Coupled fixed point theorems for rational type contractions
Anca Oprea and Gabriela Petru¸sel
Dedicated to Professor Ioan A. Rus on the occasion of his 80th anniversary
Abstract. In this paper, we will consider the coupled fixed problem inb-metric space for single-valued operators satisfying a generalized contraction condition of rational type. First part of the paper concerns with some fixed point theorems, while the second part presents a study of the solution set of the coupled fixed point problem. More precisely, we will present some existence and uniqueness theorems for the coupled fixed point problem, as well as a qualitative study of it (data dependence of the coupled fixed point set, well-posedness, Ulam-Hyers stability and the limit shadowing property of the coupled fixed point problem) under some rational type contraction assumptions on the mapping.
Mathematics Subject Classification (2010):47H05, 47H09, 47H10.
Keywords:Fixed point, ordered metric space, rational type contraction, coupled fixed point, data dependence, well-posedness, Ulam-Hyers stability, limit shad- owing property.
1. Introduction and preliminaries
The notion of b-metric spaces and discussion on the topological structure of it appeared in several papers, such as L.M. Blumenthal [2], S. Czerwik [6], N. Bourbaki [5], Heinonen [10].
On the other hand, the concept of coupled fixed point problem, was considered, for the first time, by Opoitsev in [14]-[15], but a very fruitful approach in this field was proposed by D. Guo, V. Lashmikantham [9] and T. Gnana Bhaskar and V. Lash- mikantham [7]. Later on, many results related to this kind of problem appeared (see, for example [8], [13],. . . ).
Moreover, starting with the paper of Dass and Gupta [9], several extensions of the contraction principle considered the case of self mappings satisfying some rational type contraction assumptions, see, for example, [7].
Our aim is to consider both of the above research directions. More precisely, we will prove, using some adequate fixed point theorems for monotone rational con- tractions in orderedb-metric spaces, some coupled fixed point theorems for operators T :X×X →X satisfying some rational type assumptions on comparable elements.
We shall recall some well known notions and definition of the b-metric spaces.
Definition 1.1. LetX be a set and let s ≥1 be a given real number. A functional d:X×X →R+ is said to be a b-metric if the following axioms are satisfied:
1. ifx, y∈X, thend(x, y) = 0 if and only ifx=y;
2.d(x, y) =d(y, x) for allx, y∈X;
3.d(x, z)≤s[d(x, y) +d(y, z)], for allx, y, z∈X.
A pair (X, d) with the above properties is called a b-metric space.
Let (X,≤) be a partially ordered set andda metric onX. Notice that we can endow the product spaceX×X with the partial order≤p given by
(x, y)≤p (u, v)⇔x≤u, y≥v.
Definition 1.3. Let (X,≤) be a partially ordered set and let T : X ×X → X. We say thatT has the mixed monotone property ifT(·, y) is monotone increasing for any y∈X andT(x,·) is monotone decreasing for anyx∈X.
Lemma 1.4.Let (X, d) be a b-metric space. Then the sequence (xn)n∈N⊂X is called:
i) convergent if and only if there existsx∈Xsuch thatd(xn, x)→0 asn→ ∞.
In this case we write lim
n→∞xn=x;
ii) Cauchy if and only ifd(xn, xm)→0 asn, m→ ∞.
If (X, d) is a metric space andT :X×X →Xis an operator, then by definition, a coupled fixed point forT is a pair (x∗, y∗)∈X×X satisfying
x∗=T(x∗, y∗)
y∗=T(y∗, x∗) (P1)
We will denote byCF ix(T) the coupled fixed point set forT.
The aim of this paper is to present, in the framework of complete orderedb-metric spaces, some existence and uniqueness theorems for the coupled fixed point problem, as well as, a qualitative study of this problem (data dependence of the coupled fixed point set, well-posedness, Ulam-Hyers stability and the limit shadowing property of the coupled fixed point problem) under some rational type contraction assumptions on the mapping. Our results extend and complement some theorems given in the recent literature, see e.g. [21], [22].
2. Fixed point theorems
In this part of the paper, we will present a fixed point theorems in ordered b-metric spaces for a single-valued operstor satisfying a rational type contraction condition.
Theorem 2.1.Let (X,≤) be a partially ordered set andd:X×X→R+be a complete b-metric with constants≥1. Letf :X→X be an operator which has closed graph
with respect todand it is increasing with respect to ”≤”. Suppose that there exists α, β≥0 withα+βs <1 satisfying
d(f(x), f(y))≤α·d(y, f(y))[1 +d(x, f(x))]
1 +d(x, y) +β·d(x, y), (2.1) forx, y∈X withx≤y .
If there existsx0∈X such thatx0≤f(x0), then there exists x∗∈X such that x∗=f(x∗) and (fn(x0))n∈N→x∗, as n→ ∞.
Proof.We have two cases:
Case 1.Iff(x0) =x0, then F ix(f)6=∅.
Case 2.Suppose thatx0< f(x0).
Using thatf is an increasing operator and by mathematical induction, we have x0< f(x0)≤f2(x0)≤. . .≤fn(x0)≤fn+1(x0)≤. . .
By this method we get a sequence (xn)n⊂N∈X defined by
xn+1=f(xn) =f(f(xn−1)) =f2(xn−1) =. . .=fn(x1) =fn+1(x0).
