1 Introduction

Fixed point theory plays a fundamental role in nonlinear analysis and its applications. One of the basic and important fixed point results is the Banach contraction principle. It states that if T is a contraction mapping on a complete metric (X, d), i.e., there exists \(k\in [0,1)\) such that

$$\begin{aligned} d(Tx,Ty)\le kd(x,y)\quad \forall x,y\in X, \end{aligned}$$
(1)

then T has a unique fixed point. This principle has been a powerful tool for proving the existence of solutions of differential equations, integral equations, nonlinear equations, etc., and for solving various problems in mathematical science and engineering. The Banach principle has been extended, generalized and improved in several directions. In 1962, Edelstein [8] proved the following fixed point theorem.

Theorem 1

[8] Let (X, d) be a compact metric space and \(T:X\rightarrow X\) be a self-mapping. If

$$\begin{aligned} d(Tx,Ty)<d(x,y)\quad \forall x,y\in X, x\ne y, \end{aligned}$$
(2)

then T has a unique fixed point.

Examples show that the compactness of the metric space (X, d) cannot be replaced by its completeness. The following is such an example which may be known.

Example 1

Let \(X=[2,\infty )\) with the usual metric \(d(x,y)=|x-y|\) for all \(x,y\in X\). Then, (X, d) is a complete metric space but it is not compact. We consider the mapping \(T:X\rightarrow X\) defined by \(Tx = x +\displaystyle \frac{2}{x}\) for all \(x\in X\). For all \(x,y\in X\) with \(x\ne y\), we have

$$\begin{aligned} d(Tx,Ty) = \left| x+\frac{2}{x}-y-\frac{2}{y}\right| <|x-y| = \left( 1-\frac{2}{xy}\right) d(x,y) =d(x,y). \end{aligned}$$

Thus, condition (2) is satisfied. However, T has no fixed point in X.

It is noted that if \(T:X\rightarrow X\) satisfies condition (2), then T is continuous on X. Thus, the function \(f: X\rightarrow \mathbb {R}\) defined by \(f(x) = d(x,Tx)\) is continuous on X. The standard proof of Theorem 1 is based on the fact that the continuous function f on the compact space X attains its minimum on X. Many results which generalize and extend Edelstein’s result use this proof technique and require some continuity properties of the considered mappings, see, for examples [6, 9, 13, 14, 20] and references therein.

In 2009, Suzuki [23] obtained a very interesting generalization of Edelstein’s theorem which is stated as follows.

Theorem 2

[23] Let (X, d) be a compact metric space and let \(T:X\rightarrow X\). Assume that

$$\begin{aligned} \frac{1}{2}d(x,Tx)< d(x,y) \quad \Longrightarrow \quad d(Tx,Ty)<d(x,y) \end{aligned}$$
(3)

holds for all \(x,y\in X\). Then T has a unique fixed point in X.

Note that a mapping \(T:X\rightarrow X\) satisfying condition (3) is not necessarily continuous and Theorem 2 is a real generalization of Theorem 1. The idea of the proof of Theorem 2 may be used to prove some generalizations of Edelstein’s theorem without assuming that the involving mappings are continuous (see, e.g., [7, 15, 16]). For further fixed point results involving Suzuki or Suzuki–Edelstein type contractions, we refer the reader to [2, 3, 17, 18] and references therein.

Recently, Altun et.al. [4] proved a new extension of Edelstein’s theorem.

Theorem 3

[4, Theorem 2.11] Let (X, d) be a compact metric space and \(T:X\rightarrow X\) be a P-contractive mapping, i.e.,

$$\begin{aligned} d(Tx,Ty)<d(x,y) + |d(x,Tx)-d(y,Ty)|,\quad \text{ for } \text{ all }\,\,x,y\in X, x\ne y. \end{aligned}$$
(4)

If \(f(x) = d(x,Tx)\) is lower semi-continuous, then T has a unique fixed point.

In [4], the authors also presented several examples showing that Theorem 3 and Theorem 2 are independent. We refer the reader to [5] for further results concerning P- contractive mappings.

It is worth noting that the expression

$$\begin{aligned} E(x,y): = d(x,y) + |d(x,Tx)-d(y,Ty)| \end{aligned}$$

in the right-hand side of (4) was first appeared in the papers [10, 11] by A. Fulga and A. M. Proca where they presented some fixed point theorems for a self-mapping T in a metric space (X, d) satisfying some generalized contractive conditions which involve the expression \(E(x,y) = d(x,y) + |d(x,Tx)-d(y,Ty)|\). For more interesting fixed point results for mappings satisfying conditions involving E-contractions (i.e., conditions which involve the expression \(E(x,y) = d(x,y) + |d(x,Tx)-d(y,Ty)|\)), we refer the reader to, e.g., [12, 19] and references therein.

