T. Balasubramaniam and A. Pandiarani [1] have defined the sequence spaces G(x). l (x), G(x) and have studied some topological property of their KÃƒÂ¶the-Toeplitz duals in operator form. In this paper we define the new sequence spaces G0(x, p), Ge(x, p) and GÃ‚Â¥(x,p) and have studied some topological property of their KÃƒÂ¶the-Toeplitz duals in operator form

A. Ansari2, K. B. Gupta1* and Ashfaque

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Recent Research in Science and Technology 2011, 3(7): 95-99 ISSN: 2076-5061 www.scholarjournals.org   www.recent-science.com 95  RRST-Mathematics Operator Duals of Vector Sequence Spaces  K. B. Gupta1* and Ashfaque A. Ansari2 1Department of Mathematics, St. Andrew’s College, Gorakhpur-273001 (INDIA) 2Department of Mathematics and Statistics, D.D.U. Gorakhpur University, Gorakhpur - 273009 (INDIA) Article Info  Abstract Article History  T. Balasubramaniam and A. Pandiarani [1] have defined the sequence spaces Γ(x). λ (x), G(x) and have studied some topological property of their Köthe-Toeplitz duals in operator form. In this paper we define the new sequence spaces Γ0(x, p), Γe(x, p) and Γ∞(x,p) and have studied some topological property of their Köthe-Toeplitz duals in operator form.      Received : 27-04-2011  Revised : 18-07-2011  Accepted : 27-07-2011  *Corresponding Author  Tel : +91-9415500371       Email:  dr.kbgupta77@gmail.com    ©ScholarJournals, SSR  Key Words: Analytic sequences, Entire sequences, Köthe-Toeplitz duals    Introduction The idea of dual sequence spaces was introduced by Köthe and Toeplitz [6]. In fact, the basic problems in the theory of complex sequence space or scalar valued sequence space deal with the transformation of complex sequence by infinite matrices of complex numbers. The basic results in this regard may be seen in the books of Cooke [3] ,Hardy [4] and Maddox [5].  When the infinite matrices of complex numbers operate on a complex sequence, we come across an infinite series whose convergence has given rise to the concept of β-duals also called as Köthe-Toeplitz duals. Thus if E is a set of complex sequences, then β-dual of E is denoted by Eβ and is defined as Eβ = {a = (ak) ∈ ω :  ∑∞=1kkk xa  converges for all x = (xk) ∈E, where ω is the set of all complex sequences}  The main result of Köthe and Toeplitz [2] is concerned with α-duals. If ω is the set of all complex sequence spaces, then the α-dual of E, (where E ⊂ ω) is denoted by Eα and is defined as Eα = {a = (ak) ∈ ω : ∑∞=1||kkk xa   converges for all x = (xk) ∈ E}  An account of the theory of α-duals in the scalar case may be found in G. Köthe [6]. Another dual, the γ-dual of the set E ⊂ w is denoted by Eγ and is defined as  Eγ = {a = (ak) ∈ ω : supn ∑=nkkk xa1 < ∞ for all x = < xk > ∈ E}  After Robinson’s paper in 1995 [7] where in he considered the action of matrices operators on Banach space valued sequence, a decisive break occurred. This gave the concept of generalized Köthe-Toeplitz duals as given below.  