Some generalizations of Suzuki and Edelstein type theorems

Abstract

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

MSC:54H25.

Introduction and preliminaries

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

$$d(Tx,Ty)\le rd(x,y)$$

for all $x,y\in X$.

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

Theorem 1 (Banach [1])

Let $(X,d)$ 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 [2–19].

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

$$\theta (r)=\{\begin{array}{ll}1& \mathit{\text{if }}0\le r\le (\sqrt{5}-1)/2,\\ (1-r)/r^2& \mathit{\text{if }}(\sqrt{5}-1)/2\le r\le 1/\sqrt{2},\\ 1/(1+r)& \mathit{\text{if }}1/\sqrt{2}\le r<1.\end{array}$$

(1)

Assume that there exists $r\in [0,1)$ such that $\theta (r)d(x,Tx)\le d(x,y)$ implies $d(Tx,Ty)\le rd(x,y)$ 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 [30–32] 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 $(X,d)$ be a complete metric space and $T:X\to X$ be a $(s,r)$-contractive single-valued operator:

$$x,y\in X\mathit{\text{with }}d(y,Tx)\le sd(y,x)\mathit{\text{implies}}d(Tx,Ty)\le r{M}_T(x,y),$$

where $r\in [0,1)$, $s>r$ and

$$M_T(x,y)=max\{d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Ty)+d(y,Tx)}{2}\}.$$

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 $(X,d)$ be a complete metric space, and let T be a mapping on X. Assume that there exist $r\in [0,1)$ and $s>r$ such that

$$d(y,Tx)\le sd(y,x)\mathit{\text{implies}}d(Tx,Ty)\le rd(x,y)$$

(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 $(X,d)$ be a compact metric space, and let T be a mapping on X. Assume $d(Tx,Ty)<d(x,y)$ 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 [35–38]).

Theorem 6 [34]

Let $(X,d)$ be a compact metric space, and let T be a mapping on X. Assume that $(1/2)d(x,Tx)<d(x,y)$ implies $d(Tx,Ty)<d(x,y)$ 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 $(X,d)$ be a complete metric space, and let T be a mapping on X. Assume that there exist $r\in [0,1)$, $a\in [0,1]$, $b\in [0,1)$, $(a+b)r^2+r\le 1$ if $r\in [1/2,1/\sqrt{2})$, $a+(a+b)r\le 1$ if $r\in [1/\sqrt{2},1)$ such that

$$ad(x,Tx)+bd(y,Tx)\le d(y,x)\mathit{\text{implies}}d(Tx,Ty)\le rd(x,y)$$

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 $ad(x,Tx)+bd(Tx,Tx)=ad(x,Tx)\le d(Tx,x)$ holds for every $x\in X$, by hypothesis, we get

$$d(Tx,T^{2}x)\le rd(x,Tx)$$

(3)

for all $x\in X$. We now fix $u\in X$ and define a sequence $\{u_n\}\in X$ by $u_n=T^{n}u$. Then (3) yields $d(u_n,u_{n+1})\le r^{n}d(u,Tu)$, so ${\sum }_{n=1}^{\mathrm{\infty }}d(u_n,u_{n+1})<\mathrm{\infty }$. Hence $\{u_n\}$ is a Cauchy sequence. Since X is complete, $\{u_n\}$ converges to some point $z\in X$. We next show that

$$d(Tx,z)\le rd(x,z)$$

(4)

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

Now we assume that $T^{j}z\ne z$ for every integer $j\ge 1$. Then (4) yields

$$d(T^{j+1}z,z)\le r^{j}d(Tz,z)$$

(5)

for every integer $j\ge 1$. We consider the following three cases:

• (a) $0\le r<1/2$,

• (b) $1/2\le r<1/\sqrt{2}$,

• (c) $1/\sqrt{2}\le r<1$.

In the case (a) we note that $2r<1$. Then, by (3) and (5), we have

$$d(z,Tz)\le d(z,T^{2}z)+d(Tz,T^{2}z)\le rd(z,Tz)+rd(z,Tz)=2rd(z,Tz)<d(z,Tz).$$

This is a contradiction.

