Existence theorems for single-valued and set-valued mappings with w-distances in metric spaces

Abstract

In this paper, using the concept of w-distances, and we prove existence theorems for single-valued mappings and set-valued mappings in a complete metric space which generalize Takahashi, Wong, and Yao’s theorems.

1 Introduction

Let $\ell^{\infty}$ be the Banach space of bounded sequences with supremum norm and let $(\ell^{\infty})^{*}$ be the dual space of $\ell^{\infty}$. Let μ be an element of $(\ell^{\infty})^{*}$. We denote by $\mu(f)$ the value of μ at $f=\{x_{n}\} \in\ell^{\infty}$. Sometimes, we denote by $\mu_{n}(x_{n})$ the value $\mu(f)$. A linear functional μ on $\ell^{\infty}$ is called a mean if $\mu(e)=\|\mu\|=1$, where $e=\{1, 1, 1, \dots \}$. Hasegawa et al. [1] obtained the following unique fixed point theorem on a complete metric space.

Theorem 1.1

([1])

Let $(X,d)$ be a complete metric space and let S be a mapping of X into itself. Let $\ell^{\infty}$ be the Banach space of bounded sequences with the supremum norm. Suppose that there exist a real number r with $0\leq r<1$ and an element $x\in X$ such that $\{S^{n} x\}$ is bounded and

$$\mu_{n} d \bigl(S^{n}x,Sy \bigr)\leq r\mu_{n} d \bigl(S^{n}x,y \bigr),\quad \forall y\in X $$

for some mean μ on $l^{\infty}$. Then the following hold:

• (1) S has a unique fixed point $u\in X$;

• (2) for every $z\in X$, the sequence $\{S^{n} z\}$ converges to u.

By using the idea of Caristi’s fixed point theorem [2], Chuang et al. [3] proved a unique fixed point theorem for single-valued mappings which generalizes Theorem 1.1. Furthermore, they obtained an existence theorem for set-valued mappings in a complete metric space. Using these results, Chuang et al. [3] obtained new and well-known existence theorems in a complete metric space.

On the other hand, in 1996, Kada et al. [4] introduced the concept of w-distances on a metric space.

Let $(X,d)$ be a metric space. A function $p:X\times X\to[0, \infty)$ is said to be a w-distance [4] on X if the following are satisfied:

• (1) $p(x, z)\le p(x, y)+p(y, z)$ for all $x, y, z\in X$;

• (2) for any $x\in X$, $p(x, \cdot):X\to[0, \infty)$ is lower semicontinuous;

• (3) for any $\varepsilon>0$, there exists $\delta>0$ such that $p(z, x)\le\delta$ and $p(z, y)\le\delta$ imply $d(x, y)\le\varepsilon$.

Using the concept of w-distances, they improved important results in complete metric spaces. For example, they improved Caristi’s fixed point theorem [2], Ekeland’s variational principle [5] and the nonconvex minimization theorem according to Takahashi [6]. Motivated by Chuang et al. [3], Takahashi et al. [7] improved their unique fixed point theorem for single-valued mappings by using the concept of w-distances. Furthermore, they extended Chuang et al.’s existence theorem [3] for set-valued mappings to w-distances. However, Takahashi et al. [7] assumed that w-distances are symmetric.

In this paper, without assuming that w-distances are symmetric, we prove Takahashi et al.’s unique fixed point theorems for single-valued mappings and their existence theorem for set-valued mappings in a complete metric space. Using these results, we obtained new and well-known existence theorems in a complete metric space. In particular, using this unique fixed point theorem for single-valued mappings, we obtain a unique fixed point theorem of Caristi’s type [2] with lower semicontinuous functions and w-distances. It seems that the proofs are technical and useful.

2 Preliminaries

Throughout this paper, we denote by $\mathbb {N}$ and $\mathbb {R}$ the sets of positive integers and real numbers, respectively. Let X be a metric space with metric d. Then we denote by $W(X)$ the set of all w-distances on X. A w-distance p on X is called symmetric if $p(x,y)=p(y, x)$ for all $x, y\in X$. We denote by $W_{0}(X)$ the set of all symmetric w-distances on X. Note that the metric d is an element of $W_{0}(X)$. We also know that there are many important examples of w-distances on X; see [4, 8].

The following lemma was proved by Kada et al. [4]; see also Shioji et al. [9].

Lemma 2.1

([4])

Let $(X,d)$ be a complete metric space and let p be a w-distance on X. Let $\{x_{n}\}$ and $\{y_{n}\}$ be sequences in X. Let $\{s_{n}\}$ and $\{ t_{n}\}$ be sequences in $[0,\infty)$ converging to 0, and let $x, y, z \in X$. Then the following hold:

• (1) If $p(x_{n},y)\leq s_{n}$ and $p(x_{n},z)\leq t_{n}$ for all $n\in \mathbb {N}$, then $y=z$. In particular, if $p(x,y)=0$ and $p(x,z)=0$, then $y=z$;

• (2) if $p(x_{n},y_{n})\leq s_{n}$ and $p(x_{n},z)\leq t_{n}$ for all $n\in\mathbb {N}$, then the sequence $\{y_{n}\}$ converges to z;

• (3) if $p(x_{n},x_{m})\leq s_{n}$ for all $n, m\in\mathbb {N}$ with $m>n$, then the sequence $\{x_{n}\}$ is a Cauchy sequence;

• (4) if $p(y,x_{n})\leq s_{n}$ for all $n\in\mathbb {N}$, then $\{ x_{n}\}$ is a Cauchy sequence.

Let $(X,d)$ be a metric space and let g be a function of X into $(-\infty, \infty]=\mathbb {R}\cup\{\infty\}$. Then g is proper if there exists $x\in X$ such that $g(x)<\infty$. A function g is lower semicontinuous if for any $t\in\mathbb {R}$, the set $\{x\in X: g(x)\leq t\}$ is closed. A function g is bounded below if there exists $K\in \mathbb {R}$ such that

$$K\leq g(x), \quad\forall x\in X. $$

Kada et al. [4] improved Caristi’s fixed point theorem [2] as follows; see also [8], Theorem 2.2.8.

