# Common Fixed-Point Theorems in Complete Generalized Metric Spaces

Authors: Chi-Ming Chen
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce the notions of the𝒲 function and 𝒮 function, and then we prove two common fixed point theorems in complete generalized metric spaces under contractive conditions with these two functions. Our results generalize or improve many recent common fixed point results in the literature.


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## Body

## 1. Introduction and Preliminaries

In 2000, Branciari [1] introduced the following notion of a generalized metric space where the triangle inequality of a metric space had been replaced by an inequality involving three terms instead of two. Later, many authors worked on this interesting space (e.g., [2–7]).Let(X,d) be a generalized metric space. For γ>0 and x∈X, we define that
(1.1)Bγ(x):={y∈X∣d(x,y)<γ}.
Branciari [1] also claimed that {Bγ(x):γ>0,x∈X} is a basis for a topology on X, d is continuous in each of the coordinates, and a generalized metric space is a Hausdorff space. We recall some definitions of a generalized metric space as follows.Definition 1.1 (See [1]).
LetX be a nonempty set and d:X×X→[0,∞) be a mapping such that for all x,y∈X and for all distinct point u,v∈X each of them different from x and y, one has the following:(i)
d(x,y)=0if and only if x=y,(ii)
d(x,y)=d(y,x),(iii)
d(x,y)≤d(x,u)+d(u,v)+d(v,y)(rectangular inequality). Then (X,d) is called a generalized metric space (or shortly g.m.s).We present an example to show that not every generalized metric on a setX is a metric on X.Example 1.2.
LetX={t,2t,3t,4t,5t} with t>0 be a constant, and we define d:X×X→[0,∞) by(1)
d(x,x)=0,for all x∈X,(2)
d(x,y)=d(y,x),for all x,y∈X,(3)
d(t,2t)=3γ,(4)
d(t,3t)=d(2t,3t)=γ,(5)
d(t,4t)=d(2t,4t)=d(3t,4t)=2γ,(6)
d(t,5t)=d(2t,5t)=d(3t,5t)=d(4t,5t)=(3/2)γ,where γ>0 is a constant. Then let (X,d) be a generalized metric space, but it is not a metric space, because
(1.2)d(t,2t)=3γ>d(t,3t)+d(3t,2t)=2γ.Definition 1.3 (See [1]).
Let(X,d) be a g.m.s,{xn} be a sequence in X and x∈X.We say that {xn} is g.m.s convergent to x if and only if d(xn,x)→0 as n→∞.We denote by xn→x as n→∞.Definition 1.4 (See [1]).
Let(X,d) be a g.m.s, {xn} be a sequence in X and x∈X.We say that {xn} is g.m.s Cauchy sequence if and only if, for each ε>0,there exists n0∈ℕ such that d(xm,xn)<ε for all n>m>n0.Definition 1.5 (See [1]).
Let(X,d) be a g.m.s.Then X is called complete g.m.s if every g.m.s Cauchy sequence is g.m.s convergent in X.In this paper, we also recall the concept of compatible mappings and prove two common fixed point theorems which incorporated the compatible map concept followed. In 1986, Jungck [8] introduced the below concept of compatible mappings.Definition 1.6 (See [8]).
Let(X,d) be a g.m.s, and let S,F:X→X be two single-valued functions. We say that S and F are compatible if
(1.3)lim⁡⁡n→∞d(SFxn,FSxn)=0,
whenever {xn} is a sequence in X such that limn→∞d(Fxn,Sxn)=0.
In particular,d(SFx,FSx)=0 if d(Fx,Sx)=0 by taking xn=x for all n∈ℕ.Later, many authors studied this subject (compatible mappings), and many results on fixed points and common fixed points are proved (see, e.g., [9–14]).


## 2. Main Results

In this paper, we first introduce the below concept of the𝒲 function.Definition 2.1.
We callφ:ℝ+→ℝ+ a 𝒲 function if the function φ satisfies the following conditions:(φ1)
φ(t)<tfor allt>0andφ(0)=0,(φ2)
limtn→tinfφ(tn)<tfor allt>0.Lemma 2.2.
Letφ:ℝ+→ℝ+ be a  𝒲 function. Then  limn→∞φn(t)=0 for all  t>0, where  φn(t) denotes the  nth iteration of  φ.Proof.
