A new iterative method for a common solution of fixed points for pseudo-contractive mappings and variational inequalities

Abstract

In this article, we introduce a new iterative scheme for finding a common element of the set of fixed points for a continuous pseudo-contractive mapping and the solution set of a variational inequality problem governed by continuous monotone mappings. Strong convergence for the proposed iterative scheme is proved. Our results improve and extend some recent results in the literature.

2000 Mathematics Subject Classification: 46C05; 47H09; 47H10.

1. Introduction

The theory of variational inequalities represents, in fact, a very natural generalization of the theory of boundary value problems and allows us to consider new problems arising from many fields of applied mathematics, such as mechanics, physics, engineering, the theory of convex programming, and the theory of control. While the variational theory of boundary value problems has its starting point in the method of orthogonal projection, the theory of variational inequalities has its starting point in the projection on a convex set.

Let C be a nonempty closed and convex subset of a real Hilbert space H. The classical variational inequality problem is to find a u ∈ C such that 〈v-u, Au〉 ≥ 0 for all v ∈ C, where A is a nonlinear mapping. The set of solutions of the variational inequality is denoted by VI(C, A). The variational inequality problem has been extensively studied in the literature, see [1–5] and the reference therein. In the context of the variational inequality problem, this implies that u ∈ VI(C, A) ⇔ u = P _C (u - λAu), ∀λ > 0, where P _C is a metric projection of H into C.

Let A be a mapping from C to H, then A is called monotone if and only if for each x, y ∈ C,

$$〈x-y,Ax-Ay〉\ge 0.$$

(1.1)

An operator A is said to be strongly positive on H if there exists a constant $\bar{\gamma }>0$ such that

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

A mapping A of C into itself is called L-Lipschitz continuous if there exits a positive and number L such that

$$∥Ax-Ay∥\le L∥x-y∥,\forall x,y\in C.$$

A mapping A of C into H is called α-inverse-strongly monotone if there exists a positive real number α such that

$$〈x-y,Ax-Ay〉\ge \alpha {∥Ax-Ay∥}^2,$$

for all x, y ∈ C; see [2, 6–10]. If A is an α-inverse strongly monotone mapping of C into H, then it is obvious that A is $\frac{1}{\alpha }$-Lipschitz continuous, that is, $∥Ax-Ay∥\le \frac{1}{\alpha }∥x-y∥$ for all x, y ∈ C. Clearly, the class of monotone mappings include the class of α-inverse strongly monotone mappings.

Recall that a mapping T of C into H is called pseudo-contractive if for each x, y ∈ C, we have

$$〈Tx-Ty,x-y〉\le {∥x-y∥}^2.$$

(1.2)

T is said to be a k-strict pseudo-contractive mapping if there exists a constant 0 ≤ k ≤ 1 such that

$$〈x-y,Tx-Ty〉\le {∥x-y∥}^2-k{∥\left(I-T\right)x-\left(I-T\right)y∥}^2,\text{for}\text{all}x,y\in D\left(T\right).$$

A mapping T of C into itself is called nonexpansive if ∥Tx - Ty∥ ≤ ∥x - y∥, for all x, y ∈ C. We denote by F(T) the set of fixed points of T. Clearly, the class of pseudo-contractive mappings include the class of nonexpansive and strict pseudo-contractive mappings.

For finding an element of F(T), where T is a nonexpansive mapping of C into itself, Halpern [11] was the first to study the convergence of the following scheme:

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

(1.3)

where u, x₀ ∈ C and a sequence {α _n } of real numbers in (0,1) in the framework of Hilbert spaces. Lions [12] improved the result of Halpern by proving strong convergence of {x _n } to a fixed point of T provided that the real sequence {α _n } satisfies certain mild conditions. In 2000, Moudafi [13] introduced viscosity approximation method and proved that if H is a real Hilbert space, for given x₀ ∈ C, the sequence {x _n } generated by the algorithm

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

(1.4)

where f : C → C is a contraction mapping with a constant β ∈ (0,1) and {α _n } ⊂ (0,1) satisfies certain conditions, converges strongly to fixed point of Moudafi [13] generalizes Halpern's theorems in the direction of viscosity approximations. In [14, 15], Zegeye and Shahzad extended Moudafi's result to Banach spaces which more general than Hilbert spaces. For other related results, see [16–18]. Viscosity approximations are very important because they are applied to convex optimization, linear programming, monotone inclusion and elliptic differential equations. Marino and Xu [19], studied the viscosity approximation method for nonexpansive mappings and considered the following general iterative method:

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

(1.5)

They proved that if the sequence {α _n } of parameters satisfies appropriate conditions, then the sequence {x _n } generated by (1.5) converges strongly to the unique solution of the variational inequality

$$〈\left(A-\gamma f\right)x^{*},x-x^{*}〉\ge 0,x\in C,$$

which is the optimality condition for the minimization problem

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

where h is a potential function for γf (i.e., h'(x) = γf(x) for x ∈ H).

For finding an element of F(T) ∩ VI(C, A), where T is nonexpansive and A is α-inverse strongly monotone, Takahashi and Toyoda [20] introduced the following iterative scheme:

$$x_{n+1}={\alpha }_n{x}_n+\left(1-{\alpha }_n\right)T{P}_C\left(x_n-{\lambda }_{n}A{x}_n\right),n\ge 0.$$

(1.6)

where x₀ ∈ C, {α _n } is a sequence in (0,1), and {λ _n } is a sequence in (0, 2α), and obtained weak convergence theorem in a Hilbert space H. Iiduka and Takahashi [7] proposed a new iterative scheme x₁ = x ∈ C and

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

(1.7)

and obtained strong convergence theorem in a Hilbert space.

Motivated and inspired by the work mentioned above which combined from Equations (1.5) and (1.6), in this article, we introduced a new iterative scheme (3.1) below which converges strongly to common element of the set of fixed points of continuous pseudo-contractive mappings which more general than nonexpansive mappings and the solution set of the variational inequality problem of continuous monotone mappings which more general than α-inverse strongly monotone mappings. As a consequence, we provide an iterative scheme which converges strongly to a common element of set of fixed points of finite family continuous pseudo-contractive mappings and the solutions set of finite family of variational inequality problems for continuous monotone mappings. Our theorems extend and unify most the results that have been proved for these important class of nonlinear operators.

2. Preliminaries

Let H be a nonempty closed and convex subset of a real Hilbert space H. Let A be a mapping from C into H. For every point x ∈ H, there exists a unique nearest point in C, denoted by P _C x, such that

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

PC is called the metric projection of H onto C. We know that P _C is a nonexpansive mapping of H onto C.

