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A NEW SPECTRAL THEORY FOR NONLINEAR

OPERATORS AND ITS APPLICATIONS

W. FENG

Abstract

. In this paper, by applying (p,k

)-epi mapping theory, we intro-duce a new definition of spectrum for nonlinear operators which containsall eigenvalues, as in the linear case. Properties of this spectrum are givenand comparison is made with the other definitions of spectra. We also giveapplications of the new theory.

1. Introduction

Spectral theories for nonlinear operators have been extensively studied,for example, see [1], [5]-[7] and [9]. Different attempts have been madeto define the spectrum for nonlinear operators. Clearly, a good definitionshould preserve as many properties of the spectrum for classical boundedlinear operators as possible and reduce to the familiar spectrum in the caseof linear operators. In [15], a spectrum for Lipschitz continuous operatorswas studied (we denote it by lip(f)), which is compact but may be empty(see the example in Section 5). The spectrum introduced by Furi, Martelliand Vignoli (denoted fmv(f)) has found many interesting applications (see[9]). However, it was indicated in [6] that this spectrum may be disjointfrom the eigenvalues, which is an important part of the spectrum in the

linear case. This paper is mainly based on the study of [9] and the mainaim is to introduce a new spectrum for nonlinear operators by applying the(p,k)-epi mapping theory of [10], [14], [17]. This new version of the spectrumis closed and contains all the eigenvalues as in the case of linear operators.

This paper is organized as follows. In Section 3, the definition of the newspectrum will be given and some of its special properties will be proved.For example, we will show that this spectrum is closed, bounded, upper

1991 Mathematics Subject Classification. Primary 47H12.Key words and phrases. Spectrum of a nonlinear operator, eigenvalues, (p,k)-epi map-

ping theory.Received: September 1, 1997.

c1996 Mancorp Publishing, Inc.

163

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semicontinuous and contains all the eigenvalues. In Section 4, we investigatepositively homogeneous operators where more precise results are possible.Some of the results are direct generalizations of the spectral theory for linear

operators. The results in this section will be used in Section 6 to generalizesome existence theorems. In Section 5, we will compare our spectrum (f)with the two spectra mentioned before (fmv(f) and lip(f)). We will provethat, in general,

lip(f) (f) fmv(f).

Moreover, in this section, nonemptiness of spectra is discussed. A counterex-ample shows that all these spectra may be empty, which answers one of theopen questions in [15].

In the last section, some applications of the new theory are obtained.This theory enables us to generalize the Birkoff-Kellogg theorem and theHopf theorem on spheres, which were also consequences of the theory in[9]. A non-trivial existence result for a global Cauchy problem, which wasstudied in [14], is obtained by applying the new theory.

It would be interesting to see extensions of the present theory to thecases considered by Ding and Kartsatos in [4], where the concept of a (p, 0)-epi mapping of [10] has been extended to cover perturbations of (possibly)densely defined operators.

In Section 2, we give some notations and definitions which will be used in

the sequel.

2. Preliminaries

Let E and F be complex Banach spaces and be an open bounded subsetof E. We suppose that f : E E is continuous and (A) denotes themeasure of noncompactness of a bounded set A [3]. The following notationswill be used in the sequel.

(f) = inf {k 0 : (f(A)) k(A) for every bounded A E},(f) = sup{k 0 : (f(A)) k(A) for every bounded A E},

m(f) = sup{k 0 : f(x) kx for all x E},d(f) = lim inf

x

f(x)x , |f| = lim supx

f(x)x .

Here |f| is called the quasinorm of f and f is said to be quasibounded if|f| < . Maps with (f) < 1 are k-set contractive (also condensing) withk = (f). Note that a map f satisfies (f) = 0 if and only if f is compact,

that is f(A) is compact for every bounded set A. For the properties of(f), (f) and d(f) we refer to [9].A map f : E F is said to be stably-solvable if given any compact map

h : E F with zero quasinorm, there is at least one element x of E suchthat f(x) = h(x). Spectra fmv(f) and lip(f) are defined as follows.

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SPECTRAL THEORY FOR NONLINEAR OPERATORS 165

Definition 2.1. (see [9]) f is said to be fmv-regular if it is stably-solvableand d(f) and (f) are both positive. Let

fmv(f) ={

C : I

f is f mv

regular

},

then fmv(f) = C \ fmv(f).Definition 2.2. (see [15]) Let Lip(E) be the space of all Lipschitz map-pings. For A Lip(E), the Lip-spectrum is defined by

lip(A) = { : (I A)1 does not exists or (I A)1 / Lip(E)}.The p-epi mappings were introduced by Furi, Martelli and Vignoli [10]

and then were studied and applied in [12], [14], [17]. In [17], the notion wasgeneralized to the following (p,k)-epi mappings.

Definition 2.3. A continuous mapping f : F is said to be p-admissible(p F) if f(x) = p for x .

