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Implicit eigenvalue problems for maximal monotone operators

Abstract

We study the implicit eigenvalue problem of the form

0Tx+C(λ,x),

where T is a maximal monotone multi-valued operator and the operator C satisfies condition ( S + ) or ( S ˜ + ). In a regularization method by the duality operator, we use the degree theories of Kartsatos and Skrypnik upon conditions of C as well as Browder’s degree. There are two cases to consider: One is that C is demicontinuous and bounded with condition ( S + ); and the other is that C is quasibounded and densely defined with condition ( S ˜ + ). Moreover, the eigenvalue problem 0Tx+λCx is also discussed.

1 Introduction and preliminaries

A general eigenvalue theory for maximal monotone operators has been developed in various ways with applications to partial differential equations; see [57, 1012]. A key tool was topological degrees for appropriate classes of operators in, e.g., [24, 8, 9, 1416] and the method of approach was in many cases to use regularization by means of the duality operator, while in [6, 11] the eigenvalue problem was solved by a transformed equation in terms of the approximant without using the regularization method.

Let X be a real reflexive Banach space with dual space X and T:D(T)X 2 X be a maximal monotone multi-valued operator. When the resolvents of T or the single-valued operator C are compact, the solvability of the nonlinear inclusion

0Tx+λCx

was studied in [5, 6, 10, 12] by applying the Leray-Schauder degree theory for compact operators. More generally, implicit eigenvalue problems were considered in [7, 10], where the single-valued and compact case was dealt with in [7]. In direction of [10], we study the implicit eigenvalue problem of the form

0Tx+C(λ,x),

where C:[0,Λ]×M X satisfies condition ( S + ) or ( S ˜ + ). As in [12], we adopt property (P) about the solvability of the related equation

T s x+C(λ,x)+ε J ψ x=0,

where T s is the Brezis-Crandall-Pazy approximant introduced in [1] and J ψ is the (normalized) duality operator with a gauge function ψ.

In the present paper, we divide our investigation into two cases to apply suitable degree theory. One case deals with demicontinuous bounded operators satisfying condition ( S + ) with the aid of the most elementary degree theory of Skrypnik [14], in comparison with [10]. The other case is concerned with quasibounded densely defined operators satisfying condition ( S ˜ + ), where the degree theory of Kartsatos and Skrypnik [8, 9] for densely defined operators is used. In more concrete situations, the eigenvalue problem 0Tx+λCx is discussed. We point out that Browder’s degree in [3] for the reduced simple operator T s +ε J ψ under the homotopy plays a crucial role in the proof of our results presented here.

Let X be a real Banach space with its dual X , Ω a nonempty subset of X, and Y another real Banach space. 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 . Here the symbol → () stands for strong (weak) convergence.

A multi-valued operator T:D(T)X 2 X is said to be monotone if

u v , x y 0for every x,yD(T) and every  u Tx, v Ty,

where D(T)={xX:Tx} denotes the effective domain of T. T is said to be maximal monotone if it is monotone and it follows from (x, u )X× X and

u v , x y 0for every yD(T) and every  v Ty

that xD(T) and u Tx.

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

We say that a multi-valued operator T:D(T)X 2 X satisfies condition ( S q ) on a set MD(T) if for every sequence { x n } in M with x n x 0 and every sequence { u n } in X with u n u where u n T x n , 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 its dual X are locally uniformly convex.

A function ψ:[0,)[0,) is called a gauge function if ψ is continuous, strictly increasing, ψ(0)=0 and ψ(t) as t. An operator J ψ :X X is called a duality operator with a gauge function ψ if

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

If ψ is the identity map I, then J:= J I is called a normalized duality operator. It is known in [13] that J ψ is continuous, bounded, surjective, strictly monotone, maximal monotone and satisfies condition ( S + ).

Given a maximal monotone operator T:D(T)X 2 X , xX and s>0, there exist unique elements x s D(T) and u s T x s such that

J( x s x)+s u s =0.

Two operators J s :XD(T) and T s :X X defined by

J s x:= x s and T s x:= u s for xX

are called the Brezis-Crandall-Pazy approximants. It is known in [1] that J s is continuous and bounded and T s is demicontinuous, bounded, and maximal monotone. It is easy to see that J s =Is J 1 T s and T s xT J s x for xX.

