143
Acta Math. Univ. Comenianae
Vol. LXXIX, 1(2010), pp. 143–149
THE DUAL SPACE OF THE SEQUENCE SPACE bvp (1 ≤ p < ∞)
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 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


! p1
∞


X
p
bvp = x = (xk ) ∈ ω :
|xk − xk−1 |
<∞
(1 ≤ p < ∞)


k=0
and
bv∞ =
x = (xk ) ∈ ω : sup |xk − xk−1 | < ∞
k∈n
where x−1 = 0.
Now, let
kxkbvp =
∞
X
! p1
p
|xk − xk−1 |
k=0
and
kxkbv∞ = sup |xk − xk−1 |.
k∈N
Then bvp and bv∞ are Banach spaces with these norms and except the case p = 2,
the space bvp is not a Hilbert space for 1 ≤ p ≤ ∞. If we define a sequence
Received August 26, 2009.
2000 Mathematics Subject Classification. Primary 46B10; Secondary 46B45.
Key words and phrases. dual space; sequence space; Banach space; isometrically isomorphic.
144
M. IMANINEZHAD and M. MIRI
(k)
b(k) = (bn )∞
n=0 of elements of the space bvp for every fixed k ∈ N by
0, if n < k
(k)
bn =
1, if n ≥ k
∞
then the sequence (b(k) )k=0 is a Schauder basis for bvp and any x ∈ bvp has a
unique representation of the form
x=
∞
X
λk b(k)
k=0
where λk = (xk − xk−1 ) for all k ∈ N.
2. A counterexample
In [1, Theorem 2.3] for case p = 1 suppose f = (3, −1, 0, 0, 0, . . .), i.e.,
f0 = f (e0 ) = 3,
f1 = f (e1 ) = −1,
fk = f (ek ) = 0 for all k ≥ 2.
Trivially f ∈ bv1∗ and
f (x) = f
∞
X
!
(∆x)k b
(k)
= 2(∆x)0 − (∆x)1 .
k=0
So
kf k =
(1)
sup
kxkbv1 =1
|f (x)| = P
sup
∞
i=0
|2(∆x)0 − (∆x)1 | = 2.
|(∆x)i |=1
Now inequality (2.5) in [1, Theorem 2.3] asserts that kf k ≥ sup |
k,n∈N
Pn
j=k
fj | = 3
which is a 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 we give the main theorem of the paper. Let
∞ X ∞
d∞ = a = (ak )k=0 ∈ ω : kakd∞ = sup
aj < ∞
k∈N
j=k
and
dq =
a=
(ak )∞
k=0
∈ ω : kakdq
X
q1
∞ X
∞
q
<∞ ,
=
|
aj |
(1 < q < ∞).
k=0 j=k
Theorem 3.1. d∞ is a sequence space with usual coordinatewise addition and
scalar multiplication and k·kd∞ is a norm on d∞ .
THE DUAL SPACE OF THE SEQUENCE SPACE bvp (1 ≤ p < ∞)
Proof. We only show that k·kd∞ is a norm on

1 1 1 1 1···
 0 1 1 1 1···


D =  0 0 1 1 1···
 0 0 0 1 1···

.. .. .. .. .. .. . .
...... .
145
d∞ . Let




.


Then

 P∞
aj
Pj=0
∞
 


aj 

  Pj=1


∞
 


aj 
Da = 
.
 =  Pj=2

∞

 


j=3 aj 
 


..
.
P∞
So kakd∞ = supk∈N j=k aj = supk∈N (Da)k = Dal∞ . Now, if a ∈ d∞
then kDakl∞ = 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

111
011
001
000
.. .. .. ..
....
1 1···
1 1···
1 1···
1 1···
.. .. . .
.. .

a0
a1
a2
a3
..
.

