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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. (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.

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. (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].

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. (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 \(\phi(z)< \infty\). □

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. (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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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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