On split generalised mixed equilibrium problems and fixed-point problems with no prior knowledge of operator norm

Abstract

The purpose of this paper is to introduce an iterative algorithm that does not require any knowledge of the operator norm for approximating a solution of a split generalised mixed equilibrium problem which is also a fixed point of a $\kappa$-strictly pseudocontractive mapping. Furthermore, a strong convergence theorem for approximating a common solution of a split generalised mixed equilibrium problem and a fixed-point problem for $\kappa$-strictly pseudocontractive mapping was stated and proved in the frame work of Hilbert spaces.

1 Introduction

Let H be a real Hilbert space and K a nonempty, closed and convex subset of H. A mapping $T:K\rightarrow K$ is said to be nonexpansive if

$$\begin{aligned} \Vert Tx-Ty\Vert \le \Vert x-y\Vert \quad \forall \quad x,y\in K, \end{aligned}$$

(1.1)

and $\kappa$ -strictly pseudocontractive in the sense of Browder and Petryshyn [2] if for $0\le \kappa <1$,

$$\begin{aligned} \Vert Tx-Ty\Vert ^{2} \le \Vert x-y\Vert ^{2}+\kappa \Vert (I-T)x-(I-T)y\Vert ^{2}\quad \forall \quad x,y\in K. \end{aligned}$$

(1.2)

In a Hilbert space H, we can show that (1.2) is equivalent to

$$\begin{aligned} \langle Tx-Ty,x-y\rangle \le \Vert x-y\Vert ^{2}-\dfrac{1-\kappa }{2}\Vert (I-T)x-(I-T)y\Vert ^{2}. \end{aligned}$$

(1.3)

A point $x \in K$ is called a fixed point of T if $Tx=x$. The set of fixed points of T is denoted by F(T). So a fixed-point problem for T is to find $x\in F(T)$. It is a common knowledge that if T is $\kappa$-strictly pseudocontractive and $F(T)\ne \emptyset$, then F(T) is closed and convex. See [2, 19, 20, 32] and references therein, for more details on strictly pseudocontractive mappings.

Let $g:C \times C \rightarrow \mathbb {R}$ be a bifunction, $\varphi :C\rightarrow \mathbb {R}\cup \{+\infty \}$ be a function and $B:C\rightarrow H$ be a nonlinear mapping. The Generalised mixed equilibrium problem is to find $u\in C$ such that

$$\begin{aligned} g(u, y)+\langle Bu,y-u\rangle +\varphi (y)-\varphi (u) \ge 0,~~\forall y \in C. \end{aligned}$$

(1.4)

Denote the set of solutions of the problem (1.4) by $GMEP(g,\varphi ,B).$ That is

$$\begin{aligned} GMEP(g,\varphi ,B)=\{u\in C:g(u, y)+\langle Bu,y-u\rangle +\varphi (y)-\varphi (u) \ge 0, \quad \forall y \in C\}.\nonumber \\ \end{aligned}$$

(1.5)

If $B=0,$ then the generalised mixed equilibrium problem (1.4) reduces to the following mixed equilibrium problem, find $u\in C$ such that

$$\begin{aligned} g(u, y) +\varphi (y)-\varphi (u) \ge 0, \quad \forall y \in C. \end{aligned}$$

(1.6)

If $\varphi =0,$ then the generalised mixed equilibrium problem (1.4) becomes the generalised equilibrium problem, find $u\in C$ such that

$$\begin{aligned} g(u, y)+\langle Bu,y-u\rangle \ge 0, \quad \forall y \in C. \end{aligned}$$

(1.7)

Again if $B=\varphi =0,$ then the generalised mixed equilibrium problem (1.4) becomes the equilibrium problem, find $u\in C$ such that

$$\begin{aligned} g(u, y) \ge 0, \quad \forall y \in C. \end{aligned}$$

(1.8)

Equilibrium problems and their generalisations are well known to have been important tools for solving problems arising in the fields of linear or nonlinear programming, variational inequalities, complementary problems, optimisation problems, fixed-point problems and have been widely applied to physics, structural analysis, management sciences and economics, etc. (see, for example [1, 5, 22, 23]). In solving equilibrium problem (1.8), the bifunction g is said to satisfy the following conditions:

• (A1) $g(x, x) = 0$ for all $x \in C$;

• (A2) g is monotone, i.e., $g(x, y) + g(y, x)\ge 0$ for all $x, y \in C$;

• (A3) for each $x, y \in C$, $\lim _{t\rightarrow 0}g(tz + (1 - t)x; y) \le g(x; y)$;

• (A4) for each $x \in C$; $y \mapsto g(x, y)$ is convex and lower semicontinuous. It is known (see [31]), that if g(x, y) satisfies $(A1)-(A4)$ then the function $F(x,y):=g(x,y)+\langle Bx,y-x\rangle +\varphi (y)-\varphi (x)$ satisfies $(A1)-(A4)$ and $GMEP(g,B, \varphi ,)$ is closed and convex. An interested reader may see [4, 11, 13,14,15,16,17,18, 21, 24,25,26,27, 29, 30] and the references there in for more information on equilibrium problem and its generalisations.

Let $H_{1}$ and $H_{2}$ be Hilbert spaces and C and Q nonempty, closed and convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $f_{1}:C \times C \rightarrow \mathbb {R}$, $f_{2}: Q\times Q \rightarrow \mathbb {R}$ be bifunctions, $\varphi _{1}:C\rightarrow \mathbb {R}\cup \{+\infty \}$, $\varphi _{2}:Q\rightarrow \mathbb {R}\cup \{+\infty \}$ be functions and $B_{1}:C\rightarrow H_{1}$, $B_{2}:Q\rightarrow H_{2}$ be nonlinear mappings. Let $A:H_{1} \rightarrow H_{2}$ be a bounded linear operator. Then the split generalised mixed equilibrium problem is to find $x^{*}\in C$ such that

$$\begin{aligned} f_{1}(x^{*}, x)+\langle B_{1}x^{*},x-x^{*}\rangle +\varphi _{1}(x)-\varphi _{1}(x^{*}) \ge 0, \quad \forall x \in C, \end{aligned}$$

(1.9)

and $y^{*}=Ax^{*}\in Q$ solves

$$\begin{aligned} f_{2}(y^{*}, y)+\langle B_{2}y^{*},y-y^{*}\rangle +\varphi _{2}(y)-\varphi _{2}(y^{*}) \ge 0, \quad \forall y \in Q. \end{aligned}$$

(1.10)

We shall denote the solution set of (1.9)–(1.10) by $\Omega =\{x^{*}\in GMEP(f_{1},B_{1},\varphi _{1}): Ax^{*}\in GMEP(f_{2},B_{2},\varphi _{2})\}.$ If $B_{1}=0$ and $B_{2}= 0$, then (1.9)–(1.10) reduces to the following split mixed equilibrium problem, find $x^{*}\in C$ such that

$$\begin{aligned} f_{1}(x^{*}, x) +\varphi _{1}(x)-\varphi _{1}(x^{*}) \ge 0, \quad \forall x \in C, \end{aligned}$$

(1.11)

and $y^{*}=Ax^{*}\in Q$ solves

$$\begin{aligned} f_{2}(y^{*}, y)+\varphi _{2}(y)-\varphi _{2}(y^{*}) \ge 0, \quad \forall y \in Q, \end{aligned}$$

(1.12)

with solution set $\Omega _{\varphi }=\{x^{*}\in MEP(f_{1},\varphi _{1}):Ax^{*}\in MEP(f_{2},\varphi _{2})\}.$ Again in (1.9)–(1.10) if $\varphi _{1}=\varphi _{2}=0$, we obtain the following split generalised equilibrium problem, find $x^{*}\in C$ such that

$$\begin{aligned} f_{1}(x^{*}, x)+\langle B_{1}x^{*},x-x^{*}\rangle \ge 0, \quad \forall x \in C, \end{aligned}$$

(1.13)

and $y^{*}=Ax^{*}\in Q$ solves

$$\begin{aligned} f_{2}(y^{*}, y)+\langle B_{2}y^{*},y-y^{*}\rangle \ge 0, \quad \forall y \in Q, \end{aligned}$$

(1.14)

with solution set $\Omega _{B}=\{x^{*}\in GEP(f_{1},B_{1}):Ax^{*}\in GEP(f_{2},B_{2})\}.$ Moreover, if $B_{1}=B_{2}$ and $\varphi _{1}=\varphi _{2}=0,$ we have the split equilibrium problem, find $x^{*}\in C$ such that

$$\begin{aligned} f_{1}(x^{*}, x) \ge 0, \quad \forall x \in C, \end{aligned}$$

(1.15)

and $y^{*}=Ax^{*}\in Q$ solves

$$\begin{aligned} f_{2}(y^{*}, y) \ge 0, \quad \forall y \in Q, \end{aligned}$$

(1.16)

with solution set $\Omega _{0}=\{x^{*}\in EP(f_{1}):Ax^{*}\in EP(f_{2})\}.$

Kazmi and Rizvi [12] studied the pair of equilibrium problems (1.15) and (1.16) called split equilibrium problem.