If there existsn≥1 such thatxn+1 =xn, then f(xn) =xn. So we get thatxn
is a fixed point off, which impliesF ix(f)6=∅.
Suppose thatxn+16=xn forn≥0.
Sincexn≤xn+1for any n∈N, we have
d(xn, xn+1) =d(f(xn−1), f(xn))
≤α·d(xn, f(xn))[1 +d(xn−1, f(xn−1))]
1 +d(xn−1, xn) +β·d(xn−1, xn)
= α·d(xn, xn+1)[1 +d(xn−1, xn)]
1 +d(xn−1, xn) +β·d(xn−1, xn)
=α·d(xn, xn+1) +β·d(xn−1, xn).
So we obtain
d(xn, xn+1)≤ β
1−α·d(xn−1, xn) for anyn∈N. Using mathematical induction we get that
d(xn, xn+1)≤ β
1−α·d(xn−1, xn)≤. . .≤ β
1−α n
·d(x0, x1) or
d(fn(x0), fn+1(x0))≤ β
1−α n
·d(x0, f(x0)) f or any n∈N.
Letn∈Nandp∈N∗. We will prove that (xn)n∈Ndefined by xn=fn(x0) is a Cauchy sequence inX.
d(fn(x0), fn+p(x0))≤s·d(fn(x0), fn+1(x0)) +s2·d(fn+1(x0), fn+2(x0)) +. . . +sp−1·d(fn+p−2(x0), fn+p−1(x0)) +sp−1·d(fn+p−1(x0), fn+p(x0)).
We denote
A= β
1−α.
So we obtain
d(fn(x0), fn+p(x0))≤s·An·d(x0, f(x0)) +s2·An+1·d(x0, f(x0)) +. . . +sp−1·An+p−2·d(x0, f(x0)) +sp·An+p−1·d(x0, f(x0))
=s·An[1 +s·A+. . .+ (s·A)p−1]·d(x0, f(x0)) =s·An·1−(s·A)p
1−s·A ·d(x0, f(x0)).
ButA= β 1−α < 1
s, then we get that d(fn(x0), fn+p(x0))≤s·An· 1−(s·A)p
1−s·A ·d(x0, f(x0))→0 as n→ ∞.
Hence (fn(x0))n∈N is a Cauchy sequence on X. We also know that (X, d) is a complete b-metric space. So there exists x∗ ∈ X such that (fn(x0))n∈N → x∗ as n→ ∞. Becausef has closed graph, then x∗∈F ix(f), which implies F ix(f)6=∅.
Or f is continuous, we have f(x∗) =f( lim
n→∞xn) = lim
n→∞f(xn) = lim
n→∞xn+1=x∗.
A uniqueness result concerning the fixed point equation is the following.
Theorem 2.2.Suppose that all the hypotheses of Theorem 2.1. take place. Addition- ally, suppose that the following condition holds: for all x, y∈X there exists z ∈X such thatz≤xandz≤y.
ThenF ix(f) ={x∗}.
Proof.Suppose thatx∗, y∗∈X are two fixed points off. We have two cases:
Case 1.x∗ andy∗ are comparable. Supposex∗≤y∗(ory∗≤x∗is the same) d(x∗, y∗) =d(f(x∗), f(y∗))≤ α·d(y∗, f(y∗))[1 +d(x∗, f(x∗))]
1 +d(x∗, y∗) +β·d(x∗, y∗)
=β·d(x∗, y∗).
Since β < 1, this is only possible when d(x∗, y∗) = 0. This implies x∗ = y∗, so F ix(f) ={x∗}.
Case 2.x∗ andy∗ are not comparable.
By our additional assumption, we have that there existsz∈X withz≤x∗ and z≤y∗.
Sincez≤x∗, thenfn(z)≤fn(x∗) =x∗ for anyn∈N. We obtain
d(fn(z), x∗) =d(fn(z), fn(x∗))≤α·d(fn−1(x∗), fn(x∗))[1 +d(fn−1(z), fn(z)]
1 +d(fn−1(z), fn−1(x∗)) +β·d(fn−1(z), fn−1(x∗)) =β·d(fn−1(z), fn−1(x∗)) =β·d(fn−1(z), x∗) So we have
d(fn(z), x∗)≤β·d(fn−1(z), x∗)≤β2·d(fn−2(z), x∗)≤. . .≤βn·d(z, x∗) and sinceβ <1,βn→0 then we get that
n→∞lim d(fn(z), x∗) = 0
This implies lim
n→∞fn(z) =x∗. Using a similar argument, we get that lim
n→∞fn(z) =y∗. Thenx∗=y∗.
A global version of the previous result is the following:
Theorem 2.3. Let (X, d) be a complete b- metric space with constant s ≥ 1, f : X →X be an operator ofX with the following condition: there existsα, β≥0 with max{α,1−αβ }<1s such that
d(f(x), f(y))≤α·d(y, f(y))[1 +d(x, f(x))]
1 +d(x, y) +β·d(x, y), (2.2) forx, y∈X. Thenf has a unique fixed point.
