Introduction and preliminaries

Fixed point theory constitutes an important and the core part of the subject of nonlinear functional analysis and is useful for proving the existence theorems for nonlinear differential and integral equations. The Banach contraction principle [3] is the simplest and one of the most versatile elementary results in fixed point theory, which is a very popular tool for solving existence problems in many branches of mathematical analysis. Several authors have extended the Banach’s fixed point theorem in various ways. The family of contraction mappings was introduced and studied by \(\acute{\text {C}}\)iri\(\acute{\text {c}}\) [7] and Taskovi\(\acute{\text {c}}\) [11]. Also in the process, the study of existence of common fixed point for finite and infinite family of self-mapping has been carried out by many authors. For example, one may refer [1, 2, 4–6, 12–14].

Recently, some new results for the existence and uniqueness of fixed points were presented for the cases of partially ordered metric spaces, cone metric spaces and fuzzy metric spaces (for example, see [1, 15–18]). Also, the study of common fixed points for a family of contractive type maps has been paid attention, and many interesting fixed point results have been obtained (for example, see [2, 7–11]).

The aim of this paper is to define some new conditions of common contractivity for an infinite family of mappings and give some new results on the existence and uniqueness of common fixed points in the setting of complete metric space.

Here, we state some known definitions and facts. We refer for more details to [1, 7].

Definition 1

Let X be a nonempty set and let \(\{T_{n}\}\) be a family of self-mappings on X. A point \(x_{0}\in X\) is called a common fixed point for this family iff \(T_{n}(x_{0})=x_{0},\) for each \(n\in N\).

The following interesting theorem was given by \(\acute{\text {C}}\)iri\(\acute{\text {c}}\) [7] for a family of generalized contractions.

Theorem 1

Let (X, d) be a complete metric space and let \(\{T_{\alpha }\}_{\alpha \in J}\) be a family of self-mappings of X. If there exists fixed \(\beta \in J\) such that for each \(\alpha \in J\):

$$\begin{aligned} d(T_{\alpha }x, T_{\beta }y) \le \lambda ~~ \text {max} \{ d(x,y), d(x,T_{\alpha }x), d(y,T_{\beta }y), \frac{1}{2}[d(x,T_{\beta }y)+d(y,T_{\alpha }x)]\}, \end{aligned}$$
(1)

for some \(\lambda =\lambda (\alpha )\in (0,1)\) and all x,  \(y \in X,\) then all \(T_{\alpha }\) have a unique common fixed point, which is a unique fixed point of each \(T_{\alpha },\) \(\alpha \in J.\)

Common fixed point theorems for a family of mappings

In this section, we prove existence of a unique common fixed point for a family of contractive type self-maps on a complete metric space.

Theorem 2

Let (X, d) be a complete metric space and \(0\le a_{i,j} ~~ (i,j=1,2,...)\) satisfy

  1. (i)

    for each j, \(\overline{\lim }_{i\longrightarrow \infty }a_{i, j}<1\)

  2. (ii)

    \(\sum _{n=1}^{\infty }A_{n}<\infty\) where \(A_{n}=\Pi _{i=1}^{n}\frac{a_{i,i+1}}{1-a_{i,i+1}}.\)

If \(\{T_{n}\}\) is a sequence of self-maps on X satisfying

$$\begin{aligned} d(T_{i}x,T_{j}y) \le a_{i,j} [d(x,T_{j}y)+d(y,T_{i}x)], \end{aligned}$$
(2)

for \(x, y\in X; i,j=1,2,...\) with \(x \ne y\) and \(i\ne j\) then all \(T_{n}\)s have a unique common fixed point in X.

Proof

For any \(x_{0} \in X,\) let \(x_{n}= T_{n}(x_{n-1}), n = 1, 2, \ldots\), then using (2.1) we get

$$\begin{aligned} d(x_{1},x_{2})=d(T_{1}(x_{0}),T_{2}(x_{1}))& \le a_{1,2} [d(x_{0},T_{2}(x_{1}))+d(x_{1},T_{1}(x_{0}))]\\ &\le a_{1,2} [d(x_{0},x_{2})+d(x_{1},x_{1})]\\ &= a_{1,2} d(x_{0},x_{2})\\ &\le a_{1,2} [d(x_{0},x_{1})+d(x_{1},x_{2})] \end{aligned}$$

which implies

$$\begin{aligned} (1-a_{1,2}) d(x_{1},x_{2}) \le a_{1,2} d(x_{0},x_{1}). \end{aligned}$$

