# On Best Proximity Points under theP-Property on Partially Ordered Metric Spaces

Authors: Mohamed Jleli; Erdal Karapinar; Bessem Samet
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Very recently, Abkar and Gabeleh (2013) observed that some best proximity point results under theP-property can be obtained from the same results in fixed-point theory. In this paper, motivated by this mentioned work, we show that the most best proximity point results on a metric space endowed with a partial order (under the P-property) can be deduced from existing fixed-point theorems in the literature.  We present various model examples to illustrate this point of view.


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## Body

## 1. Introduction

LetA and B be two nonempty subsets of a metric space (X,d). Through this paper, we will use the following notations:
(1)d(A,B):=inf{d(a,b):a∈A,b∈B},A0:={a∈A:d(a,b)=d(A,B)forsomeb∈B},B0:={b∈B:d(a,b)=d(A,B)forsomea∈A}.Definition 1.
An elementx*∈A is said to be a best proximity point of the nonself-mapping T:A→B if and only if it satisfies the condition that
(2)d(x*,Tx*)=d(A,B).The notion ofP-property was introduced in [1] as follows.Definition 2.
Let(A,B) be a pair of nonempty subsets of a metric space (X,d) with A0≠∅. Then the pair (A,B) is said to have the P-property if and only if
(3)d(x1,y1)=d(A,B)d(x2,y2)=d(A,B)}⟹d(x1,x2)=d(y1,y2),
where x1,x2∈A and y1,y2∈B.Various best proximity point results for different classes of nonself-mappings under theP-property were established recently (see, e.g., [1–6] and references therein). Very recently, Abkar and Gabeleh [7] observed that some best proximity point results under the P-property can be obtained from the same results in fixed-point theory. The main purpose of this paper is to show that in many cases best proximity point results (under the P-property) on partially ordered metric spaces can also be deduced from the corresponding fixed-point theorems. We present various model examples to illustrate this point of view.


