# Common best proximity points results for new proximal C-contraction mappings

## Abstract

We define a new version of proximal C-contraction and prove the existence and uniqueness of a common best proximity point for a pair of non-self functions. Then we apply our main results to get some fixed point theorems and we give an example to illustrate our results.

## 1 Introduction and preliminaries

Consider a pair $$(A,B)$$ of nonempty subsets of a metric space $$(X, d)$$. Assume that f is a function from A into B. An $$w \in A$$ is said to be a best proximity point whenever $$d(w, fw) = d(A,B)$$, where $$d(A,B) = \inf\{d(s, t): s \in A, t\in B\}$$.

Best proximity point theory of non-self functions was initiated by Fan [1] and Kirk et al. [2]; see also [313]. In this paper, we generalize some results of Kumam et al. [14] to obtain some new common best proximity point theorems. Next, by an example and some fixed point results, we support our main results and show some applications of them.

### Definition 1.1

Consider non-self functions $$f_{1}, f_{2},\ldots,f_{n}: A \rightarrow B$$. We say the a point $$s\in A$$ is a common best proximity point of $$f_{1}, f_{2},\ldots,f_{n}$$ if

$$d(s,f_{1}s) = d(s,f_{2}s) =\cdots= d(s,f_{n}s) = d(A,B).$$

### Definition 1.2

([14])

Let $$(X,d)$$ be a metric space and $$\emptyset\neq A,B\subset X$$. We say the pair $$(A,B)$$ has the V-property if for every sequence $$\{ t_{n}\}$$ of B satisfying $$d(s,t_{n})\rightarrow d(s,B)$$ for some $$s\in A$$, there exists a $$t \in B$$ such that $$d(s,t)= d(s,B)$$.

## 2 Main results

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 $$f,g: A \rightarrow B$$ non-self mappings. We say that $$(f,g)$$ is a triangular α-proximal 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 $$f,g: A \rightarrow B$$ non-self mappings. We say that $$(f,g)$$ is a generalized proximal C-contraction 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 $$f,g: A \rightarrow B$$ non-self 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 α-proximal C1-contraction 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 α-proximal C2-contraction 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 non-self functions $$f,g: A \rightarrow B$$ satisfy:

1. (i)

f, g are continuous,

2. (ii)

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

3. (iii)

$$(f,g)$$ is a generalized proximal C-contraction pair,

4. (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 non-self functions $$f,g: A \rightarrow B$$ satisfy:

1. (i)

f, g are continuous,

2. (ii)

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

3. (iii)

$$(f,g)$$ is an α-proximal C1-contraction pair or an α-proximal C2-contraction pair,

4. (iv)

$$(f,g)$$ is a triangular α-proximal admissible pair,

5. (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 α-proximal 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 α-proximal C1-contraction pair or an α-proximal C2-contraction pair, then $$(f,g)$$ is a generalized proximal C-contraction 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 α-proximal 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 α-proximal 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 α-proximal C1-contraction (C2-contraction) pair $$(f,g)$$ is a generalized proximal C-contraction 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 α-admissible pair if

1. (i)

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

2. (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:

1. (i)

f and g are continuous,

2. (ii)

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

3. (iii)

$$(f,g)$$ is a triangular α-admissible pair,

4. (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 V-property, and $$A_{0}$$ be nonempty. Moreover, suppose that the non-self mappings $$f,g: A \rightarrow B$$ satisfy:

1. (i)

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

2. (ii)

$$(f,g)$$ is a generalized proximal C-contraction pair,

3. (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 V-property, 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 C-contraction 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 V-property and $$A_{0}$$ be a nonempty set. Moreover, suppose that the non-self functions $$f,g: A \rightarrow B$$ satisfy:

1. (i)

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

2. (ii)

$$(f,g)$$ is an α-proximal C1-contraction pair or an α-proximal C2-contraction pair,

3. (iii)

$$(f,g)$$ is a triangular α-proximal admissible pair,

4. (iv)

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

5. (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 α-proximal C1-contraction pair or α-proximal C2-contraction pair, then $$(f,g)$$ is a generalized proximal C-contraction 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:

1. (i)

$$(f,g)$$ is a triangular α-admissible pair,

2. (ii)

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

3. (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 $$n \in\mathbb{N}\cup\{ 0\}$$,

4. (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 $$d(x,y)=|x-y|$$, $$A=\{-4, 0, 4\}$$, 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 α-proximal C1-contraction. 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 V-property 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 α-proximal 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 α-proximal admissible property.

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Lo’lo’, P., Vaezpour, S.M. & Saadati, R. Common best proximity points results for new proximal C-contraction mappings. Fixed Point Theory Appl 2016, 56 (2016). https://doi.org/10.1186/s13663-016-0545-0