Fixed point results for single and set-valued α-η-ψ-contractive mappings

Abstract

Samet et al. (Nonlinear Anal. 75:2154-2165, 2012) introduced α-ψ-contractive mappings and proved some fixed point results for these mappings. More recently Salimi et al. (Fixed Point Theory Appl. 2013:151, 2013) modified the notion of α-ψ-contractive mappings and established certain fixed point theorems. Here, we continue to utilize these modified notions for single-valued Geraghty and Meir-Keeler-type contractions, as well as multi-valued contractive mappings. Presented theorems provide main results of Hussain et al. (J. Inequal. Appl. 2013:114, 2013), Karapinar et al. (Fixed Point Theory Appl. 2013:34, 2013) and Asl et al. (Fixed Point Theory Appl. 2012:212, 2012) as corollaries. Moreover, some examples are given here to illustrate the usability of the obtained results.

MSC:46N40, 47H10, 54H25, 46T99.

Dedication

Dedicated to Professor Wataru Takahashi on the occasion of his seventieth birthday

1 Introduction and preliminaries

In metric fixed point theory, the contractive conditions on underlying functions play an important role for finding solution of fixed point problems. Banach contraction principle is a remarkable result in metric fixed point theory. Over the years, it has been generalized in different directions by several mathematicians (see [1–21]). In 2012, Samet et al. [15] introduced the concepts of α-ψ-contractive and α-admissible mappings and established various fixed point theorems for such mappings in complete metric spaces. Afterwards, Karapinar and Samet [13] generalized these notions to obtain fixed point results. More recently, Salimi et al. [14] modified the notions of α-ψ-contractive and α-admissible mappings and established fixed point theorems, which are proper generalizations of the recent results in [13, 15]. Here, we continue to utilize these modified notions for single-valued Geraghty and Meir-Keeler-type contractions, as well as multivalued contractive mappings. Presented theorems provide main results of Hussain et al. [9], Karapinar et al. [11] and Asl et al. [12] as corollaries. Moreover, some examples are given here to illustrate the usability of the obtained results.

Denote with Ψ the family of nondecreasing functions $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\psi }^n(t)<+\mathrm{\infty }$ for all $t>0$, where ${\psi }^n$ is the n th iterate of ψ.

The following lemma is obvious.

Lemma 1.1 If $\psi \in \mathrm{\Psi }$, then $\psi (t)<t$ for all $t>0$.

Samet et al. [15] defined the notion of α-admissible mappings as follows.

Definition 1.1 Let T be a self-mapping on X, and let $\alpha :X\times X\to [0,+\mathrm{\infty })$ be a function. We say that T is an α-admissible mapping if

$$x,y\in X,\alpha (x,y)\ge 1⟹\alpha (Tx,Ty)\ge 1.$$

Theorem 1.1 [15]

Let $(X,d)$ be a complete metric space, and let T be an α-admissible mapping. Assume that

$$\alpha (x,y)d(Tx,Ty)\le \psi (d(x,y))$$

(1.1)

for all $x,y\in X$, where $\psi \in \mathrm{\Psi }$. Also, suppose that

• (i) there exists $x_0\in X$ such that $\alpha (x_0,T{x}_0)\ge 1$;

• (ii) either T is continuous or for any sequence $\{x_n\}$ in X with $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and $x_n\to x$ as $n\to +\mathrm{\infty }$, we have $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$.

Then T has a fixed point.

Very recently Salimi et al. [14] modified the notions of α-admissible and α-ψ-contractive mappings as follows.

Definition 1.2 [14]

Let T be a self-mapping on X, and let $\alpha ,\eta :X\times X\to [0,+\mathrm{\infty })$ be two functions. We say that T is an α-admissible mapping with respect to η if

$$x,y\in X,\alpha (x,y)\ge \eta (x,y)⟹\alpha (Tx,Ty)\ge \eta (Tx,Ty).$$

Note that if we take $\eta (x,y)=1$, then this definition reduces to Definition 1.1. Also, if we take $\alpha (x,y)=1$, then we say that T is η-subadmissible mapping.

The following result properly contains Theorem 1.1 and Theorems 2.3 and 2.4 of [13].

Theorem 1.2 [14]

Let $(X,d)$ be a complete metric space, and let T be an α-admissible mapping with respect to η. Assume that

$$x,y\in X,\alpha (x,y)\ge \eta (x,y)⟹d(Tx,Ty)\le \psi (M(x,y)),$$

(1.2)

where $\psi \in \mathrm{\Psi }$ and

$$M(x,y)=max\{d(x,y),\frac{d(x,Tx)+d(y,Ty)}{2},\frac{d(x,Ty)+d(y,Tx)}{2}\}.$$

Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ such that $\alpha (x_0,T{x}_0)\ge \eta (x_0,T{x}_0)$;

• (ii) either T is continuous or for any sequence $\{x_n\}$ in X with $\alpha (x_n,x_{n+1})\ge \eta (x_n,x_{n+1})$ for all $n\in \mathbb{N}\cup \{0\}$ and $x_n\to x$ as $n\to +\mathrm{\infty }$, we have $\alpha (x_n,x)\ge \eta (x_n,x)$ for all $n\in \mathbb{N}\cup \{0\}$.

Then T has a fixed point.

2 Modified α-η-Geraghty type contractions

Our first main result of this section is concerning α-η-Geraghty-type [4] contractions.

Theorem 2.1 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping with respect to η. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1)$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$\begin{array}{rl}x,y\in X,& \alpha (x,fx)\alpha (y,fy)\ge \eta (x,fx)\eta (y,fy)\\ ⟹d(fx,fy)\le \beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}.\end{array}$$

(2.1)

Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge \eta (x_n,x_{n+1})$ for all n, then $\alpha (x,fx)\ge \eta (x,fx)$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge \eta (x_0,f{x}_0)$, then f has a fixed point.

Proof Let $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge \eta (x_0,f{x}_0)$. Define a sequence $\{x_n\}$ in X by $x_n=f^n{x}_0=f{x}_{n-1}$ for all $n\in \mathbb{N}$. If $x_{n+1}=x_n$ for some $n\in \mathbb{N}$, then $x=x_n$ is a fixed point for f, and the result is proved. Hence, we suppose that $x_{n+1}\ne x_n$ for all $n\in \mathbb{N}$. Since f is an α-admissible mapping with respect to η and $\alpha (x_0,f{x}_0)\ge \eta (x_0,f{x}_0)$, we deduce that $\alpha (x_1,x_2)=\alpha (f{x}_0,f^2{x}_0)\ge \eta (f{x}_0,f^2{x}_0)=\eta (x_1,x_2)$. By continuing this process, we get $\alpha (x_n,f{x}_n)\ge \eta (x_n,f{x}_n)$ for all $n\in \mathbb{N}\cup \{0\}$. Then,

$$\alpha (x_{n-1},f{x}_{n-1})\alpha (x_n,f{x}_n)\ge \eta (x_{n-1},f{x}_{n-1})\eta (x_n,f{x}_n).$$

Now from (2.1), we have

$$\begin{array}{rl}d(x_n,x_{n+1})& \le \beta (d(x_{n-1},x_n))max\{d(x_{n-1},x_n),min\{d(x_{n-1},f{x}_{n-1}),d(x_n,f{x}_n)\}\}\\ =\beta (d(x_{n-1},x_n))max\{d(x_{n-1},x_n),min\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}\}.\end{array}$$

Now, if $d(x_{n-1},x_n)<d(x_n,x_{n+1})$ for some $n\in \mathbb{N}$, then

$$max\{d(x_{n-1},x_n),min\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}\}=d(x_{n-1},x_n).$$

Also, if $d(x_n,x_{n+1})\le d(x_{n-1},x_n)$ for some $n\in \mathbb{N}$, then

$$max\{d(x_{n-1},x_n),min\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}\}=d(x_{n-1},x_n).$$

