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On the Krasnoselskiitype fixed point theorems for the sum of continuous and asymptotically nonexpansive mappings in Banach spaces
Journal of Inequalities and Applications volume 2011, Article number: 28 (2011)
Abstract
In this article, we prove some results concerning the Krasnoselskii theorem on fixed points for the sum A + B of a weaklystrongly continuous mapping and an asymptotically nonexpansive mapping in Banach spaces. Our results encompass a number of previously known generalizations of the theorem.
1 Introduction
As is well known, Krasnoselskii's fixed point theorem has a wide range of applications to nonlinear integral equations of mixed type (see [1]). It has also been extensively employed to address differential and functional differential equations. His theorem actually combines both the Banach contraction principle and the Schauder fixed point theorem, and is useful in establishing existence theorems for perturbed operator equations. Since then, there have appeared a large number of articles contributing generalizations or modifications of the Krasnoselskii fixed point theorem and their applications (see [2][21]).
The study of asymptotically nonexpansive mappings concerning the existence of fixed points have become attractive to the authors working in nonlinear analysis. Goebel and Kirk [22] introduced the concept of asymptotically nonexpansive mappings in Banach spaces and proved a theorem on the existence of fixed points for such mappings in uniformly convex Banach spaces. In 1971, Cain and Nashed [23] generalized to locally convex spaces a well known fixed point theorem of Krasnoselskii for a sum of contraction and compact mappings in Banach spaces. The class of asymptotically nonexpansive mappings includes properly the class of nonexpansive mappings as well as the class of contraction mappings. Recently, Vijayaraju [21] proved by using the same method some results concerning the existence of fixed points for a sum of nonexpansive and continuous mappings and also a sum of asymptotically nonexpansive and continuous mappings in locally convex spaces. Very recently, Agarwal et al. [1] proved some existence theorems of a fixed point for the sum of a weaklystrongly continuous mapping and a nonexpansive mapping on a Banach space and under Krasnoselskii, Leray Schauder, and FuriPeratype conditions.
Motivated and inspired by Agarwal et al. [1] and Vijayaraju [21], in this article we will prove some new generalized forms of the Krasnoselskii theorem on fixed points for the sum A + B of a weaklystrongly continuous mapping and an asymptotically nonexpansive mapping in Banach spaces. These results encompass a number of previously known generalizations of the theorem.
2 Preliminaries
Let M be a nonempty subset of a Banach space X and T : M → X be a mapping. We say that T is weaklystrongly continuous if for each sequence {x_{ n }} in M which converges weakly to x in M, the sequence {Tx_{ n }} converges strongly to Tx. The mapping T is nonexpansive if Tx Ty ≤ x  y for all x, y ∈ M, and T is asymptotically nonexpansive (see [22]) if there exists a sequence {k_{ n }} with k_{ n } ≥ 1 for all n and lim_{n→∞}k_{ n }= 1 such that T^{n}x  T^{n}y ≤ k_{ n }x  y for all n ≥ 1 and x, y ∈ M.
Definition 2.1. [21] If B and A map M into X, then B is called a uniformly asymptotically regular with respect to A if, for each ε > 0 there exists n_{0} ∈ ℕ, such that
for all n ≥ n_{0} and all x ∈ M.
Now, let us recall some definitions and results which will be needed in our further considerations. Let X be a Banach space, Ω(X) is the collection of all nonempty bounded subsets of X, and is the subset of Ω(X) consisting of all weak compact subsets of X. Let B_{ r } denote the closed ball in X countered at 0 with radius r > 0. In [24], De Blasi introduced the following mapping ω : Ω(X) → [0, ∞) defined by
for all M ∈ Ω(X). For completeness, we recall some properties of ω(·) needed below (for the proofs we refer the reader to [24]).
Lemma 2.2. [24] Let M_{1} and M_{2} ∈ Ω(X), then we have
(i) If M_{1} ⊆ M_{2}, then ω(M_{1}) ≤ ω(M_{2}).
(ii) ω(M_{1}) = 0 if and only if M_{1} is relatively weakly compact.
(iii) , where is the weak closure of M_{1}.
(iv) ω(λM_{1}) = λω(M_{1}) for all λ ∈ ℝ.
