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- Open Access

# Some new results on fixed and best proximity points in preordered metric spaces

- Alireza Amini-Harandi
^{1, 2}, - Majid Fakhar
^{2, 3}, - Hamid Reza Hajisharifi
^{3}and - Nawab Hussain
^{4}Email author

**2013**:263

https://doi.org/10.1186/1687-1812-2013-263

© Amini-Harandi et al.; licensee Springer. 2013

**Received: **15 July 2013

**Accepted: **9 October 2013

**Published: **7 November 2013

## Abstract

In this paper, we first introduce two new classes of $(\omega ,\delta )$-contractions of the first and second kinds and establish some related new fixed point and best proximity point theorems in preordered metric spaces. Our theorems subsume the corresponding recent results of Samet (J. Optim. Theory Appl. (2013), doi:10.1007/s10957-013-0269-9) and extend and generalize many of the well-known results in the literature. An example is also provided to support our main results.

**MSC:** 47H10, 41A65.

## Keywords

*P*-property$(\omega ,\delta )$-contractionpreordered metric space

## 1 Introduction and preliminaries

Given a metric space $(X,d)$ and a self-mapping *T* on *X*, the theory on the existence of a solution to the equation of the form $Tx=x$ has gained impetus because of its applicability to solve many interesting problems that can be formulated as ordinary differential equations, matrix equations *etc.* For some recent fixed point results, see [1–6] and references therein. Let *A* and *B* be nonempty subsets of *X*, and let $T:A\to B$ be a non-self mapping. The equation $Tx=x$ is unlikely to have a solution, because of the fact that a solution of the preceding equation demands the nonemptiness of $A\cap B$. Eventually, it is quite natural to seek an approximate solution *x* that is optimal in the sense that the distance $d(x,Tx)$ is minimum. The well-known best approximation theorem, due to Fan [7], states that if *A* is a nonempty, compact, and convex subset of a normed linear space *X* and *T* is a continuous function from *A* to *X*, then there exists a point *x* in *A* such that $\parallel x-Tx\parallel =d(Tx,A)=inf\{\parallel Tx-u\parallel :u\in A\}$. Such a point x is called a best approximant point of *T* in *A*. Many generalizations and extensions of this theorem appeared in the literature (see [8–11] and references therein).

Best proximity problem for the pairs $(A,B)$ is to find an element $x\in A$ such that $d(x,Tx)=d(A,B)$, where $d(A,B)=inf\{d(x,y):x\in A,y\in B\}$. Since $d(A,B)$ is a lower bound for the function $x\to d(x,Tx)$ on *A*, then the solutions of the best proximity problem are the minimum points of the function $x\to d(x,Tx)$ on *A*. Every solution of the best proximity problem is said to be a best proximity point of *T* in *A*. Moreover, if $A=B$ then every best proximity point of *T* is a fixed point. According to this fact, many authors by motivation of well-known fixed point results obtained sufficient conditions to solving best proximity problems; for more details, see [12–27] and the references therein.

Existence of best proximity and fixed points in partially ordered metric spaces has been considered recently by many authors (see [6, 13, 20, 28]). Recently Samet [29] studied the existence of best proximity points for a class of non-self almost $(\phi ,\theta )$-contractive mappings. In this work we define two new classes of contractions called $(\omega ,\delta )$-contractions of the first and second kind and establish some related new fixed point results in the setting of preordered metric spaces, and then we derive some new best proximity point theorems for these new classes of non-self contractive mappings. The presented theorems extend and generalize many of the well-known fixed point and best proximity point results.

