Hybrid method for equilibrium problems and variational inclusions

Abstract

By providing a new iterative method our aim is finding a common element of the set of fixed points of two nonexpansive mappings, the set of solutions to a variational inclusion and the set of solutions of a generalized equilibrium problem in a real Hilbert space. We review the strong convergence of the new iterative method in the framework of Hilbert spaces. Finally, we show that our main result is a generalization for some known theorems in this field.

1 Introduction

Let H be a real Hilbert space, let C be a nonempty closed convex subset of H, let $A: C \to H$ be a nonlinear map, and let F be a bifunction from $C \times C$ to $\mathbb{R}$. In 2008, Takahashi et al. [1] considered the generalized equilibrium problem: Finding $x\in C$ such that

$$ F(x,y)+\langle Ax,y-x\rangle \geq 0\quad (\forall y\in C). $$

(1)

The set of solutions of (1) is denoted by $\operatorname{GEP}(F,A)$. If $A=0$, then problem (1) becomes the equilibrium problem: Finding $x\in C$ such that

$$ F(x,y)\geq 0\quad (\forall y\in C). $$

(2)

Problem (2) was studied by Blum et al. [2] in 1994. The set of solutions of (2) is denoted by $\operatorname{EP}(F)$. If $F(x, y) = 0$ for all $x, y \in C$, then problem (1) becomes the variational inequality problem: Finding $x\in C$ such that

$$ \langle Ax,y-x\rangle \geq 0\quad (\forall y\in C). $$

(3)

Problem (3) was studied by Hartman et al. [3] in 1966 and has been extensively in the literature (see, e.g., [4–10]). The set of solutions of (3) is denoted by $\operatorname{VI}(C,A)$. If $C=H$, then $\operatorname{VI}(H,A)=A^{-1}(0)=\{x\in H:Ax=0\}$. Recall that a mapping $A : C\to H$ is said to be monotone if $\langle Au-Av,u-v\rangle \geq 0$ for all $u,v\in C$ [6, 7, 11]. A mapping A is said to be α-strongly monotone if there exists a positive real number α such that $\langle Au-Av,u-v\rangle \geq \alpha \Vert u-v \Vert ^{2}$ for all $u,v\in C$ [6, 7, 11]. A mapping A is said to be α-inverse strongly monotone if there exists a positive real number α such that $\langle Au-Av,u-v\rangle \geq \alpha \Vert Au-Av \Vert ^{2}$ for all $u,v\in C$ [6, 7, 11]. In such a case, A is said to be α-inverse-strongly monotone.

Let $T:C\to C$ be a mapping. We denote by $F(T)$ the fixed-point set of T, that is, $F(T)=\{x\in C: T(x)=x\}$. A mapping T is said to be L-Lipschitz if there exists $L \geq 0$ such that $\Vert Tu-Tv \Vert \leq L \Vert u-v \Vert$ for all $u,v\in C$. The mapping T is called nonexpansive if $L=1$. It is also called contraction if $L < 1$. Note that any α-inverse strongly monotone mapping A is Lipschitz and $\Vert Au-Av \Vert \leq \frac{1}{\alpha } \Vert u-v \Vert$ [6, 7, 11]. There are a lot works associated with the fixed point algorithms for nonexpansive mappings (see, e.g., [12–23]).

Let $A : H \to H$ be a single-valued nonlinear map, and let $B : H \to 2^{H}$ be a set-valued mapping. The variational inclusion is finding $p \in H$ such that

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

(4)

where θ is the zero vector in H. For $A = 0$, (4) becomes the inclusion problem introduced by Rockafellar [24]. The effective domain of B is denoted by $D(B)$, that is, $D(B) = \{x \in H : Bx \neq \emptyset \}$. 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 if $\langle x-y,f-h\rangle \geq 0$ for all $x,y\in D(B)$, $f\in Bx$, and $h\in By$ [25]. 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 [25]. Also, a monotone mapping B is maximal if and only if $\langle x-y,f-h\rangle \geq 0$ ($(x,f) \in H\times H$$(y,h)\in G(B)$) implies $f\in B x$ [25]. For a maximal monotone operator B on H and $r > 0$, we define the 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, $\langle x - y,J^{B}_{r}x - J^{B}_{r}y\rangle \geq \Vert J^{B}_{r}x-J^{B}_{r}y \Vert ^{2}$ for all $x,y\in H$ (see [13]), and that a solution of (4) is a fixed point of $J^{B}_{r}(I-rA)$ for all $r>0$ [25].

A basic problem for maximal monotone operator B is finding

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

(5)

A known method for solving problem (5) is the proximal point algorithm: $x_{1} = x\in H$, and

$$ x_{n+1} = J^{B}_{r_{n}}x_{n} \quad (n \geq 1), $$

where $J^{B}_{r_{n}} = (I + r_{n}B)^{-1}$ and $\{r_{n}\} \subset (0,\infty )$. Then Rockafellar [24, 26] proved that the sequence $\{x_{n}\}$ converges weakly to an element of $B^{-1}(0)$ (see also [27]). In the literature, there are a large number references associated with the proximal point algorithm [27–29]. In 2011, Shehu [8] suggested the following iterative sequence. Let $\{x_{n}\}$ be the sequence generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ x_{n+1}=a_{n}x_{n}+(1-a_{n})T[\beta _{n}f(x_{n})+(1-\beta _{n})J^{B}_{r_{n}}(y_{n}-r_{n} Ay_{n})] \quad (\forall n\geq 1). \end{cases}$$

Under appropriate conditions, the author proved that the sequence $\{x_{n}\}$ converges strongly to a point $P_{F(T)\cap (A +B)^{-1}(0)\cap \operatorname{GEP}(F,A)}u$ [8]. Our goal in this paper is to present an iterative method that converges strongly to a common element of the fixed point set of two nonexpansive mappings and the zero set of the sums of maximal monotone operators in Hilbert spaces. Our results extend and improve some related old results.

2 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 $\Vert x-P_{C}(x) \Vert =\inf_{y\in C} \Vert x-y \Vert =:d(x,C)$. It is well known that $P_{C}$ is nonexpansive monotone mapping from H onto C, $\langle x-P_{C}x,z-P_{C}x\rangle \leq 0$, $\Vert x-z \Vert ^{2}\geq \Vert x-P_{C}x \Vert ^{2}+ \Vert z-P_{C}x \Vert ^{2}$ for all $x\in H$ and $z\in C$, and $\langle P_{C}x-P_{C}z,x-z\rangle \geq \Vert P_{C}x-P_{C}z \Vert ^{2}$ for all $z,x\in H$ (see [13]). Let A be a monotone mapping from C into H. In the context of the variational inequality problem, it is easy to see that from the relation $\langle x-P_{C}x,z-P_{C}x\rangle \leq 0$ we have

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

For solving the equilibrium problem for a a bifunction $F: C \times C\to \mathbb{R}$, 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 semicontinuous.

Put $F (x, y) = \langle Ax, y - x\rangle$ for $x, y \in C$. Then we see that the equilibrium problem (2) is reduced to the variational inequality (3). We need the following results.

Lemma 2.1

([2, 30])

LetCbe a nonempty closed convex subset ofH, and letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})-(A_{4})$. For$r > 0$and$x \in H$, consider the map$T_{r} : H \to C$defined by$T_{r}(x)=\{z\in C: F(z,y)+\frac{1}{r}\langle y-z,z-x\rangle \geq 0\textit{ for all }y\in C\}$. Then$T_{r}(x) \neq \emptyset$for all$x \in H$, $T_{r}$is single-valued, $\operatorname{EP}(F)$is closed and convex, $F(T_{r}) = \operatorname{EP}(F)$, and$T_{r}$is firmly nonexpansive, that is, $\Vert T_{r}(x)-T_{r}(y) \Vert ^{2}\leq \langle T_{r}(x)-T_{r}(y),x-y\rangle$for all$x, y \in H$.

Lemma 2.2

([31])

LetCbe a nonempty closed convex subset ofH, and letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})-(A_{4})$. Define the multivalued mapping$A_{F}$fromHinto itself 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. Then$A_{F}$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 2.3

([32])

LetHbe a real Hilbert space, letCbe a closed convex subset ofH, and let$T : C\to C$be a nonexpansive mapping. Then$(I - T)$is demiclosed at zero, that is, if$\{x_{n}\}$is a sequence inCsuch that$x_{n}\rightharpoonup x$and$Tx_{n} -x_{n} \rightarrow 0$, then$x = T(x)$.