Let X and Y be a Banach spaces and E((X) is a set of X-valued sequences i.e. E(x) is a non-empty set of sequence x = (xk) with xk ∈ X. The several generalized Köthe-Toeplitz duals arise when the element complex sequence (ak) of α-duals and β-duals set is replaced by a sequence (Ak) of linear operators, where each Ak is a linear operator from X into Y. Thus, we define β-dual of E(X) as  Eβ(x) = {(Ak) : ∑∞=1)(kkk xA   converges in Y, for each  (xk) ∈ E (X)}    and α-dual of E (X) as   Eα(X) = {(Ak) : ∑∞=1||||kkk xA  converges for all (xk) ∈ E}  Furthermore, Maddox [8] has also given the convergence of sequence of operators which are not necessarily bounded. Notation and Terminology   We denote the set of all natural, real and complex numbers by N, R and C respectively. A sequence x = (xk) is said to be an entire sequence if 0||lim1 =∞→kkkx  and a K. B. Gupta and Ashfaque A. Ansari/Rec Res Sci Tech 2011, 3(7): 95-99     96sequence x = (xk) is said to be an analytic sequence if      { }kkx 1||      is bounded.  If (X, ||.||) is any Banach space over C, then we define }0||||,:{),(|10 =∈>=<= ∞→kpkkk xLimitXxxpx  }||||,:{),(|1lxLimitXxxpx kpkkkc =∈>=<= ∞→  }||||,:{),(|1 ∞=∈>=<=∞→∞kpkkk xLimitXxxpx  where < pk > is a sequence of positive real number such that  pk ≥ k, for each k and l is a non-negative real number.  Let X and Y are Banach spaces and < Ak > is a linear operator from X onto Y and E is non-empty set of sequence x = (xk) with xk ∈X.  Then, we define -duals, -duals (m, ) duals and (m, ) duals of E as  Eα = {< Ak > :  ∑∞=1||||kkk xA  converges for all x = < xk > ∈E}  Eβ = {< Ak > :  ∑∞=1kkk xA converges is Y for all x = < xk > ∈E}  Em,α = {< Ak > : < Ak > is a sequence of linear operator but not necessarily bounded such that for some m; ∑∞==kmkkk xA ||||   converges for all < xk > ∈ E}  Em,β = {< Ak > < Ak > is a sequence of linear operator but not necessarily bounded such that for some m; ∑∞==kmkkk xA   converges in Y for all < xk > ∈ E}  Let B (X, Y) denote the Banach spaces of the bounded linear operators from X into Y with the usual operator norm. From the definition of the spaces, .),(|,),(|,),(| 0 pXpXpX c ∞ . It is obvious that   .),(|),(|),(| 0 pXpXpX c ∞⊂⊂   Thus, if T ∈ B(X, Y), the operator norm of T is  || T || = sup {|| T (x) ||}:  x∈S}, where  S = {x ∈ X: || x || ≤ 1} is a closed unit sphere in X. Definitions: Let < Tk > = < T1, T2, T3, ........