Iterative approximation of a common solution of a split generalized equilibrium problem and a fixed point problem for nonexpansive semigroup

Abstract

Purpose

In this paper, we introduce and study an iterative method to approximate a common solution of a split generalized equilibrium problem and a fixed point problem for a nonexpansive semigroup in real Hilbert spaces.

Methods

We prove a strong convergence theorem of the iterative algorithm in Hilbert spaces under certain mild conditions.

Results

We obtain a strong convergence result for approximating a common solution of a split generalized equilibrium problem and a fixed point problem for a nonexpansive semigroup in real Hilbert spaces, which is a unique solution of a variational inequality problem. Further, we obtain some consequences of our main result.

Conclusions

The results presented in this paper are the supplement, extension, and generalization of results in the study of Plubtieng and Punpaeng and that of Cianciaruso et al. The approach of the proof given in this paper is also different.

Introduction

Throughout the paper, unless otherwise stated, let H₁ and H₂ be real Hilbert spaces with inner product 〈·,·〉 and norm ∥·∥. Let C and Q be nonempty closed convex subsets of H₁ and H₂, respectively.

A mapping f:C→C is said to be a contraction if there exists a constant α∈(0,1) such that ∥f x-f y∥≤α∥x-y∥, ∀x,y∈C. A mapping T:C→C is said to be nonexpansive if ∥T x-T y∥≤∥x-y∥, ∀x,y∈C. Fix (T) denotes the fixed point set of the nonexpansive mapping T:C→C.

Let B:H₁→H₁ be a strongly positive linear bounded operator, i.e., if there exists a constant $\bar{\gamma }>0$ such that

$$〈\mathit{\text{Bx}},x〉\ge \bar{\gamma }\parallel x{\parallel }^2,\forall x\in H_1.$$

A typical problem is to minimize a quadratic function over the set of fixed points of nonexpansive mapping T:

$$\underset{x\in \text{Fix}\left(T\right)}{min}\frac{1}{2}〈\mathit{\text{Bx}},x〉-〈x,b〉,$$

where b is a given point in H₁.

In 2006, Marino and Xu [1] considered the following iterative method:

$$x_{n+1}={\alpha }_n\mathrm{\gamma f}\left(x_n\right)+(I-{\alpha }_{n}B)T{x}_n,\forall n\ge 0,$$

with $0<\gamma <\frac{\bar{\gamma }}{\alpha }$ and proved that the sequence {x _n } converges strongly to the unique solution of the variational inequality

$$〈(B-\mathrm{\gamma f})z,x-z〉,\forall x\in \text{Fix}\left(T\right)$$

which is the optimality condition for the minimization problem

$$\underset{x\in \text{Fix}\left(T\right)}{min}\frac{1}{2}〈\mathit{\text{Bx}},x〉-h\left(x\right),$$

where h is the potential function for γ f.

A family S:={T(s):0≤s<∞} of mappings from C into itself is called a nonexpansive semigroup on C if it satisfies the following conditions:

• (i) T(0)x=x for all x∈C.

• (ii) T(s+t)=T(s)T(t) for all s,t≥0.

• (iii) ∥T(s)x-T(s)y∥≤∥x-y∥ for all x,y∈C and s≥0.

• (iv) For all x∈C, s↦T(s)x is continuous.

The set of all the common fixed points of a family S is denoted by Fix (S), i.e.,

$$\begin{array}{rl}\text{Fix}\left(S\right):& =\{x\in C:T(s)x=x,0\le s<\infty \}\\ =\bigcap _{0\le s<\infty }\text{Fix}\left(T\right(s\left)\right),\end{array}$$

where Fix(T(s)) is the set of fixed points of T(s). It is well known that Fix (S) is closed and convex.

The fixed point problem (FPP) for a nonexpansive semigroup S is:

$$\text{Find}x\in C\text{such that}x\in \text{Fix}\left(S\right).$$

(1)

In 1997, Shimizu and Takahashi [2] introduced and studied the following iterative method to prove a strong convergence theorem for FPP (1) in a real Hilbert space:

$$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n)\frac{1}{s_n}{\int }_0^{s_n}T\left(s\right)x_n\mathit{\text{ds}},\forall n\in \mathbb{N},$$

where {α _n } is a sequence in (0,1) and {s _n } is a sequence of positive real numbers which diverges to +∞. Later, Chen and Song [3] introduced and studied the following iterative method to prove a strong convergence theorem for FPP (1) in a real Hilbert space:

$$x_{n+1}={\alpha }_{n}f\left(x_n\right)+(1-{\alpha }_n)\frac{1}{s_n}{\int }_0^{s_n}T\left(s\right)x_n\mathit{\text{ds}},\forall n\in \mathbb{N},$$

where f is a contraction mapping. Recently, Plubtieng and Punpaeng [4] introduced and studied the following iterative method to prove a strong convergence theorem for FPP (1) in a real Hilbert space:

$$\begin{array}{rl}{x}_{n+1}=& {\alpha }_{n}f\left(x_n\right)+{\beta }_n{x}_n\\ +(1-{\alpha }_n-{\beta }_n)\frac{1}{s_n}{\int }_0^{s_n}T\left(s\right)x_n\mathit{\text{ds}},\forall n\in \mathbb{N},\end{array}$$

where {α _n } and {β _n } are the sequences in (0,1) and {s _n } is a positive real divergent sequence.

The equilibrium problem (EP) [5] is of finding x∈C such that

$$F(x,y)\ge 0,\forall y\in C,$$

(2)

where $F:C\times C\to \mathbb{R}$ is a bifunction. The solution set of EP (2) is denoted by EP (F).

Cianciaruso et al. [6] introduced and studied the following iterative method to prove a strong convergence theorem for FPP (1) and EP (2) in a real Hilbert space: x₀∈H₁:

$$x_{n+1}={\alpha }_n\mathrm{\gamma f}\left(x_n\right)+(1-{\alpha }_{n}B)\frac{1}{s_n}{\int }_0^{s_n}T\left(s\right)u_n\mathit{\text{ds}},\forall n\in \mathbb{N},$$

$$F(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\forall y\in H_1,$$

where {α _n } is a sequence in (0,1) and {s _n } is a positive real divergent sequence.

Recently, Moudafi [7] introduced the following split equilibrium problem (SEP):

Let $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ be nonlinear bifunctions and A:H₁→H₂ be a bounded linear operator, then the SEP is to find x^∗∈C such that

$$F_1(x^{\ast },x)\ge 0,\forall x\in C,$$

(3)

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}{F}_2(y^{\ast },y)\ge 0,\forall y\in \mathrm{Q.}$$

(4)

When looked separately, (3) is the classical EP, and we denoted its solution set by EP (F₁). SEP (3)-(4) constitutes a pair of equilibrium problems which have to be solved so that the image y^∗=A x^∗, under a given bounded linear operator A, of the solution x^∗ of EP (3) in H₁ is the solution of another EP (4) in another space H₂, and we denote the solution set of EP (4) by EP (F₂).

The solution set of SEP (3)-(4) is denoted by Ω={p∈EP(F₁):A p∈EP(F₂)}. SEP (3)-(4) includes the split variational inequality problem, split zero problem, and split feasibility problem (see, for instance, [7–12]).

In this paper, we consider a split generalized equilibrium problem (SGEP): Find x^∗∈C such that

$$F_1(x^{\ast },x)+h_1(x^{\ast },x)\ge 0,\forall x\in C,$$

(5)

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}{F}_2(y^{\ast },y)+h_2(y^{\ast },y)\ge 0,\forall y\in Q,$$

(6)

where $F_1,h_1:C\times C\to \mathbb{R}$ and $F_2,h_2:Q\times Q\to \mathbb{R}$ are nonlinear bifunctions and A:H₁→H₂ is a bounded linear operator.

We denote the solution set of generalized equilibrium problem (GEP) (5) and GEP (6) by GEP (F₁,h₁) and GEP (F₂,h₂), respectively. The solution set of SGEP (5)-(6) is denoted by Γ={p∈ GEP(F₁,h₁):A p∈GEP(F₂,h₂)}.

If h₁=0 and h₂=0, then SGEP (5)-(6) reduces to SEP (3)-(4). If h₂=0 and F₂=0, then SGEP (5)-(6) reduces to the equilibrium problem considered by Cianciaruso et al. [13].

