New
v q
-difference operator and topological
features
Abdulkadir KARAKAŞ*, Mahir Salih Abdulrahman ASSAFI
Siirt University Faculty of Arts and Sciences, Department of Mathematics, Kezer Campus, Siirt.
Geliş Tarihi (Received Date): 20.04.2020 Kabul Tarihi (Accepted Date): 14.10.2020 Abstract
We extented v by using difference operator . We generated the difference sequence vq space l p( vq) and investigated some of their properties. We showed that, if l p( vq) is supplied with an proper norm
,
.
q v
p then it will be a Banach space. We further more showed that, the sequence spaces
(
( ), . , v)
q
p p
v q
l and
(
lp, . p)
are linearly isometric. Atthe end of this studies, it was shown that ( qv) ( , v)
p p q
l l . The family of the Orlicz functions is coincides the − condition. 2
Keywords: Difference sequence spaces, isometric sequence spaces, sequence spaces.
Yeni
vq-fark operatörü ve topolojik özellikleri
Özv q
fark operatörünü kullanarak v
’yi genişlettik. ( )v p q
l fark dizi uzayını oluşturduk ve bazı topolojik özelliklerini inceledik. Eğer ( )v
p q l uygun bir , . q v p normu verilirse bunun bir Banach uzayı olacağını gösterdik. Ayrıca
(
( ), . , v)
q p p v q l ve
(
p, .)
p l diziuzaylarının lineer izometrik olduklarını gösterdik. Çalışmanın sonunda ise
( qv) ( , v)
p p q
l l olduğu gösterildi. Orlicz fonksiyonlarının ailesi , − şartı ile 2 örtüşmektedir.
* Abdulkadir KARAKAŞ, kadirkarakas21@hotmail.com, https://orcid.org/0000-0002-0630-8802
Anahtar kelimeler: Fark dizi uzayları, izometrik dizi uzayları, dizi uzayları.
1. Introduction
Let c l, and c 0 be the Banach spaces of convergent, bounded and null sequences 1
( k)
u= u respectively with complex terms, normed by
sup k
k
u = u ,
where k .
Kızmaz [1] presented the difference sequence spaces,
( ) ( k) :
U = u= u u U
for U =c, and ,l c 0 where
1
( k) ( k k )
u u u u +
= = − .
We have the norm for these Banach spaces as:
1
u = u + u .
Çolak and Et [2] have extended the spaces U ( ) to the U ( v) for U=c l, and c0. Let
U be any sequence spaces and defined
( v) ( ) : v k
U = u= u u U
where v and = vu
(
( v−1)uk)
for all k and prove that0
( v), ( v) and ( v)
c l c are Banach spaces with the norm
0 ( 1) , v v t k k t t v t u u + = =
− 1 v i v v i u u u = =
+ .Karakaş et al. [3] have defined the sequence spaces c(q), (l q) and (c0 . He also q) presented
1
( ) ( )
qu q ku quk uk+
= = −
for q . Karakaş et al. [4] have presented
v v
for U =c l, and c 0, where q v , . They showed that the spaces ( v) q U are Banach spaces by: 1 , v q v q i i v u u u = =
+ where 1 1 1 ( ) ( ) v v v v qu quk q q uk q uk − − + = = − and 0 ( ) ( 1) v v v t v t q q k k t t v q t u u −u + = = =
− .Recently, Peralta [5] has studied l and investigated the topological features of this p( v) space. In this work, we choose p [1, ) . By , we denote the space of all sequences
( )k
u= u , for u k and all kN. Taken u, describe
1 1 p p p k k u u = =
and let
( ) :
p k p l = u= u u .The linear operator q:
v
→ is presented recursively as the composition
1
q q
v v
q −
= for v and q 2 . It is obