Degree Theory

8/13/2019 Degree Theory

1/38


2/38

IntroductionMany problems in mathematics and mathematical physics can be reduced to a study of the set of solutions for an equation f (x) = y, where f is a map between appropirate spaces X and Y , andy is element of Y . Useful information about the solution set is obtained by means of topologicaldegree theory. The topological degree is mainly used in the study of ordinary and partial differentialequations, integral equations, bifurcation theory and in proving xed point theorems. Degree theoryprovides also a natural technique for proving so-called mapping theorems which include resultsconcerning the invariance of domain and generalizations of the Jordan curve theorem.

Let G be an open subset of some topological space X and f a given map of the closure of Ginto another topological space Y . For each point y of Y outside the image of boundary of G underf we assign an integer d(f ,G,y ). The integer-valued function d is said to be a topological degreeif it has certain normalizing, additivity and homotopy properties. The integer d(f ,G,y ) is a roughestimate of the number of solutions for the equation f (x) = y. In particular, if d(f ,G,y ) differsfrom zero, then there exists at least one solution of the equaiton f (x) = y.

The notion of the topological degree was rst introduced by Brouwer in 1912 [3] for continuousmappings in nite demensional spaces. The Brouwer degree can be dened either by using algebraictopology or analytically (see e.g. [11] and [19]). Leray and Schauder generalized in 1934 [18] thedegree theory for compact perturbations of identity in innite dimensional Banach spaces. Theconstruction of the Leray-Schauder degree is based on the Brouwer degree.

Since 1934, various generalizations of degree have been dened (see e.g. [19], [24]). However, itwas not untill 1972 and 1973 that F uhrer [13] and Amann and Weiss [1] proved the Brouwer degreeto be unquely determined by only a few conditions. These conditions provide a natural basis forthe formal denition of a classical topological degree (see [9], [19]).

In a series of articles [5], [6], [7], [8] and [9], F.E. Browder has recently extended the concept of atopological degree to nonlinear mappings of monotone type from a real reexive Banach spaces intoits dual space. His method is based on Galerkin approximations for which the nite dimensional

Brouwer degree is well-dened.The interest in the denition of nonlinear mappings of monotone type arises from the fact thatthese properties can be veried under concrete hypotheses for the maps between Sobolev spacesobtained from the elliptic operators of the generalized divergence form (see e.g. [9] p.18 and [17]).

In this paper we are mainly interested in the construction and uniqueness of the degree and itsdirect consequences. In applications we restrict our discussion to an abstrat level. Our constructionof the degree is based on the Leray-Schauder degree and we shall follow the approach introducedin [2].

In the rst chapter, we shall formally dene the concept of topological degree and examinethose classes of mappings for which the degree will be constructed later. We shall further extend

the notion of a degree, in a slightly modied form, to a broader class of quasimonotone mappings.In the thrid chapter, we shall study how the degree theory can be used to deduce existence andsurjectivity results for equations involving nonlinear mappings of monotone type.

In our nal chapter, we rst study the degree of a composition of mappings; a generalization of the multiplication theorem of the Leray-Schauder theory is achieved. At the end we examine brieythe connectedness of the solution set and the existence of multiple solutions.

For the general theory and applications we refer to [4], [9], [11], [19], [?] and for the denitionsand properties nonlinear mappings of monotone type to [23], [29]. Some elementary facts about theweak topology and weak convergence are needed; we refer to [25], [28].

2


3/38

1 Preliminaries

1.1 Formal Degree Theory

Denition 1.1 Let X and Y be topological spaces and let O be a class of open subsets G of X . For

each G O , we associate a class F G of maps from G into Y . For each G O , we also associate a class H G of maps [0, 1] G into Y (permissible homotopies ); if H H G , then we simply write h(t, x ) = f t (x) and speak about the homotopy f t ; 0 t 1, where t is called the parameter of the homotopy f t . For any f F G ; G O , and for any y Y \ f (G) , we associate an integer d(f ,G,y ).

The integer-valued function d is said to be a classical topological degree if the following conditions are satised:

(a) If d(f ,G,y ) = 0 , there exists a solution of the equation f (x) = y in G.

(b) (Additivity) If D G O ; D O and f F G , then the restriction f |D F D (the restricted

map is usually denoted by the same symbol ). Let G1, G2 be a pair of disjoint subsets of Gbelonging to O and suppose that y / f (G \ (G1 G2)) . Then

d(f ,G,y ) = d(f, G 1, y) + d(f, G 2, y).

(c) (Invariance under Homotopy) If f t , 0 t 0, is a homotopy in H G , then f t F G for each xed t [0, 1], and if {y(t) : t [0, 1]} is continuous curve in Y with y(t) / f t (G) for any t [0, 1], then d(f t ,G ,y(t)) is constant in t [0, 1].

(d) (Normalizing) There exists a normalizing map j : X Y such that j |G F G for each G O , and if y j (G), then

d( j, G, y ) = 1 .

The form of general theory is based on the classical degree theory in nite dimensional spacesgiven by Brouwer in 1912 [3], where exactly the properties (a) - (d) listed above guarantee theuniqueness of the degree function (see [1] or [9]). Indeed, if in Denition 1.1 we choose X = Y = R n ,O all open bounded sets, F G all continuous maps of G into R n , H G all continuous homotopies, i.e.,continuous maps of [0, 1] G into R n , and j = I ; the identity map, then we get the unique Brouwerdegree. The unique Leray-Schauder degree obtained by choosing X = Y a real innite dimensionalBanach space, O all open bounded sets, F G all continuous maps f : G X such that I f is compact, i.e., it takes bounded sets into relatively compact sets, H G the class of continuous

homotopies of the form I T , where T : [0, 1] G X is compact, and j = I the identity map.We shall now study more closely the conditions imposed on a degree function in Denition 1.1.In view of our further study it is useful to make the following simplifying assumptions.

From now on we assume that X and Y are normed spaces and O consists of open and boundedsubsets of X . Suppose that H G contains at least the affine homotopies:

f t = (1 t)f 0 + tf 1; t [0, 1],

where f 1, f 2 F G . Since every f F G can be written in the form

f = (1 t)f + tf, t [0, 1],

3


4/38

f can be regarded as a constant homotopy of class H G . Moreover, assume that for any homotopyf t in H G and for any continuous curve {y(t): t [0, 1]} in Y , also f t y(t) determines a permissiblehomotopy. In particular, f y F G whenever f F G and y Y . Let us further assume thatf F G maps closed subsets of G onto closed subsets of Y . In particular, f (G) and f (G) are closedin Y . Now we deduce three theorems which follow directly from the general structure of degreetheory. These results, much used below show characteristic features of every degree theory.

Theorem 1.1 (Boundary Dependence) Suppose that f , g F G are such that f = g on G and y Y \ f (G). Then

d(f ,G,y ) = d(g,G,y).

Proof: Consider an affine homotopy f t = (1 t)f + th ; t [0, 1]. Since f t (x) = f (x) for allt [0, 1] and x G, we have y / f t (G) for any t in [0, 1], which by the property (c) of Deni-tion 1.1 implies that d(f ,G,y ) = d(h,G,y ).

For the next theorem, recall that by an open component of an open set D we mean anequivalence class of points which can be connected by a continuous curve in D.

Theorem 1.2 Let G O and f F G be xed. Then d(f,G, ) is a function of the image point y Y \ f (G) and constant on each open component of the open set Y \ f (G) .

Proof: Let be an open component of Y \ f (G) and let y and q be two xed elements in .There is a continuous curve {y(t) : 0 t 1} in which connects y and q . By the homo-topy invariance property (c) we thus nd that d(f ,G,y (t)) is constant in t on [0, 1]. In particular,d(f ,G,y ) = d(f ,G,q ), which completes the proof.

Theorem 1.3 Let f F G and y Y \ f (G) be given. Then f y F G , 0 Y \ (f y)(G) and d(f ,G,y ) = d(f y,G, 0).

Proof: Consider an affine homotopy f t = (1 t)f + t(f y), 0 t 1, and a continuous curvey(t) = (1 t)y, 0 t 1. Clearly y(t) / f t (G) for any t [0, 1]. Hence the assertion follows fromthe homotopy invariance property (c).

In view of Theorem 1.2, it is relevant to dene the degree with respect to the component of Y \ f (G) by setting

d(f,G, ) = d(f ,G,y ), where y .

Hence the values of d(f ,G,y ) and d(f ,G,q ) can differ only if y and q are in different componentsof Y \ f (G). Thus the condition y / f (G) cannot be avoided in the denition of a degree, sinceotherwise d(f ,G,y ) would be constant for all y, which contradicts the normalizing condition ( d) of Denition 1.1.

It is useful to examine to what extent the conditions from (a) to (d) are independent. Wecan easily see that (a) follows from the additivity property of (b). Indeed, if we rst chooseG = G2 = G2 = and apply (b), then d(f, , y) = 0. If now f F G and y Y \ f (G), then bychoosing G1 = G2 = and applying (b) again we obtain d(f ,G,y ) = d(f, , y) + d(f, , y) = 0.Thus if d(f ,G,y ) = 0, then y f (G) and hence y f (G) which is precisely the condition (a).