> be a sequence in B (X, Y). Then, the group norm of (Tk) is   || (Tk)|| = sup ∑=nkkk xT1)(  where supremum is over all n ∈ N and all xk ∈ S. Main Results Lemma 1. If (Tk) is a sequence in B (X, Y) and we write Rn = (Tn, Tn+1,..), then  (i) || Tm || ≤ || Rn ||, for all m ≥ n  (ii) || Rn+1 || ≤ || Rn ||, for all n ∈ N  (iii) ∑+=pnnkkk xT  ≤ || Rn || .max {|| xk || : n ≤ k ≤ n + p} for any xk, for all n ∈ N and all non-negative integer p. Lemma 2.  If (Tk) is a sequence in B (X, Y), then   Σ || Tk || < ∞  ⇒  1sup≥k  ║Tk ║< ∞    Theorem 1.  < Ak > ∈ β,0 ),(| mpX    iff ∃ m ∈ N such that  (i) Ak ∈ B (X, Y) ;  for each  k ≥ m.  (ii) kpkmkA1||||sup≥ < ∞ Proof : Suppose the condition (i) and (ii) holds good.  Let  < xk > ∈ 0lim),(|0 =⇒ ∞→kpX   ⇒ To each ∈ > 0 ∃ +ve integer m such that  ║ xk ║1/ pk  < ∈ for all k ≥ m     ...(3.1.1)  Since, mk≥sup   ║ Ak ║1/ pk < ∞  ⇒ ∃ a real number M > 0 such that M ∈ < 1 and kpkA1||||   < M for all k ≥ m   ...(3.1.2) from (1) and (2),  ║ Ak ║1/ pk  ║ xk ║1/ pk  < M∈  ⇒ (║ Ak ║ ║ xk ║)1/ pk  < M∈  ⇒ || Ak || . || xk || < (M∈) pk  (since M ∈ < 1 and pk ≥ k ⇒ (M∈)pk ≤ (M∈)k). ⇒ || Ak || . || xk || < (M∈)k            ...(3.1.3) By result,      || Ak . (xk) ||  || Ak || . || xk ||, k ≥ m. ...(3.1.4)   From (3.1.3) and (3.1.4), || Ak . (xk) || < (M∈)k for all k ≥ m. ∑∑∑ ∞=∞=∞=∈<≤mkkmkkkmkkk MxAxA )(||||  =  ∞<∈−∈=⎥⎦⎤⎢⎣⎡∈−∈ MMMMmm1)(11)(  ⇒ ∑∞=1kkk xA  converges in Y for all < xk > ),(|0 pX    consequently, < Ak >  ∈  β.0 ),(| mpX .  Conversely, Let < Ak > ∈ β,0 ),(| mpX     But that no m exists for which Ak ∈ B (X, Y) for all k ≥ m.  ⇒ ∃ a sequence < ki > of natural no. such that ki ≥ i and Aki ∉ B (x, y), for each i  ⇒ || Aki || = ∞ where ki ≥ i,  for each ki ≥ i  ⇒ || Aki || = sup { || Aki (z) || : for each z ∈ X and || z || ≤ 1} = ∞ K. B. Gupta and Ashfaque A. Ansari/Rec Res Sci Tech 2011, 3(7): 95-99     97 ⇒ Sz∈sup  || Aki (z) || = ∞ where ki ≥ i,  for each ki ≥ i  ⇒ ∃ a sequence < zi > in S such that   || Aki (zi) || > i2pki  where ki ≥ i,  for each ki ≥ i Define    xk =    ⎪⎩⎪⎨⎧=≠ipiikkizkkik2,0 For k ≠ ki, ║ xk ║1/ pk  = 0  and for k = ki , ║ xk ║1/ pk  =  ║ xki ║1/ pk  = ║ zi ║1/ pk / i2 ≤ 1/ i2 → 0 as i → ∞, i.e. ki → ∞, i.e. k → ∞ (since || zi || ≤ 1) ⇒║ xk ║1/ pk    → 0  as k → ∞. Hence, < xk > ∈ ),(|0 pX   For k = ki,  || Ak xk || = || Aki (xki) ||   = ⎟⎟⎠⎞⎜⎜⎝⎛ikpikiizA 2        = ikpikiizA2||)(||    > 1 for large value of k.  Hence, ∑∞=mkkk xA )(  does not converge, which is contradiction to our assumption. Hence our assumption is wrong.  Therefore, ∃ m∈ N such that Ak ∈ B (X, Y), for k ≥ m. Again suppose if possible (ii) fails  ⇒ There does not exist any m ∈ N such that  ∞<≥kpkmkA1||||sup   ⇒ To each m ∈ N,  ∞=≥kpkmkA1||||sup   ⇒ kpkkA1||}{||sup1≥ = ∞  ⇒      ∃ a sequence < ki > of natural no. such that ║ Aki ║1/ pki > 2i for i ∈ N. Again since, (1/ 2k )|| Ak || ≤ || Ak || and || Ak || = sup {|| Ak (z) ||: for each  z ∈ S}  ⇒ (1/ 2k ) || Ak || ≤ sup {|| Ak (z) || : for each  z ∈ S }  ⇒ ∃ a sequence < zk > in S such