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Some eigenvalue results for perturbations of maximal monotone operators

Abstract

We study the solvability of a nonlinear eigenvalue problem for maximal monotone operators under a normalization observation. The investigation is based on degree theories for appropriate classes of operators, and a regularization method by the duality operator is used. Let X be a real reflexive Banach space with its dual X and Ω be a bounded open set in X. Suppose that T:D(T)X X is a maximal monotone operator and C:(0,)× Ω ¯ X is a bounded demicontinuous operator satisfying condition ( S + ). Applying the Browder degree theory, we solve a nonlinear eigenvalue problem of the form Tx+C(λ,x)=0. In the case where Tx+λCx=0, an eigenvalue result for generalized pseudomonotone densely defined perturbations is obtained by the Kartsatos-Skrypnik degree theory.

MSC:47J10, 47H05, 47H14, 47H11.

1 Introduction and preliminaries

Eigenvalue theory is closely related to the problem of solving nonlinear equations which was initiated by Krasnosel’skii [1] for compact operators. Regarding maximal monotone operators, it has been extensively investigated by many researchers in various aspects, with applications to evolution equations and differential equations; see, e.g., [25]. The study was mostly based on degree theories for appropriate classes of operators and the usual method of regularization by means of the duality operator; see [610].

Let X be a real reflexive Banach space with its dual X and Ω be a bounded open set in X. Suppose that T:D(T)X X is a maximal monotone operator. We consider the nonlinear eigenvalue problem

Tx+λCx=0,
(E)

where C:D(C)X X is an operator. When the operator C or the resolvents of the operator T are compact, this problem was studied, for instance, by Guan-Kartsatos [11], Kartsatos [12], and Li-Huang [13], where the method is to use the Leray-Schauder degree theory. More generally, an implicit eigenvalue problem of the form

Tx+C(λ,x)=0
(IE)

was investigated in [3, 4], where C:(0,)× Ω ¯ X is a bounded operator to be specified later. Kartsatos and Skrypnik [4] observed the above eigenvalue problem provided that the following property (P) is fulfilled: For ε>0, there exists a λ>0 such that the equation

Tx+C(λ,x)+εJx=0

has no solution in D(T)Ω, where J denotes the duality operator. This property has a close relation to the use of topological degree in eigenvalue theory by the regularization method. It is shown in [3] that two conditions about the weak closure of a certain set consisting of normalized vectors and the asymptotic behavior of the operator C at infinity of λ, called normalized conditions, are main ingredients in solving an eigenvalue problem. In this connection, we are now interested in finding eigenvectors under normalized conditions more concrete than property (P).

The purpose of this paper is to establish the existence of solutions for the above eigenvalue problems under normalized conditions, motivated by the works of Kartsatos and Skrypnik [3, 4]. We first study implicit eigenvalue problem (IE), where C is assumed to be a bounded demicontinuous operator satisfying condition ( S + ). For this, a key tool is the Browder degree given in [6]. Next, we consider two types of the operator C for eigenvalue problem (E). For the one, we apply the Browder degree theory for nonlinear operators of the form T+f with T maximal monotone and f bounded with condition ( S + ) introduced in [7]. In the other case, where C is a generalized pseudomonotone, quasibounded, and densely defined operator, we solve this problem by using the Kartsatos-Skrypnik degree theory for densely defined ( S ˜ + )-perturbations of maximal monotone operators developed in [8].

This paper is organized as follows. In Section 2, we study the solvability of implicit eigenvalue problem (IE) with normalized conditions based on the Browder degree theory. Section 3 contains an eigenvalue result for problem (E) as a special case of (IE). In Section 4, we deal with eigenvalue problem (E) for densely defined perturbations of maximal monotone operators under normalized conditions.

Let X be a real Banach space with dual space X , Ω be a nonempty subset of X, and Y be another real Banach space. Let Ω ¯ , intΩ, and Ω denote the closure, the interior, and the boundary of Ω in X, respectively. The symbol → () stands for strong (weak) convergence. An operator F:ΩY is said to be bounded if F maps bounded subsets of Ω into bounded subsets of Y. F is said to be demicontinuous if for every x 0 Ω and for every sequence { x n } in Ω with x n x 0 , we have F x n F x 0 .

