A New Equilibrium Version of Ekeland’s Variational Principle and Its Applications

Abstract

In this note, a new equilibrium version of Ekeland’s variational principle is presented. It is a modification and promotion of previous results. Subsequently, the principle is applied to discuss the equilibrium points for binary functions and the fixed points for nonlinear mappings.

1. Introduction

Ekeland’s variational principle (abbrev. $\mathbf{EVP}$), which is considered to be the basis of modern calculus of variations, was presented in 1974 (see, for instance [1,2]). It is widely used in many fields, such as differential equations, optimization, fixed point theory, etc. It is precisely the wide application of this theorem that it has attracted the attention of a large number of scholars, and has been promoted from all directions. For example, Zhong [3] extended the form of EVP in metric space; we rewrite the result as follows.

Theorem 1

(EVP of Zhong-type [3]). Let $(X,d)$ be a complete metric space and $x_0\in X$ fixed. The function $f:X\to R\cup \{+\infty \}$ is bounded from below, lower semi-continuous, and not identically $+\infty$.

If $g:[0,+\infty )\to [0,+\infty )$ is a continuous non-decreasing function such satisfying

$${\int }_0^{+\infty }\frac{1}{1+g\left(r\right)}dr=+\infty ,$$

then, for any $\epsilon >0,y\in M$ such that

$$f\left(y\right)<\underset{x\in X}{inf}f\left(x\right)+\epsilon$$

and, for any $\lambda >0$, there exists $z\in X$ satisfying

$$f\left(z\right)\le f\left(y\right)$$

$$d(z,x_0)\le \overline{r}+r_0$$

and

$$f\left(x\right)\ge f\left(z\right)-\frac{\epsilon }{\lambda (1+g\left(d(x_0,z)\right))}d(x,z)\forall x\in M$$

where $r_0=d(x_0,y)$ and $\overline{r}$ is such that

$${\int }_{r_0}^{\overline{r}+r_0}\frac{1}{1+g\left(r\right)}dr\ge \lambda .$$

Oettli and Théra [4] and Blum and Oettli [5] investigated the equilibrium versions of EVP. In [6], Bianchi et al. presented equilibrium versions of EVP as follows

Let X be an Euclidean space, $C\subseteq X$ be a closed set and $f:C\times C\to R$.

Theorem 2

([6]). Assume the following assumptions are satisfied:

(i)

$f(x,·)$ is lower bounded and lower semicontinuous, for every $x\in C$;

(ii)

$f(t,t)=0$, for every $t\in C$;

(iii)

$f(z,x)\le f(z,t)+f(t,x)$ for every $x,t,z\in C$.

Then, for every $\epsilon >0$ and for every $x_0\in C$, there exists $\overline{x}\in C$ such that

(a)

$f(x_0,\overline{x})+\epsilon \parallel x_0-\overline{x}\parallel \le 0$;

(b)

$f(\overline{x},x)+\epsilon \parallel \overline{x}-x\parallel >0,\forall x\in C,x\ne \overline{x}$.

Farkas and Molnar [7] improved the conclusion in [6], and obtained a Zhong-type variational principle for bi-functions as follows:

Theorem 3

([7]). Let $(X,d)$ be a complete metric space, $C\subset X$ be a closed set, and $f:C\times C\to {\mathbf{R}}^{+}$ be a mapping. Let $g:[0,+\infty )\to (0,+\infty )$ be a continuous nondecreasing function such that

$${\int }_0^{+\infty }\frac{1}{g\left(s\right)}ds=+\infty$$

Let $x_0\in C$ be fixed. Assume that the following assumptions be satisfied:

(i)

$f(x,·)$ is bounded from below and lower semicontinuous, for every $x\in C$;

(ii)

$f(z,z)=0$, for every $z\in C$;

(iii)

$f(z,x)\le f(z,t)+f(t,x)$ for every $x,t,z\in C$;

Then, for every $\epsilon >0$ and $y\in C$ for which we have

$$\underset{z\in C}{inf}f(y,z)>-\epsilon$$

(1)

and for every $\lambda >0$, there exists $x_{\epsilon }$ such that

(a)

$d(x_0,x_{\epsilon })<r+\overline{r}$;

(b)