Lemma 2.4

([33])

Let$\{x_{n}\}$be a sequence of nonnegative real numbers satisfying

$$ x_{n+1}\leq (1-\lambda _{n})x_{n}+\gamma _{n}, $$

where$\{\lambda _{n}\}$is a sequence in$(0,1)$, and$\gamma _{n}$is a sequence with$\sum_{n=0}^{\infty }\lambda _{n}=\infty$and$\limsup_{n\to \infty }\gamma _{n}\leq 0$or$\sum_{n=0}^{\infty } \vert \gamma _{n}\lambda _{n} \vert <\infty$. Then$\lim_{n\to \infty }x_{n}=0$.

Lemma 2.5

([34])

LetHbe a real Hilbert space, let$x_{j}\in H$, and let$a_{j}\in [0, 1]$, $j = 1, 2, 3$, be such that$a_{1} + a_{2} + a_{3} = 1$. Then we have

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

Lemma 2.6

([31])

LetBbe a maximal monotone operator on H. Then we have

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

3 Main results

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

Theorem 3.1

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap (M+B)^{-1}(0) \cap \operatorname{GEP}(F,A)\neq \emptyset$and$f: C \to C$is a contraction map with the constant$\rho \in (0,1)$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and$\{x_{n}\}$, $\{y_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}M y_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad ( \forall n\geq 1). \end{cases} $$

(6)

Suppose the following conditions hold:

$(d_{1})$:

$0< c\leq \lambda _{n}\leq d<2\beta$, $\lim_{n\to \infty } \vert \lambda _{n}-\lambda _{n-1} \vert =0$,

$(d_{2})$:

$0< a\leq r_{n}\leq b<2\alpha$, $\lim_{n\to \infty } \vert r_{n}-r_{n-1} \vert =0$,

$(d_{3})$:

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

$(d_{4})$:

$\sum_{n=1}^{\infty } \vert \mu _{n}-\mu _{n-1} \vert <\infty$, $\sum_{n=1}^{\infty } \vert a_{n}-a_{n-1} \vert <\infty$.

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

Proof

First, we show that $I - \lambda _{n}M$ is nonexpansive. Let $x,y\in C$ and $0<\lambda _{n}<2\beta$. Then

$$\begin{aligned} \bigl\Vert (I - \lambda _{n}M)x-(I - \lambda _{n}M)y \bigr\Vert ^{2} =& \bigl\Vert (x-y)-\lambda _{n}(Mx-My) \bigr\Vert ^{2} \\ \leq & \Vert x-y \Vert ^{2}-2\lambda _{n}\langle x-y,Mx-My\rangle +\lambda _{n}^{2} \Vert Mx-My \Vert ^{2} \\ \leq & \Vert x-y \Vert ^{2}-\lambda _{n}\beta \Vert Mx-My \Vert ^{2}+\lambda _{n}^{2} \Vert Mx-My \Vert ^{2} \\ =& \Vert x-y \Vert ^{2}+\lambda _{n}(\lambda _{n}-2\beta ) \Vert Mx-My \Vert ^{2} \\ \leq & \Vert x-y \Vert ^{2}. \end{aligned}$$

(7)

Thus $I - \lambda _{n}M$ is nonexpansive. Note that $y_{n}$ can be rewritten as $y_{n} = T_{r_{n}} (x_{n}-r_{n}Ax_{n})$ for $n \geq 1$. Let $q\in \varOmega$. From $(d_{2})$ and Lemma 2.1 we have

$$\begin{aligned} \Vert y_{n} - q \Vert ^{2} =& \bigl\Vert T_{r_{n}}(x_{n}-r_{n}Ax_{n})-q \bigr\Vert ^{2} \\ =& \bigl\Vert T_{r_{n}}(x_{n}-r_{n}Ax_{n})-T_{r_{n}}(q-r_{n}Aq)) \bigr\Vert ^{2} \\ \leq & \bigl\Vert (x_{n}-r_{n}Ax_{n})-(q-r_{n}Aq) \bigr\Vert ^{2} \\ =& \Vert x_{n}-q \Vert ^{2}+r_{n}(r_{n}-2 \alpha ) \Vert Ax_{n}-Aq \Vert ^{2}\leq \Vert x_{n}-q \Vert ^{2}. \end{aligned}$$

(8)

By (6) and (7), since $J_{\lambda _{n}}^{B}$ is nonexpansive, we have

$$\begin{aligned} \Vert z_{n} - q \Vert ^{2} =& \bigl\Vert \mu _{n}x_{n}+(1-\mu _{n})J_{\lambda _{n}}^{B}(y_{n}- \lambda _{n}M y_{n})-q \bigr\Vert ^{2} \\ =& \bigl\Vert \mu _{n}(x_{n}-q)+(1-\mu _{n}) \bigl(J_{\lambda _{n}}^{B}(y_{n}- \lambda _{n}M y_{n})-q\bigr) \bigr\Vert ^{2} \\ \leq &\mu _{n} \Vert x_{n}-q \Vert ^{2}+(1- \mu _{n}) \bigl\Vert J_{\lambda _{n}}^{B}(y_{n}- \lambda _{n}M y_{n})-q \bigr\Vert ^{2} \\ \leq & \mu _{n} \Vert x_{n}-q \Vert ^{2}+(1-\mu _{n}) \bigl\Vert J_{\lambda _{n}}^{B}(y_{n}- \lambda _{n}M y_{n})-J_{\lambda _{n}}^{B}(q- \lambda _{n}Mq) \bigr\Vert ^{2} \\ \leq & \mu _{n} \Vert x_{n}-q \Vert ^{2}+(1-\mu _{n}) \bigl\Vert (y_{n}-\lambda _{n}M y_{n})-(q- \lambda _{n}Mq) \bigr\Vert ^{2} \\ \leq &\mu _{n} \Vert x_{n}-q \Vert ^{2}+(1- \mu _{n}) \bigl( \Vert y_{n} - q \Vert ^{2}+ \lambda _{n}(\lambda _{n}-2\beta ) \Vert My_{n}-Mq \Vert ^{2}\bigr) \\ \leq & \Vert x_{n}-q \Vert ^{2}+(1-\mu _{n})\lambda _{n}(\lambda _{n}-2\beta ) \Vert My_{n}-Mq \Vert ^{2}\leq \Vert x_{n}-q \Vert ^{2}. \end{aligned}$$

(9)

Hence

$$\begin{aligned} \Vert x_{n+1} - q \Vert =& \bigl\Vert P_{C} \bigl[b_{n}f(x_{n})+(1-b_{n}) \bigl(a_{n}Sz_{n}+(1-a_{n}) Ty_{n} \bigr)\bigr]-P_{C}(q) \bigr\Vert \\ \leq & \bigl\Vert \bigl[b_{n}f(x_{n})+(1-b_{n}) \bigl(a_{n}Sz_{n}+(1-a_{n}) Ty_{n} \bigr)\bigr]-q \bigr\Vert \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert +(1-b_{n})\bigl[a_{n} \Vert Sz_{n}-q \Vert +(1-a_{n}) \Vert Ty_{n}-q \Vert \bigr] \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert +(1-b_{n})\bigl[a_{n} \Vert x_{n}-q \Vert +(1-a_{n}) \Vert y_{n}-q \Vert \bigr] \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert +(1-b_{n}) \Vert x_{n}-q \Vert \\ \leq & b_{n}\bigl( \bigl\Vert f(x_{n})-f(q) \bigr\Vert + \bigl\Vert f(q)-q \bigr\Vert \bigr)+(1-b_{n}) \Vert x_{n}-q \Vert \\ \leq & b_{n}\bigl(\rho \Vert x_{n}-q \Vert + \bigl\Vert f(q)-q \bigr\Vert \bigr)+(1-b_{n}) \Vert x_{n}-q \Vert \\ \leq &\bigl(1-b_{n}(1-\rho )\bigr) \Vert x_{n}-q \Vert +b_{n} \bigl\Vert f(q)-q \bigr\Vert \\ \leq &\max \biggl\{ \Vert x_{n}-q \Vert ,\frac{ \Vert f(q)-q \Vert }{(1-\rho )} \biggr\} \\ \vdots &\\ \leq &\max \biggl\{ \Vert x_{1}-q \Vert ,\frac{ \Vert f(q)-q \Vert }{(1-\rho )} \biggr\} . \end{aligned}$$

