Common Solution for a Finite Family of Equilibrium Problems, Quasi-Variational Inclusion Problems and Fixed Points on Hadamard Manifolds

Abstract

In this paper, we introduce an iterative algorithm for finding a common solution of a finite family of the equilibrium problems, quasi-variational inclusion problems and fixed point problem on Hadamard manifolds. Under suitable conditions, some strong convergence theorems are proved. Our results extend some recent results in literature.

1. Introduction

Recently, many convergence results by the proximal point algorithm have been extended from the classical linear spaces to the setting of manifolds (see, for example, refs. [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]. Li et al. [10] developed the proximal point method in the setting of Hadamard manifolds. Later, Li et al. [11] extended the Mann and Halpern iteration scheme for finding the fixed points of nonexpansive mappings from Hilbert spaces to Hadamard manifolds. Very recently, Ansari et al. [12] and Al-Homidan-Ansari-Babu [13] considered the problem of finding

$$x\in Fix\left(T\right)\bigcap {(A+B)}^{-1}\left(0\right)$$

(1)

in a Hadamard manifold, where T is a nonexpansive mapping, B is a set-valued maximal monotone mapping, and A is a single-valued continuous and monotone mapping. They proposed some Halpern-type and Mann-type iterative methods. Under suitable conditions they proved that the sequence generated by the algorithm converges strongly to a common element of the set of fixed points of the mapping T and the set of solutions of the inclusion problem.

Recently, Calao et al. [14] and Khammahawong et al. [19] studied the equilibrium problem on Hadamard manifolds. Let M be a Hadamard manifold, $TM$ be the tangent bundle of $M,K$ be a nonempty closed geodesic convex subset of M, and $F:K\times K⟶R$ be a bifunction satisfying $F(x,x)=0$, for all $x\in K$. Then the equilibrium problem on the Hadamard manifold is to find $x^{*}\in K$ such that

$$F(x^{*},y)\ge 0,\forall y\in K.$$

(2)

We denote the set of a equilibrium points of the equilibrium problem (2) by $EP\left(F\right)$.

Motivated and inspired by the above results, we consider the following common solution problem for a finite family of the equilibrium problem, quasi-variational inclusion problem and the fixed point on Hadamard manifolds: i.e., to find $x^{*}\in K$ such that

$$x^{*}\in \bigcap _{i=1}^{m}EP\left(F_i\right)\bigcap {(A+B)}^{-1}\left(0\right)\bigcap F\left(S\right).$$

(3)

In this paper, an iterative algorithm for finding a commmon solution of problem (3) is proposed. Under suitable conditions, some strong convergence is proven. Our results extend some recent results in literature.

2. Preliminaries

In this section, we recall some fundamental definitions, properties, useful results, and notations of Riemannian geometry. Readers can refer to the textbook [20].

Let M be a finite dimensional differentiable manifold, $T_{p}M$ be the tangent space of M at $p\in M$, and we denote by $TM={\bigcup }_{p\in M}{T}_{p}M$ the tangent bundle of M. An inner product ${\mathcal{R}}_p(·,·)$ on $T_{p}M$ is called a Riemannian metric on $T_{p}M$. A tensor field $\mathcal{R}(·,·)$ is said to be a Riemannian metric on M if for every $p\in M$, the tensor ${\mathcal{R}}_p(·,·)$ is a Riemannian metric on $T_{p}M$. The corresponding norm to the inner product $\mathcal{R}(·,·)$ on $T_{p}M$ is denoted by $\left|\right|·{\left|\right|}_p$. A differentiable manifold M endowed with a Riemannian metric $\mathcal{R}(·,·)$ is called a Riemannian manifold. The length of a piecewise smooth curve $\gamma :[0,1]\to M$ joining p to q (i.e., $\gamma \left(0\right)=p$ and $\gamma \left(1\right)=q$) is defined as

$$L\left(\gamma \right)={\int }_0^1\left|\right|{\gamma }^{\prime }\left(t\right)\left|\right|dt.$$

(4)

the Riemannian distance $d(p,q)$ is the minimal length over the set of all such curves joining p to q, which induces the original topology on M.

A Riemannian manifold M is complete if for any $p\in M$, all geodesics emanating from p are defined for all $t\in \mathbb{R}$. A geodesic joining p to q in M is said to be a minimal geodesic if its length is equal to $d(p,q)$. A Riemannian manifold M equipped with Riemannian distance d is a metric space $(M,d)$. By Hopf–Rinow Theorem [20], if M is complete, then any pair of points in M can be joined by a minimal geodesic. Moreover, $(M,d)$ is a complete metric space and bounded closed subsets are compact.

Definition 1.

If M is a complete Riemannian manifold, then the exponential map $ex{p}_p:T_{p}M\to M$ at $p\in M$ is defined by $ex{p}_{p}v={\gamma }_v(1,p)$ for all $v\in T_{p}M$, where ${\gamma }_v(·,p)$ is the geodesic starting from p with velocity v, that is, ${\gamma }_v(0,p)=p$ and ${\gamma }_v^{\prime }(0,p)=v.$

It is known that $ex{p}_{p}tv={\gamma }_v(t,p)$ for each real number t. It is easy to see that $ex{p}_{p}0={\gamma }_v(0,p)=p$, where 0 is the zero tangent vector. Moreover, for any $p,q\in M$ we have $d(p,q)=\left|\right|ex{p}_p^{-1}q\left|\right|$. Note that the exponential map $ex{p}_p$ is differentiable on $T_{p}M$ for any $p\in M.$

Definition 2.

A complete simply connected Riemannian manifold of non-positive sectional curvature is called a Hadamard Manifold.

