A strongly convergent algorithm for solving split equality problems beyond monotonicity

Abstract

In this paper, we focus on some split inverse problems, namely the split equality variational inequalities and common fixed point problems, and combine various operator theory techniques to establish minimum-norm strong convergence for our proposed method. We present two strong convergent results with (and without) reference to the monotonicity property of the cost operators. Our convergence analyses assume very mild conditions and thus generalize and extend recent related results in the literature. Furthermore, several numerical examples illustrate the practical potentials and advantages of our proposed algorithm.

1 Introduction

The field of optimization and fixed point theory is one of high relevance, attracting the interest of numerous researchers due to its enriching and versatile applicability in multidisciplinary areas, such as bandwidth and resource allocation, neuroscience, machine learning, image and signal recovery, optimal control problems, see e.g., Iiduka (2012), Liu et al. (2016), Taiwo et al. (2021), Tan et al. (2021),1 Zeng et al. (2020). One famous optimization problem widely studied over the years is the classical variational inequality problem (VI), introduced by Stampacchia (1968) and Fichera (1963). Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let $\langle \cdot , \cdot \rangle$ and $\Vert \cdot \Vert$ denote the inner product and induced norm on H,  respectively. Also let $\mathcal {F}:H\rightarrow H$ be a single-valued mapping. Then, the VI is defined as finding a point $x\in C$ such that

$$\begin{aligned} \langle \mathcal {F}x, z-x \rangle \ge 0, \quad \forall z\in C. \end{aligned}$$

(1.1)

We denote the solution set of VI (1.1) by $VI(C,\mathcal {F}).$

There are majorly three well known methods for solving the VI (1.1); namely, the Tseng extragradient method (TEGM) (Tseng 2000), the subgradient extragradient method (SEGM) (Censor et al. 2011a, b, 2010), and projection contraction method (PCM) (Cholamjiak et al. 2020). For interesting modifications of these methods, refer to Gibali et al. (2020), Ogwo et al. (2023), Reich et al. (2021), Reich et al. (2022), Shehu et al. (2019), Shehu et al. (2022), Oyewole and Reich (2024). The PCM is formulated as follows, see Cholamjiak et al. (2020):

$$\begin{aligned} {\left\{ \begin{array}{ll} x_1\in H, \\ y_n=P_C(x_n-\lambda \mathcal {F}x_n),\\ d(x_n,y_n):= (x_n-y_n)-\lambda (\mathcal {F}x_n-\mathcal {F}y_n),\\ z_n=x_n-\rho \xi d(x_n,y_n),\\ x_{n+1}= (1-\alpha _n-\beta _n)x_n+\beta _nz_n, \end{array}\right. } \end{aligned}$$

where $P_C$ is the projection map from H unto the feasible set C, $\rho \in (0,2), ~\lambda \in (0,\frac{1}{L}), L$ is the Lipschitz constant of the cost operator, $\xi _n:=\frac{\tau (w_n,y_n)}{\Vert d(w_n,y_n)\Vert ^2}, ~~ \tau (w_n,y_n):=\langle w_n-y_n, d(w_n,y_n) \rangle , ~~\forall n\ge 1.$ We note that while the step size ($\lambda$) in Cholamjiak et al. (2020) depends on the Lipschitz constant (which can be quite difficult to calculate), our method in this paper utilizes self adaptive step sizes that are more computationally efficient.

The split inverse problem (SIP) (Censor et al. 2012) is a viable mathematical model known for its great usefulness and applications in vast fields, such as signal processing, phase retrieval, image recovery, data compression, intensity-modulated radiation therapy, e.g. see Censor et al. (2006), Godwin et al. (2023), Eslamian et al. (2018)). The SIP model is formulated as follows:

$$\begin{aligned} \text {Find}~~ \hat{x}\in H_1 \quad \text {that solves}\ \bar{\Phi }_1 \end{aligned}$$

(1.2)

such that

$$\begin{aligned} \hat{y} := A\hat{x}\in H_2 \quad \text {solves}\ \bar{\Phi }_2, \end{aligned}$$

(1.3)

where $H_1$ and $H_2$ are real Hilbert spaces, $\bar{\Phi }_1$ denotes an inverse problem formulated in $H_1,$ $\bar{\Phi }_2$ denotes an inverse problem formulated in $H_2,$ and $A: H_1 \rightarrow H_2$ is a bounded linear operator. Since the introduction of the SIP, many researchers have proposed interesting generalizations and extensions. One of them is the split equality problem (SEP) by Moudafi’s Moudafi and Al-Shemas (2013). Let $H_1, H_2, H_3$ be real Hilbert spaces, and let C, Q be nonempty, closed, and convex subsets of $H_1$ and $H_2$ respectively. Let also $T_1, T_2$ be bounded linear operators, defined by $T_1:H_1\rightarrow H_3$ and $T_2:H_2\rightarrow H_3,$ with adjoints $T_1^*, T_2^*$ respectively. The SEP is defined as finding $u\in C$ and $v\in Q,$ such that

$$\begin{aligned} T_1u=T_2v. \end{aligned}$$

(1.4)

Also, let $\mathcal {F}:H_1\rightarrow H_1$ and $\mathcal {G}:H_2\rightarrow H_2$ be single-valued mappings, and let $C, Q, T_1, T_2$ be defined as in (1.4). The split equality variational inequality problem (SEVIP) is defined as finding

$$\begin{aligned} x\in VI(C,\mathcal {F}) ~~ \text {and}~~ y\in VI(Q,\mathcal {G}), \quad ~~ \text {such that}\quad T_1x=T_2y. \end{aligned}$$

(1.5)

We also note that if $H_2=H_3$ and $T_2=I$ (where I is the identity operator), then the SEVIP (1.5) reduces to the split variational inequality problem (SVIP) proposed by Censor et al. (2012). Recently, Kwelegano et al. (2022) proposed the following strong convergence method for solving split equality VI for pseudomonotone operators:

Algorithm 1.1

Step 0.:

Let $l\in (0,1), \mu ^*>0, \xi ^*\in (0,\frac{1}{\mu ^*}),$ choose $(x_0,y_0)\in C\times Q$ arbitrarily. Given the current iterate $(x_n,y_n),$ calculate $(x_{n+1}, y_{n+1})$ as follows;

Step 1.:

Compute: $u_n=P_C(x_n-\xi ^* \mathcal {F}x_n),~~ r_n=P_Q(y_n-\xi ^* \mathcal {G}y_n)$ and $r^*(x_n):=x_n-u_n, ~~ \kappa ^*(y_n):= y_n-r_n.$

Step 2.:

Compute: $a_n=x_n-m_nr^*(x_n)$ and $e_n=y_n-\lambda _n^* \kappa ^*(y_n)$, where $m_n=l^{j_n}, \lambda _n^*=l^{k_n},$ and $j_n$ and $k_n,$ are the smallest non-negative integers j and k,  respectively that satisfy

$$\begin{aligned}&\langle \mathcal {F}x_n-\mathcal {F}[x_n-l^jr^*(x_n)], r^*(x_n) \rangle \le \mu ^*\Vert r^*(x_n)\Vert ^2;\\&\text {and}~~ \langle \mathcal {G}y_n-\mathcal {G}[y_n-l^k \kappa ^*(y_n)], \kappa ^*(y_n) \rangle \le \mu ^*\Vert \kappa ^*(y_n)\Vert ^2. \end{aligned}$$

Step 3.:

Compute:

$$\begin{aligned} {\left\{ \begin{array}{ll} z_n=P_C[x_n-\zeta T_1^*(T_1x_n-T_2y_n)],\\ x_{n+1}=\alpha _nf(x_n)+(1-\alpha _n)[a_n^* P_{C_n}x_n+(1-a_n^*)z_n],\\ t_n=P_Q[y_n-\zeta T_2^*(T_1x_n-T_2y_n)],\\ y_{n+1}=\alpha _ng(y_n)+(1-\alpha _n)[a_n^* P_{Q_n}y_n+(1-a_n^*)t_n]. \end{array}\right. } \end{aligned}$$

where $C_n:=\{x\in C:h_n^*(x)\le 0\}, Q_n:=\{x\in Q:e_n^*(y)\le 0\},$    and   $h_n^*(x)=\langle \mathcal {F}a_n,x-a_n\rangle , e_n^*(y)=\langle \mathcal {G}e_n,y-e_n\rangle ,$

Step 4.:

Set $n:=n+1$ and go to Step 1.

where $f:C\rightarrow C$ and $g:Q\rightarrow Q$ are contraction with coefficients $\rho _1^*,\rho _2^* \in (0, \frac{1}{2})$, where $\rho ^*=\max \{\rho _1^*,\rho _2^*\},$ and $\{a_n^*\}\subset \{\pi , \pi ^*\}\subset (0,1).$

Given a nonlinear mapping $S:H\rightarrow H$, the fixed point problem (FPP) is defined as finding $p\in S$ such that

$$\begin{aligned} Sp=p. \end{aligned}$$

We denote the set of fixed points of S by F(S). Let $S_1:C\rightarrow C$ and $S_2:Q\rightarrow Q$ be nonlinear mappings such that $F(S_1)\ne \emptyset$ and $F(S_2)\ne \emptyset$ are the sets of fixed points of $S_1$ and $S_2$ respectively. Let $C, Q, T_1, T_2$ be defined as in (1.4), then the split equality fixed point problem (SEFPP) (Moudafi 2011; Reich and Tuyen 2022) is defined as finding

$$\begin{aligned} x\in F(S_1) \quad ~~ \text {and} \quad y\in F(S_2), \quad \quad \text {such that} \quad T_1x=T_2y. \end{aligned}$$

(1.6)

Boikano and Zegeye (2020) proposed the following self-adaptive algorithm for solving (1.6).

Algorithm 1.2

$$\begin{aligned} {\left\{ \begin{array}{ll} z_n=P_C[x_n-\zeta _nT_1^*(T_1x_n-T_2y_n)],\\ x_{n+1}=\alpha _nz_n+(1-\alpha _n)Y_nz_n,\\ t_n=P_Q[y_n+\zeta _nT_2^*(T_1x_n-T_2y_n)],\\ y_{n+1}=\alpha _nt_n+(1-\alpha _n)J_nt_n, \end{array}\right. } \end{aligned}$$

Where,

$$\begin{aligned}&Y_n=[(1-\eta _{n})I+\eta _{n}S_1[(1-\chi _n)I+\chi _nS_1]]\\&J_n=[(1-\eta _{n})I+\eta _{n}S_2[(1-\chi _n)I+\chi _nS_2]]. \end{aligned}$$

Where $T_1:C\rightarrow H_1$ and $T_2:Q\rightarrow H_2$ are quasi-pseudocontrative mappings, $\{\alpha _n\},\{\eta _{n}\},\{\chi _n\}$ are sequences of real numbers in (0, 1),  and $\zeta _n>0, ~~ \forall n\ge 0.$ The authors proved a strong convergence result.

The dual variational inequality (DVI) problem of the VI (1.1) is defined as finding a point $x\in C,$ such that

$$\begin{aligned} \langle \mathcal {F}z,z-x \rangle \ge 0, \quad \forall z\in C. \end{aligned}$$

(1.7)

We represent the solution set of the DVI (1.7) by $\bar{\Gamma }.$ It is known that $VI(C,\mathcal {F})=\bar{\Gamma },$ whenever $\mathcal {F}$ is pseudomonotone. However, in the case where $\mathcal {F}$ is quasimonotone, $VI(C,\mathcal {F})\nsubseteq \bar{\Gamma }.$ Hence, quasimonotone mappings are a more general class of operators than pseudomonotone, (see the references Ye and He 2015; Alakoya et al. 2022; Uzor et al. 2023).

In this work, our goal is to find a common solution to the SEVIP (1.5) and SEFPP (1.6), where $\mathcal {F}$ and $\mathcal {G}$ are quasimonotone and Lipschitz continuous. In particular, we aim to find $(x,y)\in \Omega ,$ such that

$$\begin{aligned} \Omega := \{x\in \bar{\Gamma }\cap F(S_1), ~ y\in \bar{\Gamma }_*\cap F(S_2): T_1x=T_2y\}\ne \emptyset , \end{aligned}$$

(1.8)

where $\bar{\Gamma }$ and $\bar{\Gamma }_*$ denote the solution set of the DVI involving operators $\mathcal {F}$ and $\mathcal {G},$ respectively. In recent times, authors have found interest in formulating methods that provide a common solution to certain optimization and fixed point problems like the form in (1.8). Reasons been that these common problems can be directly transformed into working models for addressing interesting sundry real-life problems that exist in communications, economics, machine learning, etc. (see Alakoya et al. 2022; Censor et al. 2006; Iiduka 2012) Eslamian (2017) proposed a method for finding a common solution of split equality variational inequality for monotone, Lipschitz continuous operators, and split equality common fixed point problems of a finite family of quasi-nonexpansive mappings. Kazmi et al. (2019) proposed a simultaneous algorithm for approximating common solution of split equality VI for monotone, Lipschitz continuous mappings, and multiple-sets split equality fixed point problem for two countable families of multivalued demicontractive mappings in real Hilbert spaces. Very recently, Izuchukwu et al. (2020) proposed an algorithm for approximating a common solution of split equality problem for finite families of $\tau -$ inverse strongly monotone variational inequalities and the set of solutions of the split equality fixed point problem of multivalued type-one demicontractive mappings in real Hilbert spaces.

Motivated by the results of Kwelegano et al. (2022), Boikano and Zegeye (2020), Eslamian (2017), Kazmi et al. (2019), and Izuchukwu et al. (2020), in this paper we propose a new self-adaptive inertial modified-Mann projection contraction method (SIMMPCM) for approximating the common solution of SEVIP for quasimonotone and Lipschitz continuous operators, and SEFPP of quasi-pseudocontractive mappings in real Hilbert spaces. Our method does not require any line-search procedure, rather it adopts a non-monotonic sequence of self-adaptive step sizes. In addition, we prove that the sequence generated by the algorithm converges strongly to a minimum-norm solution of the problem (1.8) under relaxed conditions. Some numerical experiments demonstrate the efficiency of our algorithm in comparison with other methods in the literature. Moreover, our results generalizes several recent related results in the literature.

The outline of the paper is as follows: In Sect. 2 we give useful results and definitions for our analysis. In Sects. 3 and 4, we present our algorithm and strong convergence analysis (involving and excluding monotonicity), respectively. Some numerical experiments are presented in Sect. 5 illustrating the efficiency of the proposed algorithm in comparison with existing methods in the literature. Finally, we present some concluding remarks on our work in Sect. 6.