Thus $\{x_{n}\}$ is bounded, and the sequences $\{y_{n}\}$ and $\{z_{n}\}$ are bounded as well. From $y_{n}=T_{r_{n}}(x_{n}- r_{n}Ax_{n})$ and $y_{n-1}=T_{r_{n-1}}(x_{n-1}- r_{n-1}Ax_{n-1})$ we obtain

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

(10)

and

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

(11)

By substituting $y=y_{n-1}$ into (10) and $y=y_{n}$ into (11), we find

$$ F (y_{n}, y_{n-1}) + \langle Ax_{n}, y_{n-1} - y_{n}\rangle + \frac{1}{r_{n}}\langle y_{n-1} - y_{n},y_{n} - x_{n}\rangle \geq 0 $$

and

$$ F (y_{n-1}, y_{n}) + \langle Ax_{n-1}, y_{n} - y_{n-1}\rangle + \frac{1}{r_{n-1}}\langle y_{n} - y_{n-1},y_{n-1} - x_{n-1}\rangle \geq 0. $$

Now from $(A_{2})$ we get $\langle Ax_{n-1}-Ax_{n}, y_{n} - y_{n-1}\rangle +\langle y_{n-1} - y_{n}, \frac{y_{n} - x_{n}}{r_{n}}-\frac{y_{n-1} - x_{n-1}}{r_{n-1}}\rangle \geq 0$, and so

$$\begin{aligned} 0 \leq & \biggl\langle y_{n} - y_{n-1},{r_{n}}(Ax_{n-1}-Ax_{n})+ \frac{r_{n}}{r_{n-1}}(y_{n-1} - x_{n-1})-(y_{n} - x_{n})\biggr\rangle \\ =&\biggl\langle y_{n-1} - y_{n},y_{n} - y_{n-1}+\biggl(1-\frac{r_{n}}{r_{n-1}}\biggr){y_{n-1}}+(x_{n-1}- r_{n-1}Ax_{n-1})\biggr\rangle \\ &{} -\biggl\langle y_{n+1} - y_{n},(x_{n}- r_{n}Ax_{n})+x_{n-1}- \frac{r_{n}}{r_{n-1}} x_{n-1}\biggr\rangle \\ =&\biggl\langle y_{n-1} - y_{n},y_{n} - y_{n-1}+\biggl(1-\frac{r_{n}}{r_{n-1}}\biggr) (y_{n-1}-x_{n-1}) \\ &{} +(x_{n-1}- r_{n-1}Ax_{n-1})-(x_{n}- r_{n}Ax_{n})\biggr\rangle . \end{aligned}$$

This implies that $\Vert y_{n} - y_{n-1} \Vert ^{2}\leq \Vert y_{n} - y_{n-1} \Vert [ \vert 1- \frac{r_{n}}{r_{n-1}} \vert \Vert y_{n-1}-x_{n-1} \Vert + \Vert x_{n}-x_{n-1} \Vert ]$, and so

$$ \Vert y_{n} - y_{n-1} \Vert \leq \frac{ \vert r_{n}-r_{n-1} \vert }{r_{n-1}} \Vert y_{n-1}-x_{n-1} \Vert + \Vert x_{n}-x_{n-1} \Vert . $$

(12)

Set $w_{n}=J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}My_{n})$ and $u_{n}=y_{n}-\lambda _{n}My_{n}$ for $n\geq 1$. By using Lemma 2.6 we obtain

$$\begin{aligned} \Vert w_{n} - w_{n-1} \Vert =& \bigl\Vert J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}M y_{n})-J_{ \lambda _{n-1}}^{B}(y_{n-1}-\lambda _{n-1}M y_{n-1}) \bigr\Vert \\ =& \bigl\Vert J_{\lambda _{n}}^{B}u_{n}-J_{\lambda _{n-1}}^{B}u_{n-1}+J_{ \lambda _{n}}^{B}u_{n-1}-J_{\lambda _{n}}^{B}u_{n-1} \bigr\Vert \\ \leq & \bigl\Vert (y_{n}-\lambda _{n}My_{n})-(y_{n-1}- \lambda _{n-1}My_{n-1}) \bigr\Vert \\ &{}+ \bigl\Vert J_{\lambda _{n}}^{B}u_{n-1}-J_{\lambda _{n-1}}^{B}u_{n-1} \bigr\Vert \\ \leq & \bigl\Vert (y_{n}-\lambda _{n}My_{n})-(y_{n-1}- \lambda _{n}My_{n-1})+( \lambda _{n-1}-\lambda _{n})My_{n-1} \bigr\Vert \\ &{}+\frac{ \vert \lambda _{n-1}-\lambda _{n} \vert }{\lambda _{n-1}} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert \\ \leq & \Vert y_{n}-y_{n-1} \Vert + \vert \lambda _{n-1}-\lambda _{n} \vert \Vert My_{n-1} \Vert + \frac{ \vert \lambda _{n-1}-\lambda _{n} \vert }{\lambda _{n-1}} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert \\ \leq & \Vert x_{n}-x_{n-1} \Vert +\frac{ \vert r_{n}-r_{n+1} \vert }{r_{n+1}} \Vert y_{n+1}-x_{n+1} \Vert + \vert \lambda _{n-1}-\lambda _{n} \vert \Vert My_{n-1} \Vert \\ &{}+\frac{ \vert \lambda _{n-1}-\lambda _{n} \vert }{\lambda _{n-1}} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert , \end{aligned}$$

which gives

$$\begin{aligned} \Vert z_{n} - z_{n-1} \Vert =& \bigl\Vert \bigl[\mu _{n}x_{n}+(1-\mu _{n})w_{n} \bigr] -\bigl[\mu _{n-1}x_{n-1}+(1- \mu _{n-1})w_{n-1} \bigr] \bigr\Vert \\ =& \bigl\Vert \mu _{n}(x_{n}-x_{n-1})+(1-\mu _{n}) (w_{n}-w_{n-1})+(\mu _{n}- \mu _{n-1}) (x_{n-1}-w_{n-1}) \bigr\Vert \\ \leq & \mu _{n} \Vert x_{n}-x_{n-1} \Vert + \vert \mu _{n}-\mu _{n-1} \vert \Vert x_{n-1}-w_{n-1} \Vert +(1-\mu _{n}) \Vert w_{n}-w_{n-1} \Vert \\ \leq & \Vert x_{n}-x_{n-1} \Vert + \vert \mu _{n}-\mu _{n-1} \vert \Vert x_{n-1}-w_{n-1} \Vert \\ &{} +(1-\mu _{n}) \biggl( \vert \lambda _{n-1}-\lambda _{n} \vert \Vert My_{n-1} \Vert + \frac{ \vert r_{n}-r_{n+1} \vert }{r_{n+1}} \Vert y_{n+1}-x_{n+1} \Vert \\ &{} +\frac{ \vert \lambda _{n-1}-\lambda _{n} \vert }{\lambda _{n-1}} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert \biggr). \end{aligned}$$

Set $t_{n}=a_{n}Sz_{n}+(1-a_{n}) Ty_{n}$ for $n\geq 1$. By using (12) and last inequality we have

$$\begin{aligned} \Vert t_{n}-t_{n-1} \Vert =& \bigl\Vert a_{n}(Sz_{n}-Sz_{n-1})+(1-a_{n}) (Ty_{n}-Ty_{n-1}) \\ &{}+(a_{n}-a_{n-1}) (Sz_{n-1}-Ty_{n-1}) \bigr\Vert \\ \leq &a_{n} \Vert Sz_{n}-Sz_{n-1} \Vert +(1-a_{n}) \Vert Ty_{n}-Ty_{n-1} \Vert \\ &{}+ \vert a_{n}-a_{n-1} \vert \Vert Sz_{n-1}-Ty_{n-1} \Vert \\ \leq &a_{n} \Vert z_{n}-z_{n-1} \Vert +(1-a_{n}) \Vert y_{n}-y_{n-1} \Vert + \vert a_{n}-a_{n-1} \vert \Vert Sz_{n-1}-Ty_{n-1} \Vert \\ \leq & \Vert x_{n}-x_{n-1} \Vert +\frac{ \vert r_{n}-r_{n+1} \vert }{r_{n+1}} \Vert y_{n+1}-x_{n+1} \Vert + \vert a_{n}-a_{n-1} \vert \Vert Sz_{n-1}-Ty_{n-1} \Vert \\ &{}+(1-\mu _{n})a_{n}\biggl( \vert \lambda _{n-1}-\lambda _{n} \vert \Vert My_{n-1} \Vert + \frac{ \vert \lambda _{n-1}-\lambda _{n} \vert }{\lambda _{n-1}} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert \biggr) \\ &{}+a_{n} \vert \mu _{n}-\mu _{n-1} \vert \Vert x_{n-1}-w_{n-1} \Vert , \end{aligned}$$