Let T:D(T)X X be an operator. Then T is said to be monotone if

TxTy,xy0for every x,yD(T).

T is said to be maximal monotone if it is monotone and it follows from (x, x )X× X and

x T y , x y 0for every yD(T)

that xD(T) and Tx= x .

T is said to be generalized pseudomonotone if for every sequence { x n } in D(T) with x n x 0 , T x n h and

lim sup n T x n , x n x 0 0,

we have x 0 D(T), T x 0 = h , and lim n T x n , x n =T x 0 , x 0 .

We say that T satisfies condition ( S ˜ + ) if for every sequence { x n } in D(T) with x n x 0 , T x n h and

lim sup n T x n , x n x 0 0,

we have x n x 0 , x 0 D(T), and T x 0 = h .

We say that T satisfies condition ( S + ) if for every sequence { x n } in D(T) with x n x 0 and

lim sup n T x n , x n x 0 0,

we have x n x 0 .

We say that T satisfies condition ( S 0 ) on a set MD(T) if for every sequence { x n } in M with x n x 0 and T x n h and

lim n T x n , x n = h , x 0 ,

we have x n x 0 .

We say that T satisfies condition ( S q ) on a set MD(T) if for every sequence { x n } in M with x n x 0 and T x n h , we have x n x 0 .

We say that T satisfies condition ( T ( 0 ) ) on a set MD(T) if for every sequence { x n } in M with T x n , we have T x n 1 T x n 0.

It is obvious from the definitions that ( S + ) implies ( S 0 ) and ( S 0 ) implies ( S q ). Note that if T satisfies condition ( S ˜ + ) and X is reflexive, then T is generalized pseudomonotone. For the above definitions, we refer to, e.g., [3, 4, 8], and [5].

Let C:[0,)×M X be an operator, where M is a subset of X. Then C(t,x) is said to be continuous in t uniformly with respect to xM if for every t 0 [0,) and for every sequence { t n } in [0,) with t n t 0 , we have C( t n ,x)C( t 0 ,x) uniformly with respect to xM.

We say that C satisfies condition ( S + ) if for every λ(0,) and for every sequence { x n } in M with x n x 0 and

lim sup n C ( λ , x n ) , x n x 0 0,

we have x n x 0 .

Throughout this paper, X will always be an infinite dimensional real reflexive Banach space which has been renormed so that X and X are locally uniformly convex.

An operator J ψ :X X is said to be a duality operator if

J ψ x,x=ψ ( x ) xand J ψ x=ψ ( x ) for xX,

where ψ:[0,)[0,) is continuous, strictly increasing, ψ(0)=0, and ψ(t) as t called a gauge function. If ψ is the identity map I, then J:= J I is called a normalized duality operator. It is known in [14] that J ψ is continuous, bounded, surjective, strictly monotone, maximal monotone and satisfies condition ( S + ).

The following demiclosedness property of maximal monotone operators will be frequently used; see [5].

Lemma 1.1 Let T:D(T)X X be a maximal monotone operator. Then for every sequence { x n } in D(T), x n x in X and T x n x in X imply that xD(T) and Tx= x .

2 Implicit eigenvalue problem

In this section, we establish the existence of a solution for a nonlinear implicit eigenvalue problem under normalized conditions by using the Browder degree theory for class ( S + ).

Recall that a mapping H:[0,1]× Ω ¯ X is of class ( S + ) if the following condition holds:

For any sequence { u j } in Ω ¯ with u j u 0 and any sequence { t j } in [0,1] with t j t 0 for which

lim sup j H ( t j , u j ) , u j u 0 0,

we have u j u 0 and H( t j , u j )H( t 0 , u 0 ); see [2, 6].

As mentioned in the introduction, Kartsatos and Skrypnik [4] gave the following result provided that property (P) is fulfilled. In a more concrete situation, we adopt the normalization method considered in [3].

