Proof.
Let x0,y0∈X be two arbitrary points in X. Since F(X×X)⊆g(X), we can choose x1,y1∈X such that
(31)gx1=F(x0,y0), gy1=F(y0,x0).
Continuing in this way, we can construct two sequences {xn} and {yn} in X such that
(32)gxn+1=F(xn,yn), gyn+1=F(yn,xn), ∀n≥0.
The proof is divided into 4 steps.
Step
1. Prove that {gxn} and {gyn} are Cauchy sequences.
Since T is a continuous t-representable of H-type, therefore, for any λ>0, there exists a μ>0 such that
(33)Tm-1(Ns(μ),μ)≥L*(Ns(λ),λ), ∀m∈N.
Since limt→∞MM,N(x,y,t)=1L* for all x,y∈X, there exists t0>0 such that
(34)MM,N(gx0,gx1,t0)≥L*(Ns(μ),μ),MM,N(gy0,gy1,t0)≥L*(Ns(μ),μ).
On the other hand, since ϕ∈Φ, by condition (ϕ-3), we have ∑n=1∞ϕn(t0)<∞. Then, for any t>0, there exists n0∈N such that
(35)t>∑k=n0∞ϕk(t0).
From condition (30), we have
(36)MM,N(gx1,gx2,ϕ(t0)) =MM,N(F(x0,y0),F(x1,y1),ϕ(t0)) ≥L*T(MM,N(gx0,gx1,t0),MM,N(gy0,gy1,t0)),MM,N(gy1,gy2,ϕ(t0)) =MM,N(F(y0,x0),F(y1,x1),ϕ(t0)) ≥L*T(MM,N(gy0,gy1,t0),MM,N(gx0,gx1,t0)).
Similarly, we can also get
(37)MM,N(gx2,gx3,ϕ2(t0)) =MM,N(F(x1,y1),F(x2,y2),ϕ2(t0)) ≥L*T(MM,N(gx1,gx2,ϕ(t0)),ddddddjddMM,N(gy1,gy2,ϕ(t0))) ≥L*T([MM,N(gx0,gx1,t0)]2,ddddddddd[MM,N(gy0,gy1,t0)]2),MM,N(gy2,gy3,ϕ2(t0)) =MM,N(F(y1,x1),F(y2,x2),ϕ2(t0)) ≥L*T(MM,N(gy1,gy2,ϕ(t0)),ddddddddMM,N(gx1,gx2,ϕ(t0))) ≥L*T([MM,N(gy0,gy1,t0)]2,ddddddddd[MM,N(gx0,gx1,t0)]2).
Continuing in the same way, we can get
(38)MM,N(gxn,gxn+1,ϕn(t0)) ≥L*T([MM,N(gx0,gx1,t0)]2n-1,ddddddddd[MM,N(gy0,gy1,t0)]2n-1),MM,N(gyn,gyn+1,ϕn(t0)) ≥L*T([MM,N(gy0,gy1,t0)]2n-1,ddddddddd[MM,N(gx0,gx1,t0)]2n-1).
Now, from (33), (34), and (35), for m>n≥n0, we have
(39)MM,N(gxn,gxm,t)≥L*MM,N(gxn,gxm,∑k=n0∞ϕk(t0))≥L*MM,N(gxn,gxm,∑k=nm-1ϕk(t0))≥L*Tm-n-1(M1MM,N(gxn,gxn+1,ϕn(t0)),dddddddddMM,N(gxn+1,gxn+2,ϕn+1(t0)),…,dddddddddMM,N(gxm-1,gxm,ϕm-1(t0)))≥L*Tm-n-1(T([MM,N(gx0,gx1,t0)]2n-1,dddddddddddd[MM,N(gy0,gy1,t0)]2n-1),ddddddddddT([MM,N(gx0,gx1,t0)]2n,dddddddddddd[MM,N(gy0,gy1,t0)]2n),…, ddddddddddT([MM,N(gx0,gx1,t0)]2m-2,ddddddddddddd[MM,N(gy0,gy1,t0)]2m-2))≥L*T([MM,N(gx0,gx1,t0)]2m-1-2n-1,ddddd[MM,N(gy0,gy1,t0)]2m-1-2n-1)≥L*T2m-2n-1(Ns(μ),μ)≥L*(Ns(λ),λ),
which implies that
(40)MM,N(gxn,gxm,t)>L*(Ns(λ),λ),
for all m,n∈N with m>n≥n0 and t>0. So {gxn} is a Cauchy sequence. Similarly, we can get that {gyn} is also a Cauchy sequence.
Step
2. Prove that g and F have a coupled coincidence point.
Without loss of generality, we can assume that g(X) is complete; then there exist x,y∈g(X) and a,b∈X such that
(41)limn→∞gxn=limn→∞F(xn,yn)=ga=x,limn→∞gyn=limn→∞F(yn,xn)=gb=y.
