Open Access

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

Journal of Inequalities and Applications20132013:120

https://doi.org/10.1186/1029-242X-2013-120

Received: 28 September 2012

Accepted: 28 February 2013

Published: 21 March 2013

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).

Keywords

u-distance complete metric space Ekeland’s variational principle

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 ( , + ] which is proper lower semicontinuous bounded from below. Then for u X and λ > 0 , there exists v X such that
  1. (P)
    f ( v ) f ( u ) λ d ( u , v )
    ;
     
  2. (Q)
    f ( w ) > f ( v ) λ d ( v , w )
    for every w 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 ( , + ] be proper lower semicontinuous bounded from below. Then, for all u X , λ > 0 and δ > 0 , there exists v X satisfying the following:

(P)′ f ( v ) < f ( u ) λ d ( u , v ) + δ ;
  1. (Q)
    f ( w ) > f ( v ) λ d ( v , w )
    for every w X { v } ;
     
  2. (R)

    if a sequence { x n } in X satisfies lim n ( f ( x n ) + λ 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 the set of all positive integers, by the set of real numbers, R + = [ 0 , ) . 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 × X R + is called a u-distance on X if there exists a function θ : X × X × [ 0 , ) × [ 0 , ) R + such that

(u1) p ( x , z ) p ( x , y ) + p ( y , z ) for all x , y , z X ;

(u2) θ ( x , y , 0 , 0 ) = 0 , θ ( x , y , s , t ) min { s , t } for all x , y X and s , t [ 0 , ) , and for any x X and for every ϵ > 0 , there exists δ > 0 such that | s s 0 | < δ , | t t 0 | < δ , s , s 0 , t , t 0 [ 0 , ) and y X imply
| θ ( x , y , s , t ) θ ( x , y , s 0 , t 0 ) | < ϵ ;

(u3) lim n x n = x and lim n sup { θ ( w n , z n , , p ( w n , x m ) , p ( z n , x m ) ) : m n } = 0 imply p ( y , x ) lim n inf p ( y , x n ) for all y X ;

(u4) lim n sup { p ( x n , w m ) : m n } = 0 , lim n sup { p ( y n , z m ) : m n } = 0 , lim n θ ( x n , w n , s n , t n ) = 0 and lim n θ ( y n , z n , s n , t n ) = 0 imply lim n θ ( w n , z n , s n , t n ) = 0 or lim n sup { p ( w m , x n ) : m n } = 0 , lim n sup { p ( z m , y n ) : m n } = 0 , lim n θ ( x n , w n , s n , t n ) = 0 and lim n θ ( y n , z n , s n , t n ) = 0 imply lim n θ ( w n , z n , s n , t n ) = 0 ;

(u5) lim n θ ( w n , z n , p ( w n , x n ) , p ( z n , x n ) ) = 0 and lim n θ ( w n , z n , p ( w n , y n ) , p ( z n , y n ) ) = 0 imply lim n d ( x n , y n ) = 0 or lim n θ ( a n , b n , p ( x n , a n ) , p ( x n , b n ) ) = 0 and lim n θ ( a n , b n , p ( y n , a n ) , p ( y n , b n ) ) = 0 imply lim n 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 × X R + defined by q ( x , y ) = c p ( x , y ) for every x , y 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
lim n sup { p ( x n , x m ) : m > n } = 0
or
lim n 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 p ( z , x n ) = 0 for some z X , then { x n } is a p-Cauchy sequence. Moreover, if a sequence { y n } of X also satisfies lim n p ( z , y n ) = 0 , then lim n p ( x n , y n ) = 0 . In particular, for x , y , z X , p ( z , x ) = 0 and p ( z , y ) = 0 imply x = y .

Proof Let θ be a function from X × X × [ 0 , ) × [ 0 , ) into R + satisfying (u1)-(u5). From lim n p ( z , x n ) = 0 , it follows by (u2) that lim n θ ( 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 mapping from X into itself. Suppose that there exists a u-distance p on X and r [ 0 , 1 ) such that p ( T x , T 2 x ) r p ( x , T x ) for all x X . Assume that either of the following hold:
  1. (i)

    If lim n sup { p ( x n , x m ) : m > n } = 0 , lim n p ( x n , T x n ) = 0 and lim n p ( x n , y ) = 0 , then T y = y ;

     
  2. (ii)

    if { x n } and { T x n } converge to y, then T y = y ;

     
  3. (iii)

    T is continuous.

