Open Access

# Common fixed point theorems for (ψ, φ)-weak nonlinear contraction in partially ordered sets

Fixed Point Theory and Applications20122012:62

https://doi.org/10.1186/1687-1812-2012-62

Accepted: 18 April 2012

Published: 18 April 2012

## Abstract

The purpose of this article is to present some fixed point theorems for (ψ, φ)-weak contractive mappings in a complete metric space endowed with a partial order. As an application of the main result, we give an existence theorem for the solution of a periodic boundary value problem.

2000 Mathematics Subject Classification: 47H10.

## Keywords

fixed pointweakly contractive mappingpartially ordered set

## 1. Introduction and preliminaries

It is well known that the Banach Contraction Principle has been generalized in various directions. Alber and Guerre-Delabrere [1] introduced the concept of weak contractions in Hilbert spaces and proved the corresponding fixed point result. Rhoades [2] showed that the result of Alber et al. is also valid in complete metric spaces. Weakly contractive mappings have been used in a number of subsequent articles (let us mention articles [310]) to establish various fixed point and common fixed point theorems. In particular, recent results of Zhang and Song [9] and Doric [10] are among the most general ones. Also, some applications were obtained, in particular when dealing with differential and matrix equations.

Existence of fixed point in partially ordered sets has been considered recently in [1125]. Tarski's theorem is used in [26] to show the existence of solutions for fuzzy equations and in [27] to prove existence theorems for fuzzy differential equations. In [24, 25, 27] some applications to matrix equations and to ordinary differential equations are presented. In [2832], it is proved some fixed point theorems for a mixed monotone mapping in a metric space endowed with partial order and the authors apply their results to problems of existence and uniqueness of solutions for some boundary value problems [26, 32]. We begin by stating the result of Rhoades [2] after the following definition.

A mapping f : XX, where (X, d) is a metric space, is said to be weakly contractive if
$d\left(fx,fy\right)\le d\left(x,y\right)-\phi \left(d\left(x,y\right)\right),$
(1.1)
where x, y X and ϕ : [0, ∞) → [0, ∞) is a continuous and nondecreasing function such that ϕ(t) = 0 if and only if t = 0. If one takes φ(t) = (1 - k)t, where 0 ≤ k < 1, then (1.1) reduces to the contractivity condition
$d\left(fx,fy\right)\le kd\left(x,y\right).$

Theorem 1.1. ([2]) Let (X, d) be a complete metric space, and f : XX be a weakly contractive mapping. Then f has a unique fixed point.

Weak inequalities of the above type have been used to establish fixed point results in a number of subsequent articles. For example, Zhang and Song [9] used generalized φ-weak contraction which is defined for two mappings and gave conditions for existence of a common fixed point.

Theorem 1.2. ([9]) Let (X, d) be a complete metric space, and f, g : XX be two mappings such that for all x, y X
$d\left(fx,gy\right)\le M\left(x,y\right)-\phi \left(M\left(x,y\right)\right),$
where φ : [0, ∞) → [0, ∞) is lower semi-continuous functions with φ(t) > 0 for t (0, ∞), φ(0) = 0, and
$M\left(x,y\right)=\text{max}\left\{d\left(x,y\right),d\left(x,fx\right),d\left(y,gy\right),\frac{1}{2}\left(d\left(x,gy\right)+d\left(y,fx\right)\right)\right\}.$
(1.2)

Then there exists a unique fixed point u X such that u = fu = gu.

Recently, Doric [10] extended the result of Zhang and Song using a pair of functions ψ and φ. He proved the following theorem.

Theorem 1.3. ([10]) Let (X, d) be a complete metric space, and f, g : XX be two mappings such that
$\psi \left(d\left(fx,gy\right)\right)\le \psi \left(M\left(x,y\right)\right)-\phi \left(M\left(x,y\right)\right),$

for all x,y X, where

(i) ψ : [0, ∞) → [0, ∞) is a continuous and nondecreasing function such that ψ(t) = 0 if and only if t = 0.

(ii) φ : [0, ∞) → [0, ∞) is a lower semi-continuous function such that φ(t) = 0 if and only if t = 0.

(iii) M is defined by (1.2).

Then there exists a unique fixed point u X such that u = fu = gu.

When fixed point problems in partially ordered metric spaces are concerned, first results were obtained by Ran and Reurings [25], and then by Nieto and Lopez [14]. The following two versions of the fixed point theorem were proved in these articles.

Theorem 1.4. ([14, 25]) Let (X, be a partially ordered set and let d be a metric on X such that (X, d) is a complete metric space. Let f : XX be a nondecreasing map w.r.t. . Suppose that the following conditions hold:

(i) There exists k (0,1) such that d(fx, fy) ≤ kd (x, y), for all x, y X with y x;

(ii) there exists x0 X such that x0 fx0.

