Viscosity approximation to a common solution of variational inequality problems and fixed point problems for Lipschitzian semigroup in Banach spaces

Abstract

We introduce a new iterative algorithm based on a viscosity approximation method for finding the common solution of variational inequality problems for an inverse strongly accretive operator and the solution of fixed point problems for Lipschitzian semigroup map**s in Banach spaces. In controlling suitable conditions, strong convergence theorems are proven. Our results extend and improve the recent results of some authors in the literature in this field.

MSC 2000

47H10; Secondary 47H09, 43A07, 47H20, 47J20

Introduction

Let C be a nonempty closed convex subset of a real Banach space E and E^∗ be a dual space of E with norm ∥ · ∥ and 〈·,·〉 pairing between E and E^∗. A self map** f:C→C is a contraction on C if there exists a constant α∈(0,1) and x,y∈C such that

$$\parallel f(x)-f(y)\parallel \le \alpha \parallel x-y\parallel .$$

Recall that a map** A:C→C is said to be as follows:

• 1. Lipschitzian with Lipschitz constant L>0 if

  $$\parallel \mathit{\text{Ax}}-\mathit{\text{Ay}}\parallel \le L\parallel x-y\parallel ,\forall x,y\in \mathrm{C.}$$

• 2. Nonexpansive if

  $$\parallel \mathit{\text{Ax}}-\mathit{\text{Ay}}\parallel \le \parallel x-y\parallel ,\forall x,y\in \mathrm{C.}$$

• 3. Asymptotically nonexpansive if there exists a sequence {k _n } of positive numbers, satisfying the property lim _n →∞ k _n = 1 and

  $$\parallel A^{n}x-A^{n}y\parallel \le k_n\parallel x-y\parallel ,\forall x,y\in \mathrm{C.}$$

Clearly, every nonexpansive map** A is asymptotically nonexpansive with sequence {1}. Also, every asymptotically nonexpansive map** is uniformly L-Lipschitzian with $L={sup}_{n\in \mathbb{N}}{k}_n$.

An operator A:C→E is said to be as follows:

• 1. Accretive if there exists j(x−y)∈J(x−y) such that

  $$〈\mathit{\text{Ax}}-\mathit{\text{Ay}},j(x-y)〉\ge 0,\forall x,y\in \mathrm{C.}$$

• 2. β-Strongly accretive if there exists a constant β>0 such that

  $$〈\mathit{\text{Ax}}-\mathit{\text{Ay}},j(x-y)〉\ge \beta \parallel x-y{\parallel }^2,\forall x,y\in \mathrm{C.}$$

• 3. β-Inverse strongly accretive if, for any β>0,

  $$〈\mathit{\text{Ax}}-\mathit{\text{Ay}},j(x-y)〉\ge \beta \parallel \mathit{\text{Ax}}-\mathit{\text{Ay}}{\parallel }^2,\forall x,y\in \mathrm{C.}$$

Evidently, the definition of the inverse strongly accretive operator is based on the inverse strongly monotone operator.

Let U = {x∈E:∥x∥ = 1}. A Banach space E is said to be strictly convex if, for any x,y∈U, x≠y implies $\parallel \frac{x+y}{2}\parallel \le 1$. A Banach space E is said to be uniformly convex if, for any ∊ ∈ (0,2], there exists δ> 0 such that, for any x,y∈U, ∥x−y∥≥∊ implies $\parallel \frac{x+y}{2}\parallel \le 1-\mathrm{\delta .}$ It is known that a uniformly convex Banach space is reflexive and strictly convex. A Banach space E is said to be smooth if $\underset{t\to 0}{\text{lim}}\frac{\parallel x+\mathit{\text{ty}}\parallel -\parallel x\parallel }{t}$ exists for all x,y∈U. It is also said to be uniformly smooth if the limit is attained uniformly for x,y∈U. The modulus of smoothness of E is defined by the following:

$$\begin{array}{l}\rho (\tau )=sup\{\frac{1}{2}(\parallel x+y\parallel +\parallel x-y\parallel )\\ -1:x,y\in E,\parallel x\parallel =1,\parallel y\parallel =\tau \},\end{array}$$

where ρ:[0,∞)→[0,∞) is a function. It is known that E is uniformly smooth if and only if ${\text{lim}}_{\tau \to 0}\frac{\rho (\tau )}{\tau }=0$. Let q be a fixed real number with 1 < q ≤2. A Banach space E is said to be q-uniformly smooth if there exists a constant c>0 such that ρ(τ)≤c τ^q for all τ>0. We note that E is a uniformly smooth Banach space if and only if J _q is single-valued and uniformly continuous on any bounded subset of E. Typical examples of both uniformly convex and uniformly smooth Banach spaces are L^p, where p>1. More precisely, L^p is min{p,2}-uniformly smooth for every p>1. Note also that no Banach space is q-uniformly smooth for q>2; see the work of Xu [1] for more details.

For q>1, the generalized duality map**$J_q:E\to 2^{E^{\ast }}$ is defined by the following:

$$J_q(x)=\{f\in E^{\ast }:〈x,f〉=\parallel x{\parallel }^q,\parallel f\parallel =\parallel x{\parallel }^{q-1}\}$$

for all x∈E. In particular, if q=2, the map** J₂ is called the normalized duality map** and usually write J₂=J. Further, we have the following properties of the generalized duality map** J _q :

• 1. J _q (x)=∥x∥^q−2 J ₂(x) for all x∈E with x≠0.

• 2. J _q (t x)=t ^q−1 J _q (x) for all x∈E and t∈[0,∞).

• 3. J _q (−x)=−J _q (x) for all x∈E.

Let D be a subset of C and Q:C→D, then Q is said to be sunny if

$$Q(\mathit{\text{Qx}}+t(x-\mathit{\text{Qx}}))=\mathit{\text{Qx}},$$

whenever Q x+t(x−Q x)∈C for x∈C and t≥0. A subset D of C is said to be a sunny nonexpansive retraction of C if there exists a sunny nonexpansive retraction Q of C onto D. A map** Q:C→C is called a retraction if Q²=Q. If a map** Q:C→C is a retraction, then Q z=z for all z is in the range of Q.

The following result describes a characterization of sunny nonexpansive retractions on a smooth Banach space.

Proposition 1

Let E be a smooth Banach space and let C be a nonempty subset of E. Let Q:E→ C be a retraction and let J be the normalized duality map** on E. Then, the following are equivalent [2]:

• 1. Q is sunny and nonexpansive.

• 2. ∥Q x−Q y∥²≤〈x−y,J(Q x−Q y)〉,∀x,y∈E.

• 3. 〈x−Q x,J(y−Q x)〉≤0,∀x∈E,y∈C.

Proposition 2

Let C be a nonempty closed convex subset of a uniformly convex and uniformly smooth Banach space E and let T be a nonexpansive map** of C into itself with F(T)≠∅. Then, the set F(T) is a sunny nonexpansive retract of C [3].

$$\{T(s):s\in {ℝ}^{+}\}$$

is called a strongly continuous semigroup of Lipschitzian map**s from C into itself if it satisfies the following conditions:

(i) For each s > 0, there exists a function k(·):(0,∞)→(0,∞) such that

$$\parallel T(s)x-T(s)y\parallel \le k(s)\parallel x-y\parallel ,\forall x,y\in \mathrm{C.}$$

(ii) T (0) x = x for each x∈C.