which implies that

$$\begin{aligned} \Vert x_{n+1} - x_{n} \Vert =& \bigl\Vert P_{C}\bigl[b_{n}f(x_{n})+(1-b_{n})t_{n} \bigr]-P_{C}\bigl[b_{n-1}f(x_{n-1})+(1-b_{n-1})t_{n-1} \bigr] \bigr\Vert \\ \leq & \bigl\Vert \bigl[b_{n}f(x_{n})+(1-b_{n})t_{n} \bigr]-\bigl[b_{n-1}f(x_{n-1})+(1-b_{n-1})t_{n-1} \bigr] \bigr\Vert \\ \leq & \bigl\Vert b_{n}\bigl(f(x_{n})-f(x_{n-1}) \bigr)+(1-b_{n}) (t_{n}-t_{n-1}) \bigr\Vert \\ &{} + \bigl\Vert (b_{n}-b_{n-1}) \bigl(f(x_{n-1})-t_{n-1} \bigr) \bigr\Vert \\ \leq &b_{n} \bigl\Vert f(x_{n})-f(x_{n-1}) \bigr\Vert + \vert b_{n}-b_{n-1} \vert \bigl\Vert f(x_{n-1})-t_{n-1} \bigr\Vert \\ &{} +(1-b_{n}) \Vert t_{n}-t_{n-1} \Vert \\ \leq &b_{n}\rho \Vert x_{n}-x_{n-1} \Vert + \vert b_{n}-b_{n-1} \vert \bigl\Vert f(x_{n-1})-t_{n-1} \bigr\Vert +(1-b_{n}) \Vert t_{n}-t_{n-1} \Vert \\ \leq &\bigl(1-b_{n}(1-\rho )\bigr) \Vert x_{n}-x_{n-1} \Vert + \vert b_{n}-b_{n-1} \vert \bigl\Vert f(x_{n-1})-t_{n-1} \bigr\Vert \\ &{} +\frac{ \vert r_{n}-r_{n+1} \vert }{a} \Vert y_{n+1}-x_{n+1} \Vert + \vert \mu _{n}-\mu _{n-1} \vert \Vert x_{n-1}-w_{n-1} \Vert \\ &{} + \vert \lambda _{n-1}-\lambda _{n} \vert \Vert My_{n-1} \Vert + \frac{ \vert \lambda _{n-1}-\lambda _{n} \vert }{c} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert \\ &{} + \vert a_{n}-a_{n-1} \vert \Vert Sz_{n-1}-Ty_{n-1} \Vert \\ \leq &\bigl(1-b_{n}(1-\rho )\bigr) \Vert x_{n}-x_{n-1} \Vert +L \biggl[ \vert b_{n}-b_{n-1} \vert + \vert \mu _{n}-\mu _{n-1} \vert \\ &{} +\frac{ \vert r_{n}-r_{n+1} \vert }{a}+\biggl(1+\frac{1}{c}\biggr) \vert \lambda _{n-1}-\lambda _{n} \vert + \vert a_{n}-a_{n-1} \vert \biggr], \end{aligned}$$

where L is a constant such that

$$\begin{aligned} L\geq {}&\max \Bigl\{ \sup_{n\geq 1} \Vert Sz_{n-1}-Ty_{n-1} \Vert , \sup_{n\geq 1} \Vert M y_{n} \Vert , \sup _{n\geq 1} \bigl\Vert J_{\lambda _{n-1}}^{B}u_{n-1}-u_{n-1} \bigr\Vert , \\ &{}\sup_{n\geq 1} \bigl\Vert f(x_{n-1})-t_{n-1} \bigr\Vert , \sup_{n\geq 1} \Vert y_{n+1}-x_{n+1} \Vert , \sup_{n\geq 1} \Vert x_{n-1}-w_{n-1} \Vert \Bigr\} . \end{aligned}$$

Now by using $(d_{1})$, $(d_{2})$, $(d_{3})$, $(d_{4})$, and Lemma 2.4 we get

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

(13)

From (6) we have

$$\begin{aligned} \Vert x_{n} - t_{n} \Vert \leq & \Vert x_{n} - x_{n+1} \Vert + \Vert x_{n+1} - t_{n} \Vert \\ =& \Vert x_{n} - x_{n+1} \Vert + \bigl\Vert p_{C}\bigl[b_{n}f(x_{n})+(1-b_{n})t_{n} \bigr]-P_{C}(t_{n}) \bigr\Vert \\ \leq & \Vert x_{n} - x_{n+1} \Vert + \bigl\Vert \bigl[b_{n}f(x_{n})+(1-b_{n})t_{n} \bigr]-t_{n} \bigr\Vert \\ \leq & \Vert x_{n} - x_{n+1} \Vert +b_{n} \bigl\Vert f(x_{n})-t_{n} \bigr\Vert . \end{aligned}$$

Consequently, from the condition $(d_{3})$ we obtain

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

(14)

Let $q = P_{\varOmega }f(q)$. In addition, from (6), (8), Lemma 2.5, and (9) we get

$$\begin{aligned} \Vert x_{n+1} - q \Vert ^{2} =& \bigl\Vert P_{C}\bigl[b_{n}f(x_{n})+(1-b_{n}) \bigl(a_{n}Sz_{n}+(1-a_{n}) Ty_{n} \bigr)\bigr]-P_{C}(q) \bigr\Vert ^{2} \\ \leq & \bigl\Vert \bigl[b_{n}f(x_{n})+(1-b_{n}) \bigl(a_{n}Sz_{n}+(1-a_{n}) Ty_{n} \bigr)\bigr]-q \bigr\Vert ^{2} \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+(1-b_{n})\bigl[a_{n} \Vert Sz_{n}-q \Vert ^{2}+(1-a_{n}) \Vert Ty_{n}-q \Vert ^{2} \\ &{}-a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert \bigr] \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+(1-b_{n})\bigl[a_{n} \Vert z_{n}-q \Vert ^{2}+(1-a_{n}) \Vert y_{n}-q \Vert ^{2} \\ &{}-a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert \bigr] \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+(1-b_{n}) \bigl(a_{n} \Vert x_{n}-q \Vert ^{2}-a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert \bigr) \\ &{}+(1-b_{n})a_{n}(1-\mu _{n})\lambda _{n}(\lambda _{n}-2\beta ) \Vert My_{n}-Mq \Vert ^{2} \\ &{}+(1-b_{n}) (1-a_{n}) \bigl( \Vert x_{n}-q \Vert ^{2}+r_{n}(r_{n}-2\alpha ) \Vert Ax_{n}-Aq \Vert ^{2}\bigr) \\ \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert +(1-b_{n}) \bigl( \Vert x_{n}-q \Vert -a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert \bigr) \\ &{}+(1-b_{n})a_{n}(1-\mu _{n})\lambda _{n}(\lambda _{n}-2\beta ) \Vert My_{n}-Mq \Vert ^{2} \\ &{}+(1-b_{n}) (1-a_{n})r_{n}(r_{n}-2 \alpha ) \Vert Ax_{n}-Aq \Vert ^{2}, \end{aligned}$$

which yields

$$\begin{aligned}& (1-b_{n}) (1-a_{n})r_{n}(2\alpha -r_{n})\cdot \Vert Ax_{n}-Aq \Vert ^{2} \\& \quad \leq b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+ \Vert x_{n}-q \Vert ^{2}- \Vert x_{n+1} - q \Vert ^{2} \\& \quad\quad{} +(1-b_{n})a_{n}(1-\mu _{n})\lambda _{n}(\lambda _{n}-2\beta ) \Vert My_{n}-Mq \Vert ^{2} \\& \quad\quad{} -(1-b_{n})a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert , \\& (1-b_{n}) a_{n} (1-\mu _{n})\lambda _{n}(2\beta -\lambda _{n}) \cdot \Vert My_{n}-Mq \Vert ^{2} \\& \quad \leq b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+ \Vert x_{n}-q \Vert ^{2}- \Vert x_{n+1} - q \Vert ^{2} \\& \quad\quad {} +(1-b_{n}) (1-a_{n})r_{n}(r_{n}-2 \alpha ) \Vert Ax_{n}-Aq \Vert ^{2} \\& \quad\quad{} -(1-b_{n})a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert , \end{aligned}$$

and

$$\begin{aligned} (1-b_{n})a_{n}(1-a_{n}) \Vert Sz_{n}-Ty_{n} \Vert \leq & b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+ \Vert x_{n}-q \Vert ^{2}- \Vert x_{n+1} - q \Vert ^{2} \\ &{}+(1-b_{n}) (1-a_{n})r_{n}(r_{n}-2 \alpha ) \Vert Ax_{n}-Aq \Vert ^{2} \\ &{}+(1-b_{n}) a_{n}(1-\mu _{n})\lambda _{n}(\lambda _{n}-2\beta ) \Vert My_{n}-Mq \Vert ^{2}. \end{aligned}$$

