Variational Inequalities on Geodesic Spaces

TL;DR

This paper introduces a new variational inequality problem in $CAT(0)$ spaces involving nonself multivalued nonexpansive mappings, expanding the theoretical framework of variational inequalities in metric spaces.

Contribution

It presents a novel variational inequality problem tailored for $CAT(0)$ spaces with nonself multivalued nonexpansive mappings, extending existing VIP theories.

Findings

Defines a new VIP in $CAT(0)$ spaces

Establishes foundational properties of the new VIP

Lays groundwork for future solution methods

Abstract

In this paper, we introduce a new variational inequality problem(VIP) associated with nonself multivalued nonexpansive mappings in $CAT(0)$ spaces.

Equations

$$H(A,B)=\max \{\underset{x\in A}{\sup }d(x,B),\underset{x\in B}{\sup }d(x,A)\}$$

$$H(Tx,Ty)\le d(x,y)$$

$$x_{n+1}\in T{x}_n.$$

$$x_{n+1}\in (1-{\alpha }_n)x_n+{\alpha }_{n}T{x}_n$$

$$x_{n+1}$$

$$y_n$$

$$x_{n+1}$$

$$y_n$$

$$z_n$$

$$x_{n+1}$$

$$y_n$$

$$z_n$$

$$d^2(x,(1-\lambda )y\oplus \lambda z)\le (1-\lambda )d^2(x,y)+\lambda d^2(x,z)-\frac{R}{2}\lambda (1-\lambda )d^2(y,z),$$

$$d((1-\lambda )x\oplus \lambda y,z)\le (1-\lambda )d(x,z)+\lambda d(y,z).$$

$$r(x,\{x_n\})=\underset{n\to \infty }{\mathrm{limsup}}d(x,x_n)$$

$$r(\{x_n\})=\inf \{r(x,\{x_n\});x\in X.\},$$

$$r_K(\{x_n\})=\inf \{r(x,\{x_n\});x\in K.\}$$

$$A(\{x_n\})=\{x\in X:r(x,\{x_n\})=r(\{x_n\})\}.$$

$$⟨,⟩$$

$$⟨\overrightarrow{ab},\overrightarrow{cd}⟩$$

$$⟨\overrightarrow{ab},\overrightarrow{ab}⟩$$

$$⟨\overrightarrow{ab},\overrightarrow{cd}⟩$$

$$⟨\overrightarrow{ab},\overrightarrow{ab}⟩$$

$$⟨\overrightarrow{ab},\overrightarrow{cd}⟩$$

$$⟨\overrightarrow{xu},\overrightarrow{yu}⟩\le \text{ for all }y\in K$$

$$⟨x-y,x^{\ast }-y^{\ast }⟩\ge 0$$

$$⟨u,y-x⟩\ge 0\text{, for all }y\in K$$

$$\text{Find }x\in K\text{ such that }⟨\overrightarrow{Txx},\overrightarrow{xy}⟩\ge \text{ for all }y\in K$$

$$\text{Find }(u,x{)}_{u\in Tx}\text{ such that }⟨\overrightarrow{ux},\overrightarrow{xy}⟩\ge 0\text{ for all }y\in K$$

$$P_K^{\ast }T(x)$$

$$H(P_K^{\ast }(Tx),P_K^{\ast }(Ty))$$

$$d(P_K(x_{n_k}^{\prime }),P_K(z))\le d(x_{n_k}^{\prime },z)$$

$$⟨p^{\prime }p,yp⟩\ge 0\text{ for all }y\in K$$

$$0$$

$$2(t-1)d(x,u)=0$$

Full text

1. Introduction and Preliminaries

Let $(X,d)$ be a metric space then the family of nonempty, closed and convex subsets of $X$, the family of nonempty compact and convex subsets of $X,$ the family of nonempty compact subsets of $X,$ the family of nonempty closed and bounded convex subsets of $X$ will be denoted by $C(X),KC(X),$ $K(X),$ $CB(X)$, respectively. Let $H$ be a Haussdorf Metric on $CB(X),$ defined by

[TABLE]

where $d(x,B)=\inf\{d(x,y);y\in B\}.$ A multivalued mappings $T:X\rightarrow 2^{X}$ is called nonexpansive if for all $x,y\in X$

[TABLE]

is satisfied. A point is called fixed point of $T$  if $x\in Tx$ and the set of all fixed points of $T$ is denoted by $F(T).$Many iterative processes to find a fixed point of multivalued mappings have been introduced in metric spaces and Banach spaces. One of them is defined by Nadler[1] as generalization of Picard as follows;

[TABLE]

A multivalued version of Mann and Ishikawa fixed point procedures goes as follow;

[TABLE]

and

[TABLE]

where $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are sequences in$\ [0,1]$.

Gursoy and Karakaya [17] (see also [18]) introduced Picard-S iteration as follows;

[TABLE]

where $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are  sequences in$\ [0,1]$. They have showed that it converges to fixed point of contraction mappings faster than Ishikawa, Noor, SP, CR, S and some other iterations. Also they use it to solve differential equations. Now, we define multivalued version of Picad-S iteration in $CAT(0)$ spaces as follows; Let $K\subset CAT(0)$ be a nonempty, closed and convex subset, $T:K\rightarrow C(K)$ is a mapping, $x_{0}\in K$. then for any $n\geq 0$, the proximal multivalued Picard-S iteration is defined by

[TABLE]

where $P_{K}$ is a metric projection, $\{\alpha_{n}\}$ and $\{\beta_{n}\}$ are sequences in$\ [0,1]$ with $\liminf_{n}(1-\beta_{n})\beta_{n}>0$, $u_{n}\in Ty_{n},$ $v_{n}\in Tz_{n}$ and $w_{n}\in Tx_{n}$.

