R E S E A R C H
Open Access
Common fixed points of a generalized
ordered
g
-quasicontraction in partially
ordered metric spaces
Xiaolan Liu
1,2*and Siniša Ješi´c
3*Correspondence:
1Department of Mathematics,
Sichuan Normal University, Chengdu, Sichuan 610068, China
2School of Science, Sichuan
University of Science and Engineering, Zigong, Sichuan 643000, China
Full list of author information is available at the end of the article
Abstract
The concept of a generalized orderedg-quasicontraction is introduced, and some fixed and common fixed point theorems for ag-nondecreasing generalized ordered
g-quasicontraction mapping in partially ordered complete metric spaces are proved. We also show the uniqueness of the common fixed point in the case of a generalized orderedg-quasicontraction mapping. Finally, we prove fixed point theorems for mappings satisfying the so-called weak contractive conditions in the setting of a partially ordered metric space. Presented theorems are generalizations of very recent fixed point theorems due to Golubovi´cet al.(Fixed Point Theory Appl. 2012:20, 2012). MSC: 47H10; 47N10
Keywords: G-nondecreasing; generalized orderedg-quasicontraction; coincidence point; common fixed point; comparable mappings
1 Introduction
The Banach fixed point theorem for contraction mappings has been extended in many directions (cf.[–]). Very recently Golubovićet al.[] presented some new results for ordered quasicontractions andg-quasicontractions in partially ordered metric spaces.
Recall that if (X,) is a partially ordered set andf :X→Xis such that for x,y∈X, xyimpliesfxfy, then a mappingFis said to be non-decreasing. The main result of Golubovićet al.[] is the following fixed point theorem.
Theorem .(See [], Theorem ) Let(X,d,)be a partially ordered metric space and let f,g:X→X be two self-maps on X satisfying the following conditions:
(i) fX⊂gX; (ii) gXis complete; (iii) f isg-nondecreasing;
(iv) f is an orderedg-quasicontraction; (v) there existsx∈Xsuch thatgxfx;
(vi) if{gxn}is a nondecreasing sequence that converges to somegz∈gX,thengxngz
for eachn∈Nandgzg(gz).
Then f and g have a coincidence point, i.e.,there exists z∈X such that fz=gz.If,in addition,
(vii) f andgare weakly compatible[, ],i.e.,fx=gximpliesfgx=gfxfor eachx∈X, then they have a common fixed point.
An open problem is to find sufficient conditions for the uniqueness of the common fixed point in the case of an orderedg-quasicontraction in Theorem ..
In Section of this article, we introduce generalized orderedg-quasicontractions in partially ordered metric spaces and prove the respective (common) fixed point theorems which generalize the results of Theorem ..
In Section of this article, the uniqueness of a common fixed point theorem is obtained when for allx,u∈X, there existsa∈Xsuch thatfais comparable tofxandfuin addition to the hypotheses in Theorem . of Section . Our results are an answer to finding suf-ficient conditions for the uniqueness of a common fixed point in the case of an ordered g-quasicontraction in Theorem .. Finally, two examples show that the comparability is a sufficient condition for the uniqueness of a common fixed point in the case of an ordered g-quasicontraction, so our results are extensions of known ones.
In Section of this article, we consider weak contractive conditions in the setting of a partially ordered metric space and prove respective common fixed point theorems.
2 Common fixed points of a generalized orderedg-quasicontraction
We start this section with the following definitions. Consider a partially ordered set (X,) and two mappingsf :X→Xandg:X→Xsuch thatf(X)⊂g(X).
Definition . (See []) We will say that the mappingf isg-nondecreasing (resp.,g -nonincreasing) if
gxgy ⇒ fxfy ()
(resp.,gxgy⇒fxfy) holds for eachx,y∈X.
Definition .(See []) We will say that the mappingf is an orderedg-quasicontraction if there existsα∈(, ) such that for eachx,y∈Xsatisfyinggygx, the inequality
d(fx,fy)≤α·M(x,y)
holds, where
M(x,y) =maxd(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx).
Definition . We will say that the mappingfis a generalized orderedg-quasicontraction if there is a continuous and non-decreasing functionψ: [, +∞)→[, +∞) withψ(s+t)≤
ψ(s) +ψ(t) for eachs,t> ,ψ(t)≥tfort≥ and there existsα∈(, )
ψd(fx,fy)≤αmaxψd(gx,gy),ψd(gx,fx),ψd(gy,fy),
ψd(gx,fy),ψd(gy,fx) ()
for allx,y∈Xfor whichgxgy;
For arbitraryx∈X, one can construct the so-called Jungck sequence{yn}in the follow-ing way: Denotey=fx∈f(X)⊂g(X); there existsx∈Xsuch thatgx=y=fx; now
y=fx∈f(X)⊂g(X) and there existsx∈Xsuch thatgx=y=fx and the procedure
can be continued.
