ON SOME SEMINORMED SEQUENCE SPACES DEFINED BY A MODULUS FUNCTION. Yavuz Altin, Hıfsı Altinok and Rifat Çolak

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1 121 Kragujevac J. Math ) ON SOME SEMINORMED SEQUENCE SPACES DEFINED BY A MODULUS FUNCTION Yavuz Altin, Hıfsı Altinok and Rifat Çolak Department of Mathematics, Firat University, 23119, Elazığ-TURKEY s: yaltin23@yahoo.com, hifsialtinok@yahoo.com, rcolak@firat.edu.tr) Received December 01, 2004) Abstract. In this paper we define the space BV σ p, f, q, s) on a seminormed complex linear space, by using modulus function. We give various properties and some inclusion relations on this space. Furthermore, we construct the sequence space BV σ p, f k, q, s ) and we give properties and some inclusion relations on this space. 1. INTRODUCTION Let l and c denote the Banach spaces of real bounded and convergent sequences x = x n ) normed by x = sup n x n, respectively. Let σ be a one to one mapping of the set of positive integers into itself such that σ m n) = σ σ m 1 n)), m = 1, 2,.... A continuous linear functional ϕ on l is said to be an invariant mean or a σ mean if and only if i) ϕ x) 0 when the sequence x = x n ) has x n 0 for all n, ii) ϕ e) = 1 where e = 1, 1, 1,...) and

2 122 iii) ϕ x σn) ) = ϕ xn ) for all x l. If σ is the translation mapping n n+1, a σ mean is often called a Banach limit [2], and V σ, the set of σ convergent sequences, that is, the set of bounded sequences all of whose invariant means are equal, is the set ˆf of almost convergent sequences [6]. where If x = x n ), set T x = T x n ) = x σn) ). It can be shown see Schaefer [13]) that V σ = x = x n ) : lim m t mn x) = Le uniformly in n, L = σ lim x, 1.1) t mn x) = 1 m + 1 m T j x n. The special case of 1.1), in which σ n) = n + 1 was given by Lorentz [6]. Subsequently invariant means have been studied by Ahmad and Mursaleen[1], Mursaleen [9], Raimi [11] and many others. The space BV σ = x l : m j=0 φ m,n x) <, uniformly in n were defined by Mursaleen [8], where assuming that t mn x) = 0, for m = 1. φ m,n x) = t mn x) t m 1,n x) A straightforward calculation shows that 1 m mm+1) j=1 φ m,n x) = j x σ j n) x σ n)) j 1 m 1) x n, m = 0). Note that for any sequences x, y and scalar λ, we have φ m,n x + y) = φ m,n x) + φ m,n y) and φ m,n λx) = λφ m,n x). The notion of a modulus function was introduced by Nakano [10] in We recall that a modulus f is a function from [0, ) to [0, ) such that i) fx) = 0

3 123 if and only if x = 0, ii) fx + y) fx) + fy), for all x 0, y 0, iii) f is increasing, iv) f is continuous from the right at 0. Since f x) f y) f x y ), it follows from condition iv) that f is continuous on [0, ). Furthermore, we have f nx) nf x) for all n N, from condition ii) and so f x) = f nx 1 ) x nf hence n n) 1 x ) n f x) f for all n N. n A modulus may be bounded or unbounded. For example, f x) = x p, 0 < p 1) is unbounded and f x) = x is bounded. Maddox [7] and Ruckle [12] used a modulus 1+x function to construct some sequence spaces. After then some sequence spaces, defined by a modulus function, were introduced and studied by Bhardwaj [3], Connor [5], Waszak [14], and many others. Proposition 1.1. t [0, ) Let f be a modulus and 0 < δ < 1. Then for n N and f n 1 t) > δ f n 1 t) 2f 1) δ f n 1 t) [4]. Definition 1.2. Let q 1, q 2 be seminorms on a vector space X. Then q 1 is said to be stronger than q 2 if whenever x m ) is a sequence such that q 1 x m ) 0, than also q 2 x m ) 0. If each is stronger than the others q 1 and q 2 are said to be equivalent one may refer to Wilansky [15]). Lemma 1.3. Let q 1 and q 2 be seminorms on a linear space X. Then q 1 is stronger than q 2 if and only if there exists a constant M such that q 2 x) Mq 1 x) for all x X see for instance Wilansky [15]). A sequence space E is said to be solid or normal) if α m x m ) E whenever x m ) E for all sequences α m ) of scalars with α m 1. It is well known that a sequence space E is normal implies that E is monotone. Lemma 1.4. If f is a modulus then f k is also modulus for each k = 1, 2,..., where f k = f f... f k times).