Before the results we give some definitions and lemmas about $CAT(0)$ and  $\Delta-$convergences.

Let $(X,d)$ be a metric space, $x,y\in X$ and $C\subseteq X$ nonempty subset. A geodesic path (or shortly a geodesic) joining $x$ and $y$ is a map $c:[0,t]\subseteq\mathbb{R}\rightarrow X$  such that $c(0)=x$, $c(t)=y$ and $d(c(r),c(s))=|r-s|$ for all $r,s\in[0,t]$. In particular $c$ is an isometry and $d(c(0),c(t))=t.$ The image of $c,$ $c([0,t])$ is called geodesic segment from $x$ to $y$ and it is unique (it not necessarily be unique) then it is denoted by $[x,y].$ $z\in[x,y]$ if and only if for an $\lambda\in[0,1]$ such that $d(z,x)=(1-\lambda)d(x,y)$ and $d(z,y)=\lambda d(x,y).$ The point $z$ is denoted by $z=(1-\lambda)x\oplus\lambda y.$ If for every $x,y\in X$ there is a geodesic path then $(X,d)$ called geodesic space and uniquely geodesic space if that geodesic path is unique for any pair $x,y.$ A subset $C\subseteq X$ is called convex if it contains all geodesic segment joining any pair of points in it.

In geodesic metric space $(X,d)$, a geodesic triangle $\Delta(x,y,z)$  consist of three point $x,y,z$ as vertices and three geodesic segments of any pair of these points, that is, $q\in$ $\Delta(x,y,z)$ means that $q\in[x,y]\cup[x,z]\cup[y,z].$ The triangle $\overline{\Delta}(\overline{x},\overline{y},\overline{z})$ in $(\mathbb{R}^{2},d_{2})$ is called comparison triangle for the triangle $\Delta(x,y,z)$ such that $d(x,y)=d_{2}(\overline{x},\overline{y}),d(x,z)=d_{2}(\overline{x},\overline{z})$ and $d(y,z)=d_{2}(\overline{y},\overline{z})$  A point point $\overline{z}\in[\overline{x},\overline{y}]$ called comparison point for $z\in[x,y]$ if $d(x,z)=d_{2}(\overline{x},\overline{z}).$ A geodesic triangle $\Delta(x,y,z)$ in $X$ is satisfied $CAT(0)$ inequality if$\ d(p,q)\leq d_{2}(\overline{p},\overline{q})$ for all $p,q\in$ $\Delta(x,y,z)$ where $\overline{p},\overline{q}\in\overline{\Delta}(\overline{x},\overline{y},\overline{z})$ are the comparison points of $p,q$ respectively. A geodesic space is called $CAT(0)$ space if for all geodesic triangles satisfies $CAT(0)$ inequality or alternatively: A geodesic space is called $CAT(0)$ space if and only if the inequality

[TABLE]

satisfied for every $x,y,z\in X$, $\lambda\in[0,1]$.

Proposition 1.1**.**

[10*]*Let $(X,d)$ be a $CAT(0)$ space Then, for any $x,y,z\in X$ and $\lambda\in[0,1]$, we have

[TABLE]

Let $\{x_{n}\}$ be a bounded sequence on $X$ and $x\in X$.   Then, with setting

[TABLE]

the asymptotic radius of $\{x_{n}\}$ is defined by

[TABLE]

the asymptotic radius of $\{x_{n}\}$ with respect to $K\subseteq X$ is defined by

[TABLE]

and the asymptotic center of $\{x_{n}\}$ is defined by

[TABLE]

and let $\omega_{w}(x_{n}):=\cup A(\{x_{n}\})$ where union is taken on all subsequences of $\{x_{n}\}.$

Definition 1.2**.**

[12]A sequence $\{x_{n}\}\subset X$  is said to be $\Delta-$ convergent to $x\in X$ if  $x$ is the unique asymptotic center of  all subsequence $\{u_{n}\}$ of $\{x_{n}\}$, i.e.$\omega_{w}(x_{n}):=\cup A(\{x_{n}\})$ $=\{x\}$ . In this case we write $\Delta-\lim_{n}x_{n}=x$.

Lemma 1.3**.**

[10]**

1. i)

Every bounded sequence in a complete $CAT(0)$ space has a $\Delta$-convergent subsequence 2. ii)

If $K$ is a closed convex subset of a complete $CAT(0)$ and if $\{x_{n}\}$ is a bounded sequence in $K$, then the asymptotic center of $\{x_{n}\}$ is in $K$

Lemma 1.4**.**

[10]** If $\{x_{n}\}$ is a bounded sequence in $X$ with $A(\{x_{n}\})=\{x\}$ and $\{u_{n}\}$ is a subsequence of $\{x_{n}\}$ with $A(\{u_{n}\})=u$ and the sequence $\{d(x_{n},u)\}$ converges, then $x=u$

Theorem 1.5**.**

[11*]*Let $X$ be a bounded, complete and uniformly convex metric space. If $T$ is a multivalued nonexpansive mapping which assigns to each point of $X$ a nonempty compact subset of $X$, then $T$ has a fixed point in $X$.

