An Extension in the Domain of an n-norm Defined on the Space of p-summable Sequences

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Gen. Math. Notes, Vol. 33, No., March 206, pp.-8 ISSN 229-784; Copyright c ICSRS Publication, 206 www.i-csrs.org Available free online at http://www.geman.

1 Gen. Math. Notes, Vol. 33, No., March 206, pp.-8 ISSN ; Copyright c ICSRS Publication, Available free online at An Extension in the Domain of an n-norm Defined on the Space of p-summable Sequences J.K. Srivastava Pradeep Kumar Singh 2,2 Department of Mathematics Statistics D.D.U Gorakhpur University, Gorakhpur Uttar Pradesh, India jks ddugu@yahoo.com 2 pradeep3789@gmail.com (Received: / Accepted: 7-2-6) Abstract In [6], we have already studied the space of p-summable sequences (i.e. l p ) as an n-normed space by defining a new n-norm,, p on it. In this paper, we shall extend the domain of the definition of n-norm,, p from the space l p to different vector subspaces of the vector space l containing l p. Further, we shall discuss on their derived norms also. Keywords: l p space, l space, norms, n-norms, derived norms. Introduction In [9], Gähler initialy introduced the theory of 2-norm, defined on a linear space, while that of n-norm can be found in [3] has been studied in many papers such as [, 2, 5]. Research works on sequence spaces regarded as n- normed space can be found in [, 4, 6, 7, 8]. Definition.. Let X be a vector space over K(= R or C) of dimension d n(n 2). A non-negative real valued function.,...,. defined on X n satisfying the four conditions: (N) x, x 2,, x n = 0 if only if x, x 2,, x n are linearly dependent;

2 2 J.K. Srivastava et al. (N2) x, x 2,, x n is invariant under the permutation of x, x 2,, x n ; (N3) α x, x 2,, x n = α x, x 2,, x n ; (N4) x + y, x 2,, x n x, x 2,, x n + y, x 2,, x n ; for all x, x 2,, x n, y X for all α K, is called an n-norm on X, the pair (X,.,...,. ) is called an n-normed space. Definition.2. Two n-norms,,,, 2 defined on a linear space X are said to be equivalent if only if K, K 2 > 0 such that: K x, x 2,, x n x, x 2,, x n 2 K 2 x, x 2,, x n for all x, x 2,, x n X. Definition.3. Let (X,,, ) is an n-normed space e,, e n } is a set of linearly independent vectors in X then both of the functions d d q define a norm on X (known as derived norm with respect to the set e,, e n }) they are equivalent, where. x d = max x, e t,..., e t n : t,..., t n },..., n}} 2. x d q = ( t,...,t n },...,n} x, et,..., e t n q ) /q ; q <. In this paper, we shall focus on l p l, where } l p = x = (x i ) i=0 : x i p < where x i K, for all i = 0,, 2,... i=0 l = } x = (x i ) i=0 : sup 0 i< x i <. As we know that (l p, p ) is a Banach space where x p = i=0 x i p } /p as well as (l, ) also becomes a Banach space with norm x = sup 0 i< x i ; while (l p, ) forms simply a normed space. In [6], for our convenience need we have denoted the set of whole numbers as N = 0,, 2,...}, which is also considered as a sequence N = (0,, 2,...). Further, we have denoted the sequence N = (0,, 2,...) in the form of n-consecutive terms notation as expressed as N = (0,, 2,...) = (nl, nl +,..., nl + (n )) l=0 N = (n 0 = 0, n 0 + =,..., n 0 + (n ) = n, n = n, n + = n +,..., n + (n ) = 2n,...)