So

$$\begin{aligned} d(x_{1},x_{2}) \le \frac{a_{1,2}}{1-a_{1,2}} d(x_{0},x_{1}). \end{aligned}$$

Also we have,

$$\begin{aligned} d(x_{2},x_{3})=d(T_{2}(x_{1}),T_{3}(x_{2}))\le & {} a_{2,3} [d(x_{1},T_{3}(x_{2}))+d(x_{2},T_{2}(x_{1}))]\\\le & {} a_{2,3} [d(x_{1},x_{3})+d(x_{2},x_{2})]\\= & {} a_{2,3} d(x_{1},x_{3})\\\le & {} a_{2,3} [d(x_{1},x_{2})+d(x_{2},x_{3})] \end{aligned}$$

implies

$$\begin{aligned} (1-a_{2,3}) d(x_{2},x_{3}) \le a_{2,3} d(x_{1},x_{2}). \end{aligned}$$

So

$$\begin{aligned} d(x_{2},x_{3})\le & {} \frac{a_{2,3}}{1-a_{2,3}} d(x_{1},x_{2})\\\le & {} \frac{a_{1,2}}{1-a_{1,2}}. \frac{a_{2,3}}{1-a_{2,3}}d(x_{0},x_{1}). \end{aligned}$$

In general, we get

$$\begin{aligned} d(x_{n},x_{n+1}) \le \prod _{i=1}^{n}\frac{a_{i,i+1}}{1-a_{i,i+1}}d(x_{0},x_{1})= A_{n} d(x_{0},x_{1}). \end{aligned}$$
(3)

Therefore, for \(m,n\in N\),  \(m\ge n,\) and using (2.2)

$$\begin{aligned} d(x_{n},x_{m})\le & {} \sum _{k=n}^{m-1} d(x_{k},x_{k+1})\\\le & {} \sum _{k=n}^{m-1} \prod _{i=1}^{k} \frac{a_{i,i+1}}{1-a_{i,i+1}} d(x_{0},x_{1})\\= & {} \sum _{k=n}^{m-1} A_{k}d(x_{0},x_{1}). \end{aligned}$$

Thus \(\{x_{n}\}\) is a Cauchy sequence and by completeness of X, \(\{x_{n}\}\) converges to x (say) in X.

So using (2.1), for any positive integer m we have

$$\begin{aligned} d(x,T_{m}x)\le & {} d(x,x_{n})+d(x_{n},T_{m}x)\\= & {} d(x,x_{n})+d(T_{n}x_{n-1},T_{m}x)\\\le & {} d(x,x_{n})+a_{n,m}[d(x_{n-1},T_{m}x)+d(x,T_{n}x_{n-1})]. \end{aligned}$$

Taking \(\overline{\text {lim}}\) as \(n\longrightarrow \infty ,\) we get

$$\begin{aligned} d(x,T_{m}x)\le \overline{lim}~ a_{n,m}d(x,T_{m}x), \end{aligned}$$

and it follows that \(d(x,T_{m}x)=0\) which shows that x is a common fixed of \(\{T_{m}\}.\)

Now we prove the uniqueness of the common fixed point x. Suppose that y be another common fixed point of \(\{T_{k}\}.\) Since \(\sum _{n=1}^{\infty }A_{n}<\infty\) so \({\lim}_{n\longrightarrow \infty } A_{n}=0\) and therefore there exists an \(i_{0}\in N\) such that \(a_{i_{0}, i_{0}+1}<\frac{1}{2}.\) Thus, from (2.1) we have

$$\begin{aligned} d(x,y)= & {} d(T_{i_{0}}x,T_{i_{0}+1}y)\\\le & {} a_{i_{0}, i_{0}+1}[d(x,T_{i_{0}+1}y)+d(y,T_{i_{0}}x)]\\= & {} a_{i_{0}, i_{0}+1}[d(x,y)+d(y,x)]\\< & {} d(x,y) \end{aligned}$$

which implies that \(x=y.\) So the uniqueness is proved and the proof is complete.

Corollary 1

In addition to hypotheses of Theorem 2 , suppose that for every \(n\in N\) , there exists a \(k_{n}\in N\) such that \(a_{n,k_{n}}<\frac{1}{2},\) then every \(T_{n}\) has a unique fixed point in X.