## 2. Preliminaries

As model examples, we will consider some known fixed-point theorems in the framework of partially ordered metric spaces.At first, we need to introduce two classes of real functions. Denote byΨ the class of all functions ψ:[0,+∞[→[0,+∞[ satisfying the following conditions: (Ψ1)ψ is continuous and nondecreasing,(Ψ2)ψ-1({0})={0}.We denote byΦ the set of functions φ:[0,+∞[→[0,+∞[ satisfying the following conditions: (Φ1)φ is nondecreasing,(Φ2)limn→+∞φn(t)=0 for each t>0.Through this paper,ℕ denotes the set of all natural numbers, and ℕ*=ℕ∖{0}.The following concepts will be useful later.Definition 3.
Let(X,⪯) be a partially ordered set and T:X→X be a giving mapping. We say that T is nondecreasing (with respect to ⪯) if and only if
(4)x,y∈X,x⪯y⟹Tx⪯Ty.Definition 4.
Let(X,⪯) be a partially ordered set and {xn} be a sequence in X. We say that {xn} is nondecreasing (with respect to ⪯) if and only if xn⪯xn+1 for all n∈ℕ.Definition 5.
Let(X,⪯) be a partially ordered set and d be a metric on X. We say that (X,⪯,d) is regular if and only if the following condition holds:
(5){xn}⊂Xisnondecreasinglimn→+∞d(xn,x)=0,x∈X}⟹xn⪯x,∀n∈ℕ.Definition 6.
Let(X,⪯) be a partially ordered set. We say that (X,⪯) is directed if and only if
(6)∀(x,y)∈X×X,∃z∈X∣(x⪯z)∧(y⪯z).In [8], Harjani and Sadarangani proved the following fixed point result.Theorem 7.
Let(X,⪯) be a partially ordered set and suppose that there exists a metric d on X such that (X,d) is a complete metric space. Let T:X→X be a giving mapping. Suppose that the following conditions hold: (i)
T is continuous or (X,⪯,d) is regular,(ii)
T is nondecreasing,(iii)
there existsx0∈X such that x0⪯Tx0,(iv)
for allx,y∈X such that x⪯y, we have
(7)ψ(d(Tx,Ty))≤ψ(d(x,y))-φ(d(x,y)),
where ψ,ϕ∈Ψ.
ThenT has a fixed point. Moreover, the sequence {Tnx0} converges to this fixed point. If, in addition, (X,⪯) is directed, we have uniqueness of the fixed point.In [9], Agarwal et al. established the following fixed point result.Theorem 8.
Let(X,⪯) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space. Let T:X→X be a giving mapping. Suppose that the following conditions hold: (i)
T is continuous or (X,⪯,d) is regular,(ii)
T is nondecreasing,(iii)
there existsx0∈X such that x0⪯Tx0,(iv)
for allx,y∈X such that x⪯y, we have
(8)d(Tx,Ty)≤φ(max{d(x,Ty)+d(y,Tx)2d(x,y),d(x,Tx),d(y,Ty),≤φd(x,Ty)+d(y,Tx)2}),
where φ∈Φ.
ThenT has a fixed point. Moreover, the sequence {Tnx0} converges to this fixed point.The above result was subsequently extended by Turinici in [10].Through this paper,(A,B) is a pair of nonempty subsets of a metric space (X,d), where X is endowed with a partial order ⪯. We denote
(9)d(A,B):=inf{d(a,b):a∈A,b∈B},A0:={a∈A:d(a,b)=d(A,B)forsomeb∈B},B0:={b∈B:d(a,b)=d(A,B)forsomea∈A}.Definition 9.
An elementx*∈A is said to be a best proximity point of the nonself-mapping T:A→B if and only if
(10)d(x*,Tx*)=d(A,B).The notion ofP-property was introduced in [1] as follows.Definition 10.
Let(A,B) be a pair of nonempty subsets of a metric space (X,d) with A0≠∅. Then the pair (A,B) is said to have the P-property if and only if
(11)d(x1,y1)=d(A,B)d(x2,y2)=d(A,B)}⟹d(x1,x2)=d(y1,y2),
where x1,x2∈A and y1,y2∈B.In [11], the authors introduced the concept of proximally increasing mappings.Definition 11.
A nonself-mappingT:M⊆A→B is said to be proximally increasing if and only if
(12)y1⪯y2d(x1,Ty1)=d(A,B)d(x2,Ty2)=d(A,B)}⟹x1⪯x2,
where x1,x2∈A,y1,y2∈M.Definition 12.
Let(M,N) be a pair of nonempty subsets of X. A mapping I:M→N is said to be an isometry if and only if
(13)d(Ix,Iy)=d(x,y),∀x,y∈M.The following lemmas will be useful later.Lemma 13.
Let(M,N) be a pair of nonempty subsets of X. Let I:M→N be a bijective mapping. Then I is an isometry if and only if I-1 is an isometry.Proof.
Suppose thatI is an isometry. Let z,t∈N,x=I-1z and y=I-1t. Since I is an isometry, we have
(14)d(Ix,Iy)=d(x,y),
that is,
(15)d(z,t)=d(I-1z,I-1t).
Then I-1 is an isometry.Lemma 14.
Consider two nonself-mappingsT~:A0→B0 and I:A0→B0. Suppose that the following conditions hold: (i)
T~ is proximally increasing,(ii)
I:A0→B0 is bijective,(iii)
d(x,Ix)=d(A,B) for all x∈A0.
Then the mappingI-1T~:A0→A0 is nondecreasing (with respect to ⪯).Proof.
Letx,y∈A0 such that x⪯y. We have
(16)x⪯y,d(I-1T~x,T~x)=d(I-1T~x,I(I-1T~x))=d(A,B)(from(iii)),d(I-1T~y,T~y)=d(I-1T~y,I(I-1T~y))=d(A,B).
Since T~ is proximally increasing, we get that I-1T~x⪯I-1T~y.Lemma 15.
Consider two nonself-mappingsT~:A0→B0 and I:A0→B0. Suppose that the following conditions hold: (i)
I:A0→B0 is bijective,(ii)
d(x,Ix)=d(A,B) for all x∈A0,(iii)
there existx0,x1∈A0 such that d(x1,T~x0)=d(A,B) and x0⪯x1,(iv)
the pair(A,B) satisfies the P-property.
Thenx0⪯I-1T~x0.Proof.
Using (ii) withx=x1, we have d(x1,Ix1)=d(A,B). From condition (iii), we have d(x1,T~x0)=d(A,B). It follows from the P property (condition (iv)) that d(Ix1,T~x0)=0, which implies that Ix1=T~x0. From (i) and (iii), we obtain that x0⪯x1=I-1T~x0.Lemma 16.
Let(A,B) be a pair of nonempty closed subsets of a complete metric space (X,d) such that A0 is nonempty and (A,B) has the P property. Then (A0,B0) is a closed pair of subsets of X.The proof of Lemma16 can be found in [5].Lemma 17.
Let(A,B) be a pair of nonempty closed subsets of a metric space (X,d) such that A0 is nonempty. Assume that the pair (A,B) has the P property. Then there exists a bijective isometry I:A0→B0 such that
(17)d(x,Ix)=d(A,B),∀x∈A0.The proof of Lemma17 can be found in [7].


