A new approach to $(\alpha ,\psi )$-contractive nonself multivalued mappings

Abstract

In this paper, we introduce the notions of α-admissible and α-ψ-contractive type condition for nonself multivalued mappings. We establish fixed point theorems using these new notions along with a new condition. Moreover, we have constructed examples to show that our new condition is different from the corresponding existing conditions in the literature.

MSC: 47H10, 54H25.

1 Introduction and preliminaries

In the last decades, metric fixed point theory has been appreciated by a number of authors who have extended the celebrated Banach fixed point theorem for various contractive mapping in the context of different abstract spaces; see, for example, [1–32]. Among them, we mention the interesting fixed point theorems of Samet et al. [20]. In this paper [20], the authors introduced the notions of α-ψ-contractive mappings and investigated the existence and uniqueness of a fixed point for such mappings. Further, they showed that several well-known fixed point theorems can be derived from the fixed point theorem of α-ψ-contractive mappings. Following this paper, Karapınar and Samet [21] generalized the notion α-ψ-contractive mappings and obtained a fixed point for this generalized version. On the other hand, Asl et al. [22] characterized the notions of α-ψ-contractive mapping and α-admissible mappings with the notions of ${\alpha }_{\ast }$-ψ-contractive and ${\alpha }_{\ast }$-admissible mappings to investigate the existence of a fixed point for a multivalued function. Afterward, Ali and Kamran [23] generalized the notion of ${\alpha }_{\ast }$-ψ-contractive mappings and obtained further fixed point results for multivalued mappings. Some results in this direction in the context of various abstract spaces were also given by the authors [24–28, 33–36]. The purpose of this paper is to prove fixed point theorems for nonself multivalued $(\alpha ,\psi )$-contractive type mappings using a new condition.

Let Ψ be the family of functions $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$, known in the literature as Bianchini-Grandolfi gauge functions (see, e.g., [30–32]), satisfying the following conditions:

(${\psi }_1$) ψ is nondecreasing;

(${\psi }_2$) ${\sum }_{n=1}^{+\mathrm{\infty }}{\psi }^n(t)<\mathrm{\infty }$ for all $t>0$, where ${\psi }^n$ is the n th iterate of ψ.

Notice that such functions are also known as $(c)$-comparison functions in some sources (see, e.g., [29]).

It is easily proved that if $\psi \in \mathrm{\Psi }$, then $\psi (t)<t$ for any $t>0$ and $\psi (0)=0$ for $t=0$ (see, e.g., [20, 29]). Let $(X,d)$ be a metric space. A mapping $G:X\to X$ is called α-ψ-contractive type if there exist two functions $\alpha :X\times X\to [0,\mathrm{\infty })$ and $\psi \in \mathrm{\Psi }$ such that

$$\alpha (x,y)d(Gx,Gy)\le \psi (d(x,y))$$

for each $x,y\in X$. A mapping $G:X\to X$ is called α-admissible [20] if

$$\alpha (x,y)\ge 1\Rightarrow \alpha (Gx,Gy)\ge 1.$$

We denote by $N(X)$ the space of all nonempty subsets of X and by $\mathit{CL}(X)$ the space of all nonempty closed subsets of X. For $A\in N(X)$ and $x\in X$, $d(x,A)=inf\{d(x,a):a\in A\}$. For every $A,B\in \mathit{CL}(X)$, let

$$H(A,B)=\{\begin{array}{cc}max\{{sup}_{x\in A}d(x,B),{sup}_{y\in B}d(y,A)\}\hfill & \text{if the maximum exists};\hfill \\ \mathrm{\infty }\hfill & \text{otherwise}.\hfill \end{array}$$

Such a map H is called a generalized Hausdorff metric induced by d. We use the following lemma in our results.

Lemma 1.1 [23]

Let $(X,d)$ be a metric space and $B\in \mathit{CL}(X)$. Then, for each $x\in X$ with $d(x,B)>0$ and $q>1$, there exists an element $b\in B$ such that

$$d(x,b)<qd(x,B).$$

(1.1)

Let $(X,⪯,d)$ be an ordered metric space and $A,B\subseteq X$. We say that $A{\prec }_{r}B$ if for each $a\in A$ and $b\in B$, we have $a⪯b$.

