Acta Math. Univ. Comenianae
Vol. LXXIX, 2(2010), pp. 273–280
273
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
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.
Suppose ∆ be the difference operator with matrix representation


1
0
0
0 0 0 ···
 −1 1
0
0 0 0 ··· 


 0 −1 1
0 0 0 ··· 


∆= 0
0 −1 1 0 0 · · · 


 0
0
0 −1 1 0 · · · 


..
..
..
..
.. .. . .
.
.
.
.
.
. .
∞
n
n−1
and suppose x = (xk )∞
x)
k=0 ∈ ω, then ∆x = (xk − xk−1 )k=0 and ∆ x = ∆(∆
for all n ≥ 2 where any x with negative index is zero. For every n ∈ N \ {0},
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 ).
274
M. IMANINEZHAD and M. MIRI
∆n has a triangle matrix representation, so it is invertible and
2
n
∆
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
.
.
.
.
.
.
.
.
.
.
.
.
.
.
6
.
.
.
.
.
.
.
6
6
6
.
.
6
.
.
= 6 (−1)n (n ) (−1)n−1 ( n )
.
.
−(n
1
0
n
n−1
1)
6
6
.
.
6
.
.
n
6
0
(−1)n (n
(−1)n−1 (n−1
)
.
.
−(n
1
n)
1)
6
6
.
.
6
.
.
n
n
n−1
n
6
0
0
(−1) (n )
(−1)
(n−1 )
.
.
−(n
1)
6
4
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.

∆−n




=



1
0
1
0
0
1
0
0
0
1
0
0
(n1 )
(n+1
0
(n1 )
2 )
n+2
n+1
( 3 ) ( 2 )
0
(n1 )
n+2
n+1
n
(n+3
)
(
)
(
)
1
)
(
1
4
3
2
..
..
..
..
..
.
.
.
.
.
0
0
0
0
0
..
.
···
···
···
···
···
..
.
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









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
(1)
lim kx − (α0 b0 + α1 b1 + · · · + αn bn )k = 0,
P
then (bn ) is called a Schauder basis for λ. The series
αk bk which has the
Psum x
is then called the expansion of x with respect to (bn ) and written as x =
αk bk .
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 (∆).
(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)
THE CONTINUOUS DUAL OF THE SEQUENCE SPACE lp (∆n )
275
So the sequence (∆n x(m) )∞
m=0 in lp is Chauchy and since lp is Banach, there exists
x ∈ lp such that
k∆n x(m) − xklp → 0
(5)
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 < ∞.
Proof. First we show that l2 (∆n ) is a Hilbert space. It suffices to show that
l2 (∆n ) has an inner product. Since
(7)
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 (∆). Now, if x ∈ lp (∆n ), then
∆n x ∈ lp ⊆ lp (∆). So
∆n x ∈ lp (∆) ⇒ ∆(∆n x) ∈ lp ⇒ ∆n+1 x ∈ lp ⇒ x ∈ lp (∆n+1 ).
¤
Theorem 2.6. kxklp (∆n ) ≤ 2n kxklp
Proof. Since kxklp (∆) = kxkbvp ≤ 2kxklp , kxklp (∆2 ) ≤ 2kxklp (∆) ≤ 2 · 2 · kxklp =
2 kxklp . Now by induction, we are done.
¤
2
276
M. IMANINEZHAD and M. MIRI
3. Schauder basis for space lp (∆n )
Suppose ek is a sequence whose only nonzero term is 1 in the (k + 1)th place.
n
The sequence (∆−n ek )∞
k=0 is a sequence of elements of lp (∆ ) since for all k ∈ N,
ek ∈ lp . We assert that this sequence is a Schauder basis for lp (∆n ). Suppose
Pm
Pm
x ∈ lp (∆n ), x[m] = k=0 (∆n x)k (∆−n ek ) = k=0 ∆−n ((∆n x)k ek ). Then since
x ∈ lp (∆n ), we have ∆n x ∈ lp such that
Ã
(8)
∞
X
! p1
n
p
=s<∞
|(∆ x)i |
i=0
(9)
(10)
Ã
! p1
∞
X
⇒(∀ε > 0)(∃m0 ∈ N)
n
|(∆ x)i |
p
<
i=m
n
ε
2
for all m ≥ m0
⇒kx − x[m] klp (∆n ) = k∆ x − ∆n x[m] klp
=k
∞
X
(∆n x)k ek −
k=0
(11)
≤
∞
X
k=m0
Ã
(∆n x)k ek klp =
k=m+1
Ã
(∆n x)k ek klp
k=0
∞
X
=k
m
X
|∆n x|pk
∞
X
! p1
|∆n x|pk
k=m+1
! p1
<
ε
2
P∞
((∆n x)k ek ). Now, wePshow the uniqueSo x = k=0 (∆n x)k (∆−n ek ) = k=0 ∆−nP
∞
∞
ek ) = k=0 ∆−n (µk ek ),
ness of this P
representation. Suppose x = k=0 µk (∆−nP
∞
∞
n
k
k
n
so ∆n x =
k=0 (∆ x)k e . Hence
k=0 µk e . On the other hand ∆ x =
n
µk = (∆ x)k , for all k ∈ N. So this representation is unique.
P∞
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) ¯
° (n) °
n
¯
¯


