On solving split mixed equilibrium problems and fixed point problems of hybrid-type multivalued mappings in Hilbert spaces

Abstract

In this paper, we introduce and study iterative algorithms for solving split mixed equilibrium problems and fixed point problems of λ-hybrid multivalued mappings in real Hilbert spaces and prove that the proposed iterative algorithm converges weakly to a common solution of the considered problems. We also provide an example to illustrate the convergence behavior of the proposed iteration process.

Introduction

Let H be a real Hilbert space with inner product $\langle \cdot ,\cdot \rangle$ and induced norm $\parallel \cdot \parallel$. Let C be a nonempty closed convex subset of H, $\phi :C\to \mathbb{R}$ be a function, and $F:C\times C\to \mathbb{R}$ be a bifunction. The mixed equilibrium problem is to find $x\in C$ such that

$$F(x,y)+\phi (y)-\phi (x)\ge 0,\mathrm{\forall }y\in C.$$ 1.1

The solution set of mixed equilibrium problem is denoted by $\mathit{MEP}(F,\phi )$. In particular, if $\phi =0$, this problem reduces to the equilibrium problem, which is to find $x\in C$ such that $F(x,y)\ge 0,\mathrm{\forall }y\in C$. The solution set of equilibrium problem is denoted by EP(F).

The mixed equilibrium problem is very general in the sense that it includes, as special cases, optimization problems, variational inequality problems, minimization problems, fixed point problems, Nash equilibrium problems in noncooperative games, and others; see, e.g., [1–4].

In 1994, Censor and Elfving [5] firstly introduced the following split feasibility problem in finite-dimensional Hilbert spaces: Let $H_1$, $H_2$ be two Hilbert spaces and C, Q be nonempty closed convex subsets of $H_1$ and $H_2$, respectively, and let $A:H_1\to H_2$ be a bounded linear operator. The split feasibility problem is formulated as finding a point $x^{\ast }$ with the property

$$x^{\ast }\in C\text{and}A{x}^{\ast }\in Q.$$

The split feasibility problem can extensively be applied in fields such as intensity-modulated radiation therapy, signal processing and image reconstruction, then the split feasibility problem has received so much attention by so many scholars; see [6–9].

In 2013, Kazmi and Rizvi [10] introduced and studied the following split equilibrium problem: let $C\subseteq H_1$ and $Q\subseteq H_2$. Let $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ be nonlinear bifunctions and let $A:H_1\to H_2$ be a bounded linear operator. The split equilibrium problem is to find $x^{\ast }\in C$ such that

$$F_1(x^{\ast },x)\ge 0,\mathrm{\forall }x\in C\text{ and such that }{y}^{\ast }=A{x}^{\ast }\in Q\text{ solves }{F}_2(y^{\ast },y)\ge 0,\mathrm{\forall }y\in Q.$$ 1.2

The solution set of the split equilibrium problem is denoted by

$$\mathit{SEP}(F_1,F_2):=\{x^{\ast }\in C:x^{\ast }\in \mathit{EP}(F_1)\text{ and }A{x}^{\ast }\in \mathit{EP}(F_2)\}.$$

The authors gave an iterative algorithm to find the common element of sets of solution of the split equilibrium problem and hierarchical fixed point problem; for more details refer to [11, 12].

In 2016, Suantai et al. [13] proposed the iterative algorithm to solve the problems for finding a common elements the set of solution of the split equilibrium problem and the fixed point of a nonspreading multivalued mapping in Hilbert space, given sequence $\{x_n\}$ by

$$\{\begin{array}{c}{x}_1\in C\text{arbitrarily},\hfill \\ u_n=T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n,\hfill \\ x_{n+1}\in {\alpha }_n{x}_n+(1-{\alpha }_n)S{u}_n,\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$ 1.3

where $\{{\alpha }_n\}\subset (0,1)$, $r_n\subset (0,\mathrm{\infty })$ and $\gamma \in (0,\frac{1}{L})$ such that L is the spectral radius of $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A, $C\subset H_1$, $Q\subset H_2$, $S:C\to K(C)$ is a $\frac{1}{2}$-nonspreading multivalued mapping, $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ are two bifunctions. The authors showed that under certain conditions, the sequence $\{x_n\}$ converges weakly to an element of $F(S)\cap \mathit{SEP}(F_1,F_2)$.

Several iterative algorithms have been developed for solving split feasibility problems and related split equilibrium problems; see, e.g., [14–16].

Motivated and inspired by the above results and related literature, we propose an iterative algorithm for finding a common element of the set of solutions of split mixed equilibrium problems and the set of fixed points of λ-hybrid multivalued mappings in real Hilbert spaces. Then we prove some weak convergence theorems which extend and improve the corresponding results of Kazmi and Rizvi [10] and Suantai et al. [13] and many others. We finally provide numerical examples for supporting our main result.

Preliminaries

Let C be a nonempty closed convex subset of a real Hilbert space H. We denote the strong convergence and the weak convergence of the sequence $\{x_n\}$ to a point $x\in H$ by $x_n⟶x$ and $x_n\rightharpoonup x$, respectively. It is also well known [17] that a Hilbert space H satisfies Opial’s condition, that is, for any sequence $\{x_n\}$ with $x_n\rightharpoonup x$, the inequality

$$\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-x\parallel <\underset{n⟶\mathrm{\infty }}{lim\hspace{0.17em}sup}\parallel x_n-y\parallel$$

holds for every $y\in H$ with $y\ne x$.

The following two lemmas are useful for our main results.

