 Hello, welcome to module 2 of NPTEL NOC on introductory course on point set topology, part 2. So, today we will take the topic of differentiation on Banach spaces. So, many things run parallel to what we will do in calculus of variable, one variable calculus or two variable calculus and so on. Actually, you will see that it is copied one variable calculus, but the things have to be put in a proper perspective. So, start with two Banach spaces. One can possibly do many things with just normative spaces also, but we will concentrate only on Banach spaces. Because our idea of this presentation of these things is not to do the entire thing in a very general setup, but to cover the implicit function theorem and inverse function theorem for Banach spaces as a sample, as a sample of application of topological methods. So, that is why we will concentrate only on Banach spaces. Take a subset u contained inside v open subset around a point x naught inside v is a vector space, v is a non-linear space. So, there is open subset system makes sense because we are using the norm to induce a metric to induce a topology. Start with an open subset u around a point. A function defined on this open set into another Banach space is said to be differentiable at a point x naught. If there is a continuous linear map from t, t from v to w such that this x is this limit x is equal to 0. What is this? It is f of x naught plus z minus f x naught divided by the norm h. I cannot divide by h. So, h is a vector inside this normed in space v, okay. Therefore, I cannot do just this limit, but I have to do it by subtracting t h where t is a linear map which is going to be the derivative. Then I can divide by norm h. This limit as h tends to 0 which is same thing as saying norm h tends to 0, this limit must exist. So, where is the limit existing? What is this one? f and you know f is taking values into w. t of t is also from v to w. So, t of h is also inside w. So, all this numerator is taking value inside w. It is a vector inside w, okay. So, divided by norm h, of course, that is just a scalar. So, it is a vector inside w. This limit must exist. So, the important point here is that we must have a continuously linear map t. This is the part of the definition, okay. I want to caution you that there are slightly varying definitions, you know weaker or stronger and so on. There are different definitions possible and then they will make under this condition, this will be equal to that one, that will be equal to this one and so on. We are not going to study all that in this course, okay. So, let us take this definition namely there must be a linear map which is bounded, that is continuously linear map from t to v which satisfies this equation. Limit of f of x0 plus h minus fx minus th whole thing divided by norm h, this limit must be 0, okay. As soon as such a t exists, okay, it has to be unique. You cannot have two different linear maps t1 and t2 having satisfying the same property. This is an easy consequence of just similar to what we do in a real analysis or any other different, you know, multivariable calculus and so on, okay. Uniqueness is not a difficult thing. So, that unity is called the derivative of f at the point x0 and I am using the same notation df of x0, the standard notation. The only thing is you might not have called such a thing as a fresh derivative, we are going to call it as fresh derivative, okay. This fresh A O started with Banach space calculus, okay, fresh derivative of f at the point x0. If f is differentiable at each point x inside an open set, then we say f is differentiable on the open set tube. Further, if the function which assigns to each point u its derivative, remember derivative is a bounded linear map, right, t is bounded from v to w. So, it is taking values inside bvw. If this function for each x here inside u, you take dfx, that will be denoted by df. If this itself is continuous, then we say that f is continuously differentiable on u or we can say it is a class C1, okay. There are ways of making class C2 definition C3 and so on. So, we will stop here, only class C1 here, okay. So, if you go back, the very first thing you do is the so-called increment theorem for differentiable functions or function which has a derivative at a single point. Rewriting this equation, you know, as what we said, clearing the denominator, reinterpret it. That is called increment theorem. Same thing is true here also, namely f of x0 plus h minus fx0 minus t of h0 is equal to this norm h as one on the other side. I am writing that as the remainder r of x0 depending upon h, it depends upon x0 as well as h, okay. So, you can write it as f of x0 plus h is fx0 plus dfx0 h plus some error term. This is called the increment theorem on the first approximation, linear approximation to f, okay. If we increase the value of x0 by h, the increment is roughly dfx0 of h, you can ignore this one. You can ignore the last term, that is the whole idea. Why should you ignore this one or why? I mean what makes you ignore this one? It is not always possible but because of this definition, what we have is if you divide this by h, then take the limit, it is 0. So, remainder after the first term here has the property