An algorithm for finding common solutions of various problems in nonlinear operator theory

Abstract

In this paper, it is our aim to prove strong convergence of a new iterative algorithm to a common element of the set of solutions of a finite family of classical equilibrium problems; a common set of zeros of a finite family of inverse strongly monotone operators; the set of common fixed points of a finite family of quasi-nonexpansive mappings; and the set of common fixed points of a finite family of continuous pseudocontractive mappings in Hilbert spaces on assumption that the intersection of the aforementioned sets is not empty. Moreover, the common element is shown to be the metric projection of the initial guess on the intersection of these sets.

MSC:47H06, 47H09, 47J05, 47J25.

1 Introduction

Let H be a real Hilbert space. A mapping T with domain $D(T)$ and range $R(T)$ in H is called an L-Lipschitzian mapping (or simply a Lipschitz mapping) if and only if there exists $L\ge 0$ such that for all $x,y\in D(T)$,

$$\parallel Tx-Ty\parallel \le L\parallel x-y\parallel .$$

If $L\in [0,1)$, then T is called strict contraction or simply a contraction; and if $L=1$, then T is called nonexpansive. A point $x\in D(T)$ is called a fixed point of an operator T if and only if $Tx=x$. The set of fixed points of an operator T is denoted by $Fix(T)$, that is, $Fix(T):=\{x\in D(T):Tx=x\}$.

A mapping T with domain $D(T)$ and range $R(T)$ in H is called a quasi-nonexpansive mapping if and only if $Fix(T)\ne \mathrm{\varnothing }$ and for any $x\in D(T)$, for any $u\in Fix(T)$,

$$\parallel Tx-u\parallel \le \parallel x-u\parallel .$$

Every nonexpansive mapping with a nonempty fixed point set is quasi-nonexpansive. The following examples show that the converse is not true.

Example 1.1 (see [1])

Let $E=[-\pi ,\pi ]$ be a subspace of the set of real numbers ℝ, endowed with the usual topology. Define $T:E\to E$ by $Tx=xcosx$ for all $x\in E$. Clearly, $F(T)=\{0\}$. Observe that

$$|Tx-0|=|x|\times |cosx|\le |x|=|x-0|.$$

Thus, T is quasi-nonexpansive. The mapping T is, however, not a nonexpansive mapping since for $x=\frac{\pi }{2}$ and $y=\pi$,

$$|Tx-Ty|=|\frac{\pi }{2}cos\left(\frac{\pi }{2}\right)-\pi cos\pi |=\pi .$$

But

$$|x-y|=|\frac{\pi }{2}-\pi |=\frac{\pi }{2}.$$

Example 1.2 (see [1, 2])

Let $E=\mathbb{R}$ be endowed with usual topology. Define $T:\mathbb{R}\to \mathbb{R}$ by

$$Tx=\{\begin{array}{ll}\frac{x}{2}cos(\frac{1}{x}),& x\ne 0,\\ 0,& x=0.\end{array}$$

(1.1)

It is easy to see that $F(T)=\{0\}$ since for $x\ne 0$, $Tx=x$ implies that $\frac{x}{2}cos\frac{1}{x}=x$. Thus, for any $x\ne 0$, $cos\frac{1}{x}=2$, which is not possible. So, $F(T)=\{0\}$. Next, observe that for any $x\in \mathbb{R}$,

$$|Tx-0|=\left|\frac{x}{2}\right|\times |cos\left(\frac{1}{x}\right)|\le \frac{|x|}{2}<|x|=|x-0|.$$

So, the mapping T is quasi-nonexpansive. Finally, we show that T is not nonexpansive. To see this, let $x=\frac{2}{3\pi }$ and $y=\frac{1}{\pi }$, then

$$|Tx-Ty|=|\frac{1}{3\pi }cos\left(\frac{3\pi }{2}\right)-\frac{1}{2\pi }cos\pi |=\frac{1}{2\pi }.$$

But,

$$|x-y|=|\frac{2}{3\pi }-\frac{1}{\pi }|=\frac{1}{3\pi }.$$

So,

$$|Tx-Ty|=\frac{1}{2\pi }>\frac{1}{3\pi }=|x-y|.$$

The concept of quasi-nonexpansive mappings was essentially introduced by Diaz and Metcalf [3]. Although Examples 1.1 and 1.2 guarantee the existence of a quasi-nonexpansive mapping which is not nonexpansive, we must note that a linear quasi-nonexpansive mapping defined on a subspace of a given vector space is nonexpansive on that subspace.

Another important generalization of the class of nonexpansive mappings is the class of pseudocontractive mappings. These mappings are intimately connected with the important class of nonlinear accretive operators. This connection will be made precise in what follows.

A mapping T with domain $D(T)$ and range $R(T)$ in H is called pseudocontractive if and only if for all $x,y\in D(T)$, the following inequality holds:

$$\parallel x-y\parallel \le \parallel (1+r)(x-y)-r(Tx-Ty)\parallel$$

(1.2)

for all $r>0$. As a consequence of a result of Kato [4], the pseudocontractive mappings can also be defined in terms of the normalized duality mappings as follows: the mapping T is called pseudocontractive if and only if for all $x,y\in D(T)$, we have that

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

(1.3)

It now follows trivially from (1.3) that every nonexpansive mapping is pseudocontractive. We note immediately that the class of pseudocontractive mappings is larger than that of nonexpansive mappings. For examples of pseudocontractive mappings which are not nonexpansive, the reader may see [5].

To see the connection between the pseudocontractive mappings and the monotone mappings, we introduce the following definition: a mapping A with domain $D(A)$ and range $R(A)$ in E is called monotone if and only if for all $x,y\in D(A)$, the following inequality is satisfied:

$$〈Ax-Ay,x-y〉\ge 0.$$

(1.4)

The operator A is called η-inverse strongly monotone if and only if there exists $\eta \in (0,1)$ such that for all $x,y\in D(A)$, we have that

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

(1.5)

It is easy to see from inequalities (1.3) and (1.4) that an operator A is monotone if and only if the mapping $T:=(I-A)$ is pseudocontractive. Consequently, the fixed point theory for pseudocontractive mappings is intimately connected with the zero of monotone mappings. For the importance of monotone mappings and their connections with evolution equations, the reader may consult any of the references [5, 6].