Theorem . Let(X,)be a partially ordered set and suppose there is a metric d on X such that(X,d)is a complete metric space.Let f,g:X→X be two self-maps on X satisfying the following conditions:
(i) f(X)⊂g(X); (ii) g(X)is closed;
(iii) f is ag-nondecreasing mapping;
(iv) f is a generalized orderedg-quasicontraction; (v) there exists anx∈Xwithgxfx;
(vi) {g(xn)} ⊂Xis a non-decreasing sequence withg(xn)→gzing(X),thengxngz, gzg(gz),∀nhold.
Then f and g have a coincidence point.Further,if f and g are weakly compatible,then f and g have a common fixed point.
Proof Letx∈Xbe such thatgxfx. Sincef(X)⊂g(X), we can choosex∈Xso that
gx=fx. Again fromf(X)⊂g(X), we can choosex∈Xsuch thatgx=fx. Continuing
this process, we can construct a Jungck sequence{yn}in X such that
gxn+=fxn=yn, ∀n≥. ()
Sincegxfxandgx=fx, we havegxgx. Then by (),
fxfx. ()
Thus, by (),gxgx. Again by (),
fxfx, ()
that is,gxgx. Continuing this process, we obtain
fxfxfxfx · · · fxnfxn+. ()
LetO(yk,n) ={yk,yk+, . . . ,yk+n}. We will claim that{yn}is a Cauchy sequence. To prove our claim, we follow the arguments of Das and Naik []. Fixk≥ andn∈ {, , . . .}. If diam[O(yk;n)] = , thenyk=yk+, which yields that{yn}is a constant sequence and also a Cauchy sequence. Then our claims holds. Thus we suppose thatdiam[O(yk;n)] > . Now, fori,jwith ≤i<j, by (), we have
ψd(yi,yj)
=ψd(fxi,fxj)
≤αmaxψd(gxi,gxj)
,ψd(gxi,fxi)
,ψd(gxj,fxj)
,ψd(xi,fxj)
,ψd(gxj,fxi)
=αmaxψd(yi–,yj–)
,ψd(yi–,yi)
,ψd(yj–,yj)
,ψd(yi–,yj)
,ψd(yj–,yi)
≤αψdiamO(yi–;j–i+ )
,
and so
ψd(yi,yj)
≤αψdiamO(yi–;j–i+ )
. ()
Now, for somei,jwithk≤i<j≤k+n,diam[O(yk;n)] =d(yi,yj). Ifi>kby () and (), then we have
ψdiamO(yk;n)
≤αψdiamO(yi–;j–i+ )
≤αψdiamO(yk;n)
. ()
It follows that ψ(diam[O(yk;n)]) = , as diam[O(yk;n)]≤ψ(diam[O(yk;n)]) = , then diam[O(yk;n)] = . It is a contradiction! Thus,
diamO(yk;n)
=d(yk,yj) forjwithk<j≤k+n. ()
Also, by () and (), we have
ψdiamO(yk;n)
=ψd(yk,yj)
≤αψdiamO(yk–;j–k+ )
≤αψdiamO(yk–;n+ )
. ()
Using the triangle inequality, by (), () and (), we obtain that
ψdiamO(yl;m)
=ψd(yl,yj)
≤ψd(yl,yl+) +d(yl+,yj)
≤ψd(yl,yl+)
+ψd(yl+,yj)
≤ψd(yl,yl+)
+αψdiamO(yl+;m– )
≤ψd(yl,yl+)
+αψdiamO(yl;m)
, ()
and so
ψdiamO(yl;m)
≤
–αψ
d(yl,yl+)
. ()
As a result, we have
ψdiamO(yk;n)
≤αψdiamO(yk–;n+ )
≤α·αψdiamO(yk–;n+ )
≤αkψdiamO(y;n+k)
≤ αk
–αψ
d(y,y)
Now let> , there exists an integernsuch that
αkψd(y,y)
< ( –α) for allk>n. ()
Form>n>n, we have
ψd(ym,yn)
≤ψdiamO(yn;m–n)
≤ αn
–αψ
d(y,y)
<. ()
Sinceψ(t)≥tast> , thend(ym,yn)≤ψ(d(ym,yn)) <. Therefore,{yn}is a Cauchy se-quence.
