An application of Meir-Keeler type tripled fixed point theorem

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International Journal of Advances in Applied Mathematics and Mechanics

An application of Meir-Keeler type tripled fixed point theorem

Research Article

Animesh Gupta^∗

H.No. 93/654, Ward No. 2 Gandhi Chowk, Pachmarhi-461881, Dist. Hoshangabad (M.P.), India

Visiting at : Department of Mathematics & Computer Science, Govt. Model Science College, Jabalpur (M.P.) India

Received 21 August 2014; accepted (in revised version) 13 November 2014

Abstract: Berinde and Borcut[1] introduced the concept of triple fixed point and proof some related fixed point theorem with some applications. The aim of this paper is to extend the result of Berinde and Borcut[1]. Indeed, we introduced the definition of generalized g−Meir-Keeler type contractions and prove some tripled fixed point theorems under a generalized g−Meir-Keeler type contractive condition. We also give an application of main results of this paper.

MSC: Primary 47H10• Secondary 54H25

Keywords: Tripled coincidence point• Tripled common fixed point • Mixed strict g −monotone • Mixed strict monotone

• Meir-Keeler type contraction 2014 IJAAMM all rights reserved.c

1. Introduction and preliminaries

The Banach contraction principle[2] is a classical and powerful tool in non linear analysis and has been generalized by many authors. Bhaskar and Lakshmikantham[3] introduced the concept of a coupled fixed point of mapping F : X×X → X and investigated some coupled fixed point theorems in partially ordered metric spaces. Later, various results in coupled fixed point have been obtained, see e.g. ([4]-[9]).

On the other hand, Berinde and Borcut[1] introduced the concept of triple fixed point and proof some related fixed point theorem. After this various results on tripled fixed point have been obtained. The following definitions are from[1].

Definition 1.1.

Let(X ,) be a partially ordered set, F : X³→ X mapping. The mapping F is said to have the mixed monotone property if for any x , y , z∈ X ,

(i) x₁, x₂∈ X , x1 x2⇒ F (x1, y , z)  F (x2, y , z), (ii) y₁, y₂∈ X , y1 y2⇒ F (x , y1, z)  F (x , y2, z) , (iii) z₁, z₂∈ X , z1 z2⇒ F (x , y , z1)  F (x , y, z2).

Definition 1.2.

An element(x , y, z ) ∈ X³is called a tripled fixed point of F : X³→ X if F(x , y, z ) = x , F (y, x , y ) = y, and F (z , y, x ) = z .

∗ Corresponding author.

E-mail address:[email protected]

Definition 1.3.

Let(X ,) be a partially ordered set, F : X³→ X and g : X → X two mappings. The mapping F is said to have the mixed g− monotone property if for any x , y , z ∈ X .

i. x₁, x₂∈ X , g (x1)  g (x2) ⇒ F (x1, y , z)  F (x2, y , z), ii. y₁, y₂∈ X , g (y1)  g (y2) ⇒ F (x , y1, z)  F (x , y2, z) , iii. z₁, z₂∈ X , g (z1)  g (z2) ⇒ F (x , y, z1)  F (x , y, z2).

Definition 1.4.

An element(x , y, z ) ∈ X³is called a tripled coincidence point of the mappings F : X³→ X and g : X → X if F(x , y, z ) = g x , F (y, x , y ) = g y and F (z , y, x ) = g z .

Definition 1.5.

An element(x , y, z ) ∈ X³is called a tripled common fixed point of the mappings F : X³→ X and g : X → X if F(x , y, z ) = g x = x , F (y, x , y ) = g y = y and F (z , y, x ) = g z = z .

Definition 1.6.

An element x∈ X is called a common fixed point of the mappings F : X³→ X and g : X → X if F(x , x , x ) = g x = x .

Definition 1.7.

Let X be a non empty set. The mappings F : X³→ X and g : X → X are commuting if for all x , y , z ∈ X , g(F (x , y, z )) = F (g (x ),g (y ),g (z )).

Definition 1.8.

Let(X ,d ) be a metric space. The mappings F and g where F : X³→ X and g : X → X are said to be compatible if

n→∞lim d(g (F (xn, y_n, z_n)), F (g (xn),g (yn),g (zn))) = 0

n→∞lim d(g (F (yn, xn, yn)), F (g (yn),g (xn),g (yn))) = 0 a n d

n→∞lim d(g (F (zn, y_n, x_n)), F (g (zn),g (yn),g (xn))) = 0

whenever {xn}, {yn} and {zn} are sequences in X such that limn→∞F(xn, y_n, z_n) = lim_n→∞g(xn) = x , lim_n→∞F(yn, xn, yn) = limn→∞g(yn) = y and limn→∞F(zn, yn, xn) = limn→∞g(zn) = z for some x , y, z ∈ X . In[1] Berinde and Borcut proved the following theorem.

Theorem 1.1.

Let(X ,) be a partially ordered set and (X ,d ) be a complete metric space. Let F : X³→ X be a continuous mapping having the mixed monotone property on X . Assume that there exist constants a , b , c∈ [0, 1) such that a + b + c ≺ 1 for which,

d(F (x , y, z ), F (u, v, w ))  ad (x , u) + b d (y, v ) + c d (z , w ) (1)

For all x u, y  v, z  w . Assume either, 1. F is continuous,

2. X has the following properties:

(a) if non decreasing sequence x_n→ x , then xn x for all n, (b) if non increasing sequence y_n→ y , then yn x for all n,

If there exist x₀, y₀, z₀∈ X such that

x₀ F (x0, y₀, z₀), y0 F (y0, x₀, y₀), and z0 F (z0, y₀, x₀) Then there exist x , y , z∈ X such that,

F(x , y, z ) = x , F (y, x , y ) = y,and F (z , y, x ) = z

Beside this in[10] Meir and Keeler generalized the well known Banach fixed point theorem [2] as follows;

Theorem 1.2.