dq = a ∈ ω : kakdnq = °D a° =
D
a
(12)
<
∞
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
THE CONTINUOUS DUAL OF THE SEQUENCE SPACE lp (∆n )
where
(14)

D(n)




=



1
0
0
0
0
..
.
n+4
n+3
n+2
(n1 ) (n+1
2 ) ( 3 ) ( 4 ) ( 5 )
n+1
n+2
n+3
n
1
(1 )
( 2 ) ( 3 ) ( 4 )
n+2
0
1
(n1 )
(n+1
2 ) ( 3 )
n
0
0
1
(1 )
(n+1
2 )
0
0
0
1
(n1 )
..
..
..
..
..
.
.
.
.
.
···
···
···
···
···
..
.
277









−1
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
n
Banach spaces since if (x(m) )∞
m=0 is a Chauchy sequence in dq , 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
−1
exists y in lq such that kD(n) x(m) − yklq → 0 as m → ∞. But y = D(n) D(n) y,
−1
−1
so kD(n) x(m) − D(n) D(n) yklq = kx(m) − D(n) ykdnq → 0 as m → ∞. On the
−1
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
defined
), f (e1 ), f (e2 ), f (e3 ),P
· · · ). Trivially T is linear and since
P∞by T fn = (f (e
∞
−n k
x = k=0 (∆ x)k (∆ e ) we have f (x) = k=0 (∆n x)k f (∆−n ek ). But
n+2
n+3
∆−n ek = (0, 0, · · · , 0, 1, (n1 ), (n+1
2 ), ( 3 ), ( 4 ), · · · )
| {z }
(17)
k term
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
j
If fj = f (e ), then with respect to (14), we have
f (x) =
∞ h
i
X
(n)
(n)
(n)
(∆n x)k · (Dkk fk + Dk(k+1) fk+1 + Dk(k+2) fk+2 + · · · )
k=0
=
∞
X
k=0

(∆n x)k ·
∞
X

(n)
Dkj fj 
j=k
P∞
P∞
(n)
So |f (x)| ≤ k=0 |∆n x|k · supk∈N | j=k Dkj fj | = k(f0 , f1 , f2 , · · · )kdn∞ ·kxkl1 (∆n ) .
So kf k ≤ k(f0 , f1 , f2 , · · · )kdn∞ . So T is surjective. T is injective since T (f ) = 0
278
M. IMANINEZHAD and M. MIRI
implies f = 0. Finally T is norm preserving since
(19)
|f (x)| ≤
∞
X
∞
¯X
(n) ¯
|∆n x|k · sup ¯
Dkj fj ¯ = kxkl1 (∆n ) · kT f kdn∞
k∈N
k=0
j=k
So
kf k ≤ kT f kdn∞
(20)
On the other hand,
∞
¯X
(n) ¯
Dkj fj ¯ = |f (∆−n ek )| ≤ kf k · k∆−n ek kl1 (∆n ) = kf k for all k ∈ N
(21) ¯
j=k
So
∞
¯X
(n) ¯
Dkj fj ¯ ≤ kf k
kT f kdn∞ = sup ¯
(22)
k∈N
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 < ∞,
isomorphic to dnq .
¤
1
1
+
= 1, then lp (∆n )∗ is isometrically
p
q
Proof. Let
(23)
T : lp (∆n )∗ → dnq
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) ¯¯ 
¯
|f (x)| = ¯¯
[(∆n x)k ·
Dkj fj ]¯¯ ≤
|∆n x|pk 
D
f
kj j ¯
¯
¯k=0
¯
¯
k=0 ¯j=k
j=k
k=0
= kxklp (∆n ) · k(f0 , f1 , f2 , · · · )kdnq .
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
(24)
kf k ≤ kT f kdnq .
(m)
On the other hand, let x(m) = (xk ) where
 X
∞
¯ ∞ (n) ¯q−1
¡X