Theorem 2.2

([4])

Let $(X,d)$ be a complete metric space, $p\in W(X)$, and let $\phi:X\to (\infty,\infty]$ be a proper, bounded below, and lower semicontinuous function. Let $T:X\to X$ be a mapping such that for each $x\in X$,

$$p(x,Tx)+\phi(Tx)\leq\phi(x). $$

Then there exists $z\in X$ such that $Tz=z$ and $p(z,z)=0$.

A mean μ is called a Banach limit on $\ell^{\infty}$ if $\mu _{n}(x_{n+1})=\mu_{n}(x_{n})$ for all $\{x_{n}\}\in\ell^{\infty}$. We know that there exists a Banach limit on $\ell^{\infty}$. If μ is a Banach limit on $\ell^{\infty}$, then for $f=\{x_{n}\} \in\ell^{\infty}$,

$$\liminf_{n\rightarrow\infty} x_{n} \leq\mu_{n} (x_{n}) \leq\limsup_{n\rightarrow\infty} x_{n}. $$

In particular, if $f=\{x_{n}\} \in\ell^{\infty}$ and $x_{n}\to a\in\mathbb {R}$, then we have $\mu(f)= \mu_{n} (x_{n}) = a$. For the proof of existence of a Banach limit and its other elementary properties, see [8].

3 Existence theorems for single-valued mappings

In this section, using means and w-distances, we first prove an existence theorem for mappings in metric spaces which generalizes Takahashi et al. [7].

Theorem 3.1

Let $(X,d)$ be a complete metric space, let $p\in W(X)$ and let $\{x_{n}\}$ be a sequence in X such that $\{p(x_{n}, w)\}$ and $\{p(w, x_{n})\}$ are bounded for some $w\in X$. Let μ be a mean on $\ell^{\infty}$ and let $\phi:X\to(-\infty,\infty]$ be a proper, bounded below, and lower semicontinuous function. Let $S:X\to X$ be a mapping. Suppose that there exist $l, m\in\mathbb {N}\cup\{0\}$ such that

$$ \mu_{n} p \bigl(x_{n},S^{l}y \bigr)+ \mu_{n} p \bigl(S^{m}y, x_{n} \bigr) +\phi(Sy)\leq \phi(y) $$

(3.1)

for all $y\in X$. Then there exists $x_{0}\in X$ such that

• (1) $x_{0}$ is a unique fixed point of S in $\{x\in X: \phi(x)< \infty\}$;

• (2) $x_{0}= \lim_{k\to\infty}S^{k} y$ for all $y\in X$ with $\phi (y)< \infty$;

• (3) $\phi(x_{0})=\inf_{v\in X}\phi(v)$.

Proof

Since $\{p(x_{n}, w)\}$ is bounded for some $w\in X$, we have, for any $y\in X$, $\{p(x_{n},y)\}$ is bounded. In fact, we have, for any $n\in\mathbb {N}$,

$$p(x_{n},y)\leq p(x_{n},w)+p(w,y)\leq\sup _{k\in\mathbb {N}}p(x_{k},w)+p(w,y). $$

Furthermore, since $\{p(w, x_{n})\}$ is bounded, we see that $\{p(z, x_{n})\}$ is bounded for all $z\in X$. In fact, we have, for any $n\in\mathbb {N}$,

$$p(z, x_{n})\leq p(z, w)+p(w,x_{n})\leq p(z, w)+\sup _{k\in\mathbb {N}}p(w, x_{k}). $$

We have from (3.1)

$$ \mu_{n} p \bigl(x_{n},S^{l}y \bigr) + \phi(Sy)\leq\phi(y) \quad\mbox{and}\quad \mu_{n} p \bigl(S^{m}y, x_{n} \bigr) +\phi(Sy)\leq\phi(y) $$

(3.2)

for all $y\in X$. For $y\in X$ with $\phi(y)<\infty$, we have from (3.2) $\phi(S^{k}y)< \infty$ for all $k\in \mathbb {N}\cup\{0\}$ and hence

$$ \mu_{n} p \bigl(x_{n},S^{l}S^{k}y \bigr)\leq\phi \bigl(S^{k}y \bigr)-\phi \bigl(S^{k+1}y \bigr) $$

(3.3)

and

$$ \mu_{n} p \bigl(S^{m}S^{k}y, x_{n} \bigr)\leq\phi \bigl(S^{k}y \bigr)-\phi \bigl(S^{k+1}y \bigr). $$

(3.4)

Then we see that $\{\phi(S^{k}y)\}$ is a decreasing sequence which is bounded below. Hence $\lim_{k\to\infty}\phi(S^{k}y)$ exists. Put $s=\lim_{k\to\infty}\phi(S^{k}y)$. Since

$$ \mu_{n} p \bigl(x_{n},S^{l+k}y \bigr) \leq\phi \bigl(S^{k}y \bigr)-\phi \bigl(S^{k+1}y \bigr)\leq\phi \bigl(S^{k}y \bigr)-s $$

and

$$ \mu_{n} p \bigl(S^{m+k}y, x_{n} \bigr) \leq\phi \bigl(S^{k}y \bigr)-\phi \bigl(S^{k+1}y \bigr)\leq\phi \bigl(S^{k}y \bigr)-s $$

for all $k\in\mathbb {N}$, we have

$$\limsup_{k\to\infty} \mu_{n} p \bigl(x_{n},S^{l+k}y \bigr)\leq0 \quad\mbox{and}\quad \limsup_{k\to\infty} \mu_{n} p \bigl(S^{m+k}y, x_{n} \bigr)\leq0. $$

Then we have

$$ \lim_{k\to\infty}\mu_{n} p \bigl(x_{n},S^{l+k}y \bigr)=0 \quad\mbox{and}\quad \lim _{k\to \infty}\mu_{n} p \bigl(S^{m+k}y, x_{n} \bigr)=0. $$

(3.5)

We have, for any $k, n\in\mathbb {N}$,

$$p \bigl(S^{l+m+k}y,S^{l+m+k+1}y \bigr)\leq p \bigl(S^{l+m+k}y,x_{n} \bigr)+p \bigl(x_{n},S^{l+m+k+1}y \bigr). $$