In a complete $CAT(0)$ space, the metric projection $P_{K}(x)$ of $x$ onto a nonempty, closed and convex subset $K$ is singleton and nonexpansive.

The concept of inner-product has been generalized from Hilbert space to a $CAT(0)$ space $X$ by Berg and Nikolaev [16]. as follows: For any $a,b\in X,$ with denoting $\overrightarrow{ab}$ as a vector in $X$, quasi-linearization mapping defined as

[TABLE]

for all $a,b,c,d\in X$ and satisfies following properties

[TABLE]

for all $a,b,c,d,e\in X$ The last properties is known as a Cauchy-Schwarz inequality and it is a characterization of $CAT(0)$ space: A geodesic metric space is a $CAT(0)$ if and only if it satisfies Cauchy-Schwarz inequality.

Lemma 1.6**.**

[16*]*Let $X$ be a $CAT(0)$ and $K$ be a nonempty and convex subset of $X$, $x\in X$ and $u\in K$. Then $u=P_{K}(x)$ if and only if

[TABLE]

Let $X$ be a real Hilbert space and $K\subset X$  be nonempty closed and convex. A operator $A:K\rightarrow 2^{X}$ is called monotone if and only if

[TABLE]

for all $x,y\in X,$ $x^{\ast}\in Ax,y^{\ast}\in Ay$. If $A$ is a monotone operator then the variational inequality associated with $A$ is finding $(u,x)_{u\in Ax}$ such that

[TABLE]

The VIPs associated with monotone operators have applications in applied mathematics. For interested readers can find more informations about VIPs and their applications in the book by Kinderlehrer and Stampacchia (see [2, 3] ).

Now let $X$ be a complete $CAT(0)$ space, $K\subset X$ be nonempty, closed and convex and $T:K\rightarrow X$ be a nonexpansive mapping. In 2015, Khatibzadeh, & Ranjbar [15] defined the variational inequality associated with the nonexpansive mapping $T$  as follows

[TABLE]

They prove some existence and convergence results for this problem.

In this paper, we define variational inequality associated with the a non-self multivalued nonexpansive mapping $T:K\rightarrow KC(X)$ as follows

[TABLE]

and we prove some existence and convergence theorems for this problem.

2. Existence of A Solution

In this section, it is assumed that $X$ is a complete $CAT(0)$ and $K$ is a nonempty, closed and convex subset of $X$.

Definition 2.1**.**

If $K$ is also bounded subset of $X$ and $T:K\rightarrow C(X)$. Then the projection $P_{K}T$ of multivalued mapping $T$ onto $K$ is defined by

[TABLE]

where $P_{K}$ is metric projection and $D(x^{\prime},K)=\inf_{v^{\prime}\in K}d(x^{\prime},v^{\prime}).$

Lemma 2.2**.**

$P_{K}^{\ast}T(x)$* is multivalued nonexpansive mapping from $K$ to $2^{K}$*

Proof.

Since $K$ is closed, convex and bounded, $P_{K}^{\ast}(Tx)\subset K$. We also have

[TABLE]

by the nonexpansiveness of $P_{K}$.  ∎

Lemma 2.3**.**

If $T$ is compact valued then $P_{K}^{\ast}$ is compact valued.

Proof.

Let $(v_{n})\subset P_{K}^{\ast}T(x)$ be a sequence then there is a sequence $(x_{n}^{\prime})\subset Tx$ such that for all $n\in\mathbb{N},$ there $v_{n}=P_{K}(x_{n}^{\prime}).$Since $T$ have compact values then $(x_{n}^{\prime})$ have convergent subsequence $(x_{n_{k}}^{\prime})$ with $\lim_{k\rightarrow\infty}x_{n_{k}}^{\prime}=z\in Tx$ and since for all $k\in\mathbb{N},$

[TABLE]

we get that the sequence $(v_{n})=(P_{K}(x_{n}^{\prime}))$ have convergent subsequence $(v_{n_{k}})=(P_{K}(x_{n_{k}}^{\prime}))$ with $\lim_{k\rightarrow\infty}(P_{K}(x_{n_{k}}^{\prime}))=P_{K}(z)\in P_{K}^{\ast}T(x)$ therefore $P_{K}^{\ast}Tx$ is compact. ∎

Theorem 2.4**.**

If $T:K\rightarrow KC(X)$. Then there exists a solution $(u,x)_{u\in Tx}$ of the variational inequality (1.2)

Proof.

Since $X$ is uniformly convex and $T$ is compact valued, $P_{K}T$ have fixed point $p\in P_{K}T(p)\subset K$ by Theorem 1.5. There exist $p^{\prime}\in Tp$ such that $p=P_{K}(p^{\prime})$ by definition of $P_{K}T.$ we have $\langle p^{\prime}p,yp\rangle\leq 0$ for all $y\in K$ by Lemma 1.6. Hence we have

[TABLE]

where  $p^{\prime}\in Tp$, that is $(p^{\prime},p)_{p^{\prime}\in Tp}$ is a solution of the problem (1.2). ∎

Theorem 2.5**.**

If $x\in int(K)$ and $(u,x)_{u\in Tx}$ is a solution of problem (1.2) then $x\in F(T)$, i.e., $u=x.$

Proof.

