Definition 3.1
Let \((X,d)\) be a complete metric space. Then a function \(\mathcal {S}:A\rightarrow B\), where A and B are subsets of \((X,d)\), is called an \(\mathcal{S}\)function in X if it satisfies the following hypotheses:

1.
if there exists another mapping \(F:A\rightarrow B\) in \((X,d)\), then \(d(Fx,Fy) < d(\mathcal{S}x,\mathcal{S}y)\) with \(F(A_{0})\subseteq \mathcal{S}(A_{0})\);

2.
for any \(A,B \subseteq(X,d)\), if \(A_{0}\) and \(B_{0}\) are nonempty, then \(\mathcal{S}(A_{0}) \subseteq B_{0}\);

3.
for any sequence \(\{x_{m}\}\) in A, if \(\lim_{m\rightarrow\infty }x_{m}=x \in A\), then \(\lim_{m\rightarrow\infty}\mathcal {S}x_{m}=\mathcal{S}x \in B\), where \(A\subseteq X\) and \(m\in\mathbb{N}\).
Definition 3.2
Let \((X,d)\) be a Cauchy (complete) metric space. A mapping \(T:A\rightarrow B\) with \(T(A_{0})\subseteq B_{0}\) is called an \(\mathcal{S}\)contraction in \((X,d)\) if there is an \(\mathcal{S}\)function in \((X,d)\) such that
$$d(Tx,Ty) \leq\beta\bigl(d(x,y)\bigr) d(\mathcal{S}x,\mathcal{S}y) $$
for \(x,y \in A\) and \(\beta\in\mathcal{F}\).
We denote by \(\mathcal{F}\) the collection of all mappings \(\beta: [0,\infty) \rightarrow[0,1)\) such that \(\beta(t_{n}) \to1\) implies \(t_{n} \rightarrow0\) as \(n \to\infty\). The following theorem is based on the existence of a unique common best proximity point for nonselfmaps and also furnishes fixed point results in Cauchy metric spaces.
Example 3.1
Consider \(\mathbb{R}^{2}\) with Euclidean metric. Let \(A=\{0\} \times [0,\infty)\) and \(B=\{1\} \times[0,\infty)\). Let us define \(\mathcal {S}:A\rightarrow B\) and \(T:A\rightarrow B\) as
$$T(0,y)=\biggl(1,\frac{y}{3}\biggr) $$
and
$$\mathcal{S}(0,y)=\biggl(1,\frac{y}{2}\biggr). $$
Let us take \(A_{0}=A\) and \(B_{0}=B\), where
$$\beta\bigl(d\bigl((x_{1},y_{1}),(x_{2},y_{2}) \bigr)\bigr)=\frac{1}{2} \quad\mbox{for } x_{1},x_{2} \in\{0,1 \} $$
and 0 otherwise. Then
$$\begin{aligned} d(Tx,Ty) =&d\bigl(T(0,y_{1}),T(0,y_{2})\bigr) \\ =&d\biggl(\biggl(1,\frac{y_{1}}{3}\biggr),\biggl(1,\frac{y_{2}}{3}\biggr) \biggr) \\ \leq&\biggl\frac{y_{1}y_{2}}{3}\biggr \\ \leq&\biggl\frac{y_{1}y_{2}}{2}\biggr \\ =&\frac{1}{2}y_{1}y_{2} \\ =&\beta\bigl(d\bigl((x_{1},y_{1}),(x_{2},y_{2}) \bigr)\bigr)d\bigl(\mathcal{S}(0,y_{1}),\mathcal {S}(0,y_{2})\bigr). \end{aligned}$$
Here since \(x_{1},x_{2}\in\{0,1\} \), it follows that \(0\leq\frac {1}{2}=\beta(d((x_{1},y_{1}),(x_{2},y_{2})))<1\). Therefore, \(d(T(0,y_{1}), T(0,y_{2})) \leq\beta (d((x_{1},y_{1}),(x_{2},y_{2})))d(S(0,y_{1}),S(0,y_{2}))\). The given mapping T is \(\mathcal{S}\)contraction.
Theorem 3.1
Let
A
and
B
be closed (\(A,B\neq\phi\)) be subsets of a complete metric space
\((X,d)\), and let
\(T:A\rightarrow B\)
be a continuous
\(\mathcal{S}\)contraction such that for any
\(x_{0} \in A_{0}\)
such that
\(\mathcal {S}\)
and
T
commute proximally, there exists a unique common best proximity point in
A
such that
and
$$d(x,\mathcal{S}x)=d(A,B). $$
Proof
Let us take an element \(x_{0}\in A_{0}\). Since T is an \(\mathcal {S}\)contraction, so that \(T(A_{0})\subseteq\mathcal{S}(A_{0})\), we get an element \(x_{1}\in A_{0}\) such that \(Tx_{0}=\mathcal{S}x_{1}\). Again, since \(T(A_{0})\subseteq\mathcal{S}(A_{0})\), there exists an element \(x_{2}\) in \(A_{0}\) such that \(Tx_{1}=\mathcal{S}x_{2}\). Continuing in this manner inductively, we obtain a sequence \(\{x_{n}\}\) in \(A_{0}\) with
$$Tx_{n1}=\mathcal{S}x_{n} $$
for all positive integers n by using the fact \(T(A_{0})\subseteq \mathcal{S}(A_{0})\).
