Proof of Lemma 1 : For any initial vale x(0), we consider the iteration sequence {xi(k)} and their components xi(k). We divide the proof into several steps.
(i) By (2.3) and(2.7),
bi(xi) − c(x) < 0, (3.1)
for all xi ≥ ui. Therefore
∆xi(k) = ai(x(k))[bi(xi(k)) − c(x(k))] < 0, (3.2) if xi(k) ≥ ui. Similarly, By (2.3) and (2.6),
bi(xi) − c(x) > 0, (3.3)
for all xi ≤ li. Therefore
∆xi(k) = ai(x(k))[bi(xi(k)) − c(x(k))] > 0, (3.4) if xi(k) ≤ li. We claim that for all k ∈ N0,
|bi(xi(k))| ≤ di|xi(k)| + |bi(0)|. (3.5) This follows from
bi(xi(k)) − bi(0) = b0i(·)xi(k),
where “·” means some real number between xi(k) and 0. Thus, by (2.5),
|bi(xi(k))| = |bi(0) + b0i(·)xi(k)|
≤ |bi(0)| + |b0i(·)xi(k)|
≤ |bi(0)| + di|xi(k)|.
(ii) Next, we show that for fixed constant Li, there exist some constants u0i and d0i, where u0i > 0, 0 < di < d0i < 1 such that
di|xi| + Li < d0i|xi|, if |xi| ≥ u0i. (3.6) Let us verify this. Notably,
di|xi| + Li
|xi| = di+ Li
|xi| → di < 1,
as |xi| → ∞. Therefore, there exist some u0i and d0i, where u0i > 0, 0 < di < d0i < 1 such that (di|xi| + Li)/|xi| < d0i, if |xi| ≥ u0i.
(iii)
|∆xi(k)| = |ai(x(k))[bi(xi(k)) − c(x(k))]|
≤ |bi(xi(k)) − c(x(k))| (by (2.1))
≤ |bi(xi(k))| + |c(x(k))|
≤ di|xi(k)| + |bi(0)| + |c(x(k))| (by (3.5))
≤ di|xi(k)| + |bi(0)| + M (by (2.3), (2.12)).
Hence, by (3.6), we choose |bi(0)| + M = Li, there exist some constants u0i and d0i where u0i > 0, 0 < di < d0i < 1 such that
|∆xi(k)| < d0i|xi(k)| < |xi(k)|, if |xi(k)| ≥ u0i. (3.7) (iv) Set, for each i,
qi0 := max{|ui|, |li|, u0i}. (3.8) Let Q0 := [−q10, q10] × · · · × [−q0n, q0n]. Q0 is a compact set, hence |ai(x)[bi(xi) − c(x)]|
is bounded on Q0, say
|ai(x)[bi(xi) − c(x)]| ≤ K, (3.9) for all x ∈ Q0, for all i. Set
qi := qi0+ K, (3.10)
Q := [−q1, q1] × · · · × [−qn, qn]. (3.11) We shall utilize (3.2), (3.4), (3.7), (3.8), (3.9), (3.10) in the following discus-sions.
(v) If −qi ≤ xi(0) ≤ qi, then −qi < xi(k) < qi, for all k ∈ N0.
case (a): If xi(0) ∈ [−qi, −qi0], then ∆xi(0) > 0, due to xi(0) ≤ −q0i ≤ li, and
|∆xi(0)| < |xi(0)|, due to xi(0) ≤ −u0i, hence xi(1) still stays in (−qi, −qi0], or moves into (−qi0, qi0). If the former case occurs, we consider xi(1) as case (a) again. If the latter case occurs, we consider xi(1) as in the following case (b).
case (b): If xi(0) ∈ (−qi0, qi0), then |∆xi(0)| < K, by (3.9), hence xi(1) will stay in [−qi, −q0i] or (−q0i, q0i) or [q0i, qi]. Then we can still consider xi(1) as in case (a), case (b), and case (c), respectively.
case (c): If xi(0) ∈ [qi0, qi], then ∆xi(0) < 0, by xi(0) ≥ qi0 ≥ ui, and |∆xi(0)| <
|xi(0)|, by xi(0) ≥ u0i, hence xi(1) still stays in [q0i, qi), or moves into (−qi0, qi0). If the former case occurs, we consider xi(1) as in case (c) again. If the latter case occurs, we consider xi(1) as in case (b). From the above arguments, we find that if
−qi ≤ xi(0) ≤ qi, then −qi < xi(1) < qi, and we can prove that −qi < xi(k) < qi, for all k ≥ 2, by induction.