Let(X ,) be a complete metric space and T : X × X → X be a given mapping. Suppose that, for any ε  0 then there existsδ(ε)  0 such that

ε  d (x , y ) ≺ ε + δ(ε) =⇒ d (T x ,T y ) ≺ ε (2)

for all x , y ∈ X . Then T admits a unique fixed point x0∈ X for all x ∈ X , the sequence {Tⁿ(x )} converges to x0. Definition 1.9.

Let(X ,) be a partially ordered metric space and F : X³→ X be a given mapping satisfying the following contraction condition,

ε 1

3(d (x , u) + d (y, v ) + d (z , w )) ≺ ε + δ(ε) =⇒ d (F (x , y, z ), F (u, v, w )) ≺ ε (3) for all x , y , z , u , v, w∈ X . Then F is a generalized Meir-Keeler type contraction.

Motivated by the results of Berinde and Borcut[1] and Meir-Keeler [10] we introduced the definition of g − Meir- Keeler contractive mappings and prove some tripled common fixed point theorems under the generalized g−Meir- Keeler contractive condition.

2. Main Results

We introduced the following two definitions.

Definition 2.1.

Let(X ,) be a partially ordered set and F : X³→ X and g : X → X . We say that F has the mixed strict g − monotone property if, for any x , y , z∈ X ,

x₁, x₂∈ X , g(x1) ≺ g (x2) ⇒ F (x1, y , z) ≺ F (x2, y , z) y₁, y₂∈ X , g(y1) ≺ g (y2) ⇒ F (x , y1, z)  F (x , y2, z) z₁, z₂∈ X , g(z1) ≺ g (z2) ⇒ F (x , y, z1) ≺ F (x , y, z2).







(4)

Definition 2.2.

Let(X ,) be a partially ordered set, F : X³→ X and g : X → X be two mappings. We say that F is a generalized g − Meir-Keeler type contraction if for allε  0 there exists δ(ε)  0 such that for all x , y, z , u, v, w ∈ X with g (x )  g (u), g(y )  g (v ) and g (z )  g (w ),

ε  1

3[d (g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w ))] ≺ ε + δ(ε)

⇒ d (F (x , y , z ), F (u, v, w )) ≺ ε.

(5)

Lemma 2.1.

Let(X ,) be a partially ordered set and F : X³→ X and g : X → X be two mappings. We say that F is a generalized g− Meir-Keeler type contraction then we have

d(F (x , y, z ), F (u, v, w )) ≺1

3[d (g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w ))] (6) for all x , y , z , u , v, w∈ X with g (x )  g (u), g (y )  g (v ) and g (z )  g (w ).

Proof. Let x , y , z , u , v, w∈ X such that

g(x ) ≺ g (u), g (y )  g (v ), g (z )  g (w ) or

g(x )  g (u), g (y )  g (v ), g (z )  g (w ) or

g(x )  g (u), g (y )  g (v ), g (z ) ≺ g (w ) then

d(g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w ))  0.

Since F is a generalized g− Meir-Keeler type contraction so ε =1

3 d(g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w ))

there existsδ(ε)  0 such that, for all x0, y₀, z₀, u₀, v₀, w₀∈ X with g (x0)  g (u0), g (y0)  g (v0) and g (z0)  g (w0),

ε  1

3 d(g (x0),g (u0)) + d (g (y0),g (v0)) + d (g (z0),d (w0))
≺ ε + δ(ε)

⇒ d(F (x0, y₀, z₀), F (u0, v₀, w₀)) ≺ ε (7)

therefore putting x₀= x , y0= y , z0= z , u0= u, v0= v and w0= w , we have

ε  1

3 d(g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w ))
≺ ε + δ(ε)

⇒ d(F (x , y, z ), F (u, v, w )) ≺ ε (8)

this completes the proof.

Theorem 2.1.

Let F : X³→ X and g : X → X be two mappings such that F (X³) ⊆ g (X ) also g is continuous and commutative with F . Suppose that

(a) F has the mixed strict g− monotone property;

(b) F is a generalized g− Meir-Keeler type contraction;

(c) there exist x₀, y₀, z₀∈ X such that

g(x0) ≺ F (x0, y₀, z₀), g (y0)  F (y0, x₀, y₀) and g (z0) ≺ F (z0, y₀, x₀) Then there exist x , y , z∈ X such that,

F(x , y, z ) = g (x ), F (y, x , y ) = g (y ) and F (z , y, x ) = g (z ) that is F and g have a tripled coincidence in X³.

Proof. Let x₀, y₀, z₀∈ X be such that, g(x0) ≺ F (x0, y₀, z₀),

g(y0)  F (y0, x₀, y₀), g(z0) ≺ F (z0, y₀, x₀).

Since F(X³) ⊆ g (X ), we can choose x1, y₁, z₁∈ X such that,

g(x1) = F (x0, y₀, z₀), g(y1) = F (y0, x₀, y₀), g(z1) = F (z0, y0, x0).

Again, from F(X³) ⊆ g (X ) we can choose x2, y₂, z₂∈ X such that, g(x2) = F (x1, y₁, z₁),

g(y2) = F (y1, x₁, y₁), g(z2) = F (z1, y₁, x₁).

Continuing this process we can construct the sequence{xn}, {yn} and {zn} in X such that

g(x_n+1) = F (xn, y_n, z_n), g(y_n+1) = F (yn, x_n, y_n)

g(z_n+1) = F (zn, y_n, x_n) (9)

for all n 0.

