On weak and strong convergence results for generalized equilibrium variational inclusion problems in Hilbert spaces

Abstract

We introduce a new iterative method for finding a common element of the set of fixed points of pseudo-contractive mapping, the set of solutions to a variational inclusion and the set of solutions to a generalized equilibrium problem in a real Hilbert space. We provide some results about strongly and weakly convergent of the iterative scheme sequence to a point $p\in \varOmega$ which is the unique solution of a variational inequality, where Ω is an intersection of set as given by ${\varOmega }=F(S)\cap (A+B)^{-1}(0) \cap N^{-1}(0)\cap \operatorname{GEP}(F,M)\neq \emptyset$. This gives us a common solution. Also, We show that our results extend some published recent results in this field. Finally, we provide an example to illustrate our main result.

Let H be a real Hilbert space, whose inner product and norm are denoted by $\langle \cdot,\cdot\rangle$ and $\|\cdot\|$, respectively, C a nonempty, closed and convex subset of H. Recall that a mapping $S:C\to C$ is said to be pseudo-contractive if and only if $\|Su-Sv\|^{2}\leq \|u-v\|^{2}+\|(I-S)u-(I-S)v\|^{2}$ for all $u,v\in C$. Equivalently, $\langle u-v,Su-Sv\rangle \leq \|u-v\|^{2}$ for all $u,v\in C$. A mapping $S:C\to C$ is said to be k-strictly pseudo-contractive if and only if there exists $0\leq k<1$ such that $\|Su-Sv\|^{2}\leq \|u-v\|^{2}+k\|(I-S)u-(I-S)v\|^{2}$ for all $u,v\in C$. Equivalently, $\langle u-v,Su-Sv\rangle \leq \|u-v\|^{2}-k\|(I-S)u-(I-S)v\|^{2}$ for all $u,v\in C$. A mapping L-Lipschitz if there exists $L \geq 0$ such that $\|Su-Sv\|\leq L\|u-v\|$ for all $u,v\in C$. The mapping S is called nonexpansive if $L=1$ and is called contractive if $L < 1$. A mapping S is called firmly nonexpansive if $\|Su-Sv\|^{2}\leq \|u-v\|^{2}-\|(I-S)u-(I-S)v\|^{2}$ for all $u,v\in C$. Every nonexpansive mapping is a k-strictly pseudo-contractive mapping and every k-strictly pseudo-contractive mapping is pseudo-contractive. Assume that $S:C\to C$ be a strictly pseudo-contractive. We denote by $F(S)$ the fixed point set of S, that is, $F(S)=\{x\in C: S(x)=x\}$. There is a lot of work associated with the fixed point algorithms (see for example, [1–6]). Also, there are many papers and books about iterative schemes for numerical estimations in different area of this field (see for example [7–12]).

Let $A: C \to H$ be a nonlinear mapping and F be a bi-function from $C \times C$ to $\mathbb{R}$, where $\mathbb{R}$ is the set of real numbers. The generalized equilibrium problem is to find $x^{*}\in C$ such that $F(x^{*},y)+\langle Ax^{*},y-x^{*}\rangle \geq 0$, for all $y\in C$. The set of solutions of $x^{*}$ is denoted by $\operatorname{GEP}(F,A)$ ([13]). If $A=0$, then $\operatorname{GEP}(F,A)$ is denoted by $\operatorname{EP}(F)$. If $F(x, y) = 0$ for all $x, y \in C$, then $\operatorname{GEP}(F,A)$ is denoted by $\operatorname{VI}(C,A)=\{x^{*}\in C: \langle Ax^{*},y-x^{*}\rangle \geq 0,y\in C\}$. This is the set of solutions of the variational inequality for A ([14–16]). If $C=H$, then $\operatorname{VI}(H,A)=A^{-1}(0)$ where $A^{-1}(0)=\{x\in H:Ax=0\}$. Recall that a mapping $A : C\to H$ is said to be monotone whenever $\langle Au-Av,u-v\rangle \geq 0$ for all $u,v\in C$. A mapping A is said to be α-strongly monotone whenever there exists a positive real number α such that $\langle Au-Av,u-v\rangle \geq \alpha \|u-v\|^{2}$ for all $u,v\in C$. A mapping A is said to be α-inverse strongly monotone whenever there exists a positive real number α such that $\langle Au-Av,u-v\rangle \geq \alpha \|Au-Av\|^{2}$ for all $u,v\in C$. For such a case, A is said to be α-inverse strongly monotone. Note that any α-inverse strongly monotone mapping A is Lipschitz and $\|Au-Av\|\leq \frac{1}{\alpha }\|u-v\|$. Let $A : H \to H$ be a single-valued nonlinear mapping, $B : H \to 2^{H}$ a set-valued mapping. The variational inclusion is to find $p \in H$ such that

$$ \theta \in A(p)+B(p), $$

(1)

where θ is a zero vector in H. When $A = 0$, then (1) becomes the inclusion problem introduced by Rockafellar ([17]). Let $B : H\to 2^{H}$ be a mapping. The effective domain of B is denoted by $D(B)$, namely, $D(B) = \{x \in H : Bx \neq \emptyset \}$. The graph of B is $G(B)=\{(u,v)\in H\times H :v\in Bu\}$. A set-valued mapping B is said to be monotone whenever $\langle x-y,f-h\rangle \geq 0$ for all $x,y\in D(B)$, $f\in Bx$ and $h\in By$. A monotone operator B is maximal if the graph $G(B)$ of B is not properly contained in the graph of any other monotone mapping. Also, a monotone mapping B is maximal if and only if, for $(x,f)\in H\times H$, $\langle x-y,f-h\rangle \geq 0$ for every $(y,h)\in G(B)$ implies $f\in B x$. For a maximal monotone operator B on H and $r > 0$, we define a single-valued operator $J^{B}_{r}x=(I+rB)^{-1}:H\to D(B)$, which is called the resolvent of B for r. It is well known that $J^{B}_{r}x$ is firmly nonexpansive, that is, $\langle x - y,J^{B}_{r}x - J^{B}_{r}y\rangle \geq \|J^{B}_{r}x-J^{B}_{r}y \|^{2}$ for all $x,y\in H$, and that a solution of (1) is a fixed point of $J^{B}_{r}(I-rA)$ for all $r>0$ (see[18]).

A basic problem for maximal monotone operator B is to find

$$ x \in H \quad \text{such that } 0\in Bx. $$

(2)

A well-known method for solving problem (2) is the proximal point algorithm: $x_{1} = x\in H$, and

$$ x_{n+1} = J^{B}_{r_{n}}x_{n}, \quad n = 1,2,3, \ldots , $$

where $J^{B}_{r_{n}} = (I + r_{n}B)^{-1}$ and $\{r_{n}\} \subset (0,\infty )$. For any initial guess $x^{*} \in H$, the proximal point algorithm generates an iterative sequence as $x_{n+1}=J^{B}_{r_{n}}(x_{n}+e_{n})$, where $e_{n}$ is the error sequence, then Rockafellar ([17, 19]) proved that the sequence $\{x_{n}\}$ converges weakly to an element of $B^{-1}(0)$. To ensure convergence, it is assumed that $\| e_{n+1} \| \leq \varepsilon _{n}\| x_{n+1} - x_{n} \|$ with $\sum_{n=0}^{\infty }\varepsilon _{n} < \infty$ ([17]). This criterion was then improved by Han and He as $\| e_{n+1} \| \leq \varepsilon _{n}\| x_{n+1} - x_{n} \|$ with $\sum_{n=0}^{\infty }\varepsilon ^{2}_{n} < \infty$ ([20]). Then Kamimura and Takahashi introduced the following iterative method:

$$ x_{n+1}=\lambda _{n}u+(1-\lambda _{n})J^{B}_{r_{n}}(x_{n}+e_{n}), $$

where $u \in H$ is fixed and $(\lambda _{n})$ is a real sequence ([3]). They proved that the sequence $\{x_{n}\}$ converges strongly to $x^{*}=P_{(B)^{-1}(0)}(u)$. Then Ceng, Wu and Yao obtained the norm convergence under the following conditions:

(i):

$\lim_{n\to \infty }r_{n}=\infty$,

(ii):

$\lim_{n\to \infty }\lambda _{n}=0$, $\sum_{n=0}^{\infty }\lambda _{n} = \infty$,

($iii$):

$\| e_{n+1} \| \leq \varepsilon _{n}\| x_{n+1} - x_{n} \|$ with $\sum_{n=0}^{\infty }\varepsilon ^{2}_{n} < \infty$ ([21]).

In 2013, Tian and Wang show that, if $\{r_{n}\}$ be bounded below away from zero, then the norm convergence is still guaranteed for bounded $(r_{n})$, especially for constant sequence ([22]). In the literature, there are a large number references associated with the proximal point algorithm (see for example, [20–31]).

In 2008, Takahashi and Takahashi introduced an iterative method for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions to a generalized equilibrium problem in a real Hilbert space ([13]). In 2019, Qin, Cho and Yao introduced the following iterative scheme in Banach space E:

$$ \textstyle\begin{cases} x_{0}\in C\cap D, \\ y_{n} = \beta _{n}Tx_{n} + (1-\beta _{n})x_{n}, \\ x_{n+1} = P_{C\cap D}^{E}(\alpha _{n}f(x_{n})+\delta _{n}R^{M}_{r_{n}}(x_{n}-r_{n}Nx_{n}+e_{n})+ \gamma _{n}y_{n}),\quad n\geq 0, \end{cases} $$

(3)

where $\{e_{n}\}$ is a sequence in E such that $\sum_{n=0}^{\infty }\|e_{n}\|<\infty$, C and D is two nonempty closed and convex subsets of E, $P_{C\cap D}^{E}$ is a sunny nonexpansive retraction from E onto $C\cap D$, $M : D \rightarrow 2^{E}$ is an m-accretive operator, $N :C \rightarrow E$ is an α-inverse strongly accretive operator, $R^{M}_{r}$ the resolvent of N for each $r>0$, $f :C \to E$ is a k-contraction, $T :C \rightarrow E$ is a k-strict pseudo-contraction with a nonempty fixed point set ([32]). They proved that the sequence $\{x_{n}\}$ generated by (3) converges strongly to $x^{*}=P^{C\cap D}_{F(T)\cap (N + M)^{-1}(0)}f(x^{*})$, where $x^{*}$ is the unique solution of the variational inequality $\langle f(x^{*})-x^{*},J_{q}(y-x^{*})\rangle \leq 0$, $y\in F(T)\cap (N + M)^{-1}(0)$ ([32]).

The purpose of this paper is to prove the strong and weak convergence of new algorithms under different criteria of the errors $\{e_{n}\}$. We use a new technique of argument for dealing with strong and weak convergence, also, suggest and propose the new accuracy criteria for modified approximate proximal point algorithms. Applications of the main results are also provided. In this paper, motivated by the mentioned above results, we present an iterative method which converges strong and weak to a common element of the fixed point set of pseudo-contractive mapping and the zero set of the sums of maximal monotone operators and the set of solutions to a generalized equilibrium problem in a real Hilbert space.

