Our first result is as follows.

**Theorem** **2.** Let $(P,Q)$ be a pair of non-empty subsets of a complete metric space $(M,d)$ so that ${P}_{0}$ is non-empty, $\mathcal{T}:P\to Q$ and $\alpha ,\eta :P\times P\to \left[0,\infty \right)$ be given mappings. Suppose that:

- (i)
P is closed and $\mathcal{T}\left({P}_{0}\right)\subseteq {Q}_{0}$;

- (ii)
$\mathcal{T}$ is triangular $(\alpha ,\eta )$-proximal admissible;

- (iii)
There are ${u}_{0},{u}_{1}\in {P}_{0}$ so that $d({u}_{1},\mathcal{T}{u}_{0})=d(P,Q)$ and $\alpha \left({u}_{0},{u}_{1}\right)\ge \eta \left({u}_{0},{u}_{1}\right)$;

- (iv)
$\mathcal{T}$ is a continuous Geraghty type $\mathcal{Z}$-proximal contraction.

Then, $\mathcal{T}$ has a best proximity point in $P.$ If $\alpha (u,v)\ge \eta (u,v)$ for all $u,v\in {B}_{est}\left(\mathcal{T}\right),$ then $\mathcal{T}$ has a unique best proximity point ${u}^{*}\in P.$ Moreover, for every $u\in P,$${lim}_{n\to \infty}{\mathcal{T}}^{n}u={u}^{*}$.

**Proof.** From the condition

$\left(iii\right),$ there are

${u}_{0},{u}_{1}\in {P}_{0}$ so that

Since

$\mathcal{T}\left({P}_{0}\right)\subseteq {Q}_{0}$, there is

${u}_{2}\in {P}_{0}$ so that

Since

$\mathcal{T}$ is

$(\alpha ,\eta )$-proximal admissible, we get

$\alpha \left({u}_{1},{u}_{2}\right)\ge \eta \left({u}_{1},{u}_{2}\right).$ Now, we have

Again, since

$\mathcal{T}\left({P}_{0}\right)\subseteq {Q}_{0}$, there exists

${u}_{3}\in {P}_{0}$ such that

and thus,

Since

$\mathcal{T}$ is

$(\alpha ,\eta )$-proximal admissible, this implies that

$\alpha \left({u}_{2},{u}_{3}\right)\ge \eta \left({u}_{2},{u}_{3}\right).$ Thus, we have

By repeating this process, we build a sequence

$\left\{{u}_{n}\right\}$ in

${P}_{0}\subseteq P$ so that

for all

$n\in \mathbb{N}\cup \left\{0\right\}.$ If there is

${n}_{0}$ so that

${u}_{{n}_{0}}={u}_{{n}_{0}+1},$ then

That is, ${u}_{{n}_{0}}$ is a best proximity point of $\mathcal{T}$. We should suppose that ${u}_{n}\ne {u}_{n+1},$ for all $n.$

From (

8), for all

$n\in \mathbb{N},$ we get

On the grounds that

$\mathcal{T}$ is a Geraghty type

$\mathcal{Z}$-proximal contraction, by utilizing Remark 1, we deduce that

which requires that

$d({u}_{n},{u}_{n+1})<d({u}_{n-1},{u}_{n}),$ for all

$n.$ Therefore, the sequence

$\left\{d({u}_{n},{u}_{n+1})\right\}$ is decreasing, and so there is

$\lambda \ge 0$ so that

${lim}_{n\to \infty}d\left({u}_{n},{u}_{n+1}\right)=\lambda .$ Now, we shall show that

$\lambda =0.$ On the contrary, assume that

$\lambda >0.$ Then, taking into account (

9), for any

$n\in \mathbb{N},$This yields, for any

$n\in \mathbb{N},$Taking

$n\to \infty $, we find that

and since

$\beta \in \mathsf{\Sigma},\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}$${lim}_{n\to \infty}d({u}_{n-1},{u}_{n})=0.$ This contradicts our assumption

${lim}_{n\to \infty}d({u}_{n-1},{u}_{n})=\lambda >0.$ Therefore, we get

We shall prove that

$\left\{{u}_{n}\right\}$ is Cauchy in

P. By contradiction, suppose that

$\left\{{u}_{n}\right\}$ is not a Cauchy sequence, so there is an

$\epsilon >0$ for which we can find

$\left\{{u}_{{m}_{k}}\right\}$ and

$\left\{{u}_{{n}_{k}}\right\}$ of

$\left\{{u}_{n}\right\}$ such that

${n}_{k}$ is the smallest index for which

${n}_{k}>{m}_{k}>k$ and

Taking

$k\to \infty $, by (

10), we get

By triangular inequality,

which yields that

Since

$\mathcal{T}$ is triangular

$(\alpha ,\eta )$-proximal admissible, by using (

8), we infer

Combining (

8) and (

14), for all

$k\in \mathbb{N},$ we have

Regarding the fact that

$\mathcal{T}$ is a Geraghty type

$\mathcal{Z}$-proximal contraction, from Remark 1, we deduce that

Taking the limit as

k tends to

∞ on both sides of the last inequality, and using the Equations (

12) and (

13), we get

which implies that

${lim}_{k\to \infty}\beta \left(d({u}_{{m}_{k}},{u}_{{n}_{k}})\right)=1,$ and so

${lim}_{k\to \infty}d({u}_{{m}_{k}},{u}_{{n}_{k}})=0$ which contradicts

$\epsilon >0.$ Hence,

$\left\{{u}_{n}\right\}$ is a Cauchy sequence in

P. Since

P is a closed subset of the complete metric space

$(M,d)$, there is

$p\in P$ so that

Since

$\mathcal{T}$ is continuous, we have

Combining (

8), (

15), and (

16), we get

Therefore,

$u\in P$ is a best proximity point of

$\mathcal{T}.$ Finally, we shall show that the set

${B}_{est}\left(\mathcal{T}\right)$ is a singleton. Suppose that

r is another best proximity point of

$\mathcal{T},$ that is,

$d(r,\mathcal{T}r)=d(P,Q).$ Then, by the hypothesis, we have

$\alpha (p,r)\ge \eta (p,r)$—that is,

Then, from Remark 1, we deduce

which is a contradiction. Hence, we have a unique best proximity point of

$\mathcal{T}$. □

Let us consider the following assertion in order to remove the continuity on the operator $\mathcal{T}$ in the next theorem.

