In this section, we state and prove our fixed point results.
Proof.
Since the mappings A and S are subsequentially continuous and compatible, there exist sequences {xn}, {yn} in X such that
(16)limn→∞A(xn,yn)=limn→∞Sxn=α,limn→∞A(yn,xn)=limn→∞Syn=β,
for all α,β∈X, and
(17)limn→∞M(A(Sxn,Syn),SA(xn,yn),t)=1,limn→∞M(A(Syn,Sxn),SA(yn,xn),t)=1,
that is A(α,β)=Sα and A(β,α)=Sβ. Similarly, with respect to the pair (B,T), there exist sequences {xn′}, {yn′} in X such that
(18)limn→∞B(xn′,yn′)=limn→∞Txn′=α′,limn→∞B(yn′,xn′)=limn→∞Tyn′=β′,
for all α′,β′∈X, and
(19)limn→∞M(B(Txn′,Tyn′),TB(xn′,yn′),t)=1,limn→∞M(B(Tyn′,Txn′),TB(yn′,xn′),t)=1,
that is B(α′,β′)=Tα′ and B(β′,α′)=Tβ′. Hence (α,β)∈X×X is a coupled coincidence point of the pair (A,S), whereas (α′,β′)∈X×X is a coupled coincidence point of the pair (B,T).
Now we assert that (α,β)=(α′,β′), that is, α=α′ and β=β′. Since * is a t-norm of H-type, for any λ>0, there exists an μ>0 such that
(20)(1-μ)*(1-μ)*⋯*(1-μ)︸p≥1-λ,
for all p∈ℕ.
Since M(x,y,·) is continuous and limt→∞M(x,y,t)=1 for all x,y∈X, there exists t0>0 such that M(α,α′,t0)≥1-μ and M(β,β′,t0)≥1-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 t>∑p=n0∞ϕp(t0). On using inequality (15) with x=xn, y=yn, u=xn′, and v=yn′, we have
(21)M(A(xn,yn),B(xn′,yn′),ϕ(t0)) ≥M(Sxn,Txn′,t0)*M(Syn,Tyn′,t0).
Letting n→∞, we get
(22)M(α,α′,ϕ(t0))≥M(α,α′,t0)*M(β,β′,t0).
Again using inequality (15) with x=yn, y=xn, u=yn′, and v=xn′, we have
(23)M(A(yn,xn),B(yn′,xn′),ϕ(t0)) ≥M(Syn,Tyn′,t0)*M(Sxn,Txn′,t0).
Letting n→∞, we get
(24)M(β,β′,ϕ(t0))≥M(β,β′,t0)*M(α,α′,t0).
From (22) and (24), we obtain
(25)M(α,α′,ϕ(t0))*M(β,β′,ϕ(t0)) ≥[M(α,α′,t0)]2*[M(β,β′,t0)]2.
In general, for all n∈ℕ, we have
(26)M(α,α′,ϕn(t0))*M(β,β′,ϕn(t0)) ≥[M(α,α′,ϕn-1(t0))]2*[M(β,β′,ϕn-1(t0))]2 ≥[M(α,α′,t0)]2n*[M(β,β′,t0)]2n.
Then, we have
(27)M(α,α′,t)*M(β,β′,t) ≥[M(α,α′,∑p=n0∞ϕp(t0))] *[M(β,β′,∑p=n0∞ϕp(t0))] ≥[M(α,α′,ϕn0(t0))] *[M(β,β′,ϕn0(t0))] ≥[M(α,α′,t0)]2n0*[M(β,β′,t0)]2n0 ≥(1-μ)*(1-μ)*⋯*(1-μ)︸22n0 ≥1-λ.
So for any λ>0, we have
(28)M(α,α′,t)*M(β,β′,t)≥1-λ,
for all t>0, and so α=α′ and β=β′. Therefore we have
(29)A(α,β)=Sα, A(β,α)=Sβ,B(α,β)=Tα, B(β,α)=Tβ.
Next, we show that Sα=Tα and Sβ=Tβ. Since * is a t-norm of H-type, for any λ>0, there exists an μ>0 such that
(30)(1-μ)*(1-μ)*⋯*(1-μ)︸p≥1-λ,
for all p∈ℕ.
