1.Introduction ViscosityApproximationMethodforSplitCommonNullPointProblemsbetweenBanachSpacesandHilbertSpaces

ISSN 0975-5748 (online); 0974-875X (print)

Published by RGN Publications http://www.rgnpublications.com

Research Article

Viscosity Approximation Method for Split Common Null Point Problems between Banach Spaces and Hilbert Spaces

Khanittha Promluang^1,2 and Poom Kumam^1,2,*

1KMUTT Fixed Point Research Laboratory, Department of Mathematics,

Room SCL 802 Fixed Point Laboratory, Science Laboratory Building, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok 10140, Thailand

2KMUTT-Fixed Point Theory and Applications Research Group (KMUTT-FPTA), Theoretical and Computational Science Center (TaCS), Science Laboratory Building, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok 10140, Thailand

*Corresponding author: [email protected]

Abstract. We study an iterative scheme to approximation the split common null point problems for set-valued maximal monotone operators which combine viscosity method and some fixed point technically proving method between Banach spaces and Hilbert spaces, without using the metric projection. We prove that strong convergence theorem. Also, we show that our result can be solves the split minimization problems.

Keywords. Iterative method; Viscosity approximation method; Fixed point problems; Split common null point problems; A zero point; Nonexpansive operator; (metric) resolvent operator

MSC. 47H09; 47H10; 54H25; 37C25

Received: March 28, 2016 Accepted: January 10, 2017

Copyright © 2017 Khanittha Promluang and Poom Kumam. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Many important problems in mathematics, sciences, engineering and other can be reformulation which require finding zero points or null point of nonlinear operators i.e. equation of the form 0 ∈ Ax, where x ∈ X such that X be a vector space, and A is a maximal monotone operator.

The inverse problem can be split into two inverse problems, which is called Split Inverse Problem(SIP). Let IP₁, IP₂ be two inverse problems. First one is formulated in the space X and the second one is formulated in space Y . This problem concern in a model which there are given two vector spaces X , Y , and a linear operator T : X → Y . The Split Inverse Problem is formulated as follows.

Find a point x^∗∈ X that solves IP1

and y^∗= T x^∗∈ Y that solves IP2. (1.1)

The Split Convex Feasibility Problem is the first case of the SIP, which is introduced by Censor and Elfving [1]. The two inverse problems IP₁and IP₂there are of the Convex Feasibility Problem (CFP) type. However, someone called the Split Convex Feasibility Problem is the Split Feasibility Problem (SFP). The Split (Convex) Feasibility Problem is formulated as follows.

Find a point x^∗∈ C such that T x^∗∈ Q, (1.2)

where C, Q are nonempty closed convex subspace of Hilbert space H₁, H₂, respectively. The set solution of Split Feasibility Problem is denoted byΓ:= {x^∗∈ C such that T x^∗∈ Q} = C ∩ T⁻¹(0).

Assume that the SF P has a solution, then x ∈ C solves (1.1) if and only if it solves the following fixed point equation:

x = PC(I − γT^∗(I − PQ)T)x, x ∈ C, (1.3)

where γ is any positive constant, PC is metric projections of H₁ onto C, and P_Q is metric projections of H₂ onto Q, and T^∗is the adjoint of T .

Later, many researcher have studies SIP in Hilbert spaces, for instance, [2, 3, 4] Takahash have studies SIP in Banach spaces [5].

Byrne el al. [4] introduced the Split Common Null Point Problem (SCNPP) for set- valued maximal monotone operators in Hilbert spaces. They given two set-valued operators A : H₁→ 2^H1, B : H₂→ 2^H2 and let T : H₁→ H2 be a bounded linear operator. They consider SCNPP as the following.

Find a point x^∗∈ H1that solves 0 ∈ Ax^∗

and y^∗= T x^∗∈ H2 that solves 0 ∈ B y^∗ (1.4)

where A⁻¹(0) and B⁻¹(0) are null point set of A and B, respectively. They given x₀∈ H1, define iterative scheme by the following:

x_n+1= J_λ^B1(x_n+ γA^∗(J_λ^B2− I)Axn), (1.5)

where A^∗ is the adjoint of A, L = kA^∗Ak and γ ∈ (0,_L²), andλ > 0.

In 2015, Takahashi [6] have studied SCNPP and he is the first that extended SCNPP form Hilbert space to Banach space by using metric resolvent and metric projections with applies the hybrid method. Recently, Takahashi and Yao [7] considered SCNPP in Hilbert spaces and Banach spaces by using the hybrid method as the following theorem:

Theorem 1.1 ([7]). Let H be a Hilbert space and let E be a uniformly convex and uniformly smooth of Banach spaces. Let J_Ebe the duality mapping onE. Let A : H → 2^H andB : E → 2^E∗ be maximal monotone operator such that A⁻¹(0) 6= ; and B⁻¹(0) 6= ;. Let J_λ^A be the resolvent ofA for λ > 0 and let Q^B_µ be the metric resolvent ofB forµ > 0. Let T : H → E be a bounded linear operator

such thatT 6= ; and T^∗ be an adjoint operator ofT . Suppose that A⁻¹(0) ∩ T⁻¹(B⁻¹0) 6= ;. Let x₁∈ H, and let {xn} be a sequence generated by the following.























z_n= J_λ^A_n(x_n− λnT^∗J_E(T x_n− Q^B_µnT x_n)), y_n= αnx_n+ (1 − αn)z_n,

C_n= {z ∈ H : kyn− zk ≤ kxn− zk}, Q_n= {z ∈ H : 〈xn− z, xn1 − xn〉 ≥ 0}, x_n+1= PCn∩Qnx₁; ∀ n ∈ N,

(1.6)

where αn ⊂ [0, 1], and λn,µn⊂ (0, ∞) satisfies that 0 ≤ αn≤ a < 1, 0 < b ≤ µn, and 0 < c ≤ λnkTk²≤ d < 2 for some a, b, c, d ∈ R. Then the sequence {xn} converges strongly to a point z₀∈ A⁻¹(0) ∩ T⁻¹(B⁻¹0), where z₀= P_A⁻¹_(0)∩T⁻¹_(B⁻¹₀₎x₁.

