R E S E A R C H
Open Access
Some extensions for Geragthy type
contractive mappings
Erdal Karapınar
1,2*, Hamed H Alsulami
2and Maha Noorwali
2*Correspondence: [email protected] 1Department of Mathematics, Atilim University, ˙Incek, Ankara, 06836, Turkey
2Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia
Abstract
In this paper, we establish some fixed point theorems on some extensions of Geragthy contractive type mappings in the context ofb-metric-like spaces.
MSC: 46T99; 46N40; 47H10; 54H25
Keywords: b-metric-like space; fixed point;
α-admissible; contractive mapping
1 Introduction and preliminaries
One of the interesting extensions of the notion of a metric space is the dislocated space, introduced by Hitzler []. This notion was rediscovered by Amini-Harandi [] and given the name of a metric-like space.
Definition . On a nonempty setXwe define a functionσ:X×X→[,∞) such that for allx,y,z∈X:
(σ) ifσ(x,y) = thenx=y; (σ) σ(x,y) =σ(y,x);
(σ) σ(x,y)≤σ(x,z) +σ(z,y);
and the pair (X,σ) is called a dislocated (metric-like) space.
Throughout this paper, we suppose that N=N∪ {}whereNdenotes the set of all
positive integers. Further, the symbolsR+andR+
denotes the set of positive reals and the
set of non-negative reals. First, we recall some basic concepts and notations.
The concept of ab-metric was introduced by Czerwik [] as a generalization of the met-ric (see also Bakhtin [] and Bourbaki []) to extend the celebrated Banach contraction mapping principle. Following this initial paper of Czerwik [], a number of researchers in nonlinear analysis investigated the topology of the paper and proved several fixed point theorems in the context of completeb-metric spaces (seee.g.[–] and references therein).
Definition .[] LetXbe a nonempty set ands≥ be a given real number. A mapping
d: X×X→[,∞) is said to be ab-metric if for allx,y,z∈Xthe following conditions are satisfied:
(bM) d(x,y) = if and only ifx=y;
(bM) d(x,y) =d(y,x);
(bM) d(x,z)≤s[d(x,y) +d(y,z)].
In this case, the pair (X,d) is called ab-metric space (with constants).
In what follows, we recall the notion ofb-metric-like space which is an interesting gen-eralization of bothb-metric space and metric-like space.
Definition .[] LetXbe a nonempty set ands≥ be a given real number. A mapping
d: X×X→[,∞) is said to beb-metric-like if for allx,y,z∈Xthe following conditions are satisfied:
(bML) ifd(x,y) = thenx=y;
(bML) d(x,y) =d(y,x);
(bML) d(x,z)≤s[d(x,y) +d(y,z)].
In this case, the pair (X,d) is called ab-metric-like space (with constants).
Example . LetX=C([,T]) be the set of all real continuous functions on the closed interval [,T]. Letd:X×X→R+be defined
d(f,g) =maxf(t) –g(t)p+a,
for allf,g∈X,a∈R+
, andp> . It is easy to see that (X,d) is a completeb-metric-like
space withs= p–. For more examples, seee.g.[].
Remark . Let (X,d) be ab-metric-like space with constants≥. Then it is clear that
ds(x,y) =|d(x,y) –d(x,x) –d(y,y)|satisfies the following: (S) ds(x,x) = for allx∈X.
Definition .[] Let (X,d) be ab-metric-like space. Then: () a sequence{xn}inXis called convergent tox∈Xif and only if
limn→∞d(xn,x) =d(x,x);
() a sequence{xn}inXis called Cauchy sequence if and only iflimn,m→∞d(xn,xm) exists and finite;
() (X,d)is complete if and only if every Cauchy sequence{xn}inXconverges tox∈X so that
lim
n→∞d(xn,x) =d(x,x) =m,limn→∞d(xn,xm).
Proposition .[] Let(X,d)be a b-metric-like space with constant s and let{xn}be a
sequence in X such thatlimn→∞d(xn,x) = .Then: () xis unique.
() sd(x,y)≤limn→∞d(xn,y)≤sd(x,y)for ally∈X.
Lemma .[] Let(X,d)be a b-metric-like space with constant s and{xn}a sequence in
X such that
d(xn+,xn+)≤kd(xn,xn+), n= , , . . . ,
Lemma .[] Let(X,d)be a b-metric-like space with constant s and assume that{xn}
and{yn}are sequences in X converging to x and y,respectively.Then
sd(x,y) –
sd(x,x) –d(y,y)≤lim infn→∞ d(xn,yn)≤lim supn→∞ d(xn,yn)
≤sd(x,x) +sd(y,y) +sd(x,y).
In particular,if d(x,y) = thenlimn→∞d(xn,yn) = .
Moreover,for each z∈X we have
sd(x,z) –d(x,x)≤lim infn→∞ d(xn,z)≤lim supn→∞ d(xn,z)≤sd(x,z) +sd(x,x). ()
In particular,if d(x,x) = ,then
sd(x,z)≤lim infn→∞ d(xn,z)≤lim supn→∞ d(xn,z)≤sd(x,z).
Notice that, in general, ab-metric-like mapping does not need to be continuous. The notion ofα-admissible and triangularα-admissible mappings were introduced by Sametet al.[] and Karapınaret al.[], respectively.
Definition . LetT:X→Xbe a mapping andα:X×X→[,∞) be a function. We say thatT is anα-admissible mapping if
x,y∈X, α(x,y)≥ ⇒ α(Tx,Ty)≥.
Moreover, a self-mappingTis called triangularα-admissible ifTisα-admissible and
x,y,z∈X, α(x,z)≥ andα(z,y)≥ ⇒ α(x,y)≥.
