consider the sequence of partial sums of the series, call it f 1 is u 1, f 2 is sum of u 1 and u 2 and so on, f n is u 1 plus u 2 plus etcetera plus u n. So, that is summation i goes from 1 to n u i. Now, we define definition. So, let this be definition 1. The infinite series, the infinite series summation u n, n goes from 1 to infinity is said to converge uniformly to a function say f on a b. If it is a sequence of partial sums, partial sums which we denoted by f n converges uniformly to f on the interval a b. Now, to make sure that an infinite series converges uniformly to a function f, we have the following theorem which is known as a Weistras M test. So, theorem Weistras M test. So, let m n be a sequence of positive constants such that the series of this positive constants M n, n goes from 1 to infinity converges to some number. Now, let the series of functions u n, n goes from 1 to infinity be a series of functions such that the absolute value of u x is less than or equal to absolute value of u n x is less than or equal to m n for all x in the interval a b and 4 and 4 or n is equal to 1, 2, 3, etcetera. Then the conclusion is then Weistras M test says then the series of functions then the series u n, n goes from 1 to infinity converges uniformly on the interval a b. So, each term of the series is bounded by a constant and if the series formed by that constant is a converging convergent series, then the series of functions converges uniformly on the interval a b. So, that is a test. We will see an example verifying this fact. So, example say example 1. So, consider a series that is summation n goes from 1 to infinity sin n x divided by n square and this is defined on the interval 0 less than or equal to x less than or equal to 1. So now, the sequence of numbers m n which is 1 by n square the sequence of numbers is convergent. So, we know that summation 1 by n square n goes from 1 to infinity is a convergent series. So, if you consider u n x which is by definition sin n x divided by n square and so, therefore if I take absolute value of this, this is less than or equal to 1 by n square as sin n x is less than or equal to 1 is bounded by 1 on the interval 0 to 1 and this is also true for n is equal to 1, 2, 3, etcetera. So, therefore by applying Weistras m test with the m n is equal to 1 by n square and m n n goes from 1 to infinity is finite. We conclude that this series of function sin n x divided by n square for n goes from 1 to infinity the series of function converges uniformly on the given interval 0, 1. Now, another important tool which we will be using in the existence and uniqueness theorem is Arcella-Askoli theorem. Arcella-Askoli theorem says that if you have a sequence of functions which are uniformly bounded and equicontinuous, then that sequence has a convergent subsequence. So, let me introduce what is uniformly bounded and equicontinuous sequence. So, uniformly bounded, uniformly bounded sequences and equicontinuous, uniformly bounded and equicontinuous sequence of functions. So, my definition is, say definition 2, a sequence of functions, a sequence of functions say f n defined on an interval a b is said to be uniformly bounded. A sequence of function f n defined on a b is said to be uniformly bounded if there exist a positive constant, if there exist a constant called it m greater than 0 such that f n of x, the absolute value of f n of x is less than or equal to m for all x in the interval a b and for all n is equal to 1, 2, 3, etcetera. So, there exist a constant m which is independent of n. So, uniformly there is a bound for each of the function in the sequence, then we say that the sequence of function is uniformly bounded. An example of this, example say 2, so consider a sequence of function of functions f n defined by say f n x is equal to sin n x, where x is varying in the interval 0 to 2 pi and n is equal to 1, 2, 3, etcetera, sequence of functions. So, as absolute value of f n x which is the absolute value of sin n x which is less than or equal to 1 on the interval 0 to 2 pi and for all n is equal to 1, 2, 3. So, we conclude f n x is equal to sin n x is uniformly bounded. So, this is uniformly on the given interval 0 to 2 pi. Now, we define what is known as a eke continuity of a sequence of function. So, definition say 3, a sequence of functions, a sequence of functions f n defined on a sequence of function f n defined on an interval a b defined on an interval a b is said to be eke continuous on the interval a b if, so a sequence of functions f n defined on an interval a b is said to be eke continuous on and on the same interval a b. So, it is said to be eke continuous on the interval a b if for every epsilon, every epsilon greater than 0, there exist a delta. The delta is a function of epsilon for every epsilon greater than 0, there exist a delta, that delta depends on epsilon, but is independent of n. So, does not matter from which function n it is coming. So, independent of n such that the absolute value of f n x minus f n y is less than epsilon, whenever absolute value of x minus y is less than delta. See, a sequence of function f n defined on an interval a b is said to be eke continuous if for every epsilon greater than 0, there exist a delta. So, we should be able to find a delta which is independent of n does not matter from which n which