There exists $\epsilon>0$ such that $B(x,\epsilon)\subset K.$ Let take $t\in(0,1)$ such that $tx\oplus(1-t)u\in B(x,\epsilon),$ that is, $d(x,tx\oplus(1-t)u)=(1-t)d(x,u)<\epsilon$ . Since $B(x,\epsilon)\subset K$ then $tx\oplus(1-t)u\in K$ and $d(u,tx\oplus(1-t)u)=td(u,x)$ so we have

[TABLE]

and which implies

[TABLE]

since $t\in(0,1)$ then $d(x,u)=0$. Hence $u=x\in Tx$ ∎

If $K$ is not bounded, the problem (1.2) does not always have a solution. However if $o\in X$ be arbitrary and setting $K_{r}=K\cap B(o,r)$ then if $K_{r}\neq\emptyset$ By Theorem 2.4 there is $x_{r}\in K_{r}$  such that $(u_{r},x_{r})_{u_{r}\in Tx}$ is a solution of problem

[TABLE]

Theorem 2.6**.**

The problem (1.2) have a solution if and only if there is a $r>0$ such that the solution of the problem (2.1) $(u_{r},x_{r})_{u_{r}\in Tx_{r}},x_{r}\in K_{r}$ satisfies $d(o,x_{r})<r.$

Proof.

If the problem 1.2 have a solution $(u,x)_{x\in Tx}$ then $(u,x)_{x\in Tx}$ is a solution of the problem (2.1) and $d(o,x)<r$ is satisfied. Now, let there is a $r>0$ such that the solution of the problem (2.1) $(u_{r},x_{r})_{u_{r}\in Tx_{r}}$, $x_{r}\in K_{r}$ satisfies $d(o,x_{r})<r$ and $y\in K$ be arbitrary. Then we can chose $t\in(0,1)$ such that $(1-t)x_{r}\oplus ty\in B(o,r),$ that is, $(1-t)x_{r}\oplus ty\subset K_{r}$ and $d(x_{r},(1-t)x_{r}\oplus ty)=td(x_{r},y)$. Then

[TABLE]

Hence

[TABLE]

that is $(u_{r},x_{r})_{u_{r}\in Tx_{r}}$ is a solution of the problem (1.2) . ∎

Theorem 2.7**.**

Let $T:K\rightarrow KC(X)$ and $o\in X$ be fixed. If there exist $x_{0}\in K$ and $u_{0}\in Tx_{0}$ such that

[TABLE]

where $u\in Tx$  such that $d(x,u)=d(x,Tx)$ then the problem (1.2) have a solution.

Proof.

Let $R,M\in\mathbb{R}$ such that $d(u_{0},u_{0})<M$, $d(x_{0},o)<r$  and

[TABLE]

for all $x\in K,$ $d(x,o)\geq r$. Then

[TABLE]

for $r=d(x,o)$. If $(u_{r},x_{r})_{u_{r}\in Tx_{r}}$ is a solution of the problem (2.1) then since

[TABLE]

holds so we have $d(x_{r},o)<r$. Hence by Theorem 2.6 the problem (1.2) has a solution. ∎

3. Convergence Results to The Solutions

In this section, it is assumed that $X$ is a complete $CAT(0)$ and $K$ is a nonempty, closed and convex subset of $X$.

Theorem 3.1**.**

If $T:K\rightarrow KC(X)$ is a nonexpansive mapping and $\{x_{n}\}$ is a bounded sequence in $K$ with $\Delta-\lim_{n\rightarrow\infty}x_{n}=z$ and $\lim_{n\rightarrow\infty}d(x_{n},Tx_{n})=0$ then $z\in K$ and $z\in T(z).$

Proof.

By Lemma 1.3, $z\in K.$We can find a sequence $\{y_{n}\}$ such that $y_{n}\in Tx_{n},$ $d(x_{n},y_{n})=d(x_{n},Tx_{n}),$ so we have $\lim_{n\rightarrow\infty}d(x_{n},y_{n})=0$ and we can find a sequence $\{z_{n}\}$ in $Tz$ such that $d(y_{n},z_{n})=d(y_{n},Tz).$ Then Since $Tz$ is compact, there is a convergent subsequence $\{z_{n_{i}}\}$ of $\{z_{n}\},$ say $\lim_{i\rightarrow\infty}z_{n_{i}}=u\in Tz$.

[TABLE]

implies that  $\limsup_{i\rightarrow\infty}d(x_{n_{i}},u)\leq\limsup_{i\rightarrow\infty}H(Tx_{n_{i}},Tz)$ and $\Delta-\lim_{i\rightarrow\infty}x_{n_{i}}=z$ Because of $T$ is multivalued nonexpansive mapping,

[TABLE]

which implies that

[TABLE]

which implies that $z=u\in Tz.$ ∎

Lemma 3.2**.**

If $T:K\rightarrow KC(X)$ is a nonexpansive mapping and $\{x_{n}\}$ is a bounded sequence in $K$ with $\lim_{n\rightarrow\infty}d(x_{n},Tx_{n})=0$ and $\{d(x_{n},p)\}$ converges for all $p\in F(T)$ then $\omega_{w}(x_{n})\subseteq F(T)$ and $\omega_{w}(x_{n})$ include exactly one point.

Proof.

