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# Some generalizations of Suzuki and Edelstein type theorems

Fixed Point Theory and Applications20132013:319

https://doi.org/10.1186/1687-1812-2013-319

• Accepted: 28 October 2013
• Published:

## Abstract

We prove some generalizations of Suzuki’s fixed point theorem and Edelstein’s theorem.

MSC:54H25.

## Keywords

• Banach principle
• contraction
• Suzuki’s theorem
• Edelstein’s theorem

## Introduction and preliminaries

Let $\left(X,d\right)$ be a complete metric space and T be a selfmap of X. Then T is called a contraction if there exists $r\in \left[0,1\right)$ such that
$d\left(Tx,Ty\right)\le rd\left(x,y\right)$

for all $x,y\in X$.

The following famous theorem is referred to as the Banach contraction principle.

Theorem 1 (Banach [1])

Let $\left(X,d\right)$ be a complete metric space, and let T be a contraction on X. Then T has a unique fixed point.

This theorem is a very forceful and simple, and it has become a classical tool in nonlinear analysis. It has many generalizations, see [219].

In 2008, Suzuki [20] introduced a new type of mapping and presented a generalization of the Banach contraction principle in which the completeness can also be characterized by the existence of a fixed point of these mappings.

Theorem 2 [20]

Let $\left(X,d\right)$ be a complete metric space, and let T be a mapping on X. Define a nonincreasing function θ from $\left[0,1\right)$ onto $\left(1/2,1\right]$ by
(1)

Assume that there exists $r\in \left[0,1\right)$ such that $\theta \left(r\right)d\left(x,Tx\right)\le d\left(x,y\right)$ implies $d\left(Tx,Ty\right)\le rd\left(x,y\right)$ for all $x,y\in X$. Then there exists a unique fixed point z of T. Moreover, ${lim}_{n}{T}^{n}x=z$ for all $x\in X$.

Its further outcomes by Altun and Erduran [21], Karapinar [22, 23], Kikkawa and Suzuki [24, 25], Moţ and Petruşel [26], Dhompongsa and Yingtaweesittikul [27], Popescu [28, 29], Singh and Mishra [3032] are important contributions to metric fixed point theory.

Popescu [28] introduced a new type of contractive operator and proved the following theorem.

Theorem 3 [28]

Let $\left(X,d\right)$ be a complete metric space and $T:X\to X$ be a $\left(s,r\right)$-contractive single-valued operator:
where $r\in \left[0,1\right)$, $s>r$ and
${M}_{T}\left(x,y\right)=max\left\{d\left(x,y\right),d\left(x,Tx\right),d\left(y,Ty\right),\frac{d\left(x,Ty\right)+d\left(y,Tx\right)}{2}\right\}.$

Then T has a fixed point. Moreover, if $s\ge 1$, then T has a unique fixed point.

As a direct consequence of Theorem 3, we obtain the following result.

Theorem 4 Let $\left(X,d\right)$ be a complete metric space, and let T be a mapping on X. Assume that there exist $r\in \left[0,1\right)$ and $s>r$ such that
$d\left(y,Tx\right)\le sd\left(y,x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}d\left(Tx,Ty\right)\le rd\left(x,y\right)$
(2)

for all $x,y\in X$. Then there exists a fixed point z of T. Further, if $s\ge 1$, then there exists a unique fixed point of T.

The following theorem is a well-known result in fixed point theory.

Theorem 5 (Edelstein [33])

Let $\left(X,d\right)$ be a compact metric space, and let T be a mapping on X. Assume $d\left(Tx,Ty\right) for all $x,y\in X$ with $x\ne y$. Then T has a unique fixed point.

Inspired by Theorem 2, Suzuki [34] proved a generalization of Edelstein’s fixed point theorem (see also [3538]).

Theorem 6 [34]

Let $\left(X,d\right)$ be a compact metric space, and let T be a mapping on X. Assume that $\left(1/2\right)d\left(x,Tx\right) implies $d\left(Tx,Ty\right) for all $x,y\in X$. Then T has a unique fixed point.

In this paper, we prove generalizations of Theorem 2, Theorem 4, Theorem 5 and extend Theorem 6. The direction of our extension is new, very simple and inspired by Theorem 3.

## Main results

We start this section by proving the following theorem.