Motivated by the works of Moudafi [7], Marino and Xu [1], Shimizu and Takahashi [2], Chen and Song [3], Plubtieng and Punpaeng [4], and Cianciaruso et al. [6, 13] and by the ongoing research in this direction, we introduce and study an iterative method for approximating a common solution of SGEP (5)-(6) and FPP (6) for a nonexpansive semigroup in real Hilbert spaces. The results presented in this paper extend and generalize the works of Shimizu and Takahashi [2], Chen and Song [3], Plubtieng and Punpaeng [4], and Cianciaruso et al. [6].

Now, we recall some concepts and results which are needed in sequel.

For every point x∈H₁, there exists a unique nearest point in C denoted by P _C x such that

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel ,\forall y\in \mathrm{C.}$$

(7)

P _C is called the metric projection of H₁ onto C. It is well known that P _C is a nonexpansive mapping and is characterized by the following property:

$$〈x-P_{C}x,y-P_{C}x〉\le 0.$$

(8)

Further, it is well known that every nonexpansive operator T:H₁→H₁ satisfies, for all (x,y)∈H₁×H₁, the inequality

$$\begin{array}{rl}〈(x-T(x\left)\right)& -(y-T(y\left)\right),T\left(y\right)-T\left(x\right)〉\\ \le (1/2)\parallel \left(T\right(x)-x)-\left(T\right(y)-y{\parallel }^2,\end{array}$$

(9)

and therefore, we get, for all (x,y)∈H₁×Fix(T),

$$〈x-T\left(x\right),y-T\left(x\right)〉\le (1/2)\parallel T\left(x\right)-x{\parallel }^2$$

(10)

(see, e.g., Theorem 3 in [14] and Theorem 1 in [15]).

It is also known that H₁ satisfies Opial’s condition [16], i.e., for any sequence {x _n } with $x_n\rightharpoonup x$, the inequality

$$lim\underset{n\to \infty }{inf}\parallel x_n-x\parallel <lim\underset{n\to \infty }{inf}\parallel x_n-y\parallel$$

(11)

holds for every y∈H₁ with y≠x.

Lemma 1

[[17]] Let {x _n } and {y _n } be bounded sequences in a Banach space X and {β _n } be a sequence in [0,1] with $0<lim\underset{n\to \infty }{inf}{\beta }_n\le lim\underset{n\to \infty }{sup}{\beta }_n<1$. Suppose x_n+1=(1-β _n )y _n +β _n x _n , for all integers n≥0 and $lim\underset{n\to \infty }{sup}(\parallel y_{n+1}-y_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0$. Then, $\underset{n\to \infty }{lim}\parallel y_n-x_n\parallel =0$.

Lemma 2

[[2]] Let C be a nonempty bounded closed convex subset of a Hilbert space H₁ and let S:={T(s):0≤s<∞} be a nonexpansive semigroup on C, for each x∈C and t>0. Then, for any 0≤h<∞,

$$\underset{t\to \infty }{lim}\underset{x\in C}{sup}∥\frac{1}{t}{\int }_0^{t}T\left(s\right)\mathit{\text{xds}}-T\left(h\right)\left(\frac{1}{t}{\int }_0^{t}T\left(s\right)\mathit{\text{xds}}\right)∥=0.$$

Lemma 3

[[18]] Let {a _n } be a sequence of nonnegative real numbers such that

$$a_{n+1}\le (1-{\alpha }_n)a_n+{\delta }_n,n\ge 0,$$

where {α _n } is a sequence in (0,1) and {δ _n } is a sequence in $\mathbb{R}$ such that (i)

$$\sum _{n=1}^{\infty }{\alpha }_n=\mathrm{\infty .}$$

• (ii)
  $$lim\underset{n\to \infty }{sup}\frac{{\delta }_n}{{\alpha }_n}\le 0$$

  or $\sum _{n=1}^{\infty }\mid {\delta }_n\mid <\infty$.Then,

  $$\underset{n\to \infty }{lim}{a}_n=0.$$

Lemma 4

[[1]] Assume that B is a strong positive linear bounded operator on a Hilbert space H₁ with coefficient $\bar{\gamma }>0$ and 0<ρ<∥B∥^-1. Then, $\parallel I-\mathrm{\rho B}\parallel \le 1-\rho \bar{\gamma }$.

Lemma 5

The following inequality holds in a real Hilbert space H₁:

$$\parallel x+y{\parallel }^2\le \parallel x{\parallel }^2+2〈y,x+y〉,\forall x,y\in H_1.$$

Assumption 1[19] Let $F:C\times C\to \mathbb{R}$ be a bifunction satisfying the following assumptions:

• (i) F(x,x)≥0, ∀x∈C,

• (ii) F is monotone, i.e., F(x,y)+F(y,x)≤0, ∀x∈C,

• (iii) F is upper hemicontinuous, i.e., for each x,y,z∈C,

  $$lim\underset{t\to 0}{sup}F(\mathit{\text{tz}}+(1-t)x,y)\le F(x,y),$$

• (iv) For each x∈C fixed, the function y→F(x,y) is convex and lower semicontinuous;

let $h:C\times C\to \mathbb{R}$ such that

• (i) h(x,x)≥0, ∀x∈C,

• (ii) For each y∈C fixed, the function x→h(x,y) is upper semicontinuous,

• (iii) For each x∈C fixed, the function y→h(x,y) is convex and lower semicontinuous,

and assume that for fixed r>0 and z∈C, there exists a nonempty compact convex subset K of H₁ and x∈C∩K such that

$$F(y,x)+h(y,x)+\frac{1}{r}〈y-x,x-z〉<0,\forall y\in C\setminus \mathrm{K.}$$

The proof of the following lemma is similar to the proof of Lemma 2.13 in [19] and hence omitted.

Lemma 6

Assume that $F_1,h_1:C\times C\to \mathbb{R}$ satisfying Assumption 1. Let r>0 and x∈H₁. Then, there exists z∈C such that

$$F_1(z,y)+h_1(z,y)+\frac{1}{r}〈y-z,z-x〉\ge 0,\forall y\in \mathrm{C.}$$

Lemma 7

[12]Assume that the bifunctions$F_1,h_1:C\times C\to \mathbb{R}$satisfy Assumption 1 and H₁is monotone. For r>0 and for all x∈H₁, define a mapping$T_r^{(F_1,h_1)}:H_1\to C$as follows:

$$\begin{array}{rl}{T}_r^{(F_1,h_1)}\left(x\right)=& \left\{z\in C:F_1(z,y)+h_1(z,y)\right.\\ \left.+\frac{1}{r_n}〈y-z,z-x〉\ge 0,\forall y\in C\right\}.\end{array}$$

Then, the following hold: (i)

$$T_r^{(F_1,h_1)}$$

is single-valued.(ii)

$$T_r^{(F_1,h_1)}$$

is firmly nonexpansive, i.e.,

$$\begin{array}{rl}\parallel T_r^{(F_1,h_1)}x-T_r^{(F_1,h_1)}y{\parallel }^2\le & 〈T_r^{(F_1,h_1)}x-T_r^{(F_1,h_1)}y,x-y〉,\\ \forall x,y\in H_1.\end{array}$$

(iii) Fix

$$\left(T_r^{(F_1,h_1)}\right)=\text{GEP}(F_1,h_1).$$

• (iv) GEP(F ₁,h ₁) is compact and convex.

Further, assume that $F_2,h_2:Q\times Q\to \mathbb{R}$ satisfying Assumption 1. For s>0 and for all w∈H₂, define a mapping $T_s^{(F_2,h_2)}:H_2\to Q$ as follows:

$$\begin{array}{rl}{T}_s^{(F_2,h_2)}\left(w\right)=& \left\{d\in Q:F_2(d,e)+h_2(d,e)\right.\\ \left.+\frac{1}{s}〈e-d,d-w〉\ge 0,\forall e\in Q\right\}.\end{array}$$

Then, we easily observe that $T_s^{(F_2,h_2)}$ is single-valued and firmly nonexpansive, GEP (F₂,h₂,Q) is compact and convex, and Fix$\left(T_s^{(F_2,h_2)}\right)=\text{GEP}(F_2,h_2,Q),$ where GEP (F₂,h₂,Q) is the solution set of the following generalized equilibrium problem:

Find y^∗∈Q such that F₂(y^∗,y) + h₂(y^∗,y)≥0, ∀y∈Q.

We observe that GEP (F₂,h₂)⊂ GEP (F₂,h₂,Q). Further, it is easy to prove that Γ is a closed and convex set.

Remark 1

Lemmas 6 and 7 are slight generalizations of Lemma 3.5 in[13]where the equilibrium condition F₁(x,x) = h₁(x,x) = 0 has been relaxed to F₁(x,x)≥0 and h₁(x,x)≥0 for all x∈C. Further, the monotonicity of H₁ in Lemma 6 is not required.

