THE CONTINUOUS DUAL OF THE SEQUENCE SPACE lp (∆n ),
(1 ≤ p ≤ ∞, n ∈ N)
M. IMANINEZHAD and M. MIRI
Abstract. The space lp (∆m ) consisting of all sequences whose mth order differences are p-absolutely
summable was recently studied by Altay [On the space of p-summable difference sequences of order
m, (1 ≤ p < ∞), Stud. Sci. Math. Hungar. 43(4) (2006), 387–402]. Following Altay [2], we have
found the continuous dual of the spaces l1 (∆n ) and lP (∆n ). We have also determined the norm of
the operator ∆n acting from l1 to itself and from l∞ to itself, and proved that ∆n is a bounded linear
operator on the space lp (∆n ).
1. Preliminaries, Definitions and Notations
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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
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Received February 12, 2010.
2000 Mathematics Subject Classification. Primary 46B10; Secondary 46B45.
Key words and phrases. difference operator; continuous dual; Banach space; sequence space; Schauder basis,
lp (∆n ).
bounded linear functionals on λ. Suppose ∆ be the difference operator with matrix representation





∆=



1
−1
0
0
0
..
.
0
1
−1
0
0
..
.
0
0
1
−1
0
..
.
0 0 0
0 0 0
0 0 0
1 0 0
−1 1 0
..
.. ..
.
. .
···
···
···
···
···
..
.









∞
n
n−1
and suppose x = (xk )∞
x) for all n ≥ 2 where
k=0 ∈ ω, then ∆x = (xk −xk−1 )k=0 and ∆ x = ∆(∆
n
any x with negative index is zero. For every n ∈ N \ {0}, ∆ has a triangle matrix representation,
so it is invertible and
1
0
0
0
0
0
0
6 −(n
1
0
0
0
0
0
1)
6
n
n
(2 )
−(1 )
1
0
0
0
0
6
6
n
n
n
(2 )
−(1 )
1
0
0
0
6 −(3 )
6
6
.
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6
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6
6
6
.
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6
.
.
= 6 (−1)n (n ) (−1)n−1 ( n )
.
.
−(n
1
0
n
n−1
1)
6
6
.
.
6
.
.
n
n
n−1
n
n
6
0
(−1) (n )
(−1)
(n−1 )
.
.
−(1 )
1
6
6
.
.
6
.
.
n
6
0
0
(−1)n (n
(−1)n−1 (n−1
)
.
.
−(n
n)
1)
6
4
.
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2
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∆
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n
0
0
0
0
.
.
.
···
···
···
···
···
0 ···
0 ···
1 ···
. .
. ..
.
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5

∆−n
1
0
1
0
0
1
 (n1 )
 n+1
 ( 2 )
(n1 )
 n+2
n+1
= (
(n1 )
3 ) ( 2 )
 n+3
n+2
n+1
 (
 4 ) ( 3 ) ( 2 )
..
..
..
.
.
.
0
0
0
1
0
0
0
0
(n1 ) 1
..
..
.
.
0
0
0
0
0
..
.
···
···
···
···
···
..
.