     

Then, there exists x 0 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 × X into ( , ] satisfying
  1. (1)
    ϕ ( x , z ) ϕ ( x , y ) + ϕ ( y , z )
    for all x , y , z X ;
     
  2. (2)
    ϕ ( x , ) : X ( , ]
    is lower semicontinuous for any x X ;
     
  3. (3)

    there exists an x 0 such that inf y X ϕ ( x 0 , y ) > ; and

     
  4. (4)
    ϕ ( x , y ) = ϕ ( y , x )
    .
     
Define M x = { y X : ϕ ( x , y ) + p ( x , y ) 0 } . Let u X and c R + such that ϕ ( x , u ) < for all x X , M u and c ϕ ( x , u ) inf y M u ϕ ( u , y ) . Then a function q : X × X R + defined by
q ( x , y ) = { ϕ ( u , x ) inf y M x ϕ ( u , y ) if  x M u  and  y M x , c + p ( x , y ) if  x M u  or  y M x

is a u-distance on X.

Proof Let η be a function from X × X × R + × R + into R + satisfying (u2)-(u5) for a u-distance. We note that ϕ ( x , y ) + ϕ ( y , z ) + p ( x , y ) + p ( y , z ) 0 and ϕ ( x , z ) + p ( x , z ) 0 . Thus, y M x and z M y imply z M x . If x M u and y M x , then
p ( x , y ) ϕ ( y , x ) q ( x , y ) = ϕ ( y , x ) inf y M x ϕ ( x , y ) ϕ ( x , u ) inf y M u ϕ ( x , y ) c .
Therefore, p ( x , y ) q ( y , x ) c + p ( x , y ) for all x , y X . To complete the proof, we will show (u1) q , (u3) q , η , (u4) q , η and (u5) q , η . Let x, y and z be fixed elements in X. In the case x M u , y M x , y M u and z M y , we have z M x and hence q ( x , z ) = q ( x , y ) q ( x , y ) + q ( y , z ) . In the other case, we note that
q ( x , z ) c + p ( x , z ) c + p ( x , y ) + p ( y , z ) 2 c + p ( x , y ) + p ( y , z ) = q ( x , y ) + q ( y , z ) .

This shows (u1) q .

We next suppose that lim n x n = x and lim n sup { η ( w n , z n , q ( w n , x m ) , q ( z n , x m ) ) : m n } = 0 and fix w X . Since lim n sup { θ ( w n , z n , p ( w n , x m ) , p ( z n , x m ) ) : m n } = 0 , we have p ( w , x ) lim inf n p ( w , x n ) for all y X .

In the case that w M u and there exists a subsequence { x n k } of { x n } such that x n k M w for all k N , we have
ϕ ( w , x ) + p ( w , x ) lim n inf ϕ ( w , x n ) + lim n p ( w , x n ) lim n inf ( ϕ ( w , x n ) + p ( w , x n ) ) lim k inf ( ϕ ( w , x n k ) + p ( w , x n k ) ) 0 ,
and so x M u . Hence
q ( w , x ) = ϕ ( u , w ) inf x M w ϕ ( u , x ) = lim k q ( w , x n k ) = lim n inf q ( w , x n ) .
In the other case, we obtain
q ( w , x ) c + p ( w , x ) lim n inf ( c + p ( w , x n ) ) = lim n inf q ( w , x n ) .

This shows (u3) q , η . We will show that q satisfies (u4) q , η .

Case I: Suppose that lim n sup { q ( x n , w m ) : m n } = 0 , lim n sup { q ( y n , z m ) : m n } = 0 , lim n η ( x n , w n , s n , t n ) = 0 , and lim n η ( y n , z n , s n , t n ) = 0 .