Due to the above connection, fixed point theory of pseudocontractive mappings became a flourishing area of intensive research for several authors.

Let C be a closed convex nonempty subset of a real Hilbert space H with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$. Let $f:C\times C\to \mathbb{R}$ be a bifunction. The classical equilibrium problem (EP) for a bifunction f is to find $u^{\ast }\in C$ such that

$$f(u^{\ast },y)\ge 0,\mathrm{\forall }y\in C.$$

(1.6)

The set of solutions for EP (1.6) is denoted by

$$\mathit{EP}(f)=\{u\in C:f(u,y)\ge 0,\mathrm{\forall }y\in C\}.$$

The classical equilibrium problem (EP) includes as special cases the monotone inclusion problems, saddle point problems, variational inequality problems, minimization problems, optimization problems, vector equilibrium problems, Nash equilibria in noncooperative games. Furthermore, there are several other problems, for example, the complementarity problems and fixed point problems, which can also be written in the form of the classical equilibrium problem. In other words, the classical equilibrium problem is a unifying model for several problems arising from engineering, physics, statistics, computer science, optimization theory, operations research, economics and countless other fields. For the past 20 years or so, many existence results have been published for various equilibrium problems (see, e.g., [7–10]). Approximation methods for such problems thus become a necessity.

Iterative approximation of fixed points and zeros of nonlinear mappings has been studied extensively by many authors to solve nonlinear mapping equations as well as variational inequality problems and their generalizations (see, e.g., [11–19]). Most published results on nonexpansive mappings (for example) focus on the iterative approximation of their fixed points or approximation of common fixed points of a given family of this class of mappings.

Some attempts to modify the Mann iteration method so that strong convergence is guaranteed have recently been made (we should recall that Mann iteration method only guarantees weak convergence (see, for example, Bauschke et al. [20])). Nakajo and Takahashi [16] formulated the following modification of the Mann iteration method for a nonexpansive mapping T defined on a nonempty bounded closed and convex subset C of a Hilbert space H:

$$\{\begin{array}{l}{x}_0\in C,\\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)T{x}_n,\\ C_n=\{v\in C:{\parallel y_n-v\parallel }^2\le {\parallel x_n-v\parallel }^2\},\\ Q_n=\{v\in C:〈x_n-v,x_0-x_n〉\ge 0\},\\ x_{n+1}=P_{C_n\cap Q_n}(x_0),\mathrm{\forall }n\in \mathbb{N},\end{array}$$

(1.7)

where $P_C$ denotes the metric projection from H onto a closed convex subset C of H. They proved that if the sequence ${\{{\alpha }_n\}}_{n\ge 0}$ is bounded away from 1, then ${\{x_n\}}_{n\ge 0}$ defined by (1.7) converges strongly to $P_{F(T)}(x_0)$.

Formulations similar to (1.7) for different classes of nonlinear problems had been presented by Kim and Xu [21], Nilsrakoo and Saejung [22], Ofoedu et al. [23], Yang and Su [24], Zegeye and Shahzad [25–27].

In this paper, motivated by the results of the authors mentioned above, it is our aim to prove strong convergence of a new iterative algorithm to a common element of the set of solutions of a finite family of classical equilibrium problems; a common set of zeros of a finite family of inverse strongly monotone mappings; a set of common fixed points of a finite family of quasi-nonexpansive mappings; and a set of common fixed points of a finite family of continuous pseudocontractive mappings in Hilbert spaces on assumption that the intersection of the aforementioned sets is not empty. Moreover, the common element is shown to be the metric projection of the initial guess on the intersection of these sets. Our theorems complement the results of the authors mentioned above and those of several other authors.

2 Preliminary

In what follows, we shall make use of the following lemmas.

Lemma 2.1 (see, e.g., Kopecka and Reich [28])

Let C be a nonempty closed and convex subset of a real Hilbert space. Let $x\in H$ and $P_C:H\to C$ be the metric projection of H onto C, then for any $y\in C$,

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

Lemma 2.2 Let C be a closed convex nonempty subset of a real Hilbert space H; and let $P_C:H\to C$ be the metric projection of H onto C. Let $x\in H$, then $x_0=P_{C}x$ if and only if $〈z-x_0,x-x_0〉\le 0$ for all $z\in C$.

Lemma 2.3 Let H be a real Hilbert space, then for any $x,y\in H$, $\alpha \in [0,1]$,

$${\parallel \alpha x+(1-\alpha )y\parallel }^2=\alpha {\parallel x\parallel }^2+(1-\alpha ){\parallel y\parallel }^2-\alpha (1-\alpha ){\parallel x-y\parallel }^2.$$

Lemma 2.4 (see Zegeye [29])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to H$ be a continuous pseudocontractive mapping, then for all $r>0$ and $x\in H$, there exists $z\in C$ such that

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

Lemma 2.5 (see Zegeye [29])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to C$ be a continuous pseudocontractive mapping, then for all $r>0$ and $x\in H$, define a mapping $F_r:H\to C$ by

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

then the following hold:

• (1) $F_r$ is single-valued;

• (2) $F_r$ is firmly nonexpansive type mapping, i.e., for all $x,y\in H$,

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

• (3) $Fix(F_r)$ is closed and convex; and $Fix(F_r)=Fix(T)$ for all $r>0$.

In the sequel, we shall require that the bifunction $f:C\times C\to \mathbb{R}$ satisfies the following conditions:

(A1) $f(x,x)=0$, $\mathrm{\forall }x\in C$;

(A2) f is monotone in the sense that $f(x,y)+f(y,x)\le 0$ for all $x,y\in C$;

(A3) ${lim\hspace{0.17em}sup}_{t\to 0^{+}}f(tz+(1-t)x,y)\le f(x,y)$ for all $x,y,z\in C$;

(A4) the function $y\mapsto f(x,y)$ is convex and lower semicontinuous for all $x\in C$.

