# 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.

## 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. (1)

S has a unique fixed point $$u\in X$$;

2. (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. (1)

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

2. (2)

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

3. (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.

## 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. (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. (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. (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. (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].

## 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. (1)

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

2. (2)

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

3. (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

1. (a)

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

2. (b)

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

3. (c)

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. (1)

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

2. (2)

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

3. (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 $$\phi(z)< \infty$$. □

## 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. (1)

$$x_{0}\in Sx_{0}$$;

2. (2)

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

3. (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. (1)

$$x_{0}=x$$, $$x_{1}=y$$;

2. (2)

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

3. (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. □

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## Acknowledgements

The second author was partially supported by Grant-in-Aid for Scientific Research No. 15K04906 from Japan Society for the Promotion of Science. The fourth author was partially supported by the grant MOST 103-2923-E-039-001-MY3.

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### Corresponding author

Correspondence to Jen-Chih Yao.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Kaneko, S., Takahashi, W., Wen, CF. et al. Existence theorems for single-valued and set-valued mappings with w-distances in metric spaces. Fixed Point Theory Appl 2016, 38 (2016). https://doi.org/10.1186/s13663-016-0527-2

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• DOI: https://doi.org/10.1186/s13663-016-0527-2

• 47H10
• 37C25
• 58J20

### Keywords

• complete metric space
• contractive mapping
• fixed point theorem
• generalized hybrid mapping
• w-distance