Proof

Following the proof of Theorem 2, \(\{T_{n}\}\) have a unique common fixed point \(x\in X.\) If y is another fixed point of a \(T_{m}\) then

$$\begin{aligned} d(x,y)=d(T_{m}y,T_{k_{m}}x)\le & {} a_{k_{m}, m}[d(x,T_{m}y)+d(y,T_{k_{m}}x)]\\=\, & {} a_{k_{m}, m}[d(x,y)+d(y,x)]\\< & \,{} d(x,y), \end{aligned}$$

which implies \(d(x, y) = 0.\) Therefore, \(x = y,\) which gives the desired result.

Example 1

Let \(X=[0,1]\) be a complete metric space with the distance \(d(x,y)=|x-y|\), \(x, y\in X\), and \(T_{n}:X\longrightarrow X\) be defined by

$$T_{n}(x)=\left\{ \begin{array}{ll}1, &\quad 0<x\le 1, \\ \frac{2}{3}+\frac{1}{n+2}, &\quad x=0. \end{array}\right.$$

Let \(a_{i,j}=\frac{1}{3}+\frac{1}{|i-j|+6}, i\ne j,\) then for each j,\(~\overline{\lim }_{i\longrightarrow \infty }a_{i, j}<1\) and \(A_{n}=\Pi _{i=1}^{n}\frac{a_{i,i+1}}{1-a_{i,i+1}}=\left(\frac{10}{11}\right)^{n}\), therefore \(\sum _{n=1}^{\infty }\left(\frac{10}{11}\right)^{n}<\infty .\) Now we prove that for each \(x, y \in X,\)

$$\begin{aligned} d(T_{i}x,T_{j}y) \le a_{i,j} [d(x,T_{j}y)+d(y,T_{i}x)]. \end{aligned}$$

There are three possible cases:

  1. 1.

    \(x\in (0,1],\) \(y\in (0,1].\) Then

    $$\begin{aligned} d(T_{i}x,T_{j}y)&= {}\; |T_{i}x-T_{j}y|=0\\ &\le {}\; a_{i,j}(|x-1|+|y-1|)\\&= {}\; a_{i,j} (d(x,T_{j}y)+d(y,T_{i}x)). \end{aligned}$$
  2. 2.

    \(x\in (0,1],\) \(y=0.\) Then

    $$\begin{aligned} d(T_{i}x,T_{j}y)&= {} |T_{i}x-T_{j}(0)|=\left|\frac{1}{3}-\frac{1}{j+2}\right|\le \frac{1}{3}\\ & \le {} \left(\frac{1}{3}+\frac{1}{|i-j|+6}\right) \left(\left|x-\frac{2}{3}-\frac{1}{j+2}\right|+|0-1|\right)\\&= {} a_{i,j} [d(x,T_{j}y)+d(y,T_{i}x)]. \end{aligned}$$
  3. 3.

    \(x=y=0\) , \(i<j.\) Then

    $$\begin{aligned} d(T_{i}x,T_{j}y)& = {} |T_{i}(0)-T_{j}(0)|=\left|\frac{2}{3}+\frac{1}{i+2}-\frac{2}{3}-\frac{1}{j+2}\right|=\left|\frac{1}{i+2}-\frac{1}{j+2}\right| \\ & \le {} \frac{1}{i+2}\\ & \le {} \left(\frac{1}{3}+\frac{1}{|i-j|+6}\right) \left(\left|\frac{2}{3}+\frac{1}{j+2}\right|+\left|\frac{2}{3}+\frac{1}{i+2}\right|\right)\\& = a_{i,j} [d(x,T_{j}y)+d(y,T_{i}x)]. \end{aligned}$$

So all the conditions of Theorem 2 are satisfied and note that \(x=1\) is the only fixed point for all \(T_{n}.\)

Theorem 3

Let (X, d) be a complete metric space and \(0\le a_{i,j}~~~ (i,j=1,2,...),\) satisfy

  1. (i)

    for each j, \(\overline{\lim }_{i\longrightarrow \infty }a_{i, j}<1\),

  2. (ii)

    \(\sum _{n=1}^{\infty }A_{n}<\infty\) where \(A_{n}=\Pi _{i=1}^{n}\frac{a_{i,i+1}}{1-a_{i,i+1}}.\)