## 3. Discussions and Results

In [11], the authors established the following best proximity point result.Theorem 18.
LetX be a nonempty set such that (X,⪯) is a partially ordered set and (X,d) is a complete metric space. Let (A,B) be a pair of nonempty closed subsets of the metric space (X,d) such that A0≠∅. Let T:A→B be a nonself-mapping. Suppose that the following conditions hold: (i)
T(A0)⊆B0 and (A,B) satisfy the P property,(ii)
T is a continuous on A,(iii)
T is proximally increasing,(iv)
there existx0,x1∈A0 such that
(18)d(x1,Tx0)=d(A,B),x0⪯x1,(v)
for allx,y∈A such that x⪯y, we have
(19)ψ(d(Tx,Ty))≤ψ(d(x,y))-φ(d(x,y)),
where ψ,φ∈Ψ.
Then there exists an elementx*∈A such that
(20)d(x*,Tx*)=d(A,B).
Further, the sequence {xn}, defined by
(21)d(xn+1,Txn)=d(A,B),∀n∈ℕ,
converges to the element x*.We shall prove that the best proximity point result given by Theorem18 is a consequence of the fixed point result given by Theorem 7.


### 3.1. Proof of Theorem18

Denote byT~ the restriction of the mapping T to the subset A0 of A. Since T(A0)⊆B0, we have T~:A0→B0. From Lemma 17, there exists a bijective isometry I:A0→B0 such that
(22)d(x,Ix)=d(A,B),∀x∈A0.
Consider the self-mapping S:=I-1T~:A0→A0. We shall prove that the mapping S satisfies all the conditions of Theorem 7.Claim 1.
(
A
0
,
d
) is complete.
From Lemma16, A0 is a closed subset of the complete metric space (X,d). Then (A0,d) is complete.Claim 2.
S is continuous and nondecreasing mapping.
From Lemma13, I-1 is an isometry, so I-1 is continuous on B0. Since T is continuous on A,T~ is continuous on A0. Thus, the self-mapping S=I-1T~ is continuous on A0. On the other hand, since T is proximally increasing, then T~ is also a proximally increasing mapping. Now, our claim follows immediately from Lemma 14.Claim 3.
There existsa∈A0 such that a⪯Sa.
We have(23)d(x1,Tx0)=d(A,B),d(x1,Ix1)=d(A,B).
Using the P property, we obtain that d(Tx0,Ix1)=0, which implies that T~x0=Tx0=Ix1, that is, x1=I-1T~x0=Sx0. Since x0⪯x1, our claim holds with a=x0.Claim 4.
The mappingS satisfies condition (iv) of Theorem 7.
Letx,y∈A0 such that x⪯y. Since I-1 is an isometry, it follows from condition (v) that
(24)ψ(d(Sx,Sy))=ψ(d(I-1T~x,I-1T~y))=ψ(d(Tx,Ty))≤ψ(d(x,y))-φ(d(x,y)).
This proves our claim.Now, the mappingS satisfies all the conditions of Theorem 7. We deduce that S has a fixed point x*∈A0, and the sequence {zn}⊂A0, defined by
(25)z0=x0,zn+1=Szn,∀n∈ℕ,
converges to x*.We claim thatx* is a best proximity point of T. Indeed, we have
(26)0=d(x*,Sx*)=d(x*,I-1T~x*),
which implies that x*=I-1T~x*, that is, Ix*=T~x*=Tx*. From d(x*,Ix*)=d(A,B), we obtain that d(x*,Tx*)=d(A,B). Then x* is a best proximity point of T.Let{xn}⊂A0 be the sequence defined by
(27)d(xn+1,Txn)=d(A,B),∀n∈ℕ.
Since
(28)d(xn+1,Ixn+1)=d(A,B),∀n∈ℕ,
from the P property, we get that
(29)Ixn+1=Txn,∀n∈ℕ,
which implies that
(30)xn+1=Sxn,∀n∈ℕ.
It follows from (25) and (30) that {xn}={zn}→x* as n→+∞. This makes end to the proof of Theorem 18.Remark 19.
In Theorem18, the authors considered only the continuous case. However, from Theorem 7, we can also remove the continuity assumption of T and replace it by the regularity assumption of (A0,⪯,d). Moreover, by assuming that (A0,⪯) is directed, we obtain uniqueness of the best proximity point.