That is, for all $n\in \mathbb{N}$, we have

$$max\{d(x_{n-1},x_n),min\{d(x_{n-1},x_n),d(x_n,x_{n+1})\}\}=d(x_{n-1},x_n).$$

Hence,

$$d(x_n,x_{n+1})\le \beta (d(x_{n-1},x_n))d(x_{n-1},x_n)$$

(2.2)

for all $n\in \mathbb{N}$, which implies that $d(x_n,x_{n+1})\le d(x_{n-1},x_n)$. It follows that the sequence $\{d(x_n,x_{n+1})\}$ is decreasing. Thus, there exists $d\in {\mathbb{R}}_{+}$ such that ${lim}_{n\to \mathrm{\infty }}d(x_n,x_{n+1})=d$. We shall prove that $d=0$. From (2.2), we have

$$\frac{d(x_n,x_{n+1})}{d(x_{n-1},x_n)}\le \beta (d(x_{n-1},x_n))\le 1,$$

which implies that ${lim}_{n\to \mathrm{\infty }}\beta (d(x_{n-1},x_n))=1$. Regarding the property of the function β, we conclude that

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

(2.3)

Next, we shall prove that $\{x_n\}$ is a Cauchy sequence. Suppose, to the contrary, that $\{x_n\}$ is not a Cauchy sequence. Then there is $\epsilon >0$ and sequences $\{m(k)\}$ and $\{n(k)\}$ such that for all positive integers k, we have

$$n(k)>m(k)>k,d(x_{n(k)},x_{m(k)})\ge \epsilon \text{and}d(x_{n(k)},x_{m(k)-1})<\epsilon .$$

By the triangle inequality, we derive that

$$\begin{array}{rcl}\epsilon & \le & d(x_{n(k)},x_{m(k)})\le d(x_{n(k)},x_{m(k)-1})+d(x_{m(k)-1},x_{m(k)})\\ <& \epsilon +d(x_{m(k)-1},x_{m(k)})\end{array}$$

$k\in \mathbb{N}$. Taking the limit as $k\to +\mathrm{\infty }$ in the inequality above, and regarding the limit in (2.3), we get

$$\underset{k\to +\mathrm{\infty }}{lim}d(x_{n(k)},x_{m(k)})=\epsilon .$$

(2.4)

Again, by the triangle inequality, we find that

$$d(x_{n(k)},x_{m(k)})\le d(x_{m(k)},x_{m(k)+1})+d(x_{m(k)+1},x_{n(k)+1})+d(x_{n(k)+1},x_{n(k)})$$

and

$$d(x_{n(k)+1},x_{m(k)+1})\le d(x_{m(k)},x_{m(k)+1})+d(x_{m(k)},x_{n(k)})+d(x_{n(k)+1},x_{n(k)}).$$

Taking the limit in inequality above as $k\to +\mathrm{\infty }$, together with (2.3) and (2.4), we deduce that

$$\underset{k\to +\mathrm{\infty }}{lim}d(x_{n(k)+1},x_{m(k)+1})=\epsilon .$$

(2.5)

Now, since

$$\alpha (x_{n(k)},f{x}_{n(k)})\alpha (x_{m(k)},f{x}_{m(k)})\ge \eta (x_{n(k)},f{x}_{n(k)})\eta (x_{m(k)},f{x}_{m(k)}),$$

then from (2.1), (2.4) and (2.5), we have

$$\begin{array}{r}d(x_{n(k)+1},x_{m(k)+1})\\ \le \beta (d(x_{n(k)},x_{m(k)}))max\{d(x_{n(k)},x_{m(k)}),min\{d(x_{n(k)},f{x}_{n(k)}),d(x_{m(k)},f{x}_{m(k)})\}\}\\ =\beta (d(x_{n(k)},x_{m(k)}))max\{d(x_{n(k)},x_{m(k)}),min\{d(x_{n(k)},x_{n(k)+1}),d(x_{m(k)},x_{m(k)+1})\}\}.\end{array}$$

Hence,

$$\frac{d(x_{n(k)+1},x_{m(k)+1})}{max\{d(x_{n(k)},x_{m(k)}),min\{d(x_{n(k)},x_{n(k)+1}),d(x_{m(k)},x_{m(k)+1})\}\}}\le \beta (d(x_{n(k)},x_{m(k)}))\le 1.$$

Letting $k\to \mathrm{\infty }$ in the inequality above, we get

$$\underset{n\to \mathrm{\infty }}{lim}\beta (d(x_{n(k)},x_{m(k)}))=1.$$

That is, ${lim}_{k\to \mathrm{\infty }}d(x_{n(k)},x_{m(k)})=0$, which is a contradiction. Hence $\{x_n\}$ is a Cauchy sequence. Since X is complete, then there is $z\in X$ such that $x_n\to z$. First, we suppose that f is continuous. Since f is continuous, then we have

$$fz=\underset{n\to \mathrm{\infty }}{lim}f{x}_n=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=z.$$

So z is a fixed point of f. Next, we suppose that (b) holds. Then, $\alpha (z,fz)\ge \eta (z,fz)$, and so, $\alpha (z,fz)\alpha (x_n,f{x}_n)\ge \eta (z,fz)\eta (x_n,f{x}_n)$. Now by (2.1), we have

$$d(fz,x_{n+1})\le \beta (d(z,x_n))max\{d(z,x_n),min\{d(z,fz),d(x_n,x_{n+1})\}\},$$

and hence

$$\begin{array}{rcl}d(fz,z)& \le & d(fz,x_{n+1})+d(z,x_{n+1})\\ \le & \beta (d(z,x_n))max\{d(z,x_n),min\{d(z,fz),d(x_n,x_{n+1})\}\}+d(z,x_{n+1}).\end{array}$$

Letting $n\to \mathrm{\infty }$ in the inequality above, we get $d(fz,z)=0$, that is, $z=fz$. □

If in Theorem 2.1 we take, $\eta (x,y)=1$, then we have the following corollary.

Corollary 2.1 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$\begin{array}{rl}x,y\in X,& \alpha (x,fx)\alpha (y,fy)\ge 1\\ ⟹d(fx,fy)\le \beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}.\end{array}$$

Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Corollary 2.2 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$${(d(fx,fy)+\ell )}^{\alpha (x,fx)\alpha (y,fy)}\le \beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}+\ell$$

for all $x,y\in X$, where $\ell >0$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Corollary 2.3 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$${(\alpha (x,fx)\alpha (y,fy)+1)}^{d(fx,fy)}\le 2^{\beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}}$$

for all $x,y\in X$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Corollary 2.4 Let $(X,d)$ be a metric space such that $(X,d)$ is complete and $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$\alpha (x,fx)\alpha (y,fy)d(fx,fy)\le \beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}$$

for all $x,y\in X$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,f{x}_n)\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Further, if in Theorem 2.1 we take $\alpha (x,y)=1$, then we have the following corollary.

Corollary 2.5 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be a η-subadmissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$\begin{array}{rl}x,y\in X,& \eta (x,fx)\eta (y,fy)\le 1\\ ⟹d(fx,fy)\le \beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}.\end{array}$$

Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\eta (x_n,x_{n+1})\le 1$ for all n, then $\eta (x,fx)\le 1$.

If there exists $x_0\in X$ such that $\eta (x_0,f{x}_0)\le 1$, then f has a fixed point.

Corollary 2.6 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be a η-subadmissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$d(fx,fy)+\ell \le {[\beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}+\ell ]}^{\eta (x,fx)\eta (y,fy)}$$

for all $x,y\in X$, where $\ell >0$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\eta (x_n,x_{n+1})\le 1$ for all n, then $\eta (x,fx)\le 1$.

If there exists $x_0\in X$ such that $\eta (x_0,f{x}_0)\le 1$, then f has a fixed point.