Proposition 1

([20]). Let M be a Hadamard manifold. Then for any two points $x,y\in M$, there exists a unique normalized geodesic $\gamma :[0,1]\to M$ joining $x=\gamma \left(0\right)$ to $y=\gamma \left(1\right)$, which is in fact a minimal geodesic denoted by

$$\gamma \left(t\right)=ex{p}_{x}tex{p}_x^{-1}y,\forall t\in [0,1].$$

(5)

Lemma 1

([21]). Let $▵(p,q,r)$ be a geodesic triangle in a Hadamard manifold M. Then there exist $\overline{p},\overline{q},\overline{r}\in {\mathbb{R}}^2$ such that

$$d(p,q)=\left|\right|\overline{p}-\overline{q}\left|\right|,d(q,r)=\left|\right|\overline{q}-\overline{r}\left|\right|,d(r,p)=\left|\right|\overline{r}-\overline{p}\left|\right|.$$

The triangle $▵(\overline{p},\overline{q},\overline{r})$ is called the comparison triangle of the geodesic-triangle $▵(p,q,r)$, which is unique up to isometry of $M.$

Lemma 2

([11]). Let $▵(p,q,r)$ be a geodesic triangle in a Hadamard manifold $M,▵(\overline{p},\overline{q},\overline{r})$ be its comparison triangle.

(1) 

Let $\alpha ,\beta ,\gamma$ (respectively, $\overline{\alpha },\overline{\beta },\overline{\gamma })$ be the angles of $▵(p,q,r)$ (respectively, $▵(\overline{p},\overline{q},\overline{r})$ at the vertices $p,q,r$(respectively, $\overline{p},\overline{q},\overline{r})$. Then, the following inequalities hold:

$$\overline{\alpha }\ge \alpha ,\overline{\beta }\ge \beta ,\overline{\gamma }\ge \gamma .$$

(2) 

Let z be a point on the geodesic joining p to q and $\overline{z}$ be its comparison point in the interval $[\overline{p},\overline{q}]$. suppose that $d(z,p)=\left|\right|\overline{z}-\overline{p}\left|\right|$ and $d(z,q)=\left|\right|\overline{z}-\overline{q}\left|\right|$. Then

$$d(z,r)\le \left|\right|\overline{z}-\overline{r}\left|\right|.$$

(6)

The following inequalities can be proved easily.

Lemma 3

([15]). Let M be a finite dimensional Hadamard manifold.

(i) 

Let $\gamma :[0,1]\to M$ be a geodesic joining x to y. Then we have

$$d(\gamma \left(t_1\right),\gamma \left(t_2\right))=|t_1-t_2|d(x,y),\forall t_1,t_2\in [0,1];$$

(7)

(From now on $d(x,y)$ denotes the Riemannian distance).

(ii) 

for any $x,y,z,u,w\in M$ and $t\in [0,1]$, the following inequalities hold:

$$d(ex{p}_x(1-t)ex{p}_x^{-1}y,z)\le td(x,z)+(1-t)d(y,z);$$

(8)

$$d^2(ex{p}_x(1-t)ex{p}_x^{-1}y,z)\le t{d}^2(x,z)+(1-t)d^2(y,z)-t(1-t)d^2(x,y);$$

(9)

$$d(ex{p}_x(1-t)ex{p}_x^{-1}y,ex{p}_u(1-t)ex{p}_u^{-1}w)\le td(x,u)+(1-t)d(y,w).$$

(10)

Let M be a Hadamard manifold. A subset $K\subset M$ is said to be geodesic convex if for any two points x and y in K, the geodesic joining x to y is contained in $K.$

In the sequel, unless otherwise specified, we always assume that M is a finite dimensional Hadamard manifold, and K is a nonempty, bounded, closed and geodesic convex set in M and $Fix\left(S\right)$ is the fixed point set of a mapping $S.$

A function $f:K\to (-\infty ,\infty ]$ is said to be geodesic convex if, for any geodesic $\gamma \left(\lambda \right)(0\le \lambda \le 1)$ joining $x,y\in C$, the function $f\circ \gamma$ is convex, i.e.,

$$f\left(\gamma \right(\lambda \left)\right)\le \lambda f\left(\gamma \right(0\left)\right)+(1-\lambda )f\left(\gamma \right(1\left)\right)=\lambda f\left(x\right)+(1-\lambda )f\left(y\right).$$

Definition 3.

Let X be a complete metric space and $Q\subset X$ be a nonempty set. A sequence $\left\{x_n\right\}\subset X$ is called Fej$\acute{e}$r monotone with respect to Q if for any $y\in Q$ and $n\ge 0$,

$$d(x_{n+1},y)\le d(x_n,y).$$

Lemma 4

([22,23]). Let X be a complete metric space, $Q\subset X$ be a nonempty set. If $\left\{x_n\right\}\subset X$ is Fej$\acute{e}$r monotone with respect to Q, then $\left\{x_n\right\}$ is bounded. Moreover, if a cluster point x of $\left\{x_n\right\}$ belongs to Q, then $\left\{x_n\right\}$ converges to $x.$

Definition 4.

A mapping $S:K\to K$ is said to be

(1) 

nonexpansive if

$$d(Sx,Sy)\le d(x,y),\forall x,y\in K.$$

(2) 

firmly nonexpansive [20] if for all $x,y\in K$, the function $\varphi :[0,1]\to [0,\infty ]$ defined by

$$\varphi \left(t\right):=d(ex{p}_{x}tex{p}_x^{-1}Sx,ex{p}_{y}tex{p}_y^{-1}Sy),\forall t\in [0,1]$$

is nonincreasing

Proposition 2

([24]). Let $S:K\to K$ be a mapping. Then the following statements are equivalent.

