Proof.
Let x0∈X be arbitrary. Define a sequence {xn} by x2n+1=fx2n, and x2n+2=gx2n+1 for all n≥0. If xm+1=xm for any m∈ℕ, for example, let x2n+1=x2n, then it follows from (6) that
(7)τ+F(p(x2n+1,x2n+2)) =τ+F(p(fx2n,gx2n+1)) ≤F(max{p(x,gy)+p(y,fx)2p(x2n,x2n+1),p(x2n,fx2n),p(x2n+1,gx2n+1),mmmmmmKlp(x2n,gx2n+1)+p(x2n+1,fx2n)2}) =F(max{p(x,gy)+p(y,fx)2p(x2n,x2n+1),p(x2n,x2n+1),p(x2n+1,x2n+2),mmmmmmKlp(x2n,x2n+2)+p(x2n+1,x2n+1)2}) =F(max{p(x,gy)+p(y,fx)2p(x2n,x2n),p(x2n,x2n),p(x2n,x2n+2),mmmmmmKlp(x2n,x2n+2)+p(x2n,x2n)2}) =F(p(x2n,x2n+2)),
that is, F(p(x2n,x2n+2))≤F(p(x2n,x2n+2))-τ. As, τ>0, we get a contradiction. Therefore, we must have F(p(x2n,x2n+2))=0; that is, x2n=x2n+1=x2n+2. Similarly, we obtain x2n=x2n+1=x2n+2=x2n+3=⋯, and so x2n=fx2n=gx2n. Thus, x2n is a common fixed point of f and g. Now, we assume that xm≠xm+1 for all m∈ℕ; then it follows from (6) that
(8)τ+F(p(x2n+1,x2n)) =τ+F(p(fx2n,gx2n-1)) ≤F(max{p(x,gy)+p(y,fx)2p(x2n,x2n-1),p(x2n,fx2n),p(x2n-1,gx2n-1),mmmmmmmKp(x2n,gx2n-1)+p(x2n-1,fx2n)2}) =F(max{p(x,gy)+p(y,fx)2p(x2n,x2n-1),p(x2n,x2n+1),p(x2n-1,x2n),mmmmmmmKp(x2n,x2n)+p(x2n-1,x2n+1)2}) ≤F(max{p(x,gy)+p(y,fx)2p(x2n,x2n-1),p(x2n,x2n+1),mmmmmmmKp(x2n-1,x2n)+p(x2n,x2n+1)2}) =F(max{p(x2n,x2n-1),p(x2n,x2n+1)}).
If max{p(x2n,x2n-1),p(x2n,x2n+1)}=p(x2n,x2n+1), then it follows from the above inequality that τ+F(p(x2n+1,x2n))≤F(p(x2n,x2n+1)), a contradiction (as τ>0). Therefore, we must have max{p(x2n,x2n-1),p(x2n,x2n+1)}=p(x2n,x2n-1). So, setting pn=p(xn,xn+1) from the above inequality we obtain that
(9)F(p2n)≤F(p2n-1)-τ.
Again, using (6) we obtain
(10)τ+F(p(x2n+2,x2n+1)) =τ+F(p(gx2n+1,fx2n))=τ+F(p(fx2n,gx2n+1)) ≤F(max{p(x,gy)+p(y,fx)2p(x2n,x2n+1),p(x2n,fx2n),p(x2n+1,gx2n+1),mmmmmKlmp(x2n,gx2n+1)+p(x2n+1,fx2n)2}) =F(max{p(x,gy)+p(y,fx)2p(x2n,x2n+1),p(x2n,x2n+1),p(x2n+1,x2n+2),mmmmmKlmp(x2n,x2n+2)+p(x2n+1,x2n+1)2}) ≤F(max{p(x,gy)+p(y,fx)2p(x2n,x2n+1),p(x2n+1,x2n+2),mmmmmKlmp(x2n,x2n+1)+p(x2n+1,x2n+2)2}) =F(max{p(x2n,x2n+1),p(x2n+1,x2n+2)}).
If max{p(x2n,x2n+1),p(x2n+1,x2n+2)}=p(x2n+1,x2n+2), then it follows from the above inequality that τ+F(p(x2n+2,x2n+1))≤F(p(x2n+2,x2n+1)), a contradiction (as τ>0). Therefore, we must have max{p(x2n,x2n+1),p(x2n+1,x2n+2)}=p(x2n,x2n+1). So from the above inequality we obtain that
(11)F(p2n+1)≤F(p2n)-τ.
Using (9) and (11) we obtain
(12)F(p2n)≤F(p2n-1)-τ≤F(p2n-2)-2τ≤⋯≤F(p0)-2nτ.
Similarly
(13)F(p2n+1)≤F(p2n)-τ≤F(p2n-1)-2τ≤⋯≤F(p0)-(2n+1)τ.
It follows from (12) and (13) that limn→∞F(pn)=-∞. As F∈ℱ by (F2) we have
(14)limn→∞pn=0.
Again, by (F3), there exists k∈(0,1) such that
(15)limn→∞pnkF(pn)=0.
From (12) and (13) we have
(16)p2nk[F(p2n)-F(p0)]≤-2np2nkτ≤0,p2n+1k[F(p2n+1)-F(p0)]≤-(2n+1)p2n+1kτ≤0.
Using (14) and (15) in the above inequalities we obtain
(17)limn→∞n(pn)k=0.
It follows from above that there exists n0∈ℕ such that n(pn)k<1 for all n>n0; that is,
(18)pn<1n1/k, ∀n>n0.
Let m,n∈ℕ with m>n>n0; then it follows from (18) that
(19)p(xn,xm)≤p(xn,xn+1)+p(xn+1,xn+2)+⋯+p(xm-1,xm) -[p(xn+1,xn+1)+p(xn+2,xn+2)mmll+⋯+p(xm-1,xm-1)]≤pn+pn+1+⋯≤1n1/k+1(n+1)1/k+⋯=∑i=n∞1i1/k.
As k∈(0,1), therefore the series ∑i=n∞1/i1/k converges; so it follows from the above inequality that limn→∞p(xn,xm)=0; that is, the sequence {xn} is a 0-Cauchy sequence in X. Therefore, by 0-completeness of (X,p), there exists u∈X such that
(20)limn,m→∞p(xn,xm)=limn→∞p(xn,u)=p(u,u)=0.
We will prove that u is a common fixed point of f and g. We consider two cases.

