Strong convergence theorem for amenable semigroups of nonexpansive mappings and variational inequalities

Abstract

In this paper, using strongly monotone and lipschitzian operator, we introduce a general iterative process for finding a common fixed point of a semigroup of nonexpansive mappings, with respect to strongly left regular sequence of means defined on an appropriate space of bounded real-valued functions of the semigroups and the set of solutions of variational inequality for β-inverse strongly monotone mapping in a real Hilbert space. Under suitable conditions, we prove the strong convergence theorem for approximating a common element of the above two sets.

Mathematics Subject Classification 2000: 47H09, 47H10, 43A07, 47J25

1 Introduction

Throughout this paper, we assume that H is a real Hilbert space with inner product and norm are denoted by 〈. , .〉 and || . ||, respectively, and let C be a nonempty closed convex subset of H. A mapping T of C into itself is called nonexpansive if || Tx - Ty ||≤|| x - y ||, for all x, y ∈ H. By Fix(T), we denote the set of fixed points of T (i.e., Fix(T) = {x ∈ H : Tx = x}), it is well known that Fix(T) is closed and convex. Recall that a self-mapping f : C → C is a contraction on C if there exists a constant α ∈ [0, 1) such that || f(x) - f(y) ||≤ α || x - y || for all x, y ∈ C.

Let B : C → H be a mapping. The variational inequality problem, denoted by VI(C, B), is to fined x ∈ C such that

$$〈Bx,y-x〉\ge 0,$$

(1)

for all y ∈ C. The variational inequality problem has been extensively studied in literature. See, for example, [1, 2] and the references therein.

Definition 1.1 Let B : C → H be a mapping. Then B

(1) is called η-strongly monotone if there exists a positive constant η such that

$$〈Bx-By,x-y〉\ge \eta \parallel x-y{\parallel }^2,\forall x,y\in C,$$

(2) is called k-Lipschitzian if there exist a positive constant k such that

$$\parallel Bx--By\parallel \le k\parallel x-y\parallel ,\forall x,y\in C,$$

(3) is called β-inverse strongly monotone if there exists a positive real number β > 0 such that

$$〈Bx-By,x-y〉\ge \beta \parallel Bx-By{\parallel }^2,\forall x,y\in C.$$

It is obvious that any β-inverse strongly monotone mapping B is$\frac{1}{\beta }$-Lipschitzian.

Moudafi [3] introduced the viscosity approximation method for fixed point of nonexpansive mappings (see [4] for further developments in both Hilbert and Banach spaces). Starting with an arbitrary initial x₀ ∈ H, define a sequence {x _n } recursively by

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

(2)

where α _n is sequence in (0, 1). Xu [4, 5] proved that under certain appropriate conditions on {α _n }, the sequences {x _n } generated by (2) strongly converges to the unique solution x* in Fix(T) of the variational inequality:

$$〈\left(f-I\right)x^{*},x-x^{*}〉\le 0,\forall x\in Fix\left(T\right).$$

Let A is strongly positive operator on C. That is, there is a constant $\bar{\gamma }>0$ with the property that

$$〈Ax,x〉\ge \bar{\gamma }\parallel x{\parallel }^2,\forall x\in C.$$

In [5], it is proved that the sequence {x _n } generated by the iterative method bellow with initial guess x₀ ∈ H chosen arbitrarily,

$$x_{n+1}=\left(I-{\alpha }_{n}A\right)T{x}_n+{\alpha }_{n}u,n\ge 0,$$

(3)

converges strongly to the unique solution of the minimization problem

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

where b is a given point in H.

Combining the iterative method (2) and (3), Marino and Xu [6] consider the following iterative method:

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

(4)

it is proved that if the sequence {α _n } of parameters satisfies the following conditions:

(C₁) α _n → 0,

(C₂) $\sum _{n=0}^{\infty }{\alpha }_n=\infty$,

C₃) either $\sum _{n=0}^{\infty }\mid {\alpha }_{n+1}-{\alpha }_n\mid <\infty$ or $\underset{n\to \infty }{lim}\frac{{\alpha }_{n+1}}{{\alpha }_n}=1$.

then, the sequence {x _n } generated by (4) converges strongly, as n → ∞, to the unique solution of the variational inequality:

$$〈\left(\gamma f-A\right)x^{*},x-x^{*}〉\le 0,\forall x\in Fix\left(T\right),$$

which is the optimality condition for minimization problem

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

where h is a potential function for γf (i.e., h'(x) = γf(x), for all x ∈ H). Some people also study the application of the iterative method (4) [7, 8].

Yamada [9] introduce the following hybrid iterative method for solving the variational inequality:

$$x_{n+1}=T{x}_n-\mu {\alpha }_{n}F\left(T{x}_n\right),n\in \mathbb{N},$$

(5)

where F is k-Lipschitzian and η-strongly monotone operator with k > 0, η > 0, $0<\mu <\frac{2\eta }{k^2}$, then he proved that if {α _n } satisfying appropriate conditions, then {x _n } generated by (5) converges strongly to the unique solution of the variational inequality:

$$〈F{x}^{*},x-x^{*}〉\ge 0,\forall x\in Fix\left(T\right).$$

In 2010, Tian [10] combined the iterative (4) with the iterative method (5) and considered the iterative methods:

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

(6)

and he prove that if the sequence {α _n } of parameters satisfies the conditions (C₁), (C₂), and (C₃), then the sequences {x _n } generated by (6) converges strongly to the unique solution x* ∈ Fix(T) of the variational inequality:

$$〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\forall x\in Fix\left(T\right).$$

In this paper motivated and inspired by Atsushiba and Takahashi [11], Ceng and Yao [12], Kim [13], Lau et al. [14], Lau et al [15], Marino and Xu [6], Piri and Vaezi [16], Tian [10], Xu [5] and Yamada [9], we introduce the following general iterative algorithm: Let x₀ ∈ C and

$$\left\{\begin{array}{c}{y}_n={\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}B{x}_n\right),\\ x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n,n\ge 0.\end{array}\right.$$

(7)

where P _C is a metric projection of H onto C, B is β-inverse strongly monotone, φ = {T _t : t ∈ S} is a nonexpansive semigroup on H such that the set $\mathcal{F}=Fix\left(\phi \right)\cap VI\left(C,B\right)\ne \varnothing ,$, X is a subspace of B(S) such that 1 ∈ X and the mapping t → 〈T _t x, y〉 is an element of X for each x, y ∈ H, and {μ _n } is a sequence of means on X. Our purpose in this paper is to introduce this general iterative algorithm for approximating a common element of the set of common fixed point of a semigroup of nonexpansive mappings and the set of solutions of variational inequality for β-inverse strongly monotone mapping which solves some variational inequality. We will prove that if {μ _n } is left regular and the sequences {α _n }, {β _n }, and {δ _n } of parameters satisfies appropriate conditions, then the sequences {x _n } and {y _n } generated by (7) converges strongly to the unique solution $x^{*}\in \mathcal{F}$ of the variational inequalities:

$$\left\{\begin{array}{cc}〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\hfill & \forall x\in \mathcal{F},\hfill \\ \hfill 〈B{x}^{*},y-x^{*}〉\ge 0\hfill & \hfill \forall y\in C.\hfill \end{array}\right.$$