Recently, Bnouhachem [3] stated and proved the following strong convergence result.

Theorem 1.1

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, and let $C\subset H_{1}$ and $Q\subset H_{2}$ be nonempty closed and convex subset of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a bounded linear operator. Assume that $f_{1}:C\times C\rightarrow \mathbb {R}$ and $f_{2}:Q\times Q\rightarrow \mathbb {R}$ are bifunctions satisfying $A1-A4$ and $f_{2}$ is upper semicontinuous in the first argument. Let $S,T:C\rightarrow C$ be a nonexpansive mapping such that $\Omega _{0} \cap F(T)\ne \emptyset .$ Let $f:C\rightarrow C$ be a k-Lipschitzian mapping and $\eta$-strongly monotone and let $U:C\rightarrow C$ be $\tau$-Lipschitzian mapping. For a given arbitrary $x_{0}\in C$ , let the iterative sequence $\{x_{n}\},\{u_{n}\}$ and $\{y_{n}\}$ be generated by

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}u_{n}=T_{r_{n}}^{f_{1}}(x_{n}+\gamma A^{*}(T_{r_{n}}^{f_{2}}-I)Ax_{n})\\ &{}y_{n}=\beta _{n}Sx_{n}+(1-\beta _{n})u_{n}\\ &{}x_{n+1}=P_{C}[\alpha _{n}\rho U(x_{n})+(I-\alpha _{n}\mu f)(T(y_{n}))]~~\forall n\ge 0 \end{array} \right. \end{aligned}$$

(1.17)

where $\{r_{n}\}\subset (0,2\zeta )$ and $\gamma \in (0,\frac{1}{L})$, L is the spectral radius of the operator $A^{*}A$, and $A^{*}$ is the adjoint of A. Suppose the parameters satisfy $0<\mu <\Big (\dfrac{2\eta }{k^{2}}\Big )$, $0 \le \rho \eta <\nu$, where $\nu =1-\sqrt{1-\mu (2\eta -\mu k^{2})}$ and $\{\alpha _{n}\}$ and $\{\beta _{n}\}$ are sequences in (0, 1) satisfying the following conditions:

• (a) $\lim \nolimits _{n\rightarrow \infty }\alpha _{n}=0$ and $\sum \nolimits _{n=1}^{\infty }\alpha _{n}=\infty ,$

• (b) $\lim \nolimits _{n\rightarrow \infty }(\frac{\beta _{n}}{\alpha _{n}})=0,$

• rm (c) $\sum \nolimits _{n=1}^{\infty }|\alpha _{n-1}-\alpha _{n}|<\infty$ and $\sum \nolimits _{n=1}^{\infty }|\beta _{n-1}-\beta _{n}|<\infty$

• (d) $\liminf _{n\rightarrow \infty } r_{n}<\limsup _{n\rightarrow \infty } r_{n}<2\zeta$ and $\sum \nolimits _{n=1}^{\infty }|r_{n-1}-r_{n}|<\infty .$

Then $\{x_{n}\}$ converges strongly to $z\in \Omega _{0}\cap F(T).$

Fig. 1

Errors: Case 1, $\epsilon = 10^{-4}$ (top left; 0:010 s); Case 2, $\epsilon = 10^{-4}$ (top right; 0:011  s); Case 3, $\epsilon = 10^{-4}$ (bottom; 0:013  s)

This result of Bnouhachem and other related results in literature depend on the prior knowledge of the operator norm.

Hendrickx and Oleshevsky [10] Observed that when $p = \infty$ or $p = 1$ the p-matrix norm is the largest of the row/column sums, and thus may be easily computed exactly. When $p = 2$, this problem reduces to computing an eigenvalue of $A^TA$ and thus can be solved in polynomial time in n, $log\frac{1}{\epsilon }$ and the bit-size of the entries of A.

Hendrickx and Oleshevsky [10] further stated and proved the following theorem.

Theorem 1.2

(Hendrickx and Oleshevsky [10]). Fix a rational $p \in [1,\infty )$ with $p \ne 1, 2$. Unless $P = NP$, there is no algorithm which, given input $\epsilon$ and a matrix M with entries in $\{-1, 0, 1\}$, computes $||M||_p$ to relative accuracy $\epsilon$, in time which is polynomial in $\epsilon ^{-1}$ and the dimensions of the matrix.

The result Theorem 1.2 shows that sometimes it is very difficult if not impossible to calculate or even estimate the operator norm.

It is our intention here to introduce an iterative scheme which does not require any knowledge of the operator norm and obtain a strong convergence theorem for approximating solution of split generalised mixed equilibrium problem which also solves a fixed-point problem for $\kappa$-pseudocontractive mapping.

Precisely, we consider the following problem: find $x^{*}\in F(S)$ such that

$$\begin{aligned} f_{1}(x^{*}, x)+\langle B_{1}x^{*},x-x^{*}\rangle +\varphi _{1}(x)-\varphi _{1}(x^{*}) \ge 0, \quad \forall x \in C, \end{aligned}$$

(1.18)

and $y^{*}=Ax^{*}\in Q$ solves

$$\begin{aligned} f_{2}(y^{*}, y)+\langle B_{2}y^{*},y-y^{*}\rangle +\varphi _{2}(y)-\varphi _{2}(y^{*}) \ge 0, \quad \forall y \in Q, \end{aligned}$$

(1.19)

where S is a strictly pseudocontractive mapping on C.

Many interesting practical problems (see [8]), can be formulated as fixed-point problems. The importance of equilibrium problem cannot be over emphasised as several mathematical problems (see [1]), such as optimisation problem, saddle points problem, Nash equilibria problem in noncooperative games, convex differentiable optimisation problem, variational operator inequalities problem, complementarity problems and variational inequalities with multivalued mappings can be formulated as equilibrium problems. It is easily observed that if we let $H_{1}=H_{2}$, and S, A the identity operator, then this problem we are considering reduces to the generalised mixed equilibrium problem considered by Zhang [31], which in turn generalises equilibrium problems. Our problem also complements the work of He [9] and many other related results in the literature.

2 Preliminaries

We now state some important results that are vital to the proof of the main result.

Lemma 2.1

[6, 7]. Let H be a Hilbert space and $T:H\rightarrow H$ a nonexpansive mapping, then for all $x,y\in H$,

$$\begin{aligned} \langle (x-Tx)-(y-Ty),Ty-Tx\rangle \le \dfrac{1}{2}\Vert (Tx-x)-(Ty-y)\Vert ^{2}, \end{aligned}$$

(2.1)

and consequently if $y\in F(T)$ then

$$\begin{aligned} \langle x-Tx,Ty-Tx\rangle \le \dfrac{1}{2}\Vert Tx-x\Vert ^{2}. \end{aligned}$$

(2.2)

Lemma 2.2

Let H be a real Hilbert space. Then the following result holds

$$\begin{aligned} \Vert x+y\Vert ^{2}\le \Vert x\Vert ^{2}+2\langle y,x+y\rangle , \quad \forall ~ x,y\in H. \end{aligned}$$

(2.3)

Lemma 2.3

Let H be a Hilbert space, then $\forall x,y\in H$ and $\alpha \in (0,1)$ we have

$$\begin{aligned} \Vert \alpha x + (1-\alpha )y\Vert ^{2}=\alpha \Vert x\Vert ^{2}+(1-\alpha )\Vert y\Vert ^{2}-\alpha (1-\alpha )\Vert x-y\Vert ^{2}. \end{aligned}$$

(2.4)

Lemma 2.4

(Demiclosedness principle). Let K be a nonempty, closed and convex subset of a real Hilbert space H. Let $T:K\rightarrow K$ be $\kappa$-strictly pseudocontractive mapping. Then $I-T$ is demi-closed at 0, i.e., if $x_{n} \rightharpoonup x\in K$ and $x_{n}-Tx_{n}\rightarrow 0$, then $x=Tx$.