Theorem 2.1 Let Ω be a bounded open set in X with 0Ω. Let T:D(T)X X be a maximal monotone operator such that the closure of Ω is included in the interior of D(T) with T(0)=0 and T satisfies condition ( T ( 0 ) ) on Ω ¯ . Assume that C:[0,)× Ω ¯ X is demicontinuous, bounded and satisfies condition ( S + ) such that C(0,x)=0 for all x Ω ¯ and C(t,x) is continuous in t uniformly with respect to x Ω ¯ . Further assume that

(c1) There exists a positive number N such that the weak sequential closure of the set

G= { C ( λ , x ) C ( λ , x ) : λ N , x Ω ¯ , J x + T x 2 M ( λ ) }

does not contain the origin 0, where

M(λ)=sup { C ( λ , x ) : x Ω ¯ } .

(c2) lim λ m(λ)=, where m(λ)=inf{C(λ,x):x Ω ¯ }.

Then the following statements hold:

  1. (a)

    For each ε>0, there exists a point ( λ ε , x ε ) in (0,)×Ω such that

    T x ε +C( λ ε , x ε )+εJ x ε =0.
  2. (b)

    If 0T(Ω) and T satisfies condition ( S q ) on Ω, then the implicit eigenvalue problem

    Tx+C(λ,x)=0

has a solution ( λ 0 , x 0 ) in (0,)×Ω.

Proof (a) For our aim, we use the Browder degree d B given in [6]. Let ε be any positive number. We first prove that there is a number Λ(0,) such that

d B ( T + C ( Λ , ) + ε J , Ω , 0 ) =0.
(2.1)

Assume the contrary. For a sequence { Λ n } in (0,) with Λ n , the following occurs:

For each nN, either there exists a point x n Ω such that T x n +C( Λ n , x n )+εJ x n =0, in view of d B (T+C( Λ n ,)+εJ,Ω,0)0, or there exists a point x n Ω such that T x n +C( Λ n , x n )+εJ x n =0. Thus, we get a sequence { x n } in Ω ¯ such that

T x n +C( Λ n , x n )+εJ x n =0.
(2.2)

This implies

J x n +T x n ( 1 ε ) J x n + C ( Λ n , x n ) 2M( Λ n )

for sufficiently large n. Since the sequence { C ( Λ n , x n ) 1 C( Λ n , x n )} is bounded in the reflexive Banach space X , we may suppose that C ( Λ n , x n ) 1 C( Λ n , x n ) converges weakly to some h 0 X . Using (c1), it is clear that h 0 0. It follows from (2.2) that

T x n T x n h 0 .
(2.3)

However, C( Λ n , x n ) by (c2) implies T x n , which is a contradiction to condition ( T ( 0 ) ). Hence assertion (2.1) holds.

Next, we consider a mapping H:[0,1]× Ω ¯ X given by

H(t,x)=Tx+C(tΛ,x)+εJxfor (t,x)[0,1]× Ω ¯ .

Then H is of class ( S + ). To prove this, let { u j } be any sequence in Ω ¯ with u j u 0 and { t j } be any sequence in [0,1] with t j t 0 such that

lim sup j H ( t j , u j ) , u j u 0 0.
(2.4)

Since the operators T and J are monotone, it follows from

H ( t j , u j ) , u j u 0 =T u j , u j u 0 + C ( t j Λ , u j ) , u j u 0 +εJ u j , u j u 0
(2.5)

that

H ( t j , u j ) , u j u 0 T u 0 , u j u 0 + C ( t j Λ , u j ) , u j u 0 +εJ u 0 , u j u 0 .
(2.6)

By (2.4) and (2.6), we have

lim sup j C ( t j Λ , u j ) , u j u 0 0.
(2.7)

There are two cases to consider. If t 0 =0, then C( t j Λ, u j )0 and so

lim j C ( t j Λ , u j ) , u j u 0 =0.
(2.8)

Since (2.5) implies

H ( t j , u j ) , u j u 0 T u 0 , u j u 0 + C ( t j Λ , u j ) , u j u 0 +εJ u j , u j u 0 ,

it follows from (2.4) and (2.8) that

lim sup j J u j , u j u 0 0.

