Microsoft Word - N. Pahari. _34-44_.doc N.P. Pahari / BIBECHANA 10 (2014) 20-30 : BMHSS, p.20 (Online Publication: Dec., 2013) BIBECHANA A Multidisciplinary Journal of Science, Technology and Mathematics ISSN 2091-0762 (online) Journal homepage: http://nepjol.info/index.php/BIBECHANA On 2 - normed space valued paranormed null sequence space (c0 ( S, αααα–, u– , ||. , .|| ),T ) Narayan Prasad Pahari Central Department of Mathematics, Tribhuvan University, University Campus, Kirtipur, Kathmandu, Nepal Email: nppahari @ gmail.com Article history: Received 5 September, 2013; Accepted 23 September, 2013 Abstract In this paper we introduce and study a new class c0 ( S, α–, u – , ||. , .|| ) of sequences with values in 2- normed space as a generalization of basic null sequence space c0. We investigate some conditions pertaining to the containment relations of the class c0 ( S, α–, u– , ||. , .|| ) in terms of different α– and u – and explore the linear topological structures of the space c0 ( S, α–, u– , ||. , .|| ) by endowing it with a suitable natural paranorm. Keywords: 2- normed space, 2–Banach space, Paranormed space, Sequence Space. 1. Introduction and Preliminaries Before proceeding with the main results, we recall some of the basic notations and definitions that are used in this paper. The notion of 2–normed space was initially introduced by S. GÄahler [1] as an interesting linear generalization of a normed linear space, which was further studied by J.A. White and Y.J. Cho [2], K.Iseki [3], R. Freese et al. [4,5], W.Raymond et al. [6] and many others. Recently a lot of activities have been started by many researchers to study this concept in different directions, for instances E. Savas [7], H. Gunawan and H. Mashadi [8] , J.K. Srivastava and N.P. Pahari [9] , M. Açikgöz [10] and others. Let S be a vector space of dimension d > 1 over K, the field of real or complex numbers. A 2 - norm on S is a real valued function ||. , .|| on S × S satisfying the following conditions: 2- N1: || ξ, η || ≥ 0 and || ξ, η || = 0 if and only if ξ and η are linearly dependent; 2-N2: || ξ, η || = || η, ξ ||, for all ξ, η ∈ S; 2-N3: || γξ, η || = |γ | ||ξ¸ η ||, where γ∈K and ξ, η ∈ S;and 2-N4 : || ξ1 + ξ2, η || ≤ || ξ1, η || + || ξ2, η || ,for all ξ1 , ξ2 and η∈ S. The pair (X, ||. , .||) is called a 2–normed space. Thus the notion of 2–normed space is just a two- dimensional analogue of a normed space . Recall that (X, ||. , .||) is a 2-Banach space if every Cauchy sequence in X is convergent to some x0 in X. N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.21 (Online Publication: Dec., 2013) The pair (S, ||. , .||) is called a 2–normed space. Thus the notion of 2–normed space is just a two- dimensional analogue of a normed space . Recall that (S, ||. , .