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Existence of common fixed point and best proximity point for generalized nonexpansive type maps in convex metric space
SpringerPlus volume 5, Article number: 1940 (2016)
Abstract
Here, we extend the notion of (E.A.) property in a convex metric space defined by Kumar and Rathee (Fixed Point Theory Appl 1–14, 2014) by introducing a new class of selfmaps which satisfies the common property (E.A.) in the context of convex metric space and ensure the existence of common fixed point for this newly introduced class of selfmaps. Also, we guarantee the existence of common best proximity points for this class of maps satisfying generalized nonexpansive type condition. We furnish an example in support of the proved results.
Introduction and preliminaries
In 2002, Aamri and El Moutawakil (2002) obtained the notion of (E.A.) property for a single pair of selfmaps. In the recent past, Liu et al. (2005) introduced common property (E.A.) and extend the concept of (E.A.) property defined by Aamri and El Moutawakil (2002) to two pairs of selfmaps.
Definition 1
(Liu et al. 2005) Two pairs (A, S) and (B, T) of selfmaps of a metric space (X, d) are said to satisfy the common property (E.A.) if there exist two sequences \(\{x_n\}\) and \(\{y_{n}\}\) in X such that
In 1970, Takahashi (1970) introduced the notion of convexity into metric space and proved several fixed point theorems for nonexpansive mappings in the context of convex metric space. Then after, Beg and Azam (1987), Fu and Huang (1991), Ciric (1993), and many others have obtained fixed point theorems in convex metric spaces. Very recently, Kumar and Rathee (2014) defined the concept of (E.A.) property in the setup of convex metric space and ensure the existence of common fixed point for a pair of maps satisfying this property by omitting the assumption that the range of one map is contained in other.
In the present work, we define the concept of common property (E.A.) in the context of convex metric space and extend the results of Kumar and Rathee (2014) to four selfmaps by utilizing this newly introduced concept. Further, we ensure the existence of common best proximity point for generalized nonexpansive type maps.
Before going to the main work, we recall some standard notations, known definitions and results which is required in the sequel. Throughout this paper, \({\mathbb {N}}\) and \({\mathbb {R}}\) denote the set of natural numbers and the set of real numbers, respectively.
Definition 2
(Takahashi 1970) Let (X, d) be a metric space. A continuous mapping \(W: X \times X \times [0,1] \rightarrow X\) is called a convex structure on X if, for all \(x,y \in X\) and \(\lambda \in [0,1]\), we have
for all \(u\in X.\)
A metric space (X, d) equipped with a convex structure is called a convex metric space.
Definition 3
A subset M of a convex metric space (X, d) is called a convex set (Takahashi 1970) if \(W(x,y,\lambda ) \in M\) for all \(x,y\in M\) and \(\lambda \in [0,1]\). The set M is said to be qstarshaped (Guay et al. 1982) if there exists \(q\in M\) such that \(W(x,q,\lambda ) \in M\) for all \(x\in M\) and \(\lambda \in [0,1]\).
Clearly, each convex set M is starshaped with respect to any \(q\in M\) but the converse assertion is not true. Thus, the class of starshaped set properly contains the class of convex set.
Definition 4
(Guay et al. 1982) A convex metric space (X, d) is said to satisfy the Property (I), if for all \(x, y, z \in X\) and \(\lambda \in [0,1]\),
A normed linear space X and each of its convex subset are simple examples of convex metric spaces with W given by \(W(x,y, \lambda ) = \lambda x + (1\lambda )y\) for all \(x,y\in X\) and \(0\le \lambda \le 1\). Also, Property (I) is always satisfied in a normed linear space. There are many convex metric spaces which are not normed linear space, for details (see Guay et al. 1982; Takahashi 1970).
Definition 5
Let (X, d) be a convex metric space and M be a subset of X. A mapping \(I: M \rightarrow X\) is said to be

(1)
affine (AlThagafi and Shahzad 2006; Huang and Li 1996), if M is convex and \(I(W(x,y,\lambda ))= W(Ix,Iy,\lambda )\) for all \(x,y\in M\) and \(\lambda \in [0,1]\).