In the case (b), we note that $2r^2<1$. If we assume $ad(T^{2}z,T^{3}z)+bd(z,T^{3}z)>d(z,T^{2}z)$, then we have, in view of (3) and (5),

$$\begin{array}{rcl}d(z,Tz)& \le & d(z,T^{2}z)+d(Tz,T^{2}z)\\ <& ad(T^{2}z,T^{3}z)+bd(z,T^{3}z)+d(Tz,T^{2}z)\\ \le & a{r}^{2}d(z,Tz)+b{r}^{2}d(z,Tz)+rd(z,Tz)\\ =& [(a+b)r^2+r]d(z,Tz)\\ \le & d(z,Tz).\end{array}$$

This is a contradiction. Hence $ad(T^{2}z,T^{3}z)+bd(z,T^{3}z)\le d(z,T^{2}z)$. By hypothesis and (5), we have

$$\begin{array}{rcl}d(z,Tz)& \le & d(z,T^{3}z)+d(Tz,T^{3}z)\\ \le & r^{2}d(z,Tz)+rd(z,T^{2}z)\\ \le & r^{2}d(z,Tz)+r^{2}d(z,Tz)\\ =& 2r^{2}d(z,Tz)\\ <& d(z,Tz).\end{array}$$

This is also a contradiction.

In the case (c), we assume there exists an integer $\nu \ge 1$ such that

$$ad(u_n,u_{n+1})+bd(z,u_{n+1})>d(z,u_n)$$

for all $n\ge \nu$. Then

$$\begin{array}{rcl}d(z,u_n)& <& ad(u_n,u_{n+1})+b[ad(u_{n+1},u_{n+2})+bd(z,u_{n+2})]\\ \le & (a+abr)d(u_n,u_{n+1})+b^{2}d(z,u_{n+2})\\ <& (a+abr)d(u_n,u_{n+1})+b^2[ad(u_{n+2},u_{n+3})+bd(z,u_{n+3})]\\ \le & (a+abr+a{b}^2{r}^2)d(u_n,u_{n+1})+b^{3}d(z,u_{n+3}).\end{array}$$

Continuing this process, we get

$$\begin{array}{rcl}d(z,u_n)& <& (a+abr+a{b}^2{r}^2+\cdots +a{b}^{p-1}{r}^{p-1})d(u_n,u_{n+1})+b^{p}d(z,u_{n+p})\\ \le & a\frac{1-{(br)}^p}{1-br}d(u_n,u_{n+1})+b^{p}d(z,u_{n+p})\end{array}$$

for all $n\ge \nu$, $p\ge 1$. Letting p tend to ∞, we obtain

$$d(z,u_n)\le \frac{a}{1-br}d(u_n,u_{n+1})$$

for all $n\ge \nu$. Thus,

$$d(z,u_{n+1})\le \frac{a}{1-br}d(u_{n+1},u_{n+2})\le \frac{ar}{1-br}d(u_n,u_{n+1})$$

for all $n\ge \nu$, so

$$\begin{array}{rcl}d(u_n,u_{n+1})& \le & d(z,u_n)+d(z,u_{n+1})\\ <& \frac{a}{1-br}d(u_n,u_{n+1})+\frac{ar}{1-br}d(u_n,u_{n+1})\\ =& \frac{a+ar}{1-br}d(u_n,u_{n+1})\\ \le & d(u_n,u_{n+1})\end{array}$$

for all $n\ge \nu$. This is a contradiction. Hence there exists a subsequence $\{u_{n(k)}\}$ of $\{u_n\}$ such that

$$ad(u_{n(k)},u_{n(k)+1})+bd(z,u_{n(k)+1})\le d(z,u_{n(k)})$$

for all $k\ge 1$. By hypothesis, we get $d(Tz,T{u}_{n(k)})\le rd(z,u_{n(k)})$ for all $k\ge 1$. Letting k tend to ∞, we get $d(z,Tz)=0$, 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 $d(z,Tz)=d(T^{j}z,T^{j+1}z)\le r^{j}d(z,Tz)$, so $d(z,Tz)=0$, that is, $Tz=z$.

Now we suppose that y is another fixed point of T, that is, $Ty=y$. Then

$$ad(y,Ty)+bd(z,Ty)=bd(z,y)\le d(z,y),$$

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

Remark 1 For $r\in [0,1/2)$, taking $a=1$, $b=0$, we obtain Suzuki’s condition from Theorem 2. Moreover, from our condition and the triangle inequality, we get

$$ad(x,Tx)+b[d(x,Tx)-d(y,x)]\le d(y,x),$$

that is,

$$\frac{a+b}{1+b}d(x,Tx)\le d(y,x).$$

If $r\in [1/\sqrt{2},1)$, we have

$$\frac{a+b}{1+b}=\frac{1}{1+r}=\theta (r),$$

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

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

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

$$\frac{s-r}{1+r}d(x,Tx)+d(y,Tx)\le sd(y,x)\mathit{\text{implies}}d(Tx,Ty)\le rd(x,y)$$

