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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 (,+] which is proper lower semicontinuous bounded from below. Then for uX and λ>0, there exists vX such that

  1. (P)
    f(v)f(u)λd(u,v)

    ;

  2. (Q)
    f(w)>f(v)λd(v,w)

    for every wv.

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 uX, λ>0 and δ>0, there exists vX satisfying the following:

(P)′ f(v)<f(u)λd(u,v)+δ;

  1. (Q)
    f(w)>f(v)λd(v,w)

    for every wX{v};

  2. (R)

    if a sequence { x n } in X satisfies lim n (f( x n )+λd(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,zX;

(u2) θ(x,y,0,0)=0, θ(x,y,s,t)min{s,t} for all x,yX and s,t[0,), and for any xX 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 yX 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 )):mn}=0 imply p(y,x) lim n infp(y, x n ) for all yX;

(u4) lim n sup{p( x n , w m ):mn}=0, lim n sup{p( y n , z m ):mn}=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 ):mn}=0, lim n sup{p( z m , y n ):mn}=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)=cp(x,y) for every x,yX 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 zX, 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,zX, 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(Tx, T 2 x)rp(x,Tx) for all xX. 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 Ty=y;

  2. (ii)

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

  2. (2)
    ϕ(x,):X(,]

    is lower semicontinuous for any xX;

  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 Mx={yX:ϕ(x,y)+p(x,y)0}. Let uX and c R + such that ϕ(x,u)< for all xX, Mu and cϕ(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, yMx and zMy imply zMx. If xMu and yMx, 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,yX. 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 xMu, yMx, yMu and zMy, we have zMx 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 )):mn}=0 and fix wX. Since lim n sup{θ( w n , z n ,p( w n , x m ),p( z n , x m )):mn}=0, we have p(w,x)lim inf n p(w, x n ) for all yX.

In the case that wMu and there exists a subsequence { x n k } of { x n } such that x n k Mw for all kN, 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 xMu. Hence

q(w,x)=ϕ(u,w) inf x M w ϕ(u,x)= lim k q(w, x n k )= lim n infq(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 ):mn}=0, lim n sup{q( y n , z m ):mn}=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 Mu 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 supp( x n , w m ) lim n supq( x n , w m )=0

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

Similarly, if y n Mu and z m M y n , then lim n supp( 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 Mu 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 supp( x n , w m ) lim n supq( x n , w m )=0

and therefore lim n supp( x n , w n )=0. Similarly as above, if y n Mu and z m M y n , then lim n supp( 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 ):mn}=0, lim n sup{q( z m , y n ):mn}=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 Mu 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 Mu 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 Mu 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 Mu 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,zX;

  2. (2)
    ϕ(x,):X(,]

    is lower semicontinuous for any xX;

  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 Mx={yX:ϕ(x,y)+p(x,y)0} for all xX. Then, for each uX with Mu, there exists x 0 Mu such that M x 0 { x 0 }. In particular, there exists y 0 X such that M y 0 { y 0 }.

Proof Let uX with Mu. We have u 1 Mu by ϕ(u, u 1 )<. If Mu=, the assertion holds. Suppose that M u 1 and Mx(X{x}) for all xM u 1 . Let u 2 M u 1 . We know that ϕ(x,y)0 for all xX and yMx, we define a mapping T:XX as follows: For each xM u 1 ,Tx satisfies TxMx, Txx and

ϕ( u 1 ,Tx) ϕ ( u 1 , x ) + inf y M x ϕ ( u 1 , y ) 2 .

For each xM u 1 , define Tx= 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 yMy and zMy, it follows by Lemma 3.3 that zMx. Hence TxM u 1 and MTxMx for all xM u 1 . If xM 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 xM 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 yM x n for all nN by the definition of q. Then yM u 1 and hence TyMyM x n . By Lemma 2.4 we have lim n q( x n ,Ty)= lim n 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 M u 1 Mu 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,zX;

  2. (2)
    ϕ(x,):X(,]

    is lower semicontinuous for any xX;

  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 uX, there exists vX such that ϕ(u,v)0 and ϕ(v,w)+p(v,w)>0 for all wX{v};

  2. (B)

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

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

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

Remark 3.6 By setting ϕ(x,y)=f(y)f(x), where f:XR 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,zX;

  2. (2)
    ϕ(x,):X(,]

    is lower semicontinuous for any xX;

  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 uX with p(u,u)=0. Then λ>0 and δ>0, there exists vX satisfying the following:

  1. (i)
    ϕ(u,v)0

    ;

  2. (ii)
    ϕ(u,v)+λp(u,v)<δ

    ;

  3. (iii)
    ϕ(v,w)+λp(v,w)>0

    for all wX{v};

  4. (iv)

    if a sequence { x n } in X satisfies lim n (ϕ(v, x n )+λp(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 vX. 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 vX such that ϕ(u,v)+ λ p(u,v)0 and ϕ(v,w)+ λ p(v,w)>0 for all wX{v}. Thus, we have

Therefore, ϕ(u,v)+λp(u,v)<δ. For wX{v}, we note that

ϕ(v,w)> λ p(v,w)λp(v,w).

So, ϕ(v,w)+λp(v,w)>0. Finally, we will show that (iv). Suppose that a sequence { x n } in X satisfies lim n (ϕ(v, x n )+λp(v, x n ))=0. We note that ϕ(v,w)+ λ p(v,w)0 for all wX. 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 xX. 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 vx, 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:XR 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].

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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.

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Plubtieng, S., Seangwattana, T. Generalizations of the strong Ekeland variational principle with a generalized distance in complete metric spaces. J Inequal Appl 2013, 120 (2013). https://doi.org/10.1186/1029-242X-2013-120

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Keywords

  • u-distance
  • complete metric space
  • Ekeland’s variational principle