(2)
qaffine (AlThagafi and Shahzad 2006; Kumar and Rathee 2014), if M is qstarshaped and \(I(W(x,q,\lambda ))= W(Ix,q,\lambda )\) for all \(x\in M\) and \(\lambda \in [0,1]\).
Definition 6
Let (X, d) be a metric space, M a nonempty subset of X and let I and T be selfmaps of M. A point \(x \in M\) is a coincidence point (common fixed point) of I and T if \(Ix = Tx (Ix=Tx=x)\). The pair \(\{I,T\}\) is called

(1)
commuting if \(ITx=TIx\) for all \(x \in M\).

(2)
compatible (Jungck 1986) if \(\lim _{n\rightarrow \infty }d(ITx_m,TIx_m)=0,\) whenever \(\{x_n\}\) is a sequence in X such that \(\lim _{n\rightarrow \infty }Ix_n =\lim _{n\rightarrow \infty }Tx_n=t\in X\).
For more details about these classes, one can refer to (see Agarwal et al. 2014). In 1998, Pant (1998) defined the concept of reciprocal continuity as follows.
Definition 7
(Pant 1998) Let (X, d) be a metric space and \(I,T: X\rightarrow X\). Then the pair (I, T) is said to be reciprocally continuous if
whenever \(\{x_n\}\) is a sequence in X such that \(\lim _{n\rightarrow \infty }Ix_n =\lim _{n\rightarrow \infty }Tx_n=t\in X\).
It is easy to see that if I and T are continuous, then the pair (I, T) is reciprocally continuous but the converse is not true in general (see Imdad et al. 2011, Example 2.3). Moreover, in the setting of common fixed point theorems for compatible pairs of selfmappings satisfying some contractive conditions, continuity of one of the mappings implies their reciprocal continuity.
Definition 8
(Bouhadjera and GodetThobie 2009) Let I and T be two selfmaps of a metric space (X, d). Then the pair (I, T) is said to be subcompatible if there exists a sequence \(\{x_n\}\) such that
Obviously, compatible maps which satisfy (E.A.) property are subcompatible but the converse statement does not hold in general (see Rouzkard et al. 2012, Example 2.5)
Definition 9
(Kumar and Rathee 2014) Let M be a qstarshaped subset of a convex metric space (X, d) and let \(I,T:M\rightarrow M\) with \(q\in F(I)\). The pair (I, T) is said to satisfy (E.A.) property with respect to q if there exists a sequence \(\{x_n\}\) in M such that for all \(\lambda \in [0,1]\)
where \(T_{\lambda }x = W(Tx,q,\lambda )\).
Obviously, if the pair (I, T) satisfy (E.A.) property with respect to q, then I and T satisfy (E.A.) property but converse assertion is not necessarily true (see Kumar and Rathee 2014, Example 12).
Main results
We start to this section with following definition.
Definition 10
Let M be a qstarshaped subset of a convex metric space (X, d) and let A, B, S and \(T:M\rightarrow M.\) Two pairs (A, S) and (B, T) are said to satisfy common property (E.A.) with respect to q if there exist two sequences \(\{x_n\}\) and \(\{y_n\}\) in M such that for all \(\lambda \in [0,1]\)
where \(S_{\lambda }x = W(Sx,q,\lambda )\) and \(T_{\lambda }y = W(Ty, q, \lambda )\)
Remark 11
In Definition 10, if \(A=B\) and \(S=T\), then Definition 9 can be obtained as a particular case of Definition 10. Therefore the common property (E.A.) defined here extends the notion of (E.A.) property in convex metric space defined by Kumar and Rathee (2014).
The following Lemma is particular case of the Theorem 4.1 of Chauhan and Pant (2014).
Lemma 12
Let A, B, S and T be selfmaps of a metric space (X, d). If the pairs (A, S) and (B, T) are subcompatible, reciprocally continuous and satisfy
for some \(\lambda \in (0,1)\) and all \(x,y\in X\). Then S and T have a unique common fixed point in X.
Now, we start with the following theorem.
Theorem 13
Let M be a nonempty qstarshaped subset of a convex metric space (X, d) with Property (I) and let A, B, S and T be continuous selfmaps on M such that the pair (A, S) and (B, T) satisfying common property (E.A.) w.r.t. q. Assume that A and B are qaffine, M is compact. If A, B, S and T are compatible and satisfy the inequality