Next we show that g(xn) ≺ g (x_n+1),

g(yn)  g (y_n+1),

g(zn) ≺ g (z_n+1) (10)

for all n 0. If we take n = 0 in (10) then g(x0) ≺ g (x1) = F (x0, y₀, z₀),

g(y0)  g (y1) = F (y0, x₀, y₀),

g(z0) ≺ g (z1) = F (z0, y₀, x₀). (11)

Since F has the mixed strict g− monotone property, then we have g(x0) ≺ g (x1) ⇒ F (x0, y₁, z₁) ≺ F (x1, y₁, z₁),

g(y0) ≺ g (y1) ⇒ F (y0, x₁, y₀)  F (y1, x₁, y₁),

g(z0) ≺ g (z1) ⇒ F (z0, y₁, x₁) ≺ F (z1, y₁, x₁). (12)

Continuing this process for each n 1, we have

g(x_n+2) = F (x_n+1, y_n+1, z_n+1)  F (xn, y_n+1, z_n+1)  F (xn, y_n, z_n) = g (xn+1), g(y_n+2) = F (y_n+1, x_n+1, y_n+1) ≺ F (yn, x_n+1, z_n+1) ≺ F (yn, x_n, y_n) = g (yn+1), g(zn+2) = F (z_n+1, y_n+1, x_n+1)  F (zn, y_n+1, x_n+1)  F (zn, y_n, x_n) = g (zn+1), implies that,

g(x0) ≺ g (x1) ≺ g (x2) ≺ ··· ≺ g (xn) ≺ g (x_n+1) ≺ ..., g(y0)  g (y1)  g (y2)  ···  g (yn)  g (y_n+1)  ...,

g(z0) ≺ g (z1) ≺ g (z2) ≺ ··· ≺ g (zn) ≺ g (z_n+1) ≺ .... (13)

Denote that

δn= d (g (xn),g (x_n+1)) + d (g (yn),g (y_n+1)) + d (g (zn),g (z_n+1)). (14) Since g(xn)  g (x_n−1),g (yn) ≺ g (y_n−1),g (zn)  g (z_n−1).

It follows that from (9) and Lemma2.1,

d(g (xn),g (x_n+1)) =d (F (x_n−1, y_n−1, z_n−1), F (xn, y_n, z_n))

1

3d (g (x_n−1),g (xn)) + d (g (y_n−1),g (yn))

+d (g (z_n−1),d (zn)) . (15)

In order to (15) we have

d(g (yn),g (y_n+1)) = d (F (y_n−1, z_n−1, x_n−1), F (yn, z_n, x_n))

1

3d (g (y_n−1),g (yn)) + d (g (zn−1),g (zn)) +d (g (x_n−1),d (xn))

(16)

and

d(g (zn),g (zn+1)) = d (F (z_n−1, x_n−1, y_n−1), F (zn, x_n, y_n))

1

3d (g (z_n−1),g (zn)) + d (g (x_n−1),d (xn))

+d (g (yn−1),g (yn)) . (17)

Thus it follows from (14) - (17) thatδn≺ δ_n−1. Its mean the sequence{€_δ

n 3

Š} is monotone decreasing. So there exists δ^∗ 0 such that lim_n→∞€_δ

n

3Š = δ^∗, that is ,

n→∞lim 1

3d (g (z_n−1),g (zn)) + d (g (x_n−1),d (xn)) + d (g (y_n−1),g (yn)) = δ^∗. (18) Next we show thatδ^∗= 0. Suppose that δ^∗ 0 hold. Let δ^∗= ε. Then there exists a positive integer m such that,

ε 1

3 d(g (xm),d (x_m+1)) + d (g (ym),g (y_m+1)) + d (g (zm),g (z_m+1))

= ε + δ(ε). (19)

Then, by using (10) also F is a generalized g− Meir-Keeler type contraction, so

d(F (xm, yn, zm), F (xm+1, y_m+1, z_m+1)) ≺ ε, (20)

from (9) we have

d(g (x_m+1),g (x_m+2)) ≺ ε. (21)

On the other hand, by (15), we have 1

3d (g (xm),g (x_m+1)) + d (g (ym),g (y_m+1)) + d (g (zm),d (z_m+1)) ≺ ε (22) which contradiction of (19). Thus we haveε = δ^∗= 0 so

n→∞lim 1

3d (g (xn),g (x_n+1)) + d (g (yn),g (y_n+1)) + d (g (zn),d (z_n+1))
] = 0 (23) that is,

n→∞lim δn= 0 (24)

Next we prove that{g (xn)}, {g (yn)} and {g (zn)} are Cauchy sequences in X .

Suppose that at least one of{g (xn)}, {g (yn)} or {g (zn)} are not a Cauchy sequence. Then there exist ε  0 and two subsequences{lk} and {mk} of integers mk lk k and

d(g (xlk),g (xmk)) ε 3, d(g (ylk),g (ymk)) ε

3, d(g (zlk),g (zmk)) ε

3 (25)

for all k 1. Then we have

r_k= d (g (xlk),g (xmk)) + d (g (ylk),g (ymk)) + d (g (zlk),g (zmk))  ε (26) for all k 1. Let mk be the smallest number exceeding l_k such that (26) holds. Then we have

d(g (xlk),g (xmk−1)) + d (g (ylk),g (ymk−1)) + d (g (zlk),g (zmk−1)) ≺ ε (27)

Thus from (14), (26), (27) and the triangular inequality, it follows that

ε  rk

ε d (g (xlk),g (xmk−1)) + d (g (xmk−1),g (xmk)) + d (g (ylk),g (ymk−1)) + d (g (ymk−1),g (ymk)) + d (g (zlk),g (zmk−1)) + d (g (zmk−1),g (zmk))

 ε + δmk−1 (28)

and so,

ε  lim

k→∞r_k lim

k→∞(ε + δmk−1) (29)

from (24) we have

k→∞lim r_k= ε⁺. (30)

It follow from (9),(14) and the triangle inequality that

r_k= d (g (xlk),g (xmk)) + d (g (ylk),g (ymk)) + d (g (zlk),g (zmk))

 ε

 d (g (xlk),g (xlk+1)) + d (g (xlk+1),g (xmk+1)) + d (g (xmk+1),g (xmk)) + d (g (ylk),g (ylk+1)) + d (g (ylk+1),g (ymk+1)) + d (g (ymk+1),g (ymk)) + d (g (zlk),g (zlk+1)) + d (g (zlk+1),g (zmk+1)) + d (g (zmk+1),g (zmk))

