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 Open Access
Fixed points and strict fixed points for multivalued contractions of Reich type on metric spaces endowed with a graph
 Cristian Chifu^{1},
 Gabriela Petruşel^{1}Email author and
 MonicaFelicia Bota^{2}
https://doi.org/10.1186/168718122013203
© Chifu et al.; licensee Springer 2013
 Received: 8 May 2013
 Accepted: 10 July 2013
 Published: 29 July 2013
Abstract
The purpose of this paper is to present some strict fixed point theorems for multivalued operators satisfying a Reichtype condition on a metric space endowed with a graph. The wellposedness of the fixed point problem is also studied.
MSC:47H10, 54H25.
Keywords
 fixed point
 strict fixed point
 metric space endowed with a graph
 wellposed problem
1 Preliminaries
A new approach in the theory of fixed points was recently given by Jachymski [1] and GwóźdźLukawska and Jachymski [2] by using the context of metric spaces endowed with a graph. Other recent results for singlevalued and multivalued operators in such metric spaces are given by Nicolae, O’Regan and Petruşel in [3] and by Beg, Butt and Radojevic in [4].
Let $(X,d)$ be a metric space and let Δ be the diagonal of $X\times X$. Let G be a directed graph such that the set $V(G)$ of its vertices coincides with X and $\mathrm{\Delta}\subseteq E(G)$, where $E(G)$ is the set of the edges of the graph. Assume also that G has no parallel edges and, thus, one can identify G with the pair $(V(G),E(G))$.
If x and y are vertices of G, then a path in G from x to y of length $k\in \mathbb{N}$ is a finite sequence ${({x}_{n})}_{n\in \{0,1,2,\dots ,k\}}$ of vertices such that ${x}_{0}=x$, ${x}_{k}=y$ and $({x}_{i1},{x}_{i})\in E(G)$ for $i\in \{1,2,\dots ,k\}$. Notice that a graph G is connected if there is a path between any two vertices and it is weakly connected if $\tilde{G}$ is connected, where $\tilde{G}$ denotes the undirected graph obtained from G by ignoring the direction of edges.
In particular, if ${x}_{0}\in X$ then $D({x}_{0},B):=D(\{{x}_{0}\},B)$.
In particular, we denote by $\delta (A):=\delta (A,A)$ the diameter of the set A.
Let $(X,d)$ be a metric space. If $T:X\to P(X)$ is a multivalued operator, then $x\in X$ is called a fixed point for T if and only if $x\in T(x)$. The set $Fix(T):=\{x\in X\mid x\in T(x)\}$ is called the fixed point set of T, while $SFix(T)=\{x\in X\mid \{x\}=Tx\}$ is called the strict fixed point set of T. $Graph(T):=\{(x,y)\mid y\in T(x)\}$ denotes the graph of T.
 (i)
φ is increasing;
 (ii)
${\phi}^{n}(t)\to 0$ as $n\to \mathrm{\infty}$ for all $t\in {\mathbb{R}}_{+}$;
 (iii)
${\sum}_{n=1}^{\mathrm{\infty}}{\phi}^{n}(t)<\mathrm{\infty}$ for all $t\in {\mathbb{R}}_{+}$.
In this paper, we present some fixed point and strict fixed point theorems for multivalued operators satisfying a contractive condition of Reich type involving the functional δ (see [5, 6]). The equality between $Fix(T)$ and $SFix(T)$ and the wellposedness of the fixed point problem are also studied.
Our results also generalize and extend some fixed point theorems in partially ordered complete metric spaces given in Harjani and Sadarangani [7], Nicolae et al. [3], Nieto and RodríguezLópez [8] and [9], Nieto et al. [10], O’Regan and Petruşel [11], Petruşel and Rus [12], and Ran and Reurings [13].
2 Fixed point and strict fixed point theorems
We begin this section by presenting a strict fixed point theorem for a Reich type contraction with respect to the functional δ.