## 2 Fixed point theory

**Definition 2.1**Let $(X,d)$ be a metric space, and let ${R}_{+}=[0,\mathrm{\infty})$.

- (a)Denote by Ω the family of functions $\omega :{R}_{+}\to {R}_{+}$ such that $\omega (0)=0$, $\omega (t)<t$ for each $t>0$ and for each sequence $\{{x}_{n}\}$ in
*X*with,$d({x}_{n},{x}_{n+1})\le \omega (d({x}_{n-1},{x}_{n}))\phantom{\rule{1em}{0ex}}\mathrm{\forall}n\in \mathbb{N}\Rightarrow \{{x}_{n}\}\text{is a Cauchy sequence};$ - (b)
Denote by Δ the family of functions $\delta :{R}_{+}^{4}\to R$ such that

*δ*is continuous and if ${t}_{i}=0$ for some $i\in \{1,2,3,4\}$, then $\delta ({t}_{1},{t}_{2},{t}_{3},{t}_{4})=0$; - (c)
Denote by Φ the family of non-decreasing functions $\varphi :{R}_{+}\to {R}_{+}$ such that ${\mathrm{\Sigma}}_{n=1}^{\mathrm{\infty}}{\varphi}^{n}(t)<\mathrm{\infty}$ for each $t>0$;

- (d)Denote by Σ the family of functions $\sigma :{R}_{+}\to {R}_{+}$ such that $\sigma (t)=\alpha (t)t$ for each $t>0$ and $\alpha :{R}_{+}\to [0,1)$ satisfies$\underset{s\to t}{lim\hspace{0.17em}sup}\alpha (s)<1\phantom{\rule{1em}{0ex}}\text{for each}t0;$(1)
- (e)
Denote by Ψ the family of non-decreasing functions $\psi :{R}_{+}\to {R}_{+}$ such that ${lim}_{n\to \mathrm{\infty}}{\psi}^{n}(t)=0$ for each $t>0$;

- (f)
Denote by Λ the family of non-decreasing and upper semicontinuous from the right functions $\lambda :{R}_{+}\to {R}_{+}$ such that $\lambda (t)<t$ for each $t>0$;

- (g)
Let Θ be a collection of the following functions:

$\theta ({t}_{1},{t}_{2},{t}_{3},{t}_{4})=\tau min\{{t}_{1},{t}_{2},{t}_{3},{t}_{4}\}$, $\tau >0$;

$\theta ({t}_{1},{t}_{2},{t}_{3},{t}_{4})=\tau ln(1+{t}_{1}{t}_{2}{t}_{3}{t}_{4})$, $\tau >0$;

$\theta ({t}_{1},{t}_{2},{t}_{3},{t}_{4})=\tau {t}_{1}{t}_{2}{t}_{3}{t}_{4}$, $\tau >0$.

**Lemma 2.2**

*Let*$(X,d)$

*be a metric space*.

*Then the following statements hold*:

- (i)
$\mathrm{\Phi}\subseteq \mathrm{\Omega}$,

- (ii)
$\mathrm{\Sigma}\subseteq \mathrm{\Omega}$,

- (iii)
$\mathrm{\Psi}\subseteq \mathrm{\Omega}$,

- (iv)
$\mathrm{\Lambda}\subseteq \mathrm{\Psi}\subseteq \mathrm{\Omega}$,

- (v)
$\mathrm{\Theta}\subseteq \mathrm{\Delta}$.

*Proof*Let $\{{x}_{n}\}$ be a sequence in

*X*. To prove (i), assume that $d({x}_{n},{x}_{n+1})\le \varphi (d({x}_{n-1},{x}_{n}))$ for each $n\in \mathbb{N}$, where $\varphi \in \mathrm{\Phi}$. Since

*ϕ*is non-decreasing, then by induction we get

and so $\{{x}_{n}\}$ is a Cauchy sequence.

*α*satisfying (1). Then

for each $k\in \mathbb{N}$. From the above, we obtain ${lim}_{k\to \mathrm{\infty}}\alpha (d({x}_{{m}_{k}},{x}_{{n}_{k}}))=1$. Then from (1) we get ${lim}_{k\to \mathrm{\infty}}d({x}_{{m}_{k}},{x}_{{n}_{k}})=0$, a contradiction. Therefore, $\{{x}_{n}\}$ is a Cauchy sequence.