Proof.Letx0∈X be arbitrary chosen. Using the same method as in previous proof, we can construct a sequence (xn)n∈Ngiven byxn+1=f(xn) for alln∈N, which is a Cauchy sequence.
Since (X, d) is a complete b-metric space, we get that there existsx∗∈X such that lim
n→∞xn=x∗. Then, we have
d(x∗, f(x∗))≤s·d(x∗, f(xn)) +s·d(f(xn), f(x∗))
≤s·d(x∗, f(xn)) +s· α·d(x∗, f(x∗))[1 +d(xn, f(xn))]
1 +d(xn, x∗) +s·β·d(xn, x∗)
=s·d(x∗, xn+1) +s·α·d(x∗, f(x∗))[1 +d(xn, xn+1)]
1 +d(xn, x∗) +s·β·d(xn, x∗).
Thus, we obtain
d(x∗, f(x∗))
1 +d(xn, x∗)−s·α−s·α·d(xn, xn+1) 1 +d(xn, x∗)
≤s·d(x∗, xn+1) +s·β·d(xn, x∗).
Letting n → ∞ we have d(x∗, f(x∗))(1−s·α) ≤ 0. Thus d(x∗, f(x∗)) = 0, i.e., x∗∈F ix(f).
We prove thatx∗is the unique fixed point off. Suppose thaty∗ is a fixed point off, i.e.f(y∗) =y∗. Then
d(y∗, x∗) =d(f(y∗), f(x∗))≤ α·d(x∗, f(x∗))[1 +d(y∗, f(y∗))]
1 +d(x∗, y∗) +β·d(y∗, x∗) Hence d(y∗, x∗)≤β·d(y∗, x∗) and thus y∗=x∗.
Thereforex∗ is the unique fixed point of f.
3. Coupled fixed point theorems
In this section, using the fixed point theorems proved in Section 2, we will obtain some existence and uniqueness theorems for the coupled fixed point problem.
Theorem 3.1.Let (X,≤) be a partially ordered set andd:X×X→R+be a complete b-metric onX with constants≥1. LetT :X×X →X be an operator with closed graph (or in particular, it is continuous) which has the mixed monotone property on X×X. Assume that the following conditions are satisfied:
i) Suppose that there existsα, β≥0 with 1−αβ < 1s such that d(T(x, y), T(u, v)) +d(T(y, x), T(v, u))
≤α·[d(u, T(u, v)) +d(v, T(v, u))][1 +d(x, T(x, y)) +d(y, T(y, x))]
1 +d(x, u) +d(y, v)
+β·[d(x, u) +d(y, v)], (3.1)
for all (x, y),(u, v)∈X×X withx≤u, y≥v;
ii) there exists x0, y0 ∈ X such that x0 ≤ T(x0, y0), y0 ≥ T(y0, x0), i.e.
(x0, y0)≤p(T(x0, y0), T(y0, x0)).
Then, the following conclusions hold:
a) there exists (x∗, y∗)∈X×X a solution of the coupled fixed point problem (P1), such that the sequences (xn)n∈N,(yn)n∈Nin X defined by
(xn+1=T(xn, yn),
yn+1=T(yn, xn), f or all n∈N. have the property that (xn)n∈N→x∗,(yn)n∈N→y∗ asn→ ∞.
b) in particular, ifdis a continuous b-metric onX, then d(xn, x∗) +d(yn, y∗)≤ s·An
1−s·A[d(x0, x1) +d(y0, y1)]
whereA=1−2α2β and
(x1=T(x0, y0) y1=T(y0, x0).
Proof. By ii) we have thatz0 = (x0, y0)≤p (T(x0, y0), T(y0, x0)) = (x1, y1) =z1. So we havez0≤pz1.
If we considerx2 =T(x1, y1) and y2=T(y1, x1), then we getx2=T(x1, y1) = T2(x0, y0) and y2 =T(y1, x1) = T2(y0, x0). Using the mixed monotone property of T, we get
x2=T(x1, y1)≥T(x0, y0) =x1 implies x1≤x2 y2=T(y1, x1)≤T(y0, x0) =y1 implies y1≥y2
Hencez1= (x1, y1)≤p(x2, y2) =z2.
By this approach we obtain the sequences (xn)n∈N,(yn)n∈N inX with (xn+1=T(xn, yn)
yn+1=T(yn, xn)
and by mathematical induction we obtain zn = (xn, yn) ≤p (xn+1, yn+1) = zn+1, which implies (zn)n∈Nis a monotone increasing sequence in (Z,≤p), whereZ=X×X.
Consider the metricde:Z×Z →R+ , defined by d((x, y),e (u, v)) =d(x, u) +d(y, v).
Then, deis a b-metric on Z with the same constant s≥1 and if (X, d) is complete, we have (Z,d) is complete, too.e
LetF :Z→Z be an operator defined byF(x, y) = (T(x, y), T(y, x)), ∀(x, y)∈Z.
We havezn+1=F(zn), forn≥0 wherez0= (x0, y0). Using the mixed monotone property ofT, then the operatorF is monotone increasing with respect to ”≤p” i.e.
(x, y),(u, v)∈Z, with (x, y)≤p(u, v)⇒F(x, y)≤pF(u, v).
BecauseT has a closed graph (or, in particulat it is continuous onX×X), then F has a closed graph (or respectively is continuous onZ).