Lemma 2.1

In a real Hilbert space H, the following inequalities hold:

1. ${\parallel x-y\parallel }^2\le {\parallel x\parallel }^2-{\parallel y\parallel }^2-2\langle x-y,y\rangle ,\mathrm{\forall }x,y\in H$;

2. ${\parallel x+y\parallel }^2\le {\parallel x\parallel }^2+2\langle y,x+y\rangle ,\mathrm{\forall }x,y\in H$;

3. ${\parallel tx+(1-t)y\parallel }^2=t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t){\parallel x-y\parallel }^2,\mathrm{\forall }t\in [0,1],\mathrm{\forall }x,y\in H$;

4. If $\{x_n\}$ is a sequence in H which converges weakly to $z\in H$, then

   $$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-y\parallel }^2=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{\parallel x_n-z\parallel }^2+{\parallel z-y\parallel }^2,\mathrm{\forall }y\in H.$$

Lemma 2.2

[18]

Let H be a Hilbert space and $\{x_n\}$ be a sequence in H. Let $u,v\in H$ be such that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-u\parallel$ and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-v\parallel$ exist. If $\{x_{n_k}\}$ and $\{x_{m_k}\}$ are subsequences of $\{x_n\}$ which converge weakly to u and v, respectively, then $u=v$.

A single-valued mapping $T:C⟶H$ is called δ-inverse strongly monotone [19] if there exists a positive real number δ such that

$$\langle x-y,Tx-Ty\rangle \ge \delta {\parallel Tx-Ty\parallel }^2,\mathrm{\forall }x,y\in C.$$

For each $\gamma \in (0,2\delta ]$, we see that $I-\gamma T$ is a nonexpansive single-valued mapping, that is,

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

We denote by $CB(C)$ and $K(C)$ the collection of all nonempty closed bounded subsets and nonempty compact subsets of C, respectively. The Hausdorff metric $\mathcal{H}$ on $CB(C)$ is defined by

$$\mathcal{H}(A,B):=max\{\underset{x\in A}{sup}dist(x,B),\underset{y\in B}{sup}dist(y,A)\},\mathrm{\forall }A,B\in CB(C),$$

where $dist(x,B)=inf\{d(x,y):y\in B\}$ is the distance from a point x to a subset B. Let $S:C\to CB(C)$ be a multivalued mapping. An element $x\in C$ is called a fixed point of S if $x\in Sx$. The set of all fixed points of S is denoted by $F(S)$, that is, $F(S)=\{x\in C:x\in Sx\}$. Recall that a multivalued mapping $S:C\to CB(C)$ is called

• (i)
  nonexpansive if

  $$\mathcal{H}(Sx,Sy)\le \parallel x-y\parallel ,x,y\in C;$$
• (ii)
  quasi-nonexpansive if $F(S)\ne \mathrm{\varnothing }$ and

  $$\mathcal{H}(Sx,Sp)\le \parallel x-p\parallel ,\mathrm{\forall }x\in C,\mathrm{\forall }p\in F(S);$$
• (iii)
  nonspreading [13] if

  $$2\mathcal{H}{(Sx,Sy)}^2\le dist{(y,Sx)}^2+dist{(x,Sy)}^2,\mathrm{\forall }x,y\in C;$$
• (iv)
  λ-hybrid [20] if there exists $\lambda \in \mathbb{R}$ such that

  $$(1+\lambda )\mathcal{H}{(Sx,Sp)}^2\le (1-\lambda ){\parallel x-y\parallel }^2+\lambda dist{(y,Sx)}^2+\lambda dist{(x,Sy)}^2,\mathrm{\forall }x,y\in C.$$

We note that 0-hybrid is nonexpansive, 1-hybrid is nonspreading, and if S is λ-hybrid with $F(S)\ne \mathrm{\varnothing }$, then S is quasi-nonexpansive. It is well known [20] that if S is λ-hybrid, then $F(S)$ is closed. In addition, if S satisfies the condition: $Sp=\{p\}$ for all $p\in F(S)$, then $F(S)$ is also convex.

The following result is a demiclosedness principle for λ-hybrid multivalued mapping in a real Hilbert space.

Lemma 2.3

[20]

Let C be a nonempty closed convex subset of a real Hilbert space H and $S:C\to K(C)$ be a λ-hybrid multivalued mapping. If $\{x_n\}$ is a sequence in C such that $x_n\rightharpoonup x$ and $y_n\in S{x}_n$ with $x_n-y_n\to 0$, then $x\in Sx$.

For solving the mixed equilibrium problem, we assume that the bifunction $F_1:C\times C\to \mathbb{R}$ satisfies the following assumption:

Assumption 2.4

Let C be a nonempty closed and convex subset of a Hilbert space $H_1$. Let $F_1:C\times C\to \mathbb{R}$ be the bifunction, $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ is convex and lower semicontinuous satisfies the following conditions:

1. $F_1(x,x)=0$ for all $x\in C$;

2. $F_1$ is monotone, i.e., $F_1(x,y)+F_1(y,x)\le 0,\mathrm{\forall }x,y\in C$;

3. for each $x,y,z\in C$, ${lim}_{t\downarrow 0}{F}_1(tz+(1-t)x,y)\le F_1(x,y)$;

4. for each $x\in C$, $y\mapsto F_1(x,y)$ is convex and lower semicontinuous;

5. for each $x\in H_1$ and fixed $r>0$, there exist a bounded subset $D_x\subseteq C$ and $y_x\in C$ such that, for any $z\in C\setminus D_x$,

   $$F_1(z,y_x)+\phi (y_x)-\phi (z)+\frac{1}{r}\langle y_x-z,z-x\rangle <0;$$

6. C is a bounded set.

Lemma 2.5

[21]