Lemma 2.6 (see, e.g., [7, 30])

Let C be a closed convex nonempty subset of a real Hilbert space H. Let $f:C\times C\to \mathbb{R}$ be a bifunction satisfying conditions (A1)-(A4), then for all $r>0$ and $x\in H$, there exists $u\in C$ such that

$$f(u,y)+\frac{1}{r}〈y-u,u-x〉\ge 0,\mathrm{\forall }y\in C.$$

(2.1)

Moreover, if for all $x\in H$ we define a mapping $G_r:H\to 2^C$ by

$$G_r(x)=\{u\in C:f(u,y)+\frac{1}{r}〈y-u,u-x〉\ge 0,\mathrm{\forall }y\in C\},$$

(2.2)

then the following hold:

• (1) $G_r$ is single-valued for all $r>0$;

• (2) $G_r$ is firmly nonexpansive, that is, for all $x,z\in H$,

  $${\parallel G_{r}x-G_{r}z\parallel }^2\le 〈G_{r}x-G_{r}z,x-z〉;$$

• (3) $Fix(G_r)=\mathit{EP}(f)$ for all $r>0$;

• (4) $\mathit{EP}(f)$ is closed and convex.

Lemma 2.7 (see Ofoedu [31])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T:C\to C$ be a continuous pseudocontractive mapping. For $r>0$, let $F_r:H\to C$ be the mapping in Lemma  2.5, then for any $x\in H$ and for any $p,q>0$,

$$\parallel F_{p}x-F_{q}x\parallel \le \frac{|p-q|}{p}(\parallel F_{p}x\parallel +\parallel x\parallel ).$$

Lemma 2.8 (Compare with Lemma 13 of Ofoedu [31])

Let C be a closed convex nonempty subset of a real Hilbert space H. Let $f:C\times C\to \mathbb{R}$ be a bifunction satisfying conditions (A1)-(A4). Let $r>0$ and let $G_r$ be the mapping in Lemma  2.6, then for all $p,q>0$ and for all $x\in H$, we have that

$$\parallel G_{p}x-G_{q}x\parallel \le \frac{|p-q|}{p}(\parallel G_{p}x\parallel +\parallel x\parallel ).$$

3 Main results

Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots ,T_m:C\to C$ be m continuous pseudocontractive mappings; let $S_1,S_2,\dots ,S_l:C\to C$ be l continuous quasi-nonexpansive mappings; let $A_1,A_2,\dots ,A_d:C\to H$ be d ${\gamma }_j$-inverse strongly monotone mappings with constants ${\gamma }_j\in (0,1)$, $j=1,2,\dots ,d$; let $f_1,f_2,\dots ,f_t:C\times C\to \mathbb{R}$ be t bifunctions satisfying conditions (A1)-(A4). For all $x\in E$, $i=1,2,\dots ,m$, let

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

and for all $x\in E$, $h=1,2,\dots ,t$, let

$$G_{h,r}(x)=\{u\in C:f_h(u,y)+\frac{1}{r}〈y-u,u-x〉\ge 0,\mathrm{\forall }y\in C\},$$

then in what follows we shall study the following iteration process:

$$\{\begin{array}{l}{x}_0\in C_0=C\text{chosen arbitrarily},\\ z_n=P_C(x_n-{\lambda }_n{A}_{n+1}{x}_n),\\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)S_{n+1}{z}_n,\\ w_n=\eta {\sum }_{i=1}^m{\beta }_i{F}_{i,r_n}{y}_n+(1-\eta ){\sum }_{h=1}^t{\xi }_h{G}_{h,r_n}{y}_n,\\ C_{n+1}=\{z\in C:\parallel w_n-z\parallel \le \parallel x_n-z\parallel \},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}(x_0),n\ge 0,\end{array}$$

(3.1)

where $A_n=A_{n(modd)}$, $S_n=S_{n(modl)}$; $\{r_n\}\subset (0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{r}_n=r_0>0$; ${\{{\alpha }_n\}}_{n\ge 1}$ a sequence in $(0,1)$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$; ${\{{\beta }_i\}}_{i=1}^m,{\{{\xi }_h\}}_{h=1}^t\subset (0,1)$ such that ${\sum }_{i=1}^m{\beta }_i=1={\sum }_{h=1}^t{\xi }_h$; $\eta \in (0,1)$ and $\{{\lambda }_n\}$ is a sequence in $[a,b]$ for some $a,b\in \mathbb{R}$ such that $0<a<b<2\gamma$, $\gamma ={min}_{1\le j\le d}\{{\gamma }_j\}$.

Lemma 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots ,T_m:C\to C$ be m continuous pseudocontractive mappings; let $S_1,S_2,\dots ,S_l:C\to C$ be l continuous quasi-nonexpansive mappings; let $A_1,A_2,\dots ,A_d:C\to H$ be d ${\gamma }_j$-inverse strongly monotone mappings with constants ${\gamma }_j\in (0,1)$, $j=1,2,\dots ,d$; let $f_1,f_2,\dots ,f_t:C\times C\to \mathbb{R}$ be t bifunctions satisfying conditions (A1)-(A4). Let $F:={\bigcap }_{i=1}^{m}Fix(T_i)\cap {\bigcap }_{j=1}^d{A}_j^{-1}(0)\cap {\bigcap }_{k=1}^{l}Fix(S_k)\cap {\bigcap }_{h=1}^t\mathit{EP}(f_h)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence defined by (3.1), then the sequence $\{x_n\}$ is well defined for each $n\ge 0$.

Proof We first show that $C_n$ is closed and convex for each $n\in \mathbb{N}\cup \{0\}$. From the definitions of $C_n$ it is obvious that $C_n$ is closed. Moreover, since $\parallel w_n-z\parallel \le \parallel x_n-z\parallel$ is equivalent to $2〈z,x_n-w_n〉-{\parallel x_n\parallel }^2+{\parallel w_n\parallel }^2\le 0$, it follows that $C_n$ is convex for each $n\in \mathbb{N}\cup \{0\}$. Next, we prove that $F\subset C_n$ for each $n\in \mathbb{N}\cup \{0\}$. From the assumption, we see that $F\subset C_0=C$. Suppose that $F\subset C_k$ for some $k\ge 1$, then for $p\in F$, we obtain that

$$\begin{array}{rcl}\parallel w_k-p\parallel & =& \parallel \eta \sum _{i=1}^m{\beta }_i{F}_{i,r_k}{y}_k+(1-\eta )\sum _{h=1}^m{\xi }_h{G}_{h,r_k}{y}_k-p\parallel \\ \le & \parallel y_k-p\parallel =\parallel {\alpha }_k{x}_k+(1-{\alpha }_k)S_{k+1}{z}_k-p\parallel \\ \le & {\alpha }_k\parallel x_k-p\parallel +(1-{\alpha }_k)\parallel S_{k+1}{z}_k-p\parallel \\ \le & {\alpha }_k\parallel x_k-p\parallel +(1-{\alpha }_k)\parallel z_k-p\parallel .\end{array}$$