Let take $u\in\omega_{w}(x_{n})$ then there exist subsequence $\{u_{n}\}$ of $\{x_{n}\}$ with $A(\{u_{n}\})=\{u\}.$Then by Lemma 1.3 there exist subsequence $\{v_{n}\}$ of $\{u_{n}\}$ with $\Delta-\lim_{n\rightarrow\infty}v_{n}=v\in K$ . Then by Theorem 3.1 we have $v\in F(T)$ and by Lemma 1.4 we conclude that $u=v,$ hence we get $\omega_{w}(x_{n})\subseteq F(T)$. Let take subsequence $\{u_{n}\}$ of $\{x_{n}\}$with $A(\{u_{n}\})=\{u\}$ and $A(\{x_{n}\})=\{x\}.$ Because of $v\in\omega_{w}(x_{n})\subseteq F(T)$, $\{d(x_{n},u)\}$ converges, so by Lemma 1.4 we have $x=u,$ this means that $\omega_{w}(x_{n})$ include exactly one point. ∎

Theorem 3.3**.**

If $T:K\rightarrow C(X)$ is a nonexpansive mapping with $F(T)\neq\emptyset$ and $Tp=\{p\}$ for all $p\in F(T)$ and $\{x_{n}\}$ is a sequence in $K$  defined by (1.1) with $\lim\inf_{n\rightarrow\infty}\beta_{n}(1-\beta_{n})>0$ then $\{x_{n}\}$ is bounded, $\lim_{n\rightarrow\infty}d(x_{n},Tx_{n})=0$ and $\{d(x_{n},p)\}$ converges for all $p\in F(T).$

Proof.

Let $p\in F(T)$ then for any $x\in K,$we have that

[TABLE]

since metric projection $P_{K}$ is nonexpansive and $P_{K}(p)=\{x\in K:d(p,x)=d(p,K)\}=\{p\}$ we have

[TABLE]

and

[TABLE]

Here we have $d^{2}(x_{n+1},p)\leq d^{2}(x_{n},p)$ implies that $\lim_{n\rightarrow\infty}d(x_{n},p)$ exists, it is bounded,and $d(x_{n+1},p)\leq d(y_{n},p)\leq d(x_{n},p)$ implies $\lim_{n\rightarrow\infty}[d(x_{n},p)-d(y_{n},p)]=0$. Since $\beta_{n}(1-\beta_{n})d^{2}(Tx_{n},x_{n}))\leq d^{2}(x_{n},p)-d^{2}(y_{n},p),$by assumption we have that $\lim_{n\rightarrow\infty}d^{2}(Tx_{n},x_{n})=0,$so $\lim_{n\rightarrow\infty}d(Tx_{n},x_{n})=0$ ∎

Theorem 3.4**.**

If $T:K\rightarrow KC(X)$ is a nonexpansive mapping with $F(T)\neq\emptyset$ and $Tp=\{p\}$ for all $p\in F(T)$ and $\{x_{n}\}$ is a sequence in $K$  defined by (1.1) with $\lim\inf_{n\rightarrow\infty}\beta_{n}(1-\beta_{n})>0$ then $\{x_{n}\}$ is $\Delta-$convergent to $p\in$ $F(T)$ where $(p,p)$ is a solution of the problem (1.2)

Proof.

Since we have $\lim_{n\rightarrow\infty}d(x_{n},Tx_{n})=0$, $\{d(x_{n},p)\}$ converges for all $p\in F(T)$ and $\{x_{n}\}$ is bounded by Theorem 3.3 then it follows from Lemma 3.2 that $\omega_{w}(x_{n})\subseteq F(T)$ and $\omega_{w}(x_{n})$ include exactly one point $p\in$ $F(T)$ where $(p,p)$ is a solution of the problem 1.2 ∎

Theorem 3.5**.**

Let $K$ be also compact and $T:K\rightarrow C(X)$ be a nonexpansive mapping with $F(T)\neq\emptyset$ and $Tp=\{p\}$ for all $p\in F(T)$. If $\{x_{n}\}$ is a sequence in $K$  defined by (1.1) with $\lim\inf_{n\rightarrow\infty}\beta_{n}(1-\beta_{n})>0$ then $\{x_{n}\}$ strongly converges to $q\in F(T).$where $(q,q)$ is a solution of the problem (1.2)

Proof.

By Theorem 3.3, we have that $\lim_{n\rightarrow\infty}d(Tx_{n},x_{n})=0$ and $\lim_{n\rightarrow\infty}d(x_{n},p)$ exists for all $p\in F(T)$ Since $K$ is compact there is a convergent subsequence $\{x_{n_{i}}\}$ of $\{x_{n}\},$ say $\lim_{i\rightarrow\infty}x_{{}_{n_{i}}}=q.$ Then we have

[TABLE]

and taking limit on $i,$continuity of $T$ implies that $\ q\in Tq.$ ∎

4. Common Solution of System of Variational Inequalities

Let $X$ be a $CAT(0)$ space and $K_{i}\subset X$ be a nonempty, closed and convex subsets with $\bigcap\limits_{i=1}^{N}K_{i}\neq\emptyset$. If $T_{i}:K_{i}\rightarrow C(X)$ are mappings for $i=1...N$. then the system of variational inequalities problem is

[TABLE]

It is obvious that for $N=1$ the problem is reduced the problem (1.2). The importance of studying the problem (4.1) is underlying on fact that it is unification most of the problems; for example taking if we take $T_{i}=0$ for all $i=1,...,N$  the reduce the problem (4.1) to convex feasibility problem,

[TABLE]

or if every $T_{i}$ is self operator and $K=\bigcap\limits_{i=1}^{N}F(T_{i})$ then it turn to common fixed point problem. We will show that the algorithm defined by (4.2) is convergent to common fixed point of family of non-self multivalued nonexpansive mappings $\{T_{i}\}_{i=1}^{N}$ which is also a solution of system of variational inequalities problem (4.1) Let $K=\bigcap\limits_{i=1}^{N}K_{i}\neq\emptyset$ and $x_{1}\in K$. then for any $n\geq 0$, the modified proximal multivalued Picard-S iteration is defined by

