In this section, we present new results on existence of common fixed points for auxiliary functions with two metrics endowed with a directed graph. To begin with, let us establish classes of functions which will be considered throughout this work.
Suppose that \(\phi:[0,\infty )\rightarrow [0,\infty )\) is a function which has the following properties:

ϕ is an increasing function;

ϕ is a continuous function;

\(\phi (r)=0\) if and only if \(r=0\).
From now on, we will denote the set of all such functions ϕ satisfying all the above conditions by Φ.
Inspired by [11], the next important class of functions that we shall consider is the class \(\mathcal{A}(X)\), where \((X,d)\) is a metric space. This is the class of all auxiliary functions \(h: X\times X \to [0,1]\) such that
$$\begin{aligned} \text{if}\quad\lim_{n\to \infty }h(x_{n},y_{n}) =1,\quad \text{then}\quad \lim_{n \to \infty } d(x_{n},y_{n})=0 \end{aligned}$$
for all sequences \(\{x_{n}\}\) and \(\{y_{n}\}\) in X.
As a result, we are now in a position to consider a new type of contractions, which is defined as follows.
Definition 8
Let \((X, d)\) be a metric space endowed with a directed graph \(G= (V(G),E(G) )\), and let \(f,g:X\to X\) be functions. The pair \((f,g)\) will be called an hϕcontraction with respect to d whenever the following conditions hold:

(1)
f is gedge preserving with respect to G;

(2)
There exist two functions \(h \in \mathcal{A}(X) \) and \(\phi \in \Phi \) such that, for all \(x,y\in X\) with \((gx, gy)\in E(G)\), we have
$$\begin{aligned} \phi \bigl(d(fx,fy)\bigr)\leq h(gx,gy)\phi \bigl(R(gx,gy)\bigr), \end{aligned}$$
where \(R: X\times X \to [0,\infty )\) is a function such that, for any \(x,y \in X\),
$$\begin{aligned} R(gx,gy)={}&\max \biggl\{ \frac{d(gx,fx)d(fy,gy)}{d(gx,gy)},d(gx,gy), d(gx,fx), d(gy,fy),\\ &\frac{d(gx,fy)+d(gy,fx)}{2} \biggr\} . \end{aligned}$$
The above definition allows us to generalize the result in [28] to the case of auxiliary functions. In fact, we are now ready to present and prove our main results. The following theorem involves two metrics as being motivated by [31].
Theorem 9
Let \((X,d' )\) be a complete metric space endowed with a directed graph \(G= (V(G),E(G) )\), let d be another metric on X, and let \(f,g:X\to X\) be functions. Suppose that \((f,g)\) is an hϕcontraction with respect to d, and assume further that the following conditions hold:

(1)
\(g:(X,d')\to (X,d')\) is a continuous function such that \(g(X)\) is \(d'\)closed;

(2)
\(f(X)\subseteq g(X)\);

(3)
\(E(G)\) satisfies the transitivity property;

(4)
If \(d\ngeq d'\), assume that \(f:(X,d)\to (X,d')\) is gCauchy on X;