for all \(x, y \in M\), then \(M \bigcap F(A) \bigcap F(B) \bigcap F(S) \bigcap F(T)\neq\phi\).
Proof
For each \(n \in N\), we define \(T_{n} : M \rightarrow M\) and \(S_n: M \rightarrow M\) by
for all \(x \in M\), where \(\lambda _{n}\) is a sequence in (0, 1) such that \(\lambda _{n} \rightarrow 1.\) Now we have to prove that for each \(n \in N\), the pair \((S_{n}, A)\) and \((T_{n}, B)\) are subcompatible. Since A, B, S and T are satisfying the common property (E.A.) w.r.t. q, there exist two sequences \(\left\{ x_{m}\right\}\) and \(\left\{ y_{m}\right\}\) in M such that for all \(\lambda \in [0,1]\)
for \(t \in M,\) where
Since \(\lambda _{n} \in (0,1)\), by using Eqs. (7) and (8) for each \(n \in N\), we have
Thus we have
Similarly
and so
Hence in light of Eqs. (9) and (10), we obtain
Taking in to account that A and B are qaffine and by using Property (I), we have
Similarly, we can show that
As (A, S) and (B, T) satisfy (E.A.) property w.r.t.q then these pairs also satisfy (E.A.) property and hence by using the compatibility of A, S, B,and T we get
Taking limit \(m \rightarrow \infty\) in Eqs. (12) and (13), we obtain
Taken into account Eqs. (11) and (14), it follows that \((T_{n}, B)\) and \((S_{n}, A)\) are subcompatible for each \(n \in N\). Since A, B, S and T are continuous for each \(n \in N\), the pair \((S_{n}, A)\) and \((T_{n},B)\) are reciprocally continuous.
By using equation (6) and Property (I), we get that
for each \(x, y \in M\) and \(\lambda _{n} \in (0, 1)\). By Lemma 12, for each \(n \in N\), there exists \(x_{n} \in M\) such that
Now by taking the compactness of M, we know continuous image of compact set is compact so T(M) and S(M) are compact and every compact set is sequentially compact. Therefore there exist subsequences \(\left\{ Tx_{m}\right\}\) of \(\left\{ Tx_{n}\right\}\) and \(\left\{ Sx_{m}\right\}\) of \(\left\{ Sx_{n}\right\}\) such that \(\lim _{m \rightarrow \infty }Tx_{m} =z\) and \(\lim _{m \rightarrow \infty }Sx_{m} =y.\)
Now, we have to prove that \(y =z.\)
On the contrary suppose that \(y \ne z\), then we have
This implies that the sequence \(\left\{ x_{m}\right\}\) converges to two points which is contradiction. Hence \(y=z\).
Since \(x_{m} \rightarrow z\) as \(m \rightarrow \infty\) and the mappings A, B, S and T are continuous, it follows
So, z is common fixed point of A, B, S and T.
This implies that \(M \bigcap F(A) \bigcap F(B) \bigcap F(S) \bigcap F(T) \ne \phi\). \(\square\)
Corollary 14
Let M be a nonempty qstarshaped subset of a convex metric space (X, d) with Property (I) and let A, B, S and T be continuous selfmaps on M such that the pair (A, S) and (B, T) satisfying common property (E.A.) w.r.t. q. Assume that A and B are qaffine, M is compact. If A, B, S and T are compatible and satisfy the inequality
for all \(x, y \in M\), then \(M \bigcap F(A) \bigcap F(B) \bigcap F(S) \bigcap F(T) \ne \phi.\)
Corollary 15
Let M be a nonempty qstarshaped subset of a convex metric space (X, d) with \(Property \;(I)\) and let A, B, S and T be continuous selfmaps on M such that the pair (A, S) and (B, T) satisfying common property (E.A.) w.r.t. q. Assume that A and B are qaffine, M is compact. If A, B, S and T are Rsubweakly commuting and satisfy the inequality
for all \(x, y \in M,\) then \(M \bigcap F(A) \bigcap F(B) \bigcap F(S) \bigcap F(T) \ne \phi.\)
Now we present an example in support of our theorem.
Example 16
Let \(X=R\) endowed with usual metric and let \(M= \left[ 1,\frac{2}{3}\right]\). Define A, B, S and \(T:M \rightarrow M\) by:
Then (X, d) is a convex metric space with the convex structure \(W(x, y, \lambda ) = (\lambda )x + (1\lambda )y\).
We have to check the following:
 (i):

