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Quadruple fixed point theorems in partially ordered metric spaces with mixed g-monotone property

Abstract

We prove quadruple fixed point theorems in partially ordered metric spaces with mixed g-monotone property. Also, we state some examples to show that our results are real generalization of known ones in quadruple fixed point theory.

MSC:46T99, 54H25, 47H10, 54E50.

1 Introduction

In 1987, the notion of coupled fixed point was introduced by Guo and Lakshmikantham [1]. Later, Bhaskar and Lakshmikantham [2] introduced the concept of mixed monotone property for contractive operators of the form F:X→X, where X is a partially ordered metric space, and then established some coupled fixed point theorems. They also illustrated these results by proving the existence and uniqueness of the solution for a periodic boundary value problem. Recently, Lakshmikantham and Ćirić in [3] defined a g-monotone property and proved coupled coincidence and coupled common fixed point results for nonlinear mappings satisfying certain contractive conditions in partially ordered metric spaces. They also proved related fixed point theorems. Many authors focused on coupled fixed point theory and proved remarkable results (see [4–17]).

Very recently, Berinde and Borcut [18] introduced the concept of triple fixed point and proved some tripled point theorems by virtue of mixed monotone mappings. Their contributions generalize and extend Bhaskar and Lakshmikantham’s research for nonlinear mappings. The notion of fixed point of order N≥3 was first introduced by Samet and Vetro [19]. Karapinar used the concept of quadruple fixed point and proved some fixed point theorems on the topic [20]. Following this study, a quadruple fixed point is developed and some related fixed point theorems are obtained in [21–24]. Recently, Karapinar et al. [25] have proved a number of quadruple fixed point theorems under ϕ-contractive conditions for a mapping F: X 4 →X in ordered metric spaces.

Let us recall some basic definitions from [21].

Definition 1.1 (See [21])

Let X be a nonempty set and let F: X 4 →X be a given mapping. An element (x,y,z,w)∈X×X×X×X is called a quadruple fixed point of F if

F(x,y,z,w)=x,F(y,z,w,x)=y,F(z,w,x,y)=zandF(w,x,y,z)=w.

Let (X,d) be a metric space. The mapping d ¯ : X 4 →X, given by

d ¯ ( ( x , y , z , w ) , ( u , v , h , l ) ) =d(x,y)+d(y,v)+d(z,h)+d(w,l),

defines a metric on X 4 , which will be denoted for convenience by d.

Definition 1.2 (See [21])

Let (X,≤) be a partially ordered set and let F: X 4 →X be a mapping. We say that F has the mixed monotone property if F(x,y,z,w) is monotone nondecreasing in x and z and is monotone non-increasing in y and w; that is, for any x,y,z,w∈X,

x 1 , x 2 ∈ X , x 1 ≤ x 2 implies F ( x 1 , y , z , w ) ≤ F ( x 2 , y , z , w ) , y 1 , y 2 ∈ X , y 1 ≤ y 2 implies F ( x , y 2 , z , w ) ≤ F ( x , y 1 , z , w ) , z 1 , z 2 ∈ X , z 1 ≤ z 2 implies F ( x , y , z 1 , w ) ≤ F ( x , y , z 2 , w ) and w 1 , w 2 ∈ X , w 1 ≤ w 2 implies F ( x , y , z , w 2 ) ≤ F ( x , y , z , w 1 ) .

In this article, we establish some quadruple coincidence and common fixed point theorems for F: X 4 →X and g:X→X satisfying nonlinear contractions in partially ordered metric spaces. Also, some examples are given to support our results.

2 Preliminary

We start this section with the following definitions.

Definition 2.1 Let (X,≤) be a partially ordered set. Let F: X 4 →X and g:X→X. The mapping F is said to have the mixed g-monotone property if for any x,y,z,w∈X,

x 1 , x 2 ∈ X , g x 1 ≤ g x 2 ⇒ F ( x 1 , y , z , w ) ≤ F ( x 2 , y , z , w ) , y 1 , y 2 ∈ X , g y 1 ≤ g y 2 ⇒ F ( x , y 1 , z , w ) ≥ F ( x , y 2 , z , w ) , z 1 , z 2 ∈ X , g z 1 ≤ g z 2 ⇒ F ( x , y , z 1 , w ) ≤ F ( x , y , z 2 , w ) and w 1 , w 2 ∈ X , g w 1 ≤ g w 2 ⇒ F ( x , y , z , w 1 ) ≥ F ( x , y , z , w 2 ) .

Definition 2.2 Let F: X 4 →X and g:X→X. An element (x,y,z,w) is called a quadruple coincidence point of F and g if

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw.

Definition 2.3 Let F: X 4 →X and g:X→X. An element (x,y,z,w) is called a quadruple common fixed point of F and g if

F ( x , y , z , w ) = g x = x , F ( y , z , w , x ) = g y = y , F ( z , w , x , y ) = g z = z and F ( w , x , y , z ) = g w = w .

Definition 2.4 Let X be a nonempty set. Then we say that the mappings F: X 4 →X and g:X→X are commutative if for all x,y,z,w∈X,

g ( F ( x , y , z , w ) ) =F(gx,gy,gz,gw).

Let Φ denote all the functions ϕ:[0,∞)→[0,∞) which satisfy that lim t → r ϕ(t)>0 for all r>0 and lim t → 0 + ϕ(t)=0.

Let Ψ denote all the functions ψ:[0,∞)→[0,∞) which satisfy

  1. (i)

    ψ(t)=0 if and only if t=0,

  2. (ii)

    ψ is continuous and nondecreasing,

  3. (iii)

    ψ(s+t)≤ψ(s)+ψ(t), ∀s,t∈[0,∞).

Examples of typical functions ϕ and ψ are given in [4]. The aim of this paper is to prove the following theorem.

3 Main results

Now, we present the main results of this paper.

Theorem 3.1 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose that F: X 4 →X and g:X→X are such that F is continuous and has the mixed g-monotone property. Assume also that there exist ϕ∈Φ and ψ∈Ψ such that

ψ ( d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ) ≤ 1 4 ψ ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) ) − ϕ ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) )
(1)

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), g is continuous and commutes with F. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) and g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Proof Let x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) .

Since F( X 4 )⊂g(X), then we can choose x 1 , y 1 , z 1 , w 1 ∈X such that

g x 1 = F ( x 0 , y 0 , z 0 , w 0 ) , g y 1 = F ( y 0 , z 0 , w 0 , x 0 ) , g z 1 = F ( z 0 , w 0 , x 0 , y 0 ) , g w 1 = F ( w 0 , x 0 , y 0 , z 0 ) .
(2)

Taking into account F( X 4 )⊂g(X), by continuing this process, we can construct sequences { x n }, { y n }, { z n }, and { w n } in X such that

g x n + 1 = F ( x n , y n , z n , w n ) , g y n + 1 = F ( y n , z n , w n , x n ) , g z n + 1 = F ( z n , w n , x n , y n ) , g w n + 1 = F ( w n , x n , y n , z n ) .
(3)

We shall show that

g x n ≤ g x n + 1 , g y n + 1 ≥ g y n , g z n ≤ g z n + 1 and g w n + 1 ≥ g w n for  n = 0 , 1 , 2 , … .
(4)

For this purpose, we use the mathematical induction. Since g x 0 ≤F( x 0 , y 0 , z 0 , w 0 ), g y 0 ≥F( y 0 , z 0 , w 0 , x 0 ), g z 0 ≤F( z 0 , w 0 , x 0 , y 0 ), and g w 0 ≥F( w 0 , x 0 , y 0 , z 0 ), then by (2), we get

g x 0 ≤g x 1 ,g y 1 ≥g y 0 ,g z 0 ≤g z 1 andg w 1 ≥g w 0 ,

that is, (4) holds for n=0. We presume that (4) holds for some n>0. As F has the mixed g-monotone property and g x n ≤g x n + 1 , g y n + 1 ≥g y n , g z n ≤g z n + 1 , and g w n + 1 ≥g w n , we obtain

g x n + 1 = F ( x n , y n , z n , w n ) ≤ F ( x n + 1 , y n , z n , w n ) ≤ F ( x n + 1 , y n , z n + 1 , w n ) ≤ F ( x n + 1 , y n + 1 , z n + 1 , w n ) ≤ F ( x n + 1 , y n + 1 , z n + 1 , w n + 1 ) = g x n + 2 , g y n + 2 = F ( y n + 1 , z n + 1 , w n + 1 , x n + 1 ) ≤ F ( y n + 1 , z n , w n + 1 , x n + 1 ) ≤ F ( y n , z n , w n + 1 , x n + 1 ) ≤ F ( y n , z n , w n , x n + 1 ) ≤ F ( y n , z n , w n , x n ) = g y n + 1 , g z n + 1 = F ( z n , w n , x n , y n ) ≤ F ( z n + 1 , w n , x n , y n ) ≤ F ( z n + 1 , w n , x n , y n + 1 ) ≤ F ( z n + 1 , w n , x n + 1 , y n + 1 ) ≤ F ( z n + 1 , w n + 1 , x n + 1 , y n + 1 ) = g z n + 2 , g w n + 2 = F ( w n + 1 , x n + 1 , y n + 1 , z n + 1 ) ≤ F ( w n + 1 , x n , y n + 1 , z n + 1 ) ≤ F ( w n , x n , y n + 1 , z n + 1 ) ≤ F ( w n , x n , y n , z n + 1 ) ≤ F ( w n , x n , y n , z n ) = g w n + 1 .

Thus, (4) holds for any n∈N. Assume, for some n∈N, that

g x n =g x n + 1 ,g y n =g y n + 1 ,g z n =g z n + 1 andg w n =g w n + 1 ,

then, by (3), ( x n , y n , z n , w n ) is a quadruple coincidence point of F and g. From now on, assume for any n∈N that at least g x n ≠g x n + 1 or g y n ≠g y n + 1 , or g z n ≠g z n + 1 , or g w n ≠g w n + 1 .

