 Welcome to NPTEL NOC course on point set topology part 2. This is module 1, chapter 1 on differential calculus on Banach spaces. In this introductory chapter, we shall present statements and proofs of implicit function theorem and inverse function theorem in differential calculus. Since we have developed enough background on Banach spaces in part 1, we plan to do this directly for Banach spaces rather than for Rn. Usually for Rn, one can prove inverse function theorem and then prove implicit function theorem which becomes little more transparent. In the case of Banach spaces, such a method is not possible. We have to first prove the implicit function theorem and then deduce inverse function theorem. One of the key factors which needs to be sharpened in the case of Banach spaces is the so-called weak mean value theorem. In the case of Rn, because it has a rich structure namely Hilbert's structure, the proof is much simpler. Here the proof is a little more deeper analysis involving what are called as Dini derivatives. If you have some difficulty in understanding Banach spaces, you can just replace all words Banach spaces by Rn and just think of Rn and try to follow the material. Afterwards, you can fill in your Banach spaces once you learn Banach spaces thoroughly. Let me recall these notations which I have already introduced to you. These notations will not be used for any other things during this course. So, they are kind of frozen especially these are all standard ones of course. These Euler points are C, Q, Z, N. In addition, I will also using that this Euler point I for closed interval 0 and Dn for the closed unit days less than for unit sphere. Pn and Cpn come very rarely but I have also standard notation. I do not use it for any other thing. Sometimes I need the open interval minus 1 to plus 1 raised to N for which I will use this Jn. Most important one is often I have to deal with both real and complex fields simultaneously. So, in that case I will use this notation K. The context if there is a special case then we will mention it otherwise this K will be either or a real number or complex number ok. So, let me begin with a modification of the contraction mapping theorem that we have proved in part 1. The modification is an extension actually. There we had proved it for one map. Here we will prove it for a family of maps that is the importance. So, start with a metric space Xd, a Y be any tuploid space. Take a function on Y cross X to X which is continuous ok. So, that there is a real number between 0 and 1 strictly between 0 and 1 such that this property a distance property holding a distance between f of Y x 1 and f of Y x 2 is less than it to c times it turns into x 1 x 2 for all Y and for all x 1 x 2 instead of x. So, this is some kind of lipstick condition uniform lipstick condition ok. If this condition holds then we have a conclusion. What is the conclusion? For each Y inside Y there exists the unique phi Y inside X such that f of Y phi Y is Y ok. You can think of f of Y x is x because phi Y is a point of that that X unique X will be called as phi Y because it is a unique one. So, thus we get a assignment from Y to X a function phi Y. This phi Y as the property that distance between phi Y 2 and phi Y 1 is less than or equal to 1 divided by 1 minus c that c same c here distance between phi Y 2 and f of Y 1 phi Y 2. This is a technical result that will be very helpful. In particular you can immediately see that E is continuous because of this one ok. So, this distance between Y 1 and Y 2 phi of phi of Y 1 and Y 2 is dominated by this one. So, Y 1 and Y 2 can be controlled and this can be controlled. Therefore, this difference side can be controlled is the conclusion here. If you forget about this capital Y here take one single function ok. Then we have proved this statement here. Then there is no function here it is only one. So, there is only part a namely there is a unique function unique thing and that is the contraction mapping. So, the addition here is when you have a family of functions which satisfies this uniform leftist condition we get a continuous function out of it is stronger than saying just continuous. So, that property we have put here we will use this one once again ok. So, let us go through the proof of this one alright. First of all let us recall the proof in the case of when there is a single function. I mean for each Y inside of Y let us look at f Y one single function x to given by f Y factor is to f of Y x. So, Y coordinate is fixed. So, you get one single function for this let us say what in the proof of this theorem what we have done already ok. That if Y is a singleton then this theorem is nothing but the ordinary contraction mapping principle. So, let us recall the proof. For