Eigenvalue for Densely Defined (𝑆₊)_𝐿 Perturbations of Multivalued Maximal Monotone Operators in Reflexive Banach Spaces

Abstract

Let 𝑋 be a real reflexive Banach space and let 𝑋^∗ be its dual. Let Ω⊂𝑋 be open and bounded such that 0∈Ω. Let 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ be maximal monotone with 0∈𝐷(𝑇) and 0∈𝑇(0). Using the topological degree theory developed by Kartsatos and Quarcoo we study the eigenvalue problem 𝑇𝑥+𝜆𝑃𝑥∋0, where the operator 𝑃∶𝑋⊃𝐷(𝑃)→𝑋^∗ is a single-valued of class (𝑆₊)_𝐿. The existence of continuous branches of eigenvectors of infinite length then could be easily extended to the case where the operator 𝑃∶𝑋→2^𝑋∗ is multivalued and is investigated.

1. Preliminaries

In what follows we assume that 𝑋 is a real or complex Banach space and has been renormed such that it and its dual 𝑋^∗ are locally uniformly convex. The normalized duality mapping is defined by 𝐽(𝑥)=𝑥^∗∈𝑋∶𝑥^∗=‖𝑥‖,𝑥,𝑥^∗=‖𝑥‖². (1) The mapping 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ is said to be “monotone” if for every 𝑥,𝑦∈𝐷(𝑇), 𝑢∈𝑇𝑥, and 𝑣∈𝑇𝑦 we have 𝑢−𝑣,𝑥−𝑦≥0. (2) A monotone operator 𝑇 is “maximal monotone” if 𝐺(𝑇) is maximal in 𝑋×𝑋^∗, when 𝑋×𝑋^∗ is partially ordered by inclusion. In our setting, a monotone operator 𝑇 is maximal if and only if 𝑅(𝑇+𝜆𝐽)=𝑋^∗ for every 𝜆>0. If 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ is maximal monotone operator, the operator 𝑇_𝑡≡(𝑇⁻¹+𝑡𝐽⁻¹)⁻¹∶𝑋→𝑋^∗ is called the Yosida approximant of 𝑇 and the following, which can be found in [1, page 102] is true.

Lemma 1. Let 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ be a maximal monotone operator with 0∈𝐷(𝑇) and 0∈𝑇(0). Then(i)𝑇_𝑡 is a bounded maximal monotone mapping with 0=𝑇_𝑡(0) for each 𝑡>0;(ii)𝑇_𝑡𝑥⇀𝑇^{0}𝑥 in 𝑋 as 𝑡→0 for all 𝑥∈𝐷(𝑇), where 𝑇^{0}𝑥 denotes the element 𝑦^∗∈𝑇𝑥 of minimum norm;(iii)||𝑇_𝑡𝑥||→∞ as 𝑡→0 for all 𝑥∉𝐷(𝑇).
Also 𝑇_𝑡𝑥∈𝑇𝐽_𝑡𝑥, where 𝐽_𝑡=𝐼−𝑡𝐽⁻¹𝑇_𝑡∶𝑋→𝑋 and satisfies: if 𝑥∈conv𝐷(𝑇), then 𝐽_𝑡𝑥→𝑥 in 𝑋 as 𝑡→0, where conv𝑀 denote the convex hull of the set 𝑀.

The proof of the next lemma can be found in Kartsatos and Skrypnik [2],

Lemma 2. Let 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ be maximal monotone such that 0∈𝐷(𝑇) and 0∈𝑇(0). Then the mapping (𝑡,𝑥)→𝑇_𝑡𝑥 is continuous on the set (0,∞)×𝑋.

An operator 𝑇∶𝑋⊃𝐷(𝑇)→𝑌, with 𝑌 another Banach space, is bounded if it maps bounded subsets of 𝐷(𝑇) onto bounded sets. It is compact if it is continuous and maps bounded subsets of 𝐷(𝑇) onto relatively compact subsets of 𝑌. It is demicontinuous if for every sequence {𝑥_𝑛} such that 𝑥_𝑛→𝑥₀ we have 𝑇𝑥_𝑛⇀𝑇𝑥₀.

We say that the operator 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ satisfies condition (𝑆_𝑞) on a set 𝐴⊂𝐷(𝑇) if for every sequence {𝑥_𝑛}⊂𝐴 such that 𝑥_𝑛⇀𝑥₀∈𝑋 and any 𝑦^∗_𝑛∈𝑇𝑥_𝑛, with 𝑦^∗_𝑛→ (some)𝑦^∗∈𝑋^∗, we have 𝑥_𝑛→𝑥₀. If 𝐴=𝐷(𝑇), then we say that 𝑇 satisfies (𝑆_𝑞).