= δlk+ δmk+ d (g (xlk+1),g (xmk+1))

+ d (g (ylk+1),g (ymk+1)) + d (g (zlk+1),g (zmk+1))

= δlk+ δmk+ d (F (xlk, y_lk, z_lk), F (xmk, y_mk, z_mk)) + d (F (ylk, x_lk, y_lk), F (ymk, x_mk, y_mk))

+ d (F (zlk, y_lk, x_lk), F (zmk, y_mk, x_mk)). (31)

From (13) we have g(xlk) ≺ g (xmk), g (ylk)  g (ymk) and g (zlk) ≺ g (zmk). It follows from Lemma2.1and (31) that

r_k≺ δlk+ δmk+ d (g (xlk),g (xmk))

+ d (g (ylk),g (ymk)) + d (g (zlk),g (zmk)) (32)

that is,

r_k≺ δlk+ δmk+ rk (33)

This is contradiction. Therefore,{g (xn)}, {g (yn)} and {g (zn)} are Cauchy sequences in X . Since X is complete, there exist x , y , z∈ X such that

n→∞lim g(xn) = x , lim

n→∞g(yn) = y and lim

n→∞g(zn) = z . (34)

Since{g (xn)},{g (zn)} are monotone increasing and {g (yn)} is monotone decreasing, so we have

g(xn) ≺ x , g (yn)  y and g (zn) ≺ z (35)

for all n 1. Thus it follows from (34) and the continuity of g that

n→∞lim g(xn) = g (x ), lim

n→∞g(yn) = g (y ) and lim

n→∞g(zn) = g (z ). (36)

Thus, for all m 1, there exists a positive integer nnsuch that, for all n n0,

d(g (g (xn)),g (x )) ≺ 1

4m, d(g (g (yn)),g (y )) ≺ 1

4m a n d d(g (g (zn)),g (z )) ≺ 1

4m (37)

Hence from (9), the commutativity of F and g and the triangle inequality, we have d(F (x , y, z ),g (x ))  d (F (x , y, z ),g (g (xn)))

+ d (g (g (xn)),g (x ))

= d (F (x , y, z ),g (F (x_n−1, y_n−1, z_n−1))) + d (g (g (xn)),g (x ))

= d (F (x , y, z ), F (g x_n−1, g y_n−1, g z_n−1))

+ d (g (g (xn)),g (x )). (38)

Thus, it follows from (35), (37) and Lemma2.1, that

d(F (x , y, z ),g (x )) ≺1

3[d (g (g (x_n−1)),g (x )) + d (g (g (y_n−1)),g (y )) + d (g (g (z_n−1)),g (z ))]

≺ 1

12m+ 1

12m+ 1

12m+ 1 4m 1

2m → 0 (39)

as m→ ∞. Therefore, we have F (x , y , z ) = g (x ). Similarly we can show that F (y , z , y ) = g (y ) and F (z , y , x ) = g (z ).

This means that F and g have a coupled coincidence point in X³. This completes the proof.

Corollary 2.1.

Let F : X³→ X be a mapping satisfying the following conditions:

(a) F has the mixed strict monotone property;

(b) F is a generalized Meir-Keeler type contraction;

(c) there exist x₀, y₀, z₀∈ X such that x0≺ F (x0, y₀, z₀), y0 F (y0, x₀, y₀) and z0 F (z0, y₀, x₀).

Then there exist x , y , z∈ X such that,

F(x , y, z ) = x , F (y, x , y ) = y and F (z , y, x ) = z that is, F has a tripled fixed point in X³.

Proof. The conclusion follows from Theorem2.1, by taking g= I (Identity mapping) on X .

Now we introduced the product space X³with the following partial order; for all(x , y, z ),(u, v, w ) ∈ X³,

(x , y, z )  (u, v, w ) ⇐⇒ u ≺ x , v  y, w  z . (40)

Theorem 2.2.

Suppose that all the hypothesis of Theorem 2.1 hold and further, for all (x , y, z ),(x^∗, y^∗, z^∗) ∈ X³, there is (u, v, w ) ∈ X³ such that (F (u, v, w ), F (v, u, v ), F (w, v, u)) is comparable to (F (x , y, z ), F (y, x , y ), F (z , y, x )) and (F (x^∗, y^∗, z^∗), F (y^∗, x^∗, y^∗), F (z^∗, y^∗, x^∗)). Then F and g have a unique tripled common fixed point in X³such that

F(x , y, z ) = g (x ) = x , F (y, x , y ) = g (y ) = y and F (z , y, x ) = g (z ) = z . (41) Proof. By Theorem2.1, the set of tripled coincidence of the mappings F and g is non empty. First we show that if (x , y, z ) and (x^∗, y^∗, z^∗) are tripled coincidence points of F and g , that is, if

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

F(x^∗, y^∗, z^∗) = g (x^∗), F (y^∗, x^∗, y^∗) = g (y^∗) and F (z^∗, y^∗, x^∗) = g (z^∗). (42)

then we have

g(x ) = g (x^∗) g (y ) = g (y^∗) and g (z ) = g (z^∗) (43)

Put u₀ = u, v0 = v , w0= w and choose u1, v₁, w₁ ∈ X such that g (u1) = F (u0, v₀, w₀) , g (v1) = F (v0, w₀, u₀) and g(w1) = F (w0, u₀, v₀). Similarly as in the proof of Theorem2.1, we can inductively define the sequences{g (un)}, {g (vn)} and {g (wn)} such that

g(u_n+1) = F (un, v_n, w_n), g (v_n+1) = F (vn, u_n, v_n), and g (w_n+1) = F (wn, v_n, u_n) (44) for all n 0. Also, if we set x0= x , y0= y , z0= z and x₀^∗= x^∗, y₀^∗= y^∗, z₀^∗= z^∗then we can defined the sequences {g (xn)}, {g (yn)}, {g (zn)} and {g (x_n^∗)}, {g (y_n^∗)}, {g (z_n^∗)} as follows,

g(x_n+1) = F (xn, y_n, z_n), g (y_n+1) = F (yn, x_n, y_n), and g (z_n+1) = F (zn, y_n, x_n)

g(x_n+1^∗ ) = F (x_n^∗, y_n^∗, z_n^∗), g (y_n+1^∗ ) = F (y_n^∗, x_n^∗, y_n^∗), and g (z_n+1^∗ ) = F (z_n^∗, y_n^∗, x_n^∗) (45) for all n 0. Since