Since μ is a mean on $\ell^{\infty}$, we have from (3.3) and (3.4), for any $k\in\mathbb {N}$,

$$\begin{aligned} p \bigl(S^{l+m+k}y, S^{l+m+k+1}y \bigr)&\leq \mu_{n}p \bigl(S^{l+m+k}y,x_{n} \bigr)+\mu _{n}p \bigl(x_{n},S^{l+m+k+1}y \bigr) \\ &\leq\phi \bigl(S^{l+k}y \bigr)-\phi \bigl(S^{l+k+1}y \bigr)+ \phi \bigl(S^{m+k+1}y \bigr)-\phi \bigl(S^{m+k+2}y \bigr). \end{aligned}$$

(3.6)

We have from (3.6), for any $h,k \in\mathbb {N}$ with $k> h$,

$$\begin{aligned} p \bigl(S^{l+m+h}y, S^{l+m+k}y \bigr) \leq{}& p \bigl(S^{l+m+h}y,S^{l+m+h+1}y \bigr) \\ &{} +p \bigl(S^{l+m+h+1}y,S^{l+m+h+2}y \bigr) +\cdots+ p \bigl(S^{l+m+k-1}y,S^{l+m+k}y \bigr) \\ \leq{}&\phi \bigl(S^{l+h}y \bigr)-\phi \bigl(S^{l+h+1}y \bigr)+ \phi \bigl(S^{m+h+1}y \bigr)-\phi \bigl(S^{m+h+2}y \bigr) \\ &{} + \phi \bigl(S^{l+h+1}y \bigr)-\phi \bigl(S^{l+h+2}y \bigr) + \phi \bigl(S^{m+h+2}y \bigr)-\phi \bigl(S^{m+h+3}y \bigr) + \cdots \\ &{} + \phi \bigl(S^{l+k-1}y \bigr)-\phi \bigl(S^{l+k}y \bigr) + \phi \bigl(S^{m+k}y \bigr)-\phi \bigl(S^{m+k+1}y \bigr) \\ ={}& \phi \bigl(S^{l+h}y \bigr)-\phi \bigl(S^{l+k}y \bigr)+ \phi \bigl(S^{m+h+1}y \bigr)-\phi \bigl(S^{m+k+1}y \bigr) \\ \leq{}&\phi \bigl(S^{l+h}y \bigr) -s +\phi \bigl(S^{m+h+1}y \bigr)-s \\ \leq{}&\phi \bigl(S^{l+h}y \bigr) -s +\phi \bigl(S^{m+h}y \bigr)-s \\ ={}& \alpha_{h}-s +\beta_{h}-s, \end{aligned}$$

(3.7)

where $\alpha_{h}=\phi(S^{l+h}y)$ and $\beta_{h}=\phi(S^{m+h}y)$. Since $\alpha_{h}-s +\beta_{h}-s \to0$ as $h\to\infty$, we see from Lemma 2.1 that $\{S^{l+m+k}y\}$ is a Cauchy sequence in X. Since X is complete, there exists $y_{0}\in X$ such that $\lim_{k\to\infty}S^{l+m+k} y=y_{0}$. We know from the definition of p that, for any $n\in\mathbb {N}$, $y\mapsto p(x_{n},y)$ is lower semicontinuous. Using this and following the technique of [7], we have, for any $n\in\mathbb {N}$,

$$p(x_{n},y_{0})\leq\liminf_{k\to\infty}p \bigl(x_{n},S^{l+m+k}y \bigr) $$

and hence

$$ \mu_{n} p(x_{n},y_{0})\leq \mu_{n} \Bigl(\liminf_{k\to\infty}p \bigl(x_{n},S^{l+m+k}y \bigr) \Bigr). $$

(3.8)

On the other hand, we have from (3.7), for any $h,k, n \in \mathbb {N}$ with $k> h$,

$$\begin{aligned} p \bigl(x_{n},S^{l+m+k}y \bigr)&\leq p \bigl(x_{n},S^{l+m+h}y \bigr)+p \bigl(S^{l+m+h}y, S^{l+m+k}y \bigr) \leq p \bigl(x_{n},S^{l+m+h}y \bigr)+\alpha_{h}-s + \beta_{h}-s \end{aligned}$$

and hence

$$\limsup_{k\to\infty}p \bigl(x_{n},S^{l+m+k}y \bigr) \leq p \bigl(x_{n},S^{l+m+h}y \bigr)+\alpha_{h}-s + \beta_{h}-s. $$

Applying μ to both sides of the inequality, we have

$$\mu_{n} \Bigl( \limsup_{k\to\infty}p \bigl(x_{n},S^{l+m+k}y \bigr) \Bigr) \leq\mu_{n} p \bigl(x_{n},S^{l+m+h}y \bigr)+\alpha_{h}-s +\beta_{h}-s. $$

Letting $h\to\infty$, we get from (3.5) that

$$\begin{aligned} \mu_{n} \Bigl( \limsup_{k\to\infty}p \bigl(x_{n},S^{l+m+k}y \bigr) \Bigr) &\leq\liminf _{h\to\infty}\mu_{n} p \bigl(x_{n},S^{l+m+h}y \bigr)+0 \\ &=\lim_{h\to\infty}\mu_{n} p \bigl(x_{n},S^{l+m+h}y \bigr) \\ &=0. \end{aligned}$$

(3.9)

Then we have from (3.8) and (3.9)

$$\begin{aligned} \mu_{n}p(x_{n},y_{0}) & \leq \mu_{n} \Bigl( \liminf_{k\to\infty }p \bigl(x_{n},S^{l+m+k}y \bigr) \Bigr) \\ & \leq\mu_{n} \Bigl( \limsup_{k\to\infty}p \bigl(x_{n},S^{l+m+k}y \bigr) \Bigr) \\ &\leq\lim_{k\to\infty}\mu_{n} p \bigl(x_{n},S^{l+m+k}y \bigr) \\ &=0. \end{aligned}$$

(3.10)

