1 Introduction

For more than 60 years now, there has been a lot of research activity regarding the fixed point theory of nonexpansive (that is, 1-Lipschitz) and contractive mappings. See, for instance, [2, 4, 7, 8, 11,12,13,14,15,16,17,18,, 19, 21, 24, 25] and references cited therein. This activity stems from Banach’s classical theorem [1] concerning the existence of a unique fixed point for a strict contraction on a complete metric space. It also concerns the convergence of (inexact) iterates of a nonexpansive mapping to one of its fixed points. Since that seminal result, many developments have taken place in this field including, in particular, studies of feasibility, common fixed point problems, nonlinear operator theory and variational inequalities, which find important applications in engineering, medical and the natural sciences [3, 5, 6, 22,23,24,, 25]. In particular, the study of nonexpansive and contractive mappings on complete metric spaces with graphs has recently become a rapidly growing area of research. See, for example, [9, 10, 20]. In the present paper we establish two results regarding the existence of fixed points for strict contractions on complete metric spaces with graphs.

2 Preliminaries and the first main result

Let \((X,\rho )\) be a complete metric space and let G be a (directed) graph on X. Let V(G) be the set of its vertices and let E(G) be the set of its edges. We identify the graph G with the pair (V(G), E(G)).

Denote by \({{\mathcal {M}}}\) the set of all mappings \(T: X \rightarrow X\) such that for each \(x,y \in X\) satisfying \((x,y) \in E(G)\), we have

$$\begin{aligned} (T(x),T(y)) \in E(G) \text{ and } \rho (T(x),T(y)) \le \rho (x,y). \end{aligned}$$
(2.1)

A mapping \(T \in {{\mathcal {M}}}\) is called G-nonexpansive. If \(T \in {{\mathcal {M}}}\), \(\alpha \in (0,1)\) and for each \(x,y \in X\) satisfying \((x,y) \in E(G)\), we have

$$\begin{aligned} \rho (T(x),T(y))\le \alpha \rho (x,y), \end{aligned}$$
(2.2)

then T is said to be a G-strict contraction.

Fix \(\theta \in X\). For each \(x \in X\) and each \(r>0\), set

$$\begin{aligned} B(x,r):= \{y \in X:\; \rho (x,y) \le r\}. \end{aligned}$$

In the sequel we assume that any sum over the empty set is zero and that the infimum of the empty set is \(\infty \). We also assume that \(a+\infty =\infty \) for each \(a \in R^1\). Finally, we assume that if \(x,y \in X\) satisfy \((x,y) \in E(G)\), then

$$\begin{aligned} (y,x) \in E(G). \end{aligned}$$

For each \(x,y\in X\), define

$$\begin{aligned} \rho _1(x,y):= \inf \Bigg \{\sum _{i=0}^{q-1}\rho (x_i,x_{i+1}):\; q \ge 1 \text{ is } \text{ an } \text{ integer }, \; \nonumber \\ x_i \in X,\;i=0,\dots ,q,\;x_0=x,\;x_q=y,\;(x_i,x_{i+1}) \in E(G),\;i=0,\dots ,q-1\Bigg \}. \end{aligned}$$
(2.3)

It is not difficult to see that for each \(x,y,z \in X\), we have

$$\begin{aligned} \rho _1(x,y) & \ge \rho (x,y), \nonumber \\ \rho _1(x,y)& =\rho _1(y,x), \nonumber \\ \rho _1(x,z)& \le \rho _1(x,y)+\rho _1(y,z), \end{aligned}$$
(2.4)

and if \(\rho _1(x,y)=0\), then \(x=y\). Note that it is may happen that \(\rho _1(x,y)=\infty \) for some \(x,y \in X\). The pseudometric \(\rho _1\) plays an important role in this paper because it turns out that if a mapping T is a G-strict contraction, then it is a strict contraction with respect to the pseudometric \(\rho _1\).