A viscosity approximation method is a well-known iterative method for solving a fixed point of nonexpansive mappings. Moudafi [8] is first person that proposed viscosity approximation method by combing the nonexpansive mapping and a contraction mapping. He proposed the following iterative scheme in Hilbert spaces:

x_n+1= αnf (x_n) + (1 − αn)Sx_n, (1.7)

where f is a contraction mapping and {αn} ⊆ (0,1) satisfies some conditions, he proved that {xn} converges strongly to a fixed point of S.

Further, Xu [9] developed the viscosity approximation method for solves the zero points of monotone operators in a Banach space. He proved that strong convergent theorem.

Motivated by the problems of split common null point, and result of Takahashi [7], these is interesting that the result of [7] is formulated in two different spaces. Moreover, a well-known viscosity approximation method of Moudafi [8] and Xu [9] are still effective and interesting for solving a fixed point problem. Then we consider SCNPP related between a Hilbert space and a Banach space by using the viscosity approximation method. We consider SCNPP as the following.

Find a point x^∗∈ H that solves 0 ∈ Ax^∗

and y^∗= T x^∗∈ E that solves 0 ∈ B y^∗, (1.8)

where A⁻¹(0) and B⁻¹(0) are null point set of A and B, respectively. While, we let H, E are Hilbert spaces and Banach spaces, respectively.

2. Preliminaries

Let E be a real Banach space with the norm k · kE and the dual space E^∗of E is the space of continuous linear functional on E .

The normalized duality mapping J : E → 2^E∗ is defined by J(x) = {x^∗∈ E^∗: 〈x, x^∗〉 = kxk², kxk = kx^∗k}, ∀ x ∈ E.

In the case if E is Hilbert space, then the normalized duality mapping is linear and it is just the identity mapping, i.e. J = I.

A Banach space E is said to satisfy Opial’s condition if for each sequence {x_n}^∞

n=0in E such that {x_n}converges weakly to some x in E, the inequality

lim inf

n→∞ kxn− xk < lim inf

n→∞ kxn− yk

holds for all y ∈ E with y 6= x. In fact, a Banach space with a weakly sequentially continuous duality mapping has the Opial’s condition; see [10]. It known that every Hilbert satisfies the Opial’s condition.

A Banach space E is called a strictly convex if it satisfies the following condition kxk = kyk = k(1 − λ)x + λyk ∀ x, y ∈ E and 0 < λ < 1 =⇒ x = y.

Let S(E) = {x ∈ E : kxk = 1} be the unit sphere of E. E is said to be uniformly convex if for each² ∈ (0,2] there exists a constant δ = δ(²) > 0 such that for all x, y ∈ S(E), if kx − yk ≥ ² then

°

°

°

x+y 2

°

°

° ≤1 − δ. It is well known that uniformly convex is strictly convex.

A Banach space E is said to be smooth if the limit limt→0

kx + t yk − kxk

t (2.1)

exists for all x, y ∈ S(E).

The norm of E is said to be uniformly G ˆateaux differentiable norm if for each y ∈ S(E), the limit (2.1) is attained uniformly for all x ∈ S(E) and it is said to be Fréchet differentiable if for each x ∈ S(E), the limit (2.1) is attained uniformly for all y ∈ S(E). Moreover, it is said to be uniformly smooth if the limit (2.1) is attained uniformly for all (x, y) ∈ S(E) × S(E).

The modulus of smoothness of E is the function ρ : [0,∞) → [0,∞) defined by ρ(t) = supn

1

2(kx + yk + kx − yk) − 1 : x, y ∈ E,kxk = 1,kyk = to

. A Banach space E is an uniformly smooth if and only if lim

t→0 ρ(t)

t = 0.

A Banach space E is said to be q-uniformly smooth if for 1 < q ≤ 2 be a fixed real number.

There exists a constant c > 0 such that ρ(t) ≤ ct^q for all t > 0. In the case q = 2, E is said to be 2-uniformly smooth if there exists a constant c > 0 such that ρ(t) ≤ ct² for all t > 0.

The examples of 2-uniformly smooth and uniformly convex Banach space that L_p, l_p, or the sobolev space W_m^p is







2-uniformly smooth, if p ≥ 2;

q-uniformly smooth, if 1 < p ≤ 2.

Note that no a Banach space is q-uniformly smooth for q > 2. It is known that a Hilbert space is 2-uniformly convex and 2-uniformly smooth. By [11, 12] we know that if E be a q-uniformly smooth Banach space, then for all x, y ∈ E there exists a constant c1> 0 such that kJ(x) − J(y)k ≤ c1kx − yk^q−1. Hence if E be a 2-uniformly smooth Banach space, then there exists a constant c1 > 0 such that kJ(x) − J(y)k ≤ c1kx − yk for all x, y ∈ E. If E be a q-uniformly smooth Banach space for 1 < q < 2, then there exists the constant c1 > 0 such that kJ(x) − J(y)k ≤ c1kx − yk^q−1. Then we can see that kJ(x) − J(y)k ≤ c1kx − yk too.