### 3.2. Additional Result

Using the same techniques, we can obtain a best proximity version of Theorem8.Theorem 20.
LetX be a nonempty set such that (X,⪯) is a partially ordered set and (X,d) is a complete metric space. Let (A,B) be a pair of nonempty closed subsets of the metric space (X,d) such that A0≠∅. Let T:A→B be a nonself-mapping. Suppose that the following conditions hold: (i)
T(A0)⊆B0 and (A,B) satisfy the P property,(ii)
T is continuous on A0 or (A0,⪯,d) is regular,(iii)
T is proximally increasing,(iv)
there existx0,x1∈A0 such that
(31)d(x1,Tx0)=d(A,B),x0⪯x1,(v)
for allx,y∈A such that x⪯y, we have
(32)d(Tx,Ty)≤φ(MT(A,B,x,y)),
where φ∈Φ and
(33)MT(A,B,x,y):=max{d(x,Ty)+d(y,Tx)2d(x,y),d(x,Tx)-d(A,B),d(y,Ty):=max-d(A,B),d(x,Ty)+d(y,Tx)2-d(A,B)}.
Then there exists an elementx*∈A such that
(34)d(x*,Tx*)=d(A,B).
Further, the sequence {xn}, defined by
(35)d(xn+1,Txn)=d(A,B),∀n∈ℕ,
converges to the element x*.Proof.
We continue to use the same notations as in the proof of Theorem18. Following the proof of Theorem 18, we have only to check that the self-mapping S=I-1T~:A0→A0 satisfies condition (iv) of Theorem 8. Indeed, let x,y∈A0 such that x⪯y, we have
(36)d(Sx,Sy)=d(I-1T~x,I-1T~y)=d(Tx,Ty)≤φ(MT(A,B,x,y)).
On the other hand, one has the evaluation
(37)MT(A,B,x,y)=max{d(x,Ty)+d(y,Tx)2d(x,y),d(x,Tx)-d(A,B),d(y,Ty)=max-d(A,B),d(x,Ty)+d(y,Tx)2-d(A,B)}=max{d(x,y),α,β,γ},
where we denoted
(38)α:=d(x,T~x)-d(A,B),β:=d(y,T~y)-d(A,B),γ:=d(x,T~y)+d(y,T~x)2-d(A,B).
Estimation of
α.
We have(39)α=d(x,T~x)-d(A,B)≤d(x,Sx)+d(Sx,T~x)-d(A,B)=d(x,Sx)+d(I-1T~x,T~x)-d(A,B)=d(x,Sx)+d(I-1T~x,I(I-1T~x))-d(A,B)=d(x,Sx)+d(A,B)-d(A,B)(from(22))=d(x,Sx).
Estimation of
β.
We have(40)β=d(y,T~y)-d(A,B)≤d(y,Sy)+d(Sy,T~y)-d(A,B)=d(y,Sy)+d(I-1T~y,T~y)-d(A,B)=d(y,Sy)+d(I-1T~y,I(I-1T~y))-d(A,B)=d(y,Sy).
Estimation of
γ.
We have(41)γ=d(x,T~y)+d(y,T~x)2-d(A,B)≤d(x,Sy)+d(Sy,T~y)+d(y,Sx)+d(Sx,T~x)2-d(A,B)=d(x,Sy)+d(I-1T~y,T~y)+d(y,Sx)+d(I-1T~x,T~x)2-d(A,B)=d(x,Sy)+d(A,B)+d(y,Sx)+d(A,B)2-d(A,B)=d(x,Sy)+d(y,Sx)2.
Now, we deduce that(42)MT(A,B,x,y)≤max{d(x,Sy)+d(y,Sx)2d(x,y),d(x,Sx),d(y,Sy),≤maxd(x,Sy)+d(y,Sx)2}.
Since φ is nondecreasing, we get that
(43)φ(MT(A,B,x,y))≤φ(max{d(x,Sy)+d(y,Sx)2d(x,y),d(x,Sx),d(y,Sy),≤φd(x,Sy)+d(y,Sx)2}).
Finally, it follows from (36) and (43) that
(44)d(Sx,Sy)≤φ(max{d(x,Sy)+d(y,Sx)2d(x,y),d(x,Sx),d(y,Sy),≤φd(x,Sy)+d(y,Sx)2}).
Thus, we proved that the mapping S satisfies condition (iv) of Theorem 8. Applying Theorem 8, we obtain the existence of a fixed point of S. The rest of the proof is exactly like that of Theorem 18, so we omit it.


