THE DUAL SPACE OF THE SEQUENCE SPACE bvp (1 ≤ p < 1) M. IMANINEZHAD and M. MIRI Abstract. The sequence space bvp consists of all sequences (xk) such that (xk − xk−1) belongs to the space lp. The continuous dual of the sequence space bvp has recently been introduced by Akhmedov and Basar [Acta Math. Sin. Eng. Ser., 23(10), 2007, 1757{1768]. In this paper, we show a counterexample ∗ for case p = 1 and introduce a new sequence space d1 instead of d1 and show that bv1 = d1. Also we have modified the proof for case p > 1. Our notations improve the presentation and are confirmed ∗ ∗ by last notations l1 = l1 and lp = lq. 1. Priliminaries, background and notation Let ! denote the space of all complex-valued sequences, i.e., ! = CN where N = f0; 1; 2; 3;:::g. Any vector subspace of ! which contains φ, the set of all finitely non-zero sequences, is called a sequence space. The continuous dual of a sequence space λ which is denoted by λ∗ is the set of all bounded linear functionals on λ. The space bvp is the set of all sequences of p-bounded variation and is defined by 8 1 9 JJ J I II 1 ! p < X p = Go back bvp = x = (xk) 2 ! : jxk − xk−1j < 1 (1 ≤ p < 1) : k=0 ; Full Screen Close Received August 26, 2009. 2000 Mathematics Subject Classification. Primary 46B10; Secondary 46B45. Key words and phrases. dual space; sequence space; Banach space; isometrically isomorphic. Quit and bv1 = x = (xk) 2 ! : sup jxk − xk−1j < 1 k2n where x−1 = 0. Now, let 1 1 ! p X p kxk = jx − x j bvp k k−1 k=0 and kxk = sup jx − x j: bv1 k k−1 k2N Then bvp and bv1 are Banach spaces with these norms and except the case p = 2, the space bvp (k) (k) 1 is not a Hilbert space for 1 ≤ p ≤ 1. If we define a sequence b = (bn )n=0 of elements of the space bvp for every fixed k 2 N by 0; if n < k b(k) = n 1; if n ≥ k JJ J I II (k) 1 Go back then the sequence (b )k=0 is a Schauder basis for bvp and any x 2 bvp has a unique representation of the form Full Screen 1 X (k) x = λkb Close k=0 where λk = (xk − xk−1) for all k 2 . Quit N 2. A counterexample In [1, Theorem 2.3] for case p = 1 suppose f = (3; −1; 0; 0; 0;:::), i.e., 0 1 k f0 = f(e ) = 3; f1 = f(e ) = −1; fk = f(e ) = 0 for all k ≥ 2: ∗ Trivially f 2 bv1 and 1 ! X (k) f(x) = f (∆x)kb = 2(∆x)0 − (∆x)1: k=0 So (1) kfk = sup jf(x)j = sup j2(∆x)0 − (∆x)1j = 2: P1 kxkbv1=1 i=0 j(∆x)ij=1 Pn Now inequality (2.5) in [1, Theorem 2.3] asserts that kfk ≥ sup j j=k fjj = 3 which is a k;n2N contradiction. 3. The Spaces d1 and dq (1 < q < 1) In this section, we introduce two sequence spaces and show that they are Banach spaces and then JJ J I II we give the main theorem of the paper. Let 1 1 X Go back d = a = (a ) 2 ! : kak = sup a < 1 1 k k=0 d1 j k2 N j=k Full Screen and 1 1 1 q Close 1 X X q dq = a = (ak)k=0 2 ! : kakdq = j ajj < 1 ; (1 < q < 1): k=0 j=k Quit Theorem 3.1. d1 is a sequence space with usual coordinatewise addition and scalar multipli- cation and k·k is a norm on d . d1 1 Proof. We only show that k·k is a norm on d . Let d1 1 2 1 1 1 1 1 ··· 3 6 0 1 1 1 1 ··· 7 6 7 6 0 0 1 1 1 ··· 7 D = 6 7 : 6 0 0 0 1 1 ··· 7 4 . 