Definition 3. Let 𝑋 be a separable reflexive Banach space and let 𝐿 be a dense subspace of 𝑋. A mapping 𝑃∶𝐷(𝑃)⊂𝑋→𝑋^∗ is said to be of class (𝑆₊)_𝐿 if, for any sequence of finite dimensional subspaces 𝐹_𝑛 of 𝐿 with 𝑈^∞_𝑛=1𝐹_𝑛=𝑋, ℎ∈𝑋^∗, {𝑥_𝑛}^∞_𝑛=1⊂𝐷(𝑃) with 𝑥_𝑛⇀𝑥₀ and limsup_𝑛→∞𝑃𝑥_𝑛−ℎ,𝑥_𝑛≤0,lim_𝑛→∞𝑃𝑥_𝑛−ℎ,𝑣=0 (3) for all 𝑣∈𝑈^∞_𝑛=1𝐹_𝑛, we have 𝑥_𝑛→𝑥₀, 𝑥∈𝐷(𝑃) and 𝑃𝑥₀=ℎ.
If ℎ=0, then we call 𝑃 a mapping of class (𝑆₊)_0,𝐿.

Definition 4. Let 𝑋 be a separable reflexive Banach space and let 𝐿 be a dense subspace of 𝑋. A multivalued mapping 𝑃∶𝐷(𝑃)⊂𝑋→2^𝑋∗ is said to be of class (𝑆₊)_𝐿 if it satisfies the following conditions: (i)𝑃𝑥 is bounded closed and convex for each 𝑥∈𝐷(𝑃),(ii)𝑃 is weakly upper semicontinuous in each finite dimensional space, that is, for each finite dimensional space 𝐹 of 𝐿, 𝐹∩𝐷(𝑃)≠∅, 𝑃∶𝐹∩𝐷(𝑃)→2^𝑋∗ is upper semicontinuous in the weak topology,(iii)if for any sequence of finite dimensional subspaces 𝐹_𝑛 of 𝐿 with 𝐿⊂𝑈^∞_𝑛=1𝐹_𝑛, ℎ∈𝑋^∗, {𝑥_𝑛}^∞_𝑛=1⊂𝐷(𝑃)∩𝐿 and 𝑥_𝑛⇀𝑥₀ such that limsup_𝑛→∞𝑝_𝑛−ℎ,𝑥_𝑛≤0,lim_𝑛→∞𝑝_𝑛−ℎ,𝑣=0 (4)for all 𝑣∈𝑈^∞_𝑛=1𝐹_𝑛 and some 𝑝_𝑛∈𝑃𝑥_𝑛, then we have 𝑥_𝑛→𝑥₀, 𝑥∈𝐷(𝑃) and ℎ∈𝑃𝑥₀.
If ℎ=0, then we call 𝑃 a mapping of class (𝑆₊)_0,𝐿.

We will need the following two conditions:(P1) there exist a subspace 𝐿 of 𝑋 such that 𝐿⊂𝐷(𝑃), 𝐿=𝑋 and that the operator 𝑃 satisfies condition (𝑆₊)_𝐿;(P2) there exist a function 𝜙∶ℝ₊→ℝ₊ which is nondecreasing such that 𝑝,𝑥≥−𝜙(‖𝑥‖),𝑝∈𝑃𝑥,𝑥∈𝐷(𝑃). (5)

The following lemma can be found in Zeidler [3, page 915].

Lemma 5. Let 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ be maximal monotone. Thus the following are true:(i){𝑥_𝑛}⊂𝐷(𝑇)𝑥_𝑛→𝑥₀ and 𝑇𝑥_𝑛∋𝑦_𝑛⇀𝑦₀ imply 𝑥₀∈𝐷(𝑇), and 𝑦₀∈𝑇𝑥₀;(ii){𝑥_𝑛}⊂𝐷(𝑇)𝑥_𝑛⇀𝑥₀ and 𝑇𝑥_𝑛∋𝑦_𝑛→𝑦₀ imply 𝑥₀∈𝐷(𝑇), and 𝑦₀∈𝑇𝑥₀.

We will need the following lemma from Adhikari and Kartsatos [4].

