Common fixed points of Caristi’s type mappings via wdistance
 Kittipong Sitthikul^{1} and
 Satit Saejung^{1, 2}Email author
https://doi.org/10.1186/s1366301402554
© Sitthikul and Saejung; licensee Springer 2015
Received: 18 September 2014
Accepted: 30 December 2014
Published: 23 January 2015
Abstract
Motivated by fixed point theorems of Obama and Kuroiwa (Sci. Math. Jpn. 72(1), 4148, 2010), we discuss their results with another, weaker assumption and obtain some estimating expressions. Furthermore, we also discuss the results with the lower semicontinuities of the dominated functions in place of the original orbital continuities of the mappings.
1 Introduction
In 1981, Bhakta and Basu [1] proved the following common fixed point theorem.
Theorem BB
 (BB)
\(d(Sx,Ty)\le\varphi(x)\varphi(Sx)+\psi(y)\psi(Ty)\), for all \(x,y\in X\).
 (oc)
S and T are orbitally continuous.
 (w1)
\(p(x,y)\le p(x,z)+p(z,y)\), for all \(x,y,z\in X\).
 (w2)
For each \(x\in X\), the function \(y\mapsto p(x,y)\) is lower semicontinuous.
 (w3)
For each \(\varepsilon>0\) there exists \(\delta>0\) such that \(d(y,z)\le\varepsilon\) whenever \(p(x,y)\le\delta\) and \(p(x,z)\le\delta\).
Theorem KST
([2])
 (KST)
\(p(x,Tx)\le\psi(x)\psi(Tx)\), for all \(x\in X\).
 (lsc)
ψ is lower semicontinuous.
It is clear that Theorem KST includes Caristi’s theorem as a special case.
Obama and Kuroiwa [3] used the concept of wdistance to prove the following two theorems. The first one is a generalization of Theorem BB.
Theorem OK1
 (OK1)
\(\max\{p(Sx,Ty),p(Ty,Sx)\}\le\varphi(x)\varphi(Sx)+\psi (y)\psi(Ty)\), for all \(x,y\in X\).
 (oc)
S and T are orbitally continuous.
Theorem OK2
 (OK2)
\(\max\{p(x,y),p(y,x)\}+p(x,Sx)+p(y,Ty)\le\varphi(x)\varphi (Sx)+\psi(y)\psi(Ty)\), for all \(x,y\in X\).
 (oc)
S and T are orbitally continuous.
The main tool of our result via the wdistance is based on the following lemma.
Lemma KST
([2])
 (1)
If \(p(x_{n},y_{n})\leq\alpha_{n}\) and \(p(x_{n},z)\leq\beta_{n}\), for all \(n\in\mathbb{N}\), then \(\{y_{n}\}\) converges to z. In particular, if \(p(x,y)=0\) and \(p(x,z)=0\), then \(y=z\).
 (2)
If \(p(x_{n},x_{m})\leq\alpha_{n}\), for all \(n,m\in\mathbb{N}\) with \(m>n\), then \(\{x_{n}\}\) is a Cauchy sequence.
2 Discussions on Theorems OK1 and OK2
In the appearance of the condition (OK1) or (SS) of Lemma 1 below, the problem of finding a common fixed point of S and T reduces to that of finding a fixed point of each mapping individually.
Lemma 1
 (OK1)
\(\max\{p(Sx,Ty),p(Ty,Sx)\}\leq\varphi(x)\varphi(Sx)+\psi (y)\psi(Ty)\), for all \(x,y\in X\).
 (SS)\(m(x,y)+p(x,Sx)+p(y,Ty)\leq\varphi(x)\varphi(Sx)+\psi (y)\psi(Ty)\), for all \(x,y\in X\) where$$m(x,y):=\min \left \{ \begin{array}{@{}c@{}} p(x,y), p(y,x), p(x,Ty), p(Ty,x),\\ p(y,Sx), p(Sx,y), p(Sx,Ty), p(Ty,Sx) \end{array} \right \}. $$
Proof
Let \(\widehat{x},\widehat{y}\in X\) be such that \(\widehat {x}=S\widehat{x}\) and \(\widehat{y}=T\widehat{y}\).
Remark 2
It is clear that (OK2) ⇒ (SS).
2.1 On Theorem OK1
Let \((X,d)\) be a metric space and \(S:X\to X\) be a mapping. Recall that \(G:X\to\mathbb{R}\) is Sorbitally lower semicontinuous at \(x_{0}\in X\) if \(G(\widehat{x})\le\liminf_{n\to\infty}G(x_{n})\) whenever \(\{x_{n}\}\) is a sequence in \(O(x_{0},S):=\{x_{0},Sx_{0},S^{2}x_{0},\ldots\}\) and \(\lim_{n\to\infty }x_{n}=\widehat{x}\).