||) is a 2-Banach space if every Cauchy sequence in S is convergent to some ξ0 in S. Geometrically, a 2-norm function generalizes the concept of area function of parallelogram due to the fact that, in the standard case, it represents the area of the usual parallelogram spanned by the two associated vectors.For example, consider S = R2, being equipped with || ξ, η || = | ξ1η2 – ξ2η1|, where ξ = (ξ1, ξ2) and η = (η1 , η2). Then (S, ||. , .||) forms a 2–normed space and || ξ, η || represents the area of the parallelogram spanned by the two associated vectors ξ and η. Again, let S = R 3 and define the function ||. , .|| on S × S by || ξ, η || =       Det      i j k ξ1 ξ2 ξ3 η1 η2 η3 where ξ = (ξ1, ξ2, ξ3) and y = (η1, η2, η3). Obviously (S, ||. , .||) forms a 2–normed space. Analogously, if (S , <. , .>) be a finite dimensional inner product space and ||. , .|| defined on S × S by || ξ, η || =      <ξ‚ξ > < ξ‚ η > < η‚ ξ > < η‚ η > , then we can see that (S, ||. , .||) satisfies all the conditions of 2–normed space. The notion of convergence has introduced by White and Cho [2]. A sequence (ξn) in a linear 2– normed space S is convergent if there is an ξ ∈ S such that lim n→∞ || ξn – ξ, η || = 0, for each η ∈ S .It is said to be a Cauchy if there are η and ν in S such that η and ν are linearly independent and lim m‚n→∞ ||ξm – ξn‚ η|| = 0 and lim m‚n→∞ ||ξm – ξn‚ ν|| = 0. A linear 2–normed space (S, ||.,.||) is called 2–Banach space if every Cauchy sequence in S is convergent to some ξ ∈ S. The notion of paranormed space is closely related to linear metric space , see A. Wilansky [11]. A paranormed space (S, T ) is a linear space S with zero element θ together with a function T : S → R+ (called a paranorm on S) which satisfies the following axioms: PN1: T (θ) = 0; PN2: T (ξ) = T (–ξ) , for all ξ ∈ S; PN3: T (ξ + η) ≤ T (ξ) + T (η) , for all ξ , η ∈ S; and PN4: Scalar multiplication is continuous i.e., if (γn) is a sequence of scalars with γn → γ as n → ∞ and (ξn) is a sequence of vectors with T (ξn − ξ) → 0 as n → ∞ then T (γn ξn − γξ) → 0 as n → ∞. Note that the continuity of scalar multiplication is equivalent to (i) if T (ξn) → 0 and γn → γ as n → ∞, then T(γn ξn) → 0 as n → ∞ and N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.22 (Online Publication: Dec., 2013) (ii) if γn → 0 as n → ∞ and ξ be any element in S, then T (γn ξ) → 0, see Wilansky [11]. A paranorm is called total if T (ξ) = 0 implies ξ = θ, see A. Wilansky [11]. The studies of paranorm on sequence spaces were initiated by Maddox [12] and many others. Srivastava and et al. [13-15], Pahari [16], Parashar and Choudhary [17], Bhardwaj and Bala [18] and many others further studied various types of paranormed sequence spaces and function spaces. A sequence space S is said to be normal if ξ – = (ξk) ∈ S and α– = (αk) a sequence of scalars with |αk| ≤ 1, for all k ≥ 1, then α– ξ – = (αk ξk) ∈S . 