Since \(T(A_{0})\subseteq B_{0}\), there occurs a point \(u_{n}\) in \(A_{0}\) such that
$$d(Tx_{n},u_{n})=d(A,B) $$
for any nonnegative integer n.
Since \(Tx_{n1}=\mathcal{S}x_{n}\), it follows for any \(x_{m}\) and \(x_{n}\) that
$$\begin{aligned} d(Tx_{m},Tx_{n}) \leq& \beta d(\mathcal{S}x_{m}, \mathcal{S}x_{n}) \\ =& \beta d(Tx_{m1},Tx_{n1}). \end{aligned}$$
This shows that \(\{Tx_{n}\}\) is a Cauchy sequence and thus converges to some element y in B. Similarly, the sequence \(\{\mathcal{S}x_{n}\}\) also converges to \(y\in B\). Since the sets A and B are closed, this means that if we take any point from these sets, then it will converge in the same set. For any \(u_{k}\in A\), we have a sequence \(\{u_{k}\}\) in A, and it converges to some \(u \in A\) because A is closed.
Since \(T(A_{0})\subseteq B_{0}\), there exists a point \(u_{n}\) in \(A_{0}\) such that
$$d(Tx_{n},u_{n})=d(A,B) $$
for any \(n\in\mathbb{Z^{+}}\cup\{0\}\). So, for any \(x_{n} \in A_{0}\), it follows that
$$d(\mathcal{S}x_{n},u_{n1})=d(Tx_{n1},u_{n1})=d(A,B) $$
for any nonnegative integer n. Since \(\mathcal{S}\) and T commute proximally, we obtain
$$\mathcal{S}u_{n}=Tu_{n1} $$
for any nonnegative integer n. Therefore, \(\mathcal{S}\) and T are continuous mappings, so that \(Tu=\lim_{n\rightarrow\infty}Tu_{n1}\) and \(\mathcal{S}u=\lim_{n\rightarrow\infty}\mathcal{S}u_{n}\); hence, \(\mathcal{S}u\) and Tu are identical mappings.
Since \(T(A_{0}) \subseteq B_{0}\), there is an element \(x \in A\) such that
and
$$d(x,\mathcal{S}u)=d(A,B). $$
Again, since \(\mathcal{S}\) and T both commute proximally, \(\mathcal {S}x=Tx\). Thus, we get
$$d(Tu,Tx) \leq\beta d(\mathcal{S}u,\mathcal{S}x)=\beta d(Tu,Tx), $$
which shows that \(Tu=Tx\) and hence \(\mathcal{S}u=\mathcal{S}x\). So, we have
$$\begin{aligned}& d(x,\mathcal{S}x)=d(x,\mathcal{S}u)=d(A,B), \\& d(x,Tx)=d(x,Tu)=d(A,B). \end{aligned}$$
Thus, x is a common best proximity point of the mappings T and \(\mathcal{S}\).
Now, we have to prove the uniqueness of common optimal approximate solution. Let there exists another common best proximity point \(x^{*}\) of mappings \(\mathcal{S}\) and T. Then
$$\begin{aligned}& d\bigl(x^{*},Tx^{*}\bigr)=d(A,B), \\& d\bigl(x^{*},\mathcal{S}x^{*}\bigr)=d(A,B). \end{aligned}$$
As we know, \(\mathcal{S}\) and T commute proximally; therefore, \(\mathcal{S}x=Tx\) and \(\mathcal{S}x^{*}=Tx^{*}\). We may write
$$d\bigl(Tx,Tx^{*}\bigr) \leq\beta d\bigl(\mathcal{S}x, \mathcal{S}x^{*}\bigr)= \beta d\bigl(Tx,Tx^{*}\bigr), $$
which shows that \(Tx=Tx^{*}\). Hence, to conclude we have
$$d\bigl(x,x^{*}\bigr) \leq d(x,Tx)+d\bigl(Tx,Tx^{*}\bigr)+d \bigl(Tx^{*},x^{*}\bigr)=2d(A,B), $$
which implies that \(\frac{1}{2}d(x,x^{*}) \leq d(A,B)\). If \(d(A,B)=0\), then the uniqueness is proved. If \(d(A,B)> 0\), then it is a contradiction because we know that \(d(A,B)\) itself is a minimum distance. This completes the proof. □
Our main result asserts that if we take \(\beta(t)=k \in[0,1)\) and take selfmappings in \(A=B=X\) in Theorem 3.1, by the definition of \(\mathcal {S}\)functions and \(\mathcal{S}\)contractions we get the following fixed point result of [7] and [8].
Corollary 3.1
Let
\((X,d)\)
be a complete metric space. Define an
\(\mathcal {S}\)function as a selfmapping
\(S:X \rightarrow X\)
and an
\(\mathcal {S}\)contraction as a selfmapping
\(T:X\rightarrow X\)
that obeys the following conditions:

1.
There is a nonnegative real number
\(k <1\)
such that
$$d(Tx_{n},Tx_{n+1})\leq k d(Sx_{n},Sx_{n+1}) $$
for any
\(x_{n}\)
and
\(x_{n+1}\)
in
A.