Lett>0 be fixed. If φn0(t)=0 for some n0∈ℕ, then
(2.1)φn0+1(t)=φ(φn0(t))=φ(0)=0.
It follows that φn0+k(t)=0 for all k∈ℕ, and so we get that limn→∞φn(t)=0.
Ifφn(t)>0 for all n∈ℕ, then we put αn=φn(t). Thus,
(2.2)αn+1=φn+1(t)=φ(φn(t))=φ(αn).
Since φ is a 𝒲 function, we have that αn+1=φ(αn)<αn. Therefore, the sequence {αn} is strictly decreasing and bounded from below, and so there exists an γ≥0 such that limn→∞αn=γ+. We claim that γ=0. If not, suppose that γ>0, then we have that
(2.3)γ=lim⁡n→∞αn+1=lim⁡n→∞inf⁡⁡φ(αn)=lim⁡αn→γ+inf⁡⁡φ(αn)<α,
a contradiction. So we obtain that γ=0, that is, limn→∞φn(t)=0.We now state the main common fixed-point theorem for the𝒲 function in a complete g.m.s, as follows.Theorem 2.3.
Let(X,d) be a Hausdorff and complete  g.m.s, and let  φ:ℝ+→ℝ+ be a  𝒲 function. Let  S,T,F,G:X→X be four single-valued functions such that for all  x,y∈X,
(2.4)d(Sx,Ty)≤φ(max⁡⁡{d(Fx,Gy),d(Fx,Sx),d(Gy,Ty)}).
Assume that T(X)⊂F(X) and S(X)⊂G(X), and the pairs {S,F} and {T,G} are compatible. If F or G is continuous, then S,T,F, and G have a unique common fixed point in X.Proof.
Given thatx0∈X. Define the sequence {xn} recursively as follows:
(2.5)Gx2n+1=Sx2n=z2n,Fx2n+2=Tx2n+1=z2n+1.
Step 1.
We will prove that(2.6)lim⁡n→∞d(zn,zn+1)=0.
Using (2.4), we have that for each n∈ℕ(2.7)d(z2n,z2n+1)=d(Sx2n,Tx2n+1)≤φ(max⁡⁡{d(Fx2n,Gx2n+1),d(Fx2n,Sx2n),d(Gx2n+1,Tx2n+1)})≤φ(max⁡⁡{d(z2n-1,z2n),d(z2n-1,z2n),d(z2n,z2n+1)}),
and so we can conclude that
(2.8)d(z2n,z2n+1)≤φ(d(z2n-1,z2n)).
Similarly, we also conclude that
(2.9)d(z2n+1,z2n+2)≤φ(d(z2n,z2n+1)).
Generally, we have that for each n∈ℕ(2.10)d(zn,zn+1)≤φ(d(zn-1,zn)).
By induction, we get that
(2.11)d(zn,zn+1)≤φ(d(zn-1,zn))≤φ2(d(zn-2,zn-1))≤⋯≤φn(d(z0,z1)).
By Lemma 2.2, we obtained that limn→∞d(zn,zn+1)=0.
We claim that{zn} is g.m.s Cauchy. We claim that the following result holds.Step 2.
Claim that, for everyε>0, there exists n0∈ℕ such that if m,n≥n0, then d(zm,zn)<ε.
Suppose that the above statesment is false. Then there existsε>0 such that, for any k∈ℕ, there are mk,nk∈ℕ with mk>nk≥k satisfying that(a)
mk is even and nk is odd,(b)
d(znk,zmk)≥ε,(c)
mk is the smallest even number such that the condition (b) holds. Taking into account (b) and (c), we have that
(2.12)ε≤d(znk,zmk)≤d(znk,zmk-2)+d(zmk-2,zmk-1)+d(zmk-1,zmk)≤ε+d(zmk-2,zmk-1)+d(zmk-1,zmk).
Letting k→∞, we get the following:
(2.13)lim⁡k→∞d(znk,zmk)=ε,(2.14)ε≤d(znk-1,zmk-1)≤d(znk-1,zmk-3)+d(zmk-3,zmk-2)+d(zmk-2,zmk-1)≤ε+d(zmk-3,zmk-2)+d(zmk-2,zmk-1).
