On a diffeomorphism between two Banach spaces ...

On a di¤eomorphism between two Banach spaces -simple proof and some remarks Marek Galewski January 10, 2018 Abstract We provide su¢ cient conditions for a mapping acting between two Banach spaces to be a di¤eomorphism using the approach of an auxiliary functional and also by the aid of a duality mapping corresponding to a normalization function. We simplify and generalize our previous results. Math Subject Classi…cations: 57R50, 58E05 Key Words: global di¤eomorphism; local di¤eomorphism; mountain pass lemma

1

Introduction

The aim of this work is to present a scheme allowing for a mapping between two Banach spaces which is a local di¤eomorphism to become a global one with the aid of critical point theory. The approach presented here is already well known, i.e. adding some conditions which made local di¤eomorphism a global one, and it has been started by Hadamard [13] and later developed by several authors in a number of works, out of which we recall [12] for an overview of invertibility results, [14] a general theorem with which we compare our approach in [11] showing these two do not overlap, [7] where higher order invertibility conditions are developed, [19] for the metric space setting. We mention also [26] for some general up to date version of Hadamard-Lévy Theorem (see [20], [23]) and [25] for some invertibility on a …nite dimensional space. Our inspiration lies in works [17], [16], [15] which we somewhat improve and generalize while retaining the main methodology. While we use 1

the known approaches and techniques we put emphasis on presenting the scheme which being intuitive in the case of functions of one variable becomes more involved when the dimension is enlarged to some n 2 and next to an in…nite setting. We underline that the simple remarks working on the real line can be suitable extended to cover other cases of domains (a …nite and in…nite dimensional). The notes are organized as follows. Firstly, in Section 2, we remark on the scheme leading to making a global di¤eomorphism out of a local one in the spirit of the Theorem of Hadamard. Next, in Section 3, we provide necessary background with which we can proceed to the in…nite space setting. In Sections 4 and 5 we present two approaches leading to global di¤eomorphisms between two Banach spaces with the aid of an auxiliary functional and a duality mapping. Section 6 contains some conclusions and related implicit function theorems provided without proofs since the arguments are now obvious.

2

The outline of results in a …nite dimensional setting

We start with recalling that a C 1 mapping f : Rn ! Rn is locally invert0 ible and its inverse function is also C 1 provided that det f (x) 6= 0 for any x 2 Rn . In this work we will concentrate on the case of mappings acting between Banach spaces and our aim is to make, with the aid of some additional assumption, the local invertibility condition a global one following some already developed tools. Thus we rather simplify the known proofs and provide some insight into understanding of the tools used than we introduce new notions and approaches. We still believe it is of some importance to collect in one work some results and present them in a more uniform manner. 0 Let us …rst think of such C 1 functions f : R ! R such that f (x) 6= 0 for any x 2 R. Of course such a function need not be globally invertible as a simple example of arctan shows. On the other hand a C 1 function g (x) = x3 is globally invertible while its inverse is not C 1 . This function is not locally C 1 invertible around 0. On the other function f (x) = x3 + x is locally (and globally) invertible. We see that jf (x)j ! 1 as jxj ! 1. Such a property is called coercivity of a functional and together with continuity says that a coercive functional has at least one argument of a minimum. This a version

2

of the celebrated Weierstrass Theorem since coercivity yields that for any d 2 R, the set S d is bounded, where S d = fu 2 E : jf (u)j

dg:

Since continuity provides that such sets are closed we are done. In fact it is easy to see that a locally invertible C 1 function f : R ! R 0 (i.e. f (x) 6= 0 for any x 2 R) such that jf (x)j ! 1 as jxj ! 1 is globally invertible, that is f 1 : R ! R is de…ned and C 1 . It is easy to prove this result considering an auxiliary functional g : R ! R de…ned by g (x) = jf (x)

yj2

for a …xed y 2 R. This functional is also coercive and contrary to x ! jf (x)j it is C 1 . Thus for a …xed y 2 R functional g has an argument of a minimum x which by Fermat’s rule satis…es equation (f (x)