1 Preliminaries

Let H be a real Hilbert space and let C be a nonempty closed convex subset of H. It is well known that, for any $x\in H$, there exists a unique nearest point in C, denoted by $P_{C}(x)$, such that $\|x-P_{C}(x)\|=\inf_{y\in C} \|x-y\|=:d(x,C)$. It is well known that $P_{C}$ is nonexpansive and monotone mapping of H onto C and satisfies the following:

• (1) $\langle x-P_{C}x,z-P_{C}x\rangle \leq 0$ for all $x\in H$, $z\in C$.

• (2) $\|x-z\|^{2}\geq \|x-P_{C}x\|^{2}+\|z-P_{C}x\|^{2}$ for all $x\in H$, $z\in C$.

• (3) The relation $\langle P_{C}x-P_{C}z,x-z\rangle \geq \|P_{C}x-P_{C}z\|^{2}$ holds for all $z,x\in H$.

Let A be a monotone mapping of C into H. In the context of the variational inequality problem, it is easy to see from (2) that

$$ p\in \operatorname{VI}(C,A)\quad \Leftrightarrow\quad p=P_{C}(p-\lambda Ap) \quad \text{for some } \lambda >0. $$

For solving the equilibrium problem for a bi-function $F: C \times C\to \mathbb{R}$, we assume that F satisfy the following conditions:

$(A_{1})$:

$F(x, x)=0$ for all $x \in C$,

$(A_{2})$:

F is monotone, that is, $F(x, y) + F(y, x) \leq 0$ for all $x, y \in C$,

$(A_{3})$:

for each $x, y, z \in C$, $\lim_{t\to 0} F(tz+(1-t)x, y)\leq F(x,y)$,

$(A_{4})$:

for each $x \in C$, the function $y \mapsto F(x, y)$ is convex and lower semi-continuous.

Putting $F (x, y) = \langle Ax, y - x\rangle$ for every $x, y \in C$, we see that the equilibrium problem is reduced to the variational inequality.

Lemma 1.1

([33])

Assume that B is a maximal monotone operator. The followings hold.

$(a)$:

$D(J_{r}^{B})=H$,

$(b)$:

$J_{r}^{B}$is single-valued and firmly nonexpansive

$(c)$:

$F(J_{r}^{B})=\varGamma =\{x\in D(B): 0\in B(x)\}$,

$(d)$:

its graph $G(B)$is weak-to-strong closed in $H \times H$.

Lemma 1.2

([34, 35])

Assume that $F: C \times C \to \mathbb{R}$satisfies $(A_{1})$–$(A_{4})$and C is a nonempty, closed and convex subset of H. For $r > 0$and $x \in H$, consider the map $T_{r} : H \to C$defined by

$$ T_{r}(x)=\biggl\{ z\in C: F(z,y)+\frac{1}{r}\langle y-z,z-x \rangle \geq 0 \textit{ for all } y\in C\biggr\} . $$

For each $c\in H$, we have $T_{r}(x) \neq \emptyset$, $T_{r}$is single-valued, $\operatorname{EP}(F)$is closed and convex, $F(T_{r}) = \operatorname{EP}(F)$and $T_{r}$is firmly nonexpansive, that is, $\|T_{r}(x)-T_{r}(y)\|^{2}\leq \langle T_{r}(x)-T_{r}(y),x-y\rangle$for all $x, y \in H$.

Lemma 1.3

([36])

Assume that C is a nonempty, closed and convex subset of H, F is a bi-function from $C \times C$to $\mathbb{R}$satisfying $(A_{1})$–$(A_{4})$, $A_{F}$is the multivalued mapping from H into itself defined by $A_{F}x=\{z\in C: F(z,y)\leq \langle y-x,z\rangle \textit{ for all } y\in C\}$whenever $x \in C$and $A_{F}x=\emptyset$otherwise. In this case, $A_{F}$is a maximal monotone operator with the domain $T_{r}(x)=(I+rA_{F})^{-1}x$, for all $x \in H$and $r > 0$.

Lemma 1.4

([5])

Assume that H is a real Hilbert space, C is a closed convex subset of H and $T : C\to C$is a continuous pseudo-contractive mapping. In this case, $F(T)$is a closed convex subset of C and $(I - T)$is demiclosed at zero, that is, $x = T(x)$whenever $\{x_{n}\}$is a sequence in C such that $x_{n}\rightharpoonup x$and $Tx_{n} -x_{n} \to 0$.

Lemma 1.5

([37])

If $\{x_{n}\}$, $\{a_{n}\}\subset \mathbb{R^{+}}$, $\{\lambda _{n}\}\subset (0,1)$and $\{\gamma _{n}\}\subset \mathbb{R}$are some sequences such that $x_{n+1}\leq (1-\lambda _{n})x_{n}+\lambda _{n}\gamma _{n}+a_{n}$for all $n\geq 0$, $\sum_{n=0}^{\infty }\lambda _{n}=\infty$, $\limsup_{n\to \infty }\gamma _{n}\leq 0$and $\sum_{n=0}^{\infty }a_{n}<\infty$, then $\lim_{n\to \infty }x_{n}=0$.

Lemma 1.6

([38])

Assume that H is a real Hilbert space. For each $x_{j}\in H$and $a_{j}\in [0, 1]$for $j = 1, 2, 3$with $a_{1} + a_{2} + a_{3} = 1$the following equality holds:

$$ \Vert a_{1}x_{1}+a_{2}x_{2}z+a_{3}x_{3} \Vert ^{2}= a_{1} \Vert x_{1} \Vert ^{2}+a_{2} \Vert x_{2} \Vert ^{2}+a_{3} \Vert x_{3} \Vert ^{2}- \sum_{1\leq i,j\leq 3}a_{i}a_{j} \Vert x_{i}-x_{j} \Vert ^{2}. $$

Lemma 1.7

([36])

Suppose that B is a maximal monotone operator on H. In this case, we have

$$ \frac{\lambda -r}{r}\bigl\langle J^{B}_{\lambda }x-J_{r}^{B}x,J^{B}_{ \lambda }x-x \bigr\rangle \geq \bigl\Vert J^{B}_{\lambda }x-J^{B}_{r}x \bigr\Vert ^{2} \quad \forall \lambda , r>0 \textit{ and } x\in H. $$

Lemma 1.8

([5])

Suppose that H is a real Hilbert space. For every $x, y \in H$, we have $\|x+y\|^{2}\leq \|x\|^{2}+2\langle y,x+y\rangle$.

Lemma 1.9

([39])

Assume that $\{x_{n}\}$is sequences of real numbers and there exists a subsequence $\{n_{k}\}$of $\{n\}$such that $x_{n_{k}}\leq x_{n}$for all $k\in \mathbb{N}$. There exists a nondecreasing sequence $\{t_{i}\}\subset \mathbb{N}$such that $x_{t_{i}}\leq x_{t_{i}+1}$and $x_{i}\leq x_{t_{i}+1}$for all $i\geq 1$. In fact, $t_{i}=\max \{k\leq i:x_{k}\leq x_{k+1}\}$.

2 Weak and strong convergence theorems

Now, we are ready to state and prove our main results.

Theorem 2.1

Suppose that C is a nonempty, closed and convex subset of H, F is a bi-function from $C \times C$to $\mathbb{R}$satisfying $(A_{1})$–$(A_{4})$, M is an α-inverse strongly monotone mapping from C into H, A is a β-inverse strongly monotone map from C into H, B and N are two maximal monotone operators on H such that their domains contained in C, $f : C \to C$is a ρ-contractive map with $\rho \in (0,\frac{1}{2})$and $S : C \to C$is a Lipschitz pseudo-contractive mapping with Lipschitz constants K such that ${\varOmega }=F(S)\cap (A+B)^{-1}(0) \cap N^{-1}(0)\cap \operatorname{GEP}(F,M)\neq \emptyset$. Assume that $\{b_{n}\}$, $\{\beta _{n}\}$and $\{\delta _{n}\}$are some sequences in $(0, 1)$and $\{x_{n}\}$, $\{y_{n}\}$, $\{u_{n}\}$and $\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{0}\in C, \\ F (y_{n}, y) + \langle Mx_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0, \quad \forall y\in C, \\ u_{n}=J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}Ay_{n}), \\ z_{n}=b_{n}f(x_{n})+(1-b_{n})J_{s_{n}}^{N}( u_{n}+e_{n}), \\ x_{n+1}=(1-\beta _{n})z_{n}+\beta _{n}S(\delta _{n}z_{n}+(1-\delta _{n}) Sz_{n}) \quad \forall n\geq 0. \end{cases} $$

(4)

If

$(d_{1})$:

$0< c\leq \lambda _{n}\leq d<2\beta$, $0< a\leq r_{n}\leq b<2\alpha$,

$(d_{2})$:

$0< c<\beta _{n}\leq \delta _{n}<d<\frac{1}{\sqrt{1+K^{2}}+1}$, $s_{n}>s>0$,

$(d_{3})$:

$\lim_{n\to \infty }b_{n}=0$, $\sum_{n=1}^{\infty }b_{n}=\infty$,

$(d_{4})$:

$\| e _{n}\| \leq \frac{\varepsilon _{n}}{2} \max \{\| {u}_{n} - J_{s_{n}}^{N}( u_{n}+e_{n}) \|, \|J_{s_{n}}^{N}( u_{n}+e_{n})-p\|\}$with $\sum_{n=0}^{ \infty } \varepsilon _{n} < \infty$,

then $\{x_{n}\}$converges strongly to a point $p\in \varOmega$which is the unique solution of the variational inequality $\langle (I-f)p,x-p\rangle \geq 0$for all $x \in \varOmega$.

Proof

We first show that $I - \lambda _{n}A$ is nonexpansive. For all $u,v\in C$ and $0<\lambda _{n}<2\beta$, we obtain

$$\begin{aligned} \bigl\Vert (I - \lambda _{n}A)u-(I - \lambda _{n}A)v \bigr\Vert ^{2} =& \bigl\Vert (u-v)-\lambda _{n}(Au-Av) \bigr\Vert ^{2} \\ \leq & \Vert u-v \Vert ^{2}-2\lambda _{n}\langle u-v,Au-Av\rangle +\lambda _{n}^{2} \Vert Au-Av \Vert ^{2}\quad \\ \leq & \Vert u-v \Vert ^{2}-\lambda _{n}\beta \Vert Au-Av \Vert ^{2}+\lambda _{n}^{2} \Vert Au-Av \Vert ^{2}\quad \\ =& \Vert u-v \Vert ^{2}+\lambda _{n}(\lambda _{n}-2\beta ) \Vert Au-Av \Vert ^{2} \\ \leq & \Vert u-v \Vert ^{2}. \end{aligned}$$

(5)

This proves that $I - \lambda _{n}A$ is nonexpansive. Let $p\in \varOmega$. Observe that $y_{n}$ can be re-written as $y_{n} = T_{r_{n}} (x_{n}-r_{n}Mx_{n})$, $n \geq 0$. From $(d_{2})$ and Lemma 1.2, we have

$$\begin{aligned} \Vert y_{n} - p \Vert ^{2} =& \bigl\Vert T_{r_{n}}(x_{n}-r_{n}Mx_{n})-p \bigr\Vert ^{2} \\ =&\bigl\| T_{r_{n}}(x_{n}-r_{n}Mx_{n})-T_{r_{n}}(p-r_{n}Mp)) \bigr\| ^{2} \\ \leq & \bigl\Vert (x_{n}-r_{n}Mx_{n})-(p-r_{n}Mp) \bigr\Vert ^{2} \\ =& \Vert x_{n}-p \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2} \\ \leq & \Vert x_{n}-p \Vert ^{2}. \end{aligned}$$

