The continuous dual of the sequence space $l_p(Delta^n)$, $(1leq p leq infty,ninmathbb{N})$

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 whosemth order differ-ences 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. Hun-gar. 43(4)(2006), 387–402]. Following Altay [2], we have found the continuous dual of the spacesl1(∆n_{) andl}

P(∆n). We have also determined the norm of the operator ∆n_{acting froml1to itself and froml}

∞to itself, and proved that ∆nis a bounded linear operator on the spacelp(∆n).

1. Preliminaries, Definitions and Notations

Letω denote the space of all complex-valued sequences, i.e. ω =CN

where N₌

{0,1,2,3, . . .}. Any vector subspace ofω which contains φ, the set of all finitely non-zero sequences, is called a sequence space. The continuous dual of a sequence spaceλwhich is denoted byλ∗ _{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 · · ·

..

. ... ... ... ... ... . ..



       

and supposex= (xk)∞

k=0∈ω, then ∆x= (xk−xk−1)∞k=0 and ∆nx= ∆(∆n−1x)

for all n ≥ 2 where any x with negative index is zero. For every n ∈ N_{\ {0},}

Received February 12, 2010.

2000Mathematics Subject Classification. Primary 46B10; Secondary 46B45.

∆n _{has a triangle matrix representation, so it is invertible and} ∆n = 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4

1 0 0 0 0 0 0 0 · · ·

−(n

1) 1 0 0 0 0 0 0 · · ·

(n

2) −(n1) 1 0 0 0 0 0 · · · −(n

3) (n2) −(n1) 1 0 0 0 0 · · ·

. . . . . . . . . . . . . . . . . . . . . . . . · · ·

(−1)n₍n n) (−1)

n−1

(n n−1)

. . .

. .

. −(n

1) 1 0 0 · · ·

0 (−1)n₍n

n) (−1) n−1

(n n−1)

. . .

. . . −(n

1) 1 0 · · ·

0 0 (−1)n₍n

n) (−1) n−1

(n n−1)

. . .

. . . −(n

1) 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 0 0 0 0 · · ·

(n

1) 1 0 0 0 0 · · ·

(n+1

2 ) (n1) 1 0 0 0 · · ·

(n+2

3 ) (n+12 ) (n1) 1 0 0 · · ·

(n+34 ) ( n+2

3 ) (

n+1

2 ) (n1) 1 0 · · ·

.. . ... ... ... ... ... . ..         

If a normed sequence spaceλcontains a sequence (bn) with the property that for everyx∈λ, there is a unique sequence of scalars (αn) such that

lim

n→∞kx−(α0b0+α1b1+· · ·+αnbn)k= 0,

(1)

then (bn) is called a Schauder basis forλ. The seriesPαkbk which has the sumx

is then called the expansion ofxwith respect to (bn) and written asx=Pαkbk.

2. The space lp(∆n₎

Now we introduce an apparently new sequence space and denote it bylp(∆n_{) like}

Kizmaz who defined and studiedl∞(∆),c(∆) andc0(∆).

lp(∆n_{) ={x∈ω: ∆}n_x∈lp} (2)

kxklp(∆n₎=k∆ n_xk

lp (3)

Triviallylp(∆) =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 multipli-cation 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)_kl p=kx

(r)₋

So the sequence (∆n_x(m)₎∞

m=0inlp is Chauchy and sincelpis Banach, there exists x∈lp such that

k∆nx(m)−xklp→0 as m→ ∞ (5)

Butx= (∆n_)(∆n₎−1_{x, sok∆}n_x(m)_−∆n_(∆n₎−1_xk lp=kx

(m)_−(∆n₎−1_xk

lp(∆n₎→0

asm→ ∞. Now, since (∆n₎−1_x∈lp(∆n_{) we are done. ¤}

Theorem 2.2. lp(∆n_{)is isometrically isomorphic tolp.}

Proof. Let

T :lp(∆n_)→ lp (6)

defined byT(x) = ∆n_{x. SinceTis bijective and norm preserving, we are done. ¤}

Theorem 2.3. Except the casep= 2, the spacelp(∆n_{)is not an inner product} space and hence not a Hilbert space for1≤p <∞.