By using $(d_{1})$, $(d_{2})$, $(d_{3})$, and (13) we get

$$ \lim_{n\to \infty } \Vert My_{n}-Mq \Vert =\lim_{n\to \infty } \Vert Ax_{n}-Aq \Vert = \lim _{n\to \infty } \Vert Sz_{n}-Ty_{n} \Vert =0. $$

(15)

Since $\Vert t_{n}-Ty_{n} \Vert \leq a_{n} \Vert Sz_{n}-Ty_{n} \Vert$ and $\Vert t_{n}-Sz_{n} \Vert \leq (1-a_{n}) \Vert Ty_{n}-Sz_{n} \Vert$, we obtain

$$ \lim_{n\to \infty } \Vert t_{n}-Ty_{n} \Vert =\lim_{n\to \infty } \Vert t_{n}-Sz_{n} \Vert = 0. $$

(16)

Note that

$$\begin{aligned} \Vert w_{n}-q \Vert ^{2} =& \bigl\Vert J_{\lambda _{n}}^{B}(y_{n} - \lambda _{n}My_{n})-J_{ \lambda _{n}}^{B}(q- \lambda _{n}Mq) \bigr\Vert ^{2} \\ \leq &\bigl\langle (y_{n} - \lambda _{n}My_{n})-(q- \lambda _{n}Mq),w_{n}-q \bigr\rangle \\ =&\frac{1}{2} \bigl\Vert (y_{n} - \lambda _{n}My_{n})-(q-\lambda _{n}Mq) \bigr\Vert ^{2}+ \frac{1}{2} \Vert w_{n}-q \Vert ^{2} \\ &{}-\frac{1}{2} \bigl\Vert (y_{n} - \lambda _{n}My_{n})-(q-\lambda _{n}Mq)-(w_{n}-q) \bigr\Vert ^{2} \\ \leq &\frac{1}{2} \bigl[ \Vert y_{n}-q \Vert ^{2}+ \Vert w_{n}-q \Vert ^{2}- \bigl\Vert (y_{n}-w_{n})- \lambda _{n}(My_{n}-Mq) \bigr\Vert ^{2} \bigr] \\ =& \frac{1}{2} \bigl[ \Vert y_{n}-q \Vert ^{2}+ \Vert w_{n}-q \Vert ^{2}- \Vert y_{n}-w_{n} \Vert ^{2} +2\lambda _{n}\langle y_{n}-w_{n},My_{n}-Mq \rangle \\ &{}-\lambda _{n}^{2} \Vert My_{n}-Mq \Vert ^{2} \bigr], \end{aligned}$$

and so

$$ \Vert w_{n}-q \Vert ^{2}\leq \Vert y_{n}-q \Vert ^{2}- \Vert y_{n}-w_{n} \Vert ^{2}+2\lambda _{n} \Vert y_{n}-w_{n} \Vert \Vert My_{n}-Mq \Vert . $$

(17)

By using Lemma 2.1 and (6) we have

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

It follows that

$$ \Vert y_{n}-q \Vert ^{2}\leq \Vert x_{n}-q \Vert ^{2}- \Vert x_{n}-y_{n} \Vert ^{2}+2r_{n} \langle x_{n}-y_{n},Ax_{n}-Aq \rangle . $$

(18)

From (8), (17), and (18) we have

$$\begin{aligned} \Vert t_{n}-q \Vert ^{2} =& \bigl\Vert a_{n}(Sz_{n}-q)+(1-a_{n}) (Ty_{n}-q) \bigr\Vert ^{2} \\ \leq &a_{n} \Vert Sz_{n}-q \Vert ^{2}+(1-a_{n}) \Vert Ty_{n}-q \Vert ^{2} \\ \leq &a_{n} \Vert z_{n}-q \Vert ^{2}+(1-a_{n}) \Vert y_{n}-q \Vert ^{2} \\ \leq &a_{n}\bigl(\mu _{n} \Vert x_{n}-q \Vert ^{2}+(1-\mu _{n}) \Vert w_{n}-q \Vert ^{2}\bigr) +(1-a_{n}) \Vert y_{n}-q \Vert ^{2} \\ \leq &a_{n}\bigl(\mu _{n} \Vert x_{n}-q \Vert ^{2}+(1-\mu _{n}) \bigl( \Vert y_{n}-q \Vert ^{2}- \Vert y_{n}-w_{n} \Vert ^{2}\bigr)\bigr) \\ &{} +(1-\mu _{n})a_{n}2\lambda _{n} \Vert y_{n}-w_{n} \Vert \Vert My_{n}-Mq \Vert \\ &{} +(1-a_{n}) \bigl( \Vert x_{n}-q \Vert ^{2}- \Vert x_{n}-y_{n} \Vert ^{2}+2r_{n}\langle x_{n}-y_{n},Ax_{n}-Aq \rangle \bigr) \\ \leq & \Vert x_{n}-q \Vert ^{2}-a_{n}(1-\mu _{n}) \Vert y_{n}-w_{n} \Vert ^{2} \\ &{} +(1-\mu _{n})a_{n}2\lambda _{n} \Vert y_{n}-w_{n} \Vert \Vert My_{n}-Mq \Vert \\ &{} +(1-a_{n}) \bigl(2r_{n} \Vert x_{n}-y_{n} \Vert \Vert Ax_{n}-Aq \Vert - \Vert x_{n}-y_{n} \Vert ^{2}\bigr). \end{aligned}$$

By using (6) and last inequality we see that

$$\begin{aligned} \Vert x_{n+1} - q \Vert ^{2} =& \bigl\Vert P_{C}\bigl[b_{n}f(x_{n})+(1-b_{n})t_{n} \bigr]-P_{C}(q) \bigr\Vert ^{2} \\ \leq &b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+(1-b_{n}) \Vert t_{n}-q \Vert ^{2} \\ \leq &b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+(1-b_{n}) \bigl( \Vert x_{n}-q \Vert ^{2}-a_{n}(1-\mu _{n}) \Vert y_{n}-w_{n} \Vert ^{2} \\ &{} +(1-\mu _{n})a_{n}2\lambda _{n} \Vert y_{n}-w_{n} \Vert \Vert My_{n}-Mq \Vert \\ &{} +(1-a_{n}) \bigl(2r_{n} \Vert x_{n}-y_{n} \Vert \Vert Ax_{n}-Aq \Vert - \Vert x_{n}-y_{n} \Vert ^{2}\bigr)\bigr). \end{aligned}$$

Thus

$$\begin{aligned} (1-b_{n}) (1-a_{n}) \Vert x_{n}-y_{n} \Vert ^{2} \leq &b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+ \Vert x_{n}-q \Vert ^{2}- \Vert x_{n+1}-q \Vert ^{2} \\ &{} +(1-\mu _{n})a_{n}2\lambda _{n} \Vert y_{n}-w_{n} \Vert \Vert My_{n}-Mq \Vert - \Vert y_{n}-w_{n} \Vert ^{2} \\ &{} +(1-a_{n}) \bigl(2r_{n} \Vert x_{n}-y_{n} \Vert \Vert Ax_{n}-Aq \Vert - \Vert x_{n}-y_{n} \Vert ^{2}\bigr) \end{aligned}$$

and

$$\begin{aligned} (1-b_{n}) (1-\mu _{n})a_{n} \Vert y_{n}-w_{n} \Vert ^{2} \leq &b_{n} \bigl\Vert f(x_{n})-q \bigr\Vert ^{2}+ \Vert x_{n}-q \Vert ^{2}- \Vert x_{n+1}-q \Vert ^{2} \\ &{} +(1-\mu _{n})a_{n}2\lambda _{n} \Vert y_{n}-w_{n} \Vert \Vert My_{n}-Mq \Vert \\ &{} +(1-a_{n})2r_{n} \Vert x_{n}-y_{n} \Vert \Vert Ax_{n}-Aq \Vert . \end{aligned}$$