For instance in L_p is 2-uniformly smooth Banach space for 2 ≤ p < ∞ and we know that kJ(x) − J(y)k ≤ (p − 1)kx − yk. For 1 ≤ q ≤ 2, then Lq is q-uniformly smooth Banach space. So that kJ(x) − J(y)k ≤ 2q⁻¹K_qkx − yk^q−1 and also we have kJ(x) − J(y)k ≤ 2q⁻¹K_qkx − yk, where K_q is q-uniformly smooth constant.

Remark 2.1 (See [12]). (1) If E is a uniformly smooth, then E is smooth and reflexive.

(2) If E is a uniformly convex, then E is strictly convex.

(3) If E is a smooth, reflexive and strictly convex then the normalized duality mapping J is single-valued, one-to-one and onto. Then J⁻¹: E^∗→ E is single-valued, bijective, that is the inverse mapping J⁻¹: E^∗→ (E) and also J J⁻¹= IE^∗, J⁻¹J = IE.

(4) A normed space E is reflexive, if and only if E is bounded sequence has a weakly convergent subsequence.

Next, we recall some definitions. Let C be a nonempty close convex subset of E. Let S : C → C be an operator. If there exists a coefficient k ∈ (0,1) such that kSx − S yk ≤ kkx − yk for all x, y ∈ C, then S is called k-contraction. If k = 1 that is kSx − S yk ≤ kx − yk for all x, y ∈ C, then we called nonexpansive. In this paper, fixed point of an operator S is denoted by Fix(S), i.e, Fix(S) := {x ∈ C : Sx = x}.

Let B : E → 2^E∗ be a set-value operator, B is said to be monotone if 〈x − y, u^∗− v^∗〉 ≥ 0, for all x, y ∈ D(B), u^∗∈ Bx, v^∗∈ B y, where D(B) is the domain of B. A monotone operator B on E is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on E. By [13], we known that if B be a maximal monotone operator of E into 2^E∗, then R(I + rJ⁻¹B) = E. From [14] if E be a uniformly convex and smooth Banach space, then B is maximal monotone if and only if R(J + rB) = E^∗ for r > 0, where R(J + rB) is the range of J + rB. This mean that R(I + rJ⁻¹B) = E.

For a maximal monotone operator B, we can define a nonexpansive single-valued mapping Q^B_r : R(I + rJ⁻¹B) → D(B) by Q^Br = (I + rJ⁻¹B)⁻¹ for each r > 0, which is called the metric resolvent of B. It is known that 0 ∈ B(x) ⇔ x ∈ Fix(Q^Br). The set of null point of B is generated by B⁻¹(0) = {x ∈ B : 0 ∈ Bx}. From Takahashi [15], we known that B⁻¹(0) are closed and convex.

In Hilbert space, we known that H = H^∗. For a monotone operator A : H → 2^H, we define a nonexpansive single-valued mapping J_r^A: R(I + rA) → D(A) by J_r^A= (I + r A)⁻¹for each r > 0, which is called the resolvent of A. It is known that 0 ∈ A(x) ⇔ x ∈ Fix(Jr^A).

Let C be a nonempty closed convex subset of a Banach space E and D ⊂ C, then a mapping Q : C → D is said to be sunny if Q(x + t(x − Q(x))) = Q(x) whenever Qx + t(x − Q(x)) ∈ C for all x ∈ C and t ≥ 0.

A mapping Q : C → C is called a retraction if Q²= Q. Note that if a mapping Q is a retraction, then Q z = z for all z ∈ R(Q) where R(Q) is the range of Q. A subset D of Q is called a sunny nonexpansive retract of C if there exists a sunny nonexpansive retraction from C onto D.

Lemma 2.2 ([16]). Let E be a smooth Banach space and let C be a nonempty subset of E.

Let Q : E → C be a retraction and let J be the normalized duality mapping on E. Then, the followings are equivalent:

(i) Q is sunny and nonexpansive;

(ii) kQx − Q yk²≤ 〈x − y, J(Qx − Q y)〉, ∀ x, y ∈ E;

(iii) k(x − y) − (Qx − Q y)k²≤ kx − yk²− kQx − Q yk² ; (iv) 〈x − Qx, J(y − Qx)〉 ≤ 0, ∀ x ∈ E, y ∈ C.

Lemma 2.3 ([17]). Let C be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space E and let S be a nonexpansive mapping of C into itself with Fix(S) 6= ;.

Then, the set Fix(S) is a sunny nonexpansive retract of C.

In 2004, Xu [9] studied the continuous scheme x_t = t f (xt) + (1 − t)Sxt, where f is a k- contraction and S is nonexpansive mapping. On a uniformly smooth Banach space, Xu proved that the sequence x_t∈ C be the unique fixed point of the contraction x → t f (x) + (1 − t)Sx, that is x_t= t f (xt) + (1 − t)Sxt.

Lemma 2.4 ([9]). Let E be a uniformly smooth Banach space, C be a closed convex subset of E, S : C → C be a nonexpansive mapping with Fix(S) 6= ;, and and f ∈Ξc. Then x_t defined by x_t= t f (xt) + (1 − t)Sxt converges strongly to a point in Fix(S). If we defines mapping Q :Ξc→ Fix(S), whereΞc denote the set ofk-contraction, by Q( f ) := lim

t→0x_t, thenQ( f ) solves the following variational inequality:

〈(I − f )Q( f ), J(Q( f ) − p)〉 ≤ 0, ∀ f ∈Ξc, ∀ p ∈ Fix(S).

Namely, if ¯x = QC( f ), then

〈 ¯x − f ( ¯x), J( ¯x − p)〉 ≤ 0, ∀ f ∈Ξc, ∀ p ∈ Fix(S).