Due to (1)-(4), we have

(5)
(6)
(7)
(8)

Due to (5)-(8), we conclude that

ψ ( d ( g x n + 1 , g x n + 2 ) ) + ψ ( d ( g y n + 1 , g y n + 2 ) ) + ψ ( d ( g z n + 1 , g z n + 2 ) ) + ψ ( d ( w n + 1 , w n + 2 ) ) ≤ ψ ( d ( g z n , g z n + 1 ) + d ( g w n , g w n + 1 ) + d ( g x n , g x n + 1 ) + d ( g y n , g y n + 1 ) ) − 4 ϕ ( d ( g z n , g z n + 1 ) + d ( g w n , g w n + 1 ) + d ( g x n , g x n + 1 ) + d ( g y n , g y n + 1 ) ) .
(9)

From the property (iii) of ψ, we have

ψ ( d ( g x n + 1 , g x n + 2 ) ) + d ( g y n + 1 , g y n + 2 ) + d ( g z n + 1 , g z n + 2 ) + d ( g w n + 1 , g w n + 2 ) ≤ ψ ( d ( g x n + 1 , g x n + 2 ) ) + ψ ( d ( g y n + 1 , g y n + 2 ) ) + ψ ( d ( g z n + 1 , g z n + 2 ) ) + ψ ( d ( g w n + 1 , g w n + 2 ) ) .
(10)

Combining with (9) and (10), we get that

ψ ( d ( g x n + 1 , g x n + 2 ) ) + d ( g y n + 1 , g y n + 2 ) + d ( g z n + 1 , g z n + 2 ) + d ( g w n + 1 , g w n + 2 ) ≤ ψ ( d ( g z n , g z n + 1 ) + d ( g w n , g w n + 1 ) + d ( g x n , g x n + 1 ) + d ( g y n , g y n + 1 ) ) − 4 ϕ ( d ( g z n , g z n + 1 ) + d ( g w n , g w n + 1 ) + d ( g x n , g x n + 1 ) + d ( g y n , g y n + 1 ) ) .
(11)

Set δ n =d(g x n ,g x n − 1 )+d(g y n ,g y n − 1 )+d(g z n ,g z n − 1 )+d(g w n ,g w n − 1 ). Then we have

ψ( δ n + 2 )≤ψ( δ n + 1 )−4Ï•( δ n + 1 )for all n,
(12)

which yields that ψ( δ n + 2 )≤ψ( δ n + 1 ) for all n.

Since ψ is nondecreasing, we get that δ n + 2 ≤ δ n + 1 for all n. Hence { δ n } is a non-increasing sequence. Since it is bounded below from 0, there is some δ≥0 such that

lim n → ∞ δ n =δ.
(13)

We shall show that δ=0. Suppose, on the contrary, that δ>0.

Letting n→∞ in (12) and having in mind that we suppose that lim t → r ϕ(t)>0 for all r>0 and lim t → 0 + ϕ(t)=0, we have

ψ(δ)≤ψ(δ)−4ϕ(δ)<ψ(δ),
(14)

which is a contraction. Thus, δ=0, that is,

lim n → ∞ δ n = lim n → ∞ [ d ( g x n , g x n − 1 ) + d ( g y n , g y n − 1 ) + d ( g z n , g z n − 1 ) + d ( g w n , g w n − 1 ) ] =0.
(15)

Now, we shall show that {g x n }, {g y n }, {g z n }, and {g w n } are Cauchy sequences in the metric space (X,d). Assume the contrary, that is, one of the sequences {g x n }, {g y n }, {g z n } or {g w n } is not a Cauchy sequence, that is,

lim n , m → ∞ d ( g x m , g x n ) ≠ 0 or lim n , m → ∞ d ( g y m , g y n ) ≠ 0 , or lim n , m → ∞ d ( g z m , g z n ) ≠ 0 , or lim n , m → ∞ d ( g w m , g w n ) ≠ 0 .

This means that there exists ϵ>0, for which we can find subsequences { x n ( k ) }, { x m ( k ) } of x n and { y n ( k ) }, { y m ( k ) } of y n and { z n ( k ) }, { z m ( k ) } of z n and { w n ( k ) }, { w m ( k ) } of w n with n(k)≥m(k)≥k such that

d(g x m ( k ) ,g x n ( k ) )+d(g y m ( k ) ,g y n ( k ) )+d(g z m ( k ) ,g z n ( k ) )+d(g w m ( k ) ,g w n ( k ) )≥ϵ.
(16)

In addition, by virtue of m(k), we can choose n(k) in such a way that it is the smallest integer with n(k)>m(k)≥k and satisfying (16). It follows that

d ( g x m ( k ) , g x n ( k ) − 1 ) + d ( g y m ( k ) , g y n ( k ) − 1 ) + d ( g z m ( k ) , g z n ( k ) − 1 ) + d ( g w m ( k ) , g w n ( k ) − 1 ) < ϵ .
(17)

