Binod C. Tripathy and Hemen Dutta SOME DIFFERENCE PARANORMED SEQUENCE SPACES DEFINED BY ORLICZ FUNCTIONS

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1 F A S C I C U L I M A T H E M A T I C I Nr Binod C. Tripathy and Hemen Dutta SOME DIFFERENCE PARANORMED SEQUENCE SPACES DEFINED BY ORLICZ FUNCTIONS Abstract. In this paper we introduce the difference paranormed sequence spaces c 0 (M, n m, p), c(m, n m, p) and l (M, n m, p) respectively. We study their different properties lie completeness, solidity, monotonicity, symmetricity etc. We also obtain some relations between these spaces as well as prove some inclusion results. Key words: difference sequence, Orlicz function, paranormed space, completeness, solidity, symmetricity, convergence free, monotone space. AMS Mathematics Subject Classification: 40A05, 46A45, 46E Introduction Throughout the paper w, l, c and c 0 denote the spaces of all, bounded, convergent and null sequences x = (x ) with complex terms respectively. The zero sequence is denoted by θ=(0, 0,...). The notion of difference sequence space was introduced by Kizmaz [2], who studied the difference sequence spaces l ( ), c( ) and c 0 ( ). The notion was further generalized by Et and Cola [1] by introducing the spaces l ( n ), c( n ) and c 0 ( n ). Another type of generalization of the difference sequence spaces is due to Tripathy and Esi [13], who studied the spaces l ( m ), c( m ) and c 0 ( m ). Tripathy, Esi and Tripathy [14] generalized the above notions and unified these as follows: Let m, n be non-negative integers, then for Z a given sequence space we have Z( n m) = {x = (x ) w : ( n mx ) Z}, where n mx = ( n mx ) = ( n 1 m x n 1 m x +m ) and 0 mx = x for all N, which is equivalent to the following binomial representation: n mx = n ( ( 1) v n v v=0 ) x +mv.

2 122 Binod C. Tripathy and Hemen Dutta Taing m = 1, we get the spaces l ( n ), c( n ) and c 0 ( n ) studied by Et and Cola [1]. Taing n=1, we get the spaces l ( m ), c( m ) and c 0 ( m ) studied by Tripathy and Esi [13]. Taing m=n=1, we get the spaces l ( ), c( ) and c 0 ( ) introduced and studied by Kizmaz [2]. An Orlicz function is a function M:[0, ) [0, ), which is continuous, non-decreasing and convex with M(0)=0, M(x) > 0 and M(x) as x. Lindenstrauss and Tzafriri [5] used the Orlicz function and introduced the sequence space l M as follows: l M = {(x ) w : =1 M( x ) <, for some > 0}. They proved that l M is a Banach space normed by (x ) = inf{ > 0 : =1 M( x ) 1}. Remar. An Orlicz function satisfies the inequality M(λx) λm(x) for all λ with 0 < λ < 1. The following inequality will be used throughout the article. Let p = (p ) be a positive sequence of real numbers with 0 < p sup p = G, D = max{1, 2 G 1 }. Then for all a, b C for all N, we have a + b p D( a p + b p ). The studies on paranormed sequence spaces were initiated by Naano [8] and Simons [11]. Later on it was further studied by Maddox [6], Nanda [9], Lascarides [3], Lascarides and Maddox [4], Tripathy and Sen [15] and many others. Parasar and Choudhary [10], Mursaleen, Khan and Qamaruddin [7] and many others studied paranormed sequence spaces using Orlicz functions. 2. Definitions and preliminaries A sequence space E is said to be solid (or normal) if (x ) E implies (α x ) E for all sequences of scalars (α ) with α 1 for all N. A sequence space E is said to be monotone if it contains the canonical preimages of all its step spaces. A sequence space E is said to be symmetric if (x π() ) E, where π is a permutation on N. A sequence space E is said to be convergence free if (y ) E whenever (x ) E and y = 0 whenever x = 0.