If a normed sequence space λ contains a sequence (bn ) with the property that for every x ∈ λ,
there is a unique sequence of scalars (αn ) such that
lim kx − (α0 b0 + α1 b1 + · · · + αn bn )k = 0,
P
then (bn ) is called a Schauder basis for λ. The series
αk P
bk which has the sum x is then called
the expansion of x with respect to (bn ) and written as x =
αk bk .
(1)
n→∞
2. The space lp (∆n )
Now we introduce an apparently new sequence space and denote it by lp (∆n ) like Kizmaz who
defined and studied l∞ (∆), c(∆) and c0 (∆).
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(2)
lp (∆n ) = {x ∈ ω : ∆n x ∈ lp }
(3)
kxklp (∆n ) = k∆n xklp
Trivially lp (∆) = bvp .
Theorem 2.1. lp (∆n ) is a Banach space.
Proof. Since it is a routine verification to show that lp (∆n ) is a normed space with the norm
defined by (3) and coordinate-wise addition and scalar multiplication we omit the details. To prove
the theorem, we show that every Chauchy sequence in lp (∆n ) has a limit. Suppose (x(m) )∞
m=0 is
a Chauchy sequence in lp (∆n ). So
(4)
(∀ε > 0)(∃N ∈ N)(∀r, s ≥ N )(k∆n x(r) − ∆n x(s) klp = kx(r) − x(s) klp (∆n ) < ε)
So the sequence (∆n x(m) )∞
m=0 in lp is Chauchy and since lp is Banach, there exists x ∈ lp such
that
(5)
k∆n x(m) − xklp → 0
as m → ∞
But x = (∆n )(∆n )−1 x, so k∆n x(m) − ∆n (∆n )−1 xklp = kx(m) − (∆n )−1 xklp (∆n ) → 0 as m → ∞.
Now, since (∆n )−1 x ∈ lp (∆n ) we are done.
Theorem 2.2. lp (∆n ) is isometrically isomorphic to lp .
Proof. Let
(6)
T : lp (∆n ) → lp
defined by T (x) = ∆n x. Since T is bijective and norm preserving, we are done.
Theorem 2.3. Except the case p = 2, the space lp (∆n ) is not an inner product space and hence
not a Hilbert space for 1 ≤ p < ∞.
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Proof. First we show that l2 (∆n ) is a Hilbert space. It suffices to show that l2 (∆n ) has an inner
product. Since
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(7)
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1
kxkl2 (∆n ) = k∆n xkl2 =< ∆n x, ∆n x > 2 ,
l2 (∆n ) is a Hilbert space. Now, we show that if p 6= 2, then lp (∆n ) is not Hilbert. Let
u = (∆n−1 )−1 (1, 2, 2, 2, · · · )
e = (∆n−1 )−1 (1, 0, 0, 0, · · · ).
1
Then kuklp (∆n ) = keklp (∆n ) = 2 p and ku + eklp (∆n ) = ku − eklp (∆n ) = 2. So the parallelogram
equality does not satisfy. Hence the space lp (∆n ) with p 6= 2 is not a Hilbert space.
Theorem 2.4. If 1 ≤ p < q < ∞, then lp (∆n ) ⊆ lq (∆n ) ⊆ l∞ (∆n ).
Proof. We only point out that if 1 ≤ p < q < ∞, then lp ⊆ lq ⊆ l∞ .
Theorem 2.5. lp ⊆ lp (∆) ⊆ lp (∆2 ) ⊆ lp (∆3 ) ⊆ · · ·
Proof. Since lp ⊆ bvp , it is trivial that lp ⊆ lp (∆).
∆ x ∈ lp ⊆ lp (∆). So
Now, if x ∈ lp (∆n ), then
n
∆n x ∈ lp (∆) ⇒ ∆(∆n x) ∈ lp ⇒ ∆n+1 x ∈ lp ⇒ x ∈ lp (∆n+1 ).
Theorem 2.6. kxklp (∆n ) ≤ 2n kxklp
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Proof. Since kxklp (∆) = kxkbvp ≤ 2kxklp , kxklp (∆2 ) ≤ 2kxklp (∆) ≤ 2 · 2 · kxklp = 22 kxklp . Now
by induction, we are done.
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3. Schauder basis for space lp (∆n )
Suppose ek is a sequence whose only nonzero term is 1 in the (k + 1)th place. The sequence
n
k
(∆−n ek )∞
lp . We assert that this
k=0 is a sequence of elements of lp (∆ ) since for all k ∈ N, e ∈
Pm
n
n
[m]
sequence is a Schauder basis for lp (∆ ). Suppose x ∈ lp (∆ ), x
= k=0 (∆n x)k (∆−n ek ) =
Pm
k=0
(8)
∆−n ((∆n x)k ek ). Then since x ∈ lp (∆n ), we have ∆n x ∈ lp such that
∞
X
! p1
n
p
=s<∞
|(∆ x)i |
i=0
(9)
(10)
⇒(∀ε > 0)(∃m0 ∈ N)
! p1
∞
X
n
p
|(∆ x)i |
<
i=m
n
∞
X
(∆n x)k ek −
k=0
=k
m
X
(∆n x)k ek klp
k=0
∞
X
n
k
(∆ x)k e klp =
k=m+1
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for all m ≥ m0
⇒kx − x[m] klp (∆n ) = k∆ x − ∆n x[m] klp
=k
(11)
ε
2
≤
∞
X
k=m0
! p1
|∆
n
x|pk
k=m+1
! p1
|∆
∞
X
n
x|pk
<
ε
2
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P∞
P∞
n
−n k
−n
k
So x =
e )P=
((∆n x)
of this
k e ). Now, we show the uniqueness
k=0 (∆ x)k (∆
k=0 ∆
P
P∞
∞
∞
−n k
−n
k
n
k
representation. Suppose
x
=
µ
(∆
e
)
=
∆
(µ
e
),
so
∆
x
=
µ
e
k
k=0 k
k=0
k=0 k . On
P∞
n
n
k
n
the other hand ∆ x = k=0 (∆ x)k e . Hence µk = (∆ x)k , for all k ∈ N. So this representation
is unique.
4. Continuous dual of lp (∆n )
Sequence space bvp is lp (∆) so lp (∆n ) is an extension of this space. In [1] the continuous dual of
bvp was studied. The idea was wrong. We showed a counter example and then corrected it in [4].
Now, we introduce the continuous dual of lp (∆n ).
Suppose 1 ≤ q < ∞ and let