Since J satisfies condition ( S + ), we obtain

u j u 0 ,

which implies

T u j T u 0 ,C( t j Λ, u j )C(0, u 0 ),andJ u j J u 0

on observing that T is demicontinuous on Ω ¯ . This means that H( t j , u j )H(0, u 0 ). If t 0 >0, we have

lim sup j C ( t 0 Λ , u j ) , u j u 0 lim sup j C ( t j Λ , u j ) , u j u 0 + lim sup j [ C ( t j Λ , u j ) C ( t 0 Λ , u j ) , u j u 0 ]

and hence by (2.7),

lim sup j C ( t 0 Λ , u j ) , u j u 0 lim sup j [ C ( t j Λ , u j ) C ( t 0 Λ , u j ) u j u 0 ] =0.

Since C satisfies condition ( S + ), we get u j u 0 from which T u j T u 0 , C( t j Λ, u j )C( t 0 Λ, u 0 ), and J u j J u 0 . Consequently, H( t j , u j )H( t 0 , u 0 ). We have just shown that the mapping H is of class ( S + ).

We are now ready to apply the degree theory of Browder [6, 7]. Then we have

d B ( H ( 0 , ) , Ω , 0 ) = d B (T+εJ,Ω,0)=1.

The last equality is based on Theorem 3 in [6] because the operator T+εJ is strictly monotone and demicontinuous on Ω ¯ and satisfies condition ( S + ). On the other hand, it is shown in (2.1) that

d B ( H ( 1 , ) , Ω , 0 ) = d B ( T + C ( Λ , ) + ε J , Ω , 0 ) =0.

Hence, in view of Theorem 4 in [6], there exist t 0 [0,1] and x 0 Ω such that

T x 0 +C( t 0 Λ, x 0 )+εJ x 0 =0.

It follows from the injectivity of T+εJ that t 0 >0. Consequently, if we let λ ε := t 0 Λ and x ε := x 0 , then we have λ ε (0,) and T x ε +C( λ ε , x ε )+εJ x ε =0.

  1. (b)

    Let { ε n } be a sequence in (0,) such that ε n 0. According to statement (a), there exists a sequence {( λ ε n , x ε n )} in (0,)×Ω such that

    T x ε n +C( λ ε n , x ε n )+ ε n J x ε n =0.

If we set x n := x ε n and λ n := λ ε n , it can be written in the form

T x n +C( λ n , x n )+ ε n J x n =0.
(2.9)

Without loss of generality, we may suppose that

λ n λ 0 , x n x 0 ,andC( λ n , x n ) c ,
(2.10)

where λ 0 [0,], x 0 X, and c X . Note that λ 0 belongs to (0,). In fact, if λ 0 =0, then C( λ n , x n )0 implies T x n 0. Since T satisfies condition ( S q ) on Ω, we obtain that x n x 0 Ω and therefore T x 0 =0, which contradicts the hypothesis that 0T(Ω). If λ 0 =, then (c2) implies C( λ n , x n ) and so T x n . As in (2.3), a similar argument proves that T x n 1 T x n converges weakly to some nonzero vector, which contradicts condition ( T ( 0 ) ). Thus we have shown that λ 0 (0,).

For the next aim, we now show that

lim sup n C ( λ n , x n ) , x n x 0 0.
(2.11)

Assume that (2.11) is false. Then there exists a subsequence of { x n }, denoted again by { x n }, such that

lim n C ( λ n , x n ) , x n x 0 >0.

Hence we obtain from (2.9) that

lim sup n T x n , x n x 0 <0.

Noticing by (2.9) and (2.10) that T x n c , we get

lim sup n T x n , x n < lim n T x n , x 0 = c , x 0 .
(2.12)

For every xD(T), we have by the monotonicity of T

lim inf n T x n , x n lim inf n [ T x n , x + T x , x n x ] = c , x + T x , x 0 x ,

which implies along with (2.12)

c T x , x 0 x >0.
(2.13)

By the maximal monotonicity of T, we have x 0 D(T) and T x 0 = c . Letting x= x 0 in (2.13), we get a contradiction. Therefore, (2.11) is true.

Since C( λ n , x n )C( λ 0 , x n )0, it follows from (2.11) and

C ( λ n , x n ) , x n x 0 = C ( λ n , x n ) C ( λ 0 , x n ) , x n x 0 + C ( λ 0 , x n ) , x n x 0

that

lim sup n C ( λ 0 , x n ) , x n x 0 0.