2 Main results

We begin this section with the following definition which is a modification of the notion of α-admissible.

Definition 2.1 Let $(X,d)$ be a metric space and let D be a nonempty subset of X. A mapping $G:D\to \mathit{CL}(X)$ is called α-admissible if there exists a mapping $\alpha :D\times D\to [0,\mathrm{\infty })$ such that

$$\alpha (x,y)\ge 1\Rightarrow \alpha (u,v)\ge 1$$

for each $u\in Gx\cap D$ and $v\in Gy\cap D$.

Definition 2.2 Let $(X,d)$ be a metric space and let D be a nonempty subset of X. We say that $G:D\to \mathit{CL}(X)$ is an $(\alpha ,\psi )$-contractive type mapping on D if there exist $\alpha :D\times D\to [0,\mathrm{\infty })$ and $\psi \in \mathrm{\Psi }$ satisfying the following conditions:

• (i) $Gx\cap D\ne \mathrm{\varnothing }$ for all $x\in D$,

• (ii) for each $x,y\in D$, we have

  $$\alpha (x,y)H(Gx\cap D,Gy\cap D)\le \psi (M(x,y)),$$

  (2.1)

where $M(x,y)=max\{d(x,y),\frac{d(x,Gx)+d(y,Gy)}{2},\frac{d(x,Gy)+d(y,Gx)}{2}\}$.

Note that if $\psi \in \mathrm{\Psi }$ in the above definition is a strictly increasing function, then $G:D\to \mathit{CL}(X)$ is said to be a strictly $(\alpha ,\psi )$-contractive type mapping on D.

Theorem 2.3 Let $(X,d)$ be a metric space, let D be a nonempty subset of X which is complete with respect to the metric induced by d, and let G be a strictly $(\alpha ,\psi )$-contractive type mapping on D. Assume that the following conditions hold:

• (i) G is an α-admissible map;

• (ii) there exist $x_0\in D$ and $x_1\in G{x}_0\cap D$ such that $\alpha (x_0,x_1)\ge 1$;

• (iii) G is continuous.

Then G has a fixed point.

Proof By hypothesis, there exist $x_0\in D$ and $x_1\in G{x}_0\cap D$ such that $\alpha (x_0,x_1)\ge 1$. If $x_0=x_1$, then we have nothing to prove. Let $x_0\ne x_1$. If $x_1\in G{x}_1\cap D$, then $x_1$ is a fixed point. Let $x_1\notin G{x}_1\cap D$. From (2.1) we have

$$\begin{array}{rcl}0& <& \alpha (x_0,x_1)H(G{x}_0\cap D,G{x}_1\cap D)\\ \le & \psi (max\{d(x_0,x_1),\frac{d(x_0,G{x}_0)+d(x_1,G{x}_1)}{2},\frac{d(x_0,G{x}_1)+d(x_1,G{x}_0)}{2}\})\\ \le & \psi (max\{d(x_0,x_1),d(x_1,G{x}_1)\})\end{array}$$

(2.2)

since $\frac{d(x_0,G{x}_1)}{2}\le max\{d(x_0,x_1),d(x_1,G{x}_1)\}$ and $\frac{d(x_0,G{x}_0)+d(x_1,G{x}_1)}{2}\le max\{d(x_0,x_1),d(x_1,G{x}_1)\}$. Assume that $max\{d(x_0,x_1),d(x_1,G{x}_1)\}=d(x_1,G{x}_1)$. Then from (2.2) we have

$$\begin{array}{rcl}0& <& d(x_1,G{x}_1\cap D)\le \alpha (x_0,x_1)H(G{x}_0\cap D,G{x}_1\cap D)\\ \le & \psi (d(x_1,G{x}_1))\\ <& d(x_1,G{x}_1),\end{array}$$

