Tauberian theorems for statistically convergent double sequences of fuzzy numbers

3.1 Double sequences in E ⁿ

First, we define analogues of Schmidt’s slow oscillation condition for double sequences in E ⁿ . Let λ _m  : = [λm] denote the integral part of λm and let (u _mn ) be a double sequence in E ⁿ . Then, We say that (u _mn ) is slowly oscillating in sense (1, 1) if inf λ > 1 { lim sup m , n → ∞ ( max m < j < λ m n < k < λ n D ( u j k , u m n ) ) } = 0(4)

or inf 0 < λ < 1 { lim sup m , n → ∞ ( max λ m < j < m λ n < k < n D ( u j k , u m n ) ) } = 0 ,(5) where the limit superior is defined by lim sup m , n → ∞ d m n = lim n → ∞ ( sup j , k ≥ n d j k ) .

We have (4) ⇔ (5). Now, we prove some Tauberian Theorems.

Theorem 1. Let (u _mn ) be a double sequence in E ⁿ with u m n → s t μ. If (u _mn ) is slowly oscillating in sense (1, 1), then u _mn  → μ.

Proof. Let ɛ > 0. Since (u _mn ) is slowly oscillating in sense (1, 1), there exist λ > 1 and N₁ ≥ 1 such that min ( m , n ) ≥ N 1 ⇒ max m < j < λ m n < k < λ n D ( u j k , u m n ) < ε .(6) Since S m n → s t S , there exists an N₂ ≥ 1 such that for min(M, N) ≥ N₂, 1 ( M + 1 ) ( N + 1 ) | { j ≤ M and k ≤ N : D ( u jk , μ ) ≥ ɛ } | < 1 16 ( λ - 1 λ + 1 ) 2 .(7) Define N₀ = max{ N₁, N₂, 4/(λ - 1) }. For min(m, n) ≥ N₀, we have min(λ _m , λ _n ) ≥ N₂, and so (7) implies | { j ≤ λ m and k ≤ λ n : D ( u jk , μ ) ≥ ɛ } | < 1 16 ( λ - 1 λ + 1 ) 2 ( λ m + 1 ) ( λ n + 1 ) ≤ ( λ m - m - 1 ) ( λ n - n - 1 ) . Therefore we obtain { j ≤ λ m and k ≤ λ n : D ( u jk , μ ) < ɛ } ∩ { m < j < λ m and n < k < λ n } ≠ ∅ . That is, there exists some pair (j^*, k^*) such that m < j^* < λ _m ,  n < k^* < λ _n and D (u _{j ^* k ^*} , μ) < ɛ. Combining this and (6) gives D ( u m n , μ ) ≤ D ( u j * k * , μ ) + max m < j < λ m n < k < λ n D ( u j k , u m n ) < 2 ε . Therefore, u _mn  → μ which completes the proof. □

Now we give several variants of slow oscillation condition. We say that (u _mn ) is slowly oscillating in sense (1, 0) if inf λ > 1 { lim sup m , n → ∞ ( max m < j < λ m D ( u j n , u m n ) ) } = 0 ;(8) and (u _mn ) is slowly oscillating in sense (0, 1) if inf λ > 1 { lim sup m , n → ∞ ( max n < k < λ n D ( u m k , u m n ) ) } = 0.(9)

Clearly, if (u _mn ) is slowly oscillating in senses (1, 0) and (0, 1), then (u _mn ) is slowly oscillating in sense (1, 1). So we obtain the followingconsequence.

Corollary 1. Let (u _mn ) be a double sequence in E¹ with u m n → s t μ. If (u _mn ) is slowly oscillating in senses (1, 0) and (0, 1), then u _mn  → μ.

Now we give two-sided Tauberian conditions of Hardy type for double sequences in E ⁿ . We say that (u _mn ) satisfies Hardy’s two sided condition in sense (1, 0) if there exist constants n₀ ≥ 1 and H > 0 such that jD ( u jn , u j - 1 , n ) ≤ H ( j , n > n 0 ) .(10) We claim that condition (8) follows from (10). Indeed, given ɛ > 0, chose 1 < λ < 1 + ɛ/H. Since n₀ < m < j < λ _m , by (10) we have D ( u jn , u mn ) ≤ ∑ ℓ = m + 1 j D ( u ℓ n , u ℓ - 1 , n ) ≤ ∑ ℓ = m + 1 j H ℓ ≤ H ( j - m m ) < H ( λ - 1 ) < ɛ .

Similarly, Hardy’s two sided condition in sense (0, 1) is defined as follows: kD ( u mk , u m , k - 1 ) ≤ H ( m , k > n 0 ) .(11) Since (10) ⇒ (8) and (11) ⇒ (9) and by Corollary 3.1 we get the following result.

Corollary 2. Let (u _mn ) be a double sequence in E ⁿ with u m n → s t μ. If conditions (10) and (11) are satisfied, then u _mn  → μ.

Fuzzy number space E¹ has some specific properties. E¹ is partially ordered and is not a linear space. So definitions and proofs in E¹ are quite different from those in ℝ.