A 0-admissible mapping f : F is said to be (0, k)-epi if for each k-setcontraction h : F with h(x) 0 on the equation f(x) = h(x) hasa solution in . Similarly, a p-admissible mapping f : F is said to be(p,k)-epi if the mapping f p defined by (f p)(x) = f(x) p,x , is(0, k)-epi.

It was shown in [17] that the (p,k)-epi mappings have similar propertieswith properties usually obtained via the topological degree, for example, thehomotopy property and boundary dependence property.

In the following, let Br = {x : x E, x r} and Br denote theboundary of Br. If for every x = 0, f(x) = 0, let

r(f, 0) = inf {k 0, there exists g : Br E, with (g) k,g 0 on Br s.t. f(x) = g(x) has no solutions in Br},

and (f) = inf{r(f, 0), r > 0}. We will call (f) the measure of solvabilityof f at 0. (This concept is related to the measure of unsolvability of f at

p, which was introduced in [17]). Notice that, (f) > 0 if and only if there

exists > 0, such that f(x) is (0, )-epi on every Br with r > 0.

3. A new definition of the spectrum for continuous operators

We begin with the following definition.

Definition 3.1. Suppose that f : E E is continuous, then f is said tobe regular if

(f) > 0, m(f) > 0, and (f) > 0.

For

C, if I

f is regular, is said to be in the resolvent set of f.

Let (f) denote the resolvent set of f, then the spectrum of f is defined asfollows:

(f) = { C : I f is not regular } = C\(f).Proposition 3.2. If f is a regular map, then f is surjective.

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Proof. Since f is regular, m(f) > 0. Thus for x E, f(x) asx . Also we have (f) > 0, so there exists > 0 such that f is(0, )-epi on every Br with r > 0. By Corollary 3.2 of [17], f is surjective.

The following theorem characterizes the regular maps among continuouslinear operators.

Theorem 3.3. Suppose that E is a normed linear space, f : E E is acontinuous linear operator. Then f is regular if and only if f is a linearhomeomorphism.

Proof. Assume that f is regular. By Proposition 3.2, f is surjective. Also,m(f) > 0 implies that f is one to one and f1(x) (1/m(f))x. Thusf1 is continuous, so f is a linear homeomorphism.

Conversely, suppose that f is a linear homeomorphism. Then f1 is abounded linear operator and for all x E, f(x) (1/f1)x. Thisensures that

m(f) 1/f1, (f) 1/f1.So for 0 < < 1/f1, f is (0, )-epi on every Br with r > 0 [17]. Thus(f) > > 0, and f is regular.

Remark 3.4. By Theorem 3.3, for a bounded linear operator f, the spec-trum of f in Definition 3.1 is the same as the usual one.

It is well known in linear spectral theory that (f) is closed and (f) is open.The following theorem shows that this property holds true for the spectrumof nonlinear maps given by Definition 3.1.

Theorem 3.5. For a continuous map f, (f) is an open set and (f) isclosed.

Proof. Suppose that (f), then(I f) > 0, m(I f) > 0,

and I f is (0, )-epi on every Br with r > 0 for some > 0. Now let1 = (I f)/2, 2 = m(I f)/2, 3 = /2,

and = min{1, 2, 3}. Assume that C, | | < . We shall provethat (f). Since

|(I f) (I f)| (I I) = | | < (I f)/2,we have

(I f) > (I f)/2 > 0.

For every x E,x f(x) x f(x) | |x (m(I f)/2)x,

so

m(I f) m(I f)/2.

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the equation x = h1(x)/ has a solution in Br. Thus S= and SBr = .Next we have

S

[0, 1]f(S)/ + h1(S)/.

Therefore,(S) (((f) + )/||)(S).

Hence S is compact because (f) + < || and S is closed. Let be acontinuous function such that 0 1 and

(x) =

1 for x S,0 for x r,

and let

g(x) = (x)f(x)/ + h1(x)/.

Then g is an ((f) + )/||-set contraction, g(x) 0 on the boundary ofBr. Hence x = g(x) has a solution x0 Br. Then x0 S so (x0) = 1 andh1(x0) = h(x0). Thus x0 is a solution of the equation

x f(x)/ = h(x)/.This ensures that I f is (0, )-epi on Br, so we have (f) > 0, I fis regular, and is in the resolvent set of f.

Remark 3.7. For nonlinear map f with f(0) = 0, we define the norm of fby

f = inf{k > 0 : f(x) kx}.Defining the radius of the spectrum of f by r(f) = sup{|| : (f)}, itfollows that

r(f) max{(f), f}.If f(0) = 0, for each C, either I f is not surjective, or there existsx E, x = 0, such that x f(x) = 0, and then is a eigenvalue of f. Bythe following theorem, in both these cases, (f). Thus (f) = C. Hencein what follows, unless otherwise stated, we shall assume that f(0) = 0.

Theorem 3.8. All the eigenvalues of f are in the spectrum of f.