F is a contraction in (Z,d) on all comparable elements ofe Z. Letz = (x, y)≤p
(u, v) =w∈Z, so we have
d(Fe (z), F(w)) =d((Te (x, y), T(y, x)),(T(u, v), T(v, u))
=d(T(x, y), T(u, v)) +d(T(y, x), T(v, u))
≤α·[d(u, T(u, v)) +d(v, T(v, u))][1 +d(x, T(x, y)) +d(y, T(y, x))]
1 +d(x, u) +d(y, v) +β·[d(x, u) +d(y, v)]
=α·d(w, Fe (w))[1 +d(z, Fe (z))]
1 +d(z, w)e +β·d(z, w).e The operatorF :Z →Z has the following properties:
1)F :Z →Z has a closed graph;
2)F :Z →Z is increasing onZ;
3) there existz0= (x0, y0)∈Z such thatz0≤pF(z0);
4) there existsα, β≥0 with 1−αβ <1s such that d(F(z), Fe (w))≤α·d(w, Fe (w))[1 +d(z, Fe (z))]
1 +d(z, w)e +β·d(z, w).e
We can apply the conclusion of the Theorem 2.1. and we get thatF has at least one fixed point. Hence, there existsz∗ ∈Z with F(z∗) =z∗. Letz∗ = (x∗, y∗)∈Z, so we haveF(x∗, y∗) = (x∗, y∗).
This implies
(T(x∗, y∗), T(y∗, x∗)) = (x∗, y∗)⇒
(x∗=T(x∗, y∗) y∗=T(y∗, x∗) and the sequences (xn)n∈N,(yn)n∈Nin X defined by
(xn+1=T(xn, yn)
yn+1=T(yn, xn) forn∈N have the property thatxn →x∗ andyn →y∗ asn→ ∞.
We know thatzn+1=F(zn) =F(xn, yn) forn≥0. This yields to d(ze n, zn+1) =d(F(ze n−1), F(zn))
=d((Te (xn−1, yn−1), T(yn−1, xn−1)),(T(xn, yn), T(yn, xn)))
=d(T(xn−1, yn−1), T(xn, yn)) +d(T(yn−1, xn−1), T(yn, xn))
≤α[d(xn, T(xn, yn))+d(yn, T(yn, xn))][1+d(xn−1, T(xn−1, yn−1))+d(yn−1, T(yn−1, xn−1))]
1+d(xn−1, xn)+d(yn−1, yn) +β[d(xn−1, xn) +d(yn−1, yn)]
=α·d(ze n, F(zn))[1 +d(ze n−1, F(zn−1))]
1 +d(ze n−1, zn) +β·d(ze n−1, zn)
= α·d(ze n, F(zn))[1 +d(ze n−1, zn)]
1 +d(ze n−1, zn) +β·d(ze n−1, zn) =α·d(ze n, zn+1) +β·d(ze n−1, zn).
This yields to
d(ze n, zn+1)≤ β
1−α·d(ze n−1, zn)≤ β
1−α 2
·d(ze n−2, zn−1)
≤. . .≤ β
1−α n
·d(ze 0, z1) where 1−αβ < 1s <1.
We denoteA= 1−αβ <1. Moreover, forn∈Nandp∈N∗, we have
d(ze n, zn+p)≤s·d(ze n, zn+1) +s2·d(ze n+1, zn+2) +. . .+ +sp−1·d(ze n+p−1, zn+p)
≤s·An·d(ze 0, z1) +s2·An+1·d(ze 0, z1) +. . .+sp−1·An+p−1·d(ze 0, z1)
≤s·An·[1 +s·A+. . .+ (s·A)p−1]·d(ze 0, z1)
=s·An· 1−(s·A)p−1
1−s·A ·d(ze 0, z1)≤s·An· 1
1−s·A ·d(ze 0, z1).
If the b-metric is continuous, lettingp→ ∞we obtain d(ze n, z∗)≤ s·An
1−s·A ·d(ze 0, z1).
Butzn= (xn, yn), so we get
d((xe n, yn), z∗)≤ s·An
1−s·A·d((xe 0, y0),(x1, y1)) and, by definition ofd, we finally gete
d(xn, z∗) +d(yn, z∗)≤ s·An
1−s·A ·[d(x0, x1) +d(y0, y1)].
The following theorem gives the uniqueness of the coupled fixed point.
Theorem 3.2.Consider that we have the hypotheses of Theorem 3.1. and the following condition holds:
for all (x, y),(u, v)∈X×X there exists (z, w)∈X×X such that (z, w)≤p(x, y) and (z, w)≤p(u, v).
ThenCF ix(T) ={(x∗, y∗)}.