Let C be a nonempty closed and convex subset of a Hilbert space $H_1$. Let $F_1:C\times C\to \mathbb{R}$ be a bifunction satisfies Assumption  2.4 and let $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper lower semicontinuous and convex function such that $C\cap dom\phi \ne \mathrm{\varnothing }$. For $r>0$ and $x\in H_1$. Define a mapping $T_r^{F_1}:H_1\to C$ as follows:

$$T_r^{F_1}(x)=\{z\in C:F_1(z,y)+\phi (y)-\phi (z)+\frac{1}{r}\langle y-z,z-x\rangle \ge 0,\mathrm{\forall }y\in C\},$$

for all $x\in H_1$. Assume that either (B1) or (B2) holds. Then the following conclusions hold:

1. for each $x\in H_1$, $T_r^{F_1}\ne \mathrm{\varnothing }$;

2. $T_r^{F_1}$ is single-valued;

3. $T_r^{F_1}$ is firmly nonexpansive, i.e., for any $x,y\in H_1$,

   $${\parallel T_r^{F_1}x-T_r^{F_1}y\parallel }^2\le \langle T_r^{F_1}x-T_r^{F_1}y,x-y\rangle ;$$

4. $F(T_r^{F_1})=\mathit{MEP}(F_1,\phi )$;

5. $\mathit{MEP}(F_1,\phi )$ is closed and convex.

Further, assume that $F_2:Q\times Q\to \mathbb{R}$ satisfying Assumption  2.4 and $\varphi :Q\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ is a proper lower semicontinuous and convex function such that $Q\cap dom\varphi \ne \mathrm{\varnothing }$, where Q is a nonempty closed and convex subset of a Hilbert space $H_2$. For each $s>0$ and $w\in H_2$, define a mapping $T_s^{F_2}:H_2\to Q$ as follows:

$$T_s^{F_2}(v)=\{w\in Q:F_2(w,d)+\varphi (d)-\varphi (w)+\frac{1}{r}\langle d-w,w-v\rangle \ge 0,\mathrm{\forall }d\in Q\}.$$

Then we have the following:

• (6) for each $v\in H_2$, $T_s^{F_2}\ne \mathrm{\varnothing }$;
• (7) $T_s^{F_2}$ is single-valued;
• (8) $T_s^{F_2}$ is firmly nonexpansive;
• (9) $F(T_s^{F_2})=\mathit{MEP}(F_2,\varphi )$;
• (10) $\mathit{MEP}(F_2,\varphi )$ is closed and convex.

Main results

In this section, we prove the weak convergence theorems for finding a common element of the set of solutions of split mixed equilibrium problems and the set of fixed points of λ-hybrid multivalued mappings in real Hilbert spaces and give a numerical example to support our main result.

We introduce the definition of split mixed equilibrium problems in real Hilbert spaces as follows.

Definition 3.1

Let C be a nonempty closed convex subset of a real Hilbert space $H_1$ and Q be a nonempty closed convex subset of a real Hilbert space $H_2$. Let $F_1:C\times C\to \mathbb{R}$ and $F_2:Q\times Q\to \mathbb{R}$ be nonlinear bifunctions, let $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ and $\varphi :Q\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be proper lower semicontinuous and convex functions such that $C\cap dom\phi \ne \mathrm{\varnothing }$ and $Q\cap dom\varphi \ne \mathrm{\varnothing }$, and let $A:H_1\to H_2$ be a bounded linear operator. The split mixed equilibrium problem is to find $x^{\ast }\in C$ such that

$$F_1(x^{\ast },x)+\phi (x)-\phi \left(x^{\ast }\right)\ge 0,\mathrm{\forall }x\in C,$$ 3.1

and such that

$$y^{\ast }=A{x}^{\ast }\in Q\text{solves}{F}_2(y^{\ast },y)+\varphi (y)-\varphi \left(y^{\ast }\right)\ge 0,\mathrm{\forall }y\in Q.$$ 3.2

The solution set of the split mixed equilibrium problem (3.1) and (3.2) is denoted by

$$\mathit{SMEP}(F_1,\phi ,F_2,\varphi ):=\{x^{\ast }\in C:x^{\ast }\in \mathit{MEP}(F_1,\phi )\text{ and }A{x}^{\ast }\in \mathit{MEP}(F_2,\varphi )\}.$$

We now get our main result.

Theorem 3.2

Let C be a nonempty closed convex subset of a real Hilbert space $H_1$ and Q be a nonempty closed convex subset of a real Hilbert space $H_2$. Let $A:H_1\to H_2$ be a bounded linear operator and $S:C\to K(C)$ a λ-hybrid multivalued mapping. Let $F_1:C\times C\to \mathbb{R}$, $F_2:Q\times Q\to \mathbb{R}$ be bifunctions satisfying Assumption  2.4, let $\phi :C\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ and $\varphi :Q\to \mathbb{R}\cup \{+\mathrm{\infty }\}$ be a proper lower semicontinuous and convex functions such that $C\cap dom\phi \ne \mathrm{\varnothing }$ and $Q\cap dom\varphi \ne \mathrm{\varnothing }$, respectively, and $F_2$ is upper semicontinuous in the first argument. Assume that $\mathrm{\Theta }=F(S)\cap \mathit{SMEP}(F_1,\phi ,F_2,\varphi )\ne \mathrm{\varnothing }$ and $Sp=\{p\}$ for all $p\in F(S)$. Let $\{x_n\}$ be a sequence generated by $x_1\in C$ and

$$\{\begin{array}{c}{u}_n=T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)w_n,w_n\in S{u}_n,\hfill \\ x_{n+1}={\beta }_n{w}_n+(1-{\beta }_n)z_n,z_n\in S{y}_n,\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$ 3.3

where $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, $\{r_n\}\subset (0,\mathrm{\infty })$, and $\gamma \in (0,\frac{1}{L})$ such that L is the spectral radius of $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A. Assume that the following conditions hold:

1. $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

2. $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

3. $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n$.

