 Welcome to lecture number 4 of nonlinear dynamical systems. We had just begun seeing what a Lipschitz function is. So, let us just recapitulate the definition. So, a function f from Rn to Rm is called local Lipschitz at a point x0 in Rn. If there exists a neighborhood b is a neighborhood here in this case is defined to be a ball centered around x0 with radius epsilon and the radius is greater than 0. There should exist a neighborhood and some constant L such that this inequality is satisfied for all points x1 and x2 in that neighborhood of the point x0. So, we are using a ball which is an open ball which means that the distance of every point in that ball is strictly less than epsilon from the center x0. That is why it is called an open ball. So, L is called a Lipschitz constant. Once an L is found we can see that anything larger also can be put in here and that inequality and this inequality will be satisfied with a larger L also. So, in general L depends on both x0 and on epsilon. So, some examples of Lipschitz functions fx is equal to minus 4x is locally Lipschitz at x equal to 3. We will let us see that it is locally Lipschitz everywhere. In fact, it is globally Lipschitz. This is what we will see very soon, but we had just begun drawing a graph. What is the meaning of Lipschitz? For the special case, so here is a function. So, suppose we are interested in checking if the function is Lipschitz at this point locally. To say it is locally Lipschitz at this point means we should be able to find a ball. In this case, it is an open interval such that we take any two points inside this ball and we look at the corresponding values and those values we take here. So, on one side of the inequality fx1 minus fx2 less than or equal to L times x1 minus x2. This inequality just means that if we connect those two points by a line, then the slope of this line should have absolute value at most L. So, can we find a number L such that L puts the upper bound on the absolute value of the slope? Even though this is decreasing, the slope is not positive here, but we look at the absolute value of the slope and that should be bound from above by a number L. So, we will quickly see that a discontinuous function will not satisfy Lipschitz property at the point of discontinuity. At other points, it could satisfy the property of Lipschitz, but at the point of discontinuity, suppose this is a function and at x0 we have this discontinuity. So, if at x0, how much our small neighborhood we take, we are forced to take points both from the left and the right and when we connect these two points from the left and the right, we see that this line could have a slope which becomes more and more vertical. The line connecting these two points becomes more and more vertical. It will become vertical if and only if you take the point x0 and just before it, but you cannot find there does not exist a number L greater than 0 such that fx1-fx2 is less than or equal to L times x1, x2 in which this inequality satisfied for all x1 and x2 in the ball. So, for how much our small epsilon we take, because we are forced to take points both from the left and the right, it turns out that this line connecting the point from the left and connecting the point from the right, this particular line has a slope that is not bounded in absolute value. So, because of that we see that there does not exist an L such that this inequality satisfied for all x1, x2 in the ball. For a particular x1 and x2 we might be able to find the L, but that L will work only for that x1 and x2 in that ball, but we want this inequality to be satisfied for all points x1, x2 inside the ball of radius epsilon strictly greater than 0. So, this is how we see that if it is discontinuous it cannot be locally Lipschitz at the point of discontinuity. At another point x0, suppose this is another point x0, at this point it can very well be locally Lipschitz. As long as in this case as long as we take a ball that does not contain this discontinuity we can see that the function is locally Lipschitz. So, let us see some examples. So, fx equal to minus 4x is locally Lipschitz at x equal to 3. So, take L equal to 4, take the Lipschitz constant L equal to 4 or anything larger. Consider the function fx equal to e to the power 5x. This is locally Lipschitz at point say x equal to 4. For this particular point we can take the Lipschitz constant L equal to 5 times e to the power 20 plus 1. So, notice that we are taking the slope of the function evaluated at x equal to 4 and we take something that is slightly greater. How much greater we take decides on how big the open ball around the point x equal to 4 is. But since we are interested in just an open ball of radius greater than 0 we can take the Lipschitz constant slightly more than the slope at the point x equal to 4. Consider the function fx equal to e to the power x. This is locally Lipschitz at every point x naught. Whichever point x naught we take we are able to find the Lipschitz constant L that will work for all points inside a suitable ball. The unit step is locally Lipschitz at every point x naught except the point x equal to 0. So, what do we mean by locally what do we mean by the unit step? This is like our step function. This is equal to 0 up to here then suddenly it jumps to the point 1. So, this particular function up to x equal to 0 it is equal to 0. For x greater than 0 it has jumped to 1 and this is locally Lipschitz at every point except x equal to 0. At x equal to 0 we saw that there is this discontinuity and hence there will not be a constant L that will work for that