From $(d_{3})$, (15), and $\lim_{n\to \infty } \Vert x_{n}-x_{n+1} \Vert =0$ we find $\lim_{n\to \infty } \Vert x_{n}-y_{n} \Vert =0$ and also $\lim_{n\to \infty } \Vert y_{n}-w_{n} \Vert =0$. Since $\Vert x_{n}-w_{n} \Vert \leq \Vert x_{n}-y_{n} \Vert + \Vert y_{n}-w_{n} \Vert$, we get $\Vert x_{n}-w_{n} \Vert \to 0$. Consequently, by using (6) we get $\lim_{n\to \infty } \Vert z_{n} - x_{n} \Vert = \lim_{n\to \infty }(1-\mu _{n}) \Vert w_{n} - x_{n} \Vert =0$. Moreover, from (14) and (16) we get $\Vert z_{n} - Sz_{n} \Vert \leq \Vert z_{n} - x_{n} \Vert + \Vert x_{n} - t_{n} \Vert + \Vert t_{n} - Sz_{n} \Vert \to 0$ and $\Vert y_{n} - Ty_{n} \Vert \leq \Vert y_{n} - x_{n} \Vert + \Vert x_{n} - t_{n} \Vert + \Vert t_{n} - Ty_{n} \Vert \to 0$. Hence

$$\begin{aligned} \Vert x_{n} - Sx_{n} \Vert \leq & \Vert x_{n} - z_{n} \Vert + \Vert z_{n} - Sz_{n} \Vert + \Vert Sz_{n} - Sx_{n} \Vert \\ \leq & \Vert x_{n} - z_{n} \Vert + \Vert z_{n} - Sz_{n} \Vert + \Vert z_{n} - x_{n} \Vert \to 0 \end{aligned}$$

and

$$\begin{aligned} \Vert x_{n} - Tx_{n} \Vert \leq & \Vert x_{n} - y_{n} \Vert + \Vert y_{n} - Ty_{n} \Vert + \Vert Ty_{n} - Tx_{n} \Vert \\ \leq & \Vert x_{n} - y_{n} \Vert + \Vert y_{n} - Ty_{n} \Vert + \Vert y_{n} - x_{n} \Vert \to 0, \end{aligned}$$

which implies

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

(19)

Now we show that $\limsup_{n\to \infty }\langle f(q)-q,x_{n}-q\rangle \leq 0$, where $q=P_{\varOmega }f(q)$. The existence of q is justified since $P_{\varOmega }$ is nonexpansive and f is a contraction. Hence $P_{\varOmega } \circ f$ is a contraction and so has a fixed point. We can choose a subsequence $\{x_{n_{i}}\}$ of $\{x_{n}\}$ such that

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

(20)

Since $\{x_{n_{i}}\}$ is bounded, there exists a subsequence $\{x_{n_{i_{j}}}\}$ of $\{x_{n_{i}}\}$ that converges weakly to v. Without loss of generality, we can assume that $x_{n_{i}}\rightharpoonup v$. Since $\Vert x_{n}-y_{n} \Vert \to 0$ and $\Vert x_{n}-z_{n} \Vert \to 0$, we find $y_{n_{i}}\rightharpoonup v$ and $y_{n_{i}}\rightharpoonup v$. Since $\{y_{n_{i}}\}$ and $\{z_{n_{i}}\}$ lie in C and C is closed and convex, we obtain $v\in C$. It is easy to check that $v\in F(T)$ and $v\in F(S)$. By using (19) and Lemma 2.3 we get $v\in F(T)\cap F(S)$. Now we show $v\in \operatorname{GEP}(F)$. Since $y_{n} = T_{r_{n}}(x_{n} -r_{n}Ax_{n})$, we obtain

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

From $(A_{2})$ we get $\langle Ax_{n},y-y_{n}\rangle +\frac{1}{r_{n}}\langle y-y_{n},y_{n}-x_{n} \rangle \geq F(y,y_{n})$ for all $y\in C$. Hence

$$ \langle Ax_{n_{i}},y-y_{n_{i}}\rangle +\biggl\langle y-y_{n_{i}}, \frac{y_{n_{i}}-x_{n_{i}}}{r_{n_{i}}}\biggr\rangle \geq F(y,y_{n_{i}}) $$

(21)

for all $y\in C$. For $0 < t\leq 1$ and $y \in C$, put $y_{t} = ty + (1 - t)v$. Since $y \in C$ and $v\in C$, we obtain $y_{t}\in C$. From (21) we conclude that

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

Since $\Vert y_{n_{i}}-x_{n_{i}} \Vert \to 0$, we have $\Vert Ay_{n_{i}}-Ax_{n_{i}} \Vert \to 0$. Further, from the inverse-strongly monotonicity of A we have $\langle y_{t}-y_{n_{i}},Ay_{t}-Ay_{n_{i}}\rangle \geq 0$. By using $(A_{4})$, $\frac{y_{n_{i}}-x_{n_{i}}}{r_{n_{i}}}\to 0$, and $y_{n_{i}}\rightharpoonup v$ we get $\langle y_{t}-v,Ay_{t}\rangle \geq F(y_{t},v)$. From $(A_{1})$–$(A_{4})$ we have

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

and so $0\leq F(y_{t},y)+(1-t)\langle y-v,Ay_{t}\rangle$. Thus $F(v,y)+(1-t)\langle y-v,Av\rangle \geq 0$ for all $y \in C$. This implies that $v\in \operatorname{GEP}(F, A)$. Finally, we show $v\in (M+B)^{-1}(0)$. Choose a subsequence $\{\lambda _{n_{i_{j}}}\}$ of $\{\lambda _{n_{i}}\}$ such that $\lambda _{n_{i_{j}}}\to \tilde{\lambda }\in [c,d]$. Without loss of generality, assume that $\lambda _{n_{i}}\to \tilde{\lambda }$. By using Lemma 2.6 we obtain

$$\begin{aligned} \bigl\Vert y_{n_{i}}-J_{\tilde{\lambda }}^{B}(I-\tilde{\lambda }M)y_{n_{i}} \bigr\Vert \leq & \Vert y_{n_{i}}-z_{n_{i}} \Vert + \bigl\Vert z_{n_{i}}-\bigl(\mu _{n_{i}}x_{n_{i}}+(1- \mu _{n_{i}})J_{\tilde{\lambda }}^{B}(I-\tilde{\lambda }M)y_{n_{i}}\bigr) \bigr\Vert \\ &{} + \bigl\Vert \bigl(\mu _{n_{i}}x_{n_{i}}+(1-\mu _{n_{i}})J_{\tilde{\lambda }}^{B}(I- \tilde{\lambda }M)y_{n_{i}}\bigr)-J_{\tilde{\lambda }}^{B}(I-\tilde{\lambda }M)y_{n_{i}} \bigr\Vert \\ \leq & \Vert y_{n_{i}}-z_{n_{i}} \Vert +(1-\mu _{n_{i}}) \bigl\Vert J_{{\lambda _{n_{i}}}}^{B}(I-{ \lambda _{n_{i}}}M)y_{n_{i}}-J_{\tilde{\lambda }}^{B}(I- \tilde{ \lambda }M)y_{n_{i}}) \bigr\Vert \\ &{} +\mu _{n_{i}} \bigl\Vert x_{n_{i}}-J_{\tilde{\lambda }}^{B}(I- \tilde{\lambda }M)y_{n_{i}} \bigr\Vert \\ \leq & \Vert y_{n_{i}}-z_{n_{i}} \Vert +(1-\mu _{n_{i}})\bigl[ \bigl\Vert J_{{\lambda _{n_{i}}}}^{B}(I-{ \lambda _{n_{i}}}M)y_{n_{i}}-J_{\lambda _{n_{i}}}^{B}(I- \tilde{ \lambda }M)y_{n_{i}} \bigr\Vert \\ &{} + \bigl\Vert J_{\lambda _{n_{i}}}^{B}(I-\tilde{\lambda }M)y_{n_{i}}-J_{ \tilde{\lambda }}^{B}(I-\tilde{\lambda }M)y_{n_{i}} \bigr\Vert \bigr] \\ &{} +\mu _{n_{i}} \bigl\Vert x_{n_{i}}-J_{\tilde{\lambda }}^{B}(I- \tilde{\lambda }M)y_{n_{i}} \bigr\Vert \\ \leq & \Vert y_{n_{i}}-z_{n_{i}} \Vert +(1-\mu _{n_{i}}) \vert \lambda _{n_{i}}- \tilde{\lambda } \vert \Vert My_{n_{i}} \Vert \\ &{} +(1-\mu _{n_{i}}) \biggl\vert \frac{\lambda _{n_{i}}-\tilde{\lambda }}{\tilde{\lambda }} \biggr\vert \bigl\Vert J_{ \tilde{\lambda }}^{B}(I-\tilde{\lambda }M)y_{n_{i}}-(I- \tilde{\lambda }M)y_{n_{i}} \bigr\Vert \\ &{} +\mu _{n_{i}}\bigl[ \Vert x_{n_{i}}-y_{n_{i}} \Vert + \bigl\Vert y_{n_{i}}-J_{ \tilde{\lambda }}^{B}(I-\tilde{ \lambda }M)y_{n_{i}} \bigr\Vert \bigr]. \end{aligned}$$