Since by () we have{fxn=gxn+} ⊆g(X) andg(X) is closed, then there existsz∈Xsuch that
lim
n→∞gxn=gz. ()
Now we show thatzis a coincidence point off andg. Since from condition (iv) and () we havegxngzfor alln, then by the triangle inequality and (), we have that
ψd(fz,gz)≤ψd(gz,fxn) +d(fxn,fz)
≤ψd(gz,fxn)
+ψd(fxn,fz)
≤ψd(gz,fxn)
+αmaxψd(gxn,gz)
,ψd(gxn,fxn)
,
ψd(gz,fz),ψd(gxn,fz)
,ψd(gz,fxn)
. ()
So, letting n → ∞ yields ψ(d(fz,gz))≤ αψ(d(fz,gz)). Hence ψ(d(fz,gz)) = , hence d(fz,gz) = , which yieldsfz=gz. Thus we have proved thatf andghave a coincidence point.
Suppose now thatf andgcommute atz. Setw=fz=gz. Then
fw=f(gz) =g(fz) =gw. ()
Since from (vi) we have thatgzg(gz) =gwand asfz=gzandfw=gw, from () we have that
ψd(fz,fw)≤αmaxψd(gz,gw),ψd(gz,fz),ψd(gw,fw),
ψd(gz,fw),ψd(gw,fz)
=αψd(gz,gw). ()
Hence,ψ(d(fz,fw)) = , that is,d(w,fw) = . Therefore,
fw=gw=w. ()
Accordingly, we can also obtain the results similar to Theorem in [].
Theorem . Let the conditions of Theorem.be satisfied,except that(iii), (v)and(vi) are,respectively,replaced by:
(iii) f is ag-nonincreasing mapping;
(v) there existsx∈Xsuch thatfxandgxare comparable;
(vi) if{gxn}is a sequence ing(X)which has comparable adjacent terms and that converges
to somegz∈gX,then there exists a subsequencegxnkof{gxn}having all the terms com-parable withgzandgzis comparable withggz.Then all the conclusions of Theorem. hold.
Proof Regardless of whetherfxgxorgxfx(condition (v)), Lemma of []
im-plies that the adjacent terms of the Jungck sequence{yn}are comparable. This is again sufficient to imply that{yn}is a Cauchy sequence. Hence, it converges to somegz∈gX.
By (vi), there exists a subsequenceynk =fxnk =gxnk+,k∈N, having all the terms
com-parable withgz. Hence, we can apply the contractive condition to obtain
ψd(fz,gz)≤ψd(gz,fxnk)
+ψd(fz,fxnk)
≤ψd(gz,fxnk)
+αmaxψd(gz,gxnk)
,ψd(gz,fz),
ψd(gxnk,fxnk)
,ψd(gz,fxnk)
,ψd(gxnk,fz)
.
Letting k → ∞, it yields that ψ(d(fz,gz))≤αψ(d(gz,fz)), then ψ(d(fz,gz)) = . Thus d(fz,gz) = . It follows that fz=gz=w. The rest of conclusions follow in the same way
as in Theorem ..
Corollary . (a)Let(X,)be a partially ordered set and suppose there is a metric d on X such that(X,d)is a complete metric space.Let f :X→X be a nondecreasing self-map such that for someα∈(, )
d(fx,fy)≤αmaxd(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
for all x,y∈X for which xy.Suppose also that either
(i) {xn} ⊂Xis a non-decreasing sequence withxn→uinX,thenxnu,∀nhold,or
(ii) f is continuous.
If there exists an x∈X with xfx,then f has a fixed point.
(b)The same holds if f is nonincreasing,there exists xcomparable with fxand(i)is
replaced by
(i) if a sequence{xn}converging to someu∈Xhas every two adjacent terms comparable,
then there exists a subsequence{xnk}having each term comparable withx.
Proof (a) If (i) holds, then takeψ=Iandg=I(I= the identity mapping) in Theorem .. If (ii) holds, then from () withg=I, we get
z= lim
n→∞xn+=nlim→∞fxn=f
lim
(b) Letube the limit of the Picard sequence{fnx}and letfnkx
be a subsequence having
all the terms comparable withu. Then we can apply the contractivity condition to obtain
ψd(fu,u)≤ψdu,fnk+x
+dfu,fnk+x
≤ψdu,fnk+x
+ψdfu,fnk+x
≤ψdu,fnk+x
+αmaxψdu,fnkx
,ψd(u,fu),
ψdfnkx
,fnk+x
,ψdu,fnk+x
,ψdfu,fnkx
.