(F (x , y, z ), F (y, x , y ), F (z , y, x )) = (g (x1),g (y1),g (z1)) = (g (x ),g (y ),g (z ))

(F (u, v, w ), F (v, u, v ), F (w, v, u)) = (g (u1),g (v1),g (w1)) = (g (u),g (v ),g (w )) (46) are comparable each other,so g(x ) ≺ g (u1) , g (y )  g (v1) and g (z )  g (w1). It is easy to show that (g (x ),g (y ),g (z )) and(g (un),g (vn),g (wn)) are comparable each other, that is g (x ) ≺ g (un) , g (y )  g (vn) and g (z )  g (wn) for all n  1.

Thus it follows from Lemma2.1, that

d(g (x ),g (u_n+1)) + d (g (y ),g (v_n+1)) + d (g (z ),g (w_n+1))

= d (F (x , y, z ), F (un, v_n, w_n)) + d (F (y, x , y ), F (vn, u_n, v_n)) + d (F (z , y, x ), F (wn, v_n, u_n))

≺1

3[d (g (x ),g (un)) + d (g (y ),g (vn)) + d (g (z ),g (wn))]

+1

3[d (g (y ),g (vn)) + d (g (x ),g (un)) + d (g (y ),g (vn))]

+1

3[d (g (z ),g (wn)) + d (g (x ),g (un)) + d (g (y ),g (vn))] (47)

and so 1

3[d (g (x ),g (u_n+1)) + d (g (y ),g (v_n+1)) + d (g (z ),g (w_n+1))]

≺ 1

3ⁿ[d (g (x ),g (u1)) + d (g (y ),g (v1)) + d (g (z ),g (w1))] → 0

(48)

as n→ ∞. Therefore, we have

n→∞lim d(g (x ),g (u_n+1)) = 0,

n→∞limd(g (y ),g (v_n+1)) = 0,

n→∞lim d(g (z ),g (w_n+1)) = 0.

(49)

Similarly, we can prove that

n→∞lim d(g (x^∗),g (u_n+1)) = 0,

n→∞limd(g (y^∗),g (vn+1)) = 0,

n→∞lim d(g (z^∗),g (wn+1)) = 0.

(50)

Thus by the triangle inequality, (49) and (50), we have

d(g (x ),g (x^∗))  d (g (x ),g (un+1)) + d (g (x^∗),g (un+1)) → 0, d(g (y ),g (y^∗))  d (g (y ),g (vn+1)) + d (g (y^∗),g (vn+1)) → 0, d(g (z ),g (z^∗))  d (g (z ),g (wn+1)) + d (g (z^∗),g (wn+1)) → 0

(51)

as n→ ∞, which imply that g (x ) = g (x^∗), g (y ) = g (y^∗) and g (z ) = g (z^∗).

Now, we prove that g(x ) = x , g (y ) = y and g (z ) = z .

Denote that g(x ) = θ1, g(y ) = θ2and g(z ) = θ3. Since g(x ) = F (x , y, z ), (g (y ) = F (y, z , x ) and g (z ) = F (z , x , y ), by the commutativity of F and g , we have

g(θ1) = g (g (θ1)) = g (F (x , y, z )) = F (g x ,g y,g z ) = F (θ1,θ2,θ3) (52)

g(θ2) = g (g (θ2)) = g (F (y, x , y )) = F (g y,g x ,g y ) = F (θ2,θ1,θ2) (53)

g(θ3) = g (g (θ3)) = g (F (z , y, x )) = F (g z ,g y,g x ) = F (θ3,θ2,θ1) (54) Thus,(θ1,θ2,θ3) is a tripled coincidence point of F and g .

Putting x^∗= θ1, y^∗= θ2and z^∗= θ3) in (53), it follows from (43) that θ1= g (x ) = g (x^∗) = g (θ1)

θ2= g (y ) = g (y^∗) = g (θ2) θ3= g (z ) = g (z^∗) = g (θ3)

(55)

and so, from (52) and (53) θ1= g (θ1) = F (θ1,θ2,θ3) θ2= g (θ2) = F (θ2,θ1,θ2) θ3= g (θ3) = F (θ3,θ2,θ1).

(56)

Therefore(θ1,θ2,θ3) is a tripled common fixed point of F and g .

Finally, to prove the uniqueness of the tripled common fixed point of F and g . Assume that(µ1,µ2,µ3) is another tripled common fixed point of F and g . Then by (43), we haveµ1= g (µ1) = g (θ1) = θ1,µ2= g (µ2) = g (θ2) = θ2and µ3= g (µ3) = g (θ3) = θ3. This complete the proof.

Corollary 2.2.

Suppose that all the hypothesis of corollary (2.1) hold and further for all(x , y, z ),(x^∗, y^∗, z^∗) ∈ X³then there exists (u, v, w ) ∈ X³that is comparable with(x , y, z ) and ,(x^∗, y^∗, z^∗). Then there a unique x ∈ X such that x = F (x , x , x ).

3. Application

Now we give some applications of the main results in Section 2.

Theorem 3.1.