Then the sequence $\{x_n\}$ generated by (3.3) converges weakly to $p\in \mathrm{\Theta }$.

Proof

First, we show that $A^{\ast }(I-T_{r_n}^{F_2})A$ is a $\frac{1}{L}$-inverse strongly monotone mapping. Since $T_{r_n}^{F_2}$ is firmly nonexpansive and $I-T_{r_n}^{F_2}$ is 1-inverse strongly monotone, we see that

$$\begin{array}{rl}{\parallel A^{\ast }(I-T_{r_n}^{F_2})Ax-A^{\ast }(I-T_{r_n}^{F_2})Ay\parallel }^2& =\langle A^{\ast }(I-T_{r_n}^{F_2})(Ax-Ay),A^{\ast }(I-T_{r_n}^{F_2})(Ax-Ay)\rangle \\ & =\langle (I-T_{r_n}^{F_2})(Ax-Ay),A{A}^{\ast }(I-T_{r_n}^{F_2})(Ax-Ay)\rangle \\ & \le L\langle (I-T_{r_n}^{F_2})(Ax-Ay),(I-T_{r_n}^{F_2})(Ax-Ay)\rangle \\ & =L{\parallel (I-T_{r_n}^{F_2})(Ax-Ay)\parallel }^2\\ & \le L\langle Ax-Ay,(I-T_{r_n}^{F_2})(Ax-Ay)\rangle \\ & =L\langle x-y,A^{\ast }(I-T_{r_n}^{F_2})Ax-A^{\ast }(I-T_{r_n}^{F_2})Ay\rangle \end{array}$$

for all $x,y\in H_1$. This implies that $A^{\ast }(I-T_{r_n}^{F_2})A$ is a $\frac{1}{L}$-inverse strongly monotone mapping. Since $\gamma \in (0,\frac{1}{L})$, it follows that $I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A$ is a nonexpansive mapping.

Now, we divide the proof into five steps as follows:

Step 1. Show that $\{x_n\}$ is bounded.

Let $q\in \mathrm{\Theta }$. Then we have $q=T_{r_n}^{F_1}q$ and $q=(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)q$. By nonexpansiveness of $I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A$, it implies that

$$\begin{array}{rl}\parallel u_n-q\parallel & =\parallel T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)q\parallel \\ & \le \parallel (I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)q\parallel \\ & \le \parallel x_n-q\parallel .\end{array}$$ 3.4

This implies that

$$\parallel w_n-q\parallel =dist(w_n,Sq)\le H(S{u}_n,Sq)\le \parallel u_n-q\parallel \le \parallel x_n-q\parallel ,$$ 3.5

and so

$$\begin{array}{rl}\parallel y_n-q\parallel & =\parallel {\alpha }_n{x}_n+(1-{\alpha }_n)w_n-q\parallel \\ & \le {\alpha }_n\parallel x_n-q\parallel +(1-{\alpha }_n)\parallel w_n-q\parallel \\ & =\parallel x_n-q\parallel .\end{array}$$ 3.6

It follows that

$$\parallel z_n-q\parallel =dist(z_n,Sq)\le H(S{y}_n,Sq)\le \parallel y_n-q\parallel \le \parallel x_n-q\parallel .$$ 3.7

By (3.5) and (3.7), we have

$$\begin{array}{rl}\parallel x_{n+1}-q\parallel & =\parallel {\beta }_n{w}_n+(1-{\beta }_n)z_n-q\parallel \\ & \le {\beta }_n\parallel w_n-q\parallel +(1-{\beta }_n)\parallel z_n-q\parallel \\ & =\parallel x_n-q\parallel .\end{array}$$ 3.8

This implies that $\{\parallel x_n-q\parallel \}$ is decreasing and bounded below, thus ${lim}_{n⟶\mathrm{\infty }}\parallel x_n-q\parallel$ exists for all $q\in \mathrm{\Theta }$.

Step 2. Show that ${lim}_{n\to \mathrm{\infty }}\parallel w_n-z_n\parallel =0$.

From Lemma 2.1(3), (3.5), (3.7), and $Sq=\{q\}$, we have

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2& ={\parallel {\beta }_n{w}_n+(1-{\beta }_n)z_n-q\parallel }^2\\ & \le {\beta }_n{\parallel w_n-q\parallel }^2+(1-{\beta }_n){\parallel z_n-q\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel w_n-z_n\parallel }^2\\ & \le {\parallel x_n-q\parallel }^2-{\beta }_n(1-{\beta }_n){\parallel w_n-z_n\parallel }^2.\end{array}$$ 3.9

This implies that

$${\beta }_n(1-{\beta }_n){\parallel w_n-z_n\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2.$$

From Condition (C1) and ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ exists, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel w_n-z_n\parallel =0.$$ 3.10

Step 3. Show that ${lim}_{n\to \mathrm{\infty }}\parallel u_n-x_n\parallel =0$ and ${lim}_{n\to \mathrm{\infty }}\parallel w_n-u_n\parallel =0$.