inequality for all points in a ball. How much ever small the ball may be? So, in other words the unit step is locally Lipschitz at every point x naught except the point x naught. This particular function fx is equal to x to the power 1 by 3 is locally Lipschitz at every x again except x equal to 0. This is a very important example. We will see this again and again. So, let us just draw the graph of fx is equal to x to the power 1 by 3. This particular graph we see that this is how the graph looks. So, we see that at x equal to 0 this particular curve becomes vertical almost vertical. Why? Because if we try to evaluate d by dx of f of x we get 1 by 3 times x to the power minus 2 by 3 which is nothing but 1 over 3 times x to the power 2 by 3. So, we see that as x tends to 0 this quantity becomes unbounded and hence we see that the slope is not bounded about the point 0. So, how much ever small neighborhood we take about the point 0? When we connect two points we see that it could have a slope that is unbounded. We are not able to find the number L such that the inequality in the Lipschitz condition definition that inequality will be satisfied for all points inside that ball such a number L we will not be able to find. That is why we will say this particular function is not locally Lipschitz at the point x equal to 0. At another point say here we are able to find a ball such that there will be a number L. In other words whatever ball we take as long as we do not include the point 0 when we connect this we can take the farthest and the nearest we can see where the slope is maximum and we can choose the number L accordingly that number L will work for everything. We have to see where the absolute value of the slope is maximum and we take a Lipschitz constant that is equal to that or more and that will work for all points in that ball. Hence we see that as long as we do not take the point x equal to 0 how much of a close we take if it is away from x equal to 0 we can find a ball such that there is a number L satisfying the Lipschitz inequality and hence except the point x equal to 0 there is a number L that will satisfy the Lipschitz inequality. Hence it is local Lipschitz at every point except the point x equal to 0. Some more examples so the same function fx is equal to x to the power 1 by 3 is not local Lipschitz at x equal to 0. So we could just conclude that at a particular point x0 in Rn if f is differentiable then it is indeed locally Lipschitz at that point x0. If it is locally Lipschitz at a point x0 it implies that it is continuous locally Lipschitz at a point x0 means it is continuous at that point x0. Conversely if the function f is continuous at a point x0 it does not imply that it is locally Lipschitz at that point x0 and if the function f is locally Lipschitz at the point x0 it does not mean that it is differentiable at the point x0. So we can see some examples about this just to see why differentiability is not assured by locally Lipschitz property. We will see that locally Lipschitz only requires that the slope is bounded slope between any two points if we take the unit ramp so it is continuous now. So we see that the derivative does not exist at this point because the left hand line limit of the derivative and the right hand limit of the derivative are not equal to each other hence at the point 0 the function f is not differentiable. But we see that since the slope is bounded we take any two points we connect them by this line the slope is bounded. So I will just draw a bigger figure take this particular example at the point let us say 4 we have drawn a figure such that the graph of f versus x is continuous but it is not differentiable at this point because the left hand limit of the derivative and the right hand limit of the derivative are not equal to each other. However we take any two points and we connect it we see that this line is guaranteed to have a slope that is less in absolute value than the slope of this line. So we could take any two points here maybe on the same side and the line slope in absolute value is an upper bound for the absolute value of the slopes of any two points close to the number 4. In other words we can find a ball here such that we can find the number L that will satisfy the Lipschitz inequality for all the points in that ball. In other words here is an example f that is not differentiable at x equal to 4 but locally Lipschitz at x equal to 4. In other words if somebody tells us that this particular function f is locally Lipschitz at the point x equal to 4 and the one asks us does it imply that f is differentiable at the point x equal to 4? The answer is no. So this is what the statement here says. So if the function f is differentiable it does imply that it is locally Lipschitz at the point x0. If it is locally Lipschitz at the point x0 it is also continuous at the point x0 but if it is continuous at the point x0 it does not imply that it is locally Lipschitz at x0. We saw x to the power 1 by 3 as an important example and if it is locally Lipschitz at the point x0 it does not imply that it is differentiable at the point x0. For that we saw an example now. So we will just spend one slide on the difference between locally Lipschitz and globally Lipschitz. So for this purpose we need to see to what extent the number L depends on x0 and epsilon. So consider a domain D a subset of Rn and f a map from D to Rn. So f need not be defined on the whole of Rn it is defined on a domain D. So a domain is an open and connected subset of Rn in this case. So to say that D is a domain in Rn it means D is a open