This implies that

$$\mu_{n}p(x_{n}, y_{0})=0. $$

Similarly, for another $u\in X$ with $\phi(u)<\infty$, there exists $u_{0}\in X$ such that $\lim_{k\to\infty}S^{l+m+k}u=u_{0}$ and $\mu_{n} p(x_{n},u_{0})=0$. We also have, for $k,n\in\mathbb {N}$,

$$p \bigl(S^{l+m+k}y,y_{0} \bigr)\leq p \bigl(S^{l+m+k}y, x_{n} \bigr)+p(x_{n}, y_{0}) $$

and hence

$$\begin{aligned} p \bigl(S^{l+m+k}y,y_{0} \bigr)&\leq \mu_{n}p \bigl(S^{l+m+k}y, x_{n} \bigr)+ \mu_{n} p(x_{n}, y_{0}) \\ &= \mu_{n}p \bigl(S^{l+m+k}y, x_{n} \bigr)+ 0 \\ &= \mu_{n}p \bigl(S^{l+m+k}y, x_{n} \bigr). \end{aligned}$$

(3.11)

Furthermore, we have, for $k,n\in\mathbb {N}$,

$$p \bigl(S^{l+m+k}y,u_{0} \bigr)\leq p \bigl(S^{l+m+k}y, x_{n} \bigr)+p(x_{n}, u_{0}) $$

and hence

$$\begin{aligned} p \bigl(S^{l+m+k}y,u_{0} \bigr)&\leq \mu_{n}p \bigl(S^{l+m+k}y, x_{n} \bigr)+ \mu_{n} p(x_{n}, u_{0}) \\ &= \mu_{n}p \bigl(S^{l+m+k}y, x_{n} \bigr)+ 0 \\ &= \mu_{n}p \bigl(S^{l+m+k}y, x_{n} \bigr). \end{aligned}$$

(3.12)

We know that $\mu_{n}p(S^{l+m+k}y, x_{n})\to0$ as $k\to\infty$. Thus, we have from (3.11), (3.12), and Lemma 2.1 $y_{0}=u_{0}$. Therefore we have $x_{0}=\lim_{k\to\infty}S^{k} z$ for all $z\in X$ with $\phi(z)<\infty$. Since ϕ is lower semicontinuous and $\lim_{k\to\infty}S^{k} z=x_{0}$ for all $z\in X$ with $\phi(z)<\infty$, we have

$$\phi(x_{0})\leq\liminf_{k\to\infty}\phi \bigl(S^{k}z \bigr)= \lim_{k\to\infty}\phi \bigl(S^{k} z \bigr)=\inf_{k\in\mathbb {N} \cup\{0\}}\phi \bigl(S^{k} z \bigr)\leq\phi(z). $$

This implies that

$$ \phi(x_{0})=\inf_{y\in X}\phi(y). $$

(3.13)

We finally prove that $x_{0}$ is a unique fixed point of S in $\{x\in X: \phi(x)< \infty\}$. Since, from (3.13),

$$0\leq\mu_{n} p \bigl(x_{n},S^{l}x_{0} \bigr)\leq\phi(x_{0})-\phi(Sx_{0})\leq0, $$

we have $\mu_{n} p(x_{n},S^{l}x_{0})=0$. We also know $\mu_{n} p(x_{n},x_{0})=0$. For $k,n \in\mathbb {N}$, we have

$$p \bigl(S^{k}S^{m}y, S^{l}x_{0} \bigr) \leq p \bigl(S^{k}S^{m}y, x_{n} \bigr)+p \bigl(x_{n},S^{l}x_{0} \bigr) $$

and

$$p \bigl(S^{k}S^{m}y, x_{0} \bigr)\leq p \bigl(S^{k}S^{m}y, x_{n} \bigr)+p(x_{n},x_{0} ). $$

Then, as in the above argument, we have

$$\begin{aligned} p \bigl(S^{k}S^{m}y, S^{l}x_{0} \bigr)&\leq\mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr)+\mu_{n}p \bigl(x_{n},S^{l}x_{0} \bigr) \\ &= \mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr) \end{aligned}$$

(3.14)

and

$$\begin{aligned} p \bigl(S^{k}S^{m}y, x_{0} \bigr)& \leq\mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr)+\mu_{n}p(x_{n},x_{0} ) \\ &= \mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr). \end{aligned}$$

(3.15)

We also know from (3.5) that $\mu_{n}p(S^{m+k}y, x_{n})\to0$ as $k\to\infty$. Therefore, from (3.14), (3.15), and Lemma 2.1 $S^{l}x_{0}=x_{0}$. Using $S^{l}x_{0}=x_{0}$, we have from (3.13)

$$\begin{aligned} 0&\leq\mu_{n} p(x_{n},Sx_{0})=\mu_{n} p \bigl(x_{n},S^{l+1}x_{0} \bigr) \\ &\leq\phi(Sx_{0})-\phi \bigl(S^{2}x_{0} \bigr) \\ &\leq\phi(x_{0})-\phi \bigl(S^{2}x_{0} \bigr) \leq0 \end{aligned}$$

and hence $\mu_{n} p(x_{n},Sx_{0})=0$. Since, for $k,n \in\mathbb {N}$,

$$p \bigl(S^{k}S^{m}y, Sx_{0} \bigr)\leq p \bigl(S^{k}S^{m}y, x_{n} \bigr)+p(x_{n}, Sx_{0} ), $$

we have

$$\begin{aligned} p \bigl(S^{k}S^{m}y, Sx_{0} \bigr)& \leq\mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr)+\mu_{n}p(x_{n},Sx_{0} ) \\ &= \mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr). \end{aligned}$$

(3.16)

We have from (3.15), (3.16), and Lemma 2.1 $Sx_{0}=x_{0}$. We show that $x_{0}$ is a unique fixed point of S in $\{x\in X: \phi (x)< \infty\}$. Indeed, if $z_{0}$ is a fixed point of S with $\phi(z_{0})<\infty$, then

$$0\leq\mu_{n}p(x_{n}, z_{0})=\mu_{n}p \bigl(x_{n}, S^{l}z_{0} \bigr)\leq \phi(z_{0})-\phi(Sz_{0})=\phi (z_{0})- \phi(z_{0})=0 $$

and hence $\mu_{n}p(x_{n}, z_{0})=0$. Since, for $k,n \in\mathbb {N}$,

$$p \bigl(S^{k}S^{m}y, z_{0} \bigr)\leq p \bigl(S^{k}S^{m}y, x_{n} \bigr)+p(x_{n},z_{0} ), $$

we have

$$\begin{aligned} p \bigl(S^{k}S^{m}y, z_{0} \bigr)& \leq\mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr)+\mu_{n}p(x_{n},z_{0} ) = \mu_{n}p \bigl(S^{k}S^{m}y, x_{n} \bigr). \end{aligned}$$