2 Preliminaries

Let S be a semigroup and let B(S) be the space of all bounded real-valued functions defined on S with supremum norm. For s ∈ S and f ∈ B(S), we define elements l _s f and r _s f in B(S) by

$$\left(l_{s}f\right)\left(t\right)=f\left(st\right),\left(r_{s}f\right)\left(t\right)=f\left(ts\right),\forall t\in S.$$

Let X be a subspace of B(S) containing 1 and let X* be its topological dual. An element μ of X* is said to be a mean on X if || μ || = μ(1) = 1. We often write μ _t (f(t)) instead of μ(f) for μ ∈ X* and f ∈ X. Let X be left invariant (resp. right invariant), i.e., l _s (X) ⊂ X (resp. r _s (X) ⊂ X) for each s ∈ S. A mean μ on X is said to be left invariant (resp. right invariant) if μ(l _s f) = μ(f) (resp. μ(r _s f) = μ(f)) for each s ∈ S and f ∈ X. X is said to be left (resp. right) amenable if X has a left (resp. right) invariant mean. X is amenable if X is both left and right amenable. As is well known, B(S) is amenable when S is a commutative semigroup, see [15]. A net {μ _α } of means on X is said to be strongly left regular if

$$\underset{\alpha }{lim}\parallel {l_s}^{*}{\mu }_{\alpha }-{\mu }_{\alpha }\parallel =0,$$

for each s ∈ S, where $l_s^{*}$ is the adjoint operator of l _s .

Let S be a semigroup and let C be a nonempty closed and convex subset of a reflexive Banach space E. A family φ = {T _t : t ∈ S} of mapping from C into itself is said to be a nonexpansive semigroup on C if T _t is nonexpansive and T _ts = T _t T _s for each t, s ∈ S. By Fix(φ), we denote the set of common fixed points of φ, i.e.,

$$Fix\left(\phi \right)=\bigcap _{t\in S}\left\{x\in C:T_{t}x=x\right\}.$$

Lemma 2.1[15]Let S be a semigroup and C be a nonempty closed convex subset of a reflexive Banach space E. Let φ = {T _t : t ∈ S} be a nonexpansive semigroup on H such that {T _t x : t ∈ S} is bounded for some x ∈ C, let X be a subspace of B(S) such that 1 ∈ X and the mapping t → 〈T _t x, y*〉 is an element of X for each x ∈ C and y* ∈ E*, and μ is a mean on X. If we write T _μ x instead of$\int T_{t}xd\mu \left(t\right)$, then the followings hold.

(i) T _μ is non-expansive mapping from C into C.

(ii) T _μ x = x for each x ∈ Fix(φ).

(iii)$T_{\mu }x\in \stackrel{}{co}\left\{T_{t}x:t\in S\right\}$for each x ∈ C.

Let C be a nonempty subset of a Hilbert space H and T : C → H a mapping. Then T is said to be demiclosed at v ∈ H if, for any sequence {x _n } in C, the following implication holds:

$$x_n\to u\in C,T{x}_n\to vimplyTu=v,$$

where → (resp. ⇀) denotes strong (resp. weak) convergence.

Lemma 2.2[17]Let C be a nonempty closed convex subset of a Hilbert space H and suppose that T : C → H is nonexpansive. Then, the mapping I - T is demiclosed at zero.

Lemma 2.3[18]For a given x ∈ H, y ∈ C,

$$y=P_{C}x\iff 〈y-x,z-y〉\ge 0,\forall z\in C.$$

It is well known that P _C is a firmly nonexpansive mapping of H onto C and satisfies

$$\parallel P_{C}x-P_{C}y{\parallel }^2\le 〈P_{C}x-P_{C}y,x-y〉,\forall x,y\in H$$

(8)

Moreover, P _C is characterized by the following properties: P _C x ∈ C and for all x ∈ H, y ∈ C,

$$〈x-P_{C}x,y-P_{C}x〉\le 0.$$

(9)

It is easy to see that (9) is equivalent to the following inequality

$$\parallel x-y{\parallel }^2\ge \parallel x-P_{C}x{\parallel }^2+\parallel y-P_{C}x{\parallel }^2.$$

(10)

Using Lemma 2.3, one can see that the variational inequality (24) is equivalent to a fixed point problem.

It is easy to see that the following is true:

$$u\in VI\left(C,B\right)\iff u=P_C\left(u-\lambda Bu\right),\lambda >0.$$

(11)

A set-valued mapping U : H → 2 ^H is called monotone if for all x, y ∈ H, f ∈ Ux and g ∈ Uy imply 〈x - y, f - g〉 ≥ 0. A monotone mapping U : H → 2 ^H is maximal if the graph of G(U) of U is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping U is maximal if and only if for (x, f) ∈ H × H, 〈x - y, f - g〉 ≥ 0 for every (y, g) ∈ G(U) implies that f ∈ Ux. Let B be a monotone mapping of C into H and let N _C x be the normal cone to C at x ∈ C, that is, N _C x = {y ∈ H : 〈z - x, y〉 ≤ 0, ∀z ∈ C} and define

$$Ux=\left\{\begin{array}{cc}\hfill Bx+N_{C}x,\hfill & \hfill x\in C,\hfill \\ \hfill \varnothing \hfill & \hfill x\notin C.\hfill \end{array}\right.$$

(12)

Then U is the maximal monotone and 0 ∈ Ux if and only if x ∈ VI(C, B); see [19].

The following lemma is well known.

Lemma 2.4 Let H be a real Hilbert space. Then, for all x, y ∈ H

$$\parallel x-y{\parallel }^2\le \parallel x{\parallel }^2+2〈y,x+y〉,.$$

Lemma 2.5[5]Let {a _n } be a sequence of nonnegative real numbers such that

$$a_{n+1}\le \left(1-b_n\right)a_n+b_n{c}_n,n\ge 0,$$

where {b _n } and {c _n } are sequences of real numbers satisfying the following conditions:

(i) {b _n } ⊂ (0, 1), $\sum _{n=0}^{\infty }{b}_n=\infty$,

(ii) either$\underset{n\to \infty }{limsup}{c}_n\le 0$or$\sum _{n=0}^{\infty }\mid b_n{c}_n\mid <\infty$.

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

As far as we know, the following lemma has been used implicitly in some papers; for the sake of completeness, we include its proof.

Lemma 2.6 Let H be a real Hilbert space and F be a k-Lipschitzian and η-strongly monotone operator with k > 0, η > 0. Let$0<\mu <\frac{2\eta }{k^2}$and$\tau =\mu \left(\eta -\frac{\mu k^2}{2}\right)$. Then for$t\in \left(0,min\left\{1,\frac{1}{\tau }\right\}\right)$, I - t _μ F is contraction with constant 1 - t τ.