Lemma 8

[13]Let$F_1:C\times C\to \mathbb{R}$be a bifunction satisfying Assumption 1 hold and let$T_r^{F_1}$be defined as in Lemma 4 for r>0. Let x , y∈H₁and r₁, r₂>0. Then,

$$\parallel T_{r_2}^{F_1}y-T_{r_1}^{F_1}x\parallel \le \parallel y-x\parallel +\left|\frac{r_2-r_1}{r_2}\right|\parallel T_{r_2}^{F_1}y-y\parallel .$$

Notation. Let {x _n } be a sequence in H₁, then x _n →x (respectively, $x_n\rightharpoonup x$) denotes strong (respectively, weak) convergence of the sequence {x _n } to a point x∈H₁.

Methods

In this section, we prove a strong convergence theorem based on the proposed iterative method for computing the approximate common solution of SGEP (5)-(6) and FPP (1) for a nonexpansive semigroup in real Hilbert spaces.

We assume that Γ≠∅.

Theorem 1

Let H₁and H₂be two real Hilbert spaces and let C⊆H₁and Q⊆H₂be nonempty closed convex subsets. Let A:H₁→H₂be a bounded linear operator. Assume that$F_1,h_1:C\times C\to \mathbb{R}$and$F_2,h_2:Q\times Q\to \mathbb{R}$are the bifunctions satisfying Assumption 1; h₁,h₂are monotone and F₂is upper semicontinuous in the first argument. Let S={T(s):0≤s<∞} be a nonexpansive semigroup on C such that Fix(S)∩Γ≠∅. Let f:C→C be a contraction mapping with constant α∈(0,1) and B be a strongly positive linear bounded self-adjoint operator on H₁with constant$\bar{\gamma }>0$such that$0<\gamma <\frac{\bar{\gamma }}{\alpha }<\gamma +\frac{1}{\alpha }$. Let {s _n } is a positive real sequence which diverges to + ∞. For a given x₀∈C arbitrarily, let the iterative sequences {u _n } and {x _n } be generated by iterative algorithm:

$$\begin{array}{rl}{u}_n=& T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right);\\ x_{n+1}=& {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n\\ +\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}},\end{array}$$

(12)

where r _n ⊂(0,∞) and δ∈(0,1/L), L is the spectral radius of the operator A^∗A, and A^∗ is the adjoint of A, and {α _n } and {β _n } are the sequences in (0,1) satisfying the following conditions:

• (i)
  $$\underset{n\to \infty }{lim}{\alpha }_n=0$$

  and

  $$\sum _{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$$

  (ii)

  $$0<lim\underset{n\to \infty }{inf}{\beta }_n\le lim\underset{n\to \infty }{sup}{\beta }_n<1.$$

• (iii) lim infr _n >0 and

  $$\underset{n\to \infty }{lim}|r_{n+1}-r_n|=0.$$

  (iv)

  $$\underset{n\to \infty }{lim}\frac{|s_{n+1}-s_n|}{s_{n+1}}=0$$

  .

Then, the sequence {x _n } converges strongly to z∈Fix(S)∩Γ, where z = P_Fix(S)∩Γ(I-B+γ f)z.

Proof

We note that from condition (i), we may assume without loss of generality that α _n ≤(1-β _n )∥B∥^-1 for all n. From Lemma 4, we know that if 0<ρ≤∥B∥^-1, then $\parallel I-\mathrm{\rho B}\parallel \le 1-\rho \bar{\gamma }$. We will assume that $\parallel I-B\parallel \le 1-\bar{\gamma }$. □

Since B is a positive linear bounded self-adjoint operator on H₁, then

$$\parallel B\parallel =sup\left\{\right|〈\mathit{\text{Bu}},u〉|:u\in H_1,\parallel u\parallel =1\}.$$

Observe that

$$\begin{array}{rll}〈\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)u,u〉& =& 1-{\beta }_n-{\alpha }_n〈\mathit{\text{Bu}},u〉\\ \ge & 1-{\beta }_n-{\alpha }_n\parallel B\parallel \\ \ge & 0,\end{array}$$

which implies that (1-β _n )I-α _n B is positive. It follows that

$$\begin{array}{rll}\parallel (1-{\beta }_n)I-{\alpha }_{n}B\parallel & =& sup\{〈((1-{\beta }_n)I-{\alpha }_{n}B)u,u〉:\\ u\in H_1,\parallel u\parallel =1\}\\ =& sup\{1-{\beta }_n-{\alpha }_n〈\mathit{\text{Bu}},u〉:\\ u\in H_1,\parallel u\parallel =1\}\\ \le & 1-{\beta }_n-{\alpha }_n\bar{\gamma }.\end{array}$$

Let q = P_Fix(S)∩Γ. Since f is a contraction mapping with constant α∈(0,1), it follows that

$$\begin{array}{rll}\parallel q(I-B+\mathrm{\gamma f})\left(x\right)& -& q(I-B+\mathrm{\gamma f})\left(y\right)\parallel \\ \le & \parallel (I-B+\mathrm{\gamma f})\left(x\right)-(I-B+\mathrm{\gamma f})\left(x\right)\parallel \\ \le & \parallel I-B\parallel \parallel x-y\parallel +\gamma \parallel f\left(x\right)-f\left(y\right)\parallel \\ \le & (1-\bar{\gamma })\parallel x-y\parallel +\mathrm{\gamma \alpha }\parallel x-y\parallel \\ \le & (1-(\bar{\gamma }-\mathrm{\gamma \alpha }\left)\right)\parallel x-y\parallel ,\end{array}$$

for all x,y∈H₁. Therefore, the mapping q(I-B + γ f) is a contraction mapping from H₁ into itself. It follows from the Banach contraction principle that there exists an element z∈H₁ such that z = q(I-B + γ f)z = P_Fix(S)∩Γ(I-B + γ f)(z).

Let p∈Fix(S)∩Γ, i.e., p∈Γ, and we have $p=T_{r_n}^{(F_1,h_1)}p$ and

$$\mathit{\text{Ap}}=T_{r_n}^{(F_2,h_2)}\left(\mathit{\text{Ap}}\right).$$

We estimate

$$\begin{array}{rll}\parallel u_n-p{\parallel }^2& =& \parallel T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)-p{\parallel }^2\\ =& \parallel T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)\\ -T_{r_n}^{(F_1,h_1)}p{\parallel }^2\\ \le & \parallel x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n-p{\parallel }^2\\ \le & \parallel x_n-p{\parallel }^2+{\delta }^2\parallel A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\\ +2\delta 〈x_n-p,A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉.\end{array}$$

(13)

Thus, we have

$$\begin{array}{rll}\parallel u_n-p{\parallel }^2& \le & \parallel x_n-p{\parallel }^2\\ +{\delta }^2〈(T_{r_n}^{(F_2,h_2)}-I)A{x}_n,A{A}^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\\ +2\delta 〈x_n-p,A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉.\end{array}$$

(14)

Now, we have

$$\begin{array}{l}{\delta }^2〈(T_{r_n}^{(F_2,h_2)}-I)A{x}_n,A{A}^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\\ \le L{\delta }^2〈(T_{r_n}^{(F_2,h_2)}-I)A{x}_n,(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\\ =L{\delta }^2\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2.\end{array}$$

(15)

$$\text{Denoting}\Lambda =2\delta 〈x_n-p,A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉$$

and using (10), we have

$$\begin{array}{rll}\Lambda & =& 2\delta 〈x_n-p,A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\\ =& 2\delta 〈A(x_n-p),(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\\ =& 2\delta 〈A(x_n-p)+(T_{r_n}^{(F_2,h_2)}-I)A{x}_n\\ -(T_r^{(F_2,h_2)}-I)A{x}_n,(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\\ =& 2\delta \left\{〈T_{r_n}^{(F_2,h_2)}A{x}_n-\mathit{\text{Ap}},(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\right.\\ \left.-\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\right\}\\ \le & 2\delta \left\{\frac{1}{2}\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2-\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\right\}\\ \le & -\delta \parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2.\end{array}$$

(16)

Using (14), (15), and (16), we obtain

$$\parallel u_n-p{\parallel }^2\le \parallel x_n-p{\parallel }^2+\delta (\mathrm{L\delta }-1)\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2.$$

(17)

Since $\delta \in (0,\frac{1}{L})$, we obtain

$$\parallel u_n-p{\parallel }^2\le \parallel x_n-p{\parallel }^2.$$

(18)

