A Unified Two-Step Inertial Viscosity-Type Scheme for Monotone Generalized Mixed Equilibrium and Fixed-Point Problems

Abstract

We introduce and analyze a two-step inertial viscosity-type algorithm for approximating a common solution of a generalized mixed equilibrium problem and the fixed-point problems associated with a family of quasi-nonexpansive mappings in a real Hilbert space. The generalized mixed equilibrium model contains a monotone operator and a proper convex lower semicontinuous function inside an implicit equilibrium resolvent, while a second inverse strongly monotone operator is treated explicitly through two successive forward–resolvent steps. The method further incorporates two inertial corrections, a family-dependent fixed-point step, and viscosity regularization by a contraction. We first establish boundedness of the generated sequence under summability conditions on the inertial parameters. We then derive a uniform descent inequality that simultaneously controls the two equilibrium residuals and the fixed-point residual. Using a residual-based sequence convergence criterion and an appropriate weak regularity condition for the mapping family, we prove that the iterates converge strongly to the unique point $q^{*}$ of the common solution set $\mathrm{\Omega }$ satisfying $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$. The abstract theorem is specialized to composite convex minimization, mixed variational inequalities, countable convex feasibility problems, and finite families under cyclic control. A five-dimensional example verifies the assumptions explicitly and illustrates convergence to the viscosity-selected solution.

1. Introduction

Fixed-point theory and equilibrium theory provide two closely related frameworks for formulating nonlinear problems arising in optimization, economics, engineering, game theory, and applied analysis. If $S:H\to H$ is a nonlinear mapping on a real Hilbert space, the fixed-point problem consists of finding $x\in H$ such that $Sx=x$. Contractive mappings admit a unique fixed point and the associated Picard iteration converges strongly by the classical principle of Banach [1], whereas nonexpansive and quasi-nonexpansive mappings generally require more elaborate iterative regularization. The latter classes are especially useful because metric projections, resolvents of monotone operators, feasibility operators, and many optimization algorithms can be represented as fixed-point mappings. Foundational existence results include the theorems of Brouwer [2] and Kakutani [3]; broader geometric and nonlinear-analytic treatments can be found in the monographs of Goebel and Reich [4] and Reich and Shoikhet [5]. Weak convergence and asymptotic regularity for nonexpansive-type iterations were developed, among others, by Opial [6], Bruck, Kuczumow, and Reich [7], and Akashi and Takahashi [8]. Fixed-point theory for set-valued nonexpansive mappings has likewise been studied extensively by Reich and Zaslavski [9,10,11].

Equilibrium problems form a similarly broad model. The minimax principle of Fan [12] and the formulation of Blum and Oettli [13] established the equilibrium problem as a unifying setting that includes variational inequalities, optimization problems, saddle-point systems, complementarity problems, and Nash-type equilibria. In its classical form, one seeks $q\in C$ satisfying

$$F(q,y)\ge 0\forall y\in C,$$

where C is a nonempty closed convex subset of a Hilbert space and $F:C\times C\to \mathbb{R}$ is a bifunction. The mixed and generalized equilibrium models enlarge this formulation by incorporating convex potential terms and monotone operators. Such extensions are important when the problem contains both a nonsmooth convex component and an operator component, as occurs in constrained convex minimization and mixed variational inequalities. Common-solution schemes linking equilibrium problems with variational inequalities, hierarchical constraints, nonexpansive semigroups, and split formulations were studied, for example, by Bnouhachem and Chen [14], Djafari-Rouhani, Farid, and Kazmi [15], and Kazmi and Rizvi [16].

The problem considered in this paper is the generalized mixed equilibrium problem associated with a bifunction F, a continuous monotone operator $W:C\to H$, and a proper convex lower semicontinuous function $\phi :C\to \mathbb{R}\cup \{\infty \}$. Its precise formulation is given in (1) in the Preliminaries. This formulation contains several familiar problems. If $\phi \equiv 0$, it becomes a generalized equilibrium problem; if $F\equiv 0$, it becomes a mixed variational inequality and if both $F\equiv 0$ and $\phi \equiv 0$, it reduces to the classical variational inequality for W. Appropriate choices of F and $\phi$ also recover convex minimization and constrained feasibility models.

A decisive tool in the analysis of equilibrium problems is the resolvent. Combettes and Hirstoaga [17] developed a systematic resolvent approach to equilibrium programming in Hilbert spaces. For generalized mixed equilibrium problems, Zhang [18] established well-definedness and firm nonexpansiveness properties for the corresponding resolvent under standard monotonicity, convexity, and semicontinuity conditions. These properties make the equilibrium resolvent compatible with fixed-point techniques: its fixed-point set coincides with the equilibrium solution set, and firm nonexpansiveness supplies the basic Fejér-type estimates needed in convergence arguments.

The approximation of a point that simultaneously solves an equilibrium problem and a fixed-point problem has, therefore, received sustained attention. Takahashi and Takahashi [19] introduced a strongly convergent method for a generalized equilibrium problem coupled with the fixed points of a nonexpansive mapping. Chang, Lee, and Chan [20] developed a hybrid projection method in a uniformly convex and uniformly smooth Banach space for generalized equilibria, variational inequalities, and common fixed points. Osward, Kumar, and Sangago [21] treated an asymptotically nonexpansive fixed-point constraint, while Farid [22] proposed algorithms specifically for mixed equilibrium and fixed-point problems in Hilbert spaces. The Banach-space methods in this literature rely on duality mappings and generalized projections; useful background is given by Cioranescu [23] and Takahashi [24]. These results demonstrate that the common-solution formulation is flexible enough to encode both equilibrium conditions and additional geometric or dynamical constraints.

Two mechanisms have become particularly effective in obtaining strong convergence. The first is the viscosity method, in which a contraction is combined with the principal fixed-point or equilibrium operator; see, for example, Xu [25]. Besides regularizing the iteration, the contraction selects a distinguished point of the solution set, characterized by $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$. Viscosity-like implicit methods for approximating zeros of monotone operators in Banach spaces were studied by Mendy and Shukla [26]. Maldar and Çetin [27] studied a viscosity-type method for equilibrium and fixed-point problems with data dependence. Related viscosity schemes for split-equality generalized mixed equilibrium problems and infinite families of nonexpansive mappings were considered in [28,29,30]. The second is inertial extrapolation, originating from acceleration ideas and using one or more previous increments to form the next search point. Inertial terms can improve numerical behavior, but they also introduce non-Fejér perturbations that must be controlled carefully. Husain and Asad [31], for example, combined an inertial technique with a subgradient-extragradient method for a modified variational inclusion and a mixed equilibrium problem. More recently, Kajola and Narain [32] established strong convergence of an inertial method for a common fixed-point and mixed equilibrium problem, and Sunthrayuth et al. [33] studied a two-step inertial viscosity-type method for a family of strongly quasi-nonexpansive mappings.

The present work is motivated by these developments, but its equilibrium structure and convergence mechanism are different. We separate two monotone operators according to how they are evaluated. The operator U is placed inside the generalized mixed equilibrium resolvent and is, therefore, handled implicitly. A second operator $A:H\to H$ is kept outside the resolvent and treated by explicit forward steps; accordingly, A is assumed to be $\alpha$-inverse strongly monotone, or equivalently $\alpha$-cocoercive. The combined set is

$$\begin{array}{cc}\hfill \mathrm{\Theta }& =\mathrm{GMEP}(F,U+A,\phi )\hfill \\ \hfill & =\{q\in C\cap \mathrm{dom}\phi :F(q,y)+⟨Uq+Aq,y-q⟩\hfill \\ \hfill & -\phi \left(q\right)+\phi \left(y\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi \}.\hfill \end{array}$$

The desired common-solution set is

$$\mathrm{\Omega }:=\mathrm{\Theta }\cap \bigcap _{n\ge 1}\mathrm{Fix}\left(T_n\right),$$

where $\left\{T_n\right\}$ is a family of quasi-nonexpansive mappings satisfying an appropriate weak regularity condition.

The proposed algorithm combines two-step inertial extrapolation, two successive forward–resolvent evaluations, a family-dependent fixed-point step, and viscosity regularization. The use of two forward–resolvent stages yields residual estimates for both equilibrium corrections, while the convex combination of $z_n$ and $T_n{z}_n$ produces a fixed-point residual. These three residuals are assembled into a single descent term. The final convergence argument is based on a residual convergence criterion related to Wang and Pham [34]; this avoids relying on a monotonicity assumption for the squared-distance sequence.

The main contributions of the paper are the following.

(i)

We formulate the algorithm for the domain-safe generalized mixed equilibrium problem $\mathrm{GMEP}(F,U+A,\phi )$, allowing the convex function $\phi$ to be extended-valued and hence to encode additional constraints.

(ii)

We establish boundedness of the generated sequence under explicit summability conditions on both inertial coefficients, including a second inertial term that may have a negative sign.

(iii)

We derive a uniform key inequality that controls the two forward–resolvent residuals and the fixed-point residual.

(iv)

Under the viscosity, step-size, and weak regularity assumptions, we prove strong convergence to the unique point $q^{*}\in \mathrm{\Omega }$ satisfying $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$.

(v)

We specialize the theorem to composite convex minimization, mixed variational inequalities, countable convex feasibility, and finite families under cyclic control. A fully specified five-dimensional example illustrates the assumptions and identifies the limiting point explicitly.

The remainder of the paper is organized as follows. Section 2 introduces the generalized mixed equilibrium resolvent and its coupling with the explicit operator A. Section 3 states the proposed algorithm and contains the assumptions for boundedness, the boundedness analysis, the key residual inequality, and the strong convergence theorem. Section 4 presents the applications. The cyclic finite-family specialization is developed in Section 5, and the five-dimensional illustrative example is given in Section 6. Section 7 concludes the paper.

2. Preliminaries

Let H be a real Hilbert space and $C\subseteq H$ a nonempty closed convex subset. Let $F:C\times C\to \mathbb{R}$ be a bifunction, $U:C\to H$ and $A:H\to H$ nonlinear mappings, $\phi :C\to \mathbb{R}\cup \{\infty \}$ a proper function, $f:H\to H$ a contraction, and ${\left\{T_n\right\}}_{n\ge 1}$ a family of self-mappings on H.

2.1. The Generalized Mixed Equilibrium Problem and Its Resolvent

Following the real-valued formulation of Zhang [18] and the extended-valued formulation in [35], for a nonlinear mapping $W:C\to H$ the generalized mixed equilibrium problem $\mathrm{GMEP}(F,W,\phi )$ is to find $q\in C\cap \mathrm{dom}\phi$ such that

$$F(q,y)+⟨Wq,y-q⟩-\phi \left(q\right)+\phi \left(y\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi ,$$

(1)

with solution set denoted $\mathrm{GMEP}(F,W,\phi )$ as well. This is the domain-safe interpretation of the usual extended-valued formulation: for $y\notin \mathrm{dom}\phi$ the equilibrium inequality is automatic.

Conditions (B1)–(B4) below are the standard equilibrium assumptions used in the classical resolvent literature [13,17], while (B5)–(B6) are the usual monotonicity and convexity conditions for generalized mixed equilibrium models. Zhang [18] treats the real-valued case, whereas the extended-valued setting $\phi :C\to \mathbb{R}\cup \{+\infty \}$ and its resolvent properties are covered in [35].

Assumption 1 

(Data of the GMEP).

(B1) 

$F(x,x)=0$ for all $x\in C$;

(B2) 

F is monotone: $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

(B3) 

for all $x,y,z\in C$, ${\displaystyle \underset{t\downarrow 0}{\lim \sup } F\left(tz+(1-t)x,y\right)\le F(x,y)}$;

(B4) 

for each $x\in C$, the function $y\mapsto F(x,y)$ is convex and lower semicontinuous;

(B5) 

$U:C\to H$ is continuous and monotone: $⟨Ux-Uy,x-y⟩\ge 0$ for all $x,y\in C$;

(B6) 

$\phi :C\to \mathbb{R}\cup \{\infty \}$ is proper, convex, and lower semicontinuous.