(v) ω(co(M_{1})) = w(M_{1}).
(vi) ω(M_{1} + M_{2}) ≤ ω(M_{1}) + ω(M_{2}).
(vii) If (M_{ n })_{n≥1}is a decreasing sequence of nonempty, bounded and weakly closed subsets of X with lim_{n→∞}ω(M_{ n }) = 0, then and , i.e., is relatively weakly compact.
Throughout this article, a measure of weak noncompactness will be a mapping ψ : Ω(X) → [0, ∞) which satisfies the assumptions (i)(vii) cited in Lemma 2.2.
Definition 2.3. [25] Let M be a closed subset of X and I, T : M → M be two mappings. A mapping T is said to be demiclosed at the zero, if for each sequence {x_{ n }} in M, the conditions x_{ n } → x_{0} weakly and Tx_{ n } → 0 strongly imply Tx_{0} = 0.
Lemma 2.4. [26][29] Let X be a uniformly convex Banach space, M be a nonempty closed convex subset of X, and let T : M → M be an asymptotically nonexpansive mapping with F(T) ≠ ∅. Then I  T is demiclosed at zero, i.e., for each sequence {x_{ n }} in M, if {x_{ n }} converges weakly to q ∈ M and {(I  T)x_{ n }} converges strongly to 0, then (I  T)q = 0.
Definition 2.5. [1, 13] Let X be a Banach space and let ψ be a measure of weak noncompactness on X. A mapping B : D(B) ⊆ X → X is said to be ψcontractive if it maps bounded sets into bounded sets and there is a β ∈ [0, 1) such that ψ(B(S)) ≤ βψ(S) for all bounded sets S ⊆ D(B). The mapping B : D(B) ⊆ X → X is said to be ψcondensing if it maps bounded sets into bounded sets and ψ(B(S)) < ψ(S) whenever S is a bounded subset of D(B) such that ψ(S) > 0.
Let be a nonlinear operator from into X. In the next section, we will use the following two conditions:
If (x_{ n })_{n∈ℕ}is a weakly convergent sequence in , then has a strongly convergent subsequence in X.
If (x_{ n })_{n∈ℕ}is a weakly convergent sequence in , then has a weakly convergent subsequence in X.
Remark 2.6. 1. Operators satisfying or are not necessarily weakly continuous (see [12, 19, 30]).

2.
Every wcontractive mapping satisfies .

3.
A mapping satisfies if and only if it maps relatively weakly compact sets into relatively weakly compact ones (use the EberleinŠ mulian theorem [31]).

4.
A mapping satisfies if and only if it maps relatively weakly compact sets into relatively compact ones.

5.
The condition holds true for every bounded linear operator.
The following fixed point theorems are crucial for our purposes.
Lemma 2.7. [12] Let M be a nonempty closed bounded convex subset of a Banach space X. Suppose that A : M → X and B : X → X satisfying:
(i) A is continuous, AM is relatively weakly compact and A satisfies ,
(ii) B is a strict contraction satisfying ,
(iii) Ax + By ∈ M for all x, y ∈ M.
Then, there is an x ∈ M such that Ax + Bx = x.
Lemma 2.8. [20] Let M be a nonempty closed bounded convex subset of a Banach space X. Suppose that A : M → X and B : M → X are sequentially weakly continuous such that:
(i) AM is relatively weakly compact,
(ii) B is a strict contraction,
(iii) Ax + By ∈ M for all x, y ∈ M.
Then, there is an x ∈ M such that Ax + Bx = x.
Lemma 2.9. [1] Let X be a Banach space and let ψ be measure of weak noncompactness on X. Let Q and C be closed, bounded, convex subset of X with Q ⊆ C. In addition, let U be a weakly open subset of Q with 0 ∈ U, and a weakly sequentially continuous and ψcondensing mapping. Then either
or
here ∂_{ Q }U is the weak boundary of U in Q.
Lemma 2.10. [1] Let X be a Banach space and B : X → X a kLipschitzian mapping, that is
In addition, suppose that B verifies . Then for each bounded subset S of X, we have ψ(BS) ≤ kψ(S);
here,ψ is the De Blasi measure of weak noncompactness.