Thus

$$\begin{aligned} (1-\mu _{n_{i}}) \bigl\Vert y_{n_{i}}-J_{\tilde{\lambda }}^{B}(I- \tilde{\lambda }M)y_{n_{i}} \bigr\Vert \leq & \Vert y_{n_{i}}-z_{n_{i}} \Vert +(1-\mu _{n_{i}}) \vert \lambda _{n_{i}}- \tilde{\lambda } \vert \Vert My_{n_{i}} \Vert \\ &{} +(1-\mu _{n_{i}}) \biggl\vert \frac{\lambda _{n_{i}}-\tilde{\lambda }}{\tilde{\lambda }} \biggr\vert \bigl\Vert J_{ \tilde{\lambda }}^{B}(I-\tilde{\lambda }M)y_{n_{i}}-(I- \tilde{\lambda }M)y_{n_{i}} \bigr\Vert \\ &{} +\mu _{n_{i}} \Vert x_{n_{i}}-y_{n_{i}} \Vert . \end{aligned}$$

This implies that $\lim_{k\to \infty } \Vert y_{n_{i}}-J_{\tilde{\lambda }}^{B}(I- \tilde{\lambda }M)y_{n_{i}} \Vert =0$. Since $J_{\tilde{\lambda }}^{B}(I-\tilde{\lambda }M)$ is nonexpansive, it is demiclosed, and so $v\in F(J_{\tilde{\lambda }}^{B}(I-\tilde{\lambda }M))$, that is, $v\in (M+B)^{-1}(0)$. This implies $v\in \varOmega$. By using (20) we get $\limsup_{n\to \infty } \langle f(q)-q,x_{n}-q\rangle =\lim_{i\to \infty }\langle f(q)-q,x_{n_{i}}-q\rangle =\langle f(q)-q,v-q\rangle \leq 0$. Now we show that $x_{n}\to q$. From(8) and (9) we have

$$\begin{aligned} \Vert t_{n} - q \Vert =& \bigl\Vert \bigl(a_{n}Sz_{n}+(1-a_{n}) Ty_{n}\bigr)-q \bigr\Vert \\ \leq & a_{n} \Vert Sz_{n}-q \Vert +(1-a_{n}) \Vert Ty_{n}-q \Vert \\ \leq & a_{n} \Vert z_{n}-q \Vert +(1-a_{n}) \Vert y_{n}-q \Vert \leq \Vert x_{n}-q \Vert . \end{aligned}$$

Set $v_{n}=b_{n}f(x_{n})+(1-b_{n})t_{n}$ for all $n\geq 1$. By using (6) and the property of metric projection we obtain

$$\begin{aligned} \Vert x_{n+1} - q \Vert ^{2} =&\bigl\langle P_{C}( v_{n})-v_{n}, P_{C}( v_{n}) - q \bigr\rangle +\langle v_{n}-q, x_{n+1}-q\rangle \\ \leq &\bigl\langle \bigl( b_{n}f(x_{n})+(1-b_{n})t_{n} \bigr)-q, x_{n+1} - q \bigr\rangle \\ =&b_{n}\bigl\langle f(x_{n})-q, x_{n+1} - q\bigr\rangle +(1-b_{n})\langle t_{n}-q, x_{n+1} - q \rangle \\ \leq &b_{n}\bigl\langle f(x_{n})-f(q), x_{n+1} - q \bigr\rangle +b_{n}\bigl\langle f(q)-q, x_{n+1} - q \bigr\rangle \\ &{} +(1-b_{n}) \Vert t_{n}-q \Vert \Vert x_{n+1} - q \Vert \\ \leq &b_{n}\rho \Vert x_{n} - q \Vert \Vert x_{n+1} - q \Vert +b_{n}\bigl\langle f(q)-q, x_{n+1} - q \bigr\rangle \\ &{} +(1-b_{n}) \Vert x_{n}-q \Vert \Vert x_{n+1} - q \Vert \\ \leq & \bigl(1-b_{n}(1-\rho )\bigr) \Vert x_{n}-q \Vert \Vert x_{n+1} - q \Vert +b_{n}\bigl\langle f(q)-q, x_{n+1} - q \bigr\rangle \\ \leq & \frac{(1-b_{n}(1-\rho ))}{2}\bigl( \Vert x_{n}-q \Vert ^{2}+ \Vert x_{n+1} - q \Vert ^{2} \bigr)+b_{n} \bigl\langle f(q)-q, x_{n+1} - q \bigr\rangle , \end{aligned}$$

which implies that $\Vert x_{n+1} - q \Vert ^{2}\leq (1-b_{n}(1-\rho )) \Vert x_{n}-q \Vert ^{2}+2b_{n} \langle f(q)-q, x_{n+1} - q \rangle$. Now by using $(d_{3})$ and Lemma 2.4 we get $\lim_{n\to \infty } \Vert x_{n}-q \Vert =0$. This completes the proof. □

Let $u\in C$ and $f(x) = u\in C$ for all x. By using Theorem 3.1 we obtain the following result.

Corollary 3.2

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap (M+B)^{-1}(0) \cap \operatorname{GEP}(F,A)\neq \emptyset$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$, $\{y_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}M y_{n}), \\ x_{n+1}=P_{C}[b_{n}u+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad ( \forall n\geq 1). \end{cases}$$

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

Corollary 3.3

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap (M+B)^{-1}(0) \cap \operatorname{GEP}(F,A)\neq \emptyset$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$, $\{y_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}M y_{n}), \\ x_{n+1}=P_{C}[(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad (\forall n \geq 1). \end{cases} $$

If$(d_{1})$–$(d_{4})$hold, then the sequence$\{x_{n}\}$converges strongly to a point$q =P_{\varOmega }(0)$, which is the minimum norm element in Ω.

Proof

In Theorem 3.1, put $f(x) = 0$ for all x. Note that $x_{n} \to q =P_{\varOmega }(0)$ and $P_{\varOmega }(0)$ is the minimum norm element in Ω. Since $\langle (I-f)q,x-q\rangle \geq 0$, we get $\langle q,q-x\rangle \leq 0$ for all $x\in \varOmega$, that is, $\Vert q \Vert ^{2}\leq \langle q,x\rangle \leq \Vert x \Vert \Vert q \Vert$ for all $x\in \varOmega$. Thus, the point q is the unique solution to the quadratic minimization problem $q= \arg \min_{x\in \varOmega } \Vert x \Vert ^{2}$. □

Let $I_{C}$ be the indicator function of C defined by $I_{C}(x) =0$ for $x \in C$ and $I_{C}(x)=\infty$ otherwise. Recall that the subdifferential $\partial I_{C}$ is a maximal monotone operator. Note that $I_{C}$ is a proper lower semicontinuous convex function on H. The resolvent $J_{r}^{\partial I_{C}}$ of $\partial I_{C}$ for r is $P_{C}$, and $\operatorname{VI}(C,M)=(M + \partial I_{C})^{-1}(0)$, where M is an inverse strongly monotone mapping from C into H [35].