From (30), we get
(42)MM,N(F(xn,yn),F(a,b),ϕ(t)) ≥L*T(MM,N(gxn,ga,t),MM,N(gyn,gb,t)).
Since MM,N is continuous, taking limit as n→∞, we have
(43)MM,N(ga,F(a,b),ϕ(t))=1L*,
which implies that
(44)F(a,b)=ga=x.
Similarly, we can show that
(45)F(b,a)=gb=y.
Since F and g are weakly compatible, we get that
(46)gF(a,b)=F(ga,gb), gF(b,a)=F(gb,ga),
which implies that
(47)gx=F(x,y), gy=F(y,x).
Step
3. Prove that gx=y and gy=x.
Since T is a continuous t-representable of H-type. Therefore, for any λ>0, there exists a μ>0 such that
(48)Tm-1(Ns(μ),μ)≥L*(Ns(λ),λ), ∀m∈N.
Since limt→∞MM,N(x,y,t)=1L* for all x,y∈X, there exists t0>0 such that
(49)MM,N(gx,y,t0)≥L*(Ns(μ),μ),MM,N(gy,x,t0)≥L*(Ns(μ),μ).
On the other hand, since ϕ∈Φ, by condition (ϕ-3), we have ∑n=1∞ϕn(t0)<∞. Then, for any t>0, there exists n0∈N such that
(50)t>∑k=n0∞ϕk(t0).
Since
(51)MM,N(gx,gyn+1,ϕ(t0)) =MM,N(F(x,y),F(yn,xn),ϕ(t0)) ≥L*T(MM,N(gx,gyn,t0)*MM,N(gy,gxn,t0)),
thus
(52)MM,N(gx,gyn+1,ϕ(t0)) ≥L*T(MM,N(gx,gyn,t0)*MM,N(gy,gxn,t0)).
Letting n→∞ in (52), by using (41), we get
(53)MM,N(gx,y,ϕ(t0)) ≥L*T(MM,N(gx,y,t0),MM,N(gy,x,t0)).
Similarly, we can get
(54)MM,N(gy,x,ϕ(t0)) ≥L*T(MM,N(gx,y,t0),MM,N(gy,x,t0)).
From (53) and (54), we have
(55)T(MM,N(gx,y,ϕ(t0)),MM,N(gy,x,ϕ(t0))) ≥L*T([MM,N(gx,y,t0)]2,[MM,N(gy,x,t0)]2).
By this way, we can get, for all n∈N,
(56)T(MM,N(gx,y,ϕn(t0)),MM,N(gy,x,ϕn(t0))) ≥L*T([MM,N(gx,y,ϕn-1(t0))]2,dddddddd[MM,N(gy,x,ϕn-1(t0))]2) ≥L*T([MM,N(gx,y,t0)]2n,[MM,N(gy,x,t0)]2n).
Thus,
(57)T(MM,N(gx,y,ϕn(t0)),MM,N(gy,x,ϕn(t0))) ≥L*T([MM,N(gx,y,t0)]2n,[MM,N(gy,x,t0)]2n).
Then, by (48), (49), (50), and (57), we have
(58)T(MM,N(gx,y,t),MM,N(gy,x,t)) ≥L*T(MM,N(gx,y,∑k=n0∞ϕk(t0)),DDDDDDMM,N(gy,x,∑k=n0∞ϕk(t0))) ≥L*T(MM,N(gx,y,ϕn0(t0)),MM,N(gy,x,ϕn0(t0))) ≥L*T([MM,N(gx,y,t0)]2n0,[MM,N(gy,x,t0)]2n0) ≥L*T2n0+1-1(Ns(μ),μ) ≥L*(Ns(λ),λ).
So for any λ>0 we have
(59)T(MM,N(gx,y,t),MM,N(gy,x,t))≥L*(Ns(λ),λ),ggjjg∀t>0.
We can get that gx=y and gy=x.
Step
4. Prove that x=y.
Since T is a continuous t-representable of H-type. Therefore, for any λ>0, there exists a μ>0 such that
(60)Tm-1(Ns(μ),μ)≥L*(Ns(λ),λ), ∀m∈N.
Since limt→∞MM,N(x,y,t)=1L* for all x,y∈X, there exists t0>0 such that
(61)MM,N(x,y,t0)≥L*(Ns(μ),μ).
On the other hand, since ϕ∈Φ, by condition (ϕ-3), we have ∑n=1∞ϕn(t0)<∞. Then, for any t>0, there exists n0∈N such that
(62)t>∑k=n0∞ϕk(t0).
Since, for t0>0,
(63)MM,N(gxn+1,gyn+1,ϕ(t0)) =MM,N(F(xn,yn),F(yn,xn),ϕ(t0)) ≥L*T(MM,N(gxn,gyn,t0),MM,N(gyn,gxn,t0)),
thus
(64)MM,N(gxn+1,gyn+1,ϕ(t0)) ≥L*T(MM,N(gxn,gyn,t0),ddddddddjMM,N(gyn,gxn,t0)).