3 An Extension in the Domain of an n-norm... 3 be a rearrangement of the se- Let N = ( ) m nk, m nk+,..., m nk+(n ) quence N. Then for any n vectors the n vectors x t = ( x t nl, x t nl+,..., x t nl+(n )) l=0 lp ; t =, 2,..., n x t = ( ) x t m nk, x t m nk+,, x t m nk+(n ) ; t =, 2,..., n are called parallel rearrangements of x, x 2,, x) n respectively. In [6], we have observed that (l p,,, p is an n-normed space, but not complete where = sup x, x 2,..., x n : x, x 2,..., x n are parallel rearrangements of x, x 2,, x n respectively}. () x, x 2,..., x n = Moreover, we have x x x p x 2 x 2 x 2 m det nk m nk+ m nk+(n ) x n x n x n /p. (2) n! x π p x π 2 p/ x πn p/ ; (3) where π, π 2..., π n } is any permutation of, 2,..., n} x πt p/ means either x πt p or x πt is taken freely. In [], Mal`ceski investigated that the function x i x i 2... x i n x, x 2,, x n := sup det x 2 i x 2 i 2... x 2 i n i,...,i n (4) x n i x n i 2... x i n defines an n-norm on l, where i,..., i n N. But l p is a subspace of l therefore we can show that....,. forms an n-norm on l p also. In [6, 7], we have already proved that these two n-norms.,...,. p....,. defined on l p are non-equivalent. Where as their derived norms with respect to the linearly independent set e,, e n } are equivalent equivalent to., where e t = (δ t i) i=0. For details see [7].

4 4 J.K. Srivastava et al. 2 Results Here, our aim is to investigate the possibilities of extending the domain of definition of the n-norm,, p from the domain l p to other sequence spaces containing l p. If we take any arbitrary z l define l p + [z] = x + αz : x l p α K}. then obviously l p + [z] is a subspace of (l, ). Theorem 2.. The function,, p defines an n-norm on l p + [z], for every arbitrary z l. Moreover, x + z, x 2 + z 2,..., x n + z n p n! x π p x π 2 + z π 2 x πn + z πn + n! x π 2 p x π 3 + z π 3 x πn + z πn z π (5) for every scalar multiples z, z 2,..., z n of z any permutation π, π 2..., π n } of, 2,..., n}. Proof: The proof is similar to proving that,, p defines an n-norm on l p, as we have done in [6]. Here we are going to establish the inequality only. Let x + z, x 2 + z 2,..., x n + z n l p + [z] N = ( ) m nk, m nk+,..., m nk+(n ) be a rearrangement of the sequence N. By the properties of determinant we have x + z x + z x + z m nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ) x 2 + z 2 x 2 + z 2 x 2 + z 2 m det nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ) x n + z n x n + z n x n + z n m nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ) x x x x 2 + z 2 x 2 + z 2 x 2 + z 2 m = det nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ) x n + z n x n + z n x n + z n m nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ) z z z x 2 + z 2 x 2 + z 2 x 2 + z 2 m +det nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ). x n + z n x n + z n x n + z n m nk m nk m nk+ m nk+ m nk+(n ) m nk+(n ) As we know that the expansion of a determinant of order n consists of sum of n! terms, among which each term is again a product of n terms, therefore

5 An Extension in the Domain of an n-norm... 5 breaking the last determinant along second row using Minkowski inequality; (Keeping the fact in mind that z, z 2 are linearly dependent.) by definition (2) we have x + z, x 2 + z 2,..., x n + z n n! x p x 2 + z 2 x n + z n + n! x 2 p x 2 + z 2 x n + z n z. Above is true for any rearrangement N of N breaking determinant along any row, therefore x + z, x 2 + z 2,..., x n + z n p n! x π p x π 2 + z π 2 x πn + z πn + n! x π 2 p x π 3 + z π 3 x πn + z πn z π. In general, if we take any n arbitrary vectors v, v 2,..., v n l define l p +[v ] + [v 2 ] + + [v n ] = x + α v + α 2 v α n v n : x l p α j K ; j =, 2,..., n }. It is clear that, l p + [v ] + [v 2 ] + + [v n ] is a subspace of (l, ). Corollary 2.2. The function,, p defines an n-norm on l p + [v ] + [v 2 ] + + [v n ] for v, v 2,..., v n l. Moreover, x + z, x 2 + z 2,..., x n + z n p n! x π p x π 2 + z π 2 x πn + z πn + n! x π 2 p x π 3 + z π 3 x πn + z πn z π + n! x π 3 p x π 4 + z π 4 x πn + z πn z π z π n! x πn p z π z π n (6) where each of the vectors z, z 2,..., z n is linear combination of v, v 2,..., v n. Proof: The proof can be done similar to the proof of theorem 2. by breaking the determinant successively along every row. Obviously, the function....,. defines an n-norm on l p +[v ]+[v 2 ]+ + [v n ] also. Since, the n-norms,, p....,. are non-equivalent on l p (for details see [7]), therefore they must be non-equivalent on l p +[v ]+[v 2 ]+ + [v n ] also. But, it is easy to show that their derived norms with respect to the linearly independent set e,..., e n : e t = (δi) t i=0 ; t =, 2,..., n} are equivalent equivalent to. Under some conditions, the function,, p may define an n-norm on l p + [v ] + [v 2 ] + + [v n ] for v, v 2,..., v n l, before discussing such conditions, let us consider the following theorem.

