By using the concept of F-contractions, we establish fixed point theorems for self as well as non-self mappings in complete metric spaces endowed with a graph.
Some graph theory terminologies will be presented here. Let (S,d) be metric space and △ denote the diagonal of Cartesian product S×S. Let G=(V(G),E(G)) be a directed graph such that E(G), the set of its edges consists of all loops, that is, △⊂E(G) and V(G), the vertex set coincides with S. Let G has no parallel edges (arcs). For more details of these terminologies and notations see [21] and [22].
G−1 is the converse graph of G, i.e., the edge set of G−1 is obtained by reversing the edges of G, defined as:
$$E\left(G^{-1}\right) =\left\{ \left(r,s\right) \in S\times S:\left(s,r\right) \in E\left(G\right) \right\}. $$
If s,r are vertices in the graph G, then a path from s to r of length t is a sequence \(\left \{ s_{i}\right \}_{i=1}^{t}\) of t+1 vertices of G such that s0=s,st=r and (si−1,si)∈E(G),i=1,2⋯t.
A graph G is called connected if there exist at least a path between two arbitrary vertices. If \(\tilde {G}=\left (S,E\left (\tilde {G}\right) \right) \) is the symmetric graph obtained by placing together the vertices of both G and G−1, that is,
$$E\left(\tilde{G}\right) =E\left(G\right) \cup E\left(G^{-1}\right), $$
then G is said to be weakly connected whenever \(\tilde {G}\) is connected.
If G=(V(G),E(G)) is a graph and V(G)⊃H, then the graph (H,E(G)) with
$$E\left(H\right) =E\left(G\right) \cap\left(H\times H\right) $$
is said to be the subgraph of G determined by H, denoted by GH.
Self F-contraction case
A mapping Υ:S→S is said to be defined on a metric space endowed with a graph G if it satisfies
$$ \forall\text{ }r,s\in S,\text{ }\left(r,s\right) \in E\left(G\right) \text{ implies }\left(\Upsilon r,\Upsilon s\right) \in E\left(G\right). $$
(4)
A mapping Υ:S→S defined on metric space endowed with a graph G, is said to be a GF-contraction, if there is a constant τ>0 such that ∀r,s∈S with (r,s)∈E(G), we have
$$ \left[ d\left(\Upsilon r,\Upsilon s\right) >0\Longrightarrow\tau+F\left(d\left(\Upsilon r,\Upsilon s\right) \right) \leq F\left(d\left(r,s\right) \right) \right]. $$
(5)
If Υr=r, then the element r∈S is said to be the fixed point of mapping Υ.
Theorem 2
Suppose (S,d,G) be a complete metric space endowed with a weakly connected and directed graph G such that the following property (T) holds, that is, for any sequence \(\left \{ r_{n}\right \} _{n=1}^{\infty }\subset S\) with rn→r as n→∞ and (rn,rn+1)∈E(G) for all \(n\in \mathbb {N},\) there exists a subsequence \(\left \{ r_{s_{n}}\right \}_{n=1}^{\infty }\) satisfying
$$ \left(r_{s_{n}},r\right) \in E\left(G\right),\text{ }\forall\text{ }n\in\mathbb{N}. $$
(6)
Let Υ:S→S be a GF-contraction. If the set
$$ S_{\Upsilon}=\left\{ r\in S:\left(r,\Upsilon r\right) \in E\left(G\right) \right\} $$
(7)
is nonempty, then the mapping Υ has a unique fixed point in S.