 (i)There exists $a,b,c\in {\mathbb{R}}_{+}$ with $b\ne 0$ and $a+b+c<1$ such that$\delta (T(x),T(y))\le ad(x,y)+b\delta (x,T(x))+c\delta (y,T(y))$
for all $(x,y)\in E(G)$.
 (ii)For each $x\in X$, the set$\begin{array}{rcl}{\tilde{X}}_{T}(x)& :=& \{y\in T(x):(x,y)\in E(G)\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{0.25em}{0ex}}\delta (x,T(x))\le qd(x,y)\\ \mathit{\text{for some}}\phantom{\rule{0.25em}{0ex}}q\in \phantom{\rule{0.2em}{0ex}}]1,\frac{1ac}{b}[\}\end{array}$
is nonempty.
 (a)
$Fix(T)=SFix(T)\ne \mathrm{\varnothing}$;
 (b)If we additionally suppose that${x}^{\ast},{y}^{\ast}\in Fix(T)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}({x}^{\ast},{y}^{\ast})\in E(G),$
then $Fix(T)=SFix(T)=\{{x}^{\ast}\}$.
For ${x}_{2}\in X$, we have ${\tilde{X}}_{T}({x}_{2})\ne \mathrm{\varnothing}$, and so there exists ${x}_{3}\in T({x}_{2})$ such that $\delta ({x}_{2},T({x}_{2}))\le qd({x}_{2},{x}_{3})$ and $({x}_{2},{x}_{3})\in E(G)$.
 (1)
$({x}_{n},{x}_{n+1})\in E(G)$ for each $n\in \mathbb{N}$;
 (2)
$d({x}_{n},{x}_{n+1})\le {(\frac{a+bq}{1c})}^{n}d({x}_{0},{x}_{1})$ for each $n\in \mathbb{N}$;
 (3)
$\delta ({x}_{n},T({x}_{n}))\le {(\frac{a+bq}{1c})}^{n}d({x}_{0},{x}_{1})$ for each $n\in \mathbb{N}$.
From (2) we obtain that the sequence ${({x}_{n})}_{n\in \mathbb{N}}$ is Cauchy. Since the metric space X is complete, we get that the sequence is convergent, i.e., ${x}_{n}\to {x}^{\ast}$ as $n\to \mathrm{\infty}$. By the property (P), there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ of ${({x}_{n})}_{n\in \mathbb{N}}$ such that $({x}_{{k}_{n}},{x}^{\ast})\in E(G)$ for each $n\in \mathbb{N}$.
But $d({x}^{\ast},{x}_{{k}_{n+1}})\to 0$ as $n\to \mathrm{\infty}$ and $d({x}_{{k}_{n}},{x}^{\ast})\to 0$ as $n\to \mathrm{\infty}$. Hence, $\delta ({x}^{\ast},T({x}^{\ast}))=0$, which implies that ${x}^{\ast}\in SFix(T)$. Thus $SFix(T)\ne \mathrm{\varnothing}$.
We shall prove now that $Fix(T)=SFix(T)$.
Because $SFix(T)\subset Fix(T)$, we need to show that $Fix(T)\subset SFix(T)$.
Suppose that $card(T({x}^{\ast}))>1$. This implies that $\delta (T({x}^{\ast}))>0$. Thus from (2.6) we obtain that $b+c>1$, which contradicts the hypothesis $a+b+c<1$.
Thus $\delta (T({x}^{\ast}))=0\Rightarrow T({x}^{\ast})=\{{x}^{\ast}\}$, i.e., ${x}^{\ast}\in SFix(T)$ and $Fix(T)\subset SFix(T)$.
Hence, $Fix(T)=SFix(T)\ne \mathrm{\varnothing}$.