0

y) f (x) = 0

0

and since f (x) 6= 0 for any x 2 R we see that f (x) = y which solves the problem of f being onto. Now we concentrate on the question of f being injective. This is done the help of Rolle Theorem which as Jean Mawhin points out in his celebrated book [21] serves as an introductory Mountain Pass Theorem. Thus assume that to some y there exists x1 ; x2 such that f (x1 ) = f (x2 ) = y. We see that for functional g it holds that g (x1 ) = g (x2 ) = 0. Thus, by Rolle Theorem on an interval (x1 ; x2 ) we must have 0 at least one point x3 that g (x3 ) = 0. This point, with the aid of the Weierstrass, can be chosen as a point of a maximum over [x1 ; x2 ]. Thus we see that g (x3 ) > 0 and on the other hand by the Fermat’s rule g (x3 ) = 0 0 since g (x3 ) = 0 implies g (x3 ) = 0. A contradiction …nishes this simple proof. Possibly the above argumentation is a bit to sophisticated for a one variable function. But it has the advantage of being generalized to maps between Euclidean spaces and further between Banach spaces. Moreover, the additional assumption that kf (x)k ! 1 as kxk ! 1 answers the following question for a a C 1 mapping f : Rn ! Rn : What assumptions should be imposed on the mapping f to have a global di¤eomorphism from the local one? 3

As for the result known result we recall the Hadamard’s Theorem, see Theorem 5.4 from [18]: Theorem 1. Let X; B be …nite dimensional Euclidean spaces. Assume that f : X ! B is a C 1 -mapping such that f 0 (x) is invertible for any x 2 X, kf (x)k ! 1 as kxk ! 1, then f is a di¤eomorphism. This Theorem for the proof uses the above sketched approach. Of course Rolle Theorem cannot be used here but instead a following three critical point theorem is applied Theorem 2 (Theorem 5.2 from [18] ). (Finite Dimensional MPT, Courant). Suppose that a C 1 functional f : Rn ! R is coercive and possesses two distinct strict relative minima x1 and x2 : Then f possesses a third critical point x3 distinct from x1 and x2 ; which is not a relative minimizer, that is in every neighborhood of x3 , there exists a point x such that f (x) < f (x3 ). The above theorem inspired Idczak, Skowron and Walczak in [15] to obtain a version of it for mappings between a Banach and a Hilbert space and further it inspired us to see if a Hilbert space can be replaced by a Banach space, [8], [9]. In all sources mentioned a Mountain Pass Theorem is used which requires some re…ned estimation of auxiliary functional around 0. In this work we aim to follow strictly the pattern described above. Since we see that the points x1 and x2 which we use in contradiction are in fact points of strict local minima (otherwise we arrive at a contradiction with a local di¤eomorphism), we need not use the classical Mountain Pass Theorem but a three critical point version of it due to Pucci and Serrin [24]. We would like to make some remark to the end of this section. The global inversion result which we have described has also a simple application as the unique solvability tool of nonlinear equations of the following type f (x) = y where f satis…es the mentioned assumptions. Since a solvability part is reached via a Weierstrass Theorem and requires that some auxiliary functional should have an argument of a minimum, it follows that a lower semicontinuous weakly di¤erentiable functional which has invertible weak derivative would be su¢ cient. Thus we have the following easy tool: 4

Corollary 1. Assume that f : Rn ! Rn is a weakly di¤erentiable mapping 0 such that det f (x) 6= 0 for any x 2 Rn . Assume that functional g : Rn ! R de…ned by g (x) = kf (x)k2 is lower semincontinuous and coercive. Then for any …xed y 2 Rn equation f (x) = y has at least one solution. We see that by our assumption functional g (x) := kf (x) yk2 is lower semincontinuous and coercive for any …xed y 2 Rn . Thus the assertion follows easily. This result has no meaning in case of n = 1 since both types of di¤erentiability coincide there.