The following result is motivated by the one proved by Bollenbacher and Hicks [4].
Theorem 3
 (BH)
\(\max\{p(Sx,Ty),p(Ty,Sx)\}\leq\varphi(x)\varphi(Sx)+\psi (y)\psi(Ty)\), for all \(x\in O(x_{0},S)\) and \(y\in O(y_{0},T)\).
 (cc)
Every Cauchy sequence in \(O(x_{0},S)\) converges to a point in X and every Cauchy sequence in \(O(y_{0},T)\) converges to a point in X.
 (i)There exists \(z\in X\) such that \(\lim_{n\to\infty}S^{n}x_{0}=\lim_{n\to\infty}T^{n}y_{0}=z\). Moreover,$$\lim_{n\to\infty}p\bigl(S^{n}x_{0},z\bigr)=\lim _{n\to\infty}p\bigl(T^{n}y_{0},z\bigr)=0. $$
 (ii)
\(p(S^{n+1}x_{0},z)\leq\varphi(S^{n}x_{0})+\varphi (S^{n+1}x_{0})+2\psi(T^{n}y_{0})\) and \(p(T^{n+1}y_{0},z)\leq2\varphi(S^{n+1}x_{0})+\psi(T^{n}y_{0})+\psi (T^{n+1}y_{0})\), for all \(n\ge0\).
 (iii)Define \(G(x):=p(x,Sx)\), for all \(x\in X\), and \(H(y):=p(y,Ty)\), for all \(y\in X\). Then the following statements are equivalent:
 (a)
\(z=Sz=Tz\) and \(p(z,z)=0\).
 (b)
G is Sorbitally lower semicontinuous at \(x_{0}\) and H is Torbitally lower semicontinuous at \(y_{0}\).
 (a)
Proof
Finally, we prove (iii). (a) ⇒ (b) Assume that \(z=Sz=Tz\) and \(p(z,z)=0\). Let \(\{z_{n}\}\) be a sequence in \(O(x_{0},S)\). We may assume that \(z_{n}\to z\). It follows that \(G(z)=0\leq\liminf_{n\to\infty }G(z_{n})\). Hence G is Sorbitally lower semicontinuous at \(x_{0}\). Similarly, H is Torbitally lower semicontinuous at \(y_{0}\).
Remark 4
 (1)
We replace the completeness of X with the weaker assumption (cc).
 (2)
A necessary and sufficient condition for the existence of a common fixed point of S and T is given in terms of the orbitally lower semicontinuities of \(G(x):=p(x,Sx)\) and \(H(y):=p(y,Ty)\).
 (3)
We obtain some estimating expression for the iterative sequences.
 (4)
In the presence of the conditions (BH) and (cc) of Theorem 3, it is clear that if S and T are orbitally continuous, then \(G(x):=p(x,Sx)\) is Sorbitally lower semicontinuous at \(x_{0}\) and \(H(y):=p(y,Ty)\) is Torbitally lower semicontinuous at \(y_{0}\).
2.2 On Theorem OK2
We can follow the proof of Bollenbacher and Hicks’ result [4] and the proof of Theorem 3 to obtain the following result in terms of a wdistance. This result is related to Theorem KST.
Theorem 5
 (BH)
\(p(x,Tx)\leq\Phi(x)\Phi(Tx)\), for all \(x\in O(x_{0},T)\).
 (cc)
Every Cauchy sequence in \(O(x_{0},T)\) converges to a point in X.
 (i)
There exists \(z\in X\) such that \(\lim_{n\to\infty }d(T^{n}x_{0},z)=\lim_{n\to\infty}p(T^{n}x_{0},z)=0\).
 (ii)
\(p(T^{n}x_{0},z)\leq\Phi(T^{n}x_{0})\), for all \(n\ge1\).
 (iii)
\(z=Tz\) and \(p(z,z)=0\) if and only if \(H(x):=p(x,Tx)\) is Torbitally lower semicontinuous at \(x_{0}\).
In this subsection, we give another proof of Theorem OK2 via Theorem 5 and the following lemmas. We obtain the same conclusion as Theorem 3.