By use of the triangle inequality, we have

d(g x m ( k ) ,g x n ( k ) )≤d(g x m ( k ) ,g x n ( k ) − 1 )+d(g x n ( k ) − 1 ,g x n ( k ) ).
(18)

Similarly, we get that

(19)
(20)
(21)

Adding both sides to (18), (19), (20), (21) and using (16) and (17), we have that

ϵ ≤ d ( g x m ( k ) , g x n ( k ) ) + d ( g y m ( k ) , g y n ( k ) ) + d ( g z m ( k ) , g z n ( k ) ) + d ( g w m ( k ) , g w n ( k ) ) ≤ d ( g x m ( k ) , g x n ( k ) − 1 ) + d ( g x n ( k ) − 1 , g x n ( k ) ) + d ( g y m ( k ) , g y n ( k ) − 1 ) + d ( g y n ( k ) − 1 , g y n ( k ) ) + d ( g z m ( k ) , g z n ( k ) − 1 ) + d ( g z n ( k ) − 1 , g z n ( k ) ) + d ( g w m ( k ) , g w n ( k ) − 1 ) + d ( g w n ( k ) − 1 , g w n ( k ) ) ≤ ϵ + d ( g x n ( k ) − 1 , g x n ( k ) ) + d ( g y n ( k ) − 1 , g y n ( k ) ) + d ( g z n ( k ) − 1 , g z n ( k ) ) + d ( g w n ( k ) − 1 , g w n ( k ) ) .

Letting k→∞ and by use of (15), we get

lim k → ∞ λ k = lim k → ∞ [ d ( g x m ( k ) , g x n ( k ) ) + d ( g y m ( k ) , g y n ( k ) ) + d ( g z m ( k ) , g z n ( k ) ) + d ( g w m ( k ) , g w n ( k ) ) ] = ϵ .

Again, by the triangle inequality, we have

λ k = d ( g x m ( k ) , g x n ( k ) ) + d ( g y m ( k ) , g y n ( k ) ) + d ( g z m ( k ) , g z n ( k ) ) + d ( g w m ( k ) , g w n ( k ) ) ≤ d ( g x m ( k ) , g x m ( k ) + 1 ) + d ( g x m ( k ) + 1 , g x n ( k ) + 1 ) + d ( g x n ( k ) + 1 , g x n ( k ) ) + d ( g y m ( k ) , g y m ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y n ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y n ( k ) ) + d ( g z m ( k ) , g z m ( k ) + 1 ) + d ( g z m ( k ) + 1 , g z n ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z n ( k ) ) + d ( g w m ( k ) , g w m ( k ) + 1 ) + d ( g w m ( k ) + 1 , g w n ( k ) + 1 ) + d ( g w n ( k ) + 1 , g w n ( k ) ) ≤ δ m ( k ) + 1 + δ n ( k ) + 1 + d ( g x m ( k ) + 1 , g x n ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y n ( k ) + 1 ) + d ( g z m ( k ) + 1 , g z n ( k ) + 1 ) + d ( g w m ( k ) + 1 , g w n ( k ) + 1 ) .
(22)

Since n(k)≥m(k), then

g x m ( k ) ≤ g x n ( k ) , g y m ( k ) ≥ g y n ( k ) , g z m ( k ) ≤ g z n ( k ) , g w m ( k ) ≥ g w n ( k ) .
(23)

Hence from (1), (3), and (22), we get that

(24)
(25)
(26)
(27)

Combining (22) and (24)-(27), we have that

ψ ( λ k ) ≤ ψ ( δ m ( k ) + 1 + δ n ( k ) + 1 + d ( g x m ( k ) + 1 , g x n ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y n ( k ) + 1 ) + d ( g z m ( k ) + 1 , g z n ( k ) + 1 ) + d ( g w m ( k ) + 1 , g w n ( k ) + 1 ) ) ≤ ψ ( δ m ( k ) + 1 + δ n ( k ) + 1 ) + ψ ( d ( g x m ( k ) + 1 , g x n ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y n ( k ) + 1 ) + d ( g z m ( k ) + 1 , g z n ( k ) + 1 ) + d ( g w m ( k ) + 1 , g w n ( k ) + 1 ) ) ≤ ψ ( δ m ( k ) + 1 ) + ψ ( δ n ( k ) + 1 ) + ψ ( d ( g x m ( k ) + 1 , g x n ( k ) + 1 ) ) + ψ ( d ( g y m ( k ) + 1 , g y n ( k ) + 1 ) ) + ψ ( d ( g z m ( k ) + 1 , g z n ( k ) + 1 ) ) + ψ ( d ( g w m ( k ) + 1 , g w n ( k ) + 1 ) ) ≤ ψ ( δ m ( k ) + 1 ) + ψ ( δ n ( k ) + 1 ) + ψ ( λ k ) − 4 ϕ ( λ k ) .