Letting n→∞ in the above inequality, we get
(65)MM,N(x,y,ϕ(t0)) ≥L*T(MM,N(x,y,t0),MM,N(y,x,t0)).
Thus, by (60), (61), (62), and (65), we have
(66)MM,N(x,y,t) ≥L*MM,N(x,y,∑k=n0∞ϕk(t0)) ≥L* MM,N(x,y,ϕn0(t0)) ≥L*T([MM,N(x,y,t0)]2n0-1,[MM,N(y,x,t0)]2n0-1) ≥L*T2n0+1-3(Ns(μ),μ) ≥L*(Ns(λ),λ),
which implies that x=y. Thus, we have proved that F and g have a unique common fixed point in X. This completes the proof of Theorem 31.
Example 35.
Let X={0,1,1/2,1/3,…,1/n,…} and T(a,b)= (min{a1,b1},max{a2,b2}), for all a=(a1,a2) and b=(b1,b2)∈L*. Define
(69)MM,N(x,y,t)=(tt+|x-y|,|x-y|t+|x-y|),vvvjjvvvvv∀x,y∈X, t>0.
Then (X,MM,N,T) is a modified intuitionistic fuzzy metric space. Let ϕ(t)=t/2. Let g:X→X and F:X×X→X be defined as
(70)g(x)={0,x=0,1,x=12n+1,12n+1,x=12n,F(x,y)={1(2n+1)4,(x,y)=(12n,12n),0,otherwise.
Let xn=yn=1/2n. We have
(71)gxn=12n+1⟶0,F(xn,yn)=1(2n+1)4⟶0 as n⟶∞,
but
(72)MM,N(F(gxn,gyn),gF(xn,yn),t) =MM,N(0,1,t)↛1L*,
so g and F are not compatible. From F(x,y)=g(x) and F(y,x)=g(y), we can get (x,y)=(0,0) and we have gF(0,0)=F(g0,g0), which implies that F and g are weakly compatible. The following result is easy to be verified:
(73) tX+t≥min{tY+t,tZ+t}, XX+t≤max{YY+t,ZZ+t}, ⟺X≤max{Y,Z}, ∀X,Y,Z≥0, t>0.
By the definition of MM,N,ϕ and the result above, we can get inequality (30):
(74)MM,N(F(x,y),F(u,v),ϕ(t)) ≥L*T(MM,N(gx,gu,t),MM,N(gy,gv,t)),
which is equivalent to the following:
(75)2|F(x,y)-F(u,v)|≤max{|gx-gu|,|gy-gv|}.
Now, we verify inequality (75). Let A={1/2n,n∈N},B=X-A. By the symmetry and without loss of generality, (x,y),(u,v) have six possibilities.
Case
1.
(
x
,
y
)
∈
B
×
B
,
(
u
,
v
)
∈
B
×
B
. It is obvious that (75) holds.
Case
2.
(
x
,
y
)
∈
B
×
B
,
(
u
,
v
)
∈
B
×
A
. It is obvious that (75) holds.
Case
3.
(
x
,
y
)
∈
B
×
B
,
(
u
,
v
)
∈
A
×
A
. If u≠v, (75) holds. If u=v, let u=v=1/2n, and then
(76)2|F(x,y)-F(u,v)|=2(2n+1)4,max{|gx-gu|,|gy-gv|}=2n2n+1,
which implies that (75) holds.
Case
4.
(
x
,
y
)
∈
B
×
A
,
(
u
,
v
)
∈
B
×
A
. It is obvious that (75) holds.
Case
5.
(
x
,
y
)
∈
B
×
A
,
(
u
,
v
)
∈
A
×
A
. If u≠v, (75) holds. If u=v, let x∈B, y=1/2j, u=v=1/2n, and then
(77)2|F(x,y)-F(u,v)|=2(2n+1)4,max{|gx-gu|,|gy-gv|} hh=max{12n+1,|12j+1-12n+1|},
or
(78)max{|gx-gu|,|gy-gv|} hh=max{2n2n+1,|12j+1-12n+1|}.
So, (75) holds.
Case
6.
(
x
,
y
)
∈
A
×
A
and (u,v)∈A×A. If x≠y and u≠v, (75) holds. If x≠y,u=v, let x=1/2i, y=1/2j,i≠j, and u=v=1/2n. Then
(79)2|F(x,y)-F(u,v)|=2(2n+1)4,max{|gx-gu|,|gy-gv|} hh=max{|12i+1-12n+1|,|12j+1-12n+1|}.
So, (75) holds. If x=y and u=v, let x=y=1/2i, u=v=1/2n. Then
(80)2|F(x,y)-F(u,v)|=2|1(2i+1)4-1(2n+1)4|,max{|gx-gu|,|gy-gv|} hh=12i+1-12n+1.
So, (75) holds. Then all the conditions in Theorem 31 are satisfied and 0 is the unique common fixed point of g and F.