(I) 0 ≤ kDakl∞ = kakd∞ < ∞
(II) ka + bkd∞ = kDa + Dbkl∞ ≤ kDakl∞ + kDbkl∞ = kakd∞ + kbkd∞
(III) kα · akd∞ = kα · Dakl∞ = |α| · kDakl∞ = |α| · kakd∞
k·kd∞ is a norm on d∞ .
Theorem 3.2. d∞ is a Banach space.
Proof. Let (a(n) )∞
n=0 is a Cauchy sequence in d∞ . So for each ε > 0 there exists
N ∈ N, such that for all n, m ≥ N
ka(n) − a(m) kd∞ < ε.
So
kDa(n) − Da(m) kl∞ = ka(n) − a(m) kd∞ < ε.
So the sequence (Da(n) )∞
n=0 is Cauchy in l∞ . So there exists a ∈ l∞ such that
Da(n) → a in l∞ . So kDa(n) − DD−1 akl∞ → 0 and ka(n) − D−1 akd∞ → 0
Furthermore, D−1 a ∈ d∞ since DD−1 a = a ∈ l∞ .
Theorem 3.3. bv1∗ is isometrically isomorphic to d∞ .
Proof. Define T : bv1∗ → d∞ and T f = (f (e(0) ), f (e(1) ), f (e(2) ), . . . ) where
e(k) = (0, . . . , 0, |{z}
1 , 0, . . .).
kth term
Trivially, T is linear and injective since
T f = 0 ⇒ f = 0.
146
M. IMANINEZHAD and M. MIRI
T is surjective since if g̃ = (g0 , g1 , g2 , g3 , . . .) ∈ d∞ then if we define f : bv1 → C
by
∞
∞
X
X
f (x) =
(∆x)k
gj .
k=0
Then f ∈
bv1∗ .
j=k
Trivially, since f is linear and
X
X
X
∞
∞
X
∞
∞ gj |f (x)| = (∆x)k
gj ≤
|(∆x)k | · k=0
≤
∞
X
k=0
j=k
j=k
k=0
∞ ∞
X X
|(∆x)k | sup
gj =
|(∆x)k |.kg̃kd∞
k∈N
j=k
k=0
= kg̃kd∞ .kxkbv1
and T f = g̃, so T is surjective. Now we show that T is norm preserving, we have
∞
X
∞
∞
X
X
X
∞
(j) (j) f (e )
(∆x)k
e
|f (x)| = f
= (∆x)k
k=0
≤
∞
X
k=0
j=k
k=0
j=k
X
X
X
∞
∞
∞
(j) (j) f (e )
|(∆x)k | · sup
f (e ) ≤
|(∆x)k |
k=0
j=k
k∈N
j=k
≤ kxkbv1 · kT f kd∞ .
So
kf k ≤ kT f kd∞
P∞
On the other hand, j=k f (e(j) ) = f (b(k) ) ≤ kf k · kb(k) kbv1 = kf k.
P∞
(j) j=k f (e )≤ kf k for all k ∈ N.
So
∞
X
sup |
f (e(j) )| ≤ kf k,
(∗)
k∈N
So
j=k
i.e.,
(†)
kT f kd∞ ≤ kf k
by (∗) and (†) we are done.
Theorem 3.4. dq (1 ≤ q < ∞) 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 ∈ dq ⇔ Da ∈ lq .
The continuation 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.
THE DUAL SPACE OF THE SEQUENCE SPACE bvp (1 ≤ p < ∞)
Theorem 3.6. Let 1 < p < ∞ and
isomorphic to dq .
1
p
+
1
q
147
= 1, then bvp∗ is isometrically
Proof. Define A : bvp∗ → dq by f 7→ Af = (f (e(0) ), f (e(1) ), f (e(2) ), . . .). Trivially A is linear. Additionally, since Af = 0 = (0, 0, 0, . . .) implies f = 0, A is
injective. A is surjective since if a = (ak ) ∈ dq and define f on the space bvp such
that
∞
∞
X
X
f (x) =
(∆x)k
aj .
k=0
j=k
Then f is linear. Since
∞ ∞
X
X
|f (x)| ≤
aj (∆x)k
k=0
j=k
∞
X
(∆x)k p
≤
"
1
q
∞ X
X
∞ q
aj  = kxkbvp · kakdq ,
·  # p1 
k=0 j=k
k=0
it yields to kf k ≤ 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
X
X
(j) (∆x)k
f
(e
)
f (e(j) ) ≤
|f (x)| = (∆x)k
k=0
k=0
j=k
∞
X
(∆x)k p
≤
"
k=0
# p1
j=k
1
q q
∞ X
X
∞
·  f (e(j) )  = kxkbvp · kAf kdq .

