Generalizations of the strong Ekeland variational principle with a generalized distance in complete metric spaces

Abstract

In this paper, we prove a generalization of the strong Ekeland variational principle for a generalized distance (i.e., u-distance) on complete metric spaces. The result present in this paper extends and improves the corresponding result of Georgiev (J. Math. Anal. Appl. 131:1-21, 1988) and Suzuki (J. Math. Anal. Appl. 320:788-794, 2006).

1 Introduction

In 1974, Ekeland [1] proved the following, which is called the Ekeland variational principle (for short, EVP).

Theorem 1.1 [1]

Let $(X,d)$ be a complete metric space with metric d and f be a function from X into $(-\mathrm{\infty },+\mathrm{\infty }]$ which is proper lower semicontinuous bounded from below. Then for $u\in X$ and $\lambda >0$, there exists $v\in X$ such that

• (P)
  $$f(v)\le f(u)-\lambda d(u,v)$$

  ;

• (Q)
  $$f(w)>f(v)-\lambda d(v,w)$$

  for every $w\ne v$.

Later, Takahashi [2] showed that this principle is equivalent to the Caristis fixed point theorem and nonconvex minimization theorem. In 1988, Georgiev [3] proved the following generalization of Theorem 1.1, which is called the strong Ekeland variational principle.

Theorem 1.2 [3]

Let X be a complete metric space with metric d and $f:X\to (-\mathrm{\infty },+\mathrm{\infty }]$ be proper lower semicontinuous bounded from below. Then, for all $u\in X$, $\lambda >0$ and $\delta >0$, there exists $v\in X$ satisfying the following:

(P)′ $f(v)<f(u)-\lambda d(u,v)+\delta$;

• (Q)
  $$f(w)>f(v)-\lambda d(v,w)$$

  for every $w\in X\setminus \{v\}$;

• (R) if a sequence $\{x_n\}$ in X satisfies ${lim}_{n\to \mathrm{\infty }}(f(x_n)+\lambda d(v,x_n))=f(v)$, then $\{x_n\}$ converges to v.

On the other hand, Kada et al. [4] introduced the concept of w-distance defined on a metric space and extended the Ekeland variational principle, the Kirk-Caristi fixed point theorem and the minimization theorem for w-distance. Recently, Suzuki [5, 6] introduced a more general concept than w-distance, which is called τ-distance, and established the strong Ekeland variational principle for τ-distance. Very recently, Ume [7] introduced a more generalized concept than τ-distance, which is called u-distance, and proved a new minimization and a new fixed point theorem by using u-distance on a complete metric space.

In this paper, we prove the strong Ekeland variational principle for u-distance on a complete metric space. The results of this paper extend and generalize some results in Georgiev [3], Suzuki [5], Ansari [9] and Park [10].

2 Preliminaries

Throughout the paper, we denote by ℕ the set of all positive integers, by ℝ the set of real numbers, ${\mathbb{R}}_{+}=[0,\mathrm{\infty })$. Let us recall the following well-known definition of a u-distance.

Definition 2.1 ([8] and [7])

Let X be a complete metric space with metric d. Then a function $p:X\times X\to {\mathbb{R}}_{+}$ is called a u-distance on X if there exists a function $\theta :X\times X\times [0,\mathrm{\infty })\times [0,\mathrm{\infty })\to {\mathbb{R}}_{+}$ such that

(u1) $p(x,z)\le p(x,y)+p(y,z)$ for all $x,y,z\in X$;

(u2) $\theta (x,y,0,0)=0$, $\theta (x,y,s,t)\ge min\{s,t\}$ for all $x,y\in X$ and $s,t\in [0,\mathrm{\infty })$, and for any $x\in X$ and for every $ϵ>0$, there exists $\delta >0$ such that $|s-s_0|<\delta$, $|t-t_0|<\delta$, $s,s_0,t,t_0\in [0,\mathrm{\infty })$ and $y\in X$ imply

$$|\theta (x,y,s,t)-\theta (x,y,s_0,t_0)|<ϵ;$$

(u3) ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ and ${lim}_{n\to \mathrm{\infty }}sup\{\theta (w_n,z_{n,},p(w_n,x_m),p(z_n,x_m)):m\ge n\}=0$ imply $p(y,x)\le {lim}_{n\to \mathrm{\infty }}infp(y,x_n)$ for all $y\in X$;