2.
S
and
T
commute and are continuous.

3.
\(T(X)\subseteq S(X)\).
Then the mappings
S
and
T
have a unique common fixed point.
Further, if we take \(\beta(t)=k \in[0,1)\) and add two extra conditions in Theorem 3.1, then we get the main result of [7]:
Corollary 3.2
Let
A
and
B
be nonempty closed subsets of a metric space
\((X,d)\), which is a Cauchy space such that
\(A_{0}\)
and
\(B_{0}\)
are nonempty. Let
\(\mathcal{S}\)functions
\(S:A \rightarrow B\)
and
\(T:A\rightarrow B\)
satisfy the following conditions:

1.
There is a nonnegative real number
\(\beta<1\)
such that
$$d(Tx_{n},Tx_{n+1})\leq\beta d(Sx_{n},Sx_{n+1}) $$
for any
\(x_{n}\)
and
\(x_{n+1}\)
in
A.

2.
S
and
T
commute proximally, swapped proximally and continuous.

3.
\(T(A_{0})\subseteq B_{0}\), \(S(A_{0})\).

4.
A
is approximatively compact with respect to
B.
Then we have a common best proximity point
x
in
A. In addition, if
\(x^{*}\)
is another common best proximity point of
S
and
T, then
$$d\bigl(x,x^{*}\bigr) \leq2d(A,B). $$
Proof
By adding hypotheses (4) and the condition that S and T swapped proximally in our Theorem 3.1 we obtain the above wellknown result. □
Example 3.2
Let us take \(\mathbb{R}^{2}\) with Euclidean metric. Let \(A=\{ (x,y):x\leq1\}\) and \(B=\{(x,y):x\geq2\}\). Let us define \(\mathcal {S}:A\rightarrow B\) and \(T:A\rightarrow B\) as
$$T(x,y)=\biggl(4x+6,\frac{y}{5}\biggr) $$
and
$$\mathcal{S}(x,y)=\biggl(5x+7,\frac{y}{4}\biggr). $$
We get \(d(A,B)=1\), \(A_{0}=\{(1,y): y\in[0,\infty)\}\) and \(B_{0}=\{ (2,y): y\in[0,\infty)\}\), where
$$\beta\bigl(d\bigl((x_{1},y_{1}),(x_{2},y_{2}) \bigr)\bigr)=\frac{4}{5} \quad\mbox{for } x_{1},x_{2}\leq1 $$
and 0 otherwise.
Hence, given that nonselfmappings commute proximally, we also have
$$\begin{aligned} d(Tx,Ty) =&d\bigl(T(x_{1},y_{1}),T(x_{2},y_{2}) \bigr) \\ =&d\biggl(\biggl(4x_{1}+6,\frac{y_{1}}{5}\biggr), \biggl(4x_{2}+6,\frac{y_{2}}{5}\biggr)\biggr) \\ \leq&\bigl(4x_{1}+4x_{2})\bigr+\biggl\biggl(\frac{y_{1}y_{2}}{5}\biggr)\biggr \\ \leq&\frac{4}{5}\biggl(\bigl(5x_{1}+5x_{2}+77)\bigr+\biggl\biggl( \frac{y_{1}}{4}\frac {y_{2}}{4}\biggr)\biggr\biggr) \\ =&\frac{4}{5}d\biggl(\biggl(5x_{1}+7,\frac{y_{1}}{4} \biggr),\biggl(5x_{2}+7,\frac {y_{2}}{4}\biggr)\biggr) \\ =&\beta\bigl(d\bigl((x_{1},y_{1}),(x_{2},y_{2}) \bigr)\bigr)d\bigl(\mathcal{S}(x_{1},y_{1}),\mathcal {S}(x_{2},y_{2})\bigr), \end{aligned}$$
where since \(x_{1},x_{2}\leq1\), \(0\leq\frac{4}{5}=\beta (d((x_{1},y_{1}),(x_{2},y_{2})))<1\). Therefore, \(d(T(x_{1},y_{1}),T(x_{2},y_{2})) \leq\beta (d((x_{1},y_{1}),(x_{2},y_{2})))d(S(x_{1},y_{1}),S(x_{2},y_{2}))\). The given mapping T is an \(\mathcal{S}\)contraction. Furthermore, all other hypotheses of Theorem 3.1 are also satisfied, so that there exists a unique common best proximity point for the nonselfmappings \(\mathcal{S}\) and T, which is \((1,0)\).
Remark 3.1
Replacing condition (1) in Corollary 3.2 with the condition that T dominates S proximally, we get the main result of [8].