2. The Class c0 ( X, λλλλ – , u – , ||. , .|| ) of 2-Normed Space Valued Vector Sequences Let u– = (uk) and v– = (vk) be any sequences of strictly positive real numbers and λ – = (λk) and µ–= (µk ) be the sequences of non zero complex numbers. Let ( X, ||. , .|| ) be the 2- Normed space over the field C of complex numbers and θ denotes the zero element of X. Let ω(X) denotes the linear space of all sequences x – = (ξk) and y– = (ηk ) with ξk , ηk ∈ X , k ≥ 1 with usual coordinate wise operations i.e., x – + y– = (ξk +ηk) and α x – = (αξk ) ,for each x – , y– ∈ ω (X) and α∈ C. We shall denote ω (C) by ω. Further, λ– = (λk ) ∈ ω and x – ∈ ω (X ) we shall write λ– x – = (λk ξk ) . We now introduce the following classes of 2-normed space X–valued sequences c0 ( X, λ – , u– , ||. , .|| ) = {x– = (xk) ∈ ω(X ), xk ∈ X, k ≥ 1 satisfying || λk xk‚ y|| uk → 0 as k→ ∞ , for each y ∈ X }. In fact, this class is a generalization of the familiar sequence spaces, studied in Srivastava et al. (1996) [12], Srivastava (1996), Pahari ( 2011) [16], Pahari and et al ( 2011) [11] , using 2-norm . Further, when λk = 1 for all k , then c0 ( X, λ – , u– , ||. , .|| ) will be denoted by c0 ( X, u– , ||. , .|| ) and when uk = 1 for all k, then c0 ( X, λ – , u– , ||. , .|| ) will be denoted by c0 ( X, λ – , ||. , .|| ). If uk = λk = 1 for all k , then the class c0 ( X, λ – , u– , ||. , .|| ) will be denoted by c0 ( X, ||. , .|| ) . Further, when X = C we simply write c0 ( X, λ – , u– , ||. , .|| ) as c0 (λ – , u– , ||. , .|| ). 3. Containment Relations In this section, we investigate some inclusion relations of the class c0( S, ||. , .||, α–, u– ) arising in terms of different u– and α– . Throughout, we denote sup uk =L for each k and for scalar α, A [α] = max (1, |α|). But when the sequences uk and vk occur, then to distinguish L we use the notations L(u) and L(v) respectively. Theorem 3.1: For any u – = (uk), c0 ( S, α – , u – , ||. , .|| ) ⊂⊂⊂⊂ c0 ( S, β – , u – , ||. , .|| ) if and only if lim inf k    αk βk uk > 0. Proof: N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.23 (Online Publication: Dec., 2013) For the necessity, assume that c0 ( S, α– , u– , ||. , .|| ) ⊂ c0 ( S, β – , u– , ||. , .|| ) but lim infk    αk βk uk = 0. Then we can find a sequence (k(n)) of integers such that 1 ≤ k(n) < k(n +1), n ≥ 1 for which n 2 |αk(n) | uk(n) < |βk(n)| uk(n) , for all n ≥ 1. ...( 3.1) Now, corresponding to ρ ∈ S and ρ ≠ θ, we define the sequence ξ – = ( ξk) by ξk =   αk(n) -1 n-2/uk(n) ρ‚ if k = k(n) ‚ n ≥1 and θ‚ otherwise. ...