(b) Suppose that there exist ${x}^{\ast},{y}^{\ast}\in Fix(T)=SFix(T)$ with ${x}^{\ast}\ne {y}^{\ast}$. We have that

${x}^{\ast}\in SFix(T)\Rightarrow \delta ({x}^{\ast},T({x}^{\ast}))=0$;

${y}^{\ast}\in SFix(T)\Rightarrow \delta ({y}^{\ast},T({y}^{\ast}))=0$;

$({x}^{\ast},{y}^{\ast})\in E(G)$.
Thus, $d({x}^{\ast},{y}^{\ast})\le ad({x}^{\ast},{y}^{\ast})$, which implies that $a\ge 1$, which is a contradiction.
Hence, $Fix(T)=SFix(T)=\{{x}^{\ast}\}$. □
Next we present some examples and counterexamples of multivalued operators which satisfy the hypothesis in Theorem 2.1.
Let $E(G):=\{((0,1),(0,0)),((1,0),(0,1)),((1,1),(0,0))\}\cup \mathrm{\Delta}$.
Notice that all the hypotheses in Theorem 2.1 are satisfied (the condition (i) is verified for $a=c=0.01$, $b=0.97$ and so $Fix(T)=SFix(T)=\{(0,0)\}$.
The following remarks show that it is not possible to have elements in ${F}_{T}\setminus S{F}_{T}$.
Remark 2.1 If we suppose that there exists $x\in {F}_{T}\setminus S{F}_{T}$, then, since $(x,x)\in \mathrm{\Delta}$, we get (using the condition (i) in the above theorem with $y=x$) that $\delta (T(x))\le (b+c)\delta (T(x))$, which is a contradiction with $a+b+c<1$.
Remark 2.2 If, in the previous theorem, instead of the property (P), we suppose that T has a closed graph, then we obtain again the conclusion $Fix(T)=SFix(T)\ne \mathrm{\varnothing}$.
then $Fix(T)=SFix(T)=\{{x}^{\ast}\}$.
The next result presents a strict fixed point theorem where the operator T satisfies a $(\delta ,\phi )$Gcontractive condition on $E(G)$.
 (i)
T is a $(\delta ,\phi )$Gcontraction.
 (ii)For each $x\in X$, the set${\tilde{X}}_{T}:=\{y\in T(x):(x,y)\in E(G)\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{0.25em}{0ex}}\delta (x,T(x))\le qd(x,y)\phantom{\rule{0.25em}{0ex}}\mathit{\text{for some}}\phantom{\rule{0.25em}{0ex}}q\in \phantom{\rule{0.2em}{0ex}}]1,\frac{1ac}{b}\left[\right\}$
is nonempty.
 (a)
$Fix(T)=SFix(T)\ne \mathrm{\varnothing}$;
 (b)If, in addition, the following implication holds:${x}^{\ast},{y}^{\ast}\in Fix(T)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}({x}^{\ast},{y}^{\ast})\in E(G),$
then $Fix(T)=SFix(T)=\{{x}^{\ast}\}$.
For ${x}_{1}\in X$, by the same approach as before, there exists ${x}_{2}\in T({x}_{1})$ such that $\delta ({x}_{1},T({x}_{1}))\le qd({x}_{1},{x}_{2})$ and $({x}_{1},{x}_{2})\in E(G)$.
 (1)
$({x}_{n},{x}_{n+1})\in E(G)$ for each $n\in \mathbb{N}$;
 (2)
$d({x}_{n},{x}_{n+1})\le {\phi}^{n}(d({x}_{0},{x}_{1}))$ for each $n\in \mathbb{N}$;
 (3)
$\delta ({x}_{n},T({x}_{n}))\le {\phi}^{n}(d({x}_{0},{x}_{1}))$ for each $n\in \mathbb{N}$.
By (2), using the properties of φ, we obtain that the sequence ${({x}_{n})}_{n\in \mathbb{N}}$ is Cauchy. Since the metric space is complete, we have that the sequence is convergent, i.e., ${x}_{n}\to {x}^{\ast}$ as $n\to \mathrm{\infty}$. By the property (P), we get that there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ of ${({x}_{n})}_{n\in \mathbb{N}}$ such that $({x}_{{k}_{n}},{x}^{\ast})\in E(G)$ for each $n\in \mathbb{N}$.