Theorem 3.4

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap \operatorname{VI}(C,M) \cap \operatorname{GEP}(F,A)\neq \emptyset$and$f: C \to C$is a contraction map with the constant$\rho \in (0,1)$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$, $\{y_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})P_{C}(y_{n}-\lambda _{n}M y_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad ( \forall n\geq 1). \end{cases}$$

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

Proof

If $B = \partial I_{C}$ in Theorem 3.1, then $J_{\lambda _{n}} = P_{C}$ for all $\lambda _{n} > 0$. This completes the proof. □

Note that Theorem 3.4 reduces the results of [1, 36].

Theorem 3.5

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap \operatorname{VI}(C,M)\neq \emptyset$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$and$\{z_{n}\}$are some sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})P_{C}(x_{n}-\lambda _{n}M x_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Tx_{n})] \quad ( \forall n\geq 1). \end{cases}$$

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

Proof

Put $F=A=0$, $B = \partial I_{C}$, and $r_{n}=1$ for all n in Theorem 3.1. Since $J_{\lambda _{n}} = P_{C}$ for all $\lambda _{n} > 0$, we obtain the desired result. □

We can see that Theorem 3.5 extends Theorem 11 in [37]. The next result reduces the related result of [38].

Theorem 3.6

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S: C \to C$is a nonexpansive mapping such that${\varOmega }=F(S) \cap \operatorname{VI}(C,M) \neq \emptyset$and$f: C \to C$is a contraction map with the constant$\rho \in (0,1)$. Suppose that$\{b_{n}\}$and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})P_{C}(x_{n}-\lambda _{n}M x_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})Sz_{n}] \quad (\forall n\geq 1). \end{cases}$$

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

Proof

Put $F=A=0$, $B = \partial I_{C}$, $T=I$, and $r_{n}=1$ for all n in Theorem 3.1. Since $J_{\lambda _{n}} = P_{C}$ for all $\lambda _{n} > 0$, we obtain the desired result. □

Theorem 3.7

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, and let$\psi :C\to C$be aβ-strict pseudo-contraction. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap F(\psi )\neq \emptyset$, and$f: C \to C$is a contraction map with the constant$\rho \in (0,1)$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})((1-\lambda _{n})x_{n}+\lambda _{n} \psi x_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Tx_{n})] \quad ( \forall n\geq 1). \end{cases} $$

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

Proof

Put $F=A=0$, $r_{n}=1$, and $M = I-\psi$. Then M is $\frac{1-\beta }{2}$-inverse-strongly monotone map, $F (\psi ) = V I(C, M)$, and $P_{C} (x_{n} -\lambda _{n}Mx_{n}) = (1 - \lambda _{n})x_{n} + \lambda _{n}\psi x_{n}$ for all n. Now by using Theorem 3.5 we obtain the desired result. □

Note that Theorem 3.7 reduces the result of [39].

Theorem 3.8

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letAbe anα-inverse strongly monotone mapping fromCintoH, and let$\psi :C\to C$be aβ-strict pseudo-contraction. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap F(\psi )\cap \operatorname{GEP}(F,A) \neq \emptyset$and$f: C \to C$is a contraction map with the constant$\rho \in (0,1)$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$, $\{y_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})((1-\lambda _{n})y_{n}+\lambda _{n} \psi y_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad ( \forall n\geq 1). \end{cases} $$

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

Proof

Put $M = I-\psi$. Then M is $\frac{1-\beta }{2}$-inverse-strongly monotone mapping, $F (\psi ) = V I(C, M)$, and $P_{C} (y_{n} -\lambda _{n}My_{n}) = (1 - \lambda _{n})y_{n} + \lambda _{n}\psi y_{n}$ for all n. By using Theorem 3.1 we obtain the desired result. □

We can check that Theorem 3.8 reduces the result of [40].

Theorem 3.9

LetCbe a nonempty closed convex subset ofH, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap (M+B)^{-1}(0)\neq \emptyset$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})J_{\lambda _{n}}^{B}(x_{n}-\lambda _{n}M x_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Tx_{n})]\quad ( \forall n\geq 1). \end{cases} $$

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

Proof

It is sufficient put $F=A=0$ and $r_{n}=1$ for all n in Theorem 3.1. □

We can see that Theorem 3.9 reduces the result of [31]. Let $g: H \to \mathbb{R} \cup \{+\infty \}$ be a convex lower semicontinuous proper function. Put $B = \partial g$, where ∂ denotes subdifferential of g. Then B is a maximal monotone operator, and $0 \in \partial f(x)$ is equivalent to $g(x^{\prime }) =\min_{x \in C}g(x)$ [24, 26]. Recall that the subdifferential of g at x is defined by

$$ \partial g(x) := \bigl\{ v \in H: g(y) \geq g(x) + \langle v , y - x \rangle \text{ for all } y \in H \bigr\} . $$

Theorem 3.10

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, Abe anα-inverse strongly monotone mapping fromCintoH, and let$g : H \to (-\infty , +\infty ]$be a proper convex lower semicontinuous function. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap (\partial f)^{-1}(0) \cap \operatorname{GEP}(F,A)\neq \emptyset$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$be sequences in$(0, 1)$and that$\{x_{n}\}$, $\{y_{n}\}$, $\{u_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) + \langle Ax_{n}, y - y_{n}\rangle +\frac{1}{r_{n}} \langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad (\forall y\in C), \\ u_{n}=\arg \min_{w\in H}\{g(w) + \frac{ \Vert w-y_{n} \Vert ^{2}}{2\lambda _{n}}\}, \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})u_{n}, \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad ( \forall n\geq 1). \end{cases} $$

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

Proof

Put $M = 0$. Then by Theorem 3.1 the desired result immediately follows. □

Put $A = 0$ in Theorem 3.1. Then we obtain next theorem, which reduces the result of [41].

Theorem 3.11

LetCbe a nonempty closed convex subset ofH, letFbe a bifunction from$C \times C$to$\mathbb{R}$satisfying$(A_{1})$–$(A_{4})$, letMbe aβ-inverse strongly monotone map fromCintoH, and letBbe a maximal monotone operator onHwith domain contained in C. Assume that$S,T : C \to C$are two nonexpansive mappings such that${\varOmega }=F(T)\cap F(S) \cap (M+B)^{-1}(0) \cap \operatorname{GEP}(F)\neq \emptyset$and$f: C \to C$is a contraction map with the constant$\rho \in (0,1)$. Suppose that$\{b_{n}\}$, $\{a_{n}\}$, and$\{\mu _{n}\}$are some sequences in$(0, 1)$and that$\{x_{n}\}$, $\{y_{n}\}$, and$\{z_{n}\}$are the sequences generated by

$$ \textstyle\begin{cases} x_{1}\in C, \\ F (y_{n}, y) +\frac{1}{r_{n}}\langle y - y_{n},y_{n} - x_{n}\rangle \geq 0 \quad \forall y\in C, \\ z_{n}=\mu _{n}x_{n}+(1-\mu _{n})J_{\lambda _{n}}^{B}(y_{n}-\lambda _{n}M y_{n}), \\ x_{n+1}=P_{C}[b_{n}f(x_{n})+(1-b_{n})(a_{n}Sz_{n}+(1-a_{n}) Ty_{n})] \quad ( \forall n\geq 1). \end{cases} $$

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

Here we provide an example to illustrate Theorem 3.1.

Example 3.1

Let $H = \mathbb{R}$ with Euclidean norm and usual Euclidean inner product. Put $C :=[-1,\infty )$, $Sx=\frac{x}{2}$, $Tx=\frac{x}{3}$, $Bx=\log (x+1)$, $Mx=4x$, $\beta =\frac{1}{4}$, $F(x,y)=y-x$, $\alpha =\frac{1}{3}$, and $Ax =3x-1$ for all x. It is clear that S and T are nonexpansive, M is a β-inverse strongly monotone mapping, B is a maximal monotone operator, F is a bifunction from $C \times C$ to $\mathbb{R}$ satisfying $(A_{1})$–$(A_{4})$, A is an α-inverse strongly monotone mapping, and $0\in \varOmega =F(T)\cap F(S) \cap (M+B)^{-1}(0) \cap \operatorname{GEP}(F,A)$. Now by using Theorem 3.1 the sequence $\{x_{n}\}$ converges strongly to a point $q\in \varOmega$, which is the unique solution to the variational inequality $\langle (I-f)q,x-q\rangle \geq 0$ for all $x \in \varOmega$.