For $r>0$, the resolvent $T_r^{F,U,\phi }:H\rightrightarrows C$ of $(F,U,\phi )$ is initially defined as the set-valued mapping

$$\begin{array}{ccc}\hfill T_r^{F,U,\phi }\left(x\right)& =& \left\{z\in C\cap \mathrm{dom}\phi :F(z,y)+⟨Uz,y-z⟩\right.\hfill \\ & & \left.-\phi \left(z\right)+\phi \left(y\right)+{\displaystyle \frac{1}{r}}⟨y-z,z-x⟩\ge 0\forall y\in C\cap \mathrm{dom}\phi \right\}.\hfill \end{array}$$

(2)

Lemma 1 

(Alakoya, Taiwo, and Mewomo [35] Lemma 2.5). Let Assumption 1 hold. Then, for every $r>0$ and $x\in H$:

(a) 

$T_r^{F,U,\phi }\left(x\right)$ is nonempty;

(b) 

$T_r^{F,U,\phi }$ is single-valued;

(c) 

$T_r^{F,U,\phi }$ is firmly nonexpansive, that is,

$${\parallel T_r^{F,U,\phi }\left(x\right)-T_r^{F,U,\phi }\left(y\right)\parallel }^2\le ⟨T_r^{F,U,\phi }\left(x\right)-T_r^{F,U,\phi }\left(y\right),x-y⟩\forall x,y\in H;$$

(d) 

$\mathrm{Fix}\left(T_r^{F,U,\phi }\right)=\mathrm{GMEP}(F,U,\phi )$;

(e) 

$\mathrm{GMEP}(F,U,\phi )$ is closed and convex.

Proof. 

Existence in (a) follows from the cited extended-valued resolvent result. For completeness, let $z\in T_r^{F,U,\phi }\left(x\right)$ and $w\in T_r^{F,U,\phi }\left(y\right)$. Using w in the defining inequality for z, using z in the defining inequality for w, and adding the two inequalities, the terms involving $\phi$ cancel. The monotonicity of F and U then gives

$${\parallel z-w\parallel }^2\le ⟨z-w,x-y⟩.$$

Taking $x=y$ proves uniqueness, so the resolvent is single-valued, and the same inequality proves firm nonexpansiveness. Moreover, $q=T_r^{F,U,\phi }\left(q\right)$ holds exactly when the regularization term in (2) vanishes and q satisfies (1); hence $\mathrm{Fix}\left(T_r^{F,U,\phi }\right)=\mathrm{GMEP}(F,U,\phi )$. Finally, a firmly nonexpansive mapping is nonexpansive, and its fixed-point set in a Hilbert space is closed and convex. This proves (b)–(e). □

Note that in Lemma 1 the operator entering the resolvent need only satisfy (B5); in particular the lemma applies verbatim with U replaced by $U+A$ whenever A is continuous and monotone (which holds, e.g., whenever A is $\alpha$-inverse strongly monotone, since $\alpha$-ISM implies A is monotone and $\frac{1}{\alpha }$-Lipschitz, hence continuous). We record this as a remark for later use.

Remark 1. 

If $A:H\to H$ is monotone and Lipschitz continuous on C, then $U+A$ satisfies (B5), so $\mathrm{GMEP}(F,U+A,\phi )$ is well defined, its resolvent $T_r^{F,U+A,\phi }$ is firmly nonexpansive for every $r>0$, and $\mathrm{GMEP}(F,U+A,\phi )$ is closed and convex.

2.2. Coupling with the Monotone Operator A

We now couple the (implicit) resolvent $T_r^{F,U,\phi }$ of $(F,U,\phi )$ with an explicit forward step for a second monotone operator $A:H\to H$. Define

$$\begin{array}{cc}\hfill \mathrm{\Theta }& :=\mathrm{GMEP}(F,U+A,\phi )\hfill \\ \hfill & =\{q\in C\cap \mathrm{dom}\phi :F(q,y)+⟨Uq+Aq,y-q⟩\hfill \\ \hfill & -\phi \left(q\right)+\phi \left(y\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi \}.\hfill \end{array}$$

(3)

Lemma 2 

(Characterization). Let Assumption 1 hold and let $A:H\to H$ be any mapping. For $q\in C\cap \mathrm{dom}\phi$ and $r>0$,

$$q=T_r^{F,U,\phi }(q-rAq)⟺q\in \mathrm{\Theta }.$$

Proof. 

By definition (2), $q=T_r^{F,U,\phi }(q-rAq)$ means

$$F(q,y)+⟨Uq,y-q⟩-\phi \left(q\right)+\phi \left(y\right)+{\displaystyle \frac{1}{r}}⟨y-q,q-(q-rAq)⟩\ge 0\forall y\in C\cap \mathrm{dom}\phi .$$

Since $q-(q-rAq)=rAq$, the last term equals $⟨y-q,Aq⟩=⟨Aq,y-q⟩$, and the displayed inequality becomes exactly

$$F(q,y)+⟨Uq+Aq,y-q⟩-\phi \left(q\right)+\phi \left(y\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi ,$$

i.e., $q\in \mathrm{\Theta }$ by (3). □

Lemma 3 

(Forward step and quasi-nonexpansiveness). Suppose $A:H\to H$ is α-inverse strongly monotone (α-ISM): $⟨Ax-Ay,x-y⟩\ge \alpha {\parallel Ax-Ay\parallel }^2$ for all $x,y\in H$. Then:

(a) 

for $0<r\le 2\alpha$, $I-rA$ is nonexpansive on H;

(b) 

for $0<r\le 2\alpha$, and Assumption 1 in force, the map $S_r:=T_r^{F,U,\phi }\circ (I-rA)$ is quasi-nonexpansive with respect to Θ: $\parallel S_r\left(x\right)-q\parallel \le \parallel x-q\parallel$ for all $q\in \mathrm{\Theta }$, $x\in H$.

Proof. 

(a) For $x,y\in H$,

$$\begin{array}{cc}\hfill {\parallel (I-rA)x-(I-rA)y\parallel }^2& ={\parallel x-y\parallel }^2-2r⟨Ax-Ay,x-y⟩+r^2{\parallel Ax-Ay\parallel }^2\hfill \\ \hfill & \le {\parallel x-y\parallel }^2-2r\alpha {\parallel Ax-Ay\parallel }^2+r^2{\parallel Ax-Ay\parallel }^2\hfill \\ \hfill & ={\parallel x-y\parallel }^2-r(2\alpha -r){\parallel Ax-Ay\parallel }^2\le {\parallel x-y\parallel }^2\hfill \end{array}$$

whenever $0<r\le 2\alpha$.

(b) Let $q\in \mathrm{\Theta }$. By Lemma 2, $q=T_r^{F,U,\phi }(q-rAq)$. By Lemma 1 (c), $T_r^{F,U,\phi }$ is firmly nonexpansive, hence nonexpansive; combined with (a),

$$\parallel S_r\left(x\right)-q\parallel =\parallel T_r^{F,U,\phi }\left((I-rA)x\right)-T_r^{F,U,\phi }\left((I-rA)q\right)\parallel \le \parallel (I-rA)x-(I-rA)q\parallel \le \parallel x-q\parallel .$$

□

Lemma 4 

(Minty-type lemma for $\mathrm{\Theta }$). Assume

(i) 

Assumption 1 (B1), (B3) and (B4) hold for F;

(ii) 

$U:C\to H$ is continuous;

(iii) 

$A:H\to H$ is monotone and Lipschitz continuous;

(iv) 

$\phi :C\to \mathbb{R}\cup \{\infty \}$ is proper, convex, and lower semicontinuous.

If $p\in C\cap \mathrm{dom}\phi$ satisfies the dual inequality

$$F(y,p)+⟨Uy+Ay,p-y⟩-\phi \left(y\right)+\phi \left(p\right)\le 0\forall y\in C\cap \mathrm{dom}\phi ,$$

(4)

then $p\in \mathrm{\Theta }$, i.e.,

$$F(p,y)+⟨Up+Ap,y-p⟩-\phi \left(p\right)+\phi \left(y\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi .$$

(5)

Proof. 

Define $\mathrm{\Psi }(x,y):=F(x,y)+⟨(U+A)x,y-x⟩-\phi \left(x\right)+\phi \left(y\right)$ for $x,y\in C\cap \mathrm{dom}\phi$. Fix $u\in C\cap \mathrm{dom}\phi$ and, for $t\in (0,1]$, set $z_t:=tu+(1-t)p$. Convexity of C and of $\mathrm{dom}\phi$ gives $z_t\in C\cap \mathrm{dom}\phi$. By (B1), $\mathrm{\Psi }(z_t,z_t)=F(z_t,z_t)=0$. By convexity of $y\mapsto F(z_t,y)$ (Assumption 1 (B4)) and convexity of $\phi$, the map $y\mapsto \mathrm{\Psi }(z_t,y)$ is convex (the middle term is affine), so

$$0=\mathrm{\Psi }(z_t,z_t)\le t\mathrm{\Psi }(z_t,u)+(1-t)\mathrm{\Psi }(z_t,p).$$

Applying the dual inequality (4) with $y=z_t$ gives $\mathrm{\Psi }(z_t,p)\le 0$, hence

$$0\le t\mathrm{\Psi }(z_t,u)+(1-t)\mathrm{\Psi }(z_t,p)\le t\mathrm{\Psi }(z_t,u),$$

so $\mathrm{\Psi }(z_t,u)\ge 0$ for every $t\in (0,1]$ and every $u\in C\cap \mathrm{dom}\phi$.

As $t\downarrow 0$, $z_t\to p$ strongly. By continuity of U and Lipschitz continuity (hence continuity) of A, $(U+A)z_t\to (U+A)p$; by lower semicontinuity of $\phi$ at p, ${\lim \inf }_{t\downarrow 0}\phi \left(z_t\right)\ge \phi \left(p\right)$, i.e., ${\lim \sup }_{t\downarrow 0}\left(-\phi \left(z_t\right)\right)\le -\phi \left(p\right)$ and by (B3), ${\lim \sup }_{t\downarrow 0}F(z_t,u)\le F(p,u)$. Combining these,

$$\underset{t\downarrow 0}{\lim \sup }\mathrm{\Psi }(z_t,u)\le F(p,u)+⟨(U+A)p,u-p⟩-\phi \left(p\right)+\phi \left(u\right)=\mathrm{\Psi }(p,u).$$

Since $\mathrm{\Psi }(z_t,u)\ge 0$ for all $t\in (0,1]$,

$$0\le \underset{t\downarrow 0}{\lim \inf }\mathrm{\Psi }(z_t,u)\le \underset{t\downarrow 0}{\lim \sup }\mathrm{\Psi }(z_t,u)\le \mathrm{\Psi }(p,u).$$

Hence $\mathrm{\Psi }(p,u)\ge 0$, that is, $F(p,u)+⟨Up+Ap,u-p⟩-\phi \left(p\right)+\phi \left(u\right)\ge 0$. As $u\in C$ was arbitrary in $C\cap \mathrm{dom}\phi$, $p\in \mathrm{\Theta }$. □

3. The Algorithm

Choose initial points $x_0,x_1\in H$ and set $x_{-1}:=x_0$ (so that the inertial terms are well-defined for $n=1$; in particular $\parallel x_0-x_{-1}\parallel =0$). Define ${\left\{x_n\right\}}_{n\ge 1}$ by

$$\left\{\begin{array}{cc}\hfill y_n& =x_n+{\theta }_n(x_n-x_{n-1})+{\delta }_n(x_{n-1}-x_{n-2}),\hfill \\ \hfill v_n& =T_{r_n}^{F,U,\phi }(y_n-r_{n}A\left(y_n\right)),\hfill \\ \hfill z_n& =T_{s_n}^{F,U,\phi }(v_n-s_{n}A\left(v_n\right)),\hfill \\ \hfill x_{n+1}& =a_{n}f\left(x_n\right)+{\beta }_n{z}_n+{\gamma }_n{T}_n{z}_n,\hfill \end{array}\right.$$

(6)

where $a_n,{\beta }_n,{\gamma }_n\in [0,1]$ with $a_n+{\beta }_n+{\gamma }_n=1$, $r_n,s_n>0$, ${\theta }_n\ge 0$, ${\delta }_n\le 0$. No additional initial data beyond $x_0$ and $x_1$ are required.