(3.17)

Since $\mu_{n}p(S^{m+k}y, x_{n})\to0$ as $k\to\infty$, from (3.15), (3.17), and Lemma 2.1, we have $z_{0}=x_{0}$. Therefore $x_{0}$ is a unique fixed point of S in $\{y\in X: \phi(y)<\infty\}$. This completes the proof. □

Using Theorem 3.1, we can obtain the following result proved by Takahashi et al. [7].

Theorem 3.2

([7])

Let $(X,d)$ be a complete metric space, let $p\in W_{0}(X)$ and let $\{x_{n}\}$ be a sequence in X such that $\{p(x_{n}, x)\}$ is bounded for some $x\in X$. Let μ be a mean on $\ell^{\infty}$ and let $\psi:X\to(-\infty,\infty]$ be a proper, bounded below, and lower semicontinuous function. Let $T:X\to X$ be a mapping. Suppose that there exists $m\in\mathbb {N}\cup \{0\}$ such that

$$\mu_{n} p \bigl(x_{n},T^{m}y \bigr)+ \psi(Ty)\leq\psi(y),\quad \forall y\in X. $$

(3.18)

Then there exists $\bar{x}\in X$ such that

• (a) $\bar{x}= \lim_{k\to\infty}T^{k} y$ for all $y\in X$ with $\psi (y)< \infty$;

• (b) $\psi(\bar{x})=\inf_{u\in X}\psi(u)$;

• (c) x̄ is a unique fixed point of T in $\{x\in X: \psi (x)< \infty\}$.

Proof

Since $\{x_{n}\}$ is a bounded sequence in X such that $\{p(x_{n}, x)\}$ is bounded for some $x\in X$, we see from $p\in W_{0}(X)$ that $\{p(x, x_{n})\}$ is bounded. Putting $S=T$, $l=m$, and $\phi=2\psi$ in Theorem 3.1, we have

$$ 2\mu_{n} p \bigl(T^{m}y, x_{n} \bigr) +2\psi(Ty) \leq2\psi(y),\quad \forall y\in X $$

and hence

$$ \mu_{n} p \bigl(T^{m}y, x_{n} \bigr) +\psi(Ty)\leq \psi(y), \quad\forall y\in X. $$

Thus we have the desired result from Theorem 3.1. □

Using Theorem 3.1 and the generalized Caristi’s fixed point theorem (Theorem 2.2), we also have a unique fixed point theorem of Caristi’s type [2] with lower semicontinuous functions and w-distances.

Theorem 3.3

Let $(X,d)$ be a complete metric space and let $p\in W(X)$ such that $p(x,x)=0$ for all $x\in X$. Let $\phi:X\to(-\infty, \infty]$ be a proper, bounded below, and lower semicontinuous function. Let $S:X\to X$ be a mapping. Suppose that there exists $\alpha\in\mathbb{R}$ such that

$$\begin{aligned} &\alpha \bigl(p(Sx,y)+p(y,Sx) \bigr) +(1-\alpha) \bigl(p(x,y)+p(y,x) \bigr)+\phi(Sy)\leq\phi(y),\quad \forall x,y\in X. \end{aligned}$$

(3.19)

Then there exists $x_{0}\in X$ such that

• (1) $x_{0}$ is a unique fixed point of S in $\{x\in X: \phi(x)< \infty\}$;

• (2) $x_{0}= \lim_{k\to\infty}S^{k} y$ for all $y\in X$ with $\phi (y)< \infty$;

• (3) $\phi(x_{0})=\inf_{v\in X}\phi(v)$.

Proof

Let us first consider $\alpha>0$. Putting $y=x$ in (3.19), we have from $p(x,x)=0$

$$ \alpha \bigl(p(Sx,x)+p(x,Sx) \bigr)+\phi(Sx)\leq\phi(x), \quad\forall x\in X $$

and hence

$$\alpha p(x,Sx)+\phi(Sx)\leq\phi(x),\quad \forall x\in X. $$

By Theorem 2.2, there exists $u_{0}\in X$ such that $Su_{0}=u_{0}$. Putting $x=u_{0}$ in (3.19) again, we have, for any $y\in X$,

$$ \alpha \bigl(p(Su_{0},y)+p(y,Su_{0}) \bigr)+(1-\alpha) \bigl(p(u_{0},y)+p(y,u_{0}) \bigr)+\phi(Sy)\leq \phi(y). $$

Since $Su_{0}=u_{0}$, we have, for any $y\in X$,

$$p(u_{0},y)+p(y,u_{0})+\phi(Sy)\leq\phi(y). $$

By Theorem 3.1, we see that $x_{0}$ is a unique fixed point of S in $\{x\in X: \phi(x)<\infty\}$ such that $\phi(x_{0})=\inf_{u\in X}\phi(u)$ and $x_{0}=\lim_{k\to\infty}S^{k} z$ for all $z\in X$ with $\phi(z)< \infty$.

Next let us consider the case of $\alpha=0$. Then we have

$$ p(x,y)+p(y,x)+\phi(Sy)\leq\phi(y), \quad\forall x,y\in X. $$

(3.20)

Replacing x and y by Sx and x in (3.20), respectively, we have

$$ p(Sx,x)+p(x,Sx)+\phi(Sx)\leq\phi(x),\quad \forall x\in X $$

and hence

$$p(x,Sx)+\phi(Sx)\leq\phi(x), \quad\forall x\in X. $$

We also see from Theorem 2.2 that there exists $u_{0}\in X$ such that $Su_{0}=u_{0}$. Putting $x=u_{0}$ in (3.19), we have also

$$ p(u_{0},y)+p(y,u_{0})+\phi(Sy)\leq\phi(y),\quad \forall y \in X. $$

By Theorem 3.1, we see that $x_{0}$ is a unique fixed point of S in $\{x\in X: \phi(x)<\infty\}$ such that $\phi(x_{0})=\inf_{u\in X}\phi(u)$ and $x_{0}=\lim_{k\to\infty}S^{k} z$ for all $z\in X$ with $\phi(z)< \infty$.