Some questions may arise:

Question 1

  1. (i)

    Do the conclusion of Theorem 3 still hold true if we do not require any assumption on the function f?

  2. (ii)

    Can we provide a Theorem which generalize both Theorem 2 and Theorem 3?

  3. (iii)

    Can we replace the compactness of X in Theorem 2 by other conditions on X?

Our aim is to answer these questions and provide some new fixed point of Edelstein - Suzuki type involving E-contraction mappings. Our results extend, improve and unify some known results in the literature. Several examples are also given to illustrate our results.

2 Fixed point results

Our first result is a generalization of Theorem 3.

Theorem 4

Let (X, d) be a compact metric space and let T be a self mapping on X. Assume that there exists \(C\ge 0\) such that

$$\begin{aligned}&\frac{1}{2}d(x,Tx)<d(x,y)\nonumber \\&\quad \Longrightarrow \quad d(Tx,Ty)< d(x,y) + |d(x,Tx)-d(y,Ty)| + Cd(y,Tx) \end{aligned}$$
(5)

for all \(x,y\in X\). If \(f(x) = d(x,Tx)\) is lower semi-continuous, then T has a fixed point in X.

Proof

Since \(f(x) = d(x,Tx)\) is lower semi-continuous, there exists \(x_0\in X\) such that \(f(x_0)\le f(Tx_0)\), i.e., \(d(x_0,Tx_0)\le d(Tx_0,T^2x_0)\) (see, e.g., [1]). Assume that \(x_0\ne Tx_0\). Then

$$\begin{aligned} \frac{1}{2}d(x_0,Tx_0) <d(x_0,Tx_0)\le d(Tx_0,T^2x_0). \end{aligned}$$

By (5), we have

$$\begin{aligned} d(Tx_0,T^2x_0)< & {} d(x_0,Tx_0) + |d(x_0,Tx_0)-d(Tx_0,T^2x_0)| + Cd(Tx_0,Tx_0)\\= & {} d(x_0,Tx_0) + d(Tx_0,T^2x_0) - d(x_0,Tx_0)\\= & {} d(Tx_0,T^2x_0) \end{aligned}$$

which is a contradiction. Thus, \(x_0 = Tx_0\) and T has a fixed point. \(\square \)

The following simple example showing that if \(C>0\), then T may have more than one fixed point.

Example 2

Let \(X= [0,1]\) with the usual metric \(d(x,y) = |x-y|\) for all \(x,y\in X\) and \(T:X\rightarrow X\) be defined by \(Tx = x\) for all \(x\in X\). For all \(x,y\in X\), \(x\ne y\), we have \(d(Tx,Ty) = |x-y|\) and

$$\begin{aligned} d(x,y) + |d(x,Tx)-d(y,Ty)| + Cd(y,Tx) = |x-y| + C|x-y| = (1+C)|x-y|. \end{aligned}$$

Since \(C>0\), we have

$$\begin{aligned} d(Tx,Ty)=|x-y|<(1+C)|x-y| = d(x,y) + |d(x,Tx)-d(y,Ty)| + Cd(y,Tx). \end{aligned}$$

Thus all conditions of Theorem 4 are satisfied. Of course, every \(x\in X\) is a fixed point of T.

The following corollary is a generalization of Theorem 2.11 and Theorem 2.14 in [4].

Corollary 1

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

$$\begin{aligned} \frac{1}{2}d(x,Tx)<d(x,y)\quad \Longrightarrow \quad d(Tx,Ty)< d(x,y) + |d(x,Tx)-d(y,Ty)| \end{aligned}$$

for \(x,y\in X\). If \(f(x) = d(x,Tx)\) is lower semi-continuous, then T has a unique fixed point in X.

Proof

According to Theorem 4, T has a fixed point \(z\in X\). Assume that T has another fixed point \(y\in X\) with \(y\ne z\). Then \(\displaystyle \frac{1}{2} d(z,Tz)=0 <d(z,y)\). Hence, by (6), we have

$$\begin{aligned} d(z,y) = d(Tz,Ty) <d(z,y) + L|d(z,Tz)-d(y,Ty)| = d(z,y), \end{aligned}$$

a contradiction. Therefore, T has a unique fixed point. \(\square \)

Question 2

Does the conclusion of Theorem 4 still hold true if we remove condition “\(f(x) = d(x,Tx)\) is lower semi-continuous”?