The conditions (b), (c) and (d) are independent to each other.

4


5/38

Remark 1.1 It should be noted that once a degree function is constructed for some class F G , thenit is usually possible to choose H G so simple that (c) is satised. On the other hand, the practicalvalue of a degree theory depends on the homotopies available. Thus it is important to pay attentionto the choice of permissible homotopies; at least to affine homotopies should be included.

Remark 1.2 It is useful to note that it usually suffices to construct a degree function only withrespect to the origin. Moreover, suppose that the normalizing map j : X Y is bijective and itsinverse j 1: Y X is locally bounded. Let us now suppose that a degree function d is constructedwith respect to the origin, i.e., it satises the conditions (a), (b) , (d) with y = 0 and the invarianceunder homotopy (c) with respect to the curve y(t) = 0; t [0, 1]. Moreover, we dene

d(f ,G,y ) = d(f y,g, 0) for all y Y \ f (G).(1)

Then the conditions from (a) to (d) hold also in their original form. Indeed, conditions (a), (b) and(c) are clearly satised. Too see (d), assume that y j (G) and denote j t = (1 t) j + t( j y) = j ty; t [0, 1]. Then there exists R > 0 and k > 0 such that

j t (x) Y k for all t [0, 1] and for all x X R.(2)The proof of this fact will be seen later. Hence 0 / j t (B R (0) for any t [0, 1]. The already knownproperties of d together with (3) imply that

d( j, G, y ) = d( j, B R (0), y) = d( j y, B R (0), 0) = d( j, B R (0), 0) = 1 .

1.2 On the Mappings of Monotone Type

Let X be an innite dimensional real reexive Banach space and X its dual space, and let , stand for the pairing between X and X . By the results due to Trojanski, every reexive Banachspace has an equivalence norm such that X and X are both locally uniformly convex. Thus weassume from now on that X and X are locally uniformly convex. This renorming is needed for thedenition of the duality mapping at the end of this section.

Denition 1.2 Let G be an open bounded subset of X . Let f : G X .

1. The map f is said to be monotone if for all x, u G, we have

f (x) f (u), x u 0,(3)and strictly monotone if it is monotone and equality holds in (3) only if x = u.

2. The map f is said to be of class (S )+ if for every sequence {x j } G with x j x X for which

lim sup j

f (x j ), x j x 0,

we have x j x X .

5


6/38

3. The map f is said to be pseudomonotone if for every sequence {x j } G with x j x X for which

lim sup j

f (x j ), x j x 0,

we have lim

j f (x j ), x j x = 0,

and if x G, then f (x j ) f (x) in X .

4. The map f is said to be quasimonotone if for every sequence {x j } G with x j x X ,we have

lim sup j

f (x j ), x j x 0.

Denition 1.3 Let f : X X .

1. The map f is said to be bounded if it takes bounded subsets of X into bounded subsets of X .

2. The map f is said to be locally bounded if for each x X there is a neighborhood U of xsuch that f (U ) is a bounded subset of X .

3. The map f is said to be demicontinuous if for each {x j } in X such that x j x X , we have

f (x j ) f (x) in X .

We note that every demicontinuous map is locally bounded. In fact, if there is x X such that,for each j , f (B (x, 1/j )) is not bounded, then choosing x j B(x, 1/j ) such that f (x j ) j we seethat x j x in X and the sequence {f (x j )} is not bounded in X . Hence {f (x j )} cannot convergeweakly in X so that f is not demicontinuous. Also, it is easy to see that every demicontinuous

map of class (S )+ is pseudomonotone. The class of demicontinuous quasimonotone maps is quitelarge. Indeed, a demicontinuous map f : G X is quasimonotone if for instance

(i) f is of class (S )+ or pseudomonotone;

(ii) f is monotone and bounded or dened on the weak closure of G;

(iii) f is compact;

(iv) f is continuous from the weak topology of X to the strong topology of X . This holds if f islinear and compact.

In this paper the class ( S )+ is of great importance. Denote by F G the family of all demicontinuousmaps of class (S )+ of G into X and by F G,B all bounded maps in F G . If f F G , then the preimageof any compact subset of X is compact in X . To prove this we need the boundedness of closure of G. As a special case we see that f 1(y), the solution set of the equation f (x) = y on G, is compactfor each y X . An immediate consequence is that f takes closed subsets of G into closed subsetsof X . In particular, the sets f (G) and f (G) are closed.

For a demicontinuous pseudomonotone map f : G X , f (A) is closed whenever A G isweakly closed in X , for instance closed and convex.

In view of application it is useful to know under what kind of perturbations F G remains stable.This is the content of the next lemma.

6


7/38

Lemma 1.1 Let h: G X be a demicontinuous map. Then the following conditions are equiva-lent:

(i) h is quasimonotone;

(ii) For each sequence {x j } G with x j x , we have

lim inf j

h(x j ), x j x 0.

(iii) f + h F G for all f F G .

Proof: Assume that h is quasimonotone. If (ii) does not hold, then there exists a sequence {x j } Gwith x j x such that lim inf h(x j ), x j x < 0. Then, at least for a subsequence {xk } of {x j },we have

lim supk

h(xk ), xk x = limk

h(xk ), xk x = liminf j

h(x j ), x j x < 0,

which contradicts the assumed quasimonotonicity of h. Thus (ii) follows from (i).Suppose next that (ii) holds true. Let f F G be given and let {x j } G such that x j x in

X withlim sup

j f (x j ) + h(x j ), x j x 0.

Now, in view of (ii), we have

lim sup j

f (x j ), x j x lim sup j

f (x j ) + h(x j ), x j x lim inf h(x j ), x j x 0,

which implies that x j x in X since f F G . Hence f + g F G and (iii) follows from (ii).Assume now that (iii) holds true. If h is not quasimonotone, then there exists a sequence

{x j } G such that x j x in X and

lim sup j

h(x j ), x j x = q < 0.

Passing to a subsequence, if necessary, we may assume that

lim j

h(x j ), x j x = q < 0.

Let f be a bounded map in F G . Such a map exists as we shall see later, for instance, the dualitymapping. It is clear that f is of class (S )+ for any > 0 and by (iii) also f + h F G . Sinceevery map of (S )+ is quasimonotone, we get

0 lim sup j f (x j ) + h(x j ), x j x = lim sup j f (x j ), x j x q.

But the right-hand side cannot remain nonnegative for all > 0 since f is bounded. Hence wereach a contradiction showing that h is quasimonotone and the proof is complete.

If f and g are in F G , then in view of the previous lemma

f F G for all > 0 and f + h F G .

Similarly, if f : G X and h: G X are monotone(pseudomonotone, quasimonotone), then f for all 0 and f + h are monotone(pseudomonotone, quasimonotone). Hence these classes of maps have a conical structure.

7


8/38

Remark 1.3 Let f F G be given and g: G X compact. Then f + g F G for all > 0 byLemma 1.1. However, since also g is quasimonotone, we necessarily have that

lim j

g(x j ), x j x = 0

for each sequence {x j } G with x j x in X . Thus g cannot be of the class (S )+ and, consequently,

F G contains no compact maps.Remark 1.4 If X is a Hilbert space, then certainly the identity map I is of class (S )+ . Thus thosemappings for which the Leray-Schauder degree is dened, i.e., maps of the form I g, g compact,are of class (S )+ .

We shall now dene suitable homotopies in view of degree theory. A map H : [0, 1] G X issaid to be a homotopy of class (S )+ if for every sequence {x j } G with x j x in X and forevery sequence {t j } [0, 1] with t j t in [0, 1] for which

lim sup j

H (t j , x j ), x j x 0

we have x j x in X and H (x j , t j ) H (x, t ) in X . We denote by H G (or H G,B ) those homotopiesof class (S )+ which are locally bounded(or bounded) as maps of [0, 1] G into X . Clearly everyH H G is a demicontinuous map of [0, 1] G into X and H (t, ) F G for any xed t in [0, 1]. Itis not hard to prove( see ......page 27) that H G contains all affine homotopies between any pair of maps of F G . The same conclusion holds for H G,B . Furthermore, we say that a demicontinuous mapH : [0, 1] G X is a quasimonotone homotopy if for every sequence {x j } G with x j xin X and {t j } [0, 1] with t j t in [0, 1], we have

lim sup j

H (x j , t j ), x j x 0.

It is easy to see that quasimonotone homotopies include affine homotopies between any pair of demicontinuous maps with common domain G. Clearly every H H G is also a quasimonotonehomotopy.

Remark 1.5 Pseudomonotone homotopies can also be dened. This class, although important inapplication, is somewhat unsatisfactory since it does not contain affine homotopies. However, bothpseudomonotone homotopies and affine homotopies between pseudomonotone maps are containedin the class of quasimonotone homotopies.

The following result generalizes Lemma 1.1; its proof is omitted since it is similar to that of Lemma 1.1.

Lemma 1.2 Let h: [0, 1] G X be a demicontinuous map. Then the following conditions are equivalent:

(i) h is a quasimonotone homotopy;

(ii) For every sequence {x j } G and {t j } [0, 1] with x j x and t j t in [0, 1], we have

lim inf j

h(x j , t j ), x j x 0.