In the case of $\alpha<0$, we have $1-\alpha>0$. Furthermore, replacing y by Sx in (3.19), we have from $p(Sx, Sx)=0$

$$ (1-\alpha) \bigl(p(x,Sx)+p(Sx,x) \bigr)+\phi \bigl(S^{2}x \bigr)\leq\phi(Sx), \quad\forall x\in X $$

(3.21)

and hence

$$(1-\alpha)p(x,Sx)+\phi \bigl(S^{2}x \bigr)\leq\phi(Sx),\quad \forall x \in X. $$

Take $x\in X$ with $\phi(x)<\infty$. Then we have, for any $n\in \mathbb {N}$,

$$\begin{aligned} &(1-\alpha)p(x,Sx)+\phi \bigl(S^{2}x \bigr)\leq\phi(Sx), \\ &(1-\alpha)p \bigl(Sx,S^{2}x \bigr)+\phi \bigl(S^{3}x \bigr) \leq \phi \bigl(S^{2}x \bigr), \\ & \vdots \\ &(1-\alpha)p \bigl(S^{n-1}x,S^{n}x \bigr)+\phi \bigl(S^{n+1}x \bigr)\leq\phi \bigl(S^{n}x \bigr). \end{aligned}$$

Adding these inequalities, we have

$$(1-\alpha) \bigl\{ p(x,Sx)+p \bigl(Sx,S^{2}x \bigr)+\cdots+ p \bigl(S^{n-1}x,S^{n}x \bigr) \bigr\} \leq\phi(Sx) -\phi \bigl(S^{n+1}x \bigr). $$

Since $\{\phi(S^{n}x)\}$ is a decreasing sequence and bounded below, we see that there exists $s= \lim_{n\to\infty}\phi(S^{n}x)$. Thus we have, for any $n\in\mathbb {N}$,

$$\begin{aligned} (1-\alpha)p \bigl(x,S^{n}x \bigr)& \leq(1-\alpha) \bigl\{ p(x,Sx)+p \bigl(Sx,S^{2}x \bigr)+\cdots +p \bigl(S^{n-1}x,S^{n}x \bigr) \bigr\} \\ & \leq\phi(Sx) -\phi \bigl(S^{n+1}x \bigr) \\ & \leq \phi(Sx) -s < \infty. \end{aligned}$$

Then $\{p(x,S^{n}x)\}$ is bounded. Furthermore, from (3.21) we have

$$(1-\alpha)p(Sx,x)+\phi \bigl(S^{2}x \bigr)\leq\phi(Sx),\quad \forall x \in X. $$

As in the above argument, we have, for any $n\in\mathbb {N}$,

$$(1-\alpha)p \bigl(S^{n}x, x \bigr)\leq \phi(Sx) -s < \infty. $$

Then $\{p(S^{n}x,x)\}$ is bounded. Replacing x by $S^{n}x$ in (3.19), we have, for any $n\in \mathbb {N}$,

$$\begin{aligned} &\alpha \bigl(p \bigl(S^{n+1}x,y \bigr)+p \bigl(y,S^{n+1}x \bigr) \bigr) \\ &\quad{}+(1-\alpha) \bigl(p \bigl(S^{n}x,y \bigr)+p \bigl(y,S^{n}x \bigr) \bigr)+\phi(Sy)\leq\phi(y), \quad\forall y\in X. \end{aligned}$$

Applying a Banach limit μ to the both sides of this inequality, we have

$$\begin{aligned} &\alpha \bigl(\mu_{n} p \bigl(S^{n+1}x,y \bigr)+ \mu_{n} p \bigl(y,S^{n+1}x \bigr) \bigr) \\ &\quad{}+(1-\alpha) \bigl(\mu_{n} p \bigl(S^{n}x,y \bigr)+ \mu_{n} p \bigl(y,S^{n}x \bigr) \bigr)+\phi(Sy)\leq\phi (y), \quad \forall y\in X. \end{aligned}$$

Since $\mu_{n} p(S^{n+1}x,y)+\mu_{n} p(y,S^{n+1}x) = \mu_{n} p(S^{n}x,y)+\mu _{n} p(y,S^{n}x)$, we get

$$ \mu_{n} \bigl(p \bigl(S^{n}x,y \bigr)+p \bigl(y,S^{n}x \bigr) \bigr)+\phi(Sy)\leq\phi(y), \quad\forall y\in X. $$

(3.22)

By Theorem 3.1, S has a unique fixed point $x_{0}$ in $\{x\in X: \phi(x)<\infty\}$ such that $\phi(x_{0})=\inf_{u\in X}\phi(u)$ and $x_{0}=\lim_{k\to\infty}S^{k} z$ for all $z\in X$ with $\phi(z)< \infty$. □

4 Existence theorems for set-valued mappings

Using w-distances, we have the following existence theorem for set-valued mappings in a complete metric space. Let $(X,d)$ be a metric space and let $P(X)$ be the class of all nonempty subsets of X. A mapping of X into $P(X)$ is called a set-valued mapping, or a multi-valued mapping.

Theorem 4.1

Let $(X,d)$ be a complete metric space, let $p\in W(X)$, and let $\{x_{n}\}$ be a sequence in X such that $\{p(x_{n}, w)\}$ and $\{p(w, x_{n})\}$ are bounded for some $w\in X$. Let μ be a mean on $\ell^{\infty}$ and let $\phi:X\to(-\infty, \infty]$ be a proper, bounded below, and lower semicontinuous function. Let $S:X\to P(X)$ be a set-valued mapping such that for each $x\in X$, there exists $y\in Sx$ satisfying

$$ \mu_{n} p(x_{n},x)+\mu_{n} p(x,x_{n})+ \phi(y)\leq\phi(x). $$

(4.1)

Then there exists $x_{0}\in X$ such that

• (1) $x_{0}\in Sx_{0}$;

• (2) $\phi(x_{0})=\inf_{y\in X}\phi(y)$;

• (3) for any $z\in X$ with $\phi(z)<\infty$, there exists a sequence $\{z_{m}\}\subset X$ such that $z_{m+1}\in Sz_{m}$, $m\in\mathbb {N} \cup\{0\}$ and $z_{m}\to x_{0}$ as $m\to\infty$.

