Characterization of Entire Sequences via Double Orlicz Space

Let Γ denote the space of all entire sequences and ∧ the space of all analytic sequences. This paper is a study of the characterization and general properties of entire sequences via double Orlicz space of Γ 2 M of Γ 2 establishing some inclusion relations.


Introduction
Throughout, w, Γ, and ∧ denote the classes of all, entire, and analytic scalar-valued single sequences, respectively.
We write w 2 for the set of all complex sequences (x mn ), where m,n ∈ N is the set of positive integers.Then, w 2 is a linear space under the coordinatewise addition and scalar multiplication.
We need the following inequality in the sequel of the paper.For a,b ≥ 0 and 0 < p < 1, we have (a + b) p ≤ a p + b p . (1.1) The double series ∞ m,n=1 x mn is called convergent if and only if the double sequence (S mn ) is called convergent, where S mn = m,n i, j=1 x i j (m,n = 1,2,3,...) (see [9]).A sequence x = (x mn ) is said to be double analytic if sup mn |x mn | 1/m+n < ∞.The vector space of all double analytic sequences will be denoted by ∧ 2 .A sequence x = (x mn )is called double entire sequence if |x mn | 1/m+n → 0 as m,n → ∞.The double entire sequences will be denoted by Γ 2 .Let Φ = {all finite sequences}.
with 1 in the (m,n)th position and zero otherwise.An FK-space (or a metric space) X is said to have AK property if (δ mn ) is a Schauder basis for X.Or equivalently, x [m,n] → x.
An FDK space is a double sequence space endowed with a complete metrizable; locally convex topology under which the coordinate mappings x = (x k ) → (x mn ) (m,n ∈ N) are also continuous.
Orlicz [10] used the idea of Orlicz function to construct the space (L M ).Lindenstrauss and Tzafriri [11] investigated Orlicz sequence spaces in more detail, and they proved that every Orlicz sequence space M contains a subspace isomorphic to p (1 ≤ p < ∞).Subsequently, different classes of sequence spaces were defined by Parashar and Choudhary [12], Mursaleen et al. [13], Bektas ¸and Altin [14], Tripathy et al. [15], Rao and Subramanian [16], and many others.The Orlicz sequence spaces are the special cases of Orlicz spaces studied in [17].
Let (Ω,Σ,μ) be a finite measure space.We denote by E(μ) the space of all (equivalence classes of) Σ-measurable functions x from Ω into [0,∞).Given an Orlicz function M, we define on E(μ)a convex functional I M by and an Orlicz space L M (μ) by L M (μ) = {x ∈ E(μ) : I M (λx) < +∞ for some λ > 0} (for detail, see [10,17]).Lindenstrauss and Tzafriri [11] used the idea of Orlicz function to construct Orlicz sequence space: N. Subramanian et al. 3 where w = {all complex sequences}.The space M with the norm becomes a Banach space which is called an Orlicz sequence space.For M(t , the spaces M coincide with the classical sequence space p .If X is a sequence space, we give the following definitions: Toeplitz) dual of X, γ-dual of X, and ∧-dual of X, respectively.

Definitions and preliminaries
Throughout the article, w 2 denote the spaces of all sequences.Γ 2 M and ∧ 2 M denote the Pringscheims of double Orlicz space of entire sequence and Pringscheims of double Orlicz space of bounded sequence, respectively Let w 2 denote the set of all complex double sequences x = (x mn ) ∞ m,n=1 and M : [0,∞) → [0,∞) be an Orlicz function, or a modulus function.Given a double sequence, x ∈ w 2 .Let t denote the double sequence with t mn = |x mn | 1/(m+n) for all m,n ∈ N. Define the sets (2.1) The space ∧ 2 M is a metric space with the metric and the space Γ 2 M is a metric space with the metric

Main results
Proposition 3.1.If M is a modulus function, then Γ 2 M is a linear set over the set of complex numbers C.
Proof.It is trivial.Therefore, the proof is omitted.
Proof.Let y = {y mn } be an arbitrary point in (Γ 2 M ) β .If y is not in ∧ 2 , then for each natural number p, we can find an index m p n p such that Then, z is a point of Γ 2 M .Also, M(z mn /ρ) = 0. Hence, z is in Γ 2 M ; but, by (3.3), M(z mn y mn /ρ) does not converge: =⇒ Σx mn y mn diverges. (3.5) Thus, the sequence y would not be in (Γ 2 M ) β .This contradiction proves that N. Subramanian et al. 5 If we now choose M = id, where id is the identity and y 1n = x 1n = 1 and y mn = x mn = 0 (m > 1) for all n, then obviously x mn y mn = ∞.Hence, y / ∈ Γ 2 M β . (3.7) From (3.6) and (3.7), we are granted (Γ 2 M ) β ⊂ = ∧ 2 .This completes the proof.Proposition 3.3.Γ 2 M has AK, where M is a modulus function.
has 1 in the (m,n)th position and zero otherwise, with which is a double null sequence.Hence, x mn y mn with x ∈ Γ 2 M and f ∈ (Γ 2 M ) * , where (Γ 2 M ) * is the dual space of Γ 2 M .Take x = (x mn ) = δ mn ∈ Γ 2 M .Then, Thus, (y mn ) is a double bounded sequence, and hence a double analytic sequence.In other words, y ∈ ∧ 2 .Therefore, (Γ 2 M ) * = ∧ 2 .This completes the proof.
x mn y mn converges ∀y ∈ ∧