[TABLE]

where $u_{n,i}\in T_{i}y_{n},w_{n,i}\in T_{i}x_{n},v_{n,i}\in T_{i}z_{n},\{\lambda_{n,i}\},\{\alpha_{n,i}\},\{\beta_{n,i}\}$ and $\{\gamma_{n,i}\}$ are the sequences satisfies $\sum\limits_{i=1}^{N}\lambda_{n,i}=1,\sum\limits_{i=1}^{N}(\alpha_{n,i}+\beta_{n,i})=1,\sum\limits_{i=0}^{N}\gamma_{n,i}=1$ in $[b,c]$ for some $b,c\in(0,1)$

Lemma 4.1**.**

[19]** Let $(X,d,W)$ be a uniformly convex hyperbolic space with modulus of uniform convexity $\delta$. For any $r>0,$ $\epsilon\in(0,2),$ $\lambda\in[0,1]$ and $a,x,y\in$ $X$, if $d(x,a)\leq r$, $d(y,a)\leq r$ and $d(x,y)\geq\in r$ then $d((1-\lambda)x\oplus\lambda y,z)\leq(1-2\lambda(1-\lambda)\delta(r,\in))r.$

Proposition 4.2**.**

[20]** Assume that $X$ is a $CAT(0)$ space. Then $X$ is uniformly convex and

[TABLE]

is a modulus of uniform convexity.

Lemma 4.3**.**

[9]** Let $(X,d)$ be a complete $CAT(0)$ space, $\{x_{1},x_{2},..x_{n}\}\subset X$ and $\{\lambda_{1},\lambda_{2},..,\lambda_{n}\}\subset[0,1]$ with $\sum\limits_{i=1}^{n}\lambda_{i}=1.$Then $d(\bigoplus\limits_{i=1}^{n}\lambda_{i}x_{i},z)\leq\sum\limits_{i=1}^{n}\lambda_{i}d(x_{i},z)$ for every $z\in X.$

Lemma 4.4**.**

[20]** Let $X$ be a complete $CAT(0)$ space with modulus of convexity $\delta(r,\in)$ and let $x\in E$. Suppose that $\delta(r,\in)$ increases with $r$ (for a fixed $\in$ ) and suppose $\{t_{n}\}$ is a sequence in $[b,c]$ for some $b,c\in(0,1)$, $\{x_{n}\}$ and $\{y_{n}\}$ are the sequences in $X$ such that $\limsup\nolimits_{n\rightarrow\infty}d(x_{n},x)\leq r,$ $\limsup\nolimits_{n\rightarrow\infty}d(y_{n},x)\leq r$ and $\lim\nolimits_{n\rightarrow\infty}d((1-t_{n})x_{n}\oplus t_{n}y_{n},x)=r$ for some $r\geq 0$. Then $\lim_{n\rightarrow\infty}d(x_{n},y_{n})=0.$

The following Lemma is very important to our results.

Lemma 4.5**.**

Let $X$ be a complete $CAT(0)$ space with modulus of convexity $\delta(r,\in)$ and let $x\in X$. Suppose that $\delta(r,\in)$ increases with $r$ (for a fixed $\in$ ) and suppose $\{t_{n,i}\}$ with $\sum\limits_{i=1}^{N}t_{n,i}=1$ is a sequence in $[b,c]$ for some $b,c\in(0,1)$, $\{x_{n,i}\}_{n=1}^{\infty}$ are the sequences for $i\in\{1,2,..,N\}$ in $X$ such that $\limsup\nolimits_{n\rightarrow\infty}d(x_{n,i},x)\leq r$ and $\lim\nolimits_{n\rightarrow\infty}d(\bigoplus\limits_{i=1}^{N}t_{n,i}x_{n,i},x)=r$ for some $r\geq 0$. Then $\lim\nolimits_{n\rightarrow\infty}d(x_{n,k},x_{n,l})=0$ for $k,l\in\{1,2,..,N\}.$

Proof.

If $r=0$ then it is obvious let $r>0.$Since $\limsup\nolimits_{n\rightarrow\infty}d(x_{n,i},x)\leq r$ for each $i=1,2,..N$, then, by Lemma 4.3, for every $m=1,2,..,N$,

[TABLE]

Let assume that $d(x_{n,k},x_{n,l})\nrightarrow 0$ for fixed $k,l\in\{1,2,..,N\}$ with $k\neq l$  then there is subsequence denoted by (without loss of generality) $\{x_{n,k}\}$ and $\{x_{n,l}\}$ such that $inf_{n}d(x_{n,k},x_{n,l})>0.$Since

[TABLE]

then

[TABLE]

and since $t_{n,k},t_{n,l}\in[b,c]$ and by positivity of $d$, $d(\bigoplus\limits_{i=1,i\neq k}^{N}\frac{t_{n,i}}{1-t_{n,k}}x_{n,i},x_{n,k})\nrightarrow 0.$ therefore there is subsequence again denoted by $\{x_{n,k}\}$ for some $k=1,2,..N$ such that $d(\bigoplus\limits_{i=1}^{N}\frac{t_{n,i}}{1-t_{n,k}}x_{n,i},x_{n,k})>0$ so $d(x_{n,k},x)\leq r,d(\bigoplus\limits_{i=1,i\neq k}^{N}\frac{t_{n,i}}{1-t_{n,k}}x_{n,k},x)\leq r$ and $\lim\nolimits_{n\rightarrow\infty}d(\bigoplus\limits_{i=1}^{N}t_{n,i}x_{n,i},x)=\lim\nolimits_{n\rightarrow\infty}d((1-t_{n,m})[\bigoplus\limits_{i=1,i\neq k}^{N}\frac{t_{n,i}}{1-t_{n,k}}x_{n,i}]\oplus t_{n,m}x_{n,m},x)=r$ hence we can apply Lemma 4.4. ∎