A and B are qaffine with \(q =\frac{1}{3}\)
 (ii):

The pair (A, S) and (B, T) satisfying common property (E.A.) w.r.t. \(q = \frac{1}{3}\).
 (iii):

A, B, S and T are compatible.
Proof
(i) If \(x \in \left[ 1, \frac{1}{3}\right]\), then \(W\left( x, \frac{1}{3}, \lambda \right) = (\lambda )x +(1  \lambda )\frac{1}{3} \in \left[ 1, \frac{1}{3}\right]\).
That implies \(A\left( W\left( x, \frac{1}{3}, \lambda \right) \right) = W\left( Ax, \frac{1}{3}, \lambda \right) .\)
Again, if \(x \in \left[ \frac{1}{3}, \frac{2}{3}\right]\), then \(W\left( x, \frac{1}{3}, \lambda \right) = (\lambda )x +(1  \lambda )\frac{1}{3} \in \left[ \frac{1}{3}, \frac{2}{3}\right]\), so we get
and
Thus, \(A\left( W\left( x,\frac{1}{3},\lambda \right) \right) = W\left( Ax, \frac{1}{3},\lambda \right)\) for all \(x\in M\) and hence A is qaffine with \(q=\frac{1}{3}.\)
Now we shall prove that B is qaffine with \(q = \frac{1}{3}.\)
For this, if \(x \in \left[ 1,\frac{1}{3}\right]\), then \(B\left( W\left( x, \frac{1}{3}, \lambda \right) \right) = W\left( Bx, \frac{1}{3}, \lambda \right) .\) and if \(x \in \left[ \frac{1}{3}, \frac{2}{3}\right]\), then \(W\left( x, \frac{1}{3}, \lambda \right) = (\lambda )x +(1  \lambda )\frac{1}{3} \in \left[ \frac{1}{3}, \frac{2}{3}\right]\). Therefore, we have
and
So, \(B\left( W\left( x, \frac{1}{3}, \lambda \right) \right) = W\left( Bx, \frac{1}{3}, \lambda \right)\) for each \(x\in M\). This implies that B is qaffine with \(q= \frac{1}{3}\). \(\square\)
Proof
(ii) Clearly \(A\left(\frac{1}{3}\right) =B\left(\frac{1}{3}\right) =\frac{1}{3}.\)
Consider \(x_{n} = \frac{1}{3} \frac{1}{n+2}, n \ge 1\) and \(y_{n} = \frac{1}{3} \frac{1}{3n}, n\ge 1\) then for each \(n, x_{n}\) and \(y_{n} \in [0,\frac{1}{3}]\) and for each \(\lambda \in [0,1]\), we have
This implies that the pair (A, S) and (B, T) satisfying common property (E.A.) with respect to \(q = \frac{1}{3}.\) \(\square\)
Proof
(iii) Here, we shall prove that the pairs (A, S) and (B, T) are compatible.
If \(\{x_{n}\}\) and \(\{y_{n}\}\) are two sequences in M such that
Then t and k lies in the closure of A(M), S(M) and B(M), T(M) respectively, where
So \(t=k=\frac{1}{3}.\)
Therefore, by using the continuity of A, B, S and T, we have
This implies that the pair (A, S) is compatible. Similarly we can prove that the pair (B, T) is compatible. \(\square\)
Finally we have to prove the inequality (6). There are two possibilities: (i) \(x = y\) and (ii) \(x \ne y.\)