Since $d({x}^{\ast},{x}_{{k}_{n+1}})\to 0$ as $n\to \mathrm{\infty}$ and φ is continuous in 0 with $\phi (0)=0$, we get that $\delta ({x}^{\ast},T({x}^{\ast}))=0$.
Hence, ${x}^{\ast}\in SFix(T)\Rightarrow SFix(T)\ne \mathrm{\varnothing}$.
We shall prove now that $Fix(T)=SFix(T)$.
Because $SFix(T)\subset Fix(T)$, we need to show that $Fix(T)\subset SFix(T)$.
Hence, ${x}^{\ast}\in SFix(T)$ and the proof of this conclusion is complete.
(b) Suppose that there exist ${x}^{\ast},{y}^{\ast}\in Fix(T)=SFix(T)$ with ${x}^{\ast}\ne {y}^{\ast}$. We have that

${x}^{\ast}\in SFix(T)\Rightarrow \delta ({x}^{\ast},T({x}^{\ast}))=0$;

${y}^{\ast}\in SFix(T)\Rightarrow \delta ({y}^{\ast},T({y}^{\ast}))=0$;

$({x}^{\ast},{y}^{\ast})\in E(G)$.
This is a contradiction. Hence, $Fix(T)=SFix(T)=\{{x}^{\ast}\}$. □
In the next result, the operator T satisfies another contractive condition with respect to δ on $E(G)\cap Graph(T)$.
 (i)There exist $a,b\in {\mathbb{R}}_{+}$, with $b\ne 0$ and $a+b<1$, such that$\delta (y,T(y))\le ad(x,y)+b\delta (x,T(x))\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.25em}{0ex}}(x,y)\in E(G)\cap Graph(T).$
 (ii)For each $x\in X$, the set${\tilde{X}}_{T}(x):=\{y\in T(x):(x,y)\in E(G)\phantom{\rule{0.25em}{0ex}}\mathit{\text{and}}\phantom{\rule{0.25em}{0ex}}\delta (x,T(x))\le qd(x,y)\phantom{\rule{0.25em}{0ex}}\mathit{\text{for some}}\phantom{\rule{0.25em}{0ex}}q\in \phantom{\rule{0.2em}{0ex}}]1,\frac{1a}{b}\left[\right\}$
is nonempty.
Then $Fix(T)=SFix(T)\ne \mathrm{\varnothing}$.
and $({x}_{0},{x}_{1})\in E(G)$. Since ${x}_{1}\in T({x}_{0})$, we get that $({x}_{0},{x}_{1})\in E(G)\cap Graph(T)$.
For ${x}_{1}\in X$ (since ${\tilde{X}}_{T}({x}_{1})\ne \mathrm{\varnothing}$), there exists ${x}_{2}\in T({x}_{1})$ such that $\delta ({x}_{1},T({x}_{1}))\le qd({x}_{1},{x}_{2})$ and $({x}_{1},{x}_{2})\in E(G)$. But ${x}_{2}\in T({x}_{1})$ and so $({x}_{1},{x}_{2})\in E(G)\cap Graph(T)$.
 (1)
$({x}_{n},{x}_{n+1})\in E(G)\cap Graph(T)$ for each $n\in \mathbb{N}$;
 (2)
$d({x}_{n},{x}_{n+1})\le {(a+bq)}^{n}d({x}_{0},{x}_{1})$ for each $n\in \mathbb{N}$;
 (3)
$\delta ({x}_{n},T({x}_{n}))\le {(a+bq)}^{n}d({x}_{0},{x}_{1})$ for each $n\in \mathbb{N}$.
Thus $f({x}^{\ast})=0$, which means that $\delta ({x}^{\ast},T({x}^{\ast}))=0$. Thus ${x}^{\ast}\in SFix(T)$.