(3.2)

Furthermore,

$$\begin{array}{rcl}{\parallel z_k-p\parallel }^2& =& {\parallel P_C(x_k-{\lambda }_k{A}_{k+1}{x}_k)-p\parallel }^2\\ \le & {\parallel x_k-{\lambda }_k{A}_{k+1}{x}_k-p\parallel }^2\\ =& {\parallel x_k-p-{\lambda }_k(A_{k+1}{x}_k-A_{k+1}p)\parallel }^2\\ =& {\parallel x_k-p\parallel }^2-2{\lambda }_k〈x_k-p,A_{k+1}{x}_k-A_{k+1}p〉+{\lambda }_k^2{\parallel A_{k+1}{x}_k-A_{k+1}p\parallel }^2\\ \le & {\parallel x_k-p\parallel }^2+{\lambda }_k({\lambda }_k-2\gamma ){\parallel A_{k+1}{x}_k-A_{k+1}p\parallel }^2\\ \le & {\parallel x_k-p\parallel }^2(\text{since }{\lambda }_k<2\gamma ).\end{array}$$

Thus,

$$\parallel z_k-p\parallel \le \parallel x_k-p\parallel .$$

(3.3)

Using (3.3) in (3.2) gives

$$\parallel w_k-p\parallel \le \parallel x_k-p\parallel .$$

So, $p\in C_{k+1}$. This implies, by induction, that $F\subset C_n$ so that the sequence generated by (3.1) is well defined for all $n\ge 0$. □

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots ,T_m:C\to C$ be m continuous pseudocontractive mappings; let $S_1,S_2,\dots ,S_l:C\to C$ be l continuous quasi-nonexpansive mappings; let $A_1,A_2,\dots ,A_d:C\to H$ be d ${\gamma }_j$-inverse strongly monotone mappings with constants ${\gamma }_j\in (0,1)$, $j=1,2,\dots ,d$; let $f_1,f_2,\dots ,f_t:C\times C\to \mathbb{R}$ be t bifunctions satisfying conditions (A1)-(A4). Let $F:={\bigcap }_{i=1}^{m}Fix(T_i)\cap {\bigcap }_{j=1}^d{A}_j^{-1}(0)\cap {\bigcap }_{k=1}^{l}Fix(S_k)\cap {\bigcap }_{h=1}^t\mathit{EP}(f_h)\ne \mathrm{\varnothing }$. Let $\{x_n\}$ be a sequence defined by (3.1). Then the sequence ${\{x_n\}}_{n\ge 0}$ converges strongly to the element of F nearest to $x_0$.

Proof From Lemma 3.1, we obtain that $F\subset C_n$, $\mathrm{\forall }n\ge 0$ and $x_n$ is well defined for each $n\ge 0$. From $x_n=P_{C_n}(x_0)$ and $x_{n+1}=P_{C_{n+1}}(x_0)\in C_{n+1}\subset C_n$, we obtain that

$$〈x_{n+1}-x_n,x_n-x_0〉\ge 0\text{and}\parallel x_n-x_0\parallel \le \parallel x_{n+1}-x_0\parallel .$$

Besides, by Lemma 2.1,

$${\parallel x_n-p\parallel }^2=\parallel P_{C_n}{x}_0-x_0\parallel \le {\parallel x_0-p\parallel }^2-{\parallel x_0-x_n\parallel }^2\le {\parallel x_0-p\parallel }^2.$$

Thus, the sequence ${\{\parallel x_n-x_0\parallel \}}_{n\ge 0}$ is a bounded nondecreasing sequence of real numbers. So, ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_0\parallel$ exists. Similarly, by Lemma 2.1, we have that for any positive integer μ,

$$\begin{array}{rcl}{\parallel x_{n+\mu }-x_n\parallel }^2& =& {\parallel x_{n+\mu }-P_{C_n}{x}_0\parallel }^2\\ \le & {\parallel x_{n+\mu }-x_0\parallel }^2-{\parallel P_{C_n}{x}_0-x_0\parallel }^2\\ =& {\parallel x_{n+\mu }-x_0\parallel }^2-{\parallel x_n-x_0\parallel }^2\text{for all }n\ge 0.\end{array}$$

Since ${lim}_{n\to \mathrm{\infty }}\parallel x_n-x_0\parallel$ exists, we have that ${lim}_{n\to \mathrm{\infty }}\parallel x_{n+\mu }-x_n\parallel =0$ and hence, ${\{x_n\}}_{n\ge 1}$ is a Cauchy sequence in C. Therefore, there exists $x^{\ast }\in C$ such that ${lim}_{n\to \mathrm{\infty }}{x}_n=x^{\ast }$. Since $x_{n+1}\in C_{n+1}$, we have that

$$\parallel w_n-x_{n+1}\parallel \le \parallel x_n-x_{n+1}\parallel .$$

Thus,

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

(3.4)

and hence $\parallel x_n-w_n\parallel \le \parallel x_n-x_{n+1}\parallel +\parallel x_{n+1}-w_n\parallel \to 0$ as $n\to \mathrm{\infty }$, which implies that $w_n\to x^{\ast }$ as $n\to \mathrm{\infty }$.