(iii) Notice first that $\psi (t)<t$ for each $t>0$. To see this, suppose that there exists ${t}_{0}>0$ with $\psi ({t}_{0})>{t}_{0}$, then since *ψ* is non-decreasing, we see that ${t}_{0}\le {\psi}^{n}({t}_{0})$ for all $n\in \mathbb{N}$ and it is a contradiction with ${lim}_{n\to \mathrm{\infty}}{\psi}^{n}(t)=0$ for each $t>0$. Note also that $\psi (0)=0$.

*ψ*is non-decreasing, then by induction we get

This implies that $({x}_{n})$ is Cauchy and the proof of (iii) is complete.

(iv) For each $\lambda \in \mathrm{\Lambda}$, we have ${lim}_{n\to \mathrm{\infty}}{\lambda}^{n}(t)=0$ for each $t>0$ (see Remark 2.2 in [30]). Then the conclusion follows from (iii).

(v) obviously holds. □

Let *X* be a nonempty set. A preorder ⪯ on *X* is a binary relation which is reflexive and transitive. Let $(X,\u2aaf)$ be a preordered set, and let $T:X\to X$ be a mapping. We say that *T* is non-decreasing if for each $x,y\in X$, $x\u2aafy\Rightarrow Tx\u2aafTy$.

**Definition 2.3**Let $(X,\u2aaf)$ be a preordered set and

*d*be a metric on

*X*. We say that $(X,\u2aaf,d)$ is regular if and only if the following condition holds:

**Definition 2.4**Let $(X,\u2aaf,d)$ be a preordered metric space, and let $\omega :{R}_{+}\to {R}_{+}$ and $\delta :{R}_{+}^{4}\to {R}_{+}$ be arbitrary mappings.

- (a)A mapping $T:X\to X$ is said to be $(\omega ,\delta ,\u2aaf)$-contraction of the first kind if for all $x,y\in X$ with $x\u2aafy$,$d(Tx,Ty)\le \omega (d(x,y))+\delta (d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx));$
- (b)A mapping $T:X\to X$ is said to be $(\omega ,\delta ,\u2aaf)$-contraction of the second kind if for all $x,y\in X$ with $x\u2aafy$,$d(Tx,Ty)\le \omega (M(x,y))+\delta (d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)),$
where $M(x,y)=max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\}$.

**Remark 2.5** If $\u2aaf=X\times X$, that is, $x\u2aafy$ for each $x,y\in X$, then $(\omega ,\delta ,\u2aaf)$-contractions of the first and second kind are called $(\omega ,\delta )$-contractions of the first and second kind in brief. The class of $(\omega ,\delta )$-contraction maps of the first and second kind include the mappings with condition (B) [3] and almost generalized contractions [6], respectively.

**Theorem 2.6**

*Let*$(X,\u2aaf,d)$

*be a complete preordered metric space*,

*and let*$T:X\to X$

*be a mapping*.

*Suppose that the following conditions hold*:

- (i)
*T**is continuous or*$(X,\u2aaf,d)$*is regular*, - (ii)
*T**is non*-*decreasing*, - (iii)
*there exists*${x}_{0}\in X$*such that*${x}_{0}\u2aafT{x}_{0}$, - (iv)
*T**is an*$(\omega ,\delta ,\u2aaf)$-*contraction mapping of the first kind*,*where*$\omega \in \mathrm{\Omega}$*and*$\delta \in \mathrm{\Delta}$.

*Then* *T* *has a fixed point*. *Moreover*, *the sequence* $\{{T}^{n}{x}_{0}\}$ *converges to the fixed point of* *T*.