2 Preliminaries

Throughout this paper, let C be a nonempty, closed and convex subset of a real Hilbert space H. We denote the strong and weak convergence of a sequence $\{x_n\}$ to a point $z^* \in H$ by $x_n \rightarrow z^*$ and $x_n \rightharpoonup z^*,$ respectively. We also denote the set of weak limits of $\{x_n\},$ by $\omega _*(x_n)$, defined as:

$$\begin{aligned} \omega _*(x_n):= \{z^*\in H: x_{n_k}\rightharpoonup z^*~ \text {for some subsequence}~ \{x_{n_k}\}~ \text {of} ~\{x_{n}\}\}. \end{aligned}$$

Definition 2.1

Let $S: H\rightarrow H$ be a mapping defined on a real Hilbert space H. Then, $\forall x,z\in H,$ S is said to be:

• (i) L-Lipschitz continuous, where $L>0,$ if

  $$\begin{aligned} \Vert Sx - Sz\Vert \le L\Vert x-z\Vert . \end{aligned}$$

  S is nonexpansive, if $L=1;$ and a contraction if $L\in [0,1).$

• (ii) quasi-nonexpansive on H,  if $F(S)\ne \emptyset$ and

  $$\begin{aligned} \Vert Sx-k\Vert \le \Vert x-k\Vert ,\quad \forall k\in F(S), \end{aligned}$$

• (iii) $\mu$-strictly pseudocontractive on H if

  $$\begin{aligned} \Vert Sx-Sz\Vert ^2 \le \Vert x-z\Vert ^2+\mu \Vert (I-S)x-(I-S)z\Vert ^2, \end{aligned}$$

  where $0\le \mu <1.$ If $\mu =1$, then S is said to be pseudocontractive on H; 

• (iv) $\sigma$-demicontractive if

  $$\begin{aligned} \Vert Sx-k\Vert ^2 \le \Vert x-k\Vert ^2+\sigma \Vert (I-S)x\Vert ^2, ~~\quad \forall k\in F(S), \quad ~~ \text {where}~~ 0\le \sigma <1. \end{aligned}$$

• (v) quasi-pseudocontractive if

  $$\begin{aligned} \Vert Sx-k\Vert ^2 \le \Vert x-k\Vert ^2+\Vert Sx-x\Vert ^2,~~\quad \forall k\in F(S), \end{aligned}$$

• (vii) sequentially weakly continuous, if for each sequence $\{x_n\}\subset H, ~~ x_n\rightharpoonup z^*$ implies that $Sx_n\rightharpoonup Sz^*.$

It is very clear that the class of quasi-pseudocontractive mappings are more general than the other classes of mappings discussed above, (see Alakoya et al. 2022; Taiwo et al. 2020). Below is an example to show a quasi-pseudocontractive mapping that is not a $\sigma -$ demicontractive mapping.

Example 2.2

Taiwo et al. (2021) Let H be the closed interval [0, 1] with the absolute value as a norm. Define $S:H\rightarrow H$ as follows:

$$\begin{aligned} Sx={\left\{ \begin{array}{ll} \frac{1}{2}, \quad x\in \left[ 0, \frac{1}{2}\right] ,\\ 0, \quad \quad x\in \big (\frac{1}{2},1\big ]. \end{array}\right. } \end{aligned}$$

We easily see that $F(S)=\frac{1}{2}.$ Also, for $x\in [0, \frac{1}{2}],$ we have

$$\begin{aligned} \left| Sx-\frac{1}{2}\right| ^2= \left| \frac{1}{2}-\frac{1}{2}\right| ^2=0 \le \left| x -\frac{1}{2}\right| ^2+\left| x-\frac{1}{2}\right| ^2, \end{aligned}$$

and for $(\frac{1}{2},1],$ we have that

$$\begin{aligned} \left| Sx-\frac{1}{2}\right| ^2= \left| 0-\frac{1}{2}\right| ^2 =\frac{1}{4} < \left| x-\frac{1}{2}\right| ^2+|x-0|^2. \end{aligned}$$

So, clearly, we see that for $x\in [0,1],$

$$\begin{aligned} \left| Sx-S\left( \frac{1}{2}\right) \right| ^2 \le \left| x-\frac{1}{2}\right| ^2+|x-Sx|^2. \end{aligned}$$

Thus, S is quasi-pseudocontractive. Now, we go further to show that S is not demicontractive, that is, $\not \exists ~ \sigma \in [0,1)$ such that

$$\begin{aligned} \left| Sx-S\left( \frac{1}{2}\right) \right| ^2 \le \sigma \left| x-\frac{1}{2}\right| ^2+|x-Sx|^2, \quad \quad \forall x\in [0,1]. \end{aligned}$$

By contradiction, we assume such $\sigma$ exists, then we have $\frac{1}{2}\le \frac{1}{\sigma +1}<1.$ For such $\sigma ,$ we choose x such that $\frac{1}{2}< x \le \frac{1}{\sigma +1},$ so that $\sigma <\frac{1-x}{x}$ and then, we have that

$$\begin{aligned} \left| x-\frac{1}{2}\right| ^2+\sigma |x-Sx|^2<\left| x-\frac{1}{2}\right| ^2+\frac{1-x}{x}|x-Sx|^2 = \frac{1}{4}=\left| Sx-S\left( \frac{1}{2}\right) \right| ^2. \end{aligned}$$

Hence, S is not demicontractive.

Definition 2.3

Suppose $\mathcal {F}: H\rightarrow H$ is a single-valued mapping defined on a real Hilbert space H. Then, $\forall a,b\in H,$ $\mathcal {F}$ is said to be:

• (i) $\tau$- strongly monotone, if there exists $\tau >0$ such that $\langle a-b, \mathcal {F}a-\mathcal {F}b \rangle \ge \tau \Vert a-b\Vert ^2;$

• (ii) monotone, if $\langle \mathcal {F}a - \mathcal {F}b, a-b\rangle \ge 0,$

• (iii) pseudomonotone, if $\langle \mathcal {F}a, b-a\rangle \ge 0 \implies \langle \mathcal {F}b, b-a\rangle \ge 0.$

• (iv) quasimonotone, if $\langle \mathcal {F}a, b-a\rangle > 0 \implies \langle \mathcal {F}b, b-a\rangle \ge 0.$

Lemma 2.4

(Ye and He 2015) If either,

• (i) $\mathcal {F}$ is pseudomonotone on C and $VI(C,\mathcal {F})\ne \emptyset ;$

• (ii) Suppose $\mathcal {F}$ is the gradient of $\mathcal {N}^*$, such that $\mathcal {N}^*$ is a differentiable quasiconvex function on an open set $\varPsi \supset C$ which attains its global minimum on C; 

• (iii) $\mathcal {F}$ is quasimonotone on $C, ~~ \text {int}C\ne \emptyset$ and $\exists ~x\in VI(C,\mathcal {F})$ such that $\mathcal {F}x\ne 0;$

• (iv) $\mathcal {F}$ is quasimonotone on $C, ~~ \mathcal {F}\ne 0$ on C and there is such $z\in \mathbb {N},$ that $\forall w\in C$ with $z\le \Vert w\Vert ,$ there exists $y\in C,$ such that $z\ge \Vert y\Vert$ and $\langle \mathcal {F}w, w-y \rangle \ge 0;$

• (v) $\mathcal {F}$ is quasimonotone on $C,~~ \mathcal {F}\ne 0,$ and C is bounded,

then $\bar{\Gamma }\ne \emptyset .$

Lemma 2.5

(Uzor et al. 2022a) Let H be a real Hilbert space. Then the following results hold, for all $x,y\in H$ and $\lambda \in \mathbb {R}:$

• (i) $\Vert x + y\Vert ^2 \le \Vert x\Vert ^2 + 2\langle y, x + y \rangle ;$

• (ii) $\Vert \lambda x + (1-\lambda ) y\Vert ^2 = \lambda \Vert x\Vert ^2 + (1-\lambda )\Vert y\Vert ^2 -\lambda (1-\lambda )\Vert x-y\Vert ^2;$

• (iii) $\Vert x + y\Vert ^2 = \Vert x\Vert ^2 + 2\langle x, y \rangle + \Vert y\Vert ^2.$

Definition 2.6

The metric projection $P_C: H\rightarrow C$ is defined, for each $z\in H,$ as the unique element $P_Cz\in C$ such that

$$\begin{aligned} ||z - P_Cz|| = \inf \{||z-y||: y\in C\}. \end{aligned}$$

It well-known that $P_C$ is nonexpansive (see Alakoya et al. (2022), Goebel and Reich (1984)). (See Lemma 2.7 for more characteristics of the projection mapping.)

Lemma 2.7

(Kopecká and Reich 2012) Suppose I is the identity mapping defined on a real Hilbert space H. Let C be a nonempty, closed, convex subset of H. Then, the following results hold for any $x\in H$ and $a\in C:$

• (i) $b = P_Cx \Longleftrightarrow \langle x - b, b - a\rangle \ge 0.$

• (ii) $\Vert P_Cx+P_Ca\Vert ^2=\Vert x-a\Vert ^2-\Vert x-P_Cx\Vert ^2.$

• (iii) $\langle x-a,P_Cx-P_Ca \rangle \ge \Vert P_Cx-P_Ca\Vert ^2.$

• (iv) $\Vert a-P_Cx\Vert ^2+\Vert x-P_Cx\Vert ^2 \le \Vert x-a\Vert ^2.$

• (v) $\langle (I-P_C)x-(I-P_C)a, x-a\rangle \ge \Vert (I-P_C)x-(I-P_C)a\Vert ^2.$

Lemma 2.8

(Ogwo et al. 2023) Let $S:H\rightarrow H$ be a nonlinear operator such that $F(S)\ne \emptyset .$ Then $I-S$ is said to be demiclosed at zero if for any sequence $\{x_n\}$ such that $x_n\rightharpoonup x$ and $(I-S)x_n\rightarrow 0$ implies $(I-S)x=0.$

Lemma 2.9

(Tan and Xu 1993) Let $\{\zeta _n\}$ and $\{\theta _n\}$ are two nonnegative real sequences such that

$$\begin{aligned} \zeta _{n+1}\le \zeta _n + \theta _n,\quad \forall n\ge 1. \end{aligned}$$

Suppose $\sum _{n=1}^{\infty }\theta _n<+\infty ,$ then $\lim \limits _{n\rightarrow \infty }\zeta _n$ exists.

Lemma 2.10

(Saejung and Yotkaew 2012) Let $\{p_n\}$ be a sequence of real numbers. Suppose $\{x_n\}$ is a sequence of non-negative real numbers and $\{\alpha _n\}$ is a sequence in (0, 1) with $\sum _{n=1}^\infty \alpha _n = +\infty$. Let

$$\begin{aligned} x_{n+1}\le (1 - \alpha _n)x_n + \alpha _np_n, ~~~ \text {for all}~~ n\ge 1. \end{aligned}$$

Suppose $\limsup _{k\rightarrow \infty }p_{n_k}\le 0$ for every subsequence $\{x_{n_k}\}$ of $\{x_n\}$ satisfying $\liminf _{k\rightarrow \infty }(x_{n_{k+1}} - x_{n_k})\ge 0,$ then $\lim _{n\rightarrow \infty }x_n =0.$

Lemma 2.11

(Alakoya et al. 2022; Chang et al. 2020) Let H be a real Hilbert space and $S:H\rightarrow H$ be an $L-Lipschitzian$ mapping with $L\ge 1.$ Denote $\mathcal {T}:=(1-\eta )I+\eta S[(1-\chi )I+\chi S].$ Suppose $0<\eta<\chi <\frac{1}{1+\sqrt{1+L^2}},$ then the following results hold:

• (i) $F(S)=F(S[(1-\chi )+\chi S])=F(\mathcal {T}).$

• (ii) Suppose $I-S$ is demiclosed at zero, then $I-\mathcal {T}$ is also demiclosed at zero.

• (iii) Suppose $S:H\rightarrow H$ is quasi-pseudocontractive, then the mapping $\mathcal {T}$ is quasinonexpansive.

3 Proposed method

Here, we present our algorithm: A self-adaptive inertial modified-Mann projection contraction method (SIMMPCM) for approximating the common solution of SEVIP involving quasimonotone operators, and SEFPP for quasi-pseudocontractive Lipschitz continuous mappings in real Hilbert spaces. Our results are based on some mild assumptions as follows:

Assumption 3.1

• (A1) The feasibility sets C and Q are nonempty, closed and convex subsets of the real Hilbert spaces $H_1$ and $H_2,$ respectively.

• (A2) $\mathcal {F}:H_1\rightarrow H_1$ and $\mathcal {G}:H_2\rightarrow H_2$ are quasimonotone and Lipschitz continuous mappings, with Lipschitz constants $\mathcal {D}_1,~\mathcal {D}_2$ respectively and satisfies the following property;

  • (i) whenever $\{x_n\}\subset C, ~~ x_n\rightharpoonup x^*,$ then $\Vert \mathcal {F}x^*\Vert \le \liminf _{n\rightarrow \infty }\Vert \mathcal {F}x_n\Vert ,$ and

  • (ii) whenever $\{y_n\}\subset Q, ~~ y_n\rightharpoonup y^*,$ then $\Vert \mathcal {G}y^*\Vert \le \liminf _{n\rightarrow \infty }\Vert \mathcal {G}y_n\Vert , respectively.$

• (A3) $T_1:H_1\rightarrow H_3$ and $T_2:H_2\rightarrow H_3$ are bounded linear operators, with adjoints, $T_1^*$ and $T_2^*$ respectively, $T_1\ne 0,~T_2\ne 0.$

• (A4) $S_1:H_1\rightarrow H_1$ and $S_2:H_2\rightarrow H_2$ are quasi-pseudocontractive mappings and Lipschitz continuous, with constants $L_1\ge 1$ and $L_2\ge 1,$ respectively, and $I-S_i$ is demiclosed at 0,  for each $i=1,2.$

• (A5) The solution set $\Omega := \{x\in \bar{\Gamma }\cap F(S_1), ~ y\in \bar{\Gamma }_*\cap F(S_2): T_1x=T_2y\}\ne \emptyset .$

• (A6) $\{\alpha _n\}, \{\theta _n\},$ and $\{\delta _n\},$ are nonnegative sequences such that: $\{\alpha _n\}\subset (0,1),~~\lim _{n\rightarrow \infty }\alpha _n=0,~~ \sum _{n=1}^{\infty }\alpha _n=\infty , ~~ \lim _{n\rightarrow \infty }\frac{\theta _n}{\alpha _n}=0,~~ \lim _{n\rightarrow \infty }\frac{\delta _n}{\alpha _n}=0.$

• (A7) $\{\lambda _n\}, \{\iota _n\},$ and $\{\beta _n\}$ are nonnegative sequences such that: $\sum _{n=1}^{\infty }\lambda _n<+\infty , \sum _{n=1}^{\infty }\iota _n<+\infty , \{\beta _n\}\subset [a,b]\subset (0, 1-\alpha _n),$ and $\ell \in (0,2).$

• (A8) $\{\psi _n\}\subset (0,1)$ and $\{\sigma _n\}\subset (0,1),$ such that $0<a\le \psi _n,\sigma _n\le b<1,$ and $0<\eta<\chi <\frac{1}{1+\sqrt{1+L^2}},$ where $L=\max \{L_1,L_2\}.$

Our algorithm is as follows:

Algorithm 3.2

Step 0.:

Let sequences $\{\alpha _n\}_{n=1}^\infty , \{\beta _n\}_{n=1}^\infty , \{\nu _n\}_{n=1}^\infty , \{\tau _n\}_{n=1}^\infty ,$ be chosen such Assumptions 3.1 is satisfied. Select initial point $(x_0,y_0)\in H_1\times H_2,$ let $\zeta \ge 0, \mu ,\phi \in (0,1), \xi _1>0, \kappa _1>0, \tau>0, \theta >0,$ and set $n:=1.$

