We denote by Ψ the family of all continuous functions from \([0, +\infty) \times[0, +\infty)\) to \([0, +\infty)\) such that \(\psi (u,v)=0 \) if and only if \(u=v=0\) where \(\psi\in\Psi\).
Definition 2.1
Let \((X,d)\) be a metric space, \(\emptyset\neq A,B\subset X\), \(\alpha: A \times A \rightarrow[0,\infty)\) a function and \(f,g: A \rightarrow B\) nonself mappings. We say that \((f,g)\) is a triangular αproximal admissible pair, if for all \(p,q,r,t_{1},t_{2},s_{1},s_{2} \in A\),
$$\begin{aligned}& T_{1}: \left \{ \textstyle\begin{array}{l} \alpha(t_{1},t_{2})\ge1, \\ d(s_{1},ft_{1})=d(A,B),\\ d(s_{2},gt_{2})=d(A,B) \end{array}\displaystyle \right . \quad \Longrightarrow\quad \alpha(s_{1},s_{2}) \ge1, \\& T_{2}: \left \{ \textstyle\begin{array}{l} \alpha(p,r)\ge1, \\ \alpha(r,q)\ge1 \end{array}\displaystyle \right . \quad \Longrightarrow\quad \alpha(p,q) \ge1. \end{aligned}$$
Let \((X,d)\) be a metric space and \(\emptyset\neq A,B\subset X\). We define
$$\begin{aligned}& A_{0}=\bigl\{ s \in A: d(s,t)= d(A,B) \text{ for some } t \in B\bigr\} , \\& B_{0}=\bigl\{ t \in B: d(s,t)= d(A,B) \text{ for some } s \in A\bigr\} . \end{aligned}$$
Definition 2.2
Let \((X,d)\) be a metric space, \(\emptyset\neq A,B\subset X\), and \(f,g: A \rightarrow B\) nonself mappings. We say that \((f,g)\) is a generalized proximal Ccontraction pair if, for all \(s, t, p, q \in A\),
$$ \left . \textstyle\begin{array}{l} d(s,fp)=d(A,B), \\ d(t,gq)=d(A,B) \end{array}\displaystyle \right \} \quad \Longrightarrow\quad d(s,t)\leq\frac{1}{2}\bigl(d(p,t)+d(q,s)\bigr) \psi\bigl(d(p,t),d(q,s)\bigr), $$
(1)
in which \(\psi\in\Psi\).
Definition 2.3
Let \((X,d)\) be a metric space, \(\emptyset\neq A,B\subset X\), \(\alpha: A \times A \rightarrow[0,\infty)\) a function and \(f,g: A \rightarrow B\) nonself functions. If, for all \(s, t, p, q \in A\),
$$ \left \{ \textstyle\begin{array}{l} d(s,fp)=d(A,B), \\ d(t,gq)=d(A,B) \end{array}\displaystyle \right . $$
imply
$$ \alpha(p,q) d(s,t)\leq\frac{1}{2}\bigl(d(p,t)+d(q,s)\bigr) \psi\bigl(d(p,t),d(q,s)\bigr), $$
(2)
then \((f,g)\) is said to be an αproximal C1contraction pair.
If in the definition above, we replace (2) by
$$ \bigl(\alpha(p,q)+l\bigr)^{d(s,t)}\leq(l+1)^{\frac{1}{2}(d(p,t)+d(q,s))\psi (d(p,t),d(q,s))}, $$
(3)
where \(l>0\), then \((f,g)\) is said to be an αproximal C2contraction pair.
Theorem 2.4
Let
\((X,d)\)
be a metric space and
\(\emptyset\neq A,B\subset X\). Let
A
be complete and
\(A_{0}\)
nonempty set. Moreover, assume that the nonself functions
\(f,g: A \rightarrow B\)
satisfy:

(i)
f, g
are continuous,

(ii)
\(f(A_{0})\subset B_{0}\)
and
\(g(A_{0}) \subset B_{0}\),

(iii)
\((f,g)\)
is a generalized proximal
Ccontraction pair,

(iv)
there exist
\(s_{0}, s_{1} \in A_{0}\)
such that
\(d(s_{1},fs_{0})=d(A,B)\).
Then the functions
f
and
g
have a unique common best proximity point.
Proof
From (iv) we can get \(s_{0},s_{1} \in A_{0}\) such that
$$d(s_{1},fs_{0})=d(A,B). $$
Since \(g(A_{0}) \subset B_{0}\), there exists \(s_{2} \in A_{0}\) such that \(d(s_{2},gs_{1})= d(A,B)\).
We continue this process and construct a sequence \(\{s_{n}\}\) such that
$$ \left \{ \textstyle\begin{array}{l} d(s_{2n+1},fs_{2n})=d(A,B), \\ d(s_{2n+2},gs_{2n+1})=d(A,B) \end{array}\displaystyle \right . $$
(4)
for each \(n\in\mathbb {N}\).
We divide our further derivation into four steps.
Step 1. We have
$$ \lim_{n\rightarrow\infty}d(s_{n},s_{n+1})=0. $$
(5)
Put \(s=s_{2n+1}\) and \(t=s_{2n+2}\). From (1), we get
$$\begin{aligned} d(s_{2n+1},s_{2n+2}) =&\frac {1}{2} \bigl(d(s_{2n},s_{2n+2})+d(s_{2n+1},s_{2n+1}) \bigr)\psi \bigl(d(s_{2n},s_{2n+2}),d(s_{2n+1},s_{2n+1}) \bigr) \\ =&\frac{1}{2}d(s_{2n},s_{2n+2})\psi \bigl(d(s_{2n},s_{2n+2}),0\bigr) \\ \leq&\frac{1}{2}d(s_{2n},s_{2n+2}) \\ \leq& \frac{1}{2}\bigl(d(s_{2n},s_{2n+1})+d(s_{2n+1},s_{2n+2}) \bigr), \end{aligned}$$
(6)
which implies \(d(s_{2n+1},s_{2n+2})\leq d(s_{2n},s_{2n+1})\).
Now, if we put \(d_{n}:=d(s_{n},s_{n+1})\), then we get \(d_{2n+1} \leq d_{2n}\). Also, we have \(d_{2n+2}\leq d_{2n+1}\), which implies that the sequence \(\{d_{n}\}\) is decreasing and so there is a \(d\ge0\) such that \(d_{n} \rightarrow d\) as \(n\rightarrow\infty\). Now, take \(n\rightarrow\infty\) in (6) and get
$$ d \leq\frac{1}{2}\lim_{n\rightarrow\infty}d(s_{2n},s_{2n+2}) \leq \frac{1}{2}(d+d)=d, $$
that is,
$$ \lim_{n\rightarrow\infty}d(s_{2n},s_{2n+2})=2d. $$
(7)
Take again \(n \rightarrow\infty\) in (6). By (7) and the continuity of ψ, we get
$$ d \leq d\psi(2d,0), $$
and so \(\psi(2d,0)=0\). By the properties of ψ, we get \(d=0\).
Step 2. We claim that \(\{s_{n}\}\) is a Cauchy sequence. By (5), we show that the subsequence \(\{s_{2n}\}\) of \(\{s_{n}\}\) is a Cauchy sequence in \(A_{0}\). Contrarily, let there exists an \(\epsilon>0\) for which the subsequences \(\{s_{2m(k)}\}\) and \(\{s_{2n(k)}\}\) of \(\{s_{2n}\}\) such that \(n(k)\) is the smallest integer satisfying, for all \(k > 0\),
$$ n(k)> m(k) > k \quad \text{implies} \quad d(s_{2m(k)},s_{2n(k)}) \ge\epsilon, $$
(8)
which would imply that
$$ d(s_{2m(k)},s_{2n(k)2})< \epsilon. $$
(9)