Then in each of the following two cases the mappings f has at least one fixed point.

(τ1) f is continuous, or

(τ2) if a nondecreasing sequence {x n } converges to x, then x n x for all n.

Results on weakly contractive mappings in such spaces were obtained by Harjani and Sadarangani in [7]. We state one of their results.

Theorem 1.5. ([7]) Let (X, ) be a partially ordered set and let d be a metric on X such that (X, d) is a complete metric space. Let f : XX be a nondecreasing map w.r.t. such that for every two comparable elements x, y X,
$d\left(fx,fy\right)\le d\left(x,y\right)-\phi \left(d\left(x,y\right)\right),$

(i) φ : [0,∞) → [0,∞) is similar to the control function φ of Theorem 1.3.

(ii) there exists x0 X such that x0 fx0.

Then in each of the following two cases the mappings f has at least one fixed point.

(τ1) f is continuous, or

(τ2) if a nondecreasing sequence {x n } converges to x, then x n x for all n.

## 2. Main results

Let (X, ) be a partially ordered set. A pair (f, g) is said to be weakly increasing if fx gfx and gx fgx for all x X. There are examples (see [4]) when neither of such mappings f, g is nondecreasing w.r.t. . In particular, the pair (f, i X )(i x the identy function) is weakly increasing if and only if x fx for each x X.

Theorem 2.1. Let (X, ) be a partially ordered set and let d be a metric on X such that (X, d) is a complete metric space, and let f, g : XX be two weakly increasing mappings such that for every two comparable elements x, y X,
$\psi \left(d\left(fx,gy\right)\right)\le \psi \left(M\left(x,y\right)\right)-\phi \left(\psi \left(M\left(x,y\right)\right)\right)+\theta \left(N\left(x,y\right)\right),$
(2.1)

where

(i) ψ : [0, ∞) → [0, ∞) is a continuous and nondecreasing function such that ψ(t) = 0 if and only if t = 0 and also .

(ii) φ : [0, ∞) → [0, ∞) is a lower semi-continuous function such that φ(t) = 0 if and only if t = 0 and also for any sequence {t n } with limn→∞t n = 0, there exists k (0,1) and n0 , such that φ(t n ) ≥ kt n for each nn0.

(iii) θ : [0, ∞) → [0, ∞) is a continuous function such that θ(t) = 0 if and only if t = 0.

(iv) M is defined by (1.2) and N(x, y) = min{d(y, fx), d(x, gy)}.

Then in each of the following two cases the mappings f and g have at least one common fixed point.

(τ1) f or g is continuous, or

(τ2) if a nondecreasing sequence {x n } converges to x, then x n x for all n.

Proof. Step 1. Using that the pair of functions (f, g) is weakly increasing, we can construct inductively, starting with arbitrary x0 X, a sequence {x n } such that x n xn+1. Namely, denoted:
$\begin{array}{l}{x}_{1}=f{x}_{0}⊑gf{x}_{0}=g{x}_{1},\phantom{\rule{2em}{0ex}}\\ {x}_{2}=g{x}_{1}⊑fg{x}_{1}=f{x}_{2},\phantom{\rule{2em}{0ex}}\\ {x}_{3}=f{x}_{2}⊑gf{x}_{2}=g{x}_{3},\phantom{\rule{2em}{0ex}}\\ \dots \phantom{\rule{2em}{0ex}}\end{array}$

and in general, x2n+1= fx2nand x2n+2= gx2n+1for all n .

Suppose first that ${x}_{{n}_{0}}={x}_{{n}_{0}+1}$ for some n0. Then, the sequence {x n } is constant for nn0.

Indeed, let n0 = 2k. Then x2k= x2k+1and we obtain from (2.1) that
$\begin{array}{ll}\hfill \psi \left(d\left({x}_{2k+1},{x}_{2k+2}\right)\right)& =\psi \left(d\left(f{x}_{2k},g{x}_{2k+1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(M\left({x}_{2k},{x}_{2k+1}\right)\right)-\phi \left(\psi \left(M\left({x}_{2k},{x}_{2k+1}\right)\right)\right)+\theta \left(N\left({x}_{2k},{x}_{2k+1}\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$
(2.2)
where
$\begin{array}{c}M\left({x}_{2k},{x}_{2k+1}\right)=\mathrm{max}\left\{d\left({x}_{2k},{x}_{2k+1}\right),d\left({x}_{2k},f{x}_{2k}\right),d\left({x}_{2k+1},g{x}_{2k+1}\right),\\ \frac{1}{2}\left(d\left({x}_{2k},g{x}_{2k+1}\right)+d\left({x}_{2k+1},f{x}_{2k}\right)\right)\right\}\\ =\mathrm{max}\left\{0,0,d\left({x}_{2k+1},{x}_{2k+2}\right),\frac{1}{2}\left(d\left({x}_{2k},{x}_{2k+2}\right)+0\right)\right\}\\ =\mathrm{max}\left\{d\left({x}_{2k+1},{x}_{2k+2}\right),\frac{1}{2}d\left({x}_{2k},{x}_{2k+2}\right)\right\}\\ =d\left({x}_{2k+1},{x}_{2k+2}\right),\end{array}$
and
$N\left({x}_{2k},{x}_{2k+1}\right)=\text{min}\left\{d\left(f{x}_{2k},{x}_{2k+1}\right),d\left(g{x}_{2k+1},{x}_{2k}\right)\right\}=0.$
Now from (2.2), we get that
$\psi \left(d\left({x}_{2k+1},{x}_{2k+2}\right)\right)\le \psi \left(d\left({x}_{2k+1},{x}_{2k+2}\right)\right)-\phi \left(\psi \left(d\left({x}_{2k+1},{x}_{2k+2}\right)\right)\right),$