(5)
\(f:(X,d')\to (X,d')\) is Gcontinuous, and f and g are \(d'\)compatible.
As a consequence, we get that
$$\begin{aligned} X(f,g)\neq \varnothing\quad \textit{if and only if}\quad C(f,g)\neq \varnothing. \end{aligned}$$
Proof
\((\Leftarrow ) \) This follows from Lemma 3.
\((\Rightarrow ) \) Suppose that \(X(f,g)\neq \varnothing \) and \(x_{0}\in X\) with \((gx_{0}, fx_{0})\in E(G)\). By the assumption that \(f(X)\subseteq g(X)\) and \(f(x_{0})\in X\), we may construct a sequence \(\{x_{n}\}\) in X such that \(gx_{n}=fx_{n1}\) for each number \(n\in \mathbb{N}\). If it is the case that \(gx_{n_{0}}=gx_{n_{0}1}\) for some \(n_{0}\in \mathbb{N}\), then \(x_{n_{0}1}\) must be a coincidence point of f and g. As a result, we may assume now that, for every \(n\in \mathbb{N}\), \(gx_{n}\ne gx_{n1}\).
Because \((gx_{0},fx_{0})=(gx_{0},gx_{1})\in E(G)\) and the function f is gedge preserving with respect to G, it is true that \((fx_{0},fx_{1})=(gx_{1},gx_{2})\in E(G)\). By mathematical induction, we receive \((gx_{n1},gx_{n})\in E(G)\) for any \(n\in \mathbb{N}\). Since \((f,g)\) is an hϕcontraction with respect to d, for each \(n \geq 0\),
$$\begin{aligned} \phi \bigl(d(gx_{n+1},gx_{n+2})\bigr) &= \phi \bigl(d(fx_{n},fx_{n+1})\bigr) \\ &\leq h(gx_{n},gx_{n+1})\phi \bigl(R(gx_{n},gx_{n+1}) \bigr) \\ &\leq \phi \bigl(R(gx_{n},gx_{n+1})\bigr). \end{aligned}$$
(1)
Also, a direct calculation shows that
$$\begin{aligned} R(gx_{n},gx_{n+1}) ={}& \max \biggl\{ d(gx_{n},gx_{n+1}), d(gx_{n},fx_{n}), d(gx_{n+1},fx_{n+1}), \\ &{} \frac{d(gx_{n},fx_{n+1}) + d(gx_{n+1},fx_{n})}{2} \biggr\} \\ ={}& \max \biggl\{ d(gx_{n},gx_{n+1}), d(gx_{n+1},gx_{n+2}), \frac{ d(gx_{n},gx_{n+2})}{2} \biggr\} \\ ={}& \max \bigl\{ d(gx_{n},gx_{n+1}), d(gx_{n+1},gx_{n+2}) \bigr\} . \end{aligned}$$
Then \(R(gx_{n},gx_{n+1}) = d(gx_{n},gx_{n+1})\) or \(R(gx_{n},gx_{n+1}) = d(gx_{n+1},gx_{n+2})\). In both cases, we will show that \(\lim_{n \to \infty } d(gx_{n},gx_{n+1})=0\).
If \(R(gx_{n},gx_{n+1}) = d(gx_{n+1},gx_{n+2})\), then by inequality (1) we have
$$\begin{aligned} \phi \bigl(d(gx_{n+1},gx_{n+2})\bigr) \leq h(gx_{n},gx_{n+1})\phi \bigl(d(gx_{n+1},gx_{n+2}) \bigr) \leq \phi \bigl(d(gx_{n+1},gx_{n+2})\bigr) \end{aligned}$$
for each \(n \geq 0\). By our assumption, \(gx_{n+1} \neq gx_{n+2}\) so \(d(gx_{n+1},gx_{n+2}) > 0\). As a consequence, \(\phi (d(gx_{n+1},gx_{n+2}))>0\). Hence, \(\lim_{n \to \infty } h(gx_{n},gx_{n+1})=1\). Thus, we get \(\lim_{n \to \infty } d(gx_{n},gx_{n+1})=0\).
If \(R(gx_{n},gx_{n+1}) = d(gx_{n},gx_{n+1})\), then by similar argument as in the previous case we receive
$$\begin{aligned} \phi \bigl(d(gx_{n+1},gx_{n+2})\bigr) \leq h(gx_{n},gx_{n+1})\phi \bigl(d(gx_{n},gx_{n+1}) \bigr) \leq \phi \bigl(d(gx_{n},gx_{n+1})\bigr). \end{aligned}$$