Let \(T \in {{\mathcal {M}}}\) be a G-strict contraction. It is known [10] that under certain mild assumptions, the mapping T has a unique fixed point which attracts all the iterates of T. In [20] we provided a very simple proof of this fact by using the pseudometric \(\rho _1\) under the assumption that it is always finite-valued. Then we established uniform convergence of the iterates of T on bounded subsets of X and showed that this convergence is stable in the presence of small perturbations of these iterates. We also showed there that under certain assumptions, a typical G-nonexpansive mapping has a unique fixed point which attracts all its iterates, uniformly on bounded subsets of X. In the present paper we prove the existence result of [20] under a weakened assumption. We do not assume that the pseudometric \(\rho _1\) is always finite-valued. Instead, we assume that there exists a point \(x \in X\) such that \(\rho _1(x,T(x))\) is finite. This is our first main result. It is proved in this section. In our second result we show the existence of a fixed point for a mapping defined on a closed ball in X which takes values in X.

Given a mapping \(S: X \rightarrow X\), we define \(S^0=I\), the identity self-mapping on X, \(S^1=S\), and \(S^{i+1}=S\circ S^i\) for all integers \(i \ge 0\).

Proposition 2.1

Let \(T \in {{\mathcal {M}}}\), \(\alpha \in [0,1)\) and assume that for each \(x,y \in X\) satisfying \((x,y) \in E(G)\), the inequality

$$\begin{aligned} \rho (T(x),T(y)) \le \alpha \rho (x,y) \end{aligned}$$
(2.5)

holds. Then for each \(x,y \in X\), we have

$$\begin{aligned} \rho _1(T(x),T(y)) \le \alpha \rho _1(x,y). \end{aligned}$$

Proof

Let \(x,y \in X\). We may assume without any loss of generality that \(\rho _1(x,y)<\infty \). Let \(\epsilon >0\) be given. By (2.3), there exist an integer \(q\ge 1\) and points \(x_i \in X, \; i=0,\dots ,q\), such that

$$\begin{aligned} \;x_0=x,\;x_q=y, \end{aligned}$$
(2.6)
$$\begin{aligned} (x_i,x_{i+1}) \in E(G),\; i=0,\dots ,q-1, \end{aligned}$$
(2.7)
$$\begin{aligned} \sum _{i=0}^{q-1}\rho (x_i,x_{i+1}) \le \rho _1(x,y)+\epsilon . \end{aligned}$$
(2.8)

By (2.1) and (2.5)–(2.7), we have

$$\begin{aligned} T(x)=T(x_0),\;T(y)=T(x_q), \\ (T(x_i),T(x_{i+1}))\in E(G),\;i=0,\dots ,q-1, \end{aligned}$$

and

$$\begin{aligned} \rho (T(x_i),T(x_{i+1}))\le \alpha \rho (x_i,y_i),\;\;i=0,\dots ,q-1. \end{aligned}$$

When combined with (2.3) and (2.8), these relations imply that

$$\begin{aligned} \rho _1(T(x),T(y)) & \le \sum _{i=0}^{q-1}\rho (T(x_i),T(x_{i+1})) \le \alpha \sum _{i=0}^{q-1}\rho (x_i,x_{i+1}) \\ & \le \alpha \rho _1(x,y)+\epsilon . \end{aligned}$$

Since \(\epsilon \) is any positive number, we conclude that

$$\begin{aligned} \rho _1(T(x),T(y)) \le \alpha \rho _1(x,y), \end{aligned}$$

as asserted. This completes the proof of Proposition 2.1. \(\square \)

Theorem 2.2

Let \(T \in {{\mathcal {M}}}\), \(\alpha \in [0,1)\) and assume that for each \(z,y \in X\) satisfying \((z,y) \in E(G)\),

$$\begin{aligned} \rho (T(z),T(y)) \le \alpha \rho (z,y), \end{aligned}$$
(2.9)

\(x \in X\) and \(\rho _1(x,T(x))<\infty \). Then for each integer \(k \ge 0\),

$$\begin{aligned}\sum _{i=k}^{\infty }\rho (T^{i}(x),T^{i+1}(x))\le \sum _{i=k}^{\infty }\rho _1(T^{i}(x),T^{i+1}(x))\le \alpha ^k(1-\alpha )^{-1}\rho _1(x,T(x)),\end{aligned}$$

there exists

$$\begin{aligned} x_*=\lim _{i \rightarrow \infty } T^i(x) \end{aligned}$$

in \((X,\rho )\) and

$$\begin{aligned} \rho (T^{k}(x),x_*)\le \alpha ^k(1-\alpha )^{-1}\rho _1(x,T(x)). \end{aligned}$$

Moreover, if T is continuous at \(x_*\) as a self-mapping of \((X,\rho )\), then \(T(x_*)=x_*\).

