Fixed points and strict fixed points for multivalued contractions of Reich type on metric spaces endowed with a graph

Abstract

The purpose of this paper is to present some strict fixed point theorems for multivalued operators satisfying a Reich-type condition on a metric space endowed with a graph. The well-posedness of the fixed point problem is also studied.

MSC:47H10, 54H25.

1 Preliminaries

A new approach in the theory of fixed points was recently given by Jachymski [1] and Gwóźdź-Lukawska and Jachymski [2] by using the context of metric spaces endowed with a graph. Other recent results for single-valued 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 $(x_{i-1},x_i)\in E(G)$ 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.

Denote by $G^{-1}$ the graph obtained from G by reversing the direction of edges. Thus,

Since it is more convenient to treat $\tilde{G}$ as a directed graph for which the set of its edges is symmetric, under this convention, we have that

If G is such that $E(G)$ is symmetric, then for $x\in V(G)$, the symbol ${[x]}_G$ denotes the equivalence class of the relation ℜ defined on $V(G)$ by the rule:

$$y\mathrm{\Re }z\text{if there is a path in }G\text{ from }y\text{ to }z.$$

Let us consider the following families of subsets of a metric space $(X,d)$:

$$\begin{array}{c}P(X):=\{Y\in \mathcal{P}(X)\mid Y\ne \mathrm{\varnothing }\};P_{\mathrm{b}}(X):=\{Y\in P(X)\mid Y\text{ is bounded}\};\hfill \\ P_{\mathrm{cl}}(X):=\{Y\in P(X)\mid Y\text{ is closed}\};P_{\mathrm{cp}}(X):=\{Y\in P(X)\mid Y\text{ is compact}\}.\hfill \end{array}$$

The gap functional between the sets A and B in the metric space $(X,d)$ is given by

$$D:P(X)\times P(X)\to {\mathbb{R}}_{+}\cup \{+\mathrm{\infty }\},D(A,B)=inf\{d(a,b)\mid a\in A,b\in B\}.$$

In particular, if $x_0\in X$ then $D(x_0,B):=D(\{x_0\},B)$.

The Pompeiu-Hausdorff functional is defined by

$$\begin{array}{c}H:P(X)\times P(X)\to {\mathbb{R}}_{+}\cup \{+\mathrm{\infty }\},\hfill \\ H(A,B)=max\{\underset{a\in A}{sup}D(a,B),\underset{b\in B}{sup}D(A,b)\}.\hfill \end{array}$$

The diameter generalized functional generated by d is given by

$$\begin{array}{c}\delta :P(X)\times P(X)\to {\mathbb{R}}_{+}\cup \{+\mathrm{\infty }\},\hfill \\ \delta (A,B)=sup\{d(a,b)\mid a\in A,b\in B\}.\hfill \end{array}$$

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.

Definition 1.1 Let $\phi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ be a mapping. Then φ is called a strong comparison function if the following assertions hold:

• (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 $t\in {\mathbb{R}}_{+}$.

Definition 1.2 Let $(X,d)$ be a complete metric space, let G be a directed graph, and let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. By definition, T is called a $(\delta ,\phi )$-G-contraction if there exists $\phi :{\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$, a strong comparison function, such that

$$\delta (T(x),T(y))\le \phi (d(x,y))\text{for all }(x,y)\in E(G).$$

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 well-posedness 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 Rodríguez-López [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.

Theorem 2.1 Let $(X,d)$ be a complete metric space and let G be a directed graph such that the triple $(X,d,G)$ satisfies the following property:

$$\begin{array}{rl}(P)& \begin{array}{l}\mathit{\text{for any sequence}}{(x_n)}_{n\in \mathbb{N}}\subset X\mathit{\text{with}}{x}_n\to x\mathit{\text{as}}n\to \mathrm{\infty },\\ \mathit{\text{there exists a subsequence}}{(x_{k_n})}_{n\in \mathbb{N}}\mathit{\text{of}}{(x_n)}_{n\in \mathbb{N}}\mathit{\text{such that}}(x_{k_n},x)\in E(G).\end{array}\end{array}$$

Let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that the following assertions hold:

• (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

  $$\begin{array}{rcl}{\tilde{X}}_T(x)& :=& \{y\in T(x):(x,y)\in E(G)\mathit{\text{and}}\delta (x,T(x))\le qd(x,y)\\ \mathit{\text{for some}}q\in ]1,\frac{1-a-c}{b}[\}\end{array}$$

  is nonempty.