Let F : X³→ X and g : X → X be two given mappings. Assume that there exists a function ψ : [0, +∞) → [0, +∞) satisfying the following conditions:

(a) ψ(0) = 0 and ψ(t )  0 for any t  0, (b) ψ is non decreasing and right continuous,

(c) for anyε  0 there exists δ(ε)  0 such that, for all x , y, z , u, v, w ∈ X with g (x )  g (u), g (y )  g (v ), and g(z )  g (w )

ε  ψ

1

3(d (g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w )))

‹

≺ ε + δ(ε)

⇒ ψ[d (F (x , y , z ), F (u, v, w ))] ≺ ε

Then F is a generalized g−Meir-Keeler type contraction.

Proof. for anyε  0, it follows from (a) ψ(ε)  0 and so there exists α  0 such that, for all u, v, w, u^∗, v^∗, w^∗∈ X with g(u)  g (u^∗) , g (v )  g (v^∗) and g (w )  g (w^∗),

ε  ψ

1

3(d (g (u),g (u^∗)) + d (g (v ),g (v^∗)) + d (g (w ),d (w^∗)))

‹

≺ ε + δ(ε)

⇒ ψ[d (F (u, v, w ), F (u^∗, v^∗, w^∗))] ≺ ε. (57)

From the right continuity ofψ, there exists δ  0 such that ψ(ε + δ) ≺ ψ(ε)α. For any x , y, z , u, v, w ∈ X such that g(x )  g (u), g (y )  g (v ), g (z )  g (w ) and

ε 1

3d (g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),g (w )) ≺ ε + δ (58)

sinceψ is non decreasing function, we get the following,

ψ(ε)  ψ

1

3(d (g (x ),g (u)) + d (g (y ),g (v )) + d (g (z ),d (w )))

‹

≺ ψ(ε + α)

≺ ψ(ε) + α. (59)

by (58), we have

ψ[d (F (x , y, z ), F (u, v, w ))] ≺ ψ(ε) and so

d(F (x , y, z ), F (u, v, w )) ≺ ε

Therefore, it follows that F is a generalized g−Meir-Keeler type contraction. This completes the proof.

Corollary 3.1.

Let F : X³→ X be given mapping. Assume that there exists a function ψ : [0, +∞) → [0, +∞) satisfying the following conditions:

(a) ψ(0) = 0 and ψ(t )  0 for any t  0, (b) ψ is non decreasing and right continuous,

(c) for anyε  0 there exists δ(ε)  0 such that, for all x , y, z , u, v, w ∈ X with x  u, y  v , and z  w ε  ψ1

3(d (x , u) + d (y, v ) + d (z , w ))‹

≺ ε + δ(ε)

⇒ ψ[d (F (x , y , z ), F (u, v, w ))] ≺ ε Then F is a generalized Meir-Keeler type contraction.

The following result is an immediate consequence of Theorems2.1and Theorem3.1.

Corollary 3.2.

Let F : X³→ X and g : X → X be two mappings such that F (X³) ⊆ g (x ), g is continuous and commutative with F . Also, suppose that

(a) F has the mixed strict g−monotone property;

(b) for anyε  0 there exists δ(ε)  0 such that, for all x , y, z , u, v, w ∈ X with g (x )  g (u), g (y )  g (v ), and g(z )  g (w ),

ε 

Z ¹₃(d (g (x ),g (u))+d (g (y ),g (v ))+d (g (z ),g (w ))) 0

ψ(t )d t ≺ ε + δ(ε)

⇒

Z d(F (x ,y,z ),F (u,v,w )) 0

ψ(t )d t ≺ ε (60)

whereψ is a locally integrable function from [0,+∞) into itself satisfying the following condition, Z s

0

ψ(t )d t  0, (61)

for all s 0

(c) there exist x₀, y₀, z₀∈ X such that g(x0) ≺ F (x0, y₀, z₀)

g(y0)  F (y0, x₀, y₀) g(z0) ≺ F (z0, y₀, x₀)

Then there exists(x , y, z ) ∈ X × X × X such that

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

that is, F and g have a tripled coincidence in X³. Moreover, if g(x0), g (y0) and g (z0) are comparable to each other, then F and g have a unique tripled common fixed point in X³.

Corollary 3.3.

Let F : X³→ X be a mapping satisfying the following conditions, (a) F has the mixed strict monotone property;

(b) for anyε  0 there exists δ(ε)  0 such that for all x , y, z , u, v, w ∈ X with x  u, y  v and z  w ,

ε 

Z ¹₃(d (x ,u)+d (y,v )+d (z ,w )) 0

ψ(t )d t ≺ ε + δ(ε)

⇒

Z d(F (x ,y,z ),F (u,v,w )) 0

ψ(t )d t ≺ ε (62)

whereψ is a locally integrable function from [0,+∞) into itself satisfying the following condition, Z s

0

ψ(t )d t  0, (63)

for all s 0

(c) there exist x₀, y0, z0∈ X such that x0≺ F (x0, y0, z0), y0 F (y0, x0, y0) and z0 F (z0, y0, x0).

Then there exists(x , y, z ) ∈ X × X × X such that F(x , y, z ) = x , F (y, x , y ) = y and F (z , y, x ) = z

that is F has a tripled coincidence in X³. Moreover, if x₀, y₀and z₀are comparable to each other, then F has a unique tripled fixed point in X³.

Corollary 3.4.

Let F : X³→ X and g : X → X be two mappings such that F (X³) ⊆ g (x ), g is continuous and commutative with F . Suppose that

(a) F has the mixed strict g−monotone property;

(b) for anyε  0 there exists δ(ε)  0 such that, for all x , y, z , u, v, w ∈ X with g (x )  g (u), g (y )  g (v ), and g(z )  g (w ),

Z d(F (x ,y,z ),F (u,v,w )) 0

ψ(t )d t  k

Z ¹₃(d (g (x ),g (u))+d (g (y ),g (v ))+d (g (z ),g (w ))) 0

ψ(t )d t (64)

where k∈ (0, 1) and ψ is a locally integrable function from [0, +∞) into itself satisfying the following condition, Z s

0

ψ(t )d t  0, (65)

for all s 0,

(c) there exist x₀, y₀, z₀∈ X such that g (x0) ≺ F (x0, y₀, z₀), g (y0)  F (y0, z₀, x₀) and g (z0)  F (z0, x₀, y₀).