Lemma 2.11. [15, 32] Let X be a Banach space with C ⊆ X closed and convex. Assume U is a relatively open subset of C with 0 ∈ U, bounded and a condensing mapping. Then, either F has a fixed point in or there is a point u ∈ ∂U and λ ∈ (0,1) with u = λF(u); here and ∂U denote the closure of U in C and the boundary of U in C, respectively.
Lemma 2.12. [15, 32] Let X be a Banach space and Q a closed convex bounded subset of X with 0 ∈ Q. In addition, assume F : Q → X a condensing mapping with if is a sequence in ∂Q× [0, 1] converging to (x, λ) with X = λF(x) and 0 < λ < 1, then λ_{ j }F (x_{ j }) ∈ Q for j sufficiently large, holding. Then F has a fixed point.
3 Main results
Now, we are ready to state and prove the main result of this section.
Theorem 3.1. Let M be a nonempty bounded closed convex subset of a Banach space X. Let A : M → X and B : M → M satisfy the following:
(i) A is weaklystrongly continuous, and AM is relatively weakly compact,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞) satisfying ,
(iii) if (x_{ n }) is a sequence of M such that ((I  B)x_{ n }) is weakly convergent, then the sequence (x_{ n }) has a weakly convergent subsequence,
(iv) I  B is demiclosed,
(v) B^{n}x + Ay ∈ M for all x, y ∈ M and n = 1, 2,...,
(vi) B is uniformly asymptotically regular with respect to A.
Then, there is an x ∈ M such that Ax + Bx = x.
Proof. Suppose first that 0 ∈ M and let for all n ∈ ℕ. By hypothesis (v), we have
Since B is asymptotically nonexpansive, it follows that
Hence, a_{ n }B^{n} is contraction on M. Therefore, by Lemma 2.7, there is an x_{ n } ∈ M such that
for all n ∈ ℕ. This implies that
since a_{ n } → 1 as n → ∞ and M is bounded and B^{n}x + Ay ∈ M for all x, y ∈ M. Since B is uniformly asymptotically regular with respect to A, it follows that
From (3.3) and (3.4), we obtain
Now, it is noted that
Using (3.3) and (3.5) in (3.6), we get
Using the fact that AM is weakly compact and passing eventually to a subsequence, we may assume that {Ax_{ n }} converges weakly to some y ∈ M. By (3.7), we have
By hypothesis (iii), the sequence {x_{ n }} has a subsequence which converges weakly to some x ∈ M. Since A is weaklystrongly continuous, converges strongly to Ax.
Hence, we observe that
Hence, by the demiclosedness of I  B, we have Ax + Bx = x.
To complete the proof, it remains to consider the case 0 ∉ M. In such a case, let us fix any element x_{0} ∈ M and let M_{0} = {x  x_{0}, x ∈ M }. Define two mappings A_{0} : M_{0} → X and B_{0} : M_{0} → M by and , for x ∈ M. By the result of the first case for A_{0} and B_{0}, we have an x ∈ M such that A_{0}(x  x_{0}) + B_{0}(x  x_{0}) = x  x_{0}. Hence Ax + Bx = x. □
Corollary 3.2. Let M be a nonempty bounded closed convex subset of a uniformly convex Banach space X. Let A : M → X and B : M → M satisfy the following:
(i) A is weaklystrongly continuous,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞),
(iii) B^{n}x + Ay ∈ M for all x, y ∈ M, and n = 1, 2,...,
(iv) B is uniformly asymptotically regular with respect to A.
Then, there is an x ∈ M such that Ax + Bx = x.
Our next result is the following:
Theorem 3.3. Let M be a nonempty bounded closed convex subset of a Banach space X. Suppose that A : M → X and B : M → M are two weakly sequentially continuous mappings that satisfy the following:
(i) AM is relatively weakly compact,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞),
(iii) if (x_{ n }) is a sequence of M such that ((I  B)x_{ n }) is weakly convergent, then the sequence (x_{ n }) has a weakly convergent subsequence,
(iv) B^{n}x + Ay ∈ M for all x, y ∈ M, and n = 1, 2,...,
(v) B is uniformly asymptotically regular with respect to A.
Then, there is an x ∈ M such that Ax + Bx = x.