*Proof*Let ${x}_{n}=T{x}_{n-1}$ for any $n\in \mathbb{N}$. Since ${x}_{0}\u2aafT{x}_{0}$ and

*T*is non-decreasing, then we have

*T*is an $(\omega ,\delta ,\u2aaf)$-contraction mapping of the first kind, we get

*T*. If

*T*is continuous, then from the equality ${x}_{n}=T{x}_{n-1}$, we get ${x}^{\ast}=T{x}^{\ast}$. Now assume that $(X,\u2aaf,d)$ is regular. Then, for each $n\in \mathbb{N}$, we have ${x}_{n}\u2aaf{x}^{\ast}$. On the contrary, assume that $d({x}^{\ast},T{x}^{\ast})>0$. For any $n\in \mathbb{N}$,

and so from (5) we get ${x}^{\ast}=T{x}^{\ast}$. □

**Corollary 2.7**

*Let*$(X,d)$

*be a complete metric space*,

*and let*$T:X\to X$

*be an*$(\omega ,\delta )$-

*contraction mapping of the first kind*,

*where*$\omega \in \mathrm{\Omega}$

*and*$\delta \in \mathrm{\Delta}$.

*Then*

- (i)
*T**has a unique fixed point*.*Moreover*,*for all*${x}_{0}\in X$,*the sequence*$\{{T}^{n}{x}_{0}\}$*converges to the fixed point of**T*,*that is*,*T**is the Picard operator*. - (ii)
*T**is continuous at*$Fix(T)=\{{x}^{\ast}\}$.

*Proof*(i) Let $\u2aaf=X\times X$. Then from Theorem 2.6 we deduce that

*T*has a fixed point. To prove the uniqueness, on the contrary, assume that $x,y\in X$ are distinct fixed points of

*T*. So,

a contradiction. By the uniqueness of a fixed point and from Theorem 2.6, we get that the sequence $\{{T}^{n}{x}_{0}\}$ converges to the fixed point of *T* for all ${x}_{0}\in X$.

*X*such that ${y}_{n}\to {x}^{\ast}$. Since

*T*is an $(\omega ,\delta )$-contraction mapping of the first kind, so for all $n\in \mathbb{N}$ we have

Thus $T{y}_{n}\to T{x}^{\ast}$, and so *T* is continuous at ${x}^{\ast}$. □

**Remark 2.8** Theorem 2.6 extends the main result of Babu *et al.* [3], Corollary 1 of Berinde *et al.* [4], Corollary 3.1 of Samet [29] and Theorem 2.1 of Agarwal *et al.* [30].

**Theorem 2.9**

*Let*$(X,\u2aaf,d)$

*be a complete preordered metric space*,

*and let*$T:X\to X$

*be a mapping*.

*Suppose that the following conditions hold*:

- (i)
*T**is continuous or*$(X,\u2aaf,d)$*is regular*, - (ii)
*T**is non*-*decreasing*, - (iii)
*there exists*${x}_{0}\in X$*such that*${x}_{0}\u2aafT{x}_{0}$, - (iv)
*T**is an*$(\omega ,\delta ,\u2aaf)$-*contraction mapping of the second kind*,*where*$\omega \in \mathrm{\Omega}$*and*$\delta \in \mathrm{\Delta}$.

*Then* *T* *has a fixed point*. *Moreover*, *the sequence* $\{{T}^{n}{x}_{0}\}$ *converges to the fixed point of* *T*.

*Proof*Let ${x}_{n}=T{x}_{n-1}$ for any $n\in \mathbb{N}$. If ${x}_{n-1}={x}_{n}$ for some $n\in \mathbb{N}$, then ${x}_{n-1}={x}_{n}=T{x}_{n-1}$, and so ${x}_{n-1}$ is a fixed point of

*T*, and we are finished. So, we may assume that $d({x}_{n-1},{x}_{n})>0$ for all $n\in \mathbb{N}$. Now, since ${x}_{0}\u2aafT{x}_{0}$ and

*T*is non-decreasing, so

*T*is an $(\omega ,\delta ,\u2aaf)$-contraction of the second kind, so for all $n\in \mathbb{N}$ we have

*T*. If

*T*is continuous, then from the equality ${x}_{n}=T{x}_{n-1}$, we get ${x}^{\ast}=T{x}^{\ast}$. Now, assume that $(X,\u2aaf,d)$ is regular. Then, for each $n\in \mathbb{N}$, we have ${x}_{n}\u2aaf{x}^{\ast}$. Now, on the contrary, assume that $d({x}^{\ast},T{x}^{\ast})>0$. So, for any $n\in \mathbb{N}$,

a contradiction. □

**Corollary 2.10** *Let* $(X,d)$ *be a complete metric space*, *and let* $T:X\to X$ *be an* $(\omega ,\delta )$-*contraction mapping of the second kind*, *where* $\omega \in \mathrm{\Omega}$ *and* $\delta \in \mathrm{\Delta}$. *Then* *T* *has a unique fixed point*. *Moreover*, *for all* ${x}_{0}\in X$, *the sequence* $\{{T}^{n}{x}_{0}\}$ *converges to the fixed point of* *T*, *that is*, *T* *is the Picard operator*.