(n) ¢
 ¯
Dkj fj ¯ sgn
Dkj fj
n (m)
(25) (∆ x )k =
j=k
j=k


0
0≤k≤m
k>m
279
THE CONTINUOUS DUAL OF THE SEQUENCE SPACE lp (∆n )
Then x(m) ∈ lp (∆n ) since ∆n x(m) ∈ lp . So
µX
¶
µX
¶
∞
m
f (x(m) ) = f
(∆n x(m) )k · (∆−n ek ) = f
(∆n x(m) )k · (∆−n ek )
k=0
=
=
=
m
X
k=0
(∆n x(m) )k f (∆−n ek ) =
k=0
m ¯X
∞
X
¯
¯
¯
k=0 j=k
m ¯X
∞
X
¯
¯
¯
m
X
(∆n x(m) )k
k=0
∞
X
(n)
Dkj fj
j=k
¯q−1
µX
¶µ X
¶
∞
∞
(n) ¯
(n)
(n)
Dkj fj ¯¯ sgn
Dkj fj
Dkj fj
j=k
¯q
(n) ¯¯
Dkj fj ¯ ≤ kf k · kx(m) klp (∆n ) .
j=k
k=0 j=k
So
kx(m) klp (∆n ) = k∆n x(m) klp =
µX
∞
|∆n x(m) |pk
¶ p1
=
µX
m
|∆n x(m) |pk
¶ p1
k=0
k=0
¯
¯ µX
µX
¶¯p ¶ p1
m ¯X
∞
¯ ∞ (n) ¯p(q−1) ¯
¯
(n)
¯
¯
¯
=
Dkj fj ¯
Dkj fj ¯¯
¯
¯sgn
k=0 j=k
¯ ¶1
µX
m ¯X
¯ ∞ (n) ¯q p
¯
=
Dkj fj ¯¯
¯
j=k
k=0 j=k
So
¯ ¶
¯ ¶1
µX
µX
m ¯X
m ¯X
¯ ∞ (n) ¯q 1
¯ ∞ (n) ¯q p
¯
¯
¯
¯
.
D
f
≤
kf
k
·
D
f
kj j ¯
kj j ¯
¯
¯
k=0 j=k
So
(26)
kf k ≥
k=0 j=k
¯ ¶1
µX
m ¯X
¯ ∞ (n) ¯q q
¯
¯
= kT f kdnq .
D
f
kj j ¯
¯
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 if the supremum of l1 norms of the columns of A is bounded.
P∞
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∞ ) if and only if the supremum of l1 norms of the rows of A is bounded.
P∞
In fact, kAk(l∞ ,l∞ ) = supk n=0 |ank |.
280
M. IMANINEZHAD and M. MIRI
Corollary 5.4. k∆n k(l∞ ,l∞ ) = 2n .
Lemma 5.5. Let 1 < p < ∞ and let A ∈ (l∞ , l∞ )
T
(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
T
n
5.3 ∆ ∈ (l∞ , l∞ ) (l1 , l1 ) and so by Lemma 5.5, ∆n ∈ (lp , lp ).
¤
Theorem 5.7. ∆n ∈ B(lp (∆n )).
Proof. Suppose ∆n : lp → lp and x ∈ lp . Then by Corollary 5.6, there exists
∈ 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.
¤
Mnp
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.
References
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.
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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
1Librarians of Faculty of Mathematical Sciences of Mashhad University