Lemma 2.5

[28]. Assume $\{a_{n}\}$ is a sequence of nonnegative real numbers such that

$$\begin{aligned} a_{n+1}\le (1-\gamma _{n})a_{n}+\gamma _{n}\delta _{n},~~n\ge 0, \end{aligned}$$

(2.5)

where $\{\gamma _{n}\}$ is a sequence in (0,1) and $\{\delta _{n}\}$ is a sequence in $\mathbb {R}$ such that

• (i) $\sum \nolimits _{n=0}^{\infty }\gamma _{n}=\infty ,$

• (ii) $\limsup _{n\rightarrow \infty }\delta _{n}\le 0$ or $\sum _{n=0}^{\infty }\vert \delta _{n}\gamma _{n}\vert <\infty .$

Then $\lim \nolimits _{n\rightarrow \infty }a_{n}=0.$

Lemma 2.6

([31]). Let C be nonempty closed convex subset of a Hilbert space H. Let $B : C \rightarrow H$ be a continuous and monotone mapping, $\varphi : C \rightarrow \mathbb {R}$ be a lower semicontinuous and convex function, and $f : C \times C \rightarrow \mathbb {R}$ be a bifunction that satisfies $(A1)-(A4)$. For $r > 0$ and $x\in H$; then there exists $u\in C$ such that

$$\begin{aligned} f(u; y) + \langle Bu; y - u\rangle + \varphi (y) -\varphi (u) +\frac{1}{r}\langle y-u,u-x\rangle \ge 0,~~\forall y \in C.\qquad \end{aligned}$$

(2.6)

Define a mapping $T_{r}^{f} : C \rightarrow C$ as follows:

$$\begin{aligned} T_{r}^{f}(x)= & {} \Big \{u\in C:f(u, y) + \langle Bu, y - u\rangle + \varphi (y) -\varphi (u) \nonumber \\&+\frac{1}{r}\langle y-u,u-x\rangle \ge 0, \quad \forall y \in C\Big \}. \end{aligned}$$

(2.7)

Then, the following conclusions hold:

• 1. $T_{r}^{f}$ is single-valued,

• 2. $T_{r}^{f}$ is firmly nonexpansive, i.e., for any $x, y \in H$; $||T_{r}^{f}(x) - T_{r}^{f}(y)||^{2}\le \langle T_{r}^{f}(x) - T_{r}^{f}(y),x-y\rangle$,

• 3. $F(T_{r}^{f}) = GMEP(f; B;\varphi )$,

• 4. $GMEP(F;B ;\varphi )$ is closed and convex.

3 Main results

Theorem 3.1

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, let $C\subset H_{1}$ and $Q\subset H_{2}$ be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a bounded linear operator and $A^*$ the adjoint of A. Let $f_{1}:C\times C\rightarrow \mathbb {R}$ and $f_{2}:Q\times Q\rightarrow \mathbb {R}$ be bifunctions satisfying conditions $(A1)-(A4)$ and $f_{2}$ is upper semicontinuous in first argument. Let $B_{1}:C\rightarrow H_{1}$ and $B_{2}:Q\rightarrow H_{2}$ be continuous and monotone mappings, $\varphi _{1}:C\rightarrow \mathbb {R}\cup +\infty$ and $\varphi _{2}:Q\rightarrow \mathbb {R}\cup +\infty$ be proper lower semicontinuous and convex function. Let $S:C\rightarrow C$ be a $\kappa$-strictly pseudocontraction, such that $\Omega \cap F(S)\ne \emptyset$. Let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0$; $\gamma _{n} \in (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=\gamma$, otherwise $(\gamma$ being any nonnegative real number). Then the sequence $\{w_{n}\},\{x_{n}\}$ and $\{y_{n}\}$ generated iteratively for an arbitrary $x_{0}\in C$ and a fixed $u\in C$ by

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=(1-\alpha _{n})x_{n}+\alpha _{n}u\\ &{}y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})\\ &{}x_{n+1}=(1-\beta _{n})y_{n}+\beta _{n}Sy_{n},~~\forall n\ge 0 \end{array} \right. \end{aligned}$$

(3.1)

converges strongly to a point $p\in \Omega \cap F(S),$ where $\{\alpha _{n}\}_{n=1}^{\infty }$ and $\{\beta _{n}\}_{n=1}^{\infty }$ are real sequences in (0, 1) satisfying the following conditions:

• (i) $\lim \nolimits _{n\rightarrow \infty }\alpha _{n}=0$, $\sum \nolimits _{n=1}^{\infty }\alpha _{n}=\infty$,

• (ii) $0<\liminf \beta _{n}\le \limsup \beta _{n} <1-\kappa .$

Proof

Let $p\in \Omega \cap F(S)$, then from (3.1) we have,

$$\begin{aligned} ||x_{n+1}-p||^{2}= & {} ||(1-\beta _{n})y_{n}+\beta _{n}Sy_{n}-p||^{2}\nonumber \\= & {} ||(1-\beta _{n})(y_{n}-p)+\beta _{n}(Sy_{n}-p)||^{2} \nonumber \\= & {} (1-\beta _{n})^{2}||y_{n}-p||^{2}+\beta _{n}^{2}||Sy_{n}-p||^{2} \nonumber \\&+2\beta _{n}(1-\beta _{n})\langle y_{n}-p,Sy_{n}-p\rangle \nonumber \\\le & {} (1-\beta _{n})^{2}||y_{n}-p||^{2}+\beta _{n}^{2}[||y_{n}-p||^{2}+\kappa ||y_{n}-Sy_{n}||^{2}] \nonumber \\&+2\beta _{n}(1-\beta _{n})[||y_{n}-p||^{2} \nonumber \\&-\frac{1-\kappa }{2}||y_{n}-Sy_{n}||^{2}] \nonumber \\= & {} (1-2\beta _{n}+\beta _{n}^{2})||y_{n}-p||^{2}+\beta _{n}^{2}[||y_{n}-p||^{2}+\kappa ||y_{n}-Sy_{n}||^{2}] \nonumber \\&+2\beta _{n}||y_{n}-p||^{2}-2\beta _{n}^{2}||y_{n}-p||^{2} \nonumber \\&-\beta _{n}(1-\beta _{n})(1-\kappa )||y_{n}-Sy_{n}||^{2} \nonumber \\= & {} ||y_{n}-p||^{2}+[\beta _{n}^{2}\kappa -\beta _{n}(1-\beta _{n})(1-\kappa )]||y_{n}-Sy_{n}||^{2} \nonumber \\= & {} ||y_{n}-p||^{2}+\beta _{n}[\kappa +\beta _{n}-1]||y_{n}-Sy_{n}||^{2} \nonumber \\\le & {} ||y_{n}-p||^{2}. \end{aligned}$$

(3.2)

Again from (3.1),

$$\begin{aligned} ||y_{n}-p||^{2}= & {} ||T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})-p||^{2} \nonumber \\\le & {} ||w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}-p||^{2} \nonumber \\= & {} ||w_{n}-p||^{2}+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2} \nonumber \\&+2\gamma _{n}\langle w_{n}-p,A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle , \end{aligned}$$