Step 1.:

Given the $x_{n-1}, y_{n-1}$ and $x_n,y_n$ iterates, choose $\tau _n$ such that $0\le \tau _n\le \hat{\tau }_n,$ and $\nu _n$ such that $0\le \nu _n\le \hat{\nu }_n,$ for each $n\ge 1,$ where;

$$\begin{aligned} \hat{\tau }_n = {\left\{ \begin{array}{ll} \min \Big \{\tau ,~ \frac{\theta _n}{\Vert x_n - x_{n-1}\Vert }\Big \}, \quad \text {if}~ x_n \ne x_{n-1},\\ \tau , \qquad \text {otherwise.} \end{array}\right. } \end{aligned}$$

(3.1)

Step 2.:

Compute

$$\begin{aligned}&w_n = x_n + \tau _n(x_n - x_{n-1}); \end{aligned}$$

Step 3.:

Compute:

$$\begin{aligned}&z_n=w_n-\zeta _nT_1^*(T_1w_n-T_2g_n),\\&u_n=P_C(z_n-\xi _n\mathcal {F}z_n)\\&v_n=z_n-\ell \gamma _nd_n,\\&q_n=\psi _nv_n+(1-\psi _n)Yv_n,\\&x_{n+1}=(1-\alpha _n-\beta _n)z_n+\beta _nq_n, \end{aligned}$$

where,

$$\begin{aligned}&d_n=z_n-u_n-\xi _n(\mathcal {F}z_n-\mathcal {F}u_n), \nonumber \\&\gamma _n={\left\{ \begin{array}{ll} \frac{\langle z_n-u_n,d_n \rangle }{\Vert d_n\Vert ^2}, &{}\quad \text {if}~~ d_n\ne 0,\\ 0, &{}\quad \text {otherwise}; \end{array}\right. } \nonumber \\&xi_{n+1}={\left\{ \begin{array}{ll} \min \{\frac{\mu \Vert u_n-z_n\Vert }{\Vert \mathcal {F}u_n -\mathcal {F}z_n\Vert },\xi _n+\lambda _n\}, &{}\text {if}~~ \mathcal {F}u_n-\mathcal {F}z_n\ne 0,\\ \xi _n+\lambda _n, &{} \text {otherwise,} \end{array}\right. }\nonumber \\&\text {and} \quad Y=(1-\eta )I+\eta S_1[(1-\chi )I+\chi S_1]. \end{aligned}$$

(3.2)

Step 4.:

Compute:

$$\begin{aligned} \hat{\nu }_n = {\left\{ \begin{array}{ll} \min \Big \{\nu ,~ \frac{\delta _n}{\Vert y_n - y_{n-1}\Vert }\Big \}, &{}\quad \text {if}~ y_n \ne y_{n-1},\\ \nu , &{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

(3.3)

Step 5.:

Compute

$$\begin{aligned}&g_n = y_n + \nu _n(y_n - y_{n-1}); \end{aligned}$$

Step 6.:

Compute:

$$\begin{aligned}&t_n=g_n+\zeta _nT_2^*(T_1w_n-T_2g_n),\\&r_n=P_Q(t_n-\kappa _n\mathcal {G}t_n)\\&s_n=t_n-g\rho _nf_n,\\&b_n=\sigma _ns_n+(1-\sigma _n)Js_n,\\&y_{n+1}=(1-\alpha _n-\beta _n)t_n+\beta _nb_n, \end{aligned}$$

where,

$$\begin{aligned}&f_n=t_n-r_n-\sigma _n(\mathcal {G}t_n-\mathcal {G}r_n), \nonumber \\&\rho _n={\left\{ \begin{array}{ll} \frac{\langle t_n-r_n,f_n \rangle }{\Vert f_n\Vert ^2},&{} \quad \text {if}~~ f_n\ne 0,\\ 0, &{}\quad \text {otherwise}; \end{array}\right. }\nonumber \\&\kappa _{n+1}={\left\{ \begin{array}{ll} \min \{\frac{\phi \Vert t_n-r_n\Vert }{\Vert \mathcal {G}t_n-\mathcal {G}r_n\Vert }, \kappa _n+\iota _n\}, &{}\quad \text {if} ~~\mathcal {G}t_n-\mathcal {G}r_n\ne 0,\\ \kappa _n+\iota _n,&{} \quad \text {otherwise,} \end{array}\right. }\nonumber \\&\text {and} \quad J=(1-\eta )I+\eta S_2[(1-\chi )I+\chi S_2], \end{aligned}$$

(3.4)

Update: for small enough $\epsilon >0,$ choose

$$\begin{aligned}&\zeta _n \in \Big [\epsilon , \frac{2\Vert T_1w_n-T_2g_n\Vert ^2}{\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2+\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2}-\epsilon \Big ]&\quad \text {if} ~~ T_1w_n\ne T_2g_n;\\&\quad \text {otherwise,}~~ \zeta _n=\zeta . \end{aligned}$$

Set $n:= n +1$ and go back to Step 1.

Remark 3.3

Below are some of the key features of our proposed Algorithm 3.2.

• (i) Observe that in our method, there is only a single projection onto the feasible set C and Q. Unlike the work of Kazmi et al. (2019) with three projections onto the feasible set per iteration, which is much more difficult to compute.

• (ii) Equations (3.1) and (3.3) of our Algorithm 3.2 represent the inertial iterates, which help to hasten the rate of convergence of our method. Also, note that Step 1 and Step 4 of our proposed algorithm is easily implemented, due to our prior knowledge of the estimate $\Vert x_n-x_{n-1}\Vert$ before choosing $\tau _n,$ and $\nu _n.$

• (iii) We assume our cost operators to be quasimonotone, since they are known to be more general and less restrictive than the class of pseudo(monotone), and co-coercive operators (see Alakoya et al. 2022; Uzor et al. 2023). Moreover, there are only scanty results in literature with split equality problems involving quasimonotone operators.

• (iv) Observe also that our step size $\{\zeta _n\}$ is self-adaptive and does not depend on the operator norm of $T_1$ and $T_2,$ which is an improvement over the method proposed by Moudafi and Al-Shemas (2013).

• (v) Our proposed algorithm converges strongly to a minimum-norm solution of the problem (1.8).

Remark 3.4

From condition (A6)-(A8), (3.1), and (3.3), we have that

$$\begin{aligned}&\lim _{n\rightarrow \infty }\tau _n||x_n - x_{n-1}|| = 0 \quad \text {and}\quad \lim _{n\rightarrow \infty } \frac{\tau _n}{\alpha _n}||x_n - x_{n-1}|| = 0. \end{aligned}$$

(3.5)

$$\begin{aligned}&\lim _{n\rightarrow \infty }\nu _n||y_n - y_{n-1}|| = 0 \quad \text {and}\quad \lim _{n\rightarrow \infty }\frac{\nu _n}{\alpha _n}||y_n - y_{n-1}|| = 0. \end{aligned}$$

(3.6)

Also, from the definition of $\zeta _n,$ we get that

$$\begin{aligned}&\zeta _n \le \frac{2\Vert T_1w_n-T_2g_n\Vert ^2}{\Vert T_2^* (T_1w_n-T_2g_n\Vert ^2+\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2}-\epsilon \nonumber \\&\quad \implies (\zeta _n+\epsilon )[\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2 +\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2] \le 2\Vert T_1w_n-T_2g_n\Vert ^2 \nonumber \\&\quad \implies \zeta _n \cdot \epsilon [\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2 +\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2]\nonumber \\&\quad \le \zeta _n\Big ( 2\Vert T_1w_n-T_2g_n\Vert ^2-\zeta _n[\Vert T_2^* (T_1w_n-T_2g_n)\Vert ^2+\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2]\Big ). \end{aligned}$$

(3.7)

4 Convergence analysis

In this section, we carry out the convergence analysis of our proposed algorithm. Firstly, we establish some lemmas which are needed to prove the strong convergence theorem for our proposed algorithm.

Lemma 4.1

Let $\{\xi _n\}$ and $\{\kappa _n\}$ be sequences generated by be the sequence generated by (3.2) and (3.4), whereby assumption (A7) is satisfied. Then, it follows that $\lim \limits _{n\rightarrow \infty }\xi _n=\xi ,$ where $\xi \in [\min \{\frac{\mu }{\mathcal {D}_1}, \xi _1\}, \xi _1+\Lambda ],$ $\Lambda =\sum _{n=1}^{\infty }\lambda _n,$ for some constants $\mathcal {D}_1>0,$ and $\lim \limits _{n\rightarrow \infty }\kappa _n=\kappa ,$ where $\kappa \in [\min \{\frac{\phi }{\mathcal {D}_2}, \kappa _1\}, \kappa _1+\Theta ],$ $\Theta =\sum _{n=1}^{\infty }\iota _n,$ for some constants $\mathcal {D}_2>0.$

Proof

Note that Since $\mathcal {F}$ is $\mathcal {D}_1-$ Lipschitz continuous. Thus, considering the non-trivial case $(\Vert \mathcal {F}u_n-\mathcal {F}z_n\Vert \ne 0),$ we have for all $n\ge 1$ that

$$\begin{aligned} \frac{\mu \Vert u_n-z_n\Vert }{\Vert \mathcal {F}u_n-\mathcal {F}z_n\Vert } \ge \frac{\mu \Vert u_n-z_n\Vert }{\mathcal {D}_1\Vert u_n-z_n\Vert } =\frac{\mu }{\mathcal {D}_1}, \end{aligned}$$

From the definition of $\xi _{n+1},$ we know that the sequence $\{\xi _n\}$ is bounded below and above by $\min \{\frac{\mu }{\mathcal {D}_1},\xi _1\}$ and $\xi _1 + \Lambda ,$ respectively. Applying Lemma 2.9, we see that $\lim \limits _{n\rightarrow \infty }\xi _n$ exists, and is denoted by $\xi ,$ where $\xi =\lim \limits _{n\rightarrow \infty }\xi _n,$ and $\xi \in \big [\min \{\frac{\mu }{\mathcal {D}_1},\xi _1\},\xi _1+\Lambda \big ].$

By following the same pattern above, we have that $\lim \limits _{n\rightarrow \infty }\kappa _n=\kappa ,$ where $\kappa \in [\min \{\frac{\phi }{\mathcal {D}_2}, \kappa _1\}, \kappa _1+\Theta ]$. $\square$

Lemma 4.2

Assume that Algorithm 3.2 satisfies Assumption 3.1. Then, the following inequality holds for all $(x^*,y^*)\in \Omega .$

$$\begin{aligned} \Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2\le \Vert w_n-x^*\Vert ^2+\Vert g_n-y^*\Vert ^2. \end{aligned}$$

Proof

Let $(x^*,y^*)\in \Omega .$ Then, by applying Lemma 2.5, we have

$$\begin{aligned} \Vert z_n-x^*\Vert ^2&= \Vert w_n-\zeta _n T_1^*(T_1w_n-T_2g_n)-x^*\Vert ^2\nonumber \\&=\Vert w_n-x^*\Vert ^2 +\zeta _n^2 \Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2-2\zeta _n\langle w_n-x^*,T_1^*(T_1w_n-T_2g_n) \rangle \nonumber \\&=\Vert w_n-x^*\Vert ^2 +\zeta _n^2 \Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2-2\zeta _n\langle T_1w_n-T_1x^*,T_1w_n-T_2g_n \rangle \nonumber \\&=\Vert w_n-x^*\Vert ^2 + \zeta _n^2 \Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2-\zeta _n\Vert T_1w_n-T_1x^*\Vert ^2\nonumber \\&\quad -\zeta _n\Vert T_1w_n-T_2g_n\Vert ^2+\zeta _n\Vert T_2g_n-T_1x^*\Vert ^2. \end{aligned}$$

(4.1)

Similarly, we have that

$$\begin{aligned} \Vert t_n-y^*\Vert ^2&= \Vert g_n+\zeta _n T_2^*(T_1w_n-T_2g_n)-y^*\Vert ^2\nonumber \\&=\Vert g_n-y^*\Vert ^2 +\zeta _n^2 \Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2+2\zeta _n\langle g_n-y^*,T_2^*(T_1w_n-T_2g_n) \rangle \nonumber \\&=\Vert g_n-y^*\Vert ^2 +\zeta _n^2 \Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2+2\zeta _n\langle T_2g_n-T_2y^*,T_1w_n-T_2g_n \rangle \nonumber \\&=\Vert g_n-y^*\Vert ^2 + \zeta _n^2 \Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2 -\zeta _n\Vert T_2g_n-T_2y^*\Vert ^2\nonumber \\&\quad -\zeta _n\Vert T_1w_n-T_2g_n\Vert ^2+\zeta _n\Vert T_1w_n-T_2y^*\Vert ^2. \end{aligned}$$

(4.2)

Then, by summing (4.1) and (4.2), we get

$$\begin{aligned}&\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2= \Vert w_n-x^*\Vert ^2 +\Vert g_n-y^*\Vert ^2\\&\quad +\zeta _n^2[\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2+\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2]\\&\quad -\zeta _n[\Vert T_1w_n-T_1x^*\Vert ^2+\Vert T_2g_n-T_2y^*\Vert ^2] -2\zeta _n\Vert T_1w_n-T_2g_n\Vert ^2\\&\quad +\zeta _n[\Vert T_1w_n-T_2y^*\Vert ^2+\Vert T_2g_n-T_1x^*\Vert ^2]. \end{aligned}$$

By the fact that $T_1x^*=T_2y^*,$ together with (3.7), we obtain

$$\begin{aligned}&\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2 = \Vert w_n-x^*\Vert ^2 +\Vert g_n-y^*\Vert ^2\nonumber \\&\qquad +\zeta _n[2\Vert T_1w_n-T_2g_n\Vert ^2-\zeta _n (\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2 +\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2)]\nonumber \\&\quad \le \Vert w_n-x^*\Vert ^2 +\Vert g_n-y^*\Vert ^2-\zeta _n\cdot \epsilon [\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2 +\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2]\nonumber \\&\quad \le \Vert w_n-x^*\Vert ^2 +\Vert g_n-y^*\Vert ^2. \end{aligned}$$

(4.3)

$\square$

Lemma 4.3

Assume that Assumption 3.1 is satisfied by Algorithm 3.2. Then, the following inequalities hold for all $(x^*,y^*)\in \Omega .$

$$\begin{aligned} {\left\{ \begin{array}{ll} \Vert q_n-x^*\Vert ^2\le \Vert z_n-x^*\Vert ^2-\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2-\psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert , \quad \text {and}\\ \Vert z_n-u_n\Vert ^2\le \bigg [\frac{1}{\ell } \Big (\frac{\xi _{n+1}+\mu \xi _n}{\xi _{n+1}-\mu \xi _n}\Big ) \Vert z_n-v_n\Vert \bigg ]^2. \end{array}\right. } \end{aligned}$$