and connected subset of D open connected subset of Rn. So for a function f that is a map from D to Rm there are various possibilities. f could be locally Lipschitz at a point x0 in D. We saw the definition for locally Lipschitz at a point. We could also have a situation where at every point in D f is locally Lipschitz at that point. This we will say is we will use a word f is locally Lipschitz on D. So what is the significance here? L might have to be modified depending on the point x0. If x0 is another point the slope absolute value of the slope might be larger because of which L might have to be made larger. So at each point in D f is locally Lipschitz but the Lipschitz constant L is having to be modified depending on the point x0. Perhaps L has to be modified. We will say f is locally Lipschitz on D. If it is the case that f is locally Lipschitz on D and the Lipschitz constant is independent of the point x0 in D then we will say f is Lipschitz on D. The word locally is no longer relevant since we can find one Lipschitz constant L that works for the entire domain D. Finally, when the domain D is a whole of Rn if f is Lipschitz on Rn that is there is a constant L that works for every point x0 in Rn then we will say f is globally Lipschitz. So examples of globally Lipschitz functions are sin x cos x a constant A times x. The constant function itself and the 0 function it is possible to decide for each of these functions a Lipschitz constant L that works for the entire domain of these functions. The entire domain in this case is whole of R. Examples of functions which are locally Lipschitz on R but not globally Lipschitz are x square e to the power x e to the power minus x in fact any polynomial of degree 2 or more. So these all examples are locally Lipschitz on R but not globally Lipschitz why because this x square its slope could become very large depending on the point and hence there is no one number L that works on the whole of R. Similarly, e to the power x and e to the power minus x can have slopes which are very large in absolute value. We finally come to the existence and uniqueness theorem for solution to a differential equation. So consider x0 is equal to f of x consider this differential equation where f is a map from Rn to Rn and consider the point x0 in Rn. Assume f is locally Lipschitz at x0 then there is a delta greater than 0 such that there is a unique solution x of t to the differential equation x dot is equal to f of x with the initial condition x0 equal to x0. So such a solution such a unique solution exists for the time interval t belonging to 0 to delta. So there is some number delta that is strictly positive because of which there is this interval of time 0 to delta for which we have a solution and moreover the solution is unique that is what it says with the initial condition x0 is equal to the point x0 where the function f was assumed to be locally Lipschitz. So it turns out that existence and uniqueness of a solution is being guaranteed for only an interval of time. It is possible that this interval of time is very small but it is guaranteed to be a non-zero interval of time. It is not just one point but it is an interval 0 to delta. We can ask the question is locally Lipschitz important after all we have spent analyzing the significance of Lipschitz locally Lipschitz its relation to differentiability and continuity. We could ask is this locally Lipschitz property really crucial for the existence and uniqueness of a solution to the differential equation. For this purpose we will see that the differential equation x dot is equal to x to the power 1 by 3 has at least two solutions. Why? Because x to the power 1 by 3 turned out to not be locally Lipschitz and hence uniqueness was not guaranteed by this theorem and we will see that there are two solutions. So what are these solutions? This is what we will see now. So consider this differential equation x dot is equal to x to the power 1 by 3. So we can rewrite this as x to the power minus 1 by 3 dx equal to dt and upon integrating both sides we see that we get this equality where c1 is some constant and rewriting these terms x to the power 2 by 3 is now equal to 2t by 3 plus c0. So c0 and c1 are related by the constant 2 by 3. For convenience we have renamed c1 times 2 by 3 as c0. Then upon taking suitable powers we see that x of t x of t is equal to 2t by 3 plus c0 raise to the power 3 by 2. So we see that this is also a solution to the differential equation and how do we calculate c0? We can ask a question. Suppose at t equal to 0 the solution, the differential equation was at 0, x satisfied x0 is equal to 0 at t equal to 0. By substituting this we can get c0. So we see that x of t equal to 2t by 3 to the power 3 by 2 is a solution to the differential equation. Which differential equation? x dot is equal to x to the power 1 by 3 with the initial condition x0 is equal to 0. But is this the only solution? No, because if x dot equal to 0 at x equal to 0 then we also know that x of t is also a solution of the differential equation. So we see that if x is equal to 0 at t equal to 0 then x dot is equal to 0. Why is x dot equal to 0? Because we put x equal to 0 here and cube root of 0 is nothing but 0 and hence x dot is equal to 0. And then we conclude that x is equal to 0 for all time t. This is one solution to the differential equation. But we see that x t is equal to 2t by 3 whole to the power 3 by 2 is also a solution to the differential equation with the same initial condition. So for the same initial condition we see that we have two solutions to the differential