(i) 

S is firmly nonexpansive;

(ii) 

For any $x,y\in K$ and $t\in [0,1]$

$$d(S\left(x\right),S\left(y\right))\le d(ex{p}_{x}tex{p}_x^{-1}Sx,ex{p}_{y}tex{p}_y^{-1}Sy);$$

(11)

(iii) 

For any $x,y\in K$

$$\mathcal{R}(ex{p}_{S\left(x\right)}^{-1}S\left(y\right),ex{p}_{S\left(x\right)}^{-1}x)+\mathcal{R}(ex{p}_{S\left(y\right)}^{-1}S\left(x\right),ex{p}_{S\left(y\right)}^{-1}y)\le 0.$$

(12)

Lemma 5

([16]). If $S:K\to K$ is a firmly nonexpansive mapping and $Fix\left(S\right)\ne \varnothing$, then for any $x\in K$ and $p\in Fix\left(S\right)$ the following conclusion holds:

$$d^2(Sx,p)\le d^2(x,p)-d^2(Sx,x).$$

(13)

In the sequel, we denote by $\mathsf{\Omega }\left(M\right)$ the set of all single-valued vector fields $A:M\to TM$, such that $A\left(x\right)\in T_{x}M$ for each $x\in M$, and let the domain $D\left(A\right)$ of A be defined by

$$D\left(A\right)=\{x\in M:A\left(x\right)\in T_{x}M\}.$$

Definition 5

([25]). A single-valued vector field $A\in \mathsf{\Omega }\left(M\right)$ is said to be monotone if

$$\langle A\left(x\right),ex{p}_x^{-1}y\rangle \le \langle A\left(y\right),-ex{p}_y^{-1}x\rangle ,\forall x,y\in M.$$

Let $\mathcal{X}\left(M\right)$ denote the set of all set-valued vector fields $B:M\to 2^{TM}$ such that $B\left(x\right)\subset T_{x}M$ for all $x\in M$, and let the domain $D\left(B\right)$ of B be defined by $D\left(B\right)=\{x\in M:B(x)\ne \varnothing \}.$

Definition 6

([26]). A set-valued vector field $B\in \mathcal{X}\left(M\right)$ is said to be

(1) 

monotone if for any $x,y\in D\left(B\right),$

$$\mathcal{R}(u,ex{p}_x^{-1}y)\le \mathcal{R}(v,-ex{p}_y^{-1}x),\forall u\in B\left(x\right)and\forall v\in B\left(y\right);$$

(2) 

maximal monotone if it is monotone and for all $x\in D\left(B\right)$ and $u\in T_{x}M$, the condition

$$\mathcal{R}(u,ex{p}_x^{-1}y)\le \mathcal{R}(v,-ex{p}_y^{-1}x)\forall y\in D\left(B\right)and\forall v\in B\left(y\right).$$

implies $u\in B\left(x\right).$

(3) 

For given $\lambda >0$, the resolvent of B of order $\lambda >0$ is a set-valued mapping $J_{\lambda }^B:M\to 2^{TM}$ defined by

$$J_{\lambda }^B\left(x\right):=\{z\in M:x\in ex{p}_z\lambda B\left(z\right)\},\forall x\in M.$$

Theorem 1

([24]). Let $B\in \mathcal{X}\left(M\right)$. The following assertions hold for any $\lambda >0$

(1) 

the vector field B is monotone if and only if $J_{\lambda }^B$ is single-valued and firmly nonexpansive.

(2)

if $D\left(B\right)=M$, the vector field B is maximal monotone if and only if $J_{\lambda }^B$ is single-valued, firmly nonexpansive and the domain $D\left(J_{\lambda }^B\right)=M.$

Proposition 3

([24]). Let K be a nonempty subset of M and $T:K\to M$ be a firmly nonexpansive mapping. Then

$$\mathcal{R}(ex{p}_{Ty}^{-1}x,ex{p}_{Ty}^{-1}y)\le 0$$

holds for any $x\in F\left(T\right)$ and $y\in K.$

Lemma 6

([13]). Let K be a nonempty closed subset of M and $B\in \mathcal{X}\left(M\right)$ be a maximal monotone vector field. Let $\left\{{\lambda }_n\right\}\subset (0,\infty )$ be a real sequence with lim${}_{n\to \infty }{\lambda }_n=\lambda >0$ and a sequence $\left\{x_n\right\}\subset K$ with lim${}_{n\to \infty }{x}_n=x\in K$ such that lim${}_{n\to \infty }{J}_{{\lambda }_n}^B\left(x_n\right)=y$. Then, $y=J_{\lambda }^B\left(x\right).$

Proposition 4

([12]). Let $A\in \mathsf{\Omega }\left(M\right)$ be a single-valued monotone vector field, $B\in \mathcal{X}\left(M\right)$ be a set-valued maximal monotone vector field. For any $x\in K$, the following assertions are equivalent

(1)

$x\in {(A+B)}^{-1}\left(0\right);$

(2) 

$x=J_{\lambda }^B\left(ex{p}_x(-\lambda A\left(x\right))\right),\forall \lambda >0.$

Let K be a nonempty closed geodesic convex set in M and $F:K\times K\to \mathbb{R}$ be a bifunction satisfying the following assumptions:

• $\left(A_1\right)$ for all $x\in K,F(x,x)\ge 0;$
• $\left(A_2\right)F$ is monotone, that is, for all $x,y\in K,F(x,y)+F(y,x)\le 0;$
• $\left(A_3\right)$ for every $y\in K,x\mapsto F(x,y)$ is upper semicontinuous;
• $\left(A_4\right)$ for every $x\in K,y\mapsto F(x,y)$ are geodesic convex and lower semicontinuous;
• $\left(A_5\right)x\mapsto F(x,x)$ is lower semicontinuous;
• $\left(A_6\right)$ there exists a compact set $L\subseteq M$ such that $x\in K\setminus L⟹\exists y\in K\bigcap L$ such that $F(x,y)<0.$

Definition 7

([14]). Let $F:K\times K\to \mathbb{R}$ be a bifunction. The resolvent of F is a multivalued operator $T_r^F:M\to 2^K$ defined by

$$T_r^F\left(x\right)=\{z\in K:F(z,y)-\frac{1}{r}\langle ex{p}_z^{-1}x,ex{p}_z^{-1}y\rangle \ge 0,\forall y\in K\}.$$

Theorem 2

([14]). Let $F:K\times K\to \mathbb{R}$ be a bifunction satisfying the following conditions:

(1) 

F is monotone;

(2) 

for all $r>0,T_r^F$ is properly defined, that is, the domain $D\left(T_r^F\right)\ne \varnothing .$

Then for any $r>0,$

(i) 

the resolvent $T_r^F$ is single-valued;