Proof. If f(x) = x with x = 0, then m(I f) = 0, so (f).As mentioned in Section 1, the above simple theorem represents the big

difference between fmv(f) and Definition 3.1. According to their definition,the spectrum may be disjoint with its eigenvalues [6], but it is well knownthat for a linear operator, one of the important parts of its spectrum is thepoint spectrum, the eigenvalues.

The following lemma enables us to prove the upper semicontinuity of thespectrum.

Lemma 3.9. LetA K (K = C orR) be compact with A (f) = . Thenthere exists > 0 such that for A and g : E E, a continuous mapwith f g < , (f g) < , it follows that / (g), where

f g = inf{k 0 : f(x) g(x) kx, x E}.

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Proof. For A, we have(I f) > 0, m(I f) > 0, and (I f) > 0.

Thus I f is (0, 0)-epi for some 0 > 0 on every Br. By the proof ofTheorem 3.5, there exists > 0 such that for every with | | < ,(I f) > (I f)/2, m(I f) > m(I f)/2,

and I f is (0, 0/2)-epi on Br. Let0 < < min {(I f)/2, m(I f)/2, 0/2} ,

and assume that g f < , (g f) < . Then by Proposition 3.1.3 of[9],

(I g) (I f) (f g) > 0,

and(I g)(x) x f(x) f(x) g(x) > (m(I f)/2 )x.Hence m(I g) > 0. Furthermore, for every t (0, 1], and x = 0,

x f(x) + t(f(x) g(x)) > 0.By the Continuation Principle for (0, k)-epi mappings [14], I g is (0, r0)-epi for each r0 > 0 with

r0 < min{((I f)/2) (f g), (0/2) (f g)}.This implies that (I

g) > 0, so

(g). Let

O(, ) = { K : | | < },the above discussion implies that

AO(, ) A. Since A is com-

pact, there exist a finite collection such thatni=1O(i, i) A. Let

= min{1 , 2 , , n}, and suppose that g f < , (g f) < . For A, if O(i, i), i {1, , n}, then

g f < i , (g f) < iimply that / (g).

The following theorem whose proof follows that of Theorem 8.3.2 [9] con-cerns the upper semicontinuity of the spectrum. We omit its proof here.

Theorem 3.10. Let

p(E) = {f : (f) < +, there exists M > 0 such thatf(x) Mx forx E}.

The multivalued map : p(E) K which assigns to each f its spectrum(f), is upper semicontinuous (with compact values).

Suppose that f(0) = 0, we recall that a point is called a bifurcation

point of f if there are sequences n K and xn E such that xn =0, f(xn) = nxn, n , xn 0 as n . is called an asymptoticbifurcation point of f if there are sequences n K and xn E such thatxn = 0, f(xn) = nxn, n , xn as n . The followingproposition gives the relation between the spectrum and bifurcation points.

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Proposition 3.11. Bifurcation points and asymptotic bifurcation points off are in the spectrum of f.

Proof. Suppose that n K

and xn

E are such that n

, xn

=0, xn 0 and f(xn) = nxn. Then m(I f) = 0. Otherwise theinequality xn f(xn) m(I f)xn would implies that | n| m(I f) > 0, a contradiction. Hence (f).

Similarly, if is an asymptotic bifurcation point of f, we obtain (f).

The following properties of the spectrum are easily checked.

Proposition 3.12. LetE be a normed space andf : E E be a continuousoperator. Then for every

K,

(1) (f) (f), (0) = 0, (I) = 1, (I) = .(2) (I+ f) + (f).We close this section with the following proposition devoted to the the

study of the nonlinear resolvent.

Proposition 3.13. Assume that A : E E is continuous and, (A).Let

RA() = (A I)1, RA() = (A I)1be the multivalued maps. Then

RA()x RA()(I+ ( )RA())x, x E.If I A is injective, then

RA()x = RA()(I+ ( )RA())x, x E.Proof. Let y RA()x, so that (AI)y = x. Then we can write Ayy =x + ( )y, so that

y (A I)1(x + ( )y) (A I)1(x + ( )RA()x).This implies that

RA()x RA()(I+ ( )RA())x.When I A is injective, it is easy to show that equality holds.

4. Positively homogeneous operators

According to our new definition, some special properties of eigenvalues inthe spectrum of a positively homogeneous operator can be obtained, whichwill be useful in Section 6. Firstly, we shall prove the following lemmas onthe positively homogeneous (0, k)-epi mappings from Banach space E to F,

which will be used later.

Lemma 4.1. Suppose that f : E F is a positively homogeneous mappingand f is (0, )-epi on some Br for some > 0. Thenf is (0, )-epi on everyBR with R > 0.

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Proof. f is (0, )-epi on Br and f is positively homogeneous ensure thatf(x) = 0 for all x = 0. Thus f is 0-admissible on BR. Assume that h : EF is an -set contraction with h(x) = 0 for x BR. Let

h1(x) =rR

h

Rr

x

.

