In this section, we give some applications of our main result. Throughout this section we take ϕ (t) = t^{p}, t > 0 with p > 0. There are three cases:

1.
Sublinear case when 0 < p < 1;

2.
Linear case when p = 1;

3.
Superlinear case when p > 1.
4.1 Sublinear case
Our aim in this section is to establish a sufficient, as well as a necessary and sufficient, condition for the boundedness of all solutions of (1.1) and the scalar case of (1.1), respectively.
The next result provides a sufficient condition for the boundedness of solutions of (1.1).
Theorem 4.1. Let (A), (B) be satisfied and ϕ (t) = t^{p}, t > 0, with fixed p ∈ (0,1). If (2.8) and (2.9) hold, then for any x_{0} ∈ ℝ^{d}the solution x(n; x_{0}), n ≥ 0 of (1.1) and (1.2) is bounded.
The next Lemma provides a necessary and sufficient condition for the condition (2.2) be satisfied, and will be useful in the proof of Theorem 4.1.
Lemma 4.2. Assume ϕ (t) = t^{p}, t > 0 and 0 < p < 1. Any positive constant u has property (P_{0}) if and only if (2.8) and (2.9) are satisfied.
Proof. Let the nonnegative constant u have property (P_{0}) (N = 0 in Definition 2.1). Then the condition (2.2) is satisfied for some positive v and for all n ≥ 1, so
{\alpha}_{0}:=\underset{n\ge 1}{\text{sup}}a\left(n,0\right)<\mathrm{\infty},\phantom{\rule{1em}{0ex}}{\beta}_{0}:=\underset{n\ge 1}{\text{sup}}\sum _{j=1}^{n}a\left(n,j\right)<\mathrm{\infty},
(4.1)
\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}{\gamma}_{0}:=\underset{n\ge 1}{\text{sup}}\parallel h\left(n\right)\parallel <\mathrm{\infty},
(4.2)
this imply that conditions (2.8) and (2.9) are satisfied.
Conversely, we assume (2.8), (2.9) and we prove that any positive constant u has property (P_{0}). Clearly, (2.9) is equivalent to α_{0} < ∞, β_{0} < ∞ and (2.8) implies γ_{0} < ∞.
Since p ∈ (0,1), it is clear that for an arbitrarily fixed u > 0, (2.1) and
\frac{{\alpha}_{0}{u}^{p}}{v}+{\beta}_{0}{v}^{p1}+\frac{{\gamma}_{0}}{v}\le 1,
(4.3)
are satisfied for any v large enough. From (4.3) we get
{\alpha}_{0}{u}^{p}+{\beta}_{0}{v}^{p}+{\gamma}_{0}\le v,
that (2.5) is satisfied for x_{0} = u and all n ≥ 1. Then by Definition 2.1, u has property (P_{0}).
Now we prove Theorem 4.1.
Proof. Let (A) and (B) be satisfied. By Lemma 4.2, we have that for any x_{0} ∈ ℝ^{d}, u = x_{0} has property (P_{0}) (see Remark 2.3). Thus, the conditions of Theorem 3.1 hold, and the initial vector x_{0} belongs to S, and hence the solution x(n; x_{0}) of (1.1) and (1.2) is bounded.
We consider the scalar case of Volterra difference equation
x\left(n+1\right)=\sum _{j=0}^{n}a\left(n,j\right){x}^{p}\left(j\right)+h\left(n\right),\phantom{\rule{1em}{0ex}}n\ge 0,
(4.4)
x\left(0\right)={x}_{0},
(4.5)
where x_{0} ∈ ℝ_{+}, a(n,j) ∈ ℝ_{+}, h(n) ∈ ℝ_{+}, 0 ≤ j ≤ n and p ∈ (0,1).
The following result provides a necessary and sufficient condition for the boundedness of the solution of (4.4) and (4.5). The necessary part of the next theorem was motivated by a similar result of Lipovan [18] proved for convolutiontype integral equation.