(2.3)

a contradiction to our assumption. Thus $max\{d(x_0,x_1),d(x_1,G{x}_1)\}=d(x_0,x_1)$. Then from (2.2) we have

$$0<d(x_1,G{x}_1\cap D)\le \psi (d(x_0,x_1)).$$

(2.4)

For $q>1$ by Lemma 1.1, there exists $x_2\in G{x}_1\cap D$ such that

$$0<d(x_1,x_2)<qd(x_1,G{x}_1\cap D)\le q\psi (d(x_0,x_1)).$$

(2.5)

Applying ψ in (2.5), we have

$$0<\psi (d(x_1,x_2))<\psi \left(q\psi (d(x_0,x_1))\right).$$

(2.6)

Put $q_1=\frac{\psi (q\psi (d(x_0,x_1)))}{\psi (d(x_1,x_2))}$. Then $q_1>1$. Since G is an α-admissible mapping, $\alpha (x_1,x_2)\ge 1$. If $x_2\in G{x}_2\cap D$, then $x_2$ is a fixed point. Let $x_2\notin G{x}_2\cap D$. From (2.1) we have

$$\begin{array}{rcl}0& <& \alpha (x_1,x_2)H(G{x}_1\cap D,G{x}_2\cap D)\\ \le & \psi (max\{d(x_1,x_2),\frac{d(x_1,G{x}_1)+d(x_2,G{x}_2)}{2},\frac{d(x_1,G{x}_2)+d(x_2,G{x}_1)}{2}\})\\ \le & \psi (max\{d(x_1,x_2),d(x_2,G{x}_2)\})\end{array}$$

(2.7)

since $\frac{d(x_1,G{x}_2)}{2}\le max\{d(x_1,x_2),d(x_2,G{x}_2)\}$ and $\frac{d(x_1,G{x}_1)+d(x_2,G{x}_2)}{2}\le max\{d(x_1,x_2),d(x_2,G{x}_2)\}$. Assume that $max\{d(x_1,x_2),d(x_2,G{x}_2)\}=d(x_2,G{x}_2)$. Then from (2.7) we have

$$\begin{array}{rcl}0& <& d(x_2,G{x}_2\cap D)\le \alpha (x_1,x_2)H(G{x}_1\cap D,G{x}_2\cap D)\\ \le & \psi (d(x_2,G{x}_2))\\ <& d(x_2,G{x}_2),\end{array}$$

(2.8)

a contradiction to our assumption. Thus $max\{d(x_1,x_2),d(x_2,G{x}_2)\}=d(x_1,x_2)$. Then from (2.7) we have

$$0<d(x_2,G{x}_2\cap D)\le \psi (d(x_1,x_2)).$$

(2.9)

For $q_1>1$ by Lemma 1.1, there exists $x_3\in G{x}_2\cap D$ such that

$$0<d(x_2,x_3)<q_{1}d(x_2,G{x}_2\cap D)\le q_1\psi (d(x_1,x_2))=\psi \left(q\psi (d(x_0,x_1))\right).$$

(2.10)

Applying ψ in (2.10), we have

$$0<\psi (d(x_2,x_3))<{\psi }^2\left(q\psi (d(x_0,x_1))\right).$$

(2.11)

Put $q_2=\frac{{\psi }^2(q\psi (d(x_0,x_1)))}{\psi (d(x_2,x_3))}$. Then $q_2>1$. Since G is an α-admissible mapping, $\alpha (x_2,x_3)\ge 1$. If $x_3\in G{x}_3\cap D$, then $x_3$ is a fixed point. Let $x_3\notin G{x}_3\cap D$. From (2.1) we have

$$\begin{array}{rcl}0& <& \alpha (x_2,x_3)H(G{x}_2\cap D,G{x}_3\cap D)\\ \le & \psi (max\{d(x_2,x_3),\frac{d(x_2,G{x}_2)+d(x_3,G{x}_3)}{2},\frac{d(x_2,G{x}_3)+d(x_3,G{x}_2)}{2}\})\\ \le & \psi (max\{d(x_2,x_3),d(x_3,G{x}_3)\})\end{array}$$