Using the triangular inequality, (8), and (9), we would get
$$\begin{aligned} \epsilon \leq& d(s_{2m(k)},s_{2n(k)}) \leq d(s_{2m(k)},s_{2n(k)2})+d(s_{2n(k)2},s_{2n(k)1})+d(s_{2n(k)1},s_{2n(k)}) \\ < & \epsilon+d(s_{2n(k)2},s_{2n(k)1})+d(s_{2n(k)1},s_{2n(k)}). \end{aligned}$$
Letting \(k\rightarrow\infty\) in the above inequality and using (5), we would get
$$ \lim_{k\rightarrow\infty}d(s_{2m(k)},s_{2n(k)})= \epsilon. $$
(10)
On the other hand, we have
$$ d(s_{2m(k)},s_{2n(k)})\leq d(s_{2m(k)},s_{2n(k)+1})+d(s_{2n(k)+1},s_{2n(k)}). $$
Now, we would have
$$ \epsilon\leq\lim_{n\rightarrow\infty}d(s_{2m(k)},s_{2n(k)+1}). $$
(11)
By the triangular inequality, we have
$$ d(s_{2m(k)},s_{2n(k)}) \leq d(s_{2m(k)},s_{2m(k)1})+d(s_{2m(k)1},s_{2n(k)+1})+d(s_{2n(k)+1},s_{2n(k)}). $$
Letting again \(k\rightarrow\infty\) in the above inequality and using (5) and (10), we would get
$$ \epsilon\leq\lim_{k\rightarrow\infty}d(s_{2m(k)1},s_{2n(k)+1}). $$
Also, we can get \(\lim_{k\rightarrow\infty}d(s_{2m(k)1},s_{2n(k)+1})\leq\epsilon\), and so
$$ \lim_{n\rightarrow\infty}d(s_{2m(k)1},s_{2n(k)+1})= \epsilon. $$
(12)
From (1) we have
$$\begin{aligned} d(s_{2n(k)+1},s_{2m(k)}) \leq& \frac {1}{2} \bigl(d(s_{2n(k)},s_{2m(k)})+d(s_{2n(k)+1},s_{2m(k)1}) \bigr) \\ &{}\psi\bigl(d(s_{2n(k)},s_{2m(k)}),d(s_{2n(k)+1},s_{2m(k)1}) \bigr). \end{aligned}$$
(13)
Taking \(k \rightarrow\infty\) in the above inequality and using (10), (11), (12), and the continuity of ψ, we would obtain
$$ \epsilon\leq\frac{1}{2}(\epsilon+\epsilon)\psi(\epsilon ,\epsilon) , $$
and therefore \(\psi(\epsilon,\epsilon) = 0\), which would imply \(\epsilon= 0\), a contradiction. Thus, \(\{s_{n}\}\) is a Cauchy sequence. Since A is complete, there is a \(z \in A\) such that \(s_{n} \rightarrow z\).
Step 3. Now, from
$$ d(s_{2n+1},fs_{2n})=d(A,B),\qquad d(s_{2n+2},gs_{2n+1})=d(A,B) , $$
taking \(n \rightarrow\infty\) and by continuity of f and g, we have \(d(z,fz) = d(z,gz) = d(A,B)\). So, z is a common best proximity point of the mappings f and g.
Step 4. Now, let f and g have another common best proximity point, say w, such that
$$ d(w,fw) = d(w,gw) = d(A,B). $$
From (1) we have
$$\begin{aligned} d(z,w) \leq& \frac{1}{2}\bigl(d(z,w)+d(w,z)\bigr)\psi\bigl(d(z,w),d(w,z) \bigr) \\ =& d(z,w)\psi\bigl(d(z,w),d(z,w)\bigr) \end{aligned}$$
(14)
whence \(d(z,w) = 0\), and therefore \(z=w\). □
Theorem 2.5
Let
\((X,d)\)
be a metric space and
\(\emptyset\neq A,B\subset X\)
such that
A
is complete and
\(A_{0}\)
is nonempty. Moreover, suppose that the nonself functions
\(f,g: A \rightarrow B\)
satisfy:

(i)
f, g
are continuous,

(ii)
\(f(A_{0})\subset B_{0}\)
and
\(g(A_{0}) \subset B_{0}\),

(iii)
\((f,g)\)
is an
αproximal
C1contraction pair or an
αproximal
C2contraction pair,

(iv)
\((f,g)\)
is a triangular
αproximal admissible pair,

(iv)
there exist
\(s_{0}, s_{1} \in A_{0}\)
such that
\(d(s_{1},fs_{0})=d(A,B)\), \(\alpha(s_{1},s_{0}) \ge1\).
Then
f
and
g
have a common best proximity point. Furthermore, if
\(z,w \in X\)
are common best proximity points and
\(\alpha(z,w)\ge1\), then common best proximity point is unique.
Proof
By (iv), we can find \(s_{0}, s_{1} \in A_{0}\) such that
$$d(s_{1},fs_{0})=d(A,B),\qquad \alpha(s_{1},s_{0}) \ge1. $$
Define the sequence \(\{s_{n}\}\) as in (4) of Theorem 2.4. Since \((f,g)\) is triangular αproximal admissible, we have \(\alpha(s_{n},s_{n+1}) \ge1\). Then
$$ \left \{ \textstyle\begin{array}{l} \alpha(s_{n},s_{n+1}) \ge1, \\ d(s_{2n+1},fs_{2n})=d(A,B), \\ d(s_{2n+2},gs_{2n+1})=d(A,B). \end{array}\displaystyle \right . $$
(15)
If \(s=s_{2n+1}\), \(t=s_{2n+2}\), \(p=s_{2n}\), \(q=s_{2n+1}\), and \((f,g)\) is an αproximal C1contraction pair or an αproximal C2contraction pair, then \((f,g)\) is a generalized proximal Ccontraction pair. Then Step 1 of the proof of Theorem 2.4 implies that \(\lim_{n\rightarrow\infty}d(s_{n},s_{n+1})=0\).
Now we prove that
$$ \alpha(s_{2m(k)1},s_{2n(k)})\ge1,\quad n(k)> m(k) > k. $$
(16)
Since \((f,g)\) is triangular αproximal admissible and
$$ \left \{ \textstyle\begin{array}{l} \alpha(s_{2m(k)1},s_{2m(k)})\ge1, \\ \alpha(s_{2m(k)},s_{2m(k)+1})\ge1, \end{array}\displaystyle \right . $$
from (\(T_{2}\)) of Definition 2.1, we have
$$ \alpha(s_{2m(k)1},s_{2m(k)+1})\ge1. $$
Again, since \((f,g)\) is triangular αproximal admissible and
$$ \left \{ \textstyle\begin{array}{l} \alpha(s_{2m(k)1},s_{2m(k)+1})\ge1, \\ \alpha(s_{2m(k)+1},s_{2m(k)+2})\ge1, \end{array}\displaystyle \right . $$
from (\(T_{2}\)) of Definition 2.1 again, we have
$$ \alpha(s_{2m(k)1},s_{2m(k)+2})\ge1. $$
By continuing this process, we get (16). If \(s=s_{2n(k)+1}\), \(t=s_{2m(k)}\), \(p=s_{2n(k)}\), \(q=s_{2m(k)1}\), then αproximal C1contraction (C2contraction) pair \((f,g)\) is a generalized proximal Ccontraction pair. Therefore by Step 2 of Theorem 2.4, there exists a \(z \in A\) such that \(s_{n} \rightarrow z\). Step 3 of Theorem 2.4 and continuity of f and g immediately imply that f and g have a common best proximity point z. If w is another common best proximity point of \((f,g)\), then, since \(\alpha(z,w) \ge1\), Step 4 implies that \(z=w\). □
Definition 2.6
Let \(\alpha:X \times X\rightarrow(\infty,+\infty)\) be a function and \(f,g: X \rightarrow X\) self mappings. We say that \((f,g)\) is a triangular αadmissible pair if

(i)
\(p,q \in X\), \(\alpha(p,q) \ge1 \Longrightarrow \alpha(fp,gq)\ge 1\) or \(\alpha(gp,fq)\ge1\),

(ii)
\(p, q, r \in X\), \(\bigl \{\scriptsize{ \begin{array}{l} \alpha(p,r)\ge1, \\ \alpha(r,q)\ge1, \end{array}} \bigr.\Longrightarrow\alpha(p,q) \ge1\).
The following corollary is a consequence of the last theorem.
Corollary 2.7
Let
\((X,d)\)
be a complete metric space and
\(f,g: X \rightarrow X\). Moreover, let the self functions
f
and
g
satisfy:

(i)
f
and
g
are continuous,

(ii)
there exists
\(s_{0} \in X\)
such that
\(\alpha(s_{0},fs_{0})\ge1\),

(iii)
\((f,g)\)
is a triangular
αadmissible pair,

(iv)
for all
\(p, q \in X\), \(\alpha(p,q) d(fp,gq) \leq\frac{1}{2}(d(p,gq) + d(q,fp))\psi (d(p,gq),d(q,fp))\) (or
\((\alpha(p,q)+l)^{d(fp,gq)} \leq(l+1)^{\frac{1}{2}(d(p,gq) + d(q,fp))\psi(d(p,gq),d(q,fp))} \)).
Then
f
and
g
have a common fixed point. Moreover, if
\(x,y \in X\)
are common fixed points and
\(\alpha(x,y)\ge1\), then the common fixed point of
f
and
g
is unique, that is, \(x=y\).
Now, we remove the continuity hypothesis of f and g, and get the following theorem.
Theorem 2.8
Let
\((X,d)\)
be a metric space and
\(\emptyset\neq A,B\subset X\). Let
A
be complete, the pair
\((A,B)\)
have the
Vproperty, and
\(A_{0}\)
be nonempty. Moreover, suppose that the nonself mappings
\(f,g: A \rightarrow B\)
satisfy:

(i)
\(f(A_{0})\subset B_{0}\)
and
\(g(A_{0}) \subset B_{0}\),

(ii)
\((f,g)\)
is a generalized proximal
Ccontraction pair,

(iii)
there are
\(s_{0}, s_{1} \in A_{0}\)
such that
\(d(s_{1},fs_{0})=d(A,B)\).
Then the functions
f
and
g
have unique common best proximity point.
Proof
By Theorem 2.4, there is a Cauchy sequence \(\{s_{n}\} \subset A\) and \(z\in A\) such that (4) holds and \(s_{n} \rightarrow z\). Moreover, we have
$$\begin{aligned} d(z,B) \leq& d(z,fs_{2n}) \\ \leq& d(z,s_{2n+1})+d(s_{2n+1},fs_{2n}) \\ =& d(z,s_{2n+1})+ d(A,B). \end{aligned}$$
We take \(n \rightarrow\infty\) in the above inequality, and we get
$$ \lim_{n\rightarrow\infty}d(z,fs_{n})= d(z,B)=d(A,B). $$
(17)
Since the pair \((A,B)\) has the Vproperty, there is a \(p \in B\) such that \(d(z,p) = d(A,B)\) and so \(z \in A_{0} \). Moreover, since \(f(A_{0}) \subset B_{0}\), there is a \(q \in A\) such that
$$ d(q,fz) = d(A,B). $$
(18)
Furthermore \(d(s_{2n+2},gs_{2n+1}) = d(A,B)\) for every \(n\in\mathbb{N}\).
Since \((f,g)\) is a generalized proximal Ccontraction pair, we have
$$d(q,s_{2n+2}) \leq\frac{1}{2}\bigl(d(z,s_{2n+2})+d(s_{2n+1},q) \bigr)\psi \bigl({d(z,s_{2n+2}),d(s_{2n+1},q)}\bigr). $$
Letting \(n \rightarrow\infty\) in the above inequality, we have
$$d(q,z)\leq\frac{1}{2}d(z,q)\psi\bigl(d(z,q),0\bigr). $$
Thus \(d(z,q)=0\), which implies that \(z=q\). Then, by (18), z is a best proximity point of f.
Similarly, it is easy to prove that z is a best proximity point of g. Then z is a common best proximity point of the functions f and g. By the proof of Theorem 2.4 we conclude that f and g have unique common best proximity point. □
Theorem 2.9
Let
\((X,d)\)
be a metric space and
\(\emptyset\neq A,B\subset X\). Let
A
be complete, the pair
\((A,B)\)
have the
Vproperty and
\(A_{0}\)
be a nonempty set. Moreover, suppose that the nonself functions
\(f,g: A \rightarrow B\)
satisfy:

(i)
\(f(A_{0})\subset B_{0}\)
and
\(g(A_{0}) \subset B_{0}\),

(ii)
\((f,g)\)
is an
αproximal
C1contraction pair or an
αproximal
C2contraction pair,

