 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 normal 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 an 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 of X naught divided by the norm H. I cannot divide by H. So, H is a vector inside this normed in space V. 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 X 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. 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. 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. 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 minus fx minus T H whole thing divided by norm H. This limit must be 0. As soon as such a T exists, 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 multivariate calculus and so on. 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 standard notation. The only thing is you might not have called the thing as a Frechier derivative, we are going to call it as Frechier derivative. This Frechier was started with Banach space calculus. Frechier 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 you its derivative, remember derivative is a bounded linear map, T is bounded from v to w. So, it is taking values inside b, v, w. If this function for each x here inside you, you take dfx that will be denoted by df. If this itself is continuous, then we say that f is continuously differentiable on you or we can say it is a class C1. There are ways of making class C2 definition C3 and so on. So, we will stop here only class C1 here. 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. 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 path. If you 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. Exactly as in the case of one variable calculus here. The following statements are all easy to check exactly as in the case of one variable calculus. Every constant function is differentiable everywhere. Everywhere means what on the whole of v, on the whole wherever they are different. 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. It is a translation function. I have written t upper v by u. Maybe I will forget to write this notation every time. So, 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 continuously linear map. In this case your identity of v all that you have to check is you know go back this definition f of x naught plus h minus f x naught will be what now? T v of this one will be adding v then again adding v here v v cancels off if you just think it is just like x naught plus h minus x naught. So, what should I take take a identity map then this numerator itself will be identically 0. There is no need to divide by nor match 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 u going to use identity right. So, 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 the 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 continuity. Once it is continuous it is differentiable and its derivative is again the same function t at all the points x inside. This is against standard result inside R n, R into R n if you have linear map you know that its derivative it is a linear map itself. You can directly verify it by taking t itself as in the 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 the derivative of t ok. And then this standard addition rule and scalar multiplication rule if f and g are differentiable at x naught alpha beta scalars alpha f plus beta g is differentiable at x naught ok. Indeed if alpha beta from u to k are scalar functions which are differentiable at 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 at x naught ok. Of course, you have to use Leibniz rule here ok. So, f is differentiable at x naught then it is continuous at x naught same same proof as in the case of one variable this is a one variable calculus ok. The logic the definition the 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 ok. So, far except in the definition I have started the continuous linear map this is just like one variable calculus in the in the one variable calculus you are 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 they 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 at x naught then all is directional there is sorry. So, let me let me call this one ok and jump here. 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 this is same thing as in the multivariate calculus starting with two Banach spaces V and W again opens up set U of the domain x naught belonging to U. So, now you take any vector preferably a non-zero vector even vector 0 is also valid ok take any non-zero vector let f from U to W be any function. So, the directional derivative d V d V 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 follows 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 out 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 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 word if you just look at f of x naught plus T times V this function must be differentiable as a function of T 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 exist. So, often one calls the other derivative which you have defined as total derivative ok these are direction derivatives total derivative. You can talk about partial derivatives, but then you have to fix up coordinates in Banach space is coordinate fixing is something very fishy you do not want to do that. So, let V and W be any two normed 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 that 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 this is another name they did not isomorphism inverse is 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 a 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 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 I 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 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 equal to norm of T x which you can write it as you know norm of T inverse of f x is less than equal to lambda at time lambda prime times norm of x also you can write similar way. So, both sides you get a constants 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 the identity map. So, T from W W is called a similarity of two normed spaces such as the app 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 is all set of all similarities from V to W ok. Not all transformation it is only similarities consider the function eta from A to B W V 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 W V ok, does not matter. So, it is inside B V W B W V which is a Banach space ok. So, let A be then sorry then A is an open subset of this B 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 B V W. Remember B V W itself is a Banach space ok and eta is differentiable on A. Further the derivative 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 T inverse and taking the minus. So, it is a complicated derivative here ok for every S belong to B 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 differentiable the statement is that that is differentiable and this 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 difficult 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 the 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 buffer 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 a set 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. So, 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 and T inverse cancel away S 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 ball every element here looks like T plus S where norm of S is less than 1 by K that is the ball. So, the ball 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 is a bounded linear operator why S is the variable here I am taking the right composition in the left composition by some other bounded linear operators exactly T inverse on both sides now ok. So, we have seen that composing left or right is again a bounded linear transformation you have 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 its minus sign coming here you see in the statement with minus sign. So, I have put take this as put minus V T V T equal 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 take in the inverse minus T inverse it is like f of T plus S minus f T right. It is like 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 on this one that way. So, there is a minus sign. So, here I will get a plus sign divide 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 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 this is plus sign everything will be plus here if this is minus sign here then everything will be plus here. This is plus this 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 the T plus S was this one. This 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 their 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 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 that also cancels out that is a minus sign here with the first term for 0 term this plus sign this minus sign here. So, those to cancel out what you are left out is this summation from n equal to 2 onwards ok. So, S T inverse raise to 2 comes out. So, T inverse into S T inverse raise to 2 and 0 to infinity S T inverse S T inverse powers S T inverse raise 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 as in the square term cancels out other one remains the rest of them is bounded s tends to 0 norm s tends to 0. So, this is 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 discontinuous that is phi T over right T going to T inverse is continuous multiplication is continuous light hand side is continuous and so on this is what we have seen. 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 may be 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 you have to that is that is the 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 in the course, but this is something which is very very useful for people who do Lie 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 E ok. Then this becomes a group of the order 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 requires topological group is more than topological group. Namely the multiplication mu from Bv cross Bv to Bv mu of st put st this itself is done differentiable taking inverse is differentiable we have seen but that makes it to a what is called as a differentiable group or what is called as Lie group except that the word Lie group 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 Lie groups this is the starting point. So, I am giving you this is an exercise show that mu is differentiable compute its derivative let GLV denote this opens of space Bv consisting of all isomorphisms they form a group not this Bv ok along with our theorem above exercise implies that GLV is a Lie group 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 the derivative taking value if you figure it out then you will know what is the derivative ok. So, thank you very much this is all for today.