(2.12)

since $\frac{d(x_2,G{x}_3)}{2}\le max\{d(x_2,x_3),d(x_3,G{x}_3)\}$ and $\frac{d(x_2,G{x}_2)+d(x_3,G{x}_3)}{2}\le max\{d(x_2,x_3),d(x_3,G{x}_3)\}$. Assume that $max\{d(x_2,x_3),d(x_3,G{x}_3)\}=d(x_3,G{x}_3)$. Then from (2.12) we have

$$\begin{array}{rcl}0& <& d(x_3,G{x}_3\cap D)\le \alpha (x_2,x_3)H(G{x}_2\cap D,G{x}_3\cap D)\\ \le & \psi (d(x_3,G{x}_3))\\ <& d(x_3,G{x}_3),\end{array}$$

(2.13)

a contradiction to our assumption. Thus $max\{d(x_2,x_3),d(x_3,G{x}_3)\}=d(x_2,x_3)$. Then from (2.12) we have

$$0<d(x_3,G{x}_3\cap D)\le \psi (d(x_2,x_3)).$$

(2.14)

For $q_2>1$ by Lemma 1.1, there exists $x_4\in G{x}_3\cap D$ such that

$$0<d(x_3,x_4)<q_{2}d(x_3,G{x}_3\cap D)\le q_2\psi (d(x_2,x_3))={\psi }^2\left(q\psi (d(x_0,x_1))\right).$$

(2.15)

Applying ψ in (2.15), we have

$$0<\psi (d(x_3,x_4))<{\psi }^3\left(q\psi (d(x_0,x_1))\right).$$

(2.16)

Continuing the same process, we get a sequence $\{x_n\}$ in D such that $x_{n+1}\in G{x}_n\cap D$, $x_{n+1}\ne x_n$, $\alpha (x_n,x_{n+1})\ge 1$, and

$$d(x_{n+1},x_{n+2})<{\psi }^n\left(q\psi (d(x_0,x_1))\right)\text{for each }n\in \mathbb{N}\cup \{0\}.$$

(2.17)

For $m,n\in \mathbb{N}$, we have

$$d(x_n,x_{n+m})\le \sum _{i=n}^{n+m-1}d(x_i,x_{i+1})<\sum _{i=n}^{n+m-1}{\psi }^{i-1}(d(x_0,x_1)).$$

Since $\psi \in \mathrm{\Psi }$, it follows that $\{x_n\}$ is a Cauchy sequence in D. Since D is complete, there exists $x^{\ast }\in D$ such that $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$. By the continuity of G, we have

$$d(x^{\ast },G{x}^{\ast })\le \underset{n\to \mathrm{\infty }}{lim}H(G{x}_n,G{x}^{\ast })=0.$$

 □

Theorem 2.4 Let $(X,d)$ be a metric space, D be a nonempty subset of X which is complete with respect to the metric induced by d, and let G be a strictly $(\alpha ,\psi )$-contractive type mapping on D. Assume that the following conditions hold:

• (i) G is an α-admissible map;

• (ii) there exist $x_0\in D$ and $x_1\in G{x}_0\cap D$ such that $\alpha (x_0,x_1)\ge 1$;

• (iii) either

  • (a) for any sequence $\{x_n\}$ in D such that $x_n\to x$ as $n\to \mathrm{\infty }$ and $\alpha (x_n,x_{n+1})\ge 1$ for each $n\in \mathbb{N}\cup \{0\}$, ${lim}_{n\to \mathrm{\infty }}\alpha (x_n,x)\ge 1$,

    or

  • (b) for any sequence $\{x_n\}$ in D such that $x_n\to x$ as $n\to \mathrm{\infty }$ and $\alpha (x_n,x_{n+1})\ge 1$ for each $n\in \mathbb{N}\cup \{0\}$, $\alpha (x_n,x)\ge 1$ for each $n\in \mathbb{N}\cup \{0\}$.

Then G has a fixed point.