(vi): If xi(0) < −qi, then
case (d): {xi(k)} either increases as k increases and remains bounded above by −qi, or
case (e): {xi(k)} enter [−qi, qi] at some iteration, and never leaves [−qi, qi] again.
(vii) if xi(0) > qi, then
case (f): {xi(k)} either decreases as k increases and remains bounded below by qi, or
case (g): {xi(k)} enters [−qi, qi] at some iteration, and never leaves [−qi, qi] again.
We find that no matter which case above occurs, {xi(k)} are bounded above and below for all i. Therefore, {|ai(x(k))|} are bounded below by some positive number, say 0 < ρ0i ≤ |ai(x(k))|, and {b0i(xi(k))} are bounded above by some nega-tive number, say b0i(xi(k)) ≤ −²0i < 0. In fact, it is impossible for the above case (d) and case (f) to occur. This is due to that if case (d) occurs, then
bi(xi(k)) − c(x(k)) = bi(xi(k)) − bi(li) + bi(li) − c(x(k))
Hence {xi(k)} will increase unboundedly, and this yields a contradiction. Therefore case (d) never occurs. Similarly, case (f) never occurs, either. By the arguments above, we can find that given any initial value x(0), {x(k)} will be attracted by Q.
Proof of Lemma 2: vector between x(k + 1) and x(k). Thus,
∆gi(k) = b0i(·)ai(x(k))gi(k) −
case (i) |∆gi(k)| ≤ −gi(k): It follows that ˇg(k + 1) ≤ gi(k + 1) = gi(k) + ∆gi(k) ≤ 0. vector between x(k + 1) and x(k). Thus
|∆gi(k)| = b0i(·)ai(x(k))gi(k) −
From the above arguments, we find that function ˆg may keep negative at all iterations. But once it becomes nonnegative at some iteration, it will always remain nonnegative after this iteration. Similarly, ˇg may keep positive at all iterations. But once it get nonpositive at some iteration, it will always be nonpositive after this iteration. This completes the proof of Lemma 2. With Lemma 2, we assume that ˆg(0) ≥ 0, ˇg(0) ≤ 0, without loss of generality.
Proof for Lemma 3:
We assert that limk→∞ˆb(k) exists, and denote it by ˆB; moreover, limk→∞c(x(k))= ˆB.
Case (i): There exist finitely many jumps of type-1.
In this case, there exist some K3 ∈ N, some i, say 1, such that ˆg(k) = g1(k) ≥ 0, for all k ≥ K3. Hence {x1(k)} will be non-decreasing as k increases. By Lemma 1,
{x1(k)} are bounded above. Therefore, limk→∞x1(k) exists, hence limk→∞b1(x1(k)) exists, denoted by ˆB. Restated, limk→∞ˆb(k) = ˆB.
Next, we justify that limk→∞c(x(k))= ˆB. Assume otherwise, limk→∞c(x(k)) 6=
B. It follows from ˆg(k) = gˆ 1(k) ≥ 0, for all k ≥ K3, that b1(x1(k)) ≥ c(x(k)), for all k ≥ K3. There exists some ε > 0, and subsequence {kl}∞l=1 of positive integer numbers with k1 > K3 such that |c(x(kl)) − ˆB| > ε, for all l ∈ N. Because limk→∞b1(x1(k)) = ˆB, for such ε, there exists K4 ∈ N, such that |b1(x1(k))− ˆB| ≤ ε2, for all k ≥ K4. Therefore g1(kl) = b1(x1(kl)) − c(x(kl)) > ε2, for all kl ≥ K4. We find that {x1(k)} is always increasing after K4-th iteration. In fact,
∆x1(kl) = a1(x(kl))[b1(x1(kl) − c(x(kl))] > ρε 2,
if kl ≥ K4. Hence {x1(k)} will increase unboundedly, and yields a contradiction to Lemma 1.