Also,

$$\begin{aligned} {\left\{ \begin{array}{ll} \Vert b_n-y^*\Vert ^2\le \Vert t_n-y^*\Vert ^2-\frac{2-g}{g}\Vert t_n-s_n\Vert ^2 -\kappa _n(1-\kappa _n)\Vert s_n-Js_n\Vert , \quad \text {and}\\ \Vert t_n-r_n\Vert ^2\le \bigg [\frac{1}{g}\Big (\frac{\kappa _{n+1} +\phi \kappa _n}{\kappa _{n+1}-\phi \kappa _n}\Big )\Vert t_n-s_n\Vert \bigg ]^2. \end{array}\right. } \end{aligned}$$

Proof

Let $(x^*,y^*)\in \Omega ,$ by Lemma 2.5 and the nonexpansivity of Y,  we get

$$\begin{aligned} \Vert q_n-x^*\Vert ^2&=\Vert \psi _nv_n+(1-\psi _n)Yv_n-x^*\Vert ^2 \nonumber \\&=\psi _n\Vert v_n-x^*\Vert ^2+(1-\psi _n)\Vert Yv_n-x^*\Vert ^2-\psi _n (1-\psi _n)\Vert v_n-Yv_n\Vert \nonumber \\&\le \psi _n\Vert v_n-x^*\Vert ^2+(1-\psi _n)\Vert v_n-x^*\Vert ^2-\psi _n (1-\psi _n)\Vert v_n-Yv_n\Vert \nonumber \\&=\Vert v_n-x^*\Vert ^2-\psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert . \end{aligned}$$

(4.4)

By the definition of $\{v_n\},$ we have that

$$\begin{aligned} \Vert v_n-x^*\Vert ^2=\Vert z_n-x^*\Vert ^2-2\ell \gamma _{n}\langle z_n-x^*,d_n \rangle +\ell ^2\gamma _{n}^2\Vert d_n\Vert ^2. \end{aligned}$$

(4.5)

By the definition of $\{d_n\},$ we get:

$$\begin{aligned} \langle z_n-x^*,d_n \rangle&= \langle z_n-u_n,d_n \rangle +\langle u_n-x^*,d_n \rangle \nonumber \\&= \langle z_n-u_n, d_n \rangle +\langle u_n-x^*, z_n-u_n-\xi _n(\mathcal {F}z_n-\mathcal {F}u_n) \rangle . \end{aligned}$$

(4.6)

By the definition of $\{u_n\}$ and the projection property (see Lemma 2.7), we have that

$$\begin{aligned} \langle z_n-u_n-\xi _n\mathcal {F}z_n, u_n-x^* \rangle \ge 0. \end{aligned}$$

(4.7)

Since, $x^*\in \bar{\Gamma }$ and $u_n\in C,$ we obtain $\langle \mathcal {F}u_n, u_n-x^* \rangle \ge 0,$ and by the positivity of $\{\xi _n\},$ we have

$$\begin{aligned} \xi _n \langle \mathcal {F}u_n, u_n-x^* \rangle \ge 0. \end{aligned}$$

(4.8)

Thus, we have from (4.6) that

$$\begin{aligned} \langle z_n-x^*, d_n \rangle \ge \langle z_n-u_n, d_n \rangle . \end{aligned}$$

(4.9)

From the definition of $\{v_n\},$ we have that

$$\begin{aligned} \Vert v_n-z_n\Vert =\Vert \ell \gamma _n d_n\Vert , \end{aligned}$$

By combining (4.5) and (4.9), we have

$$\begin{aligned} \Vert v_n-x^*\Vert ^2&\le \Vert z_n-x^*\Vert ^2-2\ell \gamma _{n} \langle z_n-u_n, d_n\rangle +\ell ^2\gamma _{n}^2\Vert d_n\Vert ^2, \nonumber \\&= \Vert z_n-x^*\Vert ^2-2\ell \gamma _{n}^2\Vert d_n\Vert ^2 + \ell ^2\gamma _{n}^2\Vert d_n\Vert ^2 \nonumber \\&=\Vert z_n-x^*\Vert ^2-\frac{2-\ell }{\ell } \Vert \ell \gamma _nd_n\Vert ^2 \nonumber \\&=\Vert z_n-x^*\Vert ^2-\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2. \end{aligned}$$

(4.10)

Combining (4.4), (4.10), and the conditions imposed on the control parameters, we obtain

$$\begin{aligned} \Vert q_n-x^*\Vert ^2&\le \Vert z_n-x^*\Vert ^2-\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2-\psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert \nonumber \\&\le \Vert z_n-x^*\Vert ^2. \end{aligned}$$

(4.11)

Similarly, since $y^*\in \bar{\Gamma }_*$, we obtain

$$\begin{aligned} \Vert b_n-y^*\Vert ^2&\le \Vert t_n-y^*\Vert ^2-\frac{2-g}{g} \Vert t_n-s_n\Vert ^2-\sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \nonumber \\&\le \Vert t_n-y^*\Vert ^2. \end{aligned}$$

(4.12)

Next, we have from (3.2), and the definition of $\{d_n\},$ that

$$\begin{aligned} \Vert \mathcal {F}z_n-\mathcal {F}u_n\Vert \le \frac{\mu }{\xi _{n+1}}\Vert z_n-u_n\Vert , \quad \forall n\ge 1. \end{aligned}$$

(4.13)

Consequently, we get

$$\begin{aligned} \gamma _n\Vert d_n\Vert ^2=\langle z_n-u_n, d_n \rangle&\ge \Vert z_n-u_n\Vert ^2-\xi _n\Vert \mathcal {F}z_n-\mathcal {F}u_n\Vert \Vert z_n-u_n\Vert \nonumber \\&\ge \Big (1-\frac{\mu \xi _n}{\xi _{n+1}}\Big )\Vert z_n-u_n\Vert ^2. \end{aligned}$$

(4.14)

Also,

$$\begin{aligned} \Vert d_n\Vert&\le \Vert z_n-u_n\Vert +\xi _n\Vert \mathcal {F}z_n-\mathcal {F}u_n\Vert \nonumber \\&\le \Vert z_n-u_n\Vert +\frac{\mu \xi _n}{\xi _{n+1}}\Vert z_n-u_n\Vert \nonumber \\&= \Big (1+\frac{\mu \xi _n}{\xi _{n+1}}\Big )\Vert z_n-u_n\Vert . \end{aligned}$$

(4.15)

By combining (4.14) and (4.15), we get

$$\begin{aligned} \gamma _n^2\Vert d_n\Vert ^2\ge \Big (1-\frac{\mu \xi _n}{\xi _{n+1}}\Big )^2\frac{\Vert z_n-u_n\Vert ^4}{\Vert d_n\Vert ^2}\ge \frac{\Big (1-\frac{\mu \xi _n}{\xi _{n+1}}\Big )^2}{\Big (1+\frac{\mu \xi _n}{\xi _{n+1}}\Big )^2}\Vert z_n-u_n\Vert ^2. \end{aligned}$$

Then,

$$\begin{aligned} \Vert v_n-z_n\Vert ^2=\ell ^2\gamma _n^2\Vert d_n\Vert ^2\ge \ell ^2 \frac{\Big (1-\frac{\mu \xi _n}{\xi _{n+1}}\Big )^2}{\Big (1+\frac{\mu \xi _n}{\xi _{n+1}}\Big )^2}\Vert z_n-u_n\Vert ^2, \end{aligned}$$

and

$$\begin{aligned} \Vert z_n-u_n\Vert ^2\le \bigg [\frac{1}{\ell }\Big (\frac{\xi _{n+1} +\mu \xi _n}{\xi _{n+1}-\mu \xi _n}\Big )\Vert z_n-v_n\Vert \bigg ]^2. \end{aligned}$$

$\square$

Using (3.4) and following a procedure similar to the one above, we get:

$$\begin{aligned}&\Vert \mathcal {G}t_n-\mathcal {G}r_n\Vert \le \frac{\phi }{\kappa _{n+1}}\Vert t_n-r_n\Vert , \quad ~~ \forall n\ge 1 \quad \text {and,} \nonumber \\&\Vert t_n-r_n\Vert ^2\le \bigg [\frac{1}{g} \Big (\frac{\kappa _{n+1}+\phi \kappa _n}{\kappa _{n+1} -\phi \kappa _n}\Big )\Vert t_n-s_n\Vert \bigg ]^2. \end{aligned}$$

Lemma 4.4

Suppose Algorithm 3.2 satisfies Assumption 3.1. Then, $\{(x_n,y_n)\}$ is bounded.

Proof

Let $x^*\in \Omega .$ Then, by Lemma 2.5 and the definition of $\{w_n\},$ we have

$$\begin{aligned} \Vert w_n-x^*\Vert&=\Vert (x_n+\tau _n(x_n-x_{n-1})-x^*\Vert \\&\le \Vert x_n-x^*\Vert +\tau _n\Vert x_n-x_{n-1}\Vert \\&=\Vert x_n-x^*\Vert +\alpha _n\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert . \end{aligned}$$

From Remark (), we have that $\lim _{n\rightarrow \infty }\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert =0.$ So, it implies that there exists $M_3>0$ such that

$$\begin{aligned}&\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert \le M_3,\quad \forall n\ge 1.\\&\quad \implies \Vert w_n-x^*\Vert \le \Vert x_n-x^*\Vert +\alpha _nM_3. \end{aligned}$$

Consequently,

$$\begin{aligned} \Vert w_n-x^*\Vert ^2\le \Vert x_n-x^*\Vert ^2+2\alpha _nM_3\Vert x_n-x^*\Vert +\alpha _n^2M_3^2. \end{aligned}$$

(4.16)

Similarly, using a similar procedure, we get that $\exists ~ M_4>0,$ such that

$$\begin{aligned} \Vert g_n-y^*\Vert ^2\le \Vert y_n-y^*\Vert ^2+2\alpha _nM_4\Vert y_n-y^*\Vert +\alpha _n^2M_4^2. \end{aligned}$$

(4.17)

By summing (4.16) and (4.17), we have

$$\begin{aligned}&\Vert w_n-x^*\Vert ^2+\Vert g_n-y^*\Vert ^2 \nonumber \\&\quad \le \Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2 +2\alpha _n[M_3\Vert x_n-x^*\Vert +M_4\Vert y_n-y^*\Vert ]+\alpha _n^2(M_3^2+M_4^2)\nonumber \\&\quad \le \Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2+2\alpha _n[M_3\Vert x_n-x^*\Vert +M_4\Vert y_n-y^*\Vert ] +\alpha _n(M_3^2+M_4^2) \nonumber \\&\quad =\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2+\alpha _nM_1^*,\nonumber \\&\qquad \text {where}~~ M_1^*=[2(M_3\Vert x_n-x^*\Vert +M_4\Vert y_n-y^*\Vert )+M_3^2+M_4^2]>0. \end{aligned}$$

(4.18)

By the definition of $\{x_{n+1}\},$ we get

$$\begin{aligned} \Vert x_{n+1}-x^*\Vert&=\Vert (1-\alpha _n-\beta _n)(z_n-x^*) +\beta _n(q_n-x^*)-\alpha _nx^* \Vert \nonumber \\&\le \Vert (1-\alpha _n-\beta _n)(z_n-x^*)+\beta _n (q_n-x^*)\Vert +\alpha _n\Vert x^*\Vert \end{aligned}$$

(4.19)

By (4.11) and Lemma 2.5, we obtain

$$\begin{aligned}&\Vert (1-\alpha _n-\beta _n)(z_n-x^*)+\beta _n(q_n-x^*)\Vert ^2\\&\quad =\Vert (1-\alpha _n-\beta _n)^2\Vert z_n-x^*\Vert ^2 +2(1-\alpha _n-\beta _n)\beta _n\langle z_n-x^*,q_n-x^*\rangle +\beta _n^2\Vert q_n-x^*\Vert ^2 \\&\quad \le \Vert (1-\alpha _n-\beta _n)^2\Vert z_n-x^*\Vert ^2 +2(1-\alpha _n-\beta _n)\beta _n\\&\qquad \quad \times [\Vert z_n-x^*\Vert ^2+\Vert q_n-x^*\Vert ^2] +\beta _n^2\Vert q_n-x^*\Vert ^2 \\&\quad =(1-\alpha _n-\beta _n)(1-\alpha _n)\Vert z_n-x^*\Vert ^2 +\beta _n(1-\alpha _n)\Vert q_n-x^*\Vert ^2 \\&\quad \le (1-\alpha _n-\beta _n)(1-\alpha _n) \Vert z_n-x^*\Vert ^2+\beta _n(1-\alpha _n)\Vert z_n-x^*\Vert ^2\\&\quad =(1-\alpha _n)^2\Vert z_n-x^*\Vert ^2. \end{aligned}$$

So, we have that

$$\begin{aligned} \Vert (1-\alpha _n-\beta _n)(z_n-x^*)+\beta _n(q_n-x^*)\Vert \le (1-\alpha _n)\Vert z_n-x^*\Vert . \end{aligned}$$

(4.20)

By combining (4.19) and (4.20), we obtain

$$\begin{aligned} \Vert x_{n+1}-x^*\Vert \le (1-\alpha _n)\Vert z_n-x^*\Vert +\alpha _n\Vert x^*\Vert . \end{aligned}$$

(4.21)

From (4.21), we have that

$$\begin{aligned} \Vert x_{n+1}-x^*\Vert ^2\le (1-\alpha _n)^2\Vert z_n-x^*\Vert ^2 +2\alpha _n(1-\alpha _n)\Vert x^*\Vert \Vert z_n-x^*\Vert +\alpha _n^2\Vert x^*\Vert ^2. \end{aligned}$$

(4.22)

Following, a similar procedure, we get that

$$\begin{aligned} \Vert y_{n+1}-y^*\Vert ^2\le (1-\alpha _n)^2\Vert t_n-y^*\Vert ^2 +2\alpha _n(1-\alpha _n)\Vert y^*\Vert \Vert t_n-y^*\Vert +\alpha _n^2\Vert y^*\Vert ^2. \end{aligned}$$

(4.23)

Adding (4.22) and (4.23), we obtain

$$\begin{aligned}&\Vert x_{n+1}-x^*\Vert ^2+\Vert y_{n+1}-y^*\Vert ^2 \le (1-\alpha _n)^2[\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2]\nonumber \\&\qquad +2\alpha _n(1-\alpha _n)[\Vert x^*\Vert \Vert z_n-x^*\Vert +\Vert y^*\Vert \Vert t_n-y^*\Vert ] +\alpha _n^2[\Vert x^*\Vert ^2+\Vert y^*\Vert ^2]\nonumber \\&\quad \le (1-\alpha _n) [\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2]\nonumber \\&\qquad +2\alpha _n[\Vert x^*\Vert \Vert z_n-x^*\Vert +\Vert y^*\Vert \Vert t_n-y^*\Vert ] +\alpha _n[\Vert x^*\Vert ^2+\Vert y^*\Vert ^2]\nonumber \\&\quad \le (1-\alpha _n) \Big [\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2\Big ]+ \alpha _nM_2^*,\nonumber \\&\qquad \text {where,} ~~ M_2^*= [2(\Vert x^*\Vert \Vert z_n -x^*\Vert +\Vert y^*\Vert \Vert t_n-y^*\Vert )+\Vert x^*\Vert ^2+\Vert y^*\Vert ^2]>0. \end{aligned}$$