Remark 2. 

Here $T_r^{F,U,\phi }$ is the resolvent (2) of the generalized mixed equilibrium problem $\mathrm{GMEP}(F,U,\phi )$: U is treated implicitly (inside the resolvent), while A is treated explicitly (by the forward steps $y_n-r_{n}A\left(y_n\right)$ and $v_n-s_{n}A\left(v_n\right)$), exactly as in a forward–backward splitting for the sum $U+A$. By Lemma 2, fixed points of the composite maps correspond to solutions of $\mathrm{\Theta }=\mathrm{GMEP}(F,U+A,\phi )$.

3.1. Assumptions for Boundedness

Assumption 2 

(Inverse strong monotonicity). $A:H\to H$ is α-inverse strongly monotone (α-ISM), and $0<r_n,s_n<2\alpha$ for all n.

Assumption 3 

(Contraction). $f:H\to H$ is a τ-contraction with $\tau \in (0,1)$.

Assumption 4 

(Quasi-nonexpansivity). Each $T_n:H\to H$ is quasi-nonexpansive with respect to Ω: $\parallel T_{n}x-q\parallel \le \parallel x-q\parallel$ for all $q\in \mathrm{\Omega }$, $x\in H$.

Assumption 5 

(Parameter conditions).

$$\underset{n\to \infty }{\lim }{a}_n=0,{\displaystyle \sum _{n=1}^{\infty }}{a}_n=\infty ,{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{{\theta }_n}{a_n}}<\infty ,{\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{|{\delta }_n|}{a_n}}<\infty ,0<a_n<1\mathit{for}\mathit{all}n.$$

Assumption 6 

(Non-empty solution set). $\mathrm{\Omega }:=\mathrm{\Theta }\cap {\bigcap }_{n\ge 1}\mathrm{Fix}\left(T_n\right)\ne \varnothing$, where $\mathrm{\Theta }=\mathrm{GMEP}(F,U+A,\phi )$ as in (3).

3.2. Boundedness

Lemma 5 

(Boundedness). Under Assumptions 1 and 2–6, the sequence $\left\{x_n\right\}$ generated by (6) is bounded.

Proof. 

Fix $q\in \mathrm{\Omega }\subseteq \mathrm{\Theta }$ and set $u_n:=\parallel x_n-q\parallel$, $D:={\displaystyle \frac{\parallel f\left(q\right)-q\parallel }{1-\tau }}$.

• Step 1: Basic chain of inequalities. By Lemma 3 (b) (applicable since $0<r_n,s_n<2\alpha$ by Assumption 2) and Assumption 4,

  $$\begin{array}{cc}\hfill \parallel v_n-q\parallel & \le \parallel y_n-q\parallel ,\hfill \end{array}$$

  (7)

  $$\begin{array}{cc}\hfill \parallel z_n-q\parallel & \le \parallel v_n-q\parallel \le \parallel y_n-q\parallel ,\hfill \end{array}$$

  (8)

  $$\begin{array}{cc}\hfill \parallel T_n{z}_n-q\parallel & \le \parallel z_n-q\parallel \le \parallel y_n-q\parallel .\hfill \end{array}$$

  (9)

  From the definition of $y_n$ and the triangle inequality,

  $$\parallel y_n-q\parallel \le u_n+{\theta }_n\parallel x_n-x_{n-1}\parallel +\left|{\delta }_n\right|\parallel x_{n-1}-x_{n-2}\parallel .$$

  (10)

  Using convexity of the norm, $\parallel f\left(x_n\right)-q\parallel \le \tau u_n+\parallel f\left(q\right)-q\parallel$, and (7)–(10),

  $$u_{n+1}\le a_n\left(\tau u_n+\parallel f\left(q\right)-q\parallel \right)+(1-a_n)\parallel y_n-q\parallel .$$

  Setting ${\alpha }_n:=a_n(1-\tau )$ and substituting (10),

  $$u_{n+1}\le (1-{\alpha }_n)u_n+{\alpha }_{n}D+(1-a_n)\left({\theta }_n\parallel x_n-x_{n-1}\parallel +\left|{\delta }_n\right|\parallel x_{n-1}-x_{n-2}\parallel \right).$$

  (11)
• Step 2: Choosing the cut-off index N. Since $\sum {\theta }_n/a_n<\infty$ and $\sum |{\delta }_n|/a_n<\infty$, we have ${\theta }_n/a_n\to 0$ and $|{\delta }_n|/a_n\to 0$. Fix $N\in \mathbb{N}$ such that

  $${\theta }_n\le {\displaystyle \frac{{\alpha }_n}{8}}\mathrm{and}\left|{\delta }_n\right|\le {\displaystyle \frac{{\alpha }_n}{8}}\forall n\ge N.$$
• Step 3: Finite-phase bound. The convention $x_{-1}:=x_0$ gives $\parallel x_0-x_{-1}\parallel =0$. Define $R_0:=\max \{u_0,u_1,2D\}$, so $u_n\le R_0$ and $\parallel x_n-x_{n-1}\parallel \le 2R_0$ hold for $n\in \{0,1\}$. Applying (11) iteratively for $n=1,\dots ,N-1$ (a finite computation), there exists $R_1\ge R_0$ such that

  $$u_n\le R_1\mathrm{and}\parallel x_n-x_{n-1}\parallel \le 2R_1\forall n\le N.$$

  (12)
• Step 4: Inductive phase ($\mathit{n}\mathbf{\ge }\mathit{N}$). Set $R^{*}:=\max \{R_1,2D\}$. We prove $u_n\le R^{*}$ for all $n\ge N$ by induction. The base case $n=N$ follows from (12). Suppose $u_k\le R^{*}$ for all $N\le k\le n$. Together with the finite-phase estimate (12), this gives $u_k\le R^{*}$ for all $k\le n$, and hence both $\parallel x_n-x_{n-1}\parallel \le 2R^{*}$ and $\parallel x_{n-1}-x_{n-2}\parallel \le 2R^{*}$. Inserting into (11) with ${\theta }_n\le {\alpha }_n/8$, $|{\delta }_n|\le {\alpha }_n/8$, $(1-a_n)\le 1$:

  $$u_{n+1}\le (1-{\alpha }_n)R^{*}+{\alpha }_{n}D+{\displaystyle \frac{{\alpha }_n}{8}}·2R^{*}+{\displaystyle \frac{{\alpha }_n}{8}}·2R^{*}=R^{*}-{\alpha }_n\left({\displaystyle \frac{R^{*}}{2}}-D\right).$$

  Since $R^{*}\ge 2D$, we get $u_{n+1}\le R^{*}$, completing the induction. Hence $\left\{x_n\right\}$ is bounded by $R^{*}+\parallel q\parallel$, and consequently $\left\{y_n\right\}$, $\left\{v_n\right\}$, $\left\{z_n\right\}$, $\left\{T_n{z}_n\right\}$ are all bounded. □

Remark 3 

(Choice of the constant $1/8$). The constant $1/8$ is not essential. With ${\alpha }_n=a_n(1-\tau )$, the assumptions imply ${\theta }_n/{\alpha }_n\to 0$ and $|{\delta }_n|/{\alpha }_n\to 0$. Hence, for any $c\in (0,1/4)$, one may choose N such that ${\theta }_n\le c{\alpha }_n$ and $|{\delta }_n|\le c{\alpha }_n$ for all $n\ge N$. The induction in the proof then gives

$$u_{n+1}\le R^{*}-{\alpha }_n\left((1-4c)R^{*}-D\right),$$

which closes whenever $R^{*}\ge \max \{R_1,D/(1-4c)\}$. The choice $c=1/8$ yields the convenient condition $R^{*}\ge 2D$ used above.

3.3. Augmented Assumptions and Key Inequality

For the strong convergence result, we strengthen the step-size condition and impose a joint regularity requirement on the mappings $T_n$, together with uniform lower bounds on the fixed-point weights.

Assumption 7 

(Inverse strong monotonicity and step-sizes). $A:H\to H$ is α-inverse strongly monotone (α-ISM), so A is monotone and L-Lipschitz with $L=1/\alpha$. Moreover,

$$0<r_n,s_n<\alpha ,\underset{n\to \infty }{\lim \inf }{r}_n>0,\underset{n\to \infty }{\lim \sup }\max \{r_n,s_n\}<\alpha .$$

(13)

(The condition $r_n,s_n<\alpha <2\alpha$ is compatible with Assumption 2.)

Assumption 8 

(Quasi-nonexpansivity and regularity of $\left\{T_n\right\}$). Each $T_n:H\to H$ is quasi-nonexpansive with respect to ${\bigcap }_{k\ge 1}\mathrm{Fix}\left(T_k\right)$: $\parallel T_{n}x-q\parallel \le \parallel x-q\parallel$ for all $q\in {\bigcap }_k\mathrm{Fix}\left(T_k\right)$, $x\in H$. Moreover, $\left\{T_n\right\}$ satisfies the ${\mathrm{NST}}_w$ condition (i.e., for every bounded sequence $\left\{u_n\right\}$ with ${\lim }_{n\to \infty }\parallel u_n-T_n{u}_n\parallel =0$, every weak cluster point of $\left\{u_n\right\}$ belongs to ${\bigcap }_{k\ge 1}\mathrm{Fix}\left(T_k\right)$), and there exist $\overline{\beta },\overline{\gamma }>0$ such that ${\beta }_n\ge \overline{\beta }$ and ${\gamma }_n\ge \overline{\gamma }$ for all n.

Lemma 6 

(Key inequality). Let Assumptions 1 and 7 hold. For any $q\in \mathrm{\Theta }$ and all n,

$${\parallel z_n-q\parallel }^2\le {\parallel y_n-q\parallel }^2-\left(1-{\displaystyle \frac{r_n}{\alpha }}\right){\parallel v_n-y_n\parallel }^2-\left(1-{\displaystyle \frac{s_n}{\alpha }}\right){\parallel z_n-v_n\parallel }^2.$$

(14)

Because ${\lim \sup }_n\max \{r_n,s_n\}<\alpha$, there exist $\rho >0$ and $N_0\in \mathbb{N}$ such that

$${\parallel z_n-q\parallel }^2\le {\parallel y_n-q\parallel }^2-\rho \left({\parallel v_n-y_n\parallel }^2+{\parallel z_n-v_n\parallel }^2\right)\forall n\ge N_0.$$

(15)

Proof. 