(u4) ${lim}_{n\to \mathrm{\infty }}sup\{p(x_n,w_m):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}sup\{p(y_n,z_m):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}\theta (x_n,w_n,s_n,t_n)=0$ and ${lim}_{n\to \mathrm{\infty }}\theta (y_n,z_n,s_n,t_n)=0$ imply ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,s_n,t_n)=0$ or ${lim}_{n\to \mathrm{\infty }}sup\{p(w_m,x_n):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}sup\{p(z_m,y_n):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}\theta (x_n,w_n,s_n,t_n)=0$ and ${lim}_{n\to \mathrm{\infty }}\theta (y_n,z_n,s_n,t_n)=0$ imply ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,s_n,t_n)=0$;

(u5) ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,p(w_n,x_n),p(z_n,x_n))=0$ and ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,p(w_n,y_n),p(z_n,y_n))=0$ imply ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$ or ${lim}_{n\to \mathrm{\infty }}\theta (a_n,b_n,p(x_n,a_n),p(x_n,b_n))=0$ and ${lim}_{n\to \mathrm{\infty }}\theta (a_n,b_n,p(y_n,a_n),p(y_n,b_n))=0$ imply ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$.

Proposition 2.2 [7]

Let p be a u-distance on a metric space $(X,d)$ and c be a positive real number. Then a function $q:X\times X\to {\mathbb{R}}_{+}$ defined by $q(x,y)=c\cdot p(x,y)$ for every $x,y\in X$ is also a u-distance on X.

Lemma 2.3 [7]

Let $(X,d)$ be a metric space and let p be a u-distance on X. If $\{x_n\}$ is a p-Cauchy sequence, then $\{x_n\}$ is a Cauchy sequence.

Lemma 2.4 [7]

Let $(X,d)$ be a metric space and p be a u-distance on X. Suppose that a sequence $\{x_n\}$ of X satisfies

$$\underset{n\to \mathrm{\infty }}{lim}sup\{p(x_n,x_m):m>n\}=0$$

or

$$\underset{n\to \mathrm{\infty }}{lim}sup\{p(x_m,x_n):m>n\}=0.$$

Then, $\{x_n\}$ is a p-Cauchy sequence and $\{x_n\}$ is a Cauchy sequence.

3 Main theorem

Lemma 3.1 Let X be a complete metric space and p be a u-distance on X. If a sequence $\{x_n\}$ of X satisfies ${lim}_{n\to \mathrm{\infty }}p(z,x_n)=0$ for some $z\in X$, then $\{x_n\}$ is a p-Cauchy sequence. Moreover, if a sequence $\{y_n\}$ of X also satisfies ${lim}_{n\to \mathrm{\infty }}p(z,y_n)=0$, then ${lim}_{n\to \mathrm{\infty }}p(x_n,y_n)=0$. In particular, for $x,y,z\in X$, $p(z,x)=0$ and $p(z,y)=0$ imply $x=y$.

Proof Let θ be a function from $X\times X\times [0,\mathrm{\infty })\times [0,\mathrm{\infty })$ into ${\mathbb{R}}_{+}$ satisfying (u1)-(u5). From ${lim}_{n}p(z,x_n)=0$, it follows by (u2) that ${lim}_{n\to \mathrm{\infty }}\theta (z,z,p(z,x_n),p(z,x_n))=0$. Therefore, $\{x_n\}$ is a p-Cauchy sequence. □

Theorem 3.2 Let X be a complete metric space and T be a map** from X into itself. Suppose that there exists a u-distance p on X and $r\in [0,1)$ such that $p(Tx,T^{2}x)\le r\cdot p(x,Tx)$ for all $x\in X$. Assume that either of the following hold:

• (i) If ${lim}_{n\to \mathrm{\infty }}sup\{p(x_n,x_m):m>n\}=0$, ${lim}_{n\to \mathrm{\infty }}p(x_{n,T{x}_n})=0$ and ${lim}_{n\to \mathrm{\infty }}p(x_n,y)=0$, then $Ty=y$;

• (ii) if $\{x_n\}$ and $\{T{x}_n\}$ converge to y, then $Ty=y$;

• (iii) T is continuous.

Then, there exists $x_0\in X$ such that $T{x}_0=x_0$ and $p(x_0,x_0)=0$.