3 Some difference paranormed sequence A sequence space E is said to be a sequence algebra if (x y ) E whenever (x ) E and (y ) E. Let p = (p ) be any bounded sequence of positive real numbers. Then we define the following sequence spaces for an Orlicz function M: c 0 (M, n m, p) = {x = (x ) : lim mx (M( n )) p = 0, for some > 0}, c(m, n m, p) = {x = (x ) : lim mx L (M( n )) p = 0, for some > 0 and L C}, l (M, n m, p) = {x = (x ) : sup(m( n m x )) p <, for some > 0}, 1 when p = p, a constant, for all, then c 0 (M, n m, p) = c 0 (M, n m), c(m, n m, p) = c(m, n m) and l (M, n m, p) = l (M, n m). Lemma 1. If a sequence space E is solid, then E is monotone. 3. Main results In this section we prove the results of this article. following result is easy, so omitted. The proof of the Proposition 1. The classes of sequences c 0 (M, n m, p), c(m, n m, p) and l (M, n m, p) are linear spaces. Theorem 1. For Z = l, c and c 0, the spaces Z(M, n m, p) are paranormed spaces, paranormed by where g(x) = nm =1 x + inf{ p H = max(1, sup p ). M( n mx ) 1}, Proof. Clearly g( x) = g(x), g(θ) = 0. Let (x ) and (y ) be any two sequences belong to any one of the spaces Z(M, n m, p), for Z = c 0, c and l. Then we have 1, 2 > 0 such that sup M( n mx ) 1 1 and sup M( n my ) 1. 2

4 124 Binod C. Tripathy and Hemen Dutta Let = Then by convexity of M, we have sup Hence we have, M( n m(x + y ) ) ( +( ) sup ) sup M( n mx 1 ) M( n my 2 ) 1. g(x + y) = This implies that mn =1 mn x + y + inf{ p p H x + inf{1 : sup =1 mn p H + y + inf{2 : sup =1 g(x + y) g(x) + g(y). M( n m(x + y ) ) 1} M( n mx 1 ) 1} M( n my 2 ) 1}. The continuity of the scalar multiplication follows from the following inequality: g(λx) = mn λx + inf{ p H =1 mn = λ =1 : sup x + inf{(t λ ) p M( n mλx ) 1} M( n mx ) 1}, where t = t λ. Hence the space Z(M, n m, p), for Z = c 0, c and l are paranormed spaces, paranormed by g. Theorem 2. For Z = l, c and c 0, the spaces Z(M, n m, p) are complete paranormed spaces, paranormed by g(x) = nm =1 x + inf{ p M( n mx ) 1}, where H = max(1, sup p ).

5 Some difference paranormed sequence Proof. We prove for the space l (M, n m, p) and for the other spaces it will follow on applying similar arguments. Let (x i ) be any Cauchy sequence in l (M, n m, p). Let x 0 > 0 be fixed ε and t > 0 be such that for a given 0 < ε < 1, x 0 t > 0, and x 0t 1. Then there exists a positive integer n 0 such that g(x i x j ) < Using the definition of paranorm, we get ε x 0 t, for all i, j n 0. (1) mn =1 x i xj + inf{ p M( n m(x i xj ) ) 1} < ε x 0 t, for all i, j n 0. Hence we have, This implies mn x i xj < ε, for all i, j n 0. =1 x i xj < ε, for all i, j n 0 and 1 nm. Thus (x i ) is a Cauchy sequence in C for = 1, 2,..., nm. Hence (xi ) is convergent in C for = 1, 2,..., nm. (2) Let lim i x i = x, say for = 1, 2,..., nm. Again from (1) we have, inf{ p H : sup M( n m(x i xj ) ) 1} < ε, for all i, j n 0. Hence we get sup M( n m(x i xj ) g(x i x j ) 1, for all i, j n 0. ) It follows that M( n m (xi xj ) ) 1, for each 1 and for all i, j n g(x i x j ) 0. For t > 0 with M( tx 0 2 ) 1, we have M( n m(x i xj ) g(x i x j ) M( tx 0 ) 2 ).