For $q\in \mathrm{\Theta }$, we see that

$$\begin{array}{rl}{\parallel u_n-q\parallel }^2=& {\parallel T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-T_{r_n}^{F_1}q\parallel }^2\\ \le & {\parallel (I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-q\parallel }^2\\ \le & {\parallel x_n-q\parallel }^2+{\gamma }^2{\parallel A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel }^2+2\gamma \langle q-x_n,A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \\ \le & {\parallel x_n-q\parallel }^2+{\gamma }^2\langle A{x}_n-T_{r_n}^{F_2}A{x}_n,A{A}^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \\ & +2\gamma \langle A(q-x_n),A{x}_n-T_{r_n}^{F_2}A{x}_n\rangle \\ \le & {\parallel x_n-q\parallel }^2+L{\gamma }^2\langle A{x}_n-T_{r_n}^{F_2}A{x}_n,A{x}_n-T_{r_n}^{F_2}A{x}_n\rangle \\ & +2\gamma \langle A(q-x_n)+(A{x}_n-T_{r_n}^{F_2}A{x}_n)\\ & -(A{x}_n-T_{r_n}^{F_2}A{x}_n),A{x}_n-T_{r_n}^{F_2}A{x}_n\rangle \\ \le & {\parallel x_n-q\parallel }^2+L{\gamma }^2{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2\\ & +2\gamma (\langle Ap-T_{r_n}^{F_2}A{x}_n,A{x}_n-T_{r_n}^{F_2}A{x}_n\rangle -{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2)\\ \le & {\parallel x_n-q\parallel }^2+L{\gamma }^2{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2\\ & +2\gamma (\frac{1}{2}{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2-{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2)\\ =& {\parallel x_n-q\parallel }^2+\gamma (L\gamma -1){\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2.\end{array}$$

Thus, by (3.5) and (3.7), we have

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2& \le {\beta }_n{\parallel w_n-q\parallel }^2+(1-{\beta }_n){\parallel z_n-q\parallel }^2\\ & \le {\beta }_n{\parallel u_n-q\parallel }^2+(1-{\beta }_n){\parallel x_n-q\parallel }^2\\ & \le {\beta }_n({\parallel x_n-q\parallel }^2+\gamma (L\gamma -1){\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2)+(1-{\beta }_n){\parallel x_n-q\parallel }^2\\ & \le {\parallel x_n-q\parallel }^2+\gamma (L\gamma -1){\beta }_n{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2.\end{array}$$ 3.11

Therefore, we have

$$-\gamma (L\gamma -1){\beta }_n{\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2.$$

Since $\gamma (L\gamma -1)<0$, it follows by Condition (C1) and the existence of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A{x}_n-T_{r_n}^{F_2}A{x}_n\parallel =0.$$ 3.12

Since $T_{r_n}^{F_1}$ is firmly nonexpansive and $I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A$ is nonexpansive, we have

$$\begin{array}{rl}{\parallel u_n-q\parallel }^2=& {\parallel T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-T_{r_n}^{F_1}q\parallel }^2\\ \le & \langle T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-T_{r_n}^{F_1}q,(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-q\rangle \\ =& \langle u_n-q,(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-q\rangle \\ =& \frac{1}{2}({\parallel u_n-q\parallel }^2+{\parallel (I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n-q\parallel }^2\\ & -{\parallel u_n-x_n-\gamma A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel }^2)\\ \le & \frac{1}{2}({\parallel u_n-q\parallel }^2+{\parallel x_n-q\parallel }^2-({\parallel u_n-x_n\parallel }^2+{\gamma }^2{\parallel A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel }^2\\ & -2\gamma \langle u_n-x_n,A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \left)\right),\end{array}$$

which implies that

$$\begin{array}{rl}{\parallel u_n-q\parallel }^2& \le {\parallel x_n-q\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \langle u_n-x_n,A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \\ & \le {\parallel x_n-q\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel u_n-x_n\parallel \parallel A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel .\end{array}$$ 3.13

This implies by (3.5) and (3.7) that

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & {\beta }_n{\parallel w_n-q\parallel }^2+(1-{\beta }_n){\parallel z_n-q\parallel }^2\\ \le & {\beta }_n{\parallel u_n-q\parallel }^2+(1-{\beta }_n){\parallel x_n-q\parallel }^2\\ \le & {\beta }_n({\parallel x_n-q\parallel }^2-{\parallel u_n-x_n\parallel }^2+2\gamma \parallel u_n-x_n\parallel \parallel A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel )\\ & +(1-{\beta }_n){\parallel x_n-q\parallel }^2.\end{array}$$

Therefore, we have

$$\begin{array}{rl}{\beta }_n{\parallel u_n-x_n\parallel }^2& \le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+2\gamma {\beta }_n\parallel u_n-x_n\parallel \parallel A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel \\ & \le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2+2\gamma {\beta }_{n}M\parallel A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\parallel ,\end{array}$$

where $M=sup\{\parallel u_n-x_n\parallel :n\in \mathbb{N}\}$. This implies by Condition (C1), (3.12), and the existence of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$ that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel u_n-x_n\parallel =0.$$ 3.14