Theorem 7 Let $\left(X,d\right)$ be a complete metric space, and let T be a mapping on X. Assume that there exist $r\in \left[0,1\right)$, $a\in \left[0,1\right]$, $b\in \left[0,1\right)$, $\left(a+b\right){r}^{2}+r\le 1$ if $r\in \left[1/2,1/\sqrt{2}\right)$, $a+\left(a+b\right)r\le 1$ if $r\in \left[1/\sqrt{2},1\right)$ such that
$ad\left(x,Tx\right)+bd\left(y,Tx\right)\le d\left(y,x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}d\left(Tx,Ty\right)\le rd\left(x,y\right)$

for all $x,y\in X$. Then there exists a unique fixed point z of T. Moreover, ${lim}_{n}{T}^{n}x=z$ for all $x\in X$.

Proof Since $ad\left(x,Tx\right)+bd\left(Tx,Tx\right)=ad\left(x,Tx\right)\le d\left(Tx,x\right)$ holds for every $x\in X$, by hypothesis, we get
$d\left(Tx,{T}^{2}x\right)\le rd\left(x,Tx\right)$
(3)
for all $x\in X$. We now fix $u\in X$ and define a sequence $\left\{{u}_{n}\right\}\in X$ by ${u}_{n}={T}^{n}u$. Then (3) yields $d\left({u}_{n},{u}_{n+1}\right)\le {r}^{n}d\left(u,Tu\right)$, so ${\sum }_{n=1}^{\mathrm{\infty }}d\left({u}_{n},{u}_{n+1}\right)<\mathrm{\infty }$. Hence $\left\{{u}_{n}\right\}$ is a Cauchy sequence. Since X is complete, $\left\{{u}_{n}\right\}$ converges to some point $z\in X$. We next show that
$d\left(Tx,z\right)\le rd\left(x,z\right)$
(4)

for all $x\in X$, $x\ne z$. Since ${lim}_{n}d\left({u}_{n},T{u}_{n}\right)=0$, ${lim}_{n}d\left(x,T{u}_{n}\right)={lim}_{n}d\left(x,{u}_{n}\right)=d\left(x,z\right)$, there exists a positive integer ν such that $ad\left({u}_{n},T{u}_{n}\right)+bd\left(x,T{u}_{n}\right)\le d\left(x,{u}_{n}\right)$ for all $n\ge \nu$. By hypothesis, we get $d\left(T{u}_{n},Tx\right)\le rd\left({u}_{n},x\right)$. Letting n tend to ∞, we obtain $d\left(z,Tx\right)\le rd\left(z,x\right)$. That is, we have shown (4).

Now we assume that ${T}^{j}z\ne z$ for every integer $j\ge 1$. Then (4) yields
$d\left({T}^{j+1}z,z\right)\le {r}^{j}d\left(Tz,z\right)$
(5)
for every integer $j\ge 1$. We consider the following three cases:
1. (a)

$0\le r<1/2$,

2. (b)

$1/2\le r<1/\sqrt{2}$,

3. (c)

$1/\sqrt{2}\le r<1$.

In the case (a) we note that $2r<1$. Then, by (3) and (5), we have
$d\left(z,Tz\right)\le d\left(z,{T}^{2}z\right)+d\left(Tz,{T}^{2}z\right)\le rd\left(z,Tz\right)+rd\left(z,Tz\right)=2rd\left(z,Tz\right)

In the case (b), we note that $2{r}^{2}<1$. If we assume $ad\left({T}^{2}z,{T}^{3}z\right)+bd\left(z,{T}^{3}z\right)>d\left(z,{T}^{2}z\right)$, then we have, in view of (3) and (5),
$\begin{array}{rcl}d\left(z,Tz\right)& \le & d\left(z,{T}^{2}z\right)+d\left(Tz,{T}^{2}z\right)\\ <& ad\left({T}^{2}z,{T}^{3}z\right)+bd\left(z,{T}^{3}z\right)+d\left(Tz,{T}^{2}z\right)\\ \le & a{r}^{2}d\left(z,Tz\right)+b{r}^{2}d\left(z,Tz\right)+rd\left(z,Tz\right)\\ =& \left[\left(a+b\right){r}^{2}+r\right]d\left(z,Tz\right)\\ \le & d\left(z,Tz\right).\end{array}$
This is a contradiction. Hence $ad\left({T}^{2}z,{T}^{3}z\right)+bd\left(z,{T}^{3}z\right)\le d\left(z,{T}^{2}z\right)$. By hypothesis and (5), we have
$\begin{array}{rcl}d\left(z,Tz\right)& \le & d\left(z,{T}^{3}z\right)+d\left(Tz,{T}^{3}z\right)\\ \le & {r}^{2}d\left(z,Tz\right)+rd\left(z,{T}^{2}z\right)\\ \le & {r}^{2}d\left(z,Tz\right)+{r}^{2}d\left(z,Tz\right)\\ =& 2{r}^{2}d\left(z,Tz\right)\\ <& d\left(z,Tz\right).\end{array}$

