Remarks on Suzuki-type fixed point results in relational metric spaces

Next, we show that

Since \(\omega_{n}\rightarrow z\), so for each \(\upsilon \in {X - {\{ z\}}}\) there exists \(v_{\omega} \in {\mathbb{N}}\) such that \({{d{(\omega_{n},z)}} \leq {{d{(\upsilon,z)}}/3}},\) for \(n \in {\mathbb{N}}\) with \(n \geq v_{\omega}\), and

Take \(n:={n + j}\) and \(\upsilon:={T^{j}{(z)}}\) in (3.3), we get

In view of (6) and from (2), we have

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

Thus, inequality (7) holds for all \(j \in {\mathbb{N}}\), and the sequence \(\{{T^{n}{(z)}}\}\) converges to \(z\).

We will now prove that \(z\) is a fixed point of \(T\). Let \(z \neq {T{(z)}}\). Then it follows from (9) that \({T^{j}{(z)}} \neq z\) for \(j \in {\mathbb{N}}\) and \(s \neq 0\). By the induction method, we first show that

holds for \(n \in {\mathbb{N}}\) with \(n \geq 2\). For \(n = 2\), it is straightforward to see from (5). We assume that \({d{({T^{n}{(z)}},{T{(z)}})}} \leq {sd{(z,{T{(z)}})}}\) holds for some \(n \in {\mathbb{N}}\) with \(n \geq 2\). Then by the triangle inequality,

implies

We consider two cases: In the first case, let \(0 < s < \frac{1}{\sqrt{2}}\). Then \({\vartheta{(s)}} \leq \frac{1 - s}{s^{2}}\) and in view of (5), we have

In second case, let \(\frac{1}{\sqrt{2}} \leq s < 1\). Then \({\vartheta{(s)}} = \frac{1}{1 + s}\) and by triangle inequality and the assumption, we have

We get, in both cases, \({\vartheta{(s)}d{({T^{n}{(z)}},{T^{n + 1}{(z)}})}} \leq {d{(z,{T^{n}{(z)}})}}\) for \({n \in {\mathbb{N}}},\) so by (2), we have

Making \(n\rightarrow\infty\), we get \({d{(z,{T{(z)}})}} = 0\), hence \({T{(z)}} = z\).

To prove the uniqueness of the fixed point, assume that \(\upsilon\) is another fixed point of \(T\), that is, \({T{(\upsilon)}} = \upsilon\) and \(z \neq \upsilon\). Since \({z,\upsilon} \in {T{(X)}}\) and \(T{(X)}\) is \(\mathcal{R}^{S}\)-connected by assumption, there exists a path \(\{ z_{0},z_{1},z_{2},\ldots,z_{k}\}\) of finite length \(k\) in \(\mathcal{R}\) such that \(z_{0} = z\), \(z_{k} = \upsilon\), and \({{\lbrack z_{i},z_{i + 1}\rbrack} \in \mathcal{R}},{i = {0,1,2,\ldots,{k - 1}}}\). By the \(T\)-closedness property, we get

Next, we prove that \({T^{n}{(z_{i})}}\rightarrow z\) for \(i = {0,1,\ldots,k}\). Clearly, for \(i = 0\), \({T^{n}{(z_{0})}} = z\rightarrow z\). Since \(\vartheta{(s)}d{(T^{n}{(z_{0})},T^{n + 1}{(z_{0})} = 0 \leq d{(T^{n}{(z_{0})},T^{n + 1}{(z_{1})})}}\) holds for \(n \in {\mathbb{N}}\), from (2) we have

Making \(n\rightarrow\infty\), we get \({d{({T^{n}{(z_{0})}},{T^{n}{(z_{1})}})}}\rightarrow 0\), hence \({T^{n}{(z_{i})}}\rightarrow z\) for \(i = 1\).

Assume that \({T^{n}{(z_{i})}}\rightarrow z\) for some \(i = \ell < k\). If \({T^{N}{(z_{\ell + 1})}} = z\) for some positive integer \(N\) then \({T^{n}{(z_{\ell + 1})}} = z\) for all \(n \geq N\) implies \({T^{n}{(z_{\ell + 1})}}\rightarrow z\). So, we may assume \({T^{n}{(z_{\ell + 1})}} \neq z\) for all \(n \in {\mathbb{N}}\). Since \({T^{n}{(z_{\ell})}}\rightarrow z\) and \({T^{n}{(z_{\ell + 1})}} \neq z\) for \(n \in {\mathbb{N}}\), thus in light of (3.3), for every \({T^{n}{(z_{\ell + 1})}} \neq z\) there exists \(v_{n} \in {\mathbb{N}}\) such that

and by (2), we have

which follows that

Making \(n\rightarrow\infty\), we get \({d{({T^{n}{(z_{\ell})}},{T^{n}{(z_{\ell + 1})}})}}\rightarrow 0\), that is, \({T^{n}{(z_{i})}}\rightarrow z\) for \(i = {\ell + 1}\). Hence by induction method, \({T^{n}{(z_{i})}}\rightarrow z\) for \(i = {3,4,\ldots,k}\). But \({T^{n}{(z_{k})}} = \upsilon\rightarrow\upsilon\), so by the uniqueness of the limit, we have \(\upsilon = z\).