From this point, it is assumed that $X$ is a complete $CAT(0)$ and $K=\bigcap\limits_{i=1}^{N}K_{i}$ is a nonempty, closed and convex subset of $X$ where $K_{i}\subset X$ be a nonempty, closed and convex subsets with $K=\bigcap\limits_{i=1}^{N}K_{i}\neq\emptyset$ for all $i=1,2,...,N$.

Lemma 4.6**.**

Let $\{T_{i}\}_{i=1}^{N}$ be multivalued nonexpansive mappings  from $K$ to $CC(X)$ with $F=\bigcap\limits_{i=1}^{N}F(T_{i})\neq\emptyset,$ $T_{i}p=\{p\}$ for all $p\in F$. If $\{x_{n}\}$ is the sequence defined by (4.2) then $\{x_{n}\}$ is bounded and $\lim_{n\rightarrow\infty}d(x_{n},p)$ exist for all $p\in F$.

Proof.

Let $p\in F.$Then from definition of $\{x_{n}\},$

[TABLE]

and

[TABLE]

and

[TABLE]

Hence $d(y_{n},p)\leq$ $d(x_{n},p),d(z_{n},p)\leq d(x_{n},p)$ and $d(x_{n+1},p)\leq d(x_{n},p)$ and so $\lim_{n\rightarrow\infty}d(x_{n},p)$ exists and $\{x_{n}\}$ is bounded sequence. ∎

Lemma 4.7**.**

Let $\{T_{i}\}_{i=1}^{N}$ be multivalued nonexpansive mappings from $K$ to $C(X)$ with $F=\bigcap\limits_{i=1}^{N}F(T_{i})\neq\emptyset,$ $T_{i}p=\{p\}$ for all $p\in F$. If $\{x_{n}\}$ is the sequence defined by (4.2) then $\lim_{n\rightarrow\infty}d(x_{n},T_{i}x_{n})$ exist for all $i=1,2..N$.

Proof.

Let $p\in F.$From the Lemma 4.6 $\lim_{n\rightarrow\infty}d(x_{n},p)$ exist and $\{x_{n}\}$ is bounded sequence. so let $\lim_{n\rightarrow\infty}d(x_{n},p)=c$. Since $d(y_{n},p)\leq$ $d(x_{n},p)$ and $d(u_{n,i},p)\leq d(y_{n},p),\limsup_{n\rightarrow\infty}d(y_{n},p)\leq c$ and $\limsup_{n\rightarrow\infty}d(u_{n,i},p)\leq c$ and again from Lemma 4.6 similarly $\limsup_{n\rightarrow\infty}d(z_{n},p)\leq c$ and $\limsup_{n\rightarrow\infty}d(v_{n,i},p)\leq c$ and $\limsup_{n\rightarrow\infty}d(x_{n},p)\leq c$ and $\limsup_{n\rightarrow\infty}d(w_{n,i},p)\leq c.$Moreover we have

[TABLE]

implies $\lim_{n\rightarrow\infty}d(\bigoplus\nolimits_{i=1}^{N}\lambda_{n,i}u_{n,i},p)=c$. We find that $\lim_{n\rightarrow\infty}d(u_{n,i},u_{n,j})=0$ for all $i,j=1,2,..,N$ by Lemma 4.5. Then

[TABLE]

and then $\liminf_{n\rightarrow\infty}d(u_{n,m},p)\geq c$ for all $m=1,2,..,N$. Since $\limsup_{n\rightarrow\infty}d(u_{n,i,},p)\leq c$ and $d(u_{n,i,}p)\leq d(y_{n},p)$ thus we have $\lim_{n\rightarrow\infty}d(u_{n,i},p)=c$ and $\lim_{n\rightarrow\infty}d(y_{n},p)$=$c$.

[TABLE]

which implies that $\lim_{n\rightarrow\infty}d(\bigoplus\nolimits_{i=1}^{N}\alpha_{n,i}w_{n,i}\oplus\bigoplus\nolimits_{i=1}^{N}\beta_{n,i}v_{n,i},p)=c$. Also we have $\lim_{n\rightarrow\infty}d(v_{n,i},v_{n,j})=0$, $\lim_{n\rightarrow\infty}d(v_{n,i},w_{n,j})=0$ and $\lim_{n\rightarrow\infty}d(w_{n,i},w_{n,j})=0$ for all $i,j=1,...,N$ by Lemma 4.7. Then

[TABLE]

and since $\lim_{n\rightarrow\infty}d(v_{n,i},w_{n,j})=0$ and $\lim_{n\rightarrow\infty}d(w_{n,i},w_{n,j})=0$ for all $i,j=1,...,N$ then $\liminf_{n\rightarrow\infty}d(v_{n,m},p)\geq c$ for all $m=1,2,..,N$ and since $\limsup_{n\rightarrow\infty}d(v_{n,i,},p)\leq c$ and $d(v_{n,i,}p)\leq d(z_{n},p)$ thus we have $\lim_{n\rightarrow\infty}d(v_{n,i},p)=c.$and $\lim_{n\rightarrow\infty}d(z_{n},p)$=$c$. Finally