Let C:[0,Λ]×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,Λ] and for every sequence { t n } in [0,Λ] 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,Λ] 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 .

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

lim sup n C ( λ , x n ) , x n x 0 0,

we have x n x 0 , x 0 M and C(λ, x 0 )= h 0 .

We often need the following demiclosedness property of maximal monotone operators given in [17].

Lemma 1.1 Let T:D(T)X 2 X be a maximal monotone multi-valued operator. Then for every sequence { x n } in D(T), x n x in X and u n u in X , where u n T x n , imply that xD(T) and u Tx.

2 Implicit eigenvalue problem about demicontinuous operators

In this section, we are concerned with the implicit eigenvalue problem for perturbed maximal monotone operators in reflexive Banach spaces by applying the degree theories of Skrypnik and Browder for nonlinear operators of monotone type.

In what follows, for a bounded subset Ω of X, let Ω ¯ and Ω denote the closure and the boundary of Ω in X, respectively. Following Browder [2, 3], a homotopy H:[0,1]× Ω ¯ X is said to be of class ( S + ) if the following condition holds:

For every sequence { u j } in Ω ¯ with u j u 0 and every sequence { t j } in [0,1] with t j t 0 such that

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 ).

Kartsatos and Skrypnik [10] obtained the following result by using Browder’s degree in [4] for multi-valued operators. As in [12], we adopt property (P) in terms of the Brezis-Crandall-Pazy approximant so that we can apply the most essential degree theory of Skrypnik [14] for single-valued operators to prove in a direct method.

Theorem 2.1 Let Ω be a bounded open set in X with 0Ω. Let T:D(T)X 2 X be a maximal monotone multi-valued operator with 0D(T) and 0T(0). Let Λ, s 0 , and ε 0 be three positive numbers. Suppose that C:[0,Λ]× 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 ¯ .

  1. (a)

    For a given ε>0, assume the following property:

(P) For every s(0, s 0 ), there exists a λ(0,Λ] such that the equation

T s x+C(λ,x)+εJx=0

has no solution in Ω. Then there exists a ( λ ε , x ε )(0,Λ]×(D(T)Ω) such that

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

    If 0T(D(T)Ω), T satisfies condition ( S q ) on D(T)Ω, and property (P) is fulfilled for all ε(0, ε 0 ], then there exists a ( λ 0 , x 0 )(0,Λ]×(D(T)Ω) such that

    0T x 0 +C( λ 0 , x 0 ).

Proof (a) We first claim that for any s(0, s 0 ), there exists a ( λ 0 , x 0 )(0,Λ]×Ω such that

T s x 0 +C( λ 0 , x 0 )+εJ x 0 =0.
(2.1)

Assume on the contrary that for some s(0, s 0 ) and for every λ(0,Λ], the following holds:

T s x+C(λ,x)+εJx0for all xΩ.
(2.2)

Since ( T s +εJ)(0)=0 and T s +εJ is injective by the strict monotonicity of J, we have T s x+C(0,x)+εJx0 for all xΩ. Thus, (2.2) holds for all λ[0,Λ].

Consider a homotopy H:[0,1]× Ω ¯ X defined by

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

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

lim sup j H ( t j , u j ) , u j u 0 0.
(2.3)

Since T s and J are monotone, it follows from

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

that

H ( t j , u j ) , u j u 0 T s u 0 , u j u 0 + C ( t j Λ , u j ) , u j u 0 +εJ u j , u j u 0
(2.4)

and

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

By (2.3) and (2.5), we have

lim sup j C ( t j Λ , u j ) , u j u 0 0.
(2.6)

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.
(2.7)

Using (2.3), (2.4), and (2.7), we obtain

lim sup j J u j , u j u 0 0.

Since J satisfies condition ( S + ), we have

u j u 0 ,

which implies by the demicontinuity of T s and C and the continuity of J

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

This means that H( t j , u j )H(0, u 0 ). Now, let t 0 >0. We have

C ( t j Λ , u j ) , u j u 0 = C ( t j Λ , u j ) C ( t 0 Λ , u j ) , u j u 0 + C ( t 0 Λ , u j ) , u j u 0 ,

which implies

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 (2.6)

lim sup j C ( t 0 Λ , u j ) , u j u 0 lim 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 s u j T s 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 homotopy H is of class ( S + ).