If \(\{T_{n}\}\) is a sequence of self-maps on X satisfying

$$\begin{aligned} d(T_{i}x,T_{j}y) \le a_{i,j} {\text {max}}\{d(x,y),d(x,T_{i}x),d(y,T_{j}y),d(x,T_{j}y),d(y,T_{i}x)\}, \end{aligned}$$
(4)

for all \(x, y\in X,~i,j=1,2,...\) with \(x \ne y\) and \(i\ne j\) then all \(T_{n}\) ,s have a unique common fixed point in X . Further, if \(x\in X\) be unique common fixed point of \(\{T_{n}\},\) s then x is a unique fixed point for all \(T_{n},s.\)

Proof

For any \(x_{0}\in X,\) let \(x_{n}= T_{n}(x_{n-1}),~ n = 1, 2, ...\), then using (2.3) we obtain

$$\begin{aligned} d(x_{1},x_{2})& = d(T_{1}(x_{0}),T_{2}(x_{1}))\\ & \le {} a_{1,2} {\text {max}}\{d(x_{0},x_{1}),d(x_{0},T_{1}x_{0}),d(x_{1},T_{2}x_{1}),d(x_{0},T_{2}(x_{1})),d(x_{1},T_{1}(x_{0}))\}\\ & = {} a_{1,2} {\text {max}}\{d(x_{0},x_{1}),d(x_{0},x_{1}),d(x_{1},x_{2}),d(x_{0},x_{2}),d(x_{1},x_{1})\}\\ &= a_{1,2} {\text {max}}\{d(x_{0},x_{1}),d(x_{1},x_{2}), d(x_{0},x_{2})\}\\ & \le {}\, a_{1,2} (d(x_{0},x_{1})+d(x_{1},x_{2})). \end{aligned}$$

Therefore,

$$\begin{aligned} d(x_{1},x_{2}) \le \frac{a_{1,2}}{1-a_{1,2}} d(x_{0},x_{1}). \end{aligned}$$

Also

$$\begin{aligned} d(x_{2},x_{3})& = (T_{2}(x_{1}),T_{3}(x_{2}))\\ & \le {} a_{2,3}~ {\text {max}}\{d(x_{1},x_{2}),d(x_{1},T_{2}x_{1}),d(x_{2},T_{3}x_{2}),d(x_{1},T_{3}(x_{2})),d(x_{2},T_{2}(x_{1}))\}\\ & \le a_{2,3}~ {\text {max}}\{d(x_{1},x_{2}),d(x_{2},x_{3}),d(x_{1},x_{3})\}, \end{aligned}$$

which similar to the previous case we get

$$\begin{aligned} d(x_{2},x_{3}) \le \frac{a_{2,3}}{1-a_{2,3}} d(x_{1},x_{2}). \end{aligned}$$

Hence we have

$$\begin{aligned} d(x_{2},x_{3}) \le \frac{a_{1,2}}{1-a_{1,2}}. \frac{a_{2,3}}{1-a_{2,3}}d(x_{0},x_{1}). \end{aligned}$$

In general

$$\begin{aligned} d(x_{n},x_{n+1}) \le \prod _{i=1}^{n}\frac{a_{i,i+1}}{1-a_{i,i+1}}d(x_{0},x_{1}). \end{aligned}$$
(5)

Therefore, for \(m,n\in N,\)  \(m\ge n,\) and using (2.4)

$$\begin{aligned} d(x_{n},x_{m})\le & {} \sum _{k=n}^{m-1} d(x_{k},x_{k+1})\\ \le & {} \sum _{k=n}^{m-1} \prod _{i=1}^{k} \frac{a_{i,i+1}}{1-a_{i,i+1}} d(x_{0},x_{1})\\= & {} \sum _{k=n}^{m-1} A_{k}d(x_{0},x_{1}). \end{aligned}$$

Thus \(\{x_{n}\}\) is a Cauchy sequence and by completeness of X, \(\{x_{n}\}\) converges to x (say) in X.