Let ${x}^{\ast}\in Fix(T)$. Then $({x}^{\ast},{x}^{\ast})\in Graph(T)$ and hence $({x}^{\ast},{x}^{\ast})\in E(G)\cap Graph(T)$.
So, $\delta (T({x}^{\ast}))\le b\delta (T({x}^{\ast}))$. If we suppose that $cardT({x}^{\ast})>1$, then $\delta (T({x}^{\ast}))>0$. Thus, $b\ge 1$, which contradicts the hypothesis.
Thus $\delta (T({x}^{\ast}))=0$ and so $T({x}^{\ast})=\{{x}^{\ast}\}$. The proof is now complete. □
Remark 2.4 Example 2.1 satisfies the conditions from Theorem 2.3 for $a=0.01$ and $b=0.97$.
3 Wellposedness of the fixed point problem
In this section we present some wellposedness results for the fixed point problem. We consider both the wellposedness and the wellposedness in the generalized sense for a multivalued operator T.
We begin by recalling the definition of these notions from [14] and [15].
 (i)
$SFix(T)=\{{x}^{\ast}\}$;
 (ii)
If ${({x}_{n})}_{n\in \mathbb{N}}$ is a sequence in X such that $H({x}_{n},T({x}_{n}))\to 0$ as $n\to \mathrm{\infty}$, then ${x}_{n}\stackrel{d}{\to}{x}^{\ast}$ as $n\to \mathrm{\infty}$.
 (i)
$SFixT\ne \mathrm{\varnothing}$;
 (ii)
If ${({x}_{n})}_{n\in \mathbb{N}}$ is a sequence in X such that $H({x}_{n},T({x}_{n}))\to 0$ as $n\to \mathrm{\infty}$, then there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ of ${({x}_{n})}_{n\in \mathbb{N}}$ such that ${x}_{{k}_{n}}\stackrel{d}{\to}{x}^{\ast}$ as $n\to \mathrm{\infty}$.
In our first result we will establish the wellposedness of the fixed point problem for the operator T, where T is a Reichtype δcontraction.
Theorem 3.1 Let $(X,d)$ be a complete metric space and let G be a directed graph such that the triple $(X,d,G)$ satisfies the property (P).
 (i)
conditions (i) and (ii) in Theorem 2.1 hold;
 (ii)
if ${x}^{\ast},{y}^{\ast}\in Fix(T)$, then $({x}^{\ast},{y}^{\ast})\in E(G)$;
 (iii)
for any sequence ${({x}_{n})}_{n\in \mathbb{N}}$, ${x}_{n}\in X$ with $H({x}_{n},T({x}_{n}))\to 0$ as $n\to \mathrm{\infty}$, we have $({x}_{n},{x}^{\ast})\in E(G)$.
In these conditions the fixed point problem is well posed for T with respect to H.
Hence, ${x}_{n}\to {x}^{\ast}$ as $n\to \mathrm{\infty}$. □
Remark 3.1 If we replace the property (P) with the condition that T has a closed graph, we reach the same conclusion.
The next result deals with the wellposedness of the fixed point problem in the generalized sense.
Theorem 3.2 Let $(X,d)$ be a complete metric space and let G be a directed graph such that the triple $(X,d,G)$ satisfies the property (P).
 (i)
conditions (i) and (ii) in Theorem 2.1 hold;
 (ii)
for any sequence ${({x}_{n})}_{n\in \mathbb{N}}$, ${x}_{n}\in X$ with $H({x}_{n},T({x}_{n}))\to 0$ as $n\to \mathrm{\infty}$, there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ such that $({x}_{{k}_{n}},{x}^{\ast})\in E(G)$ and $H({x}_{{k}_{n}},T({x}_{{k}_{n}}))\to 0$.
In these conditions the fixed point problem is well posed in the generalized sense for T with respect to H.