Proof Following the proof of Theorem 2.3, there exists a Cauchy sequence $\{x_n\}$ in D with $x_n\to x^{\ast }$ as $n\to \mathrm{\infty }$ and $\alpha (x_n,x_{n+1})\ge 1$ for each $n\in \mathbb{N}\cup \{0\}$. Suppose that $d(x^{\ast },G{x}^{\ast })\ne 0$. From (2.1) we have

$$\begin{array}{rcl}\alpha (x_n,x^{\ast })d(x_{n+1},G{x}^{\ast }\cap D)& \le & \alpha (x_n,x^{\ast })H(G{x}_n\cap D,G{x}^{\ast }\cap D)\\ \le & \psi (max\{d(x_n,x^{\ast }),\frac{d(x_n,G{x}_n)+d(x^{\ast },G{x}^{\ast })}{2},\\ \frac{d(x_n,G{x}^{\ast })+d(x^{\ast },G{x}_n)}{2}\left\}\right)\\ <& max\{d(x_n,x^{\ast }),\frac{d(x_n,G{x}_n)+d(x^{\ast },G{x}^{\ast })}{2},\\ \frac{d(x_n,G{x}^{\ast })+d(x^{\ast },G{x}_n)}{2}\}.\end{array}$$

(2.18)

Letting $n\to \mathrm{\infty }$ in (2.18), we have

$$\underset{n\to \mathrm{\infty }}{lim}\alpha (x_n,x^{\ast })d(x^{\ast },G{x}^{\ast }\cap D)\le \frac{d(x^{\ast },G{x}^{\ast })}{2}.$$

(2.19)

Since ${lim}_{n\to \mathrm{\infty }}\alpha (x_n,x^{\ast })\ge 1$, by condition (iii)(a), we have

$$d(x^{\ast },G{x}^{\ast }\cap D)\le \underset{n\to \mathrm{\infty }}{lim}\alpha (x_n,x^{\ast })d(x^{\ast },G{x}^{\ast }\cap D)\le \frac{d(x^{\ast },G{x}^{\ast })}{2}.$$

(2.20)

Further, it is clear that $d(x^{\ast },G{x}^{\ast })\le d(x^{\ast },G{x}^{\ast }\cap D)$. Then from (2.20) we have

$$d(x^{\ast },G{x}^{\ast })\le \frac{d(x^{\ast },G{x}^{\ast })}{2},$$

which is impossible. Thus $d(x^{\ast },G{x}^{\ast })=0$. If we use (iii)(b), then from (2.1) we have

$$\begin{array}{rcl}d(x_{n+1},G{x}^{\ast }\cap D)& \le & \alpha (x_n,x^{\ast })H(G{x}_n\cap D,G{x}^{\ast }\cap D)\\ \le & \psi (max\{d(x_n,x^{\ast }),\frac{d(x_n,G{x}_n)+d(x^{\ast },G{x}^{\ast })}{2},\\ \frac{d(x_n,G{x}^{\ast })+d(x^{\ast },G{x}_n)}{2}\left\}\right)\\ <& max\{d(x_n,x^{\ast }),\frac{d(x_n,G{x}_n)+d(x^{\ast },G{x}^{\ast })}{2},\\ \frac{d(x_n,G{x}^{\ast })+d(x^{\ast },G{x}_n)}{2}\}.\end{array}$$

(2.21)

Letting $n\to \mathrm{\infty }$ in (2.21), we have

$$d(x^{\ast },G{x}^{\ast })\le d(x^{\ast },G{x}^{\ast }\cap D)\le \frac{d(x^{\ast },G{x}^{\ast })}{2},$$

which is impossible. Thus $d(x^{\ast },G{x}^{\ast })=0$. □

Example 2.5 Let $X=(-\mathrm{\infty },-8)\cup [0,\mathrm{\infty })$ be endowed with the usual metric d, and let $D=[0,\mathrm{\infty })$. Define $G:D\to \mathit{CL}(X)$ by

$$Gx=\{\begin{array}{cc}[0,\frac{x}{4}]\hfill & \text{if }0\le x<4,\hfill \\ \{0\}\hfill & \text{if }x=4,\hfill \\ (-\mathrm{\infty },-3x]\cup [x,x^2]\hfill & \text{if }x>4\hfill \end{array}$$

and $\alpha :D\times D\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}1\hfill & \text{if }x,y\in [0,4],\hfill \\ 0\hfill & \text{otherwise}.\hfill \end{array}$$

Clearly, $Gx\cap D\ne \mathrm{\varnothing }$ for each $x\in D$. Let $\psi (t)=\frac{t}{2}$ for each $t\ge 0$. To see that G is a strictly $(\alpha ,\psi )$-contractive type mapping on D, we consider the following cases.