In the next theorem, we give a partial answer to the above question. This is also a partial answer to Question 1(ii).

Theorem 5

Let (X, d) be a compact metric space and let T be a self mapping on X. Assume that there exist \(L\in [0,1)\) and \(C\ge 0\) such that

$$\begin{aligned}&\frac{1}{2}d(x,Tx)<d(x,y)\nonumber \\&\quad \Longrightarrow \quad d(Tx,Ty)< d(x,y) + L|d(x,Tx)-d(y,Ty)| + Cd(y,Tx) \end{aligned}$$
(6)

for \(x,y\in X\). Then T has a fixed point in X.

Proof

Set

$$\begin{aligned} m = \inf \{d(x,Tx):x\in X\}. \end{aligned}$$

Then, there exists a sequence \(\{x_n\}\) in X such that

$$\begin{aligned} \lim _{n\rightarrow \infty }d(x_n,Tx_n)=m. \end{aligned}$$

Since X is compact, we may assume that

$$\begin{aligned} \lim _{n\rightarrow \infty }x_n = a,\quad \text{ and }\quad \lim _{n\rightarrow \infty }Tx_n = b \end{aligned}$$

for some \(a,b\in X\). Hence,

$$\begin{aligned} m = \lim _{n\rightarrow \infty }d(x_n,Tx_n) = \lim _{n\rightarrow \infty }d(x_n,b) = \lim _{n\rightarrow \infty }d(a,Tx_n) = d(a,b). \end{aligned}$$

We are going to show that \(m=0\). Assume to the contrary that \(m>0\). Then, there exists \(n_0\in \mathbb {N}\) such that

$$\begin{aligned} d(x_n,b)>\frac{2}{3}m \quad \text{ and }\quad d(x_n,Tx_n) <\frac{4}{3}m \end{aligned}$$

for all \(n\ge n_0\). That is,

$$\begin{aligned} \frac{1}{2}d(x_n,Tx_n) <d(x_n,b) \quad \text{ for } \text{ all }\,\,\,n\ge n_0. \end{aligned}$$

By (6), one has, for all \(n\ge n_0\), that

$$\begin{aligned} d(Tx_n,Tb) < d(x_n,b) + L|d(x_n,Tx_n)-d(b,Tb)| + Cd(b,Tx_n). \end{aligned}$$

Letting \(n\rightarrow \infty \), we get

$$\begin{aligned} d(b,Tb) \le m + L|m - d(b,Tb)| + C.0. \end{aligned}$$

By the definition of m, it follows from the latter inequality that

$$\begin{aligned} d(b,Tb) \le m + L(d(b,Tb)-m)= (1-L)m + Ld(b,Tb) \end{aligned}$$

which yields \(d(b,Tb)\le m\) since \(L\in [0,1)\). Again, by the definition of m, we obtain \(d(b,Tb)=m\). This implies that \(\displaystyle \frac{1}{2}d(b,Tb)<d(b,Tb)\). By (6), we have

$$\begin{aligned} d(Tb,T^2b) <d(b,Tb) + L|d(b,Tb)-d(Tb,T^2b)| + Cd(Tb,Tb) = m + L|m-d(Tb,T^2b)|. \end{aligned}$$

Again, by the definition of m and by the fact that \(L\in [0,1)\), we arrive at \(d(Tb,T^2b)<m\) which is a contradiction. Thus, \(m = 0\) and

$$\begin{aligned} \lim _{n\rightarrow \infty }x_n=\lim _{n\rightarrow \infty }Tx_n = a. \end{aligned}$$

We shall show that T has a fixed point in X. Assume to the contrary that T has no fixed point. Then \(d(x_n,Tx_n)>0\) for all \(n\in \mathbb {N}\). This implies that

$$\begin{aligned} \frac{1}{2}d(x_n,Tx_n)<d(x_n,Tx_n)\quad \text{ for } \text{ all }\quad n\in \mathbb {N}. \end{aligned}$$

By (6), we have for all n that

$$\begin{aligned} d(Tx_n,T^2x_n)< & {} d(x_n,Tx_n) + L|d(x_n,Tx_n) - d(Tx_n,T^2x_n)|+ Cd(Tx_n,Tx_n)\nonumber \\= & {} d(x_n,Tx_n) + L|d(x_n,Tx_n) - d(Tx_n,T^2x_n)|. \end{aligned}$$
(7)