simplicity we shall use the same notation f function f restricted to Y namely f Y in this special case ok this singleton Y ok. So, in this special case let us first prove the uniqueness. Let x 1 and x 2 be two points such that f x 1 equal to x 1 and f of x 2 equal to x 2 ok. Then distance between x 1 and x 2 will be between f x 1 and f x 2 because x 1 and f x 1 x 2 is f x 2, but this is less than c times distance between x 2 and x 2. Now, Y is suppressed here that is all f x 1 distance f x 1 f x 2 is less than 2 distance between f x 1 and x 2. So, this is a condition 1 right. So, condition 1 we will tell you we have distance between x 1 and x 2 is less than or equal to a fraction of the same distance and this is possible only if this real number is 0 ok that means x 1 is x 2. This is the way the uniqueness part was proved and we have just recorded. For the existence part what we do we follow the alteration method right. Start with any point as inductively define x 2 equal to f of x 1, x 3 equal to f of x 2 and so on, x n equal to f of x n minus 1 which is nothing but f f f operative 1 f etcetera n minus 1 times x 1 f power n minus 1 times x 1 ok. We claim that the sequence got by iterating the powers of f on x 1 namely x n is a Cauchy sequence ok and then we will appeal to that the metric space is complete to get a limit ok. That limit if I denote by x then we will show that this x is the fixed point f x is equal to x ok. So, how does one prove that this is a Cauchy sequence? Let us for the sake of simplicity put r equal to distance between x 1 and x 2. What is x 2? It is f of x 1. If f of x 1 is already x 1 then this distance will be 0 right but we have already solved this problem we do not have to go any further never mind. So, r may be 0 never mind but whatever we have if r is not 0 namely f of x 1 is not equal to x 1 then only we have to iterate right. So, we keep iterating. So, do not worry about right now that there is equal to. If it is equal to by chance if you know that we can stop there is no problem. Then d of x 2 x 1 x 2 x 3 will be d of x 2 is f x 1 and x 3 is f x 2 that is less than c times a distance between x 1 and x 2 which is less than c times r ok. Now, we repeat this one d of x 3 x 4 will be c square times r and so on distance between x n and x n plus 1 will be less than or equal to c power n minus 1 times r ok. Therefore, one more suppose suppose we assume this one we have to repeat it you will get distance and x n plus 1 x n plus c power n times r. So, same formula is there now. Therefore, distance between any x n and x n plus m now I use triangle inequality m times right. So, I go for x n to x n plus 1 x n plus 1 to x n plus 2 x n plus 2 to x n plus 3 take all these distances add them up put a less than or equal to 0 to m minus 1 distance between x n plus i and x n plus i plus 1 ok. But just now we have proved this formula this distance is r times c power n plus i minus 1 ok x n c n minus 1 comes. So, x n plus i will be c n plus i minus 1 which is the same thing as r comes out i range from n minus 1 to n plus m minus 1 c power i. What are these? These are nothing partial sums of the series 1 plus c plus c square plus c cube which is a geometric series where c is between 0 and 1. Therefore, this itself we say you know as geometric series is a partial sum of Cauchy sequences. So, now summation c i is a geometric series which converges to 1 divided by 1 minus c. So, that explains why we have got this 1 divided by 1 minus c here in this inequality ok. So, continuing with the proof of this Cauchy sequence in particular the partial sums i range to 1 to n c i this is a Cauchy sequence. So, subtract the first n terms namely the summation of n term in n plus m term what you have is this one this can be made less than epsilon given any epsilon choosing n and m properly therefore, this is a Cauchy sequence ok. So, what we have proved is x n is a Cauchy sequence it converges because x is a complete matrix. After that if you apply limit of x n is the same thing as limit of x n plus 1 is same thing as limit of f of x n plus 1 f is a continuous function. So, you can take out f. So, it will be put f of limit of x n which is equal to f of x which is equal to x ok. So, f of x is limit of f of x n is also limit of x n plus 1 because f of x n can be derived, but this limit is same thing as n all right. So, now if you take m tending to infinity in this summation distance to x n and x infinity is what now it is x ok. What is this summation what you get is distance to x n and x is less than equal to r time c power n minus 1 divided by 1 minus c the partial sums from the this is the remainder after n term ok. So, for n greater than equal to 1 this is always true all right. So, now we will do the same copy