Theorem 4.3. Assume
\underset{n\to \mathrm{\infty}}{\text{lim}\mathsf{\text{inf}}}\left(\sum _{j=0}^{n}a\left(n,j\right)\right)>0,
(4.6)
moreover for any n ≥ 0, there exists an index j_{
n
}such that
0\le {j}_{n}\le n\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}a\left(n,{j}_{n}\right)+h\left(n\right)>0.
(4.7)
For any x_{0} ∈ (0, ∞), the solution of (4.4) is bounded, if and only if (2.8) and (2.9) are satisfied.
Proof. Assume (2.8) and (2.9) are satisfied. Clearly, by Theorem 4.1 the solution of (4.4) is bounded. Conversely, let the solution x(n) = x(n;x_{0}) of (4.4) be bounded on ℝ_{+}, with x_{0} > 0. Under condition (4.7), by mathematical induction we show that x(n) > 0, n ≥ 0. For n = 0 this is clear. Suppose that required inequality is not satisfied for all n ≥ 0. Then there exists index ℓ ≥ 0 such that x(0) > 0, ..., x(ℓ) > 0 and x(ℓ+1) ≤ 0. But by condition (4.7), we get
\begin{array}{ll}\hfill x\left(\ell +1\right)& =\sum _{j=0}^{\ell}a\left(\ell ,j\right){x}^{p}\left(j\right)+h\left(\ell \right)\phantom{\rule{2em}{0ex}}\\ \ge a\left(\ell ,{j}_{\ell}\right){x}^{p}\left({j}_{\ell}\right)+h\left(\ell \right)>0,\phantom{\rule{1em}{0ex}}0\le {j}_{\ell}\le \ell ,\phantom{\rule{2em}{0ex}}\end{array}
which is a contradiction. So x(n) > 0 for all n ≥ 0. On the other hand, for any n ≥ N* ≥ 1
x\left(n+1\right)\ge \sum _{j=0}^{{N}^{*}1}a\left(n,j\right){x}^{p}\left(j\right)\ge \sum _{j=0}^{{N}^{*}1}a\left(n,j\right)\underset{0\le j\le {N}^{*}}{\text{min}}{x}^{p}\left(j\right),
hence
\underset{n\ge 0}{\text{sup}}\sum _{j=0}^{{N}^{*}1}a\left(n,j\right)<\mathrm{\infty}.
(4.8)
Since x(n + 1) ≥ h(n) for all n ≥ 0, clearly sup_{n≥0}h(n) is finite.
Define now
m=\underset{n\to \mathrm{\infty}}{\text{lim}\mathsf{\text{inf}}}x\left(n\right),
which is finite. First we show that m > 0.
Assume for the sake of contradiction that m = 0. In this case we can find a strictly increasing sequence (N_{
k
})_{k≥1}, such that
x\left({N}_{k}\right)=\underset{0\le n\le {N}_{k}}{\text{min}}x\left(n\right)>0,\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}x\left({N}_{k}\right)\to 0\mathsf{\text{as}}k\to \mathrm{\infty}\mathsf{\text{.}}
From (4.4) if n = N_{
k
} 1, we deduce
\begin{array}{ll}\hfill x\left({N}_{k}\right)& =\sum _{j=0}^{{N}_{k}1}a\left({N}_{k}1,j\right){x}^{p}\left(j\right)+h\left({N}_{k}1\right)\phantom{\rule{2em}{0ex}}\\ \ge \sum _{j=0}^{{N}_{k}1}a\left({N}_{k}1,j\right)\underset{0\le j\le {N}_{k}}{\text{min}}{x}^{p}\left(j\right)\phantom{\rule{2em}{0ex}}\\ ={x}^{p}\left({N}_{k}\right)\sum _{j=0}^{{N}_{k}1}a\left({N}_{k}1,j\right).\phantom{\rule{2em}{0ex}}\end{array}
Since x(N_{
k
}) > 0, we have that
{x}^{1p}\left({N}_{k}\right)\ge \sum _{j=0}^{{N}_{k}1}a\left({N}_{k}1,j\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}k\ge 1.