Since M(x,y,·) is continuous and limt→+∞M(x,y,t)=1 for all x,y∈X, there exists t0>0 such that M(Sα,Tα,t0)≥1-μ and M(Sβ,Tβ,t0)≥1-μ.
Since ϕ∈Φ, by condition (ϕ-3), we have ∑n=1∞ϕn(t0)<∞. Then for any t>0, there exists n0∈ℕ such that t>∑p=n0∞ϕp(t0). On using inequality (15) with x=u=α, y=v=β, we have
(31)M(A(α,β),B(α,β),ϕ(t0)) ≥M(Sα,Tα,t0)*M(Sβ,Tβ,t0),
and so
(32)M(Sα,Tα,ϕ(t0))≥M(Sα,Tα,t0)*M(Sβ,Tβ,t0).
Similarly, we can obtain
(33)M(Sβ,Tβ,ϕ(t0))≥M(Sβ,Tβ,t0)*M(Sα,Tα,t0).
From (32) and (33), we have
(34)M(Sα,Tα,ϕ(t0))*M(Sβ,Tβ,ϕ(t0)) ≥[M(Sα,Tα,t0)]2*[M(Sβ,Tβ,t0)]2.
In general, for all n∈ℕ, we get
(35)M(Sα,Tα,ϕn(t0))*M(Sβ,Tβ,ϕn(t0)) ≥[M(Sα,Tα,ϕn-1(t0))]2*[M(Sβ,Tβ,ϕn-1(t0))]2 ≥[M(Sα,Tα,t0)]2n*[M(Sβ,Tβ,t0)]2n.
Then, we have
(36)M(Sα,Tα,t)*M(Sβ,Tβ,t) ≥[M(Sα,Tα,∑p=n0∞ϕp(t0))] *[M(Sβ,Tβ,∑p=n0∞ϕp(t0))] ≥[M(Sα,Tα,ϕn0(t0))] *[M(Sβ,Tβ,ϕn0(t0))] ≥[M(Sα,Tα,t0)]2n0 *[M(Sβ,Tβ,t0)]2n0 ≥(1-μ)*(1-μ)*⋯*(1-μ)︸22n0 ≥1-λ.
So for any λ>0, we obtain
(37)M(Sα,Tα,t)*M(Sβ,Tβ,t)≥1-λ,
for all t>0, and hence Sα=Tα and Sβ=Tβ. Therefore
(38)Sα=Tα=A(α,β)=B(α,β),Sβ=Tβ=A(β,α)=B(β,α).
Now we show that Sα=α and Sβ=β. Since * is a t-norm of H-type, for any λ>0, there exists an μ>0 such that
(39)(1-μ)*(1-μ)*⋯*(1-μ)︸p≥1-λ,
for all p∈ℕ.
Since M(x,y,·) is continuous and limt→+∞M(x,y,t)=1 for all x,y∈X, there exists t0>0 such that M(Sα,α,t0)≥1-μ and M(Sβ,β,t0)≥1-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 t>∑p=n0∞ϕp(t0). On using inequality (15) with x=α, y=β, u=xn′, v=yn′, we have
(40)M(A(α,β),B(xn′,yn′),ϕ(t0)) ≥M(Sα,Txn′,t0)*M(Sβ,Tyn′,t0).
Letting n→∞, we obtain
(41)M(Sα,α,ϕ(t0))≥M(Sα,α,t0)*M(Sβ,β,t0).
Similarly, we can get
(42)M(Sβ,β,ϕ(t0))≥M(Sβ,β,t0)*M(Sα,α,t0).
Consequently, from (41) and (42), we have
(43)M(Sα,α,ϕ(t0))*M(Sβ,β,ϕ(t0)) ≥[M(Sα,α,t0)]2*[M(Sβ,β,t0)]2.
In general, for all n∈ℕ, we get
(44)M(Sα,α,ϕn(t0))*M(Sβ,β,ϕn(t0)) ≥[M(Sα,α,ϕn-1(t0))]2*[M(Sβ,β,ϕn-1(t0))]2 ≥[M(Sα,α,t0)]2n*[M(Sβ,β,t0)]2n.