function f n it is coming. So, such that f n x minus f n y is less than epsilon whenever x minus y is less than this delta. So, the eke continuity is a continuity in the uniform sense for all n. So, note that the number delta is independent of the choice of function from the sequence. So, theorem is known as Arcella Ascoli theorem. So, it states that let f n be a sequence of functions in the set of all continuous functions defined on the interval a b. So, let f n be a sequence of functions in the space C a b, C a b is the space of all continuous functions defined on a b. And if f n the sequence f n is uniformly bounded, if the sequence f n is uniformly bounded and eke continuous on the interval a b, then the conclusion of the theorem is then there is a uniformly convergent, uniformly convergent subsequence called it f n k for f n. If f n is a sequence which are functions from the space C a b and if f n is a uniformly bounded and eke continuous set of functions on a b, then there is a uniformly convergent subsequence f n k for the original sequence f n. So, this result is known as Arcella Ascoli theorem, which we will be using for proving the Exchance theorem for the initial value problem. For the proof, one may refer to any standard book on analysis or it is also given in Codingdon and Levinson, refer Codingdon and Levinson, Codingdon and Levinson's book on theory of differential equations. Now, we move on to next topic, which will be used in proving the uniqueness of solution of initial value problem, namely the Lipchitz continuity. So, Lipchitz continuity of functions. So, initially for the sake of simplicity, we will define Lipchitz continuity for a one variable function. So, definition, so this is definition 4 in this lecture, a function f from R to R, a real valued function from the real from the domain as a real space is said to be Lipchitz is said to be Lipchitz continuous, said to be Lipchitz continuous also is known as f satisfies a Lipchitz condition, we refer to this notion as Lipchitz continuity. A function f from R to R is said to be Lipchitz continuous on a subset, on a subset say D, which is a subset of R, if so if there exist a constant for it alpha strictly positive, such that a function f from R to R is said to be Lipchitz continuous on a subset D of R, if there exist a constant alpha greater than 0, such that the absolute value of f of x minus f of y is less than or equal to alpha times x minus y for all x y in the subset of in the subset D of R. So, in this case, so we now call the constant alpha as the Lipchitz constant of f. In fact, if alpha is a number satisfying this inequality, then any number larger than that will also satisfy this inequality. We take alpha the least upper bound of all such alphas and so that alpha we call as a Lipchitz constant. The least number which is the smallest number which is satisfying this inequality is known as a Lipchitz constant of f. If in case D is the all real line, so we here assume the definition that D is a subset of R, in case D is all real line, then we say that f is Lipchitz globally, so then f is the globally Lipchitz continuous, otherwise f is locally Lipchitz. So, if the condition is true, we satisfied in the old space, then the Lipchitz continuity is a global Lipchitz continuity or if it is restricted to a subset of the real line, then it is a local Lipchitz continuity. Now, we take a few examples showing the Lipchitz continuity. So, let us take examples, so example 4. So, consider a function f from R to R defined by f of x is equal to 2 x plus 3. So, this function f of x is equal to 2 x plus 3, obviously this function is not a linear function, we check the Lipchitz continuity of this function f of x minus f of y is 2 x plus 3 minus 2 y plus 3. So, if we take the absolute value, so this is less than or equal to 2 times or equal to 2 times x minus y. So, this implies that it is true for all x and y in R. So, f x is equal to 2 x plus 3 is globally Lipchitz with Lipchitz constant alpha is given by 2. Now, consider another example, so example 5, so define or consider a function, consider a function f from R to R defined by f of x is equal to f of x is equal to x square. So, f of x is equal to x square. Now, to check the Lipchitz continuity f of x minus f of y which is equal to x square minus y square which is equal to x plus y into x minus y. So, the absolute value of f of x minus f of y is equal to x plus y into x minus y. Now, this quantity, consider this quantity absolute value of x plus y, this is not a bounded quantity if x and y are allowed to vary in the ender real line. So, but if x and y are varying in a bounded set, then this modulus absolute value of x plus y is a bounded quantity. So, if x and y are varying in a bounded set, say x is less than or equal to a and y is less than or equal to b. In that case f of x minus f of y is less than or equal to, so this value x plus y could be alpha times x minus y, where alpha is a bound for x plus y, where x and y are bounded by these two constants. So, therefore, alpha is a finite quantity. Therefore, f x is equal to x square is locally Lipschitz. So, if this is locally Lipschitz, so if x and y are varying from if for example, if x is if f is defined from a set minus a to a, f x is equal to x square. Then f x is equal to x square is Lipschitz with alpha is