Proof. First we show that l2(∆n) is a Hilbert space. It suffices to show that l2(∆n) has an inner product. Since

kxkl2(∆n)=k∆

n

xkl2 =<∆

n x,∆n

x >12,

(7)

l2(∆n) is a Hilbert space. Now, we show that ifp6= 2, thenlp(∆n) is not Hilbert.

Let

u= (∆n−1₎−1_{(1,2,2,2,· · ·)} e= (∆n−1₎−1_{(1,0,0,0,· · ·).}

Thenkuklp(∆n₎=kek_lp(∆n₎= 2

1

p andku+ek

lp(∆n₎=ku−ek_lp(∆n₎= 2. So the

parallelogram equality does not satisfy. Hence the spacelp(∆n_{) withp6= 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 <∞, thenlp⊆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₎≤2nkxklp

Proof. Sincekxklp(∆)=kxkbvp≤2kxklp,kxklp(∆2)≤2kxklp(∆)≤2·2· kxklp =

22_kxk

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_ek₎∞

k=0 is a sequence of elements oflp(∆n) since for all k∈N, ek _{∈ lp. We assert that this sequence is a Schauder basis for lp(∆}n_{). Suppose} x∈ lp(∆n_{), x}[m] ₌Pm

k=0(∆ n_x)

k(∆−nek) = P m k=0∆

−n_((∆n_x)

kek). Then since x∈lp(∆n_{), we have ∆}n_{x∈lpsuch that}

̰

X

i=0

|(∆n x)i|p

!1

p

=s <∞ (8)

⇒(∀ε >0)(∃m0∈N) Ã∞

X

i=m

|(∆nx)i|p !1

p <ε

2 for allm≥m0 (9)

⇒kx−x[m]klp(∆n₎=k∆ n

x−∆nx[m]klp

(10)

=k ∞

X

k=0

(∆n

x)kek−

m X

k=0

(∆n

x)kekklp

=k ∞

X

k=m+1

(∆n

x)kekklp=

à ∞

X

k=m+1

|∆n x|p_k

!1

p

(11)

≤

à ∞

X

k=m0 |∆n

x|p_k

!1

p < ε

2

Sox=P∞k=0(∆ n_x)

k(∆−nek) =P

∞

k=0∆

−n_((∆n_x)

kek). Now, we show the

unique-ness of this representation. Supposex =P∞

k=0µk(∆

−n_ek_{) =} P∞

k=0∆

−n_(µkek_),

so ∆n_{x =} P∞

k=0µkek. On the other hand ∆nx = P∞

k=0(∆nx)kek. Hence µk= (∆n_x)

k, for allk∈N. So this representation is unique.

4. Continuous dual oflp(∆n₎

Sequence space bvp is lp(∆) so lp(∆n_{) is an extension of this space. In [1] the}

continuous dual ofbvp 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

dn q =

  

 

a∈ω:kakdn q =

° ° °D(n)a

° ° °_l

q

=





∞

X

k=0 ¯ ¯ ¯ ¯ ¯ ¯

∞

X

j=k D(_kjn)aj

¯ ¯ ¯ ¯ ¯ ¯

q



1

q

<∞

  

 

(12)

dn∞=

 



a∈ω:kakdn

∞ =

° ° °D(n)a

° ° °

l∞ = sup

k∈N

¯ ¯ ¯ ¯ ¯ ¯

∞

X

j=k D_kj(n)aj

¯ ¯ ¯ ¯ ¯ ¯

<∞

 

 ,

where

D(n)₌ 

       

1 (n

1) (n+12 ) ( n+2

3 ) (

n+3

4 ) (

n+4 5 ) · · ·

0 1 (n

1) (

n+1

2 ) (

n+2

3 ) (

n+3 4 ) · · ·

0 0 1 (n

1) (n+12 ) ( n+2

3 ) · · ·

0 0 0 1 (n

1) (

n+1 2 ) · · ·

0 0 0 0 1 (n

1) · · ·

..

. ... ... ... ... ... . ..