Proof It is the same as the proof of Theorem 1 in [5]. □

Lemma 3.3 Let X be a complete metric space, p be a u-distance on X and ϕ be a function from $X\times X$ into $(-\mathrm{\infty },\mathrm{\infty }]$ satisfying

• (1)
  $$\varphi (x,z)\le \varphi (x,y)+\varphi (y,z)$$

  for all $x,y,z\in X$;

• (2)
  $$\varphi (x,\cdot ):X\to (-\mathrm{\infty },\mathrm{\infty }]$$

  is lower semicontinuous for any $x\in X$;

• (3) there exists an $x_0$ such that ${inf}_{y\in X}\varphi (x_0,y)>-\mathrm{\infty }$; and

• (4)
  $$\varphi (x,y)=-\varphi (y,x)$$

  .

Define $Mx=\{y\in X:\varphi (x,y)+p(x,y)\le 0\}$. Let $u\in X$ and $c\in {\mathbb{R}}_{+}$ such that $\varphi (x,u)<\mathrm{\infty }$ for all $x\in X$, $Mu\ne \mathrm{\varnothing }$ and $c\ge \varphi (x,u)-{inf}_{y\in Mu}\varphi (u,y)$. Then a function $q:X\times X\to {\mathbb{R}}_{+}$ defined by

$$q(x,y)=\{\begin{array}{cc}\varphi (u,x)-{inf}_{y\in Mx}\varphi (u,y)\hfill & \mathit{\text{if }}x\in Mu\mathit{\text{ and }}y\in Mx,\hfill \\ c+p(x,y)\hfill & \mathit{\text{if }}x\notin Mu\mathit{\text{ or }}y\notin Mx\hfill \end{array}$$

is a u-distance on X.

Proof Let η be a function from $X\times X\times {\mathbb{R}}_{+}\times {\mathbb{R}}_{+}$ into ${\mathbb{R}}_{+}$ satisfying (u2)-(u5) for a u-distance. We note that $\varphi (x,y)+\varphi (y,z)+p(x,y)+p(y,z)\le 0$ and $\varphi (x,z)+p(x,z)\le 0$. Thus, $y\in Mx$ and $z\in My$ imply $z\in Mx$. If $x\in Mu$ and $y\in Mx$, then

$$\begin{array}{rcl}p(x,y)& \le & \varphi (y,x)\le q(x,y)=\varphi (y,x)-\underset{y\in Mx}{inf}\varphi (x,y)\\ \le & \varphi (x,u)-\underset{y\in Mu}{inf}\varphi (x,y)\le c.\end{array}$$

Therefore, $p(x,y)\le q(y,x)\le c+p(x,y)$ for all $x,y\in X$. To complete the proof, we will show (u1) _q , ${\text{(u3)}}_{q,\eta }$, ${\text{(u4)}}_{q,\eta }$ and ${\text{(u5)}}_{q,\eta }$. Let x, y and z be fixed elements in X. In the case $x\in Mu$, $y\in Mx$, $y\in Mu$ and $z\in My$, we have $z\in Mx$ and hence $q(x,z)=q(x,y)\le q(x,y)+q(y,z)$. In the other case, we note that

$$\begin{array}{rcl}q(x,z)& \le & c+p(x,z)\le c+p(x,y)+p(y,z)\\ \le & 2c+p(x,y)+p(y,z)\\ =& q(x,y)+q(y,z).\end{array}$$

This shows (u1) _q .

We next suppose that ${lim}_{n\to \mathrm{\infty }}{x}_n=x$ and ${lim}_{n\to \mathrm{\infty }}sup\{\eta (w_n,z_n,q(w_n,x_m),q(z_n,x_m)):m\ge n\}=0$ and fix $w\in X$. Since ${lim}_{n\to \mathrm{\infty }}sup\{\theta (w_n,z_n,p(w_n,x_m),p(z_n,x_m)):m\ge n\}=0$, we have $p(w,x)\le lim{inf}_{n\to \mathrm{\infty }}p(w,x_n)$ for all $y\in X$.