Proof From (i) we have that $SFix(T)\ne \mathrm{\varnothing}$. Let ${({x}_{n})}_{n\in \mathbb{N}}\subset X$ be a sequence which satisfies (ii). Then there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ such that $({x}_{{k}_{n}},{x}^{\ast})\in E(G)$.
Hence, ${x}_{{k}_{n}}\to {x}^{\ast}$. □
Remark 3.2 If we replace the property (P) with the condition that T has a closed graph, we reach the same conclusion.
Next we consider the case where the operator T satisfies a φcontraction condition.
Theorem 3.3 Let $(X,d)$ be a complete metric space and let G be a directed graph such that the triple $(X,d,G)$ satisfies the property (P).
 (i)
conditions (i) and (ii) in Theorem 2.2 hold;
 (ii)
the following implication holds: ${x}^{\ast},{y}^{\ast}\in Fix(T)$ implies $({x}^{\ast},{y}^{\ast})\in E(G)$;
 (iii)
the function $\psi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, given by $\psi (t)=t\phi (t)$, has the following property: if $\psi ({t}_{n})\to 0$ as $n\to \mathrm{\infty}$, then ${t}_{n}\to 0$ as $n\to \mathrm{\infty}$;
 (iv)
for any sequence ${({x}_{n})}_{n\in \mathbb{N}}\subset X$ with $H({x}_{n},T({x}_{n}))\to 0$ as $n\to \mathrm{\infty}$, we have $({x}_{n},{x}^{\ast})\in E(G)$ for all $n\in \mathbb{N}$.
In these conditions the fixed point problem is well posed for T with respect to H.
Using condition (iii), we get that $d({x}_{n},{x}^{\ast})\to 0$ as $n\to \mathrm{\infty}$. Hence, ${x}_{n}\to {x}^{\ast}$. □
Remark 3.3 If we replace the property (P) with the condition that T has a closed graph, we reach the same conclusion.
The next result gives a wellposedness (in the generalized sense) criterion for the fixed point problem.
Theorem 3.4 Let $(X,d)$ be a complete metric space and let G be a directed graph such that the triple $(X,d,G)$ satisfies the property (P).
 (i)
the conditions (i) and (ii) in Theorem 2.2 hold;
 (ii)
the function $\psi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, given $\psi (t)=t\phi (t)$, has the following property: for any sequence ${({t}_{n})}_{n\in \mathbb{N}}$, there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ such that if $\psi ({t}_{{k}_{n}})\to 0$ as $n\to \mathrm{\infty}$, then ${t}_{{k}_{n}}\to 0$ as $n\to \mathrm{\infty}$;
 (iii)
for any sequence ${({x}_{n})}_{n\in \mathbb{N}}$, ${x}_{n}\in X$ with $H({x}_{n},T({x}_{n}))\to 0$ as $n\to \mathrm{\infty}$, there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ such that $({x}_{{k}_{n}},{x}^{\ast})\in E(G)$ and $H({x}_{{k}_{n}},T({x}_{{k}_{n}}))\to 0$.
In these conditions the fixed point problem is well posed in the generalized sense for T with respect to H.
Proof From (i) we have that $SFix(T)\ne \mathrm{\varnothing}$. Let ${({x}_{n})}_{n\in \mathbb{N}}$, ${x}_{n}\in X$ be a sequence which satisfies (iii). Then there exists a subsequence ${({x}_{{k}_{n}})}_{n\in \mathbb{N}}$ such that $({x}_{{k}_{n}},{x}^{\ast})\in E(G)$.
Using condition (ii), we get that $d({x}_{{k}_{n}},{x}^{\ast})\to 0$ as $n\to \mathrm{\infty}$. Hence, ${x}_{n}\to {x}^{\ast}$. □
Remark 3.4 If we replace the property (P) with the condition that T has a closed graph, we reach the same conclusion.
Declarations
Acknowledgements
The third author is supported by a grant of the Romanian National Authority for Scientific Research, CNCS UEFISCDI, project number PNIIIDPCE201130094.
Authors’ Affiliations
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