In the case x n M u and w m M x n , we note that q ( x n , w n ) = ϕ ( u , x n ) inf w m M x n ϕ ( u , w m ) . Since ϕ ( x n , w m ) + p ( x n , w n ) 0 , it follows that
p ( x n , w m ) ϕ ( x n , w n ) = ϕ ( w m , x n ) ϕ ( w m , u ) + ϕ ( u , x n ) = ϕ ( u , x n ) ϕ ( u , w m ) ϕ ( u , x n ) inf w m M x n ϕ ( u , w m ) = q ( x n , w m ) .
Thus, we have p ( x n , w m ) q ( x n , w m ) . This implies that sup m n p ( x n , w n ) sup m n q ( x n , w m ) . Take n , so
0 lim n sup p ( x n , w m ) lim n sup q ( x n , w m ) = 0

and therefore lim n sup p ( x n , w m ) = 0 .

Similarly, if y n M u and z m M y n , then lim n sup p ( y n , z m ) = 0 .

We note that lim n θ ( x n , w n , s n , t n ) = 0 = lim n θ ( y n , z n , s n , t n ) and hence
lim n η ( w n , z n , s n , t n ) = 0 .
In the case x n M u or w m M x n , we note that p ( x n , w m ) c + p ( x n , w m ) = q ( x n , w m ) . Thus, we have p ( x n , w m ) q ( x n , w m ) . This implies that sup m n p ( x n , w m ) sup m n q ( x n , w m ) . Taking n , we obtain
0 lim n sup p ( x n , w m ) lim n sup q ( x n , w m ) = 0

and therefore lim n sup p ( x n , w n ) = 0 . Similarly as above, if y n M u and z m M y n , then lim n sup p ( y n , z m ) = 0 . We note that lim n θ ( x n , w n , s n , t n ) = 0 = lim n θ ( y n , z n , s n , t n ) and hence lim n η ( w n , z n , s n , t n ) = 0 .

Case II: Suppose that lim n sup { q ( w m , x n ) : m n } = 0 , lim n sup { q ( z m , y n ) : m n } = 0 , lim n η ( x n , w n , s n , t n ) = 0 and lim n η ( y n , z n , s n , t n ) = 0 . Similarly as in Case I, we can show that lim n η ( w n , z n , s n , t n ) = 0 . This shows (u4) q , η . We will show that q satisfies (u5) q , η .

Case I: Suppose that lim n η ( w n , z n , q ( x n , w n ) , q ( x n , z n ) ) = 0 and lim n η ( w n , z n , q ( y n , w n ) , q ( y n , z n ) ) = 0 . In the case x n M u and w n , z n M x n , we note that q ( x n , w n ) = ϕ ( u , x n ) inf w n M x n ϕ ( u , w n ) and hence q ( x n , z n ) = ϕ ( u , x n ) inf z n M x n ϕ ( u , z n ) . Thus, we have
θ ( w n , z n , p ( x n , w n ) , p ( x n , z n ) ) θ ( w n , z n , ϕ ( z n , x n ) , ϕ ( z n , x n ) ) θ ( w n , z n , ϕ ( w n , u ) + ϕ ( u , x n ) , ϕ ( z n , u ) + ϕ ( u , x n ) ) = θ ( w n , z n , ϕ ( u , x n ) ϕ ( u , w n ) , ϕ ( u , x n ) ϕ ( u , z n ) ) θ ( w n , z n , ϕ ( u , x n ) inf w n M x n ϕ ( u , w n ) , ϕ ( u , x n ) inf z n M x n ϕ ( u , z n ) ) = η ( w n , z n , q ( x n , w n ) , q ( x n , z n ) ) .
Taking n , we have
0 lim n θ ( w n , z n , p ( x n , w n ) , p ( x n , z n ) ) lim n η ( w n , z n , q ( x n , w n ) , q ( x n , z n ) ) = 0 .
Therefore lim n θ ( w n , z n , p ( x n , w n ) , p ( x n , z n ) ) = 0 . Similarly, if y n M u and z n , w n M y n , then lim n θ ( w n , z n , p ( y n , w n ) , p ( y n , z n ) ) = 0 . In the case x n M u or w n , z n 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 d ( x n , y n ) = 0 . Hence
θ ( w n , z n , c + p ( x n , w n ) , c + p ( x n , z n ) ) θ ( w n , z n , c + p ( x n , w n ) , c + p ( x n , z n ) ) η ( w n , z n , q ( x n , w n ) , q ( x n , z n ) ) .
Take n , thus
0 lim n θ ( w n , z n , p ( x n , w n ) , p ( x n , z n ) ) lim n η ( w n , z n , q ( x n , w n ) , q ( x n , z n ) ) = 0 .

