We start this section with the following theorem.

Proof.
We first prove that the coincidence point of g and F is unique if it exists. Let z and w be coincidence points of g and F. Thus, there exists some x,y∈X such that w=Fx=gx and z=Fy=gy. By (5), we derive that
(7)ψ(d(w,z))=ψ(d(Fx,Fy))≤ψ(M(gx,gy))-ϕ(M(gx,gy)),
where
(8)M(gx,gy)=max{d(gx,gy),d(gx,Fx),d(gy,Fy)}=d(gx,gy)=d(w,z).
Thus, we conclude that z=w by (7).

Remember that g and F are weakly compatible. Since z is the unique coincidence point of g and F, the point z is the unique common fixed point of g and F by Lemma 4.

Now, we will prove the existence of a coincidence point of g and F. Let x0 be an arbitrary point. Since FX⊂gX, we define two iterative sequences {xn} and {yn} in X as follows:
(9)yn=gxn+1=Fxn
for all n=0,1,2,…. If yn=yn+1 then clearly F and g have a coincidence point in X. Indeed, yn=gxn+1=Fxn=gxn+2=Fxn+1=yn+1 and xn+1 is the desired point. Thus, we assume that yn≠yn+1, that is, d(yn,yn+1)>0 for all n=0,1,…. Moreover, if yn=Fxn=Fxn+p=yn+p, then we choose xn+p+1=xn+1, for all n≥0.

We assert that
(10)limn→∞d(yn,yn+1)=0, limn→∞d(yn,yn+2)=0.
Now from (5), we have
(11)ψ(d(yn,yn+1))=ψ(d(Fxn,Fxn+1))≤ψ(M(gxn,gxn+1))-ϕ(M(gxn,gxn+1)),
where
(12)M(gxn,gxn+1) =max{d(gxn,gxn+1),d(gxn,Fxn), d(gxn+1,Fxn+1)} =max{d(gxn,gxn+1),d(gxn,gxn+1), d(gxn+1,gxn+2)} =max{d(yn-1,yn),d(yn-1,yn),d(yn,yn+1)} =max{d(yn-1,yn),d(yn,yn+1)}.
If M(gxn,gxn+1)=d(yn,yn+1), then we have
(13)ψ(d(yn,yn+1))≤ψ(d(yn,yn+1))-ϕ(d(yn,yn+1)),
which implies that ϕ(d(yn,yn+1))=0, and hence d(yn,yn+1)=0. Then yn=yn+1, which contradicts with the initial assumption. Thus, we have M(gxn,gxn+1)=d(yn-1,yn), and hence
(14) ψ(d(yn,yn+1))≤ψ(d(yn-1,yn))-ϕ(d(yn-1,yn))≤ψ(d(yn-1,yn)).
Since ψ is nondecreasing, then d(yn,yn+1)≤d(yn-1,yn) for all n≥0, that is, the sequence {d(yn,yn+1)} is nonincreasing and bounded below. Hence, it converges to a positive number, say r>0. Taking the limit as n→∞ in (14), we get
(15)ψ(r)≤ψ(r)-ϕ(r),
which leads to ϕ(r)=0, and hence r=0. Thus,
(16)limn→∞d(yn,yn+1)=0.
In this step, we show that the second limit in (5) is also 0. To prove this claim, we set x=xn-1 and y=xn+1 in (5) and get that
(17)ψ(d(yn,yn+2)) =ψ(d(Fxn,Fxn+2))≤ψ(M(gxn,gxn+2)) -ϕ(M(gxn,gxn+2)),
where
(18)M(gxn,gxn+2) =max{d(gxn,gxn+2),d(gxn,Fxn),d(gxn+2,Fxn+2)} =max{d(gxn,gxn+2),d(gxn,gxn+1),d(gxn+2,gxn+3)} =max{d(yn-1,yn+1),d(yn-1,yn),d(yn+1,yn+2)}.
We consider all possible cases for M(gxn,gxn+2). If M(gxn,gxn+2)=d(yn+1,yn+2), then we have
(19)ψ(d(yn,yn+2))≤ψ(d(yn+1,yn+2))-ϕ(d(yn+1,yn+2)).
Letting n→∞ in (19), the right hand side of (19) tends to 0. Hence, limn→∞ψ(d(yn,yn+2))=0. Since, ψ is continuous, we find that limn→∞d(yn,yn+2)=0. For the case M(gxn,gxn+2)=d(yn-1,yn), we get analogously limn→∞d(yn,yn+2)=0.