(iii) T (s₁+s₂) x = T(s₁)T (s₂) x for any $s_1,s_2\in {ℝ}^{+}$ and x∈C.

(iv) For each x∈C, the map** T(·)x from ${ℝ}^{+}$ into C is continuous.

If k(s) = L for all s>0 in (i), then $\{T(s):s\in {ℝ}^{+}\}$ is called a strongly continuous semigroup of uniformly L-Lipschitzian map**s. If k(s) = 1 for all s>0 in (i), then $\{T(s):s\in {ℝ}^{+}\}$ is called a strongly continuous semigroup of nonexpansive map**s (see the work of Sahu and O’Regan [4]). For a semigroup S, we can define a partial preordering ≺ on S by a≺b if and only if a S⊃b S. If S is a left reversible semigroup (i.e., a S∩b S≠∅ for a,b∈S), then it is a directed set. (Indeed, for every a,b∈S, applying a S∩b S≠∅, there exist a^′,b^′∈S with a a^′=b b^′; by taking c=a a^′=b b^′, we have c S⊆a S∩b S, and then a≺c and b≺c.) If a semigroup S is left amenable, then S is left reversible [5]. Let $\mathcal{S}=\{T(s):s\in S\}$ be a representation of a left reversible semigroup S as Lipschitzian map**s on C with Lipschitz constants {k(s):s∈S}. We shall say that $\mathcal{S}$ is an asymptotically nonexpansive semigroup on C if there holds the uniform Lipschitzian condition lim _s k(s)≤ 1 on the Lipschitz constants. (Note that a left reversible semigroup is a directed set.) It is worth mentioning that there is a notion of asymptotically nonexpansive defined, depending on left ideals in a semigroup in the works of Holmes and Lau [6] and Holmes [7].

In 2008, Saeidi [8] introduced the following viscosity iterative scheme:

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)x_n,\forall n\ge 1,$$

(1)

for a representation of S as Lipschitzian map**s on a compact convex subset C of a smooth Banach space E with respect to a left regular sequence {μ _n } of means defined on an appropriate invariant subspace of l^∞(S). For some related results, we refer the readers to the works of Kirk [9] and Takahashi [10]. In 2011, Katchang and Kumam [11] extended the result of Saeidi [8] and introduced a new iterative algorithm with a viscosity iteration method for approximating a common fixed point of Lipschitzian semigroups in Banach spaces as the following:

$$\left\{\begin{array}{l}{y}_n={\delta }_n{x}_n+(1-{\delta }_n)T({\mu }_n)x_n,\\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n{y}_n,\forall n\ge 1.\end{array}\right.$$

(2)

Recently, Aoyama et al. [12] first considered the following generalized variational inequality problem in a smooth Banach space. Let A be an accretive operator of C into E. Find a point x∈C such that

$$〈\mathit{\text{Ax}},j(y-x)〉\ge 0$$

(3)

for all y∈C. The set of solutions of (3) is denoted by V I(C,A). This problem is connected with the fixed point problem for nonlinear map**s, the problem of finding a zero point of an accretive operator and so on. For the problem of finding a zero point of an accretive operator by the proximal point algorithm, see Kamimura and Takahashi [13, 14]. In order to find a solution of the variational inequality (3), Aoyama et al. [12] proved the strong convergence theorem in the framework of Banach spaces which is generalized by Iiduka et al. [15] from Hilbert spaces. In 2011, Yao and Maruster [16] proved some strong convergence theorems for finding a solution of variational inequality problem (3) in Banach spaces. They defined a sequence {x _n } iteratively by the arbitrary given x₀∈C and

$$x_{n+1}={\beta }_n{x}_n+(1-{\beta }_n)Q_C[(1-{\alpha }_n)(x_n-\mathrm{\lambda A}{x}_n)],\forall n\ge 0,$$

(4)

where Q _C is a sunny nonexpansive retraction from a uniformly convex and 2-uniformly smooth Banach space X onto a nonempty closed convex subset C of X, and A is an α-inverse strongly accretive operator of C into X (in the framework of variational inequality problems, also see Katchang and Kumam [17]).

Here, motivated and inspired by the ideas of Aoyama et al. [12], Saeidi [8], and Yao and Maruster [16], we introduce the new iterative algorithm (7) and prove strong convergence theorems for finding the common solution of variational inequality problems (3) involving an inverse strongly accretive operator and the solution of fixed point problems involving a Lipschitzian semigroup map** in Banach spaces using a viscosity approximation method. Moreover, its applications are also studied.

Preliminaries

Let S be a semigroup. We denote by l^∞(S) the Banach space of all bounded real-valued functions on S with the supremum norm. For each s∈S, we define l _s and r _s on l^∞(S) by (l _s f)(t) = f(s t) and (r _s f)(t) = f(t s) for each t∈S and f∈l^∞(S). Let X be a subspace of l^∞(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 (with respect to the (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. A net {μ _α } of means on X is said to be strongly left regular if

$$\underset{\alpha }{\text{lim}}\parallel l_s^{\ast }{\mu }_{\alpha }-{\mu }_{\alpha }\parallel =0$$

for each s∈S, where $l_s^{\ast }$ is the adjoint operator of l _s . Let C be a nonempty closed and convex subset of E. Throughout this paper, S will always denote a semigroup with an identity e. S is called left reversible if any two right ideals in S have nonvoid intersection, i.e., a S∩b S≠∅ for a,b∈S. In this case, we can define a partial ordering ≺ on S by a≺b if and only if a S⊃b S. It is easy to see t≺t s,(∀t,s∈S). Furthermore, if t≺s, then p t≺p s for all p∈S. If a semigroup S is left amenable, then S is left reversible. However, the converse is not true.

$$\mathcal{S}=\{T(s):s\in S\}$$

is called a representation of S as Lipschitzian map**s on C if for each s∈S, the map** T(s) is Lipschitzian map** on C with Lipschitz constant k(s) and T(s t)=T(s)T(t) for s,t∈S. We denote by $F(\mathcal{S})$ the set of common fixed points of $\mathcal{S}$, and we denote by C _a the set of almost periodic elements in C, i.e., all x∈C such that {T(s)x:s∈S} is relatively compact in the norm topology of reflexive Banach space E. We will call a subspace X of l^∞(S), $\mathcal{S}$-stable if the functions s↦〈T(s)x,x^∗〉 and s↦∥T(s)x−y∥ on S are in X for all x,y∈C and x^∗∈E^∗. We know that if μ is a mean on X and if for each x^∗∈E^∗, the function s↦〈T(s)x,x^∗〉 is contained in X and C is weakly compact, then there exists a unique point x₀ of E such that

$${\mu }_s〈T(s)x,x^{\ast }〉=〈x_0,x^{\ast }〉$$

for each x^∗∈E^∗. We denote such a point x₀ by T(μ)x. Note that T(μ)z=z for each $z\in F(\mathcal{S})$; see related works [18–20].

We need the following lemmas to prove our main results.

Lemma 1

Let S be a left reversible semigroup and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from a nonempty weakly compact convex subset C of a Banach space E into C, with the uniform Lipschitzian condition lim _s k(s)≤ 1 on the Lipschitz constants of the map**s. Let X be a left invariant $\mathcal{S}-\mathit{\text{stable}}$ subspace of l^∞(S) containing 1, and μ be a left invariant mean on X. Then, $F(\mathcal{S})=F(T(\mu ))\cap C_a$[21].