Let $q\in \mathrm{\Theta }$. By Lemma 2, $q=T_r^{F,U,\phi }(q-rAq)$ for every $r>0$. Since $T_{r_n}^{F,U,\phi }$ is firmly nonexpansive (Lemma 1 (c)),

$$\begin{array}{cc}\hfill {\parallel v_n-q\parallel }^2& ={∥T_{r_n}^{F,U,\phi }(y_n-r_{n}A{y}_n)-T_{r_n}^{F,U,\phi }(q-r_{n}Aq)∥}^2\hfill \\ \hfill & \le {∥(y_n-r_{n}A{y}_n)-(q-r_{n}Aq)∥}^2\hfill \\ \hfill & -{∥(I-T_{r_n}^{F,U,\phi })(y_n-r_{n}A{y}_n)-(I-T_{r_n}^{F,U,\phi })(q-r_{n}Aq)∥}^2.\hfill \end{array}$$

Because $T_{r_n}^{F,U,\phi }(q-r_{n}Aq)=q$, the second term simplifies as

$$\begin{array}{cc}\hfill (I-T_{r_n}^{F,U,\phi })(y_n-r_{n}A{y}_n)& -(I-T_{r_n}^{F,U,\phi })(q-r_{n}Aq)\hfill \\ \hfill & =(y_n-r_{n}A{y}_n-v_n)-(q-r_{n}Aq-q)\hfill \\ \hfill & =(y_n-v_n)-r_n(A{y}_n-Aq).\hfill \end{array}$$

Expanding the right-hand side and cancelling the $r_n^2{\parallel A{y}_n-Aq\parallel }^2$ terms, we obtain

$${\parallel v_n-q\parallel }^2\le {\parallel y_n-q\parallel }^2-{\parallel v_n-y_n\parallel }^2-2r_n⟨v_n-q,A{y}_n-Aq⟩.$$

(16)

Now write $⟨v_n-q,A{y}_n-Aq⟩=⟨v_n-y_n,A{y}_n-Aq⟩+⟨y_n-q,A{y}_n-Aq⟩$. By $\alpha$-inverse strong monotonicity, $⟨y_n-q,A{y}_n-Aq⟩\ge \alpha {\parallel A{y}_n-Aq\parallel }^2$. Using Young’s inequality, $⟨v_n-y_n,A{y}_n-Aq⟩\ge -{\displaystyle \frac{1}{2\alpha }}\parallel v_n-y_n{\parallel }^2-{\displaystyle \frac{\alpha }{2}}{\parallel A{y}_n-Aq\parallel }^2$. Substituting these two estimates into (16) yields

$$\begin{array}{cc}\hfill {\parallel v_n-q\parallel }^2& \le {\parallel y_n-q\parallel }^2-{\parallel v_n-y_n\parallel }^2\hfill \\ \hfill & +2r_n\left({\displaystyle \frac{1}{2\alpha }}\parallel v_n-y_n{\parallel }^2+{\displaystyle \frac{\alpha }{2}}\parallel A{y}_n{-Aq\parallel }^2-\alpha {\parallel A{y}_n-Aq\parallel }^2\right)\hfill \\ \hfill & ={\parallel y_n-q\parallel }^2-\left(1-{\displaystyle \frac{r_n}{\alpha }}\right)\parallel v_n-y_n{\parallel }^2-r_n\alpha {\parallel A{y}_n-Aq\parallel }^2\hfill \\ \hfill & \le {\parallel y_n-q\parallel }^2-\left(1-{\displaystyle \frac{r_n}{\alpha }}\right){\parallel v_n-y_n\parallel }^2.\hfill \end{array}$$

An identical argument with $v_n,z_n,s_n$ replacing $y_n,v_n,r_n$ gives

$${\parallel z_n-q\parallel }^2\le {\parallel v_n-q\parallel }^2-\left(1-{\displaystyle \frac{s_n}{\alpha }}\right){\parallel z_n-v_n\parallel }^2.$$

(17)

Combining the two inequalities, we obtain (14). Finally, set

$$\rho :={\displaystyle \frac{1}{2}}\left(1-{\displaystyle \frac{1}{\alpha }}\underset{n\to \infty }{\lim \sup }\max \{r_n,s_n\}\right)>0.$$

Then choose $N_0$ so that $1-r_n/\alpha \ge \rho$ and $1-s_n/\alpha \ge \rho$ for all $n\ge N_0$; then (15) follows. □

3.4. Strong Convergence

We recall the following residual-based convergence lemma due to Wang and Pham [34] Lemma 2.3.

Lemma 7 

(Wang and Pham [34] Lemma 2.3). Let $\left\{s_k\right\}\subset [0,\infty )$ and $\left\{b_k\right\}\subset \mathbb{R}$ be bounded sequences satisfying

$$s_{k+1}\le (1-{\ell }_k)s_k+{\ell }_k{b}_k+d_k-v_k{c}_k\forall k\ge 1,$$

(18)

where

(i) 

$\left\{{\ell }_k\right\}\subset [0,1]$ and ${\sum }_{k=1}^{\infty }{\ell }_k=\infty$;

(ii) 

$\left\{d_k\right\}\subset \mathbb{R}$ and ${\sum }_{k=1}^{\infty }\left|d_k\right|<\infty$;

(iii) 

$v_k>0$ and ${\ell }_k/v_k\to 0$;

(iv) 

$c_k\ge 0$, and for every subsequence $\left\{k_j\right\}$, $c_{k_j}\to 0$ implies ${\lim \sup }_{j\to \infty }{b}_{k_j}\le 0$.

Then $s_k\to 0$.

The cited statement is formulated with $c_k>0$. Here we use its nonnegative-residual extension with $c_k\ge 0$. The proof is unchanged, because condition (iv) directly treats every subsequence along which $c_k\to 0$, including subsequences containing zero residuals.

Theorem 1 

(Strong convergence). Assume that all the conditions of Lemma 5 hold (i.e., Assumptions 1 and 2–6), with the step-size condition $0<r_n,s_n<2\alpha$ replaced by the stronger requirement $0<r_n,s_n<\alpha$, $\lim \inf r_n>0$, $\lim \sup \max \{r_n,s_n\}<\alpha$ (Assumption 7). Suppose furthermore that Assumption 8 is satisfied. Then the sequence $\left\{x_n\right\}$ generated by (6) converges strongly to the unique $q^{*}\in \mathrm{\Omega }$ satisfying $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$.

Proof. 

We first justify the existence and uniqueness of $q^{*}$. By Remark 1, $\mathrm{\Theta }$ is closed and convex. Set $K:={\bigcap }_{n\ge 1}\mathrm{Fix}\left(T_n\right)$, which is nonempty because $\mathrm{\Omega }\ne \varnothing$. We claim that K is closed and convex. If $p_m\in K$ and $p_m\to p$, then for every n,

$$\parallel T_{n}p-p_m\parallel \le \parallel p-p_m\parallel \to 0,$$

so $T_{n}p=p$ and $p\in K$. If $p_0,p_1\in K$ and $p_t:=(1-t)p_0+t{p}_1$ with $t\in (0,1)$, quasi-nonexpansivity gives

$$\parallel T_n{p}_t-p_0\parallel \le t\parallel p_1-p_0\parallel ,\parallel T_n{p}_t-p_1\parallel \le (1-t)\parallel p_1-p_0\parallel .$$

The triangle inequality forces equality throughout and hence $T_n{p}_t=p_t$; thus $p_t\in K$. Therefore, $\mathrm{\Omega }=\mathrm{\Theta }\cap K$ is nonempty, closed, and convex. The metric projection $P_{\mathrm{\Omega }}$ is consequently well defined and nonexpansive, so $P_{\mathrm{\Omega }}\circ f$ is a $\tau$-contraction. By the Banach contraction principle it has a unique fixed point $q^{*}\in \mathrm{\Omega }$.

By Lemma 5, $\left\{x_n\right\}$ is bounded; let $M>0$ be a common bound for all relevant sequences. Fix $q=q^{*}$.

• Notation. To avoid conflict with the step-size sequence $\left\{s_n\right\}$ in algorithm (6), we write

  $${\sigma }_n:={\parallel x_n-q\parallel }^2$$

  (19)

  for the squared distance from $x_n$ to q.

• Step 1. Inertial estimate. By the triangle inequality,

  $$\begin{array}{cc}\hfill \parallel y_n-q\parallel & \le \parallel x_n-q\parallel +{\theta }_n\parallel x_n-x_{n-1}\parallel \hfill \\ \hfill & +|{\delta }_n|\parallel x_{n-1}-x_{n-2}\parallel .\hfill \end{array}$$

  Squaring and using boundedness,

  $$\begin{array}{cc}\hfill {\parallel y_n-q\parallel }^2& \le {\sigma }_n+2M\left({\theta }_n\parallel x_n-x_{n-1}\parallel +\left|{\delta }_n\right|\parallel x_{n-1}-x_{n-2}\parallel \right)\hfill \\ \hfill & +{\left({\theta }_n\parallel x_n-x_{n-1}\parallel +\left|{\delta }_n\right|\parallel x_{n-1}-x_{n-2}\parallel \right)}^2\hfill \\ \hfill & \le {\sigma }_n+2M{a}_n{\displaystyle \frac{{\theta }_n\parallel x_n-x_{n-1}\parallel +\left|{\delta }_n\right|\parallel x_{n-1}-x_{n-2}\parallel }{a_n}}+{\left(2M\right)}^2({\theta }_n+|{\delta }_n{\left|\right)}^2.\hfill \end{array}$$

  Set

  $${\epsilon }_n^{\left(1\right)}:={\displaystyle \frac{{\theta }_n\parallel x_n-x_{n-1}\parallel +\left|{\delta }_n\right|\parallel x_{n-1}-x_{n-2}\parallel }{a_n}},{\epsilon }_n^{\left(2\right)}:={\displaystyle \frac{({\theta }_n+|{\delta }_n{\left|\right)}^2}{a_n}}.$$

  By Lemma 5, the difference sequences $\{x_n-x_{n-1}\}$ and $\{x_{n-1}-x_{n-2}\}$ are bounded. Moreover, Assumption 5 gives ${\theta }_n/a_n\to 0$ and $|{\delta }_n|/a_n\to 0$. Therefore, ${\epsilon }_n^{\left(1\right)}\to 0$. The same ratio limits imply ${\theta }_n=o\left(a_n\right)$; writing ${\theta }_n={\eta }_n{a}_n$ with ${\eta }_n\to 0$ gives ${\theta }_n^2/a_n={\eta }_n^2{a}_n\to 0$, and likewise for ${\delta }_n$. Hence ${\epsilon }_n^{\left(2\right)}\to 0$. Consequently, there exists a null sequence $\left\{{\epsilon }_n\right\}$ (e.g., ${\epsilon }_n=2M{\epsilon }_n^{\left(1\right)}+4M^2{\epsilon }_n^{\left(2\right)}$) such that for all large n,

  $${\parallel y_n-q\parallel }^2\le {\sigma }_n+a_n{\epsilon }_n.$$

  (20)
• Step 2. Using quasi-nonexpansivity of ${\mathit{T}}_{\mathit{n}}$. Write $x_{n+1}=a_{n}f\left(x_n\right)+(1-a_n)t_n$ where $t_n:=\left({\beta }_n{z}_n+{\gamma }_n{T}_n{z}_n\right)/(1-a_n)$, ${\xi }_n:={\beta }_n/(1-a_n)$, ${\zeta }_n:={\gamma }_n/(1-a_n)$, ${\xi }_n+{\zeta }_n=1$. By the convex-combination norm identity,

  $${\parallel t_n-q\parallel }^2={\xi }_n{\parallel z_n-q\parallel }^2+{\zeta }_n{\parallel T_n{z}_n-q\parallel }^2-{\xi }_n{\zeta }_n{\parallel z_n-T_n{z}_n\parallel }^2.$$

  (21)

  Since $q\in \mathrm{\Omega }\subseteq {\bigcap }_k\mathrm{Fix}\left(T_k\right)$, Assumption 8 gives the plain quasi-nonexpansivity bound $\parallel T_n{z}_n-q\parallel \le \parallel z_n-q\parallel$, so ${\parallel T_n{z}_n-q\parallel }^2\le {\parallel z_n-q\parallel }^2$. Substituting into (21),

  $${\parallel t_n-q\parallel }^2\le {\parallel z_n-q\parallel }^2-{\xi }_n{\zeta }_n{\parallel z_n-T_n{z}_n\parallel }^2.$$