        (14)

sinceD(n) _{is triangle, thenD}(n)−1

exists. Triviallydn

q anddn∞are normed spaces with respect to coordinate-wise addition and scalar multiplication. dn

q anddn∞are Banach spaces since if (x(m)₎∞

m=0 is a Chauchy sequence indnq, then

(∀ε >0)(∃N ∈N_{)(∀r, s > N)kD}(n)_(x(r)_−x(s)_)kl q =kx

(r)_−x(s)_k dn

q < ε (15)

so the sequence (D(n)_(x(m)₎₎∞

m=0 is Chauchy in lq and since lq is Banach, there

existsy in lq such that kD(n)_x(m)_−yk

lq →0 asm→ ∞. But y=D

(n)_D(n)−1

y, so kD(n)_x(m)_−D(n)_D(n)−1

yklq =kx

(m)_−D(n)−1

ykdn

q → 0 asm → ∞. On the

other handD(n)−1

y∈dn

q. Sodnq is Banach. In a similar waydn∞ is Banach.

Theorem 4.1. l1(∆n)∗ is isometrically isomorphic to dn∞.

Proof. Let

T :l1(∆n)∗→dn∞

(16)

defined by T f = (f(e0_{), f(e}1_{), f(e}2_{), f(e}3_{),· · ·). Trivially T is linear and since} x=P∞k=0(∆nx)k(∆−nek) we havef(x) =P

∞

k=0(∆nx)kf(∆

−n_ek_{). But}

∆−n

ek = (0,0,· · ·,0

| {z }

kterm ,1,(n

1),(n+12 ),( n+2

3 ),( n+3

4 ),· · ·)

=ek+ (n1)e

k+1_{+ (}n+1 2 )e

k+2_{+ (}n+2 3 )e

k+3_{+· · ·} (17)

so

f(x) = ∞

X

k=0 £

(∆n

x)k·(f(ek) + (n1)f(ek+1) + (n+12 )f(e

k+2_{) +· · ·)}¤ (18)

Iffj=f(ej_{), then with respect to (14), we have}

f(x) = ∞

X

k=0 h

(∆n

x)k·(D (n) kkfk+D

(n)

k(k+1)fk+1+D (n)

k(k+2)fk+2+· · ·) i

= ∞

X

k=0 

(∆nx)k·

∞

X

j=k D(_kjn)fj





So|f(x)| ≤P∞

k=0|∆nx|k·supk∈N|

P∞

j=kD (n)

kj fj|=k(f0, f1, f2,· · ·)kdn

∞·kxkl1(∆n). Sokfk ≤ k(f0, f1, f2,· · ·)kdn

impliesf = 0. FinallyT is norm preserving since

|f(x)| ≤ ∞

X

k=0

|∆n_x| k·sup

k∈N

¯ ¯

∞

X

j=k

D_kj(n)fj¯¯=kxkl1(∆n)· kT fkdn∞

(19)

So

kfk ≤ kT fkdn

∞

(20)

On the other hand,

¯ ¯

∞

X

j=k

D(_kjn)fj¯¯=|f(∆−n_ek_{)| ≤ kfk · k∆}−n_ek_k

l1(∆n₎=kfk for allk∈N (21)

So

kT fkdn

∞= sup

k∈N

¯ ¯

∞

X

j=k

D(_kjn)fj¯¯≤ kfk

(22)

From (20) and (22), we have

kT fkdn

∞ =kfk.

SoT is norm preserving and it completes the proof. ¤

Theorem 4.2. If 1 < p < ∞, 1 p +

1

q = 1, then lp(∆ n₎∗

is isometrically

isomorphic todn q.

Proof. Let

T :lp(∆n₎∗ →dn

q (23)

defined by T f = (f(e0), f(e1), f(e2), f(e3),· · ·). Trivially T is linear and (18) implies that

|f(x)|=

¯ ¯ ¯ ¯ ¯ ¯

∞

X

k=0

[(∆nx)k·

∞

X

j=k

D(_kjn)fj]

¯ ¯ ¯ ¯ ¯ ¯

≤

"∞

X

k=0

|∆nx|pk #1

p 



∞

X

k=0 ¯ ¯ ¯ ¯ ¯ ¯

∞

X

j=k D(_kjn)fj

¯ ¯ ¯ ¯ ¯ ¯

q



1

q

=kxklp(∆n₎· k(f₀, f₁, f₂,· · ·)kdn q.