*Proof*By Theorem 2.9 it is sufficient to prove the uniqueness of the fixed point. On the contrary assume that $x,y\in X$ are distinct fixed points of

*T*. Then

a contradiction. □

**Remark 2.11** Theorem 2.9 is a generalization of Theorem 2.2 and Theorem 2.3 of Agarwal *et al.* [30].

**Remark 2.12** When for all $t\in [0,\mathrm{\infty})$ we set $\omega (t)=\alpha t$ where $\alpha \in (0,1)$ and $\delta ({t}_{1},{t}_{2},{t}_{3},{t}_{4})=Lmin\{{t}_{1},{t}_{2},{t}_{3},{t}_{4}\}$ where $L\ge 0$, in Corollary 2.10, we obtain Theorem 2.4 of Berinde [5].

## 3 Best proximity point theory

*A*and

*B*be two nonempty subsets of a metric space $(X,d)$. We denote by ${A}_{0}$ and ${B}_{0}$ the following sets:

where $d(A,B)=inf\{d(x,y):x\in A,y\in B\}$.

**Definition 3.1**Let $(A,B)$ be a pair of nonempty subsets of the metric space $(X,d)$ with ${A}_{0}\ne \mathrm{\varnothing}$. Then the pair $(A,B)$ is said to have the P-property [31] if and only if

where ${x}_{1},{x}_{2}\in A$ and ${y}_{1},{y}_{2}\in B$.

The following lemma is crucial in proving our best proximity point results.

**Lemma 3.2**

*Let*$(A,B)$

*be a pair of nonempty closed subsets of a complete metric space*$(X,d)$

*such that*${B}_{0}\ne \mathrm{\varnothing}$

*and that*$(A,B)$

*satisfies the*

*P*-

*property*.

*Then there exists a mapping*$Q:{B}_{0}\to {A}_{0}$

*satisfying*

*Furthermore*, ${B}_{0}$ *is closed*.

*Proof*Let $x\in {B}_{0}$, then we show that there exists a unique $y\in {A}_{0}$ such that $d(x,y)=d(A,B)$. To prove the uniqueness, let us assume that there exists $z\in {A}_{0}$ such that $d(x,y)=d(A,B)=d(x,z)$. Since $(A,B)$ has the

*P*-property, we have $d(y,z)=d(x,x)=0$ and so $y=z$. Let $y=Qx$, then $d(x,Qx)=d(x,y)=d(A,B)$. Now, assume that $d(x,Qx)=d(A,B)=d(y,Qy)$, where $x,y\in {B}_{0}$. Then, by the

*P*-property of $(A,B)$, we get $d(x,y)=d(Qx,Qy)$. Therefore, there exists a mapping $Q:{B}_{0}\to {A}_{0}$ such that

*B*is closed). Since

*A*is a closed subset of a complete metric space, $d(Q{x}_{m},Q{x}_{n})=d({x}_{m},{x}_{n})$ for each $m,n\in \mathbb{N}$ and $\{{x}_{n}\}$ is a Cauchy sequence, we deduce that $Q{x}_{n}\to y\in A$. Since $d({x}_{n},Q{x}_{n})=d(A,B)$ for each $n\in \mathbb{N}$, we have

and so $x\in {B}_{0}$. Hence, ${B}_{0}$ is closed. □

**Remark 3.3** It is clear that the mapping *Q* in Lemma 3.2 is a bijection and for any $x\in {A}_{0}$, we have $d(x,{Q}^{-1}x)=d(Q({Q}^{-1}x),{Q}^{-1}x)=d(A,B)$.