Lemma 6. Assume that the operators 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ and 𝑇₀∶𝑋⊃𝐷(𝑇₀)→𝑋^∗ are maximal monotone, with 0∈𝐷(𝑇)∩𝐷(𝑇₀) and 0∈𝑇(0)∩𝑇₀(0). Assume further that 𝑇+𝑇₀ is maximal monotone. Assume that there is a positive sequence {𝑡_𝑛} such that 𝑡_𝑛↓0, a sequence {𝑥_𝑛}⊂𝐷(𝑇₀) and a sequence 𝑤_𝑛∈𝑇₀𝑥_𝑛 such that 𝑥_𝑛⇀𝑥₀ and 𝑇_𝑡𝑛𝑥_𝑛+𝑤_𝑛⇀𝑦^∗₀∈𝑋^∗. Then the following are true:(i)the inequality lim_𝑛→∞𝑇_𝑡𝑛𝑥_𝑛+𝑤_𝑛,𝑥_𝑛−𝑥₀<0 (6) is impossible;(ii)if lim_𝑛→∞𝑇_𝑡𝑛𝑥_𝑛+𝑤_𝑛,𝑥_𝑛−𝑥₀=0, (7) then 𝑥₀∈𝐷(𝑇+𝑇₀) and 𝑦^∗∈(𝑇+𝑇₀)𝑥₀.

The proof of the following lemma is given in [5] proof of Theorem 7 but we will repeat it here for completeness and future reference.

Lemma 7. Let 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ be a quasibounded maximal monotone operator such that 0∈𝑇(0). Let {𝑡_𝑛}⊂(0,∞) and {𝑡_𝑛}⊂𝑋 be such that 𝑢_𝑛≤𝑆,𝑇_𝑡𝑛𝑢_𝑛,𝑢_𝑛≤𝑆₁, (8) where 𝑆,𝑆₁ are positive constants. Then there exists a number 𝐾>0 such that ||𝑇_𝑡𝑛𝑢_𝑛||≤𝐾 for all 𝑛=1,2,….

Proof. Let 𝑤_𝑛=𝑇_𝑡𝑛𝑢_𝑛=𝑇⁻¹+𝑡_𝑛𝐽⁻¹⁻¹𝑢_𝑛. (9) We have that 𝑤_𝑛∈𝑇𝐽_𝑡𝑛𝑢_𝑛=𝑇𝑥_𝑛,𝑡_𝑛𝑤_𝑛=𝐽𝑢_𝑛−𝑥_𝑛, (10) where 𝑥_𝑛=𝐽_𝑡𝑛𝑢_𝑛. Thus, 𝑤_𝑛,𝑥_𝑛=𝑤_𝑛,𝑢_𝑛−𝑡_𝑛𝐽⁻¹𝑤_𝑛=𝑤_𝑛,𝑢_𝑛−𝑡_𝑛𝑤_𝑛,𝐽⁻¹𝑤_𝑛=𝑤_𝑛,𝑢_𝑛−𝑡_𝑛𝑤_𝑛²≤𝑇_𝑡𝑛𝑢_𝑛,𝑢_𝑛≤𝑆₁. (11) From (11) we obtain 𝑡_𝑛𝑤_𝑛²=𝑤_𝑛,𝑢_𝑛−𝑤_𝑛,𝑥_𝑛. (12) Now since 0∈𝑇(0) and 𝑤_𝑛∈𝑇𝑥_𝑛, we have that ⟨𝑤_𝑛,𝑥_𝑛⟩≥0, which implies 𝑡_𝑛||𝑤_𝑛||²≤𝑆₁. We claim that {𝑤_𝑛} is bounded, if not we may assume that ||𝑤_𝑛||→∞ and ||𝑤_𝑛||≤||𝑤_𝑛||² for all 𝑛. Thus 𝑡_𝑛||𝑤_𝑛||≤𝑆₁ and 𝑡_𝑛𝑤_𝑛=𝐽𝑢_𝑛−𝑥_𝑛=𝑢_𝑛−𝑥_𝑛 (13) which implies that {𝑥_𝑛} is bounded. Now since 𝑇 is strongly quasibounded, the boundedness of {𝑥_𝑛} and {⟨𝑤_𝑛,𝑥_𝑛⟩} imply the boundedness of {𝑤_𝑛}, that is, a contradiction. It follows that {𝑇_𝑡𝑛𝑢_𝑛} is bounded and the proof is complete.

We denote by 𝐽_𝜓 the duality mapping with gauge function 𝜓. The function 𝜓∶ℝ₊→ℝ₊ is continuous, strictly increasing such that 𝜓(0)=0 and 𝜓(𝑟)→∞ as 𝑟→∞. This mapping 𝐽_𝜓 is continuous, bounded, surjective, strictly and maximal monotone, and satisfies condition (𝑆₊). Also, 𝐽_𝜓𝑥,𝑥=𝜓(‖𝑥‖)‖𝑥‖,𝐽_𝜓𝑥=𝜓(‖𝑥‖),𝑥∈𝑋 (14) for these facts we refer to Petryshyn [6, pages 32-33 and 132].

2. The Eigenvalue Result

In this section we are using the topological degree developed by Kartsatos and Quarcoo in [7]. The following result will also improve Theorem 4 of Kartsatos and Skrypnik in [8] since we are no longer assuming that the perturbation is quasibounded.