and so ψ(d(x2k+1, x2k+2)) ≤ 0 and x2k+1= x2k+2. Similarly, if n0 = 2k + 1 one easily obtains that x2k+2= x2k+3and the sequence {x n } is constant (starting from some n0) and ${x}_{{n}_{0}}$ is common fixed point of f and g.

Suppose now that d(x n , xn+1) > 0 for each n. We shall show that for each n = 0, 1,...,
$d\left({x}_{2n+1},{x}_{n+2}\right)\le M\left({x}_{2n},{x}_{2n+1}\right)=d\left({x}_{2n},{x}_{2n+1}\right),$
(2.3)
and
$d\left({x}_{2n+3},{x}_{n+2}\right)\le M\left({x}_{2n+2},{x}_{2n+1}\right)=d\left({x}_{2n+2},{x}_{2n+1}\right).$
(2.4)
It is clear that N(x2n, x2n+1) = 0 and N(x2n+2, x2n+1) = 0, for all n . Using condition (2.1), since x2nand x2n+1are comparable, we obtain that
$\begin{array}{ll}\hfill \psi \left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)& =\psi \left(d\left(f{x}_{2n},g{x}_{2n+1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)-\phi \left(\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)\right)+\theta \left(N\left({x}_{2n},{x}_{2n+1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$
(2.5)
and since the control function ψ is nondecreasing, it follows that
$d\left({x}_{2n+1},{x}_{2n+2}\right)\le M\left({x}_{2n},{x}_{2n+1}\right).$
(2.6)
Hence
$\begin{array}{ll}\hfill M\left({x}_{2n},{x}_{2n+1}\right)& =\text{max}\left\{d\left({x}_{2n},{x}_{2n+1}\right),d\left({x}_{2n},f{x}_{2n}\right),d\left({x}_{2n+1},g{x}_{2n+1}\right)\right\,\phantom{\rule{2em}{0ex}}\\ \frac{1}{2}\left(d\left({x}_{2n},g{x}_{2n+1}\right)+d\left({x}_{2n+1},f{x}_{2n}\right)\right)}\phantom{\rule{2em}{0ex}}\\ =\text{max}\left\{d\left({x}_{2n},{x}_{2n+1}\right),d\left({x}_{2n},{x}_{2n+1}\right),d\left({x}_{2n+1},{x}_{2n+2}\right),\frac{1}{2}d\left({x}_{2n},{{x}_{2n}}_{+2}\right)\right\}\phantom{\rule{2em}{0ex}}\\ \le \text{max}\left\{d\left({x}_{2n},{x}_{2n+1}\right),d\left({x}_{2n+1},{x}_{2n+2}\right),\frac{1}{2}\left(d\left({x}_{2n},{x}_{2n+1}\right)+d\left({x}_{2n+1},{x}_{2n+2}\right)\right)\right\}\phantom{\rule{2em}{0ex}}\\ \le \text{max}\left\{d\left({x}_{2n},{x}_{2n+1}\right),d\left({x}_{2n+1},{x}_{2n+2}\right)\right\}.\phantom{\rule{2em}{0ex}}\end{array}$
If d(x2n+1, x2n+2) ≥ d(x2n, x2n+1), then it follows from the last inequality and (2.6) that M(x2n, x2n+1) = d(x2n+1, x2n+2) and condition (2.5) implies that
$\begin{array}{ll}\hfill \psi \left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)& =\psi \left(d\left(f{x}_{2n+},g{x}_{2n+1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)-\phi \left(\psi \left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$
which is only possible when d(x2n+1, x2n+2) = 0, it is a contradiction. Hence, d(x2n+1, x2n+2) ≤ d(x2n, x2n+1) and M(x2n, x2n+1) ≤ d(x2n, x2n+1). By definition, M(x2n, x2n+1) ≥ d(x2n, x2n+1), (2.3) is proved for d(x2n+1, x2n+2). In a Similar way, one can obtain that
$d\left({x}_{2n+3},{x}_{2n+2}\right)\le M\left({x}_{2n+2},{x}_{2n+1}\right)=d\left({x}_{2n+2},{x}_{2n+1}\right).$

So, (2.3) and (2.4) holds for each n .