We are now ready to apply the degree theory of Skrypnik [14, 15]. If d S denotes the Skrypnik degree, the homotopy invariance property of d S implies that

d S ( T s + C ( λ , ) + ε J , Ω , 0 ) = d S ( H ( t , ) , Ω , 0 ) = d S ( T s + ε J , Ω , 0 ) = 1

for all λ[0,Λ]. The last equality follows from Theorem 3 in [3], based on the fact that T s +εJ is demicontinuous, bounded, injective, and satisfies condition ( S + ) and T s x+εJx,x0 for all xΩ. For every λ(0,Λ], the existence property of d S implies that there is a point x in Ω such that

T s x+C(λ,x)+εJx=0,

which contradicts property (P). Therefore, the first claim (2.1) is true.

In view of (2.1), let { s n } be a sequence in (0, s 0 ) with s n 0 and {( λ n , x n )} be the corresponding sequence in (0,Λ]×Ω such that

T s n x n +C( λ n , x n )+εJ x n =0.
(2.8)

Without loss of generality, we may suppose that

λ n λ 0 , x n x 0 ,C( λ n , x n ) c andJ x n j ,

where λ 0 [0,Λ], x 0 X, c X and j X . To arrive at the conclusion of (a) in the next step, we may suppose that λ 0 (0,Λ]. In fact, when λ 0 =0, we know that 0Tx+C(0,x)+εJx for all xD(T)Ω because T+εJ is injective on D(T) Ω ¯ and 0(T+εJ)(D(T)Ω). For our aim, we will now show that

lim sup n C ( λ n , x n ) , x n x 0 0.
(2.9)

Assume on the contrary that 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.

Note by (2.8) and the monotonicity of J that

T s n x n , x n x 0 C ( λ n , x n ) , x n x 0 εJ x 0 , x n x 0

and so

lim sup n T s n x n , x n x 0 <0.

Hence, it follows from T s n x n c ε j that

lim sup n T s n x n , x n = lim sup n [ T s n x n , x n x 0 + T s n x n , x 0 ] < c ε j , x 0 .
(2.10)

Let xD(T) and u Tx be arbitrary. Since T is monotone, T s n x n T J s n x n , and J s n =I s n J 1 T s n , we have

T s n x n u , x n s n J 1 T s n x n x 0,

which implies

T s n x n , x n s n T s n x n 2 + T s n x n ,x+ u , x n x s n u , J 1 T s n x n .

Since { T s n x n } and J 1 are bounded and s n 0, we have

lim inf n T s n x n , x n c ε j , x + u , x 0 x .
(2.11)

Combining (2.10) and (2.11), we get

c ε j u , x 0 x >0for all xD(T) and  u Tx.
(2.12)

Since T is maximal monotone, we have x 0 D(T) and c ε j T x 0 . Letting x= x 0 D(T) in (2.12) yields a contradiction. Thus, (2.9) holds.

Since C(t,x) is continuous in t uniformly with respect to x Ω ¯ and λ n λ 0 , we have by (2.9)

lim sup n C ( λ 0 , x n ) , x n x 0 0,
(2.13)

by observing that

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 .

Since C satisfies condition ( S + ), it follows from (2.13) that

x n x 0 Ω,

which implies by (2.8)

T s n x n C( λ 0 , x 0 )εJ x 0

and as above

J s n x n = x n s n J 1 T s n x n x 0 .

Since T is maximal monotone and T s n x n T J s n x n , Lemma 1.1 implies that x 0 D(T) and

0T x 0 +C( λ 0 , x 0 )+εJ x 0 .
  1. (b)

    According to the statement (a), for a sequence { ε n } in (0, ε 0 ] with ε n 0, we can choose sequences { λ n } in (0,Λ], { x n } in D(T)Ω, and { u n } in X with u n T x n such that

    u n +C( λ n , x n )+ ε n J x n =0.
    (2.14)

We may suppose that

λ n λ 0 , x n x 0 ,C( λ n , x n ) c andJ x n j ,

where λ 0 [0,Λ], x 0 X, c X , and j X . Note that λ 0 (0,Λ]. Indeed, if λ 0 =0, then (2.14) implies u n 0 which gives by condition ( S q ) on D(T)Ω, x n x 0 Ω and therefore x 0 D(T) and 0T x 0 , which contradicts the hypothesis that 0T(D(T)Ω).

