Proof.
Define A and B by (12). For A≥0, inequality (25) is trivial. Now assume that A<0 that is, 0<p<a(1-q)<a. Since B=p-a<0, it follows from Lemma 5(ii) that f(u,v) is increasing in v, and xn≥-Aa-1 implies(27)xn+2=fxn+1,xn≥fxn+1,-Aa-1=-Aa-p>-Aa.Hence, (25) will follow if each of the subsequences {x2n}n∈N0 and {x2n+1}n∈N0 enters the interval [-Aa-1,∞). For the sake of contradiction, suppose that, for some m∈{0,1}, we have(28)xm+2n∈0,-Aa-1,n∈N0.Then, by Lemma 5(i), f(u,xm+2n) is increasing in u, and, for all n∈N0, the inequality xm+2n+1xm+2n>a implies that(29)xm+2n+2=fxm+2n+1,xm+2n>faxm+2n-1,xm+2n.Since p>0, it follows that(30)fax-1,x=xaq+ax+paq+x>x,x>0.In view of (29) and (30), (28) implies that the subsequence {xm+2n}n∈N0 is increasing and must converge to some x∗∈(0,-Aa-1] that is, (31)x∗≤1-q-pa-1<1-q.On the other hand, Lemma 6 implies that x∗>1-q, which is a contradiction. Hence, (28) cannot be true, and the proof of (25) is complete. Then, for every n>n0, we have(32)xn+2=fxn+1,xn=xn+1xn+axn+Axn+1xn+q≥xnq+xn.If xn>1-q for some n>n0, then it follows from (32) that(33)xn+2≥xnq+xn=1-qq+xn>1-q.It remains to show that both subsequences {x2n}n∈N0 and {x2n+1}n∈N0 eventually enter the interval (1-q,∞). For the sake of contradiction, suppose that, for some m∈{0,1}, we have(34)xm+2n≤1-q,n∈N0.Then, for every n∈N0 such that m+2n>n0, it follows from (32) that xm+2n+2≥xm+2n. Therefore, the subsequence {xm+2n}n∈N0 is nondecreasing and must converge to some x∗∈(0,1-q]. However, Lemma 6 implies that x∗>1-q, which is a contradiction. Hence, (34) cannot be true and the proof of (26) follows. The proof is complete.

Now we are ready to prove our main result.

Proof.
From Lemma 5(iv), it follows that there exists L>0 such that xn∈(0,L) for all n∈N0. Then, (6) implies that(35)xn+2>pxn+1q+xn>pLq+L=b>0.Thus,(36)0<b<xn<L,n≥3.First, assume that a≥p. If a-aq-p<0, then we have that f(u,v)>0 for all u,v>0 and, by Lemma 5, f(u,v) is decreasing in u and increasing in v. Therefore, in this case, the proof follows from (36), Theorem 2 with I=(0,∞), and Lemma 6. Now assume that a-aq-p≥0 that is, 0<p≤a(1-q)<a. Then q<1, and, by Lemma 7, we have xn>1-q eventually. It is easy to see that u,v∈(1-q,∞) implies f(u,v)∈(1-q,∞). From the inequalities av>a(1-q)-p and qu>q(1-q)>0≥p-a, we obtain by Lemma 5 that f(u,v) is decreasing in u and increasing in v on (1-q,∞). Hence, in this case, the proof follows from (36), Theorem 2 with I=(1-q,∞), and Lemma 6.

