Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function

Acta Univ. Sapientiae, Mathematica, 3, 20) 97 09 Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function Kuldip Raj School of Mathematics Shri Mata Vaishno Devi University Katra-82320, J&K, India email: lkuldeepraj68@rediffmail.com Sunil K. Sharma School of Mathematics Shri Mata Vaishno Devi University Katra-82320, J&K, India email: sunilksharma42@yahoo.co.in Abstract. In the present paper we introduce some sequence spaces combining lacunary sequence, invariant means in 2-normed spaces defined by Musielak-Orlicz function M = ). We study some topological properties and also prove some inclusion results between these spaces. Introduction and preliminaries The concept of 2-normed space was initially introduced by Gahler 2] as an interesting linear generalization of a normed linear space which was subsequently studied by many others see 3], 9]). Recently a lot of activities have started to study sumability, sequence spaces and related topics in these linear spaces see 4], 0]). Let X be a real vector space of dimension d, where 2 d <. A 2-norm on X is a function.,. : X X R which satisfies i) x, y = 0 if and only if x and y are linearly dependent ii) x, y = y, x iii) αx, y = α x, y, α R iv) x, y + z x, y + x for all x, y X. 200 Mathematics Subject Classification: 40A05, 40A30 Key words and phrases: Paranorm space, Difference sequence space, Orlicz function, Musielak-Orlicz function, Lacunary sequence, Invariant mean 97

98 K. Raj, S. K. Sharma The pair X,.,. ) is then called a 2-normed space see 3]. For example, we may take X = R 2 equipped with the 2-norm defined as x, y = the area of the parallelogram spanned by the vectors x and y which may be given explicitly by the formula ) x x, x 2 E = abs x 2 x 2 x 22. Then, clearly X,.,. ) is a 2-normed space. Recall that X,.,. ) is a 2-Banach space if every cauchy sequence in X is convergent to some x in X. Let σ be the mapping of the set of positive integers into itself. A continuous linear functional ϕ on l,is said to be an invariant mean or σ-mean if and only if i) ϕx) 0 when the sequence x = x k ) has x k 0 for all k, ii) ϕe) =, where e =,,,...) and iii) ϕx σk) ) = ϕx) for all x l. If x = x n ), write Tx = Tx n = x σn) ). It can be shown in ] that V σ = x l limt kn x) = l, uniformly in n, l = σ limx k }, where t kn x) = x n + x σ n +... + x σ k n. k + In the case σ is the translation mapping n n +, σ-mean is often called a Banach limit and V σ, the set of bounded sequences all of whose invariant means are equal, is the set of almost convergent sequences see 6]. By a lacunary sequence = k r ) where k 0 = 0, we shall mean an increasing sequence of non-negative integers with k r k r as r. The intervals determined by will be denoted by I r = k r, k r ]. We write = k r k r. The ratio k r k r will be denoted by q r. The space of lacunary strongly convergent sequence was defined in ]. Let X be a linear metric space. A function p : X R is called paranorm, if i) px) 0, for all x X ii) p x) = px), for all x X iii) px + y) px) + py), for all x, y X iv) if σ n ) is a sequence of scalars with σ n σ as n and x n ) is a sequence of vectors with px n x) 0 as n, then pσ n x n σx) 0 as n.

Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function 99 A paranorm p for which px) = 0 implies x = 0 is called total paranorm and the pair X, p) is called a total paranormed space. It is well known that the metric of any linear metric space is given by some total paranorm see 2], Theorem 0.4.2, P-83). An orlicz function M : 0, ) 0, ) is a continuous, non-decreasing and convex function such that M0) = 0, Mx) > 0 for x > 0 and Mx) as x. Lindenstrauss and Tzafriri 5] used the idea of Orlicz function to define the following sequence space. Let w be the space of all real or complex sequences x = x k ), then ) } xk l M = x w M < k= which is called a Orlicz sequence space. Also l M is a Banach space with the norm ) } xk x = inf > 0 M. k= Also, it was shown in 5] that every Orlicz sequence space l M contains a subspace isomorphic to l p p ). The 2 condition is equivalent to MLx) LMx), for all L with 0 < L <. An Orlicz function M can always be represented in the following integral form Mx) = x 0 ηt)dt where η is known as the kernel of M, is right differentiable for t 0, η0) = 0, ηt) > 0, η is non-decreasing and ηt) as t. A sequence M = ) of Orlicz function is called a Musielak-Orlicz function see 7], 8]). A sequence N = N k ) is called a complementary function of a Musielak-Orlicz function M N k v) = sup v u u 0 }, k =, 2,... For a given Musielak-Orlicz function M, the Musielak-Orlicz sequence space t M and its subspace h M are defined as follows t M = x w I M cx) <, for some c > 0 }, } h M = x w I M cx) <, for all c > 0,