Proof
Let r0∈SΥ. It follows from (7) that (r0,Υr0)∈E(G) and by using (4), we obtain
$$ \left(\Upsilon^{n}r_{0},\Upsilon^{n+1}r_{0}\right) \in E\left(G\right),\text{ }\forall\text{ }n\in \mathbb{N}. $$
(8)
Denote rn:=Υnr0 for all \(n\in \mathbb {N}.\) Then, by the fact that Υ is a GF-contraction and in view of (4), we get
$$ F\left(d\left(r_{n},r_{n+1}\right) \right) \leq F\left(d\left(r_{n-1},r_{n}\right) \right) -\tau, $$
(9)
for all \(n\in \mathbb {N}.\) Denote αn=d(rn,rn+1),n=0,1,…
Let rn+1≠rn, for every \(n\in \mathbb {N} \cup \left \{ 0\right \}.\) Then, αn>0 for all \(n\in \mathbb {N} \cup \left \{ 0\right \} \) and by using (2), we get
$$ F\left(\alpha_{n}\right) \leq F\left(\alpha_{n-1}\right) -\tau\leq F\left(\alpha_{n-2}\right) -2\tau\leq\cdots\leq F\left(\alpha_{0}\right) -n\tau. $$
(10)
Hence, \({\lim }_{n\rightarrow \infty }F\left (\alpha _{n}\right) =-\infty.\) By the property (f2), we obtain that αn→0 as n→∞. From (f3), there exists k∈(0,1) such that \({\lim }_{n\rightarrow \infty }\alpha _{n}^{k}F\left (\alpha _{n}\right) =0.\) By (10), the following holds for all \(n\in \mathbb {N} \)
$$ \alpha_{n}^{k}F\left(\alpha_{n}\right) -\alpha_{n}^{k}F\left(\alpha_{0}\right) \leq\alpha_{n}^{k}\left(F\left(\alpha_{0}\right) -n\tau\right) -\alpha_{n}^{k}F\left(\alpha_{0}\right) =-\alpha_{n} ^{k}n\tau. $$
(11)
Letting n→∞ in (11), we deduce \({\lim }_{n\rightarrow \infty }n\alpha _{n}^{k}=0\). From (11), we observe that there exists \(n^{\prime }\in \mathbb {N} \) such that \(n\alpha _{n}^{k}\leq 1\) for all n≥n′. Consequently, we have
$$ \alpha_{n}\leq\frac{1}{n^{1/k}}\text{ for all }n\geq n^{\prime}. $$
(12)
Choose \(m,n\in \mathbb {N} \) such that m≥n≥n′ and from (12), we have
$$d\left(r_{m},r_{n}\right) \leq\alpha_{m-1}+\cdots+\alpha_{n}<\sum\limits_{j=n}^{\infty}\alpha_{n}\leq\sum\limits_{j=n}^{\infty}\frac{1}{j^{1/k}}. $$
The convergence of the series \({\sum \nolimits }_{j=n}^{\infty }\frac {1}{j^{1/k}}\) implies that {rn} is a Cauchy sequence, hence convergent in (S,d,G). The limit of this sequence is denoted as:
$$ \underset{n\rightarrow\infty}{\lim}r_{n}=r^{^{\ast}}. $$
(13)
By using property (T) of (S,d,G), there exists a subsequence \(\left \{ r_{s_{n}}\right \} \) satisfying
$$\left(r_{s_{n}},r^{\ast}\right)\in E\left(G\right),\text{ }\forall\text{ }n\in \mathbb{N}. $$
Hence, by inequality (5) and in view of (4), we get
$$ F\left(d\left(\Upsilon r_{s_{n}},\Upsilon r^{\ast}\right) \right) \leq Fd\left(r_{s_{n}},r^{\ast}\right) -\tau<F\left(d\left(r_{s_{n}},r^{\ast}\right) \right), $$
(14)
which implies
$$ d\left(\Upsilon r_{s_{n}},\Upsilon r^{\ast}\right) \leq d\left(r_{s_{n} },r^{\ast}\right). $$
(15)
Therefore, by triangle inequality, we have
$$ \begin{aligned} d\left(r^{\ast},\Upsilon r^{\ast}\right) & \leq d\left(r^{\ast},r_{s_{n}+1}\right) +d\left(r_{s_{n}+1},\Upsilon r^{\ast}\right) \\ & =d\left(r^{\ast},r_{s_{n}+1}\right) +d\left(\Upsilon r_{s_{n}},\Upsilon r^{\ast}\right).\\ \end{aligned} $$
(16)
By using (15), inequality (16) yields
$$ d\left(r^{\ast},\Upsilon r^{\ast}\right) \leq d\left(r^{\ast},r_{s_{n} +1}\right) +d\left(r_{s_{n}},r^{\ast}\right), $$
(17)
for all n≥1. In Eq. (17), assuming n→∞ and using (13), we have d(r∗,Υr∗)=0, which implies r∗=Υr∗, i.e., r∗ a fixed point of mapping Υ.
Note that the uniqueness of r∗ follows by the GF-contraction condition (5).