From (3.5), (3.7), and the definition of $\{y_n\}$, we obtain

$$\begin{array}{rl}{\parallel x_{n+1}-q\parallel }^2\le & {\beta }_n{\parallel w_n-q\parallel }^2+(1-{\beta }_n){\parallel z_n-q\parallel }^2\\ \le & {\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n){\parallel y_n-q\parallel }^2\\ =& {\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n)({\alpha }_n{\parallel x_n-q\parallel }^2+(1-{\alpha }_n){\parallel w_n-q\parallel }^2\\ & -{\alpha }_n(1-{\alpha }_n){\parallel x_n-w_n\parallel }^2)\\ \le & {\beta }_n{\parallel x_n-q\parallel }^2+(1-{\beta }_n)({\parallel x_n-q\parallel }^2-{\alpha }_n(1-{\alpha }_n){\parallel x_n-w_n\parallel }^2)\\ =& {\parallel x_n-q\parallel }^2-{\alpha }_n(1-{\alpha }_n)(1-{\beta }_n){\parallel x_n-w_n\parallel }^2.\end{array}$$

This implies that

$${\alpha }_n(1-{\alpha }_n)(1-{\beta }_n){\parallel x_n-w_n\parallel }^2\le {\parallel x_n-q\parallel }^2-{\parallel x_{n+1}-q\parallel }^2.$$

From Conditions (C1), (C2), and the existence of ${lim}_{n\to \mathrm{\infty }}\parallel x_n-q\parallel$, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel w_n-x_n\parallel =0.$$ 3.15

By (3.14) and (3.15), we get

$$\parallel w_n-u_n\parallel \le \parallel w_n-x_n\parallel +\parallel x_n-u_n\parallel \to 0\text{as }n\to \mathrm{\infty }.$$ 3.16

Step 4. Show that ${\omega }_w(x_n)\subset \mathrm{\Theta }$, where ${\omega }_w(x_n)=\{x\in H_1:x_{n_i}\rightharpoonup x,\{x_{n_i}\}\subset \{x_n\}\}$. Since $\{x_n\}$ is bounded and $H_1$ is reflexive, ${\omega }_w(x_n)$ is nonempty. Let $p\in {\omega }_w(x_n)$ be an arbitrary element. Then there exists a subsequence $\{x_{n_i}\}\subset \{x_n\}$ converging weakly to p. From (3.14), it implies that $u_{n_i}\rightharpoonup p$ as $i\to \mathrm{\infty }$. By (3.16) and Lemma 2.3, we have $p\in F(S)$.

Next, we show that $p\in \mathit{MEP}(F_1,\phi )$. Since $u_n=T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n$, we have

$$F_1(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}\langle y-u_n,u_n-x_n-\gamma A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \ge 0,\mathrm{\forall }y\in C,$$

which implies that

$$F_1(u_n,y)+\phi (y)-\phi (u_n)+\frac{1}{r_n}\langle y-u_n,u_n-x_n\rangle -\frac{1}{r_n}\langle y-u_n,\gamma A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \ge 0,\mathrm{\forall }y\in C.$$

From Assumption 2.4(A2), we have

$$\begin{array}{rl}& \phi (y)-\phi (u_n)+\frac{1}{r_n}\langle y-u_n,u_n-x_n\rangle -\frac{1}{r_n}\langle y-u_n,\gamma A^{\ast }(I-T_{r_n}^{F_2})A{x}_n\rangle \\ & \ge -F_1(u_n,y)\ge F_1(y,u_n),\mathrm{\forall }y\in C,\end{array}$$

and hence

$$\begin{array}{rl}& \phi (y)-\phi (u_{n_i})+\frac{1}{r_{n_i}}\langle y-u_{n_i},u_{n_i}-x_{n_i}\rangle -\frac{1}{r_{n_i}}\langle y-u_{n_i},\gamma A^{\ast }(I-T_{r_{n_i}}^{F_2})A{x}_{n_i}\rangle \ge F_1(y,u_{n_i}),\\ & \mathrm{\forall }y\in C.\end{array}$$

This implies by $u_{n_i}\rightharpoonup p$, Condition (C3), (3.12), (3.14), Assumption 2.4(A2), and the proper lower semicontinuity of φ that

$$F_1(y,p)+\phi (p)-\phi (y)\le 0,\mathrm{\forall }y\in C.$$

Put $y_t=ty+(1-t)p$ for all $t\in (0,1]$ and $y\in C$. Consequently, we get $y_t\in C$ and hence $F_1(y_t,p)+\phi (p)-\phi (y_t)\le 0$. So, by Assumption 2.4(A1)-(A4), we have

$$\begin{array}{rl}0& =F_1(y_t,y_t)-\phi (y_t)+\phi (y_t)\\ & \le t{F}_1(y_t,y)+(1-t)F_1(y_t,p)+t\phi (y)+(1-t)\phi (p)-\phi (y_t)\\ & \le t(F_1(y_t,y)+\phi (y)-\phi (y_t)).\end{array}$$

Hence, we have

$$F_1(y_t,y)+\phi (y)-\phi (y_t)\ge 0,\mathrm{\forall }y\in C.$$

Letting $t\to 0$, by Assumption 2.4(A3) and the proper lower semicontinuity of φ, we have

$$F_1(p,y)+\phi (y)-\phi (p)\ge 0,\mathrm{\forall }y\in C.$$

This implies that $p\in \mathit{MEP}(F_1,\phi )$.