q  q1



∞ X
∞


X
(n) 
dnq = a ∈ ω : kakdnq = D(n) a = 
<
∞
(12)
D
a
j
kj


lq


k=0 j=k


∞


X (n) dn∞ = a ∈ ω : kakdn∞ = D(n) a = sup Dkj aj < ∞ ,
(13)


l∞
k∈N j=k
where

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(14)
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D(n)




=



1
0
0
0
0
..
.
(n1 )
1
0
0
0
..
.
−1
n+2
n+3
n+4
(n+1
2 ) ( 3 ) ( 4 ) ( 5 )
n+1
n+2
n+3
n
( 2 ) ( 3 ) ( 4 )
(1 )
n+2
1
(n1 )
(n+1
2 ) ( 3 )
n
0
1
(1 )
(n+1
2 )
0
0
1
(n1 )
..
..
..
..
.
.
.
.
···
···
···
···
···
..
.









since D(n) is triangle, then D(n) exists. Trivially dnq and dn∞ are normed spaces with respect to
coordinate-wise addition and scalar multiplication. dnq and dn∞ are Banach spaces since if (x(m) )∞
m=0
is a Chauchy sequence in dnq , then
(15)
(∀ε > 0)(∃N ∈ N)(∀r, s > N )kD(n) (x(r) − x(s) )klq = kx(r) − x(s) kdnq < ε
so the sequence (D(n) (x(m) ))∞
m=0 is Chauchy in lq and since lq is Banach, there exists y in lq such
−1
−1
that kD(n) x(m) − yklq → 0 as m → ∞. But y = D(n) D(n) y, so kD(n) x(m) − D(n) D(n) yklq =
−1
−1
kx(m) − D(n) ykdnq → 0 as m → ∞. On the other hand D(n) y ∈ dnq . So dnq is Banach. In a
similar way dn∞ is Banach.
Theorem 4.1. l1 (∆n )∗ is isometrically isomorphic to dn∞ .
Proof. Let
T : l1 (∆n )∗ → dn∞
(16)
0
1
2
3
defined
· · ). Trivially T is linear and since
P∞by T fn = (f (e−n), fk (e ), f (e ), f (e ), ·P
∞
x = k=0 (∆ x)k (∆ e ) we have f (x) = k=0 (∆n x)k f (∆−n ek ). But
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(17)
n+2
n+3
∆−n ek = (0, 0, · · · , 0, 1, (n1 ), (n+1
2 ), ( 3 ), ( 4 ), · · · )
| {z }
k term
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k+2
k+3
= ek + (n1 )ek+1 + (n+1
+ (n+2
+ ···
2 )e
3 )e
so
(18)
f (x) =
∞
X
n
k+2
(∆ x)k · (f (ek ) + (n1 )f (ek+1 ) + (n+1
) + ···)
2 )f (e
k=0
If fj = f (ej ), then with respect to (14), we have
f (x) =
∞ h
X
i
(n)
(n)
(n)
(∆n x)k · (Dkk fk + Dk(k+1) fk+1 + Dk(k+2) fk+2 + · · · )
k=0
=
∞
X