Letting k→∞, we get the following:
(2.15)lim⁡k→∞d(znk-1,zmk-1)=ε.
Using (2.4), (2.13), and (2.15), we have
(2.16)d(znk,zmk)=d(Sxnk,Txmk)≤φ(max⁡⁡{d(Fxnk,Gxmk),d(Fxnk,Sxnk),d(Gxnk,Txnk)})=φ(max⁡⁡{d(znk-1,zmk-1),d(znk-1,znk),d(zmk-1,zmk)})=φ(max⁡⁡{d(znk-1,zmk-1),cnk-1,cmk-1}),
taking limk→∞inf, we get that ε<ε, a contradiction. So {zn} is g.m.s Cauchy. Since X is complete, there exists z∈X such that limn→∞zn=z. So we have
(2.17)d(Fx2n,z)⟶0,d(Gx2n+1,z)⟶0,d(Sx2n,z)⟶0,d(Tx2n+1,z)⟶0,
as n→∞.Step 3.
We will show thatz is a common fixed point of S, T, F, and G.
Assume thatF is continuous. Then we have
(2.18)d(F2x2n,Fz)⟶0,d(FSx2n,Fz)⟶0,
as n→∞. By the rectangular property, we have
(2.19)d(SFx2n,Fz)≤d(SFx2n,FSx2n)+d(FSx2n,F2x2n)+d(F2x2n,Fz).
Since S and F are compatible and d(Sx2n,Fx2n)→0 as n→∞, we conclude that
(2.20)d(SFx2n,FSx2n)⟶0,
as n→∞. Taking into account (2.18), (2.19), and (2.20), we have that
(2.21)d(SFx2n,Fz)⟶0,
as n→∞. Since
(2.22)d(SFx2n,Tx2n+1)≤φ(max⁡⁡{d(F2x2n,Gx2n+1),d(F2x2n,SFx2n),d(Gx2n+1,Tx2n+1)}),
for each n∈ℕ. Taking limn→∞ and taking into account (2.17), (2.18), (2.19), (2.20), and (2.21), we get that
(2.23)d(Fz,z)≤φ(max⁡⁡{d(Fz,z),d(Fz,Fz),d(z,z)})=φ(max⁡⁡{d(Fz,z),0,0})<d(Fz,z),
and this is a contradiction unless d(Fz,z)=0, that is, Fz=z.
On the same way, we have that, for eachn∈ℕ,
(2.24)d(Sz,Tx2n+1)≤φ(max⁡⁡{d(Fz,Gx2n+1),d(Fz,Sz),d(Gx2n+1,Tx2n+1)}).
letting n→∞, we obtained the following:
(2.25)d(Sz,z)≤φ(max⁡⁡{d(z,z),d(z,Sz),d(z,z)})=φ(max⁡⁡{0,d(z,Sz),0})<d(Sz,z),
and this is a contradiction unless d(Sz,z)=0, that is, Sz=z.
SinceS(X)⊂G(X), put z′∈X such that Gz′=z=Sz. Then TGz′=Tz and using (2.4),
(2.26)d(z,Tz′)=d(Sz,Tz′)≤φ(max⁡⁡{d(Fz,Gz′),d(Fz,Sz),d(Gz′,Tz′)})=φ(max⁡⁡{d(z,z),d(z,z),d(z,Tz′)})<d(z,Tz′),
and this is a contradiction unless d(z,Tz′)=0, that is, Tz′=z and so d(Tz′,Gz′)=d(z,z)=0.
SinceT and G are compatible and d(Tz′,Gz′)=0, we have that
(2.27)d(Tz,Gz)=d(TGz′,GTz′)=0,
which implies that Tz=Gz. Using (2.4), we also have
(2.28)d(z,Tz)=d(Sz,Tz)×φ(max⁡⁡{d(Fz,Gz),d(Fz,Sz),d(Gz,Tz)})=φ(max⁡⁡{d(z,Tz),d(z,z),d(Tz,Tz)})<d(z,Tz),
and this is a contradiction unless d(z,Tz)=0, that is, Tz=z.
From above argument, we get that(2.29)Sz=Tz=z=Fz=Gz,
and so z is a common fixed point of S,T,F, and G.Step 4.