Next, we observe that for $p\in F$ and using Lemma 2.3,

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& =& {\parallel {\alpha }_n{x}_n+(1-{\alpha }_n)S_{n+1}{z}_n-p\parallel }^2\\ =& {\parallel {\alpha }_n(x_n-p)+(1-{\alpha }_n)(S_{n+1}{z}_n-p)\parallel }^2\\ =& {\alpha }_n\parallel x_n-p\parallel +(1-{\alpha }_n){\parallel S_{n+1}{z}_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel x_n-S_{n+1}{z}_n\parallel }^2\end{array}$$

(3.5)

$$\le {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\parallel z_n-p\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel x_n-S_{n+1}{z}_n\parallel }^2.$$

(3.6)

But

$$\begin{array}{rcl}{\parallel z_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2+{\lambda }_n({\lambda }_n-2\gamma ){\parallel A_{n+1}{x}_n-A_{n+1}p\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+{\lambda }_n({\lambda }_n-2\gamma ){\parallel A_{n+1}{x}_n\parallel }^2.\end{array}$$

(3.7)

So, using (3.7) in (3.6), we obtain that

$$\begin{array}{rcl}{\parallel y_n-p\parallel }^2& \le & {\alpha }_n{\parallel x_n-p\parallel }^2+(1-{\alpha }_n)[{\parallel x_n-p\parallel }^2+{\lambda }_n({\lambda }_n-2\gamma ){\parallel A_{n+1}{x}_n\parallel }^2]\\ -{\alpha }_n(1-{\alpha }_n){\parallel x_n-S_{n+1}{z}_n\parallel }^2\\ =& {\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\lambda }_n({\lambda }_n-2\gamma ){\parallel A_{n+1}{x}_n\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n){\parallel x_n-S_{n+1}{z}_n\parallel }^2.\end{array}$$

(3.8)

Moreover, we obtain that

$$\begin{array}{rcl}{\parallel w_n-p\parallel }^2& =& {\parallel \eta \sum _{i=1}^m{\beta }_i{F}_{i,r_n}{y}_n+(1-\eta )\sum _{h=1}^m{\xi }_h{G}_{h,r_n}{y}_n-p\parallel }^2\\ \le & {\parallel y_n-p\parallel }^2.\end{array}$$

(3.9)

Using (3.8) in (3.9) we get that

$$\begin{array}{rcl}{\parallel w_n-p\parallel }^2& \le & {\parallel x_n-p\parallel }^2+(1-{\alpha }_n){\lambda }_n({\lambda }_n-2\gamma ){\parallel A_{n+1}{x}_n\parallel }^2\\ -{\alpha }_n(1-{\alpha }_n){\parallel x_n-S_{n+1}{z}_n\parallel }^2.\end{array}$$

(3.10)

Now, using the fact that ${\lambda }_n<2\gamma$, inequality (3.10) gives (for some constant $M_0>0$) that

$${\alpha }_n(1-{\alpha }_n)\parallel x_n-S_{n+1}{z}_n\parallel \le {\parallel x_n-p\parallel }^2-{\parallel w_n-p\parallel }^2\le M_0\parallel x_n-w_n\parallel .$$

(3.11)

Hence, we obtain from inequality (3.11) that

$$\parallel x_n-S_{n+1}{z}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(3.12)

Moreover, from (3.10) we obtain that

$$(1-{\alpha }_n){\lambda }_n(2\gamma -{\lambda }_n){\parallel A_{n+1}{x}_n\parallel }^2\le {\parallel x_n-p\parallel }^2-{\parallel w_n-p\parallel }^2\le M_0\parallel x_n-w_n\parallel ,$$

which yields that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A_{n+1}{x}_n\parallel =0.$$

(3.13)

Now,

$$\begin{array}{rcl}\parallel x_n-z_n\parallel & =& \parallel x_n-P_C(x_n-{\lambda }_n{A}_{n+1}{x}_n)\parallel =\parallel P_C{x}_n-P_C(x_n-{\lambda }_n{A}_{n+1}{x}_n)\parallel \\ \le & \parallel x_n-x_n+{\lambda }_n{A}_{n+1}{x}_n\parallel ={\lambda }_n\parallel A_{n+1}{x}_n\parallel \\ \le & b\parallel A_{n+1}{x}_n\parallel .\end{array}$$

(3.14)

It follows from (3.13) and (3.14) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-z_n\parallel =0;$$

(3.15)

and hence $z_n\to x^{\ast }$ as $n\to \mathrm{\infty }$.

We now show that $x^{\ast }\in {\bigcap }_{k=1}^{l}Fix(S_k)$. Observe that from (3.12) and (3.15) we obtain that

$$\parallel S_{n+1}{z}_n-z_n\parallel \le \parallel S_{n+1}{z}_n-x_n\parallel +\parallel z_n-x_n\parallel \to 0\text{as }n\to \mathrm{\infty },$$

(3.16)

so that

$$\underset{n\to \mathrm{\infty }}{lim}{S}_{n+1}{z}_n=x^{\ast }.$$

(3.17)

Let ${\{n_{\sigma }\}}_{\sigma \ge 1}\subset \mathbb{N}$ be such that $S_{n_{\sigma }+1}=S_1$ for all $\sigma \in \mathbb{N}$, then since $z_{n_{\sigma }}\to x^{\ast }$ as $\sigma \to \mathrm{\infty }$, we obtain from (3.17), using the continuity of $S_1$, that

$$x^{\ast }=\underset{\sigma \to \mathrm{\infty }}{lim}{S}_{n_{\sigma }+1}{z}_{n_{\sigma }}=\underset{\sigma \to \mathrm{\infty }}{lim}{S}_1{z}_{n_{\sigma }}=S_1{x}^{\ast }.$$

Similarly, if ${\{n_j\}}_{j\ge 1}\subset \mathbb{N}$ is such that $S_{n_j+1}=S_2$ for all $j\in \mathbb{N}$, then we have again that

$$x^{\ast }=\underset{j\to \mathrm{\infty }}{lim}{S}_{n_j+1}{z}_{n_j}=\underset{j\to \mathrm{\infty }}{lim}{S}_2{z}_{n_j}=S_2{x}^{\ast }.$$

Continuing, we obtain that $S_k{x}^{\ast }=x^{\ast }$, $k=3,\dots ,l$. Hence, $x^{\ast }\in {\bigcap }_{k=1}^{l}F(S_k)$.

Next, we show that $x^{\ast }\in {\bigcap }_{j=1}^d{A}_j^{-1}(0)$. Since $A_j$ is γ-inverse strongly monotone for $j=1,2,\dots ,d$, we have that $A_j$ is $\frac{1}{\gamma }$-Lipschitz continuous. Thus,

$$\parallel A_{n+1}{x}_n-A_{n+1}{x}^{\ast }\parallel \le \frac{1}{\gamma }\parallel x_n-x^{\ast }\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

(3.18)

Hence, from (3.18) and (3.13), we obtain that

$$\parallel A_{n+1}{x}^{\ast }\parallel \le \parallel A_{n+1}{x}_n-A_{n+1}{x}^{\ast }\parallel +\parallel A_{n+1}{x}_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$

As a result, we get that

$$\underset{n\to \mathrm{\infty }}{lim}{A}_{n+1}{x}^{\ast }=0.$$

Let ${\{n_s\}}_{s\ge 1}\subset \mathbb{N}$ be such that $A_{n_s+1}=A_1$ for all $s\in \mathbb{N}$. Then

$$A_1{x}^{\ast }=\underset{s\to \mathrm{\infty }}{lim}{A}_{n_s+1}{x}^{\ast }=0.$$

Similarly, we have that $A_j{x}^{\ast }=0$ for $j=2,\dots ,d$. Thus, $x^{\ast }\in {\bigcap }_{j=1}^d{A}_i^{-1}(0)$.