Then for every bounded set A E ,(h1(A)) =

r

R

h

R

rA

rR

R

rA

= (A).

So h1 is an -set contraction too. Furthermore, h1(x) = 0 for x Br. Thusthe equation

f(x) =r

Rh

R

rx

has at least one solution x0 E and x0 < r. Then (R/r)x0 BR is asolution of the equation f(x) = h(x).

Remark 4.2. Lemma 4.1 and the Localization property of (0, k)-epi maps[17] show that a positively homogeneous mapping f is (0, )-epi on 1 ,where 1 f1(0) and 1 is an open bounded set of E, if and only if f is(0, )-epi on the closure of all the bounded open sets f

1

(0). This is nottrue if f is not positively homogeneous as the following example shows. Letf : R R be the function f(x) = x2 1, and let 1 = (2, 1/2) (1/2, 2).Then f is (0, k)-epi for every k 0 on 1, but f is not 0-epi on (n, n) forn > 2.

Lemma 4.3. Suppose f : E F is positively homogeneous , (f) > 0 andf is (0, )-epi on Br with r > 0. Then for each p F, there exists R > 0such that f is (p,1)-epi on BR for some 1 > 0.

Proof. Let p

F and g(x) = f(x)

p, then (g

f) = 0. Let

S = {x : f(x) + t(g(x) f(x)) = f(x) tp = 0 for some t (0, 1]}.We shall show that S is bounded. Otherwise, there would exist a sequence{xn}n=1 S with xn as n . Let tn (0, 1] be such thatf(tn) = tnp. Then

f(rxn/xn) = rtnp/xn 0 as n .Letting un = xn/xn, we would have

(f)(n=1run) (n=1f(run)) = 0.

Since (f) > 0, we have (n=1run) = 0. This implies that there exists asubsequence runk u0 and u0 = r. Thus

limk

f(runk) = f(u0) = 0.

This contradicts the fact that f is 0-admissible on Br.

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Now, let R > 0 be such that S BR. Then BR S = . By Lemma4.1, f is (0, )-epi on BR. So, the Continuation principle of (0, k)-epi maps[14] ensures that g is at least (0, 1)-epi for 0 < 1 < (f) and 1 . Thusf is (p,1)-epi on BR.

Our next theorem characterizes regular maps among positively homoge-neous maps.

Theorem 4.4. Let f be positively homogeneous. Then f is regular if andonly if

1. (f) > 0;2. There exists > 0 such that f is (0, )-epi on B1.

Proof. Clearly, we only need to prove that if f satisfies 1 and 2, then f is

regular.Suppose that f satisfies 1 and 2, then by Lemma 4.1, f is (0, )-epi onevery Br with r > 0. So (f) > 0. Assume that m(f) = 0. Then there wouldexist a sequence {xn}n=1 E, xn = 0 such that f(xn) < (1/n)xn. Letun = xn/xn, then f(un) 0 as n . Moreover,

(f)(n=1un) (n=1f(un)).Hence (n=1un) = 0. This implies that {un}n=1 has a convergent subse-quence unk u0, u0 = 1, and f(u0) = 0. This contradicts f is (0, )-epion B1. So m(f) > 0 and f is regular.

It is known that for a linear operator f, if (f) and || > (f),then is an eigenvalue of f [15]. We shall give an example later to showthat for nonlinear operators, this property is not true. But if f is positivelyhomogeneous, we have the following result on eigenvalues in the spectrum.This theorem can be used to obtain existence results for some nonlinearoperator equations as in the example which will be given later.

Theorem 4.5. Let f : E E be a positively homogeneous operator and (f) with || > (f). Then there exists t0 (0, 1] such that /t0 is aneigenvalue of f.

Proof. || > (f) ensures that (I f) || (f) > 0 (see [9]). LetS = {x E : x = 1, x tf(x) = 0 for some t (0, 1]}.

IfS = , then by the Homotopy property [17], If / is (0, r(f)/||)-epion B1 for each 1 > r > /||, since f / is a (f)/||-set contraction. Itfollows that I f is (0, r|| (f))-epi on B1. By Theorem 4.4, we knowthat I f is regular, so (f). This contradiction ensures that S = .Thus there exists t0 (0, 1] and x0 E with x0 = 1, such that

x0

t0f(x0) = 0,

so /t0 is an eigenvalue of f.

The following result, which generalizes a result of [15] (p. 85), shows thatfor odd and positively homogeneous mappings, the result on eigenvalues oflinear operators remains valid.