Proof

For each $z_{1}=z\in X$ with $\phi(z)<\infty$, there exists $z_{2}\in Sz_{1}$ such that

$$\mu_{n} p(x_{n},z_{1})+\mu_{n} p(z_{1}, x_{n})\leq\phi(z_{1})- \phi(z_{2}). $$

Repeating this process, we get a sequence $\{z_{m}\}$ in X such that $z_{m+1}\in Sz_{m}$ and

$$ \mu_{n} p(x_{n},z_{m})+ \mu_{n} p(z_{m},x_{n})\leq\phi(z_{m})- \phi(z_{m+1}) $$

(4.2)

for each $m\in\mathbb{N}$. Clearly, $\{\phi(z_{m})\}$ is a decreasing sequence which is bounded below. Hence $\lim_{m\to\infty}\phi(z_{m})$ exists. Put $s=\lim_{m\to\infty}\phi(z_{m})$. We have from (4.2)

$$ \lim_{m\to\infty}\mu_{n} p(x_{n},z_{m})=0 \quad\mbox{and}\quad \lim_{m\to\infty } \mu_{n} p(z_{m}, x_{n})=0. $$

(4.3)

We have, for any $m, n\in\mathbb {N}$,

$$p(z_{m},z_{m+1})\leq p(z_{m},x_{n})+p(x_{n},z_{m+1}). $$

Since μ is a mean on $\ell^{\infty}$, we have, for any $m\in\mathbb {N}$,

$$\begin{aligned} p(z_{m},z_{m+1})&\leq\mu_{n}p(z_{m},x_{n})+ \mu_{n}p(x_{n},z_{m+1}) \\ &\leq\phi(z_{m})-\phi(z_{m+1})+\phi(z_{m+1})- \phi(z_{m+2}) \\ &= \phi(z_{m})-\phi(z_{m+2}). \end{aligned}$$

(4.4)

We have from (4.4), for any $l,m \in\mathbb {N}$ with $m> l$,

$$\begin{aligned} p(z_{l},z_{m})\leq{}& p(z_{l},z_{l+1})+p(z_{l+1},z_{l+2}) +\cdots +p(z_{m-1},z_{m}) \\ \leq{}&\phi(z_{l})-\phi(z_{l+2})+\phi(z_{l+1})- \phi(z_{l+3}) \\ &{} +\cdots+\phi(z_{m-1})-\phi(z_{m+1}) \\ ={}& \phi(z_{l})+\phi(z_{l+1})-\phi(z_{m})- \phi(z_{m+1}) \\ \leq{}&\phi(z_{l})+\phi(z_{l+1})-s-s \\ \leq{}&\phi(z_{l})+\phi(z_{l})-s-s \\ ={}& 2\phi(z_{l})-2s \end{aligned}$$

(4.5)

and $2\phi(z_{l})-2s \to0$ as $l\to\infty$. We see from Lemma 2.1 that $\{z_{m}\}$ is a Cauchy sequence in X. Since X is complete, there exists a point $x_{0}\in X$ such that $\lim_{m\to\infty}z_{m}=x_{0}$. We know from the definition of p that, for any $n\in\mathbb {N}$, $y\mapsto p(x_{n},y)$ is lower semicontinuous. Using this and following the technique of [7], we have, for any $n\in\mathbb {N}$,

$$p(x_{n},x_{0})\leq\liminf_{m\to\infty}p(x_{n},z_{m}) $$

and hence

$$ \mu_{n}p(x_{n},x_{0})\leq \mu_{n} \Bigl( \liminf_{m\to\infty}p(x_{n},z_{m}) \Bigr). $$

(4.6)

On the other hand, we have from (4.5), for any $l,k, n \in \mathbb {N}$ with $m> l$,

$$\begin{aligned} p(x_{n},z_{m})&\leq p(x_{n},z_{l})+p(z_{l}, z_{m}) \\ &\leq p(x_{n},z_{l})+2\phi(z_{l})-2s \end{aligned}$$

and hence

$$\limsup_{m\to\infty}p(x_{n},z_{m})\leq p(x_{n},z_{l})+2\phi(z_{l})-2s. $$

Applying μ to both sides of the inequality, we have

$$\mu_{n} \Bigl( \limsup_{m\to\infty}p(x_{n},z_{m}) \Bigr) \leq\mu_{n} p(x_{n},z_{l})+2 \phi(z_{l})-2s. $$

Letting $l\to\infty$, we get

$$ \mu_{n} \Bigl( \limsup_{m\to\infty}p(x_{n},z_{m}) \Bigr) \leq\liminf_{l\to\infty}\mu_{n} p(x_{n},z_{l}). $$

(4.7)

We have from (4.3), (4.6), and (4.7)

$$\begin{aligned} \mu_{n}p(x_{n},x_{0}) & \leq \mu_{n} \Bigl( \liminf_{m\to\infty}p(x_{n},z_{m}) \Bigr) \\ & \leq\mu_{n} \Bigl( \limsup_{m\to\infty}p(x_{n},z_{m}) \Bigr) \\ &\leq\liminf_{m\to\infty}\mu_{n} p(x_{n},z_{m}) \\ &= \lim_{m\to\infty}\mu_{n} p(x_{n},z_{m})=0. \end{aligned}$$

(4.8)