8


9/38


10/38

2 Construction of Degree

2.1 Bounded Mappings of Class ( S ) +

Let X be an innite dimensional real reexive separable Banach space, renormed if necessary,

such that X and X are locally uniformly convex. Let G be an open bounded subset of X , nonemptyunless otherwise indicated. Denote by F G,B all demicontinuous maps of class ( S )+ of G into X and by H G,B all bounded homotopies of class (S )+ of [0, 1] G into X . As a normalizing map wetake the duality mapping J .

We rst recall the following embedding theorem due to Browder and Ton:Proposition 2.1 Let X be a reexive separable Banach space. Then there exists a separable Hilbert space W and a linear compact injection : W X such that (W ) is dense in X .

Using Proposition 2.1 and Frechet-Riesz theorem we dene further a map : X W by setting

((w)|v) = w, (v) for all v W, w X (4)

where (| ) stands for the inner product of W . It is easily seen that is linear and compact, andsince (W ) is dense in X , it follows that is also injective.To each f F G,B we now associate family of maps {f : > 0} dened by

f = I + 1

f , > 0,(5)

where I denotes the identity map of X . For any xed > 0, f maps G into X . Moreover, f is of type I g, g compact, a mapping for which the Leray-Schauder degree is dened.

We shall now examine how the family {f : > 0} is related to f . Since every map f F G,B isalso a constant homotopy, we focus our attention on the class H G,B .

Lemma 2.1 Assume that H H G,B and denote f t (x) = H (t, x ) for all t [0, 1] and x G. If A G is a closed set with 0 / f t (A) for any t [0, 1], then there exists such that

0 / (f t ) (A) for all t [0, 1] and 0 < < .(6)

Proof: If (6) does not hold, there exists a sequence { j }, {t j } [0, 1] and {x j } A with j 0+and (f t j ) j (x j ) = 0 for all j . At least for a subsequence we can write

x j x X, t j t [0, 1] and f t j (x j ) w X .

We thus havex j = 1

j f t j (x j ),(7)

and since {x j } is bounded, we nd that f t j (x j ) 0. On the other hand, is compact andlinear and therefore f t j (x j ) (w). By the injectivity of it can be deduced that w = 0,i.e.,

f t j (x j ) 0.(8)

Using (4) and (7) we get

f t j (x j ), x j = 1 j

f t j (x j ), f t j (x j ) = 1 j

f t j (x j ) 2W 0.

10


11/38

Thus by (8)

lim sup f t j (x j ), x j x = limsup 1 j

f t j (x j ) 2W 0,

which implies that x j x A and f t j (x j ) f t (x) in X . Since the weak limit is unique, we ndby (8) that f t (x) = 0. But this contradicts our hypothesis 0 / f t (A) for any t [0, 1]. Hence theassertion (6) is true and the proof is complete.

If we restrict ourselves with a constant homotopy, i.e., a map f F G,B , and choose A = G,then there exists > 0 such that f (x) = 0 for all x G and 0 < < , provided f (x) = 0 forall x G. This in turn means that the Leray-Schauder degree dLS (f , G, 0) is well-dened for all0 < < .

Our next lemma shows that the value of dLS (f , G, 0) remains stable as approaches zero.

Lemma 2.2 Let f F G,B and assume that f (x) = 0 for all x G. Then there exists > 0 such that f (x) = 0 for all x G and 0 < < and dLS (f , G, 0) is constant for all 0 < < .

Proof: The assertion follows from Lemma 2.1 and the homotopy invariance for the Leray-Schauderdegree. Indeed, if we take 0 < 1 < 2 < and consider and affine Leray-Schauder homotopybetween f 1 and f 2 , it is easily seen that

dLS (f 1 , G, 0) = dLS (f 2 , G, 0).

By the preceding lemma it is now relevant to give

Denition 2.1 For any f F G,B with 0 / f (G), dene

d(f,G, 0) = dLS (f , G, 0), 0 < < ,(9)

where is given by Lemma 2.2. Consequently, for any y X \ f (G),

d(f ,G,y ) = d(f y,G, 0).(10)

We are now in a position to prove that Denition 2.1 really gives us a classical topological degreein the sense of Denition 1.1.

Theorem 2.1 The integer-valued function d dened by (9) and (10) for all f F G,B satises the conditions (a) - (d) of Denition 1.1 with respect to the class H G,B of homotopies and with the duality mapping J as a normalizing map.

Proof: In view of Remark 1.2, it is sufficient to deal with the case y = 0 in (a), (b) and (d) and

y(t) = 0 , 0 t 1, in (c). We shall now verify the conditions (a)-(d) in this reduced form.(a) Assume that d(f,G, 0) = 0. If 0 / f (G), we can apply Lemma 2.1 for A = G and, consequently,we nd that there exists > 0 such that dLS (f , G, 0) = 0 for all (0, ). By (9) we getd(f,G, 0) = 0, which is a contradiction. Thus 0 f (G).(b) Let G1 and G2 be disjoint open subsets of G and assume that 0 / f (G \ (G1 G2)). ApplyingLemma 2.1 for A = G \ (G1 G2) we get > 0 such that 0 / f (A) for all (0, ). Since theadditivity property (b) holds for the Leray-Schauder degree, we have

dLS (f , G, 0) = dLS (f , G1, 0) + dLS (f , G2, 0)

for all (0, ), and thus (b) follows by (9).

11


12/38

(c) Let H H G,B and, as usual, denote f t (x) = H (t, x ). Assume that 0 / f t (G) for any t [0, 1].For each xed t [0, 1] and > 0 we have

(f t ) = I + 1

f t : G X,

and by Lemma 4 ( for A = G) there exists > 0 such that 0 / (f t ) (G) for all t [0, 1] and

(0, ). Since the homotopy H : [0, 1] D X is demicontinuous and is linear and compact,the map1

H : [0, 1] G X

is continuous for any xed > 0. Moreover, the boundedness of H together with the properties of implies that

1 H

is also compact. Thus, for a xed , (f t ) determine a permissible Leray-Schauder homotopy. Byusing Lemma 5 for each xed t [0, 1] and the Leray-Schauder homotopy invariance for each xed

(0, ), we have by (9)d(f t , G, 0) = dLS ((f t ) , G, 0) = constant

for all t [0, 1] and (0, ).(d)Assume that 0 J (G), which implies that 0 G. We consider an affine homotopy between I and J . Since

J (x), (1 t)I (x) + tJ (x) = x 2 + t

J (x) 2W > 0

for all x = 0 and 0 t 1, and since J (x) = 0 only for x = 0, we have

0 / [(1 t)I + tJ ](G) for all t [0, 1] and > 0.By the homotopy invariance for the Leray-Schauder degree we thus obtain

dLS (J , G, 0) = dLS (I,G, 0) = 1 for all > 0.

Hence (d) follows from (9). The proof is now complete.

2.2 Uniqueness of Degree

Our next aim is to prove that the degree function given by Denition 4 is unique. Suppose thereexists a degree, say d, which is dened for every map f F G,B and which satises the conditions(a) - (d) of Denition 1.1 with respect to H G,B and J . We shall prove that d = d. The proof isbased on the fact that the Leray-Schauder degree is unique. We rst give two lemmas.

Lemma 2.3 Let H = I T : [0, 1] G X be a Leray-Schauder homotopy, i.e., T is compact. Let M be an open bounded subset of X such that H (t, x ) M for all t [0, 1] and x G. For each f F M,B , we have f H H G,B .

12


13/38

Proof: Clearly f H : [0, 1] G X is bounded and demiconituous. Let sequences {t j } [0, 1]and {x j } G be such that

x j x, t j t andlim sup

j (f H )(t j , x j ), x j x 0.(11)

Since T : [0, 1] G X is compact, there exists a subsequence {T (tk , xk )} { T (t j , x j )} and anelement u X such that

T (tk , xk ) u in X.(12)Thus

H (tk , xk ) = xk T (tk , xk ) x u in X,and by denoting x u = z and H (tk , xk ) = z k we get z k z . Consequently we can write by (12)

lim supk

f (z k ), z k z = lim supk

(f H )(tk , xk ), xk T (tk , xk ) x + u

= lim supk

(f H )(tk , xk ), xk x

lim sup j

(f H )(t j, x

j), x

j x

0,

where the last inequality follows from (11). Since f is of class (S )+ , we have z k z in X and thus

xk = z k + T (tk , xk ) z + u = x.

The fact that the limit x is unique by (11) implies that the whole sequence {x j } converges to x,i.e., x j x. We further have

(f H )(t j , x j ) H (t, x ),which completes the proof.

Lemma 2.3 implies as a special case that J h F G,B whenever h: G X is a mapping of theLeray-Schauder type, i.e., I h is compact. Moreover, J h is continuous and (J h) 1(0) = h 1(0),since J (0) = 0 and the identity map is injective.

In our next lemma we use the uniqueness of the Leray-Schauder degree. We rst dene a newdegree function dLS by setting

dLS (h,G, 0) = d(J h,G, 0)(13)for all Leray-Schauder mappings h: G X with 0 / h(G). Furthermore, if y Y \ h(G)), wedene

dLS (h,G,y ) = dLS (h y,G, 0).(14)

Now we have

Lemma 2.4 dLS = dLS .