divide by norm h and take the limit, it is 0. So, this is called increment theorem, okay, exactly as in the case of one-variable calculus here, okay. The following statements are all easy to check exactly as in the case of one-variable calculus, okay. Every constant function is differentiable everywhere, everywhere means what on the whole of v, on the whole wherever they are different, okay. And its derivatives are always 0, the derivative at all the points is 0 for a constant function. For every vector v belonging to v, the translation function, see on a vector space, you have this u going to u plus v, where v is fixed, there is a translation function, I have written t upper v by u, okay. Maybe I will forget to write this notation every time, the translation function is very easy to determine. This is differentiable everywhere and its derivative is the identity function. Remember, translation is from v to v. So, the derivative will be also from v to v, it will be linear map, it is a continuous linear map. In this case, its identity of v, all that you have to check is, you know, go back this definition, f of x0 plus h minus fx0 will be what now? Tv of this one will be adding v, then again adding v here, v cancels off, this is just like x0 plus h minus x0, okay. So, what should I take? Take a identity map, then this numerator itself will be identically 0, there is no need to divide by norm h and so on. So, it will be always true, that is all. So, the translation maps, adding a constant, you know, this is a rule, you can differentiate this function, you are going to use identity, right? Identity map, derivative is identity, that v is constant, so it goes away. So, that is another way of looking at it. Every bounded linear map is also differentiable. Here, one lucky thing is that we do not have to make a further assumption. Start with a linear map, you do not know that it is continuous. So, you have to make the assumption, you know, you have to put that extra condition, continue it. Once it is continuous, it is differential and its derivative is again the same function t at all the points x inside. This is again standard result inside Rn, R into Rn, if you have linear map, you know that its derivative, it is a linear map itself, ok. You can directly verify it by taking t itself as in the slot, in the third slot here, it is a t of x naught plus x minus t of x naught minus t h is 0. So, that will give you that t itself is a derivative of t, ok. And then this standard addition rule and scalar multiplication rule, if f and g are differentiable to x naught, alpha, beta, scalar, alpha f plus beta g is differentiable to x naught, ok. Indeed, if alpha, beta from u to k are scalar functions which are differentiable to x naught, then this alpha times f, this is not a composition, this is just multiplication, right, the scalar multiplication. So, derivative of alpha f makes sense, it will be differentiable. Similarly, beta g makes sense, the sub makes sense. So, this will be also differentiable to x naught, ok. Of course, you have to use Leibniz's rule here, ok. So, f is differentiable to x naught, then it is continuous at x naught, same, same proof as in the case of one variable, this is one variable calculus, ok. The logic, the definition, you can just look at the increment theorem here, ok. You show that if f is differentiable, this rule is true now, you can show that f is continuous also, ok. So, so far, except in the definition, I have started the continuous linear map, this is just like one variable calculus. In the one variable calculus, you have just a real number, but if you think carefully, if you know that you have, you must have done it already, a real number is actually represents a linear map from R to R, ok, namely multiplication by that number. So, there is, so far there is no difference at all. So, it may be noted that if f is differentiable to x naught, then all is directional derivatives, sorry. So, let me, let me call this one, ok, and jump again. All the directional derivatives are also axial, will also exist is what I wanted to say. So, let me, what is the meaning of directional derivative? Again, it is the same thing as in the multivariate calculus, starting with two, but Banach space is V and W. Again, opens up set U of the domain, x naught belonging to U. Now, you take any vector, preferably a non-zero vector, even vector 0 is also valid, ok. Take any non-zero vector, ok. Let f from U to W be any function. So, the directional derivative dV, dV of f at x naught. So, this is derivative of f in the direction of V at the point x naught, ok. E is defined as follow, ok. Namely, it is a vector W inside W, such that in this third slot, whatever your t of h, you are writing, instead of this one now, t of W, multiplication by t here or you know the vector W is just the direction that. So, f of x naught plus t, x naught plus t times V now, we are not arbitrary h, minus f of x naught minus t of W, this entire thing is now function of a one variable t, real variable. So, you are dividing it by t itself, ok, no norm. So, see f, x naught is fixed, V is fixed, W is fixed. So, it is function of real variable, one variable. This function must be, this limit must be 0. In other