Proof. Notice that

$$\begin{array}{c}\parallel \left(I-t\mu F\right)x-\left(I-t\mu F\right)y{\parallel }^2\\ =〈\left(I-t\mu F\right)x-\left(I-t\mu F\right)y,\left(I-t\mu F\right)x-\left(I-t\mu F\right)y〉\\ =\parallel x-y{\parallel }^2+t^2{\mu }^2\parallel Fx-Fy{\parallel }^2-2t\mu 〈x-y,Fx-Fy〉\\ \le \parallel x-y{\parallel }^2+t^2{\mu }^2{k}^2\parallel x-y{\parallel }^2-2t\mu \eta \parallel x-y{\parallel }^2\\ \le \parallel x-y{\parallel }^2+t{\mu }^2{k}^2\parallel x-y{\parallel }^2-2t\mu \eta \parallel x-y{\parallel }^2\\ =\left(1-2t\mu \left(\eta -\frac{\mu k^2}{2}\right)\right)\parallel x-y{\parallel }^2\\ =\left(1-2t\tau \right)\parallel x-y{\parallel }^2\\ \le {\left(1-t\tau \right)}^2\parallel x-y{\parallel }^2.\end{array}$$

It follows that

$$\parallel \left(I-t\mu F\right)x-\left(I-t\mu F\right)y\parallel \le \left(1-t\tau \right)\parallel x-y\parallel ,$$

and hence I - tμF is contractive due to 1 - tτ ∈ (0, 1). □

Notation Throughout the rest of this paper, F will denote a k-Lipschitzian and η-strongly monotone operator of C into H with k > 0, η > 0, f is a contraction on C with coefficient 0 < α < 1. We will also always use γ to mean a number in $\left(0,\frac{\tau }{\alpha }\right)$, where $\tau =\mu \left(\eta -\frac{\mu k^2}{2}\right)$ and $0<\mu <\frac{2\eta }{k^2}$. The open ball of radius r centered at 0 is denoted by B _r and for a subset D of H, by $\stackrel{}{co}D$, we denote the closed convex hull of D. For ε > 0 and a mapping T : D → H, we let F _ε (T; D) be the set of ε-approximate fixed points of T, i.e., F _ε (T; D) = {x ∈ D : ||x - T _x || ≤ ε }. Weak convergence is denoted by ⇀ and strong convergence is denoted by →.

3 Main results

Theorem 3.1 Let S be a semigroup, C a nonempty closed convex subset of real Hilbert space H and B : C → H be a β-inverse strongly monotone. Let φ = {T _t : t ∈ S} be a nonexpansive semigroup of C into itself such that$\mathcal{F}=Fix\left(\phi \right)\cap VI\left(C,B\right)\ne \varnothing ,$, X a left invariant subspace of B(S) such that 1 ∈ X, and the function t → 〈T _t x, y〉 is an element of X for each x ∈ C and y ∈ H, {μ _n } a left regular sequence of means on X such that${\sum }_{n=1}^{\infty }\parallel {\mu }_{n+1}-{\mu }_n\parallel <\infty$. Let {α _n } and {β _n } be sequences in (0, 1) and {δ _n } be a sequence in [a, b], where 0 < a < b < 2β. Suppose the following conditions are satisfied.

(B₁) lim_n→∞α _n = 0, lim_n→∞β _n = 0,

(B₂) ${\sum }_{n=1}^{\infty }{\alpha }_n=\infty$,

(B₃) ${\sum }_{n=1}^{\infty }\mid {\alpha }_{n+1}-{\alpha }_n\mid <\infty$, ${\sum }_{n=1}^{\infty }\mid {\beta }_{n+1}-{\beta }_n\mid <\infty$, ${\sum }_{n=1}^{\infty }\mid {\delta }_{n+1}-{\delta }_n\mid <\infty$.

If {x _n } and {y _n } be generated by x₀ ∈ C and

$$\left\{\begin{array}{c}{y}_n={\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}B{x}_n\right),\\ x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n,n\ge 0.\end{array}\right.$$

Then, {x _n } and {y _n } converge strongly, as n → ∞, to$x^{*}\in \mathcal{F}$, which is a unique solution of the variational inequalities:

$$\left\{\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\hfill & \hfill \forall x\in \mathcal{F},\hfill \\ \hfill 〈B{x}^{*},y-x^{*}〉\ge 0\hfill & \hfill \forall y\in C.\hfill \end{array}\right.$$

Proof. Since {α _n } satisfies in condition (B₁), we may assume, with no loss of generality, that ${\alpha }_n<min\left\{1,\frac{1}{\tau }\right\}$. Since B is β-inverse strongly monotone and δ _n < 2β, for any x, y ∈ C, we have

$$\begin{array}{l}\left|\right|(I-{\delta }_{n}B)x-(I-{\delta }_{n}B)y|{|}^2\\ =\left|\right|(x-y)-{\delta }_n(Bx-By)|{|}^2\\ =\left|\right|x-y|{|}^2-2{\delta }_n〈x-y\mathrm{,}Bx-By〉+{\delta }_n^2||Bx-By|{|}^2\\ \le \left|\right|x-y|{|}^2-2{\delta }_n\beta ||Bx-By|{|}^2+{\delta }_n^2\left|\right|Bx-By|{|}^2\\ =\left|\right|x-y|{|}^2+{\delta }_n({\delta }_n-2\beta )||Bx-By|{|}^2\\ \le \left|\right|x-y|{|}^2.\end{array}$$

It follows that

$$\parallel \left(I-{\delta }_{n}B\right)x-\left(I-{\delta }_{n}B\right)y\parallel \le \parallel x-y\parallel .$$

(13)

Let $p\in \mathcal{F}$, in the context of the variational inequality problem, the characterization of projection (11) implies that p = P _C (p - δ _n B _p ). Using (13), we get

$$\begin{array}{ll}\hfill \parallel y_n-p\parallel & =\parallel {\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}B{x}_n\right)-p\parallel \\ =\parallel {\beta }_n\left[x_n-p\right]+\left(1-{\beta }_n\right)\left[P_C\left(x_n-{\delta }_{n}B{x}_n\right)-P_C\left(p-{\delta }_{n}Bp\right)\right]\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +\left(1-{\beta }_n\right)\parallel P_C\left(x_n-{\delta }_{n}B{x}_n\right)-P_C\left(p-{\delta }_{n}Bp\right)\parallel \\ \le {\beta }_n\parallel x_n-p\parallel +\left(1-{\beta }_n\right)\parallel x_n-p\parallel =\parallel x_n-p\parallel .\end{array}$$

(14)

We claim that {x _n } is bounded. Let $p\in \mathcal{F}$, using Lemma 2.6 and (14), we have

$$\begin{array}{ll}\hfill \parallel x_{n+1}-p\parallel & =\parallel {\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-p\parallel \\ =\parallel {\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-\left(I-{\alpha }_n\mu F\right)p-{\alpha }_n\mu Fp\parallel \\ \le {\alpha }_n\parallel \gamma f\left(x_n\right)-\mu Fp\parallel +\parallel \left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-\left(I-{\alpha }_n\mu F\right)p\parallel \\ \le {\alpha }_n\parallel \gamma f\left(x_n\right)-\gamma f\left(p\right)\parallel \\ +{\alpha }_n\parallel \gamma f\left(p\right)-\mu Fp\parallel +\left(1-{\alpha }_n\tau \right)\parallel T_{{\mu }_n}{y}_n-p\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f\left(p\right)-\mu Fp\parallel +\left(1-{\alpha }_n\tau \right)\parallel y_n-p\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f\left(p\right)-\mu Fp\parallel +\left(1-{\alpha }_n\tau \right)\parallel x_n-p\parallel \\ =\left(1-{\alpha }_n\left(\tau -\gamma \alpha \right)\right)\parallel x_n-p\parallel +{\alpha }_n\parallel \gamma f\left(p\right)-\mu Fp\parallel \\ \le max\left\{\parallel x_n-p\parallel ,{\left(\tau -\gamma \alpha \right)}^{-1}\parallel \gamma f\left(p\right)-\mu Fp\parallel \right\}.\end{array}$$

