Lemma 2:1. Suppose u(m, n), a(m, n), b(m, n) ∈ ℘+(Ω). If a(m, n) is nonincreasing in the first variable, then for (m, n) ∈ Ω,
implies
Remark 1. Lemma 2.1 is a direct variation of [[13], Lemma 2.5 (β
2)], and we note a(m, n) ≥ 0 here.
Lemma 2.2. Suppose u(m, n), a(m, n) ∈ ℘+(Ω), b(s, t, m, n) ∈ ℘+(Ω2), and a(m, n) is nonincreasing in every variable with a(m, n) > 0, while b(s, t, m, n) is nonincreasing in the third variable. φ ∈ C(ℝ+, ℝ+) is nondecreasing with φ(r) > 0 for r > 0. If for (m, n) ∈ Ω, u(m, n) satisfies the following inequality
(2)
then we have
(3)
where
(4)
Proof. Fix (m
1, n
1) ∈ Ω, and let (m, n) ∈ ([m
1, ∞] × [n
1, ∞]) ∩ Ω. Then, we have
(5)
Let the right side of (5) be v(m, n). Then,
(6)
and
that is,
(7)
On the other hand, according to the Mean Value Theorem for integrals, there exists ξ such that v(m, n) ≤ ξ ≤ v(m - 1, n), and
(8)
So, combining (7) and (8), we have
(9)
where G is defined in (4). Setting m = η in (9), and a summary with respect to η from m + 1 to ∞ yields
Noticing v(∞, n) = a(m
1, n
1), and G is increasing, it follows
(10)
Combining (6) and (10), we obtain
(11)
Setting m = m
1, n = n
1 in (11), yields
(12)
Since (m
1, n
1) is selected from Ω arbitrarily, then substituting (m
1, n
1) with (m, n) in (12), we get the desired inequality (3).
Corollary 2. 3. Under the conditions of Lemma 2.2, and furthermore assume a(m, n) ≥ 0. If for (m, n) ∈ Ω, u(m, n) satisfies the following inequality
(13)
then we have
(14)
Proof. Suppose a(m, n) > 0. By Theorem 2.1 (with ), we have
(15)
Where , z ≥ z
0
> 0. Then, a simplification of (15) yields the desired inequality (14).
If a(m, n) ≥ 0, then we can carry out the process above with a(m, n) replaced by a(m, n)+ε, where ε > 0. After letting ε → 0, we also obtain the desired inequality (14).
Lemma 2.4 [[17]]. Assume that a ≥ 0, p ≥ q ≥ 0, and p ≠ 0, then for any K > 0
Theorem 2:5. Suppose, u(m, n), w(m, n) ∈ ℘+(Ω), b
i
(s, t, m, n), c
i
(s, t, m, n) ∈ ℘+(Ω2), i = 1, 2,..., l
1, d
i
(s, t, m, n), e
i
(s, t, m, n) ∈ ℘+(Ω2), i = 1, 2,..., l
2 with b
i
, c
i
, d
i
, e
i
nonincreasing in the last two variables, and there is at least one function among d
i
, e
i
, i = 1, 2,..., l
2 not equivalent to zero. φ ∈ C(ℝ+, ℝ+) is nondecreasing with φ(r) > 0 for r > 0, and φ is submultiplicative, that is, φ(αβ ) ≤ φ(α)φ(β ) for ∀α, β ∈ ℝ+. If for (m, n) ∈ Ω, u(m, n) satisfies the following inequality
(16)
then we have
(17)
provided that 0 < μ < 1 and J is increasing, where
(18)
(19)
(20)
(21)
(22)
(23)
Proof. Denote the right side of (16) be . Then, v(m, n) is nonincreasing in every variable, and by Lemma 2.1, we obtain
(24)
where is defined in (20). Furthermore, by (24), we deduce
where and B(s, t, m, n) is defined in (22).
As we can see, H(m, n) is nonincreasing in every variable. Considering m ≥ M, n ≥ N, it follows
Since there is at least one function among d
i
, e
i
, i = 1, 2,..., l
2 not equivalent to zero, then H(M, N ) > 0.
On the other hand, as b
i
(s, t, m, n), c
i
(s, t, m, n) are nonincreasing in the last two variables, then one can see B(s, t, m, n) is also nonincreasing in the last two variables. So, a suitable application of Lemma 2.2 yields
(25)
where G, C(m, n) are defined in (18) and (21), respectively. On the other hand, we have
(26)
Then, considering v(m, n) is nonincreasing in every variable, using (25) in (26) yields
where μ is a constant defined in (23).
According to 0 < μ < 1, and G is increasing, we obtain
and
which is rewritten by
where J is defined in (19). Since J is increasing, we have
(27)
Combining (24), (25), and (27), we get the desired result.
Theorem 2.6. Suppose, u(m, n), a(m, n), w(m, n) ∈ ℘+(Ω), b
i
(s, t, m, n), c
i
(s, t, m, n) ∈ ℘+(Ω2), i = 1, 2,..., l
1, d
i
(s, t, m, n), e
i
(s, t, m, n) ∈ ℘+(Ω2), i = 1, 2,..., l
2 with b
i
, c
i
, d
i
, e
i
nonincreasing in the last two variables. q
i
, r
i
are nonnegative constants with p ≥ q
i
, p ≥ r
i
, i = 1, 2,..., l
1, while h
i
, j
i
are nonnegative constants with p ≥ h
i
, p ≥ j
i
, i = 1, 2,..., l
2. If for (m, n) ∈ Ω, u(m, n) satisfies the following inequality
(28)
then
(29)
provided that , where
(30)
(31)
(32)
(33)
(34)
(35)
(36)
Proof. Denote the right side of (28) be . Then, we have
(37)
Obviously F(m, n) is nonincreasing in the first variable. So, by Lemma 2.1, we obtain
where . Define F (m, n) = a(m, n) + v(m, n). Then
(38)
Furthermore, by (38) and Lemma 2.4, we have
where and , , , , are defined in (30)-(32), respectively. Then, using is nonincreasing in every variable, we obtain
(39)
where B(s, t, m, n) is defined in (36). Using B(s, t, m, n) is nonincreasing in the last two variables, by a suitable application of Corollary 2.3, we obtain
(40)
where is defined in (35). Furthermore, considering the definition of and (40), we have
where is defined in (34). Then, according to , we have
(41)
From (40) and (41), we deduce
(42)
Then, combining (38) and (42), we obtain the desired result.
Remark 2. As one can see, the established results above mainly deal with Volterra-Fredholm type discrete inequalities with four iterated sums and infinite sum upper limit, and they are different from the results presented in [3]. Furthermore, without considering the slight difference in infinite sum upper limit, Theorem 2.6 is generalization of [[3], Theorem 2.5]. If we take w(m, n) ≡ 0, b
i
(s, t, m, n) = b
i
(s, t), c
i
(ξ, η, m, n) = 0, i = 1, 2,..., l
1, d
i
(s, t, m, n) = d
i
(s, t), e
i
(ξ, η, m, n) ≡ 0, i = 1, 2,..., l
2, then Theorem 2.6 reduces to [[3], Theorem 2.5].