(ii) 

the resolvent $T_r^F$ is firmly nonexpansive;

(iii) 

the fixed point set of $T_r^F$ is the equilibrium point set of F,

$$F\left(T_r^F\right)=EP\left(F\right).$$

Theorem 3

([24]). Let $F:K\times K\to \mathbb{R}$ be a bifunction satisfying the assumptions $\left(A_1\right)-\left(A_3\right)$. Then $D\left(T_r^F\right)=M.$

Lemma 7

([24]). Let $F:K\times K\to \mathbb{R}$ be a bifunction satisfying the assumptions $\left(A_1\right),\left(A_3\right),\left(A_4\right),\left(A_5\right)$ and $\left(A_6\right)$. Then, there exists $z\in K$ such that

$$F(z,y)-\frac{1}{r}\langle ex{p}_z^{-1}x,ex{p}_z^{-1}y\rangle \ge 0,\forall y\in K,$$

for all $r>0$ and $x\in M.$

3. The Main Results

In the sequel, we always assume that

(1)

K is a nonempty closed bounded geodesic convex subsets of a Hadamard manifold M;

(2)

$B:K⟶2^{TM}$ is a maximal monotone setvalued vector field;

(3)

$A:K⟶TM$ is a continuous and monotone single-valued vector field satisfying the following condition [13].

(Assumption 3.1):

$$d(ex{p}_x(-\lambda A\left(x\right)),ex{p}_y(-\lambda A\left(y\right)))\le (1-\rho )d(x,y),\forall x,y\in K,\lambda >0and\rho \in [0,1].$$

(14)

(4)

$S:K\to K$ is a nonexpansive mapping;

(5)

$F_i:K\times K⟶\mathbb{R},i=1,2,\cdots ,m$, is a bifunction satisfying the conditions $\left(A_1\right)-\left(A_6\right)$, and for given $r>0$, the resolvent of $F_i$ is a multivalued operator $T_r^{F_i}:M⟶2^K$ such that for all $x\in M$

$$T_r^{F_i}\left(x\right)=\{z\in K:F_i(z,y)-\frac{1}{r}\langle ex{p}_z^{-1}x,ex{p}_z^{-1}y\rangle \ge 0,\forall y\in K\},i=1,2,\cdots ,m.$$

(6)

Denote by

$$S_r^l:=T_r^{F_l}\circ T_r^{F_{l-1}}\circ \cdots \circ T_r^{F_2}\circ T_r^{F_1},l=1,2,\cdots ,m.$$

Theorem 4.

Let $K,M,A,B,{\left\{F_i\right\}}_{i=1}^m,{\left\{S_r^l\right\}}_{l=1}^m$ and S be the same as above. Let $\left\{x_n\right\},\left\{u_n\right\},\left\{y_n\right\}$ and $\left\{z_n\right\}$ be the sequences generated by $x_0\in K$

$$\{\begin{array}{l}{u}_n=J_{{\lambda }_n}^B\left(ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right))\right),\\ y_n=ex{p}_{x_n}{\alpha }_{n}ex{p}_{x_n}^{-1}S{u}_n,\\ z_n=S_r^m\left(y_n\right),\\ x_{n+1}=ex{p}_{x_n}{\beta }_{n}ex{p}_{x_n}^{-1}{z}_n.\forall n\ge 0,\end{array}$$

(15)

where $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\lambda }_n\right\}$ are positive sequences satisfying the following conditions:

(i) 

$0<a\le {\alpha }_n,{\beta }_n\le b<1,\forall n\in \mathbb{N}$;

(ii) 

$0<\widehat{\lambda }\le {\lambda }_n\le \tilde{\lambda }<\infty ,\forall n\in \mathbb{N}$;

(iii) 

${\sum }_{n=1}^{\infty }{\alpha }_n{\beta }_n=\infty$

If $\mathsf{\Omega }={\bigcap }_{i=1}^{m}EP\left(F_i\right)\bigcap {(A+B)}^{-1}\left(0\right)\bigcap F\left(S\right)\ne \varnothing$, then $\left\{x_n\right\}$ converges strongly to a solution of problem (1.4).

Proof. 

We divide the proof of Theorem 4 into five steps.

For $n\ge 0$, let ${\gamma }_n:[0,1]\to M$ be the geodesic joining ${\gamma }_n\left(0\right)=x_n$ to ${\gamma }_n\left(1\right)=z_n$ and $\widehat{{\gamma }_n}:[0,1]\to M$ be the geodesic joining $\widehat{{\gamma }_n}\left(0\right)=x_n$ to $\widehat{{\gamma }_n}\left(1\right)=S{u}_n$. Then $\left\{x_{n+1}\right\}$ and $\left\{y_n\right\}$ can be written as

$$x_{n+1}:={\gamma }_n\left({\beta }_n\right),y_n=\widehat{{\gamma }_n}\left({\alpha }_n\right).$$

(16)

(I) We first prove that $\mathsf{\Omega }$ is closed and geodesic convex.

Because every nonexpansive mapping is continuous, therefore $F\left(S\right)$ is closed. We can prove that $F\left(S\right)$ is geodesic convex.

In fact, let $p,q\in F\left(S\right)$, we need to prove that a gecdesic $\gamma :[0,1]\to M$ joining p to q is contained in $F\left(S\right)$. It is well-known that in Hadamard manifold M, for any $p,q\in M$, and $t\in [0,1]$, there exists a unique point ${\omega }_t=\gamma \left(t\right)=ex{p}_{p}te{x}_p^{-1}q$ such that

$$d(p,q)=d(p,{\omega }_t)+d({\omega }_t,q).$$

By the nonexpansiveness of S and the geodesic convexity of Riemannian distance, we have

$$d(p,S\left({\omega }_t\right))=d(S\left(p\right),S\left({\omega }_t\right))\le d(p,{\omega }_t)=d(p,\gamma \left(t\right))\le td(p,q).$$

Similarly, we get

$$d(S\left({\omega }_t\right),q)\le (1-t)d(p,q).$$

From the above, we have

$$d(p,q)\le d(p,S\left({\omega }_t\right))+d(S\left({\omega }_t\right),q)\le d(p,q).$$

So

$$d(p,q)=(p,S\left({\omega }_t\right))+d(S\left({\omega }_t\right),q).$$

By the uniqueness of ${\omega }_t$, we have ${\omega }_t=S\left({\omega }_t\right)$. Therefore, ${\omega }_t=\gamma \left(t\right)\in F\left(S\right).$ Thus, $F\left(S\right)$ is geodesic convex.