So for any positive integer m,

$$\begin{aligned} d(x,T_{m}x)& \le {} d(x,x_{n})+d(x_{n},T_{m}x)= d(x,x_{n})+d(T_{n}x_{n-1},T_{m}x)\\& \le {} d(x,x_{n})+a_{n,m} {\text {max}} \{d(x_{n-1},x),d(x_{n-1},T_{n}x_{n-1}),d(x,T_{m}x),\\ &d(x_{n-1},T_{m}x),d(x,T_{_{n}}x_{n-1})\}\\& \le {} d(x,x_{n})+a_{n,m} {\text {max}}\{ d(x_{n-1},x),d(x_{n-1},x_{n}),d(x,T_{m}x),\\& d(x_{n-1},T_{m}x),d(x,x_{n})\}. \end{aligned}$$

Taking \(\overline{\text {lim}}\) as \(n\longrightarrow \infty ,\) we get

$$\begin{aligned} d(x,T_{m}x)\le \overline{\lim}_{n\longrightarrow \infty}a_{n,m} d(x,T_{m}x). \end{aligned}$$

From condition (i), it follows that \(d(x,T_{m}x)=0\) gives x as a common fixed point of \(\{T_{m}\}.\)

Let y be another fixed point of \(\{T_{n}\}\), then

$$\begin{aligned} d(x,y)& = {} d(T_{n}x,T_{m}y)\\ & \le {} a_{n,m} {\text {max}} \{d(x,y),d(x,T_{n}(x),d(y,T_{m}y),d(x,T_{m}y),d(y,T_{n}x)\}\\ & \le {} a_{n,m}d(x,y). \end{aligned}$$

Taking \(\overline{\text {lim}}\) as \(n\longrightarrow \infty ,\) we get

$$\begin{aligned} d(x,y)\le \overline{\lim}_{n\longrightarrow \infty}a_{n,m} d(x,y), \end{aligned}$$

which is possible only when \(x=y.\) Hence x is the unique common fixed point of \(\{T_{n}\}\). Further, if \(y\in X\) is a unique fixed point of \(T_{k},\) then according to \(\overline{\lim}_{i\longrightarrow \infty}a_{i,k} < 1,\) there exists an \(i_{k}\in N\) such that \(a_{i_{k}, k}<1.\) Thus, by (2.3) we have

$$\begin{aligned} d(x,y)& = d(T_{i_{k}}x,T_{k}y)\\ & \le {} a_{i_{k},k} {\text {max}}\{d(x,y),d(x,T_{i_{k}}x),d(y,T_{k}y),d(x,T_{k}y),d(y,T_{i_{k}}x)\}\\ & \le {} a_{i_{k},k} d(x,y), \end{aligned}$$

which implies \(d(x, y) = 0\) and hence \(x = y.\)

Theorem 4

Let (X, d) be a complete metric space and \(0\le a_{i,j}+b_{i,j} <1~~~ (i,j=1,2,...),\) satisfy

  1. (i)

    for each j, \(\overline{\lim} _{i\longrightarrow \infty}a_{i, j}<1\) and \(\overline{\lim}_{i\longrightarrow \infty}b_{i, j}<1,\)

  2. (ii)

    \(\sum _{n=1}^{\infty }A_{n}<\infty\) where \(A_{n}=\prod _{i=1}^{n}\frac{b_{i,i+1}}{1-a_{i,i+1}}.\)

If \(\{T_{n}\}\) is a sequence of self-maps on X satisfying

$$\begin{aligned} d(T_{i}x,T_{j}y) \le a_{i,j} d(y,T_{j}y) \varphi (d(x,T_{i}x),d(x,y)) +b_{i,j} d(x,y), \end{aligned}$$
(6)

for all \(x, y\in X,~ i,j=1,2,...\) with \(x \ne y\) and \(i\ne j\) where \(\varphi :[0,\infty )\times [0,\infty )\longrightarrow [0,\infty )\) is a continuous function such that \(\varphi (t,t)=1\) for all \(t \in [0,\infty )\) then, all \(T_{n}\) have a unique common fixed point in X.

Proof

For any \(x_{0}\in X,\) let \(x_{n}= T_{n}(x_{n-1}),\)  \(n = 1, 2, ...\), then from (2.5) we obtain

$$\begin{aligned} d(x_{1},x_{2})& = d(T_{1}(x_{0}),T_{2}(x_{1}))\\ & \le {} a_{1,2} d(x_{1},T_{2}x_{1}) \varphi (d(x_{0},T_{1}x_{0}),d(x_{0},x_{1}))+b_{1,2} d(x_{0},x_{1})\\ & \le {} a_{1,2} d(x_{1},x_{2})\varphi (d(x_{0},x_{1}),d(x_{0},x_{1}))+b_{1,2} d(x_{0},x_{1})\\ & \le {} a_{1,2} [d(x_{1},x_{2})+b_{1,2}d(x_{0},x_{1})], \end{aligned}$$

implies

$$\begin{aligned} (1-a_{1,2}) d(x_{1},x_{2}) \le b_{1,2} d(x_{0},x_{1}). \end{aligned}$$