(2)
So, we get that \(\{\phi (d(gx_{n},gx_{n+1}))\}\) is a nonincreasing sequence, which implies that the sequence \(\{d(gx_{n},gx_{n+1})\}\) must be nonincreasing by the definition of ϕ. Since the later sequence is bounded below, it becomes a convergent sequence. Suppose on the contrary that \(\lim_{n \to \infty } d(gx_{n},gx_{n+1})>0\). Thus, \(\lim_{n \to \infty } \phi (d(gx_{n},gx_{n+1}))>0\) by the property of ϕ. By (2), it is true that
$$\begin{aligned} 1=\lim_{n \to \infty } \frac{\phi (d(gx_{n+1},gx_{n+2}))}{\phi (d(gx_{n},gx_{n+1}))} \leq \lim_{n \to \infty } h(gx_{n},gx_{n+1}) \leq 1. \end{aligned}$$
Therefore, \(\lim_{n \to \infty } h(gx_{n},gx_{n+1}) = 1\). By the definition of auxiliary functions, \(\lim_{n \to \infty } d(gx_{n},gx_{n+1}) = 0\), which contradicts the assumption. So, the equation \(\lim_{n \to \infty } h(gx_{n},gx_{n+1}) = 0\) must be true.
Next, we will show that the sequence \(\{gx_{n}\}\) must be Cauchy. Suppose on the contrary that \(\{gx_{n}\}\) is not Cauchy. Therefore, there is \(\epsilon > 0\) such that, for all \(k \in \mathbb{N}\), there are \(n(k), m(k) \in \mathbb{N}\) such that \(n(k) > m(k) \geq k\) with the property that \(n(k)\) being the smallest number satisfies the properties as follows:
$$\begin{aligned} d(gx_{n(k)},gx_{m(k)}) \geq \epsilon \quad\text{and}\quad d(gx_{n(k)1},gx_{m(k)}) < \epsilon. \end{aligned}$$
This implies
$$\begin{aligned} \epsilon &\leq d(gx_{m(k)},gx_{n(k)}) \\ &\leq d(gx_{m(k)},gx_{n(k)1}) + d(gx_{n(k)1},gx_{n(k)}) \\ &< \epsilon + d(gx_{n(k)1},gx_{n(k)}). \end{aligned}$$
Taking \(k \to \infty \) in the above conclusion and using \(\lim_{n \to \infty } d(gx_{n},gx_{n+1}) = 0\), we receive
$$\begin{aligned} \lim_{k \to \infty }d(gx_{m(k)},gx_{n(k)}) = \epsilon > 0. \end{aligned}$$
(3)
Because \(E(G)\) has the transitivity property, we obtain that \((gx_{m(k)},gx_{n(k)})\in E(G)\) for every \(k \in \mathbb{N}\). As a consequence,
$$\begin{aligned} \phi \bigl(d(gx_{m(k)+1},gx_{n(k)+1})\bigr) &= \phi \bigl(d(fx_{m(k)},fx_{n(k)})\bigr) \\ &\leq h(gx_{m(k)},gx_{n(k)})\phi \bigl(R(gx_{m(k)},gx_{n(k)}) \bigr), \end{aligned}$$
(4)
where
$$\begin{aligned} R(gx_{m(k)},gx_{n(k)}) ={}& \max \biggl\{ \frac{d(gx_{m(k)},fx_{m(k)})d(gx_{n(k)},fx_{n(k)})}{d(gx_{m(k)},gx_{n(k)})}, \\ & d(gx_{m(k)},gx_{n(k)}),d(gx_{m(k)},fx_{m(k)}), d(gx_{n(k)},fx_{n(k)}), \\ & \frac{d(gx_{m(k)},fx_{n(k)}) + d(gx_{n(k)},fx_{m(k)})}{2} \biggr\} \\ ={}& \max \biggl\{ \frac{d(gx_{m(k)},gx_{m(k)+1})d(gx_{n(k)},gx_{n(k)+1})}{d(gx_{m(k)},gx_{n(k)})}, \\ & d(gx_{m(k)},gx_{n(k)}), d(gx_{m(k)},gx_{m(k)+1}),d(gx_{n(k)},gx_{n(k)+1}), \\ & \frac{d(gx_{m(k)},gx_{n(k)+1}) + d(gx_{n(k)},gx_{m(k)+1})}{2} \biggr\} \\ \leq{}& \max \biggl\{ \frac{d(gx_{m(k)},gx_{m(k)+1})d(gx_{n(k)},gx_{n(k)+1})}{d(gx_{m(k)},gx_{n(k)})}, \\ & d(gx_{m(k)},gx_{n(k)}), d(gx_{m(k)},gx_{m(k)+1}),d(gx_{n(k)},gx_{n(k)+1}), \\ & \frac{d(gx_{m(k)},gx_{n(k)})+d(gx_{n(k)},gx_{n(k)+1})}{2} \\ &{} + \frac{d(gx_{m(k)},gx_{n(k)})+ d(gx_{n(k)},gx_{m(k)+1})}{2} \biggr\} . \end{aligned}$$