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Theorem 4.6. Suppose f : E E is odd and positively homogeneous, (f) with || > (f). Then is an eigenvalue of f.Proof. Assume that m(I

f) > 0. Then there exists m > 0 such that

(I f)x mx for all x E.Let O1 = {x : x < 1}, since f is odd, deg(I f/, O1, 0) = 0, [3]. So fork1 satisfying 0 (f)/|| < k1 < 1, I f / is (0, k1 (f)/||)-epi ([17]Theorem 2.8). Hence, I f is (0, ||k1 (f))-epi on B1. Also we knowthat (I f) || (f) > 0. Thus, (f). This contradiction showsthat m(I f) = 0. Therefore there exists a sequence {xn}n=1 E suchthat

xn f(xn) < (1/n)xn.Letting un = xn/xn, we have

un f(un) < 1/n 0, as n .Moreover, we have

(I f)(n=1un) (n=1(I f)un) = 0.This implies (n=1un) = 0. So {un} has a convergent subsequence. Sup-pose that unk u0. Then f(u0) = u0 and u0 = 1, so is an eigenvalueof f.

The following result follows directly from Theorem 4.6, which generalizesthe result in the spectral theory for linear compact operators.

Corollary 4.7. Suppose that f is a compact, odd and homogeneous opera-tor. Then for (f), if = 0, is an eigenvalue of f.

It is known that for a continuous linear operator f, the radius of the

spectrum r(f) = limn fn 1n . The following theorem gives an estimatefor the radius of the spectrum of positively homogeneous maps.

Theorem 4.8. Let E be a Banach space over R and f : E

E be a

positively homogeneous operator with (f) < , lim infn fn1

n < . If > max

(f), lim inf

nfn 1n

,

then (f). If also x1 = x2 implies f(x1) = f(x2), thenr(f) max

(f), lim inf

nfn 1n

.(4.1)

Proof. Suppose that > max

(f), lim infn fn 1n

. Let

V =

{x : x

tf(x) = 0 for some t

(0, 1]

}.

We claim that V = {0}. Indeed, otherwise assume x0 V and x0 = 0. Lett0 (0, 1] be such that x0 t0f(x0) = 0. Then

f = supx=1

f(x) f(x0/x0) .

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Also

f2(x0/x0) = f((x0)/(t0x0)) 2/t02.(4.2)

So we have f2 2. By induction, we obtain that fn1

n . Thiscontradicts > lim infn fn 1n . Now, I f = (I f /) and f / isan (f)/||-set contraction. So, by the homotopy property [17], I f / is(0, r (f)/||)-epi on B1 for every r satisfying (f)/|| < r < 1. ThusI f is (0, r|| (f))-epi on B1. Furthermore, (I f) > 0. Theorem4.4 implies that (f).

In the case that x1 = x2 implies f(x1) = f(x2), (4.2) is also truefor < 0. So by the same proof as above, we obtain that if

|

|> max(f), lim infn

fn

1

n ,then (f). Therefore we have (4.1).

The following example shows that the estimate in Theorem 4.8 is bestpossible.

Example 4.9. Let f(x) = axe, then f is positively homogeneous andeven. a is an eigenvalue of f and fn = an. Hence liminffn 1n = a andr(f) = a.

5. Comparison and nonemptiness of spectra

We begin this section with an example concerning eigenvalues and thespectrum fmv(f).

Example 5.1. Let f : R R be the function f(x) = x3. Then fmv(f) = (see [9]). We will show that (f) = {0} {eigenvalues of f}. In fact,for every (0, ), we have f( 12 ) = 12 . Thus is an eigenvalue off, so (0, ) (f). Next, 0 (f) since m(f) = 0. Furthermore, let (, 0) , then |x f(x)| = |x x3| |x| for x R. Hencem(I f) > 0. Also, (I f) > 0 and (I f)x = x x3is (0, )-epi for all > 0. This implies that (I

f) > 0. Therefore,

(, 0) = (f), [0, ) = (f) and (f) = {0} {eigenvalues off}.The following is an interesting result of the theory.

Theorem 5.2. Suppose f : E E is continuous and f(0) = 0. Then wehave

lip(f) (f) fmv(f).Proof. We will prove that lip(f) (f) fmv(f).

(a) Assume that (f). Then (If) > 0 and m(If) > 0. Hencethere exists m > 0 such that

(I f)(x) mx for all x E.This ensures that

d(I f) = lim infx

(I f)(x)x m > 0.

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Moreover, (I f) > 0 implies that there exists > 0 such that I f is(0, )-epi on Br with r > 0. So If is stably-solvable [9]. Thus fmv(f)and therefore (f) fmv(f).

(b) Suppose that lip(f), then If is one to one, onto, and (If)1is a Lipschitz map [15]. Let L > 0 be the Lipschitz constant. Then

(5.1) (I f)x1 (I f)x2 (1/L)x1 x2 for all x1, x2 E.Letting x2 = 0, we have (I f)x1 (1/L)x1 for all x1 E. Hencem(I f) > 0. Also, by (5.1), (I f) 1/L > 0.

Let r > 0 and Or = {x : x < r}. I f : Or E is continuous,injective and (1/L)-proper [17]. Furthermore (I f)Or is open because(I f)1 is continuous. By our assumption (I f)(0) = 0. By Theorem2.3 of [17], I

f is (0, k)-epi on Br for each nonnegative k satisfying the

condition k < L. Hence (I f) > 0, (f) and lip(f) (f).The following example shows that lip(f) (f).