Lemma 2.1. Let H be a real Hilbert space. The following identity holds:

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

Lemma 2.2. Let C be a closed convex subset of a Hilbert space H. Let x ∈ H and x₀ ∈ C. Then x₀ = P _C x if and only if

$$〈z-x_0,x_0-x〉,\forall z\in C.$$

Lemma 2.3.[21]Let {a _n } be a sequence of nonnegative real numbers satisfying the following relation

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

where,

• (i) $\left\{{\gamma }_n\right\}\subset \left(0,1\right),\sum _{n=1}^{\infty }{\gamma }_n=\infty$;

• (ii) $\text{lim}\underset{n\to \infty }{\text{sup}}\frac{{\sigma }_n}{{\gamma }_n}\le 0or\sum _{n=1}^{\infty }\left|{\sigma }_n\right|<\infty$.

Then, the sequence {a _n } → 0 as n → ∞.

Lemma 2.4.[22]Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A : C → H be a continuous monotone mapping. Then, for r > 0 and x ∈ H, there exist z ∈ C such that

$$〈y-z,Az〉+\frac{1}{r}〈y-z,z-x〉\ge 0,\forall y\in C.$$

(2.1)

Moreover, by a similar argument of the proof of Lemmas 2.8 and 2.9 in[23], Zegeye[22]obtained the following lemmas:

Lemma 2.5.[22]Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A : C → H be a continuous monotone mapping. For r > 0 and x ∈ H, define a mapping F _r : H → C as follows:

$$F_{r}x:=\left\{z\in C:〈y-z,Az〉+\frac{1}{r}〈y-z,z-x〉\ge 0,\forall y\in C\right\}$$

for all x ∈ H. Then the following hold:

• (1) F _r is single-valued;

• (2) F _r is a firmly nonexpansive type mapping, i.e., for all x, y ∈ H,

  $${∥F_{r}x-F_{r}y∥}^2\le 〈F_{r}x-F_{r}y,x-y〉;$$

• (3) F(F _r ) = VI(C,A);

• (4) VI(C, A) is closed and convex.

In the sequel, we shall make use of the following lemmas:

Lemma 2.6.[22]Let C be a nonempty closed and convex subset of a real Hilbert space H. Let T : C → H be a continuous pseudo-contractive mapping. Then, for r > 0 and x ∈ H, there exist z ∈ C such that

$$〈y-z,Tz〉-\frac{1}{r}〈y-z,\left(1+r\right)z-x〉\le 0,\forall y\in C.$$

(2.2)

Lemma 2.7.[22]Let C be a nonempty closed and convex subset of a real Hilbert space H. Let T : C → C be a continuous pseudo-contractive mapping. For r > 0 and x ∈ H, define a mapping T _r : H → C as follows:

$$T_{r}x:=\left\{z\in C:〈y-z,Tz〉+\frac{1}{r}〈y-z,\left(1+r\right)z-x〉\le 0,\forall y\in C\right\}$$

for all x ∈ H. Then the following hold:

• (1) T _r is single - valued;

• (2) T _r is a firmly nonexpansive type mapping, i.e., for all x, y ∈ H,

  $${∥T_{r}x-T_{r}y∥}^2\le 〈T_{r}x-T_{r}y,x-y〉;$$

• (3) F(T _r ) = F(T);

• (4) F(T) is closed and convex.

Lemma 2.8.[19]Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient$\bar{\gamma }>0$and 0 < ρ ≤ ∥A∥^-1. Then$∥I-\rho A∥\le 1-\rho \bar{\gamma }$.

Let C be a nonempty closed and convex subset of a real Hilbert space H. Let T : C → C be a continuous pseudo-contractive mapping and A : C → H be a continuous monotone mapping. Then in what follows, $T_{r_n}$ and $F_{r_n}$ will be defined as follows: For x ∈ H and {r _n } ⊂ (0, ∞), defined

$$T_{r_n}x:=\left\{z\in C:〈y-z,Tz〉-\frac{1}{r_n}〈y-z,\left(1-r_n\right)z-x〉\le 0,\forall y\in C\right\}$$

and

$$F_{r_n}x:=\left\{z\in C:〈y-z,Az〉+\frac{1}{r_n}〈y-z,z-x〉\ge 0,\forall y\in C\right\}.$$

3. Strong convergence theorems

In this section, we will prove a strong convergence theorem for finding a common element of the set of fixed points for a continuous pseudo-contractive mapping and the solution set of a variational inequality problem governed by continuous monotone mappings.

Theorem 3.1. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let T : C → C be a continuous pseudo-contractive mapping and A : C → H be a continuous monotone mapping such that$\mathfrak{F}:=F\left(T\right)\cap VI\left(C,A\right)\ne \varnothing$. Let f be a contraction of H into itself with a contraction constant β and let B : H → H be a strongly positive linear bounded self-adjoint operator with coefficients$\bar{\beta }>0$and let {x _n } be a sequence generated by x₁ ∈ C and

$$\{\begin{array}{l}{y}_n=F_{r_n}{x}_n\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+{\delta }_n{x}_n+[(1-{\delta }_n)I-{\alpha }_{n}B]T_{r_n}{y}_n,\hfill \end{array}$$

(3.1)

where {α _n } ⊂ [0,1] and {r _n } ⊂ (0, ∞) such that

(C1) ${\text{lim}}_{n\to \infty }{\alpha }_n=0,\sum _{n=1}^{\infty }{\alpha }_n=\infty$;

(C2) $\underset{n\to \infty }{\text{lim}}{\delta }_n=0,\sum _{n=1}^{\infty }\left|{\delta }_{n+1}-{\delta }_n\right|<\infty$;

(C3) $\text{lim}\underset{n\to \infty }{\text{inf}}{r}_n>0,\sum _{n=1}^{\infty }\left|r_{n+1}-r_n\right|<\infty$.

Then, the sequence {x _n } converges strongly to$z\in \mathfrak{F}$, which is the unique solution of the variational inequality:

$$〈\left(B-\gamma f\right)z,x-z〉\ge 0,\forall x\in \mathfrak{F}.$$

(3.2)

Equivalently, $z=P_{\mathfrak{F}}\left(I-B+\gamma f\right)z$, which is the optimality condition for the minimization problem

$$\underset{x\in C}{\text{min}}\frac{1}{2}〈Az,z〉-h\left(z\right),$$

where h is a potential function for γf (i.e.,h'(z) = γf(z) for z ∈ H).