Case (ii): There exist infinitely many jumps of type-1.
We shall justify that {ˆb(k)} decreases as {k} ↑ ∞. Consider a fixed k ∈ N0. Subcase (ii-a): no jump of type-1 occurs at k-th iteration.
Suppose I(k) = I(k + 1) = i, then gi(k) ≥ 0, gi(k + 1) ≥ 0. In addition, ˆb(k + 1) = bi(xi(k + 1))
≤ bi(xi(k))
= ˆb(k),
thank to (2.5), and ∆xi(k) = ai(x(k))gi(k) ≥ 0. Thus {ˆb(k)} decreases as k in-creases.
Subcase (ii-b): jump of type-1 occurs at k-th iteration and gi(k) ≥ 0, gj(k) ≥ 0, where I(k) = i 6= I(k + 1) = j.
It follows that
ˆb(k + 1) = bj(xj(k + 1))
≤ bj(xj(k))
≤ bi(xi(k))
= ˆb(k),
due to (2.5), ∆xj((k)) = aj(x(k))gj(k) ≥ 0, and by I(k) = i 6= j.
Subcase (ii-c): jump of type-1 occurs at k-th iteration and gi(k) ≥ 0, gj(k) < 0, where I(k) = i 6= I(k + 1) = j.
Notably, we still have gj(k + 1) ≥ 0. We claim that
bj(xj(k + 1)) − bj(xj(k)) ≤ bi(xi(k)) − bj(xj(k)). (3.12) Indeed,
LHS = b0j(·)∆xj(k)
= b0j(·)aj(x(k))gj(k)
≤ b0j(·)gj(k) (by (2.1))
≤ −djgj(k) (by (2.5), and gj(k) < 0))
≤ gi(k) − gj(k) (by (1 − dj)gj(k) < 0 ≤ gi(k))
= bi(xi(k)) − bj(xj(k))
= RHS.
Herein, “ · ” is defined as before. Hence, ˆb(k + 1) = bj(xj(k + 1)) ≤ bi(xi(k)) = ˆb(k).
All these cases indicate that {ˆb(k)} decreases as {k} increases. By Lemma 1, {x(k)}
are attracted into some compact set Q contained in Rn. Therefore, {bi(xi(k))} are bounded below, and so are {ˆb(k)}. Hence {ˆb(k)} decreases and converges to some number ˆB as k tends to infinity (denoted by {ˆb(k)} ↓ ˆB).
Next, we verify that limk→∞c(x(k))= ˆB. Assume otherwise: limk→∞c(x(k)) 6=
B. There exist some positive µ, subsequence {kˆ l}∞l=1 of positive integers, such that
|c(x(kl)) − ˆB| > µ
²ρ, (3.13)
Where ², ρ are defined in (2.13) and (2.14). Because {ˆb(k)} ↓ ˆB, for µ0 := min{²ρµ, µ} >
0, there exists L ∈ N such that
B ≤ bˆ I(k)(xI(k)(k)) ≤ ˆB + µ0, (3.14)
for all k ≥ L. Moreover
ˆg(`) = bI(`)(xI(`)(`) − c(x(`)) ≥ 0, (3.15) for all ` ∈ N. Consider the kL-th iteration. Notably, kL> L. By (3.13), (3.14), and (3.15), we have
ˆg(kL) = b1(x1(kL)) − c(x(kL)) > µ
²ρ,
where, for convenience, we set I(kL)=1 without loss of generality. There are two possibilities at kL-th iteration, either jump of type-1 occurs or not. If it does not occur, then
|∆ˆb(kL)| = |ˆb(kL+ 1) − ˆb(kL)|
= |b1(x1(kL+ 1)) − b1(x1(kL))|
= |b01(·)||x1(kL+ 1) − x1(kL)|
= |b01(·)||a1(x(kL))||g1(kL)|
= |b01(·)||a1(x(kL))||ˆg(kL)|
> ²ρµ
²ρ
= µ.
But it is impossible, because of (3.14).