For more details onα-admissible and triangularα-admissible mappings, seee.g.[–]. Very recently, Popescu [] refined the notion of triangularα-orbital admissible as fol-lows.
Definition .[] LetT:X→Xbe a mapping andα:X×X→[,∞) be a function. We say thatT isα-orbital admissible if
α(x,Tx)≥ ⇒ αTx,Tx≥.
Furthermore,Tis called triangularα-orbital admissible ifTisα-orbital admissible and
α(x,y)≥ andα(y,Ty)≥ ⇒ α(x,Ty)≥.
Example . LetX={a,b,c,d,e,f,g,h}. We define a self-mappingT:X→Xsuch that
Tx=x, forx=a,dand
Tx=y for (x,y)∈(b,c), (c,b), (e,f), (f,e), (g,h), (h,g).
Moreover, we defineα:X×X→R+
, such that
α(x,y) =
if (x,y)∈ {(a,b), (a,c), (b,b), (c,c), (b,c), (c,b), (b,d), (c,d), (d,e)}, otherwise.
Note thatT isα-orbital admissible, sinceα(b,Tb) =α(b,c) = andα(c,Tc) =α(c,b) = . On the other hand, we have α(d,e) = , but α(Td,Te) =α(d,f) = . Hence,T is notα -admissible.
Lemma .[] Let T:X→X be a triangularα-orbital admissible mapping.Assume that there exists x∈X such thatα(x,Tx)≥.Define a sequence{xn}by xn+=Txnfor
each n∈N.Then we haveα(xn,xm)≥for all m,n∈Nwith n<m.
Lemma . Let T:X→X be a triangularα-orbital admissible mapping.Assume that there exists x∈X such thatα(Tx,x)≥.Define a sequence{xn}by xn+=Txnfor each
n∈N.Then we haveα(xm,xn)≥for all m,n∈Nwith n<m.
We characterize the notion ofα-regular in the setting of ab-metric-like space.
Definition .(cf.[]) Let (X,d) be ab-metric-like space,Xis said to beα-regular, if for every sequence{xn}inXsuch thatα(xn,xn+)≥ (respectively,α(xn+,xn)≥) for alln
andxn→x∈Xasn→ ∞, there exists a subsequence{xnk}of{xn}such thatα(xnk,x)≥ (respectively,α(x,xnk)≥) for allk.
In this paper, we shall prove the existence and uniqueness of a fixed point for certain operators in the setting ofb-metric-like spaces. The presented results improve, extend, and unify a number of existing results in the literature.
2 Main result forb-metric-like spaces
In this section, we shall state and prove our main results. First, we recall the following classes of auxiliary functions. Letbe the set of all increasing and continuous functions
ψ: [,∞)→[,∞) withψ–({}) ={}. LetFsbe the family of all functionsβ: [,∞)→ [,s) which satisfy the condition
lim
n→∞β(tn) =
s implies n→∞lim tn= , ()
for somes≥.
Definition . Let (X,d) be ab-metric-like space with constants≥, andT:X→Xbe a map. We say thatTis a generalized almostα-ψ-φ-Geraghty contractive type mapping if there exist a functionα:X×X→[,∞),ψ,φ∈,β∈Fs, and someL≥ such that
for allx,y∈X, where
M(x,y) =max
d(x,y),d(x,Tx),d(y,Ty),d(x,Ty) +d(y,Tx)
s and ()
N(x,y) =minds(x,Ty),ds(y,Tx),d(x,Tx),d(y,Ty). ()
Remark . Since the functions belonging toFsare strictly smaller thans, for somes≥, the expressionβ(ψ(M(x,y))) in () can be estimated from above as follows:
βψM(x,y)<
s for anyx,y∈X.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥ and T :
X→X be a generalized almostα-ψ-φ-Geraghty contractive type mapping.We suppose also that
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Proof By (ii) there existsx∈Xsuch thatα(x,Tx)≥. Define a sequence{xn} ⊂Xby
xn+=Txnfor alln∈N. AsT is triangularα-orbital admissible, by Lemma . we have α(xn,xn+)≥ for alln∈N. Throughout the proof, we suppose thatxn=xn+for alln∈
N. Indeed, if there existsnsuch thatxn=xn+, thenxnbecomes the fixed point ofT, which completes the proof.
SinceTis a generalized almostα-ψ-φ-Geraghty contractive type mapping we have
ψsd(xn+,xn+)
≤α(xn,xn+)ψ
sd(Txn,Txn+)
≤βψM(xn,xn+)
ψM(xn,xn+)
+LφN(xn,xn+)
. ()
Thus, we have
ψsd(xn+,xn+)
<
sψ
M(xn,xn+)
+LφN(xn,xn+)
, ()
whereN(xn,xn+) =min{ds(xn,xn+),ds(xn+,xn+),d(xn,xn+),d(xn+,xn+)}= , and
M(xn,xn+) =max
d(xn,xn+),d(xn,xn+),d(xn+,xn+),
d(xn,xn+) +d(xn+,xn+)
s .
Note that
d(xn,xn+) +d(xn+,xn+)
s ≤
s[d(xn,xn+) + d(xn+,xn+)]
s
=[d(xn,xn+) + d(xn+,xn+)]
≤maxd(xn,xn+),d(xn+,xn+)
Consequently, we have
M(xn,xn+) =max
d(xn,xn+),d(xn+,xn+)
.
IfM(xn,xn+) =d(xn+,xn+), then from () we have
ψsd(xn+,xn+)
<
sψ
d(xn+,xn+)
≤ψd(xn+,xn+)
.