equal to 2 a, where x can take maximum value up to a and x varies from minus a to a and y is also varying from minus a to a. So, therefore, absolute value of x plus y that can take a maximum up to a plus a that is 2 a. So, therefore, the Lipschitz constant is 2 a. So, therefore, f x is equal to x square is locally Lipschitz on a domain say d, which is set of all x such that x is less than or equal to a with Lipschitz constant alpha is equal to 2 a. Now, one can provide sufficient condition to ensure that a function is Lipschitz. See, sufficient condition to ensure that a function is Lipschitz, sufficient condition guarantee Lipschitz continuity. So, state in the form of a theorem. So, theorem 3 suppose that f is a function from d, which is a subset of r and mapping to r is differentiable on d the supremum of the bound of the derivative when x varies on d is say alpha is a finite quantity. Then the conclusion is then the function f is Lipschitz continuous on d with Lipschitz constant alpha Lipschitz constant alpha. The proof is very simple just by using the mean value theorem by the mean value theorem f of x minus f of y is equal to the derivative of the function evaluated at some point psi times x minus y, where psi is a point lies between lies between x and y. So, therefore, this implies that the absolute value of f x minus f y, which is less than or equal to soup of f prime psi into x minus y and this quantity is our alpha by hypothesis this is a this soup exist and this bounded by alpha. So, therefore, this is less than or equal to alpha times x minus y for all x y in the domain d and if for d happens to be the all real line then the Lipschitz continuity we obtain is a global Lipschitz continuity. So, global Lipschitz continuity can also be given in terms of the bound of the derivative. If the derivative of a function is a slope of a function is bounded globally then the function is Lipschitz continuous globally. If the slope is or the derivative is bounded on a bounded set then the function is Lipschitz continuous on that bounded set and we note that the condition in this theorem is just a sufficient. So, we note that so condition in the theorem is just sufficient, but not necessary for Lipschitz continuity. So, that means we can produce example where a function is Lipschitz continuous at the same time the conditions of the theorem is violated. So, example say 6. So, we take a function f x is equal to mod x and f is defined from r to r. So, obviously f x minus f y is mod x minus mod y if we take the absolute value f of x minus f of y which is absolute value of mod x minus mod y. So, it can be shown easily that this is less than or equal to absolute value of x minus y for all x y in r. So, what does it say? So, this implies that f x is equal to mod x is Lipschitz continuous with the Lipschitz constant alpha as 1. However, f does not satisfy the condition of theorem 3. The reason is f is not differentiable at 0 to verify the condition stated in the theorem. If you take another example, so example 7. So, consider the function, consider the function f x is equal to sin x and to verify whether this function is Lipschitz or not f x minus f y which is sin x minus sin y and if I want to take and I want to show if it is if f is Lipschitz then there exists a constant alpha such that it is less than or equal to that constant alpha times absolute value of x minus y. So, how to show this? It is not that straightforward provided if you use some trigonometric identities you may be able to, but if you just apply the theorem 3, the sufficient condition for Lipschitz continuity that ensures that the function f x is equal to sin x is Lipschitz continuous. So, how? Because f prime x is cos x and it is bounded. So, cos x is bounded by 1 for all x. So, therefore, by theorem 3 there is a sufficient condition condition for Lipschitz continuity implies f x is equal to sin x is Lipschitz continuous with Lipschitz constant alpha is equal to 1. So, it is a very good test. If you just recall the function which we considered f x is equal to x square the derivative is 2 x f prime x is 2 x and 2 x is not bounded globally, but whenever x is bounded 2 x is bounded. So, f prime x is bounded. So, whenever x is bounded in a bounded domain f x is equal to x square is Lipschitz or that is locally Lipschitz. If you look at the function f x is equal to sin x, f x is equal to sin x is globally Lipschitz f x is equal to x square is not globally Lipschitz, but it is locally Lipschitz. So, not globally Lipschitz. So, because of this, so this says that is locally Lipschitz. Now, we require this Lipschitz condition for functions of 2 variables that we will deal with in the next session. So, therefore, in this session we have seen, we have analyzed the uniform convergence of a series of functions and by using the Weisrass M test, we can make sure, we can test whether a given series of function is convergent uniformly or not. And also, we have defined what is a equicontinuous function and uniformly bounded functions of a sequence of functions. And finally, we stated the Arsola-Ascoli theorem which says that any every uniformly bounded and equicontinuous function defined on a bounded interval a, b has a convergent subsequence. We will deal with Lipschitz continuity for functions of 2 variables in the next session. Bye. .