Since \(\lim_{n \to \infty } d(gx_{n},gx_{n+1}) = 0\), letting \(k \to \infty \) in the above inequality implies that
$$\begin{aligned} \lim_{k \to \infty }R(gx_{m(k)},gx_{n(k)}) = \lim _{k \to \infty } d(gx_{m(k)},gx_{n(k)})= \epsilon > 0. \end{aligned}$$
By inequality (4) and the above fact, we get
$$\begin{aligned} 1=\lim_{k \to \infty } \frac{\phi (d(gx_{m(k)+1},gx_{n(k)+1}))}{\phi (d(gx_{m(k)},gx_{n(k)}))} \leq \lim_{k \to \infty } h(gx_{m(k)},gx_{n(k)}) \leq 1. \end{aligned}$$
As a result, \(\lim_{k \to \infty }h(gx_{m(k)},gx_{n(k)}) = 1\). Thus, \(\lim_{k \to \infty }d(gx_{m(k)},gx_{n(k)}) = 0\), which contradicts (3). So, it must be true that \(\{gx_{n}\}\) is Cauchy in the metric space \((X,d)\).
In the next part, we prove that \(\{gx_{n}\}\) is also Cauchy in the metric space \((X,d')\).
When \(d \geq d'\), the proof is trivial. Therefore, we consider the case \(d \ngeq d'\). Let \(\varepsilon > 0\). Because \(\{gx_{n}\}\) is Cauchy in \((X, d)\) and the function f is gCauchy on X, we obtain that \(\{fx_{n}\}\) is Cauchy in the metric space \((X, d')\). So, there is a number \(N_{0} \in \mathbb{N}\) such that
$$\begin{aligned} d'(gx_{n+1},gx_{m+1})=d'(fx_{n},fx_{m}) < \varepsilon \end{aligned}$$
for all numbers \(n,m \geq N_{0}\). Hence, the sequence \(\{gx_{n}\}\) is Cauchy in \((X,d')\).
Next, the fact that \(g(X)\) is a \(d'\)closed subset of \((X,d')\), which is complete, implies the existence of \(u = gx \in g(X)\), which satisfies
$$\begin{aligned} \lim_{n \to \infty }gx_{n} = \lim_{n \to \infty }fx_{n} = u. \end{aligned}$$
In addition, using the fact that \(f:(X,d')\to (X,d')\) is a Gcontinuous function such that f and g are \(d'\)compatible, we arrive at the conclusion that
$$\begin{aligned} \lim_{n \to \infty }d'(gfx_{n},fgx_{n}) = 0. \end{aligned}$$
(5)
Finally, consider
$$\begin{aligned} d'(gu,fu) \leq d'(gu,gfx_{n}) + d'(gfx_{n},fgx_{n}) + d'(fgx_{n},fu). \end{aligned}$$
Taking \(n \to \infty \), we obtain that \(d'(gu,fu) = 0\) because of (5), the continuity of g, and the fact that f is Gcontinuous. Therefore, \(gu = fu\), which implies that u is a coincidence point of f and g. □
In our next theorem, we consider the case when the two metrics d and \(d'\) coincide.
Theorem 10
Let \((X,d)\) be a complete metric space endowed with a directed graph \(G= (V(G),E(G) ) \), and let \(f,g:X\to X\) be functions such that \((f,g)\) is an hϕcontraction with respect to d. Assume further that the following conditions hold:

(1)
g is a continuous function such that \(g(X)\) is closed;

(2)
\(f(X)\subseteq g(X)\);

(3)
\(E(G)\) attains the transitivity property;

(4)
At least one of the following conditions is true:

(a)
f is Gcontinuous, and f and g are dcompatible;

(b)
\((X, d, G)\) has the property A.
As a consequence, we get that
$$\begin{aligned} X(f,g)\neq \varnothing \quad\textit{if and only if} \quad C(f,g)\neq \varnothing. \end{aligned}$$
Proof
From the proof of Theorem 9, it suffices to consider the only if case together with the assumption that (b) of condition \((4)\) above is true. We will adopt all the notations from the proof of Theorem 9. By the fact that the sequence \(\{gx_{n}\}\) is Cauchy and \(g(X)\) is a closed subset of X, there is an element \(u \in X\) with
$$\begin{aligned} \lim_{n \to \infty }gx_{n} = gu = \lim _{n \to \infty }fx_{n}. \end{aligned}$$
(6)
We claim that u must be a coincidence point of f and g. Suppose on the contrary that u is not a coincidence point of f and g. Therefore, \(fu \neq gu\) and hence \(d(fu,gu) > 0\). Then \((gx_{n}, gu)\in E(G)\) for every \(n\in \mathbb{N}\) because the triple \((X, d, G)\) has the property A. Also,
$$\begin{aligned} d(gu,fu) \leq d(gu,fx_{n(k)}) + d(fx_{n(k)},fu). \end{aligned}$$
As a result,
$$\begin{aligned} d(gu,fu)  d(gu,fx_{n(k)}) \leq d(fx_{n(k)},fu). \end{aligned}$$
In fact, the definition of ϕ implies that
$$\begin{aligned} \phi \bigl(d(gu,fu)  d(gu,fx_{n(k)})\bigr) & \leq \phi \bigl(d(fx_{n(k)},fu)\bigr) \\ &\leq h(gx_{n(k)},gu)\phi \bigl(R(gx_{n(k)},gu)\bigr), \end{aligned}$$
(7)
where
$$\begin{aligned} R(gx_{n(k)},gu) ={}& \max \biggl\{ \frac{d(gx_{n(k)},fx_{n(k)})d(gu,fu)}{d(gx_{n(k)},gu)}, d(gx_{n(k)},gu), \\ & d(gx_{n(k)},fx_{n(k)}), d(gu,fu), \frac{d(gx_{n(k)},fu) + d(gu,fx_{n(k)})}{2} \biggr\} . \end{aligned}$$
Letting \(k \to \infty \) in the above equation and using (6), we get
$$\begin{aligned} \lim_{k \to \infty }R(gx_{n(k)},gu) = d(gu,fu) > 0. \end{aligned}$$
Then, taking \(k \to \infty \) in (7) gives us that \(\lim_{k \to \infty }h(gx_{n(k)},gu) = 1\). This implies \(d(gu,fu)=\lim_{k \to \infty }d(gx_{n(k)},gu) =0\), which is a contradiction. Hence, \(fu = gu\) and we can deduce that f and g have u as one of their coincidence points. □
We may obtain a stronger result about the existence of a common fixed point by assuming an extra condition as in the theorem below.
Theorem 11
Let us adopt all the notations and conditions that appeared in Theorem 9. Furthermore, assume in addition that