It follows that the sequence {d(x n , xn+1)} is nondecreasing. Let limn→∞d(x n , xn+1) = d*, for some d* ≥ 0 then
$\underset{n\to \infty }{\text{lim}}M\left({x}_{2n},{x}_{2n+1}\right)={d}^{*},$
and
$\underset{n\to \infty }{\text{lim}}M\left({x}_{2n+2},{x}_{2n+1}\right)={d}^{*}.$
Suppose that d* > 0. We have
$\psi \left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)\le \psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)-\phi \left(\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)\right).$
Passing to the (upper) limit when n → ∞, it follows that
$\psi \left({d}^{*}\right)\le \psi \left({d}^{*}\right)-\underset{n\to \infty }{\text{lim inf}}\phi \left(\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)\right)\le \psi \left({d}^{*}\right)-\phi \left(\psi \left({d}^{*}\right)\right),$

i.e. ψ(d*) ≤ 0. Using the properties of control functions, we get that d* = 0 which is a contradiction. We conclude that limn→∞d(x n , xn+1) = 0.

Step 2. Now we show that {x n } is a Cauchy sequence in X. Since
$\underset{n\to \infty }{\text{lim}}\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)=0,\underset{n\to \infty }{\text{lim}}\psi \left(M\left({x}_{2n+2},{x}_{2n+1}\right)\right)=0$
then by the property of φ there exists k (0,1) and n0 such that
$\begin{array}{ll}\hfill \phi \left(\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)\right)& \ge k\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right),\phantom{\rule{2em}{0ex}}\\ \hfill \phi \left(\psi \left(M\left({x}_{2n+2},{x}_{2n+1}\right)\right)\right)& \ge k\psi \left(M\left({x}_{2n+2},{x}_{2n+1}\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$
for all nn0. For any natural number nn0, if n is even, we have
$\begin{array}{ll}\hfill \psi \left(d\left({x}_{2n+1},{x}_{2n+2}\right)\right)& =\psi \left(d\left(f{x}_{2n},g{x}_{2n+1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)-\phi \left(\psi \left(M\left({x}_{2n},{x}_{2n+1}\right)\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \left(1-k\right)\psi \left(d\left({x}_{2n},{x}_{2n+1}\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$
and similarly
$\begin{array}{ll}\hfill \psi \left(d\left({x}_{2n},{x}_{2n+1}\right)\right)& =\psi \left(d\left(f{x}_{2n},g{x}_{2n-1}\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(M\left({x}_{2n},{x}_{2n-1}\right)\right)-\phi \left(\psi \left(M\left({x}_{2n},{x}_{2n-1}\right)\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \left(1-k\right)\psi \left(d\left({x}_{2n},{x}_{2n-1}\right)\right).\phantom{\rule{2em}{0ex}}\end{array}$
Hence, for all nn0, we have
$\begin{array}{l}\psi \left(d\left({x}_{n},{x}_{n+1}\right)\right)\le \left(1-k\right)\psi \left(d\left({x}_{n},{x}_{n-1}\right)\right).\phantom{\rule{2em}{0ex}}\\ \hfill \sum _{1}^{\infty }\psi \left(d\left({x}_{n},{x}_{n+1}\right)\right)& \le \sum _{1}^{{n}_{0}}\psi \left(d\left({x}_{n},{x}_{n+1}\right)\right)+\sum _{1}^{\infty }{\left(1-k\right)}^{n}\psi \left(d\left({x}_{{n}_{0}},{x}_{{n}_{0}+1}\right)\right)<\infty .\phantom{\rule{2em}{0ex}}\end{array}$
Since

then ${\sum }_{1}^{\infty }d\left({x}_{n},{x}_{n+1}\right)<\infty$. This shows that {x n } is a Cauchy sequence. Since (X, d) is a complete metric space, then there exists a z X such that limn→∞x n = z.

Step 3. We have to prove that z is a common fixed point of f and g. We shall distinguish the case (τ1) and (τ2) of the theorem.