Lemma 6
 (i)Define \(\boldsymbol{d}:\boldsymbol{X}\times\boldsymbol{X}\to [0,\infty)\) byfor all \(x,y,z,w\in X\). Then d is a metric on X.$$\boldsymbol {d}\bigl((x,y),(z,w)\bigr)=d(x,z)+d(y,w), $$
 (ii)Define \(\boldsymbol{p}:\boldsymbol{X}\times\boldsymbol{X}\to [0,\infty)\) byfor all \(x,y,z,w\in X\). If p is a wdistance (w.r.t. d), then p is a wdistance (w.r.t. d).$$\boldsymbol{p}\bigl((x,y),(z,w)\bigr)=p(x,z)+p(y,w), $$
Lemma 7
Proof
The following result follows directly from Theorem 5, Lemmas 6, and 7.
Lemma 8
 (OK2*)
\(p(x,Sx)+p(y,Ty)\leq\varphi(x)\varphi(Sx)+\psi(y)\psi (Ty)\), for all \(x\in O(x_{0},S)\) and \(y\in O(y_{0},T)\).
 (cc)
Every Cauchy sequence in \(O(x_{0},S)\) converges to a point in X and every Cauchy sequence in \(O(y_{0},T)\) converges to a point in X.
 (i)There exist \(z,w\in X\) such that \(\lim_{n\to\infty}S^{n}x_{0}=z\) and \(\lim_{n\to\infty}T^{n}y_{0}=w\). Moreover,$$\lim_{n\to\infty}p\bigl(S^{n}x_{0},z\bigr)=\lim _{n\to\infty}p\bigl(T^{n}y_{0},w\bigr)=0. $$
 (ii)
\(p(S^{n}x_{0},z)+p(T^{n}y_{0},w)\leq\varphi(S^{n}x_{0})+\psi(T^{n}y_{0})\), for all \(n\ge1\).
 (iii)Define \(G(x):=p(x,Sx)\), for all \(x\in X\), and \(H(y):=p(y,Ty)\), for all \(y\in X\). Then the following statements are equivalent:
 (a)
\(z=Sz\) and \(w=Tw\) and \(p(z,z)=p(w,w)=0\).
 (b)
G is Sorbitally lower semicontinuous at \(x_{0}\) and H is Torbitally lower semicontinuous at \(y_{0}\).
 (a)
Proof
We finally prove the statement (iii).
(b) ⇒ (a) It follows from Theorem 5 and Lemma 7.
(a) ⇒ (b) It is similar to the proof of (iii) of Theorem 3 so we omit the proof. □
We now obtain our result related to Theorem OK2.
Theorem 9
 (OK2)
\(\max\{p(x,y),p(y,x)\}+p(x,Sx)+p(y,Ty)\leq\varphi (x)\varphi(Sx)+\psi(y)\psi(Ty)\), for all \(x\in O(x_{0},S)\) and \(y\in O(y_{0},T)\).
 (cc)
Every Cauchy sequence in \(O(x_{0},S)\) converges to a point in X and every Cauchy sequence in \(O(y_{0},T)\) converges to a point in X.
 (i)There exists \(z\in X\) such that \(\lim_{n\to\infty}S^{n}x_{0}=\lim_{n\to\infty}T^{n}y_{0}=z\). Moreover,$$\lim_{n\to\infty}p\bigl(S^{n}x_{0},z\bigr)=\lim _{n\to\infty}p\bigl(T^{n}y_{0},z\bigr)=0. $$
 (ii)
\(p(S^{n}x_{0},z)+p(T^{n}y_{0},z)\leq\varphi(S^{n}x_{0})+\psi(T^{n}y_{0})\), for all \(n\ge1\).
 (iii)Define \(G(x):=p(x,Sx)\), for all \(x\in X\) and \(H(y):=p(y,Ty)\), for all \(y\in X\). Then the following statements are equivalent:
 (a)
\(z=Sz=Tz\) and \(p(z,z)=0\).
 (b)
G is Sorbitally lower semicontinuous at \(x_{0}\) and H is Torbitally lower semicontinuous at \(y_{0}\).
 (a)
Proof
Remark 10
 (1)
We replace the completeness of X with the weaker assumption (cc).
 (2)
A necessary and sufficient condition for the existence of a common fixed point of S and T is given in terms of the orbitally lower semicontinuities of \(G(x):=p(x,Sx)\) and \(H(y):=p(y,Ty)\).
 (3)
We obtain some estimating expression for the iterative sequences.
 (4)
In the presence of the conditions (OK2) and (cc) of Theorem 9, it is clear that if S and T are orbitally continuous, then \(G(x):=p(x,Sx)\) is Sorbitally lower semicontinuous at \(x_{0}\) and \(H(y):=p(y,Ty)\) is Torbitally lower semicontinuous at \(y_{0}\).