Letting k→∞, we get a contradiction. This shows that {g x n }, {g y n }, {g z n }, and {g w n } are Cauchy sequences in the metric space (X,d). Since (X,d) is complete, there exist x,y,z,w∈X such that

lim n → ∞ g x n =x, lim n → ∞ g y n =y, lim n → ∞ g z n =zand lim n → ∞ g w n =w.
(28)

From (28) and the continuity of g, we have

lim n → ∞ g ( g x n ) = g x , lim n → ∞ g ( g y n ) = g y , lim n → ∞ g ( g z n ) = g z and lim n → ∞ g ( g w n ) = g w .
(29)

It follows from (3) and the commutativity of F and g that

(30)
(31)
(32)
(33)

Now we shall show that gx=F(x,y,z,w), gy=F(y,z,w,x), gz=F(z,w,x,y), gw=F(w,x,y,z).

By letting n→∞ in (30)-(33), by (28), (29), and the continuity of F, we obtain

(34)
(35)
(36)
(37)

We have shown that F and g have a quadruple coincidence point. □

In the following theorem, the continuity of F is removed. We state the following definition.

Definition 3.1 Let (X,≤) be a partially ordered metric space and d be a metric on X. We say that (X,d,≤) is regular if the following conditions hold:

  1. (i)

    if a nondecreasing sequence a n →a, then a n ≤a for all n,

  2. (ii)

    if a non-increasing sequence b n →b, then b≤ b n for all n.

Theorem 3.2 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d,≤) is regular. Suppose that F: X 4 →X and g:X→X are such that F has the mixed g-monotone property. Assume also that there exist ϕ∈Φ and ψ∈Ψ such that

ψ ( d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ) ≤ 1 4 ψ ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) ) − ϕ ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), (g(X),d) is a complete metric space. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Proof Proceeding exactly as in Theorem 3.1, we have that {g x n }, {g y n }, {g z n }, and {g w n } are Cauchy sequences in the complete metric space (g(X),d). Then there exist x,y,z,w∈X such that

g x n →gx,g y n →gy,g z n →gzandg w n →gw.
(38)

Since {g x n }, {g z n } are nondecreasing and {g y n }, {g w n } are non-increasing, then since (X,d,≤) is regular, we get that

g x n ≤gx,g y n ≥gy,g z n ≤gzandg w n ≥gw

for all n. If g x n =gx, g y n =gy, g z n =gz, and g w n =gw for some n>0, then gx=g x n ≤g x n + 1 ≤gx=g x n , gy≤g y n + 1 ≤g y n =gy, gz=g z n ≤g z n + 1 ≤gz=g z n , and gw≤g w n + 1 ≤g w n =gw, which implies that

g x n = g x n + 1 = F ( x n , y n , z n , w n ) , g y n = g y n + 1 = F ( y n , z n , w n , x n ) and g z n = g z n + 1 = F ( z n , w n , x n , y n ) , g w n = g w n + 1 = F ( w n , x n , y n , z n ) ,

that is, ( x n , y n , z n , w n ) is a quadruple coincidence point of F and g. Then, we suppose that (g x n ,g y n ,g z n ,g w n )≠(gx,gy,gz,gw) for all n>0. By use of (1), consider now

ψ ( d ( g x , F ( x , y , z , w ) ) ) ≤ ψ ( d ( g x , g x n + 1 ) + d ( g x n + 1 , F ( x , y , z , w ) ) ) = ψ ( d ( g x , g x n + 1 ) ) + ψ ( d ( F ( x n , y n , z n , w n ) , F ( x , y , z , w ) ) ) ≤ ψ ( d ( g x , g x n + 1 ) ) + 1 4 ψ ( d ( g x n , g x ) + d ( g y n , g y ) + d ( g z n , g z ) + d ( g w n , g w ) ) − ϕ ( d ( g x n , g x ) + d ( g y n , g y ) + d ( g z n , g z ) + d ( g w n , g w ) ) .
(39)

Letting n→∞ and by (38), then the right-hand side of (39) tends to 0, thus ψ(d(gx,F(x,y,z,w)))=0. By the property (i) of ψ, we have d(gx,F(x,y,z,w))=0. It follows that gx=F(x,y,z,w). Similarly, we can show that

gy=F(y,z,w,x),gz=F(z,w,x,y),gw=F(w,x,y,z).

Therefore, we have proved that F and g have a quadruple coincidence point. □

Corollary 3.1 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose that F: X 4 →X and g:X→X are such that F is continuous and has the mixed g-monotone property. Assume also that there exist ϕ∈Φ and ψ∈Ψ such that

ψ ( d ( F ( x , y , z , w ) , F ( u , v , , h , l ) ) ) ≤ 1 4 ψ ( max { d ( g x , g u ) , d ( g y , g v ) , d ( g z , g h ) , d ( g w , g l ) } ) − ϕ ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), g is continuous and commutes with F. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Proof Since

max { d ( g x , g u ) , d ( g y , g v ) , d ( g z , g h ) , d ( g w , g l ) } ≤d(gx,gu)+d(gy,gv)+d(gz,gh)+d(gw,gl),

then we apply Theorem 3.1, since ψ is assumed to be nondecreasing. □

Similarly, as an easy consequence of Theorem 3.2, we have the following corollary.