(3.3)

but from Lemma 2.2

$$\begin{aligned} 2\gamma _{n}\langle w_{n}-p,A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle= & {} 2\gamma _{n}\langle A(w_{n}-p) +(T_{r_{n}}^{f_{2}}-I)Aw_{n} \nonumber \\&-(T_{r_{n}}^{f_{2}}-I)Aw_{n},(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle \nonumber \\= & {} 2\gamma _{n}[\langle T_{r_{n}}^{f_{2}}Aw_{n}-Ap,(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle \nonumber \\&-||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2} \nonumber \\\le & {} 2\gamma _{n}\Big [\frac{1}{2}||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}\Big ] \nonumber \\= & {} -\gamma _{n}||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}. \end{aligned}$$

(3.4)

Therefore, from (3.3), (3.4) and the condition $\gamma _{n} \in (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon ),$ we obtain

$$\begin{aligned} ||y_{n}-p||^{2}\le & {} ||w_{n}-p||^{2}+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-\gamma _{n}||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2} \nonumber \\= & {} ||w_{n}-p||^{2}+\gamma _{n}[\gamma _{n} ||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2} \nonumber \\\le & {} ||w_{n}-p||^{2}. \end{aligned}$$

(3.5)

Thus from (3.2) and (3.5) ,

$$\begin{aligned} ||x_{n+1}-p||\le & {} ||w_{n}-p|| \nonumber \\= & {} ||(1-\alpha _{n})x_{n}+\alpha _{n}u-p|| \nonumber \\= & {} ||(1-\alpha _{n})(x_{n}-p)+\alpha _{n}(u-p)|| \nonumber \\\le & {} (1-\alpha _{n})||x_{n}-p||+\alpha _{n}||u-p|| \nonumber \\\le & {} \max \{||x_{n}-p||,||u-p||\} \nonumber \\&\vdots \nonumber \\\le & {} \max \{||x_{0}-p||,||u-p||\}. \end{aligned}$$

(3.6)

Therefore, $\{x_{n}\}$ is bounded and so also are $\{y_{n}\},\{w_{n}\}$ and $\{Sy_{n}\}$ bounded.

Since S is a $\kappa$-strictly pseudocontraction then,

$$\begin{aligned}&||Sx-p||^{2}\le ||x-p||^{2}+\kappa ||x-Sx||^{2} \nonumber \\&\quad \Rightarrow \langle Sx-p,Sx-p\rangle \le \langle x-p,x-p\rangle + \kappa ||x-Sx||^{2} \nonumber \\&\quad \Rightarrow \langle Sx-p,Sx-x\rangle +\langle Sx-p,x-p\rangle \le \langle x-p,x-p\rangle + \kappa ||x-Sx||^{2} \nonumber \\&\quad \Rightarrow \langle Sx-p,Sx-x\rangle \le \langle x-Sx,x-p\rangle + \kappa ||x-Sx||^{2} \nonumber \\&\quad \Rightarrow \langle Sx-x,Sx-x\rangle + \langle x-p,Sx-x\rangle \le \langle x-Sx,x-p\rangle +\kappa ||x-Sx||^{2} \nonumber \\&\quad \Rightarrow ||Sx-x||^{2}\le \langle x-p,x-Sx\rangle -\langle x-p,Sx-x\rangle + +\kappa ||x-Sx||^{2} \nonumber \\&\quad \Rightarrow (1-\kappa )||Sx-x||^{2}\le 2\langle x-p,x-Sx\rangle . \end{aligned}$$

(3.7)

It follows from (3.1) and (3.7) that

$$\begin{aligned} ||x_{n+1}-p||^{2}= & {} ||(1-\beta _{n})y_{n}+\beta _{n}Sy_{n}-p||^{2}\nonumber \\= & {} ||y_{n}-p+\beta _{n}(Sy_{n}-y_{n})||^{2}\nonumber \\= & {} ||y_{n}-p||^{2}+\beta _{n}^{2}||Sy_{n}-p||^{2}-2\beta _{n}\langle y_{n}-p,Sy_{n}-y_{n}\rangle \nonumber \\\le & {} ||y_{n}-p||^{2}+\beta _{n}(\beta _{n}-(1-\kappa ))||Sy_{n}-p||^{2}\nonumber \\\le & {} ||w_{n}-p||^{2}+\beta _{n}(\beta _{n}-(1-\kappa ))||Sy_{n}-p||^{2}\nonumber \\= & {} ||(1-\alpha _{n})x_{n}+\alpha _{n}u-p||^{2}+\beta _{n}(\beta _{n}-(1-\kappa ))||Sy_{n}-p||^{2}\nonumber \\= & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}+\alpha _{n}^{2}||u-p||^{2}\nonumber \\&+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&+\beta _{n}(\beta _{n}-(1-\kappa ))||Sy_{n}-p||^{2}. \end{aligned}$$

(3.8)

We now consider two cases to establish the strong convergence of $\{x_{n}\}$ to p.

Case 1. Assume that $\{||x_{n}-p||\}$ is monotonically decreasing sequence. Then $\{x_{n}\}$ is convergent and clearly

$$\begin{aligned} \lim _{n\rightarrow \infty }||x_{n}-p||=\lim _{n\rightarrow \infty }||x_{n+1}-p||. \end{aligned}$$

(3.9)

Thus, from (3.8), we have

$$\begin{aligned} \beta _{n}((1-\kappa )-\beta _{n})||Sy_{n}-y_{n}||^{2}\le & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}-||x_{n+1}-p||^{2}\nonumber \\&+\alpha _{n}^{2}||u-p||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.10)

Therefore,

$$\begin{aligned} ||Sy_{n}-y_{n}||\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.11)

From (3.1),

$$\begin{aligned} ||w_{n}-x_{n}||=\alpha _{n}||u-x_{n}||\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.12)

Again from (3.1), we obtain

$$\begin{aligned} ||x_{n+1}-p||^{2}= & {} ||(1-\beta _{n})y_{n}+\beta _{n}Sy_{n}-p||^{2}\nonumber \\= & {} (1-\beta _{n})^{2}||y_{n}-p||^{2}+\beta _{n}^{2}||Sy_{n}-p||^{2}+2\beta _{n}(1-\beta _{n})\langle y_{n}-p,Sy_{n}-p\rangle \nonumber \\\le & {} ||y_{n}-p||^{2}+\beta _{n}[-1+\kappa +\beta _{n}]||y_{n}-Sy_{n}||^{2}\nonumber \\\le & {} ||y_{n}-p||^{2}\nonumber \\= & {} ||T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})-p||^{2}\nonumber \\\le & {} ||w_{n}-p||^{2}+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-\gamma _{n}||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}\nonumber \\\le & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}+\alpha _{n}^{2}||u-p||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-\gamma _{n}||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}\nonumber \\\le & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}+\alpha _{n}^{2}||u-p||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&+\gamma _{n}[\gamma _{n} ||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}]. \end{aligned}$$

(3.13)

It then follows from (3.13) and the condition

$$\begin{aligned} \gamma _{n}\in \left( \epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \right) , \end{aligned}$$

(3.14)

that

$$\begin{aligned} ||x_{n+1}-p||^{2}\le & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}+\alpha _{n}^{2}||u-p||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&-\epsilon ||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}, \end{aligned}$$

(3.15)

which implies

$$\begin{aligned} \epsilon ||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}\le & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}-||x_{n}-p||^{2}\nonumber \\&+\alpha _{n}^{2}||u-p||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle .\qquad \quad \end{aligned}$$

(3.16)

Hence,

$$\begin{aligned} \lim _{n\rightarrow \infty }||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}=0. \end{aligned}$$

(3.17)

Furthermore, from (3.13) and (3.17)

$$\begin{aligned} \gamma _{n} ||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}\le & {} (1-\alpha _{n})^{2}||x_{n}-p||^{2}-||x_{n+1}-p||^{2}+\alpha _{n}^{2}||u-p||^{2}\nonumber \\&+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&\rightarrow 0, n\rightarrow \infty . \end{aligned}$$

(3.18)

Therefore

$$\begin{aligned} \lim _{n\rightarrow \infty }||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||=0. \end{aligned}$$

(3.19)

On the other hand, if $T_{r_{n}}^{f_{2}}Aw_{n}=Aw_{n}$, then obviously,

$$\begin{aligned} \lim _{n\rightarrow \infty }||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}=0, \end{aligned}$$