Remark 2If we use the mapping F∈Ω defined by the formula F(α)= lnα in Theorem 2, then for all k∈(0,1), we obtain the extension of [16], Theorem 2.1. □
Example 3Let (S,d) be the complete metric space and G be the complete graph on the set S, that is, E(G)=S×S. Let the mapping F∈Ω be defined as: F(α)= lnα, then the GF-contraction (5) is actually a F-contraction (2) which reduces to Banach contraction, i.e.,
$$d\left(\Upsilon r,\Upsilon s\right) \leq e^{-\tau}d\left(\Upsilon r,\Upsilon s\right),{ for\, all }\, r,s\in S,\text{ }\Upsilon r\neq\Upsilon s, $$
for any k∈(0,1) and τ>0.
Non-self F-contraction case
Let S be a Banach space, A be a nonempty, closed subset of S and Υ:A→S be a non-self mapping. We choose r∈A such that Υr∉A, then there is an element s∈∂A such that
$$s=\left(1-\mu\right) r+\mu\Upsilon r\text{ where }\mu\in\left(0,1\right), $$
which represents the fact that
$$ d\left(r,\Upsilon r\right) =d\left(r,s\right) +d\left(s,\Upsilon r\right),\text{ }s\in\partial A $$
(18)
where d(r,s)=∥r−s∥.
Caristi [17] used a condition related to (18), called inward condition, to get the generalization of Banach contraction theorem for non-self mappings. The inward condition is more general because it does not need s in (18) to belong to ∂A.
A non-self mapping Υ:A→S is said to be defined on the Banach space S endowed with a graph G, if it satisfies the property that
$$ \begin{aligned} &{for\, all }\, r,s\in A\text{ }\left(r,s\right) \in E\left(G\right) \\ &with\ \Upsilon r,\Upsilon s\in A,\text{ implies }\left(\Upsilon r,\Upsilon s\right) \in E\left(G\right) \cap\left(A\times A\right), \end{aligned} $$
(19)
for the subgraph of G induced by A.
Theorem 3
Suppose (S,d,G) be a Banach space endowed with a weakly connected and directed graph G provided that following property (T) holds, that is, for any sequence {rn}⊂S along with rn→r as n→∞ and
$$\left(r_{n},r_{n+1}\right) \in E\left(G\right),\text{ }\forall\text{ }n\in \mathbb{N}, $$
there exists a subsequence \(\left \{ r_{s_{n}}\right \} \) satisfying
$$ \left(r_{s_{n}},r\right) \in E\left(G\right),\text{ }\forall\text{ }n\in \mathbb{N}. $$
(20)
Let A be a nonempty, closed subset of S and Υ:A→S be a GAF-contraction, that is, there exists a constant τ>0 such that
$$ \tau+F\left(d\left(\Upsilon r,\Upsilon s\right) \right) \leq F\left(d\left(r,s\right) \right) \text{ for all }(r,s)\in E\left(G_{A}\right), $$
(21)
where GA is the subgraph of G determined by A. If the set
$$A_{\Upsilon}:=\left\{ r\in\partial A:\left(r,\Upsilon r\right) \in E\left(G\right) \right\} $$
is nonempty and Υ satisfies Rothe ’s boundary condition
$$ \Upsilon\left(\partial A\right) \subset A, $$
(22)
then the mapping Υ has a unique fixed point.
Proof
If Υ(A)⊂A, then Υ is a self-map of the closed set A and the conclusion follows by Theorem 2. Now, we consider the case that Υ(A)⊄A. Let r0∈AΥ. It follows that (r0,Υr0)∈E(G) and in view of equation (4), we have
$$ \left(\Upsilon^{n}r_{0},\Upsilon^{n+1}r_{0}\right)\in E\left(G\right),\text{ for all }n\in \mathbb{N}. $$
(23)
Let we denote rn:=Υnr0, for all \(n\in \mathbb {N}.\) By virtue of (22) Υr0∈A.
Consider r1≡s1=Υr0. Let Υr1∈A, set r2≡s2=Υr1. If Υr1∉A, then we can select an element r2∈∂A on the segment [r1,Υr1], that is,
$$r_{2}=\left(1-\mu\right) r_{1}+\mu\Upsilon r_{1},\text{ where }\mu \in\left(0,1\right). $$
By following the same method, we obtain two sequences {rn} and {sn} whose terms satisfy one of the succeeding properties:
-
(i)
rn≡sn=Υrn−1, if Υrn−1∈A;
-
(ii)
rn=(1−μ)rn−1+μΥrn−1∈∂A,μ∈(0,1),Υrn−1∉A.
