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

F A S C I C U L I M A T H E M A T I C I Nr 42 2009 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, 46E30. 1. 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.

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.

Some difference paranormed sequence... 123 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

124 Binod C. Tripathy and Hemen Dutta Let = 1 + 2. Then by convexity of M, we have sup Hence we have, M( n m(x + y ) ) ( +( 1 1 + 2 ) sup 2 1 + 2 ) 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 ).

Some difference paranormed sequence... 125 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 ).

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.

Some difference paranormed sequence... 127 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 = 0. 1 2 Let = 1 + 2. Then we have ((M 1 + M 2 )( n mx 1 )) p D[ M 1 ( n mx )] p 1 + 2 2 1 + D[ M 2 ( n mx )] p. 1 + 2 2

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 ) + 1 2 M( n 1 m x +m ). {( 1 2 M ( n 1 m x )) p ( ( 1 n 1 + 2 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).

Some difference paranormed sequence... 129 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.

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), 377-386.

Some difference paranormed sequence... 131 [2] Kizmaz H., On certain sequence spaces, Canad. Math. Bull., 24(2)(1981), 169-176. [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), 487-500. [4] Lascarides C.G., Maddox I.J., Matrix transformation between some classes of sequences, Proc. Camb. Phil. Soc., 68(1970), 99-104. [5] Lindenstrauss J., Tzafriri L., On Orlicz sequence spaces, Israel J. Math., 10(1971), 379-390. [6] Maddox I.J., Paranormed sequence spaces generated by infinite matrices, Proc. Camb. Phil. Soc., 64(1968), 335-340. [7] Mursaleen, Khan A., Qamaruddin, Difference sequence spaces defined by Orlicz functions, Demonstratio Mathematica, XXXII(1)(1999), 145-150. [8] Naano H., Modular sequence spaces, Proc. Japan Acad., 27(1951), 508-512. [9] Nanda S., Some sequence spaces and almost convergence, J. Austral. Math. Soc. Ser. A, 22(1976), 446-455. [10] Parasar S.D., Choudhary B., Sequence spaces defined by Orlicz functions, Indian J. Pure Appl. Math, 25(4)(1994), 419-428. [11] Simons S., The sequence spaces l(p v ) and m(p v ), Proc. London Math. Soc., 15(1965), 422-436. [12] Tripathy B.C., A class of difference sequences related to the p-normed space l p, Demonstratio Math., 36(4)(2003), 867-872. [13] Tripathy B.C., Esi A., A new type of difference sequence spaces, Internat. J. Sci. Technol., (1)(2006), 11-14. [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), 333-340. [15] Tripathy B.C., Sen M., On generalized statistically convergent sequence spaces, Indian J. Pure Appl. Math., 32(11)(2001), 1689-1694. Binod Chandra Tripathy Mathematical Sciences Division Institute of Advanced Study in Science and Technology Paschim Boragaon, Garchu, Guwahati-781035; India e-mail: tripathybc@yahoo.com or tripathybc@rediffmail.com Hemen Dutta Department of Mathematics A.D.P. COLLEGE, Nagaon-782002, Assam, India e-mail: hemen dutta@rediffmail.com Received on 26.02.2008 and, in revised form, on 13.02.2009.