that (1/ 2k )  || Ak || ≤ || Ak (zk) ||  ⇒ || Ak || ≤ 2k || Ak . (zk) || i.e. || Ak (zk) || ≥ || Ak || / 2k Define      xk   =   Zk / i pk       for k = ki      and   xk =    0  for k ≠ ki ;     For k = ki  0||||.1||||111 →== kpkpkp kpkkk ziizx  as i → ∞, i.e. ki = k → ∞ and for k ≠ ki,  ║ xk ║1/ pk  = 0 Then < xk > ∈ ),(|0 pX  But for k = ki,  || Ak xk || = ⎟⎠⎞⎜⎝⎛kpkk izA    =kk pkpkkkkkp iAiAzAi )2(||||.2||||||)(||1 ≥≥   > 1  for large value of k. ⇒ Σ Ak (xk) diverges, which is contradiction to our assumption. Hence our assumption is wrong.  Therefore, ∃ m ∈ N such that  ∞<≥kpkmkA1||||sup  Theorem : 2. < Ak > ∈ ),(0 ),(| αmpX    iff ∃ m ∈ N such that (i) Ak ∈ B (X, Y), for k ≥ m.  and (ii) ⎟⎟⎠⎞⎜⎜⎝⎛ ∑∞=1||||kkA   < ∞ Proof : Suppose the condition (i) and (ii) holds good.  Let x = (xk) ∈  ),(|0 pX   ⇒ kpkkxLimit1||||∞→ = 0  ⇒ for given ∈ > 0 ∃ a +ve integer m such that   || xk ||1/ pk < ∈ for all k ≥ m. i.e. || xk || < ∈ pk, for all k ≥ m    ...(3.2.1)  since  ⎟⎟⎠⎞⎜⎜⎝⎛ ∑∞=mkkA ||||   < ∞  ⇒ ∃ real number M > 0 such that M ∈ < 1 and  ∑∞=mkkA ||||  ≤ M, for all k ≥ m. ⇒ || Ak || < M ≤ (M pk), for all k ≥ m.         ...(3.2.2) from (3.2.1) and (3.2.2), || Ak || . || xk || < M pk . ∈ M pk = (M∈)pk ≤ (M∈)k ∞<⎥⎦⎤⎢⎣⎡∈−∈=∈<∑ ∑∞=∞=mk mkmkkk MMMxA11)()(||||(since M ∈ < 1 and pk ≥ k ⇒ (M∈) pk ≤ (M∈)k).  ⇒ ∑∞=mkkk xA ||||   converges.  consequently, < Ak > ∈ ),(0 ),(| αmPX      conversely, suppose < Ak > ∈ ),(0 ),(| αmPX    and there exist no m such that Ak ∈ B (X, Y) for k ≥ m  ⇒ ∃ a sequence < ki > of natural numbers such that Aki ∉ B (X, Y), for each i  ⇒ || Aki || = ∞, for each i  ⇒ || Aki || = sup {|| Aki (z) || : for each z ∈ X and || z || ≤ 1} = ∞  ⇒ Sz∈∀sup   || Aki (z) || = ∞ , for each z ∈ S   i.e. || z || ≤ 1  ⇒ ∃ a sequence < zi > in S such that || Aki (zi) || > i2pki  where ki ≥ i K. B. Gupta and Ashfaque A. Ansari/Rec Res Sci Tech 2011, 3(7): 95-99     98Define   xk   =   Zi / i2 pk       for k = ki      and   xk =    0  for k ≠ ki ;      For k = ki  ║ xk ║1/ pk  =  ║ Zi / i2 pk   ║1/ pk  = ║ zi ║1/ pk / i2 → 0 as i → ∞, i.e. k = ki → ∞ and   for k ≠ ki,  ║ xk ║1/ pk  = 0 Then < xk > ),(|0 PX   for k = ki ≥ m,  || Ak (xk) || = ║ Ak ( Zi / i2 pk  ) ║ = ║ Ak ( Zi) ║ (1 / i2 pk ) > 1   for every value of k.  ⇒ ∑∞=mkkk xA |||| does not converge, which is contradiction.  Hence, our assumption is wrong. Therefore, ∃ m ∈ N such that Ak ∈ B (X, Y), for each  k ≥ m.  Again suppose (ii) fails  i.e. ⎟⎟⎠⎞⎜⎜⎝⎛ ∑∞=mkkA ||||    = ∞ Then ∃ a strictly increasing sequence < ni >  such that ∑+=+=11||||iinknkkA > (2i) pk and ∃ a sequence < zk > in S such that  2k || Ak (zk) || ≥ || Ak || Define xk   =   Zk / i pk       for ni < k ≤ n(i+1)        and   xk =    0  otherwise ;     For k = ki izizxkpkpkkp kpkk111 |||||||| ==  → 0 as I → ∞ i.e. k = ki → ∞ For k ≠ ki, ║ xk ║1/ pk  = 0 Then < xk > ∈ ),(|0 PX  But ∑ ∑∑+ +++= +=+==⎟⎠⎞⎜⎝⎛=1 111 11||)(||||||iiiikiiknnknnkpkknnkpkkkk iZAizAxA    ≥    ∑ ∑+ ++= +=≥1 11 1 )2(||||,2||||i ikknnknnkpkpkkiAiA   > 1    ⇒ ∑∞=mkkk xA ||||   does not converge,  which is a contradiction.  