For the simplicity of arguments in the proof, let us denote
$$U=\left\{ r_{a}\in\left\{ r_{n}\right\} :r_{a}=s_{a}=\Upsilon r_{a-1}\right\} $$
and
$$Z=\left\{ r_{a}\in\left\{ r_{n}\right\} :r_{a}\neq\Upsilon r_{a-1}\right\}. $$
Note that {rn}⊂A for all \(n\in \mathbb {N}.\) Moreover, if ra∈Z, then both ra−1 and ra+1 belong to set U. The sequence {rn} can have consecutive terms ra and ra+1 in set U, but this assertion is not true for the set Z. First of all we have to prove that
$$r_{a}\neq\Upsilon r_{a-1}\,\, { implies}\,\, r_{a-1}=\Upsilon r_{a-2}. $$
Suppose contrary that ra−1≠Υra−2 then ra−1∈∂A. Since Υ(∂A)⊂A then Υra−1∈A. Hence, ra=Υra−1 which is a contradiction.
Here, we have three different cases to show that {rn} is Cauchy sequence which are following: □
Case 1. rn,rn+1∈U.
Since both elements belong to set U, therefore, we have rn=sn=Υrn−1 and rn+1=sn+1=Υrn. Hence,
$$d\left(r_{n+1},r_{n}\right) =d\left(s_{n+1},s_{n}\right) =d\left(\Upsilon s_{n},\Upsilon s_{n-1}\right), $$
where (sn,sn−1)∈E(G) by virtue of (23). Therefore, we have
$$d\left(\Upsilon s_{n},\Upsilon s_{n-1}\right) =d\left(\Upsilon r_{n},\Upsilon r_{n-1}\right) >0. $$
Consequently, we get the following inequality
$$ \tau+F\left(d\left(\Upsilon s_{n},\Upsilon s_{n-1}\right) \right) \leq F\left(d\left(s_{n},s_{n-1}\right) \right), $$
(24)
by using (21).
Case 2. rn∈U,rn+1∈Z.
In this case, we have rn=sn=Υrn−1, but rn+1≠sn+1=Υrn; therefore, we have
$$d\left(r_{n},\Upsilon r_{n}\right) =d\left(r_{n},r_{n+1}\right) +d\left(r_{n+1},\Upsilon r_{n}\right). $$
The above equality implies d(rn+1,Υrn)≠0 and hence
$$ d\left(r_{n},r_{n+1}\right) =d\left(r_{n},\Upsilon r_{n}\right) -d\left(r_{n+1},\Upsilon r_{n}\right) < d\left(r_{n},\Upsilon r_{n}\right) =d\left(\Upsilon r_{n-1},\Upsilon r_{n}\right), $$
(25)
since rn∈U. By using (25), we obtain
$$d\left(r_{n},r_{n+1}\right) < d\left(\Upsilon r_{n-1},\Upsilon r_{n}\right) =d\left(\Upsilon s_{n-1},\Upsilon s_{n}\right) >0. $$
We can obtain again inequality (24) by using the similar arguments to that in case 1.
Case 3. rn∈Z,rn+1∈U.