Now, setting $t_n:=\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}$ and since p∈Fix(S)∩Γ, we obtain

$$\begin{array}{lll}\parallel t_n-p\parallel & =& ∥\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-p∥\\ \le & \frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}\parallel T\left(s\right)u_n-T\left(s\right)p\parallel \mathit{\text{ds}}\\ \le & \parallel u_n-p\parallel \\ \le & \parallel x_n-p\parallel .\end{array}$$

(19)

Further, we estimate

$$\begin{array}{lll}\parallel x_{n+1}-p\parallel & =& \parallel {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n+\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)t_n-p\parallel \\ =& \parallel {\alpha }_n\left(\mathrm{\gamma f}\right(x_n)-\mathit{\text{Bp}})+{\beta }_n(x_n-p)\\ +\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)(t_n-p)\parallel \\ \le & {\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel +{\beta }_n\parallel x_n-p\parallel \\ +(1-{\beta }_n-{\alpha }_n\bar{\gamma })\parallel t_n-p\parallel \\ \le & {\alpha }_n\gamma \parallel f\left(x_n\right)-f\left(p\right)\parallel +{\alpha }_n\parallel \mathrm{\gamma f}\left(p\right)-\mathit{\text{Bp}}\parallel \\ +{\beta }_n\parallel x_n-p\parallel +(1-{\beta }_n-{\alpha }_n\bar{\gamma })\parallel x_n-p\parallel \\ \le & {\alpha }_n\mathrm{\gamma \alpha }\parallel x_n-p\parallel +{\alpha }_n\parallel \mathrm{\gamma f}\left(p\right)-\mathit{\text{Bp}}\parallel \\ +(1-{\alpha }_n\bar{\gamma })\parallel x_n-p\parallel \\ =& (1-(\bar{\gamma }-\mathrm{\gamma \alpha }\left){\alpha }_n\right)\parallel x_n-p\parallel +{\alpha }_n\parallel \mathrm{\gamma f}\left(p\right)-\mathit{\text{Bp}}\parallel \\ \le & \text{max}\left\{\parallel x_n-p\parallel ,\frac{1}{\bar{\gamma }-\mathrm{\gamma \alpha }}\parallel \mathrm{\gamma f}\left(p\right)-\mathit{\text{Bp}}\parallel \right\},n\ge 0\\ ⋮\\ \le & \text{max}\left\{\parallel x_0-p\parallel ,\frac{1}{\bar{\gamma }-\mathrm{\gamma \alpha }}\parallel \mathrm{\gamma f}\left(p\right)-\mathit{\text{Bp}}\parallel \right\}.\end{array}$$

(20)

Hence, {x _n } is bounded, and consequently, we deduce that {u _n }, {t _n }, and {f(x _n )} are bounded.

Next, we estimate

$$\begin{array}{rll}\parallel t_{n+1}-t_n\parallel & =& ∥\frac{1}{s_{n+1}}\underset{0}{\overset{s_{n+1}}{\int }}T\left(s\right)u_{n+1}\mathit{\text{ds}}-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ =& ∥\frac{1}{s_{n+1}}\underset{0}{\overset{s_{n+1}}{\int }}\left[T\right(s)u_{n+1}-T(s\left)u_n\right]\mathit{\text{ds}}\\ +\left(\frac{1}{s_{n+1}}-\frac{1}{s_n}\right)\times \underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}\\ +\frac{1}{s_{n+1}}\underset{s_n}{\overset{s_{n+1}}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ =& ∥\frac{1}{s_{n+1}}\underset{0}{\overset{s_{n+1}}{\int }}\left[T\right(s)u_{n+1}-T(s\left)u_n\right]\mathit{\text{ds}}\\ +\left(\frac{1}{s_{n+1}}-\frac{1}{s_n}\right)\times \underset{0}{\overset{s_n}{\int }}\left[T\right(s)u_n-T(s\left)p\right]\mathit{\text{ds}}\\ +\frac{1}{s_{n+1}}\underset{s_n}{\overset{s_{n+1}}{\int }}\left[T\right(s)u_n-T(s\left)p\right]\mathit{\text{ds}}∥\\ \le & \parallel u_{n+1}-u_n\parallel +\frac{|s_{n+1}-s_n|s_n}{\left(s_{n+1}\right)s_n}\parallel u_n-p\parallel \\ +\frac{|s_{n+1}-s_n|}{s_{n+1}}\parallel u_n-p\parallel \\ \le & \parallel u_{n+1}-u_n\parallel +2\frac{|s_{n+1}-s_n|}{s_{n+1}}\parallel u_n-p\parallel .\end{array}$$

(21)

Since $T_{r_{n+1}}^{(F_1,h_1)}$ and $T_{r_{n+1}}^{(F_2,h_2)}$ both are firmly nonexpansive, for $\mathrm{\delta \epsilon }(0,\frac{1}{L})$, the mapping $T_{r_{n+1}}^{(F_1,h_1)}(I+\delta A^{\ast }(T_{r_{n+1}}^{(F_2,h_2)}-I\left)A\right)$ is nonexpensive, see [7, 10]. Further, since $u_n=T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)$ and $u_{n+1}=T_{r_{n+1}}^{(F_1,h_1)}(x_{n+1}+\delta A^{\ast }(T_{r_{n+1}}^{(F_2,h_2)}-I\left)A{x}_{n+1}\right),$ it follows from Lemma 8 that

$$\begin{array}{r}\parallel u_{n+1}-u_n\parallel \\ \le & \parallel T_{r_{n+1}}^{(F_1,h_1)}\left(x_{n+1}+\delta A^{\ast }\left(T_{r_{n+1}}^{(F_2,h_2)}-I\right)A{x}_{n+1}\right)\\ -T_{r_{n+1}}^{(F_1,h_1)}\left(x_n+\delta A^{\ast }\left(T_{r_{n+1}}^{(F_2,h_2)}-I\right)A{x}_n\right)\parallel \\ +\parallel T_{r_{n+1}}^{(F_1,h_1)}\left(x_n+\delta A^{\ast }\left(T_{r_{n+1}}^{(F_2,h_2)}-I\right)A{x}_n\right)\\ -T_{r_n}^{(F_1,h_1)}\left(x_n+\delta A^{\ast }\left(T_{r_n}^{(F_2,h_2)}-I\right)A{x}_n\right)\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\parallel \left(x_n+\delta A^{\ast }\left(T_{r_{n+1}}^{(F_2,h_2)}-I\right)A{x}_n\right)\\ -\left(x_n+\delta A^{\ast }\left(T_{r_n}^{(F_2,h_2)}-I\right)A{x}_n\right)\parallel \\ +\left|1-\frac{r_n}{r_{n+1}}\right|\parallel T_{r_{n+1}}^{(F_1,h_1)}\left(x_n+\delta A^{\ast }\left(T_{r_n}^{(F_2,h_2)}-I\right)A{x}_n\right)\\ -\left(x_n+\delta A^{\ast }\left(T_{r_{n+1}}^{(F_2,h_2)}-I\right)A{x}_n\right)\parallel \\ \le & \parallel x_{n+1}-x_n\parallel +\delta \parallel A\parallel \parallel T_{r_{n+1}}^{(F_2,h_2)}A{x}_n-T_{r_n}^{(F_2,h_2)}A{x}_n\parallel +{\delta }_n\\ \le & \parallel x_{n+1}-x_n\parallel +\delta \parallel A\parallel \left|1-\frac{r_n}{r_{n+1}}\right|\parallel T_{r_{n+1}}^{(F_2,h_2)}A{x}_n-A{x}_n\parallel +{\delta }_n\\ =& \parallel x_{n+1}-x_n\parallel +\delta \parallel A\parallel {\sigma }_n+{\delta }_n\end{array}$$

(22)

where

$${\sigma }_n=\left|1-\frac{r_{n+1}}{r_n}\right|\parallel T_{r_n}^{(F_2,h_2)}A{x}_n-A{x}_n\parallel$$

and

$$\begin{array}{rl}{\delta }_n=& \left|1-\frac{r_{n+1}}{r_n}\right|\parallel T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)\\ -(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)\parallel .\end{array}$$

Using (21) and (22), we have

$$\begin{array}{rl}\parallel t_{n+1}-t_n\parallel \le & \parallel x_{n+1}-x_n\parallel +\delta \parallel A\parallel {\sigma }_n+{\delta }_n\\ +2\frac{|s_{n+1}-s_n|}{s_{n+1}}\parallel u_n-p\parallel .\end{array}$$

(23)