(6)

From (4), (5) and using the fact that $J_{\lambda _{n}}^{B}$ is nonexpansive, we have

$$\begin{aligned} \Vert u_{n} - p \Vert ^{2} =& \bigl\Vert J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}A y_{n})-p \bigr\Vert ^{2} \\ \leq & \bigl\Vert J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}A y_{n})-J_{\lambda _{n}}^{B}(p- \lambda _{n}Ap) \bigr\Vert ^{2} \\ \leq & \bigl\Vert (y_{n}-\lambda _{n}A y_{n})-(p-\lambda _{n}Ap) \bigr\Vert ^{2} \\ \leq & \Vert y_{n} - p \Vert ^{2}+\lambda _{n}(\lambda _{n}-2\beta ) \Vert Ay_{n}-Ap \Vert ^{2} \\ \leq & \Vert x_{n}-p \Vert ^{2}+\lambda _{n}(\lambda _{n}-2\beta ) \Vert Ay_{n}-Ap \Vert ^{2} \\ \leq & \Vert x_{n}-p \Vert ^{2}. \end{aligned}$$

(7)

Set $t_{n}=(1-\delta _{n})z_{n}+\delta _{n}Sz_{n}$ for all $n\geq 1$. By using Lemma 1.6, we have

$$\begin{aligned} \Vert t_{n} - p \Vert ^{2} =& \bigl\Vert (1-\delta _{n})z_{n}+\delta _{n}Sz_{n}-p \bigr\Vert ^{2} \\ \leq &(1-\delta _{n}) \Vert z_{n}-p \Vert ^{2}+\delta _{n} \Vert Sz_{n}-p \Vert ^{2}-(1- \delta _{n})\delta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq &(1-\delta _{n}) \Vert z_{n}-p \Vert ^{2}+\delta _{n}\bigl( \Vert z_{n}-p \Vert ^{2}+ \Vert z_{n}-Sz_{n} \Vert ^{2} \bigr)-(1-\delta _{n})\delta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq & \Vert z_{n}-p \Vert ^{2}+\delta _{n}^{2} \Vert z_{n}-Sz_{n} \Vert ^{2}. \end{aligned}$$

(8)

From (4) and (8), we get

$$\begin{aligned} \Vert x_{n+1} - p \Vert ^{2} =& \bigl\Vert (1-\beta _{n})z_{n}+\beta _{n}S\bigl((1- \delta _{n})z_{n}+ \delta _{n}Sz_{n} \bigr)-p \bigr\Vert ^{2} \\ =& \bigl\Vert (1-\beta _{n})z_{n}+\beta _{n}St_{n}-p \bigr\Vert ^{2} \\ \leq &(1-\beta _{n}) \Vert z_{n}-p \Vert ^{2}+\beta _{n} \Vert St_{n}-p \Vert ^{2}-(1- \beta _{n})\beta _{n} \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq &(1-\beta _{n}) \Vert z_{n}-p \Vert ^{2}+\beta _{n}\bigl( \Vert t_{n}-p \Vert ^{2}+ \Vert t_{n}-St_{n} \Vert ^{2} \bigr)-(1-\beta _{n})\beta _{n} \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq &(1-\beta _{n}) \Vert z_{n}-p \Vert ^{2}+\beta _{n} \Vert t_{n}-p \Vert ^{2}+\beta _{n} \Vert t_{n}-St_{n} \Vert ^{2}-(1-\beta _{n})\beta _{n} \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & \Vert z_{n}-p \Vert ^{2}+\delta _{n}^{2}\beta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2}+ \beta _{n} \Vert t_{n}-St_{n} \Vert ^{2} \\ &{}-(1-\beta _{n})\beta _{n} \Vert z_{n}-St_{n} \Vert ^{2}. \end{aligned}$$

(9)

Thus,

$$\begin{aligned} \Vert t_{n} - St_{n} \Vert ^{2} =& \bigl\Vert (1-\delta _{n})z_{n}+\delta _{n}Sz_{n}-St_{n} \bigr\Vert ^{2} \\ \leq &(1-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2}+\delta _{n} \Vert Sz_{n}-St_{n} \Vert ^{2} \\ &{}-(1-\delta _{n})\delta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2}. \end{aligned}$$

(10)

Since S is K-Lipschitz and $z_{n} - t_{n} = \delta _{n}(z_{n} - Sz_{n})$, by using (10) we get

$$\begin{aligned}& \Vert t_{n} - St_{n} \Vert ^{2} \\& \quad \leq (1-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2}+\delta _{n}K^{2} \Vert z_{n}-t_{n} \Vert ^{2}-(1-\delta _{n})\delta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2} \\& \quad \leq (1-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2}+\delta _{n}^{3}K^{2} \Vert z_{n}-Sz_{n} \Vert ^{2}-(1-\delta _{n})\delta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2} \\& \quad = (1-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2}-\delta _{n}\bigl(1-\delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2}. \end{aligned}$$

This together with (9) implies that

$$\begin{aligned} \Vert x_{n+1}-p \Vert ^{2} \leq& \Vert z_{n}-p \Vert ^{2}+\beta _{n}\bigl((1-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ &{}-\delta _{n}\bigl(1-\delta _{n}-\delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2}\bigr) \\ &{}+\delta _{n}^{2}\beta _{n} \Vert z_{n}-Sz_{n} \Vert ^{2}-(1-\beta _{n})\beta _{n} \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & \Vert z_{n}-p \Vert ^{2}-\delta _{n} \bigl(1-2\delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2}. \end{aligned}$$

(11)

Since $0< c<\beta _{n}\leq \delta _{n}<d<\frac{1}{\sqrt{1+K^{2}}+1}$ for all $n\geq 1$, we conclude that

$$ \Vert x_{n+1}-p \Vert ^{2}\leq \Vert z_{n}-p \Vert ^{2}.$$

(12)

Put $v_{n}=J_{\lambda _{n}}^{N}( u_{n}+e_{n})$ for all $n\geq 0$. By using Lemma 1.1, we obtain

$$\begin{aligned} \Vert v_{n}-p \Vert ^{2} \leq & \Vert u_{n}+e_{n}-p \Vert ^{2}- \Vert u_{n}+e_{n}-v_{n} \Vert ^{2} \\ =& \Vert u_{n}-p \Vert ^{2}- \Vert u_{n}-v_{n} \Vert ^{2}+2\langle e_{n},v_{n}-p\rangle \quad \\ \leq & \Vert u_{n}-p \Vert ^{2}- \Vert u_{n}-v_{n} \Vert ^{2}+2 \Vert e_{n} \Vert \Vert v_{n}-p \Vert . \end{aligned}$$

Since

$$\begin{aligned} \Vert e_{n} \Vert \leq & \frac{\varepsilon _{n}}{2}\max \bigl\{ \Vert u_{n} - v_{n} \Vert , \Vert v_{n} - p \Vert \bigr\} \\ \leq & \frac{\varepsilon _{n}}{2}\bigl( \Vert v_{n} - p \Vert + \Vert u_{n} - v_{n} \Vert \bigr), \end{aligned}$$

this implies that

$$\begin{aligned} \Vert v_{n}-p \Vert ^{2} \leq & \Vert u_{n}-p \Vert ^{2}- \Vert u_{n}-v_{n} \Vert ^{2}+2 \frac{\varepsilon _{n}}{2}\bigl( \Vert v_{n} - q \Vert + \Vert u_{n} - v_{n} \Vert \bigr) \Vert v_{n}-p \Vert . \\ \leq & \Vert u_{n} - p \Vert ^{2} -\biggl(1- \frac{\varepsilon _{n}}{2}\biggr) \Vert u_{n} - v_{n} \Vert ^{2} +2\varepsilon _{n} \Vert v_{n} - q \Vert ^{2}. \end{aligned}$$

Since $\varepsilon _{n} \rightarrow 0$, for all $n \geq m_{0}$, we see that there exists an integer $m_{0} \geq 0$ such that $1 - 2\varepsilon _{n} >0$. It follows from (7) that

$$\begin{aligned} \Vert v_{n} - p \Vert ^{2} \leq & \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert u_{n} - p \Vert ^{2}- \frac{1-\frac{\varepsilon _{n}}{2}}{1-2\varepsilon _{n}} \Vert u_{n} - v_{n} \Vert ^{2} \\ \leq & \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert u_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2} \\ \leq & \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2}. \end{aligned}$$

(13)

It follows that

$$\begin{aligned} \Vert v_{n} - p \Vert \leq & \biggl(1+ \frac{\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert . \end{aligned}$$

(14)

It follows from (4) and the last inequality that

$$\begin{aligned} \Vert z_{n} - p \Vert =& \bigl\Vert b_{n}f(x_{n})+(1-b_{n})v_{n}-p \bigr\Vert \\ \leq &b_{n} \bigl\Vert f(x_{n})-p \bigr\Vert +(1-b_{n}) \Vert v_{n}-p \Vert \\ \leq &b_{n}\bigl(\rho \Vert x_{n}-p \Vert + \bigl\Vert f(p)-p \bigr\Vert \bigr)+(1-b_{n}) \biggl(1+ \frac{\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert . \\ \leq &\biggl(1+\frac{\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \bigl(1-b_{n}(1- \rho ) \bigr) \Vert x_{n} - p \Vert +b_{n} \bigl\Vert f(p)-p \bigr\Vert . \end{aligned}$$

Now, by induction we have

$$ \Vert x_{n+1} - p \Vert \leq \prod _{i={0}}^{n}\biggl(1+ \frac{\varepsilon _{i}}{1-2\varepsilon _{i}}\biggr)\max \biggl\{ \frac{1}{1-\rho } \bigl\Vert f(p)-p \bigr\Vert , \Vert x_{0} - p \Vert \biggr\} , \quad \forall n\geq 0. $$

(15)