Theorem 8. Let Ω be open and bounded with 0∈Ω. Assume that the operator 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ is maximal monotone with 0∈𝐷(𝑇) and 0∈𝑇(0). Assume that the operator 𝑃∶𝑋⊃𝐷(𝑃)→𝑋^∗, with 𝐿⊂𝐷(𝑃), satisfies condition (𝑆₊)_𝐿 and (P2). Let 𝜖, 𝜖₀, and Λ be positive numbers. Assume that (𝒫) there exists 𝜆∈(0,Λ] such that 𝑇𝑥+𝜆𝑃𝑥+𝜖𝐽_𝜓𝑥∋0 (15) has no solution in 𝐷(𝑇+𝑃)∩Ω.
Then(i)there exists (𝜆₀,𝑥₀)∈(0,Λ]×(𝐷(𝑇+𝑃)∩𝜕Ω) such that 𝑇𝑥₀+𝜆₀𝑃𝑥₀+𝜖𝐽_𝜓𝑥₀∋0; (16)(ii)if 0∉𝑇(𝐷(𝑇)∩𝜕Ω), 𝑇 satisfies (𝑆_𝑞) on 𝜕Ω, and property (𝒫) is satisfied for every 𝜖∈(0,𝜖₀], then there exists (𝜆₀,𝑥₀)∈(0,Λ]×(𝐷(𝑇+𝑃)∩𝜕Ω) such that 𝑇𝑥₀+𝜆₀𝑃𝑥₀∋0.