Case (i) When $x,y\in [0,4)$, we have

$$\alpha (x,y)H(Gx\cap D,Gy\cap D)=|\frac{x}{4}-\frac{y}{4}|\le \frac{|x-y|}{2}=\psi (d(x,y))\le \psi (M(x,y)).$$

Case (ii) When $x\in [0,4)$ and $y=4$, we have

$$\alpha (x,y)H(Gx\cap D,Gy\cap D)=\left|\frac{x}{4}\right|\le \psi \left(\frac{d(x,Gx)+d(y,Gy)}{2}\right)\le \psi (M(x,y)).$$

Case (iii) Otherwise, we have

$$\alpha (x,y)H(Gx\cap D,Gy\cap D)=0\le \psi (M(x,y)),$$

where $M(x,y)=max\{d(x,y),\frac{d(x,Gx)+d(y,Gy)}{2},\frac{d(x,Gy)+d(y,Gx)}{2}\}$.

Thus, G is a strictly $(\alpha ,\psi )$-contractive type mapping on D. For $\alpha (x,y)\ge 1$, we have $x,y\in [0,4]$, then $Gx\cap D,Gy\cap D\subseteq [0,1]$, thus $\alpha (u,v)=1$ for each $u\in Gx\cap D$ and $v\in Gy\cap D$. Further, for any sequence $\{x_n\}$ in D such that $x_n\to x$ as $n\to \mathrm{\infty }$ and $\alpha (x_n,x_{n+1})=1$ for each $n\in \mathbb{N}\cup \{0\}$, ${lim}_{n\to \mathrm{\infty }}\alpha (x_n,x)=1$. Therefore, all the conditions of Theorem 2.4 hold and G has a fixed point.

Corollary 2.6 Let $(X,⪯,d)$ be an ordered metric space, let $(D,⪯)$ be a nonempty subset of X which is complete with respect to the metric induced by d. Let $G:D\to \mathit{CL}(X)$ be a mapping such that $Gx\cap D\ne \mathrm{\varnothing }$ for each $x\in D$ and for each $x,y\in D$ with $x⪯y$, we have

$$H(Gx\cap D,Gy\cap D)\le \psi (M(x,y)),$$

where $M(x,y)=max\{d(x,y),\frac{d(x,Gx)+d(y,Gy)}{2},\frac{d(x,Gy)+d(y,Gx)}{2}\}$ and ψ is an increasing function in  Ψ. Also, assume that the following conditions hold:

• (i) there exist $x_0\in D$ and $x_1\in G{x}_0\cap D$ such that $x_0⪯x_1$;

• (ii) if $x⪯y$ then $Gx\cap D{\prec }_{r}Gy\cap D$;

• (iii) either

  • (a) G is continuous,

    or

  • (b) for any sequence $\{x_n\}$ in D such that $x_n\to x$ as $n\to \mathrm{\infty }$ and $x_n⪯x_{n+1}$ for each $n\in \mathbb{N}\cup \{0\}$, $x_n⪯x$ as $n\to \mathrm{\infty }$,

    or

  • (c) for any sequence $\{x_n\}$ in D such that $x_n\to x$ as $n\to \mathrm{\infty }$ and $x_n⪯x_{n+1}$ for each $n\in \mathbb{N}\cup \{0\}$, $x_n⪯x$ for each $n\in \mathbb{N}\cup \{0\}$.

Then G has a fixed point.

