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 \(\lambda \in [0,1]\)
$$\begin{aligned} \lim _{n\rightarrow \infty }Ax_n = \lim _{n\rightarrow \infty }S_{\lambda }x_n =\lim _{n\rightarrow \infty }By_n = \lim _{n\rightarrow \infty }T_{\lambda }y_n = t\in M, \end{aligned}$$
(4)
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
$$\begin{aligned} d(Sx,Ty)\le \lambda \max \{d(Ax,By), d(Ax,Sx), d(By,Ty),d(Ax,Ty),d(By,Sx)\} \end{aligned}$$
(5)
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
$$\begin{aligned} d(Sx, Ty) \le \max \left\{ d(Ax, By), dist(Ax, [sx,q]),dist(By, [Ty,q]), dist(Ax, [Ty, q]), dist(By, [Sx,q])\right\} \end{aligned}$$
(6)
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
$$\begin{aligned} T_{n}(y) = W(Ty, q, \lambda _{n}) \;and \; S_{n}(x) = W(Sx, q, \lambda _{n}) \end{aligned}$$
(7)
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 \(\lambda \in [0,1]\)
$$\begin{aligned} \lim _{m \rightarrow \infty }Ax_{m} =\lim _{m \rightarrow \infty }S_{\lambda }(x_{m})= \lim _{m \rightarrow \infty }By_{m} = \lim _{m \rightarrow \infty } T_{\lambda }y_{m} =t \end{aligned}$$
(8)
for \(t \in M,\) where
$$\begin{aligned} \lim _{m \rightarrow \infty }T_{\lambda }y_{m}&= \lim _{m \rightarrow \infty }W(Ty_m, q, \lambda )\\ {\hbox{and}} \lim _{m \rightarrow \infty }S_{\lambda }x_m&= \lim _{m \rightarrow \infty }W(Sx_m, q, \lambda ) .\end{aligned}$$
Since \(\lambda _{n} \in (0,1)\), by using Eqs. (7) and (8) for each \(n \in N\), we have
$$\begin{aligned} \lim _{m \rightarrow \infty }T_{n}y_{m}&= \lim _{m \rightarrow \infty }W(Ty_{m}, q, \lambda _{n})\\&= \lim _{m \rightarrow \infty }T_{\lambda _{n}}(y_{m})\\&= t \in M. \end{aligned}$$
Thus we have
$$\begin{aligned} \lim _{m \rightarrow \infty }By_{m} = \lim _{m \rightarrow \infty }T_{n}y_{m} = t \in M. \end{aligned}$$
(9)
Similarly
$$\begin{aligned} \lim _{m \rightarrow \infty }S_{n}x_{m}&= \lim _{m \rightarrow \infty }W(Sx_{m}, q, \lambda _{n})\\&= \lim _{m \rightarrow \infty }S_{\lambda _{n}}x_{m}\\&= t \in M. \end{aligned}$$
and so
$$\begin{aligned} \lim _{m \rightarrow \infty }Ax_{m} = \lim _{m \rightarrow \infty }S_{n}x_{m} = t \in M. \end{aligned}$$
(10)
Hence in light of Eqs. (9) and (10), we obtain
$$\begin{aligned} \lim _{m \rightarrow \infty }Ax_{m} = \lim _{m \rightarrow \infty }S_{n}x_{m} = \lim _{m \rightarrow \infty }By_{m} = \lim _{m \rightarrow \infty }T_{n}y_{m} = t \in M. \end{aligned}$$
(11)
Taking in to account that A and B are qaffine and by using Property (I), we have
$$\begin{aligned} d(S_{n}Ax_{m}, AS_{n}x_{m})&= d(W(SAx_{m},q, \lambda _{n}), A(W(Sx_{m},q, \lambda _{n})))\nonumber \\&= d(W(SAx_{m},q, \lambda _{n}),W(ASx_{m},q, \lambda _{n}))\nonumber \\&\le \lambda _{n}d(SAx_{m}, ASx_{m}). \end{aligned}$$
(12)
Similarly, we can show that
$$\begin{aligned} d(T_{n}By_{m}, BT_{n}y_{m}) \le \lambda _{n}d(TBy_{m}, BTy_{m}). \end{aligned}$$