From Proposition 4, we have ${(A+B)}^{-1}\left(0\right)=F\left(J_{\lambda }^B\left(exp(-\lambda A)\right)\right)$. Because $J_{\lambda }^B$ is nonexpansive, this together with Assumption 3.1, $J_{\lambda }^B\left(exp(-\lambda A)\right)$ is nonexpansive. Therefore, ${(A+B)}^{-1}\left(0\right)$ is closed and geodesic convex in M.

From Theorem 2, we have $T_r^{F_i}$ is firmly nonexpansive and $F\left(T_r^{F_i}\right)=EP\left(F_i\right)$. Therefore, $EP\left(F_i\right)$ is closed and geodesic convex in M. Hence, $\mathsf{\Omega }$ is closed and geodesic convex.

(II) We prove that $\left\{x_n\right\}$ is Fej$\acute{e}$r monotone with respect to $\mathsf{\Omega }$.

Let $\omega \in \mathsf{\Omega }$, Then $\omega \in {(A+B)}^{-1}\left(0\right)$ and $\omega \in {\bigcap }_{i=1}^{m}EP\left(F_i\right).$ By Proposition 4, we have $\omega =J_{{\lambda }_n}^B\left(ex{p}_{\omega }(-{\lambda }_{n}A\left(\omega \right))\right)$. By nonexpansiveness of resolvent $J_{{\lambda }_n}^B$ of B and Assumption 3.1, we have

$$\begin{array}{ll}d(u_n,\omega )& =d(J_{{\lambda }_n}^B\left(ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right))\right),J_{{\lambda }_n}^B\left(ex{p}_{\omega }(-{\lambda }_{n}A\left(\omega \right))\right))\\ & \le d(ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right)),ex{p}_{\omega }(-{\lambda }_{n}A\left(\omega \right)))\\ & \le (1-\rho )d(x_n,\omega )\\ & \le d(x_n,\omega ).\end{array}$$

(17)

Since $\omega \in {\bigcap }_{i=1}^{M}EP\left(F_i\right)$, by Theorem 2, for each $i=1,2,\cdots ,m,T_r^{F_i}$ is non-expansive and $Fix\left(T_r^{F_i}\right)=EP\left(F_i\right)$, therefore $S_r^m$ is also non-expansive and $\omega \in Fix\left(S_r^m\right)$, we have

$$d(z_n,\omega )=d(S_r^m\left(y_n\right),S_r^m\left(\omega \right))\le d(y_n,\omega ).$$

(18)

Because $y_n=\widehat{{\gamma }_n}\left({\alpha }_n\right)$ for all $n\ge 0,S$ is a nonexpansive mapping on K and $\omega \in F\left(S\right)$, by using the geodesic convexity of Riemannian distance and (17), we have

$$\begin{array}{ll}d(y_n,\omega )& =d(\widehat{{\gamma }_n}\left({\alpha }_n\right),\omega )\\ & \le (1-{\alpha }_n)d(\widehat{{\gamma }_n}\left(0\right),\omega )+{\alpha }_{n}d(\widehat{{\gamma }_n}\left(1\right),\omega )\\ & \le (1-{\alpha }_n)d(x_n,\omega )+{\alpha }_{n}d(S{u}_n,\omega )\\ & \le (1-{\alpha }_n)d(x_n,\omega )+{\alpha }_{n}d(u_n,\omega )\\ & \le d(x_n,\omega ).\end{array}$$

(19)

Since $x_{n+1}:={\gamma }_n\left({\beta }_n\right)$, from (18) and (19), we have

$$\begin{array}{ll}d(x_{n+1},\omega )& =d({\gamma }_n\left({\beta }_n\right),\omega )\\ & \le (1-{\beta }_n)d({\gamma }_n\left(0\right),\omega )+{\beta }_{n}d({\gamma }_n\left(1\right),\omega )\\ & \le (1-{\beta }_n)d(x_n,\omega )+{\beta }_{n}d(z_n,\omega )\\ & \le (1-{\beta }_n)d(x_n,\omega )+{\beta }_{n}d(y_n,\omega )\\ & \le (1-{\beta }_n)d(x_n,\omega )+{\beta }_{n}d(x_n,\omega )\\ & \le d(x_n,\omega ).\end{array}$$

(20)

This shows that $\left\{d(x_n,\omega )\right\}$ is decreasing and bounded below. Hence, the limit lim${}_{n\to \infty }d(x_n,\omega )$ exists for each $\omega \in \mathsf{\Omega }$. This implies that $\left\{x_n\right\}$ is Fej$\acute{e}$r monotone with respect to $\mathsf{\Omega }$. Hence, the sequence $\left\{x_n\right\}$ is bounded. So are the sequences $\left\{ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right))\right\},\left\{S\left(u_n\right)\right\},\left\{y_n\right\}$ and $\left\{z_n\right\}$.

