We start this section with our main result.

**Theorem 8** *Let* $(X,d)$ *be a complete metric space*. *Suppose that* $(A,B)$ *is a pair of nonempty closed subsets of* *X* *and* ${A}_{0}$ *is nonempty*. *Suppose also that the pair* $(A,B)$ *has the* *P*-*property*. *If a non*-*self*-*mapping* $T:A\to B$ *is a generalized Geraghty*-*contraction satisfying* $T({A}_{0})\subseteq {B}_{0}$, *then there exists a unique best proximity point*, *that is*, *there exists* ${x}^{\ast}$ *in* *A* *such that* $d({x}^{\ast},T{x}^{\ast})=d(A,B)$.

*Proof* Let us fix an element

${x}_{0}$ in

${A}_{0}$. Since

$T{x}_{0}\in T({A}_{0})\subseteq {B}_{0}$, we can find

${x}_{1}\in {A}_{0}$ such that

$d({x}_{1},T{x}_{0})=d(A,B)$. Further, as

$T{x}_{1}\in T({A}_{0})\subseteq {B}_{0}$, there is an element

${x}_{2}$ in

${A}_{0}$ such that

$d({x}_{2},T{x}_{1})=d(A,B)$. Recursively, we obtain a sequence

$\{{x}_{n}\}$ in

${A}_{0}$ with the following property:

$d({x}_{n+1},T{x}_{n})=d(A,B)\phantom{\rule{1em}{0ex}}\text{for any}n\in \mathbb{N}.$

(2.1)

Due to the fact that the pair

$(A,B)$ has the

*P*-property, we derive that

$d({x}_{n},{x}_{n+1})=d(T{x}_{n-1},T{x}_{n})\phantom{\rule{1em}{0ex}}\text{for any}n\in \mathbb{N}.$

(2.2)

From (2.1), we get

$d({x}_{n-1},T{x}_{n-1})\le d({x}_{n-1},{x}_{n})+d({x}_{n},T{x}_{n-1})=d({x}_{n-1},{x}_{n})+d(A,B).$

On the other hand, by (2.1) and (2.2) we obtain that

$d({x}_{n},T{x}_{n})\le d({x}_{n},T{x}_{n-1})+d(T{x}_{n-1},T{x}_{n})=d({x}_{n},{x}_{n+1})+d(A,B).$

Consequently, we have

$\begin{array}{rl}M({x}_{n-1},{x}_{n})& =max\{d({x}_{n-1},{x}_{n}),d({x}_{n-1},T{x}_{n-1}),d({x}_{n},T{x}_{n})\}\\ \le max\{d({x}_{n-1},{x}_{n}),d({x}_{n},{x}_{n+1})\}+d(A,B).\end{array}$

(2.3)

If there exists

${n}_{0}\in \mathbb{N}$ such that

$d({x}_{{n}_{0}},{x}_{{n}_{0}+1})=0$, then the proof is completed. In fact, due to (2.2), we have

$0=d({x}_{{n}_{0}},{x}_{{n}_{0}+1})=d(T{x}_{{n}_{0}-1},T{x}_{{n}_{0}}),$

(2.4)

which yields that

$T{x}_{{n}_{0}-1}=T{x}_{{n}_{0}}$. Hence, equation (

2.1) implies that

$d(A,B)=d({x}_{{n}_{0}},T{x}_{{n}_{0}-1})=d({x}_{{n}_{0}},T{x}_{{n}_{0}}).$

(2.5)

For the rest of the proof, we suppose that

$d({x}_{n},{x}_{n+1})>0$ for any

$n\in \mathbb{N}$. Owing to the fact

*T* is a generalized Geraghty-contraction, we derive that

$\begin{array}{rl}d({x}_{n},{x}_{n+1})& =d(T{x}_{n-1},T{x}_{n})\\ \le \beta (M({x}_{n-1},{x}_{n}))(M({x}_{n-1},{x}_{n})-d(A,B))\\ <M({x}_{n-1},{x}_{n})-d(A,B).\end{array}$

(2.6)

Then, by (2.3) and (2.6), we deduce that

$d({x}_{n},{x}_{n+1})<M({x}_{n-1},{x}_{n})-d(A,B)\le max\{d({x}_{n-1},{x}_{n}),d({x}_{n},{x}_{n+1})\}.$

Suppose that

$max\{d({x}_{n-1},{x}_{n}),d({x}_{n},{x}_{n+1})\}=d({x}_{n},{x}_{n+1})$. Then we get that

$d({x}_{n},{x}_{n+1})<d({x}_{n},{x}_{n+1}),$

a contradiction. As a result, we conclude that

$max\{d({x}_{n-1},{x}_{n}),d({x}_{n},{x}_{n+1})\}=d({x}_{n-1},{x}_{n})$ and hence