Corollary 2.7 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be a η-subadmissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$2^{d(fx,fy)}\le {(\eta (x,fx)\eta (y,fy)+1)}^{\beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}}$$

for all $x,y\in X$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\eta (x_n,x_{n+1})\le 1$ for all n, then $\eta (x,fx)\le 1$.

If there exists $x_0\in X$ such that $\eta (x_0,f{x}_0)\le 1$, then f has a fixed point.

Corollary 2.8 Let $(X,d)$ be a metric space such that $(X,d)$ is complete, and let $f:X\to X$ be η-subadmissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$d(fx,fy)\le \eta (x,fx)\eta (y,fy)\beta (d(x,y))max\{d(x,y),min\{d(x,fx),d(y,fy)\}\}$$

for all $x,y\in X$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\eta (x_n,f{x}_n)\le 1$ for all n, then $\eta (x,fx)\le 1$.

If there exists $x_0\in X$ such that $\eta (x_0,f{x}_0)\le 1$, then f has a fixed point.

From Corollary 2.1, we can deduce the following corollary.

Corollary 2.9 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$x,y\in X,\alpha (x,fx)\alpha (y,fy)\ge 1⟹d(fx,fy)\le \beta (d(x,y))d(x,y).$$

Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Also, from the corollary above, we can deduce the following corollaries.

Corollary 2.10 (Theorem 4 of [9])

Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$${(d(fx,fy)+\ell )}^{\alpha (x,fx)\alpha (y,fy)}\le \beta (d(x,y))d(x,y)+\ell$$

for all $x,y\in X$, where $\ell \ge 1$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Corollary 2.11 (Theorem 6 of [9])

Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$${(\alpha (x,fx)\alpha (y,fy)+1)}^{d(fx,fy)}\le 2^{\beta (d(x,y))d(x,y)}$$

for all $x,y\in X$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,x_{n+1})\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Corollary 2.12 (Theorem 8 of [9])

Let $(X,d)$ be a metric space such that $(X,d)$ is complete, and let $f:X\to X$ be an α-admissible mapping. Assume that there exists a function $\beta :[0,\mathrm{\infty })\to [0,1]$ such that for any bounded sequence $\{t_n\}$ of positive reals, $\beta (t_n)\to 1$ implies that $t_n\to 0$ and

$$\alpha (x,fx)\alpha (y,fy)d(fx,fy)\le \beta (d(x,y))d(x,y)$$

for all $x,y\in X$. Suppose that either

• (a) f is continuous, or

• (b) if $\{x_n\}$ is a sequence in X such that $x_n\to x$, $\alpha (x_n,f{x}_n)\ge 1$ for all n, then $\alpha (x,fx)\ge 1$.

If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Example 2.1 Let $X=[0,\mathrm{\infty })$ be endowed with the usual metric $d(x,y)=|x-y|$ for all $x,y\in X$, and let $f:X\to X$ be defined by

$$fx=\{\begin{array}{cc}\frac{1}{4}x\hfill & \text{if }x\in [0,1],\hfill \\ ln(x^2+x+3)\hfill & \text{if }x\in (1,\mathrm{\infty }).\hfill \end{array}$$

Define also $\alpha :X\times X\to [0,+\mathrm{\infty })$ and $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ by

$$\alpha (x,y)=\{\begin{array}{cc}6\hfill & \text{if }x,y\in [0,1],\hfill \\ 0\hfill & \text{otherwise}\hfill \end{array}\text{and}\beta (t)=\frac{1}{2}.$$

We prove that Corollary 2.9 can be applied to f, but Corollaries 2.10, 2.11 and 2.12 (Theorem 4, 6 and 8 of [9]) cannot be applied to f.

Clearly, $(X,d)$ is a complete metric space. We show that f is an α-admissible mapping. Let $x,y\in X$ with $\alpha (x,y)\ge 1$, then $x,y\in [0,1]$. On the other hand, for all $x\in [0,1]$, we have $fx\le 1$. It follows that $\alpha (fx,fy)\ge 1$. Hence, the assertion holds. Also, $\alpha (0,f0)\ge 1$. Now, if $\{x_n\}$ is a sequence in X such that $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$ and $x_n\to x$ as $n\to +\mathrm{\infty }$, then $\{x_n\}\subset [0,1]$, and hence $x\in [0,1]$. This implies that $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}$.

Let $\alpha (x,y)\ge 1$. Then $x,y\in [0,1]$. We get,

$$d(fx,fy)=|fy-fx|=|\frac{1}{4}x-\frac{1}{4}y|=\frac{1}{4}|x-y|\le \frac{1}{2}|x-y|=\beta (d(x,y))d(x,y).$$

That is,

$$\alpha (x,y)\ge 1⟹d(fx,fy)\le \beta (d(x,y))d(x,y),$$

then the conditions of Corollary 2.1 hold, and f has a fixed point.

Let $x=0$, $y=1$, and let $\ell =1$, then

$${(d(f0,f1)+1)}^{\alpha (0,f0)\alpha (1,f1)}={(1/4+1)}^{36}>1/2+1=\beta (d(0,1))d(0,1)+1.$$

That is, Corollary 2.10 (Theorem 4 of [9]) cannot be applied for this example.

Let, $x=0$, and let $y=1$, then

$${(\alpha (0,f0)\alpha (1,f1)+1)}^{d(f0,f1)}=\sqrt[4]{37}>\sqrt{2}=2^{\beta (d(0,1))d(0,1)}.$$

That is, Corollary 2.11 (Theorem 6 of [9]) cannot be applied for this example.

Let, $x=0$, and let $y=1$, then

$$\alpha (0,f0)\alpha (1,f1)d(f0,f1)=9>1/2=\beta (d(0,1))d(0,1).$$

That is, Corollary 2.12 (Theorem 8 of [9]) cannot be applied for this example.

3 Modified α-ψ-Meir-Keeler contractive mappings

Recently, Karapinar et al. [11] introduced the notion of a triangular α-admissible mapping as follows.

Definition 3.1 [11]

Let $f:X\to X$, and let $\alpha :X\times X\to (-\mathrm{\infty },+\mathrm{\infty })$. We say that f is a triangular α-admissible mapping if

• (T1) $\alpha (x,y)\ge 1$ implies that $\alpha (fx,fy)\ge 1$, $x,y\in X$;

• (T2) $\{\begin{array}{l}\alpha (x,z)\ge 1,\\ \alpha (z,y)\ge 1\end{array}$ implies that $\alpha (x,y)\ge 1$.

Lemma 3.1 [11]

Let f be a triangular α-admissible mapping. Assume that there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$. Define sequence $\{x_n\}$ by $x_n=f^n{x}_0$. Then

$$\alpha (x_m,x_n)\ge 1\mathit{\text{for all }}m,n\in \mathbb{N}\mathit{\text{ with }}m<n.$$

Denote with Ψ the family of nondecreasing functions $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ continuous at $t=0$ such that

• $\psi (t)=0$ if and only if $t=0$,

• $\psi (t+s)\le \psi (t)+\psi (s)$.

Definition 3.2 [11]

Let $(X,d)$ be a metric space, and let $\psi \in \mathrm{\Psi }$. Suppose that $f:X\to X$ is a triangular α-admissible mapping satisfying the following condition:

for each $\epsilon >0$ there exists $\delta >0$ such that

$$\epsilon \le \psi (d(x,y))<\epsilon +\delta \text{implies that}\alpha (x,y)\psi (d(fx,fy))<\epsilon$$

(3.1)

for all $x,y\in X$. Then f is called an α-ψ-Meir-Keeler contractive mapping.

Now, we modify Definition 3.2 as follows.

Definition 3.3 Let $(X,d)$ be a metric space, and let $\psi \in \mathrm{\Psi }$. Suppose that $f:X\to X$ is a triangular α-admissible mapping satisfying the following condition:

for each $\epsilon >0$ there exists $\delta >0$ such that

$$\epsilon \le \psi (d(x,y))<\epsilon +\delta \text{implies that}\psi (d(fx,fy))<\epsilon$$

(3.2)

for all $x,y\in X$ with $\alpha (x,y)\ge 1$. Then f is called a modified α-ψ-Meir-Keeler contractive mapping.