Suppose that there exists \(k\in \mathbb {N}\) such that \(d(Tx_k,T^2x_k)\ge d(x_k,Tx_k)\). By (7),

$$\begin{aligned} d(Tx_k,T^2x_k)< & {} d(x_k,Tx_k)+ L(d(Tx_k,T^2x_k)-d(x_k,Tx_k) \\= & {} (1-L)d(x_k,Tx_k) + Ld(Tx_k,T^2x_k) \end{aligned}$$

which yields \(d(Tx_k,T^2x_k) < d(x_k,Tx_k)\), a contradiction. Thus,

$$\begin{aligned} d(Tx_n,T^2x_n)< d(x_n,Tx_n)\quad \text{ for } \text{ all } \quad n\in \mathbb {N}. \end{aligned}$$

We have

$$\begin{aligned} \lim _{n\rightarrow \infty }d(a,T^2x_n)\le & {} \lim _{n\rightarrow \infty }[d(a,Tx_n) + d(Tx_n,T^2x_n)]\\\le & {} \lim _{n\rightarrow \infty }[d(a,Tx_n) + d(x_n,Tx_n)] = 0. \end{aligned}$$

Hence, \(\lim _{n\rightarrow \infty }T^2x_n=a\).

Assume now that there exists \(N\in \mathbb {N}\) such that

$$\begin{aligned} \frac{1}{2}d(x_N,Tx_N)\ge d(x_N,a)\quad \text{ and }\quad \frac{1}{2}d(Tx_N,T^2x_N)\ge d(Tx_N,a). \end{aligned}$$

Then, by the triangular inequality, one has

$$\begin{aligned} d(x_N,Tx_N)\le & {} d(x_N,a)+d(a,Tx_N) \le \frac{1}{2}d(x_N,Tx_N)+ \frac{1}{2}d(Tx_N,T^2x_N\\< & {} \frac{1}{2}d(x_N,Tx_N) + \frac{1}{2}d(x_N,Tx_N) = d(x_N,Tx_N), \end{aligned}$$

a contradiction. Thus, for each \(n\in \mathbb {N}\),

$$\begin{aligned} \frac{1}{2}d(x_n,Tx_n)<d(x_n,a)\quad \text{ or }\quad \frac{1}{2}d(Tx_n,T^2x_n)<d(Tx_n,a) \end{aligned}$$

holds true. For the first case, one gets

$$\begin{aligned} d(Tx_n,Ta) <d(x_n,a) + L|d(x_n,Tx_n)-d(a,Ta)| + Cd(a,Tx_n), \end{aligned}$$
(8)

and for the second case, one gets

$$\begin{aligned} d(T^2x_n,Ta)< d(Tx_n,a) + L|d(Tx_n,T^2x_n)-d(a,Ta)| + Cd(a,T^2x_n). \end{aligned}$$
(9)

If (8) holds true for n in an infinite subset I of \(\mathbb {N}\), then passing to the limit when \(n\in I\), \(n\rightarrow \infty \), we obtain

$$\begin{aligned} d(a,Ta) \le d(a,a) + L|d(a,a)-d(a,Ta)| + Cd(a,a) = Ld(a,Ta) \end{aligned}$$

which implies \(d(a,Ta)=0\), or \(Ta=a\) since \(L\in [0,1)\).

If (9) holds for infinitely many n in \(\mathbb {N}\), then we can treat in the same way with the first case and arrive at \(Ta = a\). Hence, in any case, \(Ta=a\). This contradicts to the assumption that T has no fixed point. Therefore, there exists \(z\in X\) such that \(Tz=z\). \(\square \)

Corollary 2

Let (X, d) be a compact metric space and let T be a self mapping on X. Assume that there exist \(L\in [0,1)\) such that

$$\begin{aligned} \frac{1}{2}d(x,Tx)<d(x,y)\quad \text{ implies }\quad d(Tx,Ty)< d(x,y) + L|d(x,Tx)-d(y,Ty)| \end{aligned}$$
(10)

for \(x,y\in X\). Then T has a unique fixed point in X.

The following corollary is a partial answer to Question 1(i).

Corollary 3

Let (X, d) be a compact metric space and let T be a self mapping on X. Assume that there exist \(L\in [0,1)\) such that

$$\begin{aligned} d(Tx,Ty)< d(x,y) + L|d(x,Tx)-d(y,Ty)| \end{aligned}$$
(11)

for \(x,y\in X\). Then T has a unique fixed point in X.

Remark 1

In Corollary 2, if we let \(L=0\), then we get Theorem 2 - the Suzuki fixed point theorem. We do not know whether the conclusion of Corollary 3 hold or not if we let \(L=1\) in Corollary 3.