In this case, we have rn+1=Υrn, and rn≠sn=Υrn−1. Since rn∈Z, so we have
$$ d\left(r_{n-1},\Upsilon r_{n-1}\right) =d\left(r_{n-1},r_{n}\right) +d\left(r_{n},\Upsilon r_{n-1}\right). $$
(26)
Hence, by triangle inequality
$$ \begin{aligned} d\left(r_{n},r_{n+1}\right) & \leq d\left(r_{n},\Upsilon r_{n-1}\right) +d(\Upsilon r_{n-1},r_{n+1})\\ & =d\left(r_{n},\Upsilon r_{n-1}\right) +d(\Upsilon r_{n-1},\Upsilon r_{n})\\ & =d\left(r_{n},\Upsilon r_{n-1}\right) +d(\Upsilon s_{n-1},\Upsilon s_{n}). \end{aligned} $$
(27)
By virtue of (23) (sn−1,sn)∈E(G), and the following inequality is obtained by the contraction condition (21)
$$ F\left(d\left(\Upsilon s_{n-1},\Upsilon s_{n}\right)\right) \leq F\left(d\left(s_{n-1},s_{n}\right) \right) -\tau<F\left(d\left(s_{n-1},s_{n}\right) \right), $$
(28)
which implies
$$ d(\Upsilon s_{n-1},\Upsilon s_{n})\leq d\left(s_{n-1},s_{n}\right) =d\left(r_{n-1},r_{n}\right). $$
(29)
Thus, by using (26) and (29) in inequality (27), we have
$$\begin{aligned} d\left(r_{n},r_{n+1}\right) &\leq d\left(r_{n},\Upsilon r_{n-1}\right) +d(\Upsilon s_{n-1},\Upsilon s_{n})\\ & < d\left(r_{n},\Upsilon r_{n-1}\right) +d(r_{n-1},r_{n})\\ & =d\left(r_{n-1},\Upsilon r_{n-1}\right).\\ \end{aligned} $$
By using (23), (rn−2,rn−1)=(sn−2,sn−1)∈E(G) and by virtue of contraction condition (21), we get
$$ d\left(r_{n},r_{n+1}\right) < d\left(r_{n-1},\Upsilon r_{n-1}\right) =d\left(\Upsilon r_{n-2},\Upsilon r_{n-1}\right) \leq d\left(r_{n-2},r_{n-1}\right). $$
(30)
Now, we summarize all the above mentioned three cases. By virtue of (24) and (30), it follows that the sequence {d(rn,rn+1)} satisfies the inequality
$$ \tau+F\left(\max\left\{ d\left(r_{n-2},r_{n-1}\right),d\left(r_{n-1},r_{n}\right) \right\} \right) \leq F\left(d\left(r_{n},r_{n+1}\right) \right), $$
(31)
for all n≥2. Denote αn=d(rn,rn+1) for n=2,3,⋯.
We obtain the following inequality by simple induction for n≥2, and using (31)
$$ F\left(\alpha_{n}\right) \leq F\left(\max\left\{ \alpha_{0},\alpha_{1}\right\} \right) -\left[ \frac{n}{2}\right] \tau, $$
(32)
where \(\left [ \frac {n}{2}\right ] \) denotes the greatest integer not exceeding \(\frac {n}{2}.\)
Hence, \({\lim }_{n\rightarrow \infty }F\left (\alpha _{n}\right) =-\infty.\) By the property (f2), we obtain that αn→0 as n→∞. From (f3), there exists k∈(0,1) such that \({\lim }_{n\rightarrow \infty }\alpha _{n}^{k}F\left (\alpha _{n}\right) =0.\) Denote γ= max{α0,α1}. By (32), the following holds for all n≥2:
$$ \alpha_{n}^{k}F\left(\alpha_{n}\right) -\alpha_{n}^{k}F\left(\gamma\right) \leq\alpha_{n}^{k}\left(F\left(\gamma\right) -\left[ \frac{n}{2}\right] \tau\right) -\alpha_{n}^{k}F\left(\gamma\right) =-\alpha_{n}^{k}\left[ \frac{n}{2}\right] \tau. $$
(33)
Assuming n→∞ in (33), we deduce \({\lim }_{n\rightarrow \infty }\left [ \frac {n}{2}\right ] \alpha _{n}^{k}=0\). From (33), we observe that there exists \(n^{\prime }\in \mathbb {N} \) such that \(\left [ \frac {n}{2}\right ] \alpha _{n}^{k}< n\alpha _{n}^{k}\leq 1 \) for all n≥n′. Consequently, we have
$$ \alpha_{n}\leq\frac{1}{n^{1/k}}\text{ for all }n\geq n^{\prime}. $$
(34)
Choose \(m,n\in \mathbb {N} \) such that m≥n≥n′ and from (34), we have
$$d\left(r_{m},r_{n}\right) \leq\alpha_{m-1}+\cdots+\alpha_{n}<\sum\limits_{j=n}^{\infty}\alpha_{n}\leq\sum\limits_{j=n}^{\infty}\frac{1}{j^{1/k}}. $$