Indeed when $n = 0$, from (12) we have

$$\begin{aligned} \Vert x_{1} - p \Vert \leq &\biggl(1+ \frac{\varepsilon _{0}}{1-2\varepsilon _{0}} \biggr) \bigl(1-b_{0}(1- \rho )\bigr) \Vert x_{0} - p \Vert +b_{0} \bigl\Vert f(p)-p \bigr\Vert \\ \leq &\biggl(1+\frac{\varepsilon _{0}}{1-2\varepsilon _{0}} \biggr)\bigl[\bigl(1-b_{0}(1- \rho )\bigr) \Vert x_{0} - p \Vert +b_{0} \bigl\Vert f(p)-p \bigr\Vert \bigr] \\ \leq &\biggl(1+\frac{\varepsilon _{0}}{1-2\varepsilon _{0}} \biggr)\max \biggl\{ \frac{1}{1-\rho } \bigl\Vert f(p)-p \bigr\Vert , \Vert x_{0} - p \Vert \biggr\} , \end{aligned}$$

which implies that (15) holds for $n = 0$. Assume that (15) holds for $n \geq 1$. Then it follows that $\| x_{n} - p\|\leq \prod_{i={0}}^{n-1}(1+ \frac{\varepsilon _{i}}{1-2\varepsilon _{i}})\max \{\frac{1}{1-\rho } \|f(p)-p\|,\|x_{0} - p\|\}$. Hence, from (12) we have

$$\begin{aligned} \Vert x_{n+1} - p \Vert \leq& \biggl(1+ \frac{\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \bigl(1-b_{n}(1-\rho )\bigr) \Vert x_{n} - p \Vert +b_{n} \bigl\Vert f(p)-p \bigr\Vert \\ \leq &\biggl(1+\frac{\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)\bigl[\bigl(1-b_{n}(1- \rho )\bigr) \Vert x_{n} - p \Vert +b_{n} \bigl\Vert f(p)-p \bigr\Vert \bigr] \\ \leq &\biggl(1+\frac{\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)\prod_{i={0}}^{n-1} \biggl(1+ \frac{\varepsilon _{i}}{1-2\varepsilon _{i}}\biggr)\max \biggl\{ \frac{1}{1-\rho } \bigl\Vert f(p)-p \bigr\Vert , \Vert x_{0} - p \Vert \biggr\} \\ =&\prod_{i={0}}^{n}\biggl(1+ \frac{\varepsilon _{i}}{1-2\varepsilon _{i}}\biggr) \max \biggl\{ \frac{1}{1-\rho } \bigl\Vert f(p)-p \bigr\Vert , \Vert x_{0} - p \Vert \biggr\} . \end{aligned}$$

This indicates that (15) holds for $n+1$. Therefore, (15) holds for $n\geq 0$. We have

$$\begin{aligned} \Vert x_{n+1} - p \Vert \leq &\prod_{i={0}}^{n} \biggl(1+ \frac{\varepsilon _{i}}{1-2\varepsilon _{i}}\biggr)\max \biggl\{ \frac{1}{1-\rho } \bigl\Vert f(p)-p \bigr\Vert , \Vert x_{0} - p \Vert \biggr\} \\ \leq &\prod_{i={0}}^{\infty }\biggl(1+ \frac{\varepsilon _{i}}{1-2\varepsilon _{i}}\biggr)\max \biggl\{ \frac{1}{1-\rho } \bigl\Vert f(p)-p \bigr\Vert , \Vert x_{0} - p \Vert \biggr\} . \end{aligned}$$

Since $\sum_{n=0}^{+\infty } \varepsilon _{n}<+\infty$, it follows that $\prod_{n=m_{0}}^{+\infty }(1+ \frac{\varepsilon _{n}}{1-2\varepsilon _{n}})<+\infty$. Thus, $\{\|x_{n}-p\|\}$ is bounded. So, $\{x_{n}\}$ is bounded and so are the sequences $\{y_{n}\}$, $\{u_{n}\}$ and $\{z_{n}\}$. Let $p=P_{\varOmega }f(p)$. We have from (6), (7), (11), (14) and Lemma 1.8

$$\begin{aligned} \Vert x_{n+1} - p \Vert ^{2} \leq& \Vert z_{n}-p \Vert ^{2}-\delta _{n}\bigl(1-2 \delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ =& \bigl\Vert b_{n}f(x_{n})+(1-b_{n})v_{n}-p \bigr\Vert ^{2}-\delta _{n}\bigl(1-2\delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & \bigl\Vert b_{n}\bigl(f(x_{n})-p \bigr)+(1-b_{n}) (v_{n}-p) \bigr\Vert ^{2}- \delta _{n}\bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq &(1-b_{n}) \Vert v_{n} - p \Vert ^{2}+2b_{n}\bigl\langle f(x_{n})-p,x_{n+1} - p \bigr\rangle \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} -\delta _{n} \bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq &(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2}\biggr] \\ &{}+2b_{n}\bigl\langle f(x_{n})-p,x_{n+1} - p \bigr\rangle \\ &{}+(1-b_{n})\bigl[\lambda _{n}(\lambda _{n}-2 \beta ) \Vert Ay_{n}-Ap \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2}\bigr] \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} -\delta _{n} \bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq &(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2}\biggr] \\ &{}+2b_{n}\bigl[\bigl\langle f(x_{n})-p,x_{n} - p\bigr\rangle +\bigl\langle f(x_{n})-p,x_{n+1} - x_{n}\bigr\rangle \bigr] \\ &{}+(1-b_{n})\bigl[\lambda _{n}(\lambda _{n}-2 \beta ) \Vert Ay_{n}-Ap \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2}\bigr] \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} -\delta _{n} \bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq &(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2}\biggr] \\ &{}+2b_{n}\bigl[\bigl\langle f(x_{n})-f(p),x_{n} - p\bigr\rangle +\bigl\langle f(p)-p,x_{n} - p\bigr\rangle + \Vert x_{n+1}-x_{n} \Vert \bigl\Vert f(x_{n})-p \bigr\Vert \bigr] \\ &{}+(1-b_{n})\bigl[\lambda _{n}(\lambda _{n}-2 \beta ) \Vert Ay_{n}-Ap \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2}\bigr] \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} -\delta _{n} \bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq &(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2}\biggr] \\ &{}+2b_{n}\bigl[\rho \Vert x_{n}-p \Vert ^{2}+\bigl\langle f(p)-p,x_{n} - p\bigr\rangle + \Vert x_{n+1}-x_{n} \Vert \bigl\Vert f(x_{n})-p \bigr\Vert \bigr] \\ &{}+(1-b_{n})\bigl[\lambda _{n}(\lambda _{n}-2 \beta ) \Vert Ay_{n}-Ap \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2}\bigr] \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} -\delta _{n} \bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ \leq &\bigl(1-b_{n}(1-2\rho )\bigr) \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}-(1-b_{n}) \Vert u_{n} - v_{n} \Vert ^{2} \\ &{}+2b_{n}\bigl[\bigl\langle f(p)-p,x_{n} - p\bigr\rangle + \Vert x_{n+1}-x_{n} \Vert \bigl\Vert f(x_{n})-p \bigr\Vert \bigr] \\ &{}+(1-b_{n})\bigl[\lambda _{n}(\lambda _{n}-2 \beta ) \Vert Ay_{n}-Ap \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2}\bigr] \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} -\delta _{n} \bigl(1-2 \delta _{n}-\delta _{n}^{2}K^{2} \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2}. \end{aligned}$$

It follows that

$$\begin{aligned} \Vert x_{n+1}-p \Vert \leq &\bigl(1-b_{n}(1-2 \rho )\bigr) \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2} \\ &{}+2b_{n}\bigl[\bigl\langle f(p)-p,x_{n} - p\bigr\rangle + \Vert x_{n+1}-x_{n} \Vert \bigl\Vert f(x_{n})-p \bigr\Vert \bigr]. \end{aligned}$$

(16)

Next, we split the proof into two cases.

Case 1: Assume that there exists $n_{0}\in \mathbb{N}$ such that $\{\|x_{n}-p\|\}$ is decreasing for all $n\geq n_{0}$. Therefore, we obtain $\lim_{n\to \infty }\|x_{n}-p\|=d$. Consequently, we obtain

$$\begin{aligned}& (1-b_{n})\bigl[ \Vert u_{n} - v_{n} \Vert ^{2}+\lambda _{n}(2\beta -\lambda _{n}) \Vert Ay_{n}-Ap \Vert ^{2}+r_{n}(2\alpha -r_{n}) \Vert Mx_{n}-Mp \Vert ^{2}\bigr] \\& \qquad {}+\beta _{n}(\delta _{n}-\beta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} +\delta _{n} \bigl(2 \delta _{n}+\delta _{n}^{2}K^{2}-1 \bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\& \quad \leq \bigl(1-b_{n}(1-2\rho )\bigr) \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert x_{n+1}-p \Vert ^{2} \\& \qquad {}+2b_{n}\bigl[\bigl\langle f(p)-p,x_{n} - p\bigr\rangle + \Vert x_{n+1}-x_{n} \Vert \bigl\Vert f(x_{n})-p \bigr\Vert \bigr]. \end{aligned}$$

We find from the restrictions $(d_{1})$–$(d_{4})$ that

$$\begin{aligned} \begin{gathered} \lim_{n\to \infty } \Vert u_{n} - v_{n} \Vert =\lim_{n\to \infty } \Vert Ay_{n}-Ap \Vert , \\ \lim_{n\to \infty } \Vert Mx_{n}-Mp \Vert =\lim _{n\to \infty } \Vert z_{n}-St_{n} \Vert =\lim _{n\to \infty } \Vert z_{n}-Sz_{n} \Vert =0. \end{gathered} \end{aligned}$$

(17)

From $\|x_{n+1}-u_{n}\|\leq \|x_{n+1}-z_{n}\|+\|z_{n}-u_{n}\|$, $\|z_{n}-u_{n}\|\leq b_{n}\|f(x_{n})-u_{n}\|+(1-b_{n})\|v_{n}-u_{n} \|$, $\|x_{n+1}-z_{n}\|\leq \|z_{n}-St_{n}\|$ and the restrictions $(d_{3})$ we get

$$ \lim_{n\to \infty } \Vert x_{n+1}-z_{n} \Vert =\lim_{n\to \infty } \Vert z_{n}-u_{n} \Vert =\lim_{n\to \infty } \Vert x_{n+1}-u_{n} \Vert =0. $$

(18)

Observe that

$$\begin{aligned} \Vert u_{n}-p \Vert ^{2} =& \bigl\Vert J_{\lambda _{n}}^{B}(y_{n} - \lambda _{n}Ay_{n})-J_{ \lambda _{n}}^{B}(p- \lambda _{n}Ap) \bigr\Vert ^{2} \\ \leq &\bigl\langle (y_{n} - \lambda _{n}Ay_{n})-(P- \lambda _{n}Ap),u_{n}-p \bigr\rangle \\ =&\frac{1}{2} \bigl\Vert (y_{n} - \lambda _{n}Ay_{n})-(p-\lambda _{n}Ap) \bigr\Vert ^{2}+ \frac{1}{2} \Vert u_{n}-p \Vert ^{2} \\ &{}-\frac{1}{2} \bigl\Vert (y_{n} - \lambda _{n}Ay_{n})-(p-\lambda _{n}Ap)-(u_{n}-p) \bigr\Vert ^{2} \\ \leq &\frac{1}{2} \bigl[ \Vert y_{n}-p \Vert ^{2}+ \Vert u_{n}-p \Vert ^{2}- \bigl\Vert (y_{n}-u_{n})- \lambda _{n}(Ay_{n}-Ap) \bigr\Vert ^{2} \bigr] \\ =& \frac{1}{2} \bigl[ \Vert y_{n}-p \Vert ^{2}+ \Vert u_{n}-p \Vert ^{2}- \Vert y_{n}-u_{n} \Vert ^{2} +2\lambda _{n}\langle y_{n}-u_{n},Ay_{n}-Ap \rangle \\ &{}-\lambda _{n}^{2} \Vert Ay_{n}-Ap \Vert ^{2} \bigr], \end{aligned}$$

from which one deduces that

$$ \Vert u_{n}-p \Vert ^{2}\leq \Vert y_{n}-p \Vert ^{2}- \Vert y_{n}-u_{n} \Vert ^{2}+2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert . $$