3

Some remarks on in…nite setting

It is well known that as with the case of Rn for n 2 there are two basic concepts of di¤erentiability for operators and functionals. Let X; Y be Banach spaces, and assume that U is an open subset of X. A mapping f : U ! Y is said to be Gâteaux di¤erentiable at x0 2 U if there exists a continuous linear operator fG0 (x0 ) : X ! Y such that for every h 2 X lim t!0

f (x0 + th) t

f (x0 )

= fG0 (x0 )h:

The operator fG0 (x0 ) is called the Gâteaux derivative of f at x0 . We will denote it in the sequel by f 0 . An operator f : U ! Y is said to be Fréchet-di¤erentiable at x0 2 U if there exists a continuous linear operator f 0 (x0 ) : X ! Y such that lim

khk!0

kf (x0 + h)

f (x0 ) khk

f 0 (x0 )hk

= 0:

The operator f 0 (x0 ) is called the Fréchet derivative of operator f at x0 . When F is Fréchet-di¤erentiable it is continuous and Gâteaux di¤erentiable. A mapping f is continuously Fréchet-di¤erentiable if f 0 : X 3 x 7! f 0 (x) 2 L (X; Y ) is continuous in the respective topologies. If f is continuously Gâteaux di¤erentiable then it is also continuously Fréchet-di¤erentiable and thus it is called C 1 . It is the most common way to prove the Fréchet di¤erentiability that one shows that f is continuously Gâteaux di¤erentiable. In fact for critical point theory tools either the functional usually must be C 1 or locally Lipschitz and so it is no surprise that Gâteaux di¤erentiability is only an auxiliary tool. 5

A continuously Fréchet-di¤erentiable map f : X ! B is called a diffeomorphism if it is a bijection and its inverse f 1 : B ! X is continuously Fréchet-di¤erentiable as well. Recalling the Inverse Function Theorem a continuously Fréchet-di¤erentiable mapping f : X ! B such that for any x 2 X the derivative is surjective, i.e. f 0 (x)X = B and invertible, i.e. there exists a constant x > 0 such that kf 0 (x)hk

x

khk 0

de…nes a local di¤eomorphism. We will write that f (x) 2 Isom (X; Y ) for 0 mappings with such properties. When f (x) 2 Isom (X; Y ) for each x 2 X, it means that for each point x in X, there exists an open set U containing x, such that f (U ) is open in B and f U : U ! f (U ) is a di¤eomorphism. If f is a di¤eomorphism it obviously de…nes a local di¤eomorphism. Thus the main problem to be overcome is to make a local di¤eomorphism a global one. Prior to introducing the result by Idczak, Skowron and Walczak from [15] we give some required background on variational methods in in…nite setting. Indeed, in a in…nite dimensional setting a continuous coercive functional need not have an argument of a minimum, so we must put this issue in some proper frame. We resort to the example of a function x ! 12 kxk2 on a Hilbert space H for whom the minimizing sequence, i.e. such a sequence (xn ) H that inf n2N 12 kxn k2 = inf x2H 12 kxk2 is such that xn ! 0 and xn stands for the derivative of x ! 12 kxk2 : Now we are in position to introduce a new notion. Let E be a Banach space. A Gateaux di¤erentiable functional J : E ! R satis…es the Palais-Smale condition if every sequence (un ) such that (J(un ))is bounded and J 0 (un ) ! 0, has a convergent subsequence. The following links the Palais-Smale condition with coercivity [18]. Proposition 3. Assume that J 2 C 1 (E; R) is bounded from below and satis…es (PS) condition. Then J is coercive. The converse statement (coercivity implying the Palais-Smale condition) is valid in a …nite dimensional space only. We note that with the aid of the Palais-Smale condition, we can formulate a version of the critical point theorem which serves as a counterpart of a direct method in the calculus of variation. Namely, it can be applied when a functional is not weakly l.s.c. or else we may look at it as some generalization of the Weierstrass Theorem. Theorem 4 (Proposition 10.1 [18]). Let E be a Banach space and J : E ! R be a C 1 functional which satis…es the Palais-Smale condition. Suppose in 6

addition that J is bounded from below. Then the in…mum of J is achieved at some point u0 2 E and u0 is a critical point of J, i.e. J 0 (u0 ) = 0.