Remark 3.1 Let f be a modified α-ψ-Meir-Keeler contractive mapping. Then

$$\psi (d(fx,fy))<\psi (d(x,y))$$

for all $x,y\in X$ when $x\ne y$ and $\alpha (x,y)\ge 1$. Also, if $x=y$ and $\alpha (x,y)\ge 1$, then $d(fx,fy)=0$, i.e.,

$$\psi (d(fx,fy))\le \psi (d(x,y)).$$

Theorem 3.1 Let $(X,d)$ be a complete metric space. Suppose that f is a continuous modified α-ψ-Meir-Keeler contractive mapping, and that there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Proof Let $x_0\in X$ and define a sequence $\{x_n\}$ by $x_n=f^n{x}_0$ for all $n\in \mathbb{N}$. If $x_{n_0}=x_{n_0+1}$ for some $n_0\in \mathbb{N}\cup \{0\}$, then, obviously, f has a fixed point. Hence, we suppose that

$$x_n\ne x_{n+1}$$

(3.3)

for all $n\in \mathbb{N}\cup \{0\}$. We have $d(x_n,x_{n+1})>0$ for all $n\in \mathbb{N}\cup \{0\}$. Now, define $s_n=\psi (d(x_n,x_{n+1}))$. By Remark 3.1, we deduce that for all $n\in \mathbb{N}\cup \{0\}$ $\psi (d(x_{n+1},x_{n+2}))=\psi (d(f{x}_n,f{x}_{n+1}))<\psi (d(x_n,x_{n+1}))$. By applying Lemma 3.1 for

$$\alpha (x_m,x_n)\ge 1\text{for all }m,n\in \mathbb{N}\text{ with }m<n,$$

we have

$$\psi (d(x_{n+1},x_{n+2}))<\psi (d(x_n,x_{n+1})).$$

Hence, the sequence $\{s_n\}$ is decreasing in ${\mathbb{R}}_{+}$, and so, it is convergent to $s\in {\mathbb{R}}_{+}$. We will show that $s=0$. Suppose, to the contrary, that $s>0$. Note that

$$0<s<\psi (d(x_n,x_{n+1}))\text{for all }n\in \mathbb{N}\cup \{0\}.$$

(3.4)

Let $\epsilon =s>0$. Then by hypothesis, there exists a $\delta (\epsilon )>0$ such that (3.2) holds. On the other hand, by the definition of ε, there exists $n_0\in \mathbb{N}$ such that

$$\epsilon <s_{n_0}=\psi (d(x_{n_0},x_{n_0+1}))<\epsilon +\delta .$$

Now by (3.2), we have

$$s_{n_0+1}=\psi (d(x_{n_0+1},x_{n_0+2}))\le \psi (d(f{x}_{n_0},f{x}_{n_0+1}))<\epsilon =s,$$

which is a contradiction. Hence $s=0$, that is, ${lim}_{n\to +\mathrm{\infty }}{s}_n=0$. Now, by the continuity of ψ at $t=0$, we have ${lim}_{n\to +\mathrm{\infty }}d(x_n,x_{n+1})=0$. For given $\epsilon >0$, by the hypothesis, there exists a $\delta =\delta (\epsilon )>0$ such that (3.2) holds. Without loss of generality, we assume that $\delta <\epsilon$. Since $s=0$, then there exists $N\in \mathbb{N}$ such that

$$s_{n-1}=\psi (d(x_{n-1},x_n))<\delta \text{for all }n\ge N.$$

(3.5)

We will prove that for any fixed $k\ge N_0$,

$$\psi (d(x_k,x_{k+l}))\le \epsilon \text{for all }l\in \mathbb{N},$$

(3.6)

holds. Note that (3.6) holds for $l=1$ by (3.5). Suppose that condition (3.2) is satisfied for some $m\in \mathbb{N}$. For $l=m+1$, by (3.5), we get

$$\begin{array}{rl}\psi (d(x_{k-1},x_{k+m}))& \le \psi (d(x_{k-1},x_k)+d(x_k,x_{k+m}))\\ \le \psi (d(x_{k-1},x_k))+\psi (d(x_k,x_{k+m}))\\ <\epsilon +\delta .\end{array}$$

(3.7)

If $\psi (d(x_{k-1},x_{k+m}))\ge \epsilon$, then by (3.2), we get

$$\psi (d(x_k,x_{k+m+1}))=\psi (d(f{x}_{k-1},f{x}_{k+m}))<\epsilon ,$$

and hence (3.6) holds.

If $\psi (d(x_{k-1},x_{k+m}))<\epsilon$, by Remark 3.1, we get

$$\psi (d(x_k,x_{k+m+1}))\le \psi (d(x_{k-1},x_{k+m}))<\epsilon .$$

Consequently, (3.6) holds for $l=m+1$. Hence, $\psi (d(x_k,x_{k+l}))\le \epsilon$ for all $k\ge N_0$ and $l\ge 1$, which means

$$d(x_n,x_m)<\epsilon \text{for all }m\ge n\ge N_0.$$

(3.8)

Hence $\{x_n\}$ is a Cauchy sequence. Since $(X,d)$ is complete, there exists $z\in X$ such that $x_n\to z$ as $n\to \mathrm{\infty }$. Now, since f is continuous, then

$$fz=f\left(\underset{n\to \mathrm{\infty }}{lim}{x}_n\right)=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=z,$$

that is, f has a fixed point. □

Corollary 3.1 (Theorem 10 of [11])

Let $(X,d)$ be a complete metric space. Suppose that f is a continuous α-ψ-Meir-Keeler contractive mapping, and that there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Proof Let $\epsilon \le \psi (d(x,y))<\epsilon +\delta$, where $\alpha (x,y)\ge 1$. Then by $\epsilon \le \psi (d(x,y))<\epsilon +\delta$ and Definition 3.2, we deduce that $\alpha (x,y)\psi (d(fx,fy))<\epsilon$. On the other hand, since $\alpha (x,y)\ge 1$, then we have

$$\psi (d(fx,fy))\le \alpha (x,y)\psi (d(fx,fy))<\epsilon .$$

That is, conditions of Theorem 3.1 hold, and f has a fixed point. □

Theorem 3.2 Let $(X,d)$ be a complete metric space, and let f be a modified α-ψ-Meir-Keeler contractive mapping. If the following conditions hold:

• (i) there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$,

• (ii) if $\{x_n\}$ is a sequence in X such that $\alpha (x_n,x_{n+1})\ge 1$ for all n, and $x_n\to x$ as $n\to +\mathrm{\infty }$, then $\alpha (x_n,x)\ge 1$ for all n.

Then f has a fixed point.

Proof Following the proof of Theorem 3.1, we say that $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$, and that there exist $z\in X$ such that $x_n\to z$ as $n\to +\mathrm{\infty }$. Hence, from (ii) $\alpha (x_n,z)\ge 1$. By Remark 3.1, we have

$$\begin{array}{rl}\psi (d(fz,z))& \le \psi (d(fz,f{x}_n)+d(f{x}_n,z))\le \psi (d(fz,f{x}_n))+\psi (d(f{x}_n,z))\\ \le \psi (d(z,x_n))+\psi (d(x_{n+1},z)).\end{array}$$

By taking limit as $n\to +\mathrm{\infty }$, in the inequality above, we get $\psi (d(fz,z))\le 0$, that is, $d(fz,z)=0$. Hence $fz=z$. □

Corollary 3.2 (Theorem 11 of [11])

Let $(X,d)$ be a complete metric space, and let f be a α-ψ-Meir-Keeler contractive mapping. If the following conditions hold:

• (i) there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$,

• (ii) if $\{x_n\}$ is a sequence in X such that $\alpha (x_n,x_{n+1})\ge 1$ for all n, and $x_n\to x$ as $n\to +\mathrm{\infty }$, then $\alpha (x_n,x)\ge 1$ for all n.