(i)
If \(x =y\), then
Subcase (i): if \(x= y \in \left[ 1, \frac{1}{3}\right]\), then \(d(Sx, Tx) = 0.\) So the inequality holds trivially.
Subcase (ii): if \(x =y \in \left[ \frac{1}{3}, \frac{2}{3}\right]\), then
$$\begin{aligned} d(Sx, Tx)&= \left \frac{x}{2} + \frac{1}{6}\frac{x+1}{4}\right \\&= \left \frac{6x+23x3}{12}\right \\&= \left \frac{3x1}{12}\right \\&= \frac{1}{4}\left x\frac{1}{3}\right \\ d(Ax,Bx)&= \left \frac{5}{3} 4x1+2x \right \\&= \left \frac{1}{3} 2x\right \\&= 2\left x\frac{1}{3}\right. \end{aligned}$$That implies \(d(Sx,Tx) \le d(Ax,Bx).\)

(ii)
If \(x \ne y\), then
Subcase (i): if \(x \ne y \in [0, \frac{1}{3}]\), then \(d(Sx, Ty) = 0\). Inequality trivially holds.
Subcase (ii): if \(x \ne y \in \left[ \frac{1}{3}, \frac{2}{3}\right]\), then
$$\begin{aligned} d(Sx, Ty)&= \left \frac{x}{2} + \frac{1}{6}\frac{y+1}{4}\right \\&= \left \frac{6x+23y3}{12}\right \\&= \left \frac{6x3y1}{12}\right \\&= \frac{1}{2}\left x\frac{y}{2} \frac{1}{6}\right \\ d(Ax,By)&= \left \frac{5}{3} 4x1+2y \right \\&= \left \frac{2}{3} 4x +2y\right \\&= 4\left x\frac{y}{2}\frac{1}{6}\right. \end{aligned}$$This implies \(d(Sx,Ty) \le d(Ax, By).\)
Subcase (iii): if \(x \in \left[ 1,\frac{1}{3}\right]\) and \(y \in \left[ \frac{1}{3}, \frac{2}{3}\right]\), then
$$\begin{aligned} d(Sx, Ty)&= \left \frac{1}{3} \frac{y+1}{4}\right \\&= \left \frac{43y3}{12}\right \\&= \left \frac{13y}{12}\right \\&= \frac{1}{4}\left \frac{1}{3}y\right \\ d(Ax,By)&= \left \frac{1}{3} 1+2y \right \\&= \left 2y\frac{2}{3} \right \\&= 2\left \frac{1}{3}y\right. \end{aligned}$$Therefore, we get \(d(Sx,Ty) \le d(Ax, By).\)
Subcase (iv): if \(x \in \left[ \frac{1}{3}, \frac{2}{3}\right]\) and \(y \in \left[ 1, \frac{1}{3}\right]\), then
$$\begin{aligned} d(Sx, Ty)&= \left \frac{x}{2} \frac{1}{6}\right \\&= \frac{1}{2}\left x\frac{1}{3}\right \\ d(Ax,By)&= \left \frac{5}{3} 4x\frac{1}{3} \right \\&= 4\left \frac{1}{3}x\right. \end{aligned}$$So, we have \(d(Sx,Ty) \le d(Ax, By).\)
Thus, for each \(x,y \in M\), the mappings A, B, S and T satisfying the inequality (6). Also M is compact and A, B, S and T are continuous. Thus we conclude that A, B, S and T satisfying all the conditions of Theorem 13 and consequently
Here \(\frac{1}{3} \in M\) is such a common fixed point of A, B, S and T.
Remark 17
It is to be noted that, in Example 16, \(S(M) \not \subset A(M)\) and \(T(M)\not \subset B(M)\). Therefore all the existing common fixed point theorems which ensure the existence of common fixed point for the maps under the hypothesis that range of one set is contained in other are not applicable to Example 16 (see Chen and Li (2007), Rathee and Kumar (2014a, b), Shahzad (2001)).
Application to invariant approximation
For a nonempty subset M of a metric space (X, d) and \(p\in X\), an element \(y\in M\) is called a best approximation to p if \(d(p,y)=dist(p,M)\), where \(dist(p,M) = \inf \{d(p,z): z\in M\}\). The set of all best approximations to p is denoted by \(P_M(p)\).
As an application of Theorem 13, we present an invariant approximation theorems.
Theorem 18
Let A, B, S and T be selfmaps of a convex metric space (X, d) with Property (I), \(p\in F(S)\cap F(T)\cap F(A) \cap F(B)\), and M be a subset of X such that \(S(\delta M\cap M)\subseteq M\) and \(T(\delta M\cap M)\subseteq M\), where \(\delta M\) denotes the boundary of M. Suppose that \(P_M(p)\) is nonempty, q starshaped with \(A(P_M(p))\subset P_M(p)\) and \(B(P_M(p))\subset P_M(p)\) and also the maps A and B are qaffine and continuous on \(P_M(p)\). If the pairs (A, S) and (B, T) are compatible, satisfy the common property (E.A.) w.r.t. q and also satisfy the inequality for all \(x,y\in P_M(p)\cup \{p\}\)