Proof: The assertion follows from the uniqueness of dLS if dLS satises the conditions (a) - (d) of Denition 1.1 in the form they have in the Leray-Schauder theory. It is sufficient to deal with thecase y = 0 in (a), (b) and (d) and y(t) = 0 , 0 t 1, in (c). The verications of the conditions(a) - (d) is now effected by using the fact that d is a degree function. In the proof of (c), invarianceunder homotopy, we also use Lemma 2.3. Thus dLS satises the conditions of the Leray-Schauderdegree and, by the uniqueness argument, we necessarily have dLS = dLS .

We are now in a position to prove

13


14/38

Theorem 2.2 Suppose that d is a degree function dened on F G,B satisfying conditions (a) - (d)with respect to H G,B and J . Then d coincides with the degree d given in Denition 2.1.

Proof:

2.3 Unbounded Mappings of Class ( S ) +

In this section we extend the degree theory to all demicontinuous mappings of the class ( S )+ ,not only to bounded ones as in Section 2.1. By reducing suitably the domain of a demicontinuousmap of class (S )+ we are able to use the results of Section 2.1 and 2.2 to dene the degree uniquely.The denition is based on the following four lemmas.Lemma 2.5 For each H H G and any compact set A X , the set

K = {x G: H (t, x ) A for some t [0, 1]

is a compact subset of X .

Proof: We may suppose that K = . Let {x j } K be a given sequence. By the denition of K there exists a sequence {t j } [0, 1] such that H (t j , x j ) A for all j . For some subsequences{xk }, {tk } and {H (tk , xk )} we can write tk t [0, 1], xk x X and H (tk , xk ) w A.Hence

lim supk

H (tk , xk ), xk x = 0,

which implies that xk x and H (tk , xk ) H (t, x ) = w. Consequently, we have H (t, x ) A whichmeans that x K . Thus every sequence in K has a strongly converging subsequence with the limitin K , which proves the compactness of K .

Lemma 2.6 Let S G be any xed compact set and assume that H H G . Then there exists an open set G and R > 0 such that

(i) S G G,

(ii) H (t, x ) R for all t [0, 1] and x G.

Proof: Let us assume that S = and introduce a family {Gn } of sets, one of which satises (i)and (ii). For each n = 1, 2, . . . , denote

Dn = x X : dist(x, S ) < 1n ,

where dist( x, S ) denotes the distance of x from S . The compactness of S allows us to write Dn as

Dn = x X : x z < 1n

for some z S .

Dening Gn = Dn G we get a family {Gn } of open sets satisfying (i). If none of the sets Gnsatises (ii), there exist sequences {tn } [0, 1] and {xn }, xn Gn , with

H (tn , xn ) + as n .(15)

14


15/38

Since xn Gn , there exists z n S such that xn z n 1/n and, moreover, the sequence {z k } S has convergent subsequence, say z k z S . Thus

xk z xk z k + z k z 0,

which implies that xk z . We may assume that {tk } [0, 1] converges, say tk t [0, 1], andby the demicontinuity of H we obtain H (tk , xk ) H (t, z ), which contradicts (15), since a weaklyconvergent sequence is always bounded. Hence one of the sets {Gn }, say Gn , satises the conditions(i) and (ii) ( in fact, the same is true for any Gn , n n ), we denote G = Gn .

Remark 2.1 Lemma 2.6 actually means that by restricting the domain of H H G to [0, 1] Gwe have H H G,B .

By applying Lemma 2.6 to a constant homotopy f F G and to the compact set S = f 1(0) weget

Lemma 2.7 Assume that f F G and 0 / f (G). Then there exists an open set G X such that

(i) f 1(0) G G,

(ii) f ( G) is bounded.

In view of Lemma 2.7 the restriction of f F G to G satises f F G,B and 0 / f ( G). Hencethe degree d(f, G, 0) is well-dened.

Lemma 2.8 Let f F G be given with 0 / f (G). Then the value of d(f, G, 0) does not depend on the choice of G, provided it satises the conditions (i) and (ii) of Lemma 2.7.

Denition 2.2 For any f F G with 0 / f (G), dene d(f,G, 0) = d(f, G, 0),(16)

where G is any open set satisfying (i) and (ii) of Lemma . Consequently, if y X \ f (G), we dene

d(f ,G,y ) = d(f y,G, 0).(17)

Theorem 2.3 The integer-valued function d given by Denition 2.2 for any f F G satises the conditions (a) - (d) of Denition 1.1 with respect of H G and J .

Proof:Theorem 2.4 Let d1 be any degree function which is dened on the class F G and which satises the conditions (a) - (d) of Denition 1.1 with respect of H G and J . Then d1 = d.

Proof:

15


16/38


17/38

Denition 2.3 For any demicontinuous quasimonotone map f : G X and for any y X \ f (G)we dene

dQM (f ,G,y ) = d(f + J,G,y ), 0 < < ,(19)

where > 0 is given by Lemma 2.9 for A = A.

Theorem 2.5 The integer-valued function dQM given by Denition 2.3 is a degree function in

the weak sense on the class of all demicontinuous quasimonotone maps, which is invariant under quasimonotone homotopies and normalized by J .

Proof:

Theorem 2.6 The degree function dQM in the weak sense given by Denition 2.3 is unique.

Proof:

Remark 2.2 In the view of the previous sections, it is appropriate to denote by the same symbold all the degree functions d, d and dQM dened in this chapter.

17


18/38

3 Existence and Surjectivity Results

3.1 Noncoercive Mappings

In this section we deduce some existence and surjectivity results, which follow directly from the

basic properties of the degree function (cf. [2]). Indeed, this shows the power of the degree theoryonce constructed. Results in this direction are given, for instance, in [23] without using degreetheory.Lemma 3.1 Let f F G be given and assume that there exists x G such that

f (x), x x > f (x) x x for all x G.(20)

Then d(f,G, 0) = 1 , and there exists at least one x0 G such that f (x0) = 0 .

Proof: (Cf. [2].) It follows from (20) that 0 / f (G) and therefore d(f,G, 0) is dened. Let J stand for the mapping J (x) = J (x x) for all x X . By using (20) and affine homotopies we get

d(f,G, 0) = d( J ,G, 0) = d(J,G,J (x)) ,

where d(J,G,J (x)) = 1 since J (x) J (G). Thus d(f,G, 0) = 1 which implies that 0 f (G).Indeed, we consider the affine (S )+ homotopy

f t (x) = t J (x) + (1 t)(J (x) J (x)) , x G.

Sincef t (x), x x = t J (x x), x x + (1 t) J (x) J (x), x x

we conclude, in view of the properties of J , that

f t (x) / 0 for all x G, t [0, 1].

Otherwise, for some x G and t [0, 1], we have

0 = t x x 2 + (1 t) J (x) J (x), x x .

It is clear that t = 1. Then we have

J (x) J (x), x x = t

1 tx x 2 0

which implies that x = x (since J is strictly monotone); a contradiction. By homotopy invariance,we have

d( J ,G, 0) = d(J J (x), G, 0) = d(J,G,J (x)) .

We next consider the ( S )+ homotopy

gt (x) = t J (x) + (1 t)f (x), x G, t [0, 1].

Obviously, gt (x) = 0 for all x G with t = 0, 1. In fact, gt (x) = 0 for all x G and t [0, 1].Indeed, if gt (x) = 0 for some x G and t (0, 1), then

f (x) = t

1 tJ (x x)

18


19/38


20/38

Since {f (x j )} f (G) f (G), we conclude that 0 f (G) = f (G). Since 0 / f (G) we have0 f (G).

Also, it is not hard to show that there exists 1 > 0 such that

(1 t)(f (x) + J (x)) + th (x) = 0

for all t [0, 1], (0, 1) and x G. For if there exist sequences {t j } [0, 1], { j } with j 0+

and {x j } G, thenf (x j ) = (1 t j ) j J (x j ) j t j J (x j x) 0 as j

so that 0 f (G) which contradicts the assumption that 0 / f (G). Hence, by Denition 2.3 andthe homotopy invariance for d,

d(f,G, 0) = d(f + J,G, 0) = d(h , G, 0) = 1

for any > 0 with 0 < < 1. Thus 0 f (G), and since 0 / f (G) and f (G) is closed, we seethere to exist x0 G such that f (x0) = 0.

Remark 3.3 If y is any element of X , we can replace f by f y in Theorem 3.1. Hence thesolvability of the equation f (x) = y follows, provided there exists some x G (depending on y)such that

f (x) y, x x > f (x) y x x for all x G.

Remark 3.4 A natural application of Theorem 3.1 occurs when G X is convex and f : G X is pseudomonotone. Indeed, a closed convex set is weakly closed and the image of a weakly closedset of X under a demicontinuous pseudomonotone map is always weakly closed.(see [2]).