(τ1) Suppose that the mapping g is continuous. Since x2n+1z, we obtain that x2n+2= gx2n+1gz. On the other hand, x2n+2z (as a subsequence of {x n }). It follows that gz = z. To prove fz = z, using z z, we can put x = y = z in (2.1) and obtain that
$\begin{array}{ll}\hfill \psi \left(d\left(fz,gz\right)\right)& \le \psi \left(M\left(z,z\right)\right)-\phi \left(\psi \left(M\left(z,z\right)\right)\right)+\theta \left(N\left(z,z\right)\right),\mathsf{\text{where}}\phantom{\rule{2em}{0ex}}\\ M\left(z,z\right)=\text{max}\left\{d\left(z,z\right),d\left(z,fz\right),d\left(z,gz\right),\frac{1}{2}\left(d\left(z,gz\right)+d\left(z,fz\right)\right)\right\}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\text{max}\left\{0,d\left(z,fz\right),0,\frac{1}{2}d\left(z,fz\right)\right\}=d\left(z,fz\right),\phantom{\rule{2em}{0ex}}\end{array}$
and
$N\left(z,z\right)=\text{min}\left\{d\left(z,fz\right),d\left(z,gz\right)\right\}=0.$

Hence, ψ(d(fz, z)) ≤ ψ(d(z, fz)) - φ (ψ(d(z, fz))) it follows that z = fz.

The proof is similar if f is continuous.

(τ2) Suppose that the condition (τ2) of the theorem holds. The sequence {x n } is nondecreasing w.r.t. and it follows that x n z. Take x = x2nand y = z, which are comparable in (2.1) to obtain that
$\psi \left(d\left(f{x}_{2n},{g}_{z}\right)\right)\le \psi \left(M\left({x}_{2n},z\right)\right)-\phi \left(\psi \left(M\left({x}_{2n},z\right)\right)\right)+\theta \left(N\left({x}_{2n},z\right)\right),$
(2.7)
where
$\begin{array}{ll}\hfill M\left({x}_{2n},z\right)& =\text{max}\left\{d\left({x}_{2n},z\right),d\left({x}_{2n},f{x}_{2n}\right),d\left(z,gz\right),\frac{1}{2}\left(d\left({x}_{2n},gz\right)+d\left(z,f{x}_{2n}\right)\right)\right\}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\to \text{max}\left\{0,0,d\left(z,gz\right),\frac{1}{2}d\left(z,gz\right)\right\}=d\left(z,gz\right),\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left({x}_{2n},z\right)=\text{min}\left\{d\left({x}_{2n},gz\right),d\left(z,f{x}_{2n}\right)\right\}\to 0.\phantom{\rule{2em}{0ex}}\end{array}$
Now, passing to the limit when n → ∞ in (2.7), we get
$\psi \left(d\left(z,gz\right)\right)\le \psi \left(d\left(z,gz\right)\right)-\phi \left(\psi \left(d\left(z,gz\right)\right)\right),$

wherefrom it follows that z = gz.

To prove fz = z, using z z, we can put x = y = z in (2.1) and obtain that ψ(d(fz, gz)) ≤ ψ(M(z, z)) - φ (ψ(M(z, z))) + θ(N(z, z)), where
$\begin{array}{ll}\hfill M\left(z,z\right)& =\text{max}\left\{d\left(z,z\right),d\left(z,fz\right),d\left(z,gz\right),\frac{1}{2}\left(d\left(z,gz\right)+d\left(z,fz\right)\right)\right\}\phantom{\rule{2em}{0ex}}\\ =\text{max}\left\{0,d\left(z,fz\right),0,\frac{1}{2}d\left(z,fz\right)\right\}=d\left(z,fz\right),\phantom{\rule{2em}{0ex}}\end{array}$
and
$N\left(z,z\right)=\text{min}\left\{d\left(z,fz\right),d\left(z,gz\right)\right\}=0.$

Hence, ψ(d(fz, z)) ≤ ψ(d(z, fz)) - φ (ψ(d(z, fz))) it follows that z = fz.

Theorem 2.2. Let (X, ) be a partially ordered set and let d be a metric on X such that (X, d) is a complete metric space, and f : XX be a nondecreasing map such that x0 fx0 for some x0 X, and for every two comparable elements x, y X,
$\psi \left(d\left(fx,fy\right)\right)\le \psi \left(M\left(x,y\right)\right)-\phi \left(\psi \left(M\left(x,y\right)\right)\right),$
(2.8)

where

(i) ψ : [0, ∞) → [0, ∞) is a continuous and nondecreasing function such that ψ(t) = 0 if and only if t = 0 and also $\text{lim}\underset{s\to {0}^{+}}{\text{sup}}\frac{s}{\psi \left(s\right)}<\infty$.

(ii) φ : [0, ∞) → [0, ∞) is a lower semi-continuous function such that φ(t) = 0 if and only if t = 0 and also for any sequence {t n } with limn→∞t n = 0, there exists k (0,1) and n0 , such that φ(t n ) ≥ kt n for each nn0.