Setting x_{n + 1} = β _n x _n + (1-β _n )e _n implies from (12) that

$$e_n=\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}=\frac{{\alpha }_n\mathrm{\gamma f}\left(x_n\right)+\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)t_n}{1-{\beta }_n}.$$

Further, it follows that

$$\begin{array}{rll}{e}_{n+1}-e_n& =& \frac{{\alpha }_{n+1}\mathrm{\gamma f}\left(x_{n+1}\right)+\left(\right(1-{\beta }_{n+1})I-{\alpha }_{n+1}B)t_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_n\mathrm{\gamma f}\left(x_n\right)+\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)t_n}{1-{\beta }_n}\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\mathrm{\gamma f}\left(x_{n+1}\right)+\frac{(1-{\beta }_{n+1})t_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_{n+1}B{t}_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_n}{1-{\beta }_n}\mathrm{\gamma f}\left(x_n\right)-\frac{(1-{\beta }_n)t_n}{1-{\beta }_n}+\frac{{\alpha }_{n}B{t}_n}{1-{\beta }_n}\\ =& \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\left(\mathrm{\gamma f}\right(x_{n+1})+B{t}_{n+1})+t_{n+1}-t_n\\ +\frac{{\alpha }_n}{1-{\beta }_n}(B{t}_n-\mathrm{\gamma f}(x_n\left)\right).\end{array}$$

Using (23), we have

$$\begin{array}{rll}\parallel e_{n+1}-e_n\parallel & =& ∥\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\left(\mathrm{\gamma f}\right(x_{n+1})+B{t}_{n+1})+t_{n+1}\\ -t_n+\frac{{\alpha }_n}{1-{\beta }_n}(B{t}_n-\mathrm{\gamma f}(x_n\left)\right)∥\\ \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\left(\parallel \mathrm{\gamma f}\left(x_{n+1}\right)\parallel +\parallel B{t}_{n+1}\parallel \right)\\ +\frac{{\alpha }_n}{1-{\beta }_n}(\parallel \mathrm{\gamma f}(x_n)\parallel +\parallel B{t}_n\parallel )+\parallel t_{n+1}-t_n\parallel \\ \le & \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}\left(\parallel \mathrm{\gamma f}\left(x_{n+1}\right)\parallel +\parallel B{t}_{n+1}\parallel \right)\\ +\frac{{\alpha }_n}{1-{\beta }_n}(\parallel \mathrm{\gamma f}(x_n)\parallel +\parallel B{t}_n\parallel )+\parallel x_{n+1}-x_n\parallel \\ +\gamma \parallel A\parallel {\sigma }_n+{\delta }_n+2\left(\frac{|s_{n+1}-s_n|}{s_{n+1}}\right)\parallel u_n-p\parallel \end{array}$$

which implies that

$$\begin{array}{rl}\parallel e_{n+1}& -e_n\parallel -\parallel x_{n+1}-x_n\parallel \\ \le \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}(\parallel f(x_{n+1})\parallel +\parallel B{t}_{n+1}\parallel )+\gamma \parallel A\parallel {\sigma }_n+{\delta }_n\\ +\frac{{\alpha }_n}{1-{\beta }_n}(\parallel \mathrm{\gamma f}(x_n)\parallel +\parallel B{t}_n\parallel )\\ +2\left(\frac{|s_{n+1}-s_n|}{s_{n+1}}\right)\parallel u_n-p\parallel .\end{array}$$

Hence, it follows by conditions (i), (iii), and (iv) that

$$lim\underset{n\to \infty }{sup}\left[\parallel e_{n+1}-e_n\parallel -\parallel x_{n+1}-x_n\parallel \right]\le 0.$$

(24)

From Lemma 1 and (24), we get $\underset{n\to \infty }{lim}\parallel e_n-x_n\parallel =0$ and

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =\underset{n\to \infty }{lim}(1-{\beta }_n)\parallel e_n-x_n\parallel =0.$$

(25)

Now,

$$\begin{array}{rll}{x}_{n+1}-x_n& =& {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n+\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)t_n-x_n\\ =& {\alpha }_n\left(\mathrm{\gamma f}\right(x_n)-x_n)+\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)(t_n-x_n).\end{array}$$

Since ∥x_{n + 1}-x _n ∥→0 and α _n →0 as n→∞, we obtain

$$\underset{n\to \infty }{lim}\parallel t_n-x_n\parallel =0.$$

(26)

Next, we have

$$\begin{array}{rll}\parallel T\left(s\right)x_n-x_n\parallel & =& ∥T\left(s\right)x_n-T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}\\ +T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}\\ -\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}+\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-x_n∥\\ \le & ∥T\left(s\right)x_n-T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ +∥T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ +∥\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-x_n∥\\ \le & ∥x_n-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ +∥T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ +∥\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-x_n∥\\ \le & 2∥x_n-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥\\ +∥T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥.\end{array}$$

(27)

Since {x _n } and {f(x _n )} are bounded, let $K:=\left\{w\in C:\parallel w-p\parallel \le \parallel x_0-p\parallel ,\frac{1}{\bar{\gamma }-\mathrm{\gamma \alpha }}\parallel \mathrm{\gamma f}\left(p\right)-\mathit{\text{Bp}}\parallel \right\}$, then K is a nonempty bounded closed convex subset of C which is T(s)-invariant for each 0≤s<∞ and contains {x _n }. So, without loss of generality, we may assume that S: = {T(s):0≤s<∞} is a nonexpansive semigroup on K. By Lemma 2, we have

$$\underset{n\to \infty }{lim}∥T\left(s\right)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}-\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}}∥=0.$$

(28)

Using (21) to (23), we obtain

$$\underset{n\to \infty }{lim}\parallel T\left(s\right)x_n-x_n\parallel =0.$$

(29)

It follows from (17) and Lemma 5 that

$$\begin{array}{r}\parallel x_{n+1}-p{\parallel }^2\\ =& \parallel {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n+\left[\right(1-{\beta }_n)I-{\alpha }_{n}B]t_n-p{\parallel }^2\\ =& \parallel {\alpha }_n\left(\mathrm{\gamma f}\right(x_n)-\mathit{\text{Bp}})+{\beta }_n(x_n-t_n)+(I-{\alpha }_{n}B)(t_n-p){\parallel }^2\\ \le & \parallel (I-{\alpha }_{n}B)(t_n-p)+{\beta }_n(x_n-t_n){\parallel }^2+2{\alpha }_n〈\mathrm{\gamma f}\left(x_n\right)\\ -\mathit{\text{Bp}},x_{n+1}-p〉\\ \le & {\left[\parallel (I-{\alpha }_{n}B)(t_n-p)\parallel +{\beta }_n\parallel x_n-t_n\parallel \right]}^2+2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)\\ -\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & {\left[\parallel I-{\alpha }_{n}B\parallel \parallel u_n-p\parallel +{\beta }_n\parallel x_n-t_n\parallel \right]}^2+2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)\\ -\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ =& {(I-{\alpha }_n\bar{\gamma })}^2\parallel u_n-p{\parallel }^2+{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n\\ -p\parallel \parallel x_n-t_n\parallel +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & {(1-{\alpha }_n\bar{\gamma })}^2\left[\parallel x_n-p{\parallel }^2+\delta (\mathrm{L\delta }-1)\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\right]\\ +{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n-t_n\parallel \\ +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & \left[1-2{\alpha }_n\bar{\gamma }+{\left({\alpha }_n\bar{\gamma }\right)}^2\right]\parallel x_n-p{\parallel }^2\\ +{(1-{\alpha }_n\bar{\gamma })}^2\delta (\mathrm{L\delta }-1)\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\\ +{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n-t_n\parallel \\ +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \end{array}$$

$$\begin{array}{rl}\le & \parallel x_n-p{\parallel }^2+{\alpha }_n{\bar{\gamma }}^2\parallel x_n-p{\parallel }^2\\ +{(1-{\alpha }_n\bar{\gamma })}^2\delta (\mathrm{L\delta }-1)\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\\ +{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n-t_n\parallel \\ +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

(30)

Therefore,

$$\begin{array}{rl}{(1-{\alpha }_n\bar{\gamma })}^2& \delta (1-\mathrm{L\delta })\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\\ \le & \parallel x_n-p{\parallel }^2-\parallel x_{n+1}-p{\parallel }^2+{\beta }_n^2\parallel x_n-t_n{\parallel }^2\\ +{\alpha }_n{\bar{\gamma }}^2\parallel x_n-p{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n\\ -p\parallel \parallel x_n-t_n\parallel +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +{\beta }_n^2\parallel x_n-t_n{\parallel }^2\\ +{\alpha }_n{\bar{\gamma }}^2\parallel x_n-p{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n-t_n\parallel \\ +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