Sinceψis increasing, we derive thatsd(xn+,xn+) <d(xn+,xn+), which is a contradiction
ass≥. Thus,M(xn,xn+) =d(xn,xn+). Again by (), we find
ψsd(xn+,xn+)
<
sψ
d(xn,xn+)
≤ψd(xn,xn+)
.
Hence, we get
sd(xn+,xn+)≤d(xn,xn+) equivalently d(xn+,xn+)≤
sd(xn,xn+). ()
Case (i):s> . Since s > andss =s< , by Lemma ., the sequence{xn}is Cauchy and
lim
n,m→∞d(xn,xm) = . ()
Case (ii):s= . From (), we haved(xn+,xn+)≤d(xn,xn+) for alln. Thus, we conclude
that
lim
n→∞d(xn,xn+) =r, ()
for somer≥. We shall prove thatr= . Suppose, on the contrary, thatr> . Note that, fors= , the inequality () turns into
ψd(xn+,xn+)
<βψM(xn,xn+)
ψM(xn,xn+)
+LφN(xn,xn+)
, ()
whereN(xn,xn+) = andM(xn,xn+) =d(xn,xn+) as evaluated above. Thus, () yields
ψ(d(xn+,xn+)) ψ(d(xn,xn+)) ≤
βψd(xn,xn+)
< . ()
By taking the limit asn→ ∞in () and regarding the continuity ofψ, we get
lim
n→∞β
ψd(xn,xn+)
= .
Hence, we have
lim
n→∞ψ
d(xn,xn+)
= and so lim
Consequentlyr= . In what follows, we shall prove that{xn}is a Cauchy sequence. Indeed we will prove thatlimm,n→∞d(xn,xm) = . Suppose, on the contrary, that there existε> and corresponding subsequences{nk}and{mk}ofNsatisfyingnk>mk>kfor which
d(xmk,xnk)≥ε, ()
wherenk,mkare chosen as the smallest integers satisfying (), that is,
d(xmk,xnk–) <ε. ()
By (), (), and the triangle inequality, we easily derive that
ε≤d(xmk,xnk)≤d(xmk,xnk–) +d(xnk–,xnk) <ε+d(xnk–,xnk). ()
Using () and the squeeze theorem we get
lim
k→∞d(xmk,xnk) =ε. ()
In a similar way, we can prove thatlimk→∞d(xmk,xnk+) = ,limk→∞d(xnk,xmk+) = .
Regarding thatTis a generalized almostα-ψ-φ-Geraghty contractive type mapping, we have
ψd(xmk+,xnk+)
≤α(xmk,xnk)ψ
d(Txmk,Txnk)
≤βψM(xmk,xnk)
ψM(xmk,xnk)
+LφN(xmk,xnk)
, ()
for allx,y∈X, where
M(xmk,xnk) = max
d(xmk,xnk),d(xmk,xmk+),d(xnk,xnk+),
d(xmk,xnk+) +d(xnk,xmk+)
, ()
and
N(xmk,xnk) =min
ds(xmk,xnk+),ds(xnk,xmk+),d(xmk,xmk+),d(xnk,xnk+)
.
By taking the limit ask→ ∞in () and taking (), () into account, we get
ψ(ε)≤ lim
k→∞β
ψM(xmk,xnk)
ψ(ε). ()
Sinceβis a Geraghty function, we derive thatψ(M(xmk,xnk))→. Consequently, we have
d(xmk,xnk)→, which is a contradiction. Hence, we conclude thatlimm,n→∞d(xn,xm) = , and the sequence{xn}is Cauchy for anys≥.
By completeness of (X,d), there existsu∈Xsuch that
lim
SinceTis continuous,
Tu=T
lim
n→∞xn
= lim
n→∞Txn=n→∞limxn+=u,
anduis a fixed point forT.
In what follows, we replace the condition of continuity of the operator by the condition ofα-regularity of the space.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥ and T :
X→X be a generalized almostα-ψ-φ-Geraghty contractive type mapping.We suppose also that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Xisα-regular anddis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Proof Following the lines of the proof of Theorem ., we conclude that there existsu∈X
such that
lim
n→∞d(xn,u) =n,m→∞lim d(xn,xm) =d(u,u) = .
SinceXisα-regular,α(xn,xn+)≥ for alln. Due to the fact thatlimn→∞xn=u, there exists a subsequence{xnk}of{xn}such thatα(xnk,u)≥ for allk. To prove thatuis a fixed point forT, suppose on the contrary thatd(u,Tu) > .
Now, by using the properties ofψ and asT is a generalized almost α-ψ-φ-Geraghty contractive type mapping we have
ψd(xnk+,Tu)
≤α(xnk,u)ψ
sd(Tx
nk,Tu)
≤βψM(xnk,u)
ψM(xnk,u)
+LφN(xnk,u)
.
Thus we have
ψd(xnk+,Tu)
<
sψ
M(xnk,u)
+LφN(xnk,u)
, ()
where N(xnk,u) = min{d s(x
nk,Tu),d s(u,x
nk+),d(xnk,xnk+),d(u,Tu)}, and note that
limk→∞N(xnk,u) = . Moreover,
M(xnk,u) = max
d(xnk,u),d(xnk,xnk+),d(u,Tu),
d(xnk,Tu) +d(u,xnk+) s
≤max
d(xnk,u),d(xnk,xnk+),d(u,Tu),
d(xnk,u) +d(u,Tu) +d(u,xnk) +d(xnk,xnk+)
Hence
lim
k→∞M(xnk,u)≤max
,d(u,Tu),d(u,Tu)
=d(u,Tu),
and by the definition ofM(xnk,u) we havelimk→∞M(xnk,u) =d(u,Tu).