(19)

Using Lemma 1.2 and (4), we have

$$\begin{aligned} \Vert y_{n}-p \Vert ^{2} =& \bigl\Vert T_{r_{n}}(x_{n} - r_{n}Mx_{n})-T_{r_{n}}(p-r_{n}Mp) \bigr\Vert ^{2} \\ \leq &\bigl\langle (x_{n} - r_{n}Mx_{n})-(p-r_{n}Mp),y_{n}-p \bigr\rangle \\ =&\frac{1}{2} \bigl\Vert (x_{n} - r_{n}Mx_{n})-(p-r_{n}Mp) \bigr\Vert ^{2}+\frac{1}{2} \Vert y_{n}-p \Vert ^{2} \\ &{}-\frac{1}{2} \bigl\Vert (x_{n} - r_{n}Mx_{n})-(p-r_{n}Mp)-(y_{n}-p) \bigr\Vert ^{2} \\ \leq &\frac{1}{2} \bigl[ \Vert x_{n}-p \Vert ^{2}+ \Vert y_{n}-p \Vert ^{2}- \bigl\Vert (x_{n}-y_{n})-2r_{n}(Mx_{n}-Mp) \bigr\Vert ^{2} \bigr] \\ =& \frac{1}{2} \bigl[ \Vert x_{n}-p \Vert ^{2}+ \Vert y_{n}-p \Vert ^{2}- \Vert x_{n}-y_{n} \Vert ^{2} +2r_{n}\langle x_{n}-y_{n},Mx_{n}-Mp \rangle \\ &{}-r_{n}^{2} \Vert Mx_{n}-Mp \Vert ^{2} \bigr]. \end{aligned}$$

It follows that

$$ \Vert y_{n}-p \Vert ^{2}\leq \Vert x_{n}-p \Vert ^{2}- \Vert x_{n}-y_{n} \Vert ^{2}+2r_{n} \langle x_{n}-y_{n},Mx_{n}-Mp \rangle . $$

(20)

We have from (7), (12), (14), (19) and (20)

$$\begin{aligned} \Vert x_{n+1} - p \Vert ^{2} =& \bigl\Vert b_{n}f(x_{n})+(1-b_{n})v_{n}-p \bigr\Vert ^{2} \\ \leq &b_{n} \bigl\Vert f(x_{n})-p \bigr\Vert ^{2}+(1-b_{n}) \Vert v_{n}-p \Vert ^{2} \\ \leq &b_{n} \bigl\Vert f(x_{n})-p \bigr\Vert ^{2}+(1-b_{n}) \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)\bigl[ \Vert x_{n} - p \Vert ^{2}- \Vert y_{n}-u_{n} \Vert ^{2} \\ &{}+2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert - \Vert x_{n}-y_{n} \Vert ^{2}+2r_{n} \langle x_{n}-y_{n},Mx_{n}-Mp \rangle \bigr] \\ \leq &b_{n}\bigl( \bigl\Vert f(x_{n})-f(p) \bigr\Vert + \bigl\Vert f(p)-p \bigr\Vert \bigr)^{2}+(1-b_{n}) \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)\bigl[ \Vert x_{n} - p \Vert ^{2} \\ &{}- \Vert y_{n}-u_{n} \Vert ^{2}+2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert - \Vert x_{n}-y_{n} \Vert ^{2} \\ &{}+2r_{n}\langle x_{n}-y_{n},Mx_{n}-Mp \rangle \bigr] \\ \leq &b_{n}\bigl(\rho \Vert x_{n}-p \Vert + \bigl\Vert f(p)-p \bigr\Vert \bigr)^{2}+(1-b_{n}) \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)\bigl[ \Vert x_{n} - p \Vert ^{2} \\ &{}- \Vert y_{n}-u_{n} \Vert ^{2}+2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert - \Vert x_{n}-y_{n} \Vert ^{2} \\ &{}+2r_{n}\langle x_{n}-y_{n},Mx_{n}-Mp \rangle \bigr] \\ \leq &b_{n}\bigl(\rho ^{2} \Vert x_{n}-p \Vert ^{2}+ \bigl\Vert f(p)-p \bigr\Vert ^{2}+2\rho \Vert x_{n}-p \Vert \bigl\Vert f(p)-p \bigr\Vert \bigr) \\ &{}+(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert y_{n}-u_{n} \Vert ^{2}\\ &{}+2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert \\ &{}- \Vert x_{n}-y_{n} \Vert ^{2}+2r_{n} \langle x_{n}-y_{n},Mx_{n}-Mp\rangle \biggr] \\ \leq &b_{n}\bigl(\rho (1+\rho ) \Vert x_{n}-p \Vert ^{2}+(1+\rho ) \bigl\Vert f(p)-p \bigr\Vert ^{2}\bigr) \\ &{}+(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert y_{n}-u_{n} \Vert ^{2}\\ &{}+2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert \\ &{}- \Vert x_{n}-y_{n} \Vert ^{2}+2r_{n} \langle x_{n}-y_{n},Mx_{n}-Mp\rangle \biggr] \\ \leq &\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \bigl(1-b_{n}\bigl(1- \rho (1+\rho )\bigr)\bigr) \Vert x_{n}-p \Vert ^{2}+b_{n}(1+ \rho ) \bigl\Vert f(p)-p \bigr\Vert ^{2} \\ &{}+(1-b_{n})\biggl[\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)2 \lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert - \Vert y_{n}-u_{n} \Vert ^{2} \\ &{}+\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)2r_{n}\langle x_{n}-y_{n},Mx_{n}-Mp \rangle - \Vert x_{n}-y_{n} \Vert ^{2}\biggr]. \end{aligned}$$

Consequently, we obtain

$$\begin{aligned}& (1-b_{n}) \bigl( \Vert y_{n}-u_{n} \Vert ^{2}+ \Vert x_{n}-y_{n} \Vert ^{2} \bigr) \\& \quad \leq \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \bigl(1-b_{n}\bigl(1- \rho (1+\rho )\bigr)\bigr) \Vert x_{n}-p \Vert ^{2} \\& \qquad {}- \Vert x_{n+1}-p \Vert ^{2}+b_{n}(1+ \rho ) \bigl\Vert f(p)-p \bigr\Vert ^{2} \\& \qquad {}+\biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr)\bigl[2\lambda _{n} \Vert y_{n}-u_{n} \Vert \Vert Ay_{n}-Ap \Vert \\& \qquad {}+2r_{n}\langle x_{n}-y_{n},Mx_{n}-Mp \rangle \bigr]. \end{aligned}$$

We find from (17) and the restrictions $(d_{3})$ and $(d_{4})$ that

$$ \lim_{n\to \infty } \Vert y_{n} - u_{n} \Vert =\lim_{n\to \infty } \Vert x_{n} - y_{n} \Vert =0.$$

(21)

We have from $\|x_{n} - x_{n+1}\|\leq \|x_{n} - y_{n}\|+\|y_{n} - u_{n}\|+\|u_{n} - x_{n+1}\|$ and (18) that

$$ \lim_{n\to \infty } \Vert x_{n} - x_{n+1} \Vert =0. $$

Next, we show that

$$ \limsup_{n\to \infty }\bigl\langle f(p)-p,x_{n}-p\bigr\rangle \leq 0, $$

where $p=P_{\varOmega }f(p)$. The existence of q is justified since $P_{\varOmega }$ is nonexpansive and f is a contraction, then $P_{\varOmega }$ is a contraction so it has a fixed point. To show it, choose a subsequence $\{x_{n_{j}}\}$ of $\{x_{n}\}$ such that

$$ \limsup_{n\to \infty } \bigl\langle f(p)-p,x_{n}-p \bigr\rangle =\lim_{j\to \infty }\bigl\langle f(p)-p,x_{n_{j}}-p \bigr\rangle .$$

(22)

Since $\{x_{n_{j}}\}$ is bounded, there exists a subsequence $\{x_{n_{j_{k}}}\}$ of $\{x_{n_{j}}\}$, converges weakly to u. Without loss of generality, we assume that $x_{n_{j}}\rightharpoonup u$. Since $\|x_{n}-y_{n}\|\to 0$ as $n\to \infty$ we obtain $y_{n_{j}}\rightharpoonup u$. Since $\{y_{n_{j}}\}\subset C$ and C is closed and convex, we obtain $u\in C$. First, we show that $u\in F(S)$. Then, from (17) and Lemma 1.4, we have $u\in F(S)$. We now show $u\in \operatorname{GEP}(F,M)$. By $y_{n} = T_{r_{n}}(x_{n} -r_{n}Mx_{n})$, we know that

$$ F(y_{n},y)+\langle Mx_{n},y-y_{n}\rangle + \frac{1}{r_{n}}\langle y-y_{n},y_{n}-x_{n} \rangle \geq 0 ,\quad \forall y\in C. $$

It follows from $(A_{2})$ that

$$ \langle Mx_{n},y-y_{n}\rangle +\frac{1}{r_{n}}\langle y-y_{n},y_{n}-x_{n} \rangle \geq F(y,y_{n}) ,\quad \forall y\in C. $$

Hence,

$$ \langle Mx_{n_{j}},y-y_{n_{j}}\rangle +\biggl\langle y-y_{n_{j}}, \frac{y_{n_{j}}-x_{n_{j}}}{r_{n_{j}}}\biggr\rangle \geq F(y,y_{n_{j}}) ,\quad \forall y\in C. $$

(23)

For t with $0 < t\leq 1$ and $y \in C$, let $y_{t} = ty + (1 - t)u$. Since $y \in C$ and $u\in C$, we obtain $y_{t}\in C$. So, from (23) we have

$$\begin{aligned} \langle y_{t}-y_{n_{j}},My_{t}\rangle \geq & \langle y_{t}-y_{n_{j}},My_{t} \rangle -\langle y_{t}-y_{n_{j}},Mx_{n_{j}}\rangle \\ &{}-\biggl\langle y_{t}-y_{n_{j}},\frac{y_{n_{j}}-x_{n_{j}}}{r_{n_{j}}} \biggr\rangle + F(y_{t},y_{n_{j}}) \\ =&\langle y_{t}-y_{n_{j}},My_{t}-My_{n_{j}} \rangle +\langle y_{t}-y_{n_{j}},My_{n_{j}}-Mx_{n_{j}} \rangle \\ &{}-\biggl\langle y_{t}-y_{n_{j}},\frac{y_{n_{j}}-x_{n_{j}}}{r_{n_{j}}} \biggr\rangle + F(y_{t},y_{n_{j}}). \end{aligned}$$