(4.24)

By combining (4.24), (4.3), and (4.18), we obtain

$$\begin{aligned} \Vert x_{n+1}-x^*\Vert ^2+\Vert y_{n+1}-y^*\Vert ^2&\le (1-\alpha _n) [\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2]+ \alpha _nM_2^* \\&\le (1-\alpha _n) [\Vert w_n-x^*\Vert ^2+\Vert g_n-y^*\Vert ^2+\alpha _nM_1^*]+ \alpha _nM_2^*\\&\le (1-\alpha _n) [\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2+\alpha _nM_1^*]+ \alpha _nM_2^*\\&\le (1-\alpha _n) [\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2] +\alpha _nM_1^*+ \alpha _nM_2^* \\&= (1-\alpha _n) [\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2] +\alpha _nM_3^* \\&\le \max \{\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2,M_3^*\}\\&~~\vdots \\&\le \max \Big \{\Vert x_{n_0}-x^*\Vert ^2+\Vert y_{n_0}-y^*\Vert ^2,M_3^*\Big \}.\\&\text {where,} ~~ M_3^*=(M_1^*+M_2^*)>0. \end{aligned}$$

Hence, the sequence $\{(x_n,y_n)\}$ is bounded. Therefore, $\{z_n\},$ $\{t_n\},$ $\{q_n\},$ and $\{b_n\}$ are all also bounded. $\square$

Lemma 4.5

Assume that Algorithm 3.2 satisfies Assumption 3.1. Then, the following inequality hold for all $(x^*,y^*)\in \Omega .$

$$\begin{aligned}&\beta _n(1-\alpha _n)\Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2 + \frac{2-g}{g} \Vert t_n-s_n\Vert ^2 + \psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert \\&\qquad + \sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \Big ]\\&\qquad +(1-\alpha _n)\zeta _n\cdot \epsilon \Big [\Vert T_1^*(T_1w_n-T_2g_n)\Vert ^2+\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2\Big ]\\&\quad \le (1-\alpha _n)[\Vert x_n-x^*\Vert ^2 + \Vert y_n-y^*\Vert ^2]\\&\qquad -[\Vert x_{n+1}-x^*\Vert ^2+\Vert y_{n+1}-y^*\Vert ^2] +\alpha _nM_4^* +3M_5\alpha _n(1-\alpha _n)\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert \\&\qquad +3M_6\alpha _n(1-\alpha _n)\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert . \end{aligned}$$

Proof

Let $x^*\in \Omega ,$ then by definition of $\{w_n\}$ and Lemma 2.5, we get

$$\begin{aligned} \Vert w_n-x^*\Vert ^2&=\Vert x_n+\tau _n(x_n-x_{n-1})-x^*\Vert ^2\nonumber \\&=\Vert x_n-x^*\Vert ^2+\tau _n^2\Vert x_n-x_{n-1}\Vert ^2+2\tau _n\langle x_n-x^*,x_n-x_{n-1} \rangle \nonumber \\&\le \Vert x_n-x^*\Vert ^2+\tau _n^2\Vert x_n-x_{n-1}\Vert ^2+2\tau _n\Vert x_n-x^*\Vert \Vert x_n-x_{n-1} \Vert \nonumber \\&=\Vert x_n-x^*\Vert ^2+\tau _n\Vert x_n-x_{n-1}\Vert (\tau _n\Vert x_n-x_{n-1}\Vert +2\Vert x_n-x^*\Vert )\nonumber \\&\le \Vert x_n-x^*\Vert ^2+3M_5\tau _n\Vert x_n-x_{n-1}\Vert \nonumber \\&=\Vert x_n-x^*\Vert ^2+3M_5\alpha _n\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert ,\nonumber \\&\text {where,} ~~ M_5=\sup _{n\in \mathbb {N}}\{\Vert x_n-x^*\Vert ,\tau _n\Vert x_n-x_{n-1}\}>0. \end{aligned}$$

(4.25)

Similarly, we have that

$$\begin{aligned}&\Vert g_n-y^*\Vert ^2 \le \Vert y_n-y^*\Vert ^2+3M_6\alpha _n\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert ,\nonumber \\&\text {where,} ~~ M_6=\sup _{n\in \mathbb {N}}\{\Vert y_n-y^*\Vert ,\nu _n\Vert y_n-y_{n-1}\}>0. \end{aligned}$$

(4.26)

$\square$

Using (4.25), Lemma 2.5, and (4.11), we have

$$\begin{aligned} \Vert x_{n+1}-x^*\Vert ^2= & {} \Vert (1-\alpha _n-\beta _n)(z_n-x^*)+\beta _n(q_n-x^*)-\alpha _nx^*\Vert ^2\nonumber \\\le & {} \Vert (1-\alpha _n-\beta _n)(z_n-x^*)+\beta _n(q_n-x^*)\Vert ^2-2\alpha _n\langle x^*,x_{n+1}-x^* \rangle \nonumber \\= & {} (1-\alpha _n-\beta _n)^2\Vert z_n-x^*\Vert ^2 +\beta _n^2\Vert q_n-x^*\Vert ^2\nonumber \\{} & {} +2\beta _n(1-\alpha _n-\beta _n)\langle z_n-x^*,q_n-x^* \rangle +2\alpha _n\langle x^*,x^*-x_{n+1} \rangle \nonumber \\\le & {} (1-\alpha _n-\beta _n)^2\Vert z_n-x^*\Vert ^2 +\beta _n^2\Vert q_n-x^*\Vert ^2\nonumber \\{} & {} +\beta _n(1-\alpha _n-\beta _n)[\Vert z_n-x^*\Vert ^2+ \Vert q_n-x^* \Vert ^2] +2\alpha _n\langle x^*,x^*-x_{n+1} \rangle \nonumber \\= & {} (1-\alpha _n-\beta _n)(1-\alpha _n)\Vert z_n-x^*\Vert ^2\nonumber \\{} & {} +\beta _n(1-\alpha _n)\Vert q_n-x^*\Vert ^2+2\alpha _n\langle x^*,x^*-x_{n+1} \rangle \nonumber \\\le & {} (1-\alpha _n-\beta _n)(1-\alpha _n)\Vert z_n-x^*\Vert ^2 +2\alpha _n\langle x^*,x^*-x_{n+1} \rangle \nonumber \\{} & {} +\beta _n(1-\alpha _n)\Big [\Vert z_n-x^*\Vert ^2-\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2-\psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert \Big ] \nonumber \\= & {} (1-\alpha _n)^2\Vert z_n-x^*\Vert ^2 +2\alpha _n\langle x^*,x^*-x_{n+1} \rangle \nonumber \\{} & {} -\beta _n(1-\alpha _n)\Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2+\psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert \Big ] \nonumber \\\le & {} (1-\alpha _n)\Vert z_n-x^*\Vert ^2 +2\alpha _n\langle x^*,x^* -x_{n+1} \rangle \nonumber \\{} & {} -\beta _n(1-\alpha _n)\Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2+\psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert \Big ]. \end{aligned}$$

(4.27)

By following a procedure similar to the one above, we obtain

$$\begin{aligned} \Vert y_{n+1}-y^*\Vert ^2&\le (1-\alpha _n)\Vert t_n-y^*\Vert ^2 +2\alpha _n\langle y^*,y^*-y_{n+1} \rangle \nonumber \\&\quad -\beta _n(1-\alpha _n)\Big [\frac{2-g}{g} \Vert t_n-s_n\Vert ^2+\sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \Big ]. \end{aligned}$$

(4.28)

We sum (4.27), (4.28), and (4.3) to get

$$\begin{aligned}{} & {} \Vert x_{n+1}-x^*\Vert ^2+\Vert y_{n+1}-y^*\Vert ^2 \le (1-\alpha _n)[\Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2] \\{} & {} \qquad +2\alpha _n[\langle x^*,x^*-x_{n+1} \rangle +\langle y^*,y^*-y_{n+1} \rangle ]-\beta _n(1-\alpha _n)\\{} & {} \qquad \Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2 +\frac{2-g}{g} \Vert t_n-s_n\Vert ^2 \\{} & {} \qquad + \psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert + \sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \Big ]\\{} & {} \quad \le (1-\alpha _n)\Big [\Vert w_n-x^*\Vert ^2+\Vert g_n -y^*\Vert ^2-\zeta _n\cdot \epsilon [\Vert T_1^* (T_1w_n-T_2g_n)\Vert ^2\\{} & {} \qquad +\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2]\Big ] +2\alpha _n[\langle x^*,x^*-x_{n+1} \rangle +\langle y^*,y^*-y_{n+1} \rangle ]\\{} & {} \qquad -\beta _n(1-\alpha _n)\Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2 +\frac{2-g}{g} \Vert t_n-s_n\Vert ^2\\{} & {} \qquad + \psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert + \sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \Big ]\\{} & {} \quad \le (1-\alpha _n)\Big [\Vert x_n-x^*\Vert ^2 + \Vert y_n-y^*\Vert ^2 +3M_5\alpha _n\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert \\{} & {} \qquad +3M_6\alpha _n\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert \Big ]\\{} & {} \qquad -(1-\alpha _n)\zeta _n\cdot \epsilon \Big [\Vert T_1^* (T_1w_n-T_2g_n)\Vert ^2+\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2\Big ]\\{} & {} \qquad +2\alpha _n[\langle x^*,x^*-x_{n+1} \rangle +\langle y^*,y^*-y_{n+1} \rangle ]\\{} & {} \qquad -\beta _n(1-\alpha _n)\Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2+ \frac{2-g}{g} \Vert t_n-s_n\Vert ^2\\{} & {} \qquad + \psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert + \sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \Big ]\\{} & {} \quad \le (1-\alpha _n)[\Vert x_n-x^*\Vert ^2 + \Vert y_n-y^*\Vert ^2]+\alpha _nM_4^*\\{} & {} \qquad +3M_5\alpha _n(1-\alpha _n)\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert +3M_6\alpha _n(1-\alpha _n)\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert \\{} & {} \qquad -(1-\alpha _n)\zeta _n\cdot \epsilon \Big [\Vert T_1^* (T_1w_n-T_2g_n)\Vert ^2+\Vert T_2^*(T_1w_n-T_2g_n)\Vert ^2\Big ]\\{} & {} \qquad -\beta _n(1-\alpha _n)\Big [\frac{2-\ell }{\ell } \Vert z_n-v_n\Vert ^2+ \frac{2-g}{g} \Vert t_n-s_n\Vert ^2\\{} & {} \qquad + \psi _n(1-\psi _n)\Vert v_n-Yv_n\Vert + \sigma _n(1-\sigma _n)\Vert s_n-Js_n\Vert \Big ], \end{aligned}$$

where $M_4^*=[\Vert x^*\Vert ^2+\Vert x^*-x_{n+1}\Vert ^2+\Vert y^*\Vert ^2 +\Vert y^*-y_{n+1}\Vert ^2]>0$.

Thus, we have our required inequality.

Lemma 4.6

Assume that Assumption 3.1 is satisfied by Algorithm 3.2. Then, the following inequality hold for all $(x^*,y^*)\in \Omega :$

$$\begin{aligned}&\Vert x_{n+1}-x^*\Vert ^2 +\Vert y_{n+1}-y^*\Vert ^2 \le (1-\alpha _n)[\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2]\\&\quad +\alpha _n\Big [3M_5(1-\alpha _n) \frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert +3M_6(1-\alpha _n)\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert \\&\quad +2\beta _n \Vert z_n-q_n\Vert \Vert x_{n+1}-x^*\Vert + 2\beta _n \Vert t_n-b_n\Vert \Vert y_{n+1}-y^*\Vert \\&\quad +2\langle x^*, x^*-x_{n+1} \rangle +2\langle y^*, y^*-y_{n+1} \rangle \Big ]. \end{aligned}$$

Proof

Let $(x^*,y^*)\in \Omega ,$ and we set $l_n=(1-\beta _n)z_n+\beta _nq_n.$ By applying (4.25), (4.11), and Lemma 2.5, we get that $\square$

$$\begin{aligned} \Vert l_n-x^*\Vert ^2&=\Vert (1-\beta _n)(z_n-x^*)+\beta _n(q_n-x^*)\Vert ^2\nonumber \\&=(1-\beta _n)^2\Vert z_n-x^*\Vert ^2+\beta _n^2\Vert q_n-x^*\Vert ^2 +2\beta _n(1-\beta _n)\langle z_n-x^*,q_n-x^* \rangle \nonumber \\&\le (1-\beta _n)^2\Vert z_n-x^*\Vert ^2+\beta _n^2 \Vert q_n-x^*\Vert ^2\nonumber \\&\quad +2\beta _n(1-\beta _n)[\Vert z_n-x^*\Vert ^2+ \Vert q_n-x^* \Vert ^2] \nonumber \\&=(1-\beta _n)\Vert z_n-x^*\Vert ^2+\beta _n\Vert q_n-x^*\Vert ^2 \nonumber \\&\le (1-\beta _n)\Vert z_n-x^*\Vert ^2+\beta _n\Vert z_n-x^*\Vert ^2 \nonumber \\&=\Vert z_n-x^*\Vert ^2. \end{aligned}$$

(4.29)

Following a procedure similar to the one above, we set $p_n=(1-\alpha _n)t_n+\beta _n b_n,$ and we obtain

$$\begin{aligned} \Vert p_n-y^*\Vert ^2\le \Vert t_n-y^*\Vert ^2. \end{aligned}$$

(4.30)

By adding (4.29), (4.30), and applying (4.25) and (4.26), we have

$$\begin{aligned}&\Vert l_n-x^*\Vert ^2+\Vert p_n-y^*\Vert ^2\le \Vert z_n-x^*\Vert ^2+\Vert t_n-y^*\Vert ^2 \nonumber \\&\quad \le \Vert w_n-x^*\Vert ^2+\Vert g_n-y^*\Vert ^2 \nonumber \\&\quad \le \Vert x_n-x^*\Vert ^2+3M_5\alpha _n\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert +\Vert y_n-y^*\Vert ^2+3M_6\alpha _n\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert . \end{aligned}$$

(4.31)

Also, we have that

$$\begin{aligned} x_{n+1}&=l_n-\alpha _nz_n\nonumber \\&=(1-\alpha _n)l_n-\alpha _n(z_n-l_n)\nonumber \\&=(1-\alpha _n)l_n-\alpha _n\beta _n(z_n-q_n). \end{aligned}$$

(4.32)

Similarly, we obtain

$$\begin{aligned} y_{n+1}=(1-\alpha _n)p_n-\alpha _n\beta _n(t_n-b_n). \end{aligned}$$

(4.33)