By the continuity ofψand theb-metric-liked, taking the limit askgoes to∞on both sides of () we have
ψd(u,Tu)≤
sψ
d(u,Tu).
Thus = ψψ((dd((uu,,TuTu))))≤ s, which is a contradiction in the cases> . Henced(u,Tu) = ; thereforeTu=u. In the cases= we take the limit askgoes to∞on both sides of
ψd(xnk+,Tu)
≤βψM(xnk,u)
ψM(xnk,u)
and getlimk→∞β(ψ(M(xnk,u))) = and asβ∈F so we havelimk→∞ψ(M(xnk,u)) = .
Thus we haved(u,Tu) = ; thereforeTu=u.
For the uniqueness of a fixed point of a generalizedα-ψ-φcontractive mapping, we will consider the following hypothesis.
(H) For allx,y∈Fix(T), eitherα(x,y)≥orα(y,x)≥. Here,Fix(T) denotes the set of fixed points ofT.
Theorem . Adding condition(H)to the hypotheses of Theorem. (or Theorem.),we obtain the uniqueness of the fixed point of T.
Proof Suppose thatx∗andy∗are two fixed points ofT. Then it is obvious thatM(x∗,y∗) =
d(x∗,y∗) andN(x∗,y∗) = . So, we have
ψdx∗,y∗≤ψsdTx∗,Ty∗
≤αx∗,y∗ψsdTx∗,Ty∗
≤βψdx∗,y∗ψdx∗,y∗
<
sψ
dx∗,y∗,
which is a contradiction.
Definition . Let (X,d) be ab-metric-like space with constants≥,T :X→Xbe a map, we say thatTis a generalized rationalα-ψ-φ-Geraghty contractive mapping of type (I) if there exist a functionα:X×X→[,∞),ψ,φ∈,β∈Fs, and someL≥ such that
α(x,y)ψsd(Tx,Ty)≤βψK(x,y)ψK(x,y)+LφN(x,y), ()
for allx,y∈X, whereN(x,y) is defined as in () and
K(x,y) =max
d(x,y),d(x,Tx)d(y,Ty) +d(x,y) ,
d(x,Tx)d(y,Ty)
Definition . Let (X,d) be ab-metric-like space with constants≥,T:X→Xbe a map, we say thatT is a generalized rationalα-ψ-φ-Geraghty contractive of type (II) mapping if there exist a functionα:X×X→[,∞),ψ,φ∈,β∈Fsand someL≥ such that
α(x,y)ψsd(Tx,Ty)≤βψQ(x,y)ψQ(x,y)+LφN(x,y), ()
for allx,y∈X, whereN(x,y) is defined as in () and
Q(x,y) =max
d(x,y),d(x,Tx)d(y,Ty) +d
s(x,Ty)ds(y,Tx)
+s(d(x,Tx) +d(y,Ty)) . ()
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥and T:X→ X be a generalized rationalα-ψ-φ-Geraghty contractive mapping of type(I)such that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Proof We shall use the same techniques as in the proof of Theorem .. First of all, we shall construct a sequence{xn} ⊂Xwherexn+=Txnfor whichα(xn,xn+)≥ andxn=xn+for
alln∈N.
SinceTis generalized rationalα-ψ-φ-Geraghty contractive of type (I) we have
ψsd(xn+,xn+)
≤α(xn,xn+)ψ
sd(Txn,Txn+)
≤βψK(xn,xn+)
ψK(xn,xn+)
+LφN(xn,xn+)
. ()
SinceN(xn,xn+) = , the above inequality implies that
ψsd(xn+,xn+)
≤βψK(xn,xn+)
ψK(xn,xn+)
, ()
where
K(xn,xn+) =max
d(xn,xn+),
d(xn,xn+)d(xn+,xn+)
+d(xn,xn+)
,d(xn,xn+)d(xn+,xn+) +d(xn+,xn+)
.
On the other hand, we have
d(xn,xn+)d(xn+,xn+)
+d(xn+,xn+) ≤
d(xn,xn+)d(xn+,xn+)
d(xn+,xn+)
=d(xn,xn+)
and
d(xn,xn+)d(xn+,xn+)
+d(xn,xn+) ≤
d(xn,xn+)d(xn+,xn+)
d(xn,xn+)
=d(xn+,xn+).
Consequently, we getK(xn,xn+)≤max{d(xn,xn+),d(xn+,xn+)}.
Ifmax{d(xn,xn+),d(xn+,xn+)}=d(xn+,xn+), then from () together with Remark .,
we have
ψsd(xn+,xn+)
<
sψ
d(xn+,xn+)
This is a contradiction sinceψis increasing. Thus, we havemax{d(xn,xn+),d(xn+,xn+)}=
d(xn,xn+) and by the definition ofK(xn,xn+) we shall haveK(xn,xn+) =d(xn,xn+).
Con-sequently, the inequality () turns into
ψd(xn+,xn+)
≤ψsd(xn+,xn+)
≤βψK(xn,xn+)
ψd(xn,xn+)
. ()
By Remark ., we get
ψsd(xn+,xn+)
<ψd(xn,xn+)
and hence,d(xn+,xn+) <
sd(xn,xn+). ()
Case (i):s> . Since s > andss =s< , by Lemma ., the sequence{xn}is Cauchy and
lim
n,m→∞d(xn,xm) = . ()
Case (ii):s= . Since{d(xn,xn+)}is a decreasing sequence, there existsr≥ such that
lim
n→∞d(xn,xn+) =r, ()
for somer≥. We shall prove thatr= . Suppose, on the contrary, thatr> . By letting
n→ ∞in () we find
ψ(r)≤ lim
n→∞β
ψK(xn,xn+)
ψ(r).
