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Some remarks on ‘Multidimensional fixed point theorems for isotone mappings in partially ordered metric spaces’
Fixed Point Theory and Applications volume 2014, Article number: 245 (2014)
Abstract
The main aim of this paper is to advise researchers in the field of Fixed Point Theory against an extended mistake that can be found in some proofs. We illustrate our claim proving that theorems in the very recent paper (Wang in Fixed Point Theory Appl. 2014:137, 2014) are incorrect, and we provide different corrected versions of them.
1 Introduction and preliminaries
Let (X,d) be a metric space, let k be a positive integer number, and let {X}^{k}=X\times X\times \stackrel{(k)}{\cdots}\times X be the Cartesian product of k identical copies of X. The function {\rho}_{k}:{X}^{k}\times {X}^{k}\to [0,\mathrm{\infty}) defined, for all ({y}_{1},{y}_{2},\dots ,{y}_{k}),({v}_{1},{v}_{2},\dots ,{v}_{k})\in {X}^{k}, by
is a metric on {X}^{k}.
Let Φ denote the set of all continuous and strictly increasing functions \varphi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}), and Ψ denote the set of all functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) such that {lim}_{t\to r}\psi (t)>0, for all r>0.
Inspired by Roldán et al.’s notion of a multidimensional fixed point (see [1–3]), Wang announced the following results in [4].
Theorem 1 (Wang [[4], Theorem 3.1])
Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:{X}^{k}\to {X}^{k} be an isotone mapping for which there exist \phi \in \mathrm{\Phi} and \psi \in \mathrm{\Psi} such that, for all Y,V\in {X}^{k} with Y\u2ab0V,
where {\rho}_{k} is defined by (1). Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {Z}_{0}\in {X}^{k} such that {Z}_{0}\asymp T({Z}_{0}), then T has a fixed point.
As a consequence, she deduced the following result.
Theorem 2 (Wang [[4], Corollary 3.2])
Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi} and \psi \in \mathrm{\Psi} such that, for all y,v\in X with y\u2ab0v,
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
Remark 3 Theorems 1 and 2 are equivalent: Wang interpreted Theorem 2 as a corollary of Theorem 1 (taking k=1), but is also true that Theorem 1 is a particularization of Theorem 2 taking ({X}^{k},{\rho}_{k},\u2aaf) rather than (X,d,\u2aaf).
We claim that Theorem 2 is false, providing the following counterexample. As a consequence, all results in [4] are not correct.
2 A counterexample
Let X=\mathbb{N}=\{1,2,3,\dots \} be the set of all natural numbers endowed with the usual order ≤ of real numbers and the Euclidean metric d(x,y)=xy, for all x,y\in X. Define T:X\to X and \varphi ,\psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by
Clearly, \phi \in \mathrm{\Phi}, \psi \in \mathrm{\Psi} (in fact, notice that {lim}_{t\to r}\psi (t)=1, for all r\ge 0). Furthermore, T is a continuous, nondecreasing mapping. Moreover, it verifies condition (2) because if y,v\in X=\mathbb{N} are natural numbers, then d(y,v) is also a nonnegative integer number, so \psi (d(y,v))=0, and this proves that
(it is not necessary to assume that y\u2ab0v). Any point {z}_{0}\in X verifies {z}_{0}\u2aafT({z}_{0}). However, T does not have any fixed point.
Remark 4 Notice that Wang’s results are valid if we add the assumption \psi (t)=0\iff t=0 but, in this case, the results are not as attractive.
3 Some considerations of Wang’s paper
We write the following in order to advise researchers against the mistake in Wang’s proofs, which could be found on other papers.
In fixed point theory, given an operator T:X\to X, it is usual to consider a Picard sequence {x}_{n+1}=T{x}_{n}, for all n\in \mathbb{N}. In the context of partially ordered metric spaces, this sequence must be monotone (for instance, nondecreasing). Applying the contractivity condition (2), it follows that, for all n\in \mathbb{N},
Using the fact that φ is strictly increasing, we have d({x}_{n+1},{x}_{n+2})\le d({x}_{n},{x}_{n+1}), for all n\in \mathbb{N}, which means that {\{d({x}_{n},{x}_{n+1})\}}_{n\in \mathbb{N}} is a nonincreasing sequence of nonnegative real numbers. As a consequence, it is convergent. Let L\ge 0 be its limit. In order to prove that L=0, we reason by contradiction assuming that L>0. Taking into account that
it is usual to take the limit in the previous inequality. As φ is continuous and {lim}_{t\to L}\psi (t)>0 by hypothesis, the author tried to deduce that
which is a contradiction. However, the equality
can be false: as in the previous counterexample, the sequence {\{\psi (d({x}_{n},{x}_{n+1}))\}}_{n\in \mathbb{N}} can be identically zero but the limit {lim}_{t\to L}\psi (t) must take a positive value.
4 A suggestion to correct Wang’s paper
In this section, inspired by Wang’s paper [4], we suggest a new theorem in the context of partially ordered metric spaces. We also underline that one can easily state the same theorem in the frame of multidimensional fixed points. In fact, they are equivalent as is mentioned in Section 1.
Firstly, we give a modified version of the collection of auxiliary functions Ψ in the following way. Let {\mathrm{\Psi}}^{\prime} denote the set of all lower semicontinuous functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) such that \psi (t)=0\iff t=0. The following two lemmas will be useful in the proofs of the upcoming theorems.
Lemma 5 (See, e.g., [5])
Let (X,d) be a metric space and let \{{x}_{n}\} be a sequence in X such that