Let us consider the last case, that is, M(gxn,gxn+2)=d(yn-1,yn+1). The inequality (17) turns into
(20) 0≤ψ(d(yn,yn+2))≤ψ(d(yn-1,yn+1))-ϕ(d(yn-1,yn+1))≤ψ(d(yn-1,yn+1)).
Therefore, the sequence {d(yn,yn+2)} is non-increasing and bounded below. Hence, the sequence {d(yn,yn+2)} converges to a number, s≥0. Taking limit as n→∞ in (20), we get
(21)0≤ψ(d(s))≤ψ(s)-ϕ(s),
which implies that ϕ(s)=0, and hence s=0. In other words,
(22)limn→∞d(yn,yn+2)=0.
Suppose that yn≠ym for all m≠n and prove that {yn} is a RMS Cauchy sequence. If possible, let {yn} be not a Cauchy sequence. Then there exists ϵ>0 for which we can find subsequences {ymk} and {ynk} of {yn} with nk>mk≥k such that
(23)d(ymk,ynk)≥ϵ.
Furthermore, corresponding to mk, we can choose nk in such a way that it is the smallest integer with nk>mk and satisfying (23). Then,
(24)d(ymk,ynk-1)<ϵ.
Using (23), (24), and the rectangular inequality (RM3), we have
(25)ϵ≤d(ymk,ynk)≤d(ynk,ynk-2)+d(ynk-2,ynk-1)+d(ynk-1,ymk)≤d(ynk,ynk-2)+d(ynk-2,ynk-1)+ϵ.
Taking limit as k→∞ in (23) and using (16), (22) we get
(26)limk→∞d(ymk,ynk)=ϵ.
Again, using the rectangular inequality (RM3), we obtain
(27)d(ynk,ymk)-d(ymk,ymk-1)-d(ynk-1,ynk) ≤d(ynk-1,ymk-1) ≤d(ynk-1,ynk)+d(ynk,ymk)+d(ymk,ymk-1).
Letting k→∞ in (27), and by using (16) and (22) we get
(28)limk→∞d(ynk-1,ymk-1)=ϵ.
Now, we substitute x=xnk and y=xmk in (5). Consider
(29)ψ(d(ynk,ymk))=ψ(d(Fxnk,Fxmk))≤ψ(M(gxnk,gxmk)) -ϕ(M(gxnk,gxmk))
where
(30)M(gxnk,gxmk)=max{d(gxnk,gxmk),d(gxnk,Fxnk), d(gxmk,Fxmk)}=max{d(ynk-1,ymk-1),d(ynk-1,ynk), d(ymk-1,ymk)}.
Clearly, as k→∞, we have
(31)M(gxnk,gxmk)→max{ϵ,0,0}=ϵ.
Then letting k→∞ in (29), we have
(32)0≤ψ(ϵ)≤ψ(ϵ)-ϕ(ϵ).
This implies that
(33)ϕ(ϵ)=0, hence ϵ=0,
which contradicts the fact that ϵ>0. Thus, {yn} is a RMS Cauchy sequence. Since (gX,d) is RMS complete, there exists z∈gX such that yn→z as n→∞. Let y∈X such that gy=z. Applying the inequality (5), with x=xn, we obtain
(34)ψ(d(Fxn,Fy))≤ψ(M(gxn,gy))-ϕ(M(gxn,gy)),
where
(35)M(gxn,gy)=max{d(gxn,gy),d(gxn,Fxn),d(gy,Fy)}=max{d(gxn,gy),d(gxn,gxn+1), d(gy,Fy)}.
Now, if M(gxn,gy)=d(gxn,gy) or M(gxn,gy)=d(gxn,gxn+1), we have
(36)d(Fxn,Fy)≤d(gxn,gy) or d(Fxn,Fy)≤d(gxn,gxn+1),
since ψ is nondecreasing. In either case, letting n→∞, we get gxn+1=Fxn→Fy. Since X is Hausdorff, we deduce that gy=Fy. If, on the other hand, M(gxn,gy)=d(gy,Fy), then taking limit as n→∞ in
(37)ψ(d(Fxn,Fy))≤ψ(d(gy,Fy))-ϕ(d(gy,Fy)),
we get ϕ(d(gy,Fy))=0, hence d(gy,Fy)=0, that is, gy=Fy. Let z=gy=Fy. Then z is a point of coincidence of F and g. Suppose that there exists n,p∈ℕ such that yn=yn+p, we can choose in such a way that it is the smallest positive integer satisfying yn=yn+p. We aim to prove that p=1, then
(38)gxn+1=Fxn=Fxn+1=yn+1,
and so yn+1 is a point of coincidence of g and F. Assume that p>1. This implies that d(yn+p-1,yn+p)>0. Now we have
(39)ψ(d(yn,yn+1)) =ψ(d(yn+p,yn+p+1))=ψ(d(Fxn+p,Fxn+p+1)) ≤ψ(M(gxn+p,gxn+p+1))-ϕ(M(gxn+p,gxn+p+1)),
where
(40)M(gxn+p,gxn+p+1) =max{d(gxn+p,gxn+p+1),d(gxn+p,Fxn+p), d(gxn+p+1,Fxn+p+1)} =max{d(gxn+p,gxn+p+1),d(gxn+p,gxn+p+1), d(gxn+p+1,gxn+p+2)} =max{d(gxn+p,gxn+p+1),d(gxn+p+1,gxn+p+2)} =max{d(yn+p-1,yn+p),d(yn+p,yn+p+1)} =max{d(yn-1,yn),d(yn,yn+1)}.
If M(gxn+p,gxn+p+1)=d(yn,yn+1), then we have
(41)ψ(d(yn,yn+1))=ψ(d(yn+p,yn+p+1))<ψ(d(yn,yn+1)),
which is a contradiction. Thus p=1.