Then, we have
(45)M(Sα,α,t)*M(Sβ,β,t) ≥[M(Sα,α,∑p=n0∞ϕp(t0))] *[M(Sβ,β,∑p=n0∞ϕp(t0))] ≥[M(Sα,α,ϕn0(t0))] *[M(Sβ,β,ϕn0(t0))] ≥[M(Sα,α,t0)]2n0 *[M(Sβ,β,t0)]2n0 ≥(1-μ)*(1-μ)*⋯*(1-μ)︸22n0 ≥1-λ.
Therefore for any λ>0, we obtain
(46)M(Sα,α,t)*M(Sβ,β,t)≥1-λ,
for all t>0 and so Sα=α and Sβ=β. Thus
(47)α=Sα=Tα=A(α,β)=B(α,β),β=Sβ=Tβ=A(β,α)=B(β,α).
Finally, we assert that α=β. Since * is a t-norm of H-type, for any λ>0, there exists an μ>0 such that
(48)(1-μ)*(1-μ)*⋯*(1-μ)︸p≥1-λ,
for all p∈ℕ.
Since M(x,y,·) is continuous and limt→+∞M(x,y,t)=1 for all x,y∈X, there exists t0>0 such that M(α,β,t0)≥1-μ.
Also, since ϕ∈Φ, by condition (ϕ-3), we have ∑n=1∞ϕn(t0)<∞. Then for any t>0, there exists n0∈ℕ such that t>∑p=n0∞ϕp(t0). On using inequality (15) with x=v=α, y=u=β, we have
(49)M(A(α,β),B(β,α),ϕ(t0)) ≥M(Sα,Tβ,t0)*M(Sβ,Tα,t0),
and so
(50)M(α,β,ϕ(t0))≥M(α,β,t0)*M(β,α,t0).
Thus we have
(51)M(α,β,t)≥M(α,β,∑p=n0∞ϕp(t0))≥M(α,β,ϕn0(t0))≥[M(α,β,t0)]2n0*[M(β,α,t0)]2n0≥(1-μ)*(1-μ)*⋯*(1-μ)︸22n0≥1-λ,
which implies that α=β. Therefore, we proved that there exists α in X such that
(52)α=Sα=Tα=A(α,α)=B(α,α).
The uniqueness of such a point follows immediately from inequality (15) and so we omit the details.
Next, we illustrate our results providing the following examples.
Example 17.
Let X=[0,+∞), a*b=ab for all a,b∈[0,1] and ψ(s)=s/(s+1) for all s∈ℝ+. Then (X,M,*) is a fuzzy metric space, where
(54)M(x,y,t)=[ψ(t)]|x-y|,
for all x,y∈X and t>0. Let ϕ(s)=s/2, and let the mappings A:X×X→X, S:X→X be defined as
(55)A(x,y)={3x+3y-5,if x,y∈(1,∞),x+y6,otherwise,S(x)={3x-2, if x∈(1,∞),x6, if x∈[0,1].
In view of Definition 10, to prove compatibility, we have only to consider sequences {xn} and {yn} converging to zero from the right. In such case we have
(56)limn→∞A(xn,yn)=0=limn→∞S(xn),limn→∞A(yn,xn)=0=limn→∞S(yn).
Next, we get
(57)limn→∞A(Sxn,Syn)=0=A(0,0),limn→∞SA(xn,yn)=0=S(0),limn→∞A(Syn,Sxn)=0=A(0,0),limn→∞SA(yn,xn)=0=S(0).
Consequently
(58)limn→∞M(A(SxnSyn),SA(xn,yn),t)=1,limn→∞M(A(SynSxn),SA(yn,xn),t)=1,
for all t>0.
On the other hand, to prove subsequential continuity, in view of Definition 12, we have only to consider sequences {xn} and {yn} converging to one from the right. In such case we have
(59)limn→∞A(xn,yn)=1=limn→∞S(xn),limn→∞A(yn,xn)=1=limn→∞S(yn).
Also, note that, for the same sequences, we get
(60)limn→∞A(Sxn,Syn)=1≠A(1,1),limn→∞SA(xn,yn)=1≠S(1),limn→∞A(Syn,Sxn)=1≠A(1,1),limn→∞SA(yn,xn)=1≠S(1),
but
(61)limn→∞M(A(Sxn,Syn),SA(xn,yn),t)=1,limn→∞M(A(Syn,Sxn),SA(yn,xn),t)=1.