00 K. Raj, S. K. Sharma where I M is a convex modular defined by I M x) = x k ), x = x k ) t M. k= We consider t M equipped with the Luxemburg norm x ) } x = inf k > 0 I M k or equipped with the Orlicz norm } x 0 = inf k + I Mkx)) k > 0. Let M = ) be a Musielak-Orlicz function, X,.,. ) be a 2-normed space and p = P k ) be any sequence of strictly positive real numbers. By S2 X) we denote the space of all sequences defined over X,.,. ). We now define the following sequence spaces: w o σm, p,.,. ] = x S2 X) lim r } > 0, uniformly in n, w ] = x S2 X) lim r } > 0, uniformly in n, and w ] = x S2 X) sup r,n } for some > 0. t kn x l) = 0, =0, <, When Mx) = x for all k, the spaces w o ], w ] and w σ Mk, p,.,. ] reduces to the spaces ] σ wo p,.,., w ] σ, p,.,. and w ] respectively. If p k = for all k, the spaces w o ], w ] and w ] reduces to w o ] σ M,.,., w ] σ M,.,. and w ] σ M,.,. respectively.

Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function 0 The following inequality will be used throughout the paper. If 0 p k sup p k = H, K = max, 2 H ) then a k + b k p k K a k p k + b k p k } ) for all k and a k, b k C. Also a p k max, a H ) for all a C. In the present paper we study some topological properties of the above sequence spaces. 2 Main results Theorem Let M = ) be Musielak-Orlicz function, p = p k ) be a bounded sequence of positive real numbers, then the classes of sequences w o ], w ] and ] w are linear spaces over the field of complex numbers. Proof. Let x, y w o ] and α, β C. In order to prove the result we need to find some 3 such that t kn αx + βy) lim r 3 = 0, uniformly in n. Since x, y w o ], there exist positive, 2 such that and lim r lim r t kn y) 2 = 0, uniformly in n )] = 0, uniformly in n. Define 3 = max2 α, 2 β 2 ). Since ) is non-decreasing and convex )] t kn αx + βy) 3 )] t kn αx) 3 + t kn βy) 3 )] t kn x) + t kn y) K )] t kn x) + + K )] t kn y) 2 0 as r, uniformly in n. 2

02 K. Raj, S. K. Sharma So that αx + βy w o ]. This completes the proof. Similarly, we can prove that w ] and ] w are linear spaces. Theorem 2 Let M = ) be Musielak-Orlicz function, p = p k ) be a bounded sequence of positive real numbers. Then w o ] is a topological linear spaces paranormed by gx) = inf pr H ) H :, r =, 2,..., n =, 2,..., where H = max,sup k p k < ). Proof. Clearly gx) 0 for x = x k ) w o ]. Since 0) = 0, we get g0) = 0. Conversely, suppose that gx) = 0, then inf pr H : )] ) H, r, n = 0. This implies that for a given ǫ > 0, there exists some ǫ 0 < ǫ < ǫ) such that )] ) t kn x) H ǫ. Thus )] ) t kn x) ǫ H )] ) t kn x) H ǫ, for eac and n. Suppose that x k 0 for each k N. This implies that t kn x) 0, for each k, n N. Let ǫ 0, then. It follows that t knx) ǫ ) t kn x) ] p k ) H ǫ which is a contradiction.

Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function 03 Therefore, t kn x) = 0 for each k and thus x k = 0 for each k N. Let > 0 and 2 > 0 be such that )] ) t kn x) H and )] ) t kn y) H 2 for eac. Let = + 2. Then, we have )] ) t kn x + y) H )] ) t kn x) + t kn y) H + 2 ) t kn x) + 2 + )] 2 t kn y) ) p k H + 2 2 + 2 ) ) 2 + + 2 by Minkowski s inequality) )] ) H )] ) t kn y) H 2.

04 K. Raj, S. K. Sharma Since s are non-negative, so we have gx + y) = = inf pr )] ) t kn x) + t kn y) H H, r, n inf pr H )] ) t kn x) H, r, n + + inf pr H 2 )] ) t kn x) H 2, r, n. Therefore, gx + y) gx) + gy). Finally, we prove that the scalar multiplication is continuous. Let λ be any complex number. By definition, gλx)=inf pr H )] ) t kn λx) H, r, n. Then pr )] ) t kn x) H gλx)= inf λ t) H M k t, r, n. where t = λ. Since λ pr max, λ sup pr ), we have gλx) max, λ sup pr ) inf t pr )] ) t kn x) H H t, r, n. So, the fact that scalar multiplication is continuous follows from the above inequality. This completes the proof of the theorem.

Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function 05 Theorem 3 Let M = ) be Musielak-Orlicz function. If then sup k Mk t) ] p k < for all t > 0, w ] w ]. Proof. Let x w. By using inequality ), we have ] )] t kn x) K t kn x l) + K l. )] Since sup k Mk t) ] p k <, we can take that sup k Mk t) ] p k = T. Hence we get x w ]. Theorem 4 Let M = ) be Musielak-Orlicz function which satisfies 2 - condition for all k, then w ] w ]. Proof. Let x w ]. Then we have T r = t kn x l) as r uniformly in n, for some l. Let ǫ > 0 and choose δ with 0 < δ < such that t) < ǫ for 0 t δ for all k. So that )] t kn x l) = t kn x l) k I r t kn x l)z δ + 2 k I r t kn x l)z δ t kn x l) )]. For the first summation in the right hand side of the above equation, we have ǫ H by using continuity of for all k. For the second summation, we write t kn x l) + t knx l). δ

06 K. Raj, S. K. Sharma Since is non-decreasing and convex for all k, it follows that ) t kn x l) ) < + t kn x l) δ 2 2) + ) 2 2) t kn x l) δ. Since satisfies 2 -condition for all k, we can write ) t kn x l) 2 L t kn x l) δ 2) + 2 L t kn x l) δ 2) = L t kn x l) δ 2). So we write t kn x l) ǫ H + max, L 2))δ] H T r. Letting r, it follows that x w σ M, p,.,. ]. This completes the proof. Theorem 5 Let M = ) be Musielak-Orlicz function. Then the following statements are equivalent: i) w ] ] wo, ii) w o ] ] wo, iii) sup Mk t) ] p k < for all t > 0. r Proof. i) = ii) We have only to show that w o ] ] w. Let x w o ]. Then there exists r r o, for ǫ > 0, such that t kn x) k p < ǫ. Hence there exists H > 0 such that sup t kn x) k r,n p < H

Some sequence spaces in 2-normed spaces defined by Musielak-Orlicz function 07 for all n and r. So we get x w ]. ii) = iii) Suppose that iii) does not hold. Then for some t > 0 sup r t)] = and therefore we can find a subinterval I rm) of the set of interval I r such that ] m ) > m, m =, 2, 2) m) m) Let us define x = x k ) as follows, x k = m if k I rm) and x k = 0 if k I rm). Then x w o ] but by eqn. 2), x ] w. which contradicts ii). Hence iii) must hold. iii) = i) Suppose i) not holds, then for x w ], we have sup r,n = 3) Let t = for each k and fixed n, so that eqn. 3) becomes t knx) sup r t)] = which contradicts iii). Hence i) must hold. Theorem 6 Let M = ) be Musielak-Orlicz function. Then the following statements are equivalent: i) w o ] ] wo, ii) w o ] ] w, iii) inf Mk t) ] p k > 0 for all t > 0. r Proof. i) = ii) : It is easy to prove. ii) = iii) Suppose that iii) does not hold. Then inf r t)] = 0 for some t > 0,

08 K. Raj, S. K. Sharma and we can find a subinterval I rm) of the set of interval I r such that m)] <, m =, 2,... 4) m m) Let us define x k = m if k I rm) and x k = 0 if k I rm). Thus by eqn.4), x w o ] but x ] w which contradicts ii). Hence iii) must hold. iii) = i) It is obvious. Theorem 7 Let M = ) be Musielak-Orlicz function. Then w σ M, p,.,. ] ] wo if and only if lim r Mk t) ] p k = 5) ] p,.,. Proof. Let w ] wo σ. Suppose that eqn. 5) does not hold. Therefore there is a subinterval I rm) of the set of interval I r and a number t o > 0, where t o = for all k and n, such that m) t knx) m) t o )] M <, m =, 2,... 6) Let us define x k = t o if k I rm) and x k = 0 if k I rm). Then, by eqn. 6), x w σ, p,.,. ]. But x w o σp,.,. ]. Hence eqn. 5) must hold. Conversely, suppose that eqn. 5) hold and that x w σ, p,.,. ]. Then for eac and n M <. 7) Now suppose that x w o σp,.,. ]. Then for some number ǫ > 0 and for a subinterval I ri of the set of interval I r, there is k o such that t kn x) p k > ǫ for k k o. From the properties of sequence of Orlicz functions, we obtain ǫ )] )] which contradicts eqn.6), by using eqn. 7). This completes the proof.

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