Proof Define $\alpha :D\times D\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}1\hfill & \text{if }x⪯y,\hfill \\ 0\hfill & \text{otherwise}.\hfill \end{array}$$

By using condition (i) and the definition of α, we have $\alpha (x_0,x_1)=1$. Also, from condition (ii), we have that $x⪯y$ implies $Gx\cap D{\prec }_{r}Gy\cap D$; by using the definitions of α and ${\prec }_r$, we have that $\alpha (x,y)=1$ implies $\alpha (u,v)=1$ for each $u\in Gx\cap D$ and $v\in Gy\cap D$. Moreover, it is easy to check that G is a strictly $(\alpha ,\psi )$-contractive type mapping on D. Therefore, all the conditions of Theorem 2.3 (or Theorem 2.4 for (iii)(b), (iii)(c)) hold, hence G has a fixed point. □

Remark 2.7 Condition (a), in the statement of Theorem 2.4, was introduced by Samet et al. [20]. In Theorem 2.4 we introduce a new condition (b). The following examples show that (a) and (b) are independent conditions.

Example 2.8 Let $X=\{\frac{1}{n}:n\in \mathbb{N}\}\cup \{0\}$. Consider $x_n=\frac{1}{n+1}$ for each $n\in \mathbb{N}\cup \{0\}$, then $x_n\to 0=x^{\ast }$ as $n\to \mathrm{\infty }$. Define $\alpha :X\times X\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}max\{\frac{1}{x},\frac{1}{y}\}\hfill & \text{if }x\ne 0\text{ and }y\ne 0,\hfill \\ \frac{1}{x+y}\hfill & \text{if either }x=0\text{ or }y=0,\hfill \\ 1\hfill & \text{if }x=y=0.\hfill \end{array}$$

Now, we have $\alpha (x_n,x_{n+1})=\alpha (\frac{1}{n+1},\frac{1}{n+2})=n+2>1$ for each $n\in \mathbb{N}\cup \{0\}$ and $\alpha (x_n,x^{\ast })=\alpha (\frac{1}{n+1},0)=n+1\ge 1$ for each $n\in \mathbb{N}\cup \{0\}$. Thus condition (a) holds but ${lim}_{n\to \mathrm{\infty }}\alpha (x_n,x^{\ast })={lim}_{n\to \mathrm{\infty }}(n+1)=\mathrm{\infty }$. Thus condition (b) does not hold.

Example 2.9 Let $X=\{\frac{1}{n}:n\in \mathbb{N}\}\cup \{0\}$. Consider $x_n=\frac{1}{n+1}$ for each $n\in \mathbb{N}\cup \{0\}$, then $x_n\to 0=x^{\ast }$ as $n\to \mathrm{\infty }$. Define $\alpha :X\times X\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}max\{\frac{1}{x},\frac{1}{y}\}\hfill & \text{if }x\ne 0\text{ and }y\ne 0,\hfill \\ \frac{1}{1+(x+y)/2}\hfill & \text{if either }x=0\text{ or }y=0,\hfill \\ 1\hfill & \text{if }x=y=0.\hfill \end{array}$$

Now, we have $\alpha (x_n,x_{n+1})=\alpha (\frac{1}{n+1},\frac{1}{n+2})=n+2>1$ for each $n\in \mathbb{N}\cup \{0\}$ and $\alpha (x_n,x^{\ast })=\alpha (\frac{1}{n+1},0)=\frac{2n+2}{2n+3}$. Then ${lim}_{n\to \mathrm{\infty }}\alpha (x_n,x^{\ast })={lim}_{n\to \mathrm{\infty }}\frac{2n+2}{2n+3}=1$. Thus condition (b) holds but for $n=0$, we have $\alpha (x_n,x^{\ast })=\frac{2}{3}$; for $n=1$, we have $\alpha (x_n,x^{\ast })=\frac{4}{5}$; for $n=2$, we have $\alpha (x_n,x^{\ast })=\frac{6}{7}$, which implies that $\alpha (x_n,x)\ngeqq 1$ for each $n\in \mathbb{N}\cup \{0\}$. Thus condition (a) does not hold.