k=0 j=k
So
kf k ≤ kAf kdq .
(∗)
(n) ∞
On the other hand, suppose f ∈ bvp∗ and x(n) = xk k=0 are such that
q−1 ∞
 X
X
∞

(j)
(j)  f
(e
)
sgn
f
(e
)
, if (0 ≤ k ≤ n)
(∆x(n) )k =
j=k
j=k


0,
if k > n.
P∞
We note that j=k f (e(j) ) = f (b(k) ). So x(n) ∈ bvp since ∆x(n) ∈ lp .
Then it is clear that
∆x(n) =


∞
q−1 ∞
X
q−1 X
∞
X
X
∞

f (e(j) ) sgn
f (e(j) ) , . . . , f (e(j) ) sgn
f (e(j) ) , 0, 0, . . . .
j=0
j=0
j=n
j=n
148
M. IMANINEZHAD and M. MIRI
So

x
(n)
q−1 ∞
 ∞
∞
q−1 ∞
X
X
 X
X
(j) (j)
(j) (j)

f (e ) , b0 + f (e ) sgn
f (e ) sgn
f (e ) ,
= 
 j=0
j=0
j=1
j=1
|
{z
}
|
{z
}
b0
b1

,...,
n
X
bk
k=0


, t, t, t, . . .
.

| {z }
t=n+1th term
So if we let fk = f (e(k) ), then
f (x(n) ) = b0 f0 +
b0 f1 + b1 f1 +
b0 f2 + b1 f2 + b2 f2 +
b0 f3 + b1 f3 + b2 f3 + b3 f3 +
..
..
..
..
.
.
.
.
b0 fn + b1 fn + b2 fn + b3 fn + . . . + bn fn +
b0 fn+1 + b1 fn+1 + b2 fn+1 + b3 fn+1 + . . . + bn fn+1 +
b0 fn+2 + b1 fn+2 + b2 fn+2 + b3 fn+2 + . . . + bn fn+2 +
b0 fn+3 + b1 fn+3 + b2 fn+3 + b3 fn+3 + . . . + bn fn+3 +
..
..
..
..
..
..
.
.
.
.
.
.
n
∞
q
X X =
fj .
k=0 j=k
So

1
q p
n X
n X
∞
X
X
∞ q
fj = f (x(n) ) = |f (x(n) )| ≤ kf k · kx(n) kbvp = kf k · 
fj  .
k=0 j=k
k=0 j=k
Since
"
kx(n) kbvp = k∆x(n) klp =
∞
X
# p1
(n)
|∆xk |p
k=0
"
=
n
X
k=0

  p1
p
n X
∞
X
X
∞ q−1



=
fj sgn
fj 

k=0

=
j=k
j=k
 p1
n X
X
∞ q
fj  .
k=0 j=k
# p1
(n)
|∆xk |p
THE DUAL SPACE OF THE SEQUENCE SPACE bvp (1 ≤ p < ∞)
So
149


 p1
1
n
∞
n X
X
X
X
∞ q

|
fj  ≤ kf k · 
fj |q  .

k=0 j=k
k=0 j=k
So
1
q
n X
X
∞ q
kf k ≥ 
fj  = kAf kdq .

(†)
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. Also he would like to express his thanks to
Mr. Davoodnezhad and Mrs. Sadeghi1 for supplying him with some references.
References
1. Akhmedov A. M. , Basar F., The Fine Spectra of the Difference Operator ∆ over the Sequence
Space bvp (1 ≤ 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.
3. Goldberg S., Unbounded Linear Operators, Dover Publication Inc. New York, 1985.
4. Kreyszig E., Introductory Functional Analysis 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: imani507@gmail.com
M. Miri, Dep. of Math., University of Birjand,e-mail: mmiri@birjand.ac.ir
1Librarians of Faculty of Mathematical Sciences of Mashhad University