Remark 3.5 For d we always have y f (G) if d(f ,G,y ) = 0 and f (G) is supposed to be closed.However, the duality mapping is the only map for which we know the converse to be true, i.e.,y J (G) implies that d(J,G, 0) = 0. In view of application it is desirable to nd other such maps

for whichd(f ,G,y ) = 0 if and only if y f (G).(22)

Assume that f : X X is a demicontinuous monotone map, G is an open bounded set suchthat f (G) is closed, and y is a point of X with y / f (G). If f (x0) = y for some x0 G, then

f (x) y, x x0 0 > f (x) y x x0 for all x G,

which implies by Theorem 3.1 that d(f ,G,y ) = 1. Thus (22) holds.Our next result gives sufficient conditions for a quasimonotone map to be surjective. Let

f : X X be a given map. We say that f has the property (B) if

(B) for any y X there exists a neighborhood U of y such that f 1(U ) is bounded .

It is easy to see that the property (B) can be stated equivalently as

(B ) If {x j } X is such that f (x j ) y X , then {x j } is bounded .

It is useful to note that f has the property (B) if for instance

f (x) as x ,

which equivalently means that f 1(C ) is bounded for any bounded set C of X .

20


21/38

Theorem 3.2 Let f : X X be a demicontinuous quasimonotone map. Assume that f (B R ) is closed for each ball {x X : x < R }, R > 0, and that f has the property (B). If there exists R > 0 such that

f (x), xx

+ f (x) > 0 for all x R,(23)

then f (X ) = X , i.e., the equation f (x) = y admits a solution for any y X .

Proof: Let y X be xed. We can choose R R and k > 0 such that

f (x) ty k for all t [0, 1] and x R .

Indeed, suppose on the contrary that there exists sequences {x j } X with x j and {t j } [0, 1] such that f (x j ) t j y 0 as j . We can assume that t j t, which implies f (x j ) ty.By the property (B), {xJ } is bounded, which contradicts our assumption x j . By choosingsome 0 > 0 with 0 < (c/R ), we have for all x = R and 0 < < 0,

f (x) + J (x) ty f (x) ty J (x) c 0R > 0.

Hence (f + J )(x) = ty for all t [0, 1], 0 < < and x BR . By Denition 2.3 and thehomotopy invariance for ( S )+ -mappings we get

d(f, B R , y) = d(f + J,B R , y) = d(f + J,B R , 0) = d(f, B R , 0).

By (23) we havef (x), x > f (x) x for all x = R

and nd that the assumptions of Theorem 3.1 hold for x = 0. Thus

d(f, B R , y) = d(f, B R , 0) = 1 ,

which implies that y f (BR ). Since f (BR ) is closed, we have y f (BR ), which completes theproof.

Remark 3.6 The assumptions of Theorem 3.2 are satised if, for example, f : X X is a demi-continuous coercive pseudomonotone map, i.e.,

f (x), xx

+ as x + .

Some concrete examples of mappings satisfying the condition (23) but not coercive can be found in[27].

Example 3.1 Let f : X X be a demicontinuous monotone map which is expanding, i.e., thereexists k > 0 such that

f (x) f (u) k x u for all x, u X.(24)By (24), f (x) as x ; thus f has the property (B). Moreover, since X is a Banachspace, f (A) is closed for each closed subset A of X . Indeed, if {y j } f (A) with y j y, we canwrite y j = f (x j ), x j A, and {x j } is a Cauchy sequence by (24). Thus {x j } converges, sayx j x A, and, consequently, f (x j ) f (x) = y f (A). We can assume that f (0) = 0(if not,we shall study the map f f (0)). The monotonicity of f and property (B) imply that there existsR > 0 such that (23) holds. Hence, by Theorem 3.2, f (X ) = X (see also Section 3.3 where westudy the surjectivity of the expanding maps further).

21


22/38

3.2 Odd Mappings

In this section we shall consider the generalization of Borsuks theorem of the Leray-Schaudertheory concerning the degree of odd mappings. It can be stated as follows (see [19], [24]).

Proposition 3.1 (Borsuks theorem) Let G be an open bounded subset of X containing the

origin and symmetric, i.e., x G whenever x G and let f : G X be a map of the Leray-Schauder type, that is I f is compact. If 0 / f (G) and f ( x) = f (x) for all x G, then dLS (f,G, 0) is an odd number.

We rst consider odd mappings of class ( S )+ (cf, [23] p 223). Since we do not presuppose anyboundedness of mappings, we need the following lemma.

Lemma 3.2 Let G be an open bounded symmetric subset of X with 0 G, and f F G be given map with 0 / f (G). Assume that f ( x) = f (x) for all x G. Then there exists a symmetric open set G such that

(i) f 1(0) G G, and

(ii) f ( G) is bounded.

Proof: The solution set f 1(0) = {x G: f (x) = 0} is obviously symmetric and 0 f 1(0). Wehave already shown in the proof of Lemma 2.6 that one of the sets

Gn = x G:dist x, f 1(0) < 1n

,

n = 1, 2, . . . , satises conditions (i) and (ii). The assertion now follows provided Gn is symmetric.

Let x Gn , then there exists z f 1

(0) such that x z < 1/n , and thus ( x) ( z ) =x z < 1/n with z f 1(0), which implies that x Gn .

Let G X be as in the previous lemma, and let f 1: G X stand for the map

f 1(x) = f ( x) for all x G.

Then f 1 is demicontinuous and of class (S )+ whenever f has these properties.

Theorem 3.3 Let G be an open bounded symmetric set of X with 0 G, and let f F G be odd in G. Then there exists in G a solution of the equation f (x) = 0 and , moreover, d(f,G, 0) is an odd number whenever dened.

Proof: If 0 f (G), the assertion follows. Assume therefore that f (x) = 0 for all x G. Denea new map of class (S )+ by setting

h(x) = 12

f (x) f ( x) , x G.

Clearly h is odd on G and coincides with f on G. Hence

d(f,G, 0) = d(h,G, 0).(25)

22


23/38

According to Lemma 3.2, there exists an open symmetric subset G of G such that h 1(0) G andthe restriction h: G X is bounded. By the additivity property and Denition 2.1 we get

d(h,G, 0) = d(h, G, 0) = dLS (h , G, 0) for all 0 < < ,(26)

whereh = I +

1 h

and > 0 is given by Lemma 2.2. Since h is odd, h is odd on G. Consequently, by Proposition ?? ,

dLS (h , G, 0) = odd for all 0 < < .(27)

The relations (25), (26) and (27) allow us to deduce that d(f,G, 0) is an odd number, which certainlyimplies that 0 f (G).

Corollary 3.1 Let G be an open bounded and symmetric subset of X with 0 G and let f F Gbe such that 0 / f (G) and

f (x)f (x) = f ( x)f ( x) for all x G.(28)

Then d(f,G, 0) is odd and there exists in G a solution of the equation f (x) = 0 .

Proof: Let us consider an affine homotopy of class (S )+ given by

f t (x) = (1 t)f (x) + th (x), t [0, 1], x G,

whereh(x) =

12

f (x) f ( x) , x G.

If for some t [0, 1] and x G we have f t (x) = 0 then

1 t2

f (x) = t2

f ( x),

which contradicts (28). Thus d(f,G, 0) = d(h,G, 0) = odd, where the last equality follows from thefact that h is odd on G. In particular, d(f,G, 0) = 0, which implies that 0 f (G). Hence the proof is complete.

Our next theorem is applicable if for instance f is psuedomonotone and G is convex.

Theorem 3.4 Let G be an open bounded symmetric subset of X with 0 G and let f : G X be a demicontinuous quasimonotone map such that f (G) is closed. If F is odd on G, then there exists in G a solution of the equation f (x) = 0 and d(f,G, 0 = odd whenever dened.

Proof: If 0 f (G) f (G), the assertion follows. Thus we can assume that 0 / f (G).Since (f + J )( x) = (f + J )(x) whenever f ( x) = f (x), we conclude by Denition 2.3 andTheorem 3.3 that

d(f,G, 0) = d(f + J,G, 0) = odd for all 0 < < ,

where > 0 is given by Lemma 2.9 for A = G. Hence 0 f (G) f (G), which completes theproof.

23


24/38

3.3 Injective Mappings

It is known from the Leray-Schauder theory that for each injective map I g: G X , g compactwith y (I g)(G), we have

dLS (f ,G,y ) = 1.

The proof of this result in based on the symmetry between I g and its inverse map [19], and thereseems to be no analogous proof for maps of class (S )+ . However, by means of odd mappings weshall derive a weaker result

d(f ,G,y ) = odd ,

whenever f F G is a continuous injection and y f (G). An essential point is of course the factthat the value of degree is nonzero.

We rst give an auxiliary result which entitles us to assume that 0 G, thus simplifying theproofs. Let G be an open bounded subset of X and x0 G any xed point. Denote

s(x) = x x0 and s 1(x) = x + x0, x X.

Now if f F G , then clearly f s 1 F s (G ) . Because f (x) = ( f s 1)(s(x)) for all x G, thedegrees d(f,G, 0) and d(f s 1, s(G), 0) are both dened if and only if 0 / f (G). Hence 0 s(G),and it can be proved that

d(f,G, 0) = d(f s 1, s(G), 0).(29)

Indeed, the right-hand side of (29) determines a degree function for the class F G in the sense of Section 2.3. Thus (29) follows from the uniqueness of d.