To show that x n x 0 , we first claim that

lim sup n C ( λ n , x n ) , x n x 0 0.
(2.15)

Assume the contrary. So, we may suppose that

lim n C ( λ n , x n ) , x n x 0 >0.

Hence, it follows from (2.14) that u n c and

lim sup n u n , x n x 0 <0,

which imply

lim sup n u n , x n < c , x 0 .
(2.16)

For every xD(T) and every u Tx, we obtain from the monotonicity of T that

lim inf n u n , x n lim inf n [ u n , x + u , x n x ] c , x + u , x 0 x ,

which implies along with (2.16)

c u , x 0 x >0.
(2.17)

Since T is maximal monotone, we have x 0 D(T) and c T x 0 . Letting x= x 0 in (2.17), we have a contradiction. Thus, (2.15) is true.

As above, we can deduce from (2.15) that

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

Since C satisfies condition ( S + ) and is demicontinuous, we obtain from (2.14) that

x n x 0 and u n C( λ 0 , x 0 ).

We conclude that x 0 D(T)Ω and

0T x 0 +C( λ 0 , x 0 ).

This completes the proof. □

As a consequence of Theorem 2.1, we have the following result. When C is a compact operator, it was proved by Li and Huang [12] with the aid of the Leray-Schauder degree for compact operators.

Corollary 2.2 Let T, Ω, Λ, s 0 , and ε 0 be as in Theorem  2.1. Suppose that C: Ω ¯ X is a demicontinuous bounded operator which satisfies condition ( S + ).

  1. (a)

    For a given ε>0, assume the following property:

(P) For every s(0, s 0 ), there exists a λ(0,Λ] such that the equation

T s x+λCx+εJx=0

has no solution in Ω. Then there exists a ( λ ε , x ε )(0,Λ]×(D(T)Ω) such that

0T x ε + λ ε C x ε +εJ x ε .
  1. (b)

    If 0T(D(T)Ω), T satisfies condition ( S q ) on D(T)Ω, and property (P) is fulfilled for all ε(0, ε 0 ], then there exists a ( λ 0 , x 0 )(0,Λ]×(D(T)Ω) such that

    0T x 0 + λ 0 C x 0 .

Proof Define an operator C ˜ :[0,Λ]× Ω ¯ X by

C ˜ (λ,x)=λCxfor (λ,x)[0,Λ]× Ω ¯ .

By hypotheses on C, the operator C ˜ is obviously demicontinuous, bounded, and satisfies condition ( S + ). Moreover, C ˜ (t,x) is continuous in t uniformly with respect to x Ω ¯ because C( Ω ¯ ) is bounded. Apply Theorem 2.1 with C= C ˜ . □

3 Implicit eigenvalue problem about densely defined operators

In this section, we study the implicit eigenvalue problem for densely defined perturbations of maximal monotone operators, based on the degree theories of Kartsatos and Skrypnik.

An operator C:[0,Λ]×D(C) X is said to be uniformly quasibounded if for every S>0 there exists a constant K(S)>0 such that for all λ[0,Λ] and all uD(C) with uS and C(λ,u),u0, we have C(λ,u)K(S).

In a regularization method by the duality operator, we establish a new result on the existence of eigenvalues, by applying topological degree for densely defined operators in [8, 9].