equation. Let us just draw a graph of x versus time. So here is this x that was 0 until t equal to 0 and from here it is growing as 2 by 3 to the power 3 by 2 times t to the power 3 by 2. This is a graph of x versus time t and in addition the differential equation also has equivalently equal to 0 as a solution to the differential equation. In other words at this point t equal to 0 there is this solution that comes out of x equal to 0 and it also continues at x equal to 0 as has another solution. At this point the vector field we see that it is pointed, sorry this is something we have to plot like this. So the graph of x to the power 1 by 3 is like this. We see that of course there is this instability. But at x equal to 0 itself the arrow had length 0 because the graph of f crossed the x axis at x equal to 0 and hence if it is at the point x equal to 0 then it continues to be at the point x equal to 0. This is what our vector field diagram told us. But here we see that in addition to continuing to be at 0 there is also this possibility that it comes out. It emanates out of the equilibrium point without requiring a perturbation. While this figure says that this equilibrium point is an unstable equilibrium point we see that upon perturbation there are points that are trajectories that are going away from the equilibrium point. But here is an example because of this non-locally Lipschitz property at the point x equal to 0. Without a perturbation also we see that there is a solution that emanates out of the equilibrium point. Thus making us ask what exactly is the definition of an equilibrium point when there are solutions that can emanate out of the equilibrium point even without requiring a perturbation. For these purposes we will from now on eventually assume that a function f when studying a differential equation satisfies locally Lipschitz at every point x0. But this is an example where without locally Lipschitz property it turns out that there can be non-uniqueness of solutions to the differential equation. So we see that the solution x is differentiable. We were able to get explicitly the solution x as a function of time. We can also differentiate this as a function of time and see that it solves this differential equation. f of x is continuous but it is not locally Lipschitz at x equal to 0. That is why the previous theorem would not guarantee existence of solutions, existence and uniqueness of solutions and here is an example where uniqueness fails. So because of the importance of this result of existence and uniqueness of solutions to differential equation under locally Lipschitz property we will because of its importance we will see the proof. So the outline of the proof will be as follows. We will define an operator p that takes one estimate of the solution trajectory and gives a better estimate of the solution. So this operator p takes a trajectory which is a solution to the differential equation on the interval 0 to delta and it gives a better estimate of the solution. So this is going to be called Picard's iteration because of Picard's work in this area. So p xn is an estimate of the solution trajectory at the nth iteration. We will define p such that the desired solution will satisfy px equal to x. In other words x will be a so called fixed point. Why do we call it fixed? Because the desired solution, the solution to the differential equation takes x and gives back the same x. In other words while it takes different xn and gives back xn plus 1, possibly xn plus 1 is different from xn. While this is possible in general, we will construct p such that the desired solution x will satisfy px equal to x. In other words p fixes x. So the Lipschitz condition on f will help to prove convergence of this iteration to a unique fixed point. But this fixed point would be unique provided we are looking for a so called complete space. These are some things that we will define precisely. So for this purpose we will use a so called Banach fixed point theorem. So please note that the point in this context is a trajectory x of t for the interval 0 to delta. Possibly the interval delta, the interval 0 to delta is a very small interval. In other words delta is only slightly more than 0 but it is positive which means that the interval is not just a point but it is the interval of time of length delta. So how will we define this Picard's iterates? So define the operator p that takes a continuous function x of t and gives another continuous function y is equal to p of x. How will we define it? px is again a function of time, px at any time t is defined as x0 plus the integral from 0 to t of f of x tau d tau, where t belongs to the interval 0 to delta. So t comes in here and this integral is being added to 0. So we see that for this t equal to 0, this particular trajectory is also equal to x0. So these are different different functions of time that start from x0. So what is the significance of this operator p with our solution to the with our differential equation? We see that xt is a solution to the differential equation d by dt of x is equal to f of x with this initial condition x0 is equal to x0 if and only if xt is a solution to the integral equation to this integral equation x of t equal to x0 plus integral from 0 to t f of x tau d tau. So by differentiating this right hand side, this is something we can quickly check, y is solution to this integral equation also solution to the differential equation. We see that x appears on both sides, here is x and x is also inside this integral. So d by dt of x is equal to, derivative of x is equal to, derivative