Proof. (i) Assume that (𝒫) is true and that (16) is false. We consider the homotopy inclusion 𝐻(𝑡,𝑥)≡𝑇𝑥+𝑡Λ𝑃𝑥+𝜖𝐽_𝜓𝑥∋0,𝑡∈[0,1]. (17) It is clear that this inclusion has no solution 𝑥∈𝐷(𝐻(𝑡,⋅))∩𝜕Ω for 𝑡∈(0,1], because 𝑡Λ∈(0,Λ] and our assumption about (16). It is also true for 𝑡=0 because 0∈(𝑇+𝜖𝐽_𝜓)(0) and the operator 𝑇+𝜖𝐽_𝜓 is strictly monotone, hence one-to-one. We are going to show that 𝐻(𝑡,𝑥) is an admissible homotopy for the degree in [4]. To this end, we set 𝑇_𝑡=𝑇, 𝑃^𝑡≡𝑡Λ𝑃+𝐽_𝜓 and we recall the operators 𝑇_𝑡,𝑠=𝑇_𝑠≡(𝑇⁻¹+𝑠𝐽⁻¹)⁻¹∶𝑋→𝑋^∗, 𝑠>0, and we set 𝐽_𝑡,𝑠=𝐽_𝑠=𝐼−𝑠𝐽⁻¹𝑇_𝑡,𝑠=𝐼−𝑠𝐽⁻¹𝑇_𝑠∶𝑋→𝑋. We have 𝐷(𝐻(0,⋅))=𝐷(𝑇) and 𝐷(𝐻(𝑡,⋅))=𝐷(𝑇+𝑃), 𝑡∈(0,1]. We also set 𝐷^𝑡=𝐷(𝑡Λ𝑃)=𝐷(𝑡𝑃). We have 𝐷⁰=𝑋 and 𝐷^𝑡=𝐷(𝑃) for 𝑡∈(0,1]. Let Ω be an open and bounded subset of 𝑋. We know that the equation 𝑇^𝑡𝑥+𝑃^𝑡𝑥∋0 (18) has no solution 𝑥∈𝐷(𝐻(𝑡,⋅))∩𝜕Ω for any 𝑡∈[0,1]. Now we consider the equation 𝑇_𝑠𝑥+𝑡Λ𝑃𝑥+𝜖𝐽_𝜓=0, (19) and we show that there exists 𝑠₁>0 such that 0∉𝑇_𝑠+𝑡Λ𝑃+𝜖𝐽_𝜓𝐷^𝑡∩𝜕Ω,(𝑠,𝑡)∈0,𝑠₁×[0,1]. (20) Assume that this is not true, then there exist {𝑠_𝑛}⊂(0,∞) with 𝑠_𝑛↓0, {𝑡_𝑛}⊂[0,1], with 𝑡_𝑛→𝑡₀, {𝑥_𝑛}⊂𝜕Ω with 𝑥_𝑛⇀𝑥₀, and 𝑇_𝑠𝑛𝑥_𝑛+𝑡_𝑛Λ𝑃𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛=0. (21) Clearly we cannot have 𝑡_𝑛=0 for any 𝑛, since (𝑇_𝑠𝑛+𝐽_𝜓)(0)=0 and the operator 𝑇_𝑠𝑛+𝐽_𝜓 is strictly monotone, hence one-to-one. Thus 𝑡_𝑛>0, for all 𝑛. From (21) we have that 𝑇_𝑠𝑛𝑥_𝑛,𝑥_𝑛=−𝑡_𝑛Λ𝑃𝑥_𝑛,𝑥_𝑛−𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛≤𝑡_𝑛Λ𝜙𝑥_𝑛−𝜖𝜓𝑥_𝑛≤Λ𝜙𝑆₁+𝜖𝜓𝑆₁, (22) where 𝑆₁ is the bound of {𝑥_𝑛}. Thus we have the boundedness of ⟨𝑇_𝑠𝑛𝑥_𝑛,𝑥_𝑛⟩. Using Lemma 7, we have the boundedness of 𝑇_𝑠𝑛𝑥_𝑛. We may thus assume that 𝑇_𝑠𝑛𝑥_𝑛⇀ℎ^∗₁. From (21) we also have that the sequence {𝑃𝑥_𝑛} is bounded and we may assume that 𝑃𝑥_𝑛⇀ℎ^∗₀.
If 𝑡₀=0, then from 𝜖𝜓𝑥_𝑛𝑥_𝑛=𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛≤−𝑡_𝑛Λ𝑃𝑥_𝑛,𝑥_𝑛→0 (23) we obtain 𝑥_𝑛→0∈𝜕Ω, which is a contradiction to 0∈Ω. Hence it follows that 𝑡₀>0. Since {𝐽_𝜓𝑥_𝑛} is bounded, we may assume that 𝐽_𝜓𝑥_𝑛⇀ℎ^∗₂ and we have 𝑡₀Λℎ^∗₀=−ℎ^∗₁−𝜖ℎ^∗₂.
