THE DUAL OF THE SPACE bvp (1 ≤ p < ∞)

M. IMANINEZHAD and M. MIRI

Abstract. The sequence space bvp consists of all (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 d∞ instead of d1 and show that bv1 = d∞. Also we have modified the proof for case p > 1. Our notations improve the presentation and are confirmed ∗ ∗ by last notations l1 = l∞ and lp = lq.

1. Priliminaries, background and notation

Let ω denote the space of all complex-valued sequences, i.e., ω = CN where N = {0, 1, 2, 3,...}. Any vector subspace of ω 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

 1  JJ J I II ∞ ! p  X p  Go back bvp = x = (xk) ∈ ω : |xk − xk−1| < ∞ (1 ≤ p < ∞)  k=0  Full Screen

Close Received August 26, 2009. 2000 Subject Classification. Primary 46B10; Secondary 46B45. Key words and phrases. ; sequence space; ; isometrically isomorphic. Quit and   bv∞ = x = (xk) ∈ ω : sup |xk − xk−1| < ∞ k∈n

where x−1 = 0. Now, let 1 ∞ ! p X p kxk = |x − x | bvp k k−1 k=0 and kxk = sup |x − x |. bv∞ k k−1 k∈N

Then bvp and bv∞ are Banach spaces with these norms and except the case p = 2, the space bvp (k) (k) ∞ is not a for 1 ≤ p ≤ ∞. If we define a sequence b = (bn )n=0 of elements of the space bvp for every fixed k ∈ N by  0, if n < k b(k) = n 1, if n ≥ k JJ J I II (k) ∞ Go back then the sequence (b )k=0 is a for bvp and any x ∈ bvp has a unique representation of the form Full Screen ∞ X (k) x = λkb Close k=0

where λk = (xk − xk−1) for all k ∈ . 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 ∈ bv1 and ∞ ! X (k) f(x) = f (∆x)kb = 2(∆x)0 − (∆x)1. k=0 So

(1) kfk = sup |f(x)| = sup |2(∆x)0 − (∆x)1| = 2. P∞ kxkbv1=1 i=0 |(∆x)i|=1 Pn Now inequality (2.5) in [1, Theorem 2.3] asserts that kfk ≥ sup | j=k fj| = 3 which is a k,n∈N contradiction.

3. The Spaces d∞ and dq (1 < q < ∞)

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  ∞  ∞ X Go back d = a = (a ) ∈ ω : kak = sup a < ∞ ∞ k k=0 d∞ j k∈ N j=k Full Screen and ∞ ∞ 1    q  Close ∞ X X q dq = a = (ak)k=0 ∈ ω : kakdq = | aj| < ∞ , (1 < q < ∞). k=0 j=k Quit Theorem 3.1. d∞ is a sequence space with usual coordinatewise addition and scalar multipli- cation and k·k is a on d . d∞ ∞ Proof. We only show that k·k is a norm on d . Let d∞ ∞  1 1 1 1 1 ···   0 1 1 1 1 ···     0 0 1 1 1 ···  D =   .  0 0 0 1 1 ···   ......  ...... Then      P∞  1 1 1 1 1 ··· a0 j=0 aj P∞  0 1 1 1 1 ···   a1   j=1 aj       P∞  Da =  0 0 1 1 1 ···   a2  =  j=2 aj  .      P∞   0 0 0 1 1 ···   a3   aj       j=3  ...... P∞ So kakd∞ = sup aj = sup (Da)k = Da . Now, if a ∈ d∞ then kDakl∞ = k∈N j=k k∈N l∞ JJ J I II kakd∞ < ∞ hence Da ∈ l∞. Also if Da ∈ l∞, then kakd∞ = kDakl∞ < ∞ hence a ∈ d∞. So a ∈ d∞ if and only if Da ∈ l∞. Now since Go back (I)0 ≤ kDak = kak < ∞ l∞ d∞ Full Screen (II) ka + bk = kDa + Dbk ≤ kDak + kDbk = kak + kbk d∞ l∞ l∞ l∞ d∞ d∞ (III) kα · ak = kα · Dak = |α| · kDak = |α| · kak d∞ l∞ l∞ d∞ Close k·k is a norm on d . d∞ ∞  Quit Theorem 3.2. d∞ is a Banach space.