of this is equal to 0 and since t appears only here, this is nothing but f evaluated at the end point. So this is what our differential equation was. In other words, a solution to the integral equation is exactly a solution to this differential equation also. When we integrate this on both sides, we see that x of t also satisfies this initial condition plus integral from 0 to t of the right hand side. In other words, solution to the differential equation, when integrating both sides this differential equation, we obtain exactly the integral equation. Thus solution to the integral equation is a solution to the differential equation and solution to the differential equation is also a solution to the integral equation. However, we see that here also x appears on both sides and here also x appears on both sides and it is not clear that going from a differential equation to the integral equation is genuinely an improvement. We will see that obtaining an integral equation allows us to use Picard's iteration. So the latter, what is important is that when x satisfies this integral equation that x is also a fixed point of this operator P. Why? Because the right hand side is nothing but P of x and what x we have put in here is exactly what is also here. That is why this particular integral equation says that P of x equal to x. So we will first see the special case and f is independent of x. Maybe f is allowed to depend on time. Only here we will assume that f is a function of time explicitly and there is no dependence on x. So consider this differential equation d by dt of x is equal to f of t with this initial condition x0. So there is no dependence of f on x. Then we are able to integrate this x of t we can then define as x0 plus this integral from 0 to t of f of tau d tau. So, there is no guess or iteration required to define this. Such a definition x will always satisfy this differential equation. But when f depends on x, then we are not able to define this. Why? Because x appears both on the right hand side and left hand side of this integral equation. So the word define can no longer be used here. So only a carefully chosen x will satisfy this equation and that is indeed the solution to our integral equation. So what we will do now is we will take x1 as some function of time. It turns out that within a small neighborhood of the actual solution which solution we take will not matter. So we will take x1 of t equivalently equal to x0. So we will take the function x1 which is always equal to x0 value. x0 is a point in Rn which corresponds to our initial condition. There is one particular function which is always equal to x0 for all time t that we will take as our initial x1. Then we will define x2 as x0 plus this integral with f evaluated at x1 instead of x2. So now since we have x1 here which we know and x2 which we do not know, we are allowed to use x2 defined by this right hand side. In other words, x2 is equal to p times x1. So we can similarly define x3 as p times x2 and in general xn plus 1 as p times xn. So the question arises will this converge? Will this converge to a solution? In other words will it converge to a fixed point of the operator p? So the answer is it will converge for a carefully constructed delta that will be strictly greater than 0 and for ensuring that this delta is strictly greater than 0 and that it exists we will be using the locally lipschitz property of f at the point x0. So this is the hard and remaining part of the proof which we will proceed and do now. So please note that x0 is a point in Rn. This is the initial condition for the differential equation d by dt of x is equal to f of x. On the other hand, x1, x2 up to xn are continuous functions of time and these are iterates of the operator p. These are no longer points in Rn but they are functions which take their values in Rn for different time instance. So for this proof we need some preliminaries. So for that purpose we will need to use the Banach fixed point theorem and these are some preliminaries for that purpose. So we need the notion of a normed vector space. So vector space v is called normed f. To each vector v in the vector space V there is a real valued function which we will call as the norm and we will denote it as the norm of v by this notation such that this norm is always greater than or equal to 0. So the norm of a vector v will be a real number and it cannot be negative that is what it says. Moreover, so this is greater than or equal to 0 for all v and also it is equal to 0 only for the vector v equal to 0. This is the first condition required from the norm function. Another condition that is required is if we scale the vector v by a constant alpha then the norm also gets scaled by that same number alpha, the same number alpha if alpha is positive and absolute value of alpha in general. So for any real number alpha and a vector v this equality has to be satisfied and finally the triangular inequality is required to be satisfied. So for two vectors v1 and v2 the norm of v1 plus v2 cannot be more than norm of v1 plus norm of v2. This triangular inequality is also required to be satisfied for all vectors v1 and v2. This is called triangular inequality. This is called the linearity property of the norm function and this is the definition that the notion of length of a vector is required to satisfy. In the context of convergence we also need what is the definition of a Cauchy sequence. So in the context of convergence we like to say that elements in a sequence are going close to each other. So in general if elements in a sequence a n go