(13)
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
$$\begin{aligned} \lim _{m \rightarrow \infty }d(BTy_{m}, TBy_{m})&= 0\\ {\hbox{and}} \lim _{m \rightarrow \infty } d(ASx_{m}, SAx_{m})&= 0. \end{aligned}$$
Taking limit \(m \rightarrow \infty\) in Eqs. (12) and (13), we obtain
$$\begin{aligned} \lim _{m \rightarrow \infty }d(BT_{n}y_{m}, T_{n}By_{m}) = \lim _{m \rightarrow \infty } d(AS_{n}x_{m}, S_{n}Ax_{m}) = 0. \end{aligned}$$
(14)
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
$$\begin{aligned} d(S_{n}x, T_{n}y)&= d(W(Sx, q, \lambda _{n}),W(Ty, q, \lambda _{n})) \nonumber \\&\le \lambda _{n}d(Sx, Ty)\nonumber \\&\le \lambda _{n}\max \left\{ d(Ax, By), dist(Ax, [Sx,q]),dist(By,[Ty,q]),dist(Ax,[Ty,q]),dist(By,[Sx,q])\right\} \nonumber \\&\le \lambda _{n}\max \left\{ d(Ax, By), d(Ax, S_{n}x), d(By, T_{n}y), d(Ax, T_{n}y), d(By, S_{n}x)\right\} \end{aligned}$$
(15)
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
$$\begin{aligned} Ax_{n} = S_{n}x_{n}=Bx_{n}= T_{n}x_{n}=x_{n}. \end{aligned}$$
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
$$\begin{aligned}&x_{m} = T_{m}x_{m} = W(Tx_{m},q,\lambda _{m}) \rightarrow z\hbox { as }m \rightarrow \infty \\&\hbox {and }x_{m} = S_{m}x_{m} = W(Sx_{m},q,\lambda _{m}) \rightarrow y \hbox { as }m \rightarrow \infty. \end{aligned}$$
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
$$\begin{aligned} Az = Tz = Sz = Bz = z. \end{aligned}$$
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
$$\begin{aligned} d(Sx, Ty) \le \max \left\{ d(Ax, By), dist(Ax, [sx,q]),dist(By, [Ty,q]), \frac{1}{2}[dist(Ax, [Ty, q])+ dist(By, [Sx,q])]\right\} \end{aligned}$$
(16)
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
$$\begin{aligned} d(Sx, Ty) \le \max \left\{ d(Ax, By), dist(Ax, [sx,q]),dist(By, [Ty,q]), dist(Ax, [Ty, q]), dist(By, [Sx,q])\right\} \end{aligned}$$
(17)
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:
$$\begin{aligned} A(x)&= {\left\{ \begin{array}{ll} \frac{1}{3} &{} \quad {\text {if}}\;1\le x\le \frac{1}{3} \\ \frac{5}{3}4x &{} \quad {\text {if}}\;\frac{1}{3}\le x\le \frac{2}{3} \end{array}\right. } \quad \text {and}\quad S(x) = {\left\{ \begin{array}{ll} \frac{1}{3} &{} {\text {if}}\;1\le x \le \frac{1}{3} \\ \frac{x}{2}+\frac{1}{6} &{}\quad {\text {if}}\;\frac{1}{3}\le x\le \frac{2}{3} \end{array}\right. }\\ B(x)&= {\left\{ \begin{array}{ll} \frac{1}{3} &{}\quad {\text {if}}\;1\le x\le \frac{1}{3} \\ 12x &{}\quad {\text {if}}\;\frac{1}{3}\le x\le \frac{2}{3} \end{array}\right. } \quad \text {and}\quad T(x) = {\left\{ \begin{array}{ll} \frac{1}{3} &{}\quad {\text {if}}\;1\le x \le \frac{1}{3} \\ \frac{x}{4}+\frac{1}{4} & \quad {\text {if}}\;\frac{1}{3}\le x\le \frac{2}{3}. \end{array}\right.} \end{aligned}$$
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