By induction we have,

$$\left|\right|x_n-p\left|\right|\le \max \{(\tau -\gamma \alpha {)}^{-1}\left|\right|\gamma f(p)-\mu Fp\left|\right|\mathrm{,}\left|\right|x_0-p\left|\right|\}=M_0.$$

Hence, {x _n } is bounded and also {y _n } and {f(x _n )} are bounded. Set D = {y ∈ H : ||y - p||≤ = M₀}. We remark that D is φ-invariant bounded closed convex set and {x _n }, {y _n } ⊂ D. Now we claim that

$$\underset{n\to \infty }{limsup}\underset{y\in D}{sup}\parallel T_{{\mu }_n}y-T_t{T}_{{\mu }_n}y\parallel =0,\forall t\in S.$$

(15)

Let ε > 0. By [[20], Theorem 1.2], there exists δ > 0 such that

$$\stackrel{}{co}{F}_{\delta }\left(T_t;D\right)+B_{\delta }\subset F_{\epsilon }\left(T_t;D\right),\forall t\in S.$$

(16)

Also by [[20], Corollary 1.1], there exists a natural number N such that

$$∥\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y-T_t\left(\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}y\right)∥\le \delta ,$$

(17)

for all t, s ∈ S and y ∈ D. Let t ∈ S. Since {μ _n } is strongly left regular, there exists N₀ ∈ ℕ such that $\parallel {\mu }_n-l_{t^i}^{*}{\mu }_n\parallel \le \frac{\delta }{\left(M_0+\parallel p\parallel \right)}$ for n ≥ N₀ and i = 0, 1, 2,..., N. Then we have

$$\begin{array}{l}\underset{y\in D}{\sup }\Vert T_{{\mu }_n}y-{\displaystyle \int }\frac{1}{N+1}{\displaystyle \sum _{i=0}^N{T}_{t^{i}s}y\text{d}{\mu }_{n}s}‱\\ =\underset{y\in D}{{\displaystyle \sup }}\underset{\parallel z\parallel =1}{{\displaystyle \sup }}\left|〈T_{{\mu }_n}y\mathrm{,}z〉-〈{\displaystyle \int }\frac{1}{N+1}{\displaystyle \sum _{i=0}^N}{T}_{t^{i}s}y\text{d}{\mu }_{n}s\mathrm{,}z〉\right|\\ =\underset{y\in D}{{\displaystyle \sup }}\underset{\parallel z\parallel =1}{{\displaystyle \sup }}\left|\frac{1}{N+1}{\displaystyle \sum _{i=0}^N}{({\mu }_n)}_s〈T_{s}y\mathrm{,}z〉-\frac{1}{N+1}{\displaystyle \sum _{i=0}^N}{({\mu }_n)}_s〈T_{t^{i}s}y\mathrm{,}z〉\right|\\ \le \frac{1}{N+1}{\displaystyle \sum _{i=0}^N}\underset{y\in D}{{\displaystyle \sup }}\underset{\parallel z\parallel =1}{{\displaystyle \sup }}|({\mu }_n{)}_s〈T_{s}y\mathrm{,}z〉-{(l_{t^i}^{*}{\mu }_n)}_s〈T_{s}y\mathrm{,}z〉|\\ \le \underset{i=\mathrm{1,2,}\dots N}{\text{max}}\parallel {\mu }_n-l_{t^i}^{*}{\mu }_n\parallel (M_0+\parallel p\parallel )\le \delta \mathrm{,}\forall n\ge N_0.\end{array}$$

(18)

By Lemma 2.1, we have

$$\int \frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}yd{\mu }_{n}s\in \stackrel{}{co}\left\{\frac{1}{N+1}\sum _{i=0}^N{T}_{t^i}\left(T_{s}y\right):s\in s\right\}.$$

(19)

It follows from (16), (17), (18), and (19) that

$$\begin{array}{ll}\hfill T_{{\mu }_n}\left(y\right)& \in \stackrel{}{co}\left\{\frac{1}{N+1}\sum _{i=0}^N{T}_{t^{i}s}\left(y\right):s\in S\right\}+B_{\delta }\\ \subset \stackrel{}{co}{F}_{\delta }\left(T_t;D\right)+B_{\delta }\subset F_{\epsilon }\left(T_t;D\right),\end{array}$$

for all y ∈ D and n ≥ N₀. Therefore,

$$\underset{n\to \infty }{\mathrm{lim\; sup}}\underset{y\in D}{sup}\parallel T_t\left(T_{{\mu }_n}y\right)-T_{{\mu }_n}y\parallel \le \epsilon .$$

Since ε > 0 is arbitrary, we get (15). In this stage, we will show

$$\underset{n\to \infty }{lim}\parallel x_n-T_t{x}_n\parallel =0,\forall t\in S.$$

(20)

Let t ∈ S and ε > 0. Then, there exists δ > 0, which satisiies (16). From lim_n→∞α _n = 0 and (15) there exists N₁ ∈ ℕ such that ${\alpha }_n\le \frac{\delta }{\left(\tau +\mu k\right)M_0}$ and $T_{{\mu }_n}{y}_n\in F_{\delta }\left(T_t;D\right)$, for all n ≥ N₁. By Lemma 2.6 and (14), we have

$$\begin{array}{ll}\hfill {\alpha }_n\parallel & \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n\parallel \\ \le {\alpha }_n\left(\gamma \parallel f\left(x_n\right)-f\left(p\right)\parallel +\parallel \gamma f\left(p\right)-\mu Fp\parallel +\parallel \mu Fp-\mu F{T}_{{\mu }_n}{y}_n\parallel \right)\\ \le {\alpha }_n\left(\gamma \alpha \parallel x_n-p\parallel +\parallel \gamma f\left(p\right)-\mu Fp\parallel +\mu k\parallel y_n-p\parallel \right)\\ \le {\alpha }_n\left(\gamma \alpha M_0+\left(\tau -\gamma \alpha \right)M_0+\mu k{M}_0\right)\\ \le {\alpha }_n\left(\tau +\mu k\right)M_0\le \delta ,\end{array}$$

for all n ≥ N₁. Therefore, we have

$$\begin{array}{ll}\hfill x_{n+1}& =T_{{\mu }_n}{y}_n+{\alpha }_n\left[\gamma f\left(x_n\right)+\mu F\left(T_{{\mu }_n}{y}_n\right)\right]\\ \in F_{\delta }\left(T_t;D\right)+B_{\delta }\subset F_{\epsilon }\left(T_t;D\right),\end{array}$$

for all n ≥ N₁. This shows that

$$\parallel x_n-T_t{x}_n\parallel \le \epsilon ,\forall n\ge N_1.$$

Since ε > 0 is arbitrary, we get (20).