It well known that if E = H is a Hilbert space, then a sunny nonexpansive retraction QC is coincident with the metric projection P_C from E onto C, that is Q_C= PC.

In the sequel to give our main results, we need the following lemmas.

Lemma 2.5 ([18]). Let E be a real Banach space. Then kx + yk²≤ kxk²+ 2〈y, j(x + y)〉, j(x + y) ∈ J(x + y), where j denote a single-value.

Lemma 2.6 ([12]). Let H be a Hilbert space. Then

kx + yk²= kxk²+ 2〈y, x〉 + kyk², ∀ x, y ∈ H.

Lemma 2.7 ([20]). Let E be a real uniformly convex Banach space and B_r= {x ∈ E : kxk ≤ r}, r > 0. Then there exists a continuous, strictly increasing, and convex function g : [0,∞] → [0,∞], g(0) = 0 such that

kαx + βy + γzk²≤ αkxk²+ βkyk²+ γkzk²− αβg(kx − yk), ∀ x, y, z ∈ [0, 1], withα + β + γ = 1.

Lemma 2.8 ([21], The Resolvent Identity). For all r > 0, s > 0. Let x ∈ E and B is maximal monotone then

J_r^Bx = Js^B

³s rx +³

1 − s r

´ J_r^Bx´

.

Lemma 2.9 ([24], Demiclosed Principle). Let C be a nonempty closed convex subset of a uniformly convex Banach space E and S be a nonexpansive mapping. Then I − S is demiclosed at zero, i.e., x_n*^{x and x}n− Sxn→ 0 imply x = Sx.

Lemma 2.10 ([25]). Let {x_n} and {z_n} be bounded sequences in a Banach space E and let {βn} be a sequence in [0, 1] with

0 < liminf

n→∞ βn≤ lim sup

n→∞ βn< 1.

Suppose x_n+1= (1 − βn)z_n+ βnx_n for all integersn ≥ 0 and limsup

n→∞ (kz_n+1− znk − kx_n+1− xnk) ≤ 0.

Then lim

n→∞kzn− xnk = 0.

Lemma 2.11 ([26]). Assume that {a_n} is a sequence of nonnegative real numbers satisfying the condition

a_n+1≤ (1 − tn)a_n+ tnb_n+ cn, ∀ n ≥ 0, where {t_n} is a sequence in [0, 1] such that ^∞P

n=0

t_n= ∞, {bn} is a sequence such thatlim sup

n→∞

b_n≤ 0 and c_n≥ 0, ∀n ≥ 0 such that ^∞P

n→∞

c_n< ∞. Then, lim

n→∞a_n= 0.

Lemma 2.12([6]). Let E₁andE₂ be strictly convex, reflexive and smooth Banach space and let J_E1 andJ_E2 be the duality mapping onE₁andE, respectively. Let A : E₁→ 2^E∗1 andB : E₂→ 2^E∗2 be maximal monotone operators such that A⁻¹(0) 6= ; and B⁻¹(0) 6= ;, respectively. Let J_λ^A and Q^B_µ be the metric resolvent of be the resolvent of A forλ > 0 and B for µ > 0, respectively. Let T : E₁→ E2be a bounded linear operator such thatT 6= 0 and let T^∗ be the adjoint operator of T . Suppose that Ω:= A⁻¹(0) ∩ T⁻¹(B⁻¹(0)) 6= ;. Let λ, µ, r > 0 and p ∈ E. Then the following are equivalent:

(i) p = J_λ^A(p − rJ_E⁻¹₁T^∗J_E2(T p − Q^B_µT p));

(ii) p ∈ A⁻¹(0) ∩ T⁻¹(B⁻¹(0)).

3. Main Result

Proposition 3.1. Let H be a Hilbert space and E be a real 2-uniformly smooth Banach space with the constant c₁∈

³ 0, ¹

kTk²

´

, wherec₁ is positive constant such that kJ(x) − J(y)k ≤ c1kx − yk for all x, y ∈ E. Let JE be the duality mapping onE. Let B : E → 2^E∗ be an maximal monotone operator such thatB⁻¹(0) 6= ;. Let Q^B_µ= (I +µJ⁻¹B)⁻¹be the metric resolventB. Let T : H → E be a bounded linear operator such thatT 6= ; and T^∗: E^∗→ H be an adjoint operator of T. Assume thatT⁻¹(B⁻¹0) 6= ;. Let M := T^∗J_E(T − Q^B_µT), then M := T^∗J_E(T − Q^B_µT) is nonexpansive.

Proof. Since I and Q^B_µ are nonexpansive mappings, and we know that kJE(x)−JE( y)k ≤ c1kx− yk.

We compute

kMx − M yk = kT^∗J_E(I − Q^B_µ)T x − T^∗J_E(I − Q_µ^B)T yk = kT^∗(JE(I − Q^B_µ)T x − JE(I − Q^B_µ)T y)k

≤ kTk kJE(T − Q_µ^BT)x − JE(I − Q_µ^B)T yk ≤ c1kTk k(I − Q^B_µ)T x − (I − Q_µ^B)T xk²

≤ c1kTk kT x − T yk = c1kTk kT(x − y)k

≤ c1kTk² kx − yk.

Since c₁∈

³ 0, ¹

kTk²

´

, therefore M is a nonexpansive mapping.