In the case (c), we assume there exists an integer $\nu \ge 1$ such that
$ad\left({u}_{n},{u}_{n+1}\right)+bd\left(z,{u}_{n+1}\right)>d\left(z,{u}_{n}\right)$
for all $n\ge \nu$. Then
$\begin{array}{rcl}d\left(z,{u}_{n}\right)& <& ad\left({u}_{n},{u}_{n+1}\right)+b\left[ad\left({u}_{n+1},{u}_{n+2}\right)+bd\left(z,{u}_{n+2}\right)\right]\\ \le & \left(a+abr\right)d\left({u}_{n},{u}_{n+1}\right)+{b}^{2}d\left(z,{u}_{n+2}\right)\\ <& \left(a+abr\right)d\left({u}_{n},{u}_{n+1}\right)+{b}^{2}\left[ad\left({u}_{n+2},{u}_{n+3}\right)+bd\left(z,{u}_{n+3}\right)\right]\\ \le & \left(a+abr+a{b}^{2}{r}^{2}\right)d\left({u}_{n},{u}_{n+1}\right)+{b}^{3}d\left(z,{u}_{n+3}\right).\end{array}$
Continuing this process, we get
$\begin{array}{rcl}d\left(z,{u}_{n}\right)& <& \left(a+abr+a{b}^{2}{r}^{2}+\cdots +a{b}^{p-1}{r}^{p-1}\right)d\left({u}_{n},{u}_{n+1}\right)+{b}^{p}d\left(z,{u}_{n+p}\right)\\ \le & a\frac{1-{\left(br\right)}^{p}}{1-br}d\left({u}_{n},{u}_{n+1}\right)+{b}^{p}d\left(z,{u}_{n+p}\right)\end{array}$
for all $n\ge \nu$, $p\ge 1$. Letting p tend to ∞, we obtain
$d\left(z,{u}_{n}\right)\le \frac{a}{1-br}d\left({u}_{n},{u}_{n+1}\right)$
for all $n\ge \nu$. Thus,
$d\left(z,{u}_{n+1}\right)\le \frac{a}{1-br}d\left({u}_{n+1},{u}_{n+2}\right)\le \frac{ar}{1-br}d\left({u}_{n},{u}_{n+1}\right)$
for all $n\ge \nu$, so
$\begin{array}{rcl}d\left({u}_{n},{u}_{n+1}\right)& \le & d\left(z,{u}_{n}\right)+d\left(z,{u}_{n+1}\right)\\ <& \frac{a}{1-br}d\left({u}_{n},{u}_{n+1}\right)+\frac{ar}{1-br}d\left({u}_{n},{u}_{n+1}\right)\\ =& \frac{a+ar}{1-br}d\left({u}_{n},{u}_{n+1}\right)\\ \le & d\left({u}_{n},{u}_{n+1}\right)\end{array}$
for all $n\ge \nu$. This is a contradiction. Hence there exists a subsequence $\left\{{u}_{n\left(k\right)}\right\}$ of $\left\{{u}_{n}\right\}$ such that
$ad\left({u}_{n\left(k\right)},{u}_{n\left(k\right)+1}\right)+bd\left(z,{u}_{n\left(k\right)+1}\right)\le d\left(z,{u}_{n\left(k\right)}\right)$

for all $k\ge 1$. By hypothesis, we get $d\left(Tz,T{u}_{n\left(k\right)}\right)\le rd\left(z,{u}_{n\left(k\right)}\right)$ for all $k\ge 1$. Letting k tend to ∞, we get $d\left(z,Tz\right)=0$, that is, $z=Tz$. This is a contradiction.

Thus there exists an integer $j\ge 1$ such that ${T}^{j}z=z$. By (3) we get $d\left(z,Tz\right)=d\left({T}^{j}z,{T}^{j+1}z\right)\le {r}^{j}d\left(z,Tz\right)$, so $d\left(z,Tz\right)=0$, that is, $Tz=z$.