$f(x_{\epsilon },x_0)+\frac{\epsilon }{\lambda (1+g\left(d(x_0,x_{\epsilon })\right))}d(x_{\epsilon },x_0)\le 0$;

(c)

$f(x_{\epsilon },x)+\frac{\epsilon }{\lambda (1+g\left(d(x_0,x_{\epsilon })\right))}d(x,x_{\epsilon })>0$,$\forall x\in C,x\ne x_{\epsilon }$;

where $r_0=d(x_0,y)$ and $\overline{r}$ are chosen such that

$${\int }_{r_0}^{r_0+\overline{r}}\frac{1}{1+g\left(r\right)}dr\ge \lambda .$$

However, when proving (a), there are some errors in [7].

In the process of proving $\{d(x_0,x_n)<r_0+\overline{r}\}\left(14\right)$, they presented the following inequality,

$$\begin{array}{ccc}\hfill \sum _{n=1}^{k-1}\frac{d(x_n,x_{n+1})}{1+g\left(d(x_0,x_{n+1})\right)}& \ge & \sum _{n=1}^{k-1}\frac{d(x_0,x_{n+1})-d(x_0,x_n)}{1+g\left(d(x_0,x_{n+1})\right)}\hfill \\ & \ge & \sum _{n=1}^{k-1}{\int }_{d(x_0,x_n)}^{d(x_0,x_{n+1})}\frac{1}{1+g\left(r\right)}dr\hfill \\ & =& {\int }_{d(x_0,x_1)}^{d(x_0,x_k)}\frac{1}{1+g\left(r\right)}dr\hfill \end{array}$$

But in fact, by the continuity and monotonicity of g and the definition of $W\left(x_n\right)$, we have $d(x_0,x_n)<d(x_0,x_{n+1})$, then for $d(x_0,x_n)\le r\le d(x_0,x_{n+1})$,

$$\frac{1}{1+g\left(r\right)}\ge \frac{1}{1+g\left(d(x_0,x_{n+1})\right)}$$

Hence,

$$\begin{array}{ccc}\hfill \sum _{n=1}^{k-1}{\int }_{d(x_0,x_n)}^{d(x_0,x_{n+1})}\frac{1}{1+g\left(r\right)}dr& \ge & \sum _{n=1}^{k-1}{\int }_{d(x_0,x_n)}^{d(x_0,x_{n+1})}\frac{1}{1+g\left(d(x_0,x_{n+1})\right)}dr\hfill \\ & =& \sum _{n=1}^{k-1}\frac{d(x_0,x_{n+1})-d(x_0,x_n)}{1+g\left(d(x_0,x_{n+1})\right)},\hfill \end{array}$$

which contradicts their conclusion.

In this note, we aim at modifying the result of [7], and establish a new equilibrium form of the Ekeland’s variational principle for bi-function. Then, the conclusions are used to discuss the equilibrium point problem and fixed point problem. Some recent advances in Ekeland’s variational principles and applications can be seen in [8,9,10,11,12,13,14,15,16,17,18,19] and references therein.

This paper is organized as follows: In Section 2, we state a new version of Ekeland’s variational principle for bi-functions. In Section 3, as applications of the main result, we discuss a equilibrium problem and a fixed point problem.

2. A New Equilibrium Version of EVP

In this section, we establish a new equilibrium version of EVP.

Theorem 4.

Let $(X,d)$ be a complete metric space, $C\subset X$ be a closed set, $x_0\in C$ fixed, and $g:[0,+\infty )\to (0,+\infty )$ be a continuous nondecreasing function such that

$${\int }_0^{+\infty }\frac{1}{g\left(s\right)}ds=m,(0<m\le +\infty ).$$

If $f:C\times C\to \mathbf{R}$ satisfies:

(i)

$f(x,·)$ is bounded from below and lower semi-continuous, $\forall x\in C$;

(ii)

$f(y,y)=0$, $\forall y\in C$;

(iii)

$f(x,z)\le f(x,y)+f(y,z)$, $\forall x,y,z\in C$.