(6)
For any \(x,y\in C(f,g)\) with \(gx\neq gy\), it is true that \((gx, gy)\in E(G)\).
As a consequence, we get that
$$\begin{aligned} X(f,g)\neq \varnothing \quad\textit{if and only if}\quad Cm(f,g)\neq \varnothing. \end{aligned}$$
Proof
From the proof of Theorem 9, it suffices to consider the only if case together with the assumption that condition \((6)\) above is true. By Theorem 9, there is an element \(x\in X\) such that \(gx=fx\).
To begin with, suppose that \(y\in X\) is also a coincidence point, i.e., \(gy=fy\). We will show that \(gx = gy\). To see this, suppose on the contrary that \(gx\neq gy\). By condition (6) above, \((gx, gy)\in E(G)\), which implies
$$\begin{aligned} \phi {\bigl(d(fx,fy)\bigr)} \leq h(gx,gy)\phi \bigl(R(gx,gy)\bigr) \leq \phi \bigl(R(gx,gy)\bigr)= \phi \bigl(d(fx,fy)\bigr). \end{aligned}$$
Thus, \(h(gx,gy)=1\) by the property of ϕ. As a result, \(gx=gy\).
Next, set \(x_{0}=x\) and use condition (2) in Theorem 9 to construct a sequence \(\{x_{n}\}\) such that \(gx_{n}=fx_{n1}\) for any \(n\in \mathbb{N}\). In particular, since x is a coincidence point, we may assume it is the case that \(x_{n}=x\) so \(gx_{n}=fx\) for every \(n\in \mathbb{N}\).
Then, let \(z=gx\) so that \(gz=ggx=gfx\). Note also that \(gx_{n}=fx=fx_{n1}\) for each \(n\in \mathbb{N}\). Therefore,
$$\begin{aligned} \lim_{n\to \infty } fx_{n}=\lim_{n\to \infty } gx_{n} = fx \end{aligned}$$
in \((X,d')\). Furthermore,
$$\begin{aligned} \lim_{n\to \infty } d'(gfx_{n},fgx_{n})= 0, \end{aligned}$$
since f and g are \(d'\)compatible. This means \(gfx=fgx\). Hence, \(gz=gfx=fgx=fz\) so \(z \in C(f,g)\). By the above proof, \(fz=gz=gx=z\). Thus, \(z \in Cm(f,g)\). □
In the last part of this section, we give an example supporting our main results.
Example 1
Suppose that \(X=[0,\infty )\subseteq \mathbb{R}\), and \(d,d':X \times X \rightarrow [0,\infty )\) are such that
$$\begin{aligned} d(x,y)= \vert xy \vert \quad\text{and}\quad d'(x,y)=L \vert xy \vert \end{aligned}$$
for all \(x,y \in X\), where \(L\in (1,\infty )\) is a constant. It can be easily checked that d and \(d'\) are metrics. Furthermore, by the way we define our metrics, we clearly get that \(d< d'\).
Next, suppose
$$\begin{aligned} E(G)=\bigl\{ (x,y): x = y \text{ or } x,y \in [0, 1] \text{ with } x \leq y \bigr\} . \end{aligned}$$
In addition, let \(f:X\to X\) and \(g:X\to X\) be given by
$$\begin{aligned} gx=x^{2} \quad\text{and}\quad fx=\ln \biggl(1+\frac{x^{2}}{4} \biggr) \end{aligned}$$
for each \(x\in X\).
We will prove that requirements (1) and (2) for the pair \((f,g)\) to be an hϕcontraction with respect to d are true.
First, let \((gx, gy)\in E(G)\). Observe that if \(x=y\), then \((fx, fy)\in E(G)\). On the other hand, if \((gx,gy) \in E(G)\) and \(gx \leq gy\), then \(gx=x^{2}\), \(gy=y^{2} \in [0, 1]\) and \(x^{2} = gx \leq gy = y^{2}\). Therefore, \(fx=\ln (1+\frac{x^{2}}{4} ) \leq \ln (1+\frac{y^{2}}{4} )=fy\) and \(fx, fy\in [0, 1] \). Thus, \((fx, fy)\in E(G)\).
Second, set \(\phi (t)=\frac{t}{4}\), and define \(h: X\times X \to [0,1]\) by the following equation:
$$\begin{aligned} h(x,y)= \textstyle\begin{cases} \frac{4\ln ( 1+ \frac{ \vert xy \vert }{4} ) }{ \vert xy \vert } &\text{if } \vert xy \vert >0, \\ 0 &\text{if } \vert xy \vert =0. \end{cases}\displaystyle \end{aligned}$$