(III) Next, we prove that

$$\underset{n\to \infty }{lim}d(x_{n+1},x_n)=0$$

We fix $n\ge 0$ and let $\Delta (x_n,z_n,\omega )$ be a geodesic triangle with vertices $x_n,z_n$ and $\omega$, and $\Delta (\overline{x_n},\overline{z_n},\overline{\omega })\subseteq {\mathbb{R}}^2$ be a corresponding comparison triangle. Then, we have

$$d(x_n,\omega )=\left|\right|\overline{x_n}-\overline{\omega }\left|\right|,d(z_n,\omega )=\left|\right|\overline{z_n}-\overline{\omega }\left|\right|andd(z_n,x_n)=\left|\right|\overline{z_n}-\overline{x_n}\left|\right|.$$

Recall $x_{n+1}:=ex{p}_{x_n}{\beta }_{n}ex{p}_{x_n}^{-1}{z}_n$, so it comparison point is $\overline{x_{n+1}}=(1-{\beta }_n)\overline{x_n}+{\beta }_n\overline{z_n}$. Using (6), (18) and (19), we get

$$\begin{array}{ll}{d}^2(x_{n+1},\omega )& \le \left|\right|\overline{x_{n+1}}-\overline{\omega }{\left|\right|}^2\\ & =\left|\right|(1-{\beta }_n)\overline{x_n}+{\beta }_n\overline{z_n}-\overline{\omega }{\left|\right|}^2\\ & =\left|\right|(1-{\beta }_n)(\overline{x_n}-\overline{\omega })+{\beta }_n(\overline{z_n}-\overline{\omega }){\left|\right|}^2\\ & =(1-{\beta }_n)\left|\right|\overline{x_n}-\overline{\omega }{\left|\right|}^2+{\beta }_n\left|\right|\overline{z_n}-\overline{\omega }{\left|\right|}^2-{\beta }_n(1-{\beta }_n)\left|\right|\overline{x_n}-\overline{z_n}{\left|\right|}^2\\ & =(1-{\beta }_n)d^2(x_n,\omega )+{\beta }_n{d}^2(z_n,\omega )-{\beta }_n(1-{\beta }_n)d^2(x_n,z_n)\\ & \le (1-{\beta }_n)d^2(x_n,\omega )+{\beta }_n{d}^2(y_n,\omega )-{\beta }_n(1-{\beta }_n)d^2(x_n,z_n)\\ & \le (1-{\beta }_n)d^2(x_n,\omega )+{\beta }_n{d}^2(x_n,\omega )-{\beta }_n(1-{\beta }_n)d^2(x_n,z_n)\\ & =d^2(x_n,\omega )-{\beta }_n(1-{\beta }_n)d^2(x_n,z_n).\end{array}$$

(21)

From (21), we also obtain

$${\beta }_n(1-{\beta }_n)d^2(x_n,z_n)\le d^2(x_n,\omega )-d^2(x_{n+1},\omega ),$$

and we further have

$$\begin{array}{ll}{d}^2(x_n,z_n)& \le \frac{1}{{\beta }_n(1-{\beta }_n)}(d^2(x_n,\omega )-d^2(x_{n+1},\omega ))\\ & \le \frac{1}{a(1-b)}(d^2(x_n,\omega )-d^2(x_{n+1},\omega )).\end{array}$$

Since $\left\{x_n\right\}$ is a Fej$\acute{e}$r monotone with respect to $\mathsf{\Omega }$ which implies that lim${}_{n\to \infty }d(x_n,\omega )$ exists. By letting $n\to \infty$, we have

$$\underset{n\to \infty }{lim}d(x_n,z_n)=0.$$

(22)

Recall that $x_{n+1}:={\gamma }_n\left({\beta }_n\right)$ for all $n\in \mathbb{N}$, using the geodesic convexity of Riemannian distance, we obtain

$$\begin{array}{ll}d(x_{n+1},x_n)& =d({\gamma }_n\left({\beta }_n\right),x_n)\\ & \le (1-{\beta }_n)d({\gamma }_n\left(0\right),x_n)+{\beta }_{n}d({\gamma }_n\left(1\right),x_n)\\ & =(1-{\beta }_n)d(x_n,x_n)+{\beta }_{n}d(z_n,x_n)\\ & ={\beta }_{n}d(x_n,z_n)\\ & \le bd(x_n,z_n).\end{array}$$

Letting $n\to \infty$ and using (22), we get

$$\underset{n\to \infty }{lim}d(x_{n+1},x_n)=0.$$

(23)

(IV) Next we prove that ${lim}_{n\to \infty }d(x_n,y_n)=0$ and ${lim}_{n\to \infty }d(x_n,S{u}_n)=0$.

Because $\left\{x_n\right\}$ is bounded, there exists a constant L such that $d(x_n,\omega )\le L$ for all $n\ge 0$. By (17) and (20), Assumption 3.1 and geodesic convexity of Riemannian distance, we have

$$\begin{array}{ll}d(x_n,\omega )& \le (1-{\beta }_{n-1})d(x_{n-1},\omega )+{\beta }_{n-1}d(y_{n-1},\omega )\\ & \le (1-{\beta }_{n-1})d(x_{n-1},\omega )+{\beta }_{n-1}((1-{\alpha }_{n-1})d(x_{n-1},\omega )+{\alpha }_{n-1}d(u_{n-1},\omega ))\\ & \le (1-{\beta }_{n-1})d(x_{n-1},\omega )+{\beta }_{n-1}((1-{\alpha }_{n-1})d(x_{n-1},\omega )+{\alpha }_{n-1}(1-\rho )d(x_{n-1},\omega ))\\ & =(1-{\beta }_{n-1})d(x_{n-1},\omega )+{\beta }_{n-1}(1-\rho {\alpha }_{n-1})d(x_{n-1},\omega )\\ & =(1-\rho {\alpha }_{n-1}{\beta }_{n-1})d(x_{n-1},\omega ),\end{array}$$

where $\rho$ is the same as in Assumption 3.1. Let $0\le m\le n$, then

$$d(x_n,\omega )\le L\prod _{j=m}^{n-1}(1-\rho {\alpha }_j{\beta }_j).$$

(24)

On the other hand, by (19) and (20), we have

$$\begin{array}{ll}d(x_n,y_n)& \le d(x_n,x_{n+1})+d(x_{n+1},\omega )+d(y_n,\omega )\\ & \le d(x_n,x_{n+1})+2d(x_n,\omega ).\end{array}$$

Therefore, by using (24), the above inequality becomes

$$d(x_n,y_n)\le d(x_n,x_{n+1})+2L\prod _{j=m}^{n-1}(1-\rho {\alpha }_j{\beta }_j).$$

