We start this section by recalling the following two classes of functions.

Proof.
We will follow the lines in the proof of the main result in [13]. By injection of T, we easily check that M(Tx,Ty)=0 if and only if x=y is a common fixed point of f and g. Let x0∈X. We define two iterative sequences {xn} and {yn} in the following way:
(3.3)x2n+2=fx2n+1, x2n+1=gx2n , yn=Txn, ∀n=0,1,2,….
We prove {yn} is a Cauchy sequence. For this purpose, we first claim that lim n→∞d(yn+1,yn)=0. It follows from property of φ that if n is odd
(3.4)ψ(d(yn+1,yn)) =ψ(d(Txn+1,Txn))=ψ(d(Tfxn,Tgxn-1))≤ψ(M(Txn,Txn-1))-φ(M(Txn,Txn-1))≤ψ(M(Txn,Txn-1)),
where
(3.5)M(Txn,Txn-1)=max{d(Tgxn-1,Txn)+d(Tfxn,Txn-1)2d(Txn,Txn-1),d(Tfxn,Txn),d(Tgxn-1,Txn-1), d(Tgxn-1,Txn)+d(Tfxn,Txn-1)2}=max{d(yn,yn-1),d(yn+1,yn),d(yn,yn-1),d(yn-1,yn+1)2} ≤max{d(yn,yn-1),d(yn+1,yn),d(yn-1,yn)+d(yn,yn+1)2}.
Hence, we have (3.6)ψ(d(yn+1,yn))≤ψ(max{d(yn,yn-1),d(yn+1,yn),d(yn-1,yn)+d(yn,yn+1)2 }).
If d(yn,yn+1)>d(yn-1,yn)≥0 then M(Txn,Txn-1)=d(yn,yn+1), hence
(3.7)ψ(d(yn,yn+1))≤ψ(d(yn,yn+1))-φ(d(yn,yn+1))
and which contradicts with d(yn,yn+1)>0 and the property of φ. Thus, it follows from (3.5) that
(3.8)d(yn+1,yn)≤M(Txn,Txn-1)=d(yn,yn-1).
If n is even then by the same argument above, we obtain
(3.9)d(yn+1,yn)≤M(Txn-1,Txn)=d(yn,yn-1).
Therefore,
(3.10)d(yn+1,yn)≤M(Txn,Txn-1)=d(yn,yn-1)
for all n and {d(yn,yn+1)} is a nonincreasing sequence of nonnegative real numbers. Hence, there exists r≥0 such that
(3.11)limn→∞d(yn,yn+1)=limn→∞M(Txn,Txn-1)=r.
By the lower semicontinuity of φ, we have
(3.12)φ(r)≤limn→∞inf φ(M(Txn,Txn-1)).
Taking the upper limits as n→∞ on either side of
(3.13)ψ(d(yn,yn+1))≤ψ(M(Txn,Txn-1))-φ(M(Txn,Txn-1)),
we get
(3.14)ψ(r)≤ψ(r)-limn→∞infφ(M(Txn,Txn-1))≤ψ(r)-φ(r),
that is, φ(r)≤0. By the property of φ, this implies that φ(r)=0. It follows that r=0 and
(3.15)limn→∞d(yn,yn+1)=0.
It is implied from (3.10) that
(3.16)limn→∞M(Txn,Txn-1)=0.
Now, we claim that {yn} is a Cauchy sequence. Since lim n→∞ d(yn,yn+1)=0, it is sufficient to prove that {y2n} is a Cauchy sequence. Suppose on the contrary that {y2n} is not a Cauchy sequence. Then, there exist ɛ>0 and subsequences {y2n(k)} and {y2m(k)} of {y2n} such that n(k) is the smallest index for which
(3.17)n(k)>m(k)>k, d(y2m(k),y2n(k))>ɛ.
This means that
(3.18)d(y2m(k),y2n(k)-2)<ɛ.
From (3.18) and the triangle inequality, we get
(3.19)ɛ≤d(y2m(k),y2n(k))≤d(y2m(k),y2n(k)-2)+d(y2n(k)-2,y2n(k)-1)+d(y2n(k)-1,y2n(k))<ɛ+d(y2n(k)-2,y2n(k)-1)+d(y2n(k)-1,y2n(k)).
Letting k→∞ and using (3.15), we get
(3.20)limk→∞d(y2m(k),y2n(k))=ɛ.
By the fact
(3.21)|d(y2m(k),y2n(k)+1)-d(y2m(k),y2n(k))|≤d(y2n(k),y2n(k)+1)|d(y2m(k)-1,y2n(k))-d(y2m(k),y2n(k))|≤d(y2m(k)-1,y2m(k))
and using (3.15) and (3.20), we obtain
(3.22)limk→∞d(y2m(k)-1,y2n(k))=limk→∞d(y2m(k),y2n(k)+1)=ɛ.
Moreover, from
(3.23)|d(y2m(k)-1,y2n(k)+1)-d(y2m(k)-1,y2n(k))|≤d(y2n(k),y2n(k)+1)
and combining with (3.15) and (3.22), we conclude that
(3.24)limk→∞d(y2m(k)-1,y2n(k)+1)=ɛ.
Now, by the definition of M(Tx,Ty) and from (3.10), (3.15), and (3.20)–(3.24), we can deduce that
(3.25)limk→∞M(Tx2m(k)-1,Tx2n(k))=ɛ.
Due to (3.1), we have (3.26)ψ(d(y2m(k),y2n(k)+1))=ψ(d(Tx2m(k) ,Tx2n(k)+1))=ψ(d(Tfx2m(k)-1,Tgx2n(k)))≤ψ(M(Tx2m(k)-1,Tx2n(k)))-φ(M(Tx2m(k)-1,Tx2n(k))).
Letting k→∞ and using (3.22) and (3.25), we have
(3.27)ψ(ɛ)≤ψ(ɛ)-φ(ɛ).
It is a contradiction to φ(t)>0 for every t>0. This proves that {yn} is a Cauchy sequence.