4 124 Let p = p m ) be a sequence of strictly positive real numbers and X be a seminormed space over the field C of complex numbers with the seminorm q. We define the sequence space as follows: BV σ p, f, q, s) = x = x m ) X : m s [f q φ m,n x)))] p m <, s 0, uniformly in n, where f is a modulus function. We get the following sequence spaces from BV σ p, f, q, s) by choosing some of the special p, f, and s : For f x) = x we get BV σ p, q, s) = x = x m ) X : m s [q φ m,n x)))] pm <, s 0, uniformly in n, for p m = 1, for all m, we get BV σ f, q, s) = x = x m ) X : m s [f q φ m,n x)))] <, s 0, uniformly in n, for s = 0 we get BV σ p, f, q) = x = x m ) X : [f q φ m,n x)))] p m <, uniformly in n, for fx) = x and s = 0 we get BV σ p, q) = x = x m ) X : [q φ m,n x)))] p m <, uniformly in n, for p m = 1, for all m, and s = 0 we get BV σ f, q) = x = x m ) X : [f q φ m,n x)))] <, uniformly in n, for f x) = x, p m = 1, for all m, and s = 0 we have BV σ q) = x = x m ) X : q φ m,n x)) <, uniformly in n,

5 125 The following inequalities will be used throughout the paper. Let p = p m ) be a bounded sequence of strictly positive real numbers with 0 < p m sup p m = H, C = max 1, 2 H 1), then a m + b m p m C a m p m + b m p m, 1.2) where a m, b m C and n n n a m + b m ) i a i m + b i m 1.3) where a 1, a 2,..., a n 0, b 1, b 2,..., b n 0 and 0 < i MAIN RESULTS In this section we will prove the general results of this paper on the sequence space BV σ p, f, q, s), those characterize the structure of this space. Theorem 2.1. The sequence space BV σ p, f, q, s) is a linear space over the field C of complex numbers. Proof. Let x, y BV σ p, f, q, s) and λ, µ C. Then there exist integers M λ and N λ such that λ M λ and µ N µ. Since f is subadditive, q is a seminorm m s [f q λφ m,n x) + µφ m,n y)))] p m m s [f λ q φ m,n x))) + f µ q φ m,n y)))] pm C M λ ) H m s [f q φ m,n x)))] p m + C N µ ) H <. m s [f q φ m,n y)))] p m This proves that BV σ p, f, q, s) is a linear space. Theorem 2.2. BV σ p, f, q, s) is a paranormed need not be total paranormed) space with ) 1 M g x) = m s [f q φ m,n x)))] p m,

6 126 where M = max 1, sup p m ), H = sup m p m <. Proof. It is clear that g θ) = 0 and g x) = g x), for all x BVσ p, f, q, s), where θ = θ, θ,...). It also follows from 1.2), Minkowski s inequality and definition of f that g is subadditive and where K λ g λx) K H/M λ g x), is an integer such that λ < K λ. Therefore the function λ, x) λx is continuous at λ = 0, x = θ and that when λ is fixed, the function x λx is continuous at x = θ. If x is fixed and ε > 0, we can choose m 0 such that m=m 0 m s [f q φ m,n x)))] pm < ε 2 and δ > 0, so that λ < δ and definition of f gives m 0 m s [f q λφ m,n x)))] pm = m 0 m s [f λ q φ m,n x)))] pm < ε 2. Therefore λ < min 1, δ) implies that g λx) < ε. Thus the function λ, x) λx is continuous at λ = 0 and BV σ p, f, q, s) is a paranormed space. Theorem 2.3. Let f, f 1, f 2 be modulus functions q, q 1, q 2 seminorms and s, s 1, s 2 0. Then i) If s > 1 then BV σ p, f 1, q, s) BV σ p, f f 1, q, s), ii) BV σ p, f 1, q, s) BV σ p, f 2, q, s) BV σ p, f 1 + f 2, q, s), iii) BV σ p, f, q 1, s) BV σ p, f, q 2, s) BV σ p, f, q 1 + q 2, s), iv) If q 1 is stronger than q 2 then BV σ p, f, q 1, s) BV σ p, f, q 2, s), v) If s 1 s 2 then BV σ p, f, q, s 1 ) BV σ p, f, q, s 2 ). Proof. i) Since f is continuous at 0 from right, for ε > 0 there exists 0 < δ < 1 such that 0 c δ implies f c) < ε. If we define I 1 = m N : f 1 q φ m,n x))) δ I 2 = m N : f 1 q φ m,n x))) > δ,

7 127 then, when f 1 q φ m,n x))) > δ we get f f 1 q φ m,n x)))) 2f 1) /δ f 1 q φ m,n x))). Hence for x BV σ p, f 1, q, s) and s > 1 m s [f f 1 q φ m,n x)))] pm = m I 1 m s [f f 1 q φ m,n x)))] pm + m I 2 m s [f f 1 q φ m,n x)))] pm m I 1 m s [ε] p m + m I 2 m s [2f 1) /δ f 1 q φ m,n x)))] p m max ε h, ε H) m s + max 2f 1) /δ h, 2f 1) /δ H) m s [f 1 q φ m,n x)))] p m <. Where 0 < h = inf p m p m H = sup m p m < ). ii) The proof follows from the following inequality m s [f 1 + f 2 ) q φ m,n x)))] pm Cm s [f 1 q φ m,n x)))] pm + Cm s [f 2 q φ m,n x)))] pm. iii), iv) and v) follow easily. Corollary 2.4. Let f be a modulus function, then we have i) If s > 1, BV σ p, q, s) BV σ p, f, q, s), ii) If q 1 = equivalent to) q2, then BV σ p, f, q 1, s) = BV σ p, f, q 2, s), iii) BV σ p, f, q) BV σ p, f, q, s), iv) BV σ p, q) BV σ p, q, s), v) BV σ f, q) BV σ f, q, s). The proof is straightforward. Theorem 2.5. Suppose that 0 < p m t m < for each m N. Then