Lettingk→ ∞, we have that
ψd(fu,u)≤αmax,ψd(u,fu), , ,ψd(u,fu)
=αψd(u,fu).
It follows that ψ(d(fu,u)) = . Thusd(fu,u) = asd(fu,u)≤ψ(d(fu,u)) = . Therefore, fu=u.
Note also that instead of the completeness ofX, itsf-orbitally completeness is sufficient
to obtain the conclusion of the corollary.
3 Uniqueness of a common fixed point offandg
The following theorem gives the sufficient condition for the uniqueness of a common fixed point off andg.
Theorem . In addition to the hypotheses of Theorem.,suppose that for all x,u∈X, there exists a∈X such that
fa is comparable to fx and fu. ()
Then f and g have a unique common fixed point.
Proof Since a set of common fixed points off andg is not empty due to Theorem ., assume now thatxanduare two common fixed points off andg,i.e.,
fx=gx=x, fu=gu=u. ()
We claim thatgx=gu.
By assumption, there existsa∈Xsuch thatfais comparable tofxandfu. Define a se-quence{gan}such thata=aand
gan=fan– for alln. ()
Further, setx=xandu=uand in the same way define{gxn}and{gun}such that
Sincefx(=gx=gx) is comparable tofa(=fa=ga) andf isg-nondecreasing, it is easy
to show
gxga. ()
Recursively, we can get that
gangx for alln. ()
By (), we have that
ψd(gan+,gx)
=ψd(fan,fx)
≤αmaxψd(gan,gx)
,ψd(gan,fan)
,ψd(gx,fx),
ψd(gan,fx)
,ψd(gx,fan)
. ()
By the proof of Theorem ., we obtain that {gan} is a convergent sequence, and there existsga¯such thatgan→ga¯. Lettingn→ ∞in () andψ is continuous, we can obtain that
lim n→∞ψ
d(gan+,gx)
=ψd(ga¯,gx)
≤αmaxψd(ga¯,gx), , ,ψd(ga¯,fx),ψd(gx,ga¯)
=αψd(ga¯,gx).
Therefore, we obtain
ψd(ga¯,gx)= .
Sinceψ(t)≥tast≥, thend(ga¯,gx) = and hence
ga¯=gx. ()
Similarly, we can show that
lim n→∞ψ
d(gan+,gu)
=ψd(ga¯,gu)
≤αmaxψd(ga¯,gu), , ,ψd(ga¯,fu),ψd(gu,ga¯)
=αψd(ga¯,gu).
Therefore, we obtain
ψd(ga¯,gu)= .
Sinceψ(t)≥tast≥, thend(ga¯,gu) = and hence
Thus, from () and (), we havegx=gu. It follows that
x=fx=gx=gu=fu=u. ()
It means thatxis the unique common fixed point off andg.
Remark . Theorem . can be considered as an answer to Theorem in []. We find the sufficient conditions for the uniqueness of the common fixed point in the case of an orderedg-quasicontraction. In this paper, condition (vi) in Theorem . is weaker than the orderedg-quasicontraction in []. Whenψ=I(I= the identity mapping), our condition (vi) reduces to the orderedg-quasicontraction in [].
Example . LetX={(, ), (, )}, let (a,b)(c,d) if and only ifa≤candb≥d, and let dbe the Euclidean metric. We define the functions as follows:
f(x,y)=x, y– , g(x,y)=x,y– for all (x,y)∈X.
Letφ,ψ: [,∞)→[,∞) be given by
ψ(t) =
t for allt∈[,∞).
Obviously, for (, ) and (, )∈X, butf((, )) = (, ) is not comparable to g((, )) = (, ). However,f andghave two common fixed points (, ) and (, ) since
f(, )=g(, ) = (, ), f(, )=g(, )= (, ).
Example . LetX= [–∞, +∞) with the usual metricd(x,y) =|x–y|for allx,y∈X. Let f :X→Xandg:X→Xbe given by
f(x) = x
, g(x) = x
for allx,y,z,w∈X. Letφ,ψ: [,∞)→[,∞) be given by
ψ(t) = t for allt∈[,∞).
It is easy to check that all the conditions of Theorem . are satisfied.