Remark: (1) The variational inequality (3.2) has the unique solution; (see [19]). (2) It follows from condition (C1) that (1 - δ _n )I - α _n B is positive and $∥\left(1-{\delta }_n\right)I-{\alpha }_{n}B∥\le I-{\delta }_n-{\alpha }_n\bar{\beta }$ for all n ≥ 1; (see [24]).

Proof. We processed the proof with following four steps:

Step 1. First, we will prove that the sequence {x _n } is bounded.

Let $v\in \mathfrak{F}$ and let $u_n=T_{r_n}{y}_n$ and $y_n=F_{r_n}{x}_n$. Then, from Lemmas 2.5 and 2.7 that

$$∥u_n-v∥=∥T_{r_n}{y}_n-T_{r_n}v∥\le ∥y_n-v∥=∥F_{r_n}{x}_n-F_{r_n}v∥\le ∥x_n-v∥.$$

(3.3)

Moreover, from (3.1) and (3.2), we compute

$$\begin{array}{c}\Vert x_{n+1}-v\Vert =\Vert {\alpha }_n(\gamma (f(x_n)-Bv))+{\delta }_n(x_n-v)+[(1-{\delta }_n)I-{\alpha }_{n}B]T_{r_n}-v\Vert .\\ \le {\alpha }_n\Vert \gamma f(x_n)-Bv)\Vert +{\delta }_n\Vert (x_n-v)\Vert +\Vert (1-{\delta }_n)I-{\alpha }_{n}B\Vert \Vert T_{r_n}-v\Vert .\\ \le {\alpha }_n\beta \gamma \Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert +{\delta }_n\Vert x_n-v\Vert +(1+{\delta }_n-{\alpha }_n\overline{\beta })\Vert T_{r_n}{y}_n-v\Vert .\\ \le {\alpha }_n\beta \gamma \Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert +{\delta }_n\Vert x_n-v\Vert +(1+{\delta }_n-{\alpha }_n\overline{\beta })\Vert u_n-v\Vert .\\ \le {\alpha }_n\beta \gamma \Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert +{\delta }_n\Vert x_n-v\Vert +(1+{\delta }_n-{\alpha }_n\overline{\beta })\Vert x_n-v\Vert .\\ ={\alpha }_n\beta \gamma \Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert +{\delta }_n\Vert x_n-v\Vert +\Vert x_n-v\Vert \\ -{\delta }_n\Vert x_n-v\Vert -{\alpha }_n\overline{\beta }\Vert x_n-v\Vert .\\ ={\alpha }_n\beta \gamma \Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert +\Vert x_n-v\Vert -{\alpha }_n\overline{\beta }\Vert x_n-v\Vert .\\ \le \left({\alpha }_n\beta \gamma +1-{\alpha }_n\overline{\beta }\right)\Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert .\\ =(1-{\alpha }_n(\overline{\beta }-\beta \gamma ))\Vert x_n-v\Vert +{\alpha }_n\Vert \gamma f(v)-Bv\Vert .\\ \le \max \left\{\Vert x_n-v\Vert ,\frac{\Vert \gamma f(v)-Bv\Vert }{\overline{\beta }-\beta \gamma },\right\}\forall n\ge 1.\end{array}$$

Therefore, by the simple introduction, we have

$$∥x_n-v∥=\text{max}\left\{∥x_1-v∥,\frac{∥\gamma f\left(v\right)-Bv∥}{\bar{\beta }-\beta \gamma }\right\},\forall n\ge 1$$

which show that {x _n } is bounded, so {y _n }, {u _n }, and {f(x _n )} are bounded.

Step 2. We will show that ∥x_n+1- x _n ∥ → 0 and ∥u _n - y _n ∥ → 0 as n → ∞.

Notice that each $T_{r_n}$ and $F_{r_n}$ are firmly nonexpansive. Hence, we have

$$∥u_{n+1}-u_n∥=∥T_{r_n}{y}_{n+1}-T_{r_n}{y}_n∥\le ∥y_{n+1}-y_n∥=∥F_{r_n}{x}_{n+1}-F_{r_n}{x}_n∥\le ∥x_{n+1}-x_n∥.$$