Then, for any $\epsilon >0,0<\alpha <m$ fulfilling

$$\underset{z\in C}{inf}f(x_0,z)>-\alpha \epsilon$$

(2)

there is $x_{\epsilon }\in C$ such that

(a)

$f(x_0,x_{\epsilon })+\frac{\epsilon }{g\left(d(x_0,x_0)\right)}d(x_0,x_{\epsilon })\le 0$

(b)

$f(x_{\epsilon },x)+\frac{\epsilon }{g\left(d(x_0,x_{\epsilon })\right)}d(x_{\epsilon },x)>0$,$\forall x\in C,x\ne x_{\epsilon }$;

(c)

$d(x_0,x_{\epsilon })\le l$,

where l satisfies

$${\int }_0^l\frac{1}{g\left(s\right)}ds=\alpha .$$

Proof. 

Let

$$T\left(x\right)=\{y\in C\backslash B(x_0,d(x_0,x))|f(x_0,x)+\frac{\epsilon }{g\left(d(x_0,x)\right)}d(x_0,x)\le 0\}.$$

In the same manner as the proof of Theorem 2.1 in [7], we can construct a sequence ${\left\{x_n\right\}}_{n=0}^{\infty }\subseteq C$ such that

(1)

$x_{n+1}\in T\left(x_n\right),T\left(x_{n+1}\right)\subset T\left(x_n\right),n=0,1,2,\cdots$;

(2)

diam$T\left(x_n\right)\to 0$.

Due to the completeness of X and the closeness of C, there is a unique $x_{\epsilon }\in C$ such that

$$\underset{n\to \infty }{lim}{x}_n=x_{\epsilon },\bigcap _{n=0}^{\infty }T\left(x_n\right)=\left\{x_{\epsilon }\right\}.$$

As $x_{\epsilon }\in T\left(x_0\right)$, we have

$$f(x_0,x_{\epsilon })+\frac{\epsilon }{g\left(d(x_0,x_0)\right)}d(x_0,x_{\epsilon })\le 0.$$

This verifies assertion (a).

Due to $x_{\epsilon }\in T\left(x_n\right),n=0,1,2,\cdots$, we obtain $T\left(x_{\epsilon }\right)\subset T\left(x_n\right),n=0,1,2,\cdots$. Hence

$$T\left(x_{\epsilon }\right)\in \bigcap _{n=0}^{\infty }T\left(x_n\right).$$

and $T\left(x_{\epsilon }\right)=\left\{x_{\epsilon }\right\}$.

Therefore, the assertion

$$\left(b\right)f(x_{\epsilon },x)+\frac{\epsilon }{g\left(d(x_0,x_{\epsilon })\right)}d(x_{\epsilon },x)>0,\forall x\in C,x\ne x_{\epsilon },$$

holds.

In what follows, let us verify conclusion (c).

As $x_{n+1}\in T\left(x_n\right)$,

$$\begin{array}{c}\hfill f(x_n,x_{n+1})+\frac{\epsilon }{g\left(d(x_0,x_n)\right)}d(x_n,x_{n+1})\le 0,(n=0,1,2,\cdots )\end{array}$$

Hence,

$$\begin{array}{c}\hfill \sum _{j=0}^{n}f(x_j,x_{j+1})+\sum _{j=0}^n\frac{\epsilon }{g\left(d(x_0,x_j)\right)}d(x_j,x_{j+1})\le 0.\end{array}$$

Noting that

$$\begin{array}{c}\hfill \sum _{j=0}^{n}f(x_j,x_{j+1})\ge f(x_0,x_{n+1}),\end{array}$$

(3)

we obtain

$$\begin{array}{c}\hfill \sum _{j=0}^n\frac{\epsilon }{g\left(d(x_0,x_j)\right)}d(x_j,x_{j+1})\le -\sum _{j=0}^{n}f(x_j,x_{j+1})\le -f(x_0,x_{n+1})<\alpha \epsilon ,\end{array}$$

which means

$$\begin{array}{c}\hfill \sum _{j=0}^{\infty }\frac{1}{g\left(d(x_0,x_j)\right)}d(x_j,x_{j+1})<\alpha \end{array}$$

We assert $d(x_0,x_{\epsilon })\le l$. Contrarily, assume $d(x_0,x_{\epsilon })>l$.