In the case that $w\in Mu$ and there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k}\in Mw$ for all $k\in \mathbb{N}$, we have

$$\begin{array}{rcl}\varphi (w,x)+p(w,x)& \le & \underset{n\to \mathrm{\infty }}{lim}inf\varphi (w,x_n)+\underset{n\to \mathrm{\infty }}{lim}p(w,x_n)\\ \le & \underset{n\to \mathrm{\infty }}{lim}inf(\varphi (w,x_n)+p(w,x_n))\\ \le & \underset{k\to \mathrm{\infty }}{lim}inf(\varphi (w,x_{n_k})+p(w,x_{n_k}))\\ \le & 0,\end{array}$$

and so $x\in Mu$. Hence

$$q(w,x)=\varphi (u,w)-\underset{x\in Mw}{inf}\varphi (u,x)=\underset{k\to \mathrm{\infty }}{lim}q(w,x_{n_k})=\underset{n\to \mathrm{\infty }}{lim}infq(w,x_n).$$

In the other case, we obtain

$$\begin{array}{rcl}q(w,x)& \le & c+p(w,x)\le \underset{n\to \mathrm{\infty }}{lim}inf(c+p(w,x_n))\\ =& \underset{n\to \mathrm{\infty }}{lim}infq(w,x_n).\end{array}$$

This shows ${\text{(u3)}}_{q,\eta }$. We will show that q satisfies ${\text{(u4)}}_{q,\eta }$.

Case I: Suppose that ${lim}_{n\to \mathrm{\infty }}sup\{q(x_n,w_m):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}sup\{q(y_n,z_m):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}\eta (x_n,w_n,s_n,t_n)=0$, and ${lim}_{n\to \mathrm{\infty }}\eta (y_n,z_n,s_n,t_n)=0$.

In the case $x_n\in Mu$ and $w_m\in M{x}_n$, we note that $q(x_n,w_n)=\varphi (u,x_n)-{inf}_{w_m\in M{x}_n}\varphi (u,w_m)$. Since $\varphi (x_n,w_m)+p(x_n,w_n)\le 0$, it follows that

$$\begin{array}{rcl}p(x_n,w_m)& \le & -\varphi (x_n,w_n)=\varphi (w_m,x_n)\\ \le & \varphi (w_m,u)+\varphi (u,x_n)\\ =& \varphi (u,x_n)-\varphi (u,w_m)\\ \le & \varphi (u,x_n)-\underset{w_m\in M{x}_n}{inf}\varphi (u,w_m)=q(x_n,w_m).\end{array}$$

Thus, we have $p(x_n,w_m)\le q(x_n,w_m)$. This implies that ${sup}_{m\ge n}p(x_n,w_n)\le {sup}_{m\ge n}q(x_n,w_m)$. Take $n\to \mathrm{\infty }$, so

$$0\le \underset{n\to \mathrm{\infty }}{lim}supp(x_n,w_m)\le \underset{n\to \mathrm{\infty }}{lim}supq(x_n,w_m)=0$$

and therefore ${lim}_{n\to \mathrm{\infty }}supp(x_n,w_m)=0$.

Similarly, if $y_n\in Mu$ and $z_m\in M{y}_n$, then ${lim}_{n\to \mathrm{\infty }}supp(y_n,z_m)=0$.

We note that ${lim}_{n\to \mathrm{\infty }}\theta (x_n,w_n,s_n,t_n)=0={lim}_{n\to \mathrm{\infty }}\theta (y_n,z_n,s_n,t_n)$ and hence

$$\underset{n\to \mathrm{\infty }}{lim}\eta (w_n,z_n,s_n,t_n)=0.$$

In the case $x_n\ne Mu$ or $w_m\ne M{x}_n$, we note that $p(x_n,w_m)\le c+p(x_n,w_m)=q(x_n,w_m)$. Thus, we have $p(x_n,w_m)\le q(x_n,w_m)$. This implies that ${sup}_{m\ge n}p(x_n,w_m)\le {sup}_{m\ge n}q(x_n,w_m)$. Taking $n\to \mathrm{\infty }$, we obtain

$$0\le \underset{n\to \mathrm{\infty }}{lim}supp(x_n,w_m)\le \underset{n\to \mathrm{\infty }}{lim}supq(x_n,w_m)=0$$

and therefore ${lim}_{n\to \mathrm{\infty }}supp(x_n,w_n)=0$. Similarly as above, if $y_n\ne Mu$ and $z_m\ne M{y}_n$, then ${lim}_{n\to \mathrm{\infty }}supp(y_n,z_m)=0$. We note that ${lim}_{n\to \mathrm{\infty }}\theta (x_n,w_n,s_n,t_n)=0={lim}_{n\to \mathrm{\infty }}\theta (y_n,z_n,s_n,t_n)$ and hence ${lim}_{n\to \mathrm{\infty }}\eta (w_n,z_n,s_n,t_n)=0$.