ψd(fx,fy)= |x–y|
≤·α·
|x–y|
≤max
·
|x–y|, ·
x– x
, ·
y– y
,
· x–
y
, ·
y– x
=maxψd(gx,gy),ψd(gx,fx),ψd(gy,fy),ψd(gx,fy),ψd(gy,fx).
In addition,∀x,u∈X, there existsa∈Xsuch thatfa=a is comparable tofx=x and fu=u. So, all the conditions of Theorem . are satisfied.
By applying Theorem ., we conclude thatf andghave a unique common fixed point. In fact,f andghave only one common fixed point. It isx= .
4 Weak ordered contractions
We denote bythe set of functionsψ: [, +∞)→[, +∞) satisfying the following hy-potheses:
(ψ) ψis continuous and nondecreasing, (ψ) ψ(t) = if and only ift= .
We denote bythe set of functionsφ: [, +∞)→[, +∞) satisfying the following hy-potheses:
(φ) lims→t+φ(s) > for allt> , (φ) φ(t) = if and only ift= .
Let (X,d) be a metric space and letf,g:X→X. In the article [] (in the setting of partially ordered metric spaces), the authors obtained contractive conditions of the form
ψd(fx,fy)≤ψM(x,y)–φM(x,y), ()
where
M(x;y) =max
d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy) +d(gy,fx)
. ()
We will use here the following more general contractive condition:
M(x,y) =maxd(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx). ()
We begin with the following result.
Theorem . Let (X,d,)be a partially ordered metric space and let f and g be self-mappings of X satisfying the following conditions:
(i) f(X)⊂g(X); (ii) g(X)is complete; (iii) f isg-nondecreasing;
(iv) f andgsatisfy the following condition:
ψd(fx,fy)≤ψM(x,y)–φM(x,y) ()
for allx,y∈Xsuch thatgygx,whereψ∈,φ∈and
M(x,y) =maxd(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx). ()
Suppose that,in addition,
(vi) ψ(s+t)≤ψ(s) +ψ(t)for eachs,t> ; (vii) limt→+∞φ(t) =∞;
(viii) there existsx∈Xsuch thatgxfx;
(ix) if{gxn}is a nondecreasing sequence that converges to somegz∈gX,thengxngz
for eachn∈Nandgzg(gz).
Then f and g have a coincidence point.If,in addition,
(x) f andgare weakly compatible,then they have a common fixed point.
Further,if
(xi) for arbitraryv,w∈X,there existsy∈Xsuch thatfyis comparable tofvandfw, thenf andghave a unique common fixed point.
Proof As in the proof of Theorem ., we can construct a nondecreasing Jungck sequence
{yn}with
yn=fxn=gxn+
for alln≥. Denote
O(yk,n) ={yk,yk+,yk+, . . . ,yk+n}, ()
O(yk) ={yk,yk+,yk+, . . . ,yk+n, . . .}. ()
We will prove that the Jungck sequence{yn}is bounded, that is,
diamO(y)
=diam{y,y,y, . . . ,yn, . . .}
≤K ()
for someK∈R. Letk<nbe any fixed positive integer and letdiam(O(yk,n)) =d(yi,yj) for somei,jwithk≤i<j≤k+n. We will show that
ψdiamO(yk,n)
≤ψdiamO(yi–,j–i+ )
–φdiamO(yi–,j–i+ )
. ()
Sincediam(O(yk,n)) =d(yi,yj),yi=fxi,yj=fxjandgxigxj, then from () we have
ψdiamO(yk,n)
=ψd(fxi,fxj)
≤ψM(xi,xj)
–φM(xi,xj)
, ()
where
M(xi,xj) =max
d(gxi,gxj),d(gxi,fxi),d(gxj,fxj),d(gxi,fxj),d(gxj,fxi)
=maxd(yi–,yj–),d(yi–,yi),d(yj–,yj),d(yi–,yj),d(yj–,yi)
.
Sinceyi–,yi,yj–,yj∈O(yi–,j–i+ ), then
M(xi,xj)≤diam
{yi–,yi,yj–,yj}
≤diamO(yi–,j–i+ )
.
So, from (v),
ψM(xi,xj)
–φM(xi,xj)
≤ψdiamO(yi–,j–i+ )
–φdiamO(yi–,j–i+ )
.
Note thatφ(diam(O(yi–,j–i+ ))) > , and so from (),
diamO(yk,n)
<diamO(yi–,j–i+ )
. ()
Now we will show that ifdiam(O(yk,n)) =d(yi,yj), theni=k, that is,
diamO(yk,n)
=d(yk,yj) for somek<j≤k+n. ()
Suppose, to the contrary, thati>k. Then{yi–,yi, . . . ,yj} ⊆ {yk,yk+, . . . ,yi, . . . ,yj}and hence we conclude that
diamO(yk,n)
=d(yi,yj) =diam
O(yi–,j–i+ )
=diamO(yi,j–i)
=diamO(yk,j–k)
.