( 3.2) Then for k = k(n), n ≥ 1 ,we have || αk ξk‚ η|| uk = ||n-2/uk(n) ρ‚ η || uk(n) ≤ 1 n2 ||ρ‚ η || uk(n) ≤ 1 n2 A [ ||ρ‚ η|| L(u) ] → 0 as n→ ∞, where ||ρ, η|| uk(n) ≤ A [ || ρ, η|| L(u) ] for each n ≥ 1 is used and || αk ξk‚ η|| u k = 0, for k ≠ k(n) , n ≥ 1. This shows that ξ – ∈ c0 ( S, α–, u– , ||. , .|| ) . On the other hand, let us choose η ∈ S such that || w, η || = 1. Then for k = k(n), n ≥ 1, and in view of (3.1) and (3.2) ,we have || βk ξk ‚η|| uk = || β k(n) ξ k(n) ‚η || u k(n) = 1 n 2 || w‚ η||uk(n)    αk(n) βk (n) uk(n) ≥ 1. This shows that ξ – ∉ c0 ( S, β – , u–,||. , .|| ),a contradiction. For the sufficiency, assume that lim infk    αk βk uk > 0.Then there exists m > 0 such that m |βk| uk < |αk | uk for all sufficiently large values of k. Let ξ – = (ξk) ∈ c0 ( X, α–, u– , ||. , .|| ) . Then for each η ∈S, || αk ξk‚ η|| u k → 0 as k→ ∞ . Now for each η ∈ S, ||βk ξk‚ η|| uk ≤ |αk| uk m ||ξk ‚η|| uk ≤ 1 m || αk ξk‚ η|| uk → 0 as k → ∞. This clearly implies that ξ – ∈ c0 ( S, β – , u– , ||. , .|| ) and hence c0 ( S, α–, u– , ||. , .|| ) ⊂ c0 ( S, β – , u– , ||. , .|| ). This completes the proof . Theorem 3.2: For any u – = (uk), c0 ( S, β – , u – , ||. , .|| ) ⊂ ⊂ ⊂ ⊂ c0 ( S, α – , u – ,||. , .|| ) if and only if lim supk    αk βk uk < ∞∞∞∞. N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.24 (Online Publication: Dec., 2013) Proof: For the necessity, suppose that c0 ( S, β – , u–,||. , .|| ) ⊂ c0 ( S, α– , u– , ||. , .|| ) but lim supk    αk βk uk = ∞. Then there exists a sequence (k(n)) of positive integers with 1 ≤ k(n) < k(n + 1) , n ≥ 1 satisfying |αk(n)| uk(n) > n2 |βk(n)| uk(n) ,for each n ≥ 1. ...( 3.3) Corresponding to ρ ∈ S and ρ ≠ θ, we define a sequence ξ – = (ξk) by ξk =   βk(n) -1 n-2/uk(n) ρ‚ if k = k(n)‚ n ≥1 and θ‚ otherwise. ...( 3.4) Then for k = k(n), n ≥ 1, we have || βk ξk‚ η|| uk = || β k(n) ξ k(n) ‚η || u k(n) = ||n-2/uk(n) ρ‚ η || uk(n) ≤ 1 n2 ||ρ‚ η || uk(n) ≤ 1 n2 A [ ||ρ‚ η || L(u) ] → 0 as n→ ∞ and ||βk ξk‚ η|| uk = 0, for k ≠ k(n) , n ≥ 1. This shows that ξ – ∈ c0 ( S, β – , u–,||. , .|| ) . On the other hand, let us choose η ∈ S such that ||ρ, η|| = 1. Then for k = k(n), n ≥ 1, in view of (3.3) and (3.4) ,we have || αk ξk ‚η|| uk =    αk(n) βk (n) uk(n) . 1 n 2 ||ρ‚ η||uk(n) ≥ 1 and so ξ – ∉ c0 ( S, α– , u– , ||. , .|| ) ,a contradiction. For the sufficiency, assume that lim supk    αk βk uk < ∞. Then we can find a positive constant d such that d |βk| uk > |αk| uk for all sufficiently large values of k. Let ξ – = (ξk) ∈ c0 ( S, β – , u–,||. , .|| ) . Then we have ||βk ξk‚ η|| uk → 0 as k→ ∞ , for each η ∈S. Now we have ||αk ξk‚ η || uk ≤ d|βk| uk || ξk‚ η|| uk ≤ ||βk ξk‚ η|| uk → 0 as k→ ∞ , for each η ∈S. This clearly implies that ξ – ∈ c0 ( S, α– , u– , ||. , .|| ) and hence c0 ( S, β – , u–,||. , .|| ) ⊂ c0 ( S, α– , u– , ||. , .|| ). The proof is now complete. On combining the Theorems 3.1 and 3.2, we get: Theorem 3.3: For any u – = (uk), c0 ( S, α – , u – , ||. , .