Proof. The operator T verifies the hypotheses of Theorem 3.1. Hence there exists (x∗, y∗)∈Z:=X×X such that
(x∗=T(x∗, y∗) y∗=T(y∗, x∗)
Let (x, y)∈CF ix(T) andde:Z×Z →R+, defined by d((x, y),e (u, v)) =d(x, u) +d(y, v),
whereZ =X×X. We have two cases:
Case 1.(x∗, y∗)≤p(x, y), which implies
d((xe ∗, y∗),(x, y)) =d((Te (x∗, y∗), T(y∗, x∗)),(T(x, y), T(y, x)))
=d(T(x∗, y∗), T(x, y)) +d(T(y∗, x∗), T(y, x))
≤ α·[d(x, T(x, y)) +d(y, T(y, x))][1 +d(x∗, T(x∗, y∗)) +d(y∗, T(y∗, x∗))]
1 +d(x∗, x) +d(y∗, y)
+β·[d(x∗, x) +d(y∗, y)] =β·[d(x∗, x) +d(y∗, y)] =β·d((xe ∗, y∗),(x, y)) This yields to
d((xe ∗, y∗),(x, y))≤β·d((xe ∗, y∗),(x, y)) or
(1−β)·d((xe ∗, y∗),(x, y))≤0 (but 1−β >0) Hence, we have
(x∗, y∗) = (x, y).
Case 2.(x∗, y∗),(x, y) are not comparable.
Let F :Z →Z be defined by F(x, y) = (T(x, y), T(y, x)) ∀(x, y)∈Z. There exists (z, w)∈Z, such that (z, w)≤p(x∗, y∗), impliesFn(z, w)≤pFn(x∗, y∗) because F is an increasing operator and (z, w)≤p(x, y), impliesFn(z, w)≤pF(x, y),F is an increasing operator.
We have
d(Fe n(z, w),(x∗, y∗)) =d(Fe n(z, w), Fn(x∗, y∗)) =d(Fe (Fn−1(z, w)), F(Fn−1(x∗, y∗)))
≤ α·d(Fe n−1(x∗, y∗), Fn(x∗, y∗))[1 +d(Fe n−1(z, w), Fn(z, w))]
1 +d(Fe n−1(z, w), Fn−1(x∗, y∗)) +β·d(Fe n−1(z, w), Fn−1(x∗, y∗))
=β·d(Fe n−1(z, w), Fn−1(x∗, y∗)).
By mathematical induction we get
d(Fe n(z, w), Fn(x∗, y∗))≤β·d(Fe n−1(z, w), Fn−1(x∗, y∗))
≤. . .≤βn·d((z, w),e (x∗, y∗))→0, asn→ ∞.
Hence
n→∞lim Fn(z, w) = (x∗, y∗). (3.2) But, we also know,
(z, w)≤p(x, y) implies
Fn(z, w)≤pFn(x, y) = (x, y).
Similarly, we obtain that
d(Fe n(z, w),(x, y))≤βn·d((z, w),e (x, y))→0 as n→ ∞ Hence,
n→∞lim Fn(z, w) = (x, y). (3.3)
By (3.1)+(3.2) we obtain that
(x∗, y∗) = (x, y).
A global version of the previous results is the following.
Theorem 3.3. Let (X, d) be a complete b-metric space with constant s ≥ 1 and T :X×X →X be an operator such that there existα, β≥0 with max{α,1−αβ }< 1s such that
d(T(x, y), T(u, v)) +d(T(y, x), T(v, u))
≤α·[d(u, T(u, v)) +d(v, T(v, u))][1 +d(x, T(x, y)) +d(y, T(y, x))]
1 +d(x, u) +d(y, v)
+β·[d(x, u) +d(y, v)] for (x, y),(u, v)∈X×X.
Then, there exists an unique solution (x∗, y∗) ∈ X ×X of the coupled fixed point problem (P1), and for any initial element (x0, y0)∈X×Xthe sequencezn+1= (xn+1, yn+1) = (T(xn, yn), T(yn, xn))∈X×X converges to (x∗, y∗).
Proof.LetZ =X×X and the functionalde:Z×Z →R+, such that d((x, y),e (u, v)) =d(x, u) +d(y, v).
We know thatdeis a b-metric onZ with the same constants≥1. Moreover, if (X, d) is a complete b-metric space, then (Z,d) is a complete b-metric space too.˜
Consider the operator F : Z → Z defined by F(x, y) = (T(x, y), T(y, x)) for (x, y)∈Z.
Letz= (x, y)∈Z andw= (u, v)∈Z. We have
d(Fe (z), F(w)) =d((Te (x, y), T(y, x)),(T(u, v), T(v, u)))
=d(T(x, y), T(u, v)) +d(T(y, x), T(v, u))
≤α·[d(u, T(u, v)) +d(v, T(v, u))][1 +d(x, T(x, y)) +d(y, T(y, x))]
1 +d(x, u) +d(y, v) +β·[d(x, u) +d(y, v)]
=α·d((u, v),e (T(u, v), T(v, u)))[1 +d((x, y),e (T(x, y), T(y, x)))]
1 +d((x, y),e (u, v)) +β·d((x, y),e (u, v))
=α·d(w, Fe (w))[1 +d(z, Fe (z))]
1 +d(z, w)e
+β·d(z, w).e Therefore
d(F(z), Fe (w))≤α·d(w, Fe (w))[1 +d(z, Fe (z))]
1 +d(z, w)e +β·d(z, w).e
From Theorem 2.3. we have thatF ix(F) ={(x∗, y∗)}, so the coupled fixed point problem (P1) has a unique solution (x∗, y∗)∈Z.
An existence and uniqueness result for the fixed point ofT is given now.