$M({x}_{n-1},{x}_{n})\le max\{d({x}_{n-1},{x}_{n}),d({x}_{n},{x}_{n+1})\}+d(A,B)=d({x}_{n-1},{x}_{n})+d(A,B).$

(2.7)

By (2.6), we get

$\begin{array}{rl}d({x}_{n},{x}_{n+1})& =d(T{x}_{n-1},T{x}_{n})\\ \le \beta (M({x}_{n-1},{x}_{n}))d({x}_{n-1},{x}_{n})\\ <d({x}_{n-1},{x}_{n})\end{array}$

(2.8)

for all

$n\in \mathbb{N}$. Consequently,

$\{d({x}_{n},{x}_{n+1})\}$ is a nonincreasing sequence and bounded below. Thus, there exists

$L\ge 0$ such that

${lim}_{n\to \mathrm{\infty}}(d({x}_{n},{x}_{n+1}))=L$. We shall show that

$L=0$. Suppose, on the contrary,

$L>0$. Then, by (2.8), we have

$\frac{d({x}_{n+1},{x}_{n+2})}{d({x}_{n},{x}_{n+1})}\le \beta (M({x}_{n},{x}_{n+1}))\le 1$

for each

$n\ge 1$. In what follows,

$\underset{n\to \mathrm{\infty}}{lim}\beta (M({x}_{n},{x}_{n+1}))=1.$

On the other hand, since

$\beta \in F$, we conclude

${lim}_{n\to \mathrm{\infty}}M({x}_{n},{x}_{n+1})=0$, that is,

$L=\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},{x}_{n+1})=0.$

(2.9)

Since,

$d({x}_{n},T{x}_{n-1})=d(A,B)$ holds for all

$n\in \mathbb{N}$ and

$(A,B)$ satisfies the

*P*-property, then, for all

$m,n\in \mathbb{N}$, we can write,

$d({x}_{m},{x}_{n})=d(T{x}_{m-1},T{x}_{n-1})$. We also have

$d({x}_{l},T{x}_{l})\le d({x}_{l},{x}_{l+1})+d({x}_{l+1},T{x}_{l})=d({x}_{l},{x}_{l+1})+d(A,B)$

for all

$l\in \mathbb{N}$. It follows that

$\begin{array}{rl}M({x}_{m},{x}_{n})& =max\{d({x}_{m},{x}_{n}),d({x}_{m},T{x}_{m}),d({x}_{n},T{x}_{n})\}\\ \le max\{d({x}_{m},{x}_{n}),d({x}_{m},{x}_{m+1}),d({x}_{n},{x}_{n+1})\}+d(A,B).\end{array}$

Taking (2.9) into consideration, we find

$\underset{m,n\to \mathrm{\infty}}{lim}M({x}_{m},{x}_{n})\le \underset{m,n\to \mathrm{\infty}}{lim}d({x}_{m},{x}_{n})+d(A,B).$

(2.10)

We shall show that

$\{{x}_{n}\}$ is a Cauchy sequence. Suppose, on the contrary, that we have

$\epsilon =\underset{m,n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}d({x}_{n},{x}_{m})>0.$

(2.11)

Due to the triangular inequality, we have

$d({x}_{n},{x}_{m})\le d({x}_{n},{x}_{n+1})+d({x}_{n+1},{x}_{m+1})+d({x}_{m+1},{x}_{m}).$

(2.12)

Regarding (1.6) and (2.12), we have

$\begin{array}{rl}d({x}_{n},{x}_{m})& \le d({x}_{n},{x}_{n+1})+d(T{x}_{n},T{x}_{m})+d({x}_{m+1},{x}_{m})\\ \le d({x}_{n},{x}_{n+1})+\beta (M({x}_{n},{x}_{m}))(M({x}_{n},{x}_{m})-d(A,B))+d({x}_{m+1},{x}_{m}).\end{array}$

(2.13)

Taking (2.10), (2.13) and (2.9) into account, we derive that

$\begin{array}{rcl}\underset{m,n\to \mathrm{\infty}}{lim}d({x}_{n},{x}_{m})& \le & \underset{m,n\to \mathrm{\infty}}{lim}\beta (M({x}_{n},{x}_{m}))\underset{m,n\to \mathrm{\infty}}{lim}(M({x}_{m},{x}_{n})-d(A,B))\\ \le & \underset{m,n\to \mathrm{\infty}}{lim}\beta (M({x}_{n},{x}_{m}))\underset{m,n\to \mathrm{\infty}}{lim}d({x}_{m},{x}_{n}).\end{array}$