(∆n x)k ·
k=0

∞
X
(n)
Dkj fj 
j=k
P∞
P∞
(n)
n
n
So |f (x)| ≤
j=k Dkj fj | = k(f0 , f1 , f2 , · · · )kd∞ · kxkl1 (∆n ) . So kf k ≤
k=0 |∆ x|k · supk∈N |
k(f0 , f1 , f2 , · · · )kdn∞ . So T is surjective. T is injective since T (f ) = 0 implies f = 0. Finally T is
norm preserving since
(19)
|f (x)| ≤
∞
X
k=0
∞
X
(n) |∆n x|k · sup Dkj fj = kxkl1 (∆n ) · kT f kdn∞
k∈N
j=k
So
kf k ≤ kT f kdn∞
(20)
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On the other hand,
(21)
∞
X
(n) D fj = |f (∆−n ek )| ≤ kf k · k∆−n ek kl
kj
n
1 (∆ )
= kf k for all k ∈ N
j=k
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So
(22)
∞
X
(n) kT f kdn∞ = sup Dkj fj ≤ kf k
k∈N
Quit
j=k
From (20) and (22), we have
kT f kdn∞ = kf k.
So T is norm preserving and it completes the proof.
Theorem 4.2. If 1 < p < ∞,
1 1
+ = 1, then lp (∆n )∗ is isometrically isomorphic to dnq .
p q
Proof. Let
T : lp (∆n )∗ → dnq
(23)
defined by T f = (f (e0 ), f (e1 ), f (e2 ), f (e3 ), · · · ). Trivially T is linear and (18) implies that
"
q  q1
# p1  ∞ ∞
X
∞
∞
∞
X
X
X
X
(n)
(n) 
n
n p

|f (x)| = [(∆ x)k ·
Dkj fj ] ≤
|∆ x|k
Dkj fj k=0
k=0 j=k
j=k
k=0
= kxklp (∆n ) · k(f0 , f1 , f2 , · · · )kdnq .
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The above computations show that T is surjective. Moreover T is injective since T f = 0 implies
f = 0. T is norm preserving since |f (x)| ≤ kxklp (∆n ) · k(f0 , f1 , f2 , · · · )kdnq = kxklp (∆n ) · kT f kdnq . So
kf k ≤ kT f kdnq .
(24)
(m)
On the other hand, let x(m) = (xk ) where
 X
∞
X
∞ (n) q−1