Let

$Q=P_{\mathcal{F}}$. Then Q(I - μF + γ f) is a contraction of H into itself. In fact, we see that

$$\begin{array}{ll}\hfill \parallel Q& \left(I-\mu F+\gamma f\right)x-Q\left(I-\mu F+\gamma f\right)y\parallel \\ \le \parallel \left(I-\mu F+\gamma f\right)x-\left(I-\mu F+\gamma f\right)y\parallel \\ \le \parallel \left(I-\mu F\right)x-\left(I-\mu F\right)y\parallel +\gamma \parallel f\left(x\right)-f\left(y\right)\parallel \\ =\underset{n\to \infty }{lim}∥\left(I-\left(1-\frac{1}{n}\right)\mu F\right)x-\left(I-\left(1-\frac{1}{n}\right)\mu F\right)y∥+\gamma \parallel f\left(x\right)-f\left(y\right)\parallel \\ \le \underset{n\to \infty }{lim}\left(1-\left(1-\frac{1}{n}\right)\tau \right)\parallel x-y\parallel +\gamma \alpha \parallel x-y\parallel \\ =\left(1-\tau \right)\parallel x-y\parallel +\gamma \alpha \parallel x-y\parallel ,\end{array}$$

and hence Q(I - μF + γ f) is a contraction due to (1 - (τ -γα)) ∈ (0, 1).

Therefore, by Banachs contraction principal, $P_{\mathcal{F}}\left(I-\mu F+\gamma f\right)$ has a unique fixed point x*. Then using (9), x* is the unique solution of the variational inequality:

$$〈\left(\gamma f-\mu F\right)x^{*},x-x^{*}〉\le 0,\forall x\in \mathcal{F}.$$

(21)

We show that

$$\underset{n\to \infty }{limsup}〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_n-x^{*}〉\le 0.$$

(22)

Indeed, we can choose a subsequence $\left\{x_{n_i}\right\}$ of {x _n } such that

$$\underset{n\to \infty }{limsup}〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_n-x^{*}〉=\underset{i\to \infty }{lim}〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n_i}-x^{*}〉.$$

(23)

Because $\left\{x_{n_i}\right\}$ is bounded, we may assume that $x_{n_i}\to z$. In terms of Lemma 2.2 and (20), we conclude that z ∈ Fix (φ).

Now, let us show that z ∈ VI (C, B). Let w _n = P _C (x _n - δ _n Bx _n ), it follows from the definition of {y _n } that

$$\begin{array}{c}\parallel y_{n+1}-y_n\parallel \\ =\parallel {\beta }_{n+1}{x}_{n+1}+(1-{\beta }_{n+1})P_C(x_{n+1}-{\delta }_{n+1}B{x}_{n+1})-{\beta }_n{x}_n-(1-{\beta }_n)P_C(x_n-{\delta }_{n}B{x}_n)\parallel \\ =\parallel {\beta }_{n+1}(x_{n+1}-x_n)+({\beta }_{n+1}-{\beta }_n)x_n+(1-{\beta }_{n+1})P_C(x_{n+1}-{\delta }_{n+1}B{x}_{n+1})\\ -(1-{\beta }_{n+1})P_C(x_n-{\delta }_{n+1}B{x}_n)+(1-{\beta }_{n+1})P_C(x_n-{\delta }_{n+1}B{x}_n)-(1-{\beta }_n)P_C(x_n-{\delta }_{n}B{x}_n)\parallel \\ \le {\beta }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel x_n\parallel \\ +(1-{\beta }_{n+1})\parallel P_C(x_{n+1}-{\delta }_{n+1}B{x}_{n+1})-P_C(x_n-{\delta }_{n+1}B{x}_n)\parallel \\ +\parallel P_C(x_n-{\delta }_{n+1}B{x}_n)-P_C(x_n-{\delta }_{n}B{x}_n)\parallel \\ +\parallel {\beta }_n{P}_C(x_n-{\delta }_{n}B{x}_n)-{\beta }_{n+1}{P}_C(x_n-{\delta }_{n+1}B{x}_n)\parallel ]\\ \le {\beta }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel x_n\parallel +(1-{\beta }_{n+1})\parallel x_{n+1}-x_n\parallel \\ +|{\delta }_{n+1}-{\delta }_n|\parallel B{x}_n\parallel +\parallel {\beta }_n{P}_C(x_n-{\delta }_{n}B{x}_n)-{\beta }_n{P}_C(x_n-{\delta }_{n+1}B{x}_n)\\ +{\beta }_n{P}_C(x_n-{\delta }_{n+1}B{x}_n)-{\beta }_{n+1}{P}_C(x_n-{\delta }_{n+1}B{x}_n)\parallel \\ \le {\beta }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel x_n\parallel +(1-{\beta }_{n+1})\parallel x_{n+1}-x_n\parallel \\ +|{\delta }_{n+1}-{\delta }_n|\parallel B{x}_n\parallel +{\beta }_n|{\delta }_{n+1}-{\delta }_n|\parallel B{x}_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel P_C(x_n-{\delta }_{n+1}B{x}_n)\parallel \\ =\parallel x_{n+1}-x_n\parallel +|{\beta }_{n+1}-{\beta }_n|\parallel x_n\parallel +(1+{\beta }_n)|{\delta }_{n+1}-{\delta }_n|\parallel B{x}_n\parallel \\ +|{\beta }_{n+1}-{\beta }_n|\left|\right|P_C(x_n-{\delta }_{n+1}B{x}_n)\left|\right|.\end{array}$$

Using the last inequality, we get

$$\begin{array}{ll}\hfill \parallel & x_{n+1}-x_n\parallel \\ =\parallel {\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-{\alpha }_{n-1}\gamma f\left(x_{n-1}\right)-\left(I-{\alpha }_{n-1}\mu F\right)T_{{\mu }_{n-1}}{y}_{n-1}\parallel \\ =\parallel {\alpha }_n\gamma f\left(x_n\right)-{\alpha }_n\gamma f\left(x_{n-1}\right)+\left({\alpha }_n-{\alpha }_{n-1}\right)\gamma f\left(x_{n-1}\right)\\ +\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-\left(I-{\alpha }_n\mu F\right)T_{{\mu }_{n-1}}{y}_{n-1}\\ +\left(I-{\alpha }_n\mu F\right)T_{{\mu }_{n-1}}{y}_{n-1}-\left(I-{\alpha }_{n-1}\mu F\right)T_{{\mu }_{n-1}}{y}_{n-1}\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-x_{n-1}\parallel +\mid {\alpha }_n-{\alpha }_{n-1}\mid \gamma \parallel f\left(x_{n-1}\right)\parallel \\ +\left(1-{\alpha }_n\tau \right)\parallel T_{{\mu }_n}{y}_n-T_{{\mu }_{n-1}}{y}_{n-1}\parallel +\mid {\alpha }_n-{\alpha }_{n-1}\mid \mu \parallel F{T}_{{\mu }_{n-1}}{y}_{n-1}\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-x_{n-1}\parallel +\mid {\alpha }_n-{\alpha }_{n-1}\mid \gamma \parallel f\left(x_{n-1}\right)\parallel +\left(1-{\alpha }_n\tau \right)\parallel y_n-y_{n-1}\parallel \\ +\left(1-{\alpha }_n\tau \right)\parallel T_{{\mu }_n}{y}_{n-1}-T_{{\mu }_{n-1}}{y}_{n-1}\parallel +\mid {\alpha }_n-{\alpha }_{n-1}\mid \mu \parallel F{T}_{{\mu }_{n-1}}{y}_{n-1}\parallel \\ \le {\alpha }_n\gamma \alpha \parallel x_n-x_{n-1}\parallel +\mid {\alpha }_n-{\alpha }_{n-1}\mid \gamma \parallel f\left(x_{n-1}\right)\parallel +\left(1-{\alpha }_n\tau \right)\parallel x_n-x_{n-1}\parallel \\ +\left(1-{\alpha }_n\tau \right)\mid {\beta }_n-{\beta }_{n-1}\mid \parallel x_{n-1}\parallel +\left(1-{\alpha }_n\tau \right)\left(1+{\beta }_{n-1}\right)\mid {\delta }_n-{\delta }_{n-1}\mid \parallel B{x}_{n-1}\parallel \\ +\left(1-{\alpha }_n\tau \right)\mid {\beta }_n-{\beta }_{n-1}\mid \parallel P_C\left(x_{n-1}-{\delta }_{n}B{x}_{n-1}\right)\parallel \\ +\left(1-{\alpha }_n\tau \right)\parallel T_{{\mu }_n}{y}_{n-1}-T_{{\mu }_{n-1}}{y}_{n-1}\parallel +\mid {\alpha }_n-{\alpha }_{n-1}\mid \mu \parallel F{T}_{{\mu }_{n-1}}{y}_{n-1}\parallel .\end{array}$$