(iii)
$M\left(x,y\right)=\text{max}\left\{d\left(x,y\right),d\left(x,fx\right),d\left(y,fy\right),\frac{1}{2}\left(d\left(x,fy\right)+d\left(y,fx\right)\right)\right\}.$
(2.9)

Then in each of the following two cases the mappings f has fixed point.

(τ1) f is continuous, or

(τ2) if a nondecreasing sequence {x n } converges to x, then x n x for all n.

Theorem 2.3. Let all the conditions of Theorem 2.2 be fulfilled and let the following condition hold:

A : For arbitrary two points x, y X, there exists z X, which is comparable with both x and y and z fz. Then the fixed point of f is unique.

Proof. Let z and y be two fixed points of f. Consider the following two cases.

Case 1. z and y are comparable. Then we can apply condition (2.8) to obtain that
$\psi \left(d\left(y,z\right)\right)=\psi \left(d\left(fy,fz\right)\right)\le \psi \left(M\left(y,z\right)\right)-\phi \left(\psi \left(M\left(y,z\right)\right)\right).$
Where
$M\left(z,y\right)=\text{max}\left\{d\left(z,y\right),d\left(z,fz\right),d\left(y,fy\right),\frac{1}{2}\left(d\left(y,fz\right)+d\left(z,fy\right)\right)\right\}=d\left(z,y\right),$
and hence
$\begin{array}{ll}\hfill \psi \left(d\left(y,z\right)\right)& =\psi \left(d\left(fy,fz\right)\right)\le \psi \left(M\left(y,z\right)\right)-\phi \left(\psi \left(M\left(y,z\right)\right)\right)\phantom{\rule{2em}{0ex}}\\ =\psi \left(d\left(z,y\right)\right)-\phi \left(\psi \left(d\left(z,y\right)\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$

which implies that z = y.

Case 2. Suppose that z and y are not comparable. Choose an element x X comparable with both of them and x fx. Then also z = f n z and is comparable with f n x, for each n, since f is nondecreasing. Applying (2.8) one obtains that
$\begin{array}{ll}\hfill \psi \left(d\left(z,{f}^{n}x\right)\right)& =\psi \left(d\left(f{f}^{n-1}z,f{f}^{n-1}x\right)\right)\phantom{\rule{2em}{0ex}}\\ \le \psi \left(M\left({f}^{n-1}z,{f}^{n-1}x\right)\right)-\phi \left(\psi \left(M\left({f}^{n-1}z,{f}^{n-1}x\right)\right)\right),\phantom{\rule{2em}{0ex}}\end{array}$
where
$\begin{array}{c}M\left({f}^{n-1}z,{f}^{n-1}x\right)=\mathrm{max}\left\{d\left({f}^{n-1}z,{f}^{n-1}x\right),d\left({f}^{n-1}z,{f}^{n}z\right),d\left({f}^{n-1}x,{f}^{n}x\right),\\ \frac{1}{2}\left(d\left({f}^{n-1}z,{f}^{n}x\right)+d\left({f}^{n}z,{f}^{n-1}x\right)\right)\right\}\\ =\mathrm{max}\left\{d\left(z,{f}^{n-1}x\right),0,d\left({f}^{n-1}x,{f}^{n}x\right),\frac{1}{2}\left(d\left(z,{f}^{n}x\right)+d\left(z,{f}^{n-1}x\right)\right)\right\}\\ =\mathrm{max}\left\{d\left(z,{f}^{n-1}x\right),d\left(z,{f}^{n}x\right)\right)\right\},\end{array}$

for n sufficiently large, because d(fn-1x, f n x) → 0 when n → ∞.

Similarly as in the proof of Theorem 2.1, it can be shown that d(z, f n x) ≤ M(z, fn-1x) ≤ d(z, fn-1x). It follows that the sequence {d(z, f n x)} is non decreasing and it has a limit l ≥ 0. Assuming that l > 0 and passing to the limit in the relation
$\psi \left(d\left(z,{f}^{n}x\right)\right)\le \psi \left(M\left(z,{f}^{n-1}x\right)\right)-\phi \left(\psi \left(M\left(z,{f}^{n-1}x\right)\right)\right),$

one obtain that l = 0, a contradiction. In the same way it can be deduced that d(y, f n x) → 0 as n → ∞. Now, passing the limit in d(z, y) ≤ d(z, f n x) + d(f n x, y), it follows that z = y and the uniqueness of the fixed point is proved.