Take $\left\{n_i\right\}$ as a subsequence of $\left\{n\right\}$ such that $\left\{d(x_0,x_{n_i})\right\}$ is monotone increasing, converges to $d(x_0,x_{\epsilon })$ and

$$d(x_0,x_k)\le d(x_0,x_{n_{i-1}})(k=n_{i-1}+1,n_{i-1}+2,\cdots ,n_i-1),$$

then

$$\begin{array}{ccc}\hfill \sum _{k=n_{i-1}}^{n_i-1}\frac{d(x_k,x_{k+1})}{g\left(d(x_0,x_k)\right)}& \ge & \sum _{k=n_{i-1}}^{n_i-1}\frac{d(x_k,x_{k+1})}{g\left(d(x_0,x_{n_{i-1}})\right)}\ge \frac{d(x_{n_{i-1}},x_{n_i})}{g\left(d(x_0,x_{n_{i-1}})\right)}\hfill \\ & \ge & \frac{d(x_0,x_{n_i})-d(x_0,x_{n_{i-1}})}{g\left(d(x_0,x_{n_{i-1}})\right)}\ge {\int }_{d(x_0,x_{n_{i-1}})}^{d(x_0,x_{n_i})}\frac{1}{g\left(s\right)}ds\hfill \end{array}$$

which implies

$$\begin{array}{c}\hfill \alpha >\sum _{n=0}^{\infty }\frac{d(x_n,x_{n+1})}{g\left(d(x_0,x_n)\right)}\ge {\int }_0^{d(x_0,x_{\epsilon })}\frac{1}{g\left(s\right)}ds>{\int }_0^l\frac{1}{g\left(s\right)}ds=\alpha ,\end{array}$$

a contradiction.

This completes the proof of conclusion (c). □

If there exists $\phi :X\to {\mathbf{R}}^{+}$ such that $f(x,y)=\phi \left(y\right)-\phi \left(x\right)$, we have the following corollary.

Corollary 1.

Let $(X,d)$ be a complete metric space, $C\subset X$ be a closed set, $x_0\in C$ fixed and $\phi :C\to {\mathbf{R}}^{+}$ be a bounded from below and lower semi-continuous mapping, $g:[0,+\infty )\to (0,+\infty )$ be a continuous nondecreasing function such that

$${\int }_0^{+\infty }\frac{1}{g\left(s\right)}ds=m,(0<m\le +\infty )$$

If and $\epsilon >0$, $0<\alpha <m$ satisfy

$$\phi \left(x_0\right)\le \underset{x\in C}{inf}\phi +\alpha \epsilon ,$$

then there exists $x_{\epsilon }$ such that

(a)

$\phi \left(x_{\epsilon }\right)\le \phi \left(x_0\right)$;

(b)

$\phi \left(x\right)>\phi \left(x_{\epsilon }\right)-\frac{\epsilon }{g\left(d(x_0,x_{\epsilon })\right)}d(x,x_{\epsilon })\forall x\in Cwithx\ne x_{\epsilon }$;

(c)

$d(x_0,x_{\epsilon })\le l$;

where l satisfies

$${\int }_0^l\frac{1}{g\left(s\right)}ds=\alpha$$

Remark 1.

Corollary 1 can be seen as an extension of Theorem 2.1 in [8].

3. Applications

As applications of Theorem 4, we first discuss the existence of equilibrium point for a bi-function.

By an equilibrium problem (abbrev. EP), we understand the problem of finding

$$\overline{x}\in Xsuchthatf(\overline{x},x)\ge 0,\forall x\in C.$$

where C is a given subset of a metric space X and $f:C\times C\to \mathbb{R}$ is a given bi-function.

Theorem 5.

Let $(X,d)$ be a complete metric space, $C\subset X$ be a compact set. Assume $f:C\times C\to \mathbf{R}$ satisfies

(i)

$f(x,·)$ is bounded from below and lower semi-continuous, for every $x\in C$;

(ii)

$f(z,z)=0$, for every $z\in C$;

(iii)

$f(z,x)\le f(z,t)+f(t,x)$ for every $x,t,z\in C$;

(iv)

$f(·,y)$ is upper semi-continuous, for every $y\in C$.

Then, the equilibrium problem (EP) has a solution.

Proof. 