Example 3

Let \(X=[-1,10]\) with the usual metric and let \(T:X\rightarrow X\) be defined by

$$\begin{aligned} Tx = \left\{ \begin{matrix} \displaystyle \frac{x}{2}&{}&{}\text{ if }&{}x\in (-1,10],\\ 5&{}&{}\text{ if }&{} x = -1. \end{matrix}\right. \end{aligned}$$

Then

  1. (i)

    T has a unique fixed point.

  2. (ii)

    T satisfies condition (10) with \(L=\displaystyle \frac{39}{40}\).

  3. (iii)

    T does not satisfy Suzuki’s condition (3).

  4. (iv)

    \(x\mapsto d(x,Tx)\) is not lower semi-continuous.

Proof

(i) It is obvious that T has a unique fixed point \(x=0\).

(ii) Case 1: For \(x,y\in (-1,10]\) and \(x\ne y\), we have \(d(Tx,Ty) = |x-y|/2<|x-y| = d(x,y)\). Thus, T satisfies condition (10) for all \(x,y\in (-1,10]\).

Case 2: Let \(x=-1\) and \(y\in (-1,10]\). We have

$$\begin{aligned} \frac{1}{2}d(x,Tx) =3<1+y= d(x,y) \end{aligned}$$

when \(y>2\). In this case,

$$\begin{aligned} d(x,y) + L|d(x,Tx)-d(y,Ty)| = 1+y + \frac{39}{40}\left| 6-\frac{y}{2} \right| = \frac{137}{20}+\frac{41y}{80}>5 -\frac{y}{2} = d(Tx,Ty), \end{aligned}$$

that is, condition (10) is satisfied for \(x=-1\), \(y\in (-1,10]\).

Case 3: Let \(x\in (-1,10]\) and \(y=-1\). We have

$$\begin{aligned} \frac{1}{2}d(x,Tx) =\frac{|x|}{4}<1+x= d(x,y) \end{aligned}$$

when \(x>-\frac{4}{5}\). In this case,

$$\begin{aligned} d(Tx,Ty)< d(x,y) + \frac{39}{40} |d(x,Tx)-d(y,Ty)| \end{aligned}$$

if and only if

$$\begin{aligned} 5-\frac{x}{2} < 1+x + \frac{39}{40}\left| \frac{|x|}{2}-6\right| . \end{aligned}$$

Similar to Case 2, the latter inequality holds for \(x\ge 0\). When \( -\frac{4}{5}<x<0\), the latter inequality is equivalent to

$$\begin{aligned} 4<\frac{3}{2}x + \frac{117}{20}+\frac{39}{80}x \end{aligned}$$

or

$$\begin{aligned} x>-\frac{148}{159}. \end{aligned}$$

Thus, condition (10) is satisfied for \(x\in (-1,10]\) and \(y=-1\).

Therefore, T satisfies condition (10) with \(L=\displaystyle \frac{39}{40}\).

(iii) For \(x = 0\) and \(y=-1\), we have

$$\begin{aligned} \frac{1}{2}d(x,Tx) = 0<1 = d(x,y)\quad \text{ and }\quad d(Tx,Ty) = 5> 1 = d(x,y) \end{aligned}$$

and then (3) does not hold.

(iv) One has

$$\begin{aligned} d(x,Tx) = \left\{ \begin{matrix} \displaystyle \frac{|x|}{2}&{}&{}\text{ if }&{}x\in (-1,10],\\ 6&{}&{}\text{ if }&{} x = -1. \end{matrix}\right. \end{aligned}$$

It is easy to see that \(x\mapsto d(x,Tx)\) is not lower semi-continuous at \(x=-1\). Thus, we cannot apply Theorem 4 and Theorem 3 to this example. \(\square \)

Example 4

Let \(X = [0,1]\) with the usual metric d. Then, (X, d) is compact. We define a mapping \(T:X\rightarrow X\) as

$$\begin{aligned} Tx = \left\{ \begin{matrix} \displaystyle \frac{x}{2}&{}&{}\text{ if }&{} x\in [0,1),\\ 1&{}&{}\text{ if }&{}x=1. \end{matrix} \right. \end{aligned}$$

We first check that condition (10) is not satisfied. Assume to the contrary that (10) is satisfied. Since

$$\begin{aligned} \frac{1}{2}d(1,T1) = 0<1-y = d(1,y) \forall y \in [0,1), \end{aligned}$$

we have for all \(y\in [0,1)\) that

$$\begin{aligned} 1-\frac{y}{2} =d(T1,Ty)< d(1,y) + L|d(1,T1)-d(y,Ty)|+ Cd(y,T1) = 1-y+L\frac{y}{2}+C(1-y) \end{aligned}$$

or, equivalently,

$$\begin{aligned} (1-L)\frac{y}{2} < C(1-y). \end{aligned}$$

Letting \(y\rightarrow 1^-\) in both sides of the latter inequality, one gets \(1-L\le 0\) which contradicts to \(L<1\). Thus, condition (10) is not satisfied.