Now we have enough background to present main result from [15]. Using the Mountain Pass Lemma (which can be cited for example in the following form) Theorem 5 (Mountain Pass Theorem). Let E be a Banach space and assume that J 2 C 1 (E; R) satis…es the Palais-Smale condition. Assume that inf J(x)

kxk=r

maxfJ(0); J(e)g,

(1)

where 0 < r < kek and e 2 E. Then J has a non-zero critical point x0 . If moreover inf kxk=r J(x) > maxfJ(0); J(e)g, then also x0 6= e.

and ideas contained in the proof of Theorem 1 (see again [18] for some nice version of the proof or else our introductory remarks) the Authors in [15] proved the result concerning di¤eomorphism between a Banach and a Hilbert space:

Theorem 6. Let X be a real Banach space, H - a real Hilbert space. If f : X ! H is a C 1 -mapping such that: for any y 2 H the functional ' : X ! R given by the formula 1 ' (x) = kf (x) yk2 2 satis…es Palais-Smale condition, for any x 2 X, f 0 (x) 2 Isom (X; H) then f is a di¤eomorphism. The question aroused whether the Hilbert space H in the formulation of the above theorem could be replaced by a Banach space. This question is of some importance since one would expect di¤eomorphism to act between two Hilbert spaces or two Banach spaces rather than between a Hilbert and a Banach space. The applications given in [15] work when both X and H are Hilbert spaces. We see that given a Hilbert space H, relation x 7! 21 kxk2 can be treated as x 7! 12 hx; xi, where h ; i is the scalar product. The other point of view is to treat x 7! 12 kxk2 as a potential of a duality mapping between H and H and …nally look at the composition of identity with some C 1 functional which is 0 only at 0 and with derivative sharing the same property. 7

4

An approach by an auxiliary functional

The version of the Mountain Pass Theorem which we use is sort of a counterpart of a three critical point theorem which is actually what is applied for the proof of Theorem 1. Theorem 7. [24, Theorem 2] Let X be a Banach space and let J : X ! R be a C 1 functional satisfying Palais-Smale condition with 0X its strict local minimum. If there exists e 6= 0X such that J(e) 6 J(0X ), then there is a critical point x of J, with J(x) > J(0X ), which is not a local minimum. These observations will lead us towards obtaining the counterpart of Theorem 6 in a Banach space setting as well as related implicit function results in the spirit of introductory remarks. The application of Theorem 5 requires performing (1) while the application of Theorem 7 that some points de…ne local minima for the auxiliary functional. But the latter property is easy to be checked since this functional is nonnegative and it being zero at some set other that an isolated singleton means that the local invertibility is violated. Our main result now reads and is based on the main result from [8] which we improve by getting rid of one of the assumptions and by using a suitably shortened proof. Theorem 8. Let X; B be real Banach spaces. Assume that f : X ! B is a C 1 -mapping, : B ! R+ is a C 1 functional and that the following conditions hold 0 (c1) ( (x) = 0 () x = 0) and (x) = 0 () x = 0 ; (c2) for any y 2 B the functional ' : X ! R given by the formula ' (x) = (f (x)

y)

satis…es Palais-Smale condition; (c3) f 0 (x) 2 Isom (X; B) for any x 2 X. Then f is a di¤eomorphism. Proof. We follow some ideas applied in [1] with necessary modi…cations. In view of the remarks made above condition (c3 ) implies that f is a local di¤eomorphism. Thus it is su¢ cient to show that f is onto and one to one. Firstly we show that f is onto. Let us …x any point y 2 B. Observe that ' is a composition of two C 1 mappings, thus ' 2 C 1 (X; R). Moreover, ' is 8

bounded from below and it satis…es the Palais-Smale condition. Thus from Theorem 4 it follows that there exists at least one argument of a minimum which we denote by x. We see by the chain rule for Fréchet derivatives and by Fermat’s Principle that 0

' (x) =

0

y) f 0 (x) = 0.

(f (x)

Since by (c3) mapping f 0 (x) is invertible we see that by (c1) it follows that f (x) y = 0.