Now we suppose that y is another fixed point of T, that is, $Ty=y$. Then
$ad\left(y,Ty\right)+bd\left(z,Ty\right)=bd\left(z,y\right)\le d\left(z,y\right),$

so, by hypothesis, $d\left(y,z\right)=d\left(Ty,Tz\right)\le rd\left(y,z\right)$. Hence $d\left(y,z\right)=0$. This is a contradiction. □

Remark 1 For $r\in \left[0,1/2\right)$, taking $a=1$, $b=0$, we obtain Suzuki’s condition from Theorem 2. Moreover, from our condition and the triangle inequality, we get
$ad\left(x,Tx\right)+b\left[d\left(x,Tx\right)-d\left(y,x\right)\right]\le d\left(y,x\right),$
that is,
$\frac{a+b}{1+b}d\left(x,Tx\right)\le d\left(y,x\right).$
If $r\in \left[1/\sqrt{2},1\right)$, we have
$\frac{a+b}{1+b}=\frac{1}{1+r}=\theta \left(r\right),$

hence our condition implies Suzuki’s condition. We also note that if we take $a=\left(1-r\right)/{r}^{2}$, $b=0$ for $r\in \left[1/2,1/\sqrt{2}\right)$, we get Suzuki’s condition. Therefore, our theorem generalizes, extends and complements Suzuki’s theorem.

Example 1 Define a complete metric space X by $X=\left\{-1,0,1,2\right\}$ and a mapping T on X by $Tx=0$ if $x\in \left\{-1,0,1\right\}$ and $T2=-1$. Then T satisfies our condition from Theorem 7 for every $r\in \left[0,1/3\right)\cup \left[1/2,1\right)$, but T does not satisfy Suzuki’s condition from Theorem 2.

Proof Since $\theta \left(r\right)d\left(1,T1\right)\le 1=d\left(1,2\right)$ for every $r\in \left[0,1\right)$, and $d\left(T1,T2\right)=1=d\left(1,2\right)$, T does not satisfy Suzuki’s condition. If $r\in \left[1/2,\left(\sqrt{5}-1\right)/2\right)$, we have ${r}^{2}+r<1$, so taking $a+b=\left(1-r\right)/{r}^{2}$, we get $a+b>1$. Hence $ad\left(1,T1\right)+bd\left(1,T2\right)=a+2b>1=d\left(1,2\right)$ and $ad\left(2,T2\right)+bd\left(2,T1\right)=3a+2b>1=d\left(1,2\right)$. Now it is obvious that T satisfies our condition. If $r\in \left[\left(\sqrt{5}-1\right)/2,1\right)$, we take $b=1/2$. We have two cases: $r\in \left[\left(\sqrt{5}-1\right)/2,1/\sqrt{2}\right)$ and $r\in \left[1/\sqrt{2},1\right)$. In the first case we put $a=\left(2-2r-{r}^{2}\right)/\left(2{r}^{2}\right)$ and in the second $a=\left(2-r\right)/\left(2+2r\right)$. We have $a+2b=1+a>1$ in both cases, so T satisfies our condition. If $r\in \left[0,1/3\right)$ for $a=1$, $b=1/2$, it is obvious that T satisfies our condition. □

The following theorem is a generalization of Theorem 4.

Theorem 8 Let $\left(X,d\right)$ be a complete metric space, and let T be a mapping on X. Assume that there exist $r\in \left[0,1\right)$, $s>r$ such that
$\frac{s-r}{1+r}d\left(x,Tx\right)+d\left(y,Tx\right)\le sd\left(y,x\right)\phantom{\rule{1em}{0ex}}\mathit{\text{implies}}\phantom{\rule{1em}{0ex}}d\left(Tx,Ty\right)\le rd\left(x,y\right)$

for all $x,y\in X$. Then T has a unique fixed point. Moreover, if $s\ge 1$, then T has a unique fixed point.

Proof Let ${u}_{1}\in X$ and the sequence ${u}_{n}$ be defined by ${u}_{n+1}=T{u}_{n}$. Since
$0=d\left({u}_{n+1},T{u}_{n}\right)\le sd\left({u}_{n+1},{u}_{n}\right)-\frac{s-r}{1+r}d\left({u}_{n},T{u}_{n}\right),$
we get from hypothesis $d\left({u}_{n+1},{u}_{n+2}\right)\le rd\left({u}_{n+1},{u}_{n}\right)$ for all $n\ge 1$. Therefore, $d\left({u}_{n+1},{u}_{n+2}\right)\le {r}^{n}d\left({u}_{1},{u}_{2}\right)$ for all $n\ge 1$. Thus
$\sum _{n=1}^{\mathrm{\infty }}d\left({u}_{n+1},{u}_{n}\right)\le \sum _{n=1}^{\mathrm{\infty }}{r}^{n-1}d\left({u}_{1},{u}_{2}\right)<\mathrm{\infty }.$

Hence $\left\{{u}_{n}\right\}$ is a Cauchy sequence. Since X is complete, $\left\{{u}_{n}\right\}$ converges to some point $z\in X$.

