Proof.
Let l=max{mt,ks}. Since
(4)a=inf(u1,u2,…,us,v1,v2,…,vt)∈Es+tf(u1,u2,…,us;v1,v2,…,vt)a∈E,
we have
(5)x=f(x,x,…,x)>f(x+1,x,…,x)≥a.
Claim 1.
f
(
A
,
…
,
A
;
a
,
…
,
a
)
<
x

<
A
.
Proof of Claim 1. Assume on the contrary that x≥A. Then it follows from (H1), (H3), and (H4) that
(6)A=f(a,…,a;A,…,A)>f(x,…,x;A,…,A)=f(x,…,x;A,…,A)AA≥f(x,…,x)xA=A.
This is a contradiction. Therefore x<A. Obviously
(7)f(A,…,A;a,…,a)<f(x,…,x;x,…,x)=x.
Claim 1 is proven.
Claim 2. For any M≥A,J=[a,M] is an invariable interval of (2).
Proof of Claim 2. For any x0,x1,…,xl∈J, we have from (H4) that
(8)a≤x1=f(xk1,xk2,…,xks;xm1,xm2,…,xmt)≤f(a,…,a;M,…,M)MM≤f(a,…,a;A,…,A)AM=M.
By induction, we may show that xn∈J for any n≥1. Claim 2 is proven.
Let m0=a,M0=M≥A and for any i≥0,
(9)mi+1=f(Mi,…,Mi;mi,…,mi),Mi+1=f(mi,…,mi;Mi,…,Mi).
Claim 3. For any n≥0, we have
(10)mn≤mn+1<x<Mn+1≤Mn, limn→∞Mn=limn→∞mn=x.
Proof of Claim 3. From Claim 2, we obtain
(11)m0≤m1=f(M0,…,M0;m0,…,m0)<f(x,…,x)=x<f(m0,…,m0;M0,…,M0)=M1≤M0,m1=f(M0,…,M0;m0,…,m0)≤f(M1,…,M1;m1,…,m1)=m2<f(x,…,x)=x<f(m1,…,m1;M1,…,M1)=M2≤f(m0,…,m0;M0,…,M0)=M1.
By induction, we have that for n≥0,
(12)mn≤mn+1<x<Mn+1≤Mn.
Set
(13)β=limn→∞mn and α=limn→∞Mn.
Then
(14)β=f(α,…,α;β,…,β),α=f(β,…,β;α,…,α).
This with (H2) and (H5) implies α=β=x. Claim 3 is proven.
Claim 4. The equilibrium x of (2) is locally stable.
Proof of Claim 4. Let M=A and mn,Mn be the same as Claim 3. For any ε>0 with 0<ε<min{Ax,xa}, there exists n>0 such that
(15)xε<mn<x<Mn<x+ε.
Set 0<δ=min{xmn,Mnx}. Then for any x0,x1,…,xl∈(xδ,x+δ), we have
(16)x1=f(xk1,…,xks;xm1,…,xmt)≤f(mn,…,mn;Mn,…,Mn)=Mn+1≤Mn,x1=f(xk1,…,xks;xm1,…,xmt)≥f(Mn,…,Mn;mn,…,mn)=mn+1≥mn.
In similar fashion,we can show that for any k≥1,
(17)xk∈[mn,Mn]⊂(xε,x+ε).
Claim 4 is proven.
Claim 5.
x

is the global attractor of (2).
Proof of Claim 5. Let {xn}n=l∞ be a positive solution of (2), and let M=max{x1,…,xl+1,A} and mn,Mn be the same as Claim 3. From Claim 2, we have xn∈[m0,M0]=[a,M] for any n≥1. Moreover, we have
(18)xl+2 =f(xl+1k1,…,xl+1ks;xl+1m1,…,xl+1mt) ≤f(m0,…,m0;M0,…,M0)=M1,xl+2 =f(xl+1k1,…,xl+1ks;xl+1m1,…,xl+1mt) ≥f(M0,…,M0;m0,…,m0)=m1.
In similar fashion, we may show xn∈[m1,M1] for any n≥l+2. By induction, we obtain
(19)xn∈[mk,Mk] for n≥k(l+1)+1.
It follows from Claim 3 that limn→∞xn=x. Claim 5 is proven.
From Claims 4 and 5, Theorem 1 follows.