For \(x,y\in X\), set

$$\begin{aligned} R = d(x,y) + |d(x,Tx)-d(y,Ty)|+ Cd(y,Tx). \end{aligned}$$

If \(x,y\in [0,1)\) and \(x\ne y\), then \(d(Tx,Ty) = \displaystyle \frac{1}{2}|x-y| <|x-y| = d(x,y)\). That is, (5) is satisfied in this case.

If \(x =1 \) and \(y \in [0,1)\), then

$$\begin{aligned} R = 1-\frac{y}{2}+ C(1-y) >1-\frac{y}{2} = d(T1,Ty). \end{aligned}$$

Thus, (5) is satisfied in this case.

If \(x\in [0,1)\) and \(y=1\), then

$$\begin{aligned} R = 1-\frac{x}{2} + C\left( 1-\frac{x}{2}\right) >1-\frac{x}{2} = d(Tx,T1). \end{aligned}$$

Thus, (5) is satisfied in this case.

We have checked that condition (5) is satisfied. Moreover, since

$$\begin{aligned} d(x,Tx) = \left\{ \begin{matrix} \displaystyle \frac{x}{2}&{}&{}\text{ if }&{} x\in [0,1),\\ 0&{}&{}\text{ if }&{}x=1 \end{matrix} \right. \end{aligned}$$

the function \(f(x) = d(x,Tx)\) is lower semi-continuous on X. Thus, all conditions of Theorem 4 are satisfied and Theorem 4 is applicable to this example.

Remark 2

From Example 3 and Example 4, one can see that the results in Theorem 4 and Theorem 5 are independent.

Existence of fixed points of a mapping \(T:X\rightarrow X\) satisfying condition (5) is still guaranteed if we replace both the lower semi-continuity of \(x\mapsto d(x,Tx)\) and the compactness of X by another condition on T and the completeness of X. We recall here two results ( [22, Theorem 5] and [24, Theorem 5]) given by Suzuki in which compactness are replaced by other conditions; see [21] and references therein for more interesting results.

Theorem 6

[22] Let (X, d) be a complete metric space and T be a mapping on X. Assume that for each \(\varepsilon >0\), there exists \(\delta >0\) such that

(c1):

\(\displaystyle \frac{1}{2}d(x,Tx)<d(x,y)\quad \text{ and }\quad d(x,y)<\varepsilon +\delta \quad \text{ imply }\quad d(Tx,Ty)\le \varepsilon \) and

(c2):

\(\displaystyle \frac{1}{2}d(x,Tx)<d(x,y) \quad \text{ implies }\quad d(Tx,Ty)<d(x,y)\)

for all \(x,y\in X\). Then, there exists a unique fixed point z of T. Moreover, \(\{T^nx\}\) converges to z for any \(x\in X\).

Theorem 7

[24] Let T be a mapping on a complete metric space (X, d). Assume that

(d1):

\(\displaystyle \frac{1}{2}d(x,Tx)<d(x,y) \quad \text{ implies }\quad d(Tx,Ty)<d(x,y)\) for all \(x,y\in X\) and

(d2):

for any \(x\in X\) and \(\varepsilon >0\), there exists \(\delta >0\) such that \(d(T^ix,T^jx)<\varepsilon +\delta \) implies \(d(T^{i+1}x,T^{j+1}x) \le \varepsilon \) for any \(i,j\in \mathbb {N}\).

Then T has a unique fixed point z and \(\{T^nx\}\) converges to z for any \(x\in X\).

Similarly, we next replace the compactness of X in Theorem 5 and provide a generalization of both Theorem 6 and Theorem 7. For our aim, recalling that if \(T:X\rightarrow X\) and \(x\in X\), then the orbit of T at x is the set \(O(T,x) = \{x,Tx,T^2x,\cdots ,T^nx,\cdots \}\).