This contradicts (). Therefore,i=kand so we have proved ().
We will prove that the Jungck sequence {yn} is bounded. From () it follows that diam(O(y,n)) =d(y,yj) for someyj∈ {y,y, . . . ,yn}. By the triangle inequality,
diamO(y,n)
=d(y,yj)≤d(y,y) +d(y,yj).
Now, from (ψ) and (ψ), we get
ψdiamO(y,n)
≤ψd(y,y) +d(y,yj)
≤ψd(y,y)
+ψd(y,yj)
. ()
Sinced(y,yj) =d(fx,fxj) and asgxgxj, from () we have
ψd(y,yj)
≤ψM(x,xj)
–φM(x,xj)
,
where
M(x,xj) =max
d(y,yj–),d(y,y),d(yj–,yj),d(y,yj),d(yj–,y)
.
Clearly,M(x,xj)≤diam{y,y,yj–,yj} ≤diam(O(y,n)). Thus by (v), we get
ψM(x,xj)
–φM(x,xj)
≤ψdiamO(y,n)
–φdiamO(y,n)
.
Now, by (),
ψdiamO(y,n)
≤ψd(y,y)
+ψdiamO(y,n)
–φdiamO(y,n)
.
Hence
φdiamO(y,n)
≤ψd(y,y)
Sincediam({y,y, . . . ,yn})≤diam({y,y, . . . ,yn+}), the sequence{diam(O(y,n))}∞n= is
nondecreasing, and so there exists its limitdiam(O(y)), which is finite or infinite.
Sup-pose thatlimn→∞diam(O(y,n)) = +∞. Then (vii) implies that the left-hand side of ()
becomes unbounded whenntends to infinity, but the right-hand side is bounded, a contra-diction. Therefore,limn→∞diam(O(y,n)) =diam(O(y)) < +∞. Thus we have proved ().
Now we show that{yn}is a Cauchy sequence. For alln≥, set similarly as in (),
O(yn) ={yn,yn+, . . .}.
Clearly,O(yn+)⊂O(yn) and sodiam(O(yn+))≤diam(O(yn)). Therefore,{diam(O(yn))}∞n=
is the monotone decreasing sequence of finite nonnegative numbers and converges to someδ≥.
We will prove thatδ= . Letn≥ ands≥n+ . Sincegxn+gxs, from (),
ψd(yn+,ys)
=ψd(fxn+,fxs)
≤ψM(xn+,xs)
–φM(xn+,xs)
,
where
M(xn+,xs) =max
d(yn,ys–),d(yn,yn+),d(ys–,ys),d(yn,ys),d(ys–,yn+)
.
Sinceyn,yn+,ys–,ys∈ {yn,yn+, . . .}=O(yn), we conclude thatM(xn+,xs)≤diam(O(yn)), and so by (v), we get
ψd(yn+,ys)
≤ψdiamO(yn)
–φdiamO(yn)
. ()
Sincelims→+∞d(yn+,ys) =diam(O(yn+)) andψis continuous, we havelims→+∞ψ(d(yn+,
ys)) =ψ(diam(O(yn+))). Thus, taking the limit in () whens→+∞, we get
ψdiamO(yn+)
≤ψdiamO(yn)
–φdiamO(yn)
. ()
Suppose thatlimn→∞diam(O(yn)) =δ> . Sincediam(O(yn))→δ+ asn→ ∞, then from (φ), we havelimn→∞φ(diam(O(yn))) =q> . Therefore, taking the limits asn→+∞in () and using the continuity ofψ, we get
ψ(δ)≤ψ(δ) –q<ψ(δ),
a contradiction. Therefore,δ= and so we have proved that
lim n→∞diam
{yn,yn+, . . .}
= .
Hence we conclude that{yn}is a Cauchy sequence.
Sinceyn=fxn=gxn+, by the assumption (ii) thatg(X) is complete, there is somez∈X
such that
lim
We show thatfz=gz. Suppose, to the contrary, thatd(fz,gz) > . Condition (ix) implies thatgxngzand we can apply the contractive condition () to obtain
ψd(fz,fxn+)
≤ψM(z,xn+)
–φM(z,xn+)
, ()
where
M(z,xn+) =max
d(gz,gxn+),d(gz,fz),d(gxn+,fxn+),d(gz,fxn+),d(gxn+,fz)
=maxd(gz,fxn),d(gz,fz),d(fxn,fxn+),d(gz,fxn+),d(fxn,fz)
.