(iii)
\((f,g)\)
is a triangular
αproximal admissible pair,

(iv)
there exist
\(s_{0}, s_{1} \in A_{0}\)
such that
\(d(s_{1},fs_{0})=d(A,B)\), \(\alpha(s_{1},s_{0}) \ge1\),

(v)
if
\(\{s_{n}\}\)
is a sequence in
A
such that
\(\alpha (s_{n},s_{n+1})\ge1\)
and
\(s_{n} \rightarrow s_{0}\)
as
\(n \rightarrow\infty \), then
\(\alpha(s_{n},s_{0})\ge1\)
for all
\(n\in\mathbb{N} \cup\{0\}\).
Then
f
and
g
have a common best proximity point. Moreover, if
\(z,w \in X\)
are common best proximity points and
\(\alpha (z,w)\ge1\), then the common best proximity point is unique.
Proof
We can derive from the proof of Theorem 2.5 that there exist a sequence \(\{s_{n}\}\) and z in A such that \(s_{n} \rightarrow z\) and \(\alpha(s_{n},s_{n+1}) \ge1\). Also, by (v), \(\alpha(s_{n},z)\ge1\) for every \(n\in\mathbb{N} \cup\{ 0\}\). Let \(s=q\), \(t=s_{2n+2}\), \(p=z\), \(q=s_{2n+1}\). If \((f,g)\) is an αproximal C1contraction pair or αproximal C2contraction pair, then \((f,g)\) is a generalized proximal Ccontraction pair. Then by the proof of the last theorem, z is a common best proximity of f and g. □
The following corollary is an immediate consequence of the main theorem of this section.
Corollary 2.10
Let
\((X,d)\)
be a complete metric space and
\(f,g: X \rightarrow X\). Moreover, let the self functions
f
and
g
satisfy:

(i)
\((f,g)\)
is a triangular
αadmissible pair,

(ii)
there exists an
\(s_{0} \in X\)
such that
\(\alpha(s_{0},fs_{0})\ge1\),

(iii)
if
\(\{s_{n}\}\)
is a sequence in
A
such that
\(\alpha (s_{n},s_{n+1})\ge1\)
and
\(s_{n} \rightarrow s_{0} \in A\)
as
\(n \rightarrow \infty\), then
\(\alpha(s_{n},s_{0})\ge1\)
for all
\(n \in\mathbb{N}\cup\{ 0\}\),