Theorem 3.4. If we suppose that we have the hypotheses of Theorem 3.2., then for the unique coupled fixed point (x∗, y∗) ofT we have thatx∗=y∗ i.e.T(x∗, x∗) =x∗.
Proof.From Theorem 3.2., there exists an unique coupled fixed point ofT, (x∗, y∗)∈ X×X.
We have two cases:
Case 1.Ifx∗ andy∗ are comparable,x∗≤y∗. Then we have
d(T(x, y), T(u, v)) +d(T(y, x), T(v, u))
≤ α·[d(u, T(u, v)) +d(v, T(v, u))][1 +d(x, T(x, u)) +d(y, T(y, x))]
1 +d(x, u) +d(y, v) +β·[d(x, u) +d(y, v)].
Let
x=v=x∗ and y=u=y∗. Thus we obtain
2·d(T(x∗, y∗), T(y∗, x∗))
≤ α·[d(y∗, T(y∗, x∗)) +d(x∗, T(x∗, y∗))][1 +d(x∗, T(x∗, y∗)) +d(y∗, T(y∗, x∗))]
1 + 2d(x∗, y∗) +β·2·d(x∗, y∗).
This yields to
d(x∗, y∗)≤β·d(x∗, y∗).
So
(1−β)·d(x∗, y∗)≤0, follows thatx∗=y∗.
Case 2.Suppose thatx∗ andy∗ are not comparable.
Hence, there exists z ∈ X such that z ≤ x∗ and z ≤y∗. Thus, the following relations are satisfied:
(z, y∗)≤p(y∗, z), (z, y∗)≤p(x∗, y∗), (y∗, x∗)≤p(y∗, z).
LetF :Z→Z be defined byF(x, y) = (T(x, y), T(y, x)) ∀(x, y)∈Z. Then, d(x∗, y∗) = 1
2·d((ye ∗, x∗),(x∗, y∗)) = 1
2 ·d(Fe n(y∗, x∗), Fn(x∗, y∗))
≤ s
2 ·d(Fe n(y∗, x∗), Fn(y∗, z)) +s
2 ·d(Fe n(y∗, z), Fn(x∗, y∗))
≤ s
2·d(Fe n(y∗, x∗), Fn(y∗, z))+s2
2 ·d(Fe n(y∗, z), Fn(z, y∗))+s2
2 ·d(Fe n(z, y∗), Fn(x∗, y∗)).
But we know that
d(Fe n(y∗, x∗), Fn(y∗, z))≤βn·d((ye ∗, x∗),(y∗, z)) =βn·d(x∗, z) d(Fe n(y∗, z), Fn(z, y∗))≤βn·d((ye ∗, z),(z, y∗)) = 2βn·d(y∗, z) d(Fe n(z, y∗), Fn(x∗, y∗))≤βn·d((z, ye ∗),(x∗, y∗)) =βn·d(z, x∗).
Using this assumptions, we get that d(x∗, y∗)≤ s
2 ·βn·d(x∗, z) +s2
2 ·βn·2·d(y∗, z) +s2
2 ·βn·d(z, x∗)
=s
2 ·βn·[(1 +s)d(x∗, z) + 2·s·d(y∗, z)]→0 as n→ ∞.
Hence, we have thatx∗=T(x∗, x∗).
4. Properties of the coupled fixed point problem
This section presents data dependence, well-posedness, Ulam-Hyers stability and limit shadowing property for the coupled fixed point problem.
The following theorem is a data dependence result of a coupled fixed point problem.
Theorem 4.1. Let (X, d) be a complete b-metric space with constant s ≥ 1 and Ti:X×X →X (i∈ {1,2}) be two operators which satisfy the following conditions:
i) there existα, β≥0 with max{α,1−αβ }< 1s such that d(T1(x, y), T1(u, v)) +d(T1(y, x), T1(v, u))
≤α·[d(u, T1(u, v)) +d(v, T1(v, u))][1 +d(x, T1(x, y)) +d(y, T1(y, x))]
1 +d(x, u) +d(y, v)
+β·[d(x, u) +d(y, v)] all for (x, y),(u, v)∈X×X; ii)CF ix(T2)6=∅;
iii) there existsη >0 such thatd(T1(x, y), T2(x, y))≤η for all (x, y)∈X×X.
In the above conditions, if (x∗, y∗) ∈X×X is the unique coupled fixed point forT1, then d(x∗, x) +d(y∗, y)≤2s(1+α)1−sβ ·η, where (x, y)∈CF ix(T2).
Proof.By Theorem 3.3, there exists (x∗, y∗)∈X×X such that (x∗=T1(x∗, y∗)
y∗=T1(y∗, x∗).
Let (x, y)∈CF ix(T2), i.e.
(x=T2(x, y) y=T2(y, x).
Consider the b-metricde:Z×Z→R+, defined by d((x, y),e (u, v)) =d(x, u) +d(y, v) for (x, y),(u, v)∈Z, where Z=X×X.
Consider two operatorsFi:Z →Z defined byFi(x, y) = (Ti(x, y), Ti(y, x)), for (x, y)∈Z,i∈ {1,2}.