Owing to (2.11), we get

$1\le \underset{m,n\to \mathrm{\infty}}{lim}\beta (M({x}_{n},{x}_{m})),$

which implies

${lim}_{m,n\to \mathrm{\infty}}\beta (M({x}_{n},{x}_{m}))=1$. By the property of

*β*, we have

${lim}_{m,n\to \mathrm{\infty}}M({x}_{n},{x}_{m})=0$. Consequently, we have

${lim}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0$, a contradiction. Hence, we conclude that the sequence

$\{{x}_{n}\}$ is Cauchy. Since

*A* is a closed subset of the complete metric space

$(X,d)$ and

$\{{x}_{n}\}\subset A$, and we can find

${x}^{\ast}\in A$ such that

${x}_{n}\to {x}^{\ast}$ as

$n\to \mathrm{\infty}$. We assert that

$d({x}^{\ast},T{x}^{\ast})=d(A,B)$. Suppose, on the contrary, that

$d({x}^{\ast},T{x}^{\ast})>d(A,B)$. First, we obtain the following inequalities:

$\begin{array}{rl}d({x}^{\ast},T{x}^{\ast})& \le d({x}^{\ast},T{x}_{n})+d(T{x}_{n},T{x}^{\ast})\\ \le d({x}^{\ast},{x}_{n+1})+d({x}_{n+1},T{x}_{n})+d(T{x}_{n},T{x}^{\ast})\\ \le d({x}^{\ast},{x}_{n+1})+d(A,B)+d(T{x}_{n},T{x}^{\ast}).\end{array}$

Letting

$n\to \mathrm{\infty}$ in the inequalities above, we conclude that

$d({x}^{\ast},T{x}^{\ast})-d(A,B)\le \underset{n\to \mathrm{\infty}}{lim}d(T{x}_{n},T{x}^{\ast}).$

On the other hand, we obtain

$d({x}_{n},T{x}_{n})\le d({x}_{n},{x}_{n+1})+d({x}_{n+1},T{x}_{n})=d({x}_{n},{x}_{n+1})+d(A,B).$

Taking limit as

$n\to \mathrm{\infty}$ in the inequality above, we find

$\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},T{x}_{n})\le d(A,B).$

So, we deduce that

${lim}_{n\to \mathrm{\infty}}d({x}_{n},T{x}_{n})=d(A,B)$. As a consequence, we derive

$\underset{n\to \mathrm{\infty}}{lim}M({x}_{n},{x}^{\ast})=max\{\underset{n\to \mathrm{\infty}}{lim}d({x}^{\ast},{x}_{n}),\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},T{x}_{n}),d({x}^{\ast},T{x}^{\ast})\}=d({x}^{\ast},T{x}^{\ast}),$

and hence

$\underset{n\to \mathrm{\infty}}{lim}M({x}_{n},{x}^{\ast})-d(A,B)=d({x}^{\ast},T{x}^{\ast})-d(A,B).$

(2.14)

Combining (1.6) and (2.14), we find

$\begin{array}{rl}d({x}^{\ast},T{x}^{\ast})-d(A,B)& \le \underset{n\to \mathrm{\infty}}{lim}d(T{x}_{n},T{x}^{\ast})\\ \le \underset{n\to \mathrm{\infty}}{lim}\left[\beta \left(M({x}_{n},{x}^{\ast})\right)(M({x}_{n},{x}^{\ast})-d(A,B))\right]\\ =\underset{n\to \mathrm{\infty}}{lim}\beta \left(M({x}_{n},{x}^{\ast})\right)(d({x}^{\ast},T{x}^{\ast})-d(A,B)).\end{array}$

(2.15)

Since

$d({x}^{\ast},T{x}^{\ast})-d(A,B)>0$ together with (2.15), we get

$1\le {lim}_{n\to \mathrm{\infty}}\beta (M({x}_{n},{x}^{\ast}))$. Hence, we have

$\underset{n\to \mathrm{\infty}}{lim}\beta \left(M({x}_{n},{x}^{\ast})\right)=1,$

which yields

$\underset{n\to \mathrm{\infty}}{lim}M({x}_{n},{x}^{\ast})=d({x}^{\ast},T{x}^{\ast})=0.$

As a result, we deduce that $d({x}^{\ast},T{x}^{\ast})=0>d(A,B)$, a contradiction. So, $d({x}^{\ast},T{x}^{\ast})\le d(A,B)$ and hence $d({x}^{\ast},T{x}^{\ast})=d(A,B)$, ${x}^{\ast}$ is a best proximity point of *T*. Hence, we conclude that *T* has a best proximity point.

We claim that the best proximity point of *T* is unique.