Proof. Without loss of generality, we may assume that 0 ∈ M. Let for all n ∈ ℕ. By hypothesis (iv), we have
Since B is asymptotically nonexpansive, it follows that
Hence, a_{ n }B^{n} is a contraction on M. By Lemma 2.8, there is a x_{ n } ∈ M such that
for all n ∈ ℕ. This implies that
Since B is uniformly asymptotically regular with respect to A, it follows that
From (3.12) and (3.13), we obtain
Now, it is noted that
Using (3.12) and (3.14) in (3.15), we get
Using the fact that AM is weakly compact and passing eventually to a subsequence, we may assume that {Ax_{ n }} converges weakly to some y ∈ M. Hence, by (3.16)
By hypothesis (iii), the sequence {x_{ n }} has a subsequence which converges weakly to some x ∈ M. Since A and B are weakly sequentially continuous, converges weakly to Ax, and converges weakly to Bx. Hence, Ax + Bx = x. □
Theorem 3.4. Let Q and C be closed bounded convex subset of a Banach space X with Q ⊆ C. In addition, let U be a weakly open subset of Q with 0 ∈ U, and B : X → X are two weakly sequentially continuous mappings satisfying the following:
(i) is a relatively weakly compact,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞),
(iii) if (x_{ n }) is a sequence of M such that ((I  B)x_{ n }) is weakly convergent, then the sequence (x_{ n }) has a weakly convergent subsequence,
(iv) B^{n}x + Ay ∈ C for all , and n = 1, 2,...,
(v) B is uniformly asymptotically regular with respect to A.
Then, either
or
here, ∂_{ Q }U is the weak boundary of U in Q.
Proof. Let for all n ∈ ℕ. We first show that the mapping F_{ n } = a_{ n }A+a_{ n }B^{n} is ψcontractive with constant a_{ n }. To see that, let S be a bounded subset of . Using the homogeneity and the subadditivity of the De Blasi measure of weak noncompactness, we obtain
Keeping in mind that A is weakly compact and using Lemma 2.10, we deduce that
This proves that F_{ n } is ψcontractive with constant a_{ n }. Moreover, taking into account that 0 ∈ U and using assumption (iv), we infer that F_{ n } map into C. Next, we suppose that (3.19) does not occur, and F_{ n } does not have a fixed point on ∂_{ Q }U (otherwise we are finished since (3.18) occurs). If there exists a u ∈ ∂_{ Q }U, and λ ∈ (0, 1) with u = λF_{ n }u then u = λa_{ n }Au + λa_{ n }B^{n}u. It is impossible since λa_{ n } ∈ (0, 1). By Lemma 2.9, there exists such that
for all n ∈ ℕ. This implies that
Since B is uniformly asymptotically regular with respect to A, it follows that
From (3.20) and (3.21), we obtain
Now, it is noted that
Using (3.20) and (3.22) in (3.23), we get
Since AM is weakly compact and passing eventually to a subsequence, we may assume that {Ax_{ n }} converges weakly to some . Thus, we have
By hypothesis (iii), the sequence {x_{ n }} has a subsequence which converges weakly to some . Since A and B are weakly sequentially continuous, converges weakly to Ax, and converges weakly to Bx. Hence, Ax + Bx = x. □
Theorem 3.5. Let U be a bounded open convex set in a Banach space X with 0 ∈ U. Suppose and B : X → X are continuous mappings satisfying the following:
(i) is compact, and A is weaklystrongly continuous,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞), and I  B is demiclosed,
(iii) if (x_{ n }) is a sequence of such that ((I  B)x_{ n }) is weakly convergent, then the sequence (x_{ n }) has a weakly convergent subsequence,
(iv) B is uniformly asymptotically regular with respect to A.
Then, either
or
Proof. Suppose (3.27) does not occur and let for all n ∈ ℕ. The mapping F_{ n } := a_{ n }A + a_{ n }B^{n} is the sum of a compact and a strict contraction. This implies that F_{ n } is a condensing mapping (see [13]). By Lemma 2.11, we deduce that there is an such that
for all n ∈ ℕ. This implies that
Since B is uniformly asymptotically regular with respect to A, it follows that
From (3.28) and (3.29), we obtain
Now, it is noted that
Using (3.28) and (3.30) in (3.31), we get
Since AM is weakly compact and passing eventually to a subsequence, we may assume that {Ax_{ n }} converges weakly to some . This implies that
By hypothesis (iii), the sequence {x_{ n }} has a subsequence which converges weakly to some . Since A is weaklystrongly continuous, converges strongly to Ax.