Therefore lim n θ ( w n , z n , p ( x n , w n ) , p ( x n , z n ) ) = 0 . Similarly, if y n M u or w n , z n M y n , then lim n θ ( 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 d ( x n , y n ) = 0 .

Case II: Suppose that lim n η ( w n , z n , q ( w n , x n ) , q ( z n , x n ) ) = 0 and lim n η ( 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 d ( x n , y n ) = 0 . This shows (u5) q , η . □

Proposition 3.4 Let X be a complete metric space, p be a u-distance on X and ϕ be a function from X × X into ( , ] satisfying
  1. (1)
    ϕ ( x , z ) ϕ ( x , y ) + ϕ ( y , z )
    for all x , y , z X ;
     
  2. (2)
    ϕ ( x , ) : X ( , ]
    is lower semicontinuous for any x X ;
     
  3. (3)

    there exists an x 0 such that inf y X ϕ ( x 0 , y ) > ; and

     
  4. (4)
    ϕ ( x , y ) = ϕ ( y , x )
    .
     

Define M x = { y X : ϕ ( x , y ) + p ( x , y ) 0 } for all x X . Then, for each u X with M u , there exists x 0 M u such that M x 0 { x 0 } . In particular, there exists y 0 X such that M y 0 { y 0 } .

Proof Let u X with M u . We have u 1 M u by ϕ ( u , u 1 ) < . If M u = , the assertion holds. Suppose that M u 1 and M x ( X { x } ) for all x M u 1 . Let u 2 M u 1 . We know that ϕ ( x , y ) 0 for all x X and y M x , we define a mapping T : X X as follows: For each x M u 1 , T x satisfies T x M x , T x x and
ϕ ( u 1 , T x ) ϕ ( u 1 , x ) + inf y M x ϕ ( u 1 , y ) 2 .
For each x M u 1 , define T x = u 2 x . We also define a function q : X × X R + by
q ( x , y ) = { ϕ ( u , x ) inf y M x ϕ ( u 1 , y ) if  x M u 1  and  y M x , 2 ϕ ( u , u 1 ) 2 inf w M u 1 ϕ ( u , w ) + 1 + p ( x , y ) if  x M u 1  or  y M x .
By Lemma 3.3, we have q is a u-distance on X. Since y M y and z M y , it follows by Lemma 3.3 that z M x . Hence T x M u 1 and M T x M x for all x M u 1 . If x M u 1 , we obtain
q ( T x , T 2 x ) = ϕ ( u 1 , T x ) inf y M T x ϕ ( u 1 , y ) ϕ ( u 1 , x ) + inf y M x ϕ ( u 1 , y ) 2 inf y M x ϕ ( u 1 , y ) = q ( x , T x ) 2 .
If x M u 1 ,
q ( T x , T 2 x ) = q ( u 2 , T u 2 ) = ϕ ( u 1 , u 2 ) inf T u 2 M u 2 ϕ ( u 1 , T u 2 ) ϕ ( u , u 1 ) inf T u 1 ϕ ( u , T u 1 ) q ( x , u 2 ) 2 = q ( x , T x ) 2 .

We will show (i) in Theorem 3.2. Suppose that lim n sup { q ( x n , x m ) : m > n } = 0 and lim n q ( x n , y ) = 0 . We may assume x n M u 1 and y M x n for all n N by the definition of q. Then y M u 1 and hence T y M y M x n . By Lemma 2.4 we have lim n q ( x n , T y ) = lim n q ( x n , y ) = 0 and T y = y . Hence, by Theorem 3.2, T has a fixed point. This is a contradiction. So, there is x 0 M u 1 M u such that M x 0 { 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 × X into ( , ] satisfying
  1. (1)
    ϕ ( x , z ) ϕ ( x , y ) + ϕ ( y , z )
    for all x , y , z X ;
     
  2. (2)
    ϕ ( x , ) : X ( , ]
    is lower semicontinuous for any x X ;
     
  3. (3)

    there exists an x 0 such that inf y X ϕ ( x 0 , y ) > ; and

     
  4. (4)
    ϕ ( x , y ) = ϕ ( y , x )
    .
     
