1 Introduction

The following theorem is the key result of [3].

Theorem 1.1

If (X, d) is a complete metric space and \(T{:}\; X \rightarrow X\) is a continuous asymptotically regular mapping and if there exists \(0 \leqslant M < 1\) and \(0 \leqslant K <+ \infty \) satisfying

$$\begin{aligned} d(Tx,Ty) \leqslant Md(x, y)+K\{d(x, Tx)+d(y, Ty)\} \end{aligned}$$
(1.1)

for all \(x, y \in X\), then T has a unique fixed point \(p \in X\) and \(T^{n}x \rightarrow p\) for each \(x \in X\).

Recall that the set \(O(x;T) = \{T^{n}x{:}\; n = 0, 1, 2, \ldots \}\) is called the orbit of the self-mapping T at the point \(x \in X\).

Definition 1.1

A self-mapping T of a metric space (X, d) is said to be orbitally continuous at a point \(z \in X\) if for any sequence \(\{x_n\} \subset O(x;T)\) for some \(x \in X, x_n \rightarrow z\) implies \(Tx_n \rightarrow Tz\) as \(n \rightarrow \infty \).

Remark 1.1

Every continuous self-mapping of a metric space is orbitally continuous, but the converse need not be true (see Example 1.1 below).

Definition 1.2

[5] A self-mapping T of a metric space (X, d) is called k-continuous, \(k = 1, 2, 3,\ldots ,\) if \(T^{k}x_n \rightarrow Tz\), whenever \(\{x_n\}\) is a sequence in X such that \(T^{k-1}x_n \rightarrow z\).

Remark 1.2

It is important to note that for a self-mapping T of a metric space (X, d), the notion of 1-continuity coincides with continuity. However,

$$\begin{aligned} 1\hbox {-continuity} \Rightarrow 2\hbox {-continuity} \Rightarrow 3\hbox {-continuity} \Rightarrow \cdots , \end{aligned}$$

but not conversely. The following example illustrates this fact [5].

Example 1.1

Let \(X = [0, 4]\) and d be the usual metric on X. Define \(T{:}\; X \rightarrow X\) by

$$\begin{aligned} T(x)=2 \hbox { if } x \in [0,2], T(x)=0 \hbox { if } x \in (2, 4]. \end{aligned}$$

Then, \(Tx_n \rightarrow t \Rightarrow T^{2}x_n \rightarrow t\), since \(Tx_n \rightarrow t\) implies \(t=0\) or \(t=2\) and \(T^{2}x_{n} \rightarrow 2 =T2\) for all n. Hence, T is 2-continuous. However, T is discontinuous at \(x=2\).

In 1988, Rhoades [7] posed an open problem regarding existence of contractive definitions which yield a fixed point but the mapping need not be continuous at the fixed point. This problem was settled in the affirmative by Pant [6]. In a recent past, several new situations have been established where the existence of the fixed point is guaranteed but the mappings are discontinuous at the fixed point [1, 2, 4].

In this paper, we show that the assumption of continuity considered in Theorem 2.6 of [3] can be relaxed to some weaker notions of continuity, (orbital continuity or k-continuity) which thereby extends the scope of the study of fixed point theorems from the class of continuous mappings to a wider class of mappings which also include discontinuous mappings. As a by-product, we provide new answers to the open problem posed by Rhoades [7].

2 Main results

Theorem 2.1

If (X, d) is a complete metric space and \(T{:}\; X \rightarrow X\) is an asymptotically regular mapping and if there exists \(0 \leqslant M < 1\) and \(0 \leqslant K <+ \infty \) satisfying

$$\begin{aligned} d(Tx,Ty) \leqslant Md(x, y)+K\{d(x, Tx)+d(y, Ty)\} \end{aligned}$$
(2.1)

for all \(x, y \in X\), then T has a unique fixed point \(p \in X\) provided T is either k-continuous for \(k\geqslant 1\) or orbitally continuous.

Proof

Let \(x_0\) be any point in X. Define a sequence \(\{x_n\}\) in X given by the rule \(x_{n+1}= Tx_{n}=T^{n}x\). Then, following Theorem 2.6 of [3] we conclude that \(\{x_n\}\) is a Cauchy sequence. Since X is complete, there exists a point \(u \in X\) such that \(x_n\rightarrow u\) as \(n\rightarrow \infty \). Also, \(Tx_n \rightarrow u.\) Furthermore, for each \(k\geqslant 1\) we have \(T^{k}x_n \rightarrow u\) as \(n\rightarrow \infty \). Suppose that T is k-continuous. Since \(T^{k-1}x_n \rightarrow u\), k-continuity of T implies that \(\lim _{n \rightarrow \infty }T^{k}x_n = Tu\). This yields \(u = Tu,\) that is, u is a fixed point of T.

Finally, suppose that T is orbitally continuous. Since \(x_n\rightarrow u\), orbital continuity implies that \(\lim _{n\rightarrow \infty }Tx_n = Tu.\) This yields \(Tu = u,\) that is, u is a fixed point of T.

We now give an example to show that the condition (2.1) is strong enough to generate a fixed point but does not force the mapping to be continuous at the fixed point [6].

Example 2.1

Let \(X = [0, 2]\) and d be the usual metric on X. Define \(T{:}\;X \rightarrow X\) by

$$\begin{aligned} T\left( x\right) =\left\{ \begin{array}{l} 1,\quad \text { if } 0 \leqslant x \leqslant 1;\\ 0,\quad \text { if } 1 < x \leqslant 2. \\ \end{array}\right. \end{aligned}$$

Then, T satisfies all the conditions of Theorem 2.1 and has a unique fixed point \(x=1\) at which T is discontinuous.