Then there exists(x , y, z ) ∈ X³such that,

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

that is, F and g have a tripled coincidence in X³. Moreover, if g(x0), g (y0) and g (z0) are comparable to each other, then F and g have a unique tripled common fixed point in X³.

Corollary 3.5.

Let F : X³→ X be a mapping satisfying the following conditions, (a) F has the mixed strict monotone property;

(b) for anyε  0 there exists δ(ε)  0 such that for all x , y, z , u, v, w ∈ X with x  u, y  v , and z  w , Z d(F (x ,y,z ),F (u,v,w ))

0

ψ(t )d t  k

Z ¹₃(d (x ,u)+d (y,v )+d (z ,w )) 0

ψ(t )d t (66)

where k∈ (0, 1) and ψ is a locally integrable function from [0, +∞) into itself satisfying the following condition, Z s

0

ψ(t )d t  0 (67)

for all s 0,

(c) there exist x₀, y0, z0∈ X such that x0≺ F (x0, y0, z0), y0 F (y0, x0, y0) and z0 F (z0, y0, x0).

Then there exists(x , y, z ) ∈ X³such that

F(x , y, z ) = x , F (y, x , y ) = y and F (z , y, x ) = z

that is, F has a tripled coincidence in X³. Moreover if x₀, y₀and z₀are comparable to each other, then F has a unique tripled fixed point in X³.

Finally by using the above results, we show the existence of solutions for the following integral equation:

(x (t ), y (t ), z (t )) =

‚ZT 0

G(t , s )[f (s, x (s ) + λx (s ) − (f (s, y (s )) + λy (s ))]d s, Z T

0

G(t , s )[f (s, y (s ) + λy (s ) − (f (s, z (s )) + λz (s ))]d s, Z T

0

G(t , s )[f (s, z (s ) + λz (s ) − (f (s, x (s )) + λx (s ))]d s

Œ

(68)

where x , y , z∈ C (I , R ) where C (I , R ) is the set of continuous functions from I into R , T  0, f : I × R → R is contin- uous function and

G(t , s ) =

 _eλ(T +s −t )

e^{λT −1} i f 0 s ≺ t  T

e^{λ(s −t )}

e^{λT −1} i f 0 t ≺ s  T

Definition 3.1.

A lower solution for the integral type Eq. (68) is an element(α,β,γ) ∈ C¹(I ,R)
3

such that α⁰(t ) + λβ(t ) + λγ(t )  f (t ,α(t )) − f (t ,β(t )) − f (t ,γ(t )), α(0) ≺ α(T ),

β⁰(t ) + λγ(t ) + λα(t )  f (t ,β(t )) − f (t ,γ(t )) − f (t ,α(t )), β(0)  β(T ),

γ⁰(t ) + λα(t ) + λβ(t )  f (t ,γ(t )) − f (t ,α(t )) − f (t ,β(t )), γ(0)  γ(T ), (69) where C¹(I ,R) denotes the set of differentiable functions from I to R.

Next we prove the existence of solution for the integral Eq. (68).

Theorem 3.2.

LetΨ be the class of the functions ψ : [0,∞) → [0,∞) satisfying the following conditions:

(a) ψ is increasing,

(b) for any x 0, there exists k ∈ [0, 1) such that ψ(x ) ≺ (k /3)x .

In the integral Eq. (68) suppose that there existsλ  0 such that for all x , y ∈ R with y  x

0≺ f (t , y ) + λy − [f (t , x ) + λ(x )]  λψ(y − x ), (70)

whereψ ∈ Ψ. If a lower solution of the integral Eq. (68) exists then the solution of integral Eq. (68) exists.

Proof. Define a mapping F :(C (I ,R))³→ C (I , R ) by

F(x (t ), y (t ), z (t )) = Z T

0

G(t , s )[f (s, x (s ) + λx (s )

− (f (s , y (s )) + λy (s )) − (f (s , z (s )) + λz (s ))]d s , (71) Note that, if(x (t ), y (t ), z (t )) ∈ (C (I ,R))³is tripled fixed point of F , then(x (t ), y (t ), z (t )) is the solution of integral Eq. (68).

Now, we check the hypothesis in Corollary3.4as follows:

1. X³= (C (I ,R))³is a partially ordered set if we define the order relation in X³as follows;

(u(t ), v (t ), w (t ))  (x (t ), y (t ), z (t )) (72)

iff u(t ) ≺ x (t ), v (t )  y (t ) w (t ) ≺ z (t ), for all (u(t ), v (t ), w (t )),(x (t ), y (t ), z (t )) ∈ X³and t∈ I . 2. (X ,d ) is a complete metric space if we define a metric d as follows;

d(x (t ), y (t )) = sup

t∈I

{| x (t ) − y (t ) |: x (t ), y (t ) ∈ X }. (73)

3. The mapping F has the mixed strict monotone property. In fact by hypothesis, if x₂ x1, then we have

f(t , x2) + λx2 f (t , x1) + λx1 (74)

which implies that for any t∈ I ,

F(x2(t ), y (t ), z (t )) = ZT

0

G(t , s )[f (s, x2(s )) + λx2(s )

− (f (s , y (s )) + λy (s )) − (f (s , z (s )) + λz (s ))]d s and

F(x1(t ), y (t ), z (t )) = ZT

0

G(t , s )[f (s, x1(s )) + λx1(s )

− (f (s , y (s )) + λy (s )) − (f (s , z (s )) + λz (s ))]d s , that is,

F(x2(t ), y (t ), z (t ))  F (x1(t ), y (t ), z (t )). (75)

Similarly if y₁ y2, then we have

f(t , y2) + λy2 f (t , y1) + λy1 (76)

which implies that for any t∈ I ,

F(x (t ), y2(t ), z (t )) = Z T

0

G(t , s )[f (s, x (s )) + λx (s )

− (f (s , y2(s )) + λy2(s )) − (f (s, z (s )) + λz (s ))]d s

and

F(x (t ), y1(t ), z (t )) = Z T

0

G(t , s )[f (s, x (s )) + λx (s )

− (f (s , y1(s )) + λy1(s )) − (f (s, z (s )) + λz (s ))]d s.

that is

F(x (t ), y2(t ), z (t )) ≺ F (x (t ), y1(t ), z (t )) (77)

for any t∈ I .