**Definition 3.4**Let $(X,\u2aaf)$ be a preordered set. A non-self mapping $T:M\subseteq A\to B$ is said to be proximally non-decreasing if and only if

where ${x}_{1},{x}_{2}\in A$, ${y}_{1},{y}_{2}\in M$.

The following lemma follows from Lemma 14 in [32].

**Lemma 3.5** *Let* $(X,\u2aaf,d)$ *be a preordered metric space*, *and let* $T:A\to B$ *be a non*-*self mapping such that* $T{A}_{0}\subseteq {B}_{0}$. *Let* $(A,B)$ *and* *Q* *be as in the statement of Lemma* 3.2. *Suppose that* $T:{A}_{0}\to {B}_{0}$ *is proximally non*-*decreasing*. *Then the mapping* $S:{A}_{0}\to {A}_{0}$ *defined by* $Sx=QTx$ *for each* $x\in {A}_{0}$ *is non*-*decreasing*.

The following lemma follows from Lemma 15 in [32].

**Lemma 3.6** *Let* $(X,d,\u2aaf)$ *be a preordered metric space* $(A,B)$, *and* *Q* *be as in Lemma* 3.2 *and* $T:A\to B$ *be a non*-*self mapping such that* $T{A}_{0}\subseteq {B}_{0}$. *Suppose that there exist* ${x}_{0},{x}_{1}\in {A}_{0}$ *such that* $d({x}_{1},T{x}_{0})=d(A,B)$ *and* ${x}_{0}\u2aaf{x}_{1}$. *Let the mapping* $S:{A}_{0}\to {A}_{0}$ *be defined by* $Sx=QTx$ *for each* $x\in {A}_{0}$. *Then* ${x}_{0}\u2aafS{x}_{0}$.

Now, we are ready to establish our best proximity point theorems.

**Theorem 3.7**

*Let*$(A,B)$

*be a pair of nonempty closed subsets of a complete preordered metric space*$(X,\u2aaf,d)$

*such that*${A}_{0}\ne \mathrm{\varnothing}$.

*Let*$T:A\to B$

*be a non*-

*self mapping*.

*Suppose that the following conditions hold*:

- (i)
$T{A}_{0}\subseteq {B}_{0}$

*and*$(A,B)$*satisfy the**P*-*property*, - (ii)
*T**is continuous or*$({A}_{0},\u2aaf,d)$*is regular*, - (iii)
*T**is proximally non*-*decreasing*, - (iv)
*there exist*${x}_{0},{x}_{1}\in {A}_{0}$*such that*$d({x}_{1},T{x}_{0})=d(A,B),\phantom{\rule{1em}{0ex}}{x}_{0}\u2aaf{x}_{1},$ - (v)
*For all*$x,y\in A$*such that*$x\u2aafy$,*we have*$\begin{array}{rl}d(Tx,Ty)\le & \omega (d(x,y))+\delta (d(x,Tx)-d(A,B),d(y,Ty)-d(A,B),d(x,Ty)\\ -d(A,B),d(y,Tx)-d(A,B)),\end{array}$(18)

*where* $\omega \in \mathrm{\Omega}$, $\delta \in \mathrm{\Delta}$ *and* *δ* *is non*-*decreasing in each of its variables*.

*Then* *T* *has a best proximity point in* *A*.