[TABLE]

which implies that $\lim_{n\rightarrow\infty}[d(\gamma_{n,0}x_{n}\oplus\bigoplus\nolimits_{i=1}^{N}\gamma_{n,i}w_{n,i},p)=c$ and since $\limsup_{n\rightarrow\infty}d(x_{n},p)\leq c$ and $\limsup_{n\rightarrow\infty}d(w_{n,i},p)\leq c$ we find that $\lim_{n\rightarrow\infty}d(x_{n},w_{n,i})=0$ and $\lim_{n\rightarrow\infty}d(w_{n,i},w_{n,j})=0$ for all $i,j$ by Lemma 4.5..Hence $d(x_{n},T_{i}x_{n})\leq d(x_{n},w_{n,i})$ for all $i=1,2..N$ and $\lim_{n\rightarrow\infty}d(x_{n},T_{i}x_{n})$ $=0$ ∎

Theorem 4.8**.**

Let $\{T_{i}\}_{i=1}^{N}$ be multivalued nonexpansive mappings from $K$ to $KC(X)$ with $F=\bigcap\nolimits_{i=1}^{N}F(T_{i})\neq\emptyset,$ $T_{i}p=\{p\}$ for all $p\in F$ .  Then a sequence $\{x_{n}\}$ defined by (4.2) $\Delta$-converges to $p\in F$ where $(p,p)$ is a common solution of the problem (4.1).

Proof.

It follows from Lemma 4.6 and Lemma 4.7 that  $\lim_{n\rightarrow\infty}d(x_{n},T_{i}x_{n})=0$ for all $i\in\{1,2..N\}$, $\lim_{n\rightarrow\infty}d(x_{n},p)$ exists for all $p\in F$. Let $\omega_{w}(x_{n}):=\cup A(\{u_{n}\})$ where union take on all subsequence $\{u_{n}\}$ of $\{x_{n}\}.$ To show that $\Delta$-convergence of $\{x_{n}\}$ it is enough to show that $\omega_{w}(x_{n})\subseteq F$ and $\omega_{w}(x_{n})$ contains single point. First of all $\omega_{w}(x_{n})\subset K$ by Lemma 1.3. Let take$\ u\in\omega_{w}(x_{n})$, then there exist subsequence $\{u_{n}\}$ of $\{x_{n}\}$ such that $\ A\{u_{n}\}=\{u\}.$By Lemma 1.3 and Lemma 1.4 there exist a subsequence $(v_{n})$ of $\{u_{n}\}$ which $\Delta-$convergent to $v$. Let fix $i\in\{1,2...N\},$Since $T_{i}v$ is compact, then for each $n\geq 1$ we can pick up $z_{n,i}\in T_{i}v$ satisfies $d(v_{n},z_{n,i})=d(v_{n},T_{i}v)$ and compactness of $T_{i}v$ implies there exist a convergent subsequence $\{z_{n_{k},i}\}$ of $\{z_{n,i}\}$. Let $z_{n_{k},i}\rightarrow w_{i}\in T_{i}v$. Since $T_{i}$ is nonexpansive map we have;

[TABLE]

Hence we have

[TABLE]

which implies

[TABLE]

Hence by uniqueness of asymptotic centers, we have $w_{i}=v\in T_{i}v.$ Since $i$ was arbitrary we have $v\in$ $F=\bigcap\limits_{i=1}^{N}F(T_{i})$ so $\lim\nolimits_{n\rightarrow\infty}d(x_{n},v)$ exist by Lemma 4.6 which implies $u=v\in F$ by Lemma 1.4.Thus we have $\omega_{w}(x_{n})\subseteq F.$ If we take subsequence $\{u_{n}\}$ of $\{x_{n}\}$ with $A\{u_{n}\}=\{u\}$ and $A\{x_{n}\}=\{x\}$ then, since $\ u\in\omega_{w}(x_{n})\subseteq F$ and $\lim\nolimits_{n\rightarrow\infty}d(x_{n},v)$ exist, we have $u=x$ by Lemma 1.4. ∎

Theorem 4.9**.**

If $K$ is also compact $\{T_{i}\}_{i=1}^{N}$ are multivalued nonexpansive mappings from $K$ to $C(X)$ with $F=\bigcap\nolimits_{i=1}^{N}F(T_{i})\neq\emptyset,$ $T_{i}p=\{p\}$ for all $p\in F$ then the sequence $\{x_{n}\}$ defined by (4.2) strongly strongly converges to $p\in F$ where $(p,p)$ is a common solution of the problem (4.1)

Proof.

By Lemma 4.6 and Lemma 4.7,we have that $\lim_{n\rightarrow\infty}d(x_{n},T_{i}x_{n})=0$ for all $i\in\{1,2..N\}$, $\lim_{n\rightarrow\infty}d(x_{n},p)$ exists for all $p\in F$ Since $K$ is compact there is a convergent subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\},$ say $\lim_{i\rightarrow\infty}x_{{}_{n_{k}}}=q.$ Then for all $i\in\{1,2..N\},$ we have

[TABLE]

and taking limit on $k,$implies that $\ q\in T_{i}q$ for all $i\in\{1,2..N\}$. Hence $p\in F$ ∎