R E S E A R C H
Open Access
A fixed point problem under two
constraint inequalities
Mohamed Jleli and Bessem Samet
**Correspondence:
[email protected] Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh, 11451, Saudi Arabia
Abstract
Let (X,d) be a metric space. Suppose that the setXis equipped with two partial orders1and2. LetT,A,B,C,D:X→Xbe given operators. We provide sufficient
conditions for the existence of a fixed point ofTsatisfying the two constraint inequalities:Ax1BxandCx2Dx.
Keywords: fixed point; constraint inequalities; partial order
1 Introduction and basic definitions
Recently there have been many developments concerning the existence of fixed points for operators defined in a metric space equipped with a partial order. This trend was initiated by Turinici []. Next, Ran and Reurings [] extended the Banach contraction principle to continuous monotone operators defined in a partially ordered metric space. They also presented some applications to the existence of positive solutions to certain classes of nonlinear matrix equations. The result obtained in [] was extended and generalized by many authors in different directions (see [–] and the references therein).
Let (X,d) be a metric space. Suppose that the setXis endowed with two partial orders and. Let us consider five self-operatorsT,A,B,C,D:X→X. In this paper, we are concerned with the following problem: Findx∈Xsuch that
⎧ ⎪ ⎨ ⎪ ⎩
x=Tx,
AxBx,
CxDx.
(.)
We obtain sufficient conditions for the existence of at least one solution to (.). The following definitions will be used throughout the paper.
Definition . LetXbe a nonempty set. Letbe a binary relation onX. We say that is a partial order onXif the following conditions are satisfied:
(i) For everyx∈X, we havexx. (ii) For everyx,y,z∈X, we have
xy, yz ⇒ xz.
(iii) For everyx,y∈X, we have
xy, yx ⇒ x=y.
Definition . Let (X,d) be a metric space andbe a partial order onX. We say that the partial orderisd-regular if the following condition is satisfied: For every sequences {an},{bn} ⊂X, we have
lim
n→∞d(an,a) =nlim→∞d(bn,b) = , anbnfor alln ⇒ ab,
where (a,b)∈X×X.
Example . Let (X, · ) be a Banach space andPbe a cone onX. Let us consider the partial order onXdefined by
(x,y)∈X×X, xPy ⇐⇒ y–x∈P.
Consider the metricdonXdefined by
d(x,y) = x–y , (x,y)∈X×X.
ThenPisd-regular. In fact, suppose that{an}and{bn}are two sequences inXsuch that
anPbn for alln
and
lim
n→∞d(an,a) =nlim→∞d(bn,b) =
for some (a,b)∈X×X. Sincebn–an∈Pfor allnand the sequence{bn–an}converges tob–a, the closure of the conePyieldsb–a∈P, that is,aPb.
Definition . LetXbe a nonempty set endowed with two partial ordersand. Let
T,A,B,C,D:X→Xbe given operators. We say that the operatorTis (A,B,C,D,, )-stable, if the following condition is satisfied:
x∈X, AxBx ⇒ CTxDTx.
Example . LetX=Rand consider the standard order≤onX. LetA,B,C,D:X→X
be the operators defined by
Ax=x, Bx=x, Cx=exp(x),
Dx=expx– x+ , Tx=x+ , x∈R.
Then the operatorT is (A,B,C,D,≤,≤)-stable. In fact, letx∈Xbe such that
Ax=x≤Bx=x.
Then
which yields
exp(x+ )≤expx+ ,
that is,
CTx≤DTx.
Example . LetX=R. We consider the two partial orders
andonXdefined by
(a,b), (c,d)∈X, (a,b)(c,d) ⇐⇒ a≤c, b≤d
and
(a,b), (c,d)∈X, (a,b)(c,d) ⇐⇒ a≤c, b≥d.
Let us consider the operatorsT,A,B,C,D:X→Xdefined by
T(a,b) = (b,a+ ),
A(a,b) = (a, b),
B(a,b) = (a, b),
C(a,b) =|a|,b,
D(a,b) =
(a+ )|a|
|a|+ ,b–
for all (a,b)∈X. We claim thatTis (A,B,C,D,,)-stable. In order to prove this claim, we takex= (a,b)∈Xsuch that
AxBx,
which is equivalent (from the definition of) to
b≥.
On the other hand, under the above inequality, we have
CTx=C(b,a+ ) =|b|,a+ = (b,a+ )
and
DTx=D(b,a+ ) =
(b+ )|b|
|b|+ ,a = (b,a– ).
Clearly (from the definition of), we have
CTxDTx.
Now, we are ready to state and prove our main result.