Furthermore, we show that $x^{\ast }\in {\bigcap }_{i=1}^{m}Fix(T_i)={\bigcap }_{i=1}^{m}Fix(F_{i,r})$, $\mathrm{\forall }r>0$. Using the fact that $x_n\to x^{\ast }$, $z_n\to x^{\ast }$ as $n\to \mathrm{\infty }$, we obtain that

$$\begin{array}{rcl}\parallel F_{1,r_n}{y}_n-x^{\ast }\parallel & \le & \parallel y_n-x^{\ast }\parallel \\ \le & {\alpha }_n\parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel z_n-x^{\ast }\parallel \\ \le & \parallel x_n-x^{\ast }\parallel +\parallel z_n-x^{\ast }\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.19)

Thus, we obtain from (3.19) that

$$\underset{n\to \mathrm{\infty }}{lim}{F}_{1,r_n}{y}_n=x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{y}_n.$$

This implies that ${lim}_{n\to \mathrm{\infty }}\parallel F_{1,r_n}{y}_n-y_n\parallel =0$. But by Lemma 2.7,

$$\parallel F_{1,r_n}{y}_n-F_{1,r_0}{y}_n\parallel \le \frac{|r_n-r_0|}{r_n}(\parallel F_{1,r_n}{y}_n\parallel +\parallel y_n\parallel )\to 0\text{as }n\to \mathrm{\infty }.$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim}{F}_{1,r_0}{y}_n=\underset{n\to \mathrm{\infty }}{lim}{F}_{1,r_n}{y}_n=x^{\ast }.$$

So, the continuity of $F_{1,r_0}$ and the fact that $y_n\to x^{\ast }$ as $n\to \mathrm{\infty }$ give

$$x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{F}_{1,r_0}{y}_n=F_{1,r_0}{x}^{\ast }.$$

A similar argument gives

$$x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{F}_{i,r_0}{y}_n=F_{i,r_0}{x}^{\ast },i=2,3,\dots ,m.$$

Hence,

$$x^{\ast }\in \bigcap _{i=1}^{m}Fix(F_{i,r_0})=\bigcap _{i=1}^{m}Fix(T_i).$$

Moreover, we show that $x^{\ast }\in {\bigcap }_{h=1}^t\mathit{EP}(f_h)={\bigcap }_{h=1}^{t}Fix(G_{h,r_0})$. Observe that

$$\begin{array}{rcl}\parallel G_{1,r_n}{y}_n-x^{\ast }\parallel & \le & \parallel y_n-x^{\ast }\parallel \\ \le & {\alpha }_n\parallel x_n-x^{\ast }\parallel +(1-{\alpha }_n)\parallel z_n-x^{\ast }\parallel \\ \le & \parallel x_n-x^{\ast }\parallel +\parallel z_n-x^{\ast }\parallel \to 0\text{as }n\to \mathrm{\infty }.\end{array}$$

(3.20)

Thus, we obtain from (3.20) that

$$\underset{n\to \mathrm{\infty }}{lim}{G}_{1,r_n}{y}_n=x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{y}_n.$$

This implies that ${lim}_{n\to \mathrm{\infty }}\parallel G_{1,r_n}{y}_n-y_n\parallel =0$. But by Lemma 2.8,

$$\parallel G_{1,r_n}{y}_n-G_{1,r_0}{y}_n\parallel \le \frac{|r_n-r_0|}{r_n}(\parallel G_{1,r_n}{y}_n\parallel +\parallel y_n\parallel )\to 0\text{as }n\to \mathrm{\infty }.$$

Thus,

$$\underset{n\to \mathrm{\infty }}{lim}{G}_{1,r_0}{y}_n=\underset{n\to \mathrm{\infty }}{lim}{G}_{1,r_n}{y}_n=x^{\ast }.$$

So, the continuity of $G_{1,r_0}$ and the fact that $y_n\to x^{\ast }$ as $n\to \mathrm{\infty }$ give

$$x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{G}_{1,r_0}{y}_n=G_{1,r_0}{x}^{\ast }.$$

A similar argument gives

$$x^{\ast }=\underset{n\to \mathrm{\infty }}{lim}{G}_{h,r_0}{y}_n=G_{i,r_0}{x}^{\ast },h=2,3,\dots ,t.$$

Hence,

$$x^{\ast }\in \bigcap _{h=1}^{t}Fix(G_{h,r_0})=\bigcap _{h=1}^t\mathit{EP}(f_h).$$

Finally, we prove that $x^{\ast }=P_F(x_0)$. From $x_n=P_{C_n}(x_0)n\ge 0$, we obtain that

$$〈x_0-x_n,x_n-z〉\ge 0,\mathrm{\forall }z\in C_n.$$

Since $F\subset C_n$, we also have that

$$〈x_0-x_n,x_n-p〉\ge 0,\mathrm{\forall }p\in F.$$

(3.21)

So,

$$\begin{array}{rcl}0& \le & 〈x_0-x_n,x_n-p〉=〈x_0-x^{\ast }+x^{\ast }-x_n,x_n-x^{\ast }+x^{\ast }-p〉\\ =& 〈x_0-x^{\ast },x_n-x^{\ast }〉+〈x_0-x^{\ast },x^{\ast }-p〉\\ +〈x^{\ast }-x_n,x_n-x^{\ast }〉+〈x^{\ast }-x_n,x^{\ast }-p〉\\ \le & 〈x_0-x^{\ast }y,x^{\ast }-p〉+\parallel x_0-x^{\ast }\parallel \parallel x_n-x^{\ast }\parallel \\ +\parallel x_n-x^{\ast }\parallel \parallel x^{\ast }-p\parallel -{\parallel x_n-x^{\ast }\parallel }^2.\end{array}$$