Example 5.3. Let : R R be defined by

(x) =

x for x 1,1 for 1 < x < 2,x 1 for x 2.

Let f = I . Then for x R we have (1/2)x (x) 2x and so

f(x)

3

x

. Also, (I

f) > 0 for f is a compact map. I

f is (0, )-epifor all > 0 on every [n, n]. Hence = 1 (f).

Obviously, I f = is not one to one, so 1 lip(f). Thus (f) lip(f).

Example 5.1 shows that (f) fmv(f).

The following two results discuss the spectrum for positively homogeneousmaps and maps that are derivable at 0 respectively.

Theorem 5.4. Suppose f is a positively homogeneous map and (f) \fmv(f). Then one of the following cases occur:

1. I f is not injective ;2. (I f)1 is not continuous.

Proof. Suppose that (f) \ fmv(f) and I f is injective. We shallshow that (I f)1 is not continuous. Firstly / fmv(f) ensures that(I f) > 0 and I f is surjective. Also I f is injective impliesthat x f(x) = 0 for each x = 0 . Hence I f is (I f)-proper [17].Assume (If)1 is continuous, then If maps every open ball Or to anopen set. It follows that If is (0, k)-epi for each nonnegative k satisfyingk < 1/(I

f). By Theorem 4.4, we have

(f). This contradiction

shows that (I f)1 is not continuous.Theorem 5.5. Let f : E E be derivable at 0 with derivative T and (f) \ fmv(f). Then one of the following cases occur:

1. is an eigenvalue of f;

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Then (If)(u, v) = (x, y). Hence If is onto. Next, let g = (If)1,and for (x1, y1), (x2, y2) C2, let x = x1 x2, y = y1 y2. Then

|g(x1, y1) g(x2, y2)|2

= |1

||2 + i |2

|x + y|2

+ |1

i ||2 |2

|ix + y|2

.

Letting r = | 1||2+i

| > 0, we have|g(x1, y1) g(x2, y2)|2 r2(|| + 1)2(|x|2 + |y|2).

So, g is a Lipschitz map with constant r(|| + 1).In the case = 0, |f(x)| = |x|. Also, f is one to one, onto with |f1(x)| =

|x|. Hence, lip(f) = C and lip(f) = . By Theorem 5.2, (f) = lip(f) =fmv(f) = .

Remark 5.7. In [16], the authors showed that lip(f) is always nonemptyin the one-dimensional case and asked whether this is also true in higherdimensions. In [1] (which was seen after this part of the work had beencompleted), the authors gave a negative answer to this question by usingExample 5.6. We found Example 5.6 in [13] where it was used to showanother fact.

We close this section with the following result regarding operators whichare asymptotically linear or derivable at 0.

Proposition 5.8. Letf : E

E be continuous and f = T + R, where T isa linear operator and R satisfies one of the following conditions:

1. R(x)x 0 as x 0.2. R(x)x 0 as x .

Then (f) provided is an eigenvalue of T.Proof. Let x0 E and x0 = 0 be such that T(x0) = x0. For r R withr 0 we have

rx0 f(rx0) = R(rx0).So, in case (1), letting r 0 and in case (2) letting r , we have

rx0 f(rx0)rx0 0.

This implies that m(I f) = 0. Hence (f).

6. Applications

In this section, firstly by applying the theory, we shall study the solvabilityof a global Cauchy problem following the work of [14], we shall obtain the

existence result by different method. Then two well known theorems will begeneralized by using the theory.

We shall use the classical space C[0, 1] with the norm

x = maxt[0,1]

|x(t)|.

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178 W. FENG

We recall that a cone K in a Banach space E is a closed subset of E suchthat(1) x, y K, a, b 0 imply ax + by K;(2) x K and x K imply x = 0.

A cone K is said to be normal if there exists a constant > 0 such that

x + y xfor every x, y K.Example 6.1. We look for a non-trivial solution of the following globalCauchy problem depending on a parameter

x(t) = x2(t) + x2(1 t), x(0) = 0, t

[0, 1].(6.1)

Changing the problem into an integral equation we study the existence ofan eigenvalue and eigenvector of the operator

F x(t) =

t0

x2(s) + x2(1 s) ds.(6.2)

It is easily verified that F : C[0, 1] C[0, 1] is positively homogeneous,order preserving, and F x 2x.

Now we shall prove that I F is not surjective for every 0 < < 1/2and so [0, 1/

2]

(T). Assume it is surjective, then there exists x0

C[0, 1] such that x0 T x0 = . For every t [0, 1],(x0(t) 1) = F x0(t) 0 = x0(t) 1.

So for each natural number n, Fnx0 Fn1. On the other hand,F x0 x0, so Fnx0 nx0.