Case II: Suppose that ${lim}_{n\to \mathrm{\infty }}sup\{q(w_m,x_n):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}sup\{q(z_m,y_n):m\ge n\}=0$, ${lim}_{n\to \mathrm{\infty }}\eta (x_n,w_n,s_n,t_n)=0$ and ${lim}_{n\to \mathrm{\infty }}\eta (y_n,z_n,s_n,t_n)=0$. Similarly as in Case I, we can show that ${lim}_{n\to \mathrm{\infty }}\eta (w_n,z_n,s_n,t_n)=0$. This shows ${\text{(u4)}}_{q,\eta }$. We will show that q satisfies ${\text{(u5)}}_{q,\eta }$.

Case I: Suppose that ${lim}_{n\to \mathrm{\infty }}\eta (w_n,z_n,q(x_n,w_n),q(x_n,z_n))=0$ and ${lim}_{n\to \mathrm{\infty }}\eta (w_n,z_n,q(y_n,w_n),q(y_n,z_n))=0$. In the case $x_n\in Mu$ and $w_n,z_n\in M{x}_n$, we note that $q(x_n,w_n)=\varphi (u,x_n)-{inf}_{w_n\in M{x}_n}\varphi (u,w_n)$ and hence $q(x_n,z_n)=\varphi (u,x_n)-{inf}_{z_n\in M{x}_n}\varphi (u,z_n)$. Thus, we have

$$\begin{array}{rl}\theta (w_n,z_n,p(x_n,w_n),p(x_n,z_n))\le & \theta (w_n,z_n,\varphi (z_n,x_n),\varphi (z_n,x_n))\\ \le & \theta (w_n,z_n,\varphi (w_n,u)+\varphi (u,x_n),\varphi (z_n,u)+\varphi (u,x_n))\\ =& \theta (w_n,z_n,\varphi (u,x_n)-\varphi (u,w_n),\varphi (u,x_n)-\varphi (u,z_n))\\ \le & \theta (w_n,z_n,\varphi (u,x_n)-\underset{w_n\in M{x}_n}{inf}\varphi (u,w_n),\varphi (u,x_n)\\ -\underset{z_n\in M{x}_n}{inf}\varphi (u,z_n))\\ =& \eta (w_n,z_n,q(x_n,w_n),q(x_n,z_n)).\end{array}$$

Taking $n\to \mathrm{\infty }$, we have

$$0\le \underset{n\to \mathrm{\infty }}{lim}\theta (w_n,z_n,p(x_n,w_n),p(x_n,z_n))\le \underset{n\to \mathrm{\infty }}{lim}\eta (w_n,z_n,q(x_n,w_n),q(x_n,z_n))=0.$$

Therefore ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,p(x_n,w_n),p(x_n,z_n))=0$. Similarly, if $y_n\in Mu$ and $z_n,w_n\in M{y}_n$, then ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,p(y_n,w_n),p(y_n,z_n))=0$. In the case $x_n\ne Mu$ or $w_n,z_n\ne M{x}_n$, we have $q(x_n,w_n)=c+p(x_n,w_n)$ and $q(x_n,z_n)=c+p(x_n,z_n)$. Since p is a u-distance, we have ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$. Hence

$$\begin{array}{rcl}\theta (w_n,z_n,c+p(x_n,w_n),c+p(x_n,z_n))& \le & \theta (w_n,z_n,c+p(x_n,w_n),c+p(x_n,z_n))\\ \le & \eta (w_n,z_n,q(x_n,w_n),q(x_n,z_n)).\end{array}$$

Take $n\to \mathrm{\infty }$, thus

$$0\le \underset{n\to \mathrm{\infty }}{lim}\theta (w_n,z_n,p(x_n,w_n),p(x_n,z_n))\le \underset{n\to \mathrm{\infty }}{lim}\eta (w_n,z_n,q(x_n,w_n),q(x_n,z_n))=0.$$

Therefore ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,p(x_n,w_n),p(x_n,z_n))=0$. Similarly, if $y_n\ne Mu$ or $w_n,z_n\ne M{y}_n$, then ${lim}_{n\to \mathrm{\infty }}\theta (w_n,z_n,p(y_n,w_n),p(y_n,z_n))=0$. Since p is a u-distance, we have ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$.