Then f has a fixed point.

Example 3.1 Let $X=[0,\mathrm{\infty })$, and let $d(x,y)=|x-y|$ be a metric on X. Define $f:X\to X$ by

$$fx=\{\begin{array}{cc}\frac{x}{5}\hfill & \text{if }x\in [0,1],\hfill \\ x^{x^2+1}\hfill & x\in (1,\mathrm{\infty }),\hfill \end{array}\text{and}\alpha (x,y)=\{\begin{array}{cc}10\hfill & \text{if }x,y\in [0,1],\hfill \\ -2\hfill & \text{otherwise},\hfill \end{array}$$

and $\psi (t)=\frac{1}{4}t$. Clearly, $(X,d)$ is a complete metric space. We show that f is a triangular α-admissible mapping. Let $x,y\in X$, if $\alpha (x,y)\ge 1$, then $x,y\in [0,1]$. On the other hand, for all $x,y\in [0,1]$, we have $fx\le 1$ and $fy\le 1$. It follows that $\alpha (fx,fy)\ge 1$. Also, if $\alpha (x,z)\ge 1$ and $\alpha (z,y)\ge 1$, then $x,y,z\in [0,1]$, and hence, $\alpha (x,y)\ge 1$. Thus the assertion holds by the same arguments. Notice that $\alpha (0,f0)\ge 1$.

Now, if $\{x_n\}$ is a sequence in X such that $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$, and $x_n\to x$ as $n\to +\mathrm{\infty }$, then $\{x_n\}\subset [0,1]$, and hence $x\in [0,1]$. This implies that $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}\cup \{0\}$. Let $\alpha (x,y)\ge 1$, then $x,y\in [0,1]$. Without loss of generality, take $x\le y$. Then

$$\begin{array}{c}\psi (d(fx,fy))=\frac{y}{20}-\frac{x}{20},\hfill \\ \psi (d(x,y))=\frac{y}{4}-\frac{x}{4}.\hfill \end{array}$$

Clearly, by taking $\delta =4\epsilon$, the condition (3.2) holds. Hence, conditions of Theorem 3.2 hold, and f has a fixed point. But if $x,y\in [0,1]$ and

$$\epsilon \le d(x,y)<\epsilon +\delta ,$$

where $\epsilon >0$ and $\delta >0$. Then

$$\alpha (x,y)d(fx,fy)=2|x-y|=2d(x,y)\ge 2\epsilon .$$

That is, Corollary 3.2 (Theorem 11 of [11]) cannot be applied for this example.

Denote with ${\mathrm{\Psi }}_{\mathrm{st}}$ the family of strictly nondecreasing functions ${\psi }_{\mathrm{st}}:[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ continuous at $t=0$ such that

• ${\psi }_{\mathrm{st}}(t)=0$ if and only if $t=0$,

• ${\psi }_{\mathrm{st}}(t+s)\le {\psi }_{\mathrm{st}}(t)+{\psi }_{\mathrm{st}}(s)$.

Definition 3.4 [11]

Let $(X,d)$ be a metric space, and let ${\psi }_{\mathrm{st}}\in {\mathrm{\Psi }}_{\mathrm{st}}$. Suppose that $f:X\to X$ is a triangular α-admissible mapping satisfying the following condition:

for each $\epsilon >0$, there exists $\delta >0$ such that

$$\epsilon \le {\psi }_{\mathrm{st}}(M(x,y))<\epsilon +\delta \text{implies that}\alpha (x,y){\psi }_{\mathrm{st}}(d(fx,fy))<\epsilon$$

(3.9)

for all $x,y\in X$, where

$$M(x,y)=max\{d(x,y),d(fx,x),d(fy,y),\frac{1}{2}[d(fx,y)+d(x,fy)]\}.$$

Then f is called a generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping.

Definition 3.5 Let $(X,d)$ be a metric space, and let ${\psi }_{\mathrm{st}}\in {\mathrm{\Psi }}_{\mathrm{st}}$. Suppose that $f:X\to X$ is a triangular α-admissible mapping satisfying the following condition:

for each $\epsilon >0$ there exists $\delta >0$ such that

$$\epsilon \le {\psi }_{\mathrm{st}}(M(x,y))<\epsilon +\delta \text{implies that}{\psi }_{\mathrm{st}}(d(fx,fy))<\epsilon$$

(3.10)

for all $x,y\in X$, where $\alpha (x,y)\ge 1$ and

$$M(x,y)=max\{d(x,y),d(fx,x),d(fy,y),\frac{1}{2}[d(fx,y)+d(x,fy)]\}.$$

Then f is called a modified generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping.

Remark 3.2 Let f be a modified generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping. Then

$${\psi }_{\mathrm{st}}(d(fx,fy))<{\psi }_{\mathrm{st}}(M(x,y))$$

for all $x,y\in X$, where $\alpha (x,y)\ge 1$ when $M(x,y)>0$. Also, if $M(x,y)=0$ and $\alpha (x,y)\ge 1$, then $x=y$, which implies that $\psi (d(fx,fy))=0$, i.e.,

$${\psi }_{\mathrm{st}}(d(fx,fy))\le {\psi }_{\mathrm{st}}(M(x,y)).$$

Proposition 3.1 Let $(X,d)$ be a metric space, and let $f:X\to X$ be a modified generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping. If there exists $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then ${lim}_{n\to \mathrm{\infty }}d(f^{n+1}{x}_0,f^n{x}_0)=0$.

Proof Define a sequence $\{x_n\}$ by $x_n=f^n{x}_0$ for all $n\in \mathbb{N}$. If $x_{n_0}=x_{n_0+1}$ for some $n_0\in \mathbb{N}\cup \{0\}$, then, obviously, the conclusion holds. Hence, we suppose that

$$x_n\ne x_{n+1}$$

(3.11)

for all $n\in \mathbb{N}\cup \{0\}$. Then we have $M(x_{n+1},x_n)>0$ for every $n\ge 0$. Then by Lemma 3.1 and Remark 3.2, we have

$$\begin{array}{rcl}{\psi }_{\mathrm{st}}(d(x_{n+1},x_{n+2}))& =& {\psi }_{\mathrm{st}}(d(f{x}_n,f{x}_{n+1}))<{\psi }_{\mathrm{st}}(M(x_n,x_{n+1}))\\ =& {\psi }_{\mathrm{st}}(max\{d(x_n,x_{n+1}),d(f{x}_n,x_n),d(f{x}_{n+1},x_{n+1}),\\ \frac{1}{2}[d(f{x}_n,x_{n+1})+d(x_n,f{x}_{n+1})]\left\}\right)\\ \le & {\psi }_{\mathrm{st}}(max\{d(x_n,x_{n+1}),d(x_{n+1},x_{n+2})\}).\end{array}$$

Now, since ${\psi }_{\mathrm{st}}$ is strictly nondecreasing, then we get

$$d(x_{n+2},x_{n+1})<max\{d(x_{n+1},x_n),d(x_{n+2},x_{n+1})\}.$$

Hence the case, where

$$max\{d(x_{n+1},x_n),d(x_{n+2},x_{n+1})\}=d(x_{n+2},x_{n+1}),$$

is not possible. Therefore, we deduce that

$$d(x_{n+2},x_{n+1})<d(x_{n+1},x_n)$$

(3.12)

for all n. That is, ${\{d(x_{n+1},x_n)\}}_{n=0}^{\mathrm{\infty }}$ is a decreasing sequence in ${\mathbb{R}}_{+}$, and it converges to $\epsilon \in {\mathbb{R}}_{+}$, that is,

$$\underset{n\to \mathrm{\infty }}{lim}{\psi }_{\mathrm{st}}(d(x_{n+1},x_n))=\underset{n\to \mathrm{\infty }}{lim}{\psi }_{\mathrm{st}}(M(x_{n+1},x_n))={\psi }_{\mathrm{st}}(\epsilon ).$$