just copying put the variable y also ok. The idea of the proof is exactly same there is no change at all ok. Returning to general case since the fixed point x depends upon the function f y right we are changing y for each y I will get a different fixed point where I am going to say f phi y. So, that is the notation also this r which we have fixed as the first distance between x 1 and x 2 ok. So, this will now depend upon y because this is now the distance between here we come here distance from x n and f of y x 1. So, y is coming here. So, let us denote this r by let us replace it by this term I can write the r y or something if you want no problem. We can then rewrite this 3 as distance between you see this now x n is f n f y of x f y of n minus 1 times x 1 phi y phi y is the the the limit point here the fixed point here is less than equal distance between x 1 this r is f of y x 1 this distance c power n minus 1 by 1 minus c this is always this is independent of y there is the the constant c is independent of y remember that for every n they are taking over the 1 y in x x 1 in x 1. So, this is the inequality which we have ok. So, now given y 1 and y 2 put n equal to 1 this is very very for all n n equal to 1 y equal to y 1 x 1 equal to phi y 2 ok is above inequality to get this is n equal to 1. So, this will be g of phi y 2 phi y 1 ok. So, I am taking n equal to 1 y equal to y 1 and x 1 equal to phi y 2 this x 1 phi y 2 m power n n is 1. So, there is no operation here starting point ok. So, this will become phi y 2 phi y 1 because x 1 is phi y 2 it is less than equal to 1 by 1 minus c times the right hand side is distance between x 1 and f of y x 1 which is phi y 2 into f of y 1 phi y 2. So, this is what we wanted to prove this was conclusion of the part of theorem. Note that in this notation f of y 2 phi y 2 is phi y 2 ok f of y 1 phi y 1 phi y 1 and so on whatever f of y phi y is phi y. But continuity of f given epsilon positive we can choose a neighborhood n 1 cross n 2 of this point y 2 phi y 2 in y cross x such that distance between phi y 2 and any f of y x is less than epsilon times 1 minus c for all y and x instead n 1 cross n 2. Just continuity of y of f from y cross x to x take a point here it is going to phi y 2 and then you can find a neighborhood of this one that is epsilon neighborhood because it is a matrix space. In the domain you have y which is an arbitrary space and x of course is a matrix space. So, I am writing the neighborhood as n 1 cross n 2 instead of choosing ball neighborhoods and so on of the point y 2 f phi y 2 ok this is inside y cross x. This entire neighborhood f of this goes inside this epsilon neighborhood. So, I am making epsilon into 1 minus c times that neighborhood instead of this. So, this is you can choose epsilon prime here and then write epsilon prime is 1 minus no problem. The distance between phi y 2 and any f of y x as soon as y and x are inside this neighborhood is less than this one. So, this is continuity of f. Therefore, now using the continuity of f here y 1 inside n 1 it follows that distance between phi y 2 and phi y 1 is 1 divided by 1 minus c times instead of this I can write 1 divided by 1 minus c times this one is c epsilon times 1 minus c that 1 minus c cancels out you are left with epsilon. Just to cancel out 1 minus c I have put 1 minus c here that is one ok. So, this is last part is to derive the continuity of the function phi we have obtained ok as a solution as a fixed point alright. So, I am repeating condition 1 may be referred to as uniform contraction mapping condition why because the left hand side here the right hand side here the choice of c does not depend upon y at all for all y have this one that is why I told you that this is like uniform lips is right. So, the theorem itself can be called as uniform contraction mapping. Any questions? Let us recall a few basic facts from non-denial spaces and Banath spaces and so on. Suppose you have two non-denial spaces t from v to w is a linear map then the following conditions are equivalent. Remember on a infinite dimensional infinite dimensional vector space a linear map may not be continuous. So, therefore, these things become non-vacuous statements ok. t is continuous at 0 there exists lambda properties such that norm of t x is less than lambda times norm of x for all x inside v t is continuous uniformly on the whole of v ok. So, all these three are equivalent conditions none of them may be it is true in general when t is when v is finite dimensional this should be automatically true ok for all linear functions ok. So, let me just recall this because this is so fundamental ok 1 implies 2 that means once you assume continuity at one single point 0 ok by the way you can assume continuity at any other