Now, we will show that there exists a subsequence $\left\{{u}_{n\left(k\right)}\right\}$ of $\left\{{u}_{n}\right\}$ such that
$d\left(z,T{u}_{n\left(k\right)}\right)\le sd\left(z,{u}_{n\left(k\right)}\right)-\frac{s-r}{1+r}d\left({u}_{n\left(k\right)},T{u}_{n\left(k\right)}\right)$
for all $k\ge 1$. Arguing by contradiction, we suppose that there exists a positive integer ν such that
$d\left(z,T{u}_{n}\right)>sd\left(z,{u}_{n}\right)-\frac{s-r}{1+r}d\left({u}_{n},T{u}_{n}\right)$
for all $n\ge \nu$. Then we have
$\begin{array}{rcl}d\left(z,{u}_{n+2}\right)& >& sd\left(z,{u}_{n+1}\right)-\frac{s-r}{1+r}d\left({u}_{n+1},{u}_{n+2}\right)\\ >& {s}^{2}d\left(z,{u}_{n}\right)-s\cdot \frac{s-r}{1+r}d\left({u}_{n},{u}_{n+1}\right)-\frac{s-r}{1+r}d\left({u}_{n+1},{u}_{n+2}\right)\\ \ge & {s}^{2}d\left(z,{u}_{n}\right)-\frac{s-r}{1+r}\left[sd\left({u}_{n},{u}_{n+1}\right)+rd\left({u}_{n},{u}_{n+1}\right)\right]\\ =& {s}^{2}d\left(z,{u}_{n}\right)-\frac{s-r}{1+r}\left(s+r\right)d\left({u}_{n},{u}_{n+1}\right).\end{array}$
By induction, we get for all $n\ge \nu$, $p\ge 1$ that
$d\left(z,{u}_{n+p}\right)>{s}^{p}d\left(z,{u}_{n}\right)-\frac{s-r}{1+r}\left({s}^{p-1}+{s}^{p-2}r+\cdots +{r}^{p-1}\right)d\left({u}_{n},{u}_{n+1}\right).$
Then we have
$\begin{array}{rcl}d\left(z,{u}_{n+p}\right)& >& {s}^{p}d\left(z,{u}_{n}\right)-\frac{s-r}{1+r}\cdot {s}^{p-1}\cdot \frac{1-{\left(r/s\right)}^{p}}{1-r/s}d\left({u}_{n},{u}_{n+1}\right)\\ =& {s}^{p}\left[d\left(z,{u}_{n}\right)-\frac{s-r}{1+r}\cdot \frac{1-{\left(r/s\right)}^{p}}{s-r}d\left({u}_{n},{u}_{n+1}\right)\right].\end{array}$
Hence
${s}^{p}\left[d\left(z,{u}_{n}\right)-\frac{1-{\left(r/s\right)}^{p}}{1+r}d\left({u}_{n},{u}_{n+1}\right)\right]
(6)
On the other hand,
$\begin{array}{rcl}d\left({u}_{n+p},{u}_{n}\right)& \le & d\left({u}_{n},{u}_{n+1}\right)+d\left({u}_{n+1},{u}_{n+2}\right)+\cdots +d\left({u}_{n+p-1},{u}_{n+p}\right)\\ \le & \left(1+r+\cdots +{r}^{p-1}\right)d\left({u}_{n},{u}_{n+1}\right)\\ =& \frac{1-{r}^{p}}{1-r}d\left({u}_{n},{u}_{n+1}\right).\end{array}$
Letting $p\to \mathrm{\infty }$, we get for all $n\ge 1$ that $d\left(z,{u}_{n}\right)\le d\left({u}_{n},{u}_{n+1}\right)/\left(1-r\right)$. Thus
$d\left(z,{u}_{n+p}\right)\le d\left({u}_{n+p},{u}_{n+p+1}\right)/\left(1-r\right)\le {r}^{p}d\left({u}_{n},{u}_{n+1}\right)/\left(1-r\right).$
(7)
By (6) and (7) we have for all $n\ge \nu$, $p\ge 1$ that
$\frac{{r}^{p}}{1-r}d\left({u}_{n},{u}_{n+1}\right)>{s}^{p}\left[d\left(z,{u}_{n}\right)-\frac{1-{\left(r/s\right)}^{p}}{1+r}d\left({u}_{n},{u}_{n+1}\right)\right],$
so
$\frac{{\left(r/s\right)}^{p}}{1-r}d\left({u}_{n},{u}_{n+1}\right)>d\left(z,{u}_{n}\right)-\frac{1-{\left(r/s\right)}^{p}}{1+r}d\left({u}_{n},{u}_{n+1}\right).$
Taking the limit as $p\to \mathrm{\infty }$, we obtain that $d\left(z,{u}_{n}\right)\le d\left({u}_{n},{u}_{n+1}\right)/\left(1+r\right)$ for all $n\ge \nu$. Then we have
$d\left(z,{u}_{n+1}\right)\le d\left({u}_{n+1},{u}_{n+2}\right)/\left(1+r\right)\le rd\left({u}_{n},{u}_{n+1}\right)/\left(1+r\right)$
and
$rd\left({u}_{n},{u}_{n+1}\right)/\left(1+r\right)>sd\left(z,{u}_{n}\right)-\left(s-r\right)d\left({u}_{n},{u}_{n+1}\right)/\left(1+r\right).$
This implies $d\left(z,{u}_{n}\right) for all $n\ge \nu$. Thus,
$d\left({u}_{n},{u}_{n+1}\right)\le d\left(z,{u}_{n}\right)+d\left(z,{u}_{n+1}\right)
This is a contradiction. Therefore there exists a subsequence $\left\{{u}_{n\left(k\right)}\right\}$ of $\left\{{u}_{n}\right\}$ such that
$d\left(z,T{u}_{n\left(k\right)}\right)\le sd\left(z,{u}_{n\left(k\right)}\right)-\frac{s-r}{1+r}d\left({u}_{n\left(k\right)},T{u}_{n\left(k\right)}\right)$