*Proof*Since ${A}_{0}\ne \mathrm{\varnothing}$, so ${B}_{0}\ne \mathrm{\varnothing}$. By Lemma 3.2, ${B}_{0}$ is closed and there exists an isometry $Q:{B}_{0}\to {A}_{0}$ which satisfies (17). Let $S:{A}_{0}\to {A}_{0}$ be defined by $Sx=QTx$ for each $x\in {A}_{0}$. Let $x,y\in {A}_{0}$ and $x\u2aafy$, then from (18) we have

Thus *S* is an ordered $(\omega ,\delta ,\u2aaf)$-contraction mapping of the first kind. Now conditions (ii), (iii) and (iv) with Lemma 3.5 and Lemma 3.6 imply that *S* satisfies conditions (i), (ii) and (iii) of Theorem 2.6. Consequently, *S* has a fixed point ${x}^{\ast}\in {A}_{0}$ such that ${x}^{\ast}=S{x}^{\ast}=QT{x}^{\ast}$ and ${Q}^{-1}{x}^{\ast}=T{x}^{\ast}$. That is, $d({x}^{\ast},T{x}^{\ast})=d({x}^{\ast},{Q}^{-1}{x}^{\ast})=d(Q({Q}^{-1}{x}^{\ast}),{Q}^{-1}{x}^{\ast})=d(A,B)$. Thus ${x}^{\ast}\in {A}_{0}$ is the required best proximity point for *T*. □

**Corollary 3.8**

*Let*$(A,B)$

*be a pair of nonempty closed subsets of a complete metric space*$(X,d)$

*such that*${A}_{0}\ne \mathrm{\varnothing}$

*and*$(A,B)$

*satisfies the*

*P*-

*property*.

*Let*$T:A\to B$

*such that for all*$x,y\in A$,

*where* $\omega \in \mathrm{\Omega}$, $\delta \in \mathrm{\Delta}$ *and* *δ* *is non*-*decreasing in each of its variables*. *Moreover*, *assume that* $T{A}_{0}\subseteq {B}_{0}$. *Then* *T* *has a best proximity point in* *A*.

**Theorem 3.9**

*Let*$(A,B)$

*be a pair of nonempty closed subsets of a complete preordered metric space*$(X,\u2aaf,d)$

*such that*${A}_{0}\ne \mathrm{\varnothing}$.

*Let*$T:A\to B$

*be a non*-

*self mapping*.

*Suppose that the following conditions hold*:

- (i)
$T{A}_{0}\subseteq {B}_{0}$

*and*$(A,B)$*satisfy the**P*-*property*, - (ii)
*T**is continuous or*$({A}_{0},\u2aaf,d)$*is regular*, - (iii)
*T**is proximally non*-*decreasing*, - (iv)
*there exist*${x}_{0},{x}_{1}\in {A}_{0}$*such that*$d({x}_{1},T{x}_{0})=d(A,B),\phantom{\rule{1em}{0ex}}{x}_{0}\u2aaf{x}_{1},$ - (v)
*For all*$x,y\in A$*such that*$y\u2aafx$,*we have*$\begin{array}{rl}d(Tx,Ty)\le & \omega (max\{d(x,y),d(x,Tx)-d(A,B),d(y,Ty)-d(A,B),\\ \frac{d(x,Ty)+d(y,Tx)}{2}-d(A,B)\left\}\right)\\ +\delta (d(x,Tx)-d(A,B),d(y,Ty)-d(A,B),d(x,Ty)\\ -d(A,B),d(y,Tx)-d(A,B)),\end{array}$(21)

*where* $\omega \in \mathrm{\Omega}$ *is non*-*decreasing*, $\delta \in \mathrm{\Delta}$ *and* *δ* *is non*-*decreasing in each of its variables*.

*Then* *T* *has a best proximity point in* *A*.

*Proof*Since ${A}_{0}\ne \mathrm{\varnothing}$, so ${B}_{0}\ne \mathrm{\varnothing}$. By Lemma 3.2, ${B}_{0}$ is closed and there exists an isometry $Q:{B}_{0}\to {A}_{0}$ which satisfies (17). Let $S:{A}_{0}\to {A}_{0}$ be defined by $Sx=QTx$ for each $x\in {A}_{0}$. Let $x,y\in {A}_{0}$ and $y\u2aafx$, then from (21) we have