Hence we have

$$\begin{aligned} d(x_{1},x_{2}) \le \frac{b_{1,2}}{1-a_{1,2}} d(x_{0},x_{1}). \end{aligned}$$

Also,

$$\begin{aligned} d(x_{2},x_{3})& = (T_{2}(x_{1}),T_{3}(x_{2}))\\ & \le {} a_{2,3} d(x_{2},T_{3}x_{2})\varphi (d(x_{1},T_{2}x_{1}),d(x_{1},x_{2}))+b_{2,3} d(x_{1},x_{2})\\ & \le {} a_{2,3} d(x_{2},x_{3})\varphi (d(x_{1},x_{2}),d(x_{1},x_{2}))+b_{2,3}d(x_{1},x_{2})\\ &\le {} a_{2,3} d(x_{2},x_{3})+b_{2,3}d(x_{1},x_{2}). \end{aligned}$$

Then

$$\begin{aligned} d(x_{2},x_{3})& \le {} \frac{b_{2,3}}{1-a_{2,3}} d(x_{1},x_{2})\\ & \le {} \frac{b_{1,2}}{1-a_{1,2}}\times \frac{b_{2,3}}{1-a_{2,3}}d(x_{0},x_{1}). \end{aligned}$$

Generally we conclude that

$$\begin{aligned} d(x_{n},x_{n+1}) \le \prod _{i=1}^{n}\frac{b_{i,i+1}}{1-a_{i,i+1}}d(x_{0},x_{1})=A_{n}d(x_{0},x_{1}). \end{aligned}$$
(7)

Therefore, for \(m,n\in N\),  \(m\ge n,\) and using (2.6) we get

$$\begin{aligned} d(x_{n},x_{m})& \le {} \sum _{k=n}^{m-1} d(x_{k},x_{k+1})\\ & \le {} \sum _{k=n}^{m-1} A_{k}d(x_{0},x_{1}). \end{aligned}$$

Now passing to limit \(n,m\longrightarrow \infty\), we get \(d(x_{n},x_{m})\longrightarrow 0 .\)

Thus \(\{x_{n}\}\) is a Cauchy sequence and by completeness of X, \(\{x_{n}\}\) converges to x in X that is \(\displaystyle \lim\nolimits_{n\longrightarrow \infty }x_{n}=x\in X\).

So, for any positive integer m, 

$$\begin{aligned} d(x,T_{m}x)& \le {} d(x,x_{n})+d(x_{n},T_{m}x)= d(x,x_{n})+d(T_{n}x_{n-1},T_{m}x)\\ & \le {} d(x,x_{n})+a_{n,m} d(x,T_{m}x) \varphi (d(x_{n-1},T_{n}x_{n-1}),d(x_{n-1},x))\\ &+b_{n,m} d(x_{n-1},x)\\ & \le {} d(x,x_{n})+a_{n,m} d(x,T_{m}x)\varphi (d(x_{n-1},x_{n}),d(x_{n-1},x_{n}))\\ &+b_{n,m} d(x_{n-1},x)\\\le & {} d(x,x_{n})+a_{n,m} d(x,T_{m}x)+b_{n,m} d(x_{n-1},x). \end{aligned}$$

Taking \(\overline{\text {lim}}\) as \(n\longrightarrow \infty ,\) we get

$$\begin{aligned} d(x,T_{m}x)\le \overline{lim}~a_{n,m}d(x,T_{m}x)< d(x,T_{m}x). \end{aligned}$$

It follows that \(d(x,T_{m}x)=0\) gives x as a common fixed of \(\{T_{m}\}.\)

Let y be another common fixed point, then

$$\begin{aligned} d(x,y)& = {} d(T_{n}x,T_{m}y)\le a_{n,m} d(y,T_{m}y) \varphi (d(x,T_{n}(x)),d(x,y))+b_{n,m} d(x,y)\\ &= {} a_{n,m} d(y,y) \varphi (d(x,x),d(x,y))+b_{n,m} d(x,y)\\ & = {} b_{n,m} d(x,y). \end{aligned}$$

Taking \(\overline{lim}\) as \(n\longrightarrow \infty ,\) we get \(x=y.\) So, the uniqueness is proved. \(\square\)

Author contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.