Let $g\left(s\right)\equiv 1$. It is a continuous nondecreasing function such and

$${\int }_0^{+\infty }\frac{1}{g\left(s\right)}ds=+\infty$$

Let $x_0\in C$ be fixed, for every ${\epsilon }_n=\frac{1}{n}$ and $\alpha =n(b-1)$, where $b={inf}_{z\in C}f(x_0,z)$. Then, by Theorem 4 (b), there exists $x_n\in C$ such that

$$f(x_n,x)+\frac{1}{n}d(x_n,x)\ge 0,\forall x\in C$$

Due to compactness of C, there is a subsequence $\left\{x_{n_k}\right\}$ of $\left\{x_n\right\}$ which is convergent, i.e., there exists $\overline{x}\in C$, such that

$$\underset{k\to \infty }{lim}{x}_{n_k}=\overline{x}.$$

Hence, we have

$$f(\overline{x},x)\ge \underset{k\to \infty }{lim}sup[f(x_{n_k},x)+\frac{1}{n_k}d(x_{n_k},x)]\ge 0,\forall x\in C$$

This implies that $\overline{x}$ is a solution to the equilibrium problem (EP). □

Then, we establish the following Caristi type fixed point theorem.

Theorem 6.

Let $(X,d)$ be a complete metric space, $x_0\in X$ fixed, and $\phi :X\to {\mathbf{R}}^{+}$ be a bounded from below and lower semicontinuous mapping, $g:[0,+\infty )\to (0,+\infty )$ be a continuous nondecreasing function such that

$${\int }_0^{+\infty }\frac{1}{g\left(s\right)}ds=m$$

where $m\in {\mathbf{R}}^{+}\bigcup \{+\infty \}$.

If a mapping $K:X\to X$ satisfies: for some $\epsilon >0$,

$$\frac{\epsilon d(x,K(x\left)\right)}{g\left(d(x_0,x)\right)}\le \phi \left(x\right)-\phi \left(K\left(x\right)\right)\forall x\in X,$$

(4)

then K has a fixed point in X.

Proof. 

Let $f(x,y)=\phi \left(y\right)-\phi \left(x\right),C=X$. By the proof of Theorem 4, for each $\epsilon >0$, there exists a sequence ${\left\{x_n\right\}}_{n\in N}\subset X$ and $x_{\epsilon }\in X$, such that $x_n\to x_{\epsilon }$ as $n\to \infty$ and

$$\phi \left(x\right)>\phi \left(x_{\epsilon }\right)-\frac{\epsilon }{g\left(d(x_0,x_{\epsilon })\right)}d(x,x_{\epsilon })\forall x\in C,x\ne x_{\epsilon }$$

(5)

In what follows, we will prove that $x_{\epsilon }$ is a fixed point of K.

Conversely, suppose that $x_{\epsilon }\ne K\left(x_{\epsilon }\right)$. Let $x=K\left(x_{\epsilon }\right)$ and substitute it into (5), we find

$$\phi \left(K\left(x_{\epsilon }\right)\right)\ge \phi \left(x_{\epsilon }\right)-\frac{\epsilon }{g\left(d(x_0,x_{\epsilon })\right)}d(K\left(x_{\epsilon }\right),x_{\epsilon }).$$

(6)

Taking $x_{\epsilon }$ instead of x in (4), we have that

$$\frac{\epsilon d(x_{\epsilon },K\left(x_{\epsilon }\right))}{g\left(d(x_0,x_{\epsilon })\right)}\le \phi \left(x_{\epsilon }\right)-\phi \left(K\left(x_{\epsilon }\right)\right)$$

(7)

Combing the inequalities (6) with (7), we know

$$\frac{\epsilon d(x_{\epsilon },K\left(x_{\epsilon }\right))}{g\left(d(x_0,x_{\epsilon })\right)}\le \phi \left(x_{\epsilon }\right)-\phi \left(K\left(x_{\epsilon }\right)\right)<\frac{\epsilon d(x_{\epsilon },K\left(x_{\epsilon }\right))}{g\left(d(x_0,x_{\epsilon })\right)}$$

which is a contradiction.

Thus $x_{\epsilon }=K\left(x_{\epsilon }\right)$, i.e., $x_{\epsilon }$ is a fixed point of K. □