(3.13)

Notice that $\epsilon =inf\{d(x_n,x_{n+1}):n\in \mathbb{N}\}$. Let us prove that $\epsilon =0$. Suppose, to the contrary, that $\epsilon >0$. Then $\psi (\epsilon )>0$. Considering (3.13) together with the assumption that f is a generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping, for ${\psi }_{\mathrm{st}}(\epsilon )$, there exists $\delta >0$ and a natural number m such that

$${\psi }_{\mathrm{st}}(\epsilon )\le {\psi }_{\mathrm{st}}(M(x_m,x_{m+1}))<{\psi }_{\mathrm{st}}(\epsilon )+\delta$$

implies that

$${\psi }_{\mathrm{st}}(d(x_{m+1},x_{m+2}))={\psi }_{\mathrm{st}}(d(f{x}_m,f{x}_{m+1}))<{\psi }_{\mathrm{st}}(\epsilon ).$$

Now, since ${\psi }_{\mathrm{st}}$ is strictly nondecreasing, then we get

$$d(x_{m+2},x_{m+1})<\epsilon ,$$

which is a contradiction, since $\epsilon =inf\{d(x_n,x_{n+1}):n\in \mathbb{N}\}$. Then $\epsilon =0$, and so,

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

 □

Theorem 3.3 Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an orbitally continuous modified generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping. If there exist $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Proof Define $x_{n+1}=f^{n+1}{x}_0$ for all $n\ge 0$. We want to prove that ${lim}_{m,n\to \mathrm{\infty }}d(x_n,x_m)=0$. If this is not so, then there exist $\epsilon >0$ and a subsequence $\{x_{n(i)}\}$ of $\{x_n\}$ such that

$$d(x_{n(i)},x_{n(i+1)})>2\epsilon .$$

(3.14)

For this $\epsilon >0$, there exists $\delta >0$ such that $\epsilon \le {\psi }_{\mathrm{st}}(M(x,y))<\epsilon +\delta$ implies that $\alpha (x,y){\psi }_{\mathrm{st}}(d(fx,fy))<\epsilon$. Put $r=min\{\epsilon ,\delta \}$ and $s_n=d(x_n,x_{n+1})$ for all $n\ge 1$. From Proposition 3.1, there exists $n_0$ such that

$$s_n=d(x_n,x_{n+1})<\frac{r}{4}$$

(3.15)

for all $n\ge n_0$. Let $n(i)>n_0$. We get $n(i)\le n(i+1)-1$. If $d(x_{n(i)},x_{n(i+1)-1})\le \epsilon +\frac{r}{2}$, then

$$\begin{array}{rl}d(x_{n(i)},x_{n(i+1)})& \le d(x_{n(i)},x_{n(i+1)-1})+d(x_{n(i+1)-1},x_{n(i+1)})\\ \le d(x_{n(i)},x_{n(i+1)-1})+d(x_{n(i+1)-1},x_{n(i+1)})\\ <\epsilon +\frac{r}{2}+s_{n(i+1)-1}<\epsilon +\frac{3r}{4}<2\epsilon ,\end{array}$$

which contradicts the assumption (3.14). Therefore, there are values of k such that $n(i)\le k\le n(i+1)$ and $d(x_{n(i)},x_k)>\epsilon +\frac{r}{2}$. Now if $d(x_{n(i)},x_{n(i)+1})\ge \epsilon +\frac{r}{2}$, then

$$s_{n(i)}=d(x_{n(i)},x_{n(i)+1})\ge \epsilon +\frac{r}{2}>r+\frac{r}{2}>\frac{r}{4},$$

which is a contradiction to (3.15). Hence, there are values of k with $n(i)\le k\le n(i+1)$ such that $d(x_{n(i)},x_k)<\epsilon +\frac{r}{2}$. Choose the smallest integer k with $k\ge n(i)$ such that $d(x_{n(i)},x_k)\ge \epsilon +\frac{r}{2}$. Thus, $d(x_{n(i)},x_{k-1})<\epsilon +\frac{r}{2}$, and so,

$$\begin{array}{rl}d(x_{n(i)},x_k)& \le d(x_{n(i)},x_{k-1})+d(x_{k-1},x_k)\\ \le d(x_{n(i)},x_{k-1})+d(x_{k-1},x_k)<\epsilon +\frac{r}{2}+\frac{r}{4}=\epsilon +\frac{3r}{4}.\end{array}$$

Now, we can choose a natural number k satisfying $n(i)\le k\le n(i+1)$ such that

$$\epsilon +\frac{r}{2}\le d(x_{n(i)},x_k)<\epsilon +\frac{3r}{4}.$$

(3.16)

Therefore, we obtain

$$d(x_{n(i)},x_k)<\epsilon +\frac{3r}{4}<\epsilon +r,$$

(3.17)

$$d(x_{n(i)},x_{n(i)+1})=d_{n(i)}<\frac{r}{4}<\epsilon +r,$$

(3.18)

and

$$d(x_k,x_{k+1})=d_k<\frac{r}{4}<\epsilon +r.$$

(3.19)

Thus, we have

$$\begin{array}{rcl}\frac{1}{2}[d(x_{n(i)},x_{k+1})+d(x_{n(i)+1},x_k)]& \le & \frac{1}{2}[d(x_{n(i)},x_k)+d(x_k,x_{k+1})\\ +d(x_{n(i)+1},x_{n(i)})+d(x_{n(i)},x_k)]\\ \le & \frac{1}{2}[d(x_{n(i)},x_k)+d(x_k,x_{k+1})\\ +d(x_{n(i)+1},x_{n(i)})+d(x_{n(i)},x_k)]\\ =& d(x_{n(i)},x_k)+\frac{1}{2}[s_k+s_{n(i)}]\\ <& \epsilon +\frac{3r}{4}+\frac{1}{2}[\frac{r}{4}+\frac{r}{4}]=\epsilon +r.\end{array}$$

(3.20)

Now, inequalities (3.17)-(3.20) imply that $M(x_{n(i)},x_k)<\epsilon +r\le \epsilon +\delta$, and so, ${\psi }_{\mathrm{st}}(M(x_{n(i)},x_k))<{\psi }_{\mathrm{st}}(\epsilon +\delta )\le {\psi }_{\mathrm{st}}(\epsilon )+{\psi }_{\mathrm{st}}(\delta )$; the fact that f is a modified generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping yields that

$${\psi }_{\mathrm{st}}(d(x_{n(i)+1},x_{k+1}))<{\psi }_{\mathrm{st}}(\epsilon ).$$

Then $d(x_{n(i)+1},x_{k+1})<\epsilon$. We deduce

$$\begin{array}{rl}d(f^{n(i)}{x}_0,f^k{x}_0)\le & d(f^{n(i)}{x}_0,f^{n(i)+1}{x}_0)+d(f^{n(i)+1}{x}_0,f^k{x}_0)\\ \le & d(f^{n(i)}{x}_0,f^{n(i)+1}{x}_0)+d(f^{n(i)+1}{x}_0,f^k{x}_0)\\ \le & d(f^{n(i)}{x}_0,f^{n(i)+1}{x}_0)+d(f^{n(i)+1}{x}_0,f^{k+1}{x}_0)\\ +d(f^{k+1}{x}_0,f^k{x}_0).\end{array}$$

From (3.16), (3.18) and (3.19), we obtain

$$\begin{array}{rcl}d(x_{n(i)+1},x_{k+1})& \ge & d(x_{n(i)},x_k)-d(x_{n(i)},x_{n(i)+1})-d(x_k,x_{k+1})\\ >& \epsilon +\frac{r}{2}-\frac{r}{4}-\frac{r}{4}=\epsilon ,\end{array}$$

which is a contradiction. We obtained that ${lim}_{m,n\to \mathrm{\infty }}d(x_n,x_m)=0$, and so, $\{x_n=f^n{x}_0\}$ is a Cauchy sequence. Since X is complete, then there exists $z\in X$ such that $f^n{x}_0\to z$ as $n\to \mathrm{\infty }$. As f is orbitally continuous, so $z=fz$. □

Corollary 3.3 (Theorem 17 of [11])

Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be an orbitally continuous generalized α-${\psi }_{\mathrm{st}}$-Meir-Keeler contractive mapping. If there exist $x_0\in X$ such that $\alpha (x_0,f{x}_0)\ge 1$, then f has a fixed point.