Since $\|y_{n_{j}}-x_{n_{j}}\|\to 0$, we have $\|My_{n_{j}}-Mx_{n_{j}}\|\to 0$. Further, from the inverse strongly monotonicity of M, we have $\langle y_{t}-y_{n_{j}},My_{t}-My_{n_{j}}\rangle \geq 0$. It follows from $A_{4}$ and $\frac{y_{n_{j}}-x_{n_{j}}}{r_{n_{j}}}\to 0$ and $y_{n_{j}}\rightharpoonup u$ that we have

$$ \langle y_{t}-v,My_{t}\rangle \geq F(y_{t},u), $$

as $j\to \infty$. From $(A_{1})$, $(A_{4})$ we have

$$\begin{aligned} 0 =&F(y_{t},y_{t}) \\ =&tF(y_{t},y)+(1-t)F(y_{t},u) \\ \leq &tF(y_{t},y)+(1-t)\langle y_{t}-u,My_{t} \rangle \\ =&tF(y_{t},y)+(1-t)t\langle y-u,My_{t}\rangle , \end{aligned}$$

and hence

$$ 0\leq F(y_{t},y)+(1-t)\langle y-u,My_{t}\rangle . $$

Letting $t \to 0$, we have, for each $y \in C$,

$$ F(u,y)+(1-t)\langle y-u,Mu\rangle \geq 0. $$

This implies that $u\in \operatorname{GEP}(F, M)$. Next we show $u\in (A+B)^{-1}(0)$. Due to (a), there is a subsequence $\{\lambda _{n_{j_{k}}}\}$ of $\{\lambda _{n_{j}}\}$ such that $\lambda _{n_{j_{k}}}\to {\lambda ^{*}}\in [c,d]$. Without loss of generality, we assume that $\lambda _{n_{j}}\to {\lambda ^{*}}$. From Lemma 1.7, we have

$$\begin{aligned}& \bigl\Vert x_{n_{j}}-J_{{\lambda ^{*}}}^{B}\bigl(I-{\lambda ^{*}}A\bigr)x_{n_{j}} \bigr\Vert \\& \quad \leq \Vert x_{n_{j}}-u_{n_{j}} \Vert + \bigl\Vert J_{\lambda _{n_{j}}}^{B}(I-\lambda _{n_{j}}A)x_{n_{j}}-J_{{ \lambda ^{*}}}^{B} \bigl(I-{\lambda ^{*}}A\bigr)x_{n_{j}} \bigr\Vert \\& \quad \leq \Vert x_{n_{j}}-u_{n_{j}} \Vert + \bigl\Vert J_{{\lambda _{n_{j}}}}^{B}(I-{ \lambda _{n_{j}}}A)x_{n_{j}}-J_{\lambda _{n_{j}}}^{B} \bigl(I-{\lambda ^{*}}A\bigr)x_{n_{j}} \bigr\Vert \\& \qquad {}+ \bigl\Vert J_{\lambda _{n_{j}}}^{B}\bigl(I-{\lambda ^{*}}A\bigr)x_{n_{j}}-J_{{ \lambda ^{*}}}^{B}\bigl(I-{ \lambda ^{*}}A\bigr)x_{n_{j}} \bigr\Vert \\& \quad \leq \Vert x_{n_{j}}-u_{n_{j}} \Vert + \bigl\vert \lambda _{n_{j}}-{\lambda ^{*}} \bigr\vert \Vert Ax_{n_{j}} \Vert \\& \qquad {}+ \biggl\vert \frac{\lambda _{n_{j}}-{\lambda ^{*}}}{{\lambda ^{*}}} \biggr\vert \bigl\Vert J_{{ \lambda ^{*}}}^{B}\bigl(I-{\lambda ^{*}}A \bigr)x_{n_{j}}-\bigl(I-{\lambda ^{*}}A\bigr)x_{n_{j}} \bigr\Vert . \end{aligned}$$

This implies that

$$ \lim_{j\to \infty } \bigl\Vert x_{n_{j}}-J_{{\lambda ^{*}}}^{B} \bigl(I-{\lambda ^{*}}A\bigr)x_{n_{j}} \bigr\Vert =0. $$

Since $J_{{\lambda ^{*}}}^{B}(I-{\lambda ^{*}}M)$ is nonexpansive, the demiclosedness for a nonexpansive mapping implies that $u\in F(J_{{\lambda ^{*}}}^{B}(I-{\lambda ^{*}}A))$, that is, $u\in (A+B)^{-1}(0)$. Finally we show $u\in N^{-1}(0)$. Since $\|e_{n}\|\to 0$ and $\|x_{n}-v_{n}\|=\|u_{n}-v_{n}\|\to 0$ as $n\to \infty$, we have $v_{n_{j}}\rightharpoonup u$ and

$$ \bigl\Vert x_{n_{j}}+e_{n_{j}}-J_{{s_{n}}}^{N}(x_{n_{j}}+e_{n_{j}}) \bigr\Vert \leq \Vert x_{n_{j}}-v_{n_{j}} \Vert + \Vert e_{n_{j}} \Vert \to 0. $$

From Lemma 1.1, we have $0\in N(u)$. This implies $u\in \varOmega$. Due to (22), we arrive at

$$ \limsup_{n\to \infty } \bigl\langle f(p)-p,x_{n}-p\bigr\rangle =\lim_{j\to \infty }\bigl\langle f(p)-p,x_{n_{j}}-p\bigr\rangle =\bigl\langle f(p)-p,u-p\bigr\rangle \leq 0. $$

Since $\lim_{n\to \infty }b_{n} = 0$, $\sum_{n=0}^{\infty }b_{n}=\infty$, $\lim_{n\to \infty }\|x_{n+1}-x_{n}\|=0$ and $\sum_{n={0}}^{+\infty }\varepsilon _{n}<+\infty$, we obtain from Lemma 1.5 and (16)

$$ \lim_{n\to \infty } \Vert x_{n}-p \Vert =0. $$

Consequently, $x_{n}\to p=P_{C}f(p)$.

Case 2: Assume that there exists a subsequence $\{n_{j}\}$ of $\{n\}$ such that

$$ \Vert x_{n_{j}}-p \Vert \leq \Vert x_{n_{j}+1} -p \Vert $$

for all $j\in \mathbb{N}$. From Lemma 1.9 there exists a nondecreasing sequence $\{t_{k}\}\subset \mathbb{N}$ such that $t_{k}\to \infty$ and

$$ \Vert x_{t_{k}}-p \Vert \leq \Vert x_{t_{k}+1}-p \Vert \quad \text{and}\quad \Vert x_{k}-p \Vert \leq \Vert x_{t_{k}+1}-p \Vert $$

(24)

for all $k\in \mathbb{N}$. Since $\lim_{n\to \infty }b_{n}=0$ and $\sum_{n=0}^{+\infty } \varepsilon _{n}<+\infty$ we can obtain from (17), (18) and (21)

$$\begin{aligned} \lim_{k\to \infty } \Vert x_{t_{k}}-z_{t_{k}} \Vert &= \lim_{k\to \infty } \Vert x_{t_{k}}-y_{t_{k}} \Vert = \lim_{k\to \infty } \Vert x_{t_{k}}-v_{t_{k}} \Vert \\ &=\lim_{k\to \infty } \Vert z_{t_{k}}-Tz_{t_{k}} \Vert = \lim_{k\to \infty } \Vert x_{t_{k}+1}-x_{t_{k}} \Vert =0. \end{aligned}$$

From Case 1, we also have

$$ \limsup_{k\to \infty }\bigl\langle f(p)-p,x_{t_{k}}-p \bigr\rangle \leq 0 $$

(25)

Using (16) and following the methods used to get (16), we obtain

$$\begin{aligned} =& \Vert x_{t_{k}+1}-p \Vert ^{2} \leq (1-b_{t_{k}}) \biggl(1+ \frac{2\varepsilon _{t_{k}}}{1-2\varepsilon _{t_{k}}}\biggr) \Vert x_{t_{k}}-p \Vert ^{2}+2b_{t_{k}} \bigl\langle f(p)-p,x_{t_{k}} - p\bigr\rangle \\ &{}+2b_{t_{k}}\bigl( \Vert x_{t_{k}+1}-x_{t_{k}} \Vert \bigl\Vert f(x_{t_{k}})-p \bigr\Vert +\rho \Vert x_{t_{k}}-p \Vert ^{2}\bigr) \\ \leq &\bigl(1-b_{t_{k}}(1-2\rho )\bigr) \Vert x_{t_{k}}-p \Vert ^{2}+ \frac{2\varepsilon _{t_{k}}}{1-2\varepsilon _{t_{k}}}L+2b_{t_{k}} \bigl\langle f(p)-p,x_{t_{k}} - p\bigr\rangle \\ &{}+2b_{t_{k}} \Vert x_{t_{k}+1}-x_{t_{k}} \Vert \bigl\Vert f(x_{t_{k}})-p \bigr\Vert \end{aligned}$$

(26)

where $L>0$ is a sufficiently large number. This implies that

$$\begin{aligned} b_{t_{k}}(1-2\rho ) \Vert x_{t_{k}}-p \Vert ^{2} \leq & \Vert x_{t_{k}}-p \Vert ^{2}- \Vert x_{t_{k}+1}-p \Vert ^{2}+\frac{2\varepsilon _{t_{k}}}{1-2\varepsilon _{t_{k}}}L \\ &{}+2b_{t_{k}}\bigl\langle f(p)-p,x_{t_{k}} - p\bigr\rangle +2b_{t_{k}} \Vert x_{t_{k}+1}-x_{t_{k}} \Vert \bigl\Vert f(x_{t_{k}})-p \bigr\Vert . \end{aligned}$$

Since $b_{t_{k}}>0$, we get from (24)

$$ (1-\rho ) \Vert x_{t_{k}}-p \Vert ^{2}\leq \frac{2\varepsilon _{t_{k}}}{1-2\varepsilon _{t_{k}}}L+2\bigl\langle f(p)-p,x_{t_{k}} - p\bigr\rangle +2 \Vert x_{t_{k}+1}-x_{t_{k}} \Vert \bigl\Vert f(x_{t_{k}})-p \bigr\Vert . $$

Since $\lim_{n\to \infty }\|x_{n+1}-x_{n}\|=0$ and $\sum_{n=0}^{+\infty } \varepsilon _{n}<+\infty$, we obtain from (25) $\|x_{t_{k}}-p\|\to 0$ as ${k\to \infty }$. From (26) we have $\|x_{t_{k}+1}-p\|\to 0$ as ${k\to \infty }$. Using (24), we obtain $\lim_{k\to \infty }\|x_{k}-p\|=0$. Therefore, from the above two cases, we can conclude that $\{x_{n}\}$ converges strongly to a point $p=P_{\varOmega }f(p)$, which satisfies the variational inequality $\langle (I-f)p,x-p\rangle \geq 0$, for all $x \in \varOmega$. The proof is complete. □

If $f(x) = u\in C$ in Theorem 2.1, then we have the following result.

Corollary 2.2

Assume that C is a nonempty, closed and convex subset of H, F is a bi-function from $C \times C$to $\mathbb{R}$satisfying $(A_{1})$–$(A_{4})$, M is an α-inverse strongly monotone mapping from C into H, A is a β-inverse strongly monotone map from C into H, B and N are two maximal monotone operators on H such that their domains contained in C and $S : C \to C$is a Lipschitz pseudo-contractive mapping with Lipschitz constants K such that ${\varOmega }=F(S)\cap (A+B)^{-1}(0) \cap N^{-1}(0)\cap \operatorname{GEP}(F,M)\neq \emptyset$. Assume that $\{b_{n}\}$, $\{\beta _{n}\}$and $\{\delta _{n}\}$are some sequences in $(0, 1)$and $\{x_{n}\}$, $\{y_{n}\}$, $\{u_{n}\}$and $\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{0}\in C, \\ F (y_{n}, y) + \langle Mx_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0, \quad \forall y\in C, \\ u_{n}=J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}Ay_{n}), \\ z_{n}=b_{n}u+(1-b_{n})J_{s_{n}}^{N}( u_{n}+e_{n}), \\ x_{n+1}=(1-\beta _{n})z_{n}+\beta _{n}S(\delta _{n}z_{n}+(1-\delta _{n}) Sz_{n}), \quad \forall n\geq 0. \end{cases} $$

If the conditions $(d_{1})$–$(d_{4})$hold, then the sequence $\{x_{n}\}$converges strongly to a point $p\in \varOmega$which is the unique solution of the variational inequality $\langle p-u,x-p\rangle \geq 0$for all $x \in \varOmega$.