Thus, for some large enough constant M > 0, we have

$$\begin{array}{ll}\hfill \parallel x_{n+1}-x_n\parallel & \le \left(1-{\alpha }_n\left(\tau -\gamma \alpha \right)\right)\parallel x_n-x_{n-1}\parallel \\ +\left[\mid {\alpha }_n-{\alpha }_{n-1}\mid +\mid {\beta }_n-{\beta }_{n-1}\mid +\mid {\delta }_n-{\delta }_{n-1}\mid +\parallel {\mu }_n-{\mu }_{n-1}\parallel \right]M.\end{array}$$

Therefore, using condition B₃ and Lemma 2.5, we get

$$\underset{n\to \infty }{lim}\parallel x_{n+1}-x_n\parallel =0.$$

(24)

Let $p\in \mathcal{F}$, from (11) and deiinition of {y _n }, we have

$$\begin{array}{c}\parallel y_n-p{\parallel }^2\\ =\parallel {\beta }_n{x}_n+(1-{\beta }_n)P_C(x_n-{\delta }_{n}B{x}_n)-p{\parallel }^2\\ =\parallel {\beta }_n(x_n-p)+(1-{\beta }_n)(P_C(x_n-{\delta }_{n}B{x}_n)-P_C(p-{\delta }_{n}Bp)){\parallel }^2\\ \le {\beta }_n\parallel x_n-p{\parallel }^2+(1-{\beta }_n)\parallel (x_n-p)-{\delta }_n(B{x}_n-Bp)){\parallel }^2\\ ={\beta }_n\parallel x_n-p{\parallel }^2+(1-{\beta }_n)\parallel x_n-p{\parallel }^2+{\delta }_n^2(1-{\beta }_n)\parallel B{x}_n-Bp{\parallel }^2\\ -2{\delta }_n(1-{\beta }_n)〈x_n-p\mathrm{,}B{x}_n-Bp〉\\ \le \parallel x_n-p{\parallel }^2+{\delta }_n^2(1-{\beta }_n)\parallel B{x}_n-Bp{\parallel }^2-2{\delta }_n(1-{\beta }_n)\beta \parallel B{x}_n-Bp{\parallel }^2\\ =\parallel x_n-p{\parallel }^2+{\delta }_n(1-{\beta }_n)({\delta }_n-2\beta )\parallel B{x}_n-Bp{\parallel }^2.\end{array}$$

(25)

Using (25), we have

$$\begin{array}{ll}\hfill \parallel & x_{n+1}-p{\parallel }^2\\ =\parallel {\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-p{\parallel }^2\\ =\parallel {\alpha }_n\left(\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n\right)+\left(T_{{\mu }_n}{y}_n-p\right){\parallel }^2\\ ={\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\parallel T_{{\mu }_n}{y}_n-p\parallel \\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\parallel y_n-p{\parallel }^2\\ +2{\alpha }_n{〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉}^2\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\parallel x_n-p{\parallel }^2\\ +{\delta }_n\left(1-{\beta }_n\right)\left({\delta }_n-2\beta \right)\parallel B{x}_n-Bp{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉\\ ={\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\parallel x_n-p{\parallel }^2\\ +{\delta }_n\left({\delta }_n-2{\beta }_n\right)\parallel B{x}_n-Bp{\parallel }^2-{\delta }_n{\beta }_n\left({\delta }_n-2{\beta }_n\right)\parallel B{x}_n-Bp{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉.\end{array}$$

(26)

Therefore,

$$\begin{array}{ll}\hfill -& {\delta }_n\left({\delta }_n-2{\beta }_n\right)\parallel B{x}_n-Bp{\parallel }^2\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\left[\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel \right]\parallel x_{n+1}-x_n\parallel \\ -{\delta }_n{\beta }_n\left({\delta }_n-2{\beta }_n\right)\parallel B{x}_n-Bp{\parallel }^2+2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉.\end{array}$$

Hence, using condition B₁ and (24), we get

$$\underset{n\to \infty }{lim}\parallel B{x}_n-Bp\parallel =0.$$

(27)

From (8), we have

$$\begin{array}{c}\parallel w_n-p{\parallel }^2\\ =\parallel P_C(x_n-{\delta }_{n}B{x}_n)-P_C(p-{\delta }_{n}Bp){\parallel }^2\\ \le 〈(x_n-{\delta }_{n}B{x}_n)-(p-{\delta }_{n}Bp),w_n-p〉\\ =\frac{1}{2}[\parallel (x_n-{\delta }_{n}B{x}_n)-(p-{\delta }_{n}Bp){\parallel }^2+\parallel w_n-p{\parallel }^2\\ -\parallel (x_n-{\delta }_{n}B{x}_n)-(p-{\delta }_{n}Bp)-(w_n-p){\parallel }^2]\\ \le \frac{1}{2}[\parallel x_n-p{\parallel }^2+\parallel w_n-p{\parallel }^2\\ -\parallel (x_n-{\delta }_{n}B{x}_n)-(p-{\delta }_{n}Bp)-(w_n-p){\parallel }^2]\\ =\frac{1}{2}[\parallel x_n-p{\parallel }^2+\parallel w_n-p{\parallel }^2-\parallel x_n-w_n{\parallel }^2\\ +2{\delta }_n〈x_n-w_n\mathrm{,}B{x}_n-Bp〉-{\delta }_n^2\parallel B{x}_n-Bp{\parallel }^2].\end{array}$$

So we obtain

$$\begin{array}{ll}\hfill \parallel w_n-p{\parallel }^2& \le \parallel x_n-p{\parallel }^2-\parallel x_n-w_n{\parallel }^2\\ +2{\delta }_n〈x_n-w_n,B{x}_n-Bp〉-{\delta }_n^2\parallel B{x}_n-Bp{\parallel }^2.\end{array}$$

(28)