Now we assume that a<p. Define A and B as in (12). Since A≥B>0, from Lemma 5(v), it follows that, for all u,v>0,(37)u-Bq-1fu,v-AB-1<0provided u≠Bq-1,(38)u-Bq-1fu,v-Aqu-1>0provided u≠Bq-1,(39)fBq-1,v=AB-1.If Bq-1<u≤AB-1 and v>0, then (37) and (38) imply(40)AB-1>fu,v>Aqu-1≥Bq-1,and, in view of (39), we obtain(41)fu,v∈Bq-1,AB-1,u∈Bq-1,AB-1, v>0.In the same fashion, we obtain(42)fu,v∈AB-1,Bq-1,u∈AB-1,Bq-1, v>0.Now, assume that xm∈[Bq-1,AB-1] for some m>1. Then, by (41), we have(43)xn∈Bq-1,AB-1,n≥m.From (41), Lemma 5(i)(ii), and Theorem 2 with I=[Bq-1,AB-1], it follows that the subsequence {x2n}n∈N0 is monotone. But it is also bounded, in view of (41), and must converge to some xe∈[b,L]. Therefore, by Lemma 6, we obtain xe=x∗ and(44)limn→∞xn=x∗.Next, assume that xm∈[AB-1,Bq-1] for some m>1. Then, by (42), we have(45)xn∈AB-1,Bq-1,n≥m.In this case, I=[AB-1,Bq-1]≠∅, the function f:I×I→I defined by (7) is continuous, and, by Lemma 5(i)(ii), it is nonincreasing in each of its arguments. In order to verify the second condition of Theorem 4, we suppose for the sake of contradiction that there exist m,M∈I with f(m,m)=M, f(M,M)=m, and m≠M, which implies(46)M-mM+m+a-mM=0,mM=M+m+a.Then, in view of f(m,m)=M, we obtain(47)m+M=B-aq+1,mM=Aq+1.Thus, m and M are different positive roots of the quadratic polynomial(48)z2-B-aq+1z+Aq+1.Therefore, B-a>0 and (B-a)2-4A(q+1)>0, which contradicts (10). Since x∗ is the unique positive root of the characteristic equation (17), it follows from (42) and Theorem 4 with I=[AB-1,Bq-1] that (44) holds. It remains to prove (44) in the last case, when(49)xn∉c,d,n≥2,where(50)c=minBq-1,AB-1,d=maxBq-1,AB-1.In view of Lemma 7, (36), and (49), there exists some n0 such that(51)xn∈m,c∪d,L,n≥n0,where(52)m=maxb,1-q.Since the sequence {xn0+n}n∈N0 is also a positive solution of (6), in view of (37) and (38), we may assume without loss of generality that(53)0<m<x2n<c≤d<x2n+1<L,n∈N0.Then, by Lemma 5(i)(ii), f(x2k,v) is decreasing in v, while f(x2k+1,v) is increasing in v and f(u,v) is decreasing in u. Hence, for every n∈N0, we obtain that(54)c>x2n+2=fx2n+1,x2n>fx2n+1,m>fL,m,d<x2n+3=fx2n+2,x2n+1<fx2n+2,d<fm,d.The equation f(u,v)=z has a unique solution for u:(55)u=gv,z=av+aq+p-azq+v-vprovided z≠vq+v.Observe that the inequality f(u,v)<z is equivalent to u>g(v,z), provided z(q+v)-v>0. Since (52) and (53) imply that(56)cm>1≥1m+q,d+q>1,it follows from (54) that(57)x2n+1>gm,c,x2n+2<gd,dn∈N0.Set m1=m, c1=c, d1=d, L1=L, and, for s∈N,(58)ms+1=maxms,fLs,ms,cs+1=mincs,gds,ds,ds+1=maxds,gms,cs,Ls+1=minLs,fms,ds.Then, by induction, we obtain that(59)0<ms<x2k<cs≤ds<x2k+1<Ls,k≥s.Since the sequences {ms}, {cs}, {ds}, and {Ls} are monotone and bounded, there exist m∗,c∗,d∗,L∗>0 such that(60)lims→∞ms=m∗,lims→∞cs=c∗,lims→∞ds=d∗,lims→∞Ls=L∗.By letting s→∞ in (58) and (59), we obtain(61)m∗=maxm∗,fL∗,m∗,c∗=minc∗,gd∗,d∗,d∗=maxd∗,gm∗,c∗,L∗=minL∗,fm∗,d∗,(62)0<m∗≤c∗≤d∗≤L∗.Therefore,(63)m∗≥fL∗,m∗,d∗≥gm∗,c∗,(64)c∗≤gd∗,d∗,L∗≤fm∗,d∗.But (63) implies L∗≥g(m∗,m∗) and f(d∗,m∗)≤c∗ and, in view of (64), we obtain(65)fd∗,m∗≤gd∗,d∗,fm∗,d∗≥m∗,m∗.Finally, (65), (7), and (55) imply that(66)q-1m∗d∗-a+p≤pm∗-p-ad∗-m∗d∗2,q-1d∗m∗-a+p≥pd∗-p-am∗-d∗m∗2.Thus,(67)pd∗-p-am∗-d∗m∗2≤pm∗-p-ad∗-m∗d∗2,2p-a+d∗m∗d∗-m∗≤0.Hence d∗≤m∗ and, in view of (62), m∗=c∗=d∗. Then, from (59), it follows that the subsequence {x2n}n∈N0 converges to d∗>0, and the proof is complete by Lemma 6.