Now, we discuss weak convergence of the sequence in the new iteration.

Theorem 2.3

Assume that C is a nonempty, closed and convex subset of H, F is a bi-function from $C \times C$to $\mathbb{R}$satisfying $(A_{1})$–$(A_{4})$, M is an α-inverse strongly monotone mapping from C into H, A is a β-inverse strongly monotone map from C into H, B and N are two maximal monotone operators on H such that their domains contained in C and $S : C \to C$is a Lipschitz pseudo-contractive mapping with Lipschitz constants k such that ${\varOmega }=F(S)\cap (A+B)^{-1}(0) \cap N^{-1}(0)\cap \operatorname{GEP}(F,M)\neq \emptyset$. Assume that $\{b_{n}\}$, $\{\beta _{n}\}$and $\{\delta _{n}\}$are some sequences in $(0, 1)$and $\{x_{n}\}$, $\{y_{n}\}$, $\{u_{n}\}$and $\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{0}\in C, \\ F (y_{n}, y) + \langle Mx_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0, \quad \forall y\in C, \\ u_{n}=J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}Ay_{n}), \\ z_{n}=b_{n}x_{n}+(1-b_{n})J_{s_{n}}^{N}( u_{n}+e_{n}), \\ x_{n+1}=(1-\beta _{n})z_{n}+\beta _{n}S(\delta _{n}z_{n}+(1-\delta _{n}) Sz_{n}) \quad \forall n\geq 0. \end{cases} $$

(27)

If

$(d_{1})$:

$0< c\leq \lambda _{n}\leq d<2\beta$, $0< a\leq r_{n}\leq b<2\alpha$,

$(d_{2})$:

$0< c<\beta _{n}\leq \delta _{n}<d<\frac{1}{\sqrt{1+k^{2}}+1}$, $s_{n}>s>0$,

$(d_{3})$:

$\| e _{n}\| \leq \frac{\varepsilon _{n}}{2} \max \{\| {u}_{n} - J_{s_{n}}^{N}( u_{n}+e_{n}) \|, \|J_{s_{n}}^{N}( u_{n}+e_{n})-p\|\}$with $\sum_{n=0}^{ \infty } \varepsilon _{n} < \infty$,

then $\{x_{n}\}$converges weakly to an element $p\in \varOmega$.

Proof

Let $p\in \varOmega$. Similarly, from (6) and (7) we obtain

$$\begin{aligned} \Vert y_{n} - p \Vert ^{2} \leq & \Vert x_{n}-p \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Mx_{n}-Mp \Vert ^{2} \\ \leq & \Vert x_{n}-p \Vert ^{2} \end{aligned}$$

and

$$\begin{aligned} \Vert u_{n} - p \Vert ^{2} \leq & \Vert x_{n}-p \Vert ^{2}+\lambda _{n}(\lambda _{n}-2 \beta ) \Vert Ay_{n}-Ap \Vert ^{2} \\ \leq & \Vert x_{n}-p \Vert ^{2}. \end{aligned}$$

We also conclude from (11) and (12) that

$$\begin{aligned} \Vert x_{n+1}-p \Vert ^{2} \leq & \Vert z_{n}-p \Vert ^{2}-\delta _{n}\bigl(1-2\delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2}. \end{aligned}$$

(28)

Put $v_{n}=J_{\lambda _{n}}^{N}( u_{n}+e_{n})$ for all $n\geq 0$. From (14), we have

$$\begin{aligned} \Vert v_{n} - p \Vert ^{2} \leq & \biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2} \\ \leq & \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}. \end{aligned}$$

These have already been proved in Theorem 2.1. Since $0< c<\beta _{n}\leq \delta _{n}<d<\frac{1}{\sqrt{1+k^{2}}+1}$ for all $n\geq 1$, we conclude from (27), (28) and Lemma 1.6 that

$$\begin{aligned} =& \Vert x_{n+1}-p \Vert ^{2} \\ \leq& \Vert z_{n}-p \Vert ^{2}-\delta _{n}\bigl(1-2 \delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ =& \bigl\Vert b_{n}x_{n}+(1-b_{n})v_{n} - p \bigr\Vert ^{2} -\delta _{n}\bigl(1-2\delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & b_{n} \Vert x_{n}-p \Vert ^{2}+(1-b_{n}) \Vert v_{n} - p \Vert ^{2}-(1-b_{n})b_{n} \Vert x_{n}-v_{n} \Vert ^{2} \\ &{}-\delta _{n}\bigl(1-2\delta _{n}-\delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} +\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & b_{n} \Vert x_{n}-p \Vert ^{2}+(1-b_{n}) \biggl[\biggl(1+ \frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert u_{n} - v_{n} \Vert ^{2}\biggr] \\ &{}-(1-b_{n})b_{n} \Vert x_{n}-v_{n} \Vert ^{2}-\delta _{n}\bigl(1-2\delta _{n}- \delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} \\ &{}+\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}}(1-b_{n}) \biggr) \Vert x_{n} - p \Vert ^{2}-(1-b_{n}) \bigl( \Vert u_{n} - v_{n} \Vert ^{2}+b_{n} \Vert x_{n}-v_{n} \Vert ^{2}\bigr) \\ &{}-\delta _{n}\bigl(1-2\delta _{n}-\delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} +\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}-(1-b_{n}) \bigl( \Vert u_{n} - v_{n} \Vert ^{2}+b_{n} \Vert x_{n}-v_{n} \Vert ^{2}\bigr) \\ &{}-\delta _{n}\bigl(1-2\delta _{n}-\delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2} +\beta _{n}(\beta _{n}-\delta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\ \leq & \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}. \end{aligned}$$

(29)

For every $n= 0, 1, 2, \ldots$ , since $\sum_{n=0}^{\infty } \varepsilon _{n}^{2} < \infty$, we obtain

$$ M_{0} := \sum_{n=m_{0}}^{\infty } \frac{2\varepsilon _{n}^{2}}{1-2\varepsilon _{n}^{2}} < \infty \quad \text{and} \quad M_{1} := \prod _{n=m_{0}}^{\infty } \biggl( 1 + \frac{2\varepsilon _{n}^{2}}{1-2\varepsilon _{n}^{2}} \biggr) < \infty . $$

Hence for each integer $n \geq m_{0}$,

$$\begin{aligned} \Vert x_{n+1} - p \Vert ^{2} \leq & \biggl( 1 + \frac{2\varepsilon _{n}^{2}}{1 - 2\varepsilon _{n}^{2}} \biggr) \Vert x_{n} - p \Vert ^{2} \\ \leq & \biggl( 1 + \frac{2\varepsilon _{n}^{2}}{1 - 2\varepsilon _{n}^{2}} \biggr) \biggl( 1 + \frac{2\varepsilon _{n-1}^{2}}{1 - 2\varepsilon _{n-1}^{2}} \biggr) \Vert x_{n-1} - p \Vert ^{2} \\ \vdots &\\ \leq & \prod_{i=m_{0}}^{n} \biggl( 1 + \frac{2\varepsilon _{i}^{2}}{1-2\varepsilon _{i}^{2}} \biggr) \Vert x_{m_{0}} - p \Vert ^{2} \\ \leq & \prod_{i=m_{0}}^{\infty } \biggl( 1 + \frac{2\varepsilon _{i}^{2}}{1-2\varepsilon _{i}^{2}} \biggr) \Vert x_{m_{0}} - p \Vert ^{2} = K_{1} \Vert x_{m_{0}} - p \Vert ^{2}. \end{aligned}$$

Therefore, $\{\|x_{n}-p\|\}$ is bounded. So, $\{x_{n}\}$ is bounded and so are the sequences $\{y_{n}\}$, $\{u_{n}\}$ and $\{z_{n}\}$. Setting $K:= \sup_{n \geq 0} \| x_{n} - p \|$, we obtain from (29)

$$ \Vert x_{n+1} - p \Vert ^{2} \leq \Vert x_{n} - p \Vert ^{2} + \frac{2\varepsilon _{n}^{2}}{1 - 2\varepsilon _{n}^{2}} K^{2}, \quad \forall n\geq m_{0} . $$

Thus it follows that, for all $n,m \geq m_{0}$,

$$\begin{aligned} \Vert x_{n+m +1} - p \Vert ^{2} \leq & \Vert x_{n + m} - p \Vert ^{2} + \frac{2\varepsilon _{n+m}^{2}}{1 - 2\varepsilon _{n+m}^{2}} K^{2} \\ \leq & \Vert x_{n+m-1} - p \Vert ^{2} + \frac{2\varepsilon _{n+m-1}^{2}}{1 - 2\varepsilon _{n+m-1}^{2}} K^{2} + \frac{2\varepsilon _{n+m}^{2}}{1 - 2\varepsilon _{n+m}^{2}} K^{2} \\ \vdots &\\ \leq & \Vert x_{n} - p \Vert ^{2} + \sum _{n=m_{0}}^{\infty } \frac{2\varepsilon _{n}^{2}}{1-2\varepsilon _{n}^{2}} K^{2}. \end{aligned}$$

Since $\sum_{n=0}^{\infty } \frac{2\varepsilon _{n}^{2}}{1-2\varepsilon _{n}^{2}} < \infty$ we obtain

$$ \limsup_{m\to \infty } \Vert x_{m} - p \Vert ^{2} \leq \Vert x_{n} - p \Vert ^{2} + \sum _{i=0}^{\infty } \frac{2\varepsilon _{i}^{2}}{1-2\varepsilon _{i}^{2}} K^{2}. $$

This implies that for every $p \in \varOmega$, $\lim_{n \rightarrow \infty }\| x_{n} - p \|^{2}$ exists. From (29), we have

$$\begin{aligned}& \delta _{n}\bigl(1-2\delta _{n}-\delta _{n}^{2}K^{2}\bigr) \Vert z_{n}-Sz_{n} \Vert ^{2}+ \beta _{n}(\delta _{n}-\beta _{n}) \Vert z_{n}-St_{n} \Vert ^{2} \\& \qquad {}+(1-b_{n}) \bigl( \Vert u_{n} - v_{n} \Vert ^{2}+b_{n} \Vert x_{n}-v_{n} \Vert ^{2}\bigr) \\& \quad \leq \biggl(1+\frac{2\varepsilon _{n}}{1-2\varepsilon _{n}} \biggr) \Vert x_{n} - p \Vert ^{2}- \Vert x_{n+1} - p \Vert ^{2} . \end{aligned}$$