Theorem 3.1 Let Ω be a bounded open set in X with 0Ω and L a dense subspace of X. Let T:D(T)X 2 X be a maximal monotone multi-valued operator with 0D(T) and 0T(0). Let Λ, s 0 , and ε 0 be positive numbers. Assume that C:[0,Λ]×D(C) X is a single-valued operator with LD(C)X such that C(0,x)=0 for all xD(C) and C(t,x) is continuous in t uniformly with respect to xD(C). Assume further that

(c1) C is uniformly quasibounded,

(c2) C satisfies condition ( S ˜ + ), and

(c3) for every λ[0,Λ] and for every FF(L) and vL, the function , c(λ,F,v)(u)=C(λ,u),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 a given ε>0, assume the following property:

(P) For every s(0, s 0 ), there exists a λ(0,Λ] such that the equation

T s x+C(λ,x)+ε J ψ x=0

has no solution in D(C)Ω. Then there is a ( λ ε , x ε )(0,Λ]×(D(T)D(C)Ω) such that

0T x ε +C( λ ε , x ε )+ε J ψ x ε .
  1. (b)

    If 0T(D(T)Ω), T satisfies condition ( S q ) on D(T)Ω, and property (P) is fulfilled for all ε(0, ε 0 ], then there exists a ( λ 0 , x 0 )(0,Λ]×(D(T)D(C)Ω) such that

    0T x 0 +C( λ 0 , x 0 ).

Proof (a) First, we prove that for every s(0, s 0 ), there exists a ( λ 0 , x 0 )(0,Λ]×(D(C)Ω) such that

T s x 0 +C( λ 0 , x 0 )+ε J ψ x 0 =0.
(3.1)

Assume on the contrary that for some s(0, s 0 ) and for every λ(0,Λ], the following holds:

T s x+C(λ,x)+ε J ψ x0for all xD(C)Ω.
(3.2)

Then (3.2) holds for all λ[0,Λ]. For λ=0, the assertion is obvious.

For t[0,1], we set T t = T s and C t =C(tΛ,)+ε J ψ . To show that { T t + C t } is an admissible homotopy in the sense of Definition 2.4 in [9], we have to check the following conditions on two families { T t } and { C t }. In fact, conditions on { T t } are automatically satisfied, with T t independent of t, due to the monotonicity of T s and T s (0)=0.

( c 1 t ) { C t } is uniformly strongly quasibounded with respect to T t , i.e., for every >0 there exists a constant K()>0 such that for all uL with u and all t[0,1],

T t u + C t u , u 0implies C t u K().

This follows trivially from (c1) and the fact that T s and J ψ are monotone and J ψ is bounded.

( c 2 t ) For every pair of sequences { t j } in [0,1] and { u j } in L such that t j t 0 , u j u 0 , C t j u j h 0 and

lim sup j C t j u j , u j u 0 0, T t j u j + C t j u j , u j 0,
(3.3)

where t 0 [0,1], u 0 X, and h 0 X , we have u j u 0 , u 0 D(C) and C t 0 u 0 = h 0 .

For this, there are two cases to consider. If t 0 =0, then the second inequality in (3.3) implies

εψ ( u j ) u j =ε J ψ u j , u j T s u j , u j +ε J ψ u j , u j C ( t j Λ , u j ) , u j 0

and hence u j 0, u 0 =0D(C), and C 0 u 0 =0= h 0 . Now let t 0 >0. From the first inequality in (3.3) it follows that

C t j u j , u j u 0 C ( t j Λ , u j ) , u j u 0 +ε J ψ u 0 , u j u 0

and so

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

Combining this with

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

In view of C t j u j h 0 , we can find a subsequence of { u j }, denoted again by { u j }, such that C( t j Λ, u j ) h 1 and J ψ u j h 2 for some h 1 , h 2 X . Note that C( t 0 Λ, u j ) h 1 and h 0 = h 1 +ε h 2 . Hence (c2) implies that u j u 0 , u 0 D(C), and C( t 0 Λ, u 0 )= h 1 and so C t 0 u 0 =C( t 0 Λ, u 0 )+ε J ψ u 0 = h 0 . Thus, condition ( c 2 t ) is satisfied in both cases.

( c 3 t ) For every FF(L) and vL, the function , c ˜ (F,v)(t,u)= C t u,v, is continuous.

Actually, c ˜ (F,v) is continuous on [0,1]×F because c(tΛ,F,v) in the notation of (c3) is continuous on F and J ψ is continuous on X. Consequently, we have shown that { T t + C t }, t[0,1], is an admissible homotopy.