Using (4.33), (4.32), together with Lemma 2.5, we get

$$\begin{aligned} \Vert x_{n+1}-x^*\Vert ^2&=\Vert (1-\alpha _n)(l_n-x^*)-[\alpha _n \beta _n(z_n-q_n)+\alpha _nx^*]\Vert ^2 \nonumber \\&\le (1-\alpha _n)^2\Vert l_n-x^*\Vert ^2-2\langle \alpha _n\beta _n(z_n-q_n)+\alpha _nx^*,x_{n+1}-x^* \rangle \nonumber \\&\le (1-\alpha _n)^2\Vert l_n-x^*\Vert ^2+2\alpha _n\beta _n \langle z_n-q_n,x^*-x_{n+1}\rangle +2\alpha _n\langle x^*, x^*-x_{n+1} \rangle \nonumber \\&\le (1-\alpha _n)\Vert l_n-x^*\Vert ^2+\alpha _n[2\beta _n \Vert z_n-q_n\Vert \Vert x_{n+1}-x^*\Vert +2\langle x^*, x^*-x_{n+1} \rangle ]. \end{aligned}$$

(4.34)

Following a similar procedure, we obtain

$$\begin{aligned} \Vert y_{n+1}-y^*\Vert ^2 \le (1-\alpha _n)\Vert p_n-y^*\Vert ^2+\alpha _n[2\beta _n \Vert t_n-b_n\Vert \Vert y_{n+1}-y^*\Vert +2\langle y^*, y^*-y_{n+1} \rangle ]. \end{aligned}$$

(4.35)

Adding (4.34), (4.35), together with (4.31), we get

$$\begin{aligned}&\Vert x_{n+1}-x^*\Vert ^2 +\Vert y_{n+1}-y^*\Vert ^2 \le (1-\alpha _n)[\Vert l_n-x^*\Vert ^2+\Vert p_n-y^*\Vert ^2] \\&\qquad +\alpha _n\Big [2\beta _n \Vert z_n-q_n\Vert \Vert x_{n+1}-x^*\Vert + 2\beta _n \Vert t_n-b_n\Vert \Vert y_{n+1}-y^*\Vert \\&\qquad +2\langle x^*, x^*-x_{n+1} \rangle +2\langle y^*, y^*-y_{n+1} \rangle \Big ] \\&\quad \le (1-\alpha _n)\left[ \Vert x_n-x^*\Vert ^2+3M_5\alpha _n \frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert +\Vert y_n-y^*\Vert ^2\right. \\&\qquad \left. +3M_6\alpha _n\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert \right] \\&\qquad +\alpha _n\Big [2\beta _n \Vert z_n-q_n\Vert \Vert x_{n+1}-x^*\Vert + 2\beta _n \Vert t_n-b_n\Vert \Vert y_{n+1}-y^*\Vert \\&\qquad +2\langle x^*, x^*-x_{n+1} \rangle +2\langle y^*, y^*-y_{n+1} \rangle \Big ] \\&\quad = (1-\alpha _n)[\Vert x_n-x^*\Vert ^2+\Vert y_n-y^*\Vert ^2] +\alpha _n\Big [3M_5(1-\alpha _n)\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert \\&\qquad +3M_6(1-\alpha _n)\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert \\&\qquad +2\beta _n \Vert z_n-q_n\Vert \Vert x_{n+1}-x^*\Vert + 2\beta _n \Vert t_n-b_n\Vert \Vert y_{n+1}-y^*\Vert \\&\qquad +2\langle x^*, x^*-x_{n+1} \rangle +2\langle y^*, y^*-y_{n+1} \rangle \Big ]. \end{aligned}$$

Lemma 4.7

Suppose Assumption 3.1 holds for Algorithm 3.2. If there exists a subsequence $\{(z_{n_k}, t_{n_k})\}$ of $\{(z_n,t_n)\}$ such that $(z_{n_k}, t_{n_k})\rightharpoonup (\hat{x},\hat{y})\in H_1\times H_2,$ with $\lim _{k\rightarrow \infty }\Vert z_{n_k}-u_{n_k}\Vert =0$ and $\lim _{k\rightarrow \infty }\Vert t_{n_k}-r_{n_k}\Vert =0.$ Then, either $(\hat{x},\hat{y})\in \bar{\Gamma }\times \bar{\Gamma }_*$ or $\mathcal {F}\hat{x}=\mathcal {G}\hat{y}=0.$

Proof

We already know that $\{z_n\}$ is bounded, thus $\omega _*(z_n)\ne \emptyset$. Then for an arbitrary $\hat{x}\in \omega _* (z_n),$ there exists a subsequence $\{z_{n_k}\}$ of $\{z_n\}$ such that $z_{n_k}\rightharpoonup \hat{x},~~k\rightarrow \infty .$ By hypothesis, we have that $u_{n_k}\rightharpoonup \hat{x}\in C.$ We then consider two cases as follows.

First Case: If $\limsup _{k\rightarrow \infty }\Vert \mathcal {F}u_{n_k}\Vert =0.$ it implies that $\lim _{k\rightarrow \infty }\Vert \mathcal {F}u_{n_k}\Vert =\liminf _{k\rightarrow \infty }\Vert \mathcal {F}u_{n_k}\Vert =0.$ By condition (A2),  we have $0\le \Vert \mathcal {F}\hat{x}\Vert \le \liminf _{k\rightarrow \infty }\Vert \mathcal {F}u_{n_k}\Vert =0.$ Hence, $\mathcal {F}\hat{x}=0.$

Second Case: If $\limsup _{k\rightarrow \infty }\Vert \mathcal {F}u_{n_k}\Vert >0.$ Without loss of generality, let $\lim _{k\rightarrow \infty }\Vert \mathcal {F}u_{n_k}\Vert =\mathcal {J}_*>0.$ Thus, there exists $n_2\in \mathbb {N}$ such that $\Vert \mathcal {F}u_{n_k}\Vert >\frac{\mathcal {J}_*}{2}, ~~\forall k\ge n_2.$ By the definition of $\{u_n\}$ and the projection property, $\forall x\in C,$ we obtain

$$\begin{aligned}&\langle u_{n_k}-z_{n_k}+\xi _{n_k}\mathcal {F}z_{n_k}, x-u_{n_k} \rangle \ge 0 \implies \langle z_{n_k}-u_{n_k}, x-u_{n_k} \rangle \le \xi _{n_k} \langle \mathcal {F}z_{n_k}, x-u_{n_k} \rangle \nonumber \\&\quad \implies \frac{1}{\xi _{n_k}}\langle z_{n_k}-u_{n_k}, x-u_{n_k} \rangle \le \langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle +\langle \mathcal {F}z_{n_k}-\mathcal {F}u_{n_k}, x-u_{n_k} \rangle . \end{aligned}$$

(4.36)

Also, since $\{u_{n_k}\}$ is bounded, and $\lim _{k\rightarrow \infty }\xi _{n_k}=\xi >0,~\lim _{k \rightarrow \infty }\Vert u_{n_k}-z_{n_k}\Vert =0,$ and by the Lipschitz continuity of $\mathcal {F},$ we have

$$\begin{aligned} 0\le \liminf _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle \le \limsup _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle <+\infty . \end{aligned}$$

(4.37)

If $\limsup _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle >0,$ then there exists a subsequence $\{u_{n_{k_j}}\}$ such that $\lim _{j\rightarrow \infty } \langle \mathcal {F}u_{n_{k_j}}, x-u_{n_{k_j}} \rangle >0.$ Therefore, there exist $j_*\in \mathbb {N}$ such that $\langle \mathcal {F}u_{n_{k_j}}, x-u_{n_{k_j}} \rangle >0, ~~\forall j\ge j_*.$ By the quasimonotone property of $\mathcal {F},~ \langle \mathcal {F}x, x-u_{n_{k_j}} \rangle \ge 0,~~\forall j\ge j_*.$ Thus, $\hat{x}\in \bar{\Gamma },$ as $j\rightarrow \infty .$ However, if $\limsup _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle =0,$ then from (4.37), we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle = \liminf _{k\rightarrow \infty } \langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle =\limsup _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle =0. \end{aligned}$$

We set $\bar{\varsigma }_k=|\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle |+\frac{1}{k+1}.$ Then,

$$\begin{aligned} \langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle + \bar{\varsigma }_k>0. \end{aligned}$$

(4.38)

Let $\bar{\varkappa }_{n_k}=\frac{\mathcal {F}u_{n_k}}{\Vert \mathcal {F} u_{n_k}\Vert ^2}~\forall k\ge n_2.$ Then, we get that

$$\begin{aligned} \langle \mathcal {F}u_{n_k}, \bar{\varkappa }_{n_k} \rangle =1. \end{aligned}$$

(4.39)

Consequently from (4.38), we get $\langle \mathcal {F}u_{n_k}, x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \rangle >0,\quad \forall k\ge n_2.$ Since $\mathcal {F}$ is quasimonotone and Lipschitz continuous, we have $\langle \mathcal {F}(x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}), x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \rangle \ge 0, ~~ \forall k\ge n_2.$ Thus, $\forall k\ge n_2,$ it follows that

$$\begin{aligned} \langle \mathcal {F}x, x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \rangle&= \langle \mathcal {F}x-\mathcal {F}(x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}), x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \rangle \nonumber \\&\quad +\langle \mathcal {F}(x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}), x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \rangle \nonumber \\&\ge \langle \mathcal {F}x-\mathcal {F}(x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}), x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \rangle \nonumber \\&\ge - \Vert \mathcal {F}x-\mathcal {F}(x+\bar{\varsigma }_k \bar{\varkappa }_{n_k})\Vert \Vert x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k} \Vert \nonumber \\&\ge -\bar{\varsigma }_k\mathcal {D}_1\Vert \bar{\varkappa }_{n_k}\Vert \Vert x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k}\Vert \nonumber \\&= -\bar{\varsigma }_k\frac{\mathcal {D}_1}{\Vert \mathcal {F}u_{n_k}\Vert }\Vert x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k}\Vert \nonumber \\&\ge -\bar{\varsigma }_k\frac{2\mathcal {D}_1}{\mathcal {J}_*}\Vert x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k}\Vert . \end{aligned}$$

(4.40)

Since $\{\Vert x+\bar{\varsigma }_k \bar{\varkappa }_{n_k}-u_{n_k}\Vert \}$ is bounded and $\lim _{k\rightarrow \infty }\bar{\varsigma }_k=0.$ Then, by taking $k\rightarrow \infty$ in (4.40), we obtain $\langle \mathcal {F}x,x-\hat{x} \rangle \ge 0, ~~\forall x\in C.$ Therefore, $\hat{x}\in \bar{\Gamma }.$ Going by a similar procedure, we get that $\hat{y}\in \bar{\Gamma }_*.$ Thus, our proof is complete. $\square$

Theorem 4.8

Suppose Algorithm 3.2 satisfies Assumption 3.1 with $\mathcal {F}x\ne 0,~~\forall x\in C$ and $~\mathcal {G}y \ne 0,~\forall y\in Q.$ Then the sequence $\{(x_n,y_n)\}$ converges strongly to $(\hat{x},\hat{y})\in \Omega ,$ where $\hat{x}=P_A(0)$ and $\hat{y}=P_B(0);$ $A:=\bar{\Gamma }\cap F(S_1)$ and $B:=\bar{\Gamma }_*\cap F(S_2).$

Proof

Let $(\hat{x},\hat{y})\in \Omega ,$ where $\hat{x}=P_A(0)$ and $\hat{y}=P_B(0).$ Then we have by Lemma 4.6 that

$$\begin{aligned}&\Vert x_{n+1}-\hat{x}\Vert ^2 +\Vert y_{n+1}-\hat{y}\Vert ^2 \le (1-\alpha _n)[\Vert x_n-\hat{x}\Vert ^2+\Vert y_n-\hat{y}\Vert ^2]+\alpha _nb_n^*, \nonumber \\&\quad \text {where,}~~ b_n^*=\Big [3M_5(1-\alpha _n)\frac{\tau _n}{\alpha _n}\Vert x_n-x_{n-1}\Vert +3M_6(1-\alpha _n)\frac{\nu _n}{\alpha _n}\Vert y_n-y_{n-1}\Vert \nonumber \\&\quad +2\beta _n \Vert z_n-q_n\Vert \Vert x_{n+1}-\hat{x}\Vert + 2\beta _n \Vert t_n-b_n\Vert \Vert y_{n+1}-\hat{y}\Vert +2\langle \hat{x}, \hat{x}-x_{n+1} \rangle +2\langle \hat{y}, \hat{y}-y_{n+1} \rangle \Big ]. \end{aligned}$$

(4.41)

Hence, we claim that the sequence $\{\Vert x_n-\hat{x}\Vert +\Vert y_n-\hat{y}\Vert \}$ converges to zero. Thus, it suffices to show by Lemma 2.10 that $\lim _{k\rightarrow \infty }b_n^*\le 0,$ for every subsequence $\{\Vert x_{n_k}-\hat{x}\Vert +\Vert y_{n_k}-\hat{y}\Vert \}$ of $\{\Vert x_n-\hat{x}\Vert +\Vert y_n-\hat{y}\Vert \},$ satisfying

$$\begin{aligned} \liminf _{k\rightarrow \infty }[(\Vert x_{n_{k+1}}-\hat{x}\Vert +\Vert y_{n_{k+1}}-\hat{y}\Vert )-(\Vert x_{n_k}-\hat{x}\Vert +\Vert y_{n_k}-\hat{y}\Vert )] \ge 0. \end{aligned}$$

(4.42)

Now, we assume that $\{\Vert x_{n_k}-\hat{x}\Vert +\Vert y_{n_k}-\hat{y}\Vert \}$ is a subsequence of $\{\Vert x_n-\hat{x}\Vert +\Vert y_n-\hat{y}\Vert \},$ such that (4.42) is satisfied. And so, by Lemma 4.5, we get

$$\begin{aligned}&\beta _{n_k}(1-\alpha _{n_k})\Big [\frac{2-\ell }{\ell } \Vert z_{n_k}-v_{n_k}\Vert ^2+ \frac{2-g}{g} \Vert t_{n_k}-s_{n_k}\Vert ^2 + \psi _{n_k}(1-\psi _{n_k})\Vert v_{n_k}-Yv_{n_k}\Vert \\&\qquad + \sigma _{n_k}(1-\sigma _{n_k})\Vert s_{n_k} -Js_{n_k}\Vert \Big ]+(1-\alpha _{n_k})\zeta _{n_k}\\&\qquad \cdot \epsilon \Big [\Vert T_1^*(T_1w_{n_k}-T_2g_{n_k})\Vert ^2 +\Vert T_2^*(T_1w_{n_k}-T_2g_{n_k})\Vert ^2\Big ]\\&\quad \le (1-\alpha _{n_k})[\Vert x_{n_k}-\hat{x}\Vert ^2 +\Vert y_{n_k}-\hat{y}\Vert ^2]-[\Vert x_{{n_k}+1}-\hat{x}\Vert ^2 +\Vert y_{{n_k}+1}-\hat{y}\Vert ^2]+\alpha _{n_k}M_4^*\\&\qquad +3M_5\alpha _{n_k}(1-\alpha _{n_k}) \frac{\tau _{n_k}}{\alpha _{n_k}}\Vert x_{n_k}-x_{{n_k}-1}\Vert +3M_6\alpha _{n_k}(1-\alpha _{n_k})\frac{\nu _{n_k}}{\alpha _{n_k}}\Vert y_{n_k}-y_{{n_k}-1}\Vert . \end{aligned}$$