2 Main result
Let us denote bythe set of functionsϕ: [,∞)→[,∞) satisfying the following condi-tions:
() ϕis a lower semi-continuous function. () ϕ–({}) ={}.
Our main result in this paper is giving by the following theorem.
Theorem . Let(X,d)be a complete metric space endowed with two partial orders
and.Let T,A,B,C,D:X→X be given operators.Suppose that the following conditions
are satisfied:
(i) iisd-regular,i= , . (ii) A,B,C,andDare continuous. (iii) There existsx∈Xsuch that
AxBx.
(iv) Tis(A,B,C,D,,)-stable. (v) Tis(C,D,A,B,,)-stable. (vi) There existsϕ∈such that
AxBx, CyDy ⇒ d(Tx,Ty)≤d(x,y) –ϕ
d(x,y).
Then:
(I) The sequence{Tnx
}converges to somex∗∈Xsatisfying
Ax∗Bx∗ and Cx∗Dx∗.
(II) The pointx∗∈Xis a solution to(.).
Proof Let us prove (I). Letx∈Xbe such that
AxBx.
Such a point exists from (iii). Let us consider the sequence{xn} ⊂Xdefined by
xn=Tnx, n= , , , . . . .
SinceTis (A,B,C,D,,)-stable (see (iv)), we have
AxBx ⇒ CTxDTx,
that is,
Then we have
AxBx and CxDx.
SinceTis (C,D,A,B,,)-stable (see (v)),
CxDx ⇒ ATxBTx,
that is,
AxBx.
Again,Tis (A,B,C,D,,)-stable; this yields
AxBx ⇒ CTxDTx,
that is,
CxDx.
Thus we have
AxBx and CxDx.
By induction, we get
AxnBxn and Cxn+Dxn+, n= , , , . . . . (.)
Using (.) and (vi), by symmetry, we have
d(xn+,xn) =d(Txn,Txn–)≤d(xn–,xn) –ϕ
d(xn–,xn)
, n= , , , . . . , (.)
which yields
d(xn+,xn)≤d(xn,xn–), n= , , , . . . .
Then {d(xn+,xn)} is a decreasing sequence of positive numbers. Therefore, there exists somer≥ such that
lim
n→∞d(xn+,xn) =r. (.)
From (.), we have
d(xn+,xn) +ϕ
d(xn–,xn)
≤d(xn–,xn), n= , , , . . . ,
which yields
lim inf n→∞
d(xn+,xn) +ϕ
d(xn–,xn)
≤lim inf
Using (.) and the lower semi-continuity ofϕ, we obtain
r+ϕ(r)≤r,
which implies that
r∈ϕ–{}.
Sinceϕ–({}) ={}, we getr= ,i.e.,
lim
n→∞d(xn+,xn) = . (.)
Now, we show that{xn}is a Cauchy sequence in (X,d). Suppose that{xn}is not a Cauchy sequence. Then there exists someε> for which we find two sequences of positive inte-gers{m(k)}and{n(k)}such that, for all positive integersk,
n(k) >m(k) >k, d(xm(k),xn(k))≥ε, d(xm(k),xn(k)–) <ε. (.)
From (.), we have
ε≤d(xm(k),xn(k))
≤d(xm(k),xn(k)–) +d(xn(k)–,xn(k))
< ε+d(xn(k)–,xn(k)).
Then, for allk, we have
ε≤d(xm(k),xn(k)) <ε+d(xn(k)–,xn(k)).
Passing to the limit ask→ ∞and using (.), we get
lim
k→∞d(xm(k),xn(k)) =ε. (.)
On the other hand, we have
d(xn(k)+,xm(k)) –d(xm(k),xn(k))≤d(xn(k)+,xn(k))→ ask→ ∞(from (.)).
Then from (.), we have
lim
k→∞d(xn(k)+,xm(k)) =ε. (.)
We have also
d(xm(k),xn(k)) –d(xn(k),xm(k)–)≤d(xm(k)–,xm(k))→ ask→ ∞(from (.)).
Then from (.), we have
lim
Similarly,
d(xn(k)+,xm(k)) –d(xn(k)+,xm(k)+)≤d(xm(k),xm(k)+)→ ask→ ∞(from (.)).
Then from (.), we have
lim
k→∞d(xn(k)+,xm(k)+) =ε. (.)
Observe that, for allk, there exists a positive integer ≤i(k)≤ such that
n(k) –m(k) +i(k)≡().
From (.), for allk> , we have
Axn(k)Bxn(k) and Cxm(k)–i(k)Dxm(k)–i(k)
or
Axm(k)–i(k)Bxm(k)–i(k) and Cxn(k)Dxn(k).