Corollary 1

Let {μ _n } be an asymptotically left invariant sequence of the means on X. If z∈C _a and lim inf_n→∞∥T(μ _n )z−z∥=0, then z is a common fixed point for $\mathcal{S}$[8].

Lemma 2

Let S be a left reversible semigroup and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from a nonempty weakly compact convex subset C of a Banach space E into C, with the uniform Lipschitzian condition lim _s k(s)≤ 1 on the Lipschitz constants of the map**s. Let X be a left invariant subspace of l^∞(S) containing 1 such that the map**s s↦〈T(s)x,x^∗〉 be in X for all x∈X and x^∗∈E^∗, and {μ _n } be a strongly left regular sequence of means on X[8]. Then,

$$\underset{n\to \infty }{\text{lim sup}}\underset{x,y\in C}{sup}(\parallel T({\mu }_n)x-T({\mu }_n)y\parallel -\parallel x-y\parallel )\le 0.$$

Remark 1

Taking in Lemma 2,

$$c_n=\underset{x,y\in C}{sup}(\parallel T({\mu }_n)x-T({\mu }_n)y\parallel -\parallel x-y\parallel ),\forall n,$$

(5)

we obtain lim sup_n→∞c _n ≤0. Moreover,

$$\parallel T({\mu }_n)x-T({\mu }_n)y\parallel \le \parallel x-y\parallel +c_n,\forall x,y\in \mathrm{C.}$$

(6)

Corollary 2

Let S be a left reversible semigroup and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from a nonempty compact convex subset C of a Banach space E into C, with the uniform Lipschitzian condition lim _s k(s)≤ 1. Let X be a left invariant $\mathcal{S}-\mathit{\text{stable}}$ subspace of l^∞(S) containing 1, and μ be a left invariant mean on X. Then, T(μ) is nonexpansive and $F(\mathcal{S})\ne \varnothing$. Moreover, if E is smooth, then $F(\mathcal{S})$ is a sunny nonexpansive retract of C, and the sunny nonexpansive retraction of C onto $F(\mathcal{S})$ is unique [8].

Lemma 3

Let C be a nonempty closed convex subset of a smooth Banach space X. Let Q _C be a sunny nonexpansive retraction from X onto C and let A be an accretive operator of C into X. Then, for all λ > 0, the set V I(C,A) is coincident with the set of fixed points of Q _C (I−λ A) [12], that is,

$$\mathit{\text{VI}}(C,A)=F(Q_C(I-\mathrm{\lambda A})).$$

Lemma 4

Let C be a nonempty bounded closed convex subset of a uniformly convex Banach space X and T be the nonexpansive map** of C into itself. If {x _n } is a sequence of C such that $x_n\rightharpoonup x$ and x _n −T x _n →0, then x is a fixed point of T[22].

Lemma 5

Let {x _n } and {z _n } be bounded sequences in a Banach space X and let {β _n } be a sequence in [0,1] with 0< lim inf_n→∞β _n ≤ lim sup_n→∞β _n <1. Suppose x_n+1 = (1−β _n )z _n +β _n x _n , for all integers n≥0 and lim sup_n→∞(∥z_n+1−z _n ∥−∥x_n+1−x _n ∥)≤0, then lim _n →∞∥z _n −x _n ∥ = 0 [23].

Lemma 6

Assume that {x _n } is a sequence of nonnegative real numbers such that

$$x_{n+1}\le (1-a_n)x_n+b_n,n\ge 0,$$

where {a _n } is a sequence in (0,1) and {b _n } is a sequence in $ℝ$ such that [24](1)

$${\sum }_{n=1}^{\infty }{a}_n=\infty$$

,(2)

$${\text{lim sup}}_{n\to \infty }\frac{b_n}{a_n}\le 0$$

or

$${\sum }_{n=1}^{\infty }\left|b_n\right|<\mathrm{\infty .}$$

Then, lim _n →∞ x _n = 0.

Lemma 7

Let C be a nonempty closed convex subset of a real 2-uniformly smooth Banach space E with the best smooth constant K. Let the map** A:C→E be β-inverse strongly accretive (see Lemma 3.1 of [25], see also Lemma 2.8 of [12]. Then, we have

$$\parallel (I-\mathrm{\lambda A})x-(I-\mathrm{\lambda A})y{\parallel }^2\le \parallel x-y{\parallel }^2+2\lambda (\lambda K^2-\beta )\parallel \mathit{\text{Ax}}-\mathit{\text{Ay}}{\parallel }^2.$$

If β≥λ K², then I−λ A is nonexpansive.

Lemma 8

Let X be a real Banach space and let J be the duality map**. Then, for any given x,y∈X and j(x+y)∈J(x+y), there holds the following inequality [9, 10]:

$$\parallel x+y{\parallel }^2\le \parallel x{\parallel }^2+2〈y,j(x+y)〉.$$

Lemma 9

Let E be a real 2-uniformly smooth Banach space with the best smooth constant K. Then, the following inequality holds [1]:

$$\parallel x+y{\parallel }^2\le \parallel x{\parallel }^2+2〈y,\mathit{\text{Jx}}〉+2\parallel \mathit{\text{Ky}}{\parallel }^2,\forall x,y\in \mathrm{E.}$$

Lemma 10

Let E be a uniformly convex Banach space and B _r (0):={x∈E:∥x∥≤r} be a closed ball of E. Then, there exists a continuous strictly increasing convex function g:[0,∞)→[0,∞) with g(0)=0 such that

$$\parallel \mathrm{\lambda x}+\mathrm{\mu y}+\mathrm{\gamma z}{\parallel }^2\le \lambda \parallel x{\parallel }^2+\mu \parallel y{\parallel }^2+\gamma \parallel z{\parallel }^2-\mathrm{\lambda \mu g}(\parallel x-y\parallel )$$

for all x,y,z∈B _r (0) and λ,μ,γ∈[0,1] with λ+μ+γ=1 [26].

Lemma 11

Let r>0 and let E be a uniformly convex Banach space. Then, there exists a continuous, strictly increasing, and convex function g:[0,∞)→[0,∞) with g(0)=0 such that

$$\parallel \mathrm{\lambda x}+(1-\lambda )y{\parallel }^2\le \lambda \parallel x{\parallel }^2+(1-\lambda )\parallel y{\parallel }^2-\lambda (1-\lambda )g(\parallel x-y\parallel )$$

for all x,y∈B _r :={z∈E:∥z∥≤r} and 0≤λ≤1 [1].

We note that for a given sequence {x _n }⊂C, let ${\omega }_w(\{x_n\}):=\{x:\exists x_{n_j}\rightharpoonup x\}$ denote the weak ω-limit set of {x _n }.

Main result

In this section, we prove a strong convergence theorem in Banach spaces.