The following lemma shows how to obtain a permissible homotopy of class (S )+ between theinjection f and an odd mapping (cf. [19] p. 66). It should be noted that the continuity of f appearsnecessary.Lemma 3.3 Let B X be an open ball with center at the origin and f : B X be a continuous bounded injection of class (S )+ with f (0) = 0 . Then

H (t, x ) = f x1 + t

f tx1 + t

t [0, 1] x B

denes a bounded homotopy of class (S )+ .

Proof: The map H : [0, 1] B X is clearly bounded and continuous. Let {t j } [0, 1] and{x j } B be such that

t j t, x j x, t [0, 1], x X, and

lim sup j H (t j , x j ), x j x 0.

Denote f 1(x) = f ( x), u j = (1 + t j ) 1x j , v j = t j u j , u = (1 + t) 1x and v = tu . Now u j u andv j v .(a) Suppose rst that t = 0. Since f is bounded and t j = 0 for sufficiently large j , we can write

0 lim sup j

H (t j , x j ), x j x

= lim sup j

[ f (u j ), x j x + f 1(v j ), x j x ]

24


25/38

= lim sup j

(1 + t j ) f (u j ), u j x1 + t j

+ 1 + t j

t jf 1(v j ), v j

t j x1 + t j

= lim sup j

(1 + t j ) f (u j ), u j u + (1 + t j ) f (u j ), x1 + t

x1 + t j

+1 + t j

t jf 1(v j ), v j v +

1 + t jt j

f 1(v j ), tx1 + t j

t j x1 + t j

= lim sup j

(1 + t j ) f (u j ), u j u + 1 + t j

t jf 1(v j ), v j v

lim sup j

(1 + t j ) f (u j ), u j u + lim inf j

1 + t jt j

f 1(v j ), v j v

= lim sup j

(1 + t) f (u j ), u j u + lim inf j

1 + tt

f 1(v j ), v j v

lim sup j

(1 + t) f (u j ), u j u ,

where the last inequality follows from the fact that (see Lemma 1.1)

lim inf j

f 1(v j ), v j v 0.

Hencelim sup

j f (u j ), u j u 0,

which implies that u j u. Thus x j = (1 + t j )u j (1 + t)u = x and H (t j , x j ) H (t, x ).(b) Suppose now that t = 0. Then u = x, u j = (1 + t j ) 1x j x and v j = t j u j 0. It follows fromthe continuity of f that f 1(v j ) f 1(0) = 0. Since u j x j 0, we can write

0 lim sup j

H (t j , x j ), x j x

= lim sup j

[ f (u j ), x j x + f 1(v j ), x j x ]

= lim sup j

f (u j ), x j u j + u j x

= lim sup j

f (u j ), u j x ,

and thus u j x implying x j x. Hence the proof is complete.

Theorem 3.5 Let G be an open bounded subset of X and f F G a continuous injection. If

y f (G), then d(f ,G,y ) = odd .Proof: Since f y F G is also a continuous injection and d(f ,G,y ) = d(f y, G, 0), we mayassume without loss of generality that y = 0. Denote x0 = f 1(0) G. Since f is locally bounded,there exists an open ball B G with center at x0 such that the restriction f : B X is bounded.Moreover, we have by additivity

d(f,G, 0) = d(f,B, 0).(30)

Suppose that x0 = 0. In view of Lemma 3.3 we consider the homotopy

f t (x) = f x1 + t

f tx1 + t

, t [0, 1], x B.

25


26/38

Now f 0(x) = f (x) and f 1(x) = f (x/ 2) f ( x/ 2), which is odd everywhere on B. Consequently,f t (x) = 0 for all t [0, 1] and x B. Hence by the homotopy invariance and Theorem 3.3,

d(f,B, 0) = d(f 1, B, 0) = odd ,

which together with (30) completes the proof in the case x0 = 0.If x0 = f 1(0) = 0, we use (29) and by the earlier part of the proof we have

d(f,G, 0) = d(f s 1, s(G), 0) = odd ,

which completes the proof.

As an application of Theorem 3.5 we shall prove the invariance of domain of a continuousinjection of class (S )+ (cf, [19] p.66).

Theorem 3.6 Let D be an open subset of X ( not necessarily bounded ) and f : D X a continuous injection of class (S )+ . Then the set f (D) is open in X .

Proof: Let y f (D) be xed and denote x0 = f 1(y) D. Let B be a ball with center at x0 suchthat B D. Now

y = f (x0) F (B) f (D)

by injectivity of f , y / f (B ). This implies that d(f ,B ,y ) is well-dened, and it is known inaddition that the value of d(f,B, ) is constant on every open component of the open set X \ f (B ).Consequently, there exists r > 0 such that

d(f ,G,q ) = d(f ,B ,y ) for all q y < r.

By Theorem 3.3 we have d(f ,B ,y ) = odd, which gives

d(f ,B ,q ) = 0 for all q y < r,

and therefore we obtain xq B (for all q X , q y < r ) such that f (xq) = q . Thus y f (D)has an open neighborhood in f (D), namely,

B r (y) = {q X : q y < r } f (B ) f (D),

and since y f (D) was arbitrary, the assertion follows.

By using the open mapping theorem we get our next surjectivity result.

Theorem 3.7 Let f : X X be continuous injection of class (S )+ and assume that f has the

property (B ). Then f (X ) = X .Proof: In view of Theorem 3.6 (for D = X) it suffices to prove that f (X ) is closed in X . Indeed,if {y j } f (X ), y j y, we can write y j = f (x j ) and the sequence {x j } X is bounded. Withoutloss of generality we may assume that x j x for some x X . Hence

lim sup j

f (x j ), x j x = lim j

y j , x j x = 0,

which implies x j x in X and, consequently, y j = f (x j ) f (x) = y f (X ). Thus f (X ) is bothopen and closed, implying f (X ) = X .

26


27/38

Corollary 3.2 Let f : X X be a continuous map of class (S )+ which is expanding, i.e., there exists k > 0 such that

f (x) f (u) k x u for all x, u X.(31)

Then f (X ) = X .

Proof: By (31) the map f is injective and the inverse map f 1 is bounded because

w y k f 1(w) f 1(y) for all w, y f (X ).

The assertion now follows from Theorem 3.7.

Corollary 3.2 relates to the following Problem (P), which was stated by Nirenberg ([22] p. 175)in the special case if Hilbert space X .Problem (P): Assume that X is a Banach space and let T : X X be continuous, expanding mapwhose range contains an open set. Does T map X onto X ?

Morel and Steinlein gave in 1984 [21] a counterexample which shows that the answer is generallynegative. However, there are several partial positive answers to Problem (P), for instance in thefollowing cases ( see [21]):

(i) X is nite dimensional;

(ii) T = I g, where g is compact or contraction or, more generally, a k-net contraction;

(iii) X is a Hilbert space and T a strongly monotone map, i.e.,

(T (x) T (u)|x u) c x u 2 for all x, u X (c > 0).

Let X be a real separable Hilbert space, and let T : X X be continuous, expanding mapof class (S )+ , where we have identied X with X . In view of Corollary 3.2, T (X ) = X . Thisis a generalization of (iii), since every strongly monotone map is of class ( S )+ . Consequently, if T : X X is a monotone and expanding, then T (X ) = X by Example 3.1, even if T is onlydemicontinuous. As we showed in Example 3.1, the image of a closed set under any demicontinuousexpanding map is closed. Using this face we get

Theorem 3.8 Let X be a real separable Hilbert space and let T : X X be a continuous quasi-monotone map which is expanding. Then T (X ) = X .

Proof: Certainly, T + I is of class (S )+ for all > 0. Since T (x) T (u) k x u ,

(T + I )(x) (T + I )(u) (k ) x u for all x, u X.

Thus T + I is expanding for any 0 < < k and, by Corollary 3.2, (T + I )(X ) = X . For any y X there exists x , 0 < < k , such that T (x ) + x = y, which implies, as 0+ , that T (x ) y.Since T (X ) is closed, y T (X ) and thus T is surjective.

We close this section by some observations concerning the maps of X into X .

27


28/38

Remark 3.7 The reexivity of X implies that the map L dened by

Lw,x = w, x for all x in X and w in X

is a linear isometric homeomorphism of X onto X . Let M be an open bounded set of X andh: M X a given map. We say that h is of class (S )+ if L h is a map of class (S )+ of X into X .We can similarly dene homotopies of class (S )+ . For any such h we can dene a degree function

uniquely by settingd(h ,M,x ) = d (L h,M,Lx ),

where x X \ h(M ) and d is the degree function for demicontinuous maps of class ( S )+ of X into X . By Remark 1.7, J = L J 1 is the duality mapping of X into X . Thus the normalizingmap for d is J 1, the inverse of J .

Let now f F B,G be a continuous injection and denote M = f (G). Then, by Theorem 3.6, M is an open and bounded set of X and we clearly have M = f (G) and M = f (G). It is easyto see that f 1: M X is a continuous bounded injection of class (S )+ . Thus the degree of theinverse function, d(f 1,M,x ), is dened whenever x / f 1(M ), i.e., x / G.