Thus, by (4.42), Remark 3.4, and the conditions imposed on the control parameters, we get that

$$\begin{aligned}&\lim _{k\rightarrow \infty }\Bigg [\beta _{n_k}(1-\alpha _{n_k}) \Big [\frac{2-\ell }{\ell } \Vert z_{n_k}-v_{n_k}\Vert ^2+ \frac{2-g}{g} \Vert t_{n_k}-s_{n_k}\Vert ^2 + \psi _{n_k}(1-\psi _{n_k})\Vert v_{n_k}-Yv_{n_k}\Vert \\&\qquad + \sigma _{n_k}(1-\sigma _{n_k})\Vert s_{n_k} -Js_{n_k}\Vert \Big ]+(1-\alpha _{n_k})\zeta _{n_k}\\&\qquad \cdot \epsilon \Big [\Vert T_1^*(T_1w_{n_k}-T_2g_{n_k}) \Vert ^2+\Vert T_2^*(T_1w_{n_k}-T_2g_{n_k})\Vert ^2\Big ]\Bigg ]=0. \end{aligned}$$

Thus, we obtain

$$\begin{aligned}&\lim _{k\rightarrow \infty }\Vert z_{n_k}-v_{n_k}\Vert =0, \lim _{k\rightarrow \infty }\Vert t_{n_k}-s_{n_k}\Vert =0,\nonumber \\&\lim _{k\rightarrow \infty }\Vert v_{n_k}-Yv_{n_k}\Vert =0, \lim _{k\rightarrow \infty }\Vert s_{n_k}-Js_{n_k}\Vert =0,\nonumber \\&\lim _{k\rightarrow \infty }[\Vert T_1^*(T_1w_{n_k}-T_2g_{n_k})\Vert ^2 +\Vert T_2^*(T_1w_{n_k}-T_2g_{n_k})\Vert ^2]=0, \end{aligned}$$

(4.43)

Consequently,

$$\begin{aligned}{} & {} \lim _{k\rightarrow \infty }\Vert T_1^*(T_1w_{n_k}-T_2g_{n_k})\Vert =0, \lim _{k\rightarrow \infty }\Vert T_2^*(T_1w_{n_k}-T_2g_{n_k})\Vert =0,\\{} & {} \lim _{k\rightarrow \infty }\Vert (T_1w_{n_k}-T_2g_{n_k})\Vert =0. \end{aligned}$$

By the definition of $\{z_{n_k}\}, \{t_{n_k}\}$ and previous inequality, we see that

$$\begin{aligned}&\lim _{k\rightarrow \infty }\Vert z_{n_k}-w_{n_k}\Vert =\lim _{k\rightarrow \infty }\zeta _{n_k}\Vert T_1^*(T_1w_{n_k}-T_2g_{n_k})\Vert =0,\nonumber \\&\lim _{k\rightarrow \infty }\Vert t_{n_k}-g_{n_k}\Vert =\lim _{k\rightarrow \infty }\zeta _{n_k}\Vert T_2^*(T_1w_{n_k}-T_2g_{n_k})\Vert =0. \end{aligned}$$

(4.44)

Also, using Lemma 4.3 and (4.43), we get

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert z_{n_k}-u_{n_k}\Vert =0, \lim _{k\rightarrow \infty }\Vert t_{n_k}-r_{n_k}\Vert =0. \end{aligned}$$

(4.45)

By Remark 3.4, Step 2 and Step 5, we have

$$\begin{aligned}&\lim _{k\rightarrow \infty }\Vert w_{n_k}-x_{n_k}\Vert =\lim _{k\rightarrow \infty }\tau _{n_k}\Vert x_{n_k}-x_{n-1}\Vert =0, \quad \text {and,}\nonumber \\&\lim _{k\rightarrow \infty }\Vert g_{n_k}-y_{n_k}\Vert =\lim _{k\rightarrow \infty }\nu _{n_k}\Vert y_{n_k}-y_{n-1}\Vert =0. \end{aligned}$$

(4.46)

Then, (4.43)-(4.46), together with the onditions on the control parameters yields

$$\begin{aligned}&\lim _{k\rightarrow \infty }\Vert z_{n_k}-x_{n_k}\Vert =0; \lim _{k\rightarrow \infty }\Vert t_{n_k}-y_{n_k}\Vert =0; \lim _{k\rightarrow \infty }\Vert u_{n_k}-v_{n_k}\Vert =0; \nonumber \\&\lim _{k\rightarrow \infty }\Vert r_{n_k}-s_{n_k}\Vert =0; \lim _{k\rightarrow \infty }\Vert q_{n_k}-v_{n_k}\Vert =0; \lim _{k\rightarrow \infty }\Vert b_{n_k}-s_{n_k}\Vert =0. \end{aligned}$$

(4.47)

Using (4.43)-(4.47), we get

$$\begin{aligned}&\lim _{k\rightarrow \infty }\Vert z_{n_k}-q_{n_k}\Vert = \lim _{k\rightarrow \infty }\Vert t_{n_k}-b_{n_k}\Vert =0 \nonumber \\&\lim _{k\rightarrow \infty }\Vert u_{n_k}-x_{n_k}\Vert =\lim _{k \rightarrow \infty }\Vert r_{n_k}-x_{n_k}\Vert =0. \end{aligned}$$

(4.48)

Applying (4.21), (4.47), and the fact that $\lim _{k\rightarrow \infty }\alpha _{n_k}=0,$ we have

$$\begin{aligned}&\Vert x_{n+1}-x_{n_k}\Vert \le (1-\alpha _{n_k})\Vert z_{n_k}-x_{n_k}\Vert +\alpha _{n_k}\Vert x_{n_k}\Vert \rightarrow 0.~~ \text {as}~~ n\rightarrow \infty \nonumber \\&\quad \text {so also,}~~ \lim _{k\rightarrow \infty }\Vert y_{n+1}-y_{n_k}\Vert = \lim _{k\rightarrow \infty }\Vert x_{n+1}-w_{n_k}\Vert =0, ~~ \text {and}~~ \lim _{k\rightarrow \infty }\Vert y_{n+1}-g_{n_k}\Vert =0. \end{aligned}$$

(4.49)

To complete our proof, it suffices to show that $(\omega _* (x_n),\omega _* (y_n))\in \Omega .$ Since $\{x_n\}$ and $\{y_n\}$ are bounded, then $\omega _* (x_n)$ and $\omega _* (y_n)$ are nonempty. So, we choose $\hat{x}\in \omega _* (x_n)$ and $\hat{y}\in \omega _* (y_n)$ arbitrarily. By definition, there exists a subsequence $\{(x_{n_k},y_{n_k})\}$ of $\{(x_n,y_n)\}$ such that $(x_{n_k},y_{n_k})\rightharpoonup (\hat{x}, \hat{y}),$ as $k\rightarrow \infty .$ From (4.46) and (4.48), we have that $(\omega _* (x_n),\omega _* (y_n))=(\omega _* (w_n),\omega _* (g_n))=(\omega _* (u_n),\omega _* (r_n))$ which implies that $(w_{n_k},g_{n_k})\rightharpoonup (\hat{x},\hat{y}),$ as $k\rightarrow \infty .$ We have that $(\hat{x},\hat{y}) \in C\times Q$ since $(u_n,r_n)\in C\times Q$ and C, Q are both weakly closed. Now, since we assume that $\mathcal {F}x\ne 0,~~\forall x\in C$ and $~\mathcal {G}y \ne 0,~\forall y\in Q.$ Then, $\mathcal {F}\hat{x}\ne 0$ and $\mathcal {G}\hat{y}\ne 0.$ By applying Lemma 4.7 and (4.45), we get that $(\hat{x},\hat{y})\in \bar{\Gamma } \times \bar{\Gamma }_*.$ Hence, $(\hat{x},\hat{y})\in \Omega .$ Moreover, since $I-S_i$ is demiclosed at zero, $i=1,2,$ then by Lemma 2.11, (4.43) and (4.47), it follows that $\hat{x}=F(Y)=F(S_1),$ and $\hat{y}=F(J)=F(S_2).$ Thus, we have that $(\hat{x},\hat{y})\in \Omega .$ Again, since $(T_1\hat{x}-T_2\hat{y})\in \omega _* (T_1w_n-T_2g_n),$ we have from the weakly lower semicontinuity of the norm that

$$\begin{aligned} \Vert T_1\hat{x}-T_2\hat{y}\Vert \le \liminf _{n\rightarrow \infty }\Vert T_1w_n-T_2g_n\Vert =0. \end{aligned}$$

Therefore, $(\hat{x},\hat{y})\in \Omega .$ Now, since $\hat{x}\in \omega _* (x_n)$ and $\hat{y}\in \omega _* (y_n)$ are chosen arbitrarily, it shows that $(\omega _* (x_n),\omega _* (y_n))\in \Omega .$ Next, we show that

$$\begin{aligned} \limsup _{k\rightarrow \infty }\big (\langle \hat{x},\hat{x}-x_{n+1} \rangle +\langle \hat{y},\hat{y}-y_{n+1} \rangle \big ) \le 0. \end{aligned}$$

By the boundedness of $\{x_{n_k}\},$ there exists a subsequence $\{x_{n_{k_j}}\}$ of $\{x_{n_k}\}$ that converges weakly to some $x^\dagger \in H,$ and such that

$$\begin{aligned} \lim _{j\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_{k_j}} \rangle = \limsup _{k\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_k} \rangle . \end{aligned}$$

(4.50)

since $\hat{x}=P_A(0),$ then using (4.50), we obtain

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_k} \rangle&= \lim _{j\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_{k_j}} \rangle = \langle \hat{x},\hat{x}-x^\dagger \rangle \le 0. \end{aligned}$$

(4.51)

So, by (4.47) and (4.51), we get

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_{k+1}} \rangle = \limsup _{k\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_k} \rangle = \langle \hat{x},\hat{x}-x^\dagger \rangle \le 0. \end{aligned}$$

(4.52)

Also, since $\hat{y}=P_B(0),$ by following a similar procedure, we obtain

$$\begin{aligned} \limsup _{k\rightarrow \infty }\langle \hat{y},\hat{y}-y_{n_{k+1}} \rangle = \limsup _{k\rightarrow \infty }\langle \hat{y},\hat{y}-y_{n_k} \rangle = \langle \hat{y},\hat{y}-y^\dagger \rangle \le 0. \end{aligned}$$

(4.53)

By adding up (4.52) and (4.53), we have

$$\begin{aligned} \big (\limsup _{k\rightarrow \infty }\langle \hat{x},\hat{x}-x_{n_{k+1}} \rangle + \limsup _{k\rightarrow \infty }\langle \hat{y},\hat{y}-y_{n_{k+1}} \rangle \big )\le 0. \end{aligned}$$

(4.54)

Thus, we get from (4.47), (4.54), and (4.48) that

$$\begin{aligned} \limsup _{k\rightarrow \infty }b_n^*\le 0. \end{aligned}$$

Then, by applying Lemma 2.10 to (4.41), we see that $\{\Vert x_n-\hat{x}\Vert +\Vert y_n-\hat{y}\Vert \}$ converges to zero. Thus, $\lim _{n\rightarrow \infty }\Vert x_n-\hat{x}\Vert =0$ and $\lim _{n\rightarrow \infty }\Vert y_n-\hat{y}\Vert =0.$ Hence, $\{(x_n,y_n)\}$ converges strongly to $(\hat{x},\hat{y}).$ $\square$

Remark 4.9

Relaxing the condition $\mathcal {F}x\ne 0,~~\forall x\in C$ and $~\mathcal {G}y \ne 0,~\forall y\in Q$ has been of concern to researchers lately. We will investigate possible ways of bridging this gap in our future research.

Remark 4.10

In what follows, we present strong convergence result without the monotonicity property under the following condition A2$^*$ below:

(A2$^*$):

$\mathcal {F}:H_1\rightarrow H_1$ and $\mathcal {G}:H_2\rightarrow H_2$ are Lipschitz continuous mappings, with Lipschitz constants $\mathcal {D}_1,~\mathcal {D}_2$ respectively and satisfies the following property;

• (i) $\mathcal {F}$ and $\mathcal {G}$ are sequentially weakly continuous on C and Q,  respectively,

• (ii) If $x_n\rightharpoonup x^*$ and $\limsup _{n\rightarrow \infty }\langle \mathcal {F}x_n,x_n\rangle \le \langle \mathcal {F}x^*,x^*\rangle \implies \lim _{n\rightarrow \infty }\langle \mathcal {F}x_n,x_n\rangle = \langle \mathcal {F}x^*,x^*\rangle ,$ and

• (ii) If $y_n\rightharpoonup y^*$ and $\limsup _{n\rightarrow \infty }\langle \mathcal {G}y_n,y_n\rangle \le \langle \mathcal {G}y^*,y^*\rangle \implies \lim _{n\rightarrow \infty }\langle \mathcal {G}y_n,y_n\rangle = \langle \mathcal {G}y^*,y^*\rangle .$

Lemma 4.11

Suppose assumptions A1, A2$^*$, A3-A8 hold for Algorithm 3.2. If there exists a subsequence $\{(z_{n_k}, t_{n_k})\}$ of $\{(z_n,t_n)\}$ such that $(z_{n_k}, t_{n_k})\rightharpoonup (\hat{x},\hat{y})\in H_1\times H_2,$ with $\lim _{k\rightarrow \infty }\Vert z_{n_k}-u_{n_k}\Vert =0$ and $\lim _{k\rightarrow \infty }\Vert t_{n_k}-r_{n_k}\Vert =0.$ Then, either $(\hat{x},\hat{y})\in \bar{\Gamma }\times \bar{\Gamma }_*$ or $\mathcal {F}\hat{x}=\mathcal {G}\hat{y}=0.$

Proof

By a similar procedure as in Lemma 4.7, we fix $x\in C.$ Using the hypothesis, since $y_{n_k}\rightharpoonup \hat{x}\in C,$ and by (4.37), we have

$$\begin{aligned} \liminf _{k\rightarrow \infty }\langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle \ge 0. \end{aligned}$$

We now choose a positive sequence $\{\bar{\varsigma }_k\}$ such that $\bar{\varsigma }_k\rightarrow 0, ~~ k\rightarrow \infty ,$ and

$$\begin{aligned} \langle \mathcal {F}u_{n_k}, x-u_{n_k} \rangle + \bar{\varsigma }_k >0, \quad \forall k\in \mathbb {N}. \end{aligned}$$

Thus,

$$\begin{aligned} \langle \mathcal {F}u_{n_k}, x\rangle + \bar{\varsigma }_k > \langle \mathcal {F}u_{n_k},u_{n_k} \rangle , \quad \forall k\in \mathbb {N}. \end{aligned}$$

(4.55)

By setting $x=\hat{x}$ in (4.55), we then obtain

$$\begin{aligned} \langle \mathcal {F}u_{n_k}, \hat{x}\rangle + \bar{\varsigma }_k > \langle \mathcal {F}u_{n_k},u_{n_k} \rangle , \quad \forall k\in \mathbb {N}. \end{aligned}$$