Theorem 1

Let C be a nonempty compact convex subset of a uniformly convex and 2-uniformly smooth Banach space E with weakly sequentially continuous duality map** and the best smooth constant K, S be a left reversible semigroup, and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from C into itself, with the uniform Lipschitzian condition lim _s k(s) ≤ 1. Let X be a left invariant $\mathcal{S}$-stable subspace of l^∞(S) containing 1, {μ _n } be a strongly left regular sequence of means on X such that lim _n →∞∥μ_n+1−μ _n ∥ = 0 and {c _n } be the sequence defined by (5). Let f be a contraction of C into itself with coefficient α∈(0,1), Q _C be a sunny nonexpansive retraction from E onto C, and A:C→E be a β-inverse strongly accretive with β≥λ K² such that λ be a positive real number. Suppose $\mathcal{F}=F(\mathcal{S})\cap \mathit{\text{VI}}(C,A)\ne \varnothing$ and the sequences {α _n },{β _n },{γ _n } and {δ _n } in (0,1) satisfy α _n +β _n +γ _n = 1, n ≥ 1. The following conditions are satisfied:

(i) lim _n →∞ α _n = 0 and${\sum }_{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$

(ii) lim _n →∞ δ _n = 0.

(iii)${\text{lim sup}}_{n\to \infty }\frac{c_n}{{\alpha }_n}\le 0$, (note that, by Remark 1, lim sup_n→∞c _n ≤ 0).

(iv) 0 < lim inf_n→∞β _n ≤ lim sup_n→∞β _n <1.

If for the arbitrary given x₁∈C, the sequence {x _n } is generated by

$$\left\{\begin{array}{l}{y}_n={\delta }_n{x}_n+(1-{\delta }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n),\\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)y_n,\forall n\ge 1,\end{array}\right.$$

(7)

then {x _n } converges strongly to $x^{\ast }\in \mathcal{F}$, which is the unique solution of the variational inequality

$$〈(f-I)x^{\ast },J(z-x^{\ast })〉\le 0,\forall z\in \mathcal{F.}$$

Equivalently, we have $x^{\ast }=Q_{\mathcal{F}}f(x^{\ast })$, where Q is the unique sunny nonexpansive retraction of C onto $\mathcal{F}$.

Proof

First, we claim that for any sequence {v_n+1}∈C, ∥T(μ_n+1)v_n+1−T(μ _n )v_n+1∥→ as n→∞. Taking D= sup{∥p^∗∥:p^∗∈E^∗}, we have the following:

$$\begin{array}{l}\parallel T({\mu }_{n+1})v_{n+1}-T({\mu }_n)v_{n+1}\parallel \\ =\underset{\parallel p^{\ast }\parallel =1}{sup}|〈T({\mu }_{n+1})v_{n+1}-T({\mu }_n)v_{n+1},p^{\ast }〉|\\ =\underset{\parallel p^{\ast }\parallel =1}{sup}|{({\mu }_{n+1})}_s〈T(s)v_{n+1},p^{\ast }〉-{({\mu }_n)}_s〈T(s)v_{n+1},p^{\ast }〉|\\ \le \parallel {\mu }_{n+1}-{\mu }_n\parallel \underset{s\in S}{sup}\parallel T(s)v_{n+1}\parallel \\ \le \parallel {\mu }_{n+1}-{\mu }_n\parallel \mathrm{D.}\end{array}$$

Since lim _n →∞∥μ_n+1−μ _n ∥ = 0, therefore

$$\underset{n\to \infty }{\text{lim}}\parallel T({\mu }_{n+1})v_{n+1}-T({\mu }_n)v_{n+1}\parallel =0.$$

(8)

Next, we show that lim _n →∞∥x_n+1−x _n ∥ = 0, and by Lemma 2, we observe that

$$\begin{array}{l}\parallel y_{n+1}-y_n\parallel =\parallel {\delta }_{n+1}{x}_{n+1}+(1-{\delta }_{n+1})Q_C(x_{n+1}-\mathrm{\lambda A}{x}_{n+1})\\ -{\delta }_n{x}_n-(1-{\delta }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel \\ =\parallel {\delta }_{n+1}{x}_{n+1}-{\delta }_{n+1}{x}_n+{\delta }_{n+1}{x}_n\\ +(1-{\delta }_{n+1})Q_C(x_{n+1}-\mathrm{\lambda A}{x}_{n+1})\\ -(1-{\delta }_{n+1})Q_C(x_n-\mathrm{\lambda A}{x}_n)\\ +(1-{\delta }_{n+1})Q_C(x_n-\mathrm{\lambda A}{x}_n)\\ -{\delta }_n{x}_n-(1-{\delta }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel \\ =\parallel {\delta }_{n+1}(x_{n+1}-x_n)+({\delta }_{n+1}-{\delta }_n)x_n+(1-{\delta }_{n+1})\\ \times [Q_C(x_{n+1}-\mathrm{\lambda A}{x}_{n+1})-Q_C(x_n-\mathrm{\lambda A}{x}_n)]\\ +({\delta }_n-{\delta }_{n+1})Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel \\ \le {\delta }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\parallel x_n\parallel \\ +(1-{\delta }_{n+1})\parallel Q_C(x_{n+1}-\mathrm{\lambda A}{x}_{n+1})\\ -Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel \\ +|{\delta }_n-{\delta }_{n+1}|\parallel Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel \\ \le {\delta }_{n+1}\parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\parallel x_n\parallel \\ +(1-{\delta }_{n+1})\parallel x_{n+1}-x_n\parallel \\ +|{\delta }_n-{\delta }_{n+1}|\parallel Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel \\ =\parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\\ \times [\parallel x_n\parallel +\parallel Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel ].\end{array}$$

Setting x_n+1=(1−β _n )z _n +β _n x _n , we see that $z_n=\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}.$ Then, we compute

$$\begin{array}{l}\parallel z_{n+1}-z_n\parallel =\parallel \frac{x_{n+2}-{\beta }_{n+1}{x}_{n+1}}{1-{\beta }_{n+1}}-\frac{x_{n+1}-{\beta }_n{x}_n}{1-{\beta }_n}\parallel \\ =\parallel \frac{{\alpha }_{n+1}f(x_{n+1})+{\gamma }_{n+1}T({\mu }_{n+1})y_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_{n}f(x_n)+{\gamma }_{n}T({\mu }_n)y_n}{1-{\beta }_n}\parallel \\ =\parallel \frac{{\alpha }_{n+1}f(x_{n+1})+{\gamma }_{n+1}T({\mu }_{n+1})y_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\alpha }_{n+1}f(x_n)}{1-{\beta }_{n+1}}+\frac{{\alpha }_{n+1}f(x_n)}{1-{\beta }_{n+1}}\\ -\frac{{\gamma }_{n+1}T({\mu }_n)y_{n+1}}{1-{\beta }_{n+1}}+\frac{{\gamma }_{n+1}T({\mu }_n)y_{n+1}}{1-{\beta }_{n+1}}\\ -\frac{{\gamma }_{n}T({\mu }_n)y_{n+1}}{1-{\beta }_n}+\frac{{\gamma }_{n}T({\mu }_n)y_{n+1}}{1-{\beta }_n}\\ -\frac{{\alpha }_{n}f(x_n)+{\gamma }_{n}T({\mu }_n)y_n}{1-{\beta }_n}\parallel \\ =\parallel \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}[f(x_{n+1})-f(x_n)]\\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}[T({\mu }_{n+1})y_{n+1}-T({\mu }_n)y_{n+1}]\\ +[\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}]f(x_n)\\ +[\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}]T({\mu }_n)y_{n+1}\\ +\frac{{\gamma }_n}{1-{\beta }_n}[T({\mu }_n)y_{n+1}-T({\mu }_n)y_n]\parallel \\ \le \frac{\alpha {\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel \\ +\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}\parallel T({\mu }_{n+1})y_{n+1}-T({\mu }_n)y_{n+1}\parallel \\ +\parallel \frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\parallel f(x_n)\parallel \\ +|\frac{{\gamma }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\gamma }_n}{1-{\beta }_n}|\parallel T({\mu }_n)y_{n+1}\parallel \\ +\frac{{\gamma }_n}{1-{\beta }_n}\parallel T({\mu }_n)y_{n+1}-T({\mu }_n)y_n\parallel \\ \le \frac{\alpha {\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel \\ +\parallel T({\mu }_{n+1})y_{n+1}-T({\mu }_n)y_{n+1}\parallel \\ +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\\ \times [\parallel f(x_n)\parallel +\parallel T({\mu }_n)y_{n+1}\parallel ]\\ +\parallel y_{n+1}-y_n\parallel +c_n\\ \le \frac{\alpha {\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +\parallel T({\mu }_{n+1})y_{n+1}\\ -T({\mu }_n)y_{n+1}\parallel +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|\\ \times [\parallel f(x_n)\parallel +\parallel T({\mu }_n)y_{n+1}\parallel ]\\ +\parallel x_{n+1}-x_n\parallel +|{\delta }_{n+1}-{\delta }_n|\\ \times [\parallel x_n\parallel +\parallel Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel ]+c_n.\end{array}$$