Since C satisfies condition ( S + ) and λ 0 (0,), we have x n x 0 Ω, which implies C( λ n , x n )C( λ 0 , x 0 ). Hence we obtain from (2.9) that T x n C( λ 0 , x 0 ). By Lemma 1.1, we conclude that T x 0 +C( λ 0 , x 0 )=0. This completes the proof. □

Remark 2.2 In the proof of Theorem 2.1, the demicontinuity of T on Ω ¯ is needed to show that H is of class ( S + ). This is guaranteed under an additional condition Ω ¯ intD(T). Actually, local boundedness of T on intD(T) implies the demicontinuity of T; see, e.g., [5].

3 Eigenvalue problem with condition ( S + )

In this section, we study a multiplicative eigenvalue problem as a special case of the implicit eigenvalue problem in the previous section. As a key tool, we employ the Browder degree for nonlinear operators of the form T+f with T maximal monotone and f bounded with condition ( S + ).

We give a variant of Corollary 1 in [4] under normalized conditions.

Theorem 3.1 Let Ω be a bounded open set in X with 0Ω. Let T:D(T)X X be a maximal monotone operator with 0D(T) and T(0)=0 such that T satisfies condition ( T ( 0 ) ) on D(T) Ω ¯ . Assume that C: Ω ¯ X is a demicontinuous bounded operator which satisfies condition ( S + ) and the two additional conditions:

(c1) There is a positive number N such that the weak sequential closure of the set

G= { C x C x : λ N , x Ω ¯ , J x + T x 2 M ( λ ) }

does not contain zero vector, where

M(λ)=|λ|sup { C x : x Ω ¯ } .

(c2) inf{Cx:x Ω ¯ } is not equal to 0.

Then we have the following properties:

  1. (a)

    For each ε>0, there exists ( λ ε , x ε )(0,)×(D(T)Ω) such that

    T x ε + λ ε C x ε +εJ x ε =0.
  2. (b)

    If 0T(D(T)Ω) and T satisfies condition ( S q ) on D(T)Ω, then the eigenvalue problem

    Tx+λCx=0

has a solution ( λ 0 , x 0 ) in (0,)×(D(T)Ω).

Proof (a) Fix ε>0. Let d B denote the Browder degree in the sense of [7]. We first claim that there exists a number Λ in (0,) such that

d B (T+ΛC+εJ,Ω,0)=0.
(3.1)

Assume the contrary. As in the proof of (2.1), a similar argument establishes that for a sequence { Λ n } in (0,) with Λ n , there is a sequence { x n } in D(T) Ω ¯ such that

T x n + Λ n C x n +εJ x n =0.
(3.2)

This implies

J x n +T x n ( 1 ε ) J x n + Λ n C x n 2M( Λ n )

for sufficiently large n. By the boundedness of the sequence { C x n 1 C x n } in X , we may suppose that C x n 1 C x n h 0 for some h 0 X . It follows from (c1) and (3.2) that h 0 0 and

T x n T x n h 0 .

But (c2) implies T x n , which contradicts condition ( T ( 0 ) ). Hence assertion (3.1) holds.

Now we consider a mapping H:[0,1]×(D(T) Ω ¯ ) X given by

H(t,x):=Tx+εJx+tΛCxfor (t,x)[0,1]× ( D ( T ) Ω ¯ ) .

Using the normalization property of the Browder degree d B , e.g., Theorem 12 in [7], we have

d B ( H ( 0 , ) , Ω , 0 ) = d B (T+εJ,Ω,0)=1.

Moreover, (3.1) means that

d B ( H ( 1 , ) , Ω , 0 ) = d B (T+εJ+ΛC,Ω,0)=0.

Note that T+εJ is maximal monotone and satisfies condition ( S + ). Hence, there are t 0 [0,1] and x 0 D(T)Ω such that

T x 0 + t 0 ΛC x 0 +εJ x 0 =0.

By the injectivity of the strictly monotone operator T+εJ, we know that t 0 >0. Consequently, if we let λ ε := t 0 Λ and x ε := x 0 , then λ ε (0,) and T x ε + λ ε C x ε +εJ x ε =0.