(4.56)

Following condition A2$^*$(ii) and since $u_{n_k}\rightharpoonup \hat{x}, ~~ k\rightarrow \infty ,$ we have

$$\begin{aligned} \langle \mathcal {F}\hat{x},\hat{x} \rangle \ge \limsup _{k\rightarrow \infty } \langle \mathcal {F}u_{n_k},u_{n_k} \rangle \implies \lim _{k\rightarrow \infty } \langle \mathcal {F}u_{n_k},u_{n_k} \rangle =\langle \mathcal {F}\hat{x},\hat{x} \rangle . \end{aligned}$$

(4.57)

Hence, using (4.55)-(4.57), we obtain

$$\begin{aligned} \langle \mathcal {F}\hat{x},x \rangle&= \lim _{k\rightarrow \infty } (\langle \mathcal {F}u_{n_k},x \rangle +\bar{\varsigma }_k) \\&\ge \liminf _{k\rightarrow \infty } \langle \mathcal {F}u_{n_k},u_{n_k} \rangle \\&=\lim _{k\rightarrow \infty } \langle \mathcal {F}u_{n_k},u_{n_k} \rangle \\&=\langle \mathcal {F}\hat{x},\hat{x} \rangle . \end{aligned}$$

Therefore, $\langle \mathcal {F}\hat{x},x-\hat{x} \rangle \ge 0, \forall x\in C.$ Thus, $\hat{x}\in \bar{\Gamma }.$ Which follows that either $\hat{x}\in \bar{\Gamma },$ or $\mathcal {F}\hat{x}=0.$ Similarly, we also get that $\hat{y}\in \bar{\Gamma }_*.$ This completes our proof. $\square$

Theorem 4.12

Suppose Algorithm 3.2 satisfies assumptions A1, A2$^*$, A3-A8 hold with $\mathcal {F}x\ne 0,~~\forall x\in C$ and $~\mathcal {G}y \ne 0,~\forall y\in Q.$ Then the sequence $\{(x_n,y_n)\}$ converges strongly to $(\hat{x},\hat{y})\in \Omega ,$ where $\hat{x}=P_A(0)$ and $\hat{y}=P_B(0);$ $A:=\bar{\Gamma }\cap F(S_1)$ and $B:=\bar{\Gamma }_*\cap F(S_2).$

Proof

Applying Lemma 4.11 and following a similar procedure in Theorem 4.8, we get our desired result. $\square$

5 Numerical examples

In this section, we conduct certain numerical experiments to showcase the efficiency of our proposed Algorithm 3.2 (Proposed Alg.), in comparison with Algorithm 1.1 proposed by Kwelegano et al. (2022), Appendix 6.1 proposed by Taiwo et al. (2020), Appendix 6.2 proposed by Chang et al. (2015), and Appendix 6.3 proposed by Wang and Fang (2017). Our experiments are performed using MATLAB R2021(b). The stopping criterion used for each experiment is $\Vert x_{n+1}-x_{n}\Vert +\Vert y_{n+1}-y_{n}\Vert < 10^{-9}.$

Our numerical experiments will be conducted using the following examples below:

Example 5.1

Kwelegano et al. (2022) Let $H_1=H_2=H_3=\mathbb {R}^2$ be endowed with the norm

$$\begin{aligned} \Vert x\Vert :=\langle x,x \rangle ^{\frac{1}{2}}=\Bigl \{\sum _{n=1}^{2}\Vert x_n\Vert ^2\Bigr \}^{\frac{1}{2}}\quad \text {for all}~~ x\in \mathbb {R}^2, \end{aligned}$$

and inner product

$$\begin{aligned} \langle x,z \rangle =\sum _{n=1}^{2}x_nz_n, \quad x,z\in \mathbb {R}^2. \end{aligned}$$

Let C and Q be defined as $C=\{x\in \mathbb {R}^2:\Vert x\Vert \le 2\}$ and $Q=\{y\in \mathbb {R}^2:\Vert y\Vert \le 1\},$ respectively. Let the mapping $\mathcal {F}:\mathbb {R}^2\rightarrow \mathbb {R}^2$ and $\mathcal {G}:\mathbb {R}^2\rightarrow \mathbb {R}^2$ be defined by

$$\begin{aligned}&\mathcal {F}x=\mathcal {F}(x_1,x_2)={\left\{ \begin{array}{ll} \big (3-\sqrt{x_1^2+x_2^2}\big )(x_1,x_2), ~~ \text {if}~~ x\in C,\\ (x_1,x_2), \quad \text {if}~~ x \notin C, &{}\end{array}\right. }\\ \text {and}\\&\mathcal {G}y=\mathcal {G}(y_1,y_2)=(1-y_2,y_1), \quad \text {respectively.} \end{aligned}$$

Clearly, C and Q are nonempty, closed and convex subsets of $\mathbb {R}^2.$ Also, $\mathcal {F}$ and $\mathcal {G}$ are uniformly continuous and sequentially weakly continuous mappings on subsets of C and Q,  respectively. We also note that $\mathcal {F}$ and $\mathcal {G}$ are pseudomonotone on $\mathbb {R}^2.$ We choose $T_1x=T(x_1,x_2)=(3x_1,6x_2),$ $T_2y=T_2(y_1,y_2)=(7y_1,0),$ with adjoints $T_1^*x=T_1x = 5x,$ $T_2^*y=T_2y=2y.$ Let $Sx=-\frac{5}{4}x$ be a $\frac{5}{4}-$ Lipschitz quasi-pseudocontractive operator. We choose $S_1(x_1,x_2)=(-\frac{5}{3}x_1,x_2), ~~ S_2(y_1,y_2)=(-\frac{4}{5}y_1,y_2),$ where $S_1, S_2$ are Lipschitz quasipseudocontractive mappings (Table 1).

The choices of the starting points $x_0, x_1, y_0$ and $y_1$ are generated randomly in $\mathbb {R}^2$ as follows:

Case 1: $x_0=rand(2,1),~ x_1=rand(2,1),~y_0=rand(2,1),~y_1 = rand(2,1);$

Case 2: $x_0=rand(2,1),~ x_1=0.2\times rand(2,1),~y_0=rand(2,1),~y_1 = 0.3\times rand(2,1);$

Case 3: $x_0=3\times rand(2,1),~ x_1=rand(2,1),~y_0=2\times rand(2,1),~y_1 = rand(2,1);$

Case 4: $x_0=0.5\times rand(2,1),~ x_1=0.2\times rand(2,1),~y_0=2\times rand(2,1),~y_1 = 0.3\times rand(2,1).$

Table 1 Control Parameters for Examples 5.1- 5.2

Fig. 1

Example 5.1 Case 1

Fig. 2

Example 5.1 Case 2

Fig. 3

Example 5.1 Case 3

Fig. 4

Example 5.1 Case 4

Example 5.2

Taiwo et al. (2020) Let $H_1=H_2=H_3=(l_2(\mathbb {R}),\Vert \cdot \Vert ),$ such that $l_2(\mathbb {R}):=\{x=(x_1,x_2,...,x_n,...); x_k\in \mathbb {R}:\{\sum _{k=1}^{\infty }|x_k|^2<+\infty \}\},$ and the norm defined on $l_2(\mathbb {R})$ as follows:

$$\begin{aligned} \Big (\sum _{k=1}^{\infty }|x_k|^2\Big )^\frac{1}{2}, \quad \forall x\in l_2(\mathbb {R}). \end{aligned}$$

We define our feasibility sets C and Q as follows: $C:=\{x\in l_2(\mathbb {R}):x_k\ge 0, k\ge 1.\}$ and $Q:=\{y\in l_2(\mathbb {R}):y_j\ge 0, j\ge 1.\}.$ Let $S_1:H_1\rightarrow H_1$ and $S_2:H_2\rightarrow H_2$ be defined by $S_1(x)=-\frac{5}{4}x$ and $S_2(y)=-2y,$ respectively. Then $S_1$ is $\frac{5}{4}-$Lipschitzian quasi-pseudocontractive, and $S_2$ is $2-$Lipschitzian quasi-pseudocontractive mappings. We choose $T_1x=\frac{1}{2}x$ and $T_2y=3y.$ Also, $\mathcal {F}$ and $\mathcal {G}$ are pseudomonotone mappings defined as $\mathcal {F}=\frac{1}{3}x, ~ \mathcal {G}=\frac{1}{5}y.$ We choose different starting points as follows:

Case 1: $x_0=(1,0.1,0.01,\cdots ),~x_1 =(3,1,\frac{1}{3},\cdots ),~y_0=(2,0.2,0.02,\cdots ), ~y_1=(2,1,\frac{1}{2},\cdots );$

Case 2: $x_0=(1,\frac{1}{2},\frac{1}{4},\cdots ), ~x_1=(2,1,\frac{1}{2},\cdots ),~y_0=(\frac{3}{2}, \frac{1}{2},\frac{1}{6},\cdots ),~y_1=(3,1,\frac{1}{3},\cdots );$

Case 3: $x_0=(2,0.2,0.02,\cdots ),~x_1=(1,\frac{1}{2}, \frac{1}{4},\cdots ),~y_0=(1,\frac{1}{3},\frac{1}{9},\cdots ), ~y_1=(3,1,\frac{1}{3},\cdots );$

Case 4: $x_0=(1,\frac{1}{2},\frac{1}{4},\cdots ), ~x_1=(2,0.2,0.02,\cdots ),~y_0=(\frac{3}{2},\frac{1}{2}, \frac{1}{6},\cdots ),~y_1=(1,0.1,0.01,\cdots ).$

Fig. 5

Example 5.2 Case 1

Fig. 6

Example 5.2 Case 2

Fig. 7

Example 5.2 Case 3

Fig. 8

Example 5.2 Case 4

Table 2 Numerical Results for Example 5.1

Table 3 Numerical Results for Example 5.2

Remark 5.3

The following observations are presented from the above numerical Examples 5.1–5.2 in the following remarks:

• (1). Clearly from Figs. 1–8 and Tables 2 and 3, our proposed Algorithm 3.2 is easy to implement, efficient and accurate in handling the proposed problem and applications in both finite and infinite dimensional spaces.

• (2). From Table 2, Table 3 and Figs. 1–8, the number of iterations of our proposed method remain consistent, that is, well behaved irrespective of the starting point. Unfortunately, we did not have similar pattern in most of the existing methods that we compare our method with.

• (3). Our proposed Algorithm 3.2 is compared with it’s non-inertial version ($\tau _n = \nu _n = 0$). Clearly from the results of the numerical examples, our proposed method performs better than its non-inertial version in term of CPU time and number of iterations in all the cases considered. See Figs. 1–8 and Tables 2–3. This support the theory behind using inertial technique to speed up the rate of convergence for fixed point iterative methods.

• (4). In addition, we compared our proposed Algorithm 3.2 with some existing algorithms 1.1 proposed by Kwelegano et al. (2022), Appendix 6.1 proposed by Taiwo et al. (2020), Appendix 6.2 proposed by Chang et al, and Appendix 6.3 proposed by Wang et al. The results in 1–8 and Tables 2 and 3 indicate that our proposed method outperformed these existing methods with respect to the number of iterations and comparatively with CPU time.

6 Conclusion

In this paper, we studied the split equality variational inequality and fixed point problem involving more general class of operators; namely quasimonotone and quasi-pseudocontractive mappings in real Hilbert spaces. A self-adaptive inertial projection and contraction algorithm is presented and analyzed under some mild conditions. We proved two strong convergence theorems with and without regards to monotonicity. In addition, some numerical computations were carried out to illustrate the efficiency of our proposed algorithm in comparison with related results in the literature.

Appendices

Appendix 6.1

Algorithm 3.1 in Taiwo et al. (2020)

$$\begin{aligned} {\left\{ \begin{array}{ll} \text {choose arbitrary}~~ x_0\in H_1, y_0\in H_2,\\ z_n=x_n-\zeta _nT_1^*(T_1x_n-T_2y_n),\\ x_{n+1}=T_n(\alpha _ns_1^*(z_n)+(1-\alpha _n)z_n),\\ t_n=y_n+\zeta _nT_2^*(T_1x_n-T_2y_n),\\ y_{n+1}=U_n(\alpha _ns_2^*(t_n)+(1-\alpha _n)t_n), \end{array}\right. } \end{aligned}$$

where $T_n=\eta _{n}^*I+(1-\eta _{n}^*)S_1[\chi _n^*I+(1-\chi _n^*)S_1]$ and $U_n=\eta _{n}^*I+(1-\eta _{n}^*)S_2[\chi _n^*I+(1-\chi _n^*)S_2].$ Also, $s_1^*:H_1\rightarrow H_1$ and $s_2^*:H_2\rightarrow H_2$ are $L_1$ and $L_2$ Lipschitzian and strongly pseudocontractions with constants $\tau _1^*,\tau _2^*\in [0,1).$

Appendix 6.2

Algorithm 3.1 in Chang et al. (2015)

$$\begin{aligned} {\left\{ \begin{array}{ll} \text {Let}~~ x_0\in H_1, y_0\in H_2 ~~ \text {be chosen arbitrarily},\\ z_n=x_n-\zeta _nT_1^*(T_1x_n-T_2y_n),\\ x_{n+1}=\alpha _n x_n+(1-\alpha _n)Y_nz_n,\\ t_n=y_n+\zeta _n T_2^*(T_1x_n-T_2y_n),\\ y_{n+1}=\alpha _n y_n+(1-\alpha _n)J_nt_n, \end{array}\right. } \end{aligned}$$

where $\zeta _n\in \Big (0,\min (\frac{1}{\Vert T_1\Vert ^2},\frac{1}{\Vert T_2\Vert ^2})\Big ),~ \alpha _n \in (0,1)~~ \forall n\ge 1.$

Appendix 6.3

in Wang and Fang (2017)

$$\begin{aligned} {\left\{ \begin{array}{ll} \text {choose}~~ x_0\in H_1, y_0\in H_2 ~~ \text {arbitrarily},\\ z_n=x_n-\zeta _nT_1^*(T_1x_n-T_2y_n),\\ x_{n+1}=\alpha _nf(x_n)+(1-\alpha _n)((1-\psi _n)z_n+\psi _nS_1z_n),\\ t_n=y_n+\zeta _nT_2^*(T_1x_n-T_2y_n),\\ y_{n+1}=\alpha _ng(y_n)+(1-\alpha _n)((1-\psi _n)t_n+\psi _nS_2t_n), \end{array}\right. } \end{aligned}$$

where $\zeta _n=\Big (\epsilon , \frac{2}{\lambda _{T_1^*}+\lambda _{T_2^*}}-\epsilon \Big ),$ for small enough $\epsilon >0,$ and $\lambda _{T_1^*}$ and $\lambda _{T_2^*}$ are the spectral radius of $T_1^*T_1$ and $T_2^*T_2$ respectively, $f:H_1\rightarrow H_1$ and $g:H_2\rightarrow H_2$ are contractions with constants $\hat{\rho _1},\hat{\rho _2}\in [0,1),$ respectively, and $\alpha _n \in [0,1], \psi _n\in (0,1).$