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

$$0=d(u_{n+1},T{u}_n)\le sd(u_{n+1},u_n)-\frac{s-r}{1+r}d(u_n,T{u}_n),$$

we get from hypothesis $d(u_{n+1},u_{n+2})\le rd(u_{n+1},u_n)$ for all $n\ge 1$. Therefore, $d(u_{n+1},u_{n+2})\le r^{n}d(u_1,u_2)$ for all $n\ge 1$. Thus

$$\sum _{n=1}^{\mathrm{\infty }}d(u_{n+1},u_n)\le \sum _{n=1}^{\mathrm{\infty }}{r}^{n-1}d(u_1,u_2)<\mathrm{\infty }.$$

Hence $\{u_n\}$ is a Cauchy sequence. Since X is complete, $\{u_n\}$ converges to some point $z\in X$.

Now, we will show that there exists a subsequence $\{u_{n(k)}\}$ of $\{u_n\}$ such that

$$d(z,T{u}_{n(k)})\le sd(z,u_{n(k)})-\frac{s-r}{1+r}d(u_{n(k)},T{u}_{n(k)})$$

for all $k\ge 1$. Arguing by contradiction, we suppose that there exists a positive integer ν such that

$$d(z,T{u}_n)>sd(z,u_n)-\frac{s-r}{1+r}d(u_n,T{u}_n)$$

for all $n\ge \nu$. Then we have

$$\begin{array}{rcl}d(z,u_{n+2})& >& sd(z,u_{n+1})-\frac{s-r}{1+r}d(u_{n+1},u_{n+2})\\ >& s^{2}d(z,u_n)-s\cdot \frac{s-r}{1+r}d(u_n,u_{n+1})-\frac{s-r}{1+r}d(u_{n+1},u_{n+2})\\ \ge & s^{2}d(z,u_n)-\frac{s-r}{1+r}[sd(u_n,u_{n+1})+rd(u_n,u_{n+1})]\\ =& s^{2}d(z,u_n)-\frac{s-r}{1+r}(s+r)d(u_n,u_{n+1}).\end{array}$$

By induction, we get for all $n\ge \nu$, $p\ge 1$ that

$$d(z,u_{n+p})>s^{p}d(z,u_n)-\frac{s-r}{1+r}(s^{p-1}+s^{p-2}r+\cdots +r^{p-1})d(u_n,u_{n+1}).$$

Then we have

$$\begin{array}{rcl}d(z,u_{n+p})& >& s^{p}d(z,u_n)-\frac{s-r}{1+r}\cdot s^{p-1}\cdot \frac{1-{(r/s)}^p}{1-r/s}d(u_n,u_{n+1})\\ =& s^p[d(z,u_n)-\frac{s-r}{1+r}\cdot \frac{1-{(r/s)}^p}{s-r}d(u_n,u_{n+1})].\end{array}$$

Hence

$$s^p[d(z,u_n)-\frac{1-{(r/s)}^p}{1+r}d(u_n,u_{n+1})]<d(z,u_{n+p}).$$

(6)

On the other hand,

$$\begin{array}{rcl}d(u_{n+p},u_n)& \le & d(u_n,u_{n+1})+d(u_{n+1},u_{n+2})+\cdots +d(u_{n+p-1},u_{n+p})\\ \le & (1+r+\cdots +r^{p-1})d(u_n,u_{n+1})\\ =& \frac{1-r^p}{1-r}d(u_n,u_{n+1}).\end{array}$$

Letting $p\to \mathrm{\infty }$, we get for all $n\ge 1$ that $d(z,u_n)\le d(u_n,u_{n+1})/(1-r)$. Thus

$$d(z,u_{n+p})\le d(u_{n+p},u_{n+p+1})/(1-r)\le r^{p}d(u_n,u_{n+1})/(1-r).$$

(7)

By (6) and (7) we have for all $n\ge \nu$, $p\ge 1$ that

$$\frac{r^p}{1-r}d(u_n,u_{n+1})>s^p[d(z,u_n)-\frac{1-{(r/s)}^p}{1+r}d(u_n,u_{n+1})],$$

so

$$\frac{{(r/s)}^p}{1-r}d(u_n,u_{n+1})>d(z,u_n)-\frac{1-{(r/s)}^p}{1+r}d(u_n,u_{n+1}).$$