Since p ∈ (0,1) and x(N_{
k
}) → 0, an k → ∞, we get
\underset{k\to \mathrm{\infty}}{\text{lim}\mathsf{\text{inf}}}\left(\sum _{j=0}^{{N}_{k}1}a\left({N}_{k}1,j\right)\right)=0,
which contradicts (4.6). So m > 0 and for \frac{1}{2}m, there exists N* ≥ 0 such that
x\left(n\right)\ge \frac{1}{2}m,\phantom{\rule{1em}{0ex}}n\ge {N}^{*}.
Hence,
\begin{array}{ll}\hfill x\left(n+1\right)& =\sum _{j=0}^{n}a\left(n,j\right){x}^{p}\left(j\right)+h\left(n\right)\phantom{\rule{2em}{0ex}}\\ \ge \sum _{j={N}^{*}}^{n}a\left(n,j\right){x}^{p}\left(j\right)\ge \frac{1}{{2}^{p}}{m}^{p}\sum _{j={N}^{*}}^{n}a\left(n,j\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}n\ge {N}^{*}.\phantom{\rule{2em}{0ex}}\end{array}
But the solution x(n) is a bounded sequence, and hence
\underset{n\ge {N}^{*}}{\text{sup}}\sum _{j={N}^{*}}^{n}a\left(n,j\right)<\mathrm{\infty}.
This and (4.8) imply condition (2.9).
Remark 4.4. In general, without condition (4.7) the necessary part of Theorem 4.3 is not true. In fact if a(0, 0) = 0, and h(0) = 0, that is (4.7) does not hold for n = 0, then for any x_{0} ∈ (0, ∞) the solution of (4.4) satisfies x(1;x_{0}) = 0, and hence
x\left(n+1\right)=\sum _{\underset{j\ne 1}{j=0}}^{n}a\left(n,j\right){x}^{p}\left(j\right)+h\left(n\right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}n\ge 1.
Thus, the solution x(n; x_{0}) does not depend on the choice of the sequence (a(n, 1))_{n≥1}. This shows that the boundedness of the solutions does not imply (2.9), in general.
4.2 Linear case
Our aim in this section is to obtain sufficient condition for the boundedness of the solution of (1.1) under the initial condition (1.2), but in the linear case of Volterra difference equation.
The following result gives a sufficient condition for the boundedness.
Theorem 4.5. Assume (A), (B) are satisfied and ϕ (t) = t, t ≥ 0. Then the solution x(n;x_{0}), x_{0} ∈ S, n ≥ 0 of (1.1) and (1.2) is bounded, if (2.9) is satisfied and there exists an N ≥ 0 such that one of the following conditions holds:

(i)
condition (2.8) holds and
{\beta}_{N}=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)<1;
(4.9)

(ii)
{\beta}_{N}=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)=1,
and for anyn\in {\Gamma}_{N}^{\left(1\right)},
\sum _{j=0}^{N}a\left(n,j\right)=0,\phantom{\rule{1em}{0ex}}h\left(n\right)=0
(4.10)
hold, moreover
\underset{n\in {\Gamma}_{N}^{\left(2\right)}}{\text{sup}}{\left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{1}\sum _{j=0}^{N}a\left(n,j\right)<\mathrm{\infty},
(4.11)
and
\underset{n\in {\Gamma}_{N}^{\left(2\right)}}{\text{sup}}{\left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{1}\parallel h\left(n\right)\parallel <\mathrm{\infty},
(4.12)
where
{\Gamma}_{N}^{\left(1\right)}=\left\{n\ge N+1:\sum _{j=N+1}^{n}a\left(n,j\right)=1\right\},
and
{\Gamma}_{N}^{\left(2\right)}=\left\{n\ge N+1:\sum _{j=N+1}^{n}a\left(n,j\right)<1\right\}.