|| ) = c0 ( S, β – , u – ,||. , .|| ) if and only if 0 < lim infk    αk βk uk ≤ ≤ ≤ ≤ lim supk    αk βk uk < ∞.∞.∞.∞. Corollary 3.4: For any u – = (uk), N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.25 (Online Publication: Dec., 2013) (i) c0 ( S, α – , u – , ||. , .|| ) ⊂⊂⊂⊂ c0 ( S, u – ,||. , .|| ) if and only if lim infk |ααααk| uk > 0 ; (ii) c0 ( S, u – ,||. , .|| ) ⊂⊂⊂⊂ c0 ( S, α – , u – , ||. , .|| ) if and only if lim supk |ααααk| uk < ∞∞∞∞;and (iii) c0 ( S, α – , u – , ||. , .|| ) = c0 ( S, u – ,||. , .|| ) if and only if 0 < lim infk |ααααk| uk ≤ ≤ ≤ ≤ lim supk |ααααk| uk < ∞∞∞∞. Proof: The proof follows by putting βk = 1 for all k in Theorems 3.1, 3.2 and 3.3. Theorem 3.5: For any α– = (ααααk), c0 ( S, α – , u – , ||. , .|| ) ⊂⊂⊂⊂ c0 ( S, α – , v – ,||. , .|| ) if and only if lim infk vk uk > 0. Proof: For the necessity, suppose that the inclusion holds but lim infk vk uk = 0. Then there exists a sequence (k(n)) of positive integers such that 1 ≤ k(n) < k(n + 1), n ≥ 1 for which n vk(n) < uk(n) , for each n ≥ 1. …(3.5) Let ρ ∈ S and ρ ≠ θ. We define a sequence ξ – = (ξk) by ξk =   αk(n) -1 n -1/uk(n) ρ‚ for k = k(n) ‚ n ≥1 and θ‚ otherwise. ...( 3.6) Then for each η ∈ S, k = k(n), n ≥ 1 , we have ||αk ξk ‚ η|| u k = ||n -1/u k(n) ρ‚ η|| uk(n) = 1 n || ρ‚ η||uk(n) ≤ 1 n A [ ||ρ‚ η || L(u) ] → 0 as n →∞ and ||αk ξk ‚ η|| uk = 0, for k ≠ k(n) , n ≥ 1. This shows that ξ – ∈ c0 ( S, α– , u– , ||. , .|| ). But on the other hand, let us choose η ∈ S such that ||ρ, η|| = 1. Then for k = k(n), n ≥ 1, in view of (3.5) and (3.6) ,we have ||αk ξk ‚ η|| v k = ||α k(n) ξ k(n) ‚ η|| vk(n) = || n-1/uk(n) ρ‚ η|| vk(n) ≥ 1 n 1/n || ρ‚ η|| vk(n) ≥ 1 n 1/2 This shows that ξ – ∉ c0 ( S, α–, v–,||. , .|| ) , which contradicts our assumption. For the sufficiency of the condition, suppose that lim infk vk uk > 0. Then there exists a m > 0 such that vk > m uk for all sufficiently large values of k. Let ξ – = (ξk) ∈ c0 ( S, α–, u– , ||. , .|| ) . Then for each η ∈ S ||αk ξk ‚ η|| uk → 0 as k → ∞. Hence for a given ε > 0, if we choose 0 < δ < 1 satisfying δ m < ε satisfying N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.26 (Online Publication: Dec., 2013) ||αk ξk , η || uk < δ < 1 for each η ∈ S and for all sufficiently large values of k.Thus ||αk ξk , η || vk ≤ [||αk ξk ‚ η || uk] m ≤ δ m < ε, for each η ∈ S and for all sufficiently large values of k and consequently ξ – ∈ c0 ( S, α–, v–,||. , .|| ). Hence c0 ( S, α–, u– , ||. , .|| ) ⊂ c0 ( S, α– , v–,||. , .|| ). This completes the proof of the theorem. Theorem 3.6: For any α– = (ααααk), c0 ( S, α – , v – ,||. , .|| ) ⊂⊂⊂⊂ c0 ( S, α – , u – , ||. , .