Theorem 3.2. Let C be a nonempty closed convex subset of a Hilbert spaces H, and let E be a real uniformly convex and 2-uniformly smooth of Banach space with c₁∈

³ 0, ¹

kTk²

´ . Let J_E be the duality mapping on E. Let A : H → 2^H be maximal monotone operator such that A⁻¹(0) 6= ; and let B : E → 2^E∗ be an maximal monotone operator such that B⁻¹(0) 6= ;. Let J_λ^A

n= (I + λnA)⁻¹be the resolvent of A forλn> 0 and Q^B_µ= (I + µJ⁻¹B)⁻¹be the metric resolvent of B for µ > 0. Let f : C → C be a k−contraction mapping with k ∈ (0,1). Let T : H → E be a bounded linear operator such thatT 6= ; and T^∗: E^∗→ H be an adjoint operator of T. Assume thatΩ:= A⁻¹(0)∩T⁻¹(B⁻¹(0)) 6= ;. Let x1∈ H, and let {xn} be a sequence generate by the following

(y_n= σnx_n+ (1 − σn)J_λ^A

n(x_n− λnT^∗J_E(T x_n− Q_µ^BT x_n)),

x_n+1= αnf (x_n) + βnx_n+ γny_n, ∀ n ≥ 0, (3.1) where {σn}, {αn}, {βn}, {γn} ⊂ (0,1) and αn+βn+γn= 1. Assume that the control sequences satisfy the following conditions:

(a) lim

n→∞αn= 0 and ^∞P

n=1αn= ∞;

(b) 0 < liminf

n→∞ βn≤ lim sup

n→∞ βn< 1;

(c) 0 < liminf

n→∞ λn≤ λn≤ lim sup

n→∞ λn<_kTk²2, and lim

n→∞|λ_n+1− λn| = 0;

(d) 0 < liminf

n→∞ σn≤ σn≤ lim sup

n→∞ σn< 1, and lim

n→∞|σ_n+1− σn| = 0;

(e) 0 < liminf

n→∞ γn≤ γn.

Then {x_n} converges strongly to a point ¯x ∈Ω^{, where ¯}x = Q_Ωf ( ¯x).

Proof. We have divide the proof into five steps.

Step 1. We prove that the sequence {x_n} is bounded.

Fixed p ∈Ω:= A⁻¹(0) ∩ T⁻¹(B⁻¹(0)) 6= ;, then we have p ∈ (A)⁻¹(0) = Fix(J_λ^A_n), i.e., J_λ^A

np = p and T p = Q^B_µ(T p), and by Lemma 2.12 we have p = J_λ^A_n(p − λnT^∗J_E(T p − Q^B_µT p).

Set M := T^∗J_E(T − Q^B_µT), we see that kyn− pk²= kσnx_n+ (1 − σn)J_λ^A

n(I − λnM)x_n− pk²

≤ σnk(xn− p)k²+ (1 − σn)k[J_λ^A_n(I − λnM)x_n− p]k²

− σn(1 − σn)g(kxn− J_λ^A_n(I − λnM)x_nk)

≤ σnkxn− pk²+ (1 − σn)kxn− pk²− σn(1 − σn)g(kxn− J_λ^A_n(I − λnM)x_nk)

= kxn− pk²− σn(1 − σn)g(kxn− J_λ^A_n(I − λnM)x_nk). (3.2) Therefore kyn− pk²≤ kxn− pk², this implies that

kyn− pk ≤ kxn− pk.

Consider

kx_n+1− pk = kαnf (x_n) + βnx_n+ γny_n− pk

≤ αnk f (xn) − pk + βnkxn− pk + γnkyn− pk

≤ αnkkxn− pk + αnk f (p) − pk + βnkxn− pk + γnkyn− pk

≤ [αnk + βn+ γn]kxn− pk + αnk f (p) − pk

= [αnk + (1 − αn)]kxn− pk + αnk f (p) − pkk

≤ (1 + αnk − αn)kxn− pk + αnk f (p) − pkk

≤ (1 − αn(1 − k))kxn− pk + αnk f (p) − pk

= (1 − αn(1 − k))kxn− pk + αn(1 − k)k f (p) − pk (1 − k) . It follows by mathematical induction, we get that

kx_n+1− pk ≤ max

½

kxn− pk,k f (p) − pk 1 − k

¾ .

Therefore, this show that the sequence {x_n}is bounded. Furthermore, since kyn− pk ≤ kxn− pk then we obtain that { y_n} is bounded too. Also, we have { f (x)}, {J_λ^A

n(I − λnM)} are bounded sequence.

Step 2. We prove that the sequence {x_n} is asymptotically regular, i.e., lim

n→∞kx_n+1− xnk = 0.

Step 2.1. By use Lemma 2.10, then we set x_n+1:= βnx_n+(1−βn)z_nand we let z_n:=^{αnf (x}_(1−β^n)+γ_n)^nyn. We compute

z_n+1− zn=α_n+1f (x_n+1) + γ_n+1y_n+1

(1 − β_n+1) −αnf (x_n) + γny_n (1 − βn)

=α_n+1f (x_n+1) + y_n+1− (α_n+1+ β_n+1) y_n+1

(1 − βn+1) −αnf (x_n) + yn− (αn+ βn) y_n (1 − βn)

=α_n+1f (x_n+1) + (1 − β_n+1) y_n+1− α_n+1y_n+1

(1 − β_n+1) −αnf (x_n) + (1 − βn) y_n− αny_n (1 − βn)

=α_n+1( f (x_n+1) − y_n+1) + (1 − β_n+1) y_n+1

(1 − β_n+1) −αn[ f (x_n) − yn] + (1 − βn) y_n (1 − βn)

=α_n+1[ f (x_n+1) − y_n+1]

(1 − β_n+1) −αn[ f (x_n) − yn]

(1 − βn) + y_n+1− yn, (3.3) which implies that

kz_n+1− znk ≤ α_n+1

(1 − β_n+1)k f (x_n+1) + y_n+1k + αn

(1 − βn)k f (xn) − ynk + ky_n+1− ynk. (3.4) Step 2.2 Next, we compute kyn+1− ynk.