*ω*is non-decreasing and

*δ*is non-decreasing in each of its variables, in view of the proof of Theorem 3.7, we get

for each $x,y\in {A}_{0}$, where $M(x,y)=max\{d(x,y),d(x,Sx),d(y,Sy),\frac{d(x,Sy)+d(y,Sx)}{2}\}$. Thus *S* is an ordered $(\omega ,\delta ,\u2aaf)$-contraction mapping of the second kind. Now conditions (ii), (iii) and (iv) with Lemma 3.5 and Lemma 3.6 imply that *S* satisfies conditions (i), (ii) and (iii) of Theorem 2.9, so by Theorem 2.9 *S* has a fixed point ${x}^{\ast}\in {A}_{0}$ such that ${x}^{\ast}=S{x}^{\ast}=QT{x}^{\ast}$ and ${Q}^{-1}{x}^{\ast}=T{x}^{\ast}$. Thus $d({x}^{\ast},T{x}^{\ast})=d({x}^{\ast},{Q}^{-1}{x}^{\ast})=d(Q({Q}^{-1}{x}^{\ast}),{Q}^{-1}{x}^{\ast})=d(A,B)$, as required. □

**Corollary 3.10**

*Let*$(A,B)$

*be a pair of nonempty closed subsets of a complete metric space*$(X,d)$

*such that*${A}_{0}\ne \mathrm{\varnothing}$

*and*$(A,B)$

*satisfies the*

*P*-

*property*.

*Let*$T:A\to B$

*be such that for all*$x,y\in A$,

*where* $\omega \in \mathrm{\Omega}$ *is non*-*decreasing*, $\delta \in \mathrm{\Delta}$ *and* *δ* *is non*-*decreasing in each of its variables*. *Moreover*, *assume that* $T{A}_{0}\subseteq {B}_{0}$. *Then* *T* *has a best proximity point in* *A*.

**Remark 3.11**

From Lemma 2.2 and Corollary 3.8, we deduce the following result due to Samet [29].

**Theorem 3.12**

*Let*$(A,B)$

*be a pair of nonempty closed subsets of a complete metric space*$(X,d)$

*such that*${A}_{0}\ne \mathrm{\varnothing}$, $(A,B)$

*satisfies the*

*P*-

*property*.

*Let*$T:A\to B$

*such that for all*$x,y\in A$,

*where* $\phi \in \mathrm{\Phi}$, $\theta \in \mathrm{\Theta}$. *Moreover*, *assume that* $T{A}_{0}\subseteq {B}_{0}$. *Then* *T* *has a best proximity point in* *A*.

Now we provide the following example to show that Corollary 3.8 is an essential extension of the above mentioned theorem of Samet.

**Example 3.13** Consider the complete metric space $X=[0,2]\times [0,\mathrm{\infty})$ with the Euclidean metric. Let $A=\{(0,x):0\le x\}$ and $B=\{(2,y):0\le y\}$. Then $d(A,B)=2$, ${A}_{0}=A$, ${B}_{0}=B$ and $(A,B)$ has the *P*-property.

*T*has a best proximity point (indeed, $P=(0,0)$ is a best proximity point of

*T*). But we cannot invoke the above mentioned theorem of Samet to show that the mapping

*T*has a best proximity point in

*A*because

*T*is not an almost $(\phi ,\theta )$ contraction. On the contrary, assume that there exist $\phi \in \mathrm{\Phi}$ and $\theta \in \mathrm{\Theta}$ such that for all $x,y\in A$,

*f*on $(0,\mathrm{\infty})$ is an increasing positive function. So, we have

*φ*is non-decreasing, then from the above, we get

and so (24) holds for each $n\in \mathbb{N}$. Since ${x}_{n}>\frac{c}{n}$ for each $n\in \mathbb{N}$ and $\sum \frac{c}{n}=\mathrm{\infty}$, then we get $\sum {x}_{n}=\mathrm{\infty}$, a contradiction.

## Declarations

### Acknowledgements

The first author was partially supported by a grant from IPM (No. 92470412). The second author was partially supported by a grant from IPM (No. 92550414). The first and the second author were also partially supported by the Center of Excellence for Mathematics, University of Shahrekord, Iran. This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the fourth author acknowledges with thanks DSR, KAU for financial support.

## Authors’ Affiliations

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