Since δ(1-L δ) > 0, α _n →0, ∥x _n -t _n ∥→0 and ∥x_{n + 1}-x _n ∥→0 as n→∞, we obtain

$$\underset{n\to \infty }{lim}\parallel (T_{r_n}^{(F_2,h_2)}-I)A{x}_n\parallel =0.$$

(31)

Next, we show that ∥x _n -u _n ∥→0 as n→∞. Since p∈Fix(S)∩Γ, we obtain

$$\begin{array}{rll}\parallel u_n-p{\parallel }^2& =& \parallel T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)-p{\parallel }^2\\ =& \parallel T_{r_n}^{(F_1,h_1)}(x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I\left)A{x}_n\right)-T_{r_n}^{(F_1,h_1)}p{\parallel }^2\\ \le & 〈u_n-p,x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n-p〉\\ =& \frac{1}{2}\left\{\parallel u_n-p{\parallel }^2+\parallel x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n-p{\parallel }^2\right.\\ -\left.\parallel (u_n-p)-[x_n+\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n-p]{\parallel }^2\right\}\\ =& \frac{1}{2}\left\{\parallel u_n-p{\parallel }^2+\parallel x_n-p{\parallel }^2-\parallel u_n-x_n\right.\\ \left.-\delta A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\right\}\\ =& \frac{1}{2}\left\{\parallel u_n-p{\parallel }^2+\parallel x_n-p{\parallel }^2-\left[\parallel u_n-x_n{\parallel }^2\right.\right.\\ +{\delta }^2\parallel A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n{\parallel }^2\\ \left.\left.-2\delta 〈u_n-x_n,A^{\ast }(T_{r_n}^{(F_2,h_2)}-I)A{x}_n〉\right]\right\}.\end{array}$$

Hence, we obtain

$$\begin{array}{rl}\parallel u_n-p{\parallel }^2\le & \parallel x_n-p{\parallel }^2-\parallel u_n-x_n{\parallel }^2\\ +2\delta \parallel A(u_n-x_n)\parallel \parallel T_{r_n}^{(F_2,h_2)}-I)A{x}_n\parallel .\end{array}$$

It follows from (30) and (31) that

$$\begin{array}{rll}\parallel x_{n+1}-p{\parallel }^2& \le & {(1-{\alpha }_n\bar{\gamma })}^2\left[\parallel x_n-p{\parallel }^2-\parallel u_n-x_n{\parallel }^2\right.\\ \left.+2\delta \parallel A(u_n-x_n)\parallel \parallel \left.T_{r_n}^{(F_2,h_2)}-I\right)A{x}_n\parallel \right]\\ +{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n\\ -t_n\parallel +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & \left[1-2{\alpha }_n\bar{\gamma }+{\left({\alpha }_n\bar{\gamma }\right)}^2\right]\parallel x_n-p{\parallel }^2\\ -{(1-{\alpha }_n\bar{\gamma })}^2\parallel u_n-x_n{\parallel }^2+2{(1-{\alpha }_n\bar{\gamma })}^2\delta \parallel \\ A(u_n-x_n)\parallel \parallel \left.T_{r_n}^{(F_2,h_2)}-I\right)A{x}_n\parallel \\ +{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n\\ -t_n\parallel +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & \parallel x_n-p{\parallel }^2+{\alpha }_n{\bar{\gamma }}^2\parallel x_n-p{\parallel }^2-{(1-{\alpha }_n\bar{\gamma })}^2\parallel u_n\\ -x_n{\parallel }^2+2{(1-{\alpha }_n\bar{\gamma })}^2\delta \parallel A(u_n-x_n)\parallel \parallel T_{r_n}^{(F_2,h_2)}\\ -I\left.\right)A{x}_n\parallel +{\beta }_n^2\parallel x_n-t_n{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n\\ -p\parallel \parallel x_n-t_n\parallel +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

Therefore,

$$\begin{array}{rl}{(1-{\alpha }_n\bar{\gamma })}^2& \parallel u_n-x_n{\parallel }^2\\ \le & \parallel x_n-p{\parallel }^2-\parallel x_{n+1}-p{\parallel }^2+{\beta }_n^2\parallel x_n-t_n{\parallel }^2\\ +2{(1-{\alpha }_n\bar{\gamma })}^2\delta \parallel A(u_n-x_n)\parallel \parallel \left.T_{r_n}^{(F_2,h_2)}-I\right)A{x}_n\parallel \\ +{\alpha }_n{\bar{\gamma }}^2\parallel x_n-p{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n-t_n\parallel \\ +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel \\ \le & (\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel )\parallel x_n-x_{n+1}\parallel +{\beta }_n^2\parallel x_n-t_n{\parallel }^2\\ +2{(1-{\alpha }_n\bar{\gamma })}^2\delta \parallel A(u_n-x_n)\parallel \parallel T_{r_n}^{(F_2,h_2)}-I)A{x}_n\parallel \\ +{\alpha }_n{\bar{\gamma }}^2\parallel x_n-p{\parallel }^2+2(1-{\alpha }_n\bar{\gamma }){\beta }_n\parallel u_n-p\parallel \parallel x_n-t_n\parallel \\ +2{\alpha }_n\parallel \mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bp}}\parallel \parallel x_{n+1}-p\parallel .\end{array}$$

Since α _n →0, ∥x _n -t _n ∥→0, $\parallel T_{r_n}^{(F_2,h_2)}-I)A{x}_n\parallel \to 0$ and ∥x_{n + 1}-x _n ∥→0 as n→∞, we obtain

$$\underset{n\to \infty }{lim}\parallel u_n-x_n\parallel =0.$$

(32)

Thus, we can write

$$\begin{array}{rll}\parallel T\left(s\right)t_n-x_n\parallel & \le & \parallel T\left(s\right)t_n-T\left(s\right)x_n\parallel +\parallel T\left(s\right)x_n-x_n\parallel \\ \le & \parallel t_n-x_n\parallel +\parallel T\left(s\right)x_n-x_n\parallel \\ \to 0\text{as}n\to \mathrm{\infty .}\end{array}$$

Also, we have

$$\begin{array}{rll}\parallel T\left(s\right)t_n-t_n\parallel & \le & \parallel T\left(s\right)t_n-T\left(s\right)x_n\parallel \\ +\parallel T\left(s\right)x_n-x_n\parallel +\parallel x_n-t_n\parallel \\ \le & \parallel t_n-x_n\parallel +\parallel T\left(s\right)x_n-x_n\parallel \\ +\parallel x_n-t_n\parallel \\ \to 0\text{as}n\to \mathrm{\infty .}\end{array}$$

Next, we show that $lim\underset{n\to \infty }{sup}〈(B-\mathrm{\gamma f})z,x_n-z〉\le 0$, where z = P_Fix(S)∩Γ(I-B + γ f)z. To show this inequality, we choose a subsequence $\left\{t_{n_i}\right\}$ of {t _n }⊆K such that

$$lim\underset{n\to \infty }{sup}〈(B-\mathrm{\gamma f})z,t_n-z〉=\underset{i\to \infty }{lim}〈(B-\mathrm{\gamma f})z,t_{n_i}-z〉.$$

Since $\left\{t_{n_i}\right\}$ is bounded, there exists a subsequence $\left\{t_{n_{i_j}}\right\}$ of $\left\{t_{n_i}\right\}$ which converges weakly to some w∈C. Without loss of generality, we can assume that $t_{n_i}\rightharpoonup w$.

Now, we prove that w∈Fix(S)∩Γ. Let us first show that w∈Fix(S). Assume that w∉Fix(S). Since $t_{n_i}\rightharpoonup w$ and T(s)w≠w, from Opial’s condition (11), we have

$$\begin{array}{rll}lim\underset{i\to \infty }{inf}\parallel t_{n_i}-w\parallel & <& lim\underset{i\to \infty }{inf}\parallel t_{n_i}-T\left(s\right)w\parallel \\ \le & lim\underset{i\to \infty }{inf}\left\{\parallel t_{n_i}-T\left(s\right)t_{n_i}\parallel \right.\\ \left.+\parallel T\left(s\right)t_{n_i}-T\left(s\right)w\parallel \right\}\\ \le & lim\underset{i\to \infty }{inf}\parallel t_{n_i}-w\parallel ,\end{array}$$

which is a contradiction. Thus, we obtain w∈Fix(S).