  1. (b)

    Let { ε n } be a sequence in (0,) such that ε n 0. By (a), there exists a sequence {( λ n , x n )} in (0,)×(D(T)Ω) such that

    T x n + λ n C x n + ε n J x n =0.
    (3.3)

We may suppose that

λ n λ 0 , x n x 0 ,andC x n c ,
(3.4)

where λ 0 [0,], x 0 X, and c X . Note that λ 0 belongs to (0,). Indeed, if λ 0 =0, then it follows from T x n 0 and condition ( S q ) that x n x 0 Ω and therefore x 0 D(T) and T x 0 =0, which contradicts the hypothesis that 0T(D(T)Ω). If λ 0 =, then (c2) implies λ n C x n and so T x n . But we can show as above that T x n 1 T x n converges weakly to some nonzero vector, which contradicts condition ( T ( 0 ) ).

To prove that

lim sup n λ n C x n , x n x 0 0,
(3.5)

we assume to the contrary that there exists a subsequence of { x n }, denoted again by { x n }, such that

lim n λ n C x n , x n x 0 >0.

Hence we obtain from (3.3) that

lim sup n T x n , x n x 0 <0.

Noticing by (3.3) and (3.4) that T x n λ 0 c , we get

lim sup n T x n , x n < lim n T x n , x 0 = λ 0 c , x 0 .
(3.6)

For every xD(T), we have by the monotonicity of T

lim inf n T x n , x n λ 0 c , x +Tx, x 0 x,

which implies along with (3.6)

λ 0 c T x , x 0 x >0.

By the maximal monotonicity of T, we get a contradiction. Therefore, (3.5) holds.

It follows from (3.5) and

λ n C x n , x n x 0 =( λ n λ 0 )C x n , x n x 0 + λ 0 C x n , x n x 0

that

λ 0 lim sup n C x n , x n x 0 0.

Since C satisfies condition ( S + ) and is demicontinuous, we have x n x 0 Ω, which implies C x n C x 0 . Hence we obtain from (3.3) that T x n λ 0 C x 0 . By Lemma 1.1, we conclude that x 0 D(T)Ω and T x 0 + λ 0 C x 0 =0, what we wanted to prove. □

Remark 3.2 We point out that in Theorem 3.1, the condition Ω ¯ intD(T) is not necessary to be assumed.

4 Densely defined perturbations

This section is devoted to the eigenvalue problem for densely defined quasibounded perturbations of maximal monotone operators in reflexive Banach spaces. To do this, we apply the Kartsatos-Skrypnik degree for densely defined ( S ˜ + )-perturbations of maximal monotone operators developed in [8].

Recall that an operator C:D(C)X X is quasibounded if for every S>0 there exists a constant K(S)>0 such that for all uD(C) with uS and Cu,u0, we have CuK(S).

As in Section 3, we employ a normalization method to obtain an eigenvalue result for generalized pseudomonotone operators.

Theorem 4.1 Let Ω be a bounded open set in X with 0Ω and L be a dense subspace of X. Let T:D(T)X X be a maximal monotone operator with 0D(T) and T(0)=0 which satisfies condition ( T ( 0 ) ) on D(T) Ω ¯ . Assume that C:D(C)X X is a generalized pseudomonotone quasibounded operator with LD(C). Furthermore, assume that

(h1) There exists a positive number N such that the weak sequential closure of the set

G= { C x C x : λ N , x D ( C ) Ω ¯ , J ψ x + T x 2 M ( λ ) }

does not contain zero vector, where

M(λ)=|λ|sup { C x : x D ( C ) Ω ¯ } .

(h2) inf{Cx:xD(C) Ω ¯ } is not equal to 0.

(h3) For every FF(L) and vL, the function c(F,v):FR, c(F,v)(u)=Cu,v, is continuous on F, where F(L) denotes the set of all finite-dimensional subspaces of L.

Then the following statements hold:

  1. (a)

    For each ε>0, there exists a point ( λ ε , x ε ) in (0,)×(D(T+C)Ω) such that

    T x ε + λ ε C x ε +ε J ψ x ε =0.

Here D(T+C) denotes the intersection of D(T) and D(C).

  1. (b)

    If 0T(D(T)Ω) and T satisfies condition ( S 0 ) on D(T)Ω, then the eigenvalue problem

    Tx+λCx=0

has a solution ( λ 0 , x 0 ) in (0,)×(D(T+C)Ω).