(n) ∞ Proof. Let (a )n=0 is a in d∞. So for each ε > 0 there exists N ∈ N, such that for all n, m ≥ N ka(n) − a(m)k < ε. d∞ So kDa(n) − Da(m)k = ka(n) − a(m)k < ε. l∞ d∞ (n) ∞ (n) So the sequence (Da )n=0 is Cauchy in l∞. So there exists a ∈ l∞ such that Da → a in l∞. So kDa(n) − DD−1ak → 0 and ka(n) − D−1ak → 0 l∞ d∞ −1 −1 Furthermore, D a ∈ d∞ since DD a = a ∈ l∞. 

∗ Theorem 3.3. bv1 is isometrically isomorphic to d∞.

∗ (0) (1) (2) Proof. Define T : bv1 → d∞ 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,...) ∈ d∞ then if we define f : bv1 → C by ∞ ∞ Close X X f(x) = (∆x)k gj. k=0 j=k Quit ∗ Then f ∈ bv1 . Trivially, since f is linear and ∞ ∞ ∞ ∞ X X X X |f(x)| = (∆x)k gj ≤ |(∆x)k| · gj

k=0 j=k k=0 j=k ∞ ∞ ∞ X X X ≤ |(∆x)k| sup gj = |(∆x)k|.kg˜kd∞ k∈ k=0 N j=k k=0

= kg˜kd∞ .kxkbv1 and T f =g ˜, so T is surjective. Now we show that T is norm preserving, we have  ∞ ∞  ∞ ∞ X X (j) X X (j) |f(x)| = f (∆x)k e = (∆x)k f(e )

k=0 j=k k=0 j=k ∞ ∞ ∞ ∞ X X (j) X X (j) ≤ |(∆x)k| f(e ) ≤ |(∆x)k| · sup f(e ) k∈ k=0 j=k k=0 N j=k

≤ kxkbv1 · kT fkd∞ . So

JJ J I II (∗) kfk ≤ kT fkd∞

Go back P∞ (j) (k) (k) P∞ (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 ∈ N. So ∞ Close X sup | f(e(j))| ≤ kfk, k∈ N j=k Quit i.e.,

(†) kT fkd∞ ≤ kfk

by (∗) and (†) we are done. 

Theorem 3.4. dq (1 ≤ q < ∞) is a sequence space with usual coordinatewise addition and

and k.kdq is a norm on dq.

Proof. With notations of Theorem 3.1, kakdq = kDaklq and a ∈ dq ⇔ Da ∈ lq. The continua- tion of the proof is similar to Theorem 3.1. 

Theorem 3.5. dq (1 ≤ q < ∞) 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 < ∞ 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) ∈ dq and define f on the space bvp such that

∞ ∞ Close X X f(x) = (∆x)k aj. k=0 j=k Quit Then f is linear. Since ∞ ∞ X X |f(x)| ≤ (∆x)k aj

k=0 j=k 1 1   q " ∞ # p ∞ ∞ q X p X X ≤ (∆x)k ·  aj  = kxkbvp · kakdq ,

k=0 k=0 j=k

∗ it yields to kfk ≤ kakdq < ∞. So f ∈ bvp and Af = a. ∗ ∞ Now, we show that A is norm preserving. Let f ∈ bvp and x = (xk)k=0 ∈ bvp, then ∞ ∞ ∞ ∞ X X (j) X X (j) |f(x)| = (∆x)k f(e ) ≤ (∆x)k f(e )

k=0 j=k k=0 j=k 1 1   q " ∞ # p ∞ ∞ q X p X X (j) ≤ (∆x)k ·  f(e )  = kxkbvp · kAfkdq .