By taking limit as $n\to \infty$, we get

$$\underset{n\to \infty }{lim}d(x_n,y_n)\le \underset{n\to \infty }{lim}d(x_n,x_{n+1})+2L\underset{n\to \infty }{lim}\prod _{j=m}^{n-1}(1-\rho {\alpha }_j{\beta }_j).$$

(25)

By condition (iii), we have

$$\underset{n\to \infty }{lim}\prod _{j=m}^{n-1}(1-\rho {\alpha }_j{\beta }_j)=0.$$

This together with (23) and (25) implies that

$$\underset{n\to \infty }{lim}d(x_n,y_n)=0.$$

(26)

From (17) and (20), we have

$$\begin{array}{ll}d(x_n,S{u}_n)& \le d(x_n,x_{n+1})+d(x_{n+1},\omega )+d(S{u}_n,\omega )\\ & \le d(x_n,x_{n+1})+d(x_{n+1},\omega )+d(u_n,\omega )\\ & \le d(x_n,x_{n+1})+2d(x_n,\omega ).\end{array}$$

Similarly, we can prove that

$$\underset{n\to \infty }{lim}d(x_n,S{u}_n)=0,and\underset{n\to \infty }{lim}d(x_n,u_n)=0.$$

(27)

Since

$$\begin{array}{ll}d(S_r^m\left(x_n\right),x_n)& \le d(S_r^m\left(x_n\right),S_r^m\left(y_n\right))+d(S_r^m\left(y_n\right),x_n)\\ & \le d(x_n,y_n)+d(z_n,x_n).\end{array}$$

Letting $n\to \infty$ and using (22) and (26), we have

$$\underset{n\to \infty }{lim}d(S_r^m\left(x_n\right),x_n)=0.$$

(28)

(V) Next we prove that the cluster point $\nu$ of the sequence $\left\{x_n\right\}$ belongs to $\mathsf{\Omega }$.

We have already proved in step 1 that the sequence $\left\{x_n\right\}$ is bounded. Therefore, there exists a subsequence $\left\{x_{n_j}\right\}$ of $\left\{x_n\right\}$ which converges to a cluster point $\nu$ of $\left\{x_n\right\}$. Thus, (27) implies that $u_{n_j}\to \nu$ as $j\to \infty$. By nonexpansiveness of S, we have

$$\begin{array}{ll}d(\nu ,S\left(\nu \right))& \le d(x_{n_j},\nu )+d(x_{n_j},S\left(u_{n_j}\right))+d(S\left(u_{n_j}\right),S\left(\nu \right))\\ & \le d(x_{n_j},\nu )+d(x_{n_j},S\left(u_{n_j}\right))+d(u_{n_j},\nu ).\end{array}$$

By (27) and taking limit as $j\to \infty$, we have

$$d(\nu ,S(\nu \left)\right)=0,$$

which means that $\nu \in$Fix(S).

Next, we prove that $\nu \in {\bigcap }_{i=1}^{m}EP\left(F_i\right)$.

By (28), we have $li{m}_{j\to \infty }d(S_r^m\left(x_{n_j}\right),x_{n_j})=0$. Since $S_r^m$ is a non-expansive mapping, it is demi-closed at zero, so $\nu \in Fix\left(S_r^m\right)$. In order to prove that $\nu \in {\bigcap }_{i=1}^{m}EF\left(F_i\right)$ it should be proved that $Fix\left(S_r^m\right)={\bigcap }_{i=1}^{m}Fix\left(T_r^{F_i}\right)$.

It is obvious that ${\bigcap }_{i=1}^{m}Fix\left(T_r^{F_i}\right)\subseteq Fix\left(S_r^m\right)$. Next we prove that

$$Fix\left(S_r^m\right)\subseteq \bigcap _{i=1}^{m}Fix\left(T_r^{F_i}\right).$$

Let $q\in Fix\left(S_r^m\right)$ and $p\in {\bigcap }_{i=1}^{m}Fix\left(T_r^{F_i}\right)$, we have

$$\begin{array}{ll}d(q,p)& =d(S_r^{m}q,p)=d(T_r^{F_m}{S}_r^{m-1}q,p)\le d(S_r^{m-1}q,p)\\ & \le d(S_r^{m-2}q,p)\le \cdots \le d(S_r^{1}q,p)=d(T_r^{F_1}q,p)\le d(q,p).\end{array}$$

This implies that

$$\begin{array}{ll}d(q,p)& =d(S_r^{m}q,p)=d(S_r^{m-1}q,p)=d(S_r^{m-2}q,p)=\cdots =d(S_r^{1}q,p)\\ & =d(T_r^{F_1}q,p).\end{array}$$

(29)

It follows from (29) and Lemma 5 that for each $i=1,2,\cdots ,m$, we have

$$d^2(S_r^{i}q,p)+d^2(S_r^{i}q,S_r^{i-1}q)\le d^2(S_r^{i-1}q,p)=d^2(q,p).$$

Since $d(S_r^{i}q,p)=d(q,p)$, this implies that for each $i=1,2,\cdots ,m$

$$d(S_r^{i}q,S_r^{i-1}q)=0,i.e.,S_r^{i-1}q\in Fix\left(T_r^{F_i}\right).$$

(30)

Taking $i=1$ in (2), we have $q=T_r^{F_1}\left(q\right)$. Taking $i=2$ in (3.17), we have that

$$q=T_r^{F_1}\left(q\right)=T_r^{F_2}\left(q\right).$$

Taking $i=1,2,\cdots ,m$ in (2) we can prove that

$$q=T_r^{F_1}\left(q\right)=T_r^{F_2}\left(q\right)=\cdots =T_r^{F_{m-1}}\left(q\right)=T_r^{F_m}\left(q\right),i.e.,q\in \bigcap _{i=1}^{m}Fix\left(T_r^{F_i}\right).$$

Finally, we prove that $\nu \in {(A+B)}^{-1}\left(0\right)$.