$$\begin{aligned} A\left( W\left( x, \frac{1}{3}, \lambda \right) \right)&= \frac{5}{3}  4\left( W\left( x, \frac{1}{3}, \lambda \right) \right) \\&= \frac{5}{3} 4\lambda x4(1\lambda )\frac{1}{3}\\&= \frac{1}{3} 4\lambda x + \frac{4}{3}\lambda \\&= \frac{1}{3} + 4\lambda \left( \frac{1}{3}x\right) \\ \end{aligned}$$
and
$$\begin{aligned} W\left( Ax, \frac{1}{3}, \lambda \right)&= W\left( \frac{5}{3}  4x, \frac{1}{3}, \lambda \right) \\&= \lambda \left( \frac{5}{3} 4x\right) +(1\lambda )\frac{1}{3}\\&= \frac{5}{3}\lambda 4\lambda x + \frac{1}{3}\lambda \left( \frac{1}{3}\right) \\&= \frac{1}{3} + 4\lambda \left( \frac{1}{3}x\right) . \end{aligned}$$
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
$$\begin{aligned} B\left( W\left( x, \frac{1}{3}, \lambda \right) \right)&= 1  2\left( W\left( x, \frac{1}{3}, \lambda \right) \right) \\&= 12\lambda x2(1\lambda )\frac{1}{3}\\&= \frac{1}{3} + 2\lambda \left( \frac{1}{3}x\right) \end{aligned}$$
and
$$\begin{aligned} W\left( Bx, \frac{1}{3}, \lambda \right)&= \lambda (1  2x)+(1\lambda ) \frac{1}{3}\\&= \lambda 2\lambda x+\frac{1}{3}\frac{1}{3}\lambda \\&= \frac{1}{3} + 2\lambda \left( \frac{1}{3}x\right) . \end{aligned}$$
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
$$\begin{aligned}&\limsup _{n \rightarrow \infty }S_{\lambda }x_{n} = W \left( \frac{1}{3}, \frac{1}{3}, \lambda \right) = \frac{1}{3} = \lim _{n \rightarrow \infty }Ax_{n}\\&\hbox { and }\limsup _{n \rightarrow \infty }T_{\lambda }y_{n} = W \left( \frac{1}{3}, \frac{1}{3}, \lambda \right) = \frac{1}{3} = \lim _{n \rightarrow \infty }By_{n}. \end{aligned}$$
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
$$\begin{aligned}&\lim _{n \rightarrow \infty }Ax_{n} = \lim _{n \rightarrow \infty }Sx_{n} = t \quad {\hbox {for some}} \;t {\hbox { and }}\\&\lim _{n \rightarrow \infty }By_{n} = \lim _{n \rightarrow \infty }Ty_{n} = k \quad {\hbox {for some}} \; k. \end{aligned}$$
Then t and k lies in the closure of A(M), S(M) and B(M), T(M) respectively, where
$$\begin{aligned}&Cl(A(M)) = \left[ 1, \frac{1}{3}\right] \quad {\text {and}}\quad Cl(S(M)) = \left[ \frac{1}{3}, \frac{1}{2}\right] \\&Cl(B(M)) = \left[ \frac{1}{3}, \frac{1}{3}\right] \quad {\text {and}}\quad Cl(T(M)) = \left[ \frac{1}{3}, \frac{5}{12}\right]. \end{aligned}$$
So \(t=k=\frac{1}{3}.\)
Therefore, by using the continuity of A, B, S and T, we have
$$\begin{aligned}&\lim _{n \rightarrow \infty }ASx_{n} = A\lim _{n \rightarrow \infty }Sx_{n} = A\left( \frac{1}{3}\right) = \frac{1}{3}\\&\lim _{n \rightarrow \infty }SAx_{n} = S\lim _{n \rightarrow \infty }Ax_{n} = S\left( \frac{1}{3}\right) = \frac{1}{3}. \end{aligned}$$
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
$$\begin{aligned} M \bigcap F(A) \bigcap F(B) \bigcap F(S) \bigcap F(T) \ne \phi. \end{aligned}$$
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)).