Case II: Suppose that ${lim}_{n\to \mathrm{\infty }}\eta (w_n,z_n,q(w_n,x_n),q(z_n,x_n))=0$ and ${lim}_{n\to \mathrm{\infty }}\eta (w_n,z_n,q(w_n,y_n),q(z_n,y_n))=0$. Similarly as in Case I, we can show that ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$. This shows ${\text{(u5)}}_{q,\eta }$. □

Proposition 3.4 Let X be a complete metric space, p be a u-distance on X and ϕ be a function from $X\times X$ into $(-\mathrm{\infty },\mathrm{\infty }]$ satisfying

• (1)
  $$\varphi (x,z)\le \varphi (x,y)+\varphi (y,z)$$

  for all $x,y,z\in X$;

• (2)
  $$\varphi (x,\cdot ):X\to (-\mathrm{\infty },\mathrm{\infty }]$$

  is lower semicontinuous for any $x\in X$;

• (3) there exists an $x_0$ such that ${inf}_{y\in X}\varphi (x_0,y)>-\mathrm{\infty }$; and

• (4)
  $$\varphi (x,y)=-\varphi (y,x)$$

  .

Define $Mx=\{y\in X:\varphi (x,y)+p(x,y)\le 0\}$ for all $x\in X$. Then, for each $u\in X$ with $Mu\ne \mathrm{\varnothing }$, there exists $x_0\in Mu$ such that $M{x}_0\subset \{x_0\}$. In particular, there exists $y_0\in X$ such that $M{y}_0\subset \{y_0\}$.

Proof Let $u\in X$ with $Mu\ne \mathrm{\varnothing }$. We have $u_1\in Mu$ by $\varphi (u,u_1)<\mathrm{\infty }$. If $Mu=\mathrm{\varnothing }$, the assertion holds. Suppose that $M{u}_1\ne \mathrm{\varnothing }$ and $Mx\cap (X\{x\})\ne \mathrm{\varnothing }$ for all $x\in M{u}_1$. Let $u_2\in M{u}_1$. We know that $\varphi (x,y)\le 0$ for all $x\in X$ and $y\in Mx$, we define a map** $T:X\to X$ as follows: For each $x\in M{u}_1,Tx$ satisfies $Tx\in Mx$, $Tx\ne x$ and

$$\varphi (u_1,Tx)\le \frac{\varphi (u_1,x)+{inf}_{y\in Mx}\varphi (u_1,y)}{2}.$$

For each $x\notin M{u}_1$, define $Tx=u_2\ne x$. We also define a function $q:X\times X\to {\mathbb{R}}^{+}$ by

$$q(x,y)=\{\begin{array}{cc}\varphi (u,x)-{inf}_{y\in Mx}\varphi (u_1,y)\hfill & \text{if }x\in M{u}_1\text{ and }y\in Mx,\hfill \\ 2\varphi (u,u_1)-2{inf}_{w\in M{u}_1}\varphi (u,w)+1+p(x,y)\hfill & \mathit{\text{if }}x\notin M{u}_1\mathit{\text{ or }}y\notin Mx.\hfill \end{array}$$

By Lemma 3.3, we have q is a u-distance on X. Since $y\in My$ and $z\in My$, it follows by Lemma 3.3 that $z\in Mx$. Hence $Tx\in M{u}_1$ and $MTx\subset Mx$ for all $x\in M{u}_1$. If $x\in M{u}_1$, we obtain

$$\begin{array}{rcl}q(Tx,T^{2}x)& =& \varphi (u_1,Tx)-\underset{y\in MTx}{inf}\varphi (u_1,y)\\ \le & \frac{\varphi (u_1,x)+{inf}_{y\in Mx}\varphi (u_1,y)}{2}-\underset{y\in Mx}{inf}\varphi (u_1,y)\\ =& \frac{q(x,Tx)}{2}.\end{array}$$

If $x\notin M{u}_1$,

$$\begin{array}{rcl}q(Tx,T^{2}x)& =& q(u_2,T{u}_2)=\varphi (u_1,u_2)-\underset{T{u}_2\in M{u}_2}{inf}\varphi (u_1,T{u}_2)\\ \le & \varphi (u,u_1)-\underset{T{u}_1}{inf}\varphi (u,T{u}_1)\\ \le & \frac{q(x,u_2)}{2}=\frac{q(x,Tx)}{2}.\end{array}$$