Then we have:

• (a) $Fix(T)=SFix(T)\ne \mathrm{\varnothing }$;

• (b) If we additionally suppose that

  $$x^{\ast },y^{\ast }\in Fix(T)\Rightarrow (x^{\ast },y^{\ast })\in E(G),$$

  then $Fix(T)=SFix(T)=\{x^{\ast }\}$.

Proof (a) Let $x_0\in X$. Since ${\tilde{X}}_T(x_0)\ne \mathrm{\varnothing }$, there exists $x_1\in T(x_0)$ and $1<q<\frac{1-a-c}{b}$ such that $(x_0,x_1)\in E(G)$ and

$$\delta (x_0,T(x_0))\le qd(x_0,x_1).$$

By (i) we have that

$$\begin{array}{rl}\delta (x_1,T(x_1))& \le \delta (T(x_0),T(x_1))\le ad(x_0,x_1)+b\delta (x_0,T(x_0))+c\delta (x_1,T(x_1))\\ \le ad(x_0,x_1)+bqd(x_0,x_1)+c\delta (x_1,T(x_1)).\end{array}$$

Hence,

$$\delta (x_1,T(x_1))\le \frac{a+bq}{1-c}d(x_0,x_1).$$

(2.1)

For $x_1\in X$, since ${\tilde{X}}_T(x_1)\ne \mathrm{\varnothing }$, we get again that there exists $x_2\in T(x_1)$ such that $\delta (x_1,T(x_1))\le qd(x_1,x_2)$ and $(x_1,x_2)\in E(G)$. Then

$$d(x_1,x_2)\le \delta (x_1,T(x_1))\le \frac{a+bq}{1-c}d(x_0,x_1).$$

(2.2)

On the other hand, by (i), we have that

$$\begin{array}{rcl}\delta (x_2,T(x_2))& \le & \delta (T(x_1),T(x_2))\le ad(x_1,x_2)+b\delta (x_1,T(x_1))+c\delta (x_2,T(x_2))\\ \le & ad(x_1,x_2)+bqd(x_1,x_2)+c\delta (x_2,T(x_2)).\end{array}$$

Using (2.2) we obtain

$$\delta (x_2,T(x_2))\le \frac{a+bq}{1-c}d(x_1,x_2)\le {\left(\frac{a+bq}{1-c}\right)}^{2}d(x_0,x_1).$$

(2.3)

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 $(x_2,x_3)\in E(G)$.

Then

$$d(x_2,x_3)\le \delta (x_2,T(x_2))\le {\left(\frac{a+bq}{1-c}\right)}^{2}d(x_0,x_1).$$

(2.4)

By these procedures, we obtain a sequence ${(x_n)}_{n\in \mathbb{N}}$ with the following properties:

• (1) $(x_n,x_{n+1})\in E(G)$ for each $n\in \mathbb{N}$;

• (2) $d(x_n,x_{n+1})\le {(\frac{a+bq}{1-c})}^{n}d(x_0,x_1)$ for each $n\in \mathbb{N}$;

• (3) $\delta (x_n,T(x_n))\le {(\frac{a+bq}{1-c})}^{n}d(x_0,x_1)$ 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 $n\in \mathbb{N}$.