 (5)It is easy to see that the condition (OK2) can be replaced by the weaker condition (SS):for all \(x\in O(x_{0},S)\) and \(y\in O(y_{0},T)\). In fact, if \(m(x_{n},y_{n})\to 0\), then there is a strictly increasing sequence \(\{n_{k}\}\) on ℕ such that one of the following sequences converges to zero:$$m(x,y)+p(x,Sx)+p(y,Ty)\le\varphi(x)\varphi(Sx)+\psi(y)\psi(Ty), $$$$\begin{aligned}& \bigl\{ p(x_{n_{k}},y_{n_{k}})\bigr\} , \qquad\bigl\{ p(y_{n_{k}},x_{n_{k}})\bigr\} , \qquad\bigl\{ p(x_{n_{k}},y_{{n_{k}}+1}) \bigr\} , \qquad\bigl\{ p(y_{{n_{k}}+1},x_{n_{k}})\bigr\} , \\& \bigl\{ p(x_{{n_{k}}+1},y_{n_{k}})\bigr\} ,\qquad \bigl\{ p(y_{n_{k}},x_{{n_{k}}+1})\bigr\} ,\qquad \bigl\{ p(x_{{n_{k}}+1},y_{{n_{k}}+1}) \bigr\} ,\qquad \bigl\{ p(y_{{n_{k}}+1},x_{{n_{k}}+1})\bigr\} . \end{aligned}$$
Before moving to the next section, we give an example which satisfies our conditions in Theorems 3 and 9 but cannot be concluded from Theorems OK1 and OK2.
Example 11
3 Results on Theorems OK1 and OK2 with the lower semicontinuities of φ and ψ
As studied in Theorem KST, it is more practical to put an assumption on the dominated function ψ than to put one on the mapping T itself. In this section, we discuss Theorems OK1 and OK2 where the functions φ and ψ are assumed to be lower semicontinuous.
3.1 Results related to Theorem OK1
The following observation is obvious.
Lemma 12
 (i)
\(\widehat{p}\) is a wdistance on X.
 (ii)
If φ is lower semicontinuous, then so is \(\widehat {\varphi}\).
In the setting of Theorem OK1 with the appearance of the lower semicontinuities of φ and ψ in place of the orbital continuities of S and T, we get a partial result with some additional assumption.
Theorem 13
 (OK1)
\(\max\{p(Sx,Ty),p(Ty,Sx)\}\le\varphi(x)\varphi(Sx)+\psi (y)\psi(Ty)\), for all \(x,y\in X\).
 (lsc)
φ and ψ are lower semicontinuous.

S has a fixed point if and only if there exists an element \(\widehat{y}\in X\) such that \(\psi(\widehat{y})\le\psi(T\widehat{y})\).

T has a fixed point if and only if there exists an element \(\widehat{x}\in X\) such that \(\varphi(\widehat{x})\le\varphi(S\widehat{x})\).
Proof
 (i)
If there exists an element \(\widehat{y}\in X\) such that \(\psi (\widehat{y})\le\psi(T\widehat{y})\), then S has a fixed point.
 (ii)
If there exists an element \(\widehat{x}\in X\) such that \(\varphi(\widehat{x})\le\varphi(S\widehat{x})\), then T has a fixed point.
 (iii)
If S has a fixed point, then there exists an element \(\widehat{y}\in X\) such that \(\psi(\widehat{y})\le\psi(T\widehat{y})\).
Remark 14
In (i) of the proof of Theorem 13, we also have \(ST\widehat{y}=T\widehat{y}\). To see this, we note that \(p(\widehat{x},\widehat{x})=0\) and \(p(\widehat{x},\widehat{y})\le\psi (\widehat{y})\psi(T\widehat{y})\le0\). This gives \(\widehat{x}=\widehat {y}\) and hence \(ST\widehat{y}=S\widehat{x}=\widehat{x}=T\widehat{y}\).
Since d is a wdistance, we immediately obtain this corollary which is related to Theorem BB where the condition (oc) is replaced by the condition (lsc).
Corollary 15
 (BB)
\(d(Sx,Ty)\le\varphi(x)\varphi(Sx)+\psi(y)\psi(Ty)\), for all \(x,y\in X\).
 (lsc)
φ and ψ are lower semicontinuous.

S has a fixed point if and only if there exists an element \(\widehat{y}\in X\) such that \(\psi(\widehat{y})\le\psi(T\widehat{y})\).