0

(f (x)

y) = 0. Now

Thus f is surjective. Now we argue by contradiction that f is one to one. Suppose there are x1 and x2 , x1 6= x2 , x1 , x2 2 X, such that f (x1 ) = f (x2 ) = a 2 B. We will apply Theorem 7. Thus we put e = x1 x2 and de…ne functional : X ! R by the following formula (x) = (f (x + x2 )

a) :

By (c2 ) functional satis…es the Palais-Smale condition. Next we see by a direct calculation that (e) = (0) = 0: Moreover, 0 is a strict local minimum of , since otherwise, in any neighbourhood of 0 we would have a nonzero x with f (x + x2 ) = a and this would contradict the fact that f de…nes a local di¤eomorphism. Thus by Theorem 7 we note that has a critical point v such that (v) > 0. Since v is a critical point we have 0

(v) =

0

a) f 0 (v + x2 ) = 0:

(f (v + x2 )

0

Since f 0 (v + x2 ) is invertible, we see that (f (v + x2 ) a) = 0. So by the assumption (c1) we calculate f (v + x2 ) a = 0. This means that (v) = 0 which is impossible. We supply our result with a few of remarks. Remark 1. We see that from Theorem 8 by putting (x) = 21 kxk2 we obtain easily Theorem 6. Moreover, in [8] the following condition is additionally assumed: (c4) there exist positive constants , c, M such that (x)

c kxk

for kxk

M.

We do not need this condition now since we use a di¤erent tool for the proof the main result. Moreover, the proof becomes considerably simpler and the result is a full counterpart of the Hilbert space setting case. 9

We conclude with example of a functional satisfying our assumptions. Let us take for p 2 a uniformly convex Banach space W01;p ([0; 1]; R) := W01;p consisting of absolutely continuos functions x : [0; 1] ! R such that x(0) = x(1) = 0 and x0 2 Lp ([0; 1] ; R) considered with a usual norm jjxjjW 1;p = ( 0

Z1

p

jx0 (t)j dt)1=p ; x 2 W01;p :

0

We also refer to [2] for some background on this space. Lemma 1. Assume that 2 p < +1. A functional f : Lp ! R given by R1 the formula f (u) = ju(s)jp ds = kukpLp is continuously di¤erentiable and 0

for any u 2 Lp we have

f 0 (u)v = p

Z1

ju(t)jp

2

u(t)v(t)dt for v 2 Lp .

0

Remark 2. In case p = 2 the assertion of Lemma 1 follows from the properties of the scalar product. In case 2 < p < +1, from Lemma 1 and from the chain rule, it follows that a functional h : W01;p ! R de…ned by 1 h (u) = p

Z1

p

ju0 (t)j dt

(2)

0

is continuously di¤erentiable and for any u 2 W01;p we have h0 (u)v =

Z1

p 2

ju0 (t)j

u0 (t)v 0 (t)dt for v 2 W01;p .

0

Thus we see that both f and h above provide us with examples.

5

An approach by a duality mapping

In this section we improve results from [9] by extending them to cover the case of any duality mapping relative to some increasing function and not 10

the special function t ! jtjp 1 for p > 1, as provided there. The second improvement is the simpli…cation of the proof. A normed linear space E is called strictly convex if the unit sphere contains no line segments on its surface, i.e., condition kxk = 1; kyk = 1; x 6= y implies that 1 (x + y) < 1 2 or in other words that the unit sphere is a strictly convex set. The space E is called uniformly convex, if for each " 2 (0; 2] there exists (") > 0 such that if kxk = 1; kyk = 1 and kx yk ", then kx + yk 2 (1 (")). A uniformly convex space is necessarily strictly convex and re‡exive. We recall from [3] the notion of duality mapping from E into E relative to a normalization function. We shall write duality mapping in the sequel with the understanding that we mean a duality mapping relative to some normalization function. A continuous function ' : R+ ! R+ is called a normalization function if it is strictly increasing, '0) = 0 and '(r) ! 1 with r ! 1.. A duality mapping on E corresponding to a normalization function ' is an operator A : E ! 2E such that for all u 2 E and u 2 A (u) kA (u)k = ' (kuk) , hu ; ui = ku k kuk : We proceed to some remarks on a duality mapping which are in order especially concerning the assumptions on a duality mapping. We recall from [3] that (i) for each u 2 E, A (u) is a bounded, closed and convex subset of E ; (ii) A is monotone hu1

u2 ; u1

u2 i

(' (ku1 k)