(3.20)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}=0. \end{aligned}$$

(3.21)

Also,

$$\begin{aligned} ||y_{n}-p||^{2}= & {} ||T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})-p||^{2}\nonumber \\\le & {} \langle y_{n}-p,w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}-p\rangle \nonumber \\= & {} \frac{1}{2}[||y_{n}-p||^{2}+||w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}-p||^{2}-||y_{n}-p \nonumber \\&-(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})-p)||^{2}]\nonumber \\\le & {} \frac{1}{2}[||y_{n}-p||^{2}+||w_{n}-p||^{2}+\gamma _{n}(\gamma _{n} ||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}-|| \nonumber \\&(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2})-||y_{n}-p\nonumber \\&-(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}-p)||^{2}]\nonumber \\\le & {} \frac{1}{2}[||y_{n}-p||^{2}+||w_{n}-p||^{2}-(||y_{n}-w_{n}||^{2}+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||\nonumber \\&-2\gamma _{n}\langle y_{n}-w_{n},A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle )]\nonumber \\\le & {} \frac{1}{2}[||y_{n}-p||^{2}+||w_{n}-p||^{2}-||y_{n}-w_{n}||^{2}+\gamma _{n}^{2}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||\nonumber \\&+2\gamma _{n}|| y_{n}-w_{n}||||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||]. \end{aligned}$$

(3.22)

That is

$$\begin{aligned} ||y_{n}-p||^{2}\le & {} ||w_{n}-p||^{2}-||y_{n}-w_{n}||^{2}+2\gamma _{n}|| y_{n}-w_{n}||||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||.\nonumber \\ \end{aligned}$$

(3.23)

It then follows from (3.13) and (3.23) that

$$\begin{aligned} ||x_{n+1}-p||^{2}\le & {} ||w_{n}-p||^{2}-||y_{n}-w_{n}||^{2}+2\gamma _{n}|| y_{n}-w_{n}||||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||,\nonumber \\ \end{aligned}$$

(3.24)

which implies that

$$\begin{aligned} ||y_{n}-w_{n}||^{2}\le & {} ||w_{n}-p||^{2}-||x_{n+1}-p||^{2}+2\gamma _{n}|| y_{n}-w_{n}||||A^{*} \nonumber \\&(T_{r_{n}}^{f_{2}}-I)Aw_{n}||\nonumber \\= & {} ||(1-\alpha _{n})x_{n}+\alpha _{n}u-p||^{2}-||x_{n+1}-p||^{2}+2\gamma _{n}|| y_{n}-w_{n}||||A^{*} \nonumber \\&(T_{r_{n}}^{f_{2}}-I)Aw_{n}||\nonumber \\\le & {} ||x_{n}-p||^{2}-||x_{n+1}-p||^{2}+\alpha _{n}^{2}||u-p||^{2} \nonumber \\&+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&+2\gamma _{n}|| y_{n}-w_{n}||||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.25)

From (3.12) and (3.25), we obtain that

$$\begin{aligned} ||x_{n}-y_{n}||\le ||x_{n}-w_{n}||+||w_{n}-y_{n}||\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.26)

Let $u_{n}=w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}.$

Then

$$\begin{aligned} ||u_{n}-w_{n}||=\gamma _{n}||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.27)

Combining (3.25) and (3.27), we get

$$\begin{aligned} ||y_{n}-u_{n}||\le ||y_{n}-w_{n}||+||w_{n}-u_{n}||\rightarrow 0,n\rightarrow \infty . \end{aligned}$$

(3.28)

It follows from (3.11) and Lemma 2.4 that $\{y_{n}\}$ converges weakly to $p\in F(T)$ and consequently $\{x_{n}\}$ and $\{w_{n}\}$ converges weakly to p.

Next, we show that $p\in GMEP(f_{1},B_{1},\varphi _{1}).$ Since $y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}),$ we have

$$\begin{aligned}&f_{1}(y_{n},y)+\langle B_{1}y_{n},y-y_{n}\rangle + \varphi _{1}(y)-\varphi _{1}(y_{n})\nonumber \\&\quad +\frac{1}{r_{n}}\langle y-y_{n},y_{n}-w_{n}\rangle -\frac{1}{r_{n}}\langle y-y_{n},\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle \ge 0, \quad \forall y\in C.\nonumber \\ \end{aligned}$$

(3.29)

Thus, from the monotonicity of $F_{1}(x,y):=f_{1}(x,y)+\langle B_{1}x,y-x\rangle + \varphi _{1}(y)-\varphi _{1}(x)$, we have

$$\begin{aligned}&\frac{1}{r_{n}}\langle y-y_{n},y_{n}-w_{n}\rangle -\frac{1}{r_{n}}\langle y-y_{n},\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}\rangle \nonumber \\&\quad \ge f_{1}(y,y_{n})+\langle B_{1}y,y_{n}-y\rangle + \varphi _{1}(y_{n})-\varphi _{1}(y), \end{aligned}$$

(3.30)

which implies that

$$\begin{aligned}&\frac{1}{r_{n_{k}}}\langle y-y_{n_{k}},y_{n_{k}}-w_{n_{k}}\rangle -\frac{1}{r_{n_{k}}}\langle y-y_{n_{k}},\gamma _{n} A^{*}(T_{r_{n_{k}}}^{f_{2}}-I)Aw_{n_{k}}\rangle \ge f_{1}(y,y_{n_{k}}) \nonumber \\&\quad +\langle B_{1}y,y_{n_{k}}-y\rangle + \varphi _{1}(y_{n_{k}})-\varphi _{1}(y). \end{aligned}$$

(3.31)

Since $y_{n}\rightharpoonup p$, then it follows from (3.12), (3.19), (3.24), (3.26) and A4 that,

$$\begin{aligned} f_{1}(y,p)+\langle B_{1}y,p-y\rangle + \varphi _{1}(p)-\varphi _{1}(y)\le 0, \forall y\in C. \end{aligned}$$

(3.32)

Now, for fixed $y\in C$, let $y_{t}:=ty+(1-t)p$ for all $t\in (0,1).$ This implies that $y_{t} \in C$. Thus from A1 and A4

$$\begin{aligned} 0= & {} f_{1}(y_{t},y_{t})+\langle B_{1}y_{t},y_{t}-y_{t}\rangle + \varphi _{1}(y_{t})-\varphi _{1}(y_{t})\nonumber \\\le & {} t[f_{1}(y_{t},y)+\langle B_{1}y_{t},y-y_{t}\rangle + \varphi _{1}(y)-\varphi _{1}(y_{t})]\nonumber \\&+(1-t)[f_{1}(y_{t},p)+\langle B_{1}y_{t},p-y_{t}\rangle + \varphi _{1}(p)-\varphi _{1}(y_{t})]\nonumber \\\le & {} t[f_{1}(y_{t},y)+\langle B_{1}y_{t},y-y_{t}\rangle + \varphi _{1}(y)-\varphi _{1}(y_{t})]. \end{aligned}$$

(3.33)

Therefore

$$\begin{aligned} f_{1}(y_{t},y)+\langle B_{1}y_{t},y-y_{t}\rangle + \varphi _{1}(y)-\varphi _{1}(y_{t})\ge 0. \end{aligned}$$

(3.34)

Furthermore, from A3, we have

$$\begin{aligned} f_{1}(p,y)+\langle B_{1}p,y-p\rangle + \varphi _{1}(y)-\varphi _{1}(p)\ge 0, \end{aligned}$$

(3.35)

which implies that $p\in GMEP(f_{1},B_{1},\varphi _{1})$. Now we show that $Ap \in GMEP(f_{2},B_{2}, \varphi _{2}).$ Since $\{w_{n}\}$ is bounded and $w_{n}\rightharpoonup p$, there exists a subsequence $\{w_{n_{k}}\}$ of $\{w_{n}\}$ such that $w_{n_{k}}\rightarrow p$ and since A is a bounded linear operator, $Aw_{n_{k}}\rightarrow Ap.$