The fact that the open mapping theorem holds for continuous injections of class ( S )+ , togetherwith the observation that the situation is symmetric with respect to f and f 1, leads to the followingopen question (generalization of Jordans theorem):

If D and D are open bounded subsets of X and X , respectively, and f : D Dis a homeomorphism of class (S ) + , do X \ D and X \ D have the same number of components ?

This question arises naturally since the generalization of the Jordan curve theorem holds forthe homeomorphism of the type I g, g compact, form X into X (see [24] p. 94), and there theproof is based on the symmetry between I g and its inverse map. However, there seems to be noanalogous proof in the case of maps of class (S )+ .

3.4 Mappings of Type ( M )

Here we outline the direct application of the procedure introduced in Section 2.1, withoutconstructing the degree function. A function f : X X is said to be of type (M ) if we havef (x) = w for every sequences {x j }, x j x in X , and {f (x j )}, f (x j ) w in X , for which

lim sup j

f (x j ), x j x 0.

Maps of type (M ) are widely used in applications; see for instance [23]. Clearly every locallybounded map of type ( M ) is demicontinuous. It is not hard to see that every demicontinuous mapf : X X which is of class (S )+ or psuedomonotne is also of type (M ). However maps of type(M ) are generally not quasimonotone and no relevant degree theory seems to be available for them.

Let f : X X be abounded map of type ( M ). As in Section 2.1 the map

f = I + 1

f : G X

is of type I g, g compact, for each > 0. In order to obtain existence results we need the followingLemma 3.4 Let f : X X be a bounded map of type (M ). If there exist sequences { j }, j 0+ ,and {x j } in X , x j x , such that f j (x j ) = 0 , then f (x j ) f (x) = 0 .

28


29/38

Proof: Since {f (x j )} is bounded, there exists a weakly converging subsequence, say f (xk ) w .Thus f (xk ) (w), and since

x j = 1 j

f (x j ),(32)

we also have f (xk ) 0 and, consequently, w = 0, i.e., f (xk ) 0. This fact does not depend onthe particular sequence {f (xk )}, and hence we conclude that

f (x j ) 0.(33)

By making use of (32) and (33) we get

lim sup

f (x j ), x j x

= lim sup j

f (x j ), 1 j

f (x j )

= lim sup j

1 j

f

(x j

) 2W

0,

and since f is of type (M ), we deduce that f (x j ) f (x) = 0, which completes the proof.

As an application of Lemma we shall consider two results. The rst is a generalization of Borsuks theorem.

Theorem 3.9 Let f : X X be a bounded map of type (M ). If there exists a ball BR , R > 0 such that f is odd on the boundary BR , i.e.,

f ( x) = f (x) for all x = R,

then the equation f (x) = 0 has at least one solution x0 X with x0 R.

Proof: Clearly also f is also odd on the boundary BR . Assume rst that there exists > 00 / f (B R ) for all 0 < < . Then dLS (f , B R , 0) is well-dened and odd by Proposition 3.1.Since dLS (f , B R , 0) = 0, we can, for each with 0 < < , nd a point x BR such thatf (x ) = 0. In view of Lemma 3.4 there exists a point x0 X and such sequences that

j 0+ , x j x 0 X and f (x j ) f (x0) = 0 ,

and since any closed and convex set is weakly closed, we see that x0 BR .On the other hand, if there is no such > 0 that 0 / f (B

R) for all 0 < < , then we can

nd sequences { j }, j 0+ , and {x j } BR such that f j (x j ) = 0. Hence the assertion againfollows from Lemma 3.4.

In a similar fashion it is easy to prove the following result (cf. Section 3.1).

Theorem 3.10 Let f : X X be a bounded map of type (M ). If there exists R > 0 such that

f (x), x 0 for all x = R,

then the equation f (x) = 0 has at least one solution x0 X with x0 R.

29


30/38

4 Further Applications

4.1 Multiplication Theorem

In this section we generalize the multiplication theorem of the Leray-Schauder theory concern-

ing the computation of the degree of a composite function. It can be stated as follows (see [24] p. 93).Proposition 4.1 (Multiplication Theorem) Suppose that G is an open bounded set of a real Banach space X , f : G X a map of the Leray-Schauder type, that is, I f is compact, and M an open bounded set X containing f (G). Let = M \ f (G) and let j , ( j = 1, 2, 3, . . .), be the open components of . If h: M X is a Leray-Schauder type map and y is a point of X such that y / (h f )(G) h(M ), then

dLS (h f ,G,y ) = j

dLS (h, j , y) dLS (f,G, j ).(34)

Remark 4.1 The summation in (34) is nite since h 1

(y) is a compact set, so that, being coveredby the disjoint open sets j , it meets only a nite number of them and the other terms in (34)vanish.

Clearly the composition of two maps of X into X is not dened. However, we generalizeProposition for the case h f , where h is of class (S )+ and f is a map of Leray-Schauder type. LetG be an open bounded set of X and f = I g: G X such that g compact. Suppose that M isan open bounded set of X containing f (G) and h F M,B . According to Lemma 2.3, h f F G,B .In view of Section 2.1 we get the following maps of the Leray-Schauder type ( > 0):

h f = f + 1

(h f ): G X (35)

and(h f ) = I +

1 (h f ): G X.(36)

Let > 0 be xed and consider an affine Leray-Schauder homotopy

H (t, x ) = t(h f )(x) + (1 t)(h f ) (x), x G, 0 t 1,

which by (35) and (36) can be written in the form

H (t, x ) = ( h f ) (x) tg(x), x G, 0 t 1.(37)

The next lemma plays an essential role in this section.

Lemma 4.1 Assume that (h f )(x) = 0 for all x G. Then there exists > 0 such that

H (t, x ) = 0 for all x G, 0 t 1 and 0 < < .(38)

Proof: If this is not true, there exist sequences { j }, {t j } [0, 1] and {x j } G with 0+such that H j (t j , x j ) = 0 or, equivalently, by (36) and (37)

x j + 1

j (h f )(x j ) t j g(x j ) = 0 for all j .(39)

30


31/38


32/38

for all 0 < < . Moreover, choosing > 0 sufficiently small we have 0 / h (M ) for all 0 < < ,and, consequently,

0 / (h f )(G) h (M ) for all 0 < < .

Hence it follows from Proposition that

dLS (h f,G, 0) = j

dLS (h , j , 0) dLS (f,G, j )(45)

for all 0 < < . The right-hand side of (44) contains only a nite number of nonvanishing terms(cf. Remark 4.1), and for each component j we have

d(h, j , 0) = dLS (h , j , 0) for all 0 < < .(46)

Using (44), (45) and (46) we conclude that

d(h f,G, 0) = j

d(h, j , 0) dLS (f,G, j ),(47)

which is exactly the relation (43) if the case y = 0.Let now y X be any nonzero element satisfying (42). If we denote h() = h() y, we nd

that h F M,B and 0 / (h f )(G) h(M ). As stated above,

d(h f,G, 0) = j

d(h, j , 0) dLS (f,G, j ),

which further meansd(h f ,G,y ) =

j

d(h, j , y) dLS (f,G, j ),

and the proof is complete.As an application of Theorem 4.1 we study a simple example. Let f = I g: G X be a map

of Leray-Schauder type and z 0 X \ f (G). By choosing h = J , y = J (z 0) and an open boundedset M of X such that f (G) M and z 0 M we get

d(J f ,G,y ) = j

d(J, j , y) dLS (f,G, j ).

Since z 0 M \ f (G), there exists a unique k such that z 0 k . Thus d(J, j , y) = 0 for all j = k and

d(J f ,G,y ) = d(J, k , y) dLS (f,G, k ) = +1 dLS (f,G, k ) = dLS (f ,G,z 0).(48)

This shows that in fact the Leray-Schauder degree theory can be considered a part of the degreetheory for the maps of class (S )+ . Indeed, by (48) the value of dLS (f ,G,z 0) is always reached bymeans of d (cf. Lemma 2.4).

32


33/38

4.2 On Solution Set Structure

We shall now consider the structure of the solution set of the equation f (x) = 0. We generalizeresults obtained in [20], where the authors investigate zero-epi maps, which provide an interestingalternative to degree theory (see also [14], [15], and [16]).

Theorem 4.2 Let f F G be such that 0 / f (G) and d(f,G, 0) = 0 . Suppose that for any > 0and z f 1(0) there exists a demicontinuous quasimonotone map h : G X satisfying the following conditions:

(i) h (z ) = 0 ;

(ii) h (x) < for all x G;

(iii) the set S = {x G : f (x) + h (x) = 0} is -chained, i.e., for any two points a and b in S there exists some nite sequence of points {x1, x2, . . . , x n } S such that x1 = a, xn = b and dist( xi , x i+1 ) < 2 , i = 1, 2, . . . , n 1.

Then the set f 1(0) is a continuum, i.e., it is compact and connected.

Proof: Since d(f,G, 0) = 0, we have f 1(0) = and, moreover, f 1(0) is compact (see Lemma 2.5).Suppose that f 1(0) is not connected. Then f 1(0) = K 1 K 2, where K 1 and K 2 are compactnonempty subsets of G with K 1 K 2 = . Denote dist( K 1, K 2) = 6 1 > 0 and dene two open setsby setting

G j = {x G : dist(K j , x) < 2 1}, j = 1, 2.