' (ku2 k)) (ku1 k

ku2 k)

for each u1 ; u2 2 E, u1 2 A (u1 ) ; u2 2 A (u2 ) ; (iii) for each u 2 E, A (u) = @ (u), where @ ( ) : E ! 2E denotes the subdi¤erential in the sense of convex analysis of the functional (u) = R kuk ' (t) dt, i.e. 0 n o @ (u) = u 2 E : (y) (u) hu ; y uiE ;E for all y 2 E ;

(iv) if E is strictly convex, then card(A (u)) = 1, for all x 2 E; (v) when E is re‡exive and strictly convex, then operator A : E ! E is demicontinuous, which means if xn ! x in E then A (xn ) * A (x) in E . 11

We see by Proposition 2.8. from [22] that since A is demicontinuous (continuous "norm to weak" in the original terminology) we obtain that functional is di¤erentiable in the sense of Gâteaux and operator A provides its derivative. Lemma 2. Assume that E is a re‡exive Banach space with a strictly convex dual E . Then the duality mapping A : E ! E corresponding to a normalization function ' is single valued and the functional is di¤erentiable in the sense of Gâteaux with A being its Gâteaux derivative. Proof. The duality mapping A is now single valued and its potential, i.e. , has a single valued subdi¤erential which is demicontinous. Then the argument proceeding the formulation …nishes the proof. In view of the above Lemma, it is apparent that assuming the continuous di¤erentiability of a potential of a duality mapping is not a very restrictive condition. Indeed, in order to get some reasonable result we will have to assume that the duality mapping has the potential which is continuously di¤erentiable in order to obtain that the functional h : X ! R given by the formula h (x) = (f (x) y) is continuously di¤erentiable. Moreover, functional can be considered as a potential of a duality mapping A in case it is single-valued. In this case we write A : B ! B and by writing A : B ! B we implicitly assume that A is single valued. We will follow this observation in the sequel. We formulate our second global di¤eomorphism result: Theorem 9. Let X; B be a real Banach spaces. Let the potential of a duality mapping A : B ! B corresponding to a normalization function ' be continuously Gâteaux di¤erentiable. Assume that f : X ! B is a C 1 mapping such that: d1) for any y 2 Y the functional h : X ! R given by the formula h (x) =

(f (x)

satis…es the Palais-Smale condition, d2) f 0 (x) 2 Isom (X; B) for any x 2 X. Then f is a di¤eomorphism.

12

y)

Proof. Let us …x a point y 2 B. Functional h is a composition of two C mappings, so it is C 1 itself. Moreover, h is bounded from below and it satis…es the Palais-Smale condition by d1). Thus from Theorem 4 it follows that there exists an argument of a minimum which we denote by x. We see by the chain rule and Fermat’s Principle and by the assumptions on a duality mapping that 0 0 0 = h (x) = A (f (x) y) f (x) . 1

0

Since by d2) mapping f (x) is invertible we get that A (f (x) by the property that kA (u)k = ' (kuk) we note that kA (f (x)

y)k = ' (kf (x)

y) = 0. Now,

yk) :

So it follows since ' (0) = 0 and since ' is strictly increasing that f (x)

y = 0;

which proves the existence of x 2 X for every y 2 Y , such that f (x) = y. The uniqueness can be shown by contradiction exactly as before. Now we provide a simple example of a space and a duality mapping for which the assumptions of the above result hold. Let us de…ne a single valued operator A : W01;p ! W01;p such that hAu; vi =

Z1

p 2

ju0 (t)j

u0 (t)v 0 (t)dt:

0

From Remark 2 it follows that if 2 p < +1; then A is a potential operator and its C 1 potential is h de…ned by (2). Therefore by the cited remarks from [3] we get that a duality mapping on W01;p corresponding to a normalization function t ! tp 1 is de…ned by A.