**Theorem** **3.** Let $(P,Q)$ be a pair of non-empty subsets of a complete metric space $(M,d)$ so that ${P}_{0}$ is non-empty, $\mathcal{T}:P\to Q$ and $\alpha ,\eta :P\times P\to \left[0,\infty \right)$ be given mappings. Suppose that:

- (i)
P is closed and $\mathcal{T}\left({P}_{0}\right)\subseteq {Q}_{0}$;

- (ii)
$\mathcal{T}$ is triangular $(\alpha ,\eta )$-proximal admissible;

- (iii)
there are ${u}_{0},{u}_{1}\in {P}_{0}$ so that $d({u}_{1},\mathcal{T}{u}_{0})=d(P,Q)$ and $\alpha \left({u}_{0},{u}_{1}\right)\ge \eta \left({u}_{0},{u}_{1}\right)$;

- (iv)
the condition $\left(C\right)$ holds and $\mathcal{T}$ is a Geraghty type $\mathcal{Z}$-proximal contraction.

Then, $\mathcal{T}$ has a best proximity point in $P.$ If $\alpha (u,v)\ge \eta (u,v)$ for all $u,v\in {B}_{est}\left(\mathcal{T}\right),$ then $\mathcal{T}$ has a unique best proximity point ${u}^{*}\in P.$ Moreover, for each $u\in P,$ we have ${lim}_{n\to \infty}{\mathcal{T}}^{n}u={u}^{*}$.

**Example** **1.** Let $M={\mathbb{R}}^{2}$ be endowed with the Euclidian metric, $P=\left\{(0,u):u\ge 0\right\}$ and $Q=\left\{(1,u):u\ge 0\right\}$. Note that $d(P,Q)=1,$ ${P}_{0}=P$ and ${Q}_{0}=Q.$ Let Then, $\beta \in \mathsf{\Sigma}.$ Define $\mathcal{T}:P\to Q$ and $\alpha :P\times P\to \left[0,\infty \right)$ byand Choose $\zeta (t,s)=\frac{2}{3}s-t$ for all $t,s\in [0,\infty )$. Let $u,v,p,q\ge 0$ be such that Then, $u,v\in [0,1]$ or $u=v$.

$u,v\in [0,1]$. Here, $\mathcal{T}(0,u)=(1,\frac{u}{9})$ and $\mathcal{T}(0,v)=(1,\frac{v}{9})$. Also,that is, $p=\frac{u}{9}$ and $q=\frac{v}{9}$. So, $\alpha \left(\right(0,p),(0,q\left)\right)\ge d\left(\right(0,p),(0,q\left)\right).$ Moreover, If $u=v$, then $\beta \left(\right|u-v\left|\right)=\frac{1}{2}$ and the right-hand side of the above inequality is equal to 0.

$u=v>1$. Here, $\mathcal{T}(0,u)=(1,{u}^{2})$ and $\mathcal{T}(0,v)=(1,{v}^{2})$. Similarly, we get that $p=q={u}^{2}={v}^{2}$. So, $\alpha \left(\right(0,p),(0,q\left)\right)=0=\eta \left(\right(0,p),(0,q\left)\right).$

Also, $\zeta \left(d\right((0,p),(0,q)),\beta (d\left(\right(0,u),(0,v\left)\right)\left)d\right((0,u),(0,v)\left)\right)\ge 0$.

In each case, we get that $\mathcal{T}$ is an $(\alpha ,\eta )$-proximal admissible. It is also easy to see that $\mathcal{T}$ is triangular $(\alpha ,\eta )$-proximal admissible. Also, $\mathcal{T}$ is a Geraghty type $\mathcal{Z}$-proximal contraction. Also, if $\left\{{u}_{n}=(0,{p}_{n})\right\}$ is a sequence in P such that $\alpha \left({u}_{n},{u}_{n+1}\right)\ge \eta \left({u}_{n},{u}_{n+1}\right)$ for all n and ${u}_{n}=(0,{p}_{n})\to u=(0,p)$ as $n\to \infty ,$ then ${p}_{n}\to p$. We have ${p}_{n},{p}_{n+1}\in [0,1]$ or ${p}_{n}={p}_{n+1}$. We get that $p\in [0,1]$ or ${p}_{n}=p$. This implies that $\alpha \left({u}_{n},u\right)\ge \eta \left({u}_{n},u\right)$ for all $n.$

Moreover, there is $({u}_{0},{u}_{1})=((0,1),(0,\frac{1}{9}))\in {P}_{0}\times {P}_{0}$ so thatConsequently, all conditions of Theorem 3 are satisfied. Therefore, $\mathcal{T}$ has a unique best proximity point in P, which is $(0,0).$ On the other side, we indicate that (4) is not satisfied. In fact, for $u=(0,2),v=(0,3),$ we have