Then the following hold:
  1. (A)

    For each u X , there exists v X such that ϕ ( u , v ) 0 and ϕ ( v , w ) + p ( v , w ) > 0 for all w X { v } ;

     
  2. (B)

    For each λ > 0 and u X with p ( u , u ) = 0 , there exists v X such that ϕ ( u , v ) + λ p ( u , v ) 0 and ϕ ( v , w ) + λ p ( v , w ) > 0 for all w X { v } .

     

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

We will show that (B). By Proposition 2.2, we note that λp is a u-distance. We define M x = { y X : ϕ ( x , y ) + λ p ( x , y ) 0 } for all x X . Since p ( u , u ) = 0 , we have M u , and hence there exists v M u such that M v { v } by Proposition 3.4. Therefore v satisfies ϕ ( u , v ) + λ p ( u , v ) 0 and ϕ ( v , w ) + λ p ( v , w ) > 0 for all w X with w v . This completes the proof. □

Remark 3.6 By setting ϕ ( x , y ) = f ( y ) f ( x ) , where f : X 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 × X into ( , ] satisfying
  1. (1)
    ϕ ( x , z ) ϕ ( x , y ) + ϕ ( y , z )
    for all x , y , z X ;
     
  2. (2)
    ϕ ( x , ) : X ( , ]
    is lower semicontinuous for any x X ;
     
  3. (3)

    there exists an x 0 such that inf y X ϕ ( x 0 , y ) > ; and

     
  4. (4)
    ϕ ( x , y ) = ϕ ( y , x )
    .
     
Let u X with p ( u , u ) = 0 . Then λ > 0 and δ > 0 , there exists v X satisfying the following:
  1. (i)
    ϕ ( u , v ) 0
    ;
     
  2. (ii)
    ϕ ( u , v ) + λ p ( u , v ) < δ
    ;
     
  3. (iii)
    ϕ ( v , w ) + λ p ( v , w ) > 0
    for all w X { v } ;
     
  4. (iv)

    if a sequence { x n } in X satisfies lim n ( ϕ ( v , x n ) + λ 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 ϕ ( v , u ) = , (i) and (ii) hold for all v X . We also note that (iii) and (iv) do not depend on ϕ ( v , u ) . In the case ϕ ( v , u ) < , set λ ( 0 , λ ) satisfying
λ λ λ ( ϕ ( u , v ) inf x X ϕ ( v , x ) ) < δ .
By Theorem 3.5(B), there exists v X such that ϕ ( u , v ) + λ p ( u , v ) 0 and ϕ ( v , w ) + λ p ( v , w ) > 0 for all w X { v } . Thus, we have
Therefore, ϕ ( u , v ) + λ p ( u , v ) < δ . For w X { v } , we note that
ϕ ( v , w ) > λ p ( v , w ) λ p ( v , w ) .
So, ϕ ( v , w ) + λ p ( v , w ) > 0 . Finally, we will show that (iv). Suppose that a sequence { x n } in X satisfies lim n ( ϕ ( v , x n ) + λ p ( v , x n ) ) = 0 . We note that ϕ ( v , w ) + λ p ( v , w ) 0 for all w X . We have
lim n sup p ( v , x n ) = lim n sup ( λ λ λ λ ) p ( v , x n ) = lim n λ p ( v , x n ) λ p ( v , x n ) λ λ lim n λ p ( v , x n ) ϕ ( v , x n ) λ λ lim n λ p ( v , x n ) + ϕ ( v , x n ) λ λ = 0 .
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 X . From (u3), we have p ( v , x ) = 0 and so
ϕ ( v , x ) lim n inf ϕ ( v , x n ) lim n ( ϕ ( v , x n ) + λ p ( v , x n ) ) = 0 .
Thus, if v x , then we have
ϕ ( v , x ) > λ p ( v , x ) ϕ ( v , x ) .

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

Remark 3.8 By setting ϕ ( x , y ) = f ( y ) f ( x ) , where f : X 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].

Declarations

Acknowledgements

The authors would like to thank the Thailand Research Fund (TRF) for supporting by permit money of investment under of The Royal Golden Jubilee Ph.D. Program (RGJ-Ph.D.), Thailand.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Naresuan University

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© Plubtieng and Seangwattana; licensee Springer 2013

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