  (22)

  Because ${\beta }_n\ge \overline{\beta }>0$, ${\gamma }_n\ge \overline{\gamma }>0$ (Assumption 8) and $a_n\to 0$ (Assumption 5), there exists $N_1\in \mathbb{N}$ such that for $n\ge N_1$,

  $${\xi }_n{\zeta }_n={\displaystyle \frac{{\beta }_n{\gamma }_n}{{(1-a_n)}^2}}\ge {\displaystyle \frac{\overline{\beta }\overline{\gamma }}{{(1-a_n)}^2}}\ge \overline{\beta }\overline{\gamma }=:\kappa >0.$$

  Hence, combining with (22),

  $${\parallel t_n-q\parallel }^2\le {\parallel z_n-q\parallel }^2-\kappa {\parallel z_n-T_n{z}_n\parallel }^2\forall n\ge N_1.$$

  (23)
• Step 3. Viscosity recursion. Write $x_{n+1}-q=a_n(f\left(x_n\right)-f\left(q\right))+(1-a_n)(t_n-q)+a_n(f\left(q\right)-q)$. Using the inequality ${\parallel u+v\parallel }^2\le {\parallel u\parallel }^2+2⟨v,u+v⟩$ with $u=a_n(f\left(x_n\right)-f\left(q\right))+(1-a_n)(t_n-q)$ and $v=a_n(f\left(q\right)-q)$, we obtain

  $${\sigma }_{n+1}\le {∥a_n(f\left(x_n\right)-f\left(q\right))+(1-a_n)(t_n-q)∥}^2+2a_n⟨f\left(q\right)-q,x_{n+1}-q⟩.$$

  By convexity of ${\parallel ·\parallel }^2$,

  $$\begin{array}{cc}\hfill {∥a_n(f\left(x_n\right)-f\left(q\right))+(1-a_n)(t_n-q)∥}^2& \le a_n{\parallel f\left(x_n\right)-f\left(q\right)\parallel }^2+(1-a_n){\parallel t_n-q\parallel }^2\hfill \\ \hfill & \le a_n\tau {\sigma }_n+(1-a_n){\parallel t_n-q\parallel }^2,\hfill \end{array}$$

  because f is a $\tau$-contraction and ${\tau }^2\le \tau$. Hence

  $${\sigma }_{n+1}\le a_n\tau {\sigma }_n+(1-a_n){\parallel t_n-q\parallel }^2+2a_n⟨f\left(q\right)-q,x_{n+1}-q⟩.$$

  (24)

  Now combine the bounds for ${\parallel t_n-q\parallel }^2$. From (23) and Lemma 6 together with (20),

  $${\parallel t_n-q\parallel }^2\le {\sigma }_n+a_n{\epsilon }_n-\rho \left({\parallel v_n-y_n\parallel }^2+{\parallel z_n-v_n\parallel }^2\right)-\kappa {\parallel z_n-T_n{z}_n\parallel }^2.$$

  Substituting this into (24) gives

  $$\begin{array}{cc}\hfill {\sigma }_{n+1}& \le a_n\tau {\sigma }_n+(1-a_n)\left({\sigma }_n+a_n{\epsilon }_n-w_n\right)+2a_n⟨f\left(q\right)-q,x_{n+1}-q⟩\hfill \\ \hfill & =\left(1-a_n(1-\tau )\right){\sigma }_n+a_n(1-a_n){\epsilon }_n-(1-a_n)w_n\hfill \\ \hfill & +2a_n⟨f\left(q\right)-q,x_{n+1}-q⟩,\hfill \end{array}$$

  where we set $w_n:=\rho ({\parallel v_n-y_n\parallel }^2+{\parallel z_n-v_n\parallel }^2)+\kappa {\parallel z_n-T_n{z}_n\parallel }^2\ge 0$.

Define ${\lambda }_n:=a_n(1-\tau )$ (so ${\lambda }_n\in [0,1)$ and $\sum {\lambda }_n=\infty$). Then $2a_n={\displaystyle \frac{2{\lambda }_n}{1-\tau }}$ and $a_n(1-a_n){\epsilon }_n={\lambda }_n{\displaystyle \frac{(1-a_n){\epsilon }_n}{1-\tau }}$. Let

$$b_n:={\displaystyle \frac{2}{1-\tau }}⟨f\left(q\right)-q,x_{n+1}-q⟩+{\displaystyle \frac{(1-a_n){\epsilon }_n}{1-\tau }}.$$

We obtain, for all $n\ge \max \{N_0,N_1\}$,

$${\sigma }_{n+1}\le (1-{\lambda }_n){\sigma }_n+{\lambda }_n{b}_n-(1-a_n)w_n.$$

(25)

• Step 4. Verification of the residual condition. The sequence $\left\{b_n\right\}$ is bounded (because $\left\{x_n\right\}$ is bounded and ${\epsilon }_n\to 0$). To verify condition (iv) of Lemma 7, let $\left\{n_k\right\}$ be an arbitrary subsequence such that $w_{n_k}\to 0$. Since $\rho ,\kappa >0$, the definition of $w_n$ immediately gives

  $$\parallel v_{n_k}-y_{n_k}\parallel \to 0,\parallel z_{n_k}-v_{n_k}\parallel \to 0,\parallel z_{n_k}-T_{n_k}{z}_{n_k}\parallel \to 0.$$

  (26)

We now prove $\parallel x_{n_k+1}-x_{n_k}\parallel \to 0$. Using the last line of algorithm (6) and the definition $t_n=({\beta }_n{z}_n+{\gamma }_n{T}_n{z}_n)/(1-a_n)$ introduced in Step 2,

$$\begin{array}{cc}\hfill x_{n_k+1}-x_{n_k}& =a_{n_k}(f\left(x_{n_k}\right)-x_{n_k})+(1-a_{n_k})(t_{n_k}-x_{n_k}).\hfill \end{array}$$

Because $a_{n_k}\to 0$, it suffices to show $\parallel t_{n_k}-x_{n_k}\parallel \to 0$. Recall

$$t_{n_k}-x_{n_k}={\displaystyle \frac{{\beta }_{n_k}}{1-a_{n_k}}}(z_{n_k}-x_{n_k})+{\displaystyle \frac{{\gamma }_{n_k}}{1-a_{n_k}}}(T_{n_k}{z}_{n_k}-x_{n_k}).$$

Now

$$\parallel z_{n_k}-x_{n_k}\parallel \le \parallel z_{n_k}-v_{n_k}\parallel +\parallel v_{n_k}-y_{n_k}\parallel +\parallel y_{n_k}-x_{n_k}\parallel .$$

The first two terms tend to 0 by (26). For the third term, using $y_{n_k}-x_{n_k}={\theta }_{n_k}(x_{n_k}-x_{n_k-1})+{\delta }_{n_k}(x_{n_k-1}-x_{n_k-2})$, the boundedness of the difference sequences, and ${\theta }_n/a_n\to 0$, $|{\delta }_n|/a_n\to 0$, we obtain $\parallel y_{n_k}-x_{n_k}\parallel =o\left(a_{n_k}\right)\to 0$. Hence $\parallel z_{n_k}-x_{n_k}\parallel \to 0$. Moreover, $\parallel T_{n_k}{z}_{n_k}-x_{n_k}\parallel \le \parallel T_{n_k}{z}_{n_k}-z_{n_k}\parallel +\parallel z_{n_k}-x_{n_k}\parallel \to 0$. Therefore, $\parallel t_{n_k}-x_{n_k}\parallel \to 0$ and consequently $\parallel x_{n_k+1}-x_{n_k}\parallel \to 0$.

• Step 5. Weak cluster point belongs to $\mathbf{\Omega }$. Pass to a further subsequence, without changing notation, such that

  $$b_{n_k}\to \underset{j\to \infty }{\lim \sup }{b}_{n_j}\mathrm{and}{x}_{n_k}\rightharpoonup p$$

  for some $p\in H$; this is possible because $\left\{b_{n_k}\right\}$ is bounded and bounded sequences in a Hilbert space have weakly convergent subsequences. Since $\parallel y_{n_k}-x_{n_k}\parallel \to 0$, $\parallel v_{n_k}-y_{n_k}\parallel \to 0$, and $\parallel z_{n_k}-v_{n_k}\parallel \to 0$, we also have $y_{n_k}\rightharpoonup p$, $v_{n_k}\rightharpoonup p$, $z_{n_k}\rightharpoonup p$. Since $v_{n_k}\in C$ and a closed convex subset of a Hilbert space is weakly closed, $p\in C$.

$p\in \mathrm{\Theta }$. Recall that $v_{n_k}=T_{r_{n_k}}^{F,U,\phi }(y_{n_k}-r_{n_k}A{y}_{n_k})$. By definition of the resolvent, for every $y\in C\cap \mathrm{dom}\phi$,

$$F(v_{n_k},y)+⟨U{v}_{n_k},y-v_{n_k}⟩-\phi \left(v_{n_k}\right)+\phi \left(y\right)+{\displaystyle \frac{1}{r_{n_k}}}⟨y-v_{n_k},v_{n_k}-(y_{n_k}-r_{n_k}A{y}_{n_k})⟩\ge 0.$$

(27)

Writing $v_{n_k}-(y_{n_k}-r_{n_k}A{y}_{n_k})=(v_{n_k}-y_{n_k})+r_{n_k}A{y}_{n_k}$, (27) becomes

$$F(v_{n_k},y)+⟨U{v}_{n_k},y-v_{n_k}⟩-\phi \left(v_{n_k}\right)+\phi \left(y\right)+{\displaystyle \frac{1}{r_{n_k}}}⟨y-v_{n_k},v_{n_k}-y_{n_k}⟩+⟨A{y}_{n_k},y-v_{n_k}⟩\ge 0.$$

(28)

By monotonicity of F (Assumption 1 (B2)), $F(v_{n_k},y)\le -F(y,v_{n_k})$. Define $H(x,y):=F(x,y)+⟨Ux,y-x⟩-\phi \left(x\right)+\phi \left(y\right)$ for $x,y\in C\cap \mathrm{dom}\phi$. Since F and U are both monotone,

$$H(x,y)+H(y,x)=\left[F(x,y)+F(y,x)\right]-⟨Ux-Uy,x-y⟩\le 0\forall x,y\in C\cap \mathrm{dom}\phi ,$$

so H is monotone, i.e., $H(v_{n_k},y)\le -H(y,v_{n_k})$. Using this bound in (28) (which reads $H(v_{n_k},y)+\frac{1}{r_{n_k}}⟨y-v_{n_k},v_{n_k}-y_{n_k}⟩+⟨A{y}_{n_k},y-v_{n_k}⟩\ge 0$, and $-H(y,v_{n_k})\ge H(v_{n_k},y)$) gives

$$H(y,v_{n_k})\le {\displaystyle \frac{1}{r_{n_k}}}⟨y-v_{n_k},v_{n_k}-y_{n_k}⟩+⟨A{y}_{n_k},y-v_{n_k}⟩,$$

(29)

i.e.,

$$F(y,v_{n_k})+⟨Uy,v_{n_k}-y⟩-\phi \left(y\right)+\phi \left(v_{n_k}\right)\le {\displaystyle \frac{1}{r_{n_k}}}⟨y-v_{n_k},v_{n_k}-y_{n_k}⟩+⟨A{y}_{n_k},y-v_{n_k}⟩.$$

