A Fixed Point Theory for Quasi-Contractive Mappings in Ordered Sets - Advanced Research on Computer Education, Simulation and Modeling - page 113

2 Main Results

We first recall some necessary definitions, which are needed in our main results.

Definition 1. Let (,)

X ≤ gX → X

.We say that g be a partially ordered set and

x , y ∈ X , is monotone nondecreasing if, for all

x ≤ y ⇒ g ( x ) ≤ g ( y ).

g is monotone nonincreasing if, for all x , y ∈ X ,

x ≤ y ⇒ g ( x ) ≥ g ( y ).

g is monotone if it is monotone nonincreasing or monotone nondecreasing.

Definition 2. Let (,)

X ≤ fX → X

be a partially ordered set. is said to be

r ∈ (0,1)

quasi-contractive with constant r if there exists a constant such that

d ( f ( x ), f ( y )) ≤ rQ ( x , y ), ∀ x ≥ y or x ≤ y ,

where

Q ( x , y ) = max{ d ( x , y ), d ( x , f ( x )), d ( y , f ( y )), d ( x , f ( y )), d ( y , f ( x ))}

.

f is called to be contractive if it is quasi-contractive and satisfies that

Q ( x , y ) = d ( x , y ), ∀ x ≥ y or x ≤ y .

Theorem 1. Let (,)

X ≤

be a partially ordered set and suppose that there exists a
metric d in X such that (,)

Xd is a complete metric space. Let gX → X be a

1

k ∈ ( 0 )

. Suppose that either g is monotone quasi-contractive map with constant

2

0 xX

∈ xgx ≤

()

xgx ≥

()

, then g has a continuous. If there exists with or

0 0

x

* y are fixed points of g with

x

*

≥ y

*

fixed point. Furthermore, if and or

x

*

≤ y

*

x

*

= y

*

. , then

gx () = x Proof. If , then the proof is complete.

0 0

gx () ≠ x

Suppose that . The proof is similar to that of Theorem 2.1 in [1].

0 0

0 xX

∈ x ≤ gx

()

For the completeness, we conclude it. Let with or

0

xgx ≥

()

g

n

( x ) ≠ g

n + 1

( x ) . Without loss of generality, suppose that , i.e.,

0 0 0

d ( g

n

( x ), g

n + 1

( x )) ≠ 0 n = 0 2 ,...

0 0