Since \({({(0,0)},{(0,0)})} \in \mathcal{R}\), it follows that \({X{(T,\mathcal{R})}} \neq \varnothing\). Consider \(\omega = {(4,5)}\), \(\upsilon = {(5,4)}\), and \(z = {(0,0)}\). We have \({{({T{(\omega)}},{T{(\upsilon)}})},{({T{(\upsilon)}},{T{(z)}})}} \in \mathcal{R}\) but \({({T{(\omega)}},{T{(z)}})} \notin \mathcal{R}\), showing that \(\mathcal{R}\) is not \(T\)-transitive. For \({(\omega,\upsilon)} = \left( {(4,5)},{(5,4)} \right) \in \mathcal{R}\),

and for \({(\omega,\upsilon)} \in \mathcal{R}\) with \({(\omega,\upsilon)} \neq \left( {(4,5)},{(5,4)} \right)\), we have

Thus, \(T\) satisfies condition (2) relative to \(\mathcal{R}\). Moreover, one can easily verify that \((X,T,\mathcal{R})\) is regular and has the \(\mathcal{R}\)-SLP. Hence, all the assumptions of Theorem 3.2 are satisfied, and \(T\) has a unique fixed point at \((0,0)\).

Following the proof of Theorem 3.1, we show that for each \(\omega \in {X{(T,\mathcal{R})}}\), the sequence of iterates \(\{{T^{n}{(\omega)}}\}\) is \(\mathcal{R}\)-PS and \({T^{n}{(\omega)}}\overset{d}{\rightarrow}z \in X\). Since \(T\) is \(\mathcal{R}\)-continuous, it follows that \({T^{n + 1}{(\omega)}}\overset{d}{\rightarrow}{T{(z)}}\), and by the uniqueness of limits, we obtain \({T{(z)}} = z\). Thus, \(z\) is a fixed point of \(T\). The uniqueness of \(z\) follows directly from the arguments in Theorem 3.2.

We observe that \({X{(T,\mathcal{R})}} = {\{ 5,0,1\}} \neq \varnothing\). It is easy to verify that \(\mathcal{R}\) is \(T\)-closed and \(T\) is \(\mathcal{R}\)-continuous. For \({(\omega,\upsilon)} = {(5,7)}\) or \({(7,5)} \in \mathcal{R}\), we have

and

Also, for \({(\omega,\upsilon)} \in {\mathcal{R} \smallsetminus {\{{(5,7)},{(7,5)}\}}}\), we have

Thus, \(T\) satisfies condition (2). By Theorem 3.6, the mapping \(T\) has a unique fixed point at \(\omega = 1\).

Moreover, \(\{ 5,0,1\}\) is a \(\mathcal{R}\)-PS such that \({\{ 5,0,1\}}\overset{d}{\rightarrow}1\) but \({(5,1)} \notin \mathcal{R}\). It follows that \((X,T,\mathcal{R})\) is not a regular. Also, for \(\omega = 7\), \(\upsilon = 5\), and \(z = 0\) in \(X\), we have \({{({T{(\omega)}},{T{(\upsilon)}})},{({T{(\upsilon)}},{T{(z)}})}} \in \mathcal{R}\) but \({({T{(\omega)}},{T{(z)}})} \notin \mathcal{R}\) implies \(\mathcal{R}\) is not \(T\)-transitive.

Take \(u_{0} \in {X{(T,\mathcal{R})}}\) such that \(u_{1} = {T{(u_{0})}}\) and define a sequence \(\{ u_{n}\}\) in \(X\) by \(u_{n} = {T^{n}u_{0}}\) for each \(n \in {\mathbb{N}}\). Following the proof of Theorem 4.1 in [4], \(\{ u_{n}\}\) is \(\mathcal{R}\)-PS and converges to a point \(u \in X\).

Suppose \(T\) is \(\mathcal{R}\)-continuous. Then \(u_{n + 1} = {T{(u_{n})}}\overset{d}{\rightarrow}{T{(u)}}\). By the uniqueness of the limit, \({T{(u)}} = u\), meaning \(u\) is a fixed point of \(T\).