We estimate the right-hand side. Since A is $\alpha$-ISM and monotone,

$$\begin{array}{cc}\hfill ⟨A{y}_{n_k},y-v_{n_k}⟩& =⟨A{y}_{n_k}-Ay,y-v_{n_k}⟩+⟨Ay,y-v_{n_k}⟩\hfill \\ \hfill & =⟨A{y}_{n_k}-Ay,y_{n_k}-v_{n_k}⟩+⟨A{y}_{n_k}-Ay,y-y_{n_k}⟩+⟨Ay,y-v_{n_k}⟩.\hfill \end{array}$$

The first term tends to 0 because $y_{n_k}-v_{n_k}\to 0$ strongly and $A{y}_{n_k}-Ay$ is bounded. Monotonicity of A gives $⟨A{y}_{n_k}-Ay,y-y_{n_k}⟩=-⟨A{y}_{n_k}-Ay,y_{n_k}-y⟩\le 0$. Consequently,

$$⟨A{y}_{n_k},y-v_{n_k}⟩\le ⟨Ay,y-v_{n_k}⟩+o\left(1\right).$$

Also ${\displaystyle \frac{1}{r_{n_k}}}⟨y-v_{n_k},v_{n_k}-y_{n_k}⟩\to 0$ because ${\lim \inf }_k{r}_{n_k}>0$ and $v_{n_k}-y_{n_k}\to 0$. Therefore, for every $y\in C\cap \mathrm{dom}\phi$,

$$F(y,v_{n_k})+⟨Uy,v_{n_k}-y⟩-\phi \left(y\right)+\phi \left(v_{n_k}\right)\le ⟨Ay,y-v_{n_k}⟩+{\eta }_{n_k},$$

with ${\eta }_{n_k}\to 0$, i.e.,

$$F(y,v_{n_k})+⟨Uy+Ay,v_{n_k}-y⟩-\phi \left(y\right)+\phi \left(v_{n_k}\right)\le {\eta }_{n_k}.$$

(30)

We first show that $p\in \mathrm{dom}\phi$. Fix $y_0\in C\cap \mathrm{dom}\phi$, which exists because $\phi$ is proper. Since $v_{n_k}\rightharpoonup p$, weak lower semicontinuity of the real-valued function $F(y_0,·)$ implies that $\left\{F(y_0,v_{n_k})\right\}$ is bounded below for all sufficiently large k. The linear term in (30) converges, so that inequality with $y=y_0$ shows that $\left\{\phi \left(v_{n_k}\right)\right\}$ is bounded above. Weak lower semicontinuity of $\phi$ now gives

$$\phi \left(p\right)\le \underset{k\to \infty }{\lim \inf } \phi \left(v_{n_k}\right)<\infty .$$

Thus $p\in C\cap \mathrm{dom}\phi$.

Now fix any $y\in C\cap \mathrm{dom}\phi$. By Assumption 1 (B4), $F(y,·)$ is convex and lower semicontinuous, hence weakly lower semicontinuous, so $F(y,p)\le {\lim \inf }_{k}F(y,v_{n_k})$. Since $⟨Uy+Ay,·⟩$ is a continuous linear functional and $v_{n_k}\rightharpoonup p$, $⟨Uy+Ay,v_{n_k}-y⟩\to ⟨Uy+Ay,p-y⟩$. By weak lower semicontinuity of $\phi$ (Assumption 1 (B6)), $\phi \left(p\right)\le {\lim \inf }_k\phi \left(v_{n_k}\right)$. Taking ${\lim \inf }_{k\to \infty }$ in (30) (using $\lim \inf (a_k+b_k+c_k)\ge \lim \inf a_k+\lim b_k+\lim \inf c_k$ and ${\eta }_{n_k}\to 0$),

$$F(y,p)+⟨Uy+Ay,p-y⟩-\phi \left(y\right)+\phi \left(p\right)\le 0\forall y\in C\cap \mathrm{dom}\phi ,$$

which is exactly the dual inequality (4). By Lemma 4 (whose hypotheses hold: F satisfies (B1), (B3) and (B4); U is continuous by (B5); A is monotone and Lipschitz by Assumption 7; $\phi$ is proper, convex, lsc by (B6)), we conclude $p\in \mathrm{\Theta }$.

$p\in {\bigcap }_{n\ge 1}\mathrm{Fix}\left(T_n\right)$. We already have $\parallel z_{n_k}-T_{n_k}{z}_{n_k}\parallel \to 0$ from (26). Since the ${\mathrm{NST}}_w$ condition is stated for sequences indexed by all n, we make the following standard embedding. Choose $q_0\in \mathrm{\Omega }$ and define a bounded sequence $\left\{u_n\right\}$ by

$$u_n:=\left\{\begin{array}{cc}{z}_{n_k},\hfill & n=n_k\mathrm{for}\mathrm{some}k,\hfill \\ q_0,\hfill & n\notin \{n_k:k\ge 1\}.\hfill \end{array}\right.$$

For $n\notin \{n_k:k\ge 1\}$ we have $T_n{q}_0=q_0$, while along $n=n_k$ the residual tends to zero by (26). Hence $\parallel u_n-T_n{u}_n\parallel \to 0$. Moreover, $u_{n_k}=z_{n_k}\rightharpoonup p$, so p is a weak cluster point of $\left\{u_n\right\}$. The ${\mathrm{NST}}_w$ condition, therefore, gives $p\in {\bigcap }_{n\ge 1}\mathrm{Fix}\left(T_n\right)$. Hence $p\in \mathrm{\Omega }$.

• Step 6. Verification of $\mathbf{lim} \mathbf{sup}{\mathit{b}}_{{\mathit{n}}_{\mathit{k}}}\mathbf{\le }\mathbf{0}$. Recall $q=P_{\mathrm{\Omega }}f\left(q\right)$. Because $p\in \mathrm{\Omega }$, the variational characterisation of the projection gives $⟨f\left(q\right)-q,p-q⟩\le 0$. Now

  $$b_{n_k}={\displaystyle \frac{2}{1-\tau }}⟨f\left(q\right)-q,x_{n_k+1}-q⟩+{\displaystyle \frac{(1-a_{n_k}){\epsilon }_{n_k}}{1-\tau }}.$$

  Since $x_{n_k+1}-x_{n_k}\to 0$ and $x_{n_k}\rightharpoonup p$, we have $x_{n_k+1}\rightharpoonup p$, and therefore, $⟨f\left(q\right)-q,x_{n_k+1}-q⟩\to ⟨f\left(q\right)-q,p-q⟩\le 0$. Moreover ${\epsilon }_{n_k}\to 0$. Consequently,

  $$\underset{j\to \infty }{\lim \sup } b_{n_j}=\underset{k\to \infty }{\lim }{b}_{n_k}\le {\displaystyle \frac{2}{1-\tau }}·0+0=0.$$

  Thus condition (iv) of Lemma 7 holds with $c_n=w_n$.
• Step 7. Application of the residual convergence lemma. Apply Lemma 7 to (25), ignoring the finite initial segment, with

  $$s_n={\sigma }_n,{\ell }_n={\lambda }_n=a_n(1-\tau ),d_n=0,v_n=1-a_n,c_n=w_n.$$

  We have ${\sum }_n{\ell }_n=(1-\tau ){\sum }_n{a}_n=\infty$, ${\sum }_n\left|d_n\right|=0$, and $v_n>0$ because $a_n<1$. Moreover,

  $${\displaystyle \frac{{\ell }_n}{v_n}}={\displaystyle \frac{a_n(1-\tau )}{1-a_n}}⟶0$$

  because $a_n\to 0$. Condition (iv) was proved in Steps 4–6. Hence Lemma 7 yields ${\sigma }_n\to 0$, that is, $x_n\to q$ strongly. □

Corollary 1 

(Family of nonexpansive mappings). Let H be a real Hilbert space and let $C\subseteq H$ be nonempty, closed, and convex. Suppose that Assumptions 1, 3, 5 and 7 hold. Let ${\{T_n:H\to H\}}_{n\ge 1}$ be a family of nonexpansive mappings, and set

$$K:=\bigcap _{n\ge 1}\mathrm{Fix}\left(T_n\right),\mathrm{\Omega }:=\mathrm{\Theta }\cap K,$$

where $\mathrm{\Theta }=\mathrm{GMEP}(F,U+A,\phi )$. Assume that:

(i) 

$\mathrm{\Omega }\ne \varnothing$;

(ii) 

the family $\left\{T_n\right\}$ satisfies the ${\mathrm{NST}}_w$ condition: for every bounded sequence $\left\{u_n\right\}\subset H$ such that $\parallel u_n-T_n{u}_n\parallel \to 0$, every weak cluster point of $\left\{u_n\right\}$ belongs to K;

(iii) 

there exist constants $\overline{\beta },\overline{\gamma }>0$ such that ${\beta }_n\ge \overline{\beta }$ and ${\gamma }_n\ge \overline{\gamma }$ for every $n\ge 1$.

Then the sequence $\left\{x_n\right\}$ generated by (6) converges strongly to the unique point $q^{*}\in \mathrm{\Omega }$ satisfying

$$q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right).$$

Proof. 

Let $q\in K$. Since $q\in \mathrm{Fix}\left(T_n\right)$ for every $n\ge 1$ and $T_n$ is nonexpansive, we have, for every $x\in H$,

$$\parallel T_{n}x-q\parallel =\parallel T_{n}x-T_{n}q\parallel \le \parallel x-q\parallel .$$

Thus each $T_n$ is quasi-nonexpansive with respect to K, and hence with respect to $\mathrm{\Omega }\subseteq K$. Assumptions (ii) and (iii), therefore, verify Assumption 8.

Assumption 7 implies Assumption 2, because $0<r_n,s_n<\alpha <2\alpha$. Assumptions 3 and 5 provide the contraction and parameter conditions required in Lemma 5, while assumption (i) gives Assumption 6. Consequently, all the hypotheses of Theorem 1 are satisfied. The asserted strong convergence and the uniqueness of the viscosity-selected limit now follow directly from Theorem 1. □

4. Applications

In this section we specialize Theorem 1 to four settings: composite convex minimization, a mixed variational inequality, a countable convex feasibility problem, and a finite family under cyclic control. Throughout, equilibrium inequalities are understood over $C\cap \mathrm{dom}\phi$, as in Section 2.

4.1. Composite Convex Minimization

Let $h:H\to \mathbb{R}$ be convex and Fréchet differentiable with L-Lipschitz gradient for some $L>0$, let $\psi :C\to \mathbb{R}$ be convex and lower semicontinuous, and let $\phi :C\to \mathbb{R}\cup \{\infty \}$ be proper, convex, and lower semicontinuous. Consider

$$\underset{x\in C\cap \mathrm{dom}\phi }{\mathrm{minimize}}\left(h\left(x\right)+\psi \left(x\right)+\phi \left(x\right)\right).$$

(31)

Corollary 2. 

In algorithm (6), take

$$A:=\nabla h,U:=0,F(x,y):=\psi \left(y\right)-\psi \left(x\right)(x,y\in C).$$

Then F satisfies Assumption 1 (B1)–(B4), A is $1/L$-inverse strongly monotone, and Θ is exactly the solution set of (31). Consequently, if Assumptions 3, 5, 7 and 8 hold with $\alpha =1/L$, and

$$\mathrm{\Omega }=\mathrm{\Theta }\cap \bigcap _{n\ge 1}\mathrm{Fix}\left(T_n\right)\ne \varnothing ,$$

then $x_n\to q^{*}$ strongly, where $q^{*}$ is the unique point of Ω satisfying $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$. In particular, $q^{*}$ solves (31) and belongs to every $\mathrm{Fix}\left(T_n\right)$.

Proof. 