The convergence of the series \({\sum \nolimits }_{j=n}^{\infty }\frac {1}{j^{1/k}}\) implies that {rn} is a Cauchy sequence, hence convergent in (S,d,G). Since {rn}⊂A and A is closed, {rn} converges to some point \(r^{^{\prime }}\in A,\) i.e., \({\lim }_{n\rightarrow \infty }r_{n}=r^{^{\prime }}.\)
By property (T), there exists a subsequence \(\left \{ r_{s_{n}}\right \} \) satisfying
$$\left(r_{s_{n}},r^{^{\prime}}\right)\in E\left(G\right),\text{ for all }n\in \mathbb{N}. $$
Hence, by the F-contraction condition (21), we get
$$ d\left(\Upsilon r_{s_{n}},\Upsilon r^{^{\prime}}\right) \leq d\left(r_{s_{n}},r^{^{\prime}}\right). $$
(35)
Therefore, by triangle inequality, we have
$$\begin{aligned} d\left(r^{^{\prime}},\Upsilon r^{^{\prime}}\right) & \leq d\left(r^{^{\prime}},r_{s_{n}+1}\right) +d\left(r_{s_{n}+1},\Upsilon r^{^{\prime} }\right) \\ & =d\left(r^{^{\prime}},r_{s_{n}+1}\right) +d\left(\Upsilon r_{s_{n} },\Upsilon r^{^{\prime}}\right). \end{aligned} $$
By using (35), the above inequality yields
$$ d\left(r^{^{\prime}},\Upsilon r^{^{\prime}}\right) \leq d\left(r^{^{\prime}},r_{s_{n}+1}\right) +d\left(r_{s_{n}},r^{^{\prime}}\right), $$
(36)
for all n≥1. Taking limit n→∞ and using (36), we obtain \(d\left (r^{^{\prime }},\Upsilon r^{^{\prime }}\right) =0\) and get \(r^{^{\prime }}=\Upsilon r^{^{\prime }}\), which shows that \(r^{^{\prime }}\) is a fixed point of Υ.
The uniqueness of r∗ immediately follows by the GAF -contraction condition (21).
Remark 3
If we use the mapping F∈Ω defined by the formula F(α)= lnα in Theorem 3, then for all k∈(0,1), we obtain the extension of [16, Theorem 3.1].
Example 4
Let \(S=\mathbb {R}\) be a Banach space with the usual norm and A=(−∞,0] is a closed subset of S. Let the mapping Υ:A→S be defined as:
$$\Upsilon r=\left\{\begin{array}{lc} 0\text{ \ \ \ \ \ if }r\in\left[ -1,0\right] \\ 0.5\text{ \ \ \ if }r\in\left(-\infty,-1\right).\\ \end{array}\right. $$
Let the mapping F∈Ω be given by the formula \(F\left (\alpha \right) =\frac {-1}{\sqrt {\alpha }}\), and the edge set of graph G and the subgraph GA determined by A is defined as:
$$E\left(G\right) =\left\{ \left(r,s\right) \in S\times S:r\leq s\right\} $$
and
$$E\left(G_{A}\right) =\left\{ \left(r,s\right) \in A\times A:r\leq s\right\}, $$
respectively. It is easy to check that (19) holds, that is,
$$\begin{aligned} &{for\, all }\, r,s\in A\text{ }\left(r,s\right) \in E\left(G\right) \\ &{with }\,\, \Upsilon r,\Upsilon s\in A,\text{ implies }\left(\Upsilon r,\Upsilon s\right) \in E\left(G\right) \cap\left(A\times A\right). \end{aligned} $$
In view of (19), for t,u∈(−∞,−1) and r,s∈[−1,0], the edges (t,u),(t,r) has to be removed and for the rest of edges we have
$$\left(\Upsilon r,\Upsilon s\right) =\left(0,0\right) \in E\left(G_{A}\right). $$
Moreover, G is a weakly connected and for any k∈(0.5,1),Υ is a non-self GAF-contraction on A with \(\tau =\dfrac {1}{\sqrt {d\left (r,s\right) }},\) since
$$d\left(\Upsilon r,\Upsilon s\right) =\frac{1}{2}<\frac{1}{4}\times d\left(r,s\right) \text{ for }r\in\left(-\infty,-1\right) \text{ and }s\in\left[ -1,0\right] \text{.} $$
(for the rest of edges of E(GA), the F-contraction condition (21) is obvious, since the quantity in its left hand side is always zero). Property (T) holds with constant sequences {rn=r} satisfying the property (rn,rn+1)∈E(GA), for all \(n\in \mathbb {N}.\) Rothe’s boundary condition is also satisfied, as ∂A={0} and so Υ(∂A)⊂A. Finally, since we have AΥ={0}≠∅, all assumptions in Theorem 3 are satisfied and \(r^{^{\prime }}=0\) is the fixed point of Υ.