From (21) it follows that 𝑇_𝑠𝑛𝑥_𝑛+𝑡_𝑛Λ𝑃𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥=0,∀𝑥∈𝑈^∞_𝑛=1𝐹_𝑛. (24) Thus 𝑇_𝑠𝑛𝑥_𝑛+𝑡_𝑛Λ𝑃𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛=0,lim_𝑛→∞𝑇_𝑠𝑛𝑥_𝑛+𝑡_𝑛Λ𝑃𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑣=0,∀𝑣∈𝑈^∞_𝑛=1𝐹_𝑛. (25) Now 𝑡_𝑛Λ𝑃𝑥_𝑛+ℎ^∗₁+𝜖ℎ^∗₂,𝑥_𝑛=𝑇_𝑠𝑛𝑥_𝑛+𝑡_𝑛Λ𝑃𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛−𝑇_𝑠𝑛𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛−𝑥₀−𝑇_𝑠𝑛𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥₀+ℎ^∗₁+𝜖ℎ^∗₂,𝑥_𝑛, (26) hence limsup_𝑛→∞𝑡_𝑛Λ𝑃𝑥_𝑛+ℎ^∗₁+𝜖ℎ^∗₂,𝑥_𝑛≤limsup_𝑛→∞𝑇_𝑠𝑛𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛+𝑡_𝑛Λ𝑃𝑥_𝑛,𝑥_𝑛−liminf_𝑛→∞𝑇_𝑠𝑛𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛−𝑥₀−⟨ℎ^∗₁+𝜖ℎ^∗₂,𝑥₀⟩+ℎ^∗₁+𝜖ℎ^∗₂,𝑥₀≤−liminf_𝑛→∞𝑇_𝑠𝑛𝑥_𝑛+𝜖𝐽_𝜓𝑥_𝑛,𝑥_𝑛−𝑥₀≤0, (27) where the last inequality follows from Lemma 6. Now since 𝑃 is of class (𝑆₊)_𝐿, it follows that 𝑥_𝑛→𝑥₀, 𝑥₀∈𝐷(𝑃) and 𝑡₀Λ𝑃𝑥₀=−ℎ^∗₁−𝜖ℎ^∗₂. Also we have that 𝑇_𝑠𝑛𝑥_𝑛,𝑥_𝑛−𝑥₀=0 (28) and using Lemma 6 we get that 𝑥₀∈𝐷(𝑇) and 𝑇𝑥₀∋ℎ^∗₁=−𝑡₀Λ𝑃𝑥₀−𝜖𝐽_𝜓𝑥₀. This is a contradiction since 𝑥_𝑛→𝑥₀, we have 𝑥₀∈𝜕Ω. We have shown that 𝐻(𝑡,𝑥) is an admissible homotopy for our degree. We can now work as in Theorem 3 in [8] in order to show that 𝑑(𝐻(𝑡,⋅),Ω,0)=const.
Thus 𝑑(𝐻(𝑡,⋅),Ω,0)=𝑑𝑇_𝑠+𝜖𝐽_𝜓,Ω,0=1, (29) where the last equality follows from Theorem 3, (i) of [9].
Consequently, the inclusion 𝐻(𝑡,𝑥)∋0 has a solution in Ω for each 𝑡∈[0,1]. In particular, this says that 𝑇𝑥+𝜆𝑃𝑥+𝜖𝐽_𝜓𝑥∋0 has a solution in Ω for every 𝜆>0. This of course contradicts condition (𝒫) and finishes the proof of (i).
(ii) Let 𝜆_𝑛∈(0,Λ], 𝑥_𝑛∈𝐷(𝑃)∩𝜕Ω be such that, for some 𝑢^∗_𝑛∈𝑇𝑥_𝑛, 𝑢^∗_𝑛+𝜆_𝑛𝑃𝑥_𝑛+1𝑛𝐽_𝜓𝑥_𝑛=0 (30) again, 𝜆_𝑛=0 for any 𝑛 is not possible. Since 𝜆_𝑛>0, we have by property (𝑃2) that 𝑢^∗_𝑛,𝑥_𝑛=−𝜆_𝑛𝑃𝑥_𝑛,𝑥_𝑛−1𝑛𝐽_𝜓𝑥_𝑛,𝑥_𝑛≤𝜆_𝑛𝜙𝑥_𝑛−1𝑛𝜓𝑥_𝑛𝑥_𝑛≤Λ𝜙𝑆₁+𝜓𝑆₁𝑆₁, (31) where 𝑆₁ is the bound of {𝑥_𝑛}. This show that ⟨𝑢^∗_𝑛,𝑥_𝑛⟩ is bounded and further we obtain the boundedness of {𝑢^∗_𝑛} by Lemma 2. We may thus assume that 𝜆_𝑛→𝜆₀, 𝑥_𝑛⇀𝑥₀, 𝑢^∗_𝑛⇀ℎ^∗₀, then 𝜆_𝑛𝑃𝑥_𝑛⇀−ℎ^∗₀.
If 𝜆₀=0, then (30) implies that lim_𝑛→∞𝑢^∗_𝑛=lim_𝑛→∞−𝜆_𝑛𝑃𝑥_𝑛−1𝑛𝐽_𝜓𝑥_𝑛=0. (32) Since 𝑇 satisfies (𝑆_𝑞), it follows that 𝑥_𝑛→𝑥₀∈𝜕Ω. Now, by the demiclosedness of 𝑇 (see Lemma 6) we obtain that 𝑥₀∈𝐷(𝑇), 0∈𝑇𝑥₀ and this contradicts 0∉𝑇(𝐷(𝑇)∩𝜕Ω). Hence, 𝜆₀>0. Repeating the proof of (i) starting from (21), we get again that limsup_𝑛→∞𝜆_𝑛𝑃𝑥_𝑛+ℎ^∗₀,𝑥_𝑛≤0 (33) and since 𝑃 is of class (𝑆₊)_𝐿, we have 𝑥_𝑛→𝑥₀, −ℎ^∗₀=𝜆₀𝑃𝑥₀. Using the demiclosedness of 𝑇, we obtain 𝑥₀∈𝐷(𝑇)∩𝜕Ω, ℎ₀∈𝑇𝑥₀ and 𝑇𝑥₀+𝜆₀𝑃𝑥₀∋0. The proof is now complete.