Then K j G j G, j = 1, 2, and clearly dist( G1, G2) 2 1. By closedness of f (G \ (G1 G2)) weget

inf { f (x) : x G \ (G1 G2)} = 2 > 0.

Let 0 < < min( 1, 2). Then

dist(G1, G2) > 2 andinf { f (x) : x G \ (G1 G2)} > .

(49)

The additivity property of d implies that

d(f,G, 0) = d(f, G 1, 0) + d(f, G 2, 0),

where we may suppose that, for instance, d(f, G 1, 0) = 0. For any xed z K 2 and 0 < < there exists a demicontinuous quasimonotone map h : G X satisfying (i), (ii) and (iii). By (i),f (z ) + h (z ) = 0, and thus z S , which implies that G2 S = . Also, by (ii), we get

f (x) = h (x) < < for all x S .

Hence it follows from (49) that S G1 G2, and since S is -chained, necessarily

S G2.(50)

On the other hand, f + h F G , and we consider the homotopy given by

f t = t(f + h ) + (1 t)f = f + th , 0 t 1.

33


34/38

If f t (x) = 0 for some t [0, 1] and x G, then

f (x) = t h (x) < ,

which means that x G1 G2. Hence 0 / f t (G) for any t in [0, 1], and by homotopy invarianceproperty

d(f + h , G1, 0) = d(f, G 1, 0) = 0 .

Hence 0 (f + h )(G1), i.e., S G1 = . But this contradicts (50), and the proof is complete.

Remark 4.2 The condition (iii) of Theorem 4.2 is satised if, for instance, the equation f (x) +h (x) = 0 has at most one solution.

Remark 4.3 Theorem 4.2 implies in particular that the solution set f 1(0) contains either onepoint or is innite.

Example 4.1 Let f : G X be a demicontinuous monotone map of class ( S )+ with 0 / f (G).Suppose that there exists x G such that

f (x) > f (x) for all x G.

Then, for all x G, we have

f (x), x x = f (x) f (x), x x + f (x), x x f (x), x x f (x) x x> f (x) x x .

Then by Lemma 3.1, d(f,G, 0) = 1, and therefore f 1(0) = . For any > 0 and z f 1(0), dene

h (x) = J (x z ) for all x G,

where 0 < < (2 sup{ x : x G}) 1. Clearly

(i) h (z ) = 0;

(ii) h (x) = x z < for all x G;

(iii) since f + h is strictly monotone, S contains exactly one point, z , and therefore S is -chained.

From Theorem 4.2 we can now deduce that f 1(0) is a continuum.

34


35/38

4.3 On the Existence of Multiple Solutions

In this section we shall outline one possible way to deal with the question of multiplicity, i.e.,the number of solutions of the equation f (x) = 0. Our approach is based on the idea of seekingdistinct solutions in some maximal closed linear subspaces.

We shall rst present some background material.

Lemma 4.2 Let f F G,B be such that 0 / f (G) and d(f,G, 0) = 0 . Then

y X : y = f (x)

f (x), x G = {y X : y = 1}.

Proof: Let y X , y = 1, be given. The value of d(f,G, ) remains constant on each opencomponent of X \ f (G). Assume 0 k , where k is an open component of X \ f (G).For all sufficiently small values of > 0, y k . If, however, y k for all > 0, thend(f,G,y ) = d(f,G, 0) = 0 and y f (G) for all > 0. This is impossible since f is bounded.Hence there exists 0 > 0 such that 0y f (G), and the assertion follows.

For any closed linear subspace V of X , let jV : V X be the natural embedding for which jV (x) = x for all x V . Consequently, let jV : X V stand for the projection which is denedby

jV (w), v = w, j V (v) for all w X and v V.

Assume G is an open bounded set of X such that V G = , which is true if for instance 0 G.Clearly G V G V and V (G V ) G V , where V (G V ) denotes the boundary of G V in V . It is not hard to see that the map

f V = j V f jV : G V V

is demicontinuous and of class (S )+ . Let us denote the corresponding unique degree function by dV .Before starting the main result of this section, we recall some facts about maximal closed linear

subspaces (see e.g. [25]). A closed linear subspace V X , V = X , is said to be maximal, for anyclosed linear subspace Z X with V Z X , either Z = V or Z = X . Any maximal closedlinear subspace can be presented in the form

V = ker( w) = {x X : w, x = 0} for all w X , w = 0 .

Let V 1 and V 2 be maximal closed linear subspace given by V j = ker( w j ), w j = 0 , j = 1, 2. ThenV 1 = V 2 if and only if w1 and w2 are linearly dependent. By the reexivity of X , any maximalclosed linear subspace N of X can be given by

N = {y X : y, x = 0} for all x X, x = 0 .

We are now in a position to prove

Theorem 4.3 Let G be an open bounded subset of X with 0 G and f F G,B such that f (0) = 0 .Suppose that

(i) f (x), x > 0 for all x G and

35


36/38

(ii) for each linearly independent pair {x1, x2} G \ f 1(0), f (x1) and f (x2) are linearly inde-pendent.

Then the solution set f 1(0) is innite.

Proof : By (i), 0 / f (G), and it follows from Lemma 3.1 that d(f,G, 0) = 0. Hence there exits atleast one x1 such that

f (x1) = 0 , x1 G1, x1 = 0 .Suppose next that we have obtained a set {x1, x2, . . . , x n } G, n 1, such that

f (x j ) = 0 , j = 1, . . . , n and x j = xi for all j = i.

Certainly x j = 0 for all j . Let N 0 be some maximal closed linear subspace of X containing f (0)and denote

N i = {y X : y, x i = 0}, i = 1, 2, . . . , n .

We can obviously nd

y X , y = 1, such that y / N 0 N 1 N n .

Hence y = f (0) for all R and y, x i = 0 , i = 1, 2, . . . , n . Denote V = ker( y). In viewof Lemma 4.2, we can write V = ker( y) = ker( f (u)) for some u G. Clearly u / V sincef (u), u = 0 by (i). As stated above, the map f V = jV f jV : G V V is demicontinuous

and of class (S )+ . Moreover, by (i)

f V (v), v > 0 for all v V (G V ).

Thus dV (f V , G V,0) = 0 by Lemma 3.1, and we get a solution xn +1 G V of the equationf V (v) = 0 on V , i.e.,

f (xn +1 ), v = 0 for all v V.(51)

Since f (u), x i = 0 , i = 1, 2, . . . , n , and f (u), xn +1 = 0, we have xn +1 = xi , i = 1, 2, . . . , n . If f (xn +1 ) = 0 in X , the proof is complete. If not, then by (51) and the denition of V ,

V = ker( f (u)) = ker( f (xn +1 ),

which implies that f (xn +1 ) = f (u) for some R , = 0. By virtue of (ii), xn +1 and u arelinearly dependent, and since u / V , xn +1 V , we necessarily have xn +1 = 0. But this contradictsthe fact that f (u) = f (0) for all R . Hence f (xn +1 ) = 0 in X , and we have obtained n + 1distinct solutions {x1, x2, . . . , x n +1 } of the equation f (x) = 0. By the induction argument the proof is complete.

Remark 4.4 Since f 1(0) is compact and innite in the previous theorem, it contains at least oneaccumulation point.

36


37/38


38/38

[18] Leray, J., and J. Schauder: Topologie et equations fonctionelles , Ann. Sci. Ecole. Norm. Sup. 51(1934), 45-78.

[19] Lloyd, N.G.: Degree theory , Cambridge University Press, Cambridge, (1978).

[20] Martelli, M., and A. Vignoli: On the structure of the solution set of nonlinear equations ,Nonlinear Anal. 7 (1983), 685-693.

[21] Morel, J. -M, and H. Steinlein: On a problem of Nirenberg concerning expanding maps , J.Funct. Anal. 59 (1984), 145-150.

[22] Nirenberg, L.: Topics in nonlinear functional analysis , Lecture Notes, Courant Institute of Mathematical Sciences, New York University, New York, (1974).

[23] Pascali, D., and S. Sburlan: Nonlinear mappings of monotone type , Editura Academiei,Burkarest, (1978).

[24] Schwartz, J.T.: Nonlinear functional analysis , Gordon and Breach, New York, (1969).

[25] Taylor, A. E., and D.C. Lay: Introduction to functional analysis , John Wiley & Sons, NewYork, (1980).

[26] Trojanski, S. L.: On locally uniformly convex and differentiable norms in certain non-separable Banach spaces , Studia Math. 37 (1970/71), 173-180.

[27] Wille, F.: Monotone operatoren mit St orungen , Arch. Rational Mech. Anal. 46 (1972), 369-388.

[28] Yosida, K.: Functional analysis , Die Grundlehren der mathematischen Wissenschaften 123 ,Springer-Verlag, Berlin-Gottingen - Heidelberg, (1965).

[29] Zeilder, E.: Vorlesungen uber nichtlineare Funktionalananysis II. Monotone Operatoren ,Teubner-Texte zur mathematik, B.G. Teubner Verlagsgesellaschaft, leibzig, (1977).

Juha BerkovitsUniversity of OuluDepartment of MathematicsSF-90570 Oulu

Finland

Documents

Degree Theory