6 126 Binod C. Tripathy and Hemen Dutta This implies n mx i n mx j tx 0 2 ε tx 0 = ε 2. Hence ( n mx i ) is a Cauchy sequence in C for all N. This implies that ( n mx i ) is convergent in C for all N. Let lim i n mx i = y for each N. Let = 1. Then we have (3) lim n i mx i 1 = lim i n ( ( 1) v n v v=0 ) x i 1+mv = y 1. We have by (2) and (3) lim i x i mn+1 = x mn+1, exists. Proceeding in this way inductively, we have lim i x i = x exists for each N. Now we have for all i, j n 0, mn =1 This implies that lim { mn j x i xj + inf{ p =1 x i xj + inf{ p for all i n 0. Using the continuity of M, we have nm =1 x i x + inf{ p M( n m(x i xj ) ) 1} < ε. M( n m(x i xj ) ) 1}} < ε, M( n mx i n mx ) 1} < ε, for all i n 0. It follows that (x i x) l (M, n m, p). Since (x i ) l (M, n m, p) and l (M, n m, p) is a linear space, so we have x = x i (x i x) l (M, n m, p). This completes the proof of the Theorem. Theorem 3. If 0 < p q < for each, then Z(M, n m, p) Z(M, n m, q), for Z = c 0 and c. Proof. We prove the result for the case Z = c 0 and for the other case it will follow on applying similar arguments. Let (x ) c 0 (M, n m, p). Then there exists some > 0 such that lim mx (M( n )) p = 0.

7 Some difference paranormed sequence This implies that M( n m x ) < ε(0 < ε < 1) for sufficiently large. Hence we get lim mx (M( n )) q lim mx (M( n )) p = 0. This implies that (x ) c 0 (M, n m, q). This completes the proof. The following result is a consequence of Theorem 4. Corollary. (a) If 0 < inf p p 1, for each, then Z(M, n m, p) Z(M, n m), for Z = c 0 and c. (b)if 1 p sup p <, for each, then Z(M, n m) Z(M, n m, p), for Z = c 0 and c. Theorem 4. If M 1 and M 2 be two Orlicz functions. Then (i) Z(M 1, n m, p) Z(M 2 M 1, n m, p), (ii) Z(M 1, n m, p) Z(M 2, n m, p) Z(M 1 + M 2, n m, p), for Z = l, c and c 0. Proof. We prove this part for Z = l and the rest of the cases will follow similarly. Let (x ) l (M 1, n m, p). Then there exists 0 < U < such that (M 1 ( n mx )) p U, for all N. Let y = M 1 ( n m x ). Then y U 1 p V, say for all N. Hence we have ((M 2 M 1 )( n mx )) p = (M 2 (y )) p (M 2 (V )) p <, for all N. Hence sup ((M 2 M 1 )( n m x )) p <. Thus (x ) l (M 2 M 1, n m, p). (ii) We prove the result for the case Z = c 0 and for the other cases it will follow on applying similar arguments. Let (x ) c 0 (M 1, n m, p) c 0 (M 2, n m, p). Then there exist some 1, 2 > 0 such that lim (M 1( n mx )) p = 0 and lim (M 2( n mx )) p = Let = Then we have ((M 1 + M 2 )( n mx 1 )) p D[ M 1 ( n mx )] p D[ M 2 ( n mx )] p