## 3. Application to ordinary differential equations

We prove the existence of solution for the following first-order periodic problem
$\begin{array}{ll}\hfill {u}^{\prime }\left(t\right)& =f\left(t,u\left(t\right)\right),t\in I=\left[0,T\right]\phantom{\rule{2em}{0ex}}\\ \hfill u\left(0\right)& =u\left(T\right),\phantom{\rule{2em}{0ex}}\end{array}$
(3.1)
where T > 0 and f : I × is continuous function. Previously, we consider the space C(I)(I = [0,T]) of continuous functions defined on I. Obviously, this space with metric given by
$d\left(x,y\right)=\text{sup}\left\{\left|x\left(t\right)-y\left(t\right)\right|:t\in I\right\}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}x,y\in C\left(I\right)$
is a complete metric space. C(I) can also be equipped with partial order given by
$x,y\in C\left(I\right),\phantom{\rule{2.77695pt}{0ex}}x\le y⇔x\left(t\right)\le y\left(t\right)\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{all}}\phantom{\rule{2.77695pt}{0ex}}t\in I.$

Now, we give the following definition.

Definition 3.1. A lower solution for (3.1) is a function α C1(I) such that
$\begin{array}{ll}\hfill {\alpha }^{\prime }\left(t\right)& \le f\left(t,\alpha \left(t\right)\right),\phantom{\rule{2.77695pt}{0ex}}t\in I=\left[0,T\right]\phantom{\rule{2em}{0ex}}\\ \hfill \alpha \left(0\right)& \le \alpha \left(T\right).\phantom{\rule{2em}{0ex}}\end{array}$
Theorem 3.2. Consider problem (3.1) with f : I × continuous and suppose that there exists λ > 0 such that for x, y with yx
$0\le f\left(t,y\right)+\lambda y-\left[f\left(t,x\right)+\lambda x\right]\le \frac{\lambda }{2}\text{ln}\left(y-x+1\right).$

Then the existence of a lower solution for (3.1) provides the existence of a solution of (3.1).

Proof. Problem (3.1) is equivalent to the integral equation
$u\left(t\right)=\underset{0}{\overset{T}{\int }}G\left(s,t\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds$
where
$G\left(t,s\right)=\left\{\begin{array}{cc}\frac{{e}^{\lambda \left(T+s-t\right)}}{{e}^{\lambda T}-1},\hfill & 0\le s
Define F : C(I) → C(I) by
$\left(Fu\right)\left(t\right)=\underset{0}{\overset{T}{\int }}G\left(s,t\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds.$

Note that if u C(I) is a fixed point of F then u C1(I) is a solution of (3.1). Now, we will use Theorem 2.2.

The mapping F is increasing since for uv
$f\left(t,u\right)+\lambda u\ge f\left(t,v\right)+\lambda v$
using that G(s, t) > 0 for (s, t) I × I, we give
$\begin{array}{ll}\hfill \left(Fu\right)\left(t\right)& =\underset{0}{\overset{T}{\int }}G\left(s,t\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds\phantom{\rule{2em}{0ex}}\\ \ge \underset{0}{\overset{T}{\int }}G\left(s,t\right)\left[f\left(s,v\left(s\right)\right)+\lambda v\left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds=\left(Fv\right)\left(t\right)\phantom{\rule{2em}{0ex}}\end{array}$
for t I. Besides for uv
$\begin{array}{l}\mathrm{ln}\left(d\left(F\left(u\right),F\left(v\right)\right)+1\right)=\mathrm{ln}\left(\underset{t\in I}{\mathrm{sup}}|\left(Fu\right)\left(t\right)-\left(Fv\right)\left(t\right)|+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}=\mathrm{ln}\left(\left[\underset{t\in I}{\mathrm{sup}}\underset{0}{\overset{T}{\int }}G\left(s,t\right)\left[f\left(s,u\left(s\right)\right)+\lambda u\left(s\right)\right]-\left[f\left(s,v\left(s\right)\right)+\lambda v\left(s\right)\right]ds\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le \mathrm{ln}\left(\left[\underset{t\in I}{\mathrm{sup}}\underset{0}{\overset{T}{\int }}G\left(s,t\right).\frac{\lambda }{2}\mathrm{ln}\left(u\left(s\right)-v\left(s\right)+1\right)ds\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le \mathrm{ln}\left(\left[\frac{\lambda }{2}.\mathrm{ln}\left(d\left(u,v\right)+1\right).\underset{t\in I}{\mathrm{sup}}\underset{0}{\overset{T}{\int }}G\left(s,t\right)ds\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}=\mathrm{ln}\left(\left[\frac{\lambda }{2}.\mathrm{ln}\left(d\left(u,v\right)+1\right).\underset{t\in I}{\mathrm{sup}}\frac{1}{{e}^{\lambda T}-1}{\left(\frac{1}{\lambda }{e}^{\lambda \left(T+s-t\right)}\right]}_{0}^{t}+\frac{1}{\lambda }{e}^{\lambda \left(t-s\right)}{\right]}_{t}^{T}\right)\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}=\mathrm{ln}\left(\left[\frac{\lambda }{2}.\mathrm{ln}\left(d\left(u,v\right)+1\right).\underset{t\in I}{\mathrm{sup}}\frac{1}{\lambda \left({e}^{\lambda T}-1\right)}{e}^{\lambda T}-1\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}=\mathrm{ln}\left(\left[\frac{1}{2}\mathrm{ln}\left(d\left(u,v\right)+1\right)\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\le \mathrm{ln}\left(\left[\frac{1}{2}\mathrm{ln}\left(M\left(u,v\right)+1\right)\right]+1\right)\\ \phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\phantom{\rule{0.1em}{0ex}}=\mathrm{ln}\left(M\left(u,v\right)+1\right)-\left(\mathrm{ln}\left(M\left(u,v\right)+1\right)-\mathrm{ln}\left(\left[\frac{1}{2}\mathrm{ln}\left(M\left(u,v\right)+1\right)\right]+1\right)\right).\end{array}$