words, if you just look at f of x naught plus t times V, this function must be differentiated, ok, as a function of t, ok. And this derivative is W, that is called directional derivative, exactly same definition as in the case of usual multivariable calculus, ok. And you can immediately verify that all the directional derivatives will exist as soon as the derivative at x naught exists. So, often one calls the other derivative which you have defined as total derivative. It is a directional derivative, total derivative. You can talk about partial derivatives, but then you have to fix up coordinates. In Banach space, coordinate fixing is something very fishy, you do not want to do that. So, let V and W be any two normed in spaces. A continuous linear map T from V to W is said to be an isomorphism if it is invertible as a function and the inverse is also continuous, ok. So, here I have taken this definition for all normed in spaces also, ok. The remark here is that the inverse of a linear function is automatically linear. This is elementary algebra on any vector space, ok. If you have a continuous in a bijection which is sorry which you have just a bijection which is linear, then the inverse is automatically linear, right. It is not a problem, but inverse may not be continuous. This is what we have been telling. Even if it is a bijection, even if it is bijection it is continuous, the inverse may not be continuous. So, we have to mention it separately, ok. In the above situation, some author simply called T an invertible operator which is another name. They did not isomorphism, inverse is continuous etcetera. They will just say invertible operator. So, I will also use that term, ok. If V and W are Banach spaces, then for any invertible continuous linear transformation T from V to W automatically T inverse will be continuous, ok. See remember invertible just for me it does not mean that it is continuous. So, I want to be very careful. Invertible operator when you say by convention inverse is also continuous. A linear map may be invertible, ok. It has an inverse, but it may not be continuous. That is why I am making discussion. But if V and W are Banach spaces, no problem automatically the inverse will be continuous, ok. But this one needs a deeper theorem there namely what is called as open mapping theorem, ok. We are not going into that deep into function analysis, it is not a course on function analysis. But I am just mentioning this. I will never use this property though, ok. I will never use it because we are not going to prove inverse. I am just mentioning it as a information that is all, ok. If T is an isomorphism from definition that we have taken applied to both T and T inverse, we get two constants. This right hand side constant says T is continuous, right. Norm of T is less than or lambda x. Similarly, I must have other way around also for T inverse which will give you a lambda prime on this side. Lambda prime of norm x is less than or equal to norm of T x, which you can write it as you know norm of T inverse of x is less than or lambda prime lambda prime times norm of x also. You can write similar way. So, both sides you get a constant here. Remember this was the definition of that two functions are similar, two linear transformations are similar, ok. So, this is what we are. So, T itself is a similarity here because it is similar to the identity map. It is not identity map, but it is similar to identity map. So, T from W from W is called a similarity of two normal spaces such that we have, ok. We have studied similarities in the in the part one. Now, here is a theorem that I need to use. So, go through this carefully, start with two Banach spaces. Put A equal to A, V, W. A is a short form here when V and W are understood. What is this A, V, W? It is all set of all similarities from V to W, ok. Not all transformation is only similarities. Consider the function eta from A to B, WV, the other way around from W to V, defined by eta of T equal to T inverse. So, each element here is invertible. So, I can take the inverse. Again I am getting inside A, V, W. So, this eta is actually lands you from A to A, WV, ok. Does not matter. So, it is inside B, V, WV, B, WV which is a Banach space, ok. So, let A be then, sorry, then A is an open subset of this V, V, W and eta is differentiable on the entire of A, ok. So, I am stating a non-trivial thing here. First of all I say that this subset is an open subset of V, V, W. Remember V, V, W itself is a Banach space, ok. And eta is differentiable on A. Further, the derivative of d eta which is a map from A into B, W, V, ok. Because eta itself is the map into B, W, V. So, its derivative is of taking values there, ok. For each point A, you have a linear map. Given by d eta of d eta operating at a t will be a linear map operating on S, ok. This is nothing but, you know, pre-composing with inverse and then post-composing with inverse and taking the minus. So, it is a complicated derivative here, ok. For every S belong to V, V, W. Start with for every S. S is not invertible, S is arbitrary element, ok. Arbitrary element of linear C. A is a subset of this, right. And I am claiming A is an open subset. On an open subset, you have a function. You