If jump of type-1 occurs at kL-th iteration. Assume that I(kL+ 1)=2. Below we consider three different cases for b2(x2(kL)):
Case (a): ˆB ≤ b2(x2(kL)) < b1(x1(kL)). Then g2(kL) > ²ρµ, and |∆b2(x2(kL))| =
|b02(·)||a2(x(kL))||g2(kL)| > ²ρ²ρµ = µ. It is impossible, due to (3.14).
Case (b): ˆB > b2(x2(kL)) ≥ c(x(kL)). Then g2(kL) ≥ 0, and x2(kL+ 1) ≥ x2(kL). Thus,
ˆb(kL+ 1) = b2(x2(kL+ 1))
≤ b2(x2(kL))
< B.ˆ It is impossible, since {ˆb(k)} ↓ ˆB.
Case (c): b2(x2(kL)) < c(x(kL)). Then g2(kL) < 0, and
∆b2(x2(kL)) = b2(x2(kL+ 1)) − b2(x2(kL))
= b02(·)a2(x(kL))g2(kL)
≤ −d2g2(kL)
< −g2(kL).
Thus, b2(x2(kL+1)) = b2(x2(kL)) + ∆b2(x2(kL)) < b2(x2(kL)) − g2(kL) = c(x(kL)).
Hence ˆb(kL+ 1) = b2(x2(kL+ 1)) < c(x(kL)) < ˆB. It is impossible, since {ˆb(k)} ↓ ˆB.
From the above discussions, we conclude that limk→∞c(x(k)) = ˆB.
The second part of the lemma asserts that limk→∞ˇb(k) exists, denoted by ˇB, and limk→∞c(x(k))= ˇB. The proof for the assertion resembles the first part. Let us elaborate.
Case (i): There exist finitely many jumps of type-2.
In this case, there exists some K5 ∈ N, some i, say 1, such that ˇg(k) = g1(k) ≤ 0, for all k ≥ K5. Hence {x1(k)} will be non-increasing as k increases. By Lemma 1, {x1(k)} are bounded below. Therefore, limk→∞x1(k) exists, hence limk→∞b1(x1(k)) exists, denoted by ˇB. Restated, limk→∞ˇb(k) = ˇB.
Next,we justify that limk→∞c(x(k))= ˇB. Assume otherwise, limk→∞c(x(k)) 6=
B. It follows from ˇg(k) = gˇ 1(k) ≤ 0, for all k ≥ K5, b1(x1(k)) ≤ c(x(k)), for all k ≥ K5. There exists some ε > 0, and subsequence {kl}∞l=1 of positive integer numbers with k1 > K5 such that |c(x(kl))− ˇB| > ε, for all l ∈ N. Because limk→∞b1(x1(k)) = B, for such ε, there exists Kˇ 6 ∈ N, such that |b1(x1(k)) − ˇB| ≤ ε2, for all k ≥ K6. Therefore g1(kl) = b1(x1(kl)) − c(x(kl)) < −ε2, for all kl ≥ K6. We find that {x1(k)}
are always decreasing after K6− th iteration. In fact,
∆x1(kl) = a1(x(kl))[b1(x1(kl) − c(x(kl))] < −ρε 2,
if kl≥ K6. Hence, {x1(k)} will decrease unboundedly, and yields a contradiction to Lemma 1.
Case (ii): There exist infinitely many jumps of type-2.
We shall justify that {ˇb(k)} increases as {k} ↑ ∞. Consider a fixed k ∈ N0,
Subcase (ii-a): no jump of type-2 occurs at k-th iteration. Suppose J(k) = J(k + 1) = i, then gi(k) ≤ 0, gi(k + 1) ≤ 0. In addition,
ˇb(k + 1) = bi(xi(k + 1))
≥ bi(xi(k))
= ˇb(k)
thank to (2.5), and ∆xi((k)) = ai(x(k))gi(k) ≤ 0. Thus {ˇb(k)} increases as {k}
increases.
Subcase (ii-b): jump of type-2 occurs at k-th iteration and gi(k) ≤ 0, gj(k) ≤ 0, where J(k) = i 6= J(k + 1) = j.