Putting ψ(x) = ln(x + 1) and $\phi \left(x\right)=x-\text{ln}\left(\frac{1}{2}x+1\right)$. Obviously ψ : [0, ∞) → [0, ∞) is continuous, increasing $\left({\psi }^{\prime }\left(x\right)=\frac{1}{x+1}>0\right)$, positive in (0, ∞), ψ(0) = 0 and $\text{lim}\underset{x\to {0}^{+}}{\text{sup}}\frac{x}{\psi \left(x\right)}=1<\infty$. Also, φ : [0,∞) → [0,∞) is continuous, positive in (0, ∞) and ψ(0) = 0, now let {t n } n be sequence that t n → 0, since $\underset{n\to \infty }{\text{lim}}\frac{\phi \left({t}_{n}\right)}{{t}_{n}}={\phi }^{\prime }\left(0\right)=\frac{1}{2}$, then for $\epsilon =\frac{1}{4}$, there exists n0 such that $\left|\frac{\phi \left({t}_{n}\right)}{{t}_{n}}-\frac{1}{2}\right|<\frac{1}{4}$, for all nn0, hence, $\phi \left({t}_{n}\right)\ge \frac{1}{4}{t}_{n}$ for all nn0.

Therefore the controls function ψ and φ satisfying the conditions of Theorem 2.2. Finally, if α(t) be a lower solution for (3.1) then we will show that α.

Indeed
${\alpha }^{\prime }\left(t\right)+\lambda \alpha \left(t\right)\le f\left(t,\alpha \left(t\right)\right)+\lambda \alpha \left(t\right),\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}t\in I.$
Multiplying by eλtwe get
${\left(\alpha \left(t\right){e}^{\lambda t}\right)}^{\prime }\le \left[f\left(t,\alpha \left(t\right)\right)+\lambda \alpha \left(t\right)\right]{e}^{\lambda t},\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{for}}\phantom{\rule{2.77695pt}{0ex}}t\in I$
and this gives us
$\alpha \left(t\right){e}^{\lambda t}\le \alpha \left(0\right)+\underset{0}{\overset{t}{\int }}{e}^{\lambda s}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds$
(3.2)
which implies that
$\alpha \left(0\right){e}^{\lambda T}\le \alpha \left(T\right){e}^{\lambda T}\le \alpha \left(0\right)+\underset{0}{\overset{T}{\int }}{e}^{\lambda s}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds$
and so
$\alpha \left(0\right)\le \underset{0}{\overset{T}{\int }}\frac{{e}^{\lambda s}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds.$
From this inequality and (3.2), we obtain that
$\alpha \left(t\right){e}^{\lambda t}\le \underset{0}{\overset{t}{\int }}\frac{{e}^{\lambda \left(T+s\right)}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds+\underset{t}{\overset{T}{\int }}\frac{{e}^{\lambda s}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds$
and consequently,
$\alpha \left(t\right)\le \underset{0}{\overset{t}{\int }}\frac{{e}^{\lambda \left(T+s-t\right)}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds+\underset{t}{\overset{T}{\int }}\frac{{e}^{\lambda \left(s-1\right)}}{{e}^{\lambda T}-1}\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds.$
Hence
$\alpha \left(t\right)\le \underset{0}{\overset{T}{\int }}G\left(s,t\right)\left[f\left(s,\alpha \left(s\right)\right)+\lambda \alpha \left(s\right)\right]\phantom{\rule{2.77695pt}{0ex}}ds=\left(F\alpha \right)\left(t\right)$

for t I. Finally, it follows from Theorems 2.2 that F has a fixed point.

## Declarations

### Acknowledgements

The authors thank the referees for their valuable comments and suggestions. The third author of this study was partially supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant No. 2011-0005197).

## Authors’ Affiliations

(1)
Department of Mathematics, Semnan University, Semnan, Iran
(2)
Department of Mathematics, Semnan University, Semnan, Iran
(3)
Department of Mathematics, Kangnam University, Yongin, Gyeonggi, Korea

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