Then from (vi), by symmetry, we have
d(Txn(k),Txm(k)–i(k))≤d(xn(k),xm(k)–i(k)) –ϕ
d(xn(k),xm(k)–i(k))
, k> ,
that is,
d(xn(k)+,xm(k)–i(k)+)≤d(xn(k),xm(k)–i(k)) –ϕ
d(xn(k),xm(k)–i(k))
, k> . (.)
Set
=k> :i(k) = and =k> :i(k) = .
We consider two cases. •Case .||=∞. From (.), we get
d(xn(k)+,xm(k)+) +ϕ
d(xn(k),xm(k))
≤d(xn(k),xm(k)), k∈,
which gives us
lim inf k→∞
d(xn(k)+,xm(k)+) +ϕ
d(xn(k),xm(k))
≤lim inf
k→∞ d(xn(k),xm(k)).
Using (.), (.), and the lower semi-continuity ofϕ, we obtain
which yields
ε∈ϕ–{}.
Sinceϕ–({}) ={}, we getε= , which is a contradiction withε> . •Case .||<∞.
In this case, we have||=∞. Moreover, from (.), we have
d(xn(k)+,xm(k)) +ϕ
d(xn(k),xm(k)–)
≤d(xn(k),xm(k)–), k∈,
which gives us
lim inf k→∞
d(xn(k)+,xm(k)) +ϕ
d(xn(k),xm(k)–)
≤lim inf
k→∞ d(xn(k),xm(k)–).
Using (.), (.), and the lower semi-continuity ofϕ, we obtain
ε+ϕ(ε)≤ε,
which yieldsε∈ϕ–({}) ={}, that is, a contradiction withε> .
Therefore, we deduce that{xn}is a Cauchy sequence in (X,d). Since (X,d) is complete, there exists somex∗∈Xsuch that
lim n→∞d
xn,x∗
= . (.)
On the other hand, from (.), we have
AxnBxn, n= , , , . . . .
Using the continuity ofAandB, it follows from (.) that
lim n→∞d
Axn,Ax∗
= lim n→∞d
Bxn,Bx∗
= .
Sinceisd-regular, we get
Ax∗Bx∗. (.)
Similarly, from (.), we have
Cxn+Dxn+, n= , , , . . . .
Using the continuity ofCandD, it follows from (.) that
lim n→∞d
Cxn+,Cx∗
= lim n→∞d
Dxn+,Dx∗
= .
Sinceisd-regular, we get
Cx∗Dx∗. (.)
Now, let us prove (II). Using (.), (.), (.), and (vi), we have
dTx∗,x∗≤dTx∗,Txn
+dxn+,x∗
≤dx∗,xn
–ϕdx∗,xn
+dxn+,x∗
, n= , , , . . . ,
that is,
dTx∗,x∗+ϕdx∗,xn
≤dx∗,xn
+dxn+,x∗
, n= , , , . . . .
Then
lim inf n→∞
dTx∗,x∗+ϕdx∗,xn
≤lim inf n→∞
dx∗,xn
+dxn+,x∗
.
Using the lower semi-continuity ofϕ, the fact thatϕ() = , and (.), we obtain
dx∗,Tx∗= ,
which yields
Tx∗=x∗. (.)
Therefore, from (.), (.), and (.), we deduce thatx∗∈Xis a solution to (.). This
ends the proof.
In the next section, we present some consequences following from Theorem ..
3 Some consequences
3.1 A fixed point problem under one constraint equality
Here, we are concerned with the following problem: Findx∈Xsuch that
x=Tx,
Ax=Bx, (.)
whereT,A,B:X→Xare given operators and (X,d) is a metric space endowed with a certain partial order. Observe that (.) is equivalent to (.) with
==, C=B and D=A.
Then from Theorem ., we obtain the following result.
Corollary . Let(X,d)be a complete metric space endowed with a certain partial
or-der.Let T,A,B:X→X be three given operators.Suppose that the following conditions are satisfied:
(i) isd-regular.
(iii) There existsx∈Xsuch that
AxBx.
(iv) For allx∈X,we have
AxBx ⇒ BTxATx.
(v) For allx∈X,we have
BxAx ⇒ ATxBTx.
(vi) There existsϕ∈such that
AxBx, ByAy ⇒ d(Tx,Ty)≤d(x,y) –ϕd(x,y).
Then:
(I) The sequence{Tnx}converges to somex∗∈XsatisfyingAx∗=Bx∗. (II) The pointx∗∈Xis a solution to(.).