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,· · ·)kdn

q =kxklp(∆n)· kT fkd n q. So

kfk ≤ kT fkdn q. (24)

On the other hand, letx(m)_{= (x}(m) k ) where

(∆n

x(m))k =   

  ¯ ¯

∞

X

j=k

D_kj(n)fj¯¯q−1sgn¡ ∞

X

j=k

D(_kjn)fj¢ 0≤k≤m

0 k > m

Thenx(m)_∈lp(∆n_{) since ∆}n_x(m)_{∈lp. So}

f(x(m)) =f µ ∞

X

k=0

(∆n

x(m))k·(∆

−n ek)

¶ =f µ m X k=0 (∆n

x(m))k·(∆

−n ek)

¶

=

m X

k=0

(∆nx(m))kf(∆−nek) = m X

k=0

(∆nx(m))k

∞

X

j=k D_kj(n)fj

= m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D(_kjn)fj

¯ ¯ ¯ ¯

q−1

sgn

µ ∞

X

j=k D(_kjn)fj

¶µ ∞

X

j=k D(_kjn)fj

¶ = m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D(_kjn)fj

¯ ¯ ¯ ¯ q

≤ kfk · kx(m)_k lp(∆n₎.

So

kx(m)klp(∆n₎=k∆ n

x(m)klp= µ ∞

X

k=0

|∆n x(m)|p_k

¶1 p = µ m X k=0 |∆n x(m)|p_k

¶1 p = µ m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D_kj(n)fj

¯ ¯ ¯ ¯

p(q−1)¯

¯ ¯ ¯sgn

µ ∞

X

j=k D(_kjn)fj

¶¯ ¯ ¯ ¯

p¶1

p = µ m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D_kj(n)fj

¯ ¯ ¯ ¯

q¶1

p So µ m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D(_kjn)fj

¯ ¯ ¯ ¯

q¶1

≤ kfk ·

µ m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D(_kjn)fj

¯ ¯ ¯ ¯

q¶1

p .

So

kfk ≥

µ m X k=0 ¯ ¯ ¯ ¯ ∞ X

j=k D_kj(n)fj

¯ ¯ ¯ ¯

q¶1

q

=kT fkdnq. (26)

From (24) and (26), we have

kT fkdn q =kfk.

SoT 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 ofl1norms of the columns ofAis bounded.

In fact, kAk(l1,l1)= supn

P∞

k=0|ank|. Corollary 5.2. k∆n_k

(l1,l1)= 2

n_.

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.

In fact, kAk(l∞,l∞)= supk

P∞

Corollary 5.4. k∆n_k

(l∞,l∞)= 2

n_.

Lemma 5.5. Let 1< p <∞ and letA∈(l∞, l∞)

T

(l1, l1). Then A∈(lp, lp).

Corollary 5.6. For every integern and1< p <∞holds∆n_∈B(lp).

Proof. With respect to the matrix representation of ∆n _{and Lemma 5.1 and}

5.3 ∆n _∈(l

∞, l∞)

T

(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} Mp

n ∈ N such that k∆ n_xk

lp ≤ M p

nkxklp. So if ∆

n _{: lp(∆}n_{) → lp(∆}n_{) and x ∈} lp(∆n_{), then k∆}n_xk

lp(∆n₎=k∆n(∆nx)k_l p ≤M

p

n· k∆nxklp =M p

n· kxklp(∆n₎. So

k∆n_k

(lp(∆n_),lp(∆n₎₎≤M_np 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 onlp(∆) which confirms Theorem 5.7 in casen= 1.

Acknowledgment. We are grateful to the referee for his/her careful review-ing and suggestions. The first author is obliged to Dr. Bolbolian, Dr. Mohamma-dian and Mrs. Ramezani. Also he likes to express his thanks to Mr. Davoodnezhad and Mrs. Sadeghi1for supplying some references.

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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