Since C is bounded and lim sup_n→∞β _n <1, we have big constants M₁>0 and M₂ >0. Therefore, we observe that

$$\begin{array}{l}\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel \\ \le \frac{\alpha {\alpha }_{n+1}}{1-{\beta }_{n+1}}\parallel x_{n+1}-x_n\parallel +\parallel T({\mu }_{n+1})y_{n+1}-T({\mu }_n)y_{n+1}\parallel \\ +|\frac{{\alpha }_{n+1}}{1-{\beta }_{n+1}}-\frac{{\alpha }_n}{1-{\beta }_n}|M_1+|{\delta }_{n+1}-{\delta }_n|M_2+c_n.\end{array}$$

It follows from (i), (ii), (iv), (8), and Lemma 2 that

$$\underset{n\to \infty }{\text{lim sup}}(\parallel z_{n+1}-z_n\parallel -\parallel x_{n+1}-x_n\parallel )\le 0.$$

Applying Lemma 5, we obtain lim _n →∞∥z _n −x _n ∥=0. We also have ∥x_n+1−x _n ∥=(1−β _n )∥z _n −x _n ∥, therefore, we get

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

(9)

On the other hand, let $p\in \mathcal{F}$, and we have the following:

$$\begin{array}{l}\parallel x_{n+1}-p{\parallel }^2=\parallel {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)y_n-p{\parallel }^2\\ \le {\alpha }_n\parallel f(x_n)-p{\parallel }^2+{\beta }_n\parallel x_n-p{\parallel }^2\\ +{\gamma }_n\parallel T({\mu }_n)y_n-p{\parallel }^2\\ \le {\alpha }_n\parallel f(x_n)-p{\parallel }^2+{\beta }_n\parallel x_n-p{\parallel }^2\\ +{\gamma }_n[\parallel y_n-p\parallel +c_n{]}^2\\ ={\alpha }_n\parallel f(x_n)-p{\parallel }^2+(1-{\alpha }_n-{\gamma }_n)\parallel x_n-p{\parallel }^2\\ +{\gamma }_n[\parallel y_n-p{\parallel }^2+2c_n\parallel y_n-p\parallel +c_n^2]\\ ={\alpha }_n\parallel f(x_n)-p{\parallel }^2+(1-{\alpha }_n)\parallel x_n-p{\parallel }^2\\ -{\gamma }_n[\parallel x_n-p{\parallel }^2-\parallel y_n-p{\parallel }^2]\\ +{\gamma }_n{c}_n[2\parallel y_n-p\parallel +c_n]\\ ={\alpha }_n\parallel f(x_n)-p{\parallel }^2+(1-{\alpha }_n)\parallel x_n-p{\parallel }^2\\ -{\gamma }_n[\parallel x_n-p\parallel -\parallel y_n-p\parallel ]\\ \times [\parallel x_n-p\parallel +\parallel y_n-p\parallel ]s\\ +{\gamma }_n{c}_n[2\parallel y_n-p\parallel +c_n]\\ \le {\alpha }_n\parallel f(x_n)-p{\parallel }^2+\parallel x_n-p{\parallel }^2\\ -{\gamma }_n\parallel x_n-y_n{\parallel }^2+{\gamma }_n{c}_n[2\parallel y_n-p\parallel +c_n].\end{array}$$

It follows that

$$\begin{array}{l}{\gamma }_n\parallel x_n-y_n{\parallel }^2\le {\alpha }_n\parallel f(x_n)-p{\parallel }^2+\parallel x_n-p{\parallel }^2\\ -\parallel x_{n+1}-p{\parallel }^2+{\gamma }_n{c}_n[2\parallel y_n-p\parallel +c_n]\\ \le {\alpha }_n\parallel f(x_n)-p{\parallel }^2+\parallel x_n-x_{n+1}\parallel \\ \times [\parallel x_n-p\parallel +\parallel x_{n+1}-p\parallel ]\\ +{\gamma }_n{c}_n[2\parallel y_n-p\parallel +c_n].\end{array}$$

From conditions (i) and (iv) and by (5) and (9), we conclude that

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

(10)

We note that

$$\begin{array}{l}{x}_{n+1}-y_n={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)y_n-y_n\\ ={\alpha }_n[f(x_n)-y_n]+{\beta }_n(x_n-y_n)\\ +{\gamma }_n[T({\mu }_n)y_n-y_n].\end{array}$$

Thus, we have the following:

$$\begin{array}{l}{\gamma }_n\parallel y_n-T({\mu }_n)y_n\parallel \le {\alpha }_n\parallel f(x_n)-y_n\parallel +{\beta }_n\parallel x_n-y_n\parallel \\ +\parallel y_n-x_{n+1}\parallel \\ \le {\alpha }_n\parallel f(x_n)-y_n\parallel +{\beta }_n\parallel x_n-y_n\parallel \\ +\parallel y_n-x_n\parallel +\parallel x_n-x_{n+1}\parallel \\ ={\alpha }_n\parallel f(x_n)-y_n\parallel +(1+{\beta }_n)\parallel x_n-y_n\parallel \\ +\parallel x_n-x_{n+1}\parallel .\end{array}$$

By (i), (iv), (9), and (10), we obtain the following:

$$\underset{n\to \infty }{\text{lim}}\parallel y_n-T({\mu }_n)y_n\parallel =0.$$

(11)

We consider

$$\begin{array}{l}\parallel x_n-y_n\parallel =\parallel x_n-[{\delta }_n{x}_n+(1-{\delta }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n)]\parallel \\ =\parallel {\delta }_n(x_n-x_n)+(1-{\delta }_n)\\ \times [x_n-Q_C(x_n-\mathrm{\lambda A}{x}_n)]\parallel \\ =(1-{\delta }_n)\parallel x_n-Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel .\end{array}$$