Corollary 3.2 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d,≤) is regular. Suppose that F: X 4 →X and g:X→X are such that F has the mixed g-monotone property. Assume also that there exist ϕ∈Φ and ψ∈Ψ such that

ψ ( d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ) ≤ 1 4 ψ ( max { d ( g x , g u ) , d ( g y , g v ) , d ( g z , g h ) , d ( g w , g l ) } ) − ϕ ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), (g(X),d) is a complete metric space. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Corollary 3.3 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose that F: X 4 →X and g:X→X are such that F is continuous and has the mixed g-monotone property. Assume also that there exists k∈[0,1) such that

d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ≤ k 4 ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), g is continuous and commutes with F. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Proof It is sufficient to set ψ(t)=t and Ï•(t)= 1 − k 4 t in Theorem 3.1. □

Corollary 3.4 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d,≤) is regular. Suppose that F: X 4 →X and g:X→X are such that F has the mixed g-monotone property. Assume also that there exists k∈[0,1) such that

d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ≤ k 4 ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), (g(X),d) is a complete metric space. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Proof It is sufficient to set ψ(t)=t and Ï•(t)= 1 − k 4 t in Theorem 3.2. □

Corollary 3.5 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose that F: X 4 →X and g:X→X are such that F is continuous and has the mixed g-monotone property. Assume also that there exists k∈[0,1) such that

d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ≤ k 4 ( max { d ( g x , g u ) , d ( g y , g v ) , d ( g z , g h ) , d ( g w , g l ) } )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), g is continuous and commutes with F. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Proof It suffices to remark that

max { d ( g x , g u ) , d ( g y , g v ) , d ( g z , g h ) , d ( g w , g l ) } ≤ d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) .

Then we apply Corollary 3.3. □

Corollary 3.6 Let (X,≤) be a partially ordered set and suppose there is a metric d on X such that (X,d,≤) is regular. Suppose that F: X 4 →X and g:X→X are such that F has the mixed g-monotone property. Assume also that there exists k∈[0,1) such that

d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ≤ k 4 ( max { d ( g x , g u ) , d ( g y , g v ) , d ( g z , g h ) , d ( g w , g l ) } )

for any x,y,z,w,u,v,h,l∈X, for which gx≤gu, gv≥gy, gz≤gh, and gl≥gw. Suppose that F( X 4 )⊂g(X), (g(X),d) is a complete metric space. If there exist x 0 , y 0 , z 0 , w 0 ∈X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 , w 0 ) , g y 0 ≥ F ( y 0 , z 0 , w 0 , x 0 ) , g z 0 ≤ F ( z 0 , w 0 , x 0 , y 0 ) , g w 0 ≥ F ( w 0 , x 0 , y 0 , z 0 ) ,

then there exist x,y,z,w∈X such that

F(x,y,z,w)=gx,F(y,z,w,x)=gy,F(z,w,x,y)=gzandF(w,x,y,z)=gw,

that is, F and g have a quadruple coincidence point.

Remark 3.1 (i) Theorem 11 of Karapinar and Luong [21] is a particular case of Theorem 3.1 and Theorem 3.2 by taking g= I X , respectively. Corollary 12 of Karapinar and Luong [21] is a particular case of Theorem 3.1 and Theorem 3.2 by taking g= I X , ψ(t)=t, Ï•(t)= 1 − k 4 t.

(ii) Theorem 2.3 of Karapinar [22] is a particular case of Theorem 3.1 and Theorem 3.2 by taking g= I X and ψ(t)=t, respectively. Corollary 2.4 of Karapinar [22] is a particular case of Theorem 3.1 and Theorem 3.2 by taking g= I X , ψ(t)=t, Ï•(t)= 1 − k 4 t.

Now, we shall prove the existence and uniqueness of a quadruple common fixed point. For a product X 4 of a partial ordered set (X,≤), we define a partial ordering in the following way: For all (x,y,z,w),(u,v,r,h)∈ X 4 ,

(x,y,z,w)≤(u,v,r,h)⇒x≤u,y≥v,z≤randw≥l.
(40)

We say that (x,y,z,w) and (u,v,r,l) are comparable if

(x,y,z,w)≤(u,v,r,l)or(u,v,r,l)≥(x,y,z,w).

Also, we say that (x,y,z,w) is equal to (u,v,r,l) if and only if x=u, y=v, z=r and w=l.

Theorem 3.3 In addition to the hypotheses of Theorem  3.1, suppose that for all (x,y,z,w),(u,v,r,l)∈ X 4 , there exists (a,b,c,d)∈ X 4 such that

( F ( a , b , c , d ) , F ( b , c , d , a ) , F ( c , d , a , b ) , F ( d , a , b , c ) )

is comparable to

( F ( x , y , z , w ) , F ( y , z , w , x ) , F ( z , w , x , y ) , F ( w , x , y , z ) )

and

( F ( u , v , r , l ) , F ( v , r , l , u ) , F ( r , l , u , v ) , F ( l , u , v , r ) ) .