T has a fixed point if and only if there exists an element \(\widehat{x}\in X\) such that \(\varphi(\widehat{x})\le\varphi(S\widehat{x})\).
3.2 Results related to Theorem OK2
Lemma 16
 (OK2*)
\(p(x,Sx)+p(y,Ty)\leq\varphi(x)\varphi(Sx)+\psi(y)\psi (Ty)\), for all \(x,y\in X\).
 (lsc)
φ and ψ are lower semicontinuous.
Proof
We now obtain a result related to Theorem OK2 where the condition (oc) is replaced by the condition (lsc).
Theorem 17
 (OK2)
\(\max\{p(x,y),p(y,x)\}+p(x,Sx)+p(y,Ty)\leq\varphi (x)\varphi(Sx)+\psi(y)\psi(Ty)\), for all \(x,y\in X\).
 (lsc)
φ and ψ are lower semicontinuous.
Remark 18
With a slight modification of the condition (OK2*), we can conclude a common fixed point even if we assume that either φ or ψ is lower semicontinuous. However, the uniqueness is not guaranteed.
Theorem 19
 (SS*)
\(p(x,Sx)+p(y,Ty)\le\varphi(x)\psi(Sx)+\psi(y)\varphi (Ty)\), for all \(x,y\in X\).
 (lsc*)
Either φ or ψ is lower semicontinuous.
Proof
Remark 20
3.3 The existence of a common fixed point of S and T is equivalent to their orbital continuities
First, let us start with the following easy observation.
Lemma 21

There exists \(z\in X\) such thatfor all \(x\in X\).$$p(z,Sx)\le\varphi(x)\varphi(Sx) $$
 (i)
\(\lim_{n\to\infty}p(z,S^{n}x)=0\) for all \(x\in X\).
 (ii)
\(z=Sz\) if and only if S is orbitally continuous and \(p(z,z)=0\).
Proof
(ii) (⇒) Assume that \(z=Sz\). In particular, \(p(z,z)=p(z,Sz)\le\varphi(z)\varphi(Sz)=\varphi(z)\varphi(z)=0\). To show that S is orbitally continuous, let \(x\in X\) be such that \(S^{n_{j}}x\to w\) for some \(w\in X\). It follows from (i) and Lemma KST that \(S^{n}x\to z\). So, we have \(w=z\). Now, \(S(S^{n_{j}}x)\to z=Sz\). (⇐) Assume that S is orbitally continuous and \(p(z,z)=0\). It follows from (i) and Lemma KST that \(S^{n}z\to z\). Since S is orbitally continuous, we have \(S(S^{n}z)\to Sz\). It follows then that \(z=Sz\). □
The following result shows that the condition (oc) is not only sufficient but also necessary for the existence of a common fixed point in Theorems OK1 and OK2.
Theorem 22
 (OK1)
\(\max\{p(Sx,Ty),p(Ty,Sx)\}\le\varphi(x)\varphi(Sx)+\psi (y)\psi(Ty)\) for all \(x,y\in X\).
 (OK2)
\(\max\{p(x,y),p(y,x)\}+p(x,Sx)+p(y,Ty)\leq\varphi (x)\varphi(Sx)+\psi(y)\psi(Ty)\) for all \(x,y\in X\).
 (oc)
S and T are orbitally continuous.
Proof
We assume that S, T satisfy the condition (OK1) and \(z=Sz=Tz\). Letting \(y=z\) in the condition (OK1) gives \(p(z,Sx)\le\varphi(x)\varphi(Sx)\) for all \(x\in X\). It follows from the preceding lemma that S is orbitally continuous. Similarly, interchanging the role of S and T ensures that T is orbitally continuous as well.
We assume that S, T satisfy the condition (OK2) and \(z=Sz=Tz\). It is clear that \(p(z,z)=0\). Because of this, letting \(y=z\) in the condition (OK2) gives \(p(z,Sx)\le p(z,x)+p(x,Sx)+p(z,Tz) \le\varphi(x)\varphi (Sx)+\psi(z)\psi(Tz)=\varphi(x)\varphi(Sx)\) for all \(x\in X\). As proved in the first part, we conclude that S and T are orbitally continuous. □
Declarations
Acknowledgements
The authors would like to thank the editor and the two referees for their comments and suggestions. The first author is supported by Khon Kaen University  Integrated Multidisciplinary Research Cluster (Sciences and Technologies). The research of the corresponding author is supported by the Centre of Excellence in Mathematics, the Commission on Higher Education, Thailand.
Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Authors’ Affiliations
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