Our starting point is the following definition.

**Definition 2.1**.

*Let* (

*X*, ≤)

*be a partially ordered set and T*:

*X* →

*X a mapping. We say that T is nondecreasing if for x, y* ∈

*X*This definition coincides with the notion of a nondecreasing function in the case where *X* = ℝ and ≤ represents the usual total order in ℝ.

**Remark 2.2**. *The contractive condition given by Meir-Keeler*:

*For every ε* > 0,

*there exists δ*(

*ε*) > 0

*such that for x,y* ∈

*X**implies that the operator T*: *X* → *X is strictly nonexpansive (this means that for x, y* ∈ *X, x* ≠ *y, d*(*Tx, Ty*) < *d*(*x, y*)).

*Therefore, any operator T satisfying the Meir-Keeler condition is continuous*.

In what follows, we present the following theorem which is a version of Theorem 1.1 in the context of ordered metric spaces when the operator is nondecreasing.

**Theorem 2.3**.

*Let* (

*X*, ≤)

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

*X, d*)

*is a complete metric space*.

*Let T*:

*X* →

*X be a continuous and nondecreasing mapping such that for every ε* > 0

*there exists δ*(

*ε*) > 0

*satisfying**If there exists x*_{0} ∈ *X with x*_{0} ≤ *Tx*_{0}*, then T has a fixed point*.

**Remark 2.4**. *Condition (??) does not imply that T is continuous as it is proven with the following example*.

*Let X* = [0, 1] *be the unit interval with the usual metric and the order given by R* = {(*x, x*): *x* ∈ *X*}. *Consider the operator T*: *X* → *X given by Tx* = 0 *if x* ≠ 1 *and Tx* = 1 *if x* = 1.

*Obviously, T is not continuous and, as the elements in X are only comparable to themselves, T satisfies (??)*.

*This shows that the continuity of T in Theorem 2.3 is not redundant (compare with Remark 2.2 and Theorem 1.1)*.

**Remark 2.5**. *Condition* (1) *does not imply that T is strictly nonexpansive but d*(*Tx, Ty*) < *d*(*x, y*) *is true if x* < *y as it can be seen using ε* = *d*(*x, y*) *in* (1) *when x* < *y*.

*Proof of Theorem 2.3*. If *Tx*_{0} = *x*_{0}, then the proof is finished.

Suppose that

*x*
_{0} <

*Tx*
_{0} and

*T* is a nondecreasing mapping, we obtain by induction that

Put *x*
_{n+1}= *T*
^{
n
}
*x*
_{0}. Obviously, (*x*
_{
n
}) is a nondecreasing sequence.

For better readability, we divide the proof into several steps.

**Step 1**: lim_{n → ∞}*d*(*x*_{
n
}, *x*_{n + 1}) = 0.

In fact, if the sequence (*x*
_{
n
}) is not strictly nondecreasing, then we can find *n*
_{0} ∈ ℕ such that
and the result follows. If (*x*
_{
n
}) is strictly nondecreasing, then by Remark 2.5, (*d*(*x*
_{
n
}, *x*
_{n+1})) is strictly decreasing, and hence, it is convergent. Put *r* = lim_{n → ∞}
*d*(*x*
_{
n
}, *x*
_{n+1}) (notice that *r* = inf{*d*(*x*
_{
n
}, *x*
_{n+1}): *n* ∈ ℕ}).

Now, we will prove that *r* = 0.

Suppose that *r* > 0.

Applying condition (??) to

*r* > 0 we can find

*δ*(

*r*) > 0 such that

Since *r* = lim_{n→∞}
*d*(*x*
_{
n
}, *x*
_{n+1}), there exists *n*
_{0} ∈ ℕ such that
and, as
, we have
.

This is a contradiction because *r* = inf{*d*(*x*
_{
n
}, *x*
_{n+1}): *n* ∈ ℕ}.

Therefore,

*r* = 0. This means that

**Step 2**: (*x*_{
n
}) is a Cauchy sequence.

In fact, fix *ε* > 0 and arbitrary.

By condition (??), there exists

*δ*(

*ε*) > 0 (which can be choosen satisfying

*δ*(

*ε*) ≤

*ε*) such that

On the other hand, by (??), there exists

*n*
_{0} ∈ ℕ such that

Fix

*n* >

*n*
_{0} and in order to prove that {

*x*
_{
n
}} is a Cauchy sequence it is sufficient to show that

We will use mathematical induction.

For

*p* = 1 by (??) and the fact that

*δ*(

*ε*) ≤

*ε* we obtain

Now, we assume that (??) holds for some fixed *p*.

Then, using (??) and the inductive hypothesis, we get

Now, we consider two cases.

**Case 1:***d*(*x*_{n-1}*, x*_{n+p}) ≥ *ε*.

In this case, taking into account (??),

and, since

*x*
_{n-1}<

*x*
_{n+p}, by (??)

This proves that (??) is satisfied by *p* + 1.

**Case 2:***d*(*x*_{n-1}, *x*_{n+p}) < *ε*.

As

*d*(

*x*
_{n-1}
*, x*
_{n+p}) > 0 (because {

*x*
_{
n
}} is a nondecreasing sequence and

*d*(

*x*
_{
n
}
*, x*
_{n+1}) > 0 for any

*n* = 0,1, 2,...), applying condition (??) for

*ε*
_{0} =

*d*(

*x*
_{n-1}
*, x*
_{n+p}) we can get

(notice that *x*
_{n-1}< *x*
_{n+p}) and this proves that (??) is satisfied by *p* + 1.