6 6 J.K. Srivastava et al. Theorem 2.3. The function,, p need not be an n-norm on l p +[v ]+ [v 2 ] + + [v n ] for v, v 2,..., v n l. Moreover, the function,, p fails to be an n-norm on l. Proof: Taking n vectors v t = (vi) t i=0 l ; t =, 2,..., n where vi t, if i (t )(mod n) ; = 0, otherwise obviously, v t = therefore v t l for every t =, 2,..., n. But v,, v n p =. Thus the function,, p need not be an n-norm on l p +[v ]+[v 2 ]+ +[v n ] for v, v 2,..., v n l. Moreover, the function,, p fails to be an n-norm on l. Now we shall investigate those circumstances under which the function,, p define an n-norm on l p +[v ]+[v 2 ]+ +[v n ] for v, v 2,..., v n l. Lemma 2.4. If x, x 2 p < for x, x 2 l then for every x 3,..., x n l <. Moreover, n! 2 x3 x n x, x 2 p. Proof: Let N = ( ) m nk, m nk+,..., m nk+(n ) be a rearrangement of the sequence N. Obviously, exping the determinant x x x x 2 x 2 x 2 m det nk m nk+ m nk+(n ) x n x n x n in the form of the sum of the terms like x 3 m nk+j3 x 4 m nk+j4 x n m nk+jn x m nk+j x 2 m nk+j x m nk+j2 x 2 m nk+j2 then using the Minkowski inequality equation (2) gives x, x 2,..., x n n! = 2 x3 x n x, x 2 p.

7 An Extension in the Domain of an n-norm... 7 Hence, we have the lemma. Example: If we take x, x 2 l as follows: where x = ( x i ) i=0 x 2 = ( x 2 i ) i=0 x i = x 2 i = ; i = 0, ; i 2 i /p 0 ; i = 0 ; i =, 2 ; i 3 (i ) /p obviously, x, x 2 / l p but x, x 2 l. Whereas x, x 2 p < for, let N = ( ) m nk, m nk+,..., m nk+(n ) be a rearrangement of the sequence N then for every t N, we have ( t ( ) x det x p) /p ( t ) /p m 2k m 2k+ x 2 x 2 x m 2k m 2k+ m 2k x 2 p m 2k+ ( t ) /p + x m 2k+ x 2 p m 2k but both t x m 2k x 2 p m 2k+ 2 ( + therefore then i= ) i 2 t x m 2k+ x 2 p m 2k 2 ( + x, x 2 p <. Lemma 2.5. If for x, x 2,, x n l, < z, z 2,..., z n p < for each z t ; t =, 2,..., n is linear combination of x, x 2,, x n. Theorem 2.6. If v,, v n p < for v, v 2,..., v n l, then the function,, p defines an n-norm on l p + [v ] + [v 2 ] + + [v n ]. Proof: In view of lemma 2.5, the proof is similar to the proof of corollary 2.2. i= ) i 2

8 8 J.K. Srivastava et al. References [] A. Malćeski, l as n-normed space, Math. Bilten, 2(997), [2] A. Malćeski, Strong n-convex n-normed space, Math. Bilten, 2(997), [3] A. Misiak, n-inner product spaces, Math. Nachr., 40(989), [4] H. Gunawan, The space of p-summable sequences its natural n-norm, Bull. Austral. Math. Soc., 64(200), [5] M. Mashadi H. Gunawan, On n-normed spaces, Int. Jour. Math. Math. Sci., 27(200), [6] P.K. Singh J.K. Srivastava, A study of an n-norm on l p spaces, Int. Journal. of Math. Archive, 6(2) (205), [7] P.K. Singh J.K. Srivastava, Equivalent norms derived by nonequivalent n-norms on the space of p-summable sequences, To Appear in Proc. Indian Acad. Sci. (Math. Sci.). [8] S. Konca, H. Gunawan M. Basarir, Some remarks on l p as an n- normed space, Mathematical Sciences Applications E-Notes, 2(2) (204), [9] S. Gähler, Linear 2-normierte Räume, Math. Nachr., 28(964), -43.

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