6

Remarks on solvability of nonlinear equations and on implicit function theorems

In this section we shall concentrate on the …rst part of the proof of a diffeomorphism theorem which in fact provides the solvability of a nonlinear equation f (x) = y; 13

where f : X ! B is at least Gateaux di¤erentiable (and continuous) and y 2 B (is arbitrarily …xed). This is due to the slightly less demanding version of Theorem 4 which is given as below following [5] Theorem 10. Let E be a Banach space and J : E ! R be a Gateaux differentiable lower semi-continuous functional which satis…es the Palais-Smale condition. Suppose in addition that J is bounded from below. Then the in…mum of J is achieved at some point u0 2 E and u0 is a critical point of J, i.e. J 0 (u0 ) = 0. The above result follows directly by the Ekeland’s Variational Principle. Thus we obtain the following general solvability tool Corollary 2. Let X; B be real Banach spaces. Assume that f : X ! B is a Gateaux di¤erentiable continuous mapping, : B ! R+ is a C 1 functional and that the following conditions hold 0 (e1) ( (x) = 0 () x = 0) and (x) = 0 () x = 0 ; (e2) for any y 2 B the functional ' : X ! R given by the formula ' (x) = (f (x)

y)

satis…es Palais-Smale condition; (e3) for any x 2 X the Gateaux derivative f 0 (x) : X ! B is right invertible, that is there exists a linear mapping g (x) : B ! X such that f 0 (x) g (x) = idX : Then for any …xed y 2 B equation f (x) = y has at least one solution. It could be assumed that ' is lower-semicontinuous instead of continuity of g however this seems to be very di¢ cult to be veri…ed. Obtaining now global implicit functions theorems seems a straightforward task. The classical local version of implicit function theorem, to be found in many textbooks for example in [27], is as follows Theorem 11. Let X, Y , Z be real Banach spaces. If U X Y is an open set, F = F (x; y) : U ! Z is of class C 1 , F (a; b) = 0 and the di¤erential Fx (a; b) : X ! Z is bijective, then there exist balls B(a; r); B(b; ) and a function f : B(b; ) ! B(a; r) such that B(a; r) B(b; ) U and (i) equations F (x; y) = 0 and f (y) = x are equivalent in the set B(a; r) 14

B(b; ) (ii) function f is of class C 1 with di¤erential f 0 (y) given by f 0 (y) =

[Fx (f (y); y)]

1

(3)

Fy (f (y); y);

for y 2 B(b; ).

We shall provide two versions of a global implicit function theorem. Both depend on whether we apply a duality mapping or else if we apply an auxiliary functional. While for the applications which we mean both versions can be used, these di¤er as far as the assumption on the underlying spaces are concerned. In this case we do not need to assume anything additional about the spaces under consideration. Theorem 12. Let X; Y; Z be a real Banach spaces. Assume that F : X Y ! Z is a C 1 -mapping, : Z ! R+ is a C 1 functional such that the following conditions hold: f1)

(z) = 0 () z = 0 and

0

(z) = 0 () z = 0,

f2) for any y 2 Y the functional ' : X ! R given by the formula ' (x) = (F (x; y)) satis…es the Palais-Smale condition, f3) for any (x; y) 2 X

Y the di¤erential Fx (x; y) : X ! Z is a bijection.

Then there exists a unique function f : Y ! X such that equations F (x; y) = 0 and x = f (y) are equivalent on X Y . Moreover, f 2 C 1 (Y; X) with di¤erential given by (3). Theorem 13. Let X; Y; Z be a real Banach spaces. Let the potential of a duality mapping A : Z ! Z corresponding to a normalization function ' be continuously Gâteaux di¤erentiable. Assume that F : X Y ! Z is a C 1 mapping such that: g1) for any y 2 Y the functional h : X ! R given by the formula h (x) =

(F (x; y))

satis…es the Palais-Smale condition, g2) di¤erential Fx (x; y) : X ! Z is bijective for any (x; y) 2 X

Y.

Then there exists a unique function f : Y ! X such that equations F (x; y) = 0 and x = f (y) are equivalent in the set X Y . Moreover, f 2 C 1 (Y; X) with di¤erential given by (3). 15

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