Finally, to prove the uniqueness of the common fixed point ofS,T,F and G, let y be another common fixed point of S,T,F, and G. Then using (2.4), we have
(2.30)d(y,z)=d(Sy,Tz)×φ(max⁡⁡{d(Fy,Gz),d(Fy,Sy),d(Gz,Tz)})=φ(max⁡⁡{d(y,z),d(y,y),d(z,z)})<d(y,z),
and this is a contradiction unless d(y,z)=0, that is, y=z. Hence z is the unique common fixed point of S, T, F, and G in X.We give the following example to illustrate Theorem2.3.Example 2.4.
LetX={t,2t,3t,4t,5t} with t>0 is a constant, and we define d:X×X→[0,∞) by(1)
d(x,x)=0,for all x∈X,(2)
d(x,y)=d(y,x),for all x,y∈X,(3)
d(t,2t)=3γ,(4)
d(t,3t)=d(2t,3t)=γ,(5)
d(t,4t)=d(2t,4t)=d(3t,4t)=2γ,(6)
d(t,5t)=d(3t,5t)=γ and d(2t,5t)=d(4t,5t)=2γ,where γ>0 is a constant.
Ifφ:ℝ+→ℝ+,φ(t)=(4/5)t,then φ is a 𝒲 function. We next define S,T, F, G:X→X by
(2.31)S(x)={3tifx≠4t,5tifx=4t,T(x)={3tifx≠4t,tifx=4t,G(x)=I(x)=theidentitymapping,F(x)={3tifx=3t,tifx=t,2t,5t,2tifx=4t.
Then all conditions of Theorem 2.3 are satisfied, and 3t is a unique common fixed point of S, T, F, and G.For the caseG=F=I (the identity mapping) and S=T, we are easy to get the below fixed-point theorem.Theorem 2.5.
Let(X,d) be a Hausdorff and complete  g.m.s, and let  φ:ℝ+→ℝ+ be a  𝒲 function. Let  T:X→X be a single-valued function such that for all  x,y∈X,
(2.32)d(Tx,Ty)≤φ(max⁡⁡{d(x,y),d(x,Tx),d(y,Ty)}).
Then T has a unique fixed point in X.We next introduce the below concept of the𝒮 function.Definition 2.6.
We callϕ:ℝ+3→ℝ+ a 𝒮 function if the function ϕ satisfies the following conditions:(ϕ1)
ϕ is a strictly increasing, continuous function in each coordinate,(ϕ2)
for allt>0,ϕ(t,t,t)<t,ϕ(t,0,0)<t,ϕ(0,t,0)<t, and ϕ(0,0,t)<t.Example 2.7.
Letϕ:ℝ+3→ℝ+ denote that
(2.33)ϕ(t1,t2,t3)=k⋅max⁡⁡{t1,t2,t3},fork∈(0,1).
Then ϕ is a 𝒮 function.We now state the main common fixed point theorem for the𝒮 function in a complete g.m.s.Theorem 2.8.
Let(X,d) be a Hausdorff and complete  g.m.s, and let  φ:ℝ+3→ℝ+ be a  𝒮 function. Let  S,T,F,G:X→X be four single-valued functions such that for all  x,y∈X,
(2.34)d(Sx,Ty)≤ϕ(d(Fx,Gy),d(Fx,Sx),d(Gy,Ty)).
Assume that T(X)⊂F(X) and S(X)⊂G(X), and the pairs {S,F} and {T,G} are compatible. If F or G is continuous, then S, T, F, and G have a unique common fixed point in X.Proof.
Givenx0∈X. Define the sequence {xn} recusively as follows:
(2.35)Gx2n+1=Sx2n=z2n,Fx2n+2=Tx2n+1=z2n+1.
Step 1.
We will prove that(2.36)lim⁡n→∞d(zn,zn+1)=0,
Using (2.34) and the definition of the 𝒮 function, we have that for each n∈ℕ(2.37)d(z2n,z2n+1)=d(Sx2n,Tx2n+1)≤ϕ(d(Fx2n,Gx2n+1),d(Fx2n,Sx2n),d(Gx2n+1,Tx2n+1))≤ϕ(d(z2n-1,z2n),d(z2n-1,z2n),d(z2n,z2n+1)),
and so we can conclude that
(2.38)d(z2n,z2n+1)≤ϕ(d(z2n-1,z2n),d(z2n-1,z2n),d(z2n-1,z2n))<d(z2n-1,z2n).