Since A is a bounded linear operator, we have $A{x}_{n_i}\rightharpoonup Ap$. Then it follows from (3.12) that

$$T_{r_{n_i}}^{F_2}A{x}_{n_i}\rightharpoonup Ap\text{as }i\to \mathrm{\infty }.$$ 3.17

By the definition of $T_{r_{n_i}}^{F_2}A{x}_{n_i}$, we have

$$F_2(T_{r_{n_i}}^{F_2}A{x}_{n_i},y)+\varphi (y)-\varphi \left(T_{r_{n_i}}^{F_2}A{x}_{n_i}\right)+\frac{1}{r_{n_i}}\langle y-T_{r_{n_i}}^{F_2}A{x}_{n_i},T_{r_{n_i}}^{F_2}A{x}_{n_i}-A{x}_{n_i}\rangle \ge 0,\mathrm{\forall }y\in Q.$$

Since $F_2$ is upper semicontinuous in the first argument, it implies by (3.17) that

$$F_2(Ap,y)+\varphi (y)-\varphi (Ap)\ge 0,\mathrm{\forall }y\in Q.$$

This shows that $Ap\in \mathit{MEP}(F_2,\varphi )$. Therefore, $p\in \mathit{SMEP}(F_1,\phi ,F_2,\varphi )$ and hence $p\in \mathrm{\Theta }$.

Step 5. Show that $\{x_n\}$ converges weakly to an element of Θ. It is sufficient to show that ${\omega }_w(x_n)$ is a singleton set. Let $p,q\in {\omega }_w(x_n)$ and $\{x_{n_k}\}$, $\{x_{n_m}\}$ be two subsequences of $\{x_n\}$ such that $x_{n_k}\rightharpoonup p$ and $x_{n_m}\rightharpoonup q$. From (3.14), we also have $u_{n_k}\rightharpoonup p$ and $u_{n_m}\rightharpoonup q$. By (3.16) and Lemma 2.3, we see that $p,q\in F(S)$. Applying Lemma 2.2, we obtain $p=q$. This completes the proof. □

If $\phi =\varphi =0$ in (3.1) and (3.2), then the split mixed equilibrium problem reduces to split equilibrium problem. So, the following result can be obtained from Theorem 3.2 immediately.

Theorem 3.3

Let C be a nonempty closed convex subset of a real Hilbert space $H_1$ and Q be a nonempty closed convex subset of a real Hilbert space $H_2$. Let $A:H_1\to H_2$ be a bounded linear operator and $S:C\to K(C)$ a λ-hybrid multivalued mapping. Let $F_1:C\times C\to \mathbb{R}$, $F_2:Q\times Q\to \mathbb{R}$ be bifunctions satisfying Assumption  2.4, and $F_2$ is upper semicontinuous in the first argument. Assume that $\mathrm{\Theta }=F(S)\cap \mathit{SEP}(F_1,F_2)\ne \mathrm{\varnothing }$ and $Sp=\{p\}$ for all $p\in F(S)$. Let $\{x_n\}$ be a sequence generated by $x_1\in C$ and

$$\{\begin{array}{c}{u}_n=T_{r_n}^{F_1}(I-\gamma A^{\ast }(I-T_{r_n}^{F_2})A)x_n,\hfill \\ y_n={\alpha }_n{x}_n+(1-{\alpha }_n)w_n,w_n\in S{u}_n,\hfill \\ x_{n+1}={\beta }_n{w}_n+(1-{\beta }_n)z_n,z_n\in S{y}_n,\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$ 3.18

where $\{{\alpha }_n\}\subset (0,1)$, $\{{\beta }_n\}\subset (0,1)$, $\{r_n\}\subset (0,\mathrm{\infty })$, and $\gamma \in (0,\frac{1}{L})$ such that L is the spectral radius of $A^{\ast }A$ and $A^{\ast }$ is the adjoint of A. Assume that the following conditions hold:

1. $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\beta }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\beta }_n<1$;

2. $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_n\le {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}{\alpha }_n<1$;

3. $0<{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{r}_n$.

Then the sequence $\{x_n\}$ generated by (3.18) converges weakly to $p\in \mathrm{\Theta }$.

Remark 3.4

• (i)Theorems 3.2 and 3.3 extend the corresponding one of Suantai et al. [13] and Kazmi and Rizvi [10] to λ-hybrid multivalued mapping and to a split mixed equilibrium problem. In fact, we present a new iterative algorithm for finding a common element of the set of solutions of split mixed equilibrium problems and the set of fixed points of λ-hybrid multivalued mappings in a real Hilbert space.
• (ii)It is well known that the class of λ-hybrid multivalued mappings contains the classes of nonexpansive multivalued mappings, nonspreading multivalued mappings. Thus, Theorems 3.2 and 3.3 can be applied to these classes of mappings.

We give an example to illustrate Theorem 3.2 as follows.

Example 3.5

Let $H_1=\mathbb{R}$, $H_2=\mathbb{R}$, $C=[-3,0]$, and $Q=(-\mathrm{\infty },0]$. Let $A:H_1⟶H_2$ defined by $Ax=\frac{x}{2}$ for each $x\in H_1$. Then $A^{\ast }y=\frac{y}{2}$ for each $y\in H_2$. So, $L=\frac{1}{2}$ is the spectral radius of $A^{\ast }A$. Define a multivalued mapping $S:C⟶K(C)$ by

$$Sx=\{\begin{array}{cc}[-\frac{|x|}{|x|+1},0],\hfill & x\in [-3,-2)\text{;}\hfill \\ \{0\},\hfill & x\in [-2,0]\text{.}\hfill \end{array}$$

It easy to see that S is 1-hybrid multivalued mapping with $F(S)=\{0\}$ and $S(0)=\{0\}$. For each $x,y\in C$, define the bifunction $F_1:C\times C⟶\mathbb{R}$ by $F_1(x,y)=xy+y-x-x^2$ and define $\phi (x)=0$ for each $x\in C$. For each $u,v\in Q$, define the bifunction $F_2:Q\times Q⟶\mathbb{R}$ by $F_2(u,v)=uv+10v-10u-u^2$ and define $\varphi (u)=0$ for each $u\in Q$.