Clearly $F(x,x)=0$ and $F(x,y)+F(y,x)=0$. To verify Assumption 1 (B3), set $x_t:=tz+(1-t)x$. Convexity gives

$$\psi \left(x_t\right)\le t\psi \left(z\right)+(1-t)\psi \left(x\right),$$

while lower semicontinuity gives $\psi \left(x\right)\le {\lim \inf }_{t\downarrow 0}\psi \left(x_t\right)$. Hence $\psi \left(x_t\right)\to \psi \left(x\right)$ and, therefore, $F(x_t,y)\to F(x,y)$. Assumption 1 (B4) follows from convexity and lower semicontinuity of $\psi$.

By the Baillon–Haddad theorem, the convexity of h and the L-Lipschitz continuity of $\nabla h$ imply

$$⟨\nabla h\left(x\right)-\nabla h\left(y\right),x-y⟩\ge {\displaystyle \frac{1}{L}}{\parallel \nabla h\left(x\right)-\nabla h\left(y\right)\parallel }^2,$$

so $A=\nabla h$ is $1/L$-inverse strongly monotone.

It remains to identify $\mathrm{\Theta }$. If q minimizes the objective in (31), fix $y\in C\cap \mathrm{dom}\phi$ and put $q_t:=q+t(y-q)$. Optimality, followed by convexity of $\psi$ and $\phi$, gives

$$0\le {\displaystyle \frac{h\left(q_t\right)-h\left(q\right)}{t}}+\psi \left(y\right)-\psi \left(q\right)+\phi \left(y\right)-\phi \left(q\right).$$

Letting $t\downarrow 0$ yields

$$⟨\nabla h\left(q\right),y-q⟩+\psi \left(y\right)-\psi \left(q\right)+\phi \left(y\right)-\phi \left(q\right)\ge 0,$$

so $q\in \mathrm{\Theta }$. Conversely, if $q\in \mathrm{\Theta }$, convexity of h gives $h\left(y\right)-h\left(q\right)\ge ⟨\nabla h\left(q\right),y-q⟩$. Adding this inequality to the defining inequality for $\mathrm{\Theta }$ shows that the objective at y is not smaller than its value at q. Thus q solves (31). The convergence assertion is now Theorem 1. □

Remark 4. 

When $\psi \equiv 0$, this gives a smooth–nonsmooth composite minimization problem with the additional fixed-point constraints $q^{*}\in {\bigcap }_n\mathrm{Fix}\left(T_n\right)$. The smooth term is handled by the forward step, while the extended-valued term φ is handled implicitly by the resolvent and may encode further convex constraints.

4.2. Mixed Variational Inequality

Let $U:C\to H$ be continuous and monotone, let $A:H\to H$ be $\alpha$-inverse strongly monotone, and let $\phi :C\to \mathbb{R}\cup \{\infty \}$ be proper, convex, and lower semicontinuous. The mixed variational inequality problem is

$$\mathrm{find}q\in C\cap \mathrm{dom}\phi \mathrm{such}\mathrm{that}⟨(U+A)q,y-q⟩+\phi \left(y\right)-\phi \left(q\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi .$$

(32)

Corollary 3. 

Take $F(x,y):=0$ for all $x,y\in C$ in algorithm (6). Then F satisfies Assumption 1 (B1)–(B4), and Θ is precisely the solution set of (32). Consequently, if Assumptions 3, 5, 7 and 8 hold and

$$\mathrm{\Omega }=\mathrm{\Theta }\cap \bigcap _{n\ge 1}\mathrm{Fix}\left(T_n\right)\ne \varnothing ,$$

then $x_n\to q^{*}$ strongly, where $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)\in \mathrm{\Omega }$ is unique. In particular, $q^{*}$ solves (32).

Proof. 

All four assumptions on F are immediate for the zero bifunction. By definition,

$$\mathrm{\Theta }=\left\{q\in C\cap \mathrm{dom}\phi :⟨(U+A)q,y-q⟩+\phi \left(y\right)-\phi \left(q\right)\ge 0\forall y\in C\cap \mathrm{dom}\phi \right\},$$

which is exactly the solution set of (32). The final assertion follows from Theorem 1. □

Remark 5. 

This application separates the operator into a continuous monotone part U, treated implicitly, and a cocoercive part A, treated explicitly. The sum $U+A$ itself need not be Lipschitz continuous.

4.3. Countable Convex Feasibility

Let ${\left\{C_n\right\}}_{n\ge 1}$ be nonempty closed convex subsets of H such that

$$C_{\infty }:=\bigcap _{n\ge 1}{C}_n\ne \varnothing ,$$

and let $P_{C_n}$ be the metric projection onto $C_n$. We seek a point in

$$\mathrm{\Theta }\cap C_{\infty }.$$

(33)

Corollary 4. 

Suppose Assumptions 1 and 7 hold, and set $T_n:=P_{C_n}$. Assume the following joint regularity condition:

$$\left.\begin{array}{cc}\hfill & \left\{u_n\right\}\mathit{is}\mathit{bounded},\hfill \\ \hfill & \parallel u_n-P_{C_n}{u}_n\parallel \to 0\hfill \end{array}\right\}⟹\mathit{every}\mathit{weak}\mathit{cluster}\mathit{point}\mathit{of}\left\{u_n\right\}\mathit{belongs}\mathit{to}{C}_{\infty }.$$

(CF)

Then $T_n$ is quasi-nonexpansive, $\mathrm{Fix}\left(T_n\right)=C_n$, and the family $\left\{T_n\right\}$ satisfies the ${\mathrm{NST}}_w$ condition. Consequently, if Assumptions 3 and 5 hold, ${\beta }_n\ge \overline{\beta }>0$, ${\gamma }_n\ge \overline{\gamma }>0$, and

$$\mathrm{\Omega }=\mathrm{\Theta }\cap C_{\infty }\ne \varnothing ,$$

then $x_n\to q^{*}$ strongly, where $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)\in \mathrm{\Omega }$ is unique.

Proof. 

Metric projections are firmly nonexpansive. Hence $P_{C_n}$ is quasi-nonexpansive with respect to $C_n$, and $\mathrm{Fix}\left(P_{C_n}\right)=C_n$. Since $C_{\infty }\subseteq C_n$, it is, therefore, quasi-nonexpansive with respect to the common fixed-point set $C_{\infty }$, as required in Assumption 8. Finally, ${\bigcap }_n\mathrm{Fix}\left(T_n\right)={\bigcap }_n{C}_n=C_{\infty }$, and (CF) is exactly the ${\mathrm{NST}}_w$ condition for this family. Theorem 1, therefore, applies. □

Remark 6. 

Condition (CF) is not automatic for an arbitrary countable family. Two useful sufficient cases are as follows:

(a) 

$C_n$ converges to $C_{\infty }$ in the sense of Mosco. Indeed, if $u_{n_k}\rightharpoonup p$ and $\parallel u_n-P_{C_n}{u}_n\parallel \to 0$, then $P_{C_{n_k}}{u}_{n_k}\rightharpoonup p$, and the weak-limit condition in Mosco convergence gives $p\in C_{\infty }$;

(b) 

$C_1\supseteq C_2\supseteq \dots$ and $C_{\infty }={\bigcap }_n{C}_n$. For any weak cluster subsequence and any fixed m, the projected points eventually belong to $C_m$ and have the same weak limit; hence that limit belongs to every $C_m$.

Thus, under (CF), the algorithm simultaneously solves the generalized mixed equilibrium problem and a countable convex feasibility problem.

5. Finite Families Under Cyclic Control

Let $S_1,\dots ,S_N:H\to H$ be a finite family of mappings, where $N\ge 1$, and define the cyclic index and the corresponding sequence by

$$\iota \left(n\right):=\left((n-1)\mathrm{mod}N\right)+1,T_n:=S_{\iota \left(n\right)}(n\ge 1).$$

(34)

Thus the algorithm uses the mappings in the order $S_1,S_2,\dots ,S_N,S_1,S_2,\dots$. Set

$$K:={\displaystyle \bigcap _{i=1}^N}\mathrm{Fix}\left(S_i\right).$$

Corollary 5 

(Cyclic finite-family specialization). Suppose Assumptions 1, 3, 5 and 7 hold. Assume further that

(i) 

each $S_i$ is quasi-nonexpansive with respect to K;

(ii) 

the cyclic family satisfies the joint regularity condition

$$\left.\begin{array}{cc}\hfill & \left\{u_n\right\}\subset H\mathit{is}\mathit{bounded},\hfill \\ \hfill & \parallel u_n-S_{\iota \left(n\right)}{u}_n\parallel ⟶0\hfill \end{array}\right\}⟹\mathit{every}\mathit{weak}\mathit{cluster}\mathit{point}\mathit{of}\left\{u_n\right\}\mathit{belongs}\mathit{to}K;$$

(35)

(iii) 

there exist $\overline{\beta },\overline{\gamma }>0$ such that ${\beta }_n\ge \overline{\beta }$ and ${\gamma }_n\ge \overline{\gamma }$ for every n;

(iv) 

$\mathrm{\Omega }:=\mathrm{\Theta }\cap K\ne \varnothing$.

If the sequence $\left\{T_n\right\}$ in algorithm (6) is chosen as in (34), then $x_n\to q^{*}$ strongly, where $q^{*}$ is the unique point of Ω satisfying $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$.

Proof. 

Because every $S_i$ occurs infinitely often in the cyclic sequence,

$$\bigcap _{n\ge 1}\mathrm{Fix}\left(T_n\right)={\displaystyle \bigcap _{i=1}^N}\mathrm{Fix}\left(S_i\right)=K.$$

Quasi-nonexpansivity with respect to K, condition (35), and the lower bounds on ${\beta }_n$ and ${\gamma }_n$ verify Assumption 8 for the cyclic sequence. Moreover, Assumption 7 implies Assumption 2, while $\mathrm{\Omega }=\mathrm{\Theta }\cap K\ne \varnothing$ verifies the solution-set assumption. The conclusion, therefore, follows directly from Theorem 1. □

Remark 7 

(A sufficient cyclic regularity condition). A simple sufficient condition for (35) is that each $I-S_i$ is demiclosed at zero and

$$\mathrm{Fix}\left(S_i\right)=K(i=1,\dots ,N).$$

(36)

Indeed, let $\left\{u_n\right\}$ be bounded with $\parallel u_n-S_{\iota \left(n\right)}{u}_n\parallel \to 0$, and let $u_{n_k}\rightharpoonup p$. Since there are only finitely many cyclic indices, a further subsequence has $\iota \left(n_k\right)=i_0$ for some fixed $i_0$. Along this subsequence, $\parallel u_{n_k}-S_{i_0}{u}_{n_k}\parallel \to 0$. Demiclosedness of $I-S_{i_0}$ at zero yields $p\in \mathrm{Fix}\left(S_{i_0}\right)=K$, proving (35). Thus Corollary 5 applies automatically when the mappings share the same fixed-point set and their residual operators are demiclosed at zero.

6. A Five-Dimensional Illustrative Example

We give explicit data satisfying Corollary 5. Let $H={\mathbb{R}}^5$ with its Euclidean inner product and let

$$C:=\{x\in {\mathbb{R}}^5:\parallel x\parallel \le 5\}.$$

Take $F\equiv 0$ and let $\phi$ be the restriction to C of the indicator function ${\iota }_C$; thus $\phi \equiv 0$ on C and $\mathrm{dom}\phi =C$. Define $U\left(x\right)=M_{U}x$ and $A\left(x\right)=M_{A}x$, where

$$M_U:=\mathrm{diag}(0,0.25,0.20,0.15,0.10),M_A:=\mathrm{diag}(0,0.20,0.20,0.20,0.20).$$

(37)

The map U is continuous and monotone. Moreover, for every $d\in {\mathbb{R}}^5$,

$$⟨M_{A}d,d⟩\ge 5{\parallel M_{A}d\parallel }^2,$$

so A is $\alpha$-inverse strongly monotone with $\alpha =5$.