Observe that

kJ_λ^A_n+1(I − λ_n+1M)x_n+1− J_λ^A_n(I − λnM)x_nk

≤ kJ_λ^A_n+1(I − λ_n+1M)x_n+1− J_λ^A_n+1(I − λ_n+1M)x_nk + kJ_λ^A_n+1(I − λ_n+1M)x_n− J_λ^A_n(I − λnM)x_nk

≤ kx_n+1− xnk +

°

°

°

° J_λ^A

n+1(I − λ_n+1M)x_n− J_λ^A_n+1

·λn+1

λn (I − λnM x_n) − µ

1 −λn+1

λn

¶ J_λ^A

n(I − λnM)x_n

¸°

°

°

°

≤ kx_n+1− xnk +

°

°

°

°

(x_n− λ_n+1M)x_n−λ_n+1 λn

x_n+ λ_n+1M x_n+ µ

1 −λ_n+1 λn

¶ J_λ^A

n(I − λnM)x_n

°

°

°

°

≤ kx_n+1− xnk +

°

°

°

° µ

1 −λ_n+1 λn

¶

x_n+ (λ_n+1− λn)M x_n+ µ

1 −λ_n+1 λn

¶ J_λ^A

n(I − λnM)x_n

°

°

°

°

≤ kx_n+1− xnk +

¯

¯

¯

¯

λn− λ_n+1 λn

¯

¯

¯

¯kxnk + |λ_n+1− λn|kMxnk +

¯

¯

¯

¯

λn− λ_n+1 λn

¯

¯

¯

¯kJ_λ^A_n(xn− λnM x_n)k.

Consider that

y_n+1− yn= σ_n+1(x_n+1− xn) + (1 − σn+1)¡ J_λ^A

n+1(I − λn+1M)x_n+1− J_λ^A_n(I − λnM)xn

¢ + (σ_n+1− σn)(xn− J_λ^A_n(xn− λnM x_n).

Then, we have ky_n+1− ynk

≤ σ_n+1kx_n+1− xnk + (1 − σ_n+1)kJ_λ^A_n+1(I − λn+1M)x_n+1− J_λ^A_n(I − λnM)xnk + (σ_n+1− σn)kxn− J_λ^A_n(xn− λnM x_n)k

≤ σ_n+1kx_n+1− xnk + |σ_n+1− σn|kxn− J_λ^A_n(xn− λnM x_n)k + (1 − σn+1)

× µ

kx_n+1− xnk +

¯

¯

¯

¯

λn− λ_n+1 λn

¯

¯

¯

¯kxnk + |λ_n+1− λn|kMxnk +

¯

¯

¯

¯

λn− λ_n+1 λn

¯

¯

¯

¯kJ_λ^A_n(xn− λnM x_n)k

¶

≤ kx_n+1− xnk + |σ_n+1− σn|kxn− J_λ^A_n(x_n− λnM x_n)k + (1 − σ_n+1)

µ¯

¯

¯

¯

λn− λ_n+1 λn

¯

¯

¯

¯kxnk + |λ_n+1− λn|kMxnk +

¯

¯

¯

¯

λn− λ_n+1 λn

¯

¯

¯

¯kJ_λ^A_n(x_n− λnM x_n)k

¶

. (3.5) Step 2.3. To show that lim sup

n→∞ (kz_n+1− znk−kx_n+1− xnk) ≤ 0. From (3.4) and (3.5) we derive that kz_n+1− znk − kx_n+1− xnk

≤ α_n+1

(1 − β_n+1)k f (x_n+1) + y_n+1k − αn

(1 − βn)k f (xn) − ynk + |σ_n+1− σn|kxn− J_λ^A_n(x_n− λnM x_n)k + (1−σ_n+1)

·¯

¯

¯

¯

λn−λ_n+1 λn

¯

¯

¯

¯kxnk + |λ_n+1−λn|kMxnk +

¯

¯

¯

¯

λn−λ_n+1 λn

¯

¯

¯

¯kJ_λ^A_n(x_n−λnM x_n)k

¸ . (3.6) It follows from condition (a)-(d), we obtain

lim sup

n→∞ (kz_n+1− znk − kx_n+1− xnk) ≤ 0. (3.7)

By using Lemma 2.10, we obtain that lim sup

n→∞ kzn− xnk = 0. (3.8)

Since x_n+1− xn= (1 − βn)(z_n− xn), therefore

n→∞lim kx_n+1− xnk = 0. (3.9)

Next, we prove that kyn− xnk → 0, as n → ∞.

Since z_n=^{αnf (x}_(1−β^n)+γ_n)^nyn =_(1−β^αn_n)f (x_n) + (1 −_(1−β^αn_n)) y_n, we have kzn− ynk =

°

°

°

° αn

(1 − βn)f (x_n) + µ

1 − αn

(1 − βn)

¶

y_n− yn

°

°

°

°

=

°

°

°

° αn

(1 − βn)f (x_n) − αn

(1 − βn)y_n

°

°

°

°

≤ αn

(1 − βn)k f (xn) − ynk. (3.10)

By condition (a) we obtain that

n→∞lim kzn− ynk = 0. (3.11)

From (3.8) and (3.10), we get that

kyn− xnk ≤ kyn− znk + kzn− xnk → 0, as n → ∞. (3.12) Therefore

n→∞lim kyn− xnk = 0. (3.13)

Step 3. To show that lim

n→∞kxn− J_λ^A_n(I − λnM)x_nk = 0 and lim

n→∞kT xn− Q^BuT x_nk = 0.