We denote by z = (x∗, y∗) ∈ Z, which meansF1(z) = z and w = (x, y) ∈Z, which meansF2(w) =w. Then,
d(Fe 1(z), F1(w)) = α·d(w, Fe 1(w))[1 +d(z, Fe 1(z))]
1 +d(z, w)e +β·d(z, w)e
=α·d(w, Fe 1(w))
1 +d(z, w)e +β·d(z, w)e ≤α·d(w, Fe 1(w)) +β·d(z, w)e ≤2α·η+β·d(z, w).e Since
d(z, w) =e d(Fe 1(z), F2(w))≤s·[d(Fe 1(z), F1(w)) +d(Fe 1(w), F2(w))]
≤s·[2α·η+β·d(z, w)] + 2se ·η,
we will obtain that (1−sβ)·d(z, w)e ≤2s·(1 +α)·η.
Since max{α,1−αβ }<1s, we get that 1−sβ >0. Therefored(z, w)e ≤ 2s(1+α)1−sβ ·η and by definition of the metricd, we havee
d(x∗, x) +d(y∗, y)≤ 2s(1 +α)
1−sβ ·η.
Definition 4.2.Let (X, d) be a b-metric space with constants≥1 andT:X×X →X be an operator. By definition, the coupled fixed point problem (P1) is said to be well- posed if:
i)CF ix(T) ={(x∗, y∗)};
ii) for any sequence (xn, yn)n∈N ∈X ×X for whichd(xn, T(xn, yn)) →0 and d(yn, T(yn, xn))→ 0 asn → ∞, we have that (xn)n∈N → x∗ and (yn)n∈N → y∗ as n→ ∞.
Theorem 4.3.Assume that all the hypotheses of Theorem 3.3. take place. Then the coupled fixed problem (P1) is well-possed.
Proof.We denote byZ=X×X. By Theorem 3.3. we haveCF ix(T) ={(x∗, y∗)}.
Let (xn, yn)n∈N be a sequence on Z. We know that d(xn, T(xn, yn))→ 0 and d(yn, T(yn, xn))→0 asn→ ∞.
Consider the b-metric de: Z ×Z → R+, such that d((x, y),e (u, v)) = d(x, u) + d(y, v) for all (x, y),(u, v)∈Z.
Let F : Z → Z be an operator defined by F(x, y) = (T(x, y), T(y, x)) for all (x, y)∈Z. We know that F(x∗, y∗) = (x∗, y∗), so we have
d((xe n, yn),(x∗, y∗)) =d(xn, x∗) +d(yn, y∗)
≤s·d(xn, T(xn, yn)) +s·d(T(xn, yn), T(x∗, y∗)) +s·d(yn, T(yn, xn)) +s·d(T(yn, xn), T(y∗, x∗))
=s·[d(xn, T(xn, yn)) +d(yn, T(yn, xn))]
+s·[d(T(xn, yn), T(x∗, y∗)) +d(T(yn, xn), T(y∗, x∗))]
≤s·[d(xn, T(xn, yn)) +d(yn, T(yn, xn))]
+s·α·[d(x∗, T(x∗, y∗)) +d(y∗, T(y∗, x∗))][1 +d(xn, T(xn, yn)) +d(yn, T(yn, xn))]
1 +d(xn, x∗) +d(yn, y∗) +s·β·[1 +d(xn, x∗) +d(yn, y∗)]
≤s·[d(xn, T(xn, yn)) +d(yn, T(yn, xn))] +s·β·d((xe n, yn),(x∗, y∗)).
We obtain that
(1−sβ)·d((xe n, yn),(x∗, y∗))≤s·[d(xn, T(xn, yn)) +d(yn, T(yn, xn))]
d((xe n, yn),(x∗, y∗))≤ s
1−sβ·[d(xn, T(xn, yn)) +d(yn, T(yn, xn))]→0 as n→ ∞.
Therefore, (xn, yn)→(x∗, y∗) asn→ ∞.
Definition 4.4.Let (X, d) be a b-metric space with constants≥1 andT:X×X →X be an operator. Let ˜danyb-metric onX×Xgenerated bydBy definition, the coupled fixed point problem (P1) is said to be Ulam-Hyers stable if there exists a function
ψ:R+→R+increasing, continuous in 0 withψ(0) = 0, such that for eachε >0 and for each solution (x, y)∈X×X of the inequality
d((x, y),˜ (T(x, y), T(y, x)))≤ε,
there exists a solution (x∗, y∗)∈X×X of the coupled fixed point problem (P1) such that
d((x, y),˜ (x∗, y∗))≤ψ(ε).
Theorem 4.5.Assume that all the hypotheses of Theorem 3.3. take place. Then the coupled fixed point problem (P1) is Ulam-Hyers stable.
Proof. Let Z = X×X. By Theorem 3.3., we haveCF ix(T) = {(x∗, y∗)}. Let any ε >0 and let (x, y)∈Z such thatd(x, T(x, y)) +d(y, T(y, x))≤ε.