Theorem 8

Let (X, d) be a complete metric space and \(T:X\rightarrow X\) be a mapping. Assume that

(a):

there exists \(C\ge 0\) such that

$$\begin{aligned}&\frac{1}{2}d(x,Tx)<d(x,y)\nonumber \\&\quad \Longrightarrow \quad d(Tx,Ty)< d(x,y) + |d(x,Tx)-d(y,Ty)| + Cd(y,Tx) \end{aligned}$$
(12)

for \(x,y\in X\),

and

(b):

for any \(x\in X\) and \(\varepsilon >0\), there exists \(\delta >0\) such that

$$\begin{aligned} \frac{1}{2}d(u,Tu)<d(u,v)\quad \text{ and }\quad d(u,v)<\varepsilon +\delta \quad \text{ imply }\quad d(Tu,Tv)\le \varepsilon \end{aligned}$$

for all \(u,v\in O(T,x)\).

Then, T has a fixed point in X and for any \(x\in X\), the sequence of iterates \(\{T^nx\}\) converges to a fixed point of T.

Proof

Let \(x_0\in X\) be arbitrary but fixed and define the sequence \(\{x_n\}\) in X by \(x_n = T^nx\) for each \(n\in \mathbb {N}\). If there exists \(n_0\in \mathbb {N}\) such that \(x_{n_0} = x_{n_0+1}\), then \(x_{n_0}\) is a fixed point of T and the sequence \(\{x_n\}\) converges to \(x_{n_0}\). Assume now that \(x_n\ne x_{n+1}\) for all \(n\in \mathbb {N}\). Then, for each \(n\in \mathbb {N}\),

$$\begin{aligned} \frac{1}{2}d(x_n,Tx_n) = d(x_n,x_{n+1})< d(x_n,x_{n+1}). \end{aligned}$$

By (a), we have

$$\begin{aligned} d(x_{n+1},x_{n+2})= & {} d(Tx_n,Tx_{n+1})\\< & {} d(x_n,x_{n+1}) +|d(x_n,Tx_n)-d(x_{n+1},Tx_{n+2})| + Cd(x_{n+1},Tx_n)\\= & {} d(x_n,x_{n+1}) + |d(x_n,x_{n+1})-d(x_{n+1},x_{n+2})| \end{aligned}$$

This implies that \(d(x_{n+1},x_{n+2})<d(x_n,x_{n+1})\), i.e., \(\{d(x_n,x_{n+1})\}\) is a strictly decreasing sequence of positive real numbers. Thus there exists \(\alpha \ge 0\) such that \(\lim _{n\rightarrow \infty }d(x_n,x_{n+1}) = \alpha \). We shall show that \(\alpha =0\). Assume to the contrary that \(\alpha >0\). Then \(d(x_n,x_{n+1})>\alpha \) for each n. Moreover, there exist \(\delta >0\) and \(N\in \mathbb {N}\) such that

$$\begin{aligned} d(T^nx,T^{n+1}x)=d(x_n,x_{n+1})<\alpha +\delta \end{aligned}$$

for all \(n\ge N\). By condition (b), the latter inequality implies \(d(x_{n+1},x_{n+2}) = d(T^{n+1}x,T^{n+2}x)\le \alpha \) for \(n\ge N\). This is a contradiction. Hence, \(\alpha = 0\). We now prove that \(\{x_n\}\) is a Cauchy sequence. Let \(\varepsilon >0\) be arbitrary but fixed and let \(\delta >0\) be fixed. Since \(\lim _{n\rightarrow \infty }d(x_n,x_{n+1}) = 0\), there exists \(p\in \mathbb {N}\) such that \(d(x_n,x_{n+1})<\delta \) for all \(n\ge p\). We shall show by induction that

$$\begin{aligned} d(x_{p+1},x_{p+m})\le \varepsilon \end{aligned}$$
(13)

for all \(m\in \mathbb {N}\). It is evident that (13) holds for \(m=0\) and \(m=1\). Assume that (13) holds for some \(m\in \mathbb {N}\), \(m\ge 1\). We have

$$\begin{aligned} d(x_p,x_{p+m})\le d(x_p,x_{p+1}) + d(x_{p+1},x_{p+m}) <\varepsilon +\delta . \end{aligned}$$

Then, by condition (b), one has \(d(x_{p+1},x_{p+m+1}) = d(Tx_p,Tx_{p+m}) \le \varepsilon \). Thus, by induction, (13) holds for all \(m\in \mathbb {N}\). This implies that the sequence \(\{x_n\}\) is Cauchy. Since X is complete, the sequence \(\{x_n\}\) converges to some \(z\in X\). Our attempt is to show that z is a fixed point of T. Assume now that there exists \(k\in \mathbb {N}\) such that

$$\begin{aligned} \frac{1}{2}d(x_k,x_{k+1}) \ge d(x_n,z)\quad \text{ and }\quad \frac{1}{2}d(x_{n+1},x_{n+2}) \ge d(x_{n+1},z). \end{aligned}$$