Hence

nx0 Fnx0 Fn(1).This implies that

(2)nnx0 (2)nFn(1) 0.Also we know that K = {x(t) C[0, 1] : x(t) 0} is a normal cone inC[0, 1] and (

2)nn 0 as n . Thus we obtain that (2)nFn(1)

0 (n ). This contradicts Fn(1) (1/2)nt. So, I F is not onto.Assume that for some 1/2 < < 0, I F is onto. Then for each y

C[0, 1], there exists x C[0, 1], such that x F x = y. Hence ()(x) F(x) = y. So, F is onto. This is a contradiction since 1/2 > > 0.

By the above argument, [1/2, 1/2] (F) since the spectrum ofF is closed. Also, we know that F is a compact map [14], so (F) = 0.By Theorem 4.5, there exist 1 1/2 and 2 1/2 such that 1and 2 are eigenvalues of F. Moreover, by Theorem 4.8, we obtain that1/

2 |i| lim infn Fn1/n. Therefore, problem (6.1) has at least

two non-trivial solutions.

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Remark 6.2. It was known that for the above operator F, (1/

2) ln(1 +2) is the only positive eigenvalue of F (see [14]). So, for each [1/2,

1/

2], is not an eigenvalue ofF. This shows that Theorem 4.6 is not true

for positively homogeneous maps. It is well known that it is true for linearoperators.

The following theorem is a generalization of the well known Birkoff-Kelloggtheorem.

Theorem 6.3. LetE be an infinite dimensional Banach space andS be theunit sphere of E. Let f : S E be continuous and bounded. Assume thatinfxSf(x) > (f). Then for every complex number with = 0 thereexists r > 0 such thatr is an eigenvalue off. In particular, f has a positive

and a negative eigenvalue.Proof. Let f : E E be the positively homogeneous operator which isdefined as follows:

f(x) =

xf(x/x) ifx = 0,0 ifx = 0.

Then we have

d(f) = lim infx

f(x)

x

= inf

xSf(x), |f| = lim sup

x

f(x)

x

= sup

xSf(x),

and (f) = (f) (see [14]). Let B(f) be the set of all asymptotic bifurcation

points off. By Theorem 11.1.1 of [9], there exists > 0 such that B(f).Let {n}n=1 and {xn}n=1 E be sequences such that n , xn as n and f(xn) = nxn. Then we obtain

(I f)xnxn 0, as n .

This implies that d(I f) = 0. So, by Theorem 5.2, fmv(f) (f).Assume that (f), by our assumption (f) = (f) < d(f), so < d(f).Hence

d(I f) d(f) > 0, (see [9]).This contradicts d(I f) = 0. Thus we have > (f). By Theorem 4.5,there exists t0 (0, 1] such that /t0 is an eigenvalue of f. Let x0 E withx0 = 1 be such that f(x0) = (/t0)x0. Then f(x0) = rx0, where r = /t0.

For every complex number with || = 1, writing = ei, we haveinfxS

f(x) = infxS

f(x) > (f) = (f).

By the above argument, there exists r > (f) and x0 E with x0 = 0 suchthat f(x0) = rx0. Thus f(x0) = (r)x0. In the case || = 1, let 1 = /||.By the same argument, there exists r > 0 such that r is an eigenvalue off.

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180 W. FENG

Remark 6.4. The Birkoff-Kellogg theorem which was proved in [9] (Theo-rem 10.1.5) is the special case of Theorem 6.3 when f is compact, and theyshowed only that f has a positive eigenvalue.

The following example shows that there exists a map f to which Theorem10.1.5 of [9] does not apply but Theorem 6.3 can be used.

Example 6.5. Let B1 = {x E : x 1} and g : E B1 be the radialretraction of E onto the unit ball, that is

g(x) =

x/x ifx > 1,x if 0 x 1.

Since g(A) co(A 0), g is a 1-set contraction [3]. Let y E with y > 2and f : S E be defined by

f(x) = y + g(x), x E.Then,

infxS

f(x) = infxS

y + g(x) y supxS

g(x) = y 1 > 1.

Next,

(f) = (y + g) = (g) = 1.

So, infxS

f(x)

> (f). Furthermore, we have

supx=1

f(x) = supx=1

y + g(x) y + 1.

Hence, f satisfies the conditions of Theorem 6.3. So, for every z C , thereexists > 1 such that z is an eigenvalue of f.

Theorem 6.3 enables us to give the following generalization of Theorem10.1.2 of [9], who considered compact maps.

Theorem 6.6. Let S be the unit sphere in an infinite dimensional Banach

space E and letf : S S be a continuous strict set contraction. Then every K, || = 1, is an eigenvalue of f. In particular, f has a fixed point andan antipodal point.

Proof. Since (f) < 1, we have

infxS

f(x) = 1 > (f), supxS

f(S) < +.

By Theorem 6.3, for every complex number || = 1, there exists > 0,such that is an eigenvalue of f. Thus there exists x S such thatf(x) = x. Since

f(x)

= 1,

|

|= = 1, so is an eigenvalue of f.