Set $z_{n_{k}}=Aw_{n_{k}}-T_{r_{n_{k}}}^{f_{2}}Aw_{n_{k}}.$ Then we have that $Aw_{n_{k}}-z_{n_{k}}=T_{r_{n_{k}}}^{f_{2}}Aw_{n_{k}},$ and from (3.19), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }z_{n_{k}}=0. \end{aligned}$$

(3.36)

Therefore, from the definition of $T_{r_{n_{k}}}^{f_{2}}$, we observe that

$$\begin{aligned}&f_{2}(Aw_{n_{k}}-z_{n_{k}},y)+\langle B_{2}w_{n_{k}}-z_{n_{k}},y-w_{n_{k}}+z_{n_{k}}\rangle +\varphi _{2}(y)-\varphi _{2}(w_{n_{k}}-z_{n_{k}})\nonumber \\&\quad +\frac{1}{r_{n_{k}}}\langle y-(w_{n_{k}}-z_{n_{k}}),(w_{n_{k}}-z_{n_{k}})-w_{n_{k}}\rangle \ge 0, \forall y\in C. \end{aligned}$$

(3.37)

Since $f_{2}$ is upper semicontinuous in first argument, then $F_{2}$ defined as

$$\begin{aligned} F_{2}(x,y):=f_{2}(x,y)+\langle B_{2}x,y-x\rangle + \varphi _{2}(y)-\varphi _{2}(x) \end{aligned}$$

(3.38)

is also upper semicontinuous in first argument. Thus, taking $\limsup$ to the inequality (3.37) as $k\rightarrow \infty$ and using assumption A3, we have

$$\begin{aligned} f_{2}(Ap,y)+\langle B_{2}Ap,y-Ap\rangle + \varphi _{2}(y)-\varphi _{2}(Ap)\ge 0~~\forall y\in C, \end{aligned}$$

(3.39)

which implies $Ap \in GMEP(f_{2},B_{2},\varphi _{2}).$ Hence $p \in \Omega \cap F(S).$

We now show that $\{x_{n}\}$ converges strongly to p.

$$\begin{aligned} || x_{n+1}-p||^{2}= & {} || (1-\beta _{n})y_{n}+\beta _{n} Sy_{n}-p||^{2}\nonumber \\\le & {} || y_{n}-p||^{2}\nonumber \\\le & {} || w_{n}-p||^{2}\nonumber \\= & {} || (1-\alpha _{n}) x_{n} + \alpha _{n}u -p||^{2}\nonumber \\= & {} || (1-\alpha _{n})(x_{n}-p)+\alpha _{n}(u-p)||^{2}\nonumber \\= & {} (1-\alpha _{n})^{2}|| x_{n}-p||^{2} +\alpha _{n}^{2}||u- p||^{2}+2\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\\le & {} (1-\alpha _{n})|| x_{n}-p||^{2} +\alpha _{n}[\alpha _{n}|| u-p||^{2}+2(1-\alpha _{n})\langle x_{n}-p,u-p\rangle ].\nonumber \\ \end{aligned}$$

(3.40)

Therefore, by Lemma 2.5, we obtain $x_{n}\rightarrow p$, $n\rightarrow \infty .$

Case 2. Assume that $\{|| x_{n}-p||\}$ is not monotonically decreasing sequence. Set $\Gamma _{n}=|| x_{n}-p||^{2}$ and let $\tau :\mathbb {N}\rightarrow \mathbb {N}$ be a mapping for all $n\ge n_{0}$ (for some $n_{0}$ large enough) defined by

$$\begin{aligned} \tau (n):=\max \{k\in \mathbb {N}:k\le n,\Gamma _{k}\le \Gamma _{k+1}\}. \end{aligned}$$

(3.41)

Clearly $\tau$ is a non-decreasing sequence such that $\tau (n)\rightarrow \infty$ as $n\rightarrow \infty$ and $\Gamma _{\tau (n)}\le \Gamma _{\tau (n)+1}$, for $n\ge n_{0}$. It follows from (3.8) that

$$\begin{aligned} 0\le & {} || x_{\tau (n)+1}-p||^{2}-|| x_{\tau (n)}-p||^{2}\nonumber \\\le & {} || x_{\tau (n)+1}-p||^{2}-(1-\alpha _{\tau (n)})^{2}|| x_{\tau (n)}-p||^{2}\nonumber \\\le & {} \alpha _{n}^{2}||u- p||^{2}+\alpha _{n}(1-\alpha _{n})\langle x_{n}-p,u-p\rangle \nonumber \\&+\,\,\beta _{n}(\beta _{n}-(1-\kappa ))|| y_{n}-Sy_{n}||^{2}. \end{aligned}$$

(3.42)

That is,

$$\begin{aligned}&\beta _{\tau (n)}((1-\kappa )-\beta _{\tau (n)})|| Sy_{\tau (n)}-y_{\tau (n)}||^{2}\le \alpha _{\tau (n)}^{2}|| u-p||^{2}\nonumber \\&\quad +\,\, 2\alpha _{\tau (n)}(1-\alpha _{\tau (n)})\langle x_{\tau (n)}-p,u-p\rangle \nonumber \\&\quad \rightarrow 0,\quad as\quad n\rightarrow \infty . \end{aligned}$$

(3.43)

By the same argument as (3.11) to (3.28) in case 1, we conclude that $\{x_{\tau (n)}\}$, $\{y_{\tau (n)}\}$ and $\{w_{\tau (n)}\}$ converge weakly to $p\in F(S)\cap \Omega$. Now for all $n\ge n_{0}$,

$$\begin{aligned} 0\le & {} || x_{\tau (n)+1}-p||^{2}-|| x_{\tau (n)}-p||^{2}\nonumber \\\le & {} (1-\alpha _{\tau (n)})|| x_{\tau (n)}-p||^{2} + \alpha _{\tau (n)}^{2}||u- p||^{2}+2\alpha _{\tau (n)}(1-\alpha _{\tau (n)}\langle x_{\tau (n)}\nonumber \\&-p,u-p\rangle -|| x_{\tau (n)}-p||^{2}\nonumber \\= & {} \alpha _{\tau (n)}[\alpha _{\tau (n)}||u- p||^{2}+2\alpha _{\tau (n)}(1-\alpha _{\tau (n)})\langle x_{\tau (n)}-p,u-p\rangle -|| x_{\tau (n)}-p||^{2}].\nonumber \\ \end{aligned}$$

(3.44)

Therefore,

$$\begin{aligned} || x_{\tau (n)}-p||^{2}\le & {} \alpha _{\tau (n)}||u-p||^{2}+2\alpha _{\tau (n)}(1-\alpha _{\tau (n)})\langle x_{\tau (n)}-p,u-p\rangle \nonumber \\&\rightarrow 0,~~n\rightarrow \infty . \end{aligned}$$

(3.45)

Thus,

$$\begin{aligned} \lim _{n\rightarrow \infty }|| x_{\tau (n)}-p||^{2}=0, \end{aligned}$$

(3.46)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\Gamma _{\tau (n)}=\lim _{n\rightarrow \infty }\Gamma _{\tau (n)+1}. \end{aligned}$$

(3.47)

Furthermore, for $n\ge n_{0}$, it is observed that $\Gamma _{\tau (n)}\le \Gamma _{\tau (n)+1}$ if $n\ne \tau (n)$ (that is $\tau (n)<n$) because $\Gamma _{j}>\Gamma _{j+1}$ for $\tau (n)+1\le j\le n$. Consequently, for all $n\ge n_{0}$,

$$\begin{aligned} 0\le \Gamma _{n}\le \max \{\Gamma _{\tau (n)},\Gamma _{\tau (n)+1}\}=\Gamma _{\tau (n)+1}. \end{aligned}$$

(3.48)

So $\lim _{n\rightarrow \infty }\Gamma _{n}=0$, that is $\{x_{n}\}$,$\{y_{n}\}$ and $\{w_{n}\}$ converge strongly to $p\in F(S)\cap \Omega , ~~\forall n\ge 0$. $\square$