We have

$$\begin{array}{l}\delta (x^{\ast },T\left(x^{\ast }\right))\le d(x^{\ast },x_{k_{n+1}})+\delta (x_{k_{n+1}},T\left(x^{\ast }\right))\le d(x^{\ast },x_{k_{n+1}})+\delta (T(x_{k_n}),T\left(x^{\ast }\right))\\ \phantom{\delta (x^{\ast },T\left(x^{\ast }\right))}\le d(x^{\ast },x_{k_{n+1}})+ad(x_{k_n},x^{\ast })+b\delta (x_{k_n},T(x_{k_n}))+c\delta (x^{\ast },T\left(x^{\ast }\right))\\ \phantom{\delta (x^{\ast },T\left(x^{\ast }\right))}\le d(x^{\ast },x_{k_{n+1}})+ad(x_{k_n},x^{\ast })+b{\left(\frac{a+bq}{1-c}\right)}^{k_n}d(x_0,x_1)+c\delta (x^{\ast },T\left(x^{\ast }\right)),\\ \delta (x^{\ast },T\left(x^{\ast }\right))\le \frac{1}{1-c}d(x^{\ast },x_{k_{n+1}})+\frac{a}{1-c}d(x_{k_n},x^{\ast })+\frac{b}{1-c}{\left(\frac{a+bq}{1-c}\right)}^{k_n}d(x_0,x_1).\end{array}$$

(2.5)

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 $Fix(T)\subset SFix(T)$.

Let $x^{\ast }\in Fix(T)\Rightarrow x^{\ast }\in T(x^{\ast })$. Because $\mathrm{\Delta }\subset E(G)$, we have that $(x^{\ast },x^{\ast })\in E(G)$. Using (ii) with $x=y=x^{\ast }$, we obtain

$$\delta \left(T\left(x^{\ast }\right)\right)\le ad(x^{\ast },x^{\ast })+b\delta (x^{\ast },T\left(x^{\ast }\right))+c\delta (x^{\ast },T\left(x^{\ast }\right)).$$

So, $\delta (T(x^{\ast }))\le (b+c)\delta (x^{\ast },T(x^{\ast }))$. Because $x^{\ast }\in T(x^{\ast })$, we get that $\delta (x^{\ast },T(x^{\ast }))\le \delta (T(x^{\ast }))$. Hence, we have

$$\delta \left(T\left(x^{\ast }\right)\right)\le (b+c)\delta \left(T\left(x^{\ast }\right)\right).$$

(2.6)

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

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

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

• $(x^{\ast },y^{\ast })\in E(G)$.

Using (i) we obtain

$$d(x^{\ast },y^{\ast })=\delta (T\left(x^{\ast }\right),T\left(y^{\ast }\right))\le ad(x^{\ast },y^{\ast })+b\delta (x^{\ast },T\left(x^{\ast }\right))+c\delta (y^{\ast },T\left(y^{\ast }\right)).$$

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.

Example 2.1 Let $X:=\{(0,0),(0,1),(1,0),(1,1)\}$ and $T:X\to P_{\mathrm{cl}}(X)$ be given by

$$T(x)=\{\begin{array}{cc}\{(0,0)\},\hfill & x=(0,0),\hfill \\ \{(0,0)\},\hfill & x=(0,1),\hfill \\ \{(0,0),(0,1)\},\hfill & x=(1,0),\hfill \\ \{(0,0),(0,1)\},\hfill & x=(1,1).\hfill \end{array}$$

(2.7)

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 $Fix(T)=SFix(T)\ne \mathrm{\varnothing }$.

Remark 2.3 If, in the above remark, we additionally suppose that

$$x^{\ast },y^{\ast }\in Fix(T)\Rightarrow (x^{\ast },y^{\ast })\in E(G),$$

then $Fix(T)=SFix(T)=\{x^{\ast }\}$.

The next result presents a strict fixed point theorem where the operator T satisfies a $(\delta ,\phi )$-G-contractive condition on $E(G)$.

Theorem 2.2 Let $(X,d)$ be a complete metric space and let G be a directed graph such that the triple $(X,d,G)$ satisfies the following property:

$$\begin{array}{rl}(P)& \begin{array}{l}\mathit{\text{for any sequence}}{(x_n)}_{n\in \mathbb{N}}\subset X\mathit{\text{with}}{x}_n\to x\mathit{\text{as}}n\to \mathrm{\infty },\\ \mathit{\text{there exists a subsequence}}{(x_{k_n})}_{n\in \mathbb{N}}\mathit{\text{of}}{(x_n)}_{n\in \mathbb{N}}\mathit{\text{such that}}(x_{k_n},x)\in E(G).\end{array}\end{array}$$

Let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that the following assertions hold:

• (i) T is a $(\delta ,\phi )$-G-contraction.