Choose ${\alpha }_n=\frac{n}{5n+1}$, ${\beta }_n=\frac{n}{9n+1}$, $r_n=\frac{n}{n+1}$, and $\gamma =\frac{1}{15}$. It is easy to check that $F_1$, $F_2$, $\{{\alpha }_n\}$, $\{{\beta }_n\}$, $\{r_n\}$ satisfy all conditions in Theorem 3.2.

For each $x\in C$, we compute $T_r^{F_2}Ax$. Find z such that

$$\begin{array}{rl}0& \le F_2(z,y)+\phi (y)-\phi (z)+\frac{1}{r}\langle y-z,z-Ax\rangle \\ & =zy+10y-10z-z^2+\frac{1}{r}\langle y-z,z-\frac{x}{2}\rangle \\ & =(z+10)(y-z)+\frac{1}{r}(y-z)(z-\frac{x}{2})\\ & =(y-z)((z+10)+\frac{1}{r}(z-\frac{x}{2}))\end{array}$$

for all $y\in Q$. Thus, by Lemma 2.5(2), it follows that $z=\frac{x-20r}{2(1+r)}$. That is, $T_r^{F_2}Ax=\frac{x-20r}{2(1+r)}$ for each $x\in C$. Furthermore, we get

$$\begin{array}{rl}(I-\gamma A^{\ast }(I-T_r^{F_2})A)x& =x-\frac{1}{15}{A}^{\ast }(Ax-T_r^{F_2}Ax)\\ & =x-\frac{1}{15}{A}^{\ast }(\frac{x}{2}-\frac{x-20r}{2(1+r)})\\ & =x-\frac{1}{15}(\frac{x}{4}-\frac{x-20r}{4(1+r)})\\ & =x(1-\frac{\gamma }{60})-\frac{\gamma (x-20r)}{60(1+r)}.\end{array}$$

Next, we find $u\in C$ such that $F_1(u,v)+\phi (y)-\phi (z)+\frac{1}{r}\langle v-u,u-s\rangle \ge 0$ for all $v\in C$, where $s=(I-\gamma A^{\ast }(I-T_r^{F_2})A)x$. Note that

$$\begin{array}{rl}0\le F_1(u,v)+\phi (y)-\phi (z)+\frac{1}{r}\langle v-u,u-s\rangle & =uv+v-u-u^2+\frac{1}{r}\langle v-u,u-s\rangle \\ & =(u+1)(v-u)+\frac{1}{r}(v-u)(u-s)\\ & =(v-u)((u+1)+\frac{1}{r}(u-s)).\end{array}$$

Thus, by Lemma 2.5(2), it follows that

$$u=\frac{s-r}{1+r}=\frac{59x-60r}{60(1+r)}-\frac{x-20r}{60{(1+r)}^2}.$$

Then the algorithm (3.3) becomes

$$\{\begin{array}{c}{u}_n=\frac{59x_n-60r_n}{60(1+r_n)}-\frac{x_n-20r_n}{60{(1+r_n)}^2},r_n=\frac{n}{n+1},\hfill \\ y_n=\frac{n}{5n+1}{x}_n+(1-\frac{n}{5n+1})w_n,\hfill \\ x_{n+1}=\frac{n}{9n+1}{w}_n+(1-\frac{n}{9n+1})z_n,\mathrm{\forall }n\in \mathbb{N},\hfill \end{array}$$ 3.19

where

$$w_n=\{\begin{array}{cc}[-\frac{|u_n|}{|u_n|+1},0],\hfill & u_n\in [-3,-2)\text{;}\hfill \\ \{0\},\hfill & u_n\in [-2,0]\text{,}\hfill \end{array}{z}_n=\{\begin{array}{cc}[-\frac{|y_n|}{|y_n|+1},0],\hfill & y_n\in [-3,-2)\text{;}\hfill \\ \{0\},\hfill & y_n\in [-2,0]\text{.}\hfill \end{array}$$

We choose $w_n=-\frac{|u_n|}{|u_n|+1}$ if $u_n\in [-3,-2)$ and $z_n=-\frac{|y_n|}{|y_n|+1}$ if $y_n\in [-3,-2)$. By using SciLab, we compute the iterates of (3.19) for the initial point $x_1=-3$. The numerical experiment’s results of our iteration for approximating the point 0 are given in Table 1.

Table 1.

Numerical results of Example 3.5 for the algorithm (3.19)

n	${\mathit{x}}_{\mathit{n}}$	$\mathbf{\parallel }{\mathit{x}}_{\mathit{n}}\mathbf{-}{\mathit{x}}_{\mathit{n}\mathbf{-}\mathbf{1}}\mathbf{\parallel }$
1	−3.0000000e + 00	-
2	−6.8786127e − 02	2.9312139e + 00
3	0.0000000e + 00	6.8786127e − 02
4	0.0000000e + 00	0.0000000e + 00

Conclusions

The results presented in this paper extend and generalize the work of Suantai et al. [13] and Kazmi and Rizvi [10]. The main aim of this paper is to propose an iterative algorithm to find an element for solving a class of split mixed equilibrium problems and fixed point problems for λ-hybrid multivalued mappings under weaker conditions. Some sufficient conditions for the weak convergence of such proposed algorithm are given. Also, in order to show the significance of the considered problem, some important applications are discussed.