point also it is equivalent to that one I could have put that condition also here you can think about it take it as an exercise. So, put epsilon equal to 1 by continuity of t at 0 we get a delta positive such that norm of x less than equal to delta implies norm of t x is less than or equal to 1 x and every epsilon so I have put epsilon 1. Therefore, now you take x to be any non-zero vector ok then norm of t x is equal to norm of x divided by delta I am writing delta divided by norm of x times x here t of that and then I have to compensate for this factor which comes out and cancels out with this one. So, this all this entire thing is just norm of t x delta divided by norm of x has been multiplied inside here the same thing inverse is multiplied outside here ok. But what is the idea of this one now if norm of x is less than or equal to delta ok then what is the norm of the inside bracket here it is less than or equal to delta because x by norm x is going the way this one the delta time that is less than or equal to delta ok. So, I can apply this inequality it means that this part is less than or equal to 1. So, there is norm of x by delta here so less than or equal to norm of x. So, for all x not equal to 0 norm of t x is less than or equal to the volume of x by delta. If x is 0 left hand side is 0 right hand side is also 0 so inequality is valid ok. So, for all x this is true actually. But for writing down this proof I have to assume x is not here that is all because I have divided by norm x here. So, we can take lambda equal to 1 by delta then what do I get I get norm of t x is less than or equal to lambda times norm of x. So, that is the conclusion of 2 ok. As soon as you have such a uniform lambda uniform continuity follows right on the whole of t of x 1 minus x 2 less than or equal to lambda times norm of x 1 minus x 2 right. So, if norm of x 1 minus x 2 is less than or equal to some epsilon whatever delta you want you have to choose this appropriately by dividing out by this lambda that is all ok. So, 2 implies 3 is obvious in that way. But 3 implies 1 is obvious because this is now actually continuous on the whole of this whole continuity here also ok. Now, we make a definition here take V and W to be any two normed linear spaces. Take a linear map from V to W and we will call it bounded or a continuous linear map if there exists lambda positive such that norm of t x is less than or equal to lambda times norm of x for every x belonging to belonging to V ok. So, such a thing is called bounded linear map. This is a standard terminology in function and analysis I cannot help it. It is not the standard meaning of bounded functions, functions which taking values in a metric space bounded function as a different meaning altogether. This is a bit unfortunate terminology but you will get used to it when you are doing function analysis. The longer term is continuously linear map that is all. It is not too long if you bounded linear map if you have to say instead of that you continuously in the map you can say ok. For a bounded linear map F from V to W there is something called operator norm ok. We have introduced this one in the in park 1 also. We have in fact studied the Banach space of continuous functions on a metric space and so on. So, I mean just recalling is the operator norm norm t is defined as supremum of all norm t norm t f x where norm of x is 1. That means x is varying on the unit sphere in the domain ok domain is V. So, norm of x x must be inside V of course here. We should denote the space of all bounded linear maps from t to t from V to W by this curly B V W. We will just read it as B V W. This is clearly a vector space ok. What do you have to do? You have to show that if F and G are bounded course linearity is clear F plus G is also bounded linear alpha times F is also bounded linear that is all. So, together with operator norm as we have defined here it becomes a normed linear space. This norm has the standard properties that norm t is always not you know bigger than equal to 0. It is 0 if it only if t is identically 0 and alpha times norm t is mod alpha times norm t ok norm of sorry norm of alpha times t is mod alpha times norm of t and the triangle inequality in terms of addition norm of t plus s is less than equal to norm t plus norm s. So, these are the conditions which make a function norm ok. The following result about operator norm namely this special thing I want to not general norm as such are all very easy to derive. What are these? I have summed it up in this lemma we will keep using this again and again. So, let us go through it carefully that you understand this one completely. Take vector spaces ok V 1, V 2, V 3 they are all normed linear spaces I am not going to mention different