In particular, for = 1, there exists x S, such that f(x) = x. For = 1, there exists x S, such that f(x) = x.

Our theory also enables us to give a generalization of the Hopf theoremon spheres, Theorem 10.1.6 of [9]. Firstly, we need the following lemma. We

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shall let E denote the dual space ofE and J : E E the duality mapping,that is, for x E,

J(x) =

{x

E : x(x) =

x

2 =

x

2

}.

We recall that for a linear operator A in a Hilbert space E, the numericalrange of A is the set of values of (Ax,x) for all x with x = 1. It is knownthat the numerical range of A is a convex set. Let V(A) be the numerical

range of A, then (A) V(A).For a nonlinear operator f, we have the following.

Lemma 6.7. Suppose f : E E is a continuous operator. LetV(f) =

(f(x), x)

x

2

{0}, x E, x J(x).

Then coV(f) provided that (f) and || > (f), where co denotesthe closed convex hull.

Proof. We shall prove that || > (f) and dist(, coV(f)) > 0 implies that (f).

If || > (f) and dist(, coV(f)) > 0, then0 < d = dist(, conv V(f))

| (f(x), x)

x2 |

= |(x, x) (f(x), x)x2 |

=|(x f(x), x)|

x2

x f(x)x .

So, x f(x) dx. This implies that m(I f) > 0. Also, (If(x)) > 0 since || > (f). Let

M = {x E : x tf(x) = 0, t [0, 1]}.For x M, suppose x = tf(x) with t = 0. Then

(f(x), x)/x2 = ((/t)x, x)/x2 = /t.Hence,

= t(f(x), x)/x2 coV(f).This contradicts our assumption d = dist(, coV(f)) > 0. Thus, M = {0}.By the homotopy property [17], I f is (0, )-epi for some > 0, thus(I f) > 0, and so I f is regular. This shows that

{ : || > (f)} (f) coV(f).

Let Sn = {x Rn+1 : x = 1}. The following theorem is a generalizationof Theorem 10.1.6 of [9].

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182 W. FENG

Theorem 6.8. 1. Letf : Sn Rn+1 be continuous. Assume f(x), x =0 for all x Sn, where , is the Euclidean inner product onRn+1.Then, f vanishes at some point x Sn provided either n is an evennumber or f(Sn) is contained in a proper subspace ofRn+1.

2. Let E be an infinite dimensional Banach space and f : S E bea continuous compact mapping. Assume that (f(x), x) = 0 for allx J(x), x S. Then infxSf(x) = 0.

Proof. (1) Let

f(x) =

xf(x/x) ifx = 0,0 ifx = 0.

Then for every x

Rn+1 with x

= 0,

f(x), x = xf(x/x), x/x = 0.So, V(f) = {0}. Also, (f) = 0 since Rn+1 is finite dimensional. By Lemma6.7, we have (f) {0}.

(a) Assume that n is even. Let B(f) denote the set of all asymptotic

bifurcation points of f, then by Theorem 11.1.3 of [9], B(f) = . By Propo-sition 3.11, B(f) (f). Hence, 0 B(f). Suppose that {xn}n=1 Rn+1and n K satisfy xn , n 0 as n and nxn = f(xn). Then

f(xn/

xn

) = nxn/

xn

0.

Moreover, {xn/xn}n=1 has a convergent subsequence xnk/xnk x0 (k ). Thus x0 = 1 and f(x0) = f(x0) = 0.

(b) In the case that f(Sn) is contained in a proper subspace of Rn+1,f : Rn+1 Rn+1 can not be surjective. So 0 (f). Also, we know that(f) = since Rn+1 is finite dimensional. Firstly, if d(f) = 0, it followsthat there exists {xn}n=1 Rn+1 with xn = 1 such that

f(xn) = f(xn) 0 (n ).So, f(x0) = 0 for some x0

Rn+1 and

x0

= 1. Secondly, if d(f) > 0, then

0 K\(f) and 1 / (f). By Theorem 11.1.2 of [9], we have B(f)

(f)

separates 1 from 0. So, B(f)

(f) = . This implies that B(f) = since(f) = . By the same argument with that in (a), there exists x0 Snsuch that f(x0) = 0.

(2) Suppose E is an infinite Banach space. Let f be as in (1), then

V(f) = {0}. By Theorem 8.2.1 of [9], (f) = (according to [9], (f) ={ K : d(I f) = 0}). Again applying Lemma 6.7, we have

(f) (f) V(f).So, 0 (f). Thus there exist xn E with xn = 1 such that f(xn) =f(xn) 0. Hence infxSf(x) = 0.Remark 6.9. Theorem 10.1.6 of [9] only proved the case 1 of Theorem 6.8and n is even.

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Acknowledgment. The author wishes to thank Professor J. R. L. Webbfor valuable discussions.

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Department of Mathematics

University of South Florida

Tampa, FL 33620-5700, USA

E-mail address: [email protected]

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