(3.22)

Inequality (3.22) implies that

$$0\le 〈x_0-x^{\ast },x^{\ast }-p〉+(\parallel x_0-x^{\ast }\parallel +\parallel x^{\ast }-p\parallel )\parallel x_n-x^{\ast }\parallel .$$

(3.23)

By taking limit as $n\to \mathrm{\infty }$ in (3.23), we obtain that

$$〈x_0-x^{\ast },x^{\ast }-p〉\ge 0,\mathrm{\forall }p\in F.$$

Now, by Lemma 2.2 we have that $x^{\ast }=P_F(x_0)$. This completes the proof. □

Remark 3.3 We note that $x^{\ast }=P_F(x_0)$ makes sense since it could be easily shown that F is closed and convex. In fact, it is enough to show that the set of zeros of γ-inverse monotone mappings and a fixed point set of continuous quasi-nonexpansive mappings are convex sets. Closure of the two sets simply follows from the continuity of the mappings involved.

Remark 3.4 Several authors (see, e.g., [8, 31] and references therein) have studied the following problem: Let C be a closed convex nonempty subset of a real Hilbert space H with inner product $〈\cdot ,\cdot 〉$ and norm $\parallel \cdot \parallel$. Let $f:C\times C\to \mathbb{R}$ be a bifunction and $\mathrm{\Phi }:C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper extended real-valued function, where ℝ denotes the set of real numbers. Let $\mathrm{\Theta }:C\to H$ be a nonlinear monotone mapping. The generalized mixed equilibrium problem (abbreviated GMEP) for f, Φ and Θ is to find $u^{\ast }\in C$ such that

$$f(u^{\ast },y)+\mathrm{\Phi }(y)-\mathrm{\Phi }\left(u^{\ast }\right)+〈\mathrm{\Theta }{u}^{\ast },y-u^{\ast }〉\ge 0,\mathrm{\forall }y\in C.$$

(3.24)

The set of solutions for GMEP (3.24) is denoted by

$$\mathit{GMEP}(f,\mathrm{\Phi },\mathrm{\Theta })=\{u\in C:f(u,y)+\mathrm{\Phi }(y)-\mathrm{\Phi }(u)+〈\mathrm{\Theta }u,y-u〉\ge 0,\mathrm{\forall }y\in C\}.$$

These authors always claim that if $\mathrm{\Phi }\equiv 0\equiv \mathrm{\Theta }$ in (3.24), then (3.24) reduces to the classical equilibrium problem (abbreviated EP), that is, the problem of finding $u^{\ast }\in C$ such that

$$f(u^{\ast },y)\ge 0,\mathrm{\forall }y\in C$$

(3.25)

and $\mathit{GMEP}(f,0,0)$ is denoted by $\mathit{EP}(f)$, where

$$\mathit{EP}(f)=\{u\in C:f(u,y)\ge 0,\mathrm{\forall }y\in C\}.$$

If $f\equiv 0\equiv \mathrm{\Phi }$ in (3.24), then GMEP (1.6) reduces to the classical variational inequality problem and $\mathit{GMEP}(0,0,\mathrm{\Theta })$ is denoted by $\mathit{VI}(\mathrm{\Theta },C)$, where

$$\mathit{VI}(\mathrm{\Theta },C)=\{u\in C:〈\mathrm{\Theta }u,y-u〉\ge 0,\mathrm{\forall }y\in C\}.$$

If $f\equiv 0\equiv \mathrm{\Theta }$, then GMEP (3.24) reduces to the following minimization problem:

$$\text{find }{u}^{\ast }\in C\text{ such that }\mathrm{\Phi }(y)\ge \mathrm{\Phi }\left(u^{\ast }\right),\mathrm{\forall }y\in C;$$

and $\mathit{GMEP}(0,\mathrm{\Phi },0)$ is denoted by $Argmin(\mathrm{\Phi })$, where

$$Argmin(\mathrm{\Phi })=\{u\in C:\mathrm{\Phi }(u)\le \mathrm{\Phi }(y),\mathrm{\forall }y\in C\}.$$

If $\mathrm{\Theta }\equiv 0$, then (3.24) becomes the mixed equilibrium problem (abbreviated MEP) and $\mathit{GMEP}(f,\mathrm{\Phi },0)$ is denoted by $\mathit{MEP}(f,\mathrm{\Phi })$, where

$$\mathit{MEP}(f,\mathrm{\Phi })=\{u\in C:f(u,y)+\mathrm{\Phi }(y)-\mathrm{\Phi }(u)\ge 0,\mathrm{\forall }y\in C\}.$$

If $\mathrm{\Phi }\equiv 0$, then (1.6) reduces to the generalized equilibrium problem (abbreviated GEP) and $\mathit{GMEP}(f,0,\mathrm{\Theta })$ is denoted by $\mathit{GEP}(f,\mathrm{\Theta })$, where

$$\mathit{GEP}(f,\mathrm{\Theta })=\{u\in C:f(u,y)+〈\mathrm{\Theta }u,y-u〉\ge 0,\mathrm{\forall }y\in C\}.$$

If $f\equiv 0$, then GMEP (3.24) reduces to the generalized variational inequality problem (abbreviated GVIP) and $\mathit{GMEP}(0,\mathrm{\Phi },\mathrm{\Theta })$ is denoted by $\mathit{GVIP}(\mathrm{\Phi },\mathrm{\Theta },C)$, where

$$\mathit{GVIP}(\mathrm{\Phi },\mathrm{\Theta },C)=\{u\in K:\mathrm{\Phi }(y)-\mathrm{\Phi }(u)+〈\mathrm{\Theta }u,y-u〉\ge 0,\mathrm{\forall }y\in C\}.$$

It is worthy to note that if we define $\mathrm{\Gamma }:C\times C\to \mathbb{R}$ by

$$\mathrm{\Gamma }(x,y)=f(x,y)+\mathrm{\Phi }(y)-\mathrm{\Phi }(x)+〈\mathrm{\Theta }x,y-x〉,$$

then it could be easily checked that Γ is a bifunction and satisfies properties (A1)-(A4). Thus, the so-called generalized mixed equilibrium problem reduces to the classical equilibrium problem for the bifunction Γ. Thus, consideration of the so-called generalized mixed equilibrium problem in place of the classical equilibrium problem studied in this paper leads to no further generalization.