By (ii) and (10), we have the following:

$$\underset{n\to \infty }{\text{lim}}\parallel x_n-Q_C(x_n-\mathrm{\lambda A}{x}_n)\parallel =0.$$

(12)

Let ω({x _n }) be the ω-limit set of {x _n }. Next, we show that ω({x _n }) is a subset of $\mathcal{F}=F(\mathcal{S})\cap \mathit{\text{VI}}(C,A)$. Let z∈ω({x _n }) and $\left\{x_{n_k}\right\}$ be a subsequence of {x _n } that converges strongly to z. Since ∥x _n −y _n ∥→0, we obtain $y_{n_k}\to z$. From (11), Lemma 2 and Remark 1, we obtain the following:

$$\begin{array}{l}\underset{k\to \infty }{\text{lim sup}}\parallel z-T({\mu }_{n_k})z\parallel \\ \le \underset{k\to \infty }{\text{lim sup}}\left[\parallel z-y_{n_k}\parallel +\parallel y_{n_k}-T({\mu }_{n_k})y_{n_k}\parallel \right.\\ \left.+\parallel T({\mu }_{n_k})y_{n_k}-T({\mu }_{n_k})z\parallel \right]\\ \le \underset{k\to \infty }{\text{lim sup}}\left[\parallel y_{n_k}-T({\mu }_{n_k})y_{n_k}\parallel \right.\\ \left.+2\parallel y_{n_k}-z\parallel +c_{n_k}\right]\le 0.\end{array}$$

Moreover, we have the following:

$$\underset{k\to \infty }{\text{lim inf}}\parallel z-T({\mu }_{n_k})z\parallel \le \underset{k\to \infty }{\text{lim sup}}\parallel z-T({\mu }_{n_k})z\parallel \le 0.$$

Thus, applying Corollary 1, we get $z\in F(\mathcal{S})$. Next, we show z∈V I(C,A). From (12) and by Lemmas 3 and 4, we have z∈F(Q _C (I−λ A))=V I(C,A). Therefore $z\in \mathcal{F}$.

Next, we show that lim sup_n→∞〈(f−I)x^∗,J(x _n −x^∗)〉 ≤ 0, where $x^{\ast }=Q_{\mathcal{F}}f(x^{\ast })$. Let $\left\{x_{n_k}\right\}$ be a subsequence of {x _n } such that

$$\underset{n\to \infty }{\text{lim sup}}〈(f-I)x^{\ast },J(x_n-x^{\ast })〉=\underset{k\to \infty }{\text{lim}}〈(f-I)x^{\ast },J(x_{n_k}-x^{\ast })〉.$$

(13)

Now, from (13), Proposition 1 (iii) and the weakly sequential continuity of the duality map** J, we have the following:

$$\begin{array}{l}\underset{n\to \infty }{\text{lim sup}}〈(f-I)x^{\ast },J(x_n-x^{\ast })〉\\ =\underset{k\to \infty }{\text{lim}}〈(f-I)x^{\ast },J(x_{n_k}-x^{\ast })〉\\ =〈(f-I)x^{\ast },J(z-x^{\ast })〉\le 0.\end{array}$$

(14)

From (9), it follows that

$$\underset{n\to \infty }{\text{lim sup}}〈(f-I)x^{\ast },J(x_{n+1}-x^{\ast })〉\le 0.$$

(15)

Finally, we show that the sequence {x _n } converges strongly to $x^{\ast }=Q_{\mathcal{F}}f(x^{\ast })$. From Lemma 7 and since Q _C is a nonexpansive, we have the following:

$$\begin{array}{l}\parallel y_n-x^{\ast }\parallel =\parallel {\delta }_n{x}_n+(1-{\delta }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n)-x^{\ast }\parallel \\ \le {\delta }_n\parallel x_n-x^{\ast }\parallel +(1-{\delta }_n)\parallel Q_C(x_n-\mathrm{\lambda A}{x}_n)\\ -Q_C(x^{\ast }-\mathrm{\lambda A}{x}^{\ast })\parallel \\ \le {\delta }_n\parallel x_n-x^{\ast }\parallel +(1-{\delta }_n)\parallel (x_n-\mathrm{\lambda A}{x}_n)\\ -(x^{\ast }-\mathrm{\lambda A}{x}^{\ast })\parallel \\ \le {\delta }_n\parallel x_n-x^{\ast }\parallel +(1-{\delta }_n)\parallel x_n-x^{\ast }\parallel \\ =\parallel x_n-x^{\ast }\parallel .\end{array}$$

(16)

Using Lemmas 8 and 11 and (16), we have the following:

$$\begin{array}{l}\parallel x_{n+1}-x^{\ast }{\parallel }^2=\parallel {\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)y_n-x^{\ast }{\parallel }^2\\ =\parallel [;{\gamma }_n(T({\mu }_n)y_n-x^{\ast })+{\beta }_n(x_n-x^{\ast })]\\ +{\alpha }_n[f(x_n)-x^{\ast }]{\parallel }^2\\ \le \parallel {\gamma }_n[T({\mu }_n)y_n-x^{\ast }]+{\beta }_n(x_n-x^{\ast }){\parallel }^2\\ +2{\alpha }_n〈f(x_n)-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ \le (1-{\beta }_n)\parallel \frac{{\gamma }_n}{1-{\beta }_n}[T({\mu }_n)y_n-x^{\ast }]{\parallel }^2\\ +{\beta }_n\parallel x_n-x^{\ast }{\parallel }^2\\ +2{\alpha }_n〈f(x_n)-f(x^{\ast }),J(x_{n+1}-x^{\ast })〉\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ =\frac{{\gamma }_n^2}{1-{\beta }_n}\parallel T({\mu }_n)y_n-x^{\ast }{\parallel }^2+{\beta }_n\parallel x_n-x^{\ast }{\parallel }^2\\ +2{\alpha }_n〈f(x_n)-f(x^{\ast }),J(x_{n+1}-x^{\ast })〉\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ \le \frac{{\gamma }_n^2}{1-{\beta }_n}\parallel y_n-x^{\ast }{\parallel }^2\\ +\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}+{\beta }_n\parallel x_n-x^{\ast }{\parallel }^2\\ +2\alpha {\alpha }_n\parallel x_n-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel \\ +2{\alpha }_n\parallel f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ \le \frac{{\gamma }_n^2}{1-{\beta }_n}\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}+{\beta }_n\parallel x_n-x^{\ast }{\parallel }^2\\ +2\alpha {\alpha }_n\parallel x_n-x^{\ast }\parallel \parallel x_{n+1}-x^{\ast }\parallel \\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ \le \frac{{\gamma }_n^2}{1-{\beta }_n}\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}+{\beta }_n\parallel x_n-x^{\ast }{\parallel }^2\\ +\alpha {\alpha }_n[\parallel x_n-x^{\ast }{\parallel }^2+\parallel x_{n+1}-x^{\ast }{\parallel }^2]\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ =[\frac{{\gamma }_n^2}{1-{\beta }_n}+{\beta }_n+\alpha {\alpha }_n]\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}\\ +\alpha {\alpha }_n\parallel x_{n+1}-x^{\ast }{\parallel }^2\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ =[\frac{((1-{\beta }_n)-{\alpha }_n{)}^2}{1-{\beta }_n}+{\beta }_n+\alpha {\alpha }_n]\\ \parallel x_n-x^{\ast }{\parallel }^2+\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}+\alpha {\alpha }_n\parallel x_{n+1}-x^{\ast }{\parallel }^2\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ =[\frac{{(1-{\beta }_n)}^2-2(1-{\beta }_n){\alpha }_n+{\alpha }_n^2}{1-{\beta }_n}+{\beta }_n+\alpha {\alpha }_n]\\ \parallel x_n-x^{\ast }{\parallel }^2+\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}\\ +\alpha {\alpha }_n\parallel x_{n+1}-x^{\ast }{\parallel }^2+2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ =[1-{\beta }_n-2{\alpha }_n+\frac{{\alpha }_n^2}{1-{\beta }_n}+{\beta }_n+\alpha {\alpha }_n]\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}+\alpha {\alpha }_n\parallel x_{n+1}-x^{\ast }{\parallel }^2\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ =[(1-\alpha {\alpha }_n)+(2\alpha {\alpha }_n-2{\alpha }_n)+\frac{{\alpha }_n^2}{1-{\beta }_n}]\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\gamma }_n^2{c}_n}{1-{\beta }_n}+\alpha {\alpha }_n\parallel x_{n+1}-x^{\ast }{\parallel }^2\\ +2{\alpha }_n〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉.\end{array}$$