By the triangle inequality,
d(gz,fz)≤d(gz,fxn+) +d(fz,fxn+).
Now, from (ψ) and (ψ),
ψd(gz,fz)≤ψd(gz,fxn+) +d(fz,fxn+)
≤ψd(gz,fxn+)
+ψd(fz,fxn+)
.
Hence from () we have
ψd(gz,fz)≤ψd(gz,fxn+)
+ψM(z,xn+)
–φM(z,xn+)
. ()
Sincelimn→∞fxn=gz, for large enoughn, we have
M(z,xn+) =max
d(gz,fz),d(fxn,fz)
.
IfM(z,xn+) =d(gz,fz), then from ()
ψd(gz,fz)≤ψd(gz,fxn+)
+ψd(gz,fz)–φd(gz,fz).
Lettingntend to infinity and using the continuity ofψ, we get
ψd(gz,fz)≤ψd(gz,fz)–φd(gz,fz).
Henceφ(d(gz,fz)) = , a contradiction with (φ) and the assumptiond(gz,fz) > .
Similarly, ifM(z,xn+) =d(fxn,fz), then from ()
ψd(gz,fz)≤ψd(gz,fxn)
+ψd(fxn,fz)
–φd(fxn,fz)
.
Lettingntend to infinity and having in mind thatd(fxn,fz)→d(gz,fz)+, we obtain
ψd(gz,fz)≤ψd(gz,fz)– lim d(fxn,fz)→d(gz,fz)+φ
d(fxn,fz)
and hence we get
lim d(fxn,fz)→d(gz,fz)+
φd(fxn,fz)
a contradiction with (φ). Thus our assumptiond(gz,fz) > is wrong. Therefore, d(gz,
fz) = . Hencegz=fz, that is,zis a coincidence point off andg.
If the condition (x) is fulfilled, putw=fz=gz. We will show thatwis a common fixed point off andg. Sincefz=gzandf andgare weakly compatible, we obtain, by the defini-tion of weak compatibility, thatfgz=gfz. Thus we havefw=gw. Using the condition (ix) thatgzggz=gw, we can apply the contractive condition () to obtain
ψd(fw,fz)≤ψM(w,z)–φM(w,z),
where
M(w,z) =maxd(gw,gz),d(gw,fw),d(gz,fz),d(gw,fz),d(gz,fw)=d(fw,fz).
Thus
ψd(fw,fz)≤ψd(fw,fz)–φd(fw,fz).
Henceφ(d(fw,fz)) = , and so by (φ),d(fw,fz) = . Hencefw=fz. Therefore
w=fz=fw=ffz=gfz=gw.
Thus we showed thatwis a common fixed point off andg.
Suppose now that the condition (xi) is fulfilled. Since a set of common fixed points off andgis not empty, assume thatwandvare two common fixed points off andg,i.e.,
fw=gw=w, fv=gv=v. ()
We claim thatgw=gv.
By assumption, there existsy∈Xsuch thatfyis comparable tofwandfv. Define a
sequence{gyn}such that
gyn=fyn– for alln. ()
Further, setw=wandv=vand, in the same way, define{gwn}and{gvn}such that
gwn=fwn–, gvn=fvn– for alln. ()
From () and (), we havefw=gw=gwandfv=gv=gv. Sincefyis comparable
tofwandfv, andf isg-nondecreasing, it is easy to show
gwgy. ()
Recursively, we can get that
By (), we have that
ψd(gyn+,gw)
=ψd(fyn,fw)
≤ψmaxd(gyn,gw),d(gyn,fyn),d(gw,fw),d(gyn,fw),d(gw,fyn)
–φmaxd(gyn,gw),d(gyn,fyn),d(gw,fw),d(gyn,fw),d(gw,fyn)
. ()
Similarly as in the proof of Theorem ., we can prove that{gyn}is a convergent sequence. Thus there existsy¯∈Xsuch thatgyn→gy¯. Since alsolimn→∞fyn=gy¯, for large enough n, we have
maxd(gyn,gw),d(gyn,fyn),d(gw,fw),d(gyn,fw),d(gw,fyn)
=d(gy¯,gw).