can talk about whether it is differentiable. The statement is that that is differentiable and its derivative is this one, ok. So, that is the statement. So, let us see the proof is not all that difficult at all. First of all, what may happen is this A is empty. What is the meaning of that? There may not be any similarities between V and W, ok. So, if you want to say anything, there is no statement about this being an unempty here. But on empty space, whatever I have made, they are vacuously true. So, there is nothing no logical difficulty here. But you want to prove something, you should assume A is non-empty, that is all. Otherwise you do not have to prove anything, all right. So, assume A is non-empty, then we can follow. What is the meaning of A is non-empty? There is some similarity which means V and W are similar already. That is a non-trivial assumption, ok. The above theorem implies in particular that A is continuous because we have already remarked that any function which is differentiable at a point is continuous at that point. So, we are going to prove that D, this eta is differentiable on the whole of A. Therefore, it is continuous on the whole of A which is not stated here. But it is an easy consequence of that. So, we will use that also, ok, towards the proof of that A is an open subset. In the last lecture, we have already made the preparation for this. So, let us see how. Take a point T inside A. What is it? It is a similarity from V to W, ok. So, it has some norm, ok. I am taking K equal to norm of T inverse, ok. Then I am claiming that the ball of radius 1 by K around T is contained inside A. T is arbitrary, such a ball is contained inside A. This 1 by K is obviously, you know, it is non-zero, right. So, that will show that A is open. For each point you have a ball open before contained inside that set A. So, A is open. So, how to show this? So, this is what we want to claim, right. So, let S belong into B, V, W, ok. This ball is taking place where inside B, V, W, right. A is instead of B, V, W. It is such that norm of S is less than 1 by K. Then we know that T inverse composite S is less than or equal to norm of T inverse into norm of S is less than 1, right. That is the whole idea why I took norm of this K equal to T inverse, norm of T inverse, ok. T inverse, norm of T inverse, T inverse will cancel out though it is less than 1. Therefore, by lemma that we have proved, you know, that we have stated and indicated the proof also lemma 1.63 follows that identity plus T inverse of S is invertible. There it was identity minus, you can take minus S here and you can put, this will become plus identity, but T inverse of S is invertible, ok, because norm of this one is less than 1. But then you can write T plus S as T composite identity that is T plus T T cancel, T and T inverse cancel away as, ok. So, this is T is invertible, T is a similarity, this is invertible, so the composite is invertible. So, T plus S is an element of A, ok. So, these are the points inside the, inside this board. Every element here looks like T plus S where norm of S is less than 1 by K, that is the board. So, the board ball is contained inside A, that is all we have proved that A is open, all right. Now, we want to show the derivative, ok. Fix a T, I want to show that D is that eta is differentiable at T, ok. The map S going to T inverse composite S, composite T inverse again, it is a bounded linear operator. Why? S is the variable here, I am taking the right composition and the left composition by some other bounded linear operator, exactly T inverse on both sides now, ok. So, we have seen that composing left or right is again a bounded linear transformation, we have moved twice. What are they? They are actually L T inverse and R T inverse. So, this map S going to T inverse S T inverse is nothing, but I have a short notation S V T which is minus of R T inverse plus L T inverse, that is it is minus sign coming here, you see in the statement it is minus sign. So, I have put take this as put minus V T, 50 to minus R T inverse S T inverse. This is a linear map from B V W to B W V, ok. We want to show that this minus V T is the derivative of eta at T, ok. This is the same thing we are showing that now, remember what is eta? T plus S inverse, see eta taking the inverse minus T inverse it is like F of T plus S minus F T, right. It is like that that F is T plus S T inverse minus T inverse plus this map V T operating upon S which is T inverse S T inverse. So, V T operating upon this one. So, there is a minus sign. So, here I get a plus sign divided by the norm S tends to 0 must be equal to 0, this limit must be 0. So, this is what we have to show, ok. Then the proof is over, right. So, first of all T plus S can be written as identity plus S T inverse T. T if you push it inside T, here now I have stopped writing composition S T directly I am writing, ok. So, this is just for convenience everywhere composite-composite. I have already stopped writing compositions here. Here I wrote, here when I am doing computation I am now no, no long, it is just like multiplication now, ok. You must understand that these are compositions that is all. So, identity composite T plus S T inverse T inverse is S. So, that