close to each other, do they converge to some element a? So the answer is in general no. Just because elements a n are converging close to each other does not mean they will eventually converge to a number a. So if it does converge then we will call the sequence convergent. In order to arrive at that we will define this property called Cauchy. So a sequence a n is called Cauchy if for every epsilon greater than 0 there exists some capital N possibly depending on epsilon such that if we want that a n and a m are smaller than epsilon then that will be satisfied for all n and m that are greater than this capital N. The importance of this statement is it says that elements a m and a n are going close to each other. If you want them to be close to each other closer than a number epsilon then all we have to do is take m and n greater than some capital N. So when somebody specifies the epsilon greater than 0 we are able to find number n such that this inequality is satisfied for all n and m greater than capital N. So in general if epsilon is made smaller then capital N might have to be made larger. So this quantifies, this makes precise the notion that elements in the sequence a n are going closer and closer to each other. So every convergent sequence is Cauchy but it turns out that the converse is not true. So if a sequence is Cauchy it just means that elements are going close to each other but it does not imply that there is a number a to which it converges. So if we assume an important property called completeness then we will also be able to guarantee convergence. So what is the definition of completeness? So if a sequence is Cauchy then completeness would guarantee that there exists a number a such that a n converges to a and this latter convergence is what we will say convergent in the usual sense. So this is what we will use for the definition of complete of a normed vector space. A normed vector space v is called complete if every Cauchy sequence is convergent. Convergent in the sense of this limit. So we see that if every Cauchy sequence in other words a sequence in which all elements are going closer and closer to each other if we are going to say that it converges to a number in that particular set in that particular vector space v then this vector space v is going to be called complete. So what is a complete normed vector space? It is a normed vector space which is also complete. In other words every Cauchy convergent sequence is also convergent. Some examples of complete spaces. The set of real numbers are this is by definition of R this is complete. Rn is also complete where n is some finite number. So if we take n tuples of real numbers then this is also a complete vector space. So which norm here it turns out is not relevant. We could take for example the standard Euclidean norm. We could take the Euclidean norm which is very standard. Suppose a vector v is an Rn then we can define. This is one of the definitions v1 square plus v2 square plus v2 square plus vn square. This guarantees that it is positive and when we take square root it is also linear in v. So this is one of the norms of Rn. What is Rn? It is a n tuple. Every element in Rn is an n tuple in which n is finite. So this is one definition of norm and this is called the two norm. It is also called the Euclidean norm. We could also have taken v so called the infinity norm as the maximum for i equal to 1 up to n of the absolute value of the ith component. v1 up to vn there are n components in the vector v. For each of those vectors we take the absolute value and for each of the components we take the absolute value and we look at the maximum for i varying from 1 to n of this absolute value. That is called the infinity norm. So we will see some more notions of norm. So with respect to any of the norms it turns out that Rn is complete. What are examples of incomplete spaces? So the set q of rational numbers. This is incomplete. What is rational? q is a standard notation for the set of rational numbers. Any number q, small q in this capital Q can be written as the ratio of two integers z1 by z2. So if a number q can be written as ratio of two integers then q is said to be a rational number and capital Q is a set of all such rational numbers. So it turns out that q is not complete. For example, pi and square root of 2 these are some numbers which can be approximated arbitrarily well using rational numbers. To be able to approximate arbitrarily well for example we know that square root of 2 can have a decimal expansion. If we take more number of digits in the decimal expansion then we get a more accurate representation of the number square root of 2 and similarly for pi. But just because we can approximate square root of 2 and pi by a very good approximation using decimal expansion that does not imply that there are integers z1 and z2 such that z1 by z2 is equal to either pi or square root of 2. In fact, they do not exist z1 and z2 such that z1 by z2 is equal to pi. Similarly they do not exist integers z1 and z2 such that z1 by z2 is equal to square root of 2. In other words pi and square root of 2 are so called irrational numbers. However we know that these irrational numbers can be approximated arbitrarily well. This proves that q is not complete. Another important set, another important space that is complete is the so called space of continuous functions on this interval 0 to delta where delta is strictly greater than 0 with respect to the sup norm. With respect to the sup norm it turns out that this space of continuous functions is complete. So the space of continuous