Then,

$$\begin{aligned} d(x_n,x_{n+1})\le & {} d(x_n,z) + d(x_{n+1},z) \\\le & {} \frac{1}{2}[d(x_n,x_{n+1}) + d(x_{n+1},x_{n+2})]\\< & {} d(x_n,x_{n+1}) \end{aligned}$$

which is a contradiction. Thus, it holds either

  1. (i)

    \(\displaystyle \frac{1}{2} d(x_n,x_{n+1})<d(x_n,z)\) for all n in some infinite subset I of \( \mathbb {N}\), or

  2. (ii)

    \(\displaystyle \frac{1}{2}d(x_{n+1},x_{n+2})<d(x_{n+1},z)\) for all n in some infinite subset J of \( \mathbb {N}\).

If (i) holds, then, by Condition (a), we have for all \(n\in I\) that

$$\begin{aligned} d(x_{n+1},Tz)= & {} d(Tx_n,Tz) <d(x_n,z) + |d(x_n,Tx_{n+1})-d(z,Tz)| + Cd(z,Tx_{n})\\= & {} d(x_n,z) + d(x_n,x_{n+1}) + Cd(z,x_{n+1}). \end{aligned}$$

This yields

$$\begin{aligned} \lim _{n\in I, n\rightarrow \infty }x_{n+1} = Tz, \end{aligned}$$

that is, \(\{x_n\}\) has a subsequence converging to Tz. Similarly, if (ii) holds, then we also get that \(\{x_n\}\) has a subsequence converging to Tz. Since \(\{x_n\}\) converges to z and has a subsequence converging to Tz, we obtain \(Tz=z\). Therefore, we have shown that T has a fixed point z and the sequence \(\{T^nx\}\) converges to this fixed point. \(\square \)

We next provide a simple example for which we can apply Theorem 8 but cannot apply Theorem 6 and Theorem 7.

Example 5

Let \(X = [0,\infty )\) with the usual metric \(d(x,y)=|x-y|\) for all \(x,y\in X\). Then, (X, d) is complete but it is not compact. We define a mapping \(T:X\rightarrow X\) by

$$\begin{aligned} Tx = \left\{ \begin{matrix} 2&{}&{}\text{ if }&{}x\ge 2,\\ 0&{}&{}\text{ if }&{}x<2. \end{matrix} \right. \end{aligned}$$

First, we can see that T has two fixed points \(z=0\) and \(z=2\). Thus, we cannot apply Theorem 6 and Theorem 7. to this example. We now check that all conditions of Theorem 8 are satisfied.

We first check Condition (b). If \(x\in [0,2)\), then \(O(T,x) = \{x,0,0,\cdots ,\}\). Hence, if \(u,v\in O(T,x)\) and \(u\ne v\), then \(d(u,v) = x\) and \(d(Tu,Tv) = 0\). Similarly, if \(x\in [2,\infty )\), \(u,v\in O(T,x)\) with \(u\ne v\), then \(d(u,v)=x\) and \(d(Tu,Tv)=0\). Thus, Condition (b) is satisfied.

We now check that Condition (a) is satisfied with \(C=2\). If \(x\ne y\) and \(x,y\in [2,\infty )\) or \(x,y \in [0,2)\), then \(d(Tx,Ty) = 0<|x-y|= d(x,y)\). Thus, (12) holds for this case.

Consider \(x\in [2,\infty )\) and \(y\in [0,2)\). We have

$$\begin{aligned} RHS:= & {} d(x,y) + |d(x,Tx)-d(y,Ty)| + Cd(y,Tx)\\= & {} x-y+|x-2-y| +2(2-y)\\= & {} x+2-2y + |x-y-2| + 2>2= d(Tx,Ty). \end{aligned}$$

Thus, (12) holds.

If \(x\in [0,2)\) and \(y\in [2,\infty )\), then

$$\begin{aligned} RHS = y-x+|x-y+2| +2y>2y>2= d(Tx,Ty). \end{aligned}$$

Thus, (12) holds. Therefore, all conditions of Theorem 8 are satisfied. It is evident that if \(x\in [2,\infty )\) then \(\{T^nx\}\) converges to 2 and if \(x\in [0,2)\) then \(\{T^nx\}\) converges to 0.

The following corollary is an answer to Question 1(iii).

Corollary 4

In Theorem 8, if \(C=0\), then T has a unique fixed point in X and for any \(x\in X\), the sequence of iterates \(\{T^nx\}\) converges to the fixed point of T.