Step 3.1. We want to show that lim

n→∞kxn− J_λ^A_n(I − λnM)x_nk = 0. Consider kx_n+1− pk²= kαnf (x_n) + βnx_n+ γny_n− pk²

= kαn( f (xn) − f (p)) + βn(xn− p) + γn( yn− p) + αn( f (p) − p)k²

≤ αnk f (xn) − f (p)k²+ βnkxn− pk²+ γnkyn− pk²+ 2αn〈 f (p) − p, x_n+1− p〉

≤ αnkkxn− pk²+ βnkxn− pk²+ γnkyn− pk²+ 2αn〈 f (p) − p, x_n+1− p〉

= αnkkxn− pk²+ βnkxn− pk²+ γnkxn− pk²

− γnσn(1 − σn)g(kxn− J_λ^A_n(I − λnM)x_nk) + 2αn〈 f (p) − p, x_n+1− p〉

= [αnk + βn+ γn]kxn− pk²− γnσn(1 − σn)g(kxn− J_λ^A_n(I − λnM)x_nk) + 2αn〈 f (p) − p, x_n+1− p〉

≤ kxn− pk²− γnσn(1 − σn)g(kxn− J_λ^A_n(I − λnM)xnk) + 2αn〈 f (p) − p, x_n+1− p〉.

It follow that

γnσn(1 − σn)g(kxn− J_λ^A_n(I − λnM)x_nk)

≤ kxn− pk²− kx_n+1− pk²+ 2αn〈 f (p) − p, x_n+1− p〉

≤ kx_n+1− xnk (kxn− pk + kx_n+1− pk) + 2αn〈 f (p) − p, x_n+1− p〉.

By condition (a), (d), (e), (3.9) and by property of g, we have

n→∞lim kxn− J_λ^A_n(I − λnM)x_nk = 0. (3.14)

Step 3.2 We want to show that lim

n→∞kT xn− Q^BuT x_nk = 0. We consider kyn− pk²= kσnx_n+ (1 − σn)J_λ^A_n(I − λnM)xn− pk²

≤ σnkxn− pk²+ (1 − σn)kJ_λ^A_n(I − λnM)xn− pk²

≤ σnkxn− pk²+ (1 − σn)k(I − λnM)x_n− pk²

≤ σnkxn− pk²+ (1 − σn)k(xn− p) − λnM x_nk²

≤ σnkxn− pk²+ (1 − σn){kxn− pk²− 2λn〈xn− p, T^∗J_E(T x_n− Q^B_uT x_n)〉

+ λ²kT^∗J_E(T x_n− Q^B_uT x)k²}

= σnkxn− pk²+ (1 − σn){kxn− pk²− 2λn〈T xn− T p, JE(T x_n− Q^B_uT x_n)〉

+ λ²kT^∗J_E(T x_n− Q^B_uT x_n)k²}

= σnkxn− pk²+ (1 − σn){kxn− pk²− 2λn〈T xn− Q^B_uT x_n+ Q^B_uT x_n− T p, JE(T x_n− Q^B_uT x_n)〉

+ λ²kT^∗J_E(T x_n− Q^B_uT x_n)k²}

= σnkxn− pk²+ (1 − σn){kxn− pk²− 2λn〈T xn− Q^B_uT x_n, J_E(T x_n− Q^B_uT x_n)〉

− 2λnQ^B_uT x_n− T p, JE(T x_n− Q^B_uT x_n)〉 + λ²kTk²kT xn− Q^B_uT x_nk²}

≤ σnkxn− pk²+ (1 − σn)

{kxn− pk²− 2λn〈T xn− Q^B_uT x_n, J_E(T x_n− Q^B_uT x_n)〉 + λ²kTk²kT xn− Q^B_uT x_nk²}

≤ kxn− pk²− 2λn(1 − σn)kT xn− Q^BuT x_nk²+ λ²n(1 − σn)kTk²kT xn− Q^BuT x_nk²

≤ kxn− pk²− (1 − σn)(2λn− λ²kTk²)kT xn− Q^B_uT xk². Note that

kx_n+1− pk²= kαnf (x_n) + βnx_n+ γny_n− pk²

≤ αnk f (xn) − pk²+ βnkxn− pk²+ γnkyn− pk²

= αnk f (xn)− pk²+βnkxn− pk²+γn(kxn− pk²−(1−σn)(2λn− λ²_nkTk²)kT xn− Q^B_µT x_nk²)

= αnk f (xn) − pkk + (1 − αn)kxn− pk²− γn(1 − σn)(2λn− λ²nkTk²)kT xn− Q^B_µT x_nk²

≤ αnk f (xn) − pkk + kxn− pk²− γn(1 − σn)(2λn− λ²_nkTk²)kT xn− Q^B_µT x_nk². It follow that

γn(1 − σn)(2λn− λ²_nkTk²)kT xn− Q^B_µT x_nk²≤ αnk f (xn) − pkk + kxn− pk²− kx_n+1− pk²

= αnk f (xn) − pkk + kxn− x_n+1kk(xn− p) + (x_n+1− p)k.