For the proof of Theorem 4.4, we need the following lemma.
Lemma 4.6. Assume ϕ (t) = t, t ≥ 0. A positive constant u has property (P_{
N
}) with an integer N ≥ 0 if and only if the condition (2.9) and either (i) or (ii) are satisfied.
Proof. Necessity. We show that (P_{
N
}) implies (2.9) and either (i) or (ii). Suppose a positive constant u has property (P_{
N
}) with an integer N ≥ 0, hence (2.1) and (2.2) are satisfied for v > 0 and for any n ≥ N + 1.
From (2.1) and (2.2), it is clear that (2.8), (2.9) are satisfied and
\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)v,\phantom{\rule{1em}{0ex}}n\ge N+1.
(4.13)
Therefore
1\sum _{j=N+1}^{n}a\left(n,j\right)\ge 0,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}n\ge N+1,
and hence
{\beta}_{N}=\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)\le 1.
The latest inequality implies two cases with respect the value of β_{
N
}.

The first case β_{
N
}< 1. In this case the condition (i) is satisfied.

Consider now the second case where β_{
N
}= 1. Clearly if {\sum}_{j=N+1}^{n}a\left(n,j\right)=1, then from (4.13), we get
\sum _{j=0}^{N}a\left(n,j\right)u+h\left(n\right)=0,\phantom{\rule{1em}{0ex}}n\in {\Gamma}_{N}^{\left(1\right)},
or equivalently (4.10).
But if {\sum}_{j=N+1}^{n}a\left(n,j\right)<1, then
1\sum _{j=N+1}^{n}a\left(n,j\right)>0,\phantom{\rule{1em}{0ex}}n\in {\Gamma}_{N}^{\left(2\right)},
and (4.13) yields
{\left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{1}\left(\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \right)\le v,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}n\in {\Gamma}_{N}^{\left(2\right)}
and hence (4.11) and (4.12) are satisfied. Then condition (ii) holds.
Sufficiency. We show that if (2.9) and one of the conditions (i) and (ii) is satisfied with some u ≥ 0 and N ≥ 0, then u has property (P_{
N
}). It is easy to observe that (2.9) yields
\underset{n\ge N+1}{\text{sup}}\sum _{j=0}^{N}a\left(n,j\right)<\mathrm{\infty},\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}\underset{n\ge N+1}{\text{sup}}\sum _{j=N+1}^{n}a\left(n,j\right)<\mathrm{\infty}.
Let (i) of Theorem 4.5 be satisfied, that is β_{
N
}< 1. Then for u ≥ 0 and n ≥ N + 1, N ≥ 0, there exists v > 0 such that
\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1{\beta}_{N}\right)v,\phantom{\rule{1em}{0ex}}n\ge N+1.
It implies
\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)v,
i.e. (2.2) is satisfied and (2.1) also is satisfied for all v large enough, hence u has property P_{
N
}.
Now suppose β_{
N
}= 1, and (4.10) holds. Then, clearly (2.1) and (2.2) are satisfied for any v ≥ 0 and n\in {\Gamma}_{N}^{\left(1\right)}.
If n\in {\Gamma}_{N}^{\left(2\right)} and (4.11) and (4.12) are satisfied, then for u ≥ 0, we have
\underset{n\in {\Gamma}_{N}^{\left(2\right)}}{\text{sup}}{\left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{1}\left[\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \right]<\mathrm{\infty}.
Then there exists v > 0 large enough such that
{\left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)}^{1}\left[\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \right]\le v,\phantom{\rule{1em}{0ex}}n\in {\Gamma}_{N}^{\left(2\right)},
since 1{\sum}_{j=N+1}^{n}a\left(n,j\right)>0, for all n\in {\Gamma}_{N}^{\left(2\right)}.