for all $k\ge 1$. By hypothesis, we get $d\left(Tz,T{u}_{n\left(k\right)}\right)\le rd\left(z,{u}_{n\left(k\right)}\right)$. Letting $k\to \mathrm{\infty }$, we obtain $d\left(Tz,z\right)=0$, that is, $z=Tz$.

If $s\ge 1$, we assume that y is another fixed point of T. Then $d\left(z,Ty\right)=d\left(z,y\right)\le sd\left(z,y\right)-\left(s-r\right)d\left(y,Ty\right)/\left(1+r\right)=sd\left(z,y\right)$, so, by hypothesis, $d\left(z,y\right)=d\left(Tz,Ty\right)\le rd\left(z,y\right)$. Since $r<1$, this is a contradiction. □

## Edelstein’s theorem

The following theorem extends Theorem 6 and generalizes Theorem 5.

Theorem 9 Let $\left(X,d\right)$ be a compact metric space, and let T be a mapping on X. Assume that
$ad\left(x,Tx\right)+bd\left(y,Tx\right)
(8)

for $x,y\in X$, where $a>0$, $b>0$, $2a+b<1$. Then T has a unique fixed point.

Proof We put
$\beta =inf\left\{d\left(x,Tx\right):x\in X\right\}$
and choose a sequence $\left\{{x}_{n}\right\}$ in X such that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},T{x}_{n}\right)=\beta$. Since X is compact, without loss of generality, we may assume that $\left\{{x}_{n}\right\}$ and $\left\{T{x}_{n}\right\}$ converge to some elements $v,w\in X$, respectively. We have
$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},w\right)=\underset{n\to \mathrm{\infty }}{lim}d\left(T{x}_{n},v\right)=d\left(v,w\right)=\beta .$
We shall show $\beta =0$. Arguing by contradiction, we assume $\beta >0$. Since
$\underset{n\to \mathrm{\infty }}{lim}\left[ad\left({x}_{n},T{x}_{n}\right)+bd\left(w,T{x}_{n}\right)\right]=a\beta <\beta =\underset{n\to \mathrm{\infty }}{lim}d\left(w,{x}_{n}\right),$
we can choose a positive integer ν such that
$ad\left({x}_{n},T{x}_{n}\right)+bd\left(w,T{x}_{n}\right)
for all $n\ge \nu$. By hypothesis, $d\left(Tw,T{x}_{n}\right) holds for $n\ge \nu$. This implies
$d\left(w,Tw\right)=\underset{n\to \mathrm{\infty }}{lim}d\left(Tw,T{x}_{n}\right)\le \underset{n\to \mathrm{\infty }}{lim}d\left(w,{x}_{n}\right)=\beta .$
From the definition of β, we obtain $d\left(w,Tw\right)=\beta$. Since $ad\left(w,Tw\right)+bd\left(Tw,Tw\right), we have
$d\left(Tw,{T}^{2}w\right)