Therefore, {*x*
_{
n
}} is a Cauchy sequence.

Since

*X* is a complete metric space, there exists

*z* ∈

*X* such that lim

_{n→∞}
*x*
_{
n
}=

*z*. Finally, the continuity of

*T* implies that

and, therefore, *z* is a fixed point of *T*.

This finishes the proof.

In what follows we prove that Theorem 2.3 is still valid for

*T* not necessarily continuous, assuming the following hypothesis:

**Theorem 2.6**. *If in Theorem 2.3 we replace the continuity of T by condition (??) the result is true*.

*Proof*. Following the proof of Theorem 2.3, we only have to check that *Tz* = *z*.

As (*x*
_{
n
}) is a nondecreasing sequence in *X* with *x*
_{
n
}→ *z*, by condition (??), we can find a subsequence (*x*
_{n(k)}) such that *x*
_{n(k)}≤ *z* for all *k* ∈ ℕ.

If there exists

*k*
_{0} ∈ ℕ such that

, then the nondecreasing character of (

*x*
_{
n
}) gives us that

Particularly,
, and
is a fixed point of *T*.

Suppose that for any *k* ∈ ℕ, *x*
_{n(k)}< *z*.

Applying condition (??) to

*ε* =

*d*(

*x*
_{n(k)}
*, z*) for

*k* fixed and arbitrary, we have

As *x*
_{n(k)}→ *z*, the last inequality implies that *x*
_{n(k)+1}→ *Tz*.

As (*x*
_{n(k)+1}) is a subsequence of (*x*
_{
n
}) and *x*
_{
n
}→ *z* we have *x*
_{n(k)+1}→ *z*.

Now, the uniqueness of the limit in complete metric spaces gives us *Tz* = *z*.

This finishes the proof.

Now, we present an example where it can be appreciated that assumptions in Theorems 2.3 and 2.6 do not guarantee uniqueness of the fixed point.

Let

*X* = {(1,0), (0,1)} ⊂ ℝ

^{2} and consider the usual order

Then, (*X*, ≤) is a partially ordered set whose different elements are not comparable. Besides, (*X*, *d*
_{2}) is a complete metric space considering *d*
_{2} the Euclidean distance. The identity map *T*(*x*, *y*) = (*x*, *y*) is trivially continuous and nondecreasing and condition (??) of Theorem 2 is satisfied since elements in *X* are only comparable to themselves. Moreover, (1,0) ≤ *T*(1, 0) = (1, 0) and the operator *T* has two fixed points.

In what follows, we present a sufficient condition for the uniqueness of the fixed point in Theorems 2.3 and 2.6. This condition appears in [

15] and says:

**Theorem 2.7**. *Adding condition (??) to the hypotheses of Theorem 2.3 (or Theorem 2.6) we obtain the uniqueness of the fixed point of T*.

*Proof*. Suppose that there exist *z*, *y* ∈ *X* which are fixed points of *T* and *z* ≠ *y*.

We consider two cases.

**Case 1:** Suppose that *z* and *y* are comparable.

Without loss of generality, we suppose *z* < *y*.

Putting

*ε* =

*d*(

*z*,

*y*) and applying condition (??) of Theorem 1, we get

which is a contradiction.

**Case 2:** Suppose that *z* and *y* are not comparable.

By condition (??), there exists *x* ∈ *X* comparable to *z* and *y*.

Suppose *z* < *x* (the same argument serves for *x* < *z*).

Monotonicity of *T* implies that *T*
^{
n
}
*z* = *z* ≤ *T*
^{
n
}
*x* for *n* = 1, 2, 3,...

We consider two possibilities:

(a) Suppose that there exists *n*
_{0} ∈ ℕ such that
. Since *z* is a fixed point of *T*, *T*
^{
n
}
*x* = *z* for *n* ≥ *n*
_{0}, and, consequently, *T*
^{
n
}
*x* → *z*.

(b) Suppose that

*T*
^{
n
}
*z* =

*z* <

*T*
^{
n
}
*x* for any

*n* = 1, 2, 3,.... Applying condition (??) of Theorem 1 for

*ε* =

*d*(

*T*
^{
n
}
*z*,

*T*
^{
n
}
*x*) (where

*n* ∈ ℕ is fixed but arbitrary), we have

Thus, {

*d*(

*z*,

*T*
^{
n
}
*x*)} is a decreasing sequence of positive real numbers and, consequently, there exists

*r* ≥ 0 such that

Suppose *r* > 0.

Applying condition (??) of Theorem 2.3 for

*ε* =

*r* we find

*δ*(

*r*) > 0 such that

As lim

_{n→∞ }
*d*(

*z, T*
^{
n
}
*x*) =

*r* = inf {

*d*(

*z, T*
^{
n
}
*x*):

*n* ∈ ℕ}, there exists

*n*
_{0} ∈ ℕ such that

and, since

, (??) gives us

which contradicts to *r* = inf{*d*(*z*, *T*
^{
n
}
*x*): *n* ∈ ℕ}.

Therefore, *z* = *y*.

This finishes the proof.