5 . .. Then 2 3 2 3 2 P1 3 1 1 1 1 1 ··· a0 j=0 aj P1 6 0 1 1 1 1 ··· 7 6 a1 7 6 j=1 aj 7 6 7 6 7 6 P1 7 Da = 6 0 0 1 1 1 ··· 7 6 a2 7 = 6 j=2 aj 7 : 6 7 6 7 6 P1 7 6 0 0 0 1 1 ··· 7 6 a3 7 6 aj 7 4 5 4 5 4 j=3 5 . .. P1 So kakd1 = sup aj = sup (Da)k = Da : Now, if a 2 d1 then kDakl1 = k2N j=k k2N l1 JJ J I II kakd1 < 1 hence Da 2 l1. Also if Da 2 l1, then kakd1 = kDakl1 < 1 hence a 2 d1. So a 2 d1 if and only if Da 2 l1. Now since Go back (I)0 ≤ kDak = kak < 1 l1 d1 Full Screen (II) ka + bk = kDa + Dbk ≤ kDak + kDbk = kak + kbk d1 l1 l1 l1 d1 d1 (III) kα · ak = kα · Dak = jαj · kDak = jαj · kak d1 l1 l1 d1 Close k·k is a norm on d . d1 1 Quit Theorem 3.2. d1 is a Banach space. (n) 1 Proof. Let (a )n=0 is a Cauchy sequence in d1. So for each " > 0 there exists N 2 N, such that for all n; m ≥ N ka(n) − a(m)k < ": d1 So kDa(n) − Da(m)k = ka(n) − a(m)k < ": l1 d1 (n) 1 (n) So the sequence (Da )n=0 is Cauchy in l1. So there exists a 2 l1 such that Da ! a in l1. So kDa(n) − DD−1ak ! 0 and ka(n) − D−1ak ! 0 l1 d1 −1 −1 Furthermore, D a 2 d1 since DD a = a 2 l1. ∗ Theorem 3.3. bv1 is isometrically isomorphic to d1. ∗ (0) (1) (2) Proof. Define T : bv1 ! d1 and T f = (f(e ); f(e ); f(e );::: ) where e(k) = (0;:::; 0; 1 ; 0;:::): |{z} kthterm JJ J I II Trivially, T is linear and injective since Go back T f = 0 ) f = 0: Full Screen T is surjective since ifg ~ = (g0; g1; g2; g3;:::) 2 d1 then if we define f : bv1 ! C by 1 1 Close X X f(x) = (∆x)k gj: k=0 j=k Quit ∗ Then f 2 bv1 . Trivially, since f is linear and 1 1 1 1 X X X X jf(x)j = (∆x)k gj ≤ j(∆x)kj · gj k=0 j=k k=0 j=k 1 1 1 X X X ≤ j(∆x)kj sup gj = j(∆x)kj:kg~kd1 k2 k=0 N j=k k=0 = kg~kd1 :kxkbv1 and T f =g ~, so T is surjective. Now we show that T is norm preserving, we have 1 1 1 1 X X (j) X X (j) jf(x)j = f (∆x)k e = (∆x)k f(e ) k=0 j=k k=0 j=k 1 1 1 1 X X (j) X X (j) ≤ j(∆x)kj f(e ) ≤ j(∆x)kj · sup f(e ) k2 k=0 j=k k=0 N j=k ≤ kxkbv1 · kT fkd1 : So JJ J I II (∗) kfk ≤ kT fkd1 Go back P1 (j) (k) (k) P1 (j) On the other hand, f(e ) = f(b ) ≤ kfk · kb kbv1 = kfk. So f(e ) ≤ kfk for j=k j=k Full Screen all k 2 N. So 1 Close X sup j f(e(j))j ≤ kfk; k2 N j=k Quit i.e., (y) kT fkd1 ≤ kfk by (∗) and (y) we are done. Theorem 3.4. dq (1 ≤ q < 1) is a sequence space with usual coordinatewise addition and scalar multiplication and k.kdq is a norm on dq. Proof. With notations of Theorem 3.1, kakdq = kDaklq and a 2 dq , Da 2 lq. The continua- tion of the proof is similar to Theorem 3.1. Theorem 3.5. dq (1 ≤ q < 1) is a Banach space. Proof. The proof is similar to proof of Theorem 3.2 and we omit it. 1 1 ∗ Theorem 3.6. Let 1 < p < 1 and p + q = 1, then bvp is isometrically isomorphic to dq. JJ J I II ∗ (0) (1) (2) Proof. Define A : bv ! dq by f 7! Af = (f(e ); f(e ); f(e );:::). Trivially A is linear. Go back p Additionally, since Af = 0 = (0; 0; 0;:::) implies f = 0, A is injective. A is surjective since if Full Screen a = (ak) 2 dq and define f on the space bvp such that 1 1 Close X X f(x) = (∆x)k aj: k=0 j=k Quit Then f is linear. Since 1 1 X X jf(x)j ≤ (∆x)k aj k=0 j=k 1 1 2 3 q " 1 # p 1 1 q X p X X ≤ (∆x)k · 4 aj 5 = kxkbvp · kakdq ; k=0 k=0 j=k ∗ it yields to kfk ≤ kakdq < 1. So f 2 bvp and Af = a. ∗ 1 Now, we show that A is norm preserving. Let f 2 bvp and x = (xk)k=0 2 bvp, then 1 1 1 1 X X (j) X X (j) jf(x)j = (∆x)k f(e ) ≤ (∆x)k f(e ) k=0 j=k k=0 j=k 1 1 2 3 q " 1 # p 1 1 q X p X X (j) ≤ (∆x)k · 4 f(e ) 5 = kxkbvp · kAfkdq : k=0 k=0 j=k So JJ J I II (∗) kfk ≤ kAfkdq : Go back ∗ (n) (n)1 On the other hand, suppose f 2 bvp and x = xk k=0 are such that Full Screen 8 1 q−1 1 X (j) X (j) <> f(e ) sgn f(e ) ; if (0 ≤ k ≤ n) (n) Close (∆x )k = j=k j=k > : 0; if k > n: Quit P1 (j) (k) (n) (n) We note that j=k f(e ) = f(b ).