8 128 Binod C. Tripathy and Hemen Dutta This implies that lim ((M 1 + M 2 )( n mx )) p = 0. Thus (x ) c 0 (M 1 + M 2, n m, p). This completes the proof. Theorem 5. Z(M, n 1 m, p) Z(M, n m, p) (in general Z(M, i m, p) Z(M, n m, p), for i = 1, 2,..., n 1, for Z = l, c and c 0. Proof. We prove the result for Z = l and for the other cases it will follow on applying similar arguments. Let x = (x ) l (M, n 1 m, p). Then we can have > 0 such that (4) (M( n 1 m x )) p <, for all N On considering 2 and using the convexity of M, we have M( n mx 2 Hence we have ( ( )) n M m x p D 2 Then using (4), we have ) 1 2 M( n 1 m x ) M( n 1 m x +m ). {( 1 2 M ( n 1 m x )) p ( ( 1 n M (M( n mx )) p <, for all N. Thus l (M, n 1 m, p) l (M, n m, p). The inclusion is strict follows from the following example. m x +m )) p }. Example 1. Let m = 3, n = 2, M(x) = x 2, for all x [0, ) and p = 4 for all odd and p = 3 for all even. Consider the sequence x = (x ) = (). Then 2 3 x = 0, for all N. Hence x belongs to c 0 (M, 2 3, p). Again we have 1 3 x = 3, for all N. Hence x does not belong to c 0 (M, 1 3, p). Thus the inclusion is strict. Theorem 6. Let M be an Orlicz function. Then The inclusions are proper. c 0 (M, n m, p) c(m, n m, p) l (M, n m, p).

9 Some difference paranormed sequence Proof. It is obvious that c 0 (M, n m, p) c(m, n m, p). We shall prove that c(m, n m, p) l (M, n m, p). Let (x ) c(m, n m, p). Then there exists some > 0 and L C such that lim mx L (M( n )) p = 0. On taing 1 = 2, we have (M( n mx )) p D[ 1 mx L 1 2 (M( n ))] p + D[ 1 2 M( L )]p D( 1 2 )p [M( n mx L )] p + D( 1 2 )p max(1, (M( L ))H ), where H = max(1, sup p ). Thus we get (x ) l (M, n m, p). The inclusions are strict follow from the following examples. Example 2. Let m = 2, n = 2, M(x) = x 4, for all x [0, ) and p = 1, for all N. Consider the sequence x = (x ) = ( 2 ). Then x belongs to c(m, 2 2, p), but x does not belong to c 0(M, 2 2, p). Example 3. Let m = 2, n = 2, M(x) = x 2, for all x [0, ) and p = 2, for all odd and p = 3, for all even. Consider the sequence x = (x ) = {1, 3, 2, 4, 5, 7, 6, 8, 9, 11, 10, 12,...}. Then x belongs to l (M, 2 2, p), but x does not belong to c(m, 2 2, p). Theorem 7. The spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not monotone and as such are not solid in general. Proof. The proof follows from the following example. Example 4. Let n = 2, m = 3, p = 1, for all odd and p = 2, for all even and M(x) = x 2, for all x [0, ). Then 2 3 x = x 2x +3 + x +6, for all N. Consider the J th step space of a sequence space E defined by (x ), (y ) E J implies that y = x, for odd and y = 0, for even. Consider the sequence (x ) defined by x =, for all N. Then (x ) belongs to Z(M, 2 3, p), for Z = l, c and c 0, but its J th canonical pre-image does not belong to Z(M, 2 3, p), for Z = l, c and c 0. Hence the spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not monotone and as such are not solid in general. Theorem 8. The spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not symmetric in general. Proof. The proof follows from the following example.