Example 3.2 Let $X=[0,\mathrm{\infty })$, and let $d(x,y)=|x-y|$ be a metric on X. Define $f:X\to X$ by

$$fx=\{\begin{array}{cc}\frac{x}{7}\hfill & \text{if }x\in [0,1],\hfill \\ \frac{x}{\sqrt[3]{x}+6}\hfill & x\in (1,\mathrm{\infty })\hfill \end{array}$$

and ${\psi }_{\mathrm{st}}(t)=\frac{1}{2}t$,

$$\alpha (x,y)=\{\begin{array}{cc}28\hfill & \text{if }x,y\in [0,1],\hfill \\ -8\hfill & \text{otherwise}.\hfill \end{array}$$

Clearly, f is a triangular α-admissible mapping, and it is orbitally continuous. Let $\alpha (x,y)\ge 1$, then $x,y\in [0,1]$. Without loss of generality, take $x\le y$. Then

$$\begin{array}{c}{\psi }_{\mathrm{st}}(d(fx,fy))=\frac{y}{14}-\frac{x}{14},\hfill \\ \begin{array}{r}{\psi }_{\mathrm{st}}(M(x,y))={\psi }_{\mathrm{st}}(max\{y-x,\frac{6}{7}y,x-\frac{y}{7},y-\frac{x}{7}\})\\ \phantom{{\psi }_{\mathrm{st}}(M(x,y))}=max\{\frac{y}{2}-\frac{x}{2},\frac{6}{14}y,\frac{x}{2}-\frac{y}{14},\frac{y}{2}-\frac{x}{14}\}.\end{array}\hfill \end{array}$$

Clearly, by taking $\delta =6\epsilon$, the condition (3.10) holds. Hence, all conditions of Theorem 3.3 are satisfied, and f has a fixed point. But if $x=0$ and $y=1$

$$\epsilon \le M(0,1)<\delta +\epsilon$$

for $\delta >0$ and $\epsilon >0$, then

$$\epsilon \le 1<\delta +\epsilon ,$$

and so,

$$\alpha (0,1){\psi }_{\mathrm{st}}(d(f0,f1))=2\ge 1\ge \epsilon .$$

That is, Corollary 3.3 (Theorem 17 of [11]) cannot be applied for this example.

4 Modified α-η-contractive multifunction

Recently, Asl et al. [12] introduced the following notion.

Definition 4.1 Let $T:X\to 2^X$, and let $\alpha :X\times X\to {\mathbb{R}}_{+}$. We say that T is an ${\alpha }_{\ast }$-admissible mapping if

$$\alpha (x,y)\ge 1\text{implies that}{\alpha }_{\ast }(Tx,Ty)\ge 1,x,y\in X,$$

where

$${\alpha }_{\ast }(A,B)=\underset{x\in A,y\in B}{inf}\alpha (x,y).$$

We generalize this concept as follows.

Definition 4.2 Let $T:X\to 2^X$ be a multifunction, and let $\alpha ,\eta :X\times X\to {\mathbb{R}}_{+}$ be two functions, where η is bounded. We say that T is an ${\alpha }_{\ast }$-admissible mapping with respect to η if

$$\alpha (x,y)\ge \eta (x,y)\text{implies that}{\alpha }_{\ast }(Tx,Ty)\ge {\eta }_{\ast }(Tx,Ty),x,y\in X,$$

where

$${\alpha }_{\ast }(A,B)=\underset{x\in A,y\in B}{inf}\alpha (x,y)\text{and}{\eta }_{\ast }(A,B)=\underset{x\in A,y\in B}{sup}\eta (x,y).$$

If we take $\eta (x,y)=1$ for all $x,y\in X$, then this definition reduces to Definition 4.1. In case $\alpha (x,y)=1$ for all $x,y\in X$, then T is called an ${\eta }_{\ast }$-subadmissible mapping.

Notice that Ψ is the family of nondecreasing functions $\psi :[0,+\mathrm{\infty })\to [0,+\mathrm{\infty })$ such that ${\sum }_{n=1}^{\mathrm{\infty }}{\psi }^n(t)<+\mathrm{\infty }$ for all $t>0$, where ${\psi }^n$ is the n th iterate of ψ.

As an application of our new concept, we develop now a fixed point result for a multifunction, which generalizes Theorem 1.1.

Theorem 4.1 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\alpha }_{\ast }$-admissible, with respect to η, and closed-valued multifunction on X. Assume that for $\psi \in \mathrm{\Psi }$,

$$x,y\in X,{\alpha }_{\ast }(Tx,Ty)\ge {\eta }_{\ast }(Tx,Ty)⟹H(Tx,Ty)\le \psi (d(x,y)).$$

(4.1)

Also, suppose that the following assertions hold:

• (i) there exist $x_0\in X$ and $x_1\in T{x}_0$ such that $\alpha (x_0,x_1)\ge \eta (x_0,x_1)$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\alpha (x_n,x_{n+1})\ge \eta (x_n,x_{n+1})$ for all $n\in \mathbb{N}$, we have $\alpha (x_n,x)\ge \eta (x_n,x)$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Proof Let $x_1\in T{x}_0$ be such that $\alpha (x_0,x_1)\ge \eta (x_0,x_1)$. Since T is an ${\alpha }_{\ast }$-admissible mapping, then ${\alpha }_{\ast }(T{x}_0,T{x}_1)\ge {\eta }_{\ast }(T{x}_0,T{x}_1)$. Therefore, from (4.1), we have

$$H(T{x}_0,T{x}_1)\le \psi (d(x_0,x_1)).$$

(4.2)

If $x_0=x_1$, then $x_0$ is a fixed point of T. Hence, we assume that $x_0\ne x_1$. Also, if $x_1\in T{x}_1$, then $x_1$ is a fixed point of T. Assume that $x_1\notin T{x}_1$ and $q>1$. Then we have

$$0<d(x_1,T{x}_1)\le H(T{x}_0,T{x}_1)<qH(T{x}_0,T{x}_1),$$

and so, by (4.2), we get

$$0<d(x_1,T{x}_1)<qH(T{x}_0,T{x}_1)\le q\psi (d(x_0,x_1)).$$

This implies that there exists $x_2\in T{x}_1$ such that

$$0<d(x_1,x_2)<qH(T{x}_0,T{x}_1)\le q\psi (d(x_0,x_1)).$$

(4.3)

Note that $x_1\ne x_2$ (since $x_1\notin T{x}_1$). Also, since ${\alpha }_{\ast }(T{x}_0,T{x}_1)\ge {\eta }_{\ast }(T{x}_0,T{x}_1)$, $x_1\in T{x}_0$ and $x_2\in T{x}_1$, then $\alpha (x_1,x_2)\ge \eta (x_1,x_2)$. So ${\alpha }_{\ast }(T{x}_1,T{x}_2)\ge {\eta }_{\ast }(T{x}_1,T{x}_2)$. Therefore, from (4.1), we have

$$H(T{x}_1,T{x}_2)\le \psi (d(x_1,x_2)).$$

(4.4)