It yields =limn→∞β(ψ(K(xn,xn+))). Sinceβ∈F, we getψ(K(xn,xn+))→, which
im-plies thatd(xn,xn+)→, that is,r= .
In what follows, we shall prove that{xn}is a Cauchy sequence. Indeed we will prove that
limm,n→∞d(xn,xm) = . Suppose, on the contrary, that there existε> and corresponding
subsequences{nk}and{mk}ofNsatisfyingnk>mk>kfor which
d(xmk,xnk)≥ε, ()
wherenk,mkare chosen as the smallest integers satisfying (), that is,
d(xmk,xnk–) <ε. ()
By (), (), and the triangle inequality, we easily derive that
ε≤d(xmk,xnk)≤d(xmk,xnk–) +d(xnk–,xnk)
<ε+d(xnk–,xnk). ()
Using () and the squeeze theorem we get
lim
Regarding that T is generalized rational α-ψ-φ-Geraghty contractive mapping of type (I), we have
ψd(xmk+,xnk+)
≤α(xmk,xnk)ψ
d(Txmk,Txnk)
≤βψK(xmk,xnk)
ψK(xmk,xnk)
+LφN(xmk,xnk)
, ()
for allx,y∈X, where
K(xmk,xnk) =max
d(xmk,xnk),
d(xmk,xmk+)d(xnk,xnk+)
+d(xmk,xnk) ,
d(xmk,xmk+)d(xnk,xnk+)
+d(xmk+,xnk+)
()
and
N(xmk,xnk) =min
ds(xmk,xnk+),ds(xnk,xmk+),d(xmk,xmk+),d(xnk,xnk+)
.
It is clear that
lim
k→∞K(xmk,xnk) =ε and k→∞limN(xmk,xnk) = .
By taking the limit ask→ ∞in () and taking (), () into account, we get
ψ(ε)≤ lim
k→∞β
ψK(xmk,xnk)
lim
k→∞ψ
K(xmk,xnk)
≤ lim
k→∞β
ψK(xmk,xnk)
ψ(ε). ()
Sinceβis a Geraghty function, we derive thatψ(K(xmk,xnk))→. Consequently, we have
d(xmk,xnk)→, which is a contradiction. Hencelimm,n→∞d(xn,xm) = , and the sequence {xn}is Cauchy for anys≥.
By completeness of (X,d), there existsu∈Xsuch that
lim
n→∞d(xn,u) =d(u,u) =n,m→∞lim d(xn,xm) = .
Now, ifTis continuous, then
Tu=T
lim
n→∞xn
= lim
n→∞Txn=n→∞limxn+=u,
anduis a fixed point forT.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥and T:X→ X be a generalized rationalα-ψ-φ-Geraghty contractive of mapping type(I)such that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Xisα-regular anddis continuous.
Proof Verbatim of the proof of Theorem ., we conclude that the iterative sequence{xn} is Cauchy and converges tou∈X. SinceXisα-regular, then, as in the proof of Theo-rem ., there exists a subsequence{xnk}of{xn}such that
ψd(xnk+,Tu)
<
sψ
K(xnk,u)
+LφN(xnk,u)
, ()
whereK(xnk,u) =max{d(xnk,u),
d(xnk,xnk+)d(u,Tu)
+d(xnk,u) ,
d(xnk,xnk+)d(u,Tu)
+d(xnk+,Tu) }.
Hencelimk→∞K(xnk,u) = and as in the proof of Theorem .limk→∞N(xnk,u) = . Thus taking the limit ask→ ∞on both sides of () and keeping in mind thatψandd
are continuous we haveψ(d(u,Tu))≤. Henced(u,Tu) = ; thereforeTu=u.
Theorem . Adding condition(H)to the hypotheses of Theorem. (or Theorem.),
we obtain uniqueness of the fixed point of T.
Proof As in the proof of Theorem ., we suppose thatx∗andy∗are two fixed points ofT. Then, clearly, we haveK(x∗,y∗) =d(x∗,y∗) andN(x∗,y∗) = . So, we have
ψdx∗,y∗≤ψsdTx∗,Ty∗
≤αx∗,y∗ψsdTx∗,Ty∗
≤βψdx∗,y∗ψdx∗,y∗
<
sψ
dx∗,y∗,
which is a contradiction.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥and T :
X→X be a generalized rationalα-ψ-φ-Geraghty contractive mapping of type(II)such that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Proof Verbatim of the lines in the proof of Theorem ., we construct a sequence{xn} ⊂X wherexn+=Txnfor whichα(xn,xn+)≥ andxn=xn+for alln∈N. Moreover, by using
the fact thatT is a generalized rationalα-ψ-φ-Geraghty contractive mapping of type (II) and the property ofψwe have
ψsd(xn+,xn+)
≤α(xn,xn+)ψ
sd(Txn,Txn+)
≤βψQ(xn,xn+)
ψQ(xn,xn+)
+LφN(xn,xn+)
. ()
By using the same arguments as in the proof of Theorem ., we derive that
ψsd(xn+,xn+)
<
sψ
Q(xn,xn+)
where
Q(xn,xn+) =max
d(xn,xn+),
d(xn,xn+)d(xn+,xn+) +ds(xn,xn+)ds(xn+,xn+)
+s[d(xn,xn+) +d(xn+,xn+)]
.
Sinced(xn,xn+)≤s[d(xn,xn+) +d(xn+,xn+)], we have
d(xn,xn+)≥
d(xn,xn+)d(xn,xn+)
+d(xn,xn+)
≥ d(xn,xn+)d(xn,xn+)
+s[d(xn,xn+) +d(xn+,xn+)]
.