## 3.1. Proof of Theorem18

Denote byT~ the restriction of the mapping T to the subset A0 of A. Since T(A0)⊆B0, we have T~:A0→B0. From Lemma 17, there exists a bijective isometry I:A0→B0 such that
(22)d(x,Ix)=d(A,B),∀x∈A0.
Consider the self-mapping S:=I-1T~:A0→A0. We shall prove that the mapping S satisfies all the conditions of Theorem 7.Claim 1.
(
A
0
,
d
) is complete.
From Lemma16, A0 is a closed subset of the complete metric space (X,d). Then (A0,d) is complete.Claim 2.
S is continuous and nondecreasing mapping.
From Lemma13, I-1 is an isometry, so I-1 is continuous on B0. Since T is continuous on A,T~ is continuous on A0. Thus, the self-mapping S=I-1T~ is continuous on A0. On the other hand, since T is proximally increasing, then T~ is also a proximally increasing mapping. Now, our claim follows immediately from Lemma 14.Claim 3.
There existsa∈A0 such that a⪯Sa.
We have(23)d(x1,Tx0)=d(A,B),d(x1,Ix1)=d(A,B).
Using the P property, we obtain that d(Tx0,Ix1)=0, which implies that T~x0=Tx0=Ix1, that is, x1=I-1T~x0=Sx0. Since x0⪯x1, our claim holds with a=x0.Claim 4.
The mappingS satisfies condition (iv) of Theorem 7.
Letx,y∈A0 such that x⪯y. Since I-1 is an isometry, it follows from condition (v) that
(24)ψ(d(Sx,Sy))=ψ(d(I-1T~x,I-1T~y))=ψ(d(Tx,Ty))≤ψ(d(x,y))-φ(d(x,y)).
This proves our claim.Now, the mappingS satisfies all the conditions of Theorem 7. We deduce that S has a fixed point x*∈A0, and the sequence {zn}⊂A0, defined by
(25)z0=x0,zn+1=Szn,∀n∈ℕ,
converges to x*.We claim thatx* is a best proximity point of T. Indeed, we have
(26)0=d(x*,Sx*)=d(x*,I-1T~x*),
which implies that x*=I-1T~x*, that is, Ix*=T~x*=Tx*. From d(x*,Ix*)=d(A,B), we obtain that d(x*,Tx*)=d(A,B). Then x* is a best proximity point of T.Let{xn}⊂A0 be the sequence defined by
(27)d(xn+1,Txn)=d(A,B),∀n∈ℕ.
Since
(28)d(xn+1,Ixn+1)=d(A,B),∀n∈ℕ,
from the P property, we get that
(29)Ixn+1=Txn,∀n∈ℕ,
which implies that
(30)xn+1=Sxn,∀n∈ℕ.
It follows from (25) and (30) that {xn}={zn}→x* as n→+∞. This makes end to the proof of Theorem 18.Remark 19.
In Theorem18, the authors considered only the continuous case. However, from Theorem 7, we can also remove the continuity assumption of T and replace it by the regularity assumption of (A0,⪯,d). Moreover, by assuming that (A0,⪯) is directed, we obtain uniqueness of the best proximity point.