If u ∈ E¹, then α-level set [u]  _α of u is closed, bounded and non-empty interval and we can write [u]  _α  : = [u^- (α) , u⁺ (α)]. The partial ordering relation on E¹ is defined as follows: u ⪯ v ⇔ [ u ] α ⪯ [ v ] α ⇔ u - ( α ) ≤ v - ( α ) and u + ( α ) ≤ v + ( α ) for all α ∈ [ 0 , 1 ] .

The metric D on E¹ has following important property.

Lemma 1. [2, Lemma 6] Let u, v ∈ E¹ and ɛ > 0. The following statements are equivalent:

D (u, v) ≤ ɛ,

In this section we define the slow decrease conditions for double sequences in E¹ and prove that under these conditions, u m n → s t μ implies u _mn  → μ.

We assume that (u _mn ) is a double sequence in E¹. We say that the double sequence (u _mn ) is slowly decreasing in sense (1, 1) if for every ɛ > 0 there exist N and λ > 1 such that min ( m , n ) ≥ N ⇒ u jk ⪰ u mn - ɛ ¯ whenever n < k < λ n , m < j < λ m .(12) Clearly, if (u _mn ) is slowly decreasing in sense (1, 1) then for each α ∈ [0, 1] the double sequences ( u mn - ( α ) ) and ( u mn + ( α ) ) of real numbers satisfies the condition (1). But, the converse of this statement does not hold. The counter example is as follows.

Example 2. Let us define u mn ( x ) = { 1 , if x ∈ [ 0 , 1 2 ] , 1 2 + 1 4 ( mn + 1 ) , if x ∈ ( 1 2 , 1 ] , 0 , otherwise . Then, for α ∈ [0, 1] u n - ( α ) = 0 u n + ( α ) = { 1 , if α ∈ [ 0 , 1 2 + 1 4 ( mn + 1 ) ] , 1 2 , if α ∈ ( 1 2 + 1 4 ( mn + 1 ) , 1 ] , Clearly for each α ∈ [0, 1], u mn - ( α ) = 0 and so ( u mn - ( α ) ) satisfy the condition (1). On the other hand for each α ∈ [ 0 , 1 2 ] we have u mn + ( α ) = 1. for α ∈ ( 1 2 , 1 ] we can find N ∈ ℕ such that min ( m , n ) ≥ N ⇒ α ∈ ( 1 2 + 1 4 ( mn + 1 ) , 1 ] and u mn + ( α ) = 1 2 . Hence ( u mn + ( α ) ) satisfy the condition (1). Now we assume that 0 < ɛ < 1 2 and n < k,   m < j. Then taking α ∈ ( 1 2 + 1 4 ( jk + 1 ) , 1 2 + 1 4 ( mn + 1 ) ] , we have u mn + ( α ) = 1 and u jk + ( α ) = 1 2. So u jk + ( α ) < u mn + ( α ) - ɛ. Therefore the inequality u jk ⪰ u mn - ɛ ¯ does not hold. That is, (u _mn ) is not slowly decreasing in sense (1, 1).

Lemma 2. Let (u _mn ) be a double sequence in E¹. If (u _mn ) is slowly decreasing in sense (1, 1), then for every ɛ > 0 there exist N ≥ 1 and 0 < λ < 1 such that min ( m , n ) ≥ N ⇒ u mn ⪰ u jk - ɛ ¯ whenever λ n < k < n , λ m < j < m .(13)

Proof. We use an indirect approach to prove the Lemma. Suppose that (u _mn ) is slowly decreasing in sense (1, 1) and there exists some ɛ₀ > 0 such that for all 0 < λ < 1 and N ≥ 1 we can find integers k, j, m, n for which min(m, n) ≥ N, λ _n  < k < n,   λ _m  < j < m and u m n ≽ u j k − ε ¯. Therefore, there exists α₀ ∈ [0, 1] such that u mn - ( α 0 ) < u jk - ( α 0 ) - ɛ or u mn + ( α 0 ) < u jk + ( α 0 ) - ɛ . Without loss of generality, we can assume that u mn - ( α 0 ) < u jk - ( α 0 ) - ɛ . Clearly, { u mn - ( α 0 ) } does not satisfy the condition (2). That is, { u mn - ( α 0 ) } is not slowly decreasing. This contradicts that (u _mn ) is slowly decreasing in sense (1, 1). □

Theorem 2. Let (u _mn ) be a double sequence in E¹ with u m n → s t μ. If (u _mn ) is slowly decreasing in sense (1, 1), then u _mn  → μ.