For these data, the resolvent is the unique solution of

$$T_r^{F,U,\phi }\left(x\right)=\underset{z\in C}{\mathrm{argmin}}\left\{{\displaystyle \frac{1}{2}}⟨M_{U}z,z⟩+{\displaystyle \frac{1}{2r}}{\parallel z-x\parallel }^2\right\}.$$

(38)

It can be evaluated explicitly up to one scalar multiplier. Namely,

$$T_r^{F,U,\phi }\left(x\right)={(I+r{M}_U+\mu I)}^{-1}x,$$

(39)

where $\mu =0$ if $\parallel {(I+r{M}_U)}^{-1}x\parallel \le 5$; otherwise, $\mu >0$ is the unique solution of

$$\parallel {(I+r{M}_U+\mu I)}^{-1}x\parallel =5.$$

This multiplier formula follows from the optimality condition for (38). In particular, a projection applied after the unconstrained matrix inverse would not, in general, give the resolvent.

Let $B:=M_U+M_A$. Since B is positive semidefinite and $\ker B=\mathbb{R}{e}_1$, where $e_1={(1,0,0,0,0)}^{\top }$, the mixed variational inequality solution set is

$$\mathrm{\Theta }=\ker B\cap C=[-5,5]e_1.$$

(40)

Indeed, if $q\in \mathrm{\Theta }$, choosing $y=0$ in the variational inequality gives $-⟨Bq,q⟩\ge 0$. Positive semidefiniteness then implies $Bq=0$. Conversely, every $q\in \ker B\cap C$ satisfies the variational inequality.

• Cyclic family. Let $L:=\mathbb{R}{e}_1$, let $P_L$ be the orthogonal projection onto L, and define

  $$S_1:=P_L,S_2:={\displaystyle \frac{1}{2}}(I+P_L),S_3:={\displaystyle \frac{1}{4}}I+{\displaystyle \frac{3}{4}}{P}_L.$$

  (41)

  Each $S_i$ is a self-adjoint linear contraction whose eigenvalues belong to $[0,1]$; hence it is firmly nonexpansive. Furthermore,

  $$\mathrm{Fix}\left(S_1\right)=\mathrm{Fix}\left(S_2\right)=\mathrm{Fix}\left(S_3\right)=L.$$

  (42)

  Thus Remark 7 gives the cyclic regularity condition (35), and

  $$\mathrm{\Omega }=\mathrm{\Theta }\cap L=[-5,5]e_1.$$

  (43)

  Contraction and target point. Set

  $$f\left(x\right):=(1-{\eta }_f)x+{\eta }_{f}b,{\eta }_f=0.35,b={(2,0.5,-0.3,0.7,-0.4)}^{\top }.$$

  (44)

  The contraction modulus of f is $\tau =1-{\eta }_f=0.65$. For $q=t{e}_1\in \mathrm{\Omega }$,

  $$P_{\mathrm{\Omega }}f\left(q\right)=P_{[-5,5]}\left((1-{\eta }_f)t+2{\eta }_f\right)e_1.$$

  Consequently, the unique fixed point of $P_{\mathrm{\Omega }}\circ f$ is

  $$q^{*}=2e_1={(2,0,0,0,0)}^{\top }.$$

  (45)

  Algorithm parameters. For $n\ge 1$, choose

  $$\begin{array}{c}{r}_n=s_n=2.5,a_n={(n+1)}^{-0.9},\\ {\beta }_n={\gamma }_n={\displaystyle \frac{1-a_n}{2}},\\ {\theta }_n={\displaystyle \frac{0.80}{{(n+1)}^{2.1}}},{\delta }_n=-{\displaystyle \frac{0.40}{{(n+1)}^{2.1}}}.\end{array}$$

  (46)

  Then $0<r_n,s_n<\alpha$, ${\lim \inf }_n{r}_n>0$, and ${\lim \sup }_n\max \{r_n,s_n\}<\alpha$. Also $a_n\to 0$, ${\sum }_n{a}_n=\infty$, and $0<a_n<1$. Since $a_n$ is decreasing,

  $${\beta }_n,{\gamma }_n\ge {\displaystyle \frac{1-2^{-0.9}}{2}}>0(n\ge 1).$$

  Finally,

  $${\displaystyle \frac{{\theta }_n}{a_n}}={\displaystyle \frac{0.80}{{(n+1)}^{1.2}}},{\displaystyle \frac{|{\delta }_n|}{a_n}}={\displaystyle \frac{0.40}{{(n+1)}^{1.2}}},$$

  so both ratio series are summable. Hence all the conditions in Assumption 5 are satisfied. In every numerical run we set $x_1=x_0$ and $x_{-1}=x_0$. We use $T_n=S_{1+\left(\right(n-1\left)\mathrm{mod}3\right)}$, beginning with $S_1$ at $n=1$, and perform 400 iterations in double precision.
• Explicit verification of the assumptions. The preceding calculations verify all the hypotheses used in Section 3:

  (i)

  $H={\mathbb{R}}^5$ is a real Hilbert space and C is nonempty, closed, and convex. Since $F\equiv 0$, conditions (B1)–(B4) hold; U is continuous and monotone and $\phi \equiv 0$ on C is proper, convex, and lower semicontinuous. Thus Assumption 1 is satisfied.

  (ii)

  The operator A is 5-inverse strongly monotone, and $r_n=s_n=2.5<5$, with ${\lim \inf }_n{r}_n>0$ and ${\lim \sup }_n\max \{r_n,s_n\}<5$. Hence Assumptions 2 and 7 hold.

  (iii)

  The mapping f is a $0.65$-contraction, so Assumption 3 holds.

  (iv)

  Each cyclic mapping $S_i$ is firmly nonexpansive, hence quasi-nonexpansive, and their common fixed-point set is L. The cyclic regularity condition follows from Remark 7; therefore, Assumptions 4 and 8 are satisfied.

  (v)

  The sequences in (46) satisfy $a_n+{\beta }_n+{\gamma }_n=1$, ${\theta }_n\ge 0$, ${\delta }_n\le 0$, the required summability conditions, and uniform positive lower bounds for ${\beta }_n$ and ${\gamma }_n$. Thus Assumption 5 holds.

  (vi)

  Finally, $\mathrm{\Omega }=[-5,5]e_1\ne \varnothing$, so Assumption 6 is satisfied.

  Consequently, every hypothesis of the strong convergence theorem and its cyclic specialization is fulfilled by this example.
• Numerical behavior. The following experiments report $\parallel x_n-q^{*}\parallel$ on a logarithmic vertical scale. Figure 1 uses $x_0=x_1=0$, $x_0=x_1=5e_1$, and the seeded point

  $$x_0=x_1={(-1.23129742,0.37347927,-2.94398635,2.16689119,0.99093229)}^{\top }.$$

  The third initial point was chosen randomly. We used seed 2026 so that the same point can be generated again, and adjusted it to have norm 4. All three curves decrease toward the same viscosity-selected point $q^{*}=2e_1$, as predicted by the theorem.
  

  Figure 1. Error $\parallel x_n-q^{*}\parallel$ for three fully specified initial pairs, with $x_1=x_0$. The seeded point has norm 4 and lies in C.

Figure 2 uses $x_0=x_1=0$ and compares the following admissible choices:

$$\begin{array}{cccc}& a_n& {\theta }_n& {\delta }_n\\ ine\mathrm{Set}1& {(n+1)}^{-0.70}& 0.50{(n+1)}^{-1.80}& -0.25{(n+1)}^{-1.80}\\ \mathrm{Set}2& {(n+1)}^{-0.85}& 0.70{(n+1)}^{-2.00}& -0.35{(n+1)}^{-2.00}\\ \mathrm{Set}3& {(n+1)}^{-1.00}& 1.00{(n+1)}^{-2.30}& -0.50{(n+1)}^{-2.30}.\end{array}$$

In each case ${\beta }_n={\gamma }_n=(1-a_n)/2$ and $r_n=s_n=2.5$. The curves compare the combined finite-iteration effect of the viscosity and inertial parameters.

Figure 2. Error $\parallel x_n-q^{*}\parallel$ for three stated choices of the viscosity and inertial parameters, with $x_0=x_1=0$. Set 1 converges fastest to the target over the displayed iteration range.

Finally, Figure 3 compares the proposed scheme with the non-inertial variant ${\theta }_n={\delta }_n=0$, using $x_0=x_1=0$. To make the finite-iteration effect of the inertial correction clearly visible, the inertial run uses the alternative admissible choice

$${\theta }_n={\displaystyle \frac{17.5}{{(n+1)}^{2.1}}},{\delta }_n=-{\displaystyle \frac{2}{{(n+1)}^{2.1}}},$$

while $a_n$, ${\beta }_n$, ${\gamma }_n$, $r_n$, and $s_n$ remain as in (46). These coefficients satisfy

$${\displaystyle \frac{{\theta }_n}{a_n}}={\displaystyle \frac{17.5}{{(n+1)}^{1.2}}},{\displaystyle \frac{|{\delta }_n|}{a_n}}={\displaystyle \frac{2}{{(n+1)}^{1.2}}},$$

so both required ratio series remain summable. A direct implementation gives

$$\begin{array}{cccc}\hfill \parallel x_{400}-q^{*}\parallel & =0.019910\hfill & & \left(\mathrm{inertial}\right),\hfill \\ \hfill \parallel x_{400}-q^{*}\parallel & =0.124082\hfill & & \left(\mathrm{non}\text{-}\mathrm{inertial}\right).\hfill \end{array}$$

Thus the inertial error is of order ${10}^{-2}$ and is approximately 6.23 times smaller than the non-inertial error at iteration 400. This comparison is specific to the stated admissible parameters and is not a claim of a uniform convergence-rate advantage.

Figure 3. Reproducible comparison using $x_0=x_1=0$. The inertial run uses ${\theta }_n=17.5{(n+1)}^{-2.1}$ and ${\delta }_n=-2{(n+1)}^{-2.1}$, whereas the non-inertial run uses ${\theta }_n={\delta }_n=0$; all remaining parameters are identical. The values marked at iteration 400 were obtained directly from the stated algorithm and parameters.

All assumptions of Corollary 5 are now satisfied. Hence the sequence generated by (6), with the mappings in (41) used cyclically, converges strongly to the point $q^{*}=2e_1$ in (45).

7. Conclusions

In this paper, we introduced a unified two-step inertial viscosity-type algorithm for finding a common solution of a generalized mixed equilibrium problem and the fixed-point problems of a family of quasi-nonexpansive mappings in a real Hilbert space. The formulation separates the two monotone components of the equilibrium problem: the operator U is treated implicitly through the generalized mixed equilibrium resolvent, whereas the inverse strongly monotone operator A is evaluated explicitly through two forward–resolvent stages. The algorithm also incorporates two inertial corrections, a family-dependent fixed-point step, and viscosity regularization by a contraction.

We first established boundedness of the generated sequence under explicit summability conditions on the inertial parameters, including a second inertial coefficient that may be negative. A uniform descent inequality was then derived to control the two forward–resolvent residuals and the fixed-point residual simultaneously. Together with the weak regularity condition imposed on the mapping family and a residual-based convergence criterion, this estimate yielded strong convergence of the iterates to the unique viscosity-selected point $q^{*}\in \mathrm{\Omega }$ satisfying $q^{*}=P_{\mathrm{\Omega }}f\left(q^{*}\right)$.

The general result was applied to composite convex minimization, mixed variational inequalities, countable convex feasibility problems, and finite families under cyclic control. The five-dimensional example demonstrated that the assumptions can be verified explicitly and identified the limiting point as $q^{*}=2e_1$. The numerical plots further illustrated convergence from different initial points, sensitivity to the admissible parameter choices, and a clear acceleration of the tested inertial scheme relative to its non-inertial counterpart for the example considered.