We will show (i) in Theorem 3.2. Suppose that ${lim}_{n\to \mathrm{\infty }}sup\{q(x_n,x_m):m>n\}=0$ and ${lim}_{n\to \mathrm{\infty }}q(x_n,y)=0$. We may assume $x_n\in M{u}_1$ and $y\in M{x}_n$ for all $n\in \mathbb{N}$ by the definition of q. Then $y\in M{u}_1$ and hence $Ty\in My\subset M{x}_n$. By Lemma 2.4 we have ${lim}_{n\to \mathrm{\infty }}q(x_n,Ty)={lim}_{n\to \mathrm{\infty }}q(x_n,y)=0$ and $Ty=y$. Hence, by Theorem 3.2, T has a fixed point. This is a contradiction. So, there is $x_0\in M{u}_1\subset Mu$ such that $M{x}_0\subset \{x_0\}$. □

Theorem 3.5 Let X be a complete metric space, p be a u-distance on X and ϕ be a function from $X\times X$ into $(-\mathrm{\infty },\mathrm{\infty }]$ satisfying

• (1)
  $$\varphi (x,z)\le \varphi (x,y)+\varphi (y,z)$$

  for all $x,y,z\in X$;

• (2)
  $$\varphi (x,\cdot ):X\to (-\mathrm{\infty },\mathrm{\infty }]$$

  is lower semicontinuous for any $x\in X$;

• (3) there exists an $x_0$ such that ${inf}_{y\in X}\varphi (x_0,y)>-\mathrm{\infty }$; and

• (4)
  $$\varphi (x,y)=-\varphi (y,x)$$

  .

Then the following hold:

• (A) For each $u\in X$, there exists $v\in X$ such that $\varphi (u,v)\le 0$ and $\varphi (v,w)+p(v,w)>0$ for all $w\in X\setminus \{v\}$;

• (B) For each $\lambda >0$ and $u\in X$ with $p(u,u)=0$, there exists $v\in X$ such that $\varphi (u,v)+\lambda p(u,v)\le 0$ and $\varphi (v,w)+\lambda p(v,w)>0$ for all $w\in X\setminus \{v\}$.

Proof We will show that (A). For each $x\in X$, we define Mx as in Proposition 3.4. If $Mu=\mathrm{\varnothing }$, we have u that satisfies $\varphi (u,w)+p(u,w)>0$ for all $w\in X$ with $w\ne u$. If $Mu\ne \mathrm{\varnothing }$ and there exists $v\in Mu$, then it follows by Proposition 3.4 that $Mv\subset \{v\}$. Since $v\in Mu$ implies $\varphi (u,v)\le 0$ and $Mv\subset \{v\}$, this shows that $\varphi (v,w)+p(v,w)>0$ for all $w\in X$ with $w\ne v$.

We will show that (B). By Proposition 2.2, we note that λp is a u-distance. We define $Mx=\{y\in X:\varphi (x,y)+\lambda p(x,y)\le 0\}$ for all $x\in X$. Since $p(u,u)=0$, we have $Mu\ne \mathrm{\varnothing }$, and hence there exists $v\in Mu$ such that $Mv\subseteq \{v\}$ by Proposition 3.4. Therefore v satisfies $\varphi (u,v)+\lambda p(u,v)\le 0$ and $\varphi (v,w)+\lambda p(v,w)>0$ for all $w\in X$ with $w\ne v$. This completes the proof. □

Remark 3.6 By setting $\varphi (x,y)=f(y)-f(x)$, where $f:X\to \mathbb{R}$ is lower semicontinuous bounded below, and letting p be a τ-distance in Theorem 3.5, we obtain the Ekeland variational principle proved by Suzuki [5].

Theorem 3.7 Let X be a complete metric space, p be a u-distance on X and ϕ be a function from $X\times X$ into $(-\mathrm{\infty },\mathrm{\infty }]$ satisfying

• (1)
  $$\varphi (x,z)\le \varphi (x,y)+\varphi (y,z)$$

  for all $x,y,z\in X$;

• (2)
  $$\varphi (x,\cdot ):X\to (-\mathrm{\infty },\mathrm{\infty }]$$

  is lower semicontinuous for any $x\in X$;

• (3) there exists an $x_0$ such that ${inf}_{y\in X}\varphi (x_0,y)>-\mathrm{\infty }$; and

• (4)
  $$\varphi (x,y)=-\varphi (y,x)$$

  .