Since X is a complete metric space, there exists u∈X such that lim n→∞yn=u. Since T is sequentially convergent, we can deduce that {xn} converges to v∈X. By the continuity of T, we infer that
(3.28)u=limn→∞yn=limn→∞Txn=Tv.

We will show that v=fv=gv. Indeed, suppose that v≠fv, since T is injective, we have u=Tv≠Tfv. Hence, d(Tv,Tfv)>0. Since
(3.29)limn→∞y2n+1=limn→∞y2n=u,limn→∞d(y2n,y2n+1)=0,
we can seek N0∈ℕ such that for any n≥N0(3.30)d(y2n+1,u)<d(Tv,Tfv)4, d(y2n,u)<d(Tv,Tfv)4, d(y2n,y2n+1)<d(Tv,Tfv)4.
Then, we have
(3.31)d(Tv,Tfv)≤M(Tv,Tx2n)=max{d(Tv,Tgx2n)+d(Tx2n,Tfv)2d(Tv,Tx2n),d(Tv,Tfv),d(Tx2n,Tgx2n), d(Tv,Tgx2n)+d(Tx2n,Tfv)2}=max{d(u,y2n+1)+d(y2n,Tfv)2d(u,y2n),d(Tv,Tfv),d(y2n,y2n+1), d(u,y2n+1)+d(y2n,Tfv)2}≤max{d(u,y2n+1)+d(y2n,Tv)+d(Tv,Tfv)2d(u,y2n),d(Tv,Tfv),d(y2n,y2n+1), d(u,y2n+1)+d(y2n,Tv)+d(Tv,Tfv)2}≤max{d(Tv,Tfv)4,d(Tv,Tfv),d(Tv,Tfv)4, d(Tv,Tfv)/4+d(Tv,Tfv)/4+d(Tv,Tfv)2}≤max{d(Tv,Tfv),34 d(Tv,Tfv)}=d(Tv,Tfv).
Therefore, M(Tv,Tx2n)=d(Tv,Tfv) for every n≥N0. Since
(3.32)ψ(d(Tfv,y2n+1))=ψ(d(Tfv,Tx2n+1))=ψ(d(Tfv,Tgx2n))≤ψ(M(Tv,Tx2n))-φ(M(Tv,Tx2n))=ψ(d(Tv,Tfv))-φ(d(Tv,Tfv))
and letting n→∞, we arrive at
(3.33)ψ(d(Tfv,Tv))≤ψ(d(Tv,Tfv))-φ(d(Tv,Tfv)).
We get a contradiction. Hence, v=fv. By the same argument, we get v=gv.

Let w∈X such that w=fw=gw. Then, we have
(3.34)M(Tv,Tw)=max{d(Tv,Tw),d(Tv,Tfv),d(Tw,Tgw),d(Tv,Tgw)+d(Tfv,Tw)2}=max{d(Tv,Tw),d(Tv,Tw)+d(Tv,Tw)2}=d(Tw,Tv).
Thus
(3.35)ψ(d(Tv,Tw))=ψ(d(Tfv,Tgw))≤ψ(M(Tv,Tw))-φ(M(Tv,Tw))=ψ(d(Tv,Tw))-φ(d(Tv,Tw)).
This implies that d(Tv,Tw)=0, or Tv=Tw. Since T is injective, we have w=v. The theorem is proved.

Example 3.3.
Let X=[1,+∞) and d be the usual metric in X. Consider the maps f(x)=g(x)=4x. It is easy to see that 16 is the unique fixed point of f and g. We claim that f and g are not generalized φ-weak contraction. Indeed, if there exist lower semicontinuous functions ψ,φ:[0,∞)→[0,∞) with ψ(t)>0, φ(t)>0 for t∈(0,∞) and φ(0)=ψ(0)=0, such that
(3.36)ψ(d(fx,gy))≤ψ(M(x,y))-φ(M(x,y)), ∀x,y∈X,
then
(3.37)ψ(4|x-y|)≤ψ(M(x,y))-φ(M(x,y)), ∀x,y∈X,
where M(x,y)=max {d(x,y),d(fx,x),d(gy,y),(1/2)[d(gx,y)+d(fy,x)]}. For x=4 and y=1, we obtain
(3.38)M(x,y)=max{3,4,3,72}=4.
It follows from (3.37) that
(3.39)ψ(4)≤ψ(4)-φ(4).
Hence, φ(4)≤0. We arrive at a contradiction with φ(t)>0 for t∈(0,∞).

Consider the map Tx=ln x+1, for all x∈X. It is easy to see that T is injective, continuous, and sequentially convergent. Let ψ(t)=t and φ(t)=t/3, for all t∈[0,+∞). Now, we show that f and g are generalized φ-weak T-contractions. It reduces to check the following inequality:
(3.40)|ln4x-ln4y|≤23M(Tx,Ty), ∀x,y∈[1,+∞).
We have
(3.41)|ln4x-ln4y|=12|lnxy|(3.42)M(Tx,Ty)=max{|ln4y-lnx|+|ln4x-lny|2|lnx-lny|,|ln4x-lnx|,|ln4y-lny| |ln4y-lnx|+|ln4x-lny|2}≥|lnx-lny|=|lnxy|.
It follows from (3.41) and (3.42) that
(3.43)|ln4x-ln4y|≤12M(Tx,Ty)
for every x,y∈X. This proves that (3.40) is true.