Following Kartsatos and Skrypnik [8, 9], we can use the homotopy invariance property of the degree d with respect to the bounded open set Ω as follows:

d ( T s + C ( λ , ) + ε J ψ , Ω , 0 ) = d ( T s + ε J ψ , Ω , 0 ) = 1

for all λ[0,Λ]. The latter follows from Theorem 3 in [3], noticing that T s +ε J ψ is demicontinuous, bounded, strictly monotone and satisfies condition ( S + ). For every λ(0,Λ], the existence property of the degree implies that

T s x+C(λ,x)+ε J ψ x=0for some xD(C)Ω,

which contradicts property (P). Therefore, the first assertion (3.1) is true.

According to assertion (3.1), let { s n } be a sequence in (0, s 0 ) with s n 0 and {( λ n , x n )} be a sequence in (0,Λ]×(D(C)Ω) such that

T s n x n +C( λ n , x n )+ε J ψ x n =0.
(3.4)

Without loss of generality, we may suppose that

λ n λ 0 , x n x 0 ,C( λ n , x n ) c and J ψ x n j ,

where λ 0 [0,Λ], x 0 X, c X and j X . We may consider the case that λ 0 (0,Λ] as the conclusion of (a) does not hold for λ 0 =0. In order to apply the condition ( S ˜ + ), we first show that

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

Assume on the contrary that 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.

Note by (3.4) and the monotonicity of J ψ that

T s n x n , x n x 0 C ( λ n , x n ) , x n x 0 ε J ψ x 0 , x n x 0

and so

lim sup n T s n x n , x n x 0 <0.

Hence, it follows from T s n x n c ε j that

lim sup n T s n x n , x n < c ε j , x 0 .
(3.6)

For every xD(T) and every u Tx, it is shown as in the proof of part (a) of Theorem 2.1 that

lim inf n T s n x n , x n c ε j , x + u , x 0 x ,

which implies in view of (3.6)

c ε j u , x 0 x >0.
(3.7)

Since T is maximal monotone, we have x 0 D(T) and c ε j T x 0 . Letting x= x 0 in (3.7), we have a contradiction. Thus, (3.5) holds.

As before, we can deduce from (3.5) that

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

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

x n x 0 , x 0 D(C),andC( λ 0 , x 0 )= c .
(3.8)

Combining (3.4) and (3.8), we obtain

T s n x n C( λ 0 , x 0 )ε J ψ x 0

and

J s n x n x 0 .

Since T is maximal monotone and T s n x n T J s n x n , Lemma 1.1 implies that x 0 D(T)D(C)Ω and

0T x 0 +C( λ 0 , x 0 )+ε J ψ x 0 .
  1. (b)

    In view of (a), for a sequence { ε n } in (0, ε 0 ] with ε n 0, we take sequences { λ n } in (0,Λ], { x n } in D(T)D(C)Ω and { u n } in X with u n T x n such that

    u n +C( λ n , x n )+ ε n J ψ x n =0.
    (3.9)

We may suppose that

λ n λ 0 , x n x 0 ,C( λ n , x n ) c and J ψ x n j ,

where λ 0 [0,Λ], x 0 X, c X and j X . Note that λ 0 (0,Λ]. As in the proof of part (b) of Theorem 2.1, a similar argument shows that

lim sup n C ( λ n , x n ) , x n x 0 0
(3.10)

and so in a usual way

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

Hence, it follows from (c2) that

x n x 0 Ω, x 0 D(C),andC( λ 0 , x 0 )= c

which implies by (3.9)

u n C( λ 0 , x 0 ).

Consequently, we obtain from the maximal monotonicity of T that x 0 D(T) and

0T x 0 +C( λ 0 , x 0 ).

This completes the proof. □

Remark 3.2 In an analogous way to Theorem 3.1, we can observe the eigenvalue problem 0Tx+λCx for quasibounded perturbations of maximal monotone operators; see [10].

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Acknowledgements

This research was supported by 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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Kim, I. Implicit eigenvalue problems for maximal monotone operators. Fixed Point Theory Appl 2012, 178 (2012). https://doi.org/10.1186/1687-1812-2012-178

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Keywords

  • Satisfy Condition
  • Compact Operator
  • Regularization Method
  • Maximal Monotone
  • Real Banach Space