|| ) if and only if lim supk vk uk < ∞∞∞∞. Proof: For the necessity, suppose that the inclusion holds but lim supk vk uk = ∞ . Then there exists a sequence (k(n)) of positive integers such that 1 ≤ k(n) < k(n+ 1), n ≥ 1 for which vk(n) > n uk(n) ,for all n ≥ 1. ...( 3.7) Corresponding to ρ ∈ S and ρ ≠ θ , we define a sequence ξ – = (ξk) by ξk =   αk(n) -1 n -1/vk(n) ρ‚ for k = k(n) ‚ n ≥1 and θ‚ otherwise. ...( 3.8) Then for each η ∈ S, k = k(n), n ≥ 1 , we have ||αk ξk ‚η || vk = ||n-1/vk(n) ρ‚η|| vk(n) = 1 n || ρ‚ η||vk(n) ≤ 1 n A [ ||ρ‚ η || L(v) ] → 0 as n →∞ and ||αk ξk ‚ η|| v k = 0, for k ≠ k(n) , n ≥ 1. This shows that ξ – ∈ c0 ( S, α– , v–,||. , .|| ). But on the other hand, let us choose η ∈ S such that || ρ, η|| = 1. Then for k = k(n), n ≥ 1, in view of (3.7) and (3.8) ,we have ||αk ξk ‚ η|| u k = || n-1/vk(n) ρ‚ η|| uk(n) ≥ 1 n 1/n || ρ‚ η|| uk(n) ≥ 1 n 1/2 This shows that ξ – ∉ c0 ( S, α– , u– , ||. , .|| ) , a contradiction. For the sufficiency of the condition, assume that lim supk vk uk < ∞. Hence there exists d > 0 such that vk < d uk for all sufficiently large values of k. Let ξ – = (ξk) ∈ c0 ( S, α– , v–,||. , .|| ). Then for each η ∈ S ||αk ξk ‚ η|| v k → 0 as k → ∞. Hence for a given ε > 0, if we choose 0 < δ < 1 satisfying δ 1/d < ε satisfying ||αk ξk , η || vk < δ < 1 for each η ∈ S and for all sufficiently large values of k.Thus ||αk ξk , η || uk ≤ [||αk ξk ‚ η|| v k] 1/d N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.27 (Online Publication: Dec., 2013) ≤ δ 1/d < ε, for each η ∈ S and for all sufficiently large values of k and consequently ξ – ∈ c0 ( S, α– , u–,||. , .|| ). Hence c0 ( S, α– , v–,||. , .|| ) ⊂ c0 ( S, α– , u– , ||. , .|| ) . This completes the proof. On combining the Theorems 3.5 and 3.6, one obtain Theorem 3.7: For any α– = (ααααk),c0 ( S, α – , u – , ||. , .|| ) = c0 ( S, α – , v – ,||. , .|| ) if and only if 0 < lim infk vk uk ≤≤≤≤ lim supk vk uk < ∞.∞.∞.∞. Corollary 3.8: For any α– = (ααααk), (i) c0 ( S, α – , ||. , .|| ) ⊂ ⊂ ⊂ ⊂ c0 ( S, α – , u – , ||. , .|| ) if and only if lim inf k uk > 0; (ii) c0 ( S, α – , u – , ||. , .|| ) ⊂⊂⊂⊂ c0 ( S, α – , ||. , .|| ) if and only if lim sup k uk < ∞∞∞∞; and (iii) c0 ( S, α – , u – , ||. , .|| ) = c0 ( S, α – , ||. , .|| ) if and only if 0 < lim infk uk ≤≤≤≤ lim supk uk < ∞∞∞∞. Proof: Proof follows by taking uk = 1 for all k and replacing v– by u– in Theorems 3.5 , 3.6 and 3.7. Theorem 3.9: For any α– = (ααααk), β – = (ββββk), u – = (uk) and v – = (vk), c0 ( S, α – , u – , ||. , .|| ) ⊂ ⊂ ⊂ ⊂ c0 ( S, β – , v – , ||. , .|| ) if and only if (i) lim infk    αk βk uk > 0 ; and (ii) lim infk vk uk > 0. Proof: Proof of the theorem follows immediately from the Theorems 3.1 and 3.5. In the following example, c0 (S, α–, u– , ||. , .|| ) may strictly be contained in c0 ( S, β – , v–, ||. , .|| ) in spite of the satisfaction of the conditions (i) and (ii) of Theorem 3.9. Example 3.10: Let ( S, ||. , .