It follows from (26) and (28) that

$$\begin{array}{ll}\hfill \parallel & x_{n+}-p{\parallel }^2\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\parallel y_n-p{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\parallel {\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}B{x}_n\right)-p{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+{\beta }_n\parallel x_n-p{\parallel }^2+\left(1-{\beta }_n\right)\parallel w_n-p{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉\\ \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+{\beta }_n\parallel x_n-p{\parallel }^2+\left(1-{\beta }_n\right)\parallel x_n-p{\parallel }^2\\ -\left(1-{\beta }_n\right)\parallel x_n-w_n{\parallel }^2+2{\delta }_n\left(1-{\beta }_n\right)〈x_n-w_n,B{x}_n-Bp〉-{\delta }_n^2\left(1-{\beta }_n\right)\parallel B{x}_n-Bp{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉.\end{array}$$

Which implies that

$$\begin{array}{ll}\hfill \parallel x_n-w_n{\parallel }^2& \le {\alpha }_n^2\parallel \gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n{\parallel }^2+\left[\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel \right]\parallel x_{n+1}-x_n\parallel \\ +{\beta }_n\parallel x_n-w_n{\parallel }^2+2{\delta }_n\left(1-{\beta }_n\right)\parallel x_n-w_n\parallel \parallel B{x}_n-Bp\parallel \\ -{\delta }_n^2\left(1-{\beta }_n\right)\parallel B{x}_n-Bp{\parallel }^2+2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{T}_{{\mu }_n}{y}_n,T_{{\mu }_n}{y}_n-p〉.\end{array}$$

Therefore, using condition B₁, (24) and (27), we get

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

(29)

Let U : H →2 ^H be a set-valued mapping is defined by

$$Ux=\left\{\begin{array}{cc}\hfill Ax+N_{C}x,\hfill & \hfill x\in C,\hfill \\ \hfill \varnothing ,\hfill & \hfill x\notin C.\hfill \end{array}\right.$$

where N _C x is the normal cone to C at x ∈ C. Since B is relaxed, β-inverse strongly monotone. Thus, U is maximal monotone see [19]. Let (x, y) ∈ G (U), hence y - Bx ∈ N _C x. Since w _n = P _C (x _n - ζ _n Bx _n ) therefore, 〈x - w _n , y - Bx〉 ≥ 0. On the other hand from w _n = P _C (x _n - ζ _n Bx _n ), we have

$$〈x-w_n,w_n-\left(x_n-{\delta }_{n}B{x}_n\right)〉\ge 0,$$

that is

$$〈x-w_n,\frac{w_n-x_n}{{\delta }_n}+B{x}_n〉\ge 0.$$

Therefore, we have

$$\begin{array}{c}〈x-w_{n_i}\mathrm{,}y〉\\ \ge 〈x-w_{n_i}\mathrm{,}Bx〉\\ \ge 〈x-w_{n_i}\mathrm{,}Bx〉-〈x-w_{n_i}\mathrm{,}\frac{w_{n_i}-x_{n_i}}{{\delta }_{n_i}}+B{x}_{n_i})〉\\ =〈x-w_{n_i}\mathrm{,}Bx-\frac{w_{n_i}-x_{n_i}}{{\delta }_{n_i}}-B{x}_{n_i}〉\\ =〈x-w_{n_i}\mathrm{,}Bx-B{w}_{n_i}〉+〈x-w_{n_i}\mathrm{,}B{w}_{n_i}-B{x}_{n_i}〉-〈x-w_{n_i}\mathrm{,}\frac{w_{n_i}-x_{n_i}}{{\delta }_{n_i}}〉\\ \ge 〈x-w_{n_i}\mathrm{,}B{w}_{n_i}-B{x}_{n_i}〉-〈x-w_{n_i}\mathrm{,}\frac{w_{n_i}-x_{n_i}}{{\delta }_{n_i}}〉\\ \ge 〈x-w_{n_i}\mathrm{,}B{w}_{n_i}-B{x}_{n_i}〉-\parallel x-w_{n_i}\left|\right|\Vert \frac{w_{n_i}-x_{n_i}}{{\delta }_{n_i}}‱.\end{array}$$

Noting that ${lim}_{i\to \infty }\parallel w_{n_i}-x_{n_i}\parallel =0,x_{n_i}\to z$, $x_{n_i}\to z$ and B is $\frac{1}{\beta }$-lipschitzian, we obtain

$$〈x-z,y〉\ge 0.$$

Since U is maximal monotone, we have z ∈ U^-10, and hence z ∈ VI(C, B). Therefore, $z\in \mathcal{F}$.

Since $x_{n_i}\to z$ from (21) and (23), we have

$$\underset{n\to \infty }{limsup}〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_n-x^{*}〉\le 0.$$

Finally, we prove that x _n → x* as n → ∞. By Lemmas 2.4, 2.6, and (14), we have

$$\begin{array}{ll}\hfill \parallel & x_{n+1}-x^{*}{\parallel }^2\\ =\parallel {\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-x^{*}{\parallel }^2\\ =\parallel {\alpha }_n\gamma f\left(x_n\right)-{\alpha }_n\mu F{x}^{*}+\left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n\\ -\left(I-{\alpha }_n\mu F\right)x^{*}{\parallel }^2\\ \le \parallel \left(I-{\alpha }_n\mu F\right)T_{{\mu }_n}{y}_n-\left(I-{\alpha }_n\mu F\right)x^{*}{\parallel }^2\\ +2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉\\ \le {\left(1-{\alpha }_n\tau \right)}^2\parallel y_n-x^{*}{\parallel }^2+2{\alpha }_n〈\gamma f\left(x_n\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉\\ \le {\left(1-{\alpha }_n\tau \right)}^2\parallel y_n-x^{*}{\parallel }^2+2{\alpha }_n〈\gamma f\left(x_n\right)-\gamma f\left(x^{*}\right),x_{n+1}-x^{*}〉\\ +2{\alpha }_n〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉.\\ \le {\left(1-{\alpha }_n\tau \right)}^2\parallel y_n-x^{*}{\parallel }^2+{\alpha }_n\gamma \alpha \left[\parallel x_n-x^{*}{\parallel }^2+\parallel x_{n+1}-x^{*}{\parallel }^2\right]\\ +2{\alpha }_n〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉.\\ \le {\left(1-{\alpha }_n\tau \right)}^2\parallel x_n-x^{*}{\parallel }^2+{\alpha }_n\gamma \alpha \left[\parallel x_n-x^{*}{\parallel }^2+\parallel x_{n+1}-x^{*}{\parallel }^2\right]\\ +2{\alpha }_n〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉.\end{array}$$

(30)

So from (30), we reach the following

$$\begin{array}{ll}\hfill \parallel x_{n+1}-x^{*}{\parallel }^2& \le \frac{1+{\alpha }^2{\tau }^2-2{\alpha }_n\tau +{\alpha }_n\gamma \alpha }{1-{\alpha }_n\gamma \alpha }\parallel x_n-x^{*}{\parallel }^2\\ +\frac{2{\alpha }_n}{1-{\alpha }_n\gamma \alpha }〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉\\ \le \left(1-{\alpha }_n\frac{2\left(\tau -\gamma \alpha \right)-{\alpha }_n{\tau }^2}{1-{\alpha }_n\gamma \alpha }\right)\parallel x_n-x^{*}{\parallel }^2\\ +{\alpha }_n\frac{2\left(\tau -\gamma \alpha \right)-{\alpha }_n{\tau }^2}{1-{\alpha }_n\gamma \alpha }\frac{2}{2\left(\tau -\gamma \alpha \right)-{\alpha }_n{\tau }^2}〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉\end{array}$$