• (ii) For each $x\in X$, the set

  $${\tilde{X}}_T:=\{y\in T(x):(x,y)\in E(G)\mathit{\text{and}}\delta (x,T(x))\le qd(x,y)\mathit{\text{for some}}q\in ]1,\frac{1-a-c}{b}\left[\right\}$$

is nonempty.

Then we have:

• (a) $Fix(T)=SFix(T)\ne \mathrm{\varnothing }$;

• (b) If, in addition, the following implication holds:

  $$x^{\ast },y^{\ast }\in Fix(T)\Rightarrow (x^{\ast },y^{\ast })\in E(G),$$

then $Fix(T)=SFix(T)=\{x^{\ast }\}$.

Proof (a) Let $x_0\in X$. Then, since ${\tilde{X}}_T(x_0)$ is nonempty, there exist $x_1\in T(x_0)$ and $q\in ]1,\frac{1-a-c}{b}[$ such that $(x_0,x_1)\in E(G)$ and

$$\delta (x_0,T(x_0))\le qd(x_0,x_1).$$

By (i) we have that

$$\delta (x_1,T(x_1))\le \delta (T(x_0),T(x_1))\le \phi (d(x_0,x_1)).$$

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 $(x_1,x_2)\in E(G)$.

We have

$$d(x_1,x_2)\le \delta (x_1,T(x_1))\le \delta (T(x_0),T(x_1))\le \phi (d(x_0,x_1)).$$

(2.8)

On the other hand, by (i) we have that

$$\delta (x_2,T(x_2))\le \delta (T(x_1),T(x_2))\le \phi (d(x_1,x_2))\le {\phi }^2(d(x_0,x_1)).$$

By the same procedure, for $x_2\in X$ there exists $x_3\in T(x_2)$ such that $\delta (x_2,T(x_2))\le qd(x_2,x_3)$ and $(x_2,x_3)\in E(G)$. Thus

$$d(x_2,x_3)\le \delta (x_2,T(x_2))\le {\phi }^2(d(x_0,x_1)).$$

(2.9)

We have

$$\delta (x_3,T(x_3))\le \delta (T(x_2),T(x_3))\le \phi (d(x_2,x_3))\le {\phi }^3(d(x_0,x_1)).$$

By these procedures, we obtain a sequence ${(x_n)}_{n\in \mathbb{N}}$ with the following properties:

• (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 $n\in \mathbb{N}$.

We shall prove now that $x^{\ast }\in SFix(T)$. We have

$$\begin{array}{rcl}\delta (x^{\ast },T\left(x^{\ast }\right))& \le & d(x^{\ast },x_{k_{n+1}})+\delta (x_{k_{n+1}},T\left(x^{\ast }\right))\le d(x^{\ast },x_{k_{n+1}})+\delta (T(x_{k_n}),T\left(x^{\ast }\right))\\ \le & d(x^{\ast },x_{k_{n+1}})+\phi \left(d(x_{k_n},x^{\ast })\right).\end{array}$$

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 $Fix(T)\subset SFix(T)$.

Let $x^{\ast }\in Fix(T)$. Because $\mathrm{\Delta }\subset E(G)$, we have that $(x^{\ast },x^{\ast })\in E(G)$. Using (i) with $x=y=x^{\ast }$, we obtain

$$\delta \left(T\left(x^{\ast }\right)\right)\le \phi \left(d(x^{\ast },x^{\ast })\right)=0.$$

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

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

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

• $(x^{\ast },y^{\ast })\in E(G)$.

Using (i) we obtain

$$d(x^{\ast },y^{\ast })=\delta (T\left(x^{\ast }\right),T\left(y^{\ast }\right))\le \phi \left(d(x^{\ast },y^{\ast })\right)<d(x^{\ast },y^{\ast }).$$

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 $E(G)\cap Graph(T)$.