|| ) be a 2- normed space and consider a sequence ξ – = (ξk) defined by ξk = k –k ρ, if k = 1, 2, 3, …, where ρ ∈ S and ρ ≠ θ. Further, let uk = k –1, if k is odd integer, uk = k –2, if k is even integer, vk = k –1 for all values of k, αk = 3k, βk = 2k for all values of k. Then    αk βk uk = 3 2 or     3 2 1/k according as k is odd or even integer and hence lim infk    αk βk uk > 0. Further, vk uk = 1, if k is odd integer, vk uk = k, if k is even integer. Therefore lim infk vk uk > 0. Hence the conditions (i) and (ii) of Theorem 3.9 are satisfied. Now for each η ∈ S, we have N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.28 (Online Publication: Dec., 2013) ||βk ξk ‚ η|| v k = ||2k k –k ρ ‚ η|| 1/k ≤ 1 k 2 || ρ ‚ η|| 1/k ≤ 1 k 2A [ || ρ‚ η || ] → 0 as k → ∞, and it shows that ξ – ∈ c0 ( S, β – , v–, ||. , .|| ). But on the other hand, let us choose η ∈ S such that || ρ, η || = 1. Then for each even integer k, we have || αk ξk ‚ η|| uk = ||3k k –k ρ ‚ η|| 1/k 2 = (3/k)1/ k || ρ ‚ η|| 1/k 2 > 1 2 . This implies that ξ – ∉ c0 ( S, α– , u– , ||. , .|| ).Thus the containment of c0 ( S, α– , u– , ||. , .|| ) in c0 ( S, β – , v–, ||. , .|| ) is strict inspite of the satisfaction of the conditions (i) and (ii) of the Theorem 3.9. 4. Linear Topological Structures of c0 ( S, α – , u – , ||. , .|| ) In this section, we shall investigate some results that characterize the linear topological structure of the class c0 ( S, α–, u– , ||. , .|| ) by endowing it with suitable natural paranorm. Throughout we take coordinatewise operations of sequences over the field C of complex numbers i.e.,for ξ – = (ξk ) and η– = (ηk ) and scalar γ, ξ – + η– = (ξk + ηk ) and γ ξ – = (γ ξk ) and we see below that each of these classes forms a complete paranormed linear space over C. Moreover, we use frequently | ξ + η | uk ≤ S {|ξ| uk + |η| uk }, where ξ , η ∈ C, 0 < uk ≤ supk uk = L < ∞ and A(α) = max (1, |α| ). Theorem 4.1: The space c0 ( S, α – , u – , ||. , .|| ) forms a linear space over C if and only if supk uk is bounded above . Proof: For the necessity of the conditions, suppose that c0 ( S, α–, u– , ||. , .|| ) is a linear space over C but supk uk = ∞. Then there exists a sequence (k(n)) of positive integers satisfying 1 ≤ k(n) < k(n + 1), n ≥ 1 for which uk(n) > n ,for each n ≥ 1 … .(4.1) Now, corresponding to ρ ∈ S and ρ ≠ θ, we define the sequence ξ – = (ξk) by ξk =   αk(n) -1 n-2/uk(n) ρ‚ if k = k(n) ‚ n ≥1 and θ‚ otherwise. … .(4.2) Then for k = k(n), n ≥ 1, we have ||αk ξk‚ η|| uk = ||n-2/uk(n) ρ‚ η || uk(n)) ≤ ||ρ‚ η || uk(n) n2 → 0 as n→ ∞, N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.29 (Online Publication: Dec., 2013) and ||αk ξk‚ η|| uk = 0, for k ≠ k(n) , n ≥ 1 showing that ξ – ∈ c0 ( S, α– , u– , ||. , .