Consequently
By the demiclosedness of I  B, we have Ax + Bx = x. □
Corollary 3.6. Let U be a bounded open convex set in a uniformly convex Banach space X with 0 ∈ U. Suppose and B : X → X are continuous mappings satisfying the following.
(i) is compact, and A is weaklystrongly continuous,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞),
(iii) B is uniformly asymptotically regular with respect to A.
Then, either
or
Theorem 3.7. Let Q be a closed convex bounded set in a Banach space X with 0 ∈ Q. Suppose A : Q → X and B : X → X are continuous mappings satisfying the following:
(i) A(Q) is compact, and A is weaklystrongly continuous,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞), and I  B is demiclosed,
(iii) if (x_{ n }) is a sequence of such that ((I  B)x_{ n }) is weakly convergent, then the sequence (x_{ n }) has a weakly convergent subsequence,
(iv) if is a sequence of ∂Q × [0, 1] converging to (x, λ) with X = λAx + λB^{n}x and 0 ≤ λ < 1, then λ_{ j }Ax_{ j } + λ_{ j }B^{n}x_{ j } ∈ Q for j sufficiently large,
(v) B is uniformly asymptotically regular with respect to A.
Then, A + B has a fixed point in Q.
Proof. We first define F_{ n } := a_{ n }A + a_{ n }B^{n}, where for all n ∈ ℕ. Since F_{ n } is the sum of a compact mapping and a strict contraction mapping, it follows that F_{ n } is a condensing mapping. For any let fixed n, we have is a sequence of ∂Q × [0, 1] converging to (y, λ) with y = λF_{ n }(y) and 0 ≤ λ < 1. Then y = a_{ n }λAy + a_{ n }λB^{n}y. From assumption (iv), it follows that a_{ n }λ_{ j }Ay_{ j } + a_{ n }λ_{ j }B^{n}y_{ j } ∈ Q for j sufficiently large. Applying Lemma 2.12 to F_{ n }, we deduce that there is an x_{ n } ∈ Q such that
As in Theorem 3.5 this implies that
By hypothesis (iii), the sequence {x_{ n }} has a subsequence which converges weakly to some x ∈ Q. Since A is weaklystrongly continuous, converges strongly to Ax.
It follows that
Hence, by the demiclosedness of I  B, we have Ax + Bx = x. □
Corollary 3.8. Let Q be a closed convex bounded set in a uniformly convex Banach space X with 0 ∈ Q. Suppose A : Q → X and B : X → X are continuous mappings satisfying the following:
(i) A(Q) is compact and A is weaklystrongly continuous,
(ii) B is an asymptotically nonexpansive mapping with a sequence (k_{ n }) ⊂ [1, ∞),
(iii) if is a sequence of ∂Q × [0, 1] converging to (x, λ) with X = λAx + λB^{n}x and 0 ≤ λ < 1, then λ_{ j }Ax_{ j } + λ_{ j }B^{n}x_{ j } ∈ Q for j sufficiently large,
(iv) B is uniformly asymptotically regular with respect to A.
Then, A + B has a fixed point in Q.
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Acknowledgements
The authors would like to thank the referee for the insightful comments and suggestions. The first author would like to thanks The Thailand Research Fund for financial support and the second author is also supported by the Royal Golden Jubilee Program under Grant PHD/0282/2550, Thailand. Moreover, the second author the Thailand Research Fund for financial support under Grant BRG5280016.
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The work presented here was carried out in collaboration between all authors. SP and AA defined the research theme. SP designed theorems and methods of proof and interpreted the results. AA proved the theorems, interpreted the results and wrote the paper. All authors have contributed to, seen and approved the manuscript.
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Arunchai, A., Plubtieng, S. On the Krasnoselskiitype fixed point theorems for the sum of continuous and asymptotically nonexpansive mappings in Banach spaces. J Inequal Appl 2011, 28 (2011). https://doi.org/10.1186/1029242X201128
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DOI: https://doi.org/10.1186/1029242X201128