k=0 k=0 j=k So JJ J I II (∗) kfk ≤ kAfkdq . Go back ∗ (n) (n)∞ On the other hand, suppose f ∈ bvp and x = xk k=0 are such that Full Screen  ∞ q−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 P∞ (j) (k) (n) (n) We note that j=k f(e ) = f(b ). So x ∈ bvp since ∆x ∈ lp. Then it is clear that ∆x(n) =   ∞ q−1  ∞  ∞ q−1  ∞  X (j) X (j) X (j) X (j)  f(e ) sgn f(e ) ,..., f(e ) sgn f(e ) , 0, 0,... .

j=0 j=0 j=n j=n So   ∞ q−1  ∞  ∞ q−1  ∞  (n)  X (j) X (j) X (j) X (j) x =  f(e ) sgn f(e ) , b0 + f(e ) sgn f(e ) ,   j=0 j=0 j=1 j=1 | {z } | {z } b0 b1  n  X  ,..., bk , t, t, t, . . . .  k=0  | {z } JJ J I II t=n+1thterm So if we let f = f(e(k)), then Go back k (n) f(x ) = b0f0 + Full Screen b0f1 + b1f1+ b0f2 + b1f2 + b2f2+ Close b0f3 + b1f3 + b2f3 + b3f3+ ...... Quit b0fn + b1fn + b2fn + b3fn + ... + bnfn+ b0fn+1 + b1fn+1 + b2fn+1 + b3fn+1 + ... + bnfn+1+ b0fn+2 + b1fn+2 + b2fn+2 + b3fn+2 + ... + bnfn+2+ b0fn+3 + b1fn+3 + b2fn+3 + b3fn+3 + ... + bnfn+3+ ...... n ∞ q X X = fj .

k=0 j=k So

1   p n ∞ q n ∞ q X X (n) (n) (n) X X fj = f(x ) = |f(x )| ≤ kfk · kx kbvp = kfk ·  fj  .

k=0 j=k k=0 j=k

Since

1 1 " ∞ # p " n # p (n) (n) X (n) p X (n) p kx kbvp = k∆x klp = |∆xk | = |∆xk | JJ J I II k=0 k=0 1     p n ∞ q−1 ∞ p Go back X X X =  fj sgn  fj 

Full Screen k=0 j=k j=k 1   p n ∞ q Close X X =  fj  .

k=0 j=k Quit So 1  1   p n ∞ q n ∞ X X X X q  fj  ≤ kfk ·  | fj|  .

k=0 j=k k=0 j=k So 1   q n ∞ q X X (†) kfk ≥  fj  = kAfkdq .

k=0 j=k

∗ Therefore, by combinig the results (∗) and (†), A is norm preserving. Hence bvp is isometrically isomorphic to dq.  Acknowledgment. The authors would like to express their indebtedness to A. M. Akhmedov and F. Basar since they were the source of inspiration. The first author thanks to several of colleagues. Particularly, he is obliged to Dr. Bolbolian, Dr. Mohammadian and Dr. Roozbeh. JJ J I II Also he would like to express his thanks to Mr. Davoodnezhad and Mrs. Sadeghi1 for supplying him with some references.

Go back

1. Akhmedov A. M. , Basar F., The Fine Spectra of the Difference ∆ over the Sequence Space bvp(1 ≤ Full Screen p < ∞) , Acta Math. Sin. Eng. Ser., 23(10) (2007), 1757–1768. 2. Basar F., Altay B., On the Space of Sequences of p-Bounded Variation and Related Matrix Mappings, Ukrainian Math. J.,55(1) (2003), 136–147. Close 3. Goldberg S., Unbounded Linear Operators, Dover Publication Inc. New York, 1985.

1Librarians of Faculty of Mathematical Sciences of Mashhad University Quit 4. Kreyszig E., Introductory with Applications, John Wiley & Sons Inc. New York-Chichester- Brisbane-Toronto, 1978. 5. Wilansky A., Summability through Functional Analysis, North-Holland Mathematics Studies, Amsterdam-New York-Oxford, 1984.

M. Imaninezhad, Islamic Azad University of Birjand, Iran,e-mail: [email protected]

M. Miri, Dep. of Math., University of Birjand,e-mail: [email protected]

JJ J I II

Go back

Full Screen

Close

Quit