Put $t_0=d(x_0,x_1)$. Then from (4.3), we have $d(x_1,x_2)<q\psi (t_0)$, where $t_0>0$. Now, since ψ is strictly increasing, then $\psi (d(x_1,x_2))<\psi (q\psi (t_0))$. Put

$$q_1=\frac{\psi (q\psi (t_0))}{\psi (d(x_1,x_2))},$$

and so $q_1>1$. If $x_2\in T{x}_2$, then $x_2$ is a fixed point of T. Hence, we suppose that $x_2\notin T{x}_2$. Then

$$0<d(x_2,T{x}_2)\le H(T{x}_1,T{x}_2)<q_{1}H(T{x}_1,T{x}_2).$$

So there exists $x_3\in T{x}_2$ such that

$$0<d(x_2,x_3)<q_{1}H(T{x}_1,T{x}_2),$$

and then from (4.4), we get

$$0<d(x_2,x_3)<q_{1}H(T{x}_1,T{x}_2)\le q_1\psi (d(x_1,x_2))=\psi (q\psi (t_0)).$$

Again, since ψ is strictly increasing, then $\psi (d(x_2,x_3))<\psi (\psi (q\psi (t_0)))$. Put

$$q_2=\frac{\psi (\psi (q\psi (t_0)))}{\psi (d(x_2,x_3))}.$$

So, $q_2>1$. If $x_3\in T{x}_3$, then $x_3$ is a fixed point of T. Hence, we assume that $x_3\notin T{x}_3$. Then

$$0<d(x_3,T{x}_3)\le H(T{x}_2,T{x}_3)<q_{2}H(T{x}_2,T{x}_3),$$

and so, there exists $x_4\in T{x}_3$ such that

$$0<d(x_3,x_4)\le H(T{x}_2,T{x}_3)<q_{2}H(T{x}_2,T{x}_3).$$

(4.5)

Clearly, $x_2\ne x_3$. Also again, since ${\alpha }_{\ast }(T{x}_1,T{x}_2)\ge {\eta }_{\ast }(T{x}_1,T{x}_2)$, $x_2\in T{x}_1$ and $x_3\in T{x}_2$, then $\alpha (x_2,x_3)\ge \eta (x_2,x_3)$, and so, ${\alpha }_{\ast }(T{x}_2,T{x}_3)\ge {\eta }_{\ast }(T{x}_2,T{x}_3)$. Then from (4.1), we have

$$H(T{x}_2,T{x}_3)\le \psi (d(x_2,x_3)),$$

and so, from (4.5), we deduce that

$$d(x_3,x_4)<q_{2}H(T{x}_2,T{x}_3)\le q_2\psi (d(x_2,x_3))=\psi \left(\psi (q\psi (t_0))\right).$$

By continuing this process, we obtain a sequence $\{x_n\}$ in X such that $x_n\in T{x}_{n-1}$, $x_n\ne x_{n-1}$, ${\alpha }_{\ast }(x_n,x_{n+1})\ge {\eta }_{\ast }(x_n,x_{n+1})$ and $d(x_n,x_{n+1})\le {\psi }^{n-1}(q\psi (t_0))$ for all $n\in \mathbb{N}$. Now, for all $m>n$, we can write

$$d(x_n,x_m)\le \sum _{k=n}^{m-1}d(x_k,x_{k+1})\le \sum _{k=n}^{m-1}{\psi }^{k-1}(q\psi (t_0)).$$

Therefore, $\{x_n\}$ is a Cauchy sequence. Since $(X,d)$ is a complete metric space, then there exists $z\in X$ such that $x_n\to z$ as $n\to \mathrm{\infty }$. Now, since $\alpha (x_n,z)\ge \eta (x_n,z)$ for all $n\in \mathbb{N}$, then ${\alpha }_{\ast }(T{x}_n,Tz)\ge {\eta }_{\ast }(T{x}_n,Tz)$, and so, from (4.1), we have

$$d(z,Tz)\le H(T{x}_n,Tz)+d(x_n,z)\le \psi (d(x_n,z))+d(x_n,z)$$

for all $n\in \mathbb{N}$. Taking limit as $n\to \mathrm{\infty }$ in the inequality above, we get $d(z,Tz)=0$, i.e., $z\in Tz$. □

If in Theorem 4.1 we take $\eta (x,y)=1$, we have the following corollary.

Corollary 4.1 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\alpha }_{\ast }$-admissible and closed-valued multifunction on X. Assume that

$$x,y\in X,{\alpha }_{\ast }(Tx,Ty)\ge 1⟹H(Tx,Ty)\le \psi (d(x,y)).$$

Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\alpha (x_0,x_1)\ge 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

If in Theorem 4.1 we take $\alpha (x,y)=1$, then we have the following result.

Corollary 4.2 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\eta }_{\ast }$-subadmissible and closed-valued multifunction on X. Assume that

$$x,y\in X,{\eta }_{\ast }(Tx,Ty)\le 1⟹H(Tx,Ty)\le \psi (d(x,y)).$$

Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\eta (x_0,x_1)\le 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\eta (x_n,x_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta (x_n,x)\le 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Corollary 4.3 (Theorem 2.1 and 2.3 of [12])

Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\alpha }_{\ast }$-admissible and closed-valued multifunction on X. Assume that

$${\alpha }_{\ast }(Tx,Ty)H(Tx,Ty)\le \psi (d(x,y))$$

(4.6)

for all $x,y\in X$. Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\alpha (x_0,x_1)\ge 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Proof Suppose that ${\alpha }_{\ast }(Tx,Ty)\ge 1$ for $x,y\in X$. Then by (4.6), we have

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

That is, conditions of Corollary 4.1 hold, and T has a fixed point. □

Similarly, we can deduce the following corollaries.

Corollary 4.4 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\alpha }_{\ast }$-admissible and closed-valued multifunction on X. Assume that

$${({\alpha }_{\ast }(Tx,Ty)+1)}^{H(Tx,Ty)}\le 2^{\psi (d(x,y))}$$

for all $x,y\in X$. Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\alpha (x_0,x_1)\ge 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Corollary 4.5 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\alpha }_{\ast }$-admissible and closed-valued multifunction on X. Assume that

$${(H(Tx,Ty)+\ell )}^{{\alpha }_{\ast }(Tx,Ty)}\le \psi (d(x,y))+\ell$$

for all $x,y\in X$, where $\ell >0$. Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\alpha (x_0,x_1)\ge 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\alpha (x_n,x_{n+1})\ge 1$ for all $n\in \mathbb{N}$, we have $\alpha (x_n,x)\ge 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Corollary 4.6 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\eta }_{\ast }$-subadmissible and closed-valued multifunction on X. Assume that

$$H(Tx,Ty)\le {\eta }_{\ast }(Tx,Ty)\psi (d(x,y))$$

for all $x,y\in X$. Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\eta (x_0,x_1)\le 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\eta (x_n,x_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta (x_n,x)\le 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Corollary 4.7 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\eta }_{\ast }$-subadmissible and closed-valued multifunction on X. Assume that

$$2^{H(Tx,Ty)}\le {({\eta }_{\ast }(Tx,Ty)+1)}^{\psi (d(x,y))}$$

for all $x,y\in X$. Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\eta (x_0,x_1)\le 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\eta (x_n,x_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta (x_n,x)\le 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.

Corollary 4.8 Let $(X,d)$ be a complete metric space, and let $T:X\to 2^X$ be an ${\alpha }_{\ast }$-admissible and closed-valued multifunction on X. Assume that

$$H(Tx,Ty)+\ell \le {(\psi (d(x,y))+\ell )}^{{\eta }_{\ast }(Tx,Ty)}$$

for all $x,y\in X$, where $\ell >0$. Also, suppose that the following assertions hold:

• (i) there exists $x_0\in X$ and $x_1\in T{x}_0$ such that $\eta (x_0,x_1)\le 1$;

• (ii) for a sequence $\{x_n\}\subset X$ converging to $x\in X$ and $\eta (x_n,x_{n+1})\le 1$ for all $n\in \mathbb{N}$, we have $\eta (x_n,x)\le 1$ for all $n\in \mathbb{N}$.

Then T has a fixed point.