norms are here I am not going to mention that. This is the standard practice just like we keep saying x 1 is at a logical space without mentioning the topology there right. So, V 1, V 2, V 3 etcetera in a normed linear space. Take a bounded linear map from V 1 to V 2 another from V 2 to V 3 ok. Now, I am defining in operation LT on B 1, B V 1, V 2 ok LT of s is T composite s. You can view it as from the operation from this side namely operating via s on t. So, you can think of this R s of t what is this? S LT of s is composing T on the left R s of t is composing with s on the right ok. So, if you vary s this will be a map from B V 1, V 2 to B V 1, V 3. If you vary T namely keep s as it then it will be a map from B what vary T I am varying now ok B V 2, V 3 to B V 1, V 3 ok. So, this is the interesting thing that I am I am interested in namely left multiplication or right multiplication you can say. Obviously, these functional compositions are non commutative. So, that is why you have to worry about this whether right composition or left composition separately. In any case the first thing is very simple norm of T composite s is less than little norm t into norm s. This is a very fundamental property which makes the normed space into what is called as normed algebras ok. Now, LT is again a bounded linear map from B 1, V 2 to B V 1, V 3. Similarly, RT is a bounded linear map from V 2, V 3 to B V 1, V 3 ok. They are linear maps they are bounded linear maps because of you keep using this one equation in equation one this inequality one will tell you that. Now, you vary T itself ok for each T you have a bounded linear map. So, L itself from B V 2, V 3 to B of B of this ok. Similarly, R from B V 1, V 2 to the bounded linear maps from the bounded linear maps to bounded linear maps. So, they are themselves continuous. Continuity is what I wanted to emphasize all of them follow from this one single in equation here. Linearity is obvious for all of them ok. If you add your s 1 plus s 2 here that will be T composite s 1 plus T composite s 2 or if you add T 1 plus T 2 here R s of T 1 plus T 2 is R s of T 1 plus R s of T 2. So, linearity etcetera is not a problem. Continuity follows from just this one ok. Indeed you can talk about the operator norm of L T and operator norm of L and operator norm of R and find out all these things is easy exercise very easy. Do that so that you definitely understand these things completely ok. So, I will have one more important lemma here namely now I am assuming it is a W is a Banach space ok. Banach space is nothing but a normal linear space in which the induced metric is complete every Cauchy sequence is convergent right. So, take B as this codomain as a Banach space then the set of all bounded linear functions from V to W with the operator norm becomes a Banach space itself. This is what we have studied last time in the part 1 ok. Let T belong into B V. Now, B V is a short notation for B V comma V when V and W are same instead of writing B V V I am just writing B V. So, these are self self maps from V to V linear maps ok, linear operator bounded linear operator. If T is an element of that where V itself is a Banach space such that norm T is less than 1 then identity minus T or you can take identity plus T also the same thing because I can change T to minus T is invertible with the inverse given by the convergent series identity minus T inverse what is that it is given by summation 0 to infinity t power n the geometric series 1 1 is identity plus T plus T square plus T cube and so on ok. Why this is convergent because the partial sums they are bounded by 1 plus you know this is a norm T norm let us call it as T 1 plus C plus C square and so on this norm T summation that becomes a geometric series. So, that is a proof ok all right. So, if you look at identity map that is invertible you can take a bound we can take a ball of radius less than 1 actually equal to 1 ok the open ball if you take all the elements in that are invertible. So, this is the hypothesis this is the conclusion here ok in any Banach space ok the unit ball centered around the identity represents all invertible elements ok there may be more invertible elements of course, but this is definitely all these are invertible elements this is the meaning of this all right. So, here are some elementary exercises other than working out whatever I told you some of these without proof and so on that is the first lesson you have to do ok. So, let me just read out these please show that formula 6 defines a norm on BVW and just given the norm we have to show we have to verify those three conditions prove that 1.5 and 1.6 these lemmas in 0.6 this one this is what we have to do that is the exercise for you. Thank you we will meet next time.