3. Continuous Branches of Eigenvectors

In this section we are interested in showing that the results obtained in previous sections could be used in order to obtain the existence of continuous branches of eigenvectors. We need the following definition.

Definition 9. Let 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗, 𝑃∶𝑋⊃𝐷(𝑃)→𝑋^∗ be given and consider the problem 𝑇𝑥+𝜆𝑃𝑥∋0. (34) An eigenvector 𝑥 is solution of (34) for some eigenvalue 𝜆 with 𝑥∈(𝐷(𝑇)∩𝐷(𝑃)). We say that the nonzero eigenvectors of the problem (34) form a continuous branch of infinite length if there exists 𝑟₀>0 such that, for every 𝑟≥𝑟₀, the sphere 𝜕𝐵_𝑟(0) contains at least one nonzero eigenvector of (34).

Theorem 10. Assume that the operator 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ is maximal monotone with 0∈𝐷(𝑇) and 0∈𝑇(0). Assume that the operator 𝑃∶𝑋⊃𝐷(𝑃)→𝑋^∗ with 𝐿⊂𝐷(𝑃) is of class (𝑆₊)_𝐿 and satisfies (𝑃2). Let Λ be a positive number. Assume that 𝑇𝑥∋0 implies 𝑥=0, 𝑇 satisfies (𝑆_𝑞) and (𝒫1) there exists 𝛼>0 and 𝜆∈(0,Λ] such that |𝑇𝑥+𝜆𝑃𝑥|≥𝛼,𝑥∈𝐷(𝑇+𝑃). (35)Then the nonzero eigenvectors of problem (34) form a continuous branch of infinite length with corresponding eigenvalues 𝜆∈(0,Λ].

Proof. Let 𝑟₀>0 be given. Let 𝜖₀>0 be so small that 𝜖₀𝑟₀<𝛼. Then ||≥||−||‖𝑥‖𝑇𝑥+𝜆𝑃𝑥+𝜖𝐽𝑥𝑇𝑥+𝜆𝑃𝑥𝜖𝐽𝑥≥𝛼−𝜖≥𝛼−𝜖₀𝑟₀, (36)𝑥∈𝐷(𝑇+𝑃), which implies that the inclusion 𝑇𝑥+𝜆𝑃𝑥+𝜖𝐽𝑥∋0 (37) has no solution 𝑥∈𝐷(𝑇+𝑃)∩𝐵_𝑟(0) for any 𝜖∈(0,𝜖₀]. Since 0∉𝑇(𝜕𝐵_𝑟0(0)) and 𝑇 is of type (𝑆_𝑞), Theorem 8 implies the existence of a solution 𝑥_𝜆0∈𝐷(𝑇+𝑃)∩𝜕𝐵_𝑟0(0), for some 𝜆₀∈(0,Λ]. The same argument can be repeated for any 𝑟>𝑟₀. This complete the proof.

Remark 11. This result is also true when 𝑃∶𝑋⊃𝐷(𝑃)→2^𝑋∗ is assumed to be multivalued. We have the following theorem without proof.

Theorem 12. Assume that the operator 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ is maximal monotone with 0∈𝐷(𝑇) and 0∈𝑇(0). Assume that the operator 𝑃∶𝑋⊃𝐷(𝑃)→2^𝑋∗ with 𝐿⊂𝐷(𝑃) is of class (𝑆₊)_𝐿 and satisfies (𝑃2). Let Λ be a positive number. Assume that 𝑇𝑥∋0 implies 𝑥=0, 𝑇 satisfies (𝑆_𝑞), and
(𝒫1) there exists 𝛼>0 and 𝜆∈(0,Λ] such that |𝑇𝑥+𝜆𝑃𝑥|≥𝛼,𝑥∈𝐷(𝑇+𝑃). (38) Then the nonzero eigenvectors of problem (34) form a continuous branch of infinite length with corresponding eigenvalues 𝜆∈(0,Λ].

In the following result, we assume that the operator 𝑇 is defined and bounded on all of 𝑋. In this case we demonstrate the fact that the assumption |𝑇𝑥+𝑃𝑥|≥0 may be replaced by the assumption ||𝑃𝑥||≥0 on 𝐷(𝑃).

Theorem 13. Assume that the operator 𝑇∶𝑋⊃𝐷(𝑇)→2^𝑋∗ is maximal monotone and bounded with 0∈𝐷(𝑇) and 0∈𝑇(0). Assume that the operator 𝑃∶𝑋⊃𝐷(𝑃)→𝑋^∗, with 𝐿⊂𝐷(𝑃) satisfies condition (𝑆₊)_𝐿 and (𝑃2). Assume that 𝑇𝑥∋0 implies 𝑥=0, 𝑇 satisfies (𝑆_𝑞) and there exist 𝛼>0 such that ‖𝑃𝑥‖≥𝛼,𝑥∈𝐷(𝑃). (39) Then the nonzero eigenvectors of the problem (34) form a continuous branch of infinite length.