which contradicts the definition of β. Therefore we obtain $\beta =0$. We have ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},w\right)={lim}_{n\to \mathrm{\infty }}d\left(T{x}_{n},v\right)={lim}_{n\to \mathrm{\infty }}d\left(T{x}_{n},{x}_{n}\right)=d\left(v,w\right)=0$, so $v=w$. Thus, ${lim}_{n\to \mathrm{\infty }}{x}_{n}={lim}_{n\to \mathrm{\infty }}T{x}_{n}=w$.

We next show that T has a fixed point. Arguing by contradiction, we assume that T does not have a fixed point. Since $ad\left({x}_{n},T{x}_{n}\right)+bd\left(T{x}_{n},T{x}_{n}\right) for all $n\ge 1$, we get $d\left({T}^{2}{x}_{n},T{x}_{n}\right), so ${lim}_{n\to \mathrm{\infty }}{T}^{2}{x}_{n}=w$. By induction, we obtain that $d\left({T}^{p}{x}_{n},{T}^{p+1}{x}_{n}\right) and ${lim}_{n\to \mathrm{\infty }}{T}^{p}{x}_{n}=w$ for all integers $p\ge 1$. If there exist an integer $p\ge 1$ and a subsequence $\left\{{x}_{n\left(k\right)}\right\}$ of $\left\{{x}_{n}\right\}$ such that
$ad\left({T}^{p-1}{x}_{n\left(k\right)},{T}^{p}{x}_{n\left(k\right)}\right)+bd\left(w,{T}^{p}{x}_{n\left(k\right)}\right)
for all $k\ge 1$, by hypothesis we get $d\left(Tw,{T}^{p}{x}_{n\left(k\right)}\right). Taking the limit as $k\to \mathrm{\infty }$, we obtain $d\left(w,Tw\right)=0$, that is, $Tw=w$, which is a contradiction. Hence, we can assume that for every $m\ge 1$, there exists an integer $n\left(m\right)\ge 1$ such that
$ad\left({T}^{m-1}{x}_{n},{T}^{m}{x}_{n}\right)+bd\left(w,{T}^{m}{x}_{n}\right)\ge d\left(w,{T}^{m-1}{x}_{n}\right)$
(9)
for all $n\ge n\left(m\right)$. Since
$\underset{p\to \mathrm{\infty }}{lim}\frac{p{b}^{p}}{1-{b}^{p}}=0,$
and
$\frac{2a}{1-b}<1,$
we can choose p satisfying
$\frac{p{b}^{p}}{1-{b}^{p}}+\frac{\left(p-1\right){b}^{p-1}}{1-{b}^{p-1}}+\frac{2a}{1-b}<1.$
(10)
We put $\nu =max\left\{n\left(1\right),n\left(2\right),\dots ,n\left(p\right)\right\}$. Then by (9) we have
$\begin{array}{rcl}d\left(w,{x}_{n}\right)& \le & ad\left({x}_{n},T{x}_{n}\right)+bd\left(w,T{x}_{n}\right)\\ \le & ad\left({x}_{n},T{x}_{n}\right)+b\left[ad\left(T{x}_{n},{T}^{2}{x}_{n}\right)+bd\left(w,{T}^{2}{x}_{n}\right)\right]\\ =& ad\left({x}_{n},T{x}_{n}\right)+abd\left(T{x}_{n},{T}^{2}{x}_{n}\right)+{b}^{2}d\left(w,{T}^{2}{x}_{n}\right)\\ \le & \cdots \\ \le & ad\left({x}_{n},T{x}_{n}\right)+abd\left(T{x}_{n},{T}^{2}{x}_{n}\right)+\cdots \\ +a{b}^{p-1}d\left({T}^{p-2}{x}_{n},{T}^{p-1}{x}_{n}\right)+{b}^{p}d\left(w,{T}^{p}{x}_{n}\right)\\ \le & \left(a+ab+\cdots +a{b}^{p-1}\right)d\left({x}_{n},T{x}_{n}\right)+{b}^{p}d\left(w,{T}^{p}{x}_{n}\right)\\ \le & \left[a\left(1-{b}^{p}\right)/\left(1-b\right)\right]d\left({x}_{n},T{x}_{n}\right)+{b}^{p}d\left(w,{T}^{p}{x}_{n}\right)\end{array}$
for all $n\ge \nu$. Since
$\begin{array}{rcl}d\left(w,{T}^{p}{x}_{n}\right)& \le & d\left(w,{x}_{n}\right)+d\left({x}_{n},T{x}_{n}\right)+\cdots +d\left({T}^{p-1}{x}_{n},{T}^{p}{x}_{n}\right)\\ <& d\left(w,{x}_{n}\right)+pd\left({x}_{n},T{x}_{n}\right),\end{array}$
we get
$d\left(w,{x}_{n}\right)<\left[a\left(1-{b}^{p}\right)/\left(1-b\right)\right]d\left({x}_{n},T{x}_{n}\right)+{b}^{p}\left[d\left(w,{x}_{n}\right)+pd\left({x}_{n},T{x}_{n}\right)\right],$
so
$d\left(w,{x}_{n}\right)<\left(\frac{a}{1-b}+\frac{p{b}^{p}}{1-{b}^{p}}\right)d\left({x}_{n},T{x}_{n}\right)$
(11)
for all $n\ge \nu$. Similarly, we can obtain
$\begin{array}{rcl}d\left(w,T{x}_{n}\right)& <& \left[\frac{a}{1-b}+\frac{\left(p-1\right){b}^{p-1}}{1-{b}^{p-1}}\right]d\left(T{x}_{n},{T}^{2}{x}_{n}\right)\\ <& \left[\frac{a}{1-b}+\frac{\left(p-1\right){b}^{p-1}}{1-{b}^{p-1}}\right]d\left({x}_{n},T{x}_{n}\right)\end{array}$
for all $n\ge \nu$. Using (11), we get
$d\left({x}_{n},T{x}_{n}\right)\le d\left(w,{x}_{n}\right)+d\left(w,T{x}_{n}\right)<\left[\frac{2a}{1-b}+\frac{p{b}^{p}}{1-{b}^{p}}+\frac{\left(p-1\right){b}^{p-1}}{1-{b}^{p-1}}\right]d\left({x}_{n},T{x}_{n}\right)$