functions on this interval functions from this interval to Rn is also denoted by this notation. So the 0 here means that it is just continuous, differentiability is not being guaranteed. So I should emphasize that it is important that with respect to the sup norm only c0 this one is complete. There could be some other norms with respect to which it is complete but here it is no longer the case that the norm is not relevant. Which norm we are taking that will very crucially decide whether this space of continuous functions is complete or not. That statement depends on the norm with respect to which we are asking the question. The space of continuous functions from the interval 0 to delta to Rn is also not these are functions which are not necessarily differentiable but they are just continuous and hence 0 has been put here. They take their domain is from 0 to delta, it takes its values, time varies from 0 to delta and for each time instant in this interval it gives a vector in Rn. So this is the space of continuous functions from this interval to Rn and we are saying sup norm. What is the sup norm? If f is an element of this set then the sup norm we are going to define as the maximum then p varies from 0 to delta of f of t. So at each time instant f of t is an element of Rn. This is the meaning of the term f is an element of continuous functions from this interval to Rn. So at each time instant t this is some vector and for that vector we defined already the Euclidean norm, the two norm and this two norm is defined for each time t in this interval and we will look at the maximum of this particular function as t varies from 0 to delta. This maximum is said to be the sup norm of the function f. So sup norm is no longer relevant to a particular time instant even though f of t to norm was relevant to which time instant the two norm was being taken. So the sup norm is a norm on the space of continuous functions from this interval to Rn. So with respect to this sup norm it turns out that this is a complete space but with respect to which norm is it not complete for example we can also define the so called L2 norm. So what is the L2 norm? For the same space of functions we can define f L2 norm as integral from 0 to delta of, so we can take the two norm of the vector f of t at a time instant t and the two norm we take the square and we integrate from 0 to delta upon integrating from 0 to delta we get some number that is no longer dependent on the time t and its square root is said to be the L2 norm of the function f where f is an element of this. So this L2 norm also is no longer dependent on t but f of t before we took the two norm here is indeed a function of time t and after we integrate from 0 to delta this is no longer dependent on time t. So this is a L2 norm, so with respect to with respect to L2 norm this same space of functions f of t is the same space of is not complete just because elements are going very close to each other does not mean that it will also converge to a function that is continuous and close to each other for that we are using the notion of L2 norm. So with respect to the sup norm however two functions if they are going close to each other they will eventually converge to a continuous function again inside the space. So finally using these notions we can prove we can state the Banach fixed point theorem so for that we just need the notion of a closed subset and a contractive mapping. So for a set x a subset s a subset s of x is said to be a closed subset of x if the boundary of s is within x this is how we understand the closed subset more precisely s is a closed subset of x if and only if elements of x which are arbitrarily close to s are also within s. If we take an element of x which is very close to elements of s which is close to one or more elements of s then it is not necessary that this elements of x should also be within s but if it is indeed within s then s is said to be a closed subset of a set x and what is contractor? So if x is a normed vector space a map p from x to x is said to be contractor if there exists a number rho that is strictly less than 1 such that this inequality is satisfied for all x1, x2 in x what is this inequality? p x1 minus p x2 is at most rho times x1 minus x2. So what is the importance of this inequality? x1 minus x2 can be interpreted as a vector from x2 to x1 and p x2 to p x1 is a vector upon action under p. So this vector is getting contracted under the action of p. Why contraction? Because this rho is some numbers strictly less than 1. So we will say this map is contractive if there exists such a number rho such that this inequality is satisfied for all x1 and x2 in x. The number rho should not have to be modified depending on which x1 and x2 we take in capital x. For such a map p we will say it is contractive if such a number rho exists. So we are ready to state the Banach fixed point theorem using the notion we have just now defined. So let x be a Banach space and let s be a closed subset of x. Then let t a map from s to s let t be contractive. Then first of all there exists a unique fixed point x star in s fixed point for this operator p. Moreover that fixed point x star can be found quite easily. What is easily? It can be found by successively operating t on any x1 in s. We take any x1 then we take t times x1 then we take t times t of x1. When we do this iteration we will converge to that unique fixed point. We will use this Banach fixed point theorem for the proof of the existence and uniqueness theorem of the solution to the differential equation in the next lecture. Thank you.