Suppose, on the contrary, that

${x}^{\ast}$ and

${y}^{\ast}$ are two distinct best proximity points of

*T*. Thus, we have

$d({x}^{\ast},T{x}^{\ast})=d(A,B)=d({y}^{\ast},T{y}^{\ast}).$

(2.16)

By using the

*P*-property, we find

$d({x}^{\ast},{y}^{\ast})=d(T{x}^{\ast},T{y}^{\ast})$

(2.17)

and

$\begin{array}{rcl}M({x}^{\ast},{y}^{\ast})& =& max\{d({x}^{\ast},{y}^{\ast}),d({x}^{\ast},T{x}^{\ast}),d({y}^{\ast},T{y}^{\ast})\}\\ =& max\{d({x}^{\ast},{y}^{\ast}),d(A,B),d(A,B)\}=d({x}^{\ast},{y}^{\ast}).\end{array}$

Due to the fact that

*T* is a generalized Geraghty-contraction, we have

$\begin{array}{rcl}d({x}^{\ast},{y}^{\ast})& =& d(T{x}^{\ast},T{y}^{\ast})\\ \le & \beta \left(M({x}^{\ast},{y}^{\ast})\right)(M({x}^{\ast},{y}^{\ast})-d(A,B))\\ =& \beta \left(d({x}^{\ast},{y}^{\ast})\right)(d({x}^{\ast},{y}^{\ast})-d(A,B))\\ \le & \beta \left(d({x}^{\ast},{y}^{\ast})\right)d({x}^{\ast},{y}^{\ast})<d({x}^{\ast},{y}^{\ast}),\end{array}$

a contradiction. This completes the proof. □

**Remark 9** Let $(X,d)$ be a metric space and *A* be any nonempty subset of *X*. It is evident that a pair $(A,A)$ satisfies the *P*-property.

**Corollary 10** *Suppose that* $(X,d)$ *is a complete metric space and* *A* *is a nonempty closed subset of* *X*. *If a self*-*mapping* $T:A\to A$ *is a generalized Geraghty*-*contraction*, *then it has a unique fixed point*.

*Proof* Taking Remark 9 into consideration, we conclude the desired result by applying Theorem 8 with $A=B$. □

In order to illustrate our main result, we present the following example.

**Example 11** Suppose that

$X={\mathbb{R}}^{2}$ with the metric

$d((x,y),({x}^{\prime},{y}^{\prime}))=max\{|x-{x}^{\prime}|,|y-{y}^{\prime}|\},$

and consider the closed subsets

$\begin{array}{c}A=\{(x,0):0\le x\le 1\},\hfill \\ B=\{(x,0):-1\le x\le 0\},\hfill \end{array}$

and let

$T:A\to B$ be the mapping defined by

$T((x,0))=(\frac{-x}{1+x},0).$

Since $d(A,B)=0$, the pair $(A,B)$ has the *P*-property.

Notice that ${A}_{0}=(0,0)$ and ${B}_{0}=(0,0)$ and $T({A}_{0})\subseteq {B}_{0}$.

Moreover,

$d(T(x,0),T({x}^{\prime},0))=d((\frac{-x}{1+x},0),(\frac{-{x}^{\prime}}{1+{x}^{\prime}},0))=|\frac{-x}{1+x}+\frac{{x}^{\prime}}{1+{x}^{\prime}}|=\frac{|{x}^{\prime}-x|}{(1+x)(1+{x}^{\prime})},$

and, as

$(1+x)(1+{x}^{\prime})\ge 1+|x-{x}^{\prime}|$, we have

$d(T(x,0),T({x}^{\prime},0))=\frac{|{x}^{\prime}-x|}{(1+x)(1+{x}^{\prime})}\le \frac{|{x}^{\prime}-x|}{1+|x-{x}^{\prime}|}=\beta \left(|x-{x}^{\prime}|\right)=\beta \left(d((x,0),({x}^{\prime},0))\right),$

where $\beta :[0,\mathrm{\infty})\to [0,1)$ is defined as $\beta (t)=\frac{t}{1+t}$.

Notice that *β* is nondecreasing since ${\beta}^{\prime}(t)=\frac{1}{{(1+t)}^{2}}$.

Therefore,

$\begin{array}{rcl}d(T(x,0),T({x}^{\prime},0))& \le & \beta \left(d((x,0),({x}^{\prime},0))\right)\le \beta \left(M((x,0),({x}^{\prime},0))\right)\\ =& \frac{\beta (M((x,0),({x}^{\prime},0)))}{M((x,0),({x}^{\prime},0))}\cdot M((x,0),({x}^{\prime},0))\end{array}$

and it is easily seen that the function $\gamma (t)=\frac{\beta (t)}{t}=\frac{1}{1+t}$ belongs to *F*.

Therefore, since the assumptions of Theorem 8 are satisfied, by Theorem 8 there exists a unique

$({x}^{\ast},0)\in A$ such that

$d(({x}^{\ast},0),T({x}^{\ast},0))=0=d(A,B).$

The point $({x}^{\ast},0)\in A$ is $(0,0)\in A$.