Next, we show that w∈GEP(F₁,h₁). Since $u_n=T_{r_n}^{(F_1,h_1)}{x}_n$, we have

$$F_1(u_n,y)+h_1(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge 0,\forall y\in \mathrm{C.}$$

It follows from the monotonicity of F₁ that

$$h_1(u_n,y)+\frac{1}{r_n}〈y-u_n,u_n-x_n〉\ge F_1(y,u_n),$$

and hence,

$$h_1(u_{n_i},y)+〈y-u_{n_i},\frac{u_{n_i}-x_{n_i}}{r_n}〉\ge F_1(y,u_{n_i}).$$

Since ∥u _n -x _n ∥→0, we get $u_{n_i}\rightharpoonup w$ and $\frac{u_{n_i}-x_{n_i}}{r_n}\to 0$. It follows by Assumption 1 (iv) that 0≥F₁(y,w), ∀w∈C. For t with 0<t≤1 and y∈C, let y _t = t y + (1-t)w. Since y∈C, w∈C, we get y _t ∈C, and hence, F₁(y _t ,w)≤0. So, from Assumption 1 (i) and (iv), we have

$$\begin{array}{rl}0& =F_1(y_t,y_t)+h_1(y_t,y_t)\le t\left[F_1\right(y_t,y)+h_1(y_t,y\left)\right]\\ +(1-t)\left[F_1\right(y_t,w)+h_1(y_t,w\left)\right]\le t\left[F_1\right(y_t,y)+h_1(y_t,y\left)\right]\\ +(1-t)\left[F_1\right(w,y_t)+h_1(w,y_t\left)\right]\le \left[F_1\right(y_t,y)+h_1(y_t,y\left)\right].\end{array}$$

Therefore, 0≤F₁(y _t ,y) + h₁(y _t ,y). From Assumption 1 (iii), we have 0≤F₁(w,y) + h₁(w,y). This implies that w∈GEP(F₁,h₁).

Next, we show that A w∈GEP(F₂,h₂). Since $\parallel u_n-x_n\parallel \to 0,u_n\rightharpoonup w$ as n→∞ and {x _n } is bounded, there exists a subsequence $\left\{x_{n_k}\right\}$ of {x _n } such that $x_{n_k}\rightharpoonup w$, and since A is a bounded linear operator, so $A{x}_{n_k}\rightharpoonup \mathit{\text{Aw}}$.

Now, setting $v_{n_k}=A{x}_{n_k}-T_{r_{n_k}}^{F_2}A{x}_{n_k}$. It follows from (31) that $\underset{k\to \infty }{lim}{v}_{n_k}=0$ and $A{x}_{n_k}-v_{n_k}=T_{r_{n_k}}^{F_2}A{x}_{n_k}$.

Therefore, from Lemma 7, we have

$$\begin{array}{rl}{F}_2(A{x}_{n_k}-v_{n_k},z)& +h_2(A{x}_{n_k}-v_{n_k},z)\\ +\frac{1}{r_{n_k}}〈z-(A{x}_{n_k}-v_{n_k}),(A{x}_{n_k}-v_{n_k})\\ -A{x}_{n_k}〉\ge 0,\forall z\in \mathrm{Q.}\end{array}$$

Since F₂ and H₂ are upper semicontinuous in the first argument, taking lim sup to above inequality as k→∞ and using condition (iii), we obtain

$$F_2(\mathit{\text{Aw}},z)+h_2(\mathit{\text{Aw}},z)\ge 0,\forall z\in Q,$$

which means that A w∈GEP(F₂,h₂), and hence, w∈Γ.

Next, we claim that $lim\underset{n\to \infty }{sup}〈f\left(z\right)-z,x_n-z〉\le 0$, where z = P_Fix(S)∩Γ(I-B + γ f)z. Now, from (8), we have

$$\begin{array}{rl}lim\underset{n\to \infty }{sup}& 〈(B-\mathrm{\gamma f})z-z,x_n-z〉\\ =lim\underset{n\to \infty }{sup}〈(B-\mathrm{\gamma f})z-z,t_n-z〉\\ \le lim\underset{i\to \infty }{sup}〈(B-\mathrm{\gamma f})z-z,t_{n_i}-z〉\\ =〈(B-\mathrm{\gamma f})z-z,w-z〉\\ \le 0.\end{array}$$

(33)

Finally, we show that x _n →z:

$$\begin{array}{rll}\parallel x_{n+1}-z{\parallel }^2& =& \parallel {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n\\ +\left[\right(1-{\beta }_n)I-{\alpha }_{n}B]t_n-z{\parallel }^2\\ =& \parallel {\alpha }_n\left(\mathrm{\gamma f}\right(x_n)-\mathit{\text{Bz}})+{\beta }_n(x_n-z)\\ +\left[\right(1-{\beta }_n)I-{\alpha }_{n}B](t_n-z){\parallel }^2\\ \le & \parallel {\beta }_n(x_n-z)+\left[\right(1-{\beta }_n)I-{\alpha }_{n}B](t_n-z){\parallel }^2\\ +2{\alpha }_n〈\mathrm{\gamma f}\left(x_n\right)-\mathit{\text{Bz}},x_{n+1}-z〉\\ \le & \left[\parallel \left(\right(1-{\beta }_n)I-{\alpha }_{n}B)(t_n-z)\parallel \right.\\ {\left.+\parallel {\beta }_n(x_n-z)\parallel \right]}^2+2{\alpha }_n\gamma 〈f\left(x_n\right)\\ -f\left(z\right),x_{n+1}-z〉+2{\alpha }_n〈\mathrm{\gamma f}\left(z\right)\\ -\mathit{\text{Bz}},x_{n+1}-z〉\\ \le & {\left[\left.(1-{\beta }_n)-{\alpha }_n\bar{\gamma }\right)\parallel x_n-z\parallel +{\beta }_n\parallel x_n-z\parallel \right]}^2\\ +2{\alpha }_n\mathrm{\gamma \alpha }\parallel x_n-z\parallel \parallel x_{n+1}-z\parallel \\ +2{\alpha }_n〈\mathrm{\gamma f}\left(z\right)-\mathit{\text{Bz}},x_{n+1}-z〉\\ \le & {(1-{\alpha }_n\bar{\gamma })}^2\parallel x_n-z{\parallel }^2+{\alpha }_n\mathrm{\gamma \alpha }\{\parallel x_n-z{\parallel }^2\\ +\parallel x_{n+1}-z{\parallel }^2\}+2{\alpha }_n〈\mathrm{\gamma f}(z)-\mathit{\text{Bz}},x_{n+1}-z〉\\ \le & {(1-{\alpha }_n\bar{\gamma })}^2\parallel x_n-z{\parallel }^2+{\alpha }_n\mathrm{\gamma \alpha }\parallel x_n-z{\parallel }^2\\ +\mathrm{\gamma \alpha }\parallel x_{n+1}-z{\parallel }^2+2{\alpha }_n〈\mathrm{\gamma f}\left(z\right)\\ -\mathit{\text{Bz}},x_{n+1}-z〉.\end{array}$$

This implies that

$$\begin{array}{rll}\parallel x_{n+1}-z{\parallel }^2& \le & \frac{1-2{\alpha }_n\bar{\gamma }+{\left({\alpha }_n\bar{\gamma }\right)}^2+{\alpha }_n\mathrm{\gamma \alpha }}{1-\mathrm{\gamma \alpha }}\parallel x_n-z{\parallel }^2\\ +\frac{2{\alpha }_n}{1-\mathrm{\gamma \alpha }}〈\mathrm{\gamma f}\left(z\right)-\mathit{\text{Bz}},x_{n+1}-z〉\\ =& \left[1-\frac{2(\bar{\gamma }-\mathrm{\gamma \alpha }){\alpha }_n}{1-\mathrm{\gamma \alpha }}\right]\parallel x_n-z{\parallel }^2+\frac{{\left({\alpha }_n\bar{\gamma }\right)}^2}{1-\mathrm{\gamma \alpha }}\parallel x_n-z{\parallel }^2\\ +\frac{2{\alpha }_n}{1-\mathrm{\gamma \alpha }}〈\mathrm{\gamma f}\left(z\right)-\mathit{\text{Bz}},x_{n+1}-z〉\\ \le & \left[1-\frac{2(\bar{\gamma }-\mathrm{\gamma \alpha }){\alpha }_n}{1-\mathrm{\gamma \alpha }}\right]\parallel x_n-z{\parallel }^2+\frac{2(\bar{\gamma }-\mathrm{\gamma \alpha }){\alpha }_n}{1-\mathrm{\gamma \alpha }}\\ \times \left\{\frac{\left({\alpha }_n{\bar{\gamma }}^2\right)M}{2(\bar{\gamma }-\mathrm{\gamma \alpha })}+\frac{1}{\bar{\gamma }-\mathrm{\gamma \alpha }}〈\mathrm{\gamma f}\left(z\right)-\mathit{\text{Bz}},x_{n+1}-z〉\right\}\\ =& (1-{\delta }_n)\parallel x_n-z{\parallel }^2+{\delta }_n{\sigma }_n,\end{array}$$