Because $\widehat{\lambda }\le {\lambda }_n\le \tilde{\lambda }$ then we can choose $\lambda >0$ such that the subsequence $\left\{{\lambda }_{n_j}\right\}$ of $\left\{{\lambda }_n\right\}$ converges to $\lambda$. Because $u_n=J_{{\lambda }_n}^B\left(ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right))\right).$ Then, by (27) and Lemma 6, we have

$$\begin{array}{ll}0& =\underset{j\to \infty }{lim}d(x_{n_j},u_{n_j})\\ & =\underset{j\to \infty }{lim}d(x_{n_j},J_{{\lambda }_{n_j}}^B(ex{p}_{x_{n_j}}(-{\lambda }_{n_j}A(x_{n_j}))))\\ & =d(\nu ,J_{\lambda }^B\left(ex{p}_{\nu }(-\lambda A\left(\nu \right))\right)).\end{array}$$

From Proposition 4, we have $\nu \in {(A+B)}^{-1}\left(0\right)$. Hence, $\nu \in \mathsf{\Omega }$. This completes the proof. □

In Theorem 4, take $F_i\equiv 0(i=1,2,\cdots ,m)$, then the following Corollary can be obtained from Theorem 4 immediately.

Corollary 1.

Let $K,M,A,B$ and S be the same as in Theorem 4. Let $\left\{x_n\right\},\left\{u_n\right\}$ and $\left\{y_n\right\}$ be sequences generated by $x_0\in K$ and

$$\{\begin{array}{l}{u}_n=J_{{\lambda }_n}^B\left(ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right))\right),\\ y_n=ex{p}_{x_n}{\alpha }_{n}ex{p}_{x_n}^{-1}S\left(u_n\right),\\ x_{n+1}=ex{p}_{x_n}{\beta }_{n}ex{p}_{x_n}^{-1}{y}_n.\forall n\ge 0,\end{array}$$

(31)

where $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\lambda }_n\right\}$ are positive sequences satisfying the conditions (i), (ii) and (iii) in Theorem 4. If ${\mathsf{\Omega }}_1={(A+B)}^{-1}\left(0\right)\bigcap F\left(S\right)\ne \varnothing$, then $\left\{x_n\right\}$ converges strongly to some point $\nu \in {\mathsf{\Omega }}_1$.

In Theorem 4 take $F_i\equiv 0(i=1,2,\cdots ,m)$ and $S=I$ (identity mapping on K), then the following corollary can be obtained from Theorem 4.

Corollary 2.

Let $K,M,A$ and B be the same as in Theorem 4. Let $\left\{x_n\right\},\left\{u_n\right\}$ and $\left\{y_n\right\}$ be the sequences generated by $x_0\in K$ and

$$\{\begin{array}{l}{u}_n=J_{{\lambda }_n}^B\left(ex{p}_{x_n}(-{\lambda }_{n}A\left(x_n\right))\right),\\ y_n=ex{p}_{x_n}{\alpha }_{n}ex{p}_{x_n}^{-1}{u}_n,\\ x_{n+1}=ex{p}_{x_n}{\beta }_{n}ex{p}_{x_n}^{-1}{y}_n.\forall n\ge 0,\end{array}$$

(32)

where $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\lambda }_n\right\}$ are positive sequences satisfying the conditions (i), (ii) and (iii) in Theorem 4. If ${\mathsf{\Omega }}_2={(A+B)}^{-1}\left(0\right)\ne \varnothing$, then $\left\{x_n\right\}$ converges strongly to some point $\nu \in {\mathsf{\Omega }}_2$.

In Theorem 4, take $A\equiv 0$ and $S=I$ (the identity mapping on K), then the following result can be obtained from Theorem 4 immediately.

Corollary 3.

Let $K,M,B,{\left\{F_i\right\}}_{i=1}^m$ and ${\left\{S_{r_n}^l\right\}}_{l=1}^m$ be the same as in Theorem 4. Let $\left\{x_n\right\},\left\{u_n\right\},\left\{y_n\right\}$ and $\left\{z_n\right\}$ be the sequences generated by $x_0\in K$:

$$\{\begin{array}{l}{u}_n=J_{{\lambda }_n}^B\left(x_n\right),\\ y_n=ex{p}_{x_n}{\alpha }_{n}ex{p}_{x_n}^{-1}{u}_n,\\ z_n=S_r^m\left(y_n\right),\\ x_{n+1}=ex{p}_{x_n}{\beta }_{n}ex{p}_{x_n}^{-1}{z}_n.\forall n\ge 0,\end{array}$$

(33)

where $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\lambda }_n\right\}$ are positive sequences satisfying the the conditions (i), (ii) and (iii) in Theorem 4. If ${\mathsf{\Omega }}_3={\bigcap }_{i=1}^{m}EP\left(F_i\right)\bigcap B^{-1}\left(0\right)\ne \varnothing$, then $\left\{x_n\right\}$ converges strongly to some point $\nu \in {\mathsf{\Omega }}_3$.

In Theorem 4 take $A\equiv 0,B\equiv 0$ and $S=I$ (identity mapping on K), then the following result can be obtained from Theorem 4 immediately.

Corollary 4.

Let $K,M,{\left\{F_i\right\}}_{i=1}^m$ and ${\left\{S_{r_n}^l\right\}}_{l=1}^m$ be the same as in Theorem 4. Let $\left\{x_n\right\}$ and $\left\{z_n\right\}$ be sequences generated by $x_0\in K$ and

$$\{\begin{array}{l}{z}_n=S_r^m\left(x_n\right),\\ x_{n+1}=ex{p}_{x_n}{\beta }_{n}ex{p}_{x_n}^{-1}{z}_n.\forall n\ge 0,\end{array}$$

(34)

where $\left\{{\alpha }_n\right\},\left\{{\beta }_n\right\}$ and $\left\{{\lambda }_n\right\}$ are positive sequences satisfying the conditions (i), (ii) and (iii) in Theorem 4. If ${\mathsf{\Omega }}_4={\bigcap }_{i=1}^{m}EP\left(F_i\right)\ne \varnothing$, then $\left\{x_n\right\}$ converges to some point $\nu \in {\mathsf{\Omega }}_4$.