Sinceσn∈ [0, 1), and by condition (a), (c) and (3.9), then we have

n→∞lim kT xn− Q^B_uT x_nk = 0. (3.15)

Step 3.3. To show that lim

n→∞kT^∗J_E(T x_n− Q^B_ux_n)k = 0. Since T is bounded linear and T^∗ is adjoint operator of T , we have

kT^∗J_E(T x_n− Q^Bux_n)k²≤ kTk²kJE(T x_n− Q^BuT x_n)k²

= kTk²kT xn− Q^B_uT x_nk². (3.16) By (3.15) then

n→∞lim kT^∗J_E(T x_n− Q^BuT x_n)k = 0. (3.17)

Step 4. We want to show thatlim sup

n→∞ 〈 f ( ¯x) − ¯x, xn− ¯x)〉 ≤ 0, ∀ xn∈Ω, where ¯x = Q_Ωf ( ¯x).

Step 4.1. Set W :=

h

I +^γ_(1−α^n(1−σ_nⁿ₎⁾J^A

λn(I − λnM)i

, where M := T^∗J_E(T − Q^B_µT). Then W is nonexpansive mapping. We see that x_n+1:= αnf (xn) + (αn)W xn.

By apply Lemma 2.4, then we let x_t be a unique solution of equation x_t= t f (xt) + (1 − t)W xt,

∀ t ∈ (0, 1). Namely, xt be a fixed point of contraction mapping which is unique fixed point.

Putting ¯x = Q_Ωf ( ¯x) = lim

t→0x_t, where Q_Ωf ( ¯x) is an unique sunny nonexpansive retraction from C onto Fix(W), as t → 0.

Since we know that Fix(W) = Fix(J_λ^A_n)∩Fix(Q^B_µ(T)) =Ω. Thus, we know that x_t→ ¯x = Q_Ωf ( ¯x).

By Lemma 2.4, then there exist 〈f ( ¯x) − ¯x, q − ¯x〉 ≤ 0, ∀ f ∈Ξc, ∀ q ∈Ω^.

Step 4.2 To show that x_n→ q ∈ A⁻¹(0).

Next, we show that there exist q such that 0 ∈ Aq. Let sn:= J_λ^A_n(I − λnT^∗J_E(T − Q^B_µT)x_n. We obtain

(I − λnT^∗J_E(T − Q^B_µT)x_n∈ (I + λnA)s_n x_n− λnT^∗J_E(T x_n− Q^B_µT x_n) ∈ sn+ λnAs_n x_n− λnT^∗J_E(T x_n− Q^B_µT x_n) ∈ sn+ λnAs_n x_n− sn− λnT^∗J_E(T x_n− Q^B_µT x_n) ∈ λnAs_n

1 λn

(x_n− sn− λnT^∗J_E(T x_n− Q_µ^BT x_n)) ∈ Asn. (3.18) Next, we show that lim

n→∞kxn− sn− λnT^∗J_E(T x_n− Q^B_µT x_n)k = 0, we observe that kxn− sn− λnT^∗J_E(T xn− Q^B_µT x_n)k = k(xn− sn) − λnT^∗J_E(T xn− Q^B_µT x_n)k

≤ kxn− snk + 1 λn

kT^∗J_E(T xn− Q^B_µT x_n)k.

From (3.12),we get kxn− snk → 0, as n → ∞ and (3.14), we derive that

n→∞lim kxn− sn− λnT^∗J_E(T xn− Q^B_µT x_n)k = 0. (3.19) By (3.18) we obtain that 0 ∈ Asn.

Since lim

n→∞kxn−snk = 0, and the boundedness of {xn}such that {x_n}has a weakly convergence subsequence, i.e., x_ni*q. From 0 ∈ Asn, there exist q solve 0 ∈ Aq.

Step 4.3. To show that x_n→ q ∈ T⁻¹(B⁻¹(0)). Namely, we show that for some q such that T q solves 0 ∈ B(T q). Since Q^B_µ is nonexpansive, then we apply the demiclose principle, i.e., (I − Q^B_µ) is demiclose at zero. Since T is linear and bounded, then we have T x_ni*T q and from kT xn− Q^B_µT x_nk → 0, as n → ∞, this implies that T q = Q_µ^BT q. Therefore q ∈Ω^.

Next, we show that {x_nj} be another subsequence of {x_n} such that x_nj * ^{q0. That is} q⁰∈ Fix(W) = Fix(J_r^A) ∩ Fix(Q^B_µ(T)) =Ω. We want to show that q = q⁰. Assume that q 6= q⁰, by Opial’s condition we have

n→∞lim kxn− qk = lim inf

i→∞ kxni− qk < lim inf

i→∞ kxni− q⁰k

= lim

n→∞kxn− q⁰k = lim

j→∞kxn_j− q⁰k

≤ lim inf

j→∞ kxn_j− qk = lim

n→∞kxn− qk.

This contradiction. Thus, we have q = q⁰. This implies that x_n*q ∈Ω^.

Step 4.4. From above step we have q ∈Ω. From step 4.1 there exist 〈f ( ¯x) − ¯x, q − ¯x〉 ≤ 0, ∀q ∈Ω^, where ¯x = Q_Ωf ( ¯x).

Next, we show that lim sup

n→∞ 〈 f ( ¯x) − ¯x, xn− ¯x)〉 ≤ 0, ∀ xn∈Ω, where ¯x = Q_Ωf ( ¯x).

From step 4.3 we have 〈f ( ¯x) − ¯x, q − ¯x〉 ≤ 0, ∀f ∈Ξc, ∀q ∈Ω. To show this, we can choose a subsequence {x_ni}of {x_n} such that

lim sup

n→∞ 〈 f ( ¯x) − ¯x, xn− ¯x)〉 = lim

i→∞〈 f ( ¯x) − ¯x, xni− ¯x)〉 (3.20)

= 〈 f ( ¯x) − ¯x, q − ¯x〉 ≤ 0. (3.21)