4 Application (convex differentiable optimization)

In Section 1, we defined a Lipschitz continuous mapping and an inverse strongly monotone mapping. Inverse strongly monotone mappings arise in various areas of optimization and nonlinear analysis (see, for example, [32–38]). It follows from the Cauchy-Schwarz inequality that if a mapping $A:D(A)\subseteq H\to R(A)\subseteq H$ is $\frac{1}{L}$-inverse strongly monotone, then A is L-Lipschitz continuous. The converse of this statement, however, fails to be true. To see this, take for instance $A=-I$, where I is the identity mapping on H, then A is L-Lipschitz continuous (with $L=1$) but not $\frac{1}{L}$-inverse strongly monotone (that is, not firmly nonexpansive in this case).

Baillon and Haddad [39] showed in 1977 that if $D(A)=H$ and A is the gradient of a convex functional on H, then A is $\frac{1}{L}$-inverse strongly monotone if and only if A is L-Lipschitz continuous. This remarkable result, which has important applications in optimization theory (see, for example, [40–42]), has become known as the Baillon-Haddad theorem. In fact, we have the following theorem.

Theorem 4.1 (Baillon-Haddad) (see Corollary 10 of [39])

Let $\varphi :H\to \mathbb{R}$ be a convex Fréchet-differentiable functional on H such that ∇ϕ is L-Lipschitz continuous for some $L\in (0,+\mathrm{\infty })$, then ∇ϕ is a $\frac{1}{L}$-inverse strongly monotone mapping (where ∇ϕ denotes the gradient of the functional ϕ).

Now, let us turn to the problem of minimizing a continuously Fréchet-differentiable convex functional with minimum norm in Hilbert spaces.

Let K be a closed convex subset of a real Hilbert space H, consider the minimization problem given by

$$\underset{x\in K}{min}\varphi (x),$$

(4.1)

where ϕ is a Fréchet-differentiable convex functional. Let $\mathrm{\Omega }\subseteq K$, the solution set of (4.1), be nonempty. It is known that a point $z\in \mathrm{\Omega }$ if and only if the following optimality condition holds:

$$z\in K,〈\mathrm{\nabla }\varphi (z),x-z〉\ge 0,x\in K.$$

(4.2)

It is easy to see that if $K=H$, then optimality condition (4.2) is equivalent to $z\in \mathrm{\Omega }$ if and only if $z\in {(\mathrm{\nabla }\varphi )}^{-1}(0)$.

Thus, we obtain the following as a corollary of Theorem 3.2.

Theorem 4.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let $T_1,T_2,\dots ,T_m:C\to C$ be m continuous pseudocontractive mappings; let $S_1,S_2,\dots ,S_l:C\to C$ be l continuous quasi-nonexpansive mappings; let ${\varphi }_1,{\varphi }_2,\dots ,{\varphi }_d:H\to H$ be d convex and Fréchet-differentiable functionals on H such that ${(\mathrm{\nabla }\varphi )}_j$ is $L_j$-Lipschitz continuous for some $L_j\in (0,+\mathrm{\infty })$, $j=1,2,\dots ,d$; let $f_1,f_2,\dots ,f_t:C\times C\to \mathbb{R}$ be t bifunctions satisfying conditions (A1)-(A4). Let $F:={\bigcap }_{i=1}^{m}Fix(T_i)\cap {\bigcap }_{j=1}^d{(\mathrm{\nabla }{\varphi }_j)}^{-1}(0)\cap {\bigcap }_{k=1}^{l}Fix(S_k)\cap {\bigcap }_{h=1}^t\mathit{EP}(f_h)\ne \mathrm{\varnothing }$. Let ${\{x_n\}}_{n\ge 0}$ be a sequence defined by

$$\{\begin{array}{l}{x}_0\in C_0=C\mathit{\text{chosen arbitrarily}},\\ z_n=P_C(x_n-{\lambda }_n{(\mathrm{\nabla }\varphi )}_{n+1}{x}_n),\\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)S_{n+1}{z}_n,\\ w_n=\eta {\sum }_{i=1}^m{\beta }_i{F}_{i,r_n}{y}_n+(1-\eta ){\sum }_{h=1}^t{\xi }_h{G}_{h,r_n}{y}_n,\\ C_{n+1}=\{z\in C_n:\parallel w_n-z\parallel \le \parallel x_n-z\parallel \},\\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}(x_0),n\ge 0,\end{array}$$

where ${(\mathrm{\nabla }\varphi )}_n={(\mathrm{\nabla }\varphi )}_{n(modd)}$, $S_n=S_{n(modl)}$; $\{r_n\}\subset (0,\mathrm{\infty })$ such that ${lim}_{n\to \mathrm{\infty }}{r}_n=r_0>0$; ${\{{\alpha }_n\}}_{n\ge 1}$ a sequence in $(0,1)$ such that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n(1-{\alpha }_n)>0$; ${\{{\beta }_i\}}_{i=1}^m,{\{{\xi }_h\}}_{h=1}^t\subset (0,1)$ such that ${\sum }_{i=1}^m{\beta }_i=1={\sum }_{h=1}^t{\xi }_h$; $\eta \in (0,1)$ and $\{{\lambda }_n\}$ is a sequence in $[a,b]$ for some $a,b\in \mathbb{R}$ such that $0<a<b<\frac{2}{L}$, $L={max}_{1\le j\le d}\{L_j\}$. Then the sequence ${\{x_n\}}_{n\ge 0}$ converges strongly to the element of F nearest to $x_0$.

Proof Since, by our hypothesis, ${(\mathrm{\nabla }\varphi )}_j$ is $L_j$-Lipschitz continuous for some $L_j\in (0,+\mathrm{\infty })$, $j=1,2,\dots ,d$, we obtain from Theorem 4.1 that ${(\mathrm{\nabla }\varphi )}_j$ is $\frac{1}{L_j}$-inverse strongly monotone, $j=1,2,\dots ,d$; and since $L={max}_{1\le j\le d}\{L_j\}$, it is then easy to see that ${(\mathrm{\nabla }\varphi )}_j$ is $\frac{1}{L}$-inverse strongly monotone, $j=1,2,\dots ,d$. The rest, therefore, follows as in the proof of Theorem 3.2 with $\gamma =\frac{1}{L}$. This completes the proof. □