Similarly, we also conclude that
(2.39)d(z2n+1,z2n+2)≤ϕ(d(z2n,z2n+1),d(z2n,z2n+1),d(z2n,z2n+1))<d(z2n,z2n+1).
Generally, we have that for each n∈ℕ(2.40)d(zn,zn+1)≤ϕ(d(zn-1,zn),d(zn-1,zn),d(zn-1,zn))<d(zn-1,zn).
Now, for each m∈ℕ, if we denote cm=d(zm,zm+1), then {cm} is a strictly decreasing sequence. Thus, it must converge to some c with c≥0. We claim that c=0. If not, suppose that c>0, then
(2.41)c≤cm+1≤ϕ(cm,cm,cm).
Passing to the limit, as m→∞, we have that c≤c<ϕ(c,c,c)<c, which is a contradiction. So we get c=0.
We claim that{zn} is g.m.s Cauchy. We claim that the following result holds.Step 2.
Claim that, for everyε>0, there exists n0∈ℕ such that if m,n≥n0, then d(zm,zn)<ε.
Suppose that the above statesment is false. Then there existsε>0 such that, for any k∈ℕ, there are mk,nk∈ℕ with mk>nk≥k satisfying that(d)
mk is even and nk is odd,(e)
d(znk,zmk)≥ε,(f)
mk is the smallest even number such that the condition (e) holds. Taking into account (e) and (f), we have that
(2.42)ε≤d(znk,zmk)≤d(znk,zmk-2)+d(zmk-2,zmk-1)+d(zmk-1,zmk)≤ε+d(zmk-2,zmk-1)+d(zmk-1,zmk).
Letting k→∞, we get the following:
(2.43)lim⁡k→∞d(znk,zmk)=ε,(2.44)ε≤d(znk-1,zmk-1)≤d(znk-1,zmk-3)+d(zmk-3,zmk-2)+d(zmk-2,zmk-1)≤ε+d(zmk-3,zmk-2)+d(zmk-2,zmk-1).
Letting k→∞, we get the following:
(2.45)lim⁡k→∞d(znk-1,zmk-1)=ε.
Using (2.34), (2.43), and (2.45), we have
(2.46)d(znk,zmk)=d(Sxnk,Txmk)≤ϕ(d(Fxnk,Gxmk),d(Fxnk,Sxnk),d(Gxnk,Txnk))=ϕ(d(znk-1,zmk-1),d(znk-1,znk),d(zmk-1,zmk))=ϕ(d(znk-1,zmk-1),cnk-1,cmk-1),
taking k→∞, we get that ε≤ϕ(ε,0,0)<ε, a contradiction. So {zn} is g.m.s Cauchy. Since X is complete, there exists z∈X such that limn→∞zn=z. So we have
(2.47)d(Fx2n,z)⟶0,d(Gx2n+1,z)⟶0,d(Sx2n,z)⟶0,d(Tx2n+1,z)⟶0,
as n→∞.Step 3.
We will show thatz is a common fixed point of S, T, F, and G.
Assume thatF is continuous. Then, we have
(2.48)d(F2x2n,Fz)⟶0,d(FSx2n,Fz)⟶0,
as n→∞. By the rectangular property, we have
(2.49)d(SFx2n,Fz)≤d(SFx2n,FSx2n)+d(FSx2n,F2x2n)+d(F2x2n,Fz).
Since S and F are compatible and d(Sx2n,Fx2n)→0 as n→∞, we conclude that
(2.50)d(SFx2n,FSx2n)⟶0,
as n→∞. Taking into account (2.48), (2.49), and (2.50), we have that
(2.51)d(SFx2n,Fz)⟶0,
as n→∞. Since
(2.52)d(SFx2n,Tx2n+1)≤ϕ(d(F2x2n,Gx2n+1),d(F2x2n,SFx2n),d(Gx2n+1,Tx2n+1)),
for each n∈ℕ. Letting n→∞ and taking into account (2.47), (2.48), (2.49), (2.50), and (2.51), we get that
(2.53)d(Fz,z)≤ϕ(d(Fz,z),d(Fz,Fz),d(z,z))=ϕ(d(Fz,z),0,0)<d(Fz,z),
and this is a contradiction unless d(Fz,z)=0, that is, Fz=z.