If \{{x}_{n}\} is not a Cauchy sequence in (X,d), then there exist \epsilon >0 and two sequences \{n(k)\} and \{m(k)\} of positive integers such that, for all k\in \mathbb{N},
Furthermore,
Lemma 6 Let \phi \in \mathrm{\Phi}, \psi \in {\mathrm{\Psi}}^{\prime}, and let \{{t}_{n}\},\{{s}_{n}\},\{{a}_{n}\}\subset [0,\mathrm{\infty}) be three sequences of nonnegative real numbers such that
If there exists \alpha \in [0,\mathrm{\infty}) such that \{{t}_{n}\}\to \alpha, \{{s}_{n}\}\to \alpha, \{{a}_{n}\}\to 0, and {s}_{n}>\alpha, for all n\in \mathbb{N}, then \alpha =0.
Proof Notice that
As \{{t}_{n}\}\to \alpha, \{{s}_{n}\}\to \alpha, \{{a}_{n}\}\to 0, and ψ is continuous, we deduce that
As {s}_{n}>\alpha, for all n\in \mathbb{N}, and ψ is lower semicontinuous, we deduce that
Therefore, \psi (\alpha )=0. As \psi \in {\mathrm{\Psi}}^{\prime}, we conclude that \alpha =0. □
Theorem 7 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi}, \psi \in {\mathrm{\Psi}}^{\prime} and L\ge 0 such that
for all x,y\in X with x\u2aafy, where
and
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
Proof By assumption, there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}). Without loss of generality, we assume that {z}_{0}\u2aafT{z}_{0} (the case {z}_{0}\u2ab0T{z}_{0} can be treated analogously). Define an iterative sequence \{{z}_{n}\} as {z}_{n+1}=T{z}_{n} for each n\in \{0,1,2,\dots \}. If {z}_{{n}_{0}}=T{z}_{{n}_{0}} for some {n}_{0}, then we conclude the assertion of the theorem. So, assume that {z}_{n+1}\ne {z}_{n} for each n\in \{0,1,2,\dots \}. Since T is nondecreasing, {z}_{0}\u2aafT{z}_{0}={z}_{1} implies that {z}_{1}=T{z}_{0}\u2aafT{z}_{1}={z}_{2}. Recursively, we derive that
Letting x={z}_{n} and y={z}_{n+1} in (3) we get, for all n\in \mathbb{N},
By elementary calculation we find that {N}_{d}({z}_{n},{z}_{n+1})=0, and
If there exists some n such that {M}_{d}({z}_{n},{z}_{n+1})=d({z}_{n+1},{z}_{n+2}), then (4) turns into
which is a contradiction because d({z}_{n+1},{z}_{n+2})>0. Hence, (4) is equivalent to
for all n\in \mathbb{N}. Since φ is a strictly increasing function, we have, for all n\in \mathbb{N},
Thus, \{d({z}_{n},{z}_{n+1})\} is a decreasing sequence that is bounded from below. Thus, there exists \delta \ge 0 such that {lim}_{n\to \mathrm{\infty}}d({z}_{n},{z}_{n+1})=\delta. We assert that \delta =0. Suppose, on the contrary, that \delta >0. Letting n\to \mathrm{\infty} in (5) and regarding the properties of the auxiliary functions φ, ϕ, we derive that
Hence, \psi (\delta )=0 and, as a result, \delta =0. Thus,
In what follows, we prove that \{{z}_{n}\} is a Cauchy sequence. Suppose, on the contrary, that \{{z}_{n}\} is not Cauchy. By Lemma 5, there exist \epsilon >0 and two sequences \{n(k)\} and \{m(k)\} of positive integers such that, for all k\in \mathbb{N},
and
As \{{z}_{n}\} is nondecreasing, {z}_{n(k)}\preccurlyeq {z}_{m(k)}, for all k\in \mathbb{N}. Applying the contractivity condition (3), it follows, for all k\in \mathbb{N}, that
where
and
As {\{d({z}_{n(k)},{z}_{n(k)+1})\}}_{k\in \mathbb{N}}\to 0,
Furthermore, by (6), we have
Using the sequences
Lemma 6 guarantees that \epsilon =\alpha =0, which is a contradiction. As a consequence, we must admit that \{{z}_{n}\} is a Cauchy sequence in (X,d). As it is complete, there exists z\in X such that
If T is continuous, it is clear that z is a fixed point of T.
Suppose that condition (b) holds. Hence, as \{{z}_{n}\} is nondecreasing and converges to z, we have {z}_{n}\u2aafz, for all n=\{0,1,2,\dots \}. Due to (3), we have, for all n\in \mathbb{N},
where
and
Letting n\to \mathrm{\infty} in (7), we get
so \psi (d(z,Tz))=0 and d(z,Tz)=0. As a consequence, Tz=z, which completes the proof. □
The following corollary follows from Theorem 7 using L=0.
Corollary 8 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi}, \psi \in {\mathrm{\Psi}}^{\prime} such that
for all x,y\in X with x\u2aafy, where
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
In the next result, we take φ as the identity mapping on [0,\mathrm{\infty}), which clearly belongs to Φ.
Corollary 9 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exists \psi \in {\mathrm{\Psi}}^{\prime} such that
for all x,y\in X with x\u2aafy, where
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
The following result is not a direct consequence of Theorem 7, but its proof is verbatim the proof of Theorem 7. Thus, we skip it.
Theorem 10 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi}, \psi \in {\mathrm{\Psi}}^{\prime}, and L\ge 0 such that
for all x,y\in X with x\u2aafy, where
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
If L=0 in the previous theorem, we obtain the following result.
Corollary 11 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi}, \psi \in {\mathrm{\Psi}}^{\prime} such that
for all x,y\in X with x\u2aafy. Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
If we take φ as the identity mapping on [0,\mathrm{\infty}), we derive the following result.
Corollary 12 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exists \psi \in {\mathrm{\Psi}}^{\prime} such that
for all x,y\in X with x\u2aafy. Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
In the sequel we suggest another theorem by changing the properties of the auxiliary functions in the following way.
Let {\mathrm{\Psi}}^{\u2033} denote the set of all functions \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) such that