Hence, we getQ(xn,xn+) =d(xn,xn+).
By using () we get ψ(sd(x
n+,xn+))≤ ψ(d(xn,xn+)). Since ψ is increasing we
haved(xn+,xn+) < sd(xn,xn+). Ifs> then, as in the proof of Theorem ., by using
Lemma ., we conclude that{xn} is a Cauchy sequence andlimn,m→∞d(xn,xm) = . If
s= , by verbatim of the proof of Theorem ., we deduce that{xn}is a Cauchy sequence. Since (X,d) is complete, there exists u ∈ X such that = limn,m→∞d(xn,xm) =
limn→∞d(xn,u) =d(u,u). Now, sinceTis continuous,Tu=T(limn→∞xn) =limn→∞Txn=
limn→∞xn+=uanduis a fixed point forT.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥and T:
X→X be a generalized rationalα-ψ-φ-Geraghty contractive mapping of type(II)such that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Xisα-regular anddis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Proof By following the proof of Theorem . line by line, we see that{xn}converges to
u∈X. Due to the fact thatXisα-regular and by following the lines of the proof of Theo-rem . there exists a subsequence{xnk}of{xn}such that
ψd(xnk+,Tu)
<
sψ
Q(xnk,u)
+LφN(xnk,u)
, ()
where
Q(xnk,u) =max
d(xnk,u),
d(xnk,xnk+)d(u,Tu) +ds(xnk,Tu)ds(u,xnk+)
+s[d(xnk,xnk+) +d(u,Tu)]
.
Note thatlimk→∞Q(xnk,u) = andlimk→∞N(xnk,u) = . Thus taking the limit asn→ ∞ on both sides of () and keeping in mind that ψ and d are continuous we have
ψ(d(u,Tu)) = , and sod(u,Tu) = . Thusu=Tu.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥and T:
X→X be a mapping. Suppose that there exist a functionα:X×X→[,∞),ψ ∈,
β∈Fssuch that
for all x,y∈X.Suppose also that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Proof By (ii) there existsx∈Xsuch thatα(x,Tx)≥. Define a sequence{xn} ⊂Xby
xn+=Txnfor alln∈N. AsT is triangularα-orbital admissible, by Lemma . we have α(xn,xn+)≥ for alln∈N. Notice that if there exists a natural numbernsuch that
xn=xn+, then the proof is complete. To avoid this trivial case, from now on, we assume thatxn=xn+for alln∈N.
SinceTsatisfies () we have
ψsd(xn+,xn+)
≤α(xn,xn+)ψ
sd(Txn,Txn+)
≤βψd(xn,xn+)
ψd(xn,xn+)
.
Thus, we have
ψsd(xn+,xn+)
<
sψ
d(xn,xn+)
<ψd(xn,xn+)
. ()
Sinceψis increasing, we haved(xn+,xn+) <sd(xn,xn+).
Case (i): Ifs> , then, since s > andss =s < , by Lemma .,{xn}is a Cauchy se-quence andlimn,m→∞d(xn,xm) = .
Case (ii): If s = , then as d(xn+,xn+) <d(xn,xn+), there exists r≥ such that
limn→∞d(xn,xn+) =r. Ifr> then taking the limit asn→ ∞on both sides of
ψd(xn+,xn+)
≤βψd(xn,xn+)
ψd(xn,xn+)
we getlimn→∞β(ψ(d(xn,xn+))) = and asβis a Geraghty function, we derive that
lim
n→∞d(xn,xn+) = . ()
That is,r= . In what follows we shall prove that{xn}is a Cauchy sequence. Indeed we will prove thatlimm,n→∞d(xn,xm) = . Suppose, on the contrary, that there existε> and
corresponding subsequences{nk}and{mk}ofNsatisfyingnk>mk>kfor which
d(xmk,xnk)≥ε, ()
wherenk,mkare chosen as the smallest integers satisfying (), that is,
d(xmk,xnk–) <ε. ()
By (), (), and the triangle inequality, we easily derive that
Using () and the squeeze theorem we get
lim
k→∞d(xmk,xnk) =ε. ()
AsTsatisfies (), we have
ψd(xmk+,xnk+)
≤α(xmk,xnk)ψ
d(Txmk,Txnk)
≤βψd(xmk,xnk)
ψd(xmk,xnk)
. ()
Taking the limit ask→ ∞for () we getlimk→∞β(ψ(d(xmk,xnk))) = . Thus
lim
k→∞d(xmk,xnk) = .
Therefore,{xn}is a Cauchy sequence fors≥. By completeness of (X,d), there existsu∈X such that =limn,m→∞d(xn,xm) =limn→∞d(xn,u) =d(u,u).
SinceT is continuous,Tu=T(limn→∞xn) =limn→∞Txn=limn→∞xn+=uanduis a
fixed point forT.
Theorem . Let(X,d)be a complete b-metric-like space with constant s≥and T :
X→X be a mapping.Suppose that there exist a functionα:X×X→[,∞),ψ∈,and β∈Fssuch that
α(x,y)ψsd(Tx,Ty)≤βψd(x,y)ψd(x,y), ()
for all x,y∈X.Suppose also that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Xisα-regular anddis continuous.
Then T has a fixed point,u∈X with d(u,u) = .
Theorem . Adding condition(H)to the hypotheses of Theorem. (or Theorem.),
we obtain the uniqueness of the fixed point of T.
Remark . Notice that we get several corollaries by replacing the auxiliary functions
ψ andβin a proper way. In particular, by takingψ(t) =twe find the extended version of several existing results.