Let $u\in X$ with $p(u,u)=0$. Then $\lambda >0$ and $\delta >0$, there exists $v\in X$ satisfying the following:

• (i)
  $$\varphi (u,v)\le 0$$

  ;

• (ii)
  $$\varphi (u,v)+\lambda p(u,v)<\delta$$

  ;

• (iii)
  $$\varphi (v,w)+\lambda p(v,w)>0$$

  for all $w\in X\setminus \{v\}$;

• (iv) if a sequence $\{x_n\}$ in X satisfies ${lim}_n(\varphi (v,x_n)+\lambda p(v,x_n))=0$, then $\{x_n\}$ is p-Cauchy, ${lim}_n{x}_n=v$ and $p(v,v)={lim}_{n}p(v,x_n)=0$.

Proof In the case $\varphi (v,u)=\mathrm{\infty }$, (i) and (ii) hold for all $v\in X$. We also note that (iii) and (iv) do not depend on $\varphi (v,u)$. In the case $\varphi (v,u)<\mathrm{\infty }$, set ${\lambda }^{\prime }\in (0,\lambda )$ satisfying

$$\frac{\lambda -{\lambda }^{\prime }}{{\lambda }^{\prime }}(\varphi (u,v)-\underset{x\in X}{inf}\varphi (v,x))<\delta .$$

By Theorem 3.5(B), there exists $v\in X$ such that $\varphi (u,v)+{\lambda }^{\prime }p(u,v)\le 0$ and $\varphi (v,w)+{\lambda }^{\prime }p(v,w)>0$ for all $w\in X\setminus \{v\}$. Thus, we have

Therefore, $\varphi (u,v)+\lambda p(u,v)<\delta$. For $w\in X\setminus \{v\}$, we note that

$$\varphi (v,w)>-{\lambda }^{\prime }p(v,w)\ge -\lambda p(v,w).$$

So, $\varphi (v,w)+\lambda p(v,w)>0$. Finally, we will show that (iv). Suppose that a sequence $\{x_n\}$ in X satisfies ${lim}_n(\varphi (v,x_n)+\lambda p(v,x_n))=0$. We note that $\varphi (v,w)+{\lambda }^{\prime }p(v,w)\ge 0$ for all $w\in X$. We have

$$\begin{array}{rcl}\underset{n\to \mathrm{\infty }}{lim}supp(v,x_n)& =& \underset{n\to \mathrm{\infty }}{lim}sup\left(\frac{\lambda -{\lambda }^{\prime }}{\lambda -{\lambda }^{\prime }}\right)p(v,x_n)\\ =& \underset{n\to \mathrm{\infty }}{lim}\frac{\lambda p(v,x_n)-{\lambda }^{\prime }p(v,x_n)}{\lambda -{\lambda }^{\prime }}\\ \le & \underset{n\to \mathrm{\infty }}{lim}\frac{\lambda p(v,x_n)-\varphi (v,x_n)}{\lambda -{\lambda }^{\prime }}\\ \le & \underset{n\to \mathrm{\infty }}{lim}\frac{\lambda p(v,x_n)+\varphi (v,x_n)}{\lambda -{\lambda }^{\prime }}\\ =& 0.\end{array}$$

By Lemma 3.1, $\{x_n\}$ is a p-Cauchy sequence. From Lemma 2.3, therefore $\{x_n\}$ is a Cauchy sequence. By the completeness of X, $\{x_n\}$ converges to some point $x\in X$. From (u3), we have $p(v,x)=0$ and so

$$\begin{array}{rcl}\varphi (v,x)& \le & \underset{n\to \mathrm{\infty }}{lim}inf\varphi (v,x_n)\\ \le & \underset{n\to \mathrm{\infty }}{lim}(\varphi (v,x_n)+\lambda p(v,x_n))=0.\end{array}$$

Thus, if $v\ne x$, then we have

$$\varphi (v,x)>-{\lambda }^{\prime }p(v,x)\ge \varphi (v,x).$$

This is a contradiction. Hence, we obtain $v=x$. □

Remark 3.8 By setting $\varphi (x,y)=f(y)-f(x)$, where $f:X\to \mathbb{R}$ is lower semicontinuous bounded below. Let p be a τ-distance in Theorem 3.7, we obtain the strong Ekeland variational principle proved by Suzuki [6].