Proof

Proposition 2.1 and (2.1), (2.9) imply that for each integer \(i \ge 0\),

$$\begin{aligned} \rho _1(T^{i+1}(x),T^{i+2}(x))\le \alpha \rho _1(T^{i}(x),T^{i+1}(x)) \end{aligned}$$

and

$$\begin{aligned} \rho _1(T^{i}(x),T^{i+1}(x)) \le \alpha ^i \rho _1(x,T(x)). \end{aligned}$$

This implies that for each integer \(k \ge 0\),

$$\begin{aligned} \sum _{i=k}^{\infty }\rho (T^{i}(x),T^{i+1}(x)) \le \rho _1(x,T(x))\sum _{i=k}^{\infty }\alpha ^i\le \alpha ^k(1-\alpha )^{-1}\rho _1(x,T(x)). \end{aligned}$$
(2.10)

In view of (2.10), \(\{T^i(x)\}_{i=0}^{\infty }\) is a Cauchy sequence in \((X,\rho )\). Therefore there exists

$$\begin{aligned} x_*=\lim _{i \rightarrow \infty } T^i(x) \end{aligned}$$

in \((X,\rho )\) and for each integer \(k \ge 0\),

$$\begin{aligned} \rho (T^{k}(x),x_*) & =\lim _{i \rightarrow \infty }\rho (T^k(x),T^{i+k}(x)) \\ & \le \sum _{i=k}^{\infty }\rho (T^{i}(x),T^{i+1}(x)) \le \rho _1(x,T(x))\sum _{i=k}^{\infty }\alpha ^i \le \alpha ^k(1-\alpha )^{-1}\rho _1(x,T(x)), \end{aligned}$$

as asserted. This completes the proof of Theorem 2.2. \(\square \)

The following proposition easily implies the well-posedness of our fixed point problem.

Proposition 2.3

Let \(T \in {{\mathcal {M}}}\), \(\alpha \in [0,1)\) and assume that (2.9) holds for each \((z,y) \in E(G)\). Let \(x,y \in X\),

$$\begin{aligned} \rho _1(x,y),\;\rho _1(x,T(x)),\; \rho _1(y,T(y))<\infty . \end{aligned}$$

Then

$$\begin{aligned} \rho (x,y)\le \rho _1(x,y) \le (1-\alpha )^{-1}(\rho _1(x,T(x))+ \rho _1(y,T(y))). \end{aligned}$$

Proof

Proposition 2.1 and (2.1) imply that

$$\begin{aligned} \rho _1(x,y) & \le \rho _1(x,T(x))+ \rho _1(T(x),T(y))+\rho _1(T(y),y) \\ & \le \rho _1(x,T(x))+ \alpha \rho _1(x,y)+\rho _1(T(y),y), \\ (1-\alpha )\rho _1(x,y) & \le \rho _1(x,T(x))+\rho _1(T(y),y) \end{aligned}$$

and

$$\begin{aligned} \rho (x,y) \le \rho _1(x,y) \le (1-\alpha )^{-1}(\rho _1(x,T(x))+ \rho _1(y,T(y))). \end{aligned}$$

This completes the proof of Proposition 2.3. \(\square \)

3 The second main result

Assume that \(x \in X\), \(r >0\), \(\alpha \in (0,1)\) and that for each \((z,y) \in E(G) \cap (B(x,r) \times B(x,r))\), \(T: B(x,r) \rightarrow X\) satisfies

$$\begin{aligned} (T(z),T(y)) \in E(G), \; \rho (T(z),T(y)) \le \alpha \rho (z,y).\end{aligned}$$
(3.1)

Theorem 3.1

Assume that

$$\begin{aligned} \rho _1(x,T(x)) < r(1-\alpha ). \end{aligned}$$
(3.2)

Then

$$\begin{aligned} T^k(x) \in B(x,r) \end{aligned}$$

for each integer \(k \ge 1\),

$$\begin{aligned} \sum _{i=0}^{\infty }\rho (T^{i}(x),T^{i+1}(x))\le (1-\alpha )^{-1}\rho _1(x,T(x)) \end{aligned}$$

and there exists

$$\begin{aligned} x_*=\lim _{i \rightarrow \infty } T^i(x) \end{aligned}$$

in \((X,\rho )\). Moreover, if T is continuous at \(x_*\) with respect to \(\rho \), then \(T(x_*)=x_*\).