10 130 Binod C. Tripathy and Hemen Dutta Example 5. Let n = 2, m = 2, p = 2, for all odd and p = 3, for all even and M(x) = x 2, for all x [0, ). Then 2 2 x = x 2x +2 + x +4, for all N. Consider the sequence (x ) defined by x =, for odd and x = 0, for even. Then 2 2 x = 0, for all N. Hence (x ) belongs to Z(M, 2 2, p), for Z = l, c and c 0. Consider the rearranged sequence, (y ) of (x ) defined by (y ) = (x 1, x 3, x 2, x 4, x 5, x 7, x 6, x 8, x 9, x 11, x 10, x 12,...). Then (y ) does not belong to Z(M, 2 2, p), for Z = l, c and c 0. Hence the spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not symmetric in general. Theorem 9. The spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not convergence free in general. Proof. The proof follows from the following example. Example 6. Let m = 3, n = 1, p = 6, for all N and M(x) = x 3, for all x [0, ). Then 1 3 x = x x +3, for all N. Consider the sequences (x ) and (y ) defined by x = 4 for all N and y = 2, for all N. Then (x ) belongs to Z(M, 1 3, p), but (y ) does not belong to Z(M, 1 3, p), for Z = l, c and c 0. Hence the spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not convergence free in general. Theorem 10. The spaces l (M, n m, p), c(m, n m, p) and c 0 (M, n m, p) are not sequence algebra in general. Proof. The proof follows from the following examples. Example 7. Let n = 2, m = 1, p = 1, for all N and M(x) = x 3, for all x [0, ). Then 2 1 x = x 2x +1 + x +2, for all N. Let x = () and y = ( 2 ). Then x, y both belong to Z(M, 2 1, p), for Z = l and c, but xy does not belong to Z(M, 2 1, p), for Z = l and c. Hence the spaces l (M, n m, p) and c(m, n m, p) are not sequence algebra in general. Example 8. Let n = 2, m = 1, p = 7, for all N and M(x) = x 7, for all x [0, ). Then 2 1 x = x 2x +1 + x +2, for all N. Let x = () and y = (). Then x, y both belong to c 0 (M, 2 1, p), but xy does not belong to c 0 (M, 2 1, p). Hence the space c 0(M, n m, p) is not sequence algebra in general. References [1] Et M., Colo R., On generalized difference sequence spaces, Soochow J. Math., 21(4)(1995),

11 Some difference paranormed sequence [2] Kizmaz H., On certain sequence spaces, Canad. Math. Bull., 24(2)(1981), [3] Lascarides C.G., A study of certain sequence spaces of Maddox and generalization of a theorem of Iyer, Pacific Jour. Math., 38(2)(1971), [4] Lascarides C.G., Maddox I.J., Matrix transformation between some classes of sequences, Proc. Camb. Phil. Soc., 68(1970), [5] Lindenstrauss J., Tzafriri L., On Orlicz sequence spaces, Israel J. Math., 10(1971), [6] Maddox I.J., Paranormed sequence spaces generated by infinite matrices, Proc. Camb. Phil. Soc., 64(1968), [7] Mursaleen, Khan A., Qamaruddin, Difference sequence spaces defined by Orlicz functions, Demonstratio Mathematica, XXXII(1)(1999), [8] Naano H., Modular sequence spaces, Proc. Japan Acad., 27(1951), [9] Nanda S., Some sequence spaces and almost convergence, J. Austral. Math. Soc. Ser. A, 22(1976), [10] Parasar S.D., Choudhary B., Sequence spaces defined by Orlicz functions, Indian J. Pure Appl. Math, 25(4)(1994), [11] Simons S., The sequence spaces l(p v ) and m(p v ), Proc. London Math. Soc., 15(1965), [12] Tripathy B.C., A class of difference sequences related to the p-normed space l p, Demonstratio Math., 36(4)(2003), [13] Tripathy B.C., Esi A., A new type of difference sequence spaces, Internat. J. Sci. Technol., (1)(2006), [14] Tripathy B.C., Esi A., Tripathy B.K., On a new type of generalized difference Cesàro sequence spaces, Soochow J. Math., 31(3)(2005), [15] Tripathy B.C., Sen M., On generalized statistically convergent sequence spaces, Indian J. Pure Appl. Math., 32(11)(2001), Binod Chandra Tripathy Mathematical Sciences Division Institute of Advanced Study in Science and Technology Paschim Boragaon, Garchu, Guwahati ; India tripathybc@yahoo.com or tripathybc@rediffmail.com Hemen Dutta Department of Mathematics A.D.P. COLLEGE, Nagaon , Assam, India hemen dutta@rediffmail.com Received on and, in revised form, on