It follows that

$$\parallel x_{n+1}-x^{*}{\parallel }^2\le \left(1-b_n\right)\parallel x_n-x^{*}{\parallel }^2+b_n{c}_n,$$

(31)

where

$$b_n={\alpha }_n\frac{2\left(\tau -\gamma \alpha \right)-{\alpha }_n{\tau }^2}{1-{\alpha }_n\gamma \alpha },c_n=\frac{2}{2\left(\tau -\gamma \alpha \right)-{\alpha }_n{\tau }^2}〈\gamma f\left(x^{*}\right)-\mu F{x}^{*},x_{n+1}-x^{*}〉$$

Since α _n → o and $\sum _{n=0}^{\infty }{\alpha }_n=\infty$, we have $\sum _{n=0}^{\infty }{b}_n=\infty$ and by (22), we get lim sup_n→∞c _n ≤ 0. Consequently, applying Lemma 2.5, to (31), we conclude that xⁿ→ x*. Since || y _n - x* || ≤ || x _n - x* ||, we have yⁿ →x*. □

Corollary 3.2 Let {α _n }, {β _n }, {δ _n } and B be as in Theorem 3.1. Let T a nonexpansive mapping of C into C such that$\mathcal{F}=Fix\left(T\right)\cap VI\left(C,B\right)\ne \varnothing$. Suppose x₀ ∈ H and {x _n } and {y _n } be generated by the iteration algorithm

$$\left\{\begin{array}{c}\hfill y_n={\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}B{x}_n\right),\hfill \\ \hfill x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\mu F\right)\frac{n}{n-1}{\int }_0^{\frac{n-1}{n}}T\left(t\right)y_{n}dt,n\ge 0.\hfill \end{array}\right.$$

Then {x _n } and {y _n } convergence strongly to x* which is the unique solution of the systems of variational inequalities:

$$\left\{\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\hfill & \hfill \forall x\in \mathcal{F},\hfill \\ \hfill 〈B{x}^{*},y-x^{*}〉\ge 0\hfill & \hfill \forall y\in C,\hfill \end{array}\right.$$

Proof. Take ${\lambda }_n=\frac{n-1}{n}$, for n ∈ ℕ, we define ${\mu }_n\left(f\right)=\frac{1}{{\lambda }_n}{\int }_0^{{\lambda }_n}f\left(t\right)dt$ for each f ∈ C(ℝ₊), where C(ℝ₊) denotes the space of all real-valued bounded continuous functions on R⁺ with supremum norm. Then, {μ _n } is regular sequence of means on C(ℝ₊) such that

$$\parallel {\mu }_{n+1}-{\mu }_n\parallel \le 2\left(1-\frac{{\lambda }_n}{{\lambda }_{n+1}}\right)$$

for more details, see [21]. Further, for each y ∈ C, we have

$$T_{{\mu }_n}y=\frac{1}{{\lambda }_n}\underset{0}{\overset{{\lambda }_n}{\int }}T\left(t\right)ydt.$$

On the other hand

$$\begin{array}{ll}\hfill \sum _{n=1}^{\infty }∥{\mu }_{n+1}-{\mu }_n∥& \le 2\sum _{n=1}^{\infty }\left(\frac{{\lambda }_{n+1}-{\lambda }_n}{{\lambda }_{n+1}}\right)\\ =2\sum _{n=1}^{\infty }\left(\frac{\frac{n}{n+1}-\frac{n-1}{n}}{\frac{n}{n+1}}\right)\\ =2\sum _{n=1}^{\infty }\frac{1}{n^2}<\infty \end{array}$$

Now, apply Theorem 3.1 to conclude the result. □

Corollary 3.3 Let S, φ, X, {μ _n }, $\mathcal{F}$, {α _n }, {β _n }, and {δ _n } be as in Theorem 3.1. Let A be a strongly positive bounded linear operator with coefficient$\bar{\gamma }>0$, ζ a number in$\left(0,\frac{\bar{\tau }}{\alpha }\right)$, where$\bar{\tau }=\bar{\mu }\left(\bar{\gamma }-\frac{\bar{\mu }\parallel A{\parallel }^2}{2}\right)$and$0<\bar{\mu }<\frac{2\bar{\gamma }}{\parallel A{\parallel }^2}$. If {x _n } and {y _n } are generated by x₀ ∈ C and

$$\left\{\begin{array}{c}\hfill y_n={\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}A{x}_n\right),\hfill \\ \hfill x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+\left(I-{\alpha }_n\bar{\mu }A\right)T_{{\mu }_n}{y}_n,n\ge 0.\hfill \end{array}\right.$$

Then, {x _n } and {y _n } converge strongly, as n → ∞, to$x^{*}\in \mathcal{F}$, which is a unique solution of the variational inequalities:

$$\left\{\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\hfill & \hfill \forall x\in \mathcal{F},\hfill \\ \hfill 〈A{x}^{*},y-x^{*}〉\ge 0\hfill & \hfill \forall y\in C.\hfill \end{array}\right.$$

Proof. Because A is strongly positive bounded linear operator on H with coefficient $\bar{\gamma }$, we have

$$〈Ax-Ay,x-y〉\ge \bar{\gamma }\parallel x-y{\parallel }^2.$$

Therefore, A is $\bar{\gamma }$-strongly monotone.

On the other hand

$$\parallel Ax-Ay\parallel \le \parallel A\parallel \parallel x-y\parallel .$$

Therefore,

$$\frac{\bar{\gamma }}{\parallel A{\parallel }^2}\parallel Ax-Ay{\parallel }^2\le 〈Ax-Ay,x-y〉.$$

Hence, A is $\frac{\bar{\gamma }}{\parallel A{\parallel }^2}$-inverse strongly monotone. Now apply Theorem 3.1 to conclude the result. □

Corollary 3.4 Let {α _n }, {β _n } and B be as in Theorem 3.1. Let u, x₀ ∈ C and {x _n } and {y _n } be generated by the iterative algorithm

$$\left\{\begin{array}{c}\hfill y_n={\beta }_n{x}_n+\left(1-{\beta }_n\right)P_C\left(x_n-{\delta }_{n}B{x}_n\right),\hfill \\ \hfill x_{n+1}={\alpha }_{n}u+\left(I-{\alpha }_n\mu F\right)y_n,n\ge 0.\hfill \end{array}\right.$$

Then {x _n } and {y _n } convergence strongly to x* which is the unique solution of the systems of variational inequalities:

$$\left\{\begin{array}{cc}\hfill 〈\left(\mu F-\gamma f\right)x^{*},x-x^{*}〉\ge 0,\hfill & \hfill \forall x\in \mathcal{F},\hfill \\ \hfill 〈B{x}^{*},y-x^{*}〉\ge 0\hfill & \hfill \forall y\in C.\hfill \end{array}\right.$$

Proof. It is sufficient to take $f=\frac{1}{\gamma }u$ and φ = {I} in Theorem 3.1. □