Therefore
\sum _{j=0}^{N}a\left(n,j\right)u+\parallel h\left(n\right)\parallel \le \left(1\sum _{j=N+1}^{n}a\left(n,j\right)\right)v,
i.e. for all v large enough the conditions (2.2) and (2.1) are satisfied. Hence, u has property (P_{
N
}).
The following lemma is extracted from [2] (Lemma 5.3) and will be needed in this section.
Lemma 4.7. Assume (A), (B) are satisfied and ϕ (t) = t, t ≥ 0. For every integer N > 0, there exists a nonnegative constant K_{1}(N) independent of the sequence (h(n))_{n≥0}and x_{0}, such that the solution (x(n))_{n≥0}of (1.1) and (1.2), satisfies
\parallel x\left(n\right)\parallel \le {K}_{1}\left(N\right)\left(\underset{0\le m\le N}{\text{max}}\parallel h\left(m\right)\parallel +\parallel {x}_{0}\parallel \right),\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}0\le n\le N.
(4.14)
Now we give the proof of Theorem 4.5.
Proof. Let (A), (B), (2.9) and either (i) or (ii) in Theorem 4.5 be satisfied. By Lemma 4.7 the solution of (1.1) and (1.2) satisfies (4.14) for all 0 ≤ n ≤ N. This means that, there exists a nonnegative constant u such that
\parallel x\left(n\right)\parallel \le u,\phantom{\rule{1em}{0ex}}0\le n\le N.
By Lemma 4.6 we have u has property (P_{
N
}). Then the conditions of Theorem 3.1 hold for the initial vector x_{0} belonging to S, and hence the solution x(n; x_{0}) of (1.1) and (1.2) is bounded.
4.3 Superlinear case
Our aim in this section is to obtain sufficient condition for the boundedness in the superlinear case.
Theorem 4.8. Assume that conditions (A) and (B) are satisfied with the function ϕ(t) = t^{p}, t > 0, where p > 1. Suppose also (2.8) and (2.9) hold. Then the solution x(n;x_{0}) of (1.1) and (1.2) is bounded for some x_{0} ∈ ℝ^{d}if there existsv\in \left[{\left(\frac{1}{p{\beta}_{0}}\right)}^{\frac{1}{p1}},{\left(\frac{1}{{\beta}_{0}}\right)}^{\frac{1}{p1}}\right]such that
a\left(0,0\right)\parallel {x}_{0}{\parallel}^{p}+\parallel h\left(0\right)\parallel \le v,
(4.15)
and
{\alpha}_{0}\parallel {x}_{0}{\parallel}^{p}+{\gamma}_{0}\le v{\beta}_{0}{v}^{p},
(4.16)
where α_{0}, β_{0}and γ_{0}are defined in (4.1) and (4.2).
Proof. Assume (2.8), (2.9), (4.15), and (4.16) are satisfied. The case β_{0} = 0 is clear. So we assume that β_{0} > 0. In this case, clearly, v{\beta}_{0}{v}^{p}\ge 0 if
v\in \left[0,{\left(\frac{1}{{\beta}_{0}}\right)}^{\frac{1}{p1}}\right],
and the maximum value of the function g\left(v\right)=v{\beta}_{0}{v}^{p}\mathsf{\text{is}}{\left(\frac{1}{p{\beta}_{0}}\right)}^{1/\left(p1\right)}.
Then there exists v such that
v\in \left[{\left(\frac{1}{p{\beta}_{0}}\right)}^{\frac{1}{p1}},{\left(\frac{1}{{\beta}_{0}}\right)}^{\frac{1}{p1}}\right],
and the conditions (2.4) and (2.5) hold. By Remark 2.3 we get that under conditions (4.15) and (4.16), u = x_{0} has property (P_{0}) and x_{0} belongs to S. Then the conditions of Theorem 3.1 hold, so the solution of (1.1) with the initial condition (1.2) is bounded.