Corollary 3.2

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, let $C\subset H_{1}$ and $Q\subset H_{2}$ be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a bounded linear operator and $A^*$ the adjoint of A. Let $f_{1}:C\times C\rightarrow \mathbb {R}$ and $f_{2}:Q\times Q\rightarrow \mathbb {R}$ be bifunctions satisfying conditions $(A1)-(A4)$ and $f_{2}$ is upper semicontinuous in first argument. Let $B_{1}:H_{1}\rightarrow H_{1}$ and $B_{2}:H_{2}\rightarrow H_{2}$ be continuous and monotone mappings, $\varphi _{1}:C\rightarrow \mathbb {R}\cup +\infty$ and $\varphi _{2}:Q\rightarrow \mathbb {R}\cup +\infty$ be proper lower semicontinuous and convex function. Let $S:C\rightarrow C$ be a nonexpansive mapping, such that $\Omega \cap F(S)\ne \emptyset$. Let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0$; $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=\gamma$ otherwise ($\gamma$ being any nonnegative real number). Then the sequence $\{w_{n}\},\{x_{n}\}$ and $\{y_{n}\}$ generated iteratively for an arbitrary $x_{0}\in C$ and a fixed $u\in C$ by

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=(1-\alpha _{n})x_{n}+\alpha _{n}u\\ &{}y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})\\ &{}x_{n+1}=(1-\beta _{n})y_{n}+\beta _{n}Sy_{n},~~\forall n\ge 0 \end{array} \right. \end{aligned}$$

(3.49)

converges strongly to a point $p\in \Omega \cap F(S)$ where $\{\alpha _{n}\}_{n=1}^{\infty }$ and $\{\beta _{n}\}_{n=1}^{\infty }$ are real sequences in (0, 1) satisfying the following conditions:

• (i) $\lim _{n\rightarrow \infty }\alpha _{n}=0$, $\sum \nolimits _{n=1}^{\infty }\alpha _{n}=\infty$,

• (ii) $0<\liminf \beta _{n}\le \limsup \beta _{n} <1.$

Corollary 3.3

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, let $C\subset H_{1}$ and $Q\subset H_{2}$ be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a bounded linear operator and $A^*$ the adjoint of A. Let $f_{1}:C\times C\rightarrow \mathbb {R}$ and $f_{2}:Q\times Q\rightarrow \mathbb {R}$ be bifunctions satisfying conditions $(A1)-(A4)$ and $f_{2}$ is upper semicontinuous in first argument. Let $B_{1}:H_{1}\rightarrow H_{1}$ and $B_{2}:H_{2}\rightarrow H_{2}$ be continuous and monotone mappings. Let $S:C\rightarrow C$ be a $\kappa$ strictly pseudocontraction, such that $\Omega _{B} \cap F(S)\ne \emptyset$. Let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0$; $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=\gamma$, otherwise ($\gamma$ being any nonnegative real number). Then the sequence $\{w_{n}\},\{x_{n}\}$ and $\{y_{n}\}$ generated iteratively for an arbitrary $x_{0}\in C$ and a fixed $u\in C$ by

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=(1-\alpha _{n})x_{n}+\alpha _{n}u\\ &{}y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})\\ &{}x_{n+1}=(1-\beta _{n})y_{n}+\beta _{n}Sy_{n},~~\forall n\ge 0 \end{array} \right. \end{aligned}$$

(3.50)

converges strongly to a point $p\in \Omega _{B} \cap F(S)$ where $\{\alpha _{n}\}_{n=1}^{\infty }$ and $\{\beta _{n}\}_{n=1}^{\infty }$ are real sequences in (0, 1) satisfying the following conditions:

• (i) $\lim \nolimits _{n\rightarrow \infty }\alpha _{n}=0$, $\sum \nolimits _{n=1}^{\infty }\alpha _{n}=\infty$,

• (ii) $0<\liminf \beta _{n}\le \limsup \beta _{n} <1-\kappa .$

Corollary 3.4

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, let $C\subset H_{1}$ and $Q\subset H_{2}$ be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a bounded linear operator and $A^*$ the adjoint of A. Let $f_{1}:C\times C\rightarrow \mathbb {R}$ and $f_{2}:Q\times Q\rightarrow \mathbb {R}$ be bifunctions satisfying conditions $(A1)--(A4)$ and $f_{2}$ is upper semicontinuous in first argument. Let $\varphi _{1}:C\rightarrow \mathbb {R}\cup +\infty$ and $\varphi _{2}:Q\rightarrow \mathbb {R}\cup +\infty$ be proper lower semicontinuous and convex function. Let $S:C\rightarrow C$ be a $\kappa$ strictly pseudocontraction, such that $\Omega _{\varphi } \cap F(S)\ne \emptyset$. Let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0$; $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=\gamma$, otherwise $(\gamma$ being any nonnegative real number). Then the sequence $\{w_{n}\},\{x_{n}\}$ and $\{y_{n}\}$ generated iteratively for an arbitrary $x_{0}\in C$ and a fixed $u\in C$ by

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=(1-\alpha _{n})x_{n}+\alpha _{n}u\\ &{}y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})\\ &{}x_{n+1}=(1-\beta _{n})y_{n}+\beta _{n}Sy_{n}, \quad \forall n\ge 0 \end{array} \right. \end{aligned}$$

(3.51)

converges strongly to a point $p\in \Omega _{\varphi } \cap F(S)$ where $\{\alpha _{n}\}_{n=1}^{\infty }$ and $\{\beta _{n}\}_{n=1}^{\infty }$ are real sequences in (0, 1) satisfying the following conditions

• (i) $\lim \nolimits _{n\rightarrow \infty }\alpha _{n}=0$, $\sum \nolimits _{n=1}^{\infty }\alpha _{n}=\infty$,

• (ii) $0<\liminf \beta _{n}\le \limsup \beta _{n} <1-\kappa .$

Corollary 3.5

Let $H_{1}$ and $H_{2}$ be two real Hilbert spaces, let $C\subset H_{1}$ and $Q\subset H_{2}$ be nonempty closed convex subsets of $H_{1}$ and $H_{2}$, respectively. Let $A:H_{1}\rightarrow H_{2}$ be a bounded linear operator and $A^*$ the adjoint of A. Let $f_{1}:C\times C\rightarrow \mathbb {R}$ and $f_{2}:Q\times Q\rightarrow \mathbb {R}$ be bifunctions satisfying conditions $(A1)-(A4)$ and $f_{2}$ is upper semicontinuous in first argument. Let $S:C\rightarrow C$ be a $\kappa$ strictly pseudocontraction, such that $\Omega _{0} \cap F(S)\ne \emptyset$. Let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0$; $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=\gamma$, otherwise $(\gamma$ being any nonnegative real number). Then the sequence $\{w_{n}\},\{x_{n}\}$ and $\{y_{n}\}$ generated iteratively for an arbitrary $x_{0}\in C$ and a fixed $u\in C$ by

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=(1-\alpha _{n})x_{n}+\alpha _{n}u\\ &{}y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})\\ &{}x_{n+1}=(1-\beta _{n})y_{n}+\beta _{n}Sy_{n},~~\forall n\ge 0 \end{array} \right. \end{aligned}$$

(3.52)

converges strongly to a point $p\in \Omega _{0} \cap F(S)$ where $\{\alpha _{n}\}_{n=1}^{\infty }$ and $\{\beta _{n}\}_{n=1}^{\infty }$ are real sequences in (0, 1) satisfying the following conditions

• (i) $\lim \nolimits _{n\rightarrow \infty }\alpha _{n}=0$, $\sum \nolimits _{n=1}^{\infty }\alpha _{n}=\infty$,

• (ii) $0<\liminf \beta _{n}\le \limsup \beta _{n} <1-\kappa .$

4 Numerical example and application

We present here in this section an example, a numerical result and an application to split convex minimisation problem .