where M:= sup{∥x _n -z∥²:n≥1}, ${\delta }_n=\frac{2(\bar{\gamma }-\mathrm{\gamma \alpha }){\alpha }_n}{1-\mathrm{\gamma \alpha }}$, and ${\sigma }_n=\frac{\left({\alpha }_n{\bar{\gamma }}^2\right)M}{2(\bar{\gamma }-\mathrm{\gamma \alpha })}+\frac{1}{\bar{\gamma }-\mathrm{\gamma \alpha }}〈\mathrm{\gamma f}\left(z\right)-\mathit{\text{Bz}},x_{n+1}-z〉.$ Since $\underset{n\to \infty }{lim}{\alpha }_n=0$ and $\sum _{n=0}^{\infty }{\alpha }_n=\infty$, it is easy to see that $\underset{n\to \infty }{lim}{\delta }_n=0$, $\sum _{n=0}^{\infty }{\delta }_n=\infty$, and $lim\underset{n\to \infty }{sup}{\sigma }_n\le 0$. Hence, from (33), (34), and Lemma 3, we deduce that x _n →z. This completes the proof.

We have the following consequences of Theorem 1.

Corollary 1

Let H₁ and H₂ be two real Hilbert spaces and let C⊆H₁ and Q⊆H₂ be nonempty closed convex subsets. Let A:H₁→H₂ be a bounded linear operator. Assume that $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are the bifunctions satisfying Assumption 1 and F₂ is upper semicontinuous in the first argument. Let S = {T(s):0≤s<∞} be a nonexpansive semigroup on C such that Fix(S)∩Ω≠∅. Let f:C→C be a contraction mapping with constant α∈(0,1) and B be a strongly positive linear bounded self-adjoint operator on H₁ with constant $\bar{\gamma }>0$ such that $0<\gamma <\frac{\bar{\gamma }}{\alpha }<\gamma +\frac{1}{\alpha }$. Let {s _n } is a positive real sequence which diverges to +∞. For a given x₀∈C arbitrarily, let the iterative sequences {u _n } and {x _n } be generated by

$$\begin{array}{rl}{u}_n=& T_{r_n}^{F_1}(x_n+\delta A^{\ast }(T_{r_n}^{F_2}-I\left)A{x}_n\right);\\ x_{n+1}=& {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n\\ +\left(\right(1-{\beta }_n)I-{\alpha }_{n}B)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}},\end{array}$$

where r _n ⊂(0,∞) and δ∈(0,1/L), L is the spectral radius of the operator A^∗A, and A^∗ is the adjoint of A, and {α _n } and {β _n } are the sequences in (0,1) satisfying the following conditions:

(i)

$$\underset{n\to \infty }{lim}{\alpha }_n=0$$

and

$$\sum _{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$$

(ii)

$$0<lim\underset{n\to \infty }{inf}{\beta }_n\le lim\underset{n\to \infty }{sup}{\beta }_n<1.$$

• (iii)
  $$lim\underset{n\to \infty }{inf}{r}_n>0$$

  ,

  $$\sum _{n=1}^{\infty }|r_{n+1}-r_n|<\mathrm{\infty .}$$

  (iv)

  $$\underset{n\to \infty }{lim}\frac{|s_{n+1}-s_n|}{s_{n+1}}=0$$

  .

Then, the sequence {x _n } converges strongly to z∈Fix(S)∩Ω, where z = P_Fix(S)∩Ω(I-B + γ f)z.

Proof

Taking h₁ = h₂ = 0 in Theorem 1, then the conclusion of Corollary 1 is obtained. □

Corollary 2

[6]Let H be a real Hilbert space and let C⊆H be a nonempty closed convex subset. Assume that$F:C\times C\to \mathbb{R}$is a bifunction satisfying Assumption 1 for F only. Let S = {T(s):0≤s<∞} be a nonexpansive semigroup on C such that Fix(S)∩EP(F)=∅. Let f:C→C be a contraction mapping with constant α∈(0,1) and B be a strongly positive linear bounded self-adjoint operator on H with constant$\bar{\gamma }>0$such that$0<\gamma <\frac{\bar{\gamma }}{\alpha }<\gamma +\frac{1}{\alpha }$. Let {s _n } is a positive real sequence which diverges to +∞. For a given x₀∈C arbitrarily, let the iterative sequences {u _n } and {x _n } be generated by

$$\begin{array}{rl}{u}_n=& T_{r_n}^F{x}_n;\\ x_{n+1}=& {\alpha }_n\mathrm{\gamma f}\left(x_n\right)\\ +(1-{\alpha }_{n}B)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)u_n\mathit{\text{ds}},\end{array}$$

where r _n ⊂(0,∞) and {α _n } is a sequence in (0,1) satisfying

• (i)
  $$\underset{n\to \infty }{lim}{\alpha }_n=0$$

  and

  $$\sum _{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$$
• (ii)
  $$lim\underset{n\to \infty }{inf}{r}_n>0$$

  ,

  $$\sum _{n=1}^{\infty }|r_{n+1}-r_n|<\mathrm{\infty .}$$

  (iii)

  $$\underset{n\to \infty }{lim}\frac{|s_{n+1}-s_n|}{s_{n+1}}=0$$

  .

Then, the sequence {x _n } converges strongly to z∈PFix(S)∩EP(F), where z = P_{Fix(S)∩EP(F)}(I-B + γ f)z.

Proof

Taking F₁ = F₂ = F, H₁ = H₂ = H, h₁ = h₂ = 0, {β _n } = 0, and A = 0 in Theorem 1, then the conclusion of Corollary 2 is obtained. □

Corollary 3

[4]Let H be a real Hilbert space and let C⊆H be a nonempty closed convex subset. Let S = {T(s):0≤s<∞} be a nonexpansive semigroup on C such that Fix(S)≠∅. Let f:C→C be a contraction mapping with constant α∈(0,1). Let {s _n } be a positive real sequence which diverges to +∞. For a given x₀∈C arbitrarily, let the iterative sequence {x _n } be generated by

$$\begin{array}{rll}{x}_{n+1}& =& {\alpha }_n\mathrm{\gamma f}\left(x_n\right)+{\beta }_n{x}_n\\ +(1-{\alpha }_n-{\beta }_n)\frac{1}{s_n}\underset{0}{\overset{s_n}{\int }}T\left(s\right)x_n\mathit{\text{ds}},\end{array}$$

where {α _n } and {β _n } are the sequences in (0,1) satisfying the following conditions:

• (i)
  $$\underset{n\to \infty }{\text{lim}}{\alpha }_n=0$$

  and

  $$\sum _{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$$

  (ii)

  $$0<lim\underset{n\to \infty }{inf}{\beta }_n\le lim\underset{n\to \infty }{sup}{\beta }_n<1.$$

  (iii)

  $$\underset{n\to \infty }{lim}\frac{|s_{n+1}-s_n|}{s_{n+1}}=0$$

  .

Then, the sequence {x _n } converges strongly to z∈Fix(S), where z = P_Fix(S)f(z).

Proof

Taking H₁ = H₂ = H, u _n = x _n , F₁ = F₂ = h₁ = h₂ = 0, and B = I in Theorem 1, then the conclusion of Corollary 3 is obtained. □

Results and discussion

We introduce and study an iterative method for approximating a common solution of split generalized equilibrium problem and fixed point problem for a nonexpansive semigroup in real Hilbert spaces. We obtain a strong convergence result for approximating a common solution of split generalized equilibrium problem and fixed point problem for a nonexpansive semigroup in real Hilbert spaces. Further, we obtain some consequences of our main result.

Conclusions

The results presented in this paper extend and generalize the works of Shimizu and Takahashi [2], Chen and Song [3], Plubtieng and Punpaeng [4], and Cianciaruso et al. [6]. The algorithm considered in Theorem 1 is different from those considered in [7–10] in the sense that the variable sequence {r _n } has been taken in place of fixed r. Further, the approach of the proof presented in this paper is also different. The use of the iterative method presented in this paper for the split monotone variational inclusions considered in Moudafi [9] needs further research effort.