Proof. We show that the problem (34) has eigenvectors on the set 𝜕𝐵_𝑟(0) for every 𝑟>0. To this end, let fix 𝑟>0, 𝜖>0 and show the existence of 𝜆>0 such that 𝑑𝑇+𝜆𝑃+𝜖𝐽,𝐵_𝑟(0),0=0. (40) If this is not true, then there exist a sequence of {𝜆_𝑛}⊂(0,∞) such that 𝜆_𝑛→∞ and one of the following is true:(i)the degree 𝑑(𝑇+𝜆_𝑛𝑃+𝜖𝐽,𝐵_𝑟(0),0) is not well-defined; (ii)𝑑(𝑇+𝜆_𝑛𝑃+𝜖𝐽,𝐵_𝑟(0),0)≠0.
In the case (i) there exist eigenvectors 𝑥_𝑛∈𝜕𝐵_𝑟(0) such that 𝑇𝑥_𝑛+𝜆_𝑛𝑃𝑥_𝑛+𝜖𝐽𝑥_𝑛∋0. (41) In the case (ii) there exist eigenvectors 𝑥_𝑛∈𝐵_𝑟(0) such that (41) holds. Thus in either case, there exist a sequence {𝑥_𝑛}⊂𝐵_𝑟(0) such that (41) is true. But this leads to a contradiction because ||𝜆_𝑛𝑃𝑥_𝑛+𝜖𝐽𝑥_𝑛||≥𝛼𝜆_𝑛−𝜖𝑟→∞, while the set 𝑇𝑥_𝑛 lie in a bounded set. Thus, (40) is true for some 𝜆>0. We consider the homotopy 𝐻(𝑡,𝑥)≡𝑇𝑥+𝑡𝜆𝑃𝑥+𝜖𝐽𝑥,𝑡∈[0,1],𝑥∈𝐷(𝐻(𝑡,⋅)), (42) either there exist 𝑡₀∈[0,1] and 𝑥_𝑡0∈𝜕𝐵_𝑟(0) such that 𝑇𝑥_𝑡0+𝑡₀𝜆𝑃𝑥_𝑡0+𝜖𝐽𝑥_𝑡0∋0, (43) or 𝑑(𝐻(𝑡,⋅))=𝑑(𝐻(1,⋅))=0=𝑑(𝐻(0,⋅))=1,𝑡∈[0,1] (44) that is a contradiction. The last equality in (44) follows from Theorem 3, (i) in [2]. It follows that (43) is true. Naturally, we must have 𝑡₀≠0 in (43) because otherwise 0∈(𝑇+𝜖𝐽)(𝜕𝐵_𝑟(0)). This cannot happen because we already have 0∈(𝑇+𝜖𝐽)(0) and 𝑇+𝜖𝐽 is one-to-one. From (43) we obtain sequences 𝜆_𝑛∈(0,∞), {𝑥_𝑛}⊂𝜕𝐵_𝑟(0) such that 𝑇𝑥_𝑛+𝜆_𝑛𝑃𝑥_𝑛+1𝑛𝐽𝑥_𝑛∋0. (45) We may assume that 𝑥_𝑛⇀𝑥₀. Again, {𝜆_𝑛} cannot contain a subsequence {𝜆_𝑛𝑘} such that 𝜆_𝑛𝑘→∞ as 𝑘→∞ because the sequence 𝑇𝑥_𝑛𝑘+(1/𝑛_𝑘)𝐽𝑥_𝑛𝑘 lies in a bounded set and 𝜆_𝑛𝑘||𝑃𝑥_𝑛𝑘||→∞ as 𝑘→∞. Also we have that the sequence {𝑃𝑥_𝑛} is bounded. Thus we may assume that 𝑃𝑥_𝑛⇀𝑝^∗∈𝑋^∗. Since the sequence {𝜆_𝑛} is bounded, we may assume that 𝜆_𝑛→𝜆₀. Again, 𝜆₀≠0 otherwise 𝜆_𝑛𝑃𝑥_𝑛+(1/𝑛)𝐽𝑥_𝑛→0 and the (𝑆_𝑞) property of 𝑇 would imply that 𝑥_𝑛→𝑥₀∈𝜕Ω. Since 𝑇 is demiclosed, Lemma 6 would imply 𝑇𝑥₀∋0. This is a contradiction to our assumption that 𝑇𝑥∋0 implies 𝑥=0. It follows that 𝜆₀>0 and we may also assume that 𝜆_𝑛>0 for all 𝑛.
Now proceeding as in the proof of Theorem 8, it is easy to see that limsup_𝑛→∞𝜆_𝑛𝑃𝑥_𝑛+𝜆₀𝑝^∗,𝑥_𝑛≤0,lim_𝑛→∞𝜆_𝑛𝑃𝑥_𝑛+𝜆₀𝑝^∗,𝑣=0,∀𝑣∈𝑈^∞_𝑛=1𝐹_𝑛. (46) Now since 𝑃 is of class (𝑆₊)_𝐿, we conclude that 𝑥_𝑛→𝑥₀, 𝑥₀∈𝐷(𝑃) and 𝑃𝑥₀=𝑝^∗. Thus since also 𝑦^∗_𝑛⇀−𝜆₀𝑃𝑥₀, 𝑥_𝑛→𝑥₀ and the demiclosedness of 𝑇, we obtain 𝑥₀∈𝐷(𝑃)∩𝜕𝐵_𝑟(0) and 𝑇𝑥₀+𝜆₀𝑃𝑥₀∋0. Since 𝑟>0 is arbitrary, the nonzero eigenvectors of problem (34) form a continuous branch of infinite length.