Consider the b-metricde:Z×Z→R+ given by
d((x, y),e (u, v)) =d(x, u) +d(y, v), ∀(x, y),(u, v)∈Z
andF :Z →Z an operator defined byF(x, y) = (T(x, y), T(y, x)) for all (x, y)∈Z. We have
d((x, y),e (x∗, y∗)) =d(x, x∗) +d(y, y∗) =d(x, T(x∗, y∗)) +d(y, T(y∗, x∗))
≤s·[d(x, T(x, y)) +d(T(x, y), T(x∗, y∗))] +s·[d(y, T(y, x)) +d(T(y, x), T(y∗, x∗))]
≤s·[d(x, T(x, y)) +d(y, T(y, x))]
+s·α·[d(x∗, T(x∗, y∗)) +d(y∗, T(y∗, x∗))][1 +d(x, T(x, y)) +d(y, T(y, x))]
1 +d(x, x∗) +d(y, y∗) +s·β·[d(x, x∗) +d(y, y∗)].
Thus
d((x, y),e (x∗, y∗))≤ s
1−sβ ·[d(x, T(x, y)) +d(y, T(y, x))]≤ s 1−sβ ·ε.
Therefore the coupled fixed point problem (P1) is Ulam-Hyers stable, with a mapping ψ:R+→R+, ψ(t) :=ct, wherec=1−sβs >0.
Definition 4.6. Let (X, d) be a b-metric space with constant s ≥ 1 and T : X ×X → X be an operator. By definition, the coupled fixed point problem (P1) has the limit shadowing property, if for any sequence (xn, yn)n∈N ∈ X ×X for which d(xn+1, T(xn, yn)) → 0 and respectively d(yn+1, T(yn, xn)) → 0 as n → ∞, there exists a sequence (Tn(x, y), Tn(y, x))n∈N such that d(xn, Tn(x, y)) → 0 and d(yn, Tn(y, x))→0 asn→ ∞.
Theorem 4.7. Assume that the hypotheses from Theorem 3.3. take place. Then the coupled fixed point problem (P1) forT has the limit shadowing property.
Proof. By Theorem 3.3, we have CF ix(T) = {(x∗, y∗)} and for any initial point (x, y)∈X×X the sequencezn+1= (Tn(x, y), Tn(y, x))∈X×X converge to (x∗, y∗) asn→ ∞.
Let (xn, yn)n∈Nbe a sequence onZ =X×X such thatd(xn+1, T(xn, yn))→0 andd(yn+1, T(yn, xn))→0 asn→ ∞.
We consider the b-metricde:Z×Z →R+, defined by
d((x, y),e (u, v)) =d(x, u) +d(y, v) for all (x, y),(u, v)∈Z.
Let F : Z → Z be an operator defined by F(u, v) = (T(u, v), T(v, u)) for all (u, v)∈Z. We know thatF(x∗, y∗) = (x∗, y∗). Then for every (x, y)∈Z we have:
d((xe n+1, yn+1),(Tn+1(x, y), Tn+1(y, x)))
≤s·[d((xe n+1, yn+1),(x∗, y∗)) +d((xe ∗, y∗),(Tn+1(x, y), Tn+1(y, x)))]) But
d((xe n+1, yn+1),(x∗, y∗))≤s·[d((xe n+1, yn+1), F(xn, yn)) +d(Fe (xn, yn), F(x∗, y∗))]
≤s·d((xe n+1, yn+1), F(xn, yn))+s·α·d((xe ∗, y∗), F(x∗, y∗))[1 +d((xe n, yn), F(xn, yn))]
1 +d((xe n, yn),(x∗, y∗)) +s·β·d((xe n, yn),(x∗, y∗))
=s·[d(xn+1, T(xn, yn)) +d(yn+1, T(yn, xn))] +s·β·d((xe n, yn),(x∗, y∗)).
This yields to
d((xe n+1, yn+1),(x∗, y∗))≤s·[d(xn+1, T(xn, yn)) +d(yn+1, T(yn, xn))]
+s·β·{s·[d(xn, T(xn−1, yn−1))+d(yn, T(yn−1, xn−1))]+s·β·d((xe n−1, yn−1),(x∗, y∗))}.
Therefore,
d((xe n+1, yn+1),(x∗, y∗))≤s·[d(xn+1, T(xn, yn)) +d(yn+1, T(yn, xn))]
+s·(s·β)·[d(xn, T(xn−1, yn−1))+d(yn, T(yn−1, xn−1))]+(s·β)2·d((xe n−1, yn−1),(x∗, y∗))
≤. . .≤(s·β)n+1·d((xe 0, y0),(x∗, y∗)) +s·
" n X
p=0
(s·β)n−p·d((xe p+1, yp+1), F(xp, yp))
# .
From Cauchy’s Lemma we haved((xe n+1, yn+1),(x∗, y∗))→0 asn→ ∞.
Thusd((xe n+1, yn+1),(Tn+1(x, y), Tn+1(y, x)))→0 as n→ ∞, so there exists a sequence (Tn(x, y), Tn(y, x))∈Z with
d((xe n, yn),(Tn(x, y), Tn(y, x))) =d(xn, Tn(x, y))+d(yn, Tn(y, x))→0 as n→ ∞.
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Anca Oprea
Babe¸s-Bolyai University, Faculty of Mathematics and Computer Sciences 400084 Cluj-Napoca, Romania
e-mail:[email protected] Gabriela Petru¸sel
Babe¸s-Bolyai University, Faculty of Business Str. Horea nr. 7, Cluj-Napoca, Romania e-mail:[email protected]