Theorem 2.3 Let $(X,d)$ be a complete metric space, let G be a directed graph, and let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that $f:X\to {\mathbb{R}}_{+}$ defined $f(x):=\delta (x,T(x))$ is a lower semicontinuous mapping. Suppose that the following assertions hold:

• (i) There exist $a,b\in {\mathbb{R}}_{+}$, with $b\ne 0$ and $a+b<1$, such that

  $$\delta (y,T(y))\le ad(x,y)+b\delta (x,T(x))\mathit{\text{for all}}(x,y)\in E(G)\cap Graph(T).$$

• (ii) For each $x\in X$, the set

  $${\tilde{X}}_T(x):=\{y\in T(x):(x,y)\in E(G)\mathit{\text{and}}\delta (x,T(x))\le qd(x,y)\mathit{\text{for some}}q\in ]1,\frac{1-a}{b}\left[\right\}$$

  is nonempty.

Then $Fix(T)=SFix(T)\ne \mathrm{\varnothing }$.

Proof Let $x_0\in X$. Then, since ${\tilde{X}}_T(x_0)$ is nonempty, there exist $x_1\in T(x_0)$ and $1<q<\frac{1-a}{b}$ such that

$$\delta (x_0,T(x_0))\le qd(x_0,x_1)$$

and $(x_0,x_1)\in E(G)$. Since $x_1\in T(x_0)$, we get that $(x_0,x_1)\in E(G)\cap Graph(T)$.

By (i), taking $y=x_1$ and $x=x_0$, we have that

$$\begin{array}{rcl}\delta (x_1,T(x_1))& \le & ad(x_0,x_1)+b\delta (x_0,T(x_0))\\ \le & ad(x_0,x_1)+bqd(x_0,x_1).\end{array}$$

Hence,

$$\delta (x_1,T(x_1))\le (a+bq)d(x_0,x_1).$$

(2.10)

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 $(x_1,x_2)\in E(G)\cap Graph(T)$.

Then

$$d(x_1,x_2)\le \delta (x_1,T(x_1))\le (a+bq)d(x_0,x_1).$$

(2.11)

By (i), taking $y=x_2$ and $x=x_1$, we have that

$$\begin{array}{rcl}\delta (x_2,T(x_2))& \le & ad(x_1,x_2)+b\delta (x_1,T(x_1))\\ \le & ad(x_1,x_2)+bqd(x_1,x_2)=(a+bq)d(x_1,x_2)\le {(a+bq)}^{2}d(x_0,x_1).\end{array}$$

By these procedures, we obtain a sequence ${(x_n)}_{n\in \mathbb{N}}$ with the following properties:

• (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 $n\in \mathbb{N}$.

From (2) we obtain that the sequence ${(x_n)}_{n\in \mathbb{N}}$ is Cauchy. Since the metric space X is complete, we have that the sequence is convergent, i.e., $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$. Now, by the lower semicontinuity of the function f, we have

$$0\le f\left(x^{\ast }\right)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}f(x_n)=0.$$

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 $(x^{\ast },x^{\ast })\in E(G)\cap Graph(T)$.

Using (i) with $x=y=x^{\ast }$, we obtain

$$\delta \left(T\left(x^{\ast }\right)\right)=\delta (x^{\ast },T\left(x^{\ast }\right))\le ad(x^{\ast },x^{\ast })+b\delta (x^{\ast },T\left(x^{\ast }\right)).$$

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 Well-posedness of the fixed point problem

In this section we present some well-posedness results for the fixed point problem. We consider both the well-posedness and the well-posedness in the generalized sense for a multivalued operator T.

We begin by recalling the definition of these notions from [14] and [15].

Definition 3.1 Let $(X,d)$ be a metric space and let $T:X\to P(X)$ be a multivalued operator. By definition, the fixed point problem is well posed for T with respect to H if:

• (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 $n\to \mathrm{\infty }$.