Then F and g have a unique quadruple common fixed point (x,y,z,w) such that

x = g x = F ( x , y , z , w ) , y = g y = F ( y , z , w , x ) , z = g z = F ( z , w , x , y ) , w = g w = F ( w , x , y , z ) .

Proof The set of quadruple coincidence points of F and g is not empty due to Theorem 3.1. Assume now that (x,y,z,w) and (u,v,r,l) are two quadruple coincidence points of F and g, i.e.,

F ( x , y , z , w ) = g x , F ( u , v , r , l ) = g u , F ( y , z , w , x ) = g y , F ( v , r , l , u ) = g v , F ( z , w , x , y ) = g z , F ( r , l , u , v ) = g r , F ( w , x , y , z ) = g w , F ( l , u , v , r ) = g l .

We shall show that (gx,gy,gz,gw) and (gu,gv,gr,gl) are equal. By assumption, there exists (a,b,c,d)∈ X 4 such that (F(a,b,c,d),F(b,c,d,a),F(c,d,a,b),F(d,a,b,c)) is comparable to (F(x,y,z,w),F(y,z,w,x),F(z,w,x,y),F(w,x,y,z)) and (F(u,v,r,l),F(v,r,l,u),F(r,l,u,v),F(l,u,v,r)). Define sequences {g a n }, {g b n }, {g c n }, and {g d n } such that a 0 =a, b 0 =b, c 0 =c, d 0 =d, and for any n≥1,

g a n = F ( a n − 1 , b n − 1 , c n − 1 , d n − 1 ) , g b n = F ( b n − 1 , c n − 1 , d n − 1 , a n − 1 ) , g c n = F ( c n − 1 , d n − 1 , a n − 1 , b n − 1 ) , g d n = F ( d n − 1 , a n − 1 , b n − 1 , c n − 1 )
(41)

for all n. Further, set x 0 =x, y 0 =y, z 0 =z, w 0 =w and u 0 =u, v 0 =v, r 0 =r, l 0 =l, and in the same way define the sequences {g x n }, {g y n }, {g z n }, {g w n } and {g u n }, {g v n }, {g r n }, {g l n }. Then it is easy to see that

g x n = F ( x n − 1 , y n − 1 , z n − 1 , w n − 1 ) , g u n = F ( u n − 1 , v n − 1 , r n − 1 , l n − 1 ) , g y n = F ( y n − 1 , z n − 1 , w n − 1 , x n − 1 ) , g v n = F ( v n − 1 , r n − 1 , l n − 1 , u n − 1 ) , g z n = F ( z n − 1 , w n − 1 , x n − 1 , y n − 1 ) , g r n = F ( r n − 1 , l n − 1 , u n − 1 , v n − 1 ) , g w n = F ( w n − 1 , x n − 1 , y n − 1 , z n − 1 ) , g l n = F ( l n − 1 , u n − 1 , v n − 1 , r n − 1 )
(42)

for all n≥1.

Since (F(x,y,z,w),F(y,z,w,x),F(z,w,x,y),F(w,x,y,z))=(g x 1 ,g y 1 ,g z 1 ,g w 1 )=(gx,gy,gz,gw) is comparable to

( F ( a , b , c , d ) , F ( b , c , d , a ) , F ( c , d , a , b ) , F ( d , a , b , c ) ) =(g a 1 ,g b 1 ,g c 1 ,g d 1 ),

then it is easy to show (gx,gy,gz,gw)≥(g a 1 ,g b 1 ,g c 1 ,g d 1 ). Recursively, we get that

(43)
(44)
(45)
(46)

and

ψ ( d ( g w , g d n + 1 ) ) = 1 4 d ( F ( w , x , y , z ) , F ( d n , a n , b n , c n ) ) ≤ ψ ( d ( g w , g d n ) + d ( g x , g a n ) + d ( g y , g b n ) + d ( g z , g c n ) ) − ϕ ( d ( g w , g d n ) + d ( g x , g a n ) + d ( g y , g b n ) + d ( g z , g c n ) ) .
(47)

From (44)-(47), it follows that

ψ ( d ( g a n + 1 , g x ) ) + ψ ( d ( g y , g b n + 1 ) ) + ψ ( d ( g c n + 1 , g z ) ) + ψ ( d ( g w , g d n + 1 ) ) ≤ ψ ( d ( g w , g d n ) + d ( g x , g a n ) + d ( g y , g b n ) + d ( g z , g c n ) ) − 4 ϕ ( d ( g w , g d n ) + d ( g x , g a n ) + d ( g y , g b n ) + d ( g z , g c n ) ) .

By the property (iii) of ψ, we obtain that

(48)

Set σ n =d(g a n ,gx)+d(gy,g b n )+d(g c n ,gz)+d(gw,g d n ). Then due to (48), we have

ψ( σ n + 1 )≤ψ( σ n )−4Ï•( σ n )for all n,
(49)

which implies that ψ( σ n + 1 )≤ψ( σ n ). By the property of ψ, we obtain that σ n + 1 ≤ σ n . Thus, the sequence { σ n } is decreasing and bounded below from 0. Therefore, there exists σ≥0 such that

lim n → ∞ σ n =σ.