Alternatively, assume that \(\mathcal{R}\) is \(d\)-self-closed. Then there exists a subsequence \(\{ u_{n_{k}}\}\) of \(\{ u_{n}\}\) such that \({\lbrack u_{n_{k}},u\rbrack} \in \mathcal{R}\) for all \(k \in {{\mathbb{N}} \cup {\{ 0\}}}\). Since condition (e) implies that \({d{({T{(\omega)}},{T{(\upsilon)}})}} < {d{(\omega,\upsilon)}}\) whenever \(\omega \neq \upsilon\), we assert that either

holds for each \(k \in {{\mathbb{N}} \cup {\{ 0\}}}\).

Assume, on the contrary, that for some \(k\),

Then, by the triangle inequality,

which is a contradiction. Hence, by a standard argument, there exists a subsequence \(\{ n_{k_{j}}\}\) of \(\{ n_{k}\}\) such that

and applying condition (e), we get

In view of property \((F_{1})\), we conclude that

Letting \(j\rightarrow\infty\), we deduce that \({d{(u,{T{(u)}})}} = 0\), and hence \({T{(u)}} = u\). Thus, \(u\) is a fixed point of \(T\).

Suppose \(u\) and \(v\) are two fixed points of \(T\), i.e., \({T{(u)}} = u\), \({T{(v)}} = v\), and \(u \neq v\). Since \(T{(X)}\) is \(\mathcal{R}\)-connected, there exists a finite sequence \(\{\xi_{0},\xi_{1},\ldots,\xi_{k}\}\) with \(\xi_{0} = u\), \(\xi_{k} = v\), and \({\lbrack\xi_{i},\xi_{i + 1}\rbrack} \in \mathcal{R}\) for all \(i = {0,1,\ldots,{k - 1}}\). By the \(T\)-closedness of \(\mathcal{R}\), we have

We will show that \({T^{n}{(\xi_{i})}}\rightarrow u\) for each \(i = {0,1,\ldots,k}\). Clearly, \({T^{n}{(\xi_{0})}} = u\) for all \(n\), hence \({T^{n}{(\xi_{0})}}\rightarrow u\). Assume by induction that \({T^{n}{(\xi_{\ell})}}\rightarrow u\) for some \(\ell < k\). We aim to show \({T^{n}{(\xi_{\ell + 1})}}\rightarrow u\). Suppose \({T^{n}{(\xi_{\ell + 1})}} \neq u\) for all \(n\). Then, there exists \(v_{n} \in {\mathbb{N}}\) such that for all \(n \geq v_{n}\),

Applying condition (e), we get

Set \(\Delta_{n}:={d{({T^{n}{(\xi_{\ell})}},{T^{n}{(\xi_{\ell + 1})}})}}\). Then (10) becomes

Define

Then from (11), it follows

Since \(\{\Delta_{n}\}\) is decreasing and bounded below, there exists \(\Delta \geq 0\) such that \({\lim_{n\rightarrow\infty}\Delta_{n}} = \Delta\). If \(\Delta > 0\), then \({\operatorname{lim\ inf}_{n\rightarrow\infty}{\varphi{(\Delta_{n}^{p})}}} > 0\), implying \({F{(\Delta_{n})}}\rightarrow{- \infty}\) as \(n\rightarrow\infty\), a contradiction. Therefore, \(\Delta = 0\), hence \({T^{n}{(\xi_{\ell + 1})}}\rightarrow u\). By induction, \({T^{n}{(\xi_{i})}}\rightarrow u\) for each \(i = {0,1,\ldots,k}\). But \({T^{n}{(\xi_{k})}} = v\) for all \(n\), so \(v = u\) by uniqueness of the limit. Thus, the fixed point is unique.

Let \(u\) and \(v\) be two distinct fixed points of \(T\), that is, \({T{(u)}} = u\), \({T{(v)}} = v\), and \(u \neq v\). Since \(\gamma{(u,v,\mathcal{R}^{S})}\) is nonempty, there exists a finite path \(\{\xi_{0},\xi_{1},\xi_{2},\ldots,\xi_{k}\}\) of finite length \(k\) such that \(\xi_{0} = u\), \(\xi_{k} = v\), and \({\lbrack\xi_{i},\xi_{i + 1}\rbrack} \in \mathcal{R}\) for each \(i = {0,1,\ldots,{k - 1}}\). By the \(T\)-closedness property, we have

Since \(\mathcal{R}\) is \(T\)-transitive, it follows that \({(u,v)} \in \mathcal{R}\). Then, by applying condition (d), we get

a contradiction unless \(u = v\). Thus, \(T\) has a unique fixed point.