1.
\psi (t)=0\iff t=0 and {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}\psi ({t}_{n})>0 whenever {lim}_{n\to \mathrm{\infty}}{t}_{n}=t>0.

2.
\psi (t)>\phi (t)\phi (t) for any t>0, where \phi (t) is the left limit of φ at t, where \phi \in \mathrm{\Phi}.
Theorem 13 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:X\to X be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi} and \psi \in {\mathrm{\Psi}}^{\u2033}, L\ge 0 such that
for all y,v\in X with y\u2aafv, where
and
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {z}_{0}\in X such that {z}_{0}\asymp T({z}_{0}), then T has a fixed point.
The proof is analog to the proof of Theorem 7 and, hence, we skip it.
Remark 14 As in Theorem 13, by changing the property of the auxiliary function, we get various results (see, e.g., [6–9] and related references therein).
5 Fixed point theorems from (one dimensional) fixed point to multidimensional fixed point
As discussed in Remark 3, multidimensional fixed point theorems are equivalent to (one dimensional) fixed point theorems. Thus, Theorem 7 can be translated in a multidimensional case as follows.
Theorem 15 Let (X,\u2aaf) be a partially ordered set and suppose that there is a metric d on X such that (X,d) is a complete metric space. Let T:{X}^{k}\to {X}^{k} be a nondecreasing mapping for which there exist \phi \in \mathrm{\Phi}, \psi \in {\mathrm{\Psi}}^{\prime}, and L\ge 0 such that
for all V,Y\in X with V\u2aafY, where
and
Suppose either