It follows that

$$\begin{array}{l}\parallel x_{n+1}-x^{\ast }{\parallel }^2\le [1-\frac{2{\alpha }_n(1-\alpha )}{1-\alpha {\alpha }_n}]\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\alpha }_n^2}{(1-\alpha {\alpha }_n)(1-{\beta }_n)}\parallel x_n-x^{\ast }{\parallel }^2\\ +\frac{{\alpha }_n{\gamma }_n^2}{(1-\alpha {\alpha }_n)(1-{\beta }_n)}\left(\frac{c_n}{{\alpha }_n}\right)\\ +\frac{2{\alpha }_n}{1-\alpha {\alpha }_n}〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\\ \le [1-\frac{2{\alpha }_n(1-\alpha )}{1-\alpha {\alpha }_n}]\parallel x_n-x^{\ast }{\parallel }^2+\frac{{\alpha }_n}{1-\alpha {\alpha }_n}\\ \times \left[\frac{{\alpha }_n}{1-{\beta }_n}\parallel x_n-x^{\ast }{\parallel }^2+\frac{{\gamma }_n^2}{1-{\beta }_n}\left(\frac{c_n}{{\alpha }_n}\right)\right.\\ \left.+2〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\right]\\ =(1-a_n)\parallel x_n-x^{\ast }{\parallel }^2+b_n,\end{array}$$

where

$$a_n=\frac{2{\alpha }_n(1-\alpha )}{1-\alpha {\alpha }_n}$$

and

$$\begin{array}{l}{b}_n=\frac{{\alpha }_n}{1-\alpha {\alpha }_n}\left[\frac{{\alpha }_n}{1-{\beta }_n}\parallel x_n-x^{\ast }{\parallel }^2+\frac{{\gamma }_n^2}{1-{\beta }_n}\left(\frac{c_n}{{\alpha }_n}\right)\right.\\ \left.+2〈f(x^{\ast })-x^{\ast },J(x_{n+1}-x^{\ast })〉\right].\end{array}$$

Now, from (i), (iii), (iv), and (15) and Lemma 6, we get ∥x _n −x^∗∥→0 as n→∞. This completes the proof. □

Example 1

We can choose for an example of the control condition α _n in Theorem 1 as follows:

$${\alpha }_n=\left\{\begin{array}{l}\frac{1}{n+1}+\sqrt{c_n}\text{if}{c}_n\ge 0,\\ \frac{1}{n+1}\text{if}{c}_n<0.\end{array}\right.$$

(17)

Corollary 3

Let C be a nonempty compact convex subset of a uniformly convex and 2-uniformly smooth Banach space E with weakly sequentially continuous duality map** and the best smooth constant K, S be a left reversible semigroup and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from C into itself, with the uniform Lipschitzian condition lims k(s) ≤ 1. Let X be a left invariant $\mathcal{S}$-stable subspace of l^∞(S) containing 1, {μ _n } be a strongly left regular sequence of means on X such that limn→∞∥μ_n+1−μ _n ∥=0 and {c _n } be the sequence defined by (5). Let f be a contraction of C into itself with coefficient α∈(0,1), Q _C be a sunny nonexpansive retraction from E onto C and A:C→E be a β-inverse-strongly accretive with β≥λ K² such that λ be a positive real number. Suppose $\mathcal{F}=F(\mathcal{S})\cap \mathit{\text{VI}}(C,A)\ne \varnothing$ and the sequences {α _n }, {β _n } and {γ _n } in (0,1) satisfy α _n +β _n +γ _n =1, n≥1. The sequence {x _n } is generated by x₁∈C and

$$x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n),\forall n\ge 1.$$

(18)

If the sequence {x _n } satisfy the conditions (i) to (iv) in Theorem 1 then {x _n } converges strongly to $x^{\ast }\in \mathcal{F}$.

Proof

Taking δ _n =0 in Theorem 1, we can conclude the desired conclusion easily. This completes the proof. □

Remark 2

Our result extends and improves the results of Saeidi [8], Katchang and Kumam [11], and Yao and Maruster [16].

Applications

Application to the other form of semigroups

Theorem 2

Let C be a nonempty compact convex subset of a uniformly convex and 2-uniformly smooth Banach space E with weakly sequentially continuous duality map** and the best smooth constant K, and let $\mathcal{S}=\{T(t):t\in {ℝ}^{+}\}$ be a strongly continuous semigroup of Lipschitzian map**s from C into itself, with the uniform Lipschitzian condition lims k(s) ≤ 1 and {t _n } be increasing sequence in (0,∞) such that limn→∞ t _n =∞ and $\underset{n\to \infty }{\text{lim}}\frac{t_n}{t_{n+1}}=1$. Let f be a contraction of C into itself with coefficient α∈(0,1), Q _C be a sunny nonexpansive retraction from E onto C and A:C→E be a β-inverse strongly accretive with β≥λ K² such that λ be a positive real number.

Suppose $\mathcal{F}=F(\mathcal{S})\cap \mathit{\text{VI}}(C,A)\ne \varnothing$ and the sequences {α _n },{β _n },{γ _n }, and {δ _n } in (0,1) satisfy α _n +β _n +γ _n = 1, n ≥1, the following conditions are satisfied:

(i) lim _n →∞ (iii) _n = 0 and ${\sum }_{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$

(ii) lim _n →∞ δ _n = 0.

(iii)

$${\text{lim sup}}_{n\to \infty }\frac{c_n}{{\alpha }_n}\le 0$$

, where,$c_n=\underset{x,y\in C}{sup}\{\parallel \frac{1}{t_n}{\int }_0^{t_n}T(s)\mathit{\text{xds}}-\frac{1}{t_n}{\int }_0^{t_n}T(s)\mathit{\text{yds}}\parallel -\parallel x-y\parallel \}.$

(iv) 0 < lim inf_n→∞β _n ≤ lim sup_n→∞β _n < 1.