Proof

By (2.4), we have

$$\begin{aligned} \rho (x,T(x))\le \rho _1(x,T(x)) < r(1-\alpha ). \end{aligned}$$
(3.3)

Assume that \(k \ge 1\) is an integer,

$$\begin{aligned} T^i(x) \in B(x,r),\;i=0,\dots ,k, \end{aligned}$$
(3.4)

and that for each \(i \in \{0,\dots ,k-1\}\),

$$\begin{aligned} \rho _1(T^{i}(x),T^{i+1}(x)) \le \alpha ^i \rho _1(x,T(x)). \end{aligned}$$
(3.5)

(In view of (3.3), our assumption does hold for \(k=1\).) By (3.5), we have

$$\begin{aligned} \rho _1(T^{k-1}(x),T^{k}(x))\le \alpha ^{k-1} \rho _1(x,T(x)). \end{aligned}$$
(3.6)

Let

$$\begin{aligned} \epsilon \in (0, r-\rho _1(x,T(x))(1-\alpha )^{-1}). \end{aligned}$$
(3.7)

It follows from (2.3) that there exist an integer \(q\ge 1\) and points \(x_i \in X, \; i=0,\dots ,q\), such that

$$\begin{aligned} x_0=T^{k-1}(x),\;x_q=T^k(x), \end{aligned}$$
(3.8)
$$\begin{aligned} (x_i,x_{i+1}) \in E(G),\;i=0,\dots ,q-1, \end{aligned}$$
(3.9)
$$\begin{aligned} \sum _{i=0}^{q-1}\rho (x_i,x_{i+1}) \le \rho _1(T^{k-1}(x),T^k(x))+\epsilon . \end{aligned}$$
(3.10)

There are two cases: \(k=1\) and \(k>1\). Assume first that \(k=1\). In view of (3.7), (3.8) and (3.10),

$$\begin{aligned} x_0=x \end{aligned}$$

and for each \(i=0,\dots ,q\),

$$\begin{aligned} \rho (x,x_i) \le \sum _{j=0}^{q-1}\rho (x_j,x_{j+1}) \le \rho _1(x,T(x)) + \epsilon < r \end{aligned}$$

and

$$\begin{aligned} x_i \in B(x,r). \end{aligned}$$
(3.11)

By (3.8) and (3.11), for each \(i=0,\dots ,q-1\), we have

$$\begin{aligned} (T(x_i),T(x_{i+1}))\in E(G) \end{aligned}$$

and

$$\begin{aligned} \rho _1(T(x_i),T)x_{i+1}))\le \alpha \rho (x_i,x_{i+1}). \end{aligned}$$

It follows from (3.8), (3.10) and the above relations that

$$\begin{aligned} \rho _1(T(x),T^2(x)) & =\rho _1(T(x_0),T(x_q))\le \sum _{i=0}^{q-1}\rho (T(x_i),T(x_{i+1})) \\ & \le \alpha \sum _{i=0}^{q-1}\rho (x_i,x_{i+1}) \le \alpha \rho _1(x,T(x))+\alpha \epsilon . \end{aligned}$$

Since \(\epsilon \) is any positive number, we conclude that

$$\begin{aligned} \rho _1(T(x),T^2(x)) & \le \alpha \rho _1(x,T(x)), \\ \rho (x,T^2(x)) & \le \rho _1(x,T^2(x)) \le \rho _1(x,T(x))+\rho _1(T(x),T^2(x)) \\ & \le \rho _1(x,T(x))(1+\alpha ) < r \end{aligned}$$

by (3.2). Thus

$$\begin{aligned} T^2(x) \in B(x,r) \end{aligned}$$

and our assumption holds for \(k+1=2\).