(iv)
for all
\(x,y \in X\), \(\alpha(p,q) d(fp,gq) \leq\frac{1}{2}(d(p,gq) + d(q,fp))\psi (d(p,gq),d(q,fp))\) (or
\((\alpha(p,q)+l)^{d(fp,gq)} \leq(l+1)^{\frac{1}{2}(d(p,gq) + d(q,fp))\psi(d(p,gq),d(q,fp))} \)).
Then
f
and
g
have a common fixed point. Moreover, if
\(x,y \in X\)
are common fixed points and
\(\alpha(x,y)\ge1\), then the common fixed point of
f
and
g
is unique, that is, \(x=y\).
In order to illustrate our results, we present the following example.
Example 2.11
Consider \(X=\mathbb{R}\) with the usual metric \(d(x,y)=xy\), \(A=\{4, 0, 4\}\), and \(B=\{2, 1, 2\}\). Then A and B are nonempty closed subsets of X with \(d(A,B)=1\), \(A_{0}=\{0\}\), and \(B_{0}=\{1\}\). We define \(f,g: A \rightarrow B\) by
$$f(0)=1,\qquad f(10)=2,\qquad f(10)=2\quad \text{and}\quad g(x)=1\quad \forall x\in A, $$
and \(\psi: [0,\infty) \times[0,\infty) \rightarrow[0,\infty)\) by \(\psi(s,t)=st\).
It is immediate to see that \(f(A_{0})\subset B_{0}\) and \(g(A_{0})\subset B_{0}\). Also, if
$$ \left \{ \textstyle\begin{array}{l} d(u,fp)=d(A,B)=1, \\ d(v,gq)=d(A,B)=1, \end{array}\displaystyle \right . $$
then \(u=v=p=0\) and \(q \in A\) and therefore (1) is satisfied. Hence all the conditions of Theorem 2.4 hold for this example and clearly 0 is the unique common best proximity of f and g.
Example 2.12
Let \(X = [0,1]\times[0,1]\) and d be the Euclidean metric. Let
$$A:=\bigl\{ (0,m): 0 \leq m \leq1\bigr\} , \qquad B:= \bigl\{ (1,n): 0 \leq n \leq1 \bigr\} . $$
Then \(d(A,B)=1\), \(A_{0} = A\), and \(B_{0} =B\). We define \(f,g: A \rightarrow B\) by
$$f(0,m)=(1,m),\qquad g(0,m)=(1,1). $$
Define \(\alpha: A \times A \rightarrow[0,\infty)\) by
$$ \alpha(p,q)= \left \{ \textstyle\begin{array}{l@{\quad}l} 2, &\text{if }p,q\in(0,1)\times\{(0,0),(0,1)\}, \\ 0, &\text{otherwise}, \end{array}\displaystyle \right . $$
and \(\psi: [0,\infty) \times[0,\infty) \rightarrow[0,\infty)\) by
$$\psi(s,t)=\frac{1}{2}(s+t) \quad \mbox{for all }s,t \in X. $$
Then \(f(A_{0})\subset B_{0}\), \(g(A_{0})\subset B_{0}\). Assume that
$$ \left \{ \textstyle\begin{array}{l} d(u,fp)=d(A,B)=1, \\ d(v,gq)=d(A,B)=1. \end{array}\displaystyle \right . $$
Hence, \(u = p\) and \(v = (0,1)\). If \(p=(0,1)\), then \(u=v\) and (2) holds. If \(p \neq(0,1)\), then \(\alpha(p,q) = 0\) and (2) holds, which implies that \((f,g)\) is an αproximal C1contraction. Hence, all the hypotheses of Theorem 2.5 are satisfied. Moreover, if \(\{s_{n}\}\) is a sequence such that \(\alpha(s_{n},s_{n+1})\ge 1\) for every \(n \in{\mathbb{N}\cup\{0\}}\) and \(s_{n} \rightarrow s_{0}\), then \(s_{n} = (0,1)\) for all \(n \in{\mathbb{N}\cup\{0\}}\) and hence \(s_{0}=(0,1)\). Then \(\alpha(s_{n},s_{0}) \ge1\) for every \(n \in{\mathbb{N}\cup\{0\}}\). Clearly, \((A,B)\) has the Vproperty and then all the conclusions of Theorem 2.9 hold. Clearly \((0,1)\) is the unique common best proximity of f and g.
The following example shows that the triangular αproximal admissible condition for \((f,g)\) cannot be relaxed from Theorem 2.9.
Example 2.13
Let \(X = [0,1]\times[0,1]\) and d be the Euclidean metric. Let
$$A:=\bigl\{ (0,m): 0 \leq m \leq1\bigr\} , \qquad B:= \bigl\{ (1,n): 0 \leq n \leq1 \bigr\} . $$
Then \(d(A,B)=1\), \(A_{0} = A\), and \(B_{0} =B\). We define \(f,g: A \rightarrow B\) by
$$ f(0,m)= \left \{ \textstyle\begin{array}{l@{\quad}l} (1,1),& m=\frac{1}{2}, \\ (1,\frac{m}{2}),& m \ne\frac{1}{2}, \end{array}\displaystyle \right . $$
and \(g(0,m)=(1,1)\). Also we define \(\alpha: A \times A \rightarrow [0,\infty)\) by
$$ \alpha(p,q)= \left \{ \textstyle\begin{array}{l@{\quad}l} 2, &\text{if }p,q\in\{(0,\frac{1}{2})\}\times A, \\ 0, &\text{otherwise}, \end{array}\displaystyle \right . $$
and \(\psi: [0,\infty) \times[0,\infty) \rightarrow[0,\infty)\) by
$$\psi(s,t)=\frac{1}{2}(s+t)\quad \mbox{for all }s,t \in X. $$
It is easy to see that all the required hypotheses of Theorem 2.9 are satisfied unless (iii). Clearly f and g do not have a common best proximity point. It is worth noting that the pair \((f, g)\) does not have the triangular αproximal admissible property.