If for arbitrary given x₁∈C, the sequence {x _n } is generated by

$$\left\{\begin{array}{l}{y}_n={\delta }_n{x}_n+(1-{\delta }_n)Q_C(x_n-\mathrm{\lambda A}{x}_n),\\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_n\frac{1}{t_n}{\int }_0^{t_n}T(s)y_n\mathit{\text{ds}},\forall n\ge 1,\end{array}\right.$$

(19)

then {x _n } converges strongly to $x^{\ast }\in \mathcal{F}$, which is the unique solution of the following variational inequality:

$$〈(f-I)x^{\ast },J(z-x^{\ast })〉\le 0,\forall z\in \mathcal{F.}$$

Equivalently, we have $x^{\ast }=Q_{\mathcal{F}}f(x^{\ast })$, where Q is the unique sunny nonexpansive retraction of C onto $\mathcal{F}$.

Proof

For n ≥ 1, define ${\mu }_n(g)=\frac{1}{t_n}{\int }_0^{t_n}g(t)\mathit{\text{dt}}$ for each $g\in C({ℝ}^{+})$, where $C({ℝ}^{+})$ is the space of all real-valued bounded continuous functions on ${ℝ}^{+}$ with the supremum norm. Then, {μ _n } is a strongly regular sequence of means and lim _n →∞∥μ_n+1−μ _n ∥ = 0 (see [27]). Furthermore, for each x∈C, we have $T({\mu }_n)x=\frac{1}{t_n}{\int }_0^{t_n}T(s)\mathit{\text{xds}}$. Therefore, we apply Theorem 1 to conclude the result. □

Application to the strongly accretive and Lipschitz continuous operators

Now, we prove a strong convergence theorem for strongly accretive operators.

Theorem 3

Let C be a nonempty compact convex subset of a uniformly convex and 2-uniformly smooth Banach space E with weakly sequentially continuous duality map** and the best smooth constant K, S be a left reversible semigroup, and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from C into itself, with the uniform Lipschitzian condition lim _s k(s)≤ 1. Let X be a left invariant $\mathcal{S}$-stable subspace of l^∞(S) containing 1, {μ _n } be a strongly left regular sequence of means on X such that lim _n →∞∥μ_n+1−μ _n ∥ = 0, and {c _n } be the sequence defined by (5). Let f be a contraction of C into itself with coefficient α∈(0,1), Q _C be a sunny nonexpansive retraction from E onto C, and A be an β-strongly accretive and L-Lipschitz continuous operator of C into E with β≥λ K²L² such that λ be a positive real number. Suppose $\mathcal{F}=F(\mathcal{S})\cap \mathit{\text{VI}}(C,A)\ne \varnothing$ and the sequences {α _n },{β _n },{γ _n } and {δ _n } in (0,1) satisfy α _n +β _n +γ _n =1, n≥1. If the sequence {x _n } is generated by x₁∈C and (7) such that they satisfy conditions (i) to (iv), then {x _n } converges strongly to $x^{\ast }\in \mathcal{F}$, which is the unique solution of the following variational inequality:

$$〈(f-I)x^{\ast },J(z-x^{\ast })〉\le 0,\forall z\in \mathcal{F.}$$

Equivalently, we have $x^{\ast }=Q_{\mathcal{F}}f(x^{\ast })$, where Q is the unique sunny nonexpansive retraction of C onto $\mathcal{F}$.

Proof

Since A is a β-strongly accretive and L-Lipschitz continuous operator of C into E, we have the following:

$$〈\mathit{\text{Ax}}-\mathit{\text{Ay}},j(x-y)〉\ge \beta \parallel x-y{\parallel }^2\ge \frac{\beta }{L^2}\parallel \mathit{\text{Ax}}-\mathit{\text{Ay}}{\parallel }^2,\forall x,y\in \mathrm{C.}$$

Therefore, A is $\frac{\beta }{L^2}$-inverse strongly accretive. Using Theorem 1, we can obtain that {x _n } converges strongly to X^∗. This completes the proof. □

Application to Hilbert spaces

Let C be a closed convex subset of a real Hilbert space H. Let A:C→H be a map**. The classical variational inequality problem is to find x∈C such that

$$〈\mathit{\text{Ax}},y-x〉\ge 0$$

(20)

for all y∈C.

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

$$\parallel x-P_{C}x\parallel \le \parallel x-y\parallel \text{for all}y\in \mathrm{C.}$$

P _C is called the metric projection of H onto C. It is well known that P _C is a nonexpansive map** of H onto C and satisfies the following:

$$〈x-y,P_{C}x-P_{C}y〉\ge \parallel P_{C}x-P_{C}y{\parallel }^2,$$

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

$$\begin{array}{l}〈x-P_{C}x,y-P_{C}x〉\le 0,\end{array}$$

(22)

$$\begin{array}{l}\parallel x-y{\parallel }^2\ge \parallel x-P_{C}x{\parallel }^2+\parallel y-P_{C}x{\parallel }^2\end{array}$$

(23)

for all x∈H,y∈C.

It is well known in Hilbert spaces that the smooth constant $K=\frac{\sqrt{2}}{2}$ and J=I (identity map**). From Theorem 1, we can obtain the following result immediately.

Theorem 4

Let C be a nonempty compact convex subset of a real Hilbert space H, S be a left reversible semigroup, and $\mathcal{S}=\{T(s):s\in S\}$ be a representation of S as Lipschitzian map**s from C into itself, with the uniform Lipschitzian condition lim _s k(s)≤ 1. Let X be a left invariant $\mathcal{S}$-stable subspace of l^∞(S) containing 1, {μ _n } be a strongly left regular sequence of means on X such that lim _n →∞∥μ_n+1−μ _n ∥ = 0 and {c _n } be the sequence defined by (5). Let f be a contraction of C into itself with coefficient α∈(0,1), P _C be a nonexpansive map** of H onto C, and A:C→H be a β-inverse strongly monotone with λ∈(0,2β). Suppose $\mathcal{F}=F(\mathcal{S})\cap \mathit{\text{VI}}(C,A)\ne \varnothing$ and the sequences {α _n },{β _n },{γ _n } and {δ _n } in (0,1) satisfy α _n +β _n +γ _n = 1, n≥1, the following conditions are satisfied:

(i) lim _n →∞ α _n = 0 and ${\sum }_{n=0}^{\infty }{\alpha }_n=\mathrm{\infty .}$

(ii) lim _n →∞ δ _n = 0.

(iii)${\text{lim sup}}_{n\to \infty }\frac{c_n}{{\alpha }_n}\le 0$, (note that, by Remark 1, lim sup_n→∞c _n ≤0).

(iv) 0< lim inf_n→∞β _n ≤ lim sup_n→∞β _n <1.

If for the arbitrary given x₁∈C the sequence {x _n } is generated by

$$\left\{\begin{array}{l}{y}_n={\delta }_n{x}_n+(1-{\delta }_n)P_C(x_n-\mathrm{\lambda A}{x}_n),\\ x_{n+1}={\alpha }_{n}f(x_n)+{\beta }_n{x}_n+{\gamma }_{n}T({\mu }_n)y_n,\forall n\ge 1,\end{array}\right.$$

(24)

then {x _n } converges strongly to $x^{\ast }=P_{\mathcal{F}}f(x^{\ast })\in \mathcal{F}$, which is the unique solution of the following variational inequality:

$$〈(f-I)x^{\ast },z-x^{\ast }〉\le 0,\forall z\in \mathcal{F.}$$