8 128 i) BV σ f, p, q) BV σ f, t, q), ii) BV σ f, q) BV σ f, q, s). Proof. i) Let x BV σ f, p, q). This implies that [f q φ i,n x)))] p m 1 for sufficiently large values of i, say i m 0 for some fixed m 0 N. Since f is non decreasing, we have Hence x BV σ f, t, q). m=m 0 [f q φ m,n x)))] tm The proof of ii) is trivial. m=m 0 [f q φ m,n x)))] pm <. The following result is a consequence of the above result. Corollary 2.6. i) If 0 < p m 1 for each m, then BV σ p, f, q) BV σ f, q). ii) If p m 1 for all m, then BV σ f, q) BV σ p, f, q). Theorem 2.7. Let x BV σ p, f, q, s) i.e The sequence space BV σ p, f, q, s) is solid. m s [f q φ k,n x)))] p m <. Let α m ) be sequence of scalars such that α m 1 for all m N. Then the result follows from the following inequality. m s [f q α m φ k,n x)))] p k m s [f q φ k,n x)))] p m Corollary 2.8. The sequence space BV σ p, f, q, s) is monotone. Proposition 2.9. For any two sequences p = p k ) and t = t k ) of positive real numbers and any two seminorms q 1 and q 2 we have BV σ p, f, q 1, r) BV σ t, f, q 1, s) φ for all r > 0, s > 0.

9 SOME RELATIONS ON BV σ p, f k, q, s ) If we replace f k by f in the definition of BV σ p, f, q, s) from Lemma 1.4), then we have BV σ p, f k, q, s ) = x = x k ) l : m s [ f k q φ m,n x))) ] p m <, s 0 uniformly in n. The all results obtained for BV σ p, f, q, s) also hold for BV σ p, f k, q, s ). Theorem 3.1. If s > 1 and < k 2 then BV σ p, f, q, s ) BV σ p, f k 2, q, s ). Proof. The proof can be proved by using mathematical induction. Let k 2 = r. So r 1. Now we show that the assertion is true for r = 1. That is, BV σ p, f, q, s ) BV σ p, f +1, q, s ). By the continuity of f, for ε > 0, there exists 0 < δ < 1 such that 0 c δ implies f c) < ε. Let I 1 = m N : f q φ m,n x))) δ, I 2 = m N : f q φ m,n x))) > δ. Hence for x BV σ p, f, q, s ) and s > 1, m [ s f k1+1 q φ m,n x))) ] p m = m [ s f f q φ m,n x))) )] p m + m [ s f f k1 q φ m,n x))) )] pm m I 1 m I 2 m s [ε] p m + m [ s 2f 1) /δ f q φ m,n x))) ] p m m I 1 m I 2 max ε h, ε H) m s + max a 1, a 2). m [ s f q φ m,n x))) ] p m where a 1 = 2f 1) /δ h, a 2 = 2f 1) /δ H. Thus x BV σ p, f +1, q, s ). Now assume that the assertion is true for any r, that is BV σ p, f, q, s ) BV σ p, f +r, q, s ). 3.1)

10 130 We show that it is also true for r + 1, that is, BV σ p, f, q, s ) BV σ p, f +r+1, q, s ). But from 3.1) it suffices to show that BV σ p, f +r, q, s ) BV σ p, f +r+1, q, s ). This can be easily done as in the proof for r = 1. Where 0 < h = inf p m p m H = sup m p m < ). Corollary 3.2. Let s > 1 and k N, then i) BV σ p, f, q, s) BV σ p, f k, q, s ), ii) BV σ p, q, s) BV σ p, f k, q, s ). Theorem 3.3. Let, k 2 N and < k 2, then i) If f c) < c for all c [0, ), then BV σ p, q, s) BV σ p, f, q, s ) BV σ p, f k 2, q, s ) ii) If f c) c for all c [0, ), then BV σ p, f k 2, q, s ) BV σ p, f, q, s ) BV σ p, q, s). Proof. Since f c) < c and f is increasing we have f k 2 c) f k 2 1 c)... f c)... f c) < c. Thus for each m and p m > 0, the proof follows from m s [ f k 2 q φ m,n x)) ] p m m s [ f k 2 1 q φ m,n x)) ] p m... m s [ f q φ m,n x)) ] p m... m s [f q φ m,n x)))] pm < m s q φ m,n x)) pm. ii) Omitted.

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