Thus, the mappings A and S are compatible as well as subsequentially continuous but not reciprocally continuous. Next, by a routine calculation, one can verify that condition (53) holds true. For instance, for all t>0 and x,y,u,v∈[0,1], we have
(62)M(A(x,y),A(u,v),ϕ(t)) =M(A(x,y),A(u,v),(t2)) =[ψ(t2)]|x-u+y-v|/6 ≥[ψ(t2)]|x-u|/6·[ψ(t2)]|y-v|/6 =[tt+2]|x-u|/6·[tt+2]|y-v|/6 ≥[tt+1]|x-u|/6·[tt+1]|y-v|/6 =[ψ(t)]|x-u|/6·[ψ(t)]|y-v|/6 =M(Sx,Su,t)·M(Sy,Sv,t) =M(Sx,Su,t)*M(Sy,Sv,t).
Example 18.
In the setting of Example 17 (besides retaining the rest), let X=(-∞,∞), and let the mappings A:X × X→X, S:X→X be defined as
(63)A(x,y)={x+y4,if x,y∈(-∞,1),3x+3y-5,if x,y∈[1,∞),x-y4,if x∈(-∞,1), y∈[1,∞),S(x)={x+1,if x∈(-∞,1),3x-2,if x∈[1,∞).
In view of Definitions 11 and 13, to prove reciprocal continuity and subcompatibility, we have only to consider sequences {xn} and {yn} converging to one from the right. For such sequences, we get
(64)limn→∞A(xn,yn)=1=limn→∞S(xn),limn→∞A(yn,xn)=1=limn→∞S(yn).
Also, we deduce that
(65)limn→∞A(Sxn,Syn)=1=A(1,1),limn→∞SA(xn,yn)=1=S(1),limn→∞A(Syn,Sxn)=1=A(1,1),limn→∞SA(yn,xn)=1=S(1).
Therefore, we have
(66)limn→∞M(A(Sxn,Syn),SA(xn,yn),t)=1,limn→∞M(A(Syn,Sxn),SA(yn,xn),t)=1,
for all t>0. Finally, to show that the mappings A and S are not compatible, it suffices to consider the particular sequences {xn}={1/n-2}n∈ℕ and {yn}={1/3n-2}n∈ℕ in X. In fact, in such case, we have
(67)limn→∞A(xn,yn)=-1=limn→∞S(xn),limn→∞A(yn,xn)=-1=limn→∞S(yn).
Next, we deduce that
(68)limn→∞A(Sxn,Syn)=-12=A(-1,-1),limn→∞SA(xn,yn)=0=S(-1),limn→∞A(Syn,Sxn)=-12=A(-1,-1),limn→∞SA(yn,xn)=0=S(-1).
Consequently, we obtain
(69)limn→∞M(A(Sxn,Syn),SA(xn,yn),t)≠1,limn→∞M(A(Syn,Sxn),SA(yn,xn),t)≠1,
for all t>0. Thus, the mappings A and S are reciprocally continuous as well as subcompatible but not compatible. Next, by a routine calculation, one can verify that condition (53) holds true. For instance, for all t>0 and x,y,u,v∈[1,∞), we have
(70)M(A(x,y),A(u,v),ϕ(t)) =M(A(x,y),A(u,v),(t2)) =[ψ(t2)]|3x-3u+3y-3v| ≥[ψ(t2)]3|x-u|·[ψ(t2)]3|y-v| =[tt+2]3|x-u|·[tt+2]3|y-v| ≥[tt+1]3|x-u|.[tt+1]3|y-v| =[ψ(t)]3|x-u|·[ψ(t)]3|y-v| =M(Sx,Su,t)·M(Sy,Sv,t) =M(Sx,Su,t)*M(Sy,Sv,t).
Therefore, all the conditions of Corollary 16 are satisfied, and (1, 1) is the unique common fixed point of the pair (A,S). It is also noted that this example cannot be covered by those fixed point theorems which involve compatibility and reciprocal continuity both.