It is straightforward to prove that \(\phi \in \Phi \) and \(h \in \mathcal{A}(X)\). Next, let \(x,y \in X\) such that \((gx, gy)\in E(G)\). If \(gx=gy\), then \(x=y\) so that requirement (2) is satisfied. On the other hand, if \(gx=x^{2}\), \(gy=y^{2} \in [0,1] \) and \(x^{2} = gx < gy = y^{2}\), then it follows that
$$\begin{aligned} \phi \bigl(d(fx,fy)\bigr)&= \frac{d(fx,fy)}{4}= \frac{ \vert \ln (1+\frac{x^{2}}{4} )\ln (1+\frac{y^{2}}{4} ) \vert }{4} \\ &= \frac{\ln (1+\frac{y^{2}}{4} )\ln (1+\frac{x^{2}}{4} ) }{4} \\ &= \frac{\ln (\frac{1+\frac{y^{2}}{4} }{1+\frac{x^{2}}{4} } )}{4} \\ &= \frac{\ln ( 1+\frac{ \frac{y^{2}}{4}  \frac{x^{2}}{4} }{1+\frac{x^{2}}{4} } ) }{4} \\ &\leq \frac{\ln ( 1+ \vert \frac{x^{2}}{4}\frac{y^{2}}{4} \vert ) }{4} \\ &= \frac{4 \ln (1 + \frac{1}{4} \vert x^{2}y^{2} \vert ) }{ \vert x^{2}y^{2} \vert } \frac{ \vert x^{2}y^{2} \vert }{4} \\ &=\frac{ 4\ln (1 + \frac{1}{4}d(gx,gy) ) }{d(gx,gy) } \phi \bigl(d(gx,gy)\bigr) \\ &=h(gx,gy) \phi \bigl(d(gx,gy)\bigr). \end{aligned}$$
Therefore, condition (2) holds for the pair \((f,g)\).
In the last part of this example, we will prove that conditions (1)–(5) of Theorem 9 are satisfied.
(1) \(g:(X,d') \rightarrow (X,d')\) is clearly a continuous function such that \(g(X)=[0,\infty )\) is also \(d'\)closed;
(2) It is not hard to see that \(f(X) = g(X) = X\);
(3) \(E(G)\) has the transitivity property;
(4) From the fact that \(d< d'\), we will show that \(f:(X,d) \rightarrow (X,d')\) is gCauchy. Given \(\epsilon >0 \) and a sequence \(\{x_{n}\}\) in X with \(\{gx_{n}\}\) being Cauchy in \((X,d)\), there is \(N \in \mathbb{N}\) such that \(d(gx_{n},gx_{m}) < \frac{\epsilon }{L}\) for all \(n,m \geq N\). Therefore,
$$\begin{aligned} d'(fx_{n},fx_{m}) &=L \vert fx_{n}fy_{m} \vert \\ &=L \biggl\vert \ln \biggl( 1+\frac{(x_{n})^{2}}{4} \biggr) \ln \biggl( 1+ \frac{(x_{m})^{2}}{4} \biggr) \biggr\vert \\ &=L \biggl\vert \ln \biggl( \frac{1+\frac{(x_{m})^{2}}{4} }{1+\frac{(x_{n})^{2}}{4} } \biggr) \biggr\vert \\ &= L \biggl\vert \ln \biggl( 1+ \frac{ \frac{(x_{m})^{2}}{4}  \frac{(x_{n})^{2}}{4} }{1+\frac{(x_{n})^{2}}{4} } \biggr) \biggr\vert \\ &\leq L \biggl[ \ln \biggl( 1+ \biggl\vert \frac{(x_{n})^{2}}{4} \frac{(x_{m})^{2}}{4} \biggr\vert \biggr) \biggr] \\ &\leq L \biggl[ \frac{4 \ln (1 + \frac{1}{4} \vert (x_{n})^{2}(x_{m})^{2} \vert ) }{ \vert (x_{n})^{2}(x_{m})^{2} \vert } \bigl\vert (x_{n})^{2}(x_{m})^{2} \bigr\vert \biggr] \\ &< L \bigl\vert (x_{n})^{2}(x_{m})^{2} \bigr\vert \\ &=Ld(gx_{n},gx_{m}) \\ &< L \biggl( \frac{\epsilon }{L} \biggr) \\ & = \epsilon. \end{aligned}$$
This amounts to saying that \(f:(X,d) \rightarrow (X,d')\) is gCauchy;
(5) \(f:(X,d') \rightarrow (X,d')\) is clearly Gcontinuous. Moreover, f and g are \(d'\)compatible since for any sequence \(\{x_{n}\}\) in X with
$$\begin{aligned} \lim_{n\to \infty }gx_{n}=\lim_{n\to \infty }fx_{n}=x, \end{aligned}$$
it results that \(\ln ( 1+\frac{x}{4} ) =x\). This implies \(x=0\). Also, as \(n\to \infty \),
$$\begin{aligned} d'(gfx_{n},fgx_{n})=L \biggl\vert \biggl( \ln \biggl( 1+ \frac{(x_{n})^{2}}{2} \biggr) \biggr) ^{2} \ln \biggl( 1+ \frac{(x_{n})^{4}}{2} \biggr) \biggr\vert \to 0. \end{aligned}$$
Finally, observe that \((g0,f0)=(0,0)\in E(G)\) so \(X(f,g)\) is nonempty. By Theorem 9, \(C(f,g) \neq \varnothing \). In fact, it can be easily seen that \(0 \in C(f,g)\).