This implies that

$$\mu_{n}p(x_{n}, x_{0})=0. $$

Doing the same argument as above for each $y_{1}=y\in X$ with $\phi (y)<\infty$, we can construct a sequence $\{y_{m}\}$ in X such that $\{\phi(y_{m})\}$ is a decreasing sequence, $\lim_{m\to\infty}y_{m}=y_{0}$ for some $y_{0}\in X$, and $\mu_{n} p(x_{n},y_{0})=0$. We show that $x_{0}=y_{0}$. We have, for any $m,n \in\mathbb {N}$,

$$p(z_{m}, x_{0})\leq p(z_{m}, x_{n})+p(x_{n},x_{0} ). $$

Then, we have

$$\begin{aligned} p(z_{m}, x_{0})&\leq\mu_{n}p(z_{m}, x_{n})+\mu_{n}p(x_{n},x_{0} ) \\ &= \mu_{n}p(z_{m}, x_{n}). \end{aligned}$$

(4.9)

Furthermore, we have, for any $m,n \in\mathbb {N}$,

$$p(z_{m}, y_{0})\leq p(z_{m}, x_{n})+p(x_{n},y_{0} ) $$

and hence

$$\begin{aligned} p(z_{m}, y_{0})&\leq\mu_{n}p(z_{m}, x_{n})+\mu_{n}p(x_{n},y_{0} ) \\ &= \mu_{n}p(z_{m}, x_{n}). \end{aligned}$$

(4.10)

We know from (4.3) that $\mu_{n}p(z_{m}, x_{n})\to0$ as $m\to\infty$. Therefore, from (4.9), (4.10), and Lemma 2.1 $x_{0}=y_{0}$. Since ϕ is lower semicontinuous,

$$\phi(x_{0})=\phi(y_{0})\leq\liminf_{m\to\infty} \phi(y_{m})= \lim_{m\to\infty}\phi(y_{m})=\inf _{m\in\mathbb{N}}\phi(y_{m})\leq\phi(y_{1}). $$

Since $y_{1}$ is any point of X with $\phi(y_{1})<\infty$, we have

$$ \phi(x_{0})=\inf_{y\in X}\phi(y). $$

(4.11)

Using (4.1), we have $u_{0}\in X$ such that $u_{0}\in Sx_{0}$ and

$$ \mu_{n} p(x_{n},x_{0})+ \mu_{n} p(x_{0},x_{n})\leq\phi(x_{0})- \phi(u_{0}). $$

(4.12)

Furthermore, repeating this process, we have $v_{0}\in X$ such that $v_{0}\in Su_{0}$ and

$$ \mu_{n} p(x_{n},u_{0})+\mu_{n} p(u_{0},x_{n})\leq\phi(u_{0})- \phi(v_{0}). $$

Using (4.11), we have

$$ \mu_{n} p(x_{n},u_{0})+ \mu_{n} p(u_{0},x_{n})\leq\phi(u_{0})- \phi(v_{0})\leq\phi (u_{0})-\phi(x_{0}). $$

(4.13)

Then we have from (4.12) and (4.13)

$$ \mu_{n} p(x_{n},u_{0})+\mu_{n} p(u_{0},x_{n})+\mu_{n} p(x_{n},u_{0})+ \mu_{n} p(u_{0},x_{n})\leq0. $$

This implies that

$$\mu_{n} p(x_{n},u_{0})=0. $$

Since $p(z_{m}, u_{0})\leq p(z_{m}, x_{n})+p(x_{n},u_{0} )$ for $m,n\in\mathbb {N}$, we have

$$\begin{aligned} p(z_{m}, u_{0})&\leq\mu_{n}p(z_{m}, x_{n})+\mu_{n}p(x_{n},u_{0} ) \\ &= \mu_{n}p(z_{m}, x_{n}). \end{aligned}$$

(4.14)

We know from (4.3) that $\mu_{n}p(z_{m}, x_{n})\to0$ as $m\to\infty$. Therefore, from (4.9), (4.14), and Lemma 2.1 $x_{0}=u_{0}$. Since $u_{0}\in Sx_{0}$, we have $x_{0}\in Sx_{0}$. This completes the proof. □

Let $(X,d)$ be a metric space. Then $S:X\to P(X)$ is called a multi-valued weakly Picard operator [10] if for each $x\in X$ and each $y\in Sx$, there exists a sequence $\{x_{n}\}$ in X such that

• (1) $x_{0}=x$, $x_{1}=y$;

• (2) $x_{n+1}\in Sx_{n}$, $n\in\mathbb {N} \cup\{0\}$;

• (3) $\{x_{n}\}$ is convergent and its limit is a fixed point of S.

Using Theorem 4.1, we can get the following result proved by Takahashi et al. [7].

Theorem 4.2

([7])

Let $(X,d)$ be a complete metric space, let $p\in W_{0}(X)$ and let $\{x_{n}\}$ be a sequence in X such that $\{p(x_{n}, x)\}$ is bounded for some $x\in X$. Let μ be a mean on $\ell^{\infty}$ and let $\psi:X\to(-\infty, \infty)$ be a bounded below and lower semicontinuous function. Let $T:X\to P(X)$ be a set-valued mapping such that for each $u\in X$, there exists $v\in Tu$ satisfying

$$\mu_{n} p(x_{n},u)+ \psi(v)\leq\psi(u). $$

Then T is a multi-valued weakly Picard operator.

Proof

Putting $S=T$ and $\phi=2\psi$ in Theorem 4.1, we see that, for each $x\in X$, there exists $y\in Tx$ such that

$$ 2\mu_{n} p(x_{n},x) +2\psi(y)\leq2\psi(x) $$

and hence

$$ \mu_{n} p(x_{n},x) +\psi(y)\leq\psi(x). $$

For each $x\in X$ and each $y\in Tx$, put $u_{0}=x$ and $u_{1}=y$. Then we can take $u_{2}\in Tu_{1}$ such that

$$\mu_{n} p(x_{n},u_{1})+ \psi(u_{2})\leq \psi(u_{1}). $$

Repeating this process, we get a sequence $\{u_{m}\}$ in X such that $u_{m+1}\in Tu_{m}$ and

$$ \mu_{n} p(x_{n},u_{m})\leq \psi(u_{m})-\psi(u_{m+1}) $$

(4.15)

for each $m\in\mathbb{N}\cup\{0\}$. Thus we have the desired result from Theorem 4.1. □