On the same way, we have that, for eachn∈ℕ,
(2.54)d(Sz,Tx2n+1)≤ϕ(d(Fz,Gx2n+1),d(Fz,Sz),d(Gx2n+1,Tx2n+1)).
Taking limn→∞, we obtained that
(2.55)d(Sz,z)≤ϕ(d(Fz,z),d(Fz,Sz),d(z,z))=ϕ(d(0,d(z,Sz),0))<d(Sz,z),
and this is a contradiction unless d(Sz,z)=0, that is, Sz=z.
SinceS(X)⊂G(X), put z′∈X such that Gz′=z=Sz. Then TGz′=Tz and using (2.34),
(2.56)d(z,Tz′)=d(Sz,Tz′)≤ϕ(d(Fz,Gz′),d(Fz,Sz),d(Gz′,Tz′))=ϕ(d(z,z),d(z,z),d(z,Tz′))<d(z,Tz′),
and this is a contradiction unless d(z,Tz′)=0, that is, Tz′=z and so d(Tz′,Gz′)=d(z,z)=0.
SinceT and G are compatible and d(Tz′,Gz′)=0, we have that
(2.57)d(Tz,Gz)=d(TGz′,GTz′)=0,
which implies that Tz=Gz. Using (2.34), we also have
(2.58)d(z,Tz)=d(Sz,Tz)≤ϕ(d(Fz,Gz),d(Fz,Sz),d(Gz,Tz))=ϕ(d(z,Tz),d(z,z),d(Tz,Tz))<d(z,Tz),
and this is a contradiction unless d(z,Tz)=0, that is, Tz=z.
From above argument, we get that(2.59)Sz=Tz=z=Fz=Gz,
and so z is a common fixed point of S, T, F, and G.Step 4.
Finally, to prove the uniqueness of the common fixed point ofS, T, F, and G, let y be another common fixed point of S, T, F, and G. Then using (2.34), we have
(2.60)d(y,z)=d(Sy,Tz)≤ϕ(d(Fy,Gz),d(Fy,Sy),d(Gz,Tz))=ϕ(d(y,z),d(y,y),d(z,z))<d(y,z),
and this is a contradiction unless d(y,z)=0, that is, y=z. Hence z is the unique common fixed point of S, T, F, and G in X.Using Example2.4, we get the following example to illustrate Theorem 2.8.Example 2.9.
LetX={t,2t,3t,4t,5t} with t>0 be a constant, and we define d:X×X→[0,∞) by(1)
d(x,x)=0,for all x∈X,(2)
d(x,y)=d(y,x),for all x,y∈X,(3)
d(t,2t)=3γ,(4)
d(t,3t)=d(2t,3t)=γ,(5)
d(t,4t)=d(2t,4t)=d(3t,4t)=2γ,(6)
d(t,5t)=d(3t,5t)=γand d(2t,5t)=d(4t,5t)=2γ,     where γ>0 is a constant.
Ifϕ:ℝ+3→ℝ+,ϕ(t)=(8/9)·max{t1,t2,t3}, then ϕ is a 𝒮 function. We next define S,T,F,G:X→X by
(2.61)S(x)={3tifx≠4t,5tifx=4t,T(x)={3tifx≠4t,tifx=4t,G(x)=I(x)=the identity mapping,F(x)={3tifx=3t,tifx=t,2t,5t,2tifx=4t.
Then all conditions of Theorem 2.8 are satisfied, and 3t is a unique common fixed point of S, T, F, and G.For the caseG=F=I (the identity mapping) and S=T, we are easy to get the below fixed-point theorem.Theorem 2.10.
Let(X,d) be a Hausdorff and complete  g.m.s, and let  ϕ:ℝ+3→ℝ+ be a  𝒮 function. Let  T:X→X be a single-valued function such that for all  x,y∈X,
(2.62)d(Tx,Ty)≤ϕ(d(x,y),d(x,Tx),d(y,Ty)).
Then T has a unique fixed point in X.



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*Source: 945915-2012-05-29.xml*