3.2 A common fixed point problem
Let us consider the following problem: Findx∈Xsuch that
x=Tx,
x=Bx, (.)
whereT,B:X→Xare given operators and (X,d) is a metric space endowed with a certain partial order. Observe that (.) is equivalent to (.) withA=IX, the identity mapping onX. So, takeA=IXin Corollary ., We obtain the following result.
Corollary . Let(X,d)be a complete metric space endowed with a certain partial or-der.Let T,B:X→X be two given operators.Suppose that the following conditions are satisfied:
(i) isd-regular. (ii) Bis continuous.
(iii) There existsx∈Xsuch that
xBx.
(iv) For allx∈X,we have
xBx ⇒ BTxTx.
(v) For allx∈X,we have
(vi) There existsϕ∈such that
xBx, Byy ⇒ d(Tx,Ty)≤d(x,y) –ϕd(x,y).
Then:
(I) The sequence{Tnx
}converges to somex∗∈Xsatisfyingx∗=Bx∗. (II) The pointx∗∈Xis a solution to(.).
Next, we present an example that illustrates the above result.
Example . LetX⊂Rbe the set defined by
X=(, ), (, ), (, ), (, ), (, ).
Letbe the partial order onXdefined by
(x,y), (z,w)∈X, (x,y)(z,w) ⇐⇒ x≤z, y≤w.
We endowXwith the metricddefined by
d(x,y), (z,w)=(x–z)+ (y–w), (x,y), (z,w)∈X.
LetT,B:X→Xbe the mappings defined by
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
T(, ) = (, ),
T(, ) = (, ),
T(, ) = (, ),
T(, ) = (, ),
T(, ) = (, ) and ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
B(, ) = (, ),
B(, ) = (, ),
B(, ) = (, ),
B(, ) = (, ),
B(, ) = (, ).
Observe that
u∈X, uBu ⇐⇒ u∈(, ), (, ), (, ).
However,
v∈X, Bvv ⇐⇒ v∈(, ), (, ).
Letu∈Xbe such thatuBu. Foru= (, ), we have
BTu=B(, ) = (, ) =Tu.
Foru= (, ), we have
Foru= (, ), we have
BTu=B(, ) = (, ) =Tu.
Then we have
u∈X, uBu ⇒ BTuTu.
Letv∈Xbe such thatBvv. Forv= (, ), we have
Tv=T(, ) = (, ) =BTv.
Forv= (, ), we have
Tv=T(, ) = (, ) = (, ) =BTv.
Then we have
v∈X, Bvv ⇒ TvBTv.
Now, let (u,v)∈X×Xbe such that
uBu and Bvv.
We have
(u,v)∈(, ), (, ),(, ), (, ),(, ), (, ),
(, ), (, ),(, ), (, ),(, ), (, ).
For (u,v) = ((, ), (, )), we have
d(Tu,Tv) =d(, ), (, )= .
For (u,v) = ((, ), (, )), we have
d(Tu,Tv) =d(, ), (, )= =d(u,v) .
For (u,v) = ((, ), (, )), we have
d(Tu,Tv) =d(, ), (, )= =d(u,v) .
For (u,v) = ((, ), (, )), we have
For (u,v) = ((, ), (, )), we have
d(Tu,Tv) =d(, ), (, )= .
For (u,v) = ((, ), (, )), we have
d(Tu,Tv) =d(, ), (, )= < =
d(u,v) .
Then we have
uBu, Bvv ⇒ d(Tu,Tv)≤d(u,v) –ϕd(u,v),
whereϕ(t) =t,t≥. So, all the required conditions of Corollary . are satisfied. Observe thatx∗= (, ) is a common fixed point ofTandB. Observe also that
dT(, ),T(, )=d(, ), (, )= >√ =d(, ), (, ),
which shows thatTis not a contraction onX.
TakingB=Tin Corollary ., we obtain the following fixed point result.
Corollary . Let(X,d)be a complete metric space endowed with a certain partial or-der.Let T:X→X be a given operator.Suppose that the following conditions are satis-fied:
(i) isd-regular. (ii) Tis continuous.
(iii) There existsx∈Xsuch that
xTx.
(iv) For allx∈X,we have
xTx ⇒ TxTx.
(v) For allx∈X,we have
Txx ⇒ TxTx.
(vi) There existsϕ∈such that
xTx, Tyy ⇒ d(Tx,Ty)≤d(x,y) –ϕd(x,y).
Then the sequence{Tnx}converges to a fixed point of T.
4 Conclusion
technique can also be adapted for any finite number of constraint inequalities and other contractive conditions. An interesting question is the existence of a best proximity point of a certain operator under constraint inequalities. Such a question will be studied in a future work.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Acknowledgements
The authors extend their appreciation to Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia).
Received: 3 December 2015 Accepted: 25 January 2016
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