|| ). But on the other hand, let us choose η ∈ S such that ||ρ, η|| = 1. Then for such η and scalar γ = 4, for k = k(n), n ≥ 1, in view of (4.1) and (4.2), we have || γ αk ξk‚ η || uk = || αk(n) γ ξk(n ) ‚ η || uk(n) = || 4 n–2/ uk(n) ρ ‚η || uk(n) = 4 uk(n) n2 ≥ sup n 4 n n 2 ≥ 1 . This shows that γ ξ – ∉ c0 ( S, α–, u– , ||. , .|| ) , a contradiction. For the sufficiency of the condition, assume that supk uk < ∞. Let ξ – , η– ∈ c0 ( S, α–, u– , ||. , .|| ) and γ, ρ ∈ C. Then for each η ∈ S, we have ||αk ξk‚ η|| uk → 0 and ||αk ηk‚ η|| uk →0, as k → ∞. We now setting S = max ( 1,2L-1 ), then we have ||αk (γξk + ρηk)‚ η || uk = (|| γαk ξk‚ η || + || ρ αk ηk‚ η || ) uk ≤ ( S || γ αk ξk‚ η || uk + S || ρ αk ηk‚ η || uk ) = S |γ| uk || αk ξk‚ η || uk + S |ρ| uk || αk ηk‚ η || uk which tends to 0 as k → ∞, for each η ∈ S and hence αξ – + ρ η– ∈ c0 ( S, α–, u– , ||. , .|| ). This implies that c0 ( S, α–, u– , ||. , .|| ) forms a linear space over C. This completes the proof. Let u– ∈ ℓ∞ i.e., supk uk < ∞ , M = max (1, supk uk ) and consider a real valued function T on c0 ( S, α–, u– , ||. , .|| ) defined by T (x–) ={ sup k || αk ξk ‚ η || u k /M , for each η ∈ S },for ξ – ∈ ℓ∞ ( S, α–, u– , ||. , .|| ). … (4.3) We prove below that c0 ( S, α–, u– , ||. , .|| ) with respect to T forms a paranormed space. Theorem: 4.2 (c0 ( S, α – , u – , ||. , .|| ), T ) forms a total paranormed space. Proof: Since T(ξ – ) ≥ 0 ; T(ξ – ) = 0 if and only if ξ – = θ–; T (–ξ – ) = T (ξ – ) ; and T (ξ – + η–) ≤ T (ξ – ) + T (η –) easily follow. So PN1 , PN2 and PN3 are obvious. For PN4 i.e., the continuity of scalar multiplication, it suffices to show that (a) if ξ – (n) → θ– in T and γn → γ as n → ∞,then γn ξ – (n) → θ– in T; and (b) if γn → 0 and ξ – ∈ c0 ( S, α–, u– , ||. , .|| ) , then γn ξ – → θ– in T . Now (a) is easily proved if we suppose |γn| ≤ L for all n ≥ 1 and in view of (4.3) , we have T (γn ξ – (n)) = { sup k || γn αk ξk (n) ‚ η || u k /M , for each η ∈ S } ≤ { sup k |γn| u k /M sup k || αk ξk (n) ‚ η || u k /M , for each η ∈ S } N.P. Pahari / BIBECHANA 10 (2014) 20-30: BMHSS, p.30 (Online Publication: Dec., 2013) = A(L) { sup k || αk ξk (n) ‚ η || u k /M , for each η ∈ S } ≤ A(L) T (ξ – (n)), implies that T (γn ξ – (n)) → 0, as T (ξ – (n)) → 0 as n → ∞. To prove (b), let ξ – ∈ c0 ( S, α–, u– , ||. , .|| ), |γn| < 1 for all n ≥ N1 and ε > 0. Then there exists a positive integer K such that ||αk ξk ‚ η || u k < ε M for all k ≥ K and for each η ∈ S and hence for all k ≥ K and n ≥ N1, we have || γn αk ξk ‚ η || u k = |γn| u k || αk ξk ‚ η || u k < εM, for each η ∈ S. Now choose N2 such that for all k = 1, 2, …, K – 1 and n ≥ N2, || γn αk ξk ‚ η || u k = |γn| u k || αk ξk ‚ η || u k < ε M, for each η ∈ S. 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