It follows that
ˇb(k + 1) = bj(xj(k + 1))
≥ bj(xj(k))
≥ bi(xi(k))
= ˇb(k)
due to (2.5), ∆xj((k)) = aj(x(k))gj(k) ≤ 0 and J(k) = i 6= j.
Subcase (ii-c): jump of type-2 occurs at k-th iteration and gi(k) ≤ 0, gj(k) > 0, where J(k) = i 6= J(k + 1) = j.
Notably, we still have gj(k + 1) ≤ 0. We claim that
bj(xj(k + 1)) − bj(xj(k)) ≥ bi(xi(k)) − bj(xj(k)). (3.16) Indeed,
LHS = b0j(·)∆xj(k)
= b0j(·)aj(x(k))gj(k)
≥ b0j(·)gj(k) (by (2.1))
≥ −djgj(k) (by (2.5), and gj(k) > 0))
≥ gi(k) − gj(k) (by (1 − dj)gj(k) > 0 ≥ gi(k))
= bi(xi(k)) − bj(xj(k))
= RHS.
Herein, “ · ” is defined as before. Hence, ˇb(k + 1) = bj(xj(k + 1)) ≥ bi(xi(k)) = ˇb(k).
All these cases indicate that {ˇb(k)} increase as {k} increases. By Lemma 1, {x(k)}
are attracted into some compact set Q contained in Rn. Therefore, {bi(xi(k))} are bounded above, and so are {ˇb(k)}. Hence {ˇb(k)} increase and converge to some number, say ˇB as {k} tend to infinity (denoted by ˇb(k)} ↑ ˇB).
Next, we verify that limk→∞c(x(k))= ˇB. Assume otherwise: limk→∞c(x(k)) 6=
B. There exist some positive µ, subsequence {kˇ l}∞l=1 of positive integers, such that
|c(x(kl)) − ˇB| > µ
²ρ. (3.17)
Where ², ρ are defined in (2.13) and (2.14). Because ˇb(k)} ↑ ˇB, for µ0 := min{²ρµ, µ} >
0, there exists L ∈ N, such that
B ≥ bˇ J(k)(xJ(k)(k)) ≥ ˇB − µ0, (3.18)
for all k ≥ L. Moreover
ˇg(`) = bJ(`)(xJ(`)(`) − c(x(`)) ≤ 0, (3.19) for all ` ∈ N. Consider the kL-th iteration. Notably, kL > L. By (3.17), (3.18),and (3.19), we have
ˇg(kL) = b1(x1(kL)) − c(x(kL)) < −µ
²ρ,
where, for convenience, we set J(kL)=1 without loss of generality. There are two possibilities at kL− th iteration, either jump of type-2 occurs or not. If it dose not occur, then
|∆ˇb(kL)| = |ˇb(kL+ 1) − ˇb(kL)|
= |b1(x1(kL+ 1)) − b1(x1(kL))|
= |b01(·)||x1(kL+ 1)) − (x1(kL))|
= |b01(·)||a1(x(kL))||g1(kL)|
= |b01(·)||a1(x(kL))||ˇg(kL)|
> ²ρµ
²ρ
= µ.
But it is impossible, because (3.18).
If jump of type-2 occurs at kL− th iteration. Assume that J(kL+ 1)=2. Below we consider three different cases for b2(x2(kL)):
Case (a): ˇB ≥ b2(x2(kL)) > b1(x1(kL)). Then g2(kL) < −²ρµ, and |∆b2(x2(kL))| =
|b02(·)||a2(x(kL))||g2(kL)| > ²ρ²ρµ = µ. It is impossible, due to (3.18).
Case (b): ˇB < b2(x2(kL)) ≤ c(x(kL)). Then g2(kL) ≤ 0, and x2(kL+ 1) ≤ x2(kL). Thus
ˇb(kL+ 1) = b2(x2(kL+ 1))
≥ b2(x2(kL))
> B.ˇ It is impossible, since {ˇb(k)} ↑ ˇB.
Case (c): b2(x2(kL)) > c(x(kL)). Then g2(kL) > 0, and
From the above discussions, we conclude that limk→∞c(x(k)) = ˇB.