(a)
T is continuous or

(b)
(X,d,\u2aaf) is regular.
If there exists {Z}_{0}\in {X}^{k} such that {Z}_{0}\asymp T({Z}_{0}), then T has a fixed point.
In similar way, we may state the analog of Corollary 8, Corollary 9, Theorem 10, Corollary 11, Corollary 12.
6 Applications
In this section, based on the results in [10], we propose an application to our results. Consider the integral equation
where T>0. We introduce the following space:
equipped with the metric
It is clear that (C[0,T],d) is a complete metric space. Furthermore, C[0,T] can be equipped with the partial order ⪯ as follows:
Due to [11], we know that (C[0,T],d,\u2aaf) is regular.
In what follows we state the main result of this section.
Theorem 16 We assume that the following hypotheses hold:

(i)
K:[0,T]\times [0,T]\times \mathbb{R}\to \mathbb{R} and g:\mathbb{R}\to \mathbb{R} are continuous,

(ii)
for all s,t\in [0,T] and u,v\in C[0,T] with v\u2aafu, we have
K(t,s,v(s))\le K(t,s,u(s)), 
(iii)
there exists a continuous function G:[0,T]\times [0,T]\to [0,\mathrm{\infty}) such that
K(t,s,x)K(t,s,y)\le G(t,s)\sqrt{\frac{{(xy)}^{4}}{1+{(xy)}^{2}}},
for all s,t\in [0,T] and x,y\in \mathbb{R} with x\ge y,

(iv)
{sup}_{t\in [0,T]}{\int}_{0}^{T}G{(t,s)}^{2}\phantom{\rule{0.2em}{0ex}}ds\le \frac{1}{T}.
Then the integral (8) has a solution {u}^{\ast}\in C[0,T].
Proof We, first, define T:C[0,T]\to C[0,T] by
We first prove that T is nondecreasing. Assume that v\u2aafu. From (ii), for all s,t\in [0,T], we have K(t,s,v(s))\le K(t,s,u(s)). Thus, we get
Now, for all u,v\in C[0,T] with v\u2aafu, due to (iii), we derive that
On account of the CauchySchwarz inequality in the last integral above, we find that
Taking (iv) into account, we estimate the first integral in (9) as follows:
For the second integral in (9) we proceed in the following way:
By combining the above estimation, we conclude that
It yields
or equivalently,
Now, if choose \phi (t)={t}^{2} and \psi (t)={t}^{2}\frac{{t}^{2}}{1+{t}^{2}}, then we get
for all u,v\in C[0,T] with v\u2aafu. Hence, all hypotheses of Corollary 11 are satisfied. Thus, T has a fixed point {u}^{\ast}\in C[0,T] which is a solution of (8). □
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Acknowledgements
AntonioFrancisco RoldánLópezdeHierro has been partially supported by Junta de Andalucía by project FQM268 of the Andalusian CICYE.
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Agarwal, R.P., Karapınar, E. & RoldánLópezdeHierro, AF. Some remarks on ‘Multidimensional fixed point theorems for isotone mappings in partially ordered metric spaces’. Fixed Point Theory Appl 2014, 245 (2014). https://doi.org/10.1186/168718122014245
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DOI: https://doi.org/10.1186/168718122014245