Hence our assumption is wrong Therefore, ∃ m ∈ N such that  ∑∞=mkkA ||||  < ∞. Theorem : 3. < Ak > ∈   ),(),(| β∞ mpX    iff  (i) < Ak > ∈   ),(0 ),(| βmpX      (ii) || Rn || 1/n → 0 as n → ∞. Proof : Suppose (i) and (ii) holds good.  Let x = < xk > ∈ ),(| pX∞     ⇒ ∃ a real number M > 0 such that || xk ||1/ pk < M. since,  || Rn || 1/n → 0, and for every ∈ > 0,   || Rk || 1/k < ∈/M   for sufficiently large k. we have the identity by the Lemma (1), ∑+=pnnkkk xA )(  ≤ || Rn || .max {|| xk || : n ≤ k ≤ n + p} ≤ ∈k for sufficiently large k. Hence Σ Ak xk converges. consequently, (Ak) ∈ ),(),(| β∞ mpX     conversely, Let < Ak > ∈ ),(),(| β∞ mpX      since  ),(| pX∞   ⊂ ),(| pX∞     ⇒ ),(),(| β∞ mpX  ⊂ ),(),(| β∞ mpX     and therefore (Ak) ∈ ),(),(| β∞ mpX     ⇒ (Ak) ∈ ),(),(| β∞mpX     which proves (1)  Now suppose, if possible that (ii) fails  Let   limn sup || Rn || 1/n = 3p > 0.  Then, there exist natural numbers n1 > m1 > m and zm, .........., zn in S such that  ∑=11nmkkk zA      > p choose m2 > n1 such that || Rm2 || 1/m2 > 2p.  Then there exist n2  m2 and zm2, ...... ,zn2 in S such that ∑22)(nmkk zA    > p. Proceeding in this way, we and define   xk =  0  , k ≤m1       = zk  , m1≤ k≤ n1      =  0  ,  n1≤ k≤ m2      =  zk  , m2≤ k≤ n2  and so on.  Then  x = (xk) ∈ ),(| pX∞    But Σ Ak xk diverges which is contradiction to the fact that (Ak) ∈ ),(),(| β∞ mpX     Hence our assumption is wrong. Therefore  || Rn || 1/n → 0 as n → ∞. References [1] T. Balasubramanim and A. Pandiarari, 2009: Duals of vector sequence spaces; Int. Journal of Math. Analysis, Vol. 3, No. 16, 779-785. [2] Köthe G. and Toeplitz O., 1934 : Lineare Raime mit unendlichvielence koordinaten and Ringe unendlicher materizen J. F. Reine U. Angew Math., 174 193-226. K. B. Gupta and Ashfaque A. Ansari/Rec Res Sci Tech 2011, 3(7): 95-99     99[3] R. G. Cooke 1949: Infinite matrices and sequence spaces Macmillan and Co. London.  [4] G. H. Hardy 1949: Divergent series, Oxford University Press, [5] J. Maddox, 1970: Element of functional Analysis. Combridge University Press. [6] G. Köthe, 1969: Topological vector space I (English); Springer-Verlag. [7] A.  R. Robinson 1995: On functional transformations and summability, Poc. London Math. Soc. 52,132-160 [8] J. Maddox 1986: Infinite matrices of operators, Lecture notes in mathematics, Springer-verlag, Barlin.    

Operator Duals of Vector Sequence Spaces

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Operator Duals of Vector Sequence Spaces

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