Case 1. Suppose f is continuous. Using continuity of f, (20), and Lemma 7 we obtain
(21)p(u,fu)=limn→∞p(x2n+1,fu)=limn→∞p(fx2n,fu)=p(fu,fu).
We claim that p(fu,fu)=0. Suppose p(fu,fu)>0. If for each n∈ℕ there exists kn such that p(xkn+1,fu)=0 and kn>kn-1 with k0=1, then by Lemma 7 and (20) we have fu=u, and so, by Lemma 9, u is a unique common fixed point of f and g.

Now suppose there exists n1∈ℕ such that p(fu,gxn)>0 for all n≥n1. Then, since p(fu,fu)>0, there exists ε>0 such that p(fu,fu)>ε.

For any n≥n1 we have
(22)τ+F(p(fu,gx2n+1)) ≤F(max{p(x,gy)+p(y,fx)2p(u,x2n+1),p(u,fu),p(x2n+1,gx2n+1),mmmmmmmp(u,gx2n+1)+p(x2n+1,fu)2}) =F(max{p(x,gy)+p(y,fx)2p(u,x2n+1),p(fu,fu),p(x2n+1,x2n+2),mmmmmmmp(u,x2n+2)+p(x2n+1,fu)2}).
Using (20) and (21), there exists n2∈ℕ such that p(u,x2n+1),p(x2n+1,x2n+2)<ε/2 and p(x2n+1,fu)<p(fu,fu)+ε/2 for all n>n2; therefore, it follows from the above inequality that
(23)τ+F(p(fu,gx2n+1)) ≤F(max{ε/2+p(fu,fu)+ε/22ε2,p(fu,fu),ε2,mmmmKmmmε/2+p(fu,fu)+ε/22}),F(p(fu,gx2n+1))<F(p(fu,fu)).
As F∈ℱ by (F1) we have p(fu,gx2n+1)<p(fu,fu) for all n>max{n1,n2}, a contradiction. Therefore, we must have p(fu,fu)=p(u,fu)=0; that is, fu=u, and by Lemma 9, u is a unique common fixed point of f and g. Similarly, if g is continuous, then u is a unique common fixed point of f and g.

Case 2. Now suppose that F is continuous. We can assume that there exists n3 such that p(xn,fu)>0 for all n≥n3; otherwise we get fu=u (similar as in previous case).

For any n≥n3, we obtain from (6) that
(24)τ+F(p(fu,x2n+2)) =τ+F(p(fu,gx2n+1)) ≤F(max{p(x,gy)+p(y,fx)2p(u,x2n+1),p(u,fu),p(x2n+1,gx2n+1),mmmmKkKmp(u,gx2n+1)+p(x2n+1,fu)2}) ≤F(max{p(x,gy)+p(y,fx)2p(u,x2n+1),p(u,fu),p(x2n+1,x2n+2),mmmmKkKmp(u,x2n+2)+p(x2n+1,u)+p(u,fu)2}).
From (20), there exists n4∈ℕ such that
(25)max{p(x,gy)+p(y,fx)2p(u,x2n+1),p(u,fu),p(x2n+1,x2n+2),mllllmp(u,x2n+2)+p(x2n+1,u)+p(u,fu)2}=p(u,fu),
for all n>n4. Therefore, it follows from (24) that
(26)τ+F(p(fu,x2n+2))≤F(p(u,fu)), ∀n>max{n3,n4}.
Using continuity of F and letting n→∞ in the above inequality we obtain
(27)τ+F(p(fu,u))≤F(p(u,fu)),
a contradiction (as τ>0). Therefore we must have p(u,fu)=0; that is, fu=u. Again by Lemma 9 u is a unique common fixed point of f and g.

The following are some examples which illustrate the above results and that the generalizations are proper.