From (3.1), we note that

$$\begin{array}{ll}\hfill ∥x_{n+1}-x_n∥& =∥{\alpha }_n\gamma f\left(x_n\right)+{\delta }_n{x}_n+\left[\left(1-{\delta }_n\right)I-{\alpha }_{n}B\right]T_{r_n}{y}_n\\ -{\alpha }_{n-1}\gamma f\left(x_{n-1}\right)-{\delta }_{n-1}{x}_{n-1}-\left[\left(1-{\delta }_{n-1}\right)I-{\alpha }_{n-1}B\right]T_{r_n}{y}_{n-1}∥.\\ =∥{\alpha }_n\gamma f\left(x_n\right)+{\delta }_n{x}_n+\left(I-{\delta }_n-{\alpha }_{n}B\right)u_n\\ -{\alpha }_{n-1}\gamma f\left(x_{n-1}\right)-{\delta }_{n-1}{x}_{n-1}-\left(I-{\delta }_{n-1}-{\alpha }_{n-1}B\right)u_{n-1}∥.\\ \le ∥{\alpha }_n\gamma f\left(x_n\right)-{\alpha }_n\gamma f\left(x_{n-1}\right)+{\alpha }_n\gamma f\left(x_{n-1}\right)+{\delta }_n{x}_n-{\delta }_{n-1}{x}_{n-1}-{\alpha }_{n-1}\gamma f\left(x_{n-1}\right)\\ +\left(I-{\delta }_n-{\alpha }_{n}B\right)u_n-\left(I-{\delta }_n-{\alpha }_{n}B\right)u_{n-1}+\left(I-{\delta }_n-{\alpha }_{n}B\right)u_{n-1}\\ -\left(I-{\delta }_{n-1}-{\alpha }_{n-1}B\right)u_{n-1}∥.\\ \le ∥{\alpha }_n\gamma f\left(x_n\right)-{\alpha }_n\gamma f\left(x_{n-1}\right)∥+∥{\alpha }_n\gamma f\left(x_{n-1}\right)-{\alpha }_{n-1}\gamma f\left(x_{n-1}\right)∥+∥{\delta }_n{x}_n-{\delta }_{n-1}{x}_{n-1}∥\\ +∥\left(I-{\delta }_n-{\alpha }_{n}B\right)u_n-\left(I-{\delta }_n-{\alpha }_{n}B\right)u_{n-1}∥+∥\left(I-{\delta }_n-{\alpha }_{n}B\right)u_{n-1}\\ -\left(I-{\delta }_{n-1}-{\alpha }_{n-1}B\right)u_{n-1}∥.\\ ={\alpha }_n\gamma ∥f\left(x_n\right)-f\left(x_{n-1}\right)∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|∥\gamma f\left(x_{n-1}\right)∥+{\delta }_n{x}_n-{\delta }_{n-1}{x}_{n-1}+{\delta }_n{x}_{n-1}-{\delta }_n{x}_{n-1}\\ +\left(I-{\delta }_n-{\alpha }_{n}B\right)∥u_n-u_{n-1}∥+∥\left(I-{\delta }_n-{\alpha }_{n}B-I+{\delta }_{n-1}+{\alpha }_{n-1}B\right)u_{n-1}∥.\\ ={\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|\gamma ∥f\left(x_{n-1}\right)∥+{\delta }_n∥x_n-x_{n-1}∥+\left|{\delta }_n-{\delta }_{n-1}\right|∥x_{n-1}∥\\ +\left|I-{\delta }_n-{\alpha }_{n}B\right|∥u_n-u_{n-1}∥+\left|{\delta }_{n-1}-{\delta }_n+{\alpha }_{n-1}B+{\alpha }_{n}B\right|∥u_{n-1}∥.\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|\gamma ∥f\left(x_{n-1}\right)∥+{\delta }_n∥x_n-x_{n-1}∥+\left|{\delta }_n-{\delta }_{n-1}\right|∥x_{n-1}∥\\ +\left|I-{\delta }_n-{\alpha }_{n}B\right|∥x_n-x_{n-1}∥+\left|{\delta }_{n-1}-{\delta }_n\right|∥u_{n-1}∥+\left|{\alpha }_{n-1}B+{\alpha }_{n}B\right|∥u_{n-1}∥.\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|\gamma ∥f\left(x_{n-1}\right)∥+{\delta }_n∥x_n-x_{n-1}∥+\left|{\delta }_n-{\delta }_{n-1}\right|∥x_{n-1}∥\\ +\left|I-{\delta }_n-{\alpha }_{n}B\right|∥x_n-x_{n-1}∥+\left|{\delta }_{n-1}-{\delta }_n\right|∥x_{n-1}∥-\left|{\alpha }_{n-1}-{\alpha }_n\right|B∥x_{n-1}∥.\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|\gamma ∥f\left(x_{n-1}\right)∥+{\delta }_n∥x_n-x_{n-1}∥\\ +\left|I-{\delta }_n-{\alpha }_{n}B\right|∥x_n-x_{n-1}∥-\left|{\alpha }_{n-1}-{\alpha }_n\right|B∥x_{n-1}∥.\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|\gamma ∥f\left(x_{n-1}\right)∥+\left|{\delta }_n+I-{\delta }_n-{\alpha }_{n}B\right|∥x_n-x_{n-1}∥\\ -\left|{\alpha }_{n-1}-{\alpha }_n\right|B∥x_{n-1}∥.\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|\gamma ∥f\left(x_{n-1}\right)∥+\left|I-{\alpha }_{n}B\right|∥x_n-x_{n-1}∥\\ -\left|{\alpha }_n-{\alpha }_{n-1}\right|B∥x_{n-1}∥.\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|∥\gamma f\left(x_{n-1}\right)-B{x}_{n-1}∥+\left|I-{\alpha }_{n}B\right|∥x_n-x_{n-1}∥\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|∥\gamma f\left(x_{n-1}\right)-B{x}_{n-1}∥+\left|I-{\alpha }_{n}B\right|∥y_n-y_{n-1}∥\\ \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|K+\left|I-{\alpha }_{n}B\right|∥y_n-y_{n-1}∥,\end{array}$$

(3.4)

where K = ∥γf(x_n-1) - Bx_n-1∥ = 2 sup{∥f(x _n ) ∥ + ∥u _n ∥:n ∈ N}. Moreover, since $y_n=F_{r_n}{x}_n$ and $y_{n+1}=F_{r_{n+1}}{x}_{n+1}$, we get

$$〈y-y_n,A{y}_n〉+\frac{1}{r_n}〈y-y_n,y_n-x_n〉\ge 0,\forall y\in C$$

(3.5)

and

$$〈y-y_{n+1},A{y}_{n+1}〉+\frac{1}{r_{n+1}}〈y-y_{n+1},y_{n+1}-x_{n+1}〉\ge 0,\forall y\in C.$$

(3.6)

Putting y = y_n+1in (3.5) and y = y _n in (3.6), we obtain

$$〈y_{n+1}-y_n,A{y}_n〉+\frac{1}{r_n}〈y_{n+1}-y_n,y_n-x_n〉\ge 0$$

(3.7)

and

$$〈y_n-y_{n+1},A{y}_{n+1}〉+\frac{1}{r_{n+1}}〈y_n-y_{n+1},y_{n+1}-x_{n+1}〉\ge 0.$$

(3.8)

Adding (3.7) and (3.8), we have

$$〈y_{n+1}-y_n,A{y}_n-A{y}_{n+1}〉+〈y_{n+1}-y_n,\frac{y_n-x_n}{r_n}-\frac{y_{n+1}-x_{n+1}}{r_{n+1}}〉\ge 0$$

which implies that

$$-〈y_{n+1}-y_n,A{y}_{n+1}-A{y}_n〉+〈y_{n+1}-y_n,\frac{y_n-x_n}{r_n}-\frac{y_{n+1}-x_{n+1}}{r_{n+1}}〉\ge 0.$$

Using the fact that A is monotone, we get

$$〈y_{n+1}-y_n,\frac{y_n-x_n}{r_n}-\frac{y_{n+1}-x_{n+1}}{r_{n+1}}〉\ge 0.$$

and hence

$$〈y_{n+1}-y_n,y_n-y_{n+1}+y_{n+1}-y_n-\frac{r_n}{r_{n+1}}\left(y_{n+1}-x_{n+1}\right)〉\ge 0.$$

We observe that

$$\begin{array}{ll}\hfill {∥y_{n+1}-y_n∥}^2& \le 〈y_{n+1}-y_n,x_{n+1}-x_n\left(1-\frac{r_n}{r_{n+1}}\right)\left(y_{n+1}-x_{n+1}\right)〉\\ \le ∥y_{n+1}-y_n∥\left\{∥x_{n+1}-x_n∥+\left|\left(1-\frac{r_n}{r_{n+1}}\right)\right|∥\left(y_{n+1}-x_{n+1}\right)∥\right\}.\end{array}$$

(3.9)