is T plus S, ok. So, since we can choose norm S to be less than 1 by k, ok. So, that norm of S T inverse is less than 1, then we have this identity plus S T inverse can be inverted. Identity plus S T inverse, inverse will be nothing but 0 to infinity minus minus 1 raise to n because there is a minus sign, ok. If it is plus sign everything will be plus here. If it is minus sign here, then everything will be plus here. This plus is alternative minus 1 minus 1 raise to n. S T inverse raise to n. This is geometric series, 0 to infinity. Therefore, T plus S inverse minus T inverse plus T inverse S T inverse, I am computing this one, ok. So, T plus S inverse can be written as now T plus S was this one. Its inverse will be T inverse into identity plus S T inverse. Identity plus S T inverse is this summation. So, I am writing that T inverse into that summation minus T inverse plus T inverse S T inverse as it is, they are terms, ok. So, from here to here what I have come? I have just used this T plus S inverse is this way. I have written inverse of that will be first T inverse, then inverse of this, but this is like this. This is a geometric series. So, I have substituted it here. Now, look at the first term here n equal to 0. So, it is just T inverse that cancels as this one. What is the first term here? n equal to 1. What is this? S T inverse into T inverse on the left. So, that is this term. It also cancels out. That is a minus sign here with the first term. For the 0 term, this is plus sign, this is minus sign. So, those to cancel out, what you are left out is this summation from n equal to 2 onwards, ok. So, S T inverse raised to 2 comes out. So, T inverse into S T inverse raised to 2 and 0 to infinity S T inverse, S T inverse powers, S T inverse raised to n. So, up till here you have computed. Now, what we have to do? We have to divide it by norm s and then take the limit. When you take the norm of this part, the norm s comes here with a power twice, right. So, one power cancels out when you divide by norm s. Other power norm s remains. Rest of the terms are anyway bounded function there. They are just some bounded functions. When you take the norm of the whole thing, right. Therefore, as norm s tends to 0, this will be 0, ok. The claim follows. Only one norm s comes in the square term, cancels out. The other one remains. The rest of them is bounded. S tends to 0, norm s tends to 0. So, this is 0, ok. So, we have shown that eta is differentiable. From an earlier remark, it is continuous. I have already told you, I am repeating it. Ok, eta is continuous. How to show that d eta is continuous? Now, we have formula for d eta. Look at the formula. Formula says d eta is caught by T inverse phi T inverse. Sorry, minus T inverse on the left, T inverse on the right, ok. So, there is a minus sign does not matter. So, is this continuous? That is phi T inverse, right. T inverse is continuous. Multiplication is continuous. Light hand side is continuous and so on. This is what we are doing. Therefore, continuity of eta, d eta also follows, ok. Continuity of d eta follows because of, first of all, eta is continuous and phi is continuous, ok. The phi is what? Taking the left multiplication and right multiplication, maybe taking minus sign also, whatever. So, all that I am heavily using 1.53 here, right. This part I am using that L and R are continuous. L corresponds to left multiplication, R corresponds to right multiplication. Before that, you have to take inverse. T going to T inverse is continuous also, that is that is eta itself. Combining this, what we get is the derivative d eta is continuous at all the points and the whole of A. So, here is an exercise. I am not going to use it in the course, but this is something which is very, very useful for people who do leaguer groups and so on. So, that is precisely what it is. Take any Banach space that B will denote the space of all bounded linear operators from T to V, ok. Then this becomes a group out on auto obviously, right. You can compose, you can take the inverse and this is a group. What is more important is that this is a differentiable group or what we call as topological group. It is more than a topological group, namely the multiplication mu from B V cross B V to B V to mu of S T put S T. This itself is done differentiable. Taking inverse is differentiable we have seen. So, that makes it to a what is called as a differentiable group or what is called as leagroup, except that the word leagroup is used only for manifolds. This is what is called as a Banach manifold, ok, modeled on a Banach space. So, if you want to study Banach leagroups, this is the starting point. So, I am giving you this an exercise. Show that mu is differentiable, compute its derivative. Let G L V denote the open subspace B V consisting of all isomorphisms. They form a group, not this B V, ok. Along with our theorem above exercise implies that G L V is a leagroup modeled on Banach space. So, that is just an exercise. The only thing is you will see how to differentiate this one, where is the derivative and where is the derivative taking values. If you figure it out, then you will know what is the derivative, ok. So, thank you very much. This is all for today.