Taking the limit as $p\to \mathrm{\infty }$, we obtain that $d(z,u_n)\le d(u_n,u_{n+1})/(1+r)$ for all $n\ge \nu$. Then we have

$$d(z,u_{n+1})\le d(u_{n+1},u_{n+2})/(1+r)\le rd(u_n,u_{n+1})/(1+r)$$

and

$$rd(u_n,u_{n+1})/(1+r)>sd(z,u_n)-(s-r)d(u_n,u_{n+1})/(1+r).$$

This implies $d(z,u_n)<d(u_n,u_{n+1})/(1+r)$ for all $n\ge \nu$. Thus,

$$d(u_n,u_{n+1})\le d(z,u_n)+d(z,u_{n+1})<d(u_n,u_{n+1})/(1+r)+rd(u_n,u_{n+1})/(1+r)=d(u_n,u_{n+1}).$$

This is a contradiction. Therefore there exists a subsequence $\{u_{n(k)}\}$ of $\{u_n\}$ such that

$$d(z,T{u}_{n(k)})\le sd(z,u_{n(k)})-\frac{s-r}{1+r}d(u_{n(k)},T{u}_{n(k)})$$

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

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

Edelstein’s theorem

The following theorem extends Theorem 6 and generalizes Theorem 5.

Theorem 9 Let $(X,d)$ be a compact metric space, and let T be a mapping on X. Assume that

$$ad(x,Tx)+bd(y,Tx)<d(y,x)\mathit{\text{implies}}d(Tx,Ty)<d(x,y)$$

(8)

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

Proof We put

$$\beta =inf\{d(x,Tx):x\in X\}$$

and choose a sequence $\{x_n\}$ in X such that ${lim}_{n\to \mathrm{\infty }}d(x_n,T{x}_n)=\beta$. Since X is compact, without loss of generality, we may assume that $\{x_n\}$ and $\{T{x}_n\}$ converge to some elements $v,w\in X$, respectively. We have

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,w)=\underset{n\to \mathrm{\infty }}{lim}d(T{x}_n,v)=d(v,w)=\beta .$$

We shall show $\beta =0$. Arguing by contradiction, we assume $\beta >0$. Since

$$\underset{n\to \mathrm{\infty }}{lim}[ad(x_n,T{x}_n)+bd(w,T{x}_n)]=a\beta <\beta =\underset{n\to \mathrm{\infty }}{lim}d(w,x_n),$$

we can choose a positive integer ν such that

$$ad(x_n,T{x}_n)+bd(w,T{x}_n)<d(w,x_n)$$

for all $n\ge \nu$. By hypothesis, $d(Tw,T{x}_n)<d(w,x_n)$ holds for $n\ge \nu$. This implies

$$d(w,Tw)=\underset{n\to \mathrm{\infty }}{lim}d(Tw,T{x}_n)\le \underset{n\to \mathrm{\infty }}{lim}d(w,x_n)=\beta .$$

From the definition of β, we obtain $d(w,Tw)=\beta$. Since $ad(w,Tw)+bd(Tw,Tw)<d(Tw,w)$, we have

$$d(Tw,T^{2}w)<d(w,Tw)=\beta ,$$

which contradicts the definition of β. Therefore we obtain $\beta =0$. We have ${lim}_{n\to \mathrm{\infty }}d(x_n,w)={lim}_{n\to \mathrm{\infty }}d(T{x}_n,v)={lim}_{n\to \mathrm{\infty }}d(T{x}_n,x_n)=d(v,w)=0$, 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 $ad(x_n,T{x}_n)+bd(T{x}_n,T{x}_n)<d(T{x}_n,x_n)$ for all $n\ge 1$, we get $d(T^2{x}_n,T{x}_n)<d(T{x}_n,x_n)$, so ${lim}_{n\to \mathrm{\infty }}{T}^2{x}_n=w$. By induction, we obtain that $d(T^p{x}_n,T^{p+1}{x}_n)<d(T^{p-1}{x}_n,T^p{x}_n)<\cdots <d(x_n,T{x}_n)$ 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 $\{x_{n(k)}\}$ of $\{x_n\}$ such that

$$ad(T^{p-1}{x}_{n(k)},T^p{x}_{n(k)})+bd(w,T^p{x}_{n(k)})<d(w,T^{p-1}{x}_{n(k)})$$