Then A, B, S and T have a common fixed point in \(P_M(p)\), provided that \(P_M(p)\) is compact and the maps T and S are continuous on \(P_M(p)\).
Proof
Let \(x\in P_M(p)\). Then for all \(\lambda \in (0,1)\), we have
Therefore \(W(x,p,\lambda )\notin M\) for any \(\lambda \in (0,1)\) and hence \(x\in \delta M\cap M\). Thus, as \(S(\delta M\cap M)\subseteq M\) and \(T(\delta M\cap M)\subseteq M\), we have \(Tx\in M\) and \(Sx\in M\). Also, since \(Ax\in P_M(p)\) and \(p\in F(S)\cap F(T)\cap F(A) \cap F(B)\), by using Eq. (18), we get
and
Thus, \(Tx\in P_M(p)\) and \(Sx\in P_M(p)\). So A, B, S and T are selfmaps on \(P_M(p)\). In view of Theorem 13, we can say that A, B, S and T have a common fixed point in \(P_M(p)\). \(\square\)
Define \(D= P_M(p)\cap C_M^{A,B}(p)\), where \(C_M^{A,B}(p)=\{x\in M : Ax\in P_M(p)\quad \text {and}\quad Bx\in P_M(p)\}\)
Theorem 19
Let A, B, S and T be selfmaps of a convex metric space (X, d) with Property (I), \(p\in F(S)\cap F(T)\cap F(A) \cap F(B)\), and M be a subset of X such that \(S(\delta M\cap M)\subseteq M\) and \(T(\delta M\cap M)\subseteq M\), where \(\delta M\) denotes the boundary of M. Suppose that D is nonempty, qstarshaped with \(A(D)\subset D\) and \(B(D)\subset D\) and also the maps A and B are qaffine and nonexpansive on D. If the pairs (A, S) and (B, T) are compatible, satisfy the common property (E.A.) w.r.t. q and also satisfy the inequality for all \(x,y\in D\cup \{p\}\)
Then A, B, S and T have a common fixed point in \(P_M(p)\), provided that D is compact and the maps T and S are continuous on D.
Proof
Let \(x\in D\). Then by following the steps as we have done in Theorem 18, we get that \(Tx\in P_M(p)\) and \(Sx\in P_M(p)\). Since the maps A and B are nonexpansive and \(p\in F(S)\cap F(T)\cap F(A) \cap F(B)\), by using Eq. 19, we have
and
That imply ATx and \(BTx \in P_M(p)\) and hence \(Tx\in C_M^{A,B}(p)\). Similarly we can show that \(Sx\in C_M^{A,B}(p)\). Thus we can say A, B, S and T are selfmaps on D and so Theorem 13 guarantees the existence of \(z\in P_M(p)\) such that z is a common fixed point of A, B, S and T. \(\square\)
Best proximity point
First we discuss the concept of best proximity. Let \(T:A \rightarrow B\) be a map where A and B are two nonempty subsets of a metric space (X, d) and let A and B are disjoint subsets of a metric space then the equation \(Tx=x\) might have no solution. Therefore in case of nonselfmaps we are not sure about the existence of fixed point. In such a case we try to minimize the distance d(x, Tx) and a point x for which d(x, Tx) is minimum is called a best proximity point. In the recent years there have been many interesting best proximity point theorems are proved, for example, see De la Sen et al. (2013), Eldred and Veeramani (2006), Prolla (1983), Reich (1978), and Sankar Raj (2011), Sehgal and Singh (1988). In the present section we prove a new best proximity theorem for four maps but before this we recall some definitions which are required in the sequel.
Definition 20
Let (X, d) be a convex metric space and A, B be two nonempty subsets of X. A mapping \(f:A \rightarrow B\) is called pqaffine if
 (i):