Also if z₁≺ z2, then we have

f(t , z2) + λz2 f (t , z1) + λz1 (78)

F(x (t ), y (t ), z2(t )) = Z T

0

G(t , s )[f (s, x (s )) + λx (s )

− (f (s , y (s )) + λy (s )) − (f (s , z2(s )) + λz2(s ))]d s and

F(x (t ), y (t ), z1(t )) = Z T

0

G(t , s )[f (s, x (s )) + λx (s )

− (f (s , y (s )) + λy (s )) − (f (s , z1(s )) + λz1(s ))]d s that is

F(x (t ), y (t ), z2(t ))  F (x (t ), y (t ), z1(t )) (79)

In fact, let(x , y, z )  (u, v, w ) and t ∈ I then we have

d(F (x (t ), y (t ), z (t )), F (u(t ), v (t ), u(t )))

= sup{| F (x (t ), y (t ), z (t )) − F (u(t ), v (t ), u(t )) |: t ∈ I }

= sup

t∈I

{|

Z T 0

G(t , s )[(f (s, x (s )) + λx (s )) − (f (s, y (s )) + λy (s ))

− (f (s , z (s )) + λz (s ))]d s

− ZT

0

G(t , s )[f (s, u(s )) + λu(s ) − (f (s, v (s )) + λv (s ))

− (f (s , w (s )) + λw (s ))]d s |: t ∈ I }

 sup

t∈I

{|

Z T 0

G(t , s )(f (s, x (s )) + λx (s )) − (f (s, y (s )) + λy (s ))

− (f (s , z (s )) + λz (s )) − (f (s , u(s )) + λu(s ))

−(f (s , v (s )) + λv (s )) − (f (s , w (s )) + λw (s )) d s. (80)

Since the functionψ(x ) is non decreasing and (x , y, z )  (u, v, w ), we have ψ(x (s ) − u(s ))  ψ(d (x (s ), u(s )))

ψ(y (s ) − v (s ))  ψ(d (y (s ), v (s ))) ψ(z (s ) − w (s ))  ψ(d (z (s ), w (s )))

(81)

d(F (x (t ), y (t ), z (t )), F (u(t ), v (t ), u(t )))

 sup

t∈I

ZT 0

G(t , s ) | λψ(x (s ) − u(s )) + λψ(y (s ) − v (s )) + λψ(z (s ) − w (s )) | d s

 λ sup

t∈I

Z T 0

G(t , s ) | ψ(d (x (s ), u(s ))) + ψ(d (y (s ), v (s ))) + ψ(d (z (s ), w (s ))) | d s

= λ(ψ(d (x (s ), u(s ))) + ψ(d (y (s ), v (s ))) + ψ(d (z (s ), w (s ))))sup

t∈I

ZT 0

G(t , s )d s

= λ(ψ(d (x (s ), u(s ))) + ψ(d (y (s ), v (s ))) + ψ(d (z (s ), w (s ))))sup

t∈I

1 e^λT− 1

•1

λeλ(T +s −t )˜t 0

+

•1 λe^{λ(s −t )}

˜T t

= λ(ψ(d (x (s ), u(s ))) + ψ(d (y (s ), v (s ))) + ψ(d (z (s ), w (s ))))

≺k

3(d (x (s ), u(s )) + d (y (s ), v (s )) + d (z (s ), w (s )))

k

3sup{| x (t ) − u(t ) |: t ∈ I } + k

3sup{| y (t ) − v (t ) |: t ∈ I } +k

3sup{| z (t ) − w (t ) |: t ∈ I }

=k

3(d (x (t ), u(t )) + d (y (t ), v (t )) + d (z (t ), w (t ))) (82)

Then by the Lemma6, F is generalized Meir-Keeler type contraction.

Finally, let(α(t ),β(t ),γ(t )) ∈ (C¹(I ,R))³be a lower solution for the integral Eq. (68) then we show that

α ≺ F (α,β,γ), β  F (β,α,β), γ  F (γ,β,α) (83)

Indeed, we have

α⁰(t ) + λβ(t ) + λγ(t )  f (t ,α(t )) − f (t ,β(t )) − f (t ,γ(t )) for any t∈ I and so

α⁰(t ) + λα(t )  f (t ,α(t )) − f (t ,β(t )) − f (t ,γ(t )) + λα(t ) − λβ(t ) − λγ(t ) (84) for any t∈ I .

Multiplying by e^λtin (84), we get the following:

(α(t ) + λα(t ))e^λt
0

 [(f (t , α(t )) + λα(t ))

− (f (t , β (t )) − λβ (t )) − (f (t , γ(t )) − λγ(t ))]e^λt (85) for any t∈ I , which implies that

α(t )e^λt α(0) + Z t

0

[(f (s,α(s )) + λα(s ))

− (f (s , β (s )) − λβ (s )) − (f (s , γ(s )) − λγ(s ))]e^λsd s (86)

for any t∈ I , this implies that α(0)e^λt≺ α(T )e^λT

 α(0) + ZT

0

[(f (s,α(s )) + λα(s )) − (f (s,β(s )) − λβ(s ))

− (f (s , γ(s )) − λγ(s ))]e^λsd s (87)

and so α(0) ≺

Z T 0

e^λs

e^λT− 1[(f (s,α(s )) + λα(s ))

− (f (s , β (s )) − λβ (s )) − (f (s , γ(s )) − λγ(s ))]d s (88)