Thus from (), for large enoughn,
ψd(gyn+,gw)
≤ψd(gy¯,gw)–φd(g¯y,gw). ()
Lettingn→ ∞in (), by (ψ) we get
lim n→∞ψ
d(gyn+,gw)
=ψd(gy¯,gw)≤ψd(g¯y,gw)–φd(gy¯,gw).
Hence we obtain
ψd(gy¯,gw)= .
Then by (ψ),d(gy¯,gw) = and hence
gy¯=gw. ()
Similarly, we can show that
lim n→∞ψ
d(gyn+,gv)
=ψd(gy¯,gv)≤ψd(gy¯,gv)–φd(gy¯,gv),
and hence we obtain
gy¯=gv. ()
Therefore, from () and (), we havegw=gv. It follows that
w=fw=gw=gv=fv=v. ()
Corollary . Let(X,d,)be a complete partially ordered metric space and let f be a self-mapping of X satisfying the following condition:
d(fx,fy)≤m(x,y) –φm(x,y)
for all x,y∈X such that gygx,where
m(x,y) =maxd(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)
andφ∈.Suppose that,in addition,t–φ(t)is non-decreasing,limt→+∞φ(t) =∞,there exists x∈X such that xfxand if{fxn}is a nondecreasing sequence such that it con-verges to some z∈X,then fxnz.Then f has a unique fixed point.
Proof Takingψ(t) =tandg(t) =tin the proof of Theorem ., we obtain Corollary ..
Remark . Theorem . extends Theorem due to Berinde [], Theorems . and . due to Beg and Abbas [] and Theorem . due to Song [].
We present an example to show that our result is a real generalization of the recent result of Golubovićet al.[] as well as of the existing results in the literature.
Example . LetX= [,] be the closed interval with the usual metric and letf,g:X→X andψ,φ: [, +∞)→[, +∞) be mappings defined as follows:
f(x) =x–x for allx∈X,
g(x) =x for allx∈X,
ψ(t) =t for allx∈X,
φ(t) =t for ≤t≤ ,
φ(t) =
t fort> .
Letx,yinXbe arbitrary. We say thatxyifx≤y. For anyx,y∈Xsuch thatxy, we have
M(x,y) =maxdg(x),g(y),dg(x),f(x),dg(y),f(y),dg(x),f(y),dg(y),f(x)
=dg(y),f(x),
ψdg(y),f(x)=dg(y),f(x)=y–x –x
=y–x –x.
Sincey≥y–x( –x) for allx∈[,
], it follows that
Thus we have
ψdf(x),f(y)=y–y–x+x=y–x –x–y
≤y–x –x–y–x –x
=dg(y),f(x)–dg(y),f(x)
=ψM(x,y)–φM(x,y).
Therefore,fandgsatisfy (). Also, it is easy to see that the mappingsψ(t) andφ(t) possess all properties (ψ), (ψ) and (φ), (φ) respectively, as well as hypotheses (v), (vi) and (vii)
in Theorem .. Thus we can apply our Theorem . and Corollary ..
On the other hand, forx= and eachy> , the contractive condition in Theorems and of Golubovićet al.[]:
d(fx,fy)≤λ·M(x,y), ()
where <λ< and
M(x;y) =maxd(gx;gy);d(gx;fx);d(gy;fy);d(gx;fy),d(gy;fx),
is not satisfied. Indeed,
M(;y) =maxdg();g(y);dg();f();dg(y);f(y);dg();f(y),dg(y);f()
=maxy; ;y;y–y,y=y.
Thus, for any fixedλ; <λ< , we have, forx= and eachy∈Xwith <y<√ –λ,
df(),f(y)=y–y= –yy>λ·y
=λ·dg(y),g()=λ·M(,y).
Thus, f does not satisfy (). Therefore, the theorems of Jungck and Hussain [], Al-Thagafi and Shahzad [] and Das and Naik [] also cannot be applied.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Both authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Sichuan Normal University, Chengdu, Sichuan 610068, China.2School of Science, Sichuan
University of Science and Engineering, Zigong, Sichuan 643000, China.3Department of Applied Mathematics, Faculty of
Electrical Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, Belgrade, Serbia.
Acknowledgements
Siniša Ješi´c was supported by the Ministry of Education, Science and Technological Development of the Republic of Serbia, Project grant number 174032. Xiaolan Liu was supported by Scientific Research Fund of Sichuan Provincial Education Department (12ZA098), Scientific Research Fund of Sichuan University of Science and Engineering (2012KY08), and Scientific Research Fund of School of Science SUSE (10LXYB03).
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