Definition 3.2 Let $(X,d)$ be a metric space and let $T:X\to P(X)$ be a multivalued operator. By definition, the fixed point problem is well posed in the generalized sense for T with respect to H if:

• (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 well-posedness of the fixed point problem for the operator T, where T is a Reich-type δ-contraction.

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).

Let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that

• (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.

Proof From (i) and (ii) we obtain that $SFix(T)=\{x^{\ast }\}$. Let ${(x_n)}_{n\in \mathbb{N}}\subset X$ be a sequence which satisfies (iii). It is obvious that $H(x_n,T(x_n))=\delta (x_n,T(x_n))$,

$$\begin{array}{rcl}d(x_n,x^{\ast })& \le & \delta (x_n,T\left(x^{\ast }\right))\le \delta (x_n,T(x_n))+\delta (T(x_n),T\left(x^{\ast }\right))\\ \le & \delta (x_n,T(x_n))+ad(x_n,x^{\ast })+b\delta (x_n,T(x_n))+c\delta (x^{\ast },T\left(x^{\ast }\right)).\end{array}$$

Thus

$$d(x_n,x^{\ast })\le \frac{1+b}{1-a}\delta (x_n,T(x_n))\to 0\text{as }n\to \mathrm{\infty }.$$

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 well-posedness 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).

Let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that

• (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 $(x_{k_n},x^{\ast })\in E(G)$.

We have $H(x_{k_n},T(x_{k_n}))=\delta (x_{k_n},T(x_{k_n}))$,

$$\begin{array}{rcl}d(x_{k_n},x^{\ast })& \le & \delta (x_{k_n},T\left(x^{\ast }\right))\le \delta (x_{k_n},T(x_{k_n}))+\delta (T(x_{k_n}),T\left(x^{\ast }\right))\\ \le & \delta (x_{k_n},T(x_{k_n}))+ad(x_{k_n},x^{\ast })+b\delta (x_{k_n},T(x_{k_n}))+c\delta (x^{\ast },T\left(x^{\ast }\right)).\end{array}$$

Thus

$$d(x_{k_n},x^{\ast })\le \frac{1+b}{1-a}\delta (x_{k_n},T(x_{k_n}))\to 0\text{as }n\to \mathrm{\infty }.$$

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 φ-contraction 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).

Let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that

• (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.

Proof From (i) and (ii) we obtain that $SFix(T)=\{x^{\ast }\}$. Let ${(x_n)}_{n\in \mathbb{N}}$, $x_n\in X$ be a sequence which satisfies (iv). It is obvious that $H(x_n,T(x_n))=\delta (x_n,T(x_n))$,

$$\begin{array}{rcl}d(x_n,x^{\ast })& \le & \delta (x_n,T\left(x^{\ast }\right))\le \delta (x_n,T(x_n))+\delta (T(x_n),T\left(x^{\ast }\right))\\ \le & \delta (x_n,T(x_n))+\phi \left(d(x_n,x^{\ast })\right).\end{array}$$

Thus

$$d(x_n,x^{\ast })-\phi \left(d(x_n,x^{\ast })\right)\le \delta (x_n,T(x_n))\to 0\text{as }n\to \mathrm{\infty }.$$

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 well-posedness (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).

Let $T:X\to P_{\mathrm{b}}(X)$ be a multivalued operator. Suppose that

• (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 $(x_{k_n},x^{\ast })\in E(G)$.

We have $H(x_{k_n},T(x_{k_n}))=\delta (x_{k_n},T(x_{k_n}))$,

$$\begin{array}{rcl}d(x_{k_n},x^{\ast })& \le & \delta (x_{k_n},T\left(x^{\ast }\right))\le \delta (x_{k_n},T(x_{k_n}))+\delta (T(x_{k_n}),T\left(x^{\ast }\right))\\ \le & \delta (x_{k_n},T(x_{k_n}))+\phi \left(d(x_{k_n},x^{\ast })\right).\end{array}$$

Thus

$$d(x_{k_n},x^{\ast })-\phi \left(d(x_{k_n},x^{\ast })\right)\le \delta (x_{k_n},T(x_{k_n}))\to 0\text{as }n\to \mathrm{\infty }.$$

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.