We find from the restrictions $(d_{1})$–$(d_{3})$ that

$$ \lim_{n\to \infty } \Vert u_{n} - v_{n} \Vert =\lim_{n\to \infty } \Vert x_{n}-v_{n} \Vert =\lim_{n\to \infty } \Vert z_{n}-St_{n} \Vert =\lim_{n\to \infty } \Vert z_{n}-Sz_{n} \Vert =0.$$

(30)

From $\|x_{n+1}-u_{n}\|\leq \|x_{n+1}-z_{n}\|+\|z_{n}-u_{n}\|$, $\|z_{n}-u_{n}\|\leq b_{n}\|x_{n}-u_{n}\|+(1-b_{n})\|v_{n}-u_{n}\|$, $\|x_{n+1}-z_{n}\|\leq \|z_{n}-St_{n}\|$ and $\|x_{n}-u_{n}\|\leq \|x_{n}-v_{n}\|+\|v_{n}-u_{n}\|$ we get

$$ \lim_{n\to \infty } \Vert x_{n+1}-z_{n} \Vert =\lim_{n\to \infty } \Vert z_{n}-u_{n} \Vert =\lim_{n\to \infty } \Vert x_{n+1}-u_{n} \Vert =\lim_{n\to \infty } \Vert x_{n}-u_{n} \Vert =0 .$$

(31)

Also from $\|x_{n+1}-x_{n}\|\leq \|x_{n+1}-z_{n}\|+\|z_{n}-x_{n}\|$ and $\|x_{n}-z_{n}\|\leq \|x_{n}-u_{n}\|+\|u_{n}-z_{n}\|$ we obtain

$$ \lim_{n\to \infty } \Vert x_{n}-z_{n} \Vert =\lim_{n\to \infty } \Vert x_{n+1}-x_{n} \Vert =0. $$

(32)

Since $\{x_{n}\}$ is bounded, there exists a subsequence $\{x_{n_{j}}\}$ of $\{x_{n}\}$ converging weakly to u. Since $\|x_{n}-y_{n}\|\to 0$ as $n\to \infty$ we obtain $y_{n_{j}}\rightharpoonup u$. Since $\{y_{n_{j}}\}\subset C$ and C is closed and convex, we obtain $u\in C$. First, we show that $u\in F(S)$. Then, from (30) and Lemma 1.4, we have $u\cap F(S)$. Using the same argument we had in Theorem 2.1, we get $u\in \operatorname{GEP}(F,M)$ and $u\in (A+B)^{-1}(0)$. In a similar way, we have $0\in N(u)$. This implies $u\in \varOmega$.

Let us consider the uniqueness of the weak cluster point of $\{x_{n}\}$. Suppose there exist two weak cluster points û and ū of the sequence $\{ x_{n}\}$, then û and ū belong to Ω and the sequences $\{ \| \hat{u} - x_{n}\| \}$ and $\{ \| \bar{u} - x_{n}\| \}$ converge; i.e., there exist $\hat{\beta }, \bar{\beta } \in \mathbb{R}^{+}$ such that

$$ \lim_{n \to +\infty } \Vert \hat{u} - x_{n} \Vert = \hat{\beta },\qquad \lim_{n \to +\infty } \Vert \hat{u} - x_{n} \Vert = \bar{\beta } .$$

(33)

Since

$$ \Vert \hat{u} - x_{n} \Vert ^{2} = \Vert \bar{u} - x_{n} \Vert ^{2} + 2\langle x_{n} - \hat{u} , \hat{u} - \bar{u} \rangle + \Vert \hat{u} - \bar{u} \Vert ^{2}, $$

from (33), we have

$$ \lim_{n \to +\infty }\langle x_{n} - \hat{u} , \hat{u} - \bar{u} \rangle = \frac{1}{2}\bigl(\bar{\beta }^{2} - \hat{\beta }^{2} - \Vert \hat{u} - \bar{u} \Vert ^{2} \bigr). $$

(34)

Because û is a weak cluster point of $\{ x_{n}\}$, which implies that

$$ \bar{\beta }^{2} - \hat{\beta }^{2} = \Vert \hat{u} - \bar{u} \Vert ^{2}. $$

(35)

Reversing the roles of p̄ and p̂, hence $\hat{\beta }^{2} - \bar{\beta }^{2} = \| \hat{u} - \bar{u} \|^{2}$, Combining this with (35), we have $\| \hat{u} - \bar{u} \| = 0$, i.e., $\hat{u} = \bar{u}$, which is a contradiction. Therefore, there exists an unique weak cluster point of $\{x_{n}\}$.Then $\{ x_{n}\}$ is weakly convergent to an element of Ω, and this completes the proof of Theorem 2.3. □

Remark 2.1

Theorem 2.1 and Theorem 2.3 improves and extends the result in Ceng, Wu, Yao ([21]), Han, He ([20]) and Tian, Wang ([22]).

Let $I_{C}$ be the indicator function of C defined by $I_{C}(x) =0$ whenever $x \in C$ and $I_{C}(x)=\infty$ otherwise. Recall that the subdifferential $\partial I_{C}$ of $I_{C}$ is a maximal monotone operator since $I_{C}$ is a proper lower semi-continuous convex function on H. The resolvent $J_{r}^{\partial I_{C}}$ of $\partial I_{C}$ for r is $P_{C}$ and $\operatorname{VI}(C,A)=(A + \partial I_{C})^{-1}(0)$, where A is an inverse strongly monotone mapping of C into H ([40]). We obtain the following result.

Theorem 2.4

Suppose that C is a nonempty, closed and convex subset of H, F is a bi-function from $C \times C$to $\mathbb{R}$satisfying $(A_{1})$–$(A_{4})$, M is an α-inverse strongly monotone mapping from C into H, A is a β-inverse strongly monotone map from C into H, B and N are two maximal monotone operators on H such that their domains contained in C, $f : C \to C$is a ρ-contractive map with $\rho \in (0,\frac{1}{2})$and $S : C \to C$a Lipschitz pseudo-contractive mapping with Lipschitz constants K such that ${\varOmega }=F(S)\cap \operatorname{VI}(C,A) \cap N^{-1}(0)\cap \operatorname{GEP}(F,M)\neq \emptyset$. Assume that $\{b_{n}\}$, $\{\beta _{n}\}$and $\{\delta _{n}\}$are some sequences in $(0, 1)$and $\{x_{n}\}$, $\{y_{n}\}$, $\{u_{n}\}$and $\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{0}\in C, \\ F (y_{n}, y) + \langle Mx_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0, \quad \forall y\in C, \\ u_{n}=P_{C}(y_{n}-\lambda _{n}Ay_{n}), \\ z_{n}=b_{n}f(x_{n})+(1-b_{n})J_{s_{n}}^{N}( u_{n}+e_{n}), \\ x_{n+1}=(1-\beta _{n})z_{n}+\beta _{n}S(\delta _{n}z_{n}+(1-\delta _{n}) Sz_{n}) \quad \forall n\geq 0. \end{cases} $$

If the conditions $(d_{1})$–$(d_{4})$hold, then, $\{x_{n}\}$converges strongly to a point $p\in \varOmega$which is the unique solution of the variational inequality $\langle (I-f)p,x-p\rangle \geq 0$for all $x \in \varOmega$.

Proof

Putting $B = \partial I_{C}$ in Theorem 2.1, we know that $J_{\lambda _{n}} = P_{C}$ for all $\lambda _{n} > 0$, we obtain the desired result. □

Remark 2.2

Theorem 2.4 improves and extends the result in Takahashi, Takahashi ([13]) and Su, Shang, Qin ([41]).

Theorem 2.5

Suppose that C is a nonempty, closed and convex subset of H, F is a bi-function from $C \times C$to $\mathbb{R}$satisfying $(A_{1})$–$(A_{4})$, M is an α-inverse strongly monotone mapping from C into H, $\psi :C\to C$is a β-strict pseudo-contraction, N is a maximal monotone operator on H such that its domains contained in C, $f : C \to C$is a ρ-contractive map with $\rho \in (0,\frac{1}{2})$and $S : C \to C$is a Lipschitz pseudo-contractive mapping with Lipschitz constants K such that ${\varOmega }=F(S)\cap F(\psi )\cap N^{-1}(0)\cap \operatorname{GEP}(F,M)\neq \emptyset$. Assume that $\{b_{n}\}$, $\{\beta _{n}\}$and $\{\delta _{n}\}$are some sequences in $(0, 1)$and $\{x_{n}\}$, $\{y_{n}\}$, $\{u_{n}\}$and $\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{0}\in C, \\ F (y_{n}, y) + \langle Mx_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0, \quad \forall y\in C, \\ u_{n}=(1-\lambda _{n})y_{n}+\lambda _{n}\psi y_{n}, \\ z_{n}=b_{n}f(x_{n})+(1-b_{n})J_{s_{n}}^{N}( u_{n}+e_{n}), \\ x_{n+1}=(1-\beta _{n})z_{n}+\beta _{n}S(\delta _{n}z_{n}+(1-\delta _{n}) Sz_{n}) \quad \forall n\geq 0. \end{cases} $$

If the conditions $(d_{1})$–$(d_{4})$hold, $0< c<\lambda _{n}<d<1-\beta$, then $\{x_{n}\}$converges strongly to a point $p\in \varOmega$which is the unique solution of the variational inequality $\langle (I-f)p,x-p\rangle \geq 0$for all $x \in \varOmega$.

Proof

Putting $B = \partial I_{C}$, $A = I-\psi$, we see that A is $\frac{1-\beta }{2}$-inverse strongly monotone. We also have $J_{\lambda _{n}} = P_{C}$ for all $\lambda _{n} > 0$, $F (\psi ) = V I(C, A)$ and $P_{C} (y_{n} -\lambda _{n}Ay_{n}) = (1 - \lambda _{n})y_{n} + \lambda _{n}\psi y_{n}$, by Theorem 2.1 we obtain the desired result. □

Now, we provide an example to illustrate our first result.

Example 2.1

Let $H = \mathbb{R}$ with Euclidean norm and usual Euclidean inner product. Let $C :=(-\infty ,1]$, $Sx=\frac{x}{x-2}$, $Bx=\log (1-x)$, $Ax=2x$, $\beta \leq \frac{1}{2}$, $F(x,y)=y-x$, $N(x)=\log (1-x^{3})$, $\alpha \leq 1$ and $Mx =x-1$. Clearly, S is a Lipschitz pseudo-contractive mapping with Lipschitz constants $K\leq \frac{1}{10}$, A a β-inverse strongly monotone mapping, B, N maximal monotone operators, F a bi-function from $C \times C$ to $\mathbb{R}$ satisfying $(A_{1})$–$(A_{4})$, M an α-inverse strongly monotone mapping and $0\in N^{-1}(0)\cap F(S) \cap (A+B)^{-1}(0) \cap \operatorname{GEP}(F,M)$.

3 Conclusion

As is well known, many things need to be optimized. Numerous techniques and methods have been used to optimize a variety of issues. This has even been used to solve some differential equations. In this work, we introduced a new iterative method for finding a common element of the set of fixed points of a pseudo-contractive mapping, the set of solutions to a variational inclusion and the set of solutions to a generalized equilibrium problem in a real Hilbert space. We provided some strong and weak convergence results as regards the common solutions. Finally, we provided an example to illustrate our first main result.