Proof (a) We will apply the Kartsatos-Skrypnik degree d S given in [8]. Let ε be an arbitrary positive number. Since T satisfies condition ( T ( 0 ) ) on D(T) Ω ¯ and J ψ is bounded, we can prove as in Theorem 3.1 that under hypotheses (h1) and (h2), there is a positive number Λ such that

d S (T+ΛC+ε J ψ ,Ω,0)=0.
(4.1)

For t[0,1], we set T t :=T and C t :=tΛC+ε J ψ , where D( T t ) and D( C t ) denote the domain of T t and C t , respectively. In this case, D( T t )=D(T) for t[0,1], D( C 0 )=X for t=0 and D( C t )=D(C) for t(0,1]. Notice that the operators C 0 =ε J ψ and C 1 =ΛC+ε J ψ satisfy condition ( S ˜ + ), based on the facts that C is generalized pseudomonotone and J ψ is bounded and satisfies condition ( S + ).

In the sense of Definition 4.2 in [8], we check the following conditions on two families { T t } and { C t }. In fact, conditions on { T t } are obviously satisfied, with T t independent of t, due to maximal monotonicity of T, 0D(T), and T(0)=0.

( c 1 t ) Since J ψ is monotone and bounded, it follows from the quasiboundedness of C that { C t } is uniformly quasibounded.

( c 2 t ) Let { t n } be any sequence in [0,1] and { u n } be any sequence in L such that t n t 0 , u n u 0 , C t n u n h and

lim sup n C t n u n , u n u 0 0, C t n u n , u n 0,
(4.2)

where t 0 [0,1], u 0 X, and h X . If t 0 =0, then the second inequality in(4.2) implies

εψ ( u n ) u n =ε J ψ u n , u n t n ΛC u n , u n 0

and hence u n 0, u 0 =0D( C 0 ), and C 0 u 0 = h . Now let t 0 >0. From the first inequality in (4.2) and the following inequality

C t n u n , u n u 0 t n ΛC u n , u n u 0 +ε J ψ u 0 , u n u 0 ,

we obtain that

lim sup n C u n , u n u 0 0.
(4.3)

In view of C t n u n h , there exists a subsequence of { u n }, denoted again by { u n }, such that C u n h 1 and J ψ u n h 2 for some h 1 , h 2 X . Since C is generalized pseudomonotone, we obtain from (4.3) that u 0 D(C), C u 0 = h 1 , and C u n , u n C u 0 , u 0 . Thus,

lim n C u n , u n u 0 =C u 0 , u 0 h 1 , u 0 =0.

Hence it follows from the first inequality in (4.2) that

ε lim sup n J ψ u n , u n u 0 = lim sup n t n ΛC u n +ε J ψ u n , u n u 0 0.

Since J ψ satisfies condition ( S + ), we have u n u 0 and so J ψ u n J ψ u 0 . Consequently, u 0 D( C t 0 ) and C t 0 u 0 = t 0 Λ h 1 +ε h 2 = h . Therefore, condition ( c 2 t ) is satisfied.

( c 3 t ) For every FF(L) and vL, the function c ˜ (F,v):[0,1]×FR, c ˜ (F,v)(t,u)= C t u,v, is continuous on [0,1]×F because c(F,v) is continuous on F and J ψ is continuous on X.

We can now consider a mapping H:[0,1]×(D(T+C) Ω ¯ ) X given by

H(t,x):=Tx+tΛCx+ε J ψ x.

By Theorem 4.4 in [8] and (4.1), we have

d S ( H ( 0 , ) , Ω , 0 ) = d S (T+ε J ψ ,Ω,0)=1

and

d S ( H ( 1 , ) , Ω , 0 ) = d S (T+ΛC+ε J ψ ,Ω,0)=0.

According to Theorem 4.3 in [8], there exist t 0 [0,1] and x 0 D(T+C)Ω such that

T x 0 + t 0 ΛC x 0 +ε J ψ x 0 =0.

The injectivity of T+ε J ψ implies that t 0 >0. If we let λ ε := t 0 Λ and x ε := x 0 , then the conclusion follows.