for all $n\ge \nu$. Thus, by (10), we obtain $d\left({x}_{n},T{x}_{n}\right), which is a contradiction. Therefore there exists $z\in X$ such that $Tz=z$. Fix $y\in X$ with $y\ne x$. Then since $ad\left(x,Tx\right)+bd\left(y,Tx\right)=bd\left(y,x\right), we have $d\left(Ty,x\right)=d\left(Ty,Tx\right) and hence y is not a fixed point of T. Therefore, the fixed point of T is unique. □

Remark 2 The proof of Theorem 9 is available for $a=1/2$, $b=0$. In this case we obtained Theorem 6. We do not know if Theorem 9 is still correct for $a=0$, $b=1$, or, more generally, for $2a+b=1$. This is an open question.

Example 2 Define a complete metric space X by $X=\left\{A,B,C,D,E\right\}$ such that $d\left(A,B\right)=d\left(A,C\right)=d\left(B,D\right)=d\left(C,D\right)=2$, $d\left(A,D\right)=d\left(B,C\right)=3$, $d\left(A,E\right)=d\left(C,E\right)=5/2$, $d\left(B,E\right)=d\left(D,E\right)=1$ and a mapping T on X by $TA=B$, $TB=E$, $TC=D$, $TD=E$, $TE=E$. Then T satisfies our condition from Theorem 9 for $a=1/8$, $b=2/3$, but T does not satisfy Suzuki’s condition from Theorem 6.

Proof We have $d\left(A,C\right)=2=d\left(TA,TC\right)$ and $\left(1/2\right)d\left(A,TA\right)=1, so T does not satisfy Suzuki’s condition from Theorem 6. Moreover, we have $ad\left(A,TA\right)+bd\left(C,TA\right)=ad\left(C,TC\right)+bd\left(A,TC\right)=2a+3b=9/4>d\left(A,C\right)$. It is now obvious that T satisfies our condition from Theorem 9. □

## Declarations

### Acknowledgements

The author is highly indebted to the referees for their careful reading of the manuscript and valuable suggestions.

## Authors’ Affiliations

(1)
Department of Mathematics and Computer Sciences, Transilvania University, Iuliu Maniu, Brasov, Romania

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