3 Expected consequences
In this section, we shall consider some immediate consequences of our main results. The following result is obtained by lettingL= in Theorem . or ..
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥and T:X→ X be a mapping.Suppose that there existα:X×X→[,∞),ψ∈,β∈Fssuch that
for all x,y∈X,where
M(x,y) =max
d(x,y),d(x,Tx),d(y,Ty),d(x,Ty) +d(y,Tx)
s . ()
Suppose also that:
(i) Tis triangularα-orbital admissible; (ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous or(iii)Xisα-regular anddis continuous.
Then T has a fixed point.
Adding condition (H) to the hypothesis of Corollary ., we guarantee the uniqueness of the fixed point.
Again by lettingL= in Theorem . and Theorem . we get two more corollaries as Corollary .. We skip the details regarding the volume of the paper.
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥,T:X→X
be a map andα:X×X→[,∞)be a function.Suppose that T satisfies at least one of the following conditions:
(a) α(x,y)d(Tx,Ty)≤sM(x,y); (b) α(x,y)d(Tx,Ty)≤
sK(x,y);
where M(x,y),K(x,y)are defined as in(), ().Suppose also that: (i) Tis triangularα-orbital admissible;
(ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous or(iii)Xisα-regular anddis continuous.
Then T has a unique fixed point u∈X with d(u,u) = .
Proof It is sufficient to takeL= ,ψ(t) =t, andβ(t) =sin Theorem . and Theorem . (and thus, Theorem . or Theorem ., Theorem . or Theorem ., respectively).
Adding condition (H) to the hypothesis of Corollary ., we guarantee the uniqueness of the fixed point.
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥,T:X→X be a map,andα:X×X→[,∞)be a function.Suppose that T satisfies at least one of the following conditions:
(c) α(x,y)d(Tx,Ty)≤sQ(x,y),
where M(x,y),K(x,y),Q(x,y)are defined as in().Suppose also that: (i) Tis triangularα-orbital admissible;
(ii) there existsx∈Xsuch thatα(x,Tx)≥;
(iii) Tis continuous or(iii)Xisα-regular anddis continuous.
Then T has a fixed point u∈X with d(u,u) = .
Proof It is sufficient to takeL= ,ψ(t) =t, andβ(t) =
sin Theorem . or Theorem .,
3.1 For standardb-metric-like spaces
If we setα(x,y) = for allx,y∈Xin Theorem ., then we derive the following results.
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥such that d is continuous and T:X→X be a mapping.Suppose that there existψ,φ∈,β∈Fs,and
some L≥such that
ψsd(Tx,Ty)≤βψM(x,y)ψM(x,y)+LφN(x,y), ()
for all x,y∈X,where
M(x,y) =max
d(x,y),d(x,Tx),d(y,Ty),d(x,Ty) +d(y,Tx)
s and ()
N(x,y) =minds(x,Ty),ds(y,Tx),d(x,Tx),d(y,Ty). ()
Then T has a unique fixed point u∈X with d(u,u) = .
If we setα(x,y) = for allx,y∈Xin Theorem ., then we derive the following results.
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥such that d is continuous and T:X→X be a mapping.Suppose that there existψ,φ∈,β∈Fs,and
some L≥such that
ψsd(Tx,Ty)≤βψK(x,y)ψK(x,y)+LφN(x,y), ()
for all x,y∈X,where
K(x,y) =max
d(x,y),d(x,Tx)d(y,Ty) +d(x,y) ,
d(x,Tx)d(y,Ty)
+d(Tx,Ty) and ()
N(x,y) =minds(x,Ty),ds(y,Tx),d(x,Tx),d(y,Ty). ()
Then T has a unique fixed point u∈X with d(u,u) = .
If we setα(x,y) = for allx,y∈Xin Theorem ., then we derive the following results.
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥such that d is continuous and T:X→X be a mapping.Suppose that there existψ,φ∈,β∈Fs,and
some L≥such that
ψsd(Tx,Ty)≤βψQ(x,y)ψQ(x,y)+LφN(x,y), ()
for all x,y∈X,where
Q(x,y) =max
d(x,y),d(x,Tx)d(y,Ty) +d
s(x,Ty)ds(y,Tx)
+s(d(x,Tx) +d(y,Ty)) and ()
N(x,y) =minds(x,Ty),ds(y,Tx),d(x,Tx),d(y,Ty). ()
If takeL= in Corollaries .-., we get three more consequences. Regarding the vol-ume of the paper, we skip the details.
Corollary . Let(X,d)be a complete b-metric-like space with constant s≥such that d
is continuous and T:X→X be mapping.Suppose that there existψ∈andβ∈Fssuch
that
ψsd(Tx,Ty)≤βψd(x,y)ψd(x,y), ()
for all x,y∈X.Then T has a unique fixed point u∈X with d(u,u) = .
Proof It follows from Theorem . byα(x,y) = for allx,y∈X.
3.2 Forb-metric-like spaces endowed with a partial order
In this section, from our main results, we shall derive easily various fixed point results on ab-metric-like space endowed with a partial order. We, first, recall some notions.
Definition . Let (X,) be a partially ordered set andT:X→Xbe a given mapping. We say thatT is nondecreasing with respect toif
x,y∈X, xy ⇒ TxTy.
Definition . Let (X,) be a partially ordered set. A sequence{xn} ⊂Xis said to be nondecreasing (respectively, nonincreasing) with respect toifxnxn+(respectively,
xn+xnfor alln).