4.1 Example

Let $H_{1}=H_{2}=L^{2}([0,1])$ with inner product given as $\langle f, g\rangle = \int _{0}^{1}f(t)g(t)\mathrm{{d}}t$. Now take $f_{1}(x,y):=||y||_{L^{2}}-||x||_{L^{2}}$; $B_{1}x :=2x$; $\varphi _{1}(x)=||x||_{L^{2}}$ and $Sx=x.$ Suppose $A : L^2([0, 1]) \rightarrow L^2([0, 1])$ is defined by

$$\begin{aligned} A x(s) =\int _{0}^{1}V(s,t)x(t)\mathrm{{d}}t, \forall x\in L^{2}([0,1]), \end{aligned}$$

where $V : [0, 1] \times [0, 1] \rightarrow \mathbb {R}$ is continuous. Then A is a bounded linear operator and the adjoint $A^*$ of A is defined by

$$\begin{aligned} A^* x(s)=\int _{0}^{1}V(t,s)x(t)\mathrm{{d}}t, \quad \forall x\in L^{2}([0,1]). \end{aligned}$$

Here we take $V(s,t)=e^{st}$. Finally take $f_{2}(x,y):=||y||_{L^{2}}^{2}-||x||_{L^{2}}^{2}$; $B_{2}x :=3x$; $\varphi _{2}(x)=||x||_{L^{2}}^{2}.$ We consider the problem; find $x^{*}\in H_{1}$ such that

$$\begin{aligned}&\displaystyle Sx^{*}=x^*,\end{aligned}$$

(4.1)

$$\begin{aligned}&\displaystyle f_{1}(x^{*}, x)+\langle B_{1}x^{*},x-x^{*}\rangle +\varphi _{1}(x)-\varphi _{1}(x^{*}) \ge 0,~~\forall x \in H_{1}, \end{aligned}$$

(4.2)

and $y^{*}=Ax^{*}\in H_{2}$ solves

$$\begin{aligned} f_{2}(y^{*}, y)+\langle B_{2} y^{*},y-y^{*}\rangle +\varphi _{2}(y)-\varphi _{2}(y^{*}) \ge 0,~~\forall y \in H_{2}. \end{aligned}$$

(4.3)

The set of solutions of problem (4.1 4.2)–(4.3) is nonempty (since $x(t)=0$, a.e. is in the set of solutions). Take $\alpha _{n}=\frac{1}{n+3}$, $\beta _{n}=\frac{1}{2}(1-\frac{1}{n+2})$ and let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0,$ $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=\gamma$, otherwise ($\gamma$ being any nonnegative real number) in iterative scheme (3.1) to obtain

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=\left( 1-\frac{1}{n+3}\right) x_{n}+\frac{1}{n+3}u\\ &{}y_{n}=T_{r_{n}}^{f_{1}}(w_{n}+\gamma _{n} A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n})\\ &{}x_{n+1}=\left( 1-\frac{1}{2}\left( 1-\frac{1}{n+2}\right) \right) y_{n}+\frac{1}{2}\left( 1-\frac{1}{n+2}\right) y_{n}, \quad \forall n\ge 0. \end{array} \right. \end{aligned}$$

(4.4)

4.2 Example with numerical computation

Let $H_{1}=H_{2}=\mathbb {R}$ and $C = Q = \mathbb {R}$. Let $f_{1}(x,y)=-5x^{2}+xy+4y^{2},$ $\phi _{1}(x)=x^{2}$ and $B_{1}(x)=4x$, then $T_{r}^{f_{1}}(x)=\dfrac{x}{15r+1}.$ Also Let $f_{2}(x,y)=-3x^{2}+xy+2y^{2},$ $\phi _{2}(x)=2x^{2}$ and $B_{2}(x)=2x$, then $T_{r}^{f_{2}}(x)=\dfrac{x}{11r+1}.$ Furthermore, let $Ax=8x$, $A^{*}x=8x$ and $S(x)=-2x.$ We make difference choices of $x_{0}, u$ and use $\frac{||x_{n+1}-x_{n}||}{||x_{1}-x_{0}||}<10^{-4}$ for stopping criterion. Take $\alpha _{n}=\frac{1}{n+2}$, $\beta _{n}=\frac{1}{6}(1-\frac{1}{n+2}),$ $r_{n}=\dfrac{n}{n+1}$ and let the step size $\gamma _n$ be chosen in such a way that for some $\epsilon > 0$, $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}$ any positive real number otherwise, in iterative scheme (3.1) to obtain

$$\begin{aligned} \left\{ \begin{array}{lllll} &{}w_{n}=\left( 1-\frac{1}{n+2}\right) x_{n}+\frac{1}{n+2}u\\ &{}y_{n}=\dfrac{w_{n}+\gamma _{n} 8\left( \frac{-11\left( \frac{n}{n+1}\right) w_{n}}{11\left( \frac{n}{n+1}\right) +1}\right) }{15\left( \frac{n}{n+1}\right) +1}\\ &{}x_{n+1}=\left( 1-\frac{1}{6}\left( 1-\frac{1}{n+2}\right) \right) y_{n}-\frac{1}{3}\left( 1-\frac{1}{n+2}\right) y_{n}, \quad \forall n\ge 0. \end{array} \right. \end{aligned}$$

(4.5)

Case 1. $x_{0}=2$, $u=1$ and $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=0.0000021$ otherwise.

Case 2. $x_{0}=6$, $u=3$ and $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=0.0000222$ otherwise.

Case 3. $x_{0}=1$, $u=8$ and $\gamma _{n} \in \Big (\epsilon ,\dfrac{||(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}{||A^{*}(T_{r_{n}}^{f_{2}}-I)Aw_{n}||^{2}}-\epsilon \Big )$ for $T_{r_{n}}^{f_{2}}Aw_{n}\ne Aw_{n}$ and $\gamma _{n}=0.0003$ otherwise.

The Mathlab version used is R2014a and the execution times are as follows:

• (1) (case 1, $\varepsilon = 10^{-4})$ and execution time is 0.010 s.

• (2) (case 2, $\varepsilon = 10^{-4})$ and execution time is 0.011 s.

• (3) (case 3, $\varepsilon = 10^{-2})$ and execution time is 0.013 s (Fig. 1).

4.3 Applications to split convex minimisation problem

Here, we apply our result to study the following split convex minimisation problem: find

$$\begin{aligned} x^{*}\in F(S)~~\mathrm{such~~that~~} x^{*}=\arg \min _{x\in C}(h_{1}(x)+ \phi _{1}(x)+\varphi _{1}(x)), \end{aligned}$$

(4.6)

and such that

$$\begin{aligned} Ax^{*}=\arg \min _{y\in Q}(h_{2}(x)+ \phi _{2}(x)+\varphi _{2}(x)), \end{aligned}$$

(4.7)

where C and Q are nonempty closed and convex subset of $H_{1}$ and $H_{2}$. Also $h_{1}, \varphi _{1} :C\rightarrow \mathbb {R}$ and $h_{2}, \varphi _{2} :Q\rightarrow \mathbb {R}$ are four convex and lower semi-continuous functionals. Furthermore, $\phi _{1}: C\rightarrow \mathbb {R}$ and $\phi _{2}:Q\rightarrow \mathbb {R}$ are convex continuously differentiable functions and $A :H_{1}\rightarrow H_{2}$ a bounded linear operator.

Let $f_{i}(x,y)=h_{i}(y)-h_{i}(x)$ and $B_{i}=\nabla \phi _{i},~~i=1,2$ and $\nabla \phi$ denote the gradient of $\phi$.

Then the split convex minimisation problem (4.6)–(4.7) can be formulated as the following split generalised mixed equilibrium problem: find $x^{*}\in F(S)$ such that

$$\begin{aligned} h_{1}(x)-h_{1}(x^{*})+\langle \nabla \phi _{1}x^{*},x-x^{*}\rangle +\varphi _{1}(x)-\varphi _{1}(x^{*}) \ge 0, \quad \forall x \in C, \end{aligned}$$

(4.8)

and $y^{*}=Ax^{*}\in Q$ solves

$$\begin{aligned} h_{2}(y)-h_{2}(y^{*})+\langle \nabla \phi _{2}y^{*},y-y^{*}\rangle +\varphi _{2}(y)-\varphi _{2}(y^{*}) \ge 0, \quad \forall y \in Q. \end{aligned}$$

(4.9)

Thus Theorem 3.1 provides a strong convergence theorem for solving split convex minimisation problem (4.6)–(4.7).