for all $k\ge 1$, by hypothesis we get $d(Tw,T^p{x}_{n(k)})<d(w,T^{p-1}{x}_{n(k)})$. Taking the limit as $k\to \mathrm{\infty }$, we obtain $d(w,Tw)=0$, that is, $Tw=w$, which is a contradiction. Hence, we can assume that for every $m\ge 1$, there exists an integer $n(m)\ge 1$ such that

$$ad(T^{m-1}{x}_n,T^m{x}_n)+bd(w,T^m{x}_n)\ge d(w,T^{m-1}{x}_n)$$

(9)

for all $n\ge n(m)$. 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

$$\frac{p{b}^p}{1-b^p}+\frac{(p-1)b^{p-1}}{1-b^{p-1}}+\frac{2a}{1-b}<1.$$

(10)

We put $\nu =max\{n(1),n(2),\dots ,n(p)\}$. Then by (9) we have

$$\begin{array}{rcl}d(w,x_n)& \le & ad(x_n,T{x}_n)+bd(w,T{x}_n)\\ \le & ad(x_n,T{x}_n)+b[ad(T{x}_n,T^2{x}_n)+bd(w,T^2{x}_n)]\\ =& ad(x_n,T{x}_n)+abd(T{x}_n,T^2{x}_n)+b^{2}d(w,T^2{x}_n)\\ \le & \cdots \\ \le & ad(x_n,T{x}_n)+abd(T{x}_n,T^2{x}_n)+\cdots \\ +a{b}^{p-1}d(T^{p-2}{x}_n,T^{p-1}{x}_n)+b^{p}d(w,T^p{x}_n)\\ \le & (a+ab+\cdots +a{b}^{p-1})d(x_n,T{x}_n)+b^{p}d(w,T^p{x}_n)\\ \le & [a(1-b^p)/(1-b)]d(x_n,T{x}_n)+b^{p}d(w,T^p{x}_n)\end{array}$$

for all $n\ge \nu$. Since

$$\begin{array}{rcl}d(w,T^p{x}_n)& \le & d(w,x_n)+d(x_n,T{x}_n)+\cdots +d(T^{p-1}{x}_n,T^p{x}_n)\\ <& d(w,x_n)+pd(x_n,T{x}_n),\end{array}$$

we get

$$d(w,x_n)<[a(1-b^p)/(1-b)]d(x_n,T{x}_n)+b^p[d(w,x_n)+pd(x_n,T{x}_n)],$$

so

$$d(w,x_n)<(\frac{a}{1-b}+\frac{p{b}^p}{1-b^p})d(x_n,T{x}_n)$$

(11)

for all $n\ge \nu$. Similarly, we can obtain

$$\begin{array}{rcl}d(w,T{x}_n)& <& [\frac{a}{1-b}+\frac{(p-1)b^{p-1}}{1-b^{p-1}}]d(T{x}_n,T^2{x}_n)\\ <& [\frac{a}{1-b}+\frac{(p-1)b^{p-1}}{1-b^{p-1}}]d(x_n,T{x}_n)\end{array}$$

for all $n\ge \nu$. Using (11), we get

$$d(x_n,T{x}_n)\le d(w,x_n)+d(w,T{x}_n)<[\frac{2a}{1-b}+\frac{p{b}^p}{1-b^p}+\frac{(p-1)b^{p-1}}{1-b^{p-1}}]d(x_n,T{x}_n)$$

for all $n\ge \nu$. Thus, by (10), we obtain $d(x_n,T{x}_n)<d(x_n,T{x}_n)$, which is a contradiction. Therefore there exists $z\in X$ such that $Tz=z$. Fix $y\in X$ with $y\ne x$. Then since $ad(x,Tx)+bd(y,Tx)=bd(y,x)<d(y,x)$, we have $d(Ty,x)=d(Ty,Tx)<d(y,x)$ 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 $X=\{A,B,C,D,E\}$ such that $d(A,B)=d(A,C)=d(B,D)=d(C,D)=2$, $d(A,D)=d(B,C)=3$, $d(A,E)=d(C,E)=5/2$, $d(B,E)=d(D,E)=1$ 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 $d(A,C)=2=d(TA,TC)$ and $(1/2)d(A,TA)=1<d(A,C)=2$, so T does not satisfy Suzuki’s condition from Theorem 6. Moreover, we have $ad(A,TA)+bd(C,TA)=ad(C,TC)+bd(A,TC)=2a+3b=9/4>d(A,C)$. It is now obvious that T satisfies our condition from Theorem 9. □