Definition . Let (X,) be a partially ordered set anddbe ab-metric-like onX. We say that (X,,d) is regular if for every nondecreasing (respectively, nonincreasing) sequence {xn} ⊂Xsuch thatxn→x∈Xasn→ ∞, there exists a subsequence{xnk}of{xn}such thatxnkx(respectively,xnkx) for allk.
We have the following result.
Corollary . Let(X,)be a partially ordered set(which does not contain an infinite totally unordered subset)and d be a b-metric-like on X with constant s≥such that(X,d)
is complete.Let T:X→X be a nondecreasing mapping with respect to.Suppose that
there existψ∈,β∈Fssuch that
ψsd(Tx,Ty)≤βψM(x,y)ψM(x,y), ()
for all x,y∈X with xy or yx where M(x,y)is defined as in().Suppose also that the following conditions hold:
(i) there existsx∈Xsuch thatxTx;
(ii) T is continuous or(ii)(X,,d)is regular anddis continuous.
Proof Define the mappingα:X×X→[,∞) by
α(x,y) =
ifxyorxy, otherwise.
Clearly,Tsatisfies (), that is,
α(x,y)ψsd(Tx,Ty)≤βψM(x,y)ψM(x,y),
for allx,y∈X. From condition (i), we haveα(x,Tx)≥. Moreover, for allx,y∈X, from
the monotone property ofT, we have
α(x,y)≥ ⇒ xyorxy ⇒ TxTyorTxTy ⇒ α(Tx,Ty)≥.
Hence, the self-mappingTisα-admissible. Similarly, we can prove thatTis triangularα -admissible and so triangularα-orbital admissible. Now, ifT is continuous, the existence of a fixed point follows from Corollary .. Suppose now that (X,,d) is regular. Let{xn} be a sequence inXsuch thatα(xn,xn+)≥ for allnandxn→x∈Xasn→ ∞. From the regularity hypothesis and asXdoes not contain an infinite totally unordered subset, there exists a subsequence{xnk}of{xn}such thatxnkxorxxnkfor allk. This implies from the definition ofαthatα(xnk,x)≥ for allk. In this case, the existence of a fixed point
follows again from Corollary ..
In an analogous way, we derive the following results from Theorem . and Theo-rem ., respectively.
Corollary . Let(X,)be a partially ordered set(which does not contain an infinite totally unordered subset)and d be a b-metric-like on X with constant s≥such that(X,d)
is complete.Let T:X→X be a nondecreasing mapping with respect to.Suppose that
there existψ∈,β∈Fssuch that
ψsd(Tx,Ty)≤βψK(x,y)ψK(x,y), ()
for all x,y∈X with xy or yx where M(x,y)is defined as in().Suppose also that the following conditions hold:
(i) there existsx∈Xsuch thatxTx;
(ii) T is continuous or(ii)(X,,d)is regular anddis continuous.
Then T has a fixed point u∈X with d(u,u) = .
Corollary . Let(X,)be a partially ordered set(which does not contain an infinite totally unordered subset)and d be a b-metric-like on X with constant s≥such that(X,d)
is complete.Let T:X→X be a nondecreasing mapping with respect to.Suppose that
there existψ∈,β∈Fssuch that
ψsd(Tx,Ty)≤βψQ(x,y)ψQ(x,y), ()
(i) there existsx∈Xsuch thatxTx;
(ii) T is continuous or(ii)(X,,d)is regular anddis continuous.
Then T has a fixed point u∈X with d(u,u) = .
Corollary . Let(X,)be a partially ordered set(which does not contain an infinite totally unordered subset)and d be b-metric-like on X with constant s≥such that(X,d)
is complete.Let T:X→X be a nondecreasing mapping with respect to.Suppose that
there existψ∈,β∈Fssuch that
ψsd(Tx,Ty)≤βψd(x,y)ψd(x,y), ()
for all x,y∈X with xy or yx.Suppose also that the following conditions hold: (i) there existsx∈Xsuch thatxTx;
(ii) T is continuous or(ii)(X,,d)is regular anddis continuous.
Then T has a fixed point u∈X with d(u,u) = .
Example . LetX= [,∞) defined:X×X→[,∞) byd(x,y) =|x–y|. Then (X,d)
is completeb-metric (sob-metric-like) space with constants= . DefineT:X→Xand
α(x,y) :X×X→[,∞) as follows:
Tx=
x
ifx∈[, ],
x
x+ ifx∈(,∞),
and
α(x,y) =
ifx,y∈[, ], otherwise.
It is clear thatTis triangularα-orbital admissible and we haveα(,T)≥. Moreover,X
isα-regular anddis continuous.
Letψ(t) =t,β(t) = then clearlyψ∈andβ∈F. Moreover,T satisfies () for the following reason: ifx,y∈[, ], then
α(x,y)ψd(Tx,Ty)=
x
–
y
=(x–y)
=β
ψd(x,y)ψ(d(x,y).
Otherwise,
α(x,y)ψd(Tx,Ty)= ≤βψd(x,y)ψ(d(x,y).
Therefore, by Theorem .,Thas a fixed pointx= .
Example . LetX={, , }defined:X×X→[,∞) byd(x,y) = (max{x,y}). Then (X,d) is completeb-metric-like space with constants= –= such thatdis continu-ous. DefineT:X→XbyT={(, ), (, ), (, )}.
Letψ(t) =t,β(t) =
e
–t or β(t) =
+t, then clearly
ψ ∈ andβ∈F
. Note that
4 Conclusion
It is clear that we can list several more results by replacing theb-metric-like space, with some other abstract space, such as ab-metric space, a metric space, a metric-like space, a partial metric space, and so on.
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.
Received: 16 July 2015 Accepted: 15 September 2015
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