Now, we shall show that σ=0. Suppose to the contrary that σ>0. Letting n→∞ in (49), we obtain that

ψ(σ)≤ψ(σ)−4 lim n → ∞ ϕ( σ n )<ψ(σ),

which is a contradiction. It yields that σ=0. That is, lim n → ∞ σ n =0.

Consequently, we have

lim n → ∞ d ( g a n , g x ) = 0 , lim n → ∞ d ( g y , g b n ) = 0 , lim n → ∞ d ( g c n , g z ) = 0 , lim n → ∞ d ( g w , g d n ) = 0 .
(50)

Similarly, we can prove that

lim n → ∞ d ( g a n , g u ) = 0 , lim n → ∞ d ( g v , g b n ) = 0 , lim n → ∞ d ( g c n , g r ) = 0 , lim n → ∞ d ( g l , g d n ) = 0 .
(51)

Combining (50) and (51) yields that (gx,gy,gz,gw) and (gu,gv,gr,gl) are equal.

Since gx=F(x,y,z,w), gy=F(y,z,w,x), gz=F(z,w,x,y), and gw=F(w,x,y,z), by the commutativity of F and g, we obtain that

g x ′ = g ( g x ) = g ( F ( x , y , z , w ) ) = F ( g x , g y , g w , g z ) = F ( x ′ , y ′ , z ′ , w ′ ) , g y ′ = g ( g y ) = g ( F ( y , z , w , x ) ) = F ( g y , g z , g w , g x ) = F ( y ′ , z ′ , w ′ , x ′ ) , g z ′ = g ( g z ) = g ( F ( z , w , x , y ) ) = F ( g z , g w , g x , g y ) = F ( z ′ , w ′ , x ′ , y ′ ) , g w ′ = g ( g z ) = g ( F ( w , x , y , z ) ) = F ( g w , g x , g y , g z ) = F ( w ′ , x ′ , y ′ , z ′ ) ,

where gx= x ′ , gy= y ′ , gz= z ′ , and gw= w ′ . Thus, ( x ′ , y ′ , z ′ , w ′ ) is a quadruple coincidence point of F and g. Therefore, (g x ′ ,g y ′ ,g z ′ ,g w ′ ) and (gx,gy,gz,gw) are equal. We obtain that

g x ′ =gx= x ′ ,g y ′ =gy= y ′ ,g z ′ =gz= z ′ ,g w ′ =gw= w ′ .

Thus, ( x ′ , y ′ , z ′ , w ′ ) is a quadruple common fixed point of F and g. Its uniqueness follows from contraction (1). □

Example 3.1 Let X=R with the metric d(x,y)=|x−y| for all x,y∈X and the usual ordering. Let F: X 4 →X and g:X→X be given by

g(x)= 3 4 x,F(x,y,z,w)= x − y + z − w 16 for all x,y,z,w∈X.

Let ψ,ϕ:[0,∞)→[0,∞) be given by

ψ(t)= 1 5 tandÏ•(t)= t 60 for all t∈[0,∞).

We will check that the condition (1) is satisfied for all x,y,z,w,u,v,h,l∈X satisfying gx≤gu, gv≤gy, gz≤gh, gl≤gw. In this case, we have

ψ ( d ( F ( x , y , z , w ) , F ( u , v , h , l ) ) ) = 1 5 [ u − x 16 + y − v 16 + h − z 16 + w − l 16 ] = 1 4 { 1 5 [ 3 ( u − x ) 4 + 3 ( y − v ) 4 + 3 ( h − z ) 4 + 3 ( w − l ) 4 ] } − 1 30 [ 3 ( u − x ) 4 + 3 ( y − v ) 4 + 3 ( h − z ) 4 + 3 ( w − l ) 4 ] ≤ 1 4 { 1 5 [ 3 ( u − x ) 4 + 3 ( y − v ) 4 + 3 ( h − z ) 4 + 3 ( w − l ) 4 ] } − 1 60 [ 3 ( u − x ) 4 + 3 ( y − v ) 4 + 3 ( h − z ) 4 + 3 ( w − l ) 4 ] = 1 4 ψ [ d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) ] − ϕ [ d ( g x , g u ) + d ( g y , g v ) + d ( g z , g h ) + d ( g w , g l ) ] .

It is easy to check that all the conditions of Theorem 3.3 are satisfied and (0,0,0,0) is the unique quadruple fixed point of F and g.

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Acknowledgements

This work is partially supported by the Scientific Research Fund of Sichuan Provincial Education Department (12ZA098), Scientific Research Fund of Sichuan University of Science and Engineering (2012KY08).

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Correspondence to Xiao-lan Liu.

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Liu, Xl. Quadruple fixed point theorems in partially ordered metric spaces with mixed g-monotone property. Fixed Point Theory Appl 2013, 147 (2013). https://doi.org/10.1186/1687-1812-2013-147

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