  1. (b)

    Let { ε n } be a sequence in (0,) such that ε n 0. According to (a), there are sequences { λ n } in (0,) and { x n } in D(T+C)Ω such that

    T x n + λ n C x n + ε n J ψ x n =0.
    (4.4)

Then it follows from the monotonicity of T with T(0)=0 that

C x n , x n ε n λ n ψ ( x n ) x n 0.

Hence the quasiboundedness of C implies that {C x n } is bounded. Without loss of generality, we may suppose that

λ n λ 0 , x n x 0 ,andC x n c ,
(4.5)

where λ 0 [0,], x 0 X, and c X . Since T satisfies conditions ( S q ) and ( T ( 0 ) ) and since 0T(D(T)Ω), it is easily verified that λ 0 belongs to (0,).

As in the proof of Theorem 3.1, we can show that

lim sup n λ n C x n , x n x 0 0.
(4.6)

Then it follows from (4.6) and λ n λ 0 that

lim sup n C x n , x n x 0 0.

Since C is generalized pseudomonotone, we have by (4.5)

x 0 D(C),C x 0 = c ,and lim n C x n , x n =C x 0 , x 0 .

Hence we obtain from (4.4) that T x n λ 0 C x 0 and

lim n T x n , x n = lim n λ n C x n , x n = λ 0 C x 0 , x 0 .

Since T satisfies condition ( S 0 ) on D(T)Ω, we have x n x 0 . By the maximal monotonicity of T, we conclude that x 0 D(T+C)Ω and T x 0 + λ 0 C x 0 =0. This completes the proof. □

As a consequence of Theorem 4.1, we get another eigenvalue result for densely defined operators satisfying condition ( S ˜ + ) in comparison with Theorem 4 in [4].

Corollary 4.2 Let T, Ω, and L be as in Theorem  4.1. Assume that C:D(C)X X is a quasibounded operator with LD(C) and satisfies condition ( S ˜ + ). Furthermore, assume that conditions (h1), (h2), and (h3) in Theorem  4.1 are satisfied. Then:

  1. (a)

    For each ε>0, there exists ( λ ε , x ε )(0,)×(D(T+C)Ω) such that T x ε + λ ε C x ε +ε J ψ x ε =0.

  2. (b)

    If 0T(D(T)Ω) and T satisfies condition ( S q ) on D(T)Ω, then there exists ( λ 0 , x 0 )(0,)×(D(T+C)Ω) such that T x 0 + λ 0 C x 0 =0.

Proof Statement (a) follows from part (a) of Theorem 4.1 by noting that if C satisfies condition ( S ˜ + ), then C is generalized pseudomonotone.

  1. (b)

    Let { ε n } be a sequence in (0,) such that ε n 0. In view of (a), we can choose a sequence {( λ n , x n )} in (0,)×(D(T+C)Ω) such that

    T x n + λ n C x n + ε n J ψ x n =0.

We may suppose that

λ n λ 0 , x n x 0 ,andC x n c ,

where λ 0 [0,], x 0 X, and c X . Obviously, λ 0 (0,). As before, the same argument shows that

lim sup n C x n , x n x 0 0.

Since C satisfies condition ( S ˜ + ), we have

x n x 0 , x 0 D(C),andC x 0 = c .

Combining this with T x n λ 0 C x 0 , we obtain that x 0 D(T+C)Ω and T x 0 + λ 0 C x 0 =0 as required. □

Remark 4.3 When C satisfies condition ( S ˜ + ), condition ( S 0 ) on T appearing in Theorem 4.1 may be replaced by weaker condition ( S q ); see the proof of Theorem 4.1.

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Acknowledgements

The authors would like to thank anonymous referees for valuable comments and suggestions. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (NRF-2011-0021-829).

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BJ participated in the sequence alignment and coordination. KI conceived of the study and drafted the manuscript. All authors read and approved the final manuscript.

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Kim, IS., Bae, JH. Some eigenvalue results for perturbations of maximal monotone operators. J Inequal Appl 2013, 72 (2013). https://doi.org/10.1186/1029-242X-2013-72

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Keywords

  • eigenvalue
  • maximal monotone operator
  • perturbation
  • degree theory