(n)  Dkj fj sgn
Dkj fj
n (m)
(25)
(∆ x )k =
j=k
j=k


0
0≤k≤m
k>m
Then x(m) ∈ lp (∆n ) since ∆n x(m) ∈ lp . So
X
X
∞
m
(m)
n (m)
−n k
n (m)
−n k
f (x ) = f
(∆ x )k · (∆ e ) = f
(∆ x )k · (∆ e )
k=0
=
=
=
m
X
k=0
(∆n x(m) )k f (∆−n ek ) =
k=0
m X
∞
X
k=0
q
(n) Dkj fj ≤ kf k · kx(m) klp (∆n ) .
k=0 j=k
m X
∞
X
j=k
n (m)
klp (∆n ) = k∆ x
klp =
X
∞
j=k
|∆n x(m) |pk
p1
|∆n x(m) |pk
k=0
k=0 j=k
j=k
1
X
m X
∞ (n) q p
=
Dkj fj Go back
k=0 j=k
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So
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=
X
m
X
X
p p1
m X
∞
∞ (n) p(q−1) (n)
=
Dkj fj Dkj fj sgn
I II
Close
(n)
Dkj fj
j=k
k=0
JJ J
∞
X
k=0 j=k
kx
(∆n x(m) )k
q−1 X
X
∞
∞
(n) (n)
(n)
Dkj fj sgn
Dkj fj
Dkj fj
So
(m)
m
X
1
X
X
m X
m X
∞ (n) q 1
∞ (n) q p
Dkj fj ≤ kf k ·
Dkj fj .
k=0 j=k
k=0 j=k
p1
So
(26)
1
X
m X
∞ (n) q q
Dkj fj kf k ≥
= kT f kdnq .
k=0 j=k
From (24) and (26), we have
kT f kdnq = kf k.
So T is norm preserving and this completes the proof.
5. Continuity of ∆n on some sequence spaces
Lemma 5.1. The matrix A = (ank ) gives rise to a bounded linear operator T ∈ B(l1 ) if and only
P∞
if the supremum of l1 norms of the columns of A is bounded. In fact, kAk(l1 ,l1 ) = supn k=0 |ank |.
Corollary 5.2. k∆n k(l1 ,l1 ) = 2n .
Lemma 5.3. The matrix A = (ank ) gives rise to a bounded linear operator T ∈ B(l∞P
) if and only
∞
if the supremum of l1 norms of the rows of A is bounded. In fact, kAk(l∞ ,l∞ ) = supk n=0 |ank |.
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Corollary 5.4. k∆n k(l∞ ,l∞ ) = 2n .
T
Lemma 5.5. Let 1 < p < ∞ and let A ∈ (l∞ , l∞ ) (l1 , l1 ). Then A ∈ (lp , lp ).
Corollary 5.6. For every integer n and 1 < p < ∞ holds ∆n ∈ B(lp ).
Proof. With respect to the matrix representation of ∆n and Lemma 5.1 and 5.3
T
∆ ∈ (l∞ , l∞ ) (l1 , l1 ) and so by Lemma 5.5, ∆n ∈ (lp , lp ).
n
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Theorem 5.7. ∆n ∈ B(lp (∆n )).
Proof. Suppose ∆n : lp → lp and x ∈ lp . Then by Corollary 5.6, there exists Mnp ∈ N such
that k∆n xklp ≤ Mnp kxklp . So if ∆n : lp (∆n ) → lp (∆n ) and x ∈ lp (∆n ), then k∆n xklp (∆n ) =
k∆n (∆n x)klp ≤ Mnp · k∆n xklp = Mnp · kxklp (∆n ) . So k∆n k(lp (∆n ),lp (∆n )) ≤ Mnp and it completes
the proof.
In [1, Theorem 3.2] claims that the norm of operator Delta is 2 i.e. ∆ is a bounded operator
on lp (∆) which confirms Theorem 5.7 in case n = 1.
Acknowledgment. We are grateful to the referee for his/her careful reviewing and suggestions. The first author is obliged to Dr. Bolbolian, Dr. Mohammadian and Mrs. Ramezani.
Also he likes to express his thanks to Mr. Davoodnezhad and Mrs. Sadeghi1 for supplying some
references.
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1. Akhmedov A. M. and Basar F., The Fine Spectra of the Difference Operator ∆ over the Sequense Space bvp (1 ≤
p < ∞), Acta Math. Sin. Eng. Ser. 23(10) (2007), 1757–1768.
2. Altay B., On the space of p-summable difference sequences of order m, (1 ≤ p < ∞), Stud. Sci. Math. Hungar.
43(4) (2006), 387–402.
3. Basar F. and Altay B., On the Space of Sequences of p-Bounded Variation and Related Matrix Mappings,
Ukrainian Math. J. 55(1) (2003), 136–147.
4. Imaninezhad M. and Miri M., The Dual Space of the Sequence Space bvp (1 ≤ p < ∞), Acta Math. Univ.
Comenianae 79(1)(2010), 143–149.
5. Goldberg S., Unbounded Linear Operators, Dover Publication Inc. New York, 1985.
6. Kizmaz H., On Certain Sequence Spaces, Canad. Math. Bull. 24(2)(1981), 169–176.
1Librarians of Faculty of Mathematical Sciences of Mashhad University
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7. Kreyszig E., Introductory Functional Analysis with Applications, John Wiley & Sons Inc. New York-ChichesterBrisbane-Toronto, 1978.
8. Maddox I. J., Elements of Functional Analysis, University Press, Cambridge, 1988.
9. Wilansky A., Summability through Functional Analysis, North-Holland Mathematics Studies, Amsterdam-New
York-Oxford, 1984.
M. Imaninezhad, Islamic Azad University-Ferdows Branch, Iran, e-mail: imani507@gmail.com
M. Miri, Dep. of Math., University of Birjand, Iran, e-mail: mmiri@birjand.ac.ir
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