A is pstarshaped set and B is qstarshaped set;
 (ii):

\(f(W(x,p, \lambda )) = W(fx,q,\lambda )\).
Definition 21
Let A and B be two nonempty subsets of convex metric space (X, d). Let A be a pstarshaped set and B be a q starshaped set. Let f, g, S, and T be four nonselfmaps from A to B. Two pairs (f, S) and (g, T) are said to satisfy common property (E.A.) with respect to q if there exists two sequences \(\{x_{n}\}\) and \(\{y_{n}\}\) in A such that for all \(\lambda \in [0,1]\)
where \(S_{\lambda }x = W(Sx,q,\lambda )\) and \(T_{\lambda }y = W(Ty,q,\lambda ).\)
Definition 22
Let (X, d) be a convex metric space and A and B be two nonempty subsets of X such that B is qstarshaped set. A pair (f, S) of two nonselfmaps from A to B is said to be proximally commuting if for some \(\lambda \in [0,1]\) whenever \(d(x,W(Su,q,\lambda ))=d(y,fu)=d(A,B) \Longrightarrow W(Sy,q, \lambda ) = fx\).
If A and B are two nonempty subsets of a metric space (X, d), we define the following two sets.
Definition 23
(Sankar Raj, preprint) If \(A_{0} \ne \phi,\) then the pair (A, B) is said to have Pproperty if and only if for any \(x_{1},x_{2} \in A_{0}\) and \(y_{1},y_{2} \in B_{0}\)
Now we presents a best proximity point theorem:
Theorem 24
Let (A, B) be a pair of nonempty, closed subsets of a convex metric space (X, d). Suppose that A is pstarshaped and B is qstasrshaped set with Property (I). Also suppose that \(A_{0}\) is closed. Let f, g, S and, T be continuous nonself maps from A to B satisfying the conditions:

(i)
Two pairs (f, S) and (g, T) satisfying common property (E.A.) w.r.t q and proximally commuting;

(ii)
\(T(A) \subseteq f(A), S(A) \subseteq g(A), f(A_{0}) \subseteq B_{0}, g(A_{0}) \subseteq B_{0}\);

(iii)
The pair (A, B) has Pproperty;

(iv)
f, g, S and, T satisfying the condition \(d(Sx,Ty) \le \max \{d(fx,gy),dist(fx,[Sx,q]),dist(gy,[Ty,q]), \frac{1}{2}[dist(fx,[Ty,q])+d(gy,[Sx,q])]\};\)