## 3.2. Additional Result

Using the same techniques, we can obtain a best proximity version of Theorem8.Theorem 20.
LetX be a nonempty set such that (X,⪯) is a partially ordered set and (X,d) is a complete metric space. Let (A,B) be a pair of nonempty closed subsets of the metric space (X,d) such that A0≠∅. Let T:A→B be a nonself-mapping. Suppose that the following conditions hold: (i)
T(A0)⊆B0 and (A,B) satisfy the P property,(ii)
T is continuous on A0 or (A0,⪯,d) is regular,(iii)
T is proximally increasing,(iv)
there existx0,x1∈A0 such that
(31)d(x1,Tx0)=d(A,B),x0⪯x1,(v)
for allx,y∈A such that x⪯y, we have
(32)d(Tx,Ty)≤φ(MT(A,B,x,y)),
where φ∈Φ and
(33)MT(A,B,x,y):=max{d(x,Ty)+d(y,Tx)2d(x,y),d(x,Tx)-d(A,B),d(y,Ty):=max-d(A,B),d(x,Ty)+d(y,Tx)2-d(A,B)}.
Then there exists an elementx*∈A such that
(34)d(x*,Tx*)=d(A,B).
Further, the sequence {xn}, defined by
(35)d(xn+1,Txn)=d(A,B),∀n∈ℕ,
converges to the element x*.Proof.
We continue to use the same notations as in the proof of Theorem18. Following the proof of Theorem 18, we have only to check that the self-mapping S=I-1T~:A0→A0 satisfies condition (iv) of Theorem 8. Indeed, let x,y∈A0 such that x⪯y, we have
(36)d(Sx,Sy)=d(I-1T~x,I-1T~y)=d(Tx,Ty)≤φ(MT(A,B,x,y)).
On the other hand, one has the evaluation
(37)MT(A,B,x,y)=max{d(x,Ty)+d(y,Tx)2d(x,y),d(x,Tx)-d(A,B),d(y,Ty)=max-d(A,B),d(x,Ty)+d(y,Tx)2-d(A,B)}=max{d(x,y),α,β,γ},
where we denoted
(38)α:=d(x,T~x)-d(A,B),β:=d(y,T~y)-d(A,B),γ:=d(x,T~y)+d(y,T~x)2-d(A,B).
Estimation of
α.
We have(39)α=d(x,T~x)-d(A,B)≤d(x,Sx)+d(Sx,T~x)-d(A,B)=d(x,Sx)+d(I-1T~x,T~x)-d(A,B)=d(x,Sx)+d(I-1T~x,I(I-1T~x))-d(A,B)=d(x,Sx)+d(A,B)-d(A,B)(from(22))=d(x,Sx).
Estimation of
β.
We have(40)β=d(y,T~y)-d(A,B)≤d(y,Sy)+d(Sy,T~y)-d(A,B)=d(y,Sy)+d(I-1T~y,T~y)-d(A,B)=d(y,Sy)+d(I-1T~y,I(I-1T~y))-d(A,B)=d(y,Sy).
Estimation of
γ.
We have(41)γ=d(x,T~y)+d(y,T~x)2-d(A,B)≤d(x,Sy)+d(Sy,T~y)+d(y,Sx)+d(Sx,T~x)2-d(A,B)=d(x,Sy)+d(I-1T~y,T~y)+d(y,Sx)+d(I-1T~x,T~x)2-d(A,B)=d(x,Sy)+d(A,B)+d(y,Sx)+d(A,B)2-d(A,B)=d(x,Sy)+d(y,Sx)2.
Now, we deduce that(42)MT(A,B,x,y)≤max{d(x,Sy)+d(y,Sx)2d(x,y),d(x,Sx),d(y,Sy),≤maxd(x,Sy)+d(y,Sx)2}.
Since φ is nondecreasing, we get that
(43)φ(MT(A,B,x,y))≤φ(max{d(x,Sy)+d(y,Sx)2d(x,y),d(x,Sx),d(y,Sy),≤φd(x,Sy)+d(y,Sx)2}).
Finally, it follows from (36) and (43) that
(44)d(Sx,Sy)≤φ(max{d(x,Sy)+d(y,Sx)2d(x,y),d(x,Sx),d(y,Sy),≤φd(x,Sy)+d(y,Sx)2}).
Thus, we proved that the mapping S satisfies condition (iv) of Theorem 8. Applying Theorem 8, we obtain the existence of a fixed point of S. The rest of the proof is exactly like that of Theorem 18, so we omit it.



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*Source: 150970-2013-06-25.xml*