Proof. Given ɛ > 0. Since (u _mn ) is slowly decreasing in sense (1, 1), there exist λ > 1 and N₁ ≥ 1 such that min ( m , n ) ≥ N 1 ⇒ u jk ⪰ u mn - ɛ ¯ whenever n < k < λ n , m < j < λ m .(14) Let N₂ and N₀ be defined as in the proof of Theorem 1. Take min(m, n) ≥ N₀. From the proof of Theorem 1, there exist (j^*, k^*) such that m < j * < λ m , n < k * < λ n and D ( u j * k * , μ ) < ɛ . By Lemma 1, we have μ - ɛ ¯ ⪯ u j * k * ⪯ μ + ɛ ¯. Combining this and (14) gives u mn ⪯ u j * k * + ɛ ¯ ⪯ μ + 2 ɛ ¯ .(15) To replace (12) by (13), as a similar proof to the above we can find N₃ such that min ( m , n ) ≥ N 3 ⇒ μ - 2 ɛ ¯ ⪯ u mn .(16) Take N₄ = max(N₀, N₃). Then, (15) and (16) lead us to the consequence that min ( m , n ) ≥ N 4 ⇒ μ - 2 ɛ ¯ ⪯ u mn ⪯ μ + 2 ɛ ¯ which yields with Lemma 1 that min ( m , n ) ≥ N 4 ⇒ D ( u mn , μ ) ≤ 2 ɛ . This means that u _mn  → u which completes the proof. □

Now we give several variants of slow decrease condition. We say that (u _mn ) is slowly decreasing in sense (1, 0) if for every ɛ > 0 there exist N and λ > 1 such that min ( m , n ) ≥ N ⇒ u jn ⪰ u mn - ɛ ¯ whenever m < j < λ m . (u _mn ) is slowly decreasing in sense (0, 1) if for every ɛ > 0 there exist N and λ > 1 such that min ( m , n ) ≥ N ⇒ u mk ⪰ u mn - ɛ ¯ whenever n < k < λ n . Clearly, if (u _mn ) is slowly decreasing in senses (1, 0) and (0, 1), then (u _mn ) is slowly decreasing in sense (1, 1). In view of the above facts, we formulate the following corollary.

Corollary 3. Let (u _mn ) be a double sequence in E¹ with u m n → s t μ . If (u _mn ) is slowly decreasing in senses (1, 0) and (0, 1), then u _mn  → μ.

Now, we define the Landau’s one-sided Tauberian condition for double sequences in E¹. We say that (u _mn ) in E¹ satisfies Landau’s condition in sense (1, 0) if there exist constants n₀ ≥ 1 and H > 0 such that j u jn ⪰ j u j - 1 , n - H ¯ ( j , n > n 0 ) ; and (u _mn ) satisfies Landau’s condition in sense (0, 1) if there exist constants n₀ ≥ 1 and H > 0 such that k u mk ⪰ k u m , k - 1 - H ¯ ( m , k > n 0 ) .

Lemma 3. Let (u _mn ) be a double sequence in E¹. If (u _mn ) satisfies Landau’s condition in sense (γ, β), then (u _mn ) is slowly decreasing in sense (γ, β) where (γ, β) = (1, 0)  or  (0, 1).

Proof. In case of (0, 1) since the proof is obtained by the similar way, we prove only (1, 0). Assume that (u _mn ) satisfies Landau’s condition in sense (1, 0). Then there exist constants n₀ ≥ 1 and H > 0 such that

Therefore, for all α ∈ [0, 1] we have

If n₀ ≤ m < j < λ _m and n ≥ n₀, we obtain u jn - ( α ) - u mn - ( α ) ≥ ∑ ℓ = m + 1 j [ u ℓ n - ( α ) - u ℓ - 1 , n - ( α ) ] ≥ ∑ j = m + 1 j - H ℓ ≥ - H ln ( j m ) > - H ln λ . Hence, for each ɛ > 0 if we choose λ so that 1 < λ ≤ e^(ɛ/H), then u jn - ( α ) - u mn - ( α ) > - H ln λ(18) > - H ln λ ≥ - ɛ .

Similarly, we can obtain

u jn + ( α ) - u mn + ( α ) > - ɛ .(19) Combining (18) and (19), we can see that u jn ⪰ u mn - ɛ ¯ which proves that (u _mn ) is slowly decreasing in sense (1, 0). □

From Lemma 3 and Theorem 2 we have the following consequence.

Corollary 4. Let (u _mn ) be a double sequence in E¹ with u m n → s t μ . If (u _mn ) satisfies Landau’s conditions in senses (1, 0) and (0, 1), then u _mn  → μ.

Now we give an example to show that both the conditions in Corollary 4 must be satisfied.

Example 3. u jk ( x ) = { x - k , for k ≤ x ≤ k + 1 - x + k + 2 , for k + 1 ≤ x ≤ k + 2 0 , otherwise } if k is square , for all j xk + 1 , for - 1 k ≤ x ≤ 0 - xk + 1 , for 0 ≤ x ≤ 1 k 0 , otherwise } otherwise

and we obtain α-level sets [u _jk ]  _α as [ u jk ] α = { [ k + α , k + 2 - α ] if k is square , for all j [ α - 1 k , 1 - α k ] otherwise . We get D ( u jk , 0 ¯ ) = { k + 2 if k is square , for all j , 1 k otherwise . Then clearly u j k → s t 0 ¯ . Also since u _jk  = u_j-1,k for all k, the sequence (u _jk ) satisfies Landau’s condition in sense (1, 0). However for (k - 1) is square, we get u jk - ( 0 ) - u j , k - 1 - ( 0 ) = - 1 k - ( k - 1 ) .

We can not find H such that

So, (u _jk ) does not satisfy Landau’s condition in sense (0, 1). Further (u _jk ) does not convergent.