Without loss of generality, let k be a real number such that r _n > k > 0 for all n ∈ N. Then, we have

$$\begin{array}{ll}\hfill ∥y_{n+1}-y_n∥& \le ∥x_{n+1}-x_n∥+\frac{1}{r_{n+1}}\left|r_{n+1}-r_n\right|∥y_{n+1}-x_{n+1}∥\\ \le ∥x_{n+1}-x_n∥+\frac{1}{k}\left|r_{n+1}-r_n\right|M,\end{array}$$

(3.10)

where M = sup{∥y _n - x _n ∥: n ∈ N}. Furthermore, from (3.4) and (3.10), we have

$$\begin{array}{ll}\hfill ∥x_{n+1}-x_n∥& \le {\alpha }_n\gamma \beta ∥x_n-x_{n-1}∥+∥{\alpha }_n-{\alpha }_{n-1}∥K+\left(1-{\alpha }_n\right)\left(∥x_n-x_{n-1}∥+\frac{1}{k}\left|r_n-r_{i-1}\right|M\right).\\ =\left(1-{\alpha }_n+{\alpha }_n\gamma \beta \right)∥x_n-x_{n-1}∥+\left|{\alpha }_n-{\alpha }_{n-1}\right|K+\frac{1}{k}\left|r_n-r_{n-1}\right|M.\\ =\left(1-{\alpha }_n\left(1-\gamma \beta \right)\right)∥x_n-x_{n-1}∥+K\left|{\alpha }_n-{\alpha }_{n-1}\right|+\frac{M}{k}\left|r_n-r_{n-1}\right|.\end{array}$$

Using Lemma 2.3, and by the conditions (C1) and (C3), we have

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

Consequently, from (3.10), we obtain

$$\underset{n\to \infty }{\text{lim}}∥y_{n+1}-y_n∥=0.$$

(3.11)

Since $u_n=T_{r_n}{y}_n$ and $u_{n+1}=T_{r_{n+1}}{y}_{n+1}$, we have

$$〈y-u_n,T{u}_n〉-\frac{1}{r_n}〈y-u_n,\left(1-r_n\right)u_n-y_n〉\le 0,\forall y\in C$$

(3.12)

and

$$〈y-u_{n+1},T{u}_{n+1}〉-\frac{1}{r_{n+1}}〈y-u_{n+1},\left(1-r_{n+1}\right)u_{n+1}-y_{n+1}〉\le 0,\forall y\in C.$$

(3.13)

Putting y := u_n+1in (3.12) and y := u _n in (3.13), we get

$$〈u_{n+1}-u_n,T{u}_n〉-\frac{1}{r_n}〈u_{n+1}-u_n,\left(1-r_n\right)u_n-y_n〉\le 0.$$

(3.14)

and

$$〈u_n-u_{n+1},T{u}_{n+1}〉-\frac{1}{r_{n+1}}〈u_n-u_{n+1},\left(1-r_{n+1}\right)u_{n+1}-y_{n+1}〉\le 0.$$

(3.15)

Adding (3.14) and (3.15), we have

$$〈u_{n+1}-u_n,T{u}_n-T{u}_{n+1}〉-〈u_{n+1}-u_n,\frac{\left(1-r_n\right)u_n-y_n}{r_n}-\frac{\left(1-r_{n+1}\right)u_{n+1}-y_{n+1}}{r_{n+1}}〉\le 0.$$

Using the fact that T is pseudo-contractive, we get

$$〈u_{n+1}-u_n,\frac{u_n-y_n}{r_n}-\frac{u_{n+1}-y_{n+1}}{r_{n+1}}〉\ge 0$$

and hence

$$〈u_{n+1}-u_n,u_n-u_{n+1}+u_{n+1}-y_n-\frac{r_n}{r_{n+1}}\left(u_{n+1}+y_{n+1}\right)〉\ge 0.$$

Thus, using the methods in (3.9) and (3.10), we can obtain

$$∥u_{n+1}-u_n∥\le ∥y_{n+1}-y_n∥+\frac{1}{r_{n+1}}\left|r_{n+1}+r_n\right|M_1,$$

(3.16)

where M₁ = sup{∥u _n - y _n ∥: n ∈ N}. Therefore, from (3.11) and property of {r _n }, we get

$$\underset{n\to \infty }{\text{lim}}∥u_{n+1}-u_n∥=0.$$

Furthermore, since $x_n={\alpha }_{n-1}\gamma f\left(x_{n+1}\right)+{\delta }_{n-1}{x}_{n-1}+\left[\left(1-{\delta }_{n-1}\right)I-{\alpha }_{n-1}B\right]T_{r_n}{y}_{n-1}$, we have

$$\begin{array}{ll}\hfill ∥x_n-u_n∥& \le ∥x_n-u_{n-1}∥+∥u_{n-1}-u_n∥\\ =∥{\alpha }_{n-1}\gamma f\left(x_{n-1}\right)+{\delta }_{n-1}{x}_{n-1}+\left[\left(1-{\delta }_{n-1}\right)I-{\alpha }_{n-1}B\right]T_{r_n}{y}_{n-1}-u_{n-1}∥+∥u_{n-1}-u_n∥\\ ={\alpha }_{n-1}\gamma f\left(x_{n-1}\right)+{\delta }_{n-1}{x}_{n-1}+\left(I-{\delta }_{n-1}-{\alpha }_{n-1}B\right)u_{n-1}-u_{n-1}∥+∥u_{n-1}-u_n∥\\ ={\alpha }_{n-1}\gamma f\left(x_{n-1}\right)+{\delta }_{n-1}{x}_{n-1}+u_{n-1}-{\delta }_{n-1}{u}_{n-1}-{\alpha }_{n-1}B{u}_{n-1}-u_{n-1}∥+∥u_{n-1}-u_n∥\\ \le {\alpha }_{n-1}\gamma f\left(x_{n-1}\right)-{\alpha }_{n-1}B{u}_{n-1}+{\delta }_{n-1}{x}_{n-1}-{\delta }_{n-1}{u}_{n-1}∥+∥u_{n-1}-u_n∥\\ \le {\alpha }_{n-1}∥\gamma f\left(x_{n-1}\right)-B{u}_{n-1}∥+{\delta }_{n-1}∥x_{n-1}-u_{n-1}∥+∥u_{n-1}-u_n∥.\end{array}$$

Thus, by (C1) and (C2), we obtain

$$∥x_n-u_n∥\to 0,n\to \infty .$$

(3.17)

For $v\in \mathfrak{F}$, using Lemma 2.5, we obtain

$$\begin{array}{ll}\hfill {∥y_n-v∥}^2& ={∥F_{r_n}{y}_n-F_{r_{n}v}∥}^2\\ \le 〈F_{r_n}{x}_n-F_{r_n}v,x_n-v〉\\ \le 〈y_n-v,x_n-v〉\\ =\frac{1}{2}\left({∥y_n-v∥}^2+{∥x_n-v∥}^2-{∥x_n-y_n∥}^2\right)\end{array}$$

and

$${∥y_n-v∥}^2\le {∥x_n-v∥}^2-{∥x_n-y_n∥}^2.$$

(3.18)