Assume now that \(k>1\). Proposition 2.1 and (3.5) imply that

$$\begin{aligned} \rho _1(x,T^{k-1}(x)) \le \sum _{i=0}^{k-2}\rho _1(T^i(x),T^{i+1}(x))\le \rho _1(x,T(x))\sum _{i=0}^{k-2}\alpha ^i. \end{aligned}$$
(3.12)

By (3.8), (3.10) and (3.12), for \(i=1,\dots ,q\), we have

$$\begin{aligned} \rho (x_i,T^{k-1}(x))& \le \sum _{j=0}^{q-1}\rho (x_j,x_{j+1}) \\ & \le \rho _1(T^{k-1}(x),T^k(x))+\epsilon \end{aligned}$$

and

$$\begin{aligned} \rho (x_i,x) & \le \rho (x_i,T^{k-1}(x))+\rho (T^{k-1}(x),x) \\ & \le \rho _1(T^{k-1}(x),T^k(x))+\epsilon +\rho _1(x,T(x)) \sum _{i=0}^{k-2} \alpha ^i \\ & \le \rho _1(x,T(x))\sum _{i=0}^{k-1} \alpha ^i+\epsilon \\ & \le \rho _1(x,T(x))(1-\alpha )^{-1}+\epsilon <r. \end{aligned}$$

Thus

$$\begin{aligned} x_i \in B(x,r), \; i=0,\dots ,q. \end{aligned}$$
(3.13)

By (3.1), (3.9) and (3.13), for \(i=0,\dots ,q-1\), we have

$$\begin{aligned} (T(x_i),T(x_{i+1})) \in E(G) \end{aligned}$$

and

$$\begin{aligned} \rho (T(x_i),T(x_{i+1})) \le \alpha \rho (x_i,x_{i+1}). \end{aligned}$$

When combined with (3.8) and (3.10), the above relations imply that

$$\begin{aligned} \rho _1(T^k(x),T^{k+1}(x))& \le \sum _{i=0}^{q-1}\rho (T(x_i),T(x_{i+1})) \le \sum _{i=0}^{q-1}\alpha \rho (x_i,x_{i+1}) \\ & \le \alpha \rho _1(T^{k-1}(x),T^k(x))+\epsilon . \end{aligned}$$

Since \(\epsilon \) is any positive number, using (3.6), we conclude that

$$\begin{aligned} \rho _1(T^k(x),T^{k+1}(x)) \le \alpha \rho _1(T^{k-1}(x),T^k(x))\le \alpha ^k\rho _1(x,T(x)), \\ \rho _1(x,T^{k+1}(x))\le \sum _{i=0}^{k}\rho _1(T^i(x),T^{i+1}(x)) \\ \le \rho _1(x,T(x))\sum _{i=1}^k \alpha ^i \le (1-\alpha )^{-1}\rho _1(x,T(x))<r \end{aligned}$$

and

$$\begin{aligned} T^{k+1}(x) \in B(x,r). \end{aligned}$$

Thus our assumption holds for \(k+1\) too and by mathematical induction it is now seen to hold for each integer \(k \ge 1\). This implies that

$$\begin{aligned} \sum _{i=0}^{\infty }\rho (T^{i}(x),T^{i+1}(x)) \le \rho _1(x,T(x))(1-\alpha )^{-1} \end{aligned}$$

and so \(\{T^i(x)\}_{i=0}^{\infty }\) is a Cauchy sequence in \((X,\rho )\). Therefore there exists

$$\begin{aligned} x_*=\lim _{i \rightarrow \infty } T^i(x) \end{aligned}$$

in \((X,\rho )\).

Let \(k \ge 1\) be an integer. Then

$$\begin{aligned} \rho (T^{k}(x),x_*) & =\lim _{i \rightarrow \infty }\rho (T^k(x),T^{i+k}(x)) \\ & \le \sum _{i=k}^{\infty }\rho (T^{i}(x),T^{i+1}(x)) \le \rho _1(x,T(x))\sum _{i=k}^{\infty }\alpha ^i \le \alpha ^k(1-\alpha )^{-1}\rho _1(x,T(x)) \rightarrow 0 \end{aligned}$$

as \(k \rightarrow \infty \). Clearly, if T is continuous at \(x_*\), then \(x_*=T(x_*)\). This completes the proof of Theorem 3.1. \(\square \)