(v)
Two mappings S and T are pqaffine.
then f, g, S and T have a best proximity point.
Proof
For each \(n \in N\), we define sequences \(T_{n}: A \rightarrow B\) and \(S_{n}: A \rightarrow B\) by \(T_{n}y = W(Ty,q, \lambda _{n})\) and \(S_{n}x =W(Sx,q, \lambda _{n})\) for all \(x,y \in A\) and \(\lambda _{n}\) is a sequence in (0, 1) such that \(\lambda _{n} \rightarrow 1\)
Consider
By using Property (I) for the set B
\(\Longrightarrow d(S_{n}x, T_{n}y) \le \lambda _{n}\max \{d(fx,gy), dist(fx,S_{n}x),dist(gy,T_{n}y), \frac{1}{2}[dist(fx,T_{n}y)+ dist(gy,S_{n}x)]\}\) Now \(T(A) \subseteq f(A)\) we can prove that \(T_{n}(A) \subseteq f(A)\). For this purpose, consider
T is pqaffine and A is pstarshaped set
Similarly it can be proved that \(S_{n}(A) \subseteq f(A)\). Now \(T_{n}(A)\subseteq f(A)\) so for fixed \(x_{0} \in A\), there exists an element \(x_{1} \in A\) such that \(T_{n}x_{0} = fx_{1}\) similarly a point \(x_{2} \in A\) can be chosen such that \(S_{n}x_{1} = gx_{2}\), continuing in process, we can obtain a sequence \(\{x_{2n}\} \in A\) such that
Since \(f(A_{0}) \subseteq B_{0}\) and \(g(A_{0}) \subseteq B_{0}\), there exists \(\{u_{n}\} \in A_{0}\) such that
As the pair (A, B) has Pproperty then by Eq. (21) \(d(u_{2n},u_{2n+1}) = d(fx_{2n+1},gx_{2n+2}).\)
By using Eq. (20), we have \(d(u_{2n},u_{2n+1}) = d(T_{n}x_{2n}, S_{n}x_{2n+1}). \text{Using Property (I) and the condition (iv), we can write the expression }\)
\(\text{Hence} \hspace{0.5cm} d(u_{2n},u_{2n+1}) \le \lambda _{n}\max \{d(u_{2n},u_{2n1}), d(u_{2n},u_{2n+1}), \frac{1}{2}d(u_{2n1},u_{2n+1})\}\)
Similarly
From Eqs. (22) and (23), we obtain
This implies
Let \(m,n \in \mathbb {N}\) and \(m < n\), we have
By using Eq. (25), we come across
\(\Longrightarrow d(u_{m}, u_{n}) \rightarrow 0\) when \(m \rightarrow \infty\) this implies \(\{u_{n}\}\) is a Cauchy sequence. Since \(\{u_{n}\} \subset A_{0}\) and \(A_{0}\) is closed subset of the complete metric space (X, d), we can find \(u \in A_{0}\) such that \(\lim _{n \rightarrow \infty }u_{n} = u\).
Since (f, S) and (g, T) have common property (E.A.) with respect to q so there exists a sequence \(\{u_{m}\}\) in A such that
Since \(\lambda _{n} \in (0,1)\), we have
Thus, we have
Similarly
and so
Hence in light of Eqs. (26) and (27), we obtain
Since f, g, S and, T are continuous so \(f,g,S_{n}\) and \(T_{n}\) are continuous and \(u_{n} \rightarrow u\). Then from Eq. (28)
Since \(f(A_{0}) \subseteq B_{0}\), there exists \(x \in A_{0}\) such that
As \((f,S_{n})\) and \((g,T_{n})\) proximally commuting, so
Taking limit \(n \rightarrow \infty\) in Eqs. (29) and (31) we have
Since \(f(A_{0}) \subseteq B_{0}\), there exists \(z \in A_{0}\) such that
Because the pair (A, B) has Pproperty so \(d(x,z) = d(S_{n}u, T_{n}x)\)
This implies that \((1\lambda _{n})d(x,z) \le 0\). So, \(x=z\) and hence
Suppose that y is another best proximity point of the mappings f, g, S and T such that
Using Eqn. (29) and Pproperty for the pair (A, B), we get that \(x=y\). \(\square\)
Conclusion
In this note, we defined the common property (E.A.) in the context of convex metric space that means here we assign the algebraic structure to the common property (E.A.) that is already exists in metric space. Due to this, we have been able to obtained a set of common fixed point theorems in which to ensure the existence of common fixed points the condition of range of one set is contained in other is not required. Thus, this newly introduced concept plays a great role in solving many kinds of physical sciences problems which can be recast in terms of common fixed point problems.
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Rathee, S., Dhingra, K. & Kumar, A. Existence of common fixed point and best proximity point for generalized nonexpansive type maps in convex metric space. SpringerPlus 5, 1940 (2016). https://doi.org/10.1186/s4006401633993
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DOI: https://doi.org/10.1186/s4006401633993