Therefore, from (3.1), the convexity of ∥·∥², (3.2) and (3.18), we get

$$\begin{array}{ll}\hfill {∥x_{n+1}-v∥}^2& ={∥{\alpha }_n\gamma f\left(x_n\right)+{\delta }_n{x}_n+\left[\left(1-{\delta }_n\right)I-{\alpha }_{n}B\right]T_{r_n}{y}_n-v∥}^2\\ ={∥\left(1-{\delta }_n\right)\left(T_{r_n}{y}_n-v\right)+{\delta }_n\left(x_n-v\right)+{\alpha }_n\left(\gamma f\left(x_n\right)-B{T}_{r_n}{y}_n\right)∥}^2\\ \le {∥\left(1-{\delta }_n\right)\left(T_{r_n}{y}_n-v\right)+{\delta }_n\left(x_n-v\right)∥}^2+2{\alpha }_n〈\gamma f\left(x_n\right)-B{T}_{r_n}{y}_n,x_{n+1}-v〉\\ \le \left(1-{\delta }_n\right){∥\left(y_n-v\right)∥}^2+{\delta }_n{∥\left(x_n-v\right)∥}^2+2{\alpha }_n{L}^2\end{array}$$

and hence

$$\left(1-{\delta }_n\right){∥\left(y_n-v\right)∥}^2\le {\delta }_n{∥\left(x_n-v\right)∥}^2-{∥\left(x_{n+1}-v\right)∥}^2+2{\alpha }_n{L}^2.$$

(3.19)

So, we have ∥y _n - v∥ → 0 as n → ∞. Consequently, from (3.16) and (3.18), we obtain

$$∥u_n-y_n∥\le ∥u_n-x_n∥+∥x_n-y_n∥\to 0\text{as}n\to \infty .$$

Step 3. We will show that

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈\left(\gamma f-B\right)z,x_n-z〉\le 0.$$

(3.20)

Let $Q=P_{\mathfrak{F}}$, and since, Q(I - B + γf) is contraction on H into C (see also [[25], pp. 18]) and H is complete. Thus, by Banach Contraction Principle, then there exist a unique element z of H such that z = Q(I - B + γf)z.

We choose subsequence $\left\{x_{n_i}\right\}$ of {x _n } such that

$$\underset{n\to \infty }{\text{lim}\text{sup}}〈\left(\gamma f-B\right)z,x_n-z〉=\underset{n\to \infty }{\text{lim}}〈\gamma fz-Bz,x_{n_i}-z〉$$

Since $\left\{x_{n_i}\right\}$ is bounded, there exists a sequence $\left\{x_{n_{ij}}\right\}$ of $\left\{x_{n_i}\right\}$ and y ∈ C such that $\left\{x_{n_{ij}}\right\}\rightharpoonup y$. Without loss of generality, we may assume that $x_{n_i}\rightharpoonup y$. Since C is closed and convex it is weakly closed and hence y ∈ C. Since x _n - y _n → 0 as n → ∞ we have that $y_{n_i}\rightharpoonup y$. Now, we show that $y\in \mathfrak{F}$. Since $y_n=F_{r_n}$, Lemma 2.5 and using (3.5), we get

$$〈y-y_n,A{y}_n〉+〈y-y_n,\frac{y_n-x_n}{r_n}〉\ge 0,\forall y\in C.$$

(3.21)

and

$$〈y-y_{n_i},A{y}_{n_i}〉+〈y-y_{n_i},\frac{y_{n_i}-x_{n_i}}{r_{n_i}}〉\ge 0,\forall y\in C.$$

(3.22)

Set v _t = tv + (1- t)y for all t ∈ (0,1] and v ∈ C. Consequently, we get v _i ∈ C. From (3.22), it follows that

$$\begin{array}{ll}\hfill 〈v_t-y_{n_i}〉& \ge 〈v_t-y_{n_i},A{v}_t〉-〈v_t-y_{n_i},A{v}_t〉-〈v_t-y_{n_i},\frac{y_{n_i}-x_{n_i}}{r_n}〉\\ =〈v_t-y_{n_i},A{v}_t-A{y}_{n_i}〉-〈v_t-y_{n_i},\frac{y_{n_i}-x_{n_i}}{r_n}〉,\end{array}$$

from the fact that $y_{n_i}-x_{n_i}\to 0$ as i → ∞, we obtain that $\frac{u_{n_i}-x_{n_i}}{r_n}\to 0$ as i → ∞. Since A is monotone, we also have that $〈v_t-y_{n_i},A{v}_t-A{y}_{n_i}〉\ge 0$. Thus, if follows that

$$0\le \underset{i\to \infty }{\text{lim}}〈v_t-y_{n_i},A{v}_t〉=〈v_t-w,A{v}_t〉,$$

and hence $〈v-y,A{v}_t〉\ge 0,\forall v\in C$.

If t → 0, the continuity of A gives that

$$〈v-y,Ay〉\ge 0,\forall v\in C.$$

This implies that y ∈ VI(C, A).

Furthermore, since $u_n=T_{r_n}{y}_n$, Lemma 2.5 and using (3.12), we get

$$〈y-u_{n_i},T{u}_{n_i}〉-\frac{1}{r_n}〈y-u_{n_i},\left(r_{n_i}+1\right)u_{n_i}-y_{n+1}〉\le 0,\forall y\in C.$$

(3.23)

Put z _t = t(v) + (1 - t)y for all t ∈ (0,1] and v ∈ C. Then, z _t ∈ C and from (3.23) and pseudo-contractivity of T, we get

$$\begin{array}{ll}\hfill ∥u_{n_i}-z_t,T{z}_t∥& =〈u_{n_i}-z_t,T{z}_t〉+〈z_t-u_{n_i},T{u}_i〉-\frac{1}{r_n}〈z_t-u_{n_i},\left(1+r_{n_i}\right)u_{n_i}-y_{n_i}〉\\ =-〈z_t-u_{n_i},T{z}_t〉-\frac{1}{r_{n_i}}〈z_t-u_{n_i},u_{n_i}-y_{n_i}〉-〈z_t-u_{n_i},u_{n_i}〉\\ \ge {∥z_t-u_{n_i}∥}^2-\frac{1}{r_{n_i}}〈z_t-u_{n_i},u_{n_i}-y_{n_i}〉-〈z_t-u_{n_i},u_{n_i}〉\\ =-〈z_t-u_{n_i},z_t〉-〈z_t-u_{n_i},\frac{u_{n_i}-y_{n_i}}{r_{n_i}}〉.\end{array}$$

Thus, since u _n - y _n → 0, as n → ∞ we obtain that $\frac{u_{n_i}-y_{n_i}}{r_{n_i}}\to 0$ as i → ∞. Therefore, as i → ∞, it follows that

$$〈y-z_t,T{z}_t〉\ge 〈y-z_t,z_t〉$$

and hence

$$-〈v-y,T{z}_t〉\ge -〈v-y,z_t〉,\forall v\in C.$$

Taking t → 0 and since T is continuous we obtain

$$-〈v-y,Ty〉\ge -〈v-y,y〉,\forall v\in C.$$

Now, we get v = Ty. Then we obtain that y = Ty and hence y ∈ F(T). Therefore, y ∈ F(T) ∩ VI(C, A) and since $z=P_{\mathfrak{F}}\left(I-B+\gamma f\right)z$, Lemma 2.2 implies that

$$\begin{array}{ll}\hfill \underset{n\to \infty }{\text{lim}\text{sup}}〈\left(\gamma f-B\right)z,x_n-z〉& =\underset{i\to \infty }{\text{lim}}〈\left(I-B+\gamma f\right)z-z,x_{n_i}-z〉\\ =〈\left(\gamma f-B\right)z,y-z〉\le 0.\end{array}$$

(3.24)

Step 4. Finally, we will show that x _n → z as n → ∞, where $z=P_{\mathfrak{F}}\left(I-B+rf\right)z$.

From (3.1) and (3.2) we observe that

$$\begin{array}{ll}\hfill {∥x_{n+1}-z∥}^2& =〈{\alpha }_n\gamma f\left(x_n\right)+{\delta }_n{x}_n+\left[\left(1-{\delta }_n\right)I-{\alpha }_{n}B\right]T_{r_n}{y}_n-z,x_{n+1}-z〉\\ ={\alpha }_n〈\gamma f\left(x_n\right)-Bz,x_{n+1}-z〉+{\delta }_n〈x_n-z,x_{n+1}-z〉\\ +〈\left[\left(1-{\delta }_n\right)I-{\alpha }_{n}B\right]\left(T_{r_n}-z\right),x_{n+1}-z〉\\ \le {\alpha }_n\gamma 〈f\left(x_n\right)-f\left(z\right),x_{n+1}-z〉+{\alpha }_n〈\gamma f\left(z\right)-Bz,x_{n+1}-z〉\\ +{\delta }_n∥x_n-z∥∥x_{n+1}-z∥+\left(1-{\delta }_n-{\alpha }_n\bar{\beta }\right)∥z_n-z∥∥x_{n+1}-z∥\\ \le {\alpha }_n\gamma K∥x_n-z∥∥x_{n+1}-z∥+{\alpha }_n〈\gamma f\left(z\right)-Bz,x_{n+1}-z〉\\ +{\delta }_n∥x_n-z∥∥x_{n+1}-z∥+\left(1-{\delta }_n-{\alpha }_n\bar{\beta }\right)∥z_n-z∥∥x_{n+1}-z∥\\ ={\alpha }_n\gamma K∥x_n-z∥∥x_{n+1}-z∥+{\alpha }_n〈\gamma f\left(z\right)-Bz,x_{n+1}-z〉\\ +\left(1-{\alpha }_n\bar{\beta }\right)∥x_n-z∥∥x_{n+1}-z∥\\ \le \frac{\gamma k}{2}{\alpha }_n\left({∥x_n-z∥}^2+{∥x_{n+1}-z∥}^2\right)+{\alpha }_n〈\gamma f\left(z\right)-Bz,x_{n+1}-z〉\\ +\left(1-{\alpha }_n\bar{\beta }\right)\left(∥x_n-z∥∥x_{n+1}-z∥\right)\\ \le \frac{\gamma k}{2}{\alpha }_n\left({∥x_n-z∥}^2+{∥x_{n+1}-z∥}^2\right)+{\alpha }_n〈\gamma f\left(z\right)-Bz,x_{n+1}-z〉\\ +\frac{\left(1-{\alpha }_n\bar{\beta }\right)}{2}\left({∥x_n-z∥}^2+{∥x_{n+1}-z∥}^2\right)\\ \le \frac{1-{\alpha }_n\left(\bar{\beta }-k\gamma \right)}{2}{∥x_n-z∥}^2+\frac{1}{2}{∥x_{n+1}-z∥}^2+{\alpha }_n〈\gamma f\left(x\right)-Bz,x_{n+1}-z〉,\end{array}$$

which implies that

$${∥x_{n+1}+z∥}^2\le \left[1-{\alpha }_n\left(\bar{\beta }-k\gamma \right)\right]{∥x_n-z∥}^2+2{\alpha }_n〈\gamma f\left(z\right)-Bz,x_{n+1}-z〉.$$

By the condition (C1), (3.24) and using Lemma 2.3, we see that lim_n→∞∥x _n - z∥ = 0. This complete to proof. □

If we take f(x) = u, ∀x ∈ H and γ = 1, then by Theorem 3.1, we have the following corollary:

Corollary 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C → C be a continuous pseudo-contractive mapping and A : C → H be a continuous monotone mapping such that$\mathfrak{F}:=F\left(T\right)\cap VI\left(C,A\right)\ne \varnothing$. let B : H → H be a strongly positive linear bounded self-adjoint operator with coefficients$\bar{\beta }>0$and let {x _n } be a sequence generated by x₁ ∈ H and

$$\{\begin{array}{l}{y}_n=F_{r_n}{x}_n\hfill \\ x_{n+1}={\alpha }_{n}u+{\delta }_n{x}_n+[(1-{\delta }_n)I-{\alpha }_{n}B]T_{r_n}{y}_n,\hfill \end{array}$$

(3.25)

where {α _n } ⊂ [0,1] and {r _n } ⊂ (0, ∞) such that

(C1) $\underset{n\to \infty }{\text{lim}}{\alpha }_n=0,\sum _{n=1}^{\infty }{\alpha }_n=\infty$;

(C2) $\underset{n\to \infty }{\text{lim}}{\delta }_n=0,\sum _{n=1}^{\infty }\left|{\delta }_{n+1}-{\delta }_n\right|<\infty$;

(C3) $\underset{n\to \infty }{\text{lim}\text{inf}}{r}_n>0,\sum _{n=1}^{\infty }\left|r_{n+1}-r_n\right|<\infty$.

Then, the sequence {x _n } converges strongly to$z\in \mathfrak{F}$, which is the unique solution of the variational inequality:

$$〈\left(B-f\right)z,x-z〉\ge 0,\forall x\in \mathfrak{F}.$$

(3.26)

Equivalently, $z=P_{\mathfrak{F}}\left(I-B+f\right)z$.

If we take T ≡ 0, then $T_{r_n}\equiv I$ (the identity map on C). So by Theorem 3.1, we obtain the following corollary.

Corollary 3.3. Let C be a nonempty closed convex subset of a real Hilbert space H. Let A : C → H be a continuous monotone mapping such that$\mathfrak{F}:=VI\left(C,A\right)\ne \varnothing$. Let f be a contraction of H into itself and let B : H → H be a strongly positive linear bounded self-adjoint operator with coefficients$\bar{\beta }>0$and let {x _n } be a sequence generated by x₁ ∈ H and

$$x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+{\delta }_n{x}_n+\left[\left(1-{\delta }_n\right)I-{\alpha }_{n}B\right]F_{r_n}{x}_n,$$

(3.27)

where {α _n } ⊂ [0,1] and {r _n } ⊂ (0, ∞) such that

(C1) $\underset{n\to \infty }{\text{lim}}{\alpha }_n=0,\sum _{n=1}^{\infty }{\alpha }_n=\infty$;

(C2) $\underset{n\to \infty }{\text{lim}}{\delta }_n=0,\sum _{n=1}^{\infty }\left|{\delta }_{n+1}-{\delta }_n\right|<\infty$;

(C3) $\underset{n\to \infty }{\text{lim}\text{inf}}{r}_n>0,\sum _{n=1}^{\infty }\left|r_{n+1}-r_n\right|<\infty$.

Then, the sequence {x _n } converges strongly to$z\in \mathfrak{F}$, which is the unique solution of the variational inequality:

$$〈\left(B-\gamma f\right)z,x-z〉\ge 0,\forall x\in \mathfrak{F}$$

(3.28)

Equivalently, $z=P_{\mathfrak{F}}\left(I-B+\gamma f\right)z$.

If we take A ≡ 0, then $F_{r_n}\equiv I$ (the identity map on C). So by Theorem 3.1, we obtain the following corollary.

Corollary 3.4. Let C be a nonempty closed convex subset of a real Hilbert space H. Let T : C → C be a continuous pseudo-contractive mapping such that$\mathfrak{F}:=F\left(T\right)\ne \varnothing$. Let f be a contraction of H into itself and let B : H → H be a strongly positive linear bounded self-adjoint operator with coefficients$\bar{\beta }>0$and let {x _n } be a sequence generated by x₁ ∈ H and

$$x_{n+1}={\alpha }_n\gamma f\left(x_n\right)+{\delta }_n{x}_n+\left[\left(1-{\delta }_n\right)I-{\alpha }_{n}B\right]T_{r_n}{x}_n,$$

(3.29)

where {α _n } ⊂ [0,1] and {r _n } ⊂ (0, ∞) such that

(C1) $\underset{n\to \infty }{\text{lim}}{\alpha }_n=0,\sum _{n=1}^{\infty }{\alpha }_n=\infty$;

(C2) $\underset{n\to \infty }{\text{lim}}{\delta }_n=0,\sum _{n=1}^{\infty }\left|{\delta }_{n+1}-{\delta }_n\right|<\infty$;

(C3) $\underset{n\to \infty }{\text{lim}\text{inf}}{r}_n>0,\sum _{n=1}^{\infty }\left|r_{n+1}-r_n\right|<\infty$.

Then, the sequence {x _n }_n≥1converges strongly to$z\in \mathfrak{F}$, which is the unique solution of the variational inequality:

$$〈\left(B-\gamma f\right)z,x-z〉\ge 0,\forall x\in \mathfrak{F}.$$

(3.30)

Equivalently, $z=P_{\mathfrak{F}}\left(I-B+\gamma f\right)z$.

If we take C ≡ H in Theorem 3.1, then we obtain the following corollary.

Corollary 3.5. Let H be a real Hilbert space. Let T _n : H → H be a continuous pseudo-contractive mapping and A : H → H be a continuous monotone mapping such that$\mathfrak{F}:=F\left(T\right)\cap A^{-1}\left(0\right)\ne \varnothing$. Let f be a contraction of C into itself and let B : H → H be a strongly positive linear bounded self-adjoint operator with coefficients$\bar{\beta }>0$and let {x _n } be a sequence generated by x₁ ∈ H and

$$\{\begin{array}{l}{y}_n=F_{r_n}{x}_n\hfill \\ x_{n+1}={\alpha }_n\gamma f(x_n)+{\delta }_n{x}_n+[(1-{\delta }_n)I-{\alpha }_{n}B]T_{r_n}{y}_n\hfill \end{array}$$

(3.31)

where {α _n } ⊂ [0,1] and {r _n } ⊂ (0, ∞) such that

(C1) $\underset{n\to \infty }{\text{lim}}{\alpha }_n=0,\sum _{n=1}^{\infty }{\alpha }_n=\infty$;

(C2) $\underset{n\to \infty }{\text{lim}}{\delta }_n=0,\sum _{n=1}^{\infty }\left|{\delta }_{n+1}-{\delta }_n\right|<\infty$;

(C3) $\underset{n\to \infty }{\text{lim}\text{inf}}{r}_n>0,\sum _{n=1}^{\infty }\left|r_{n+1}-r_n\right|<\infty$.

Then, the sequence {x _n } converges strongly to$z\in \mathfrak{F}$, which is the unique solution of the variational inequality:

$$〈\left(B-\gamma f\right)z,x-z〉\ge 0,\forall x\in \mathfrak{F}$$

(3.32)

Equivalently, $z=P_{\mathfrak{F}}\left(I-B+\gamma f\right)z$.

Proof. Since D(A) = H, we note that VI(H, A) = A^-1(0). So, by Theorem 3.1, we obtain the desired result. □

Remark 3.6. Our results extend and unify most of the results that have been proved for these important classes of nonlinear operators. In particular, Theorem 3.1 extends Theorem 3.1 of Iiduka and Takahashi [7] and Zegeye et al. [26], Corollary 3.2 of Su et al. [27] in the sense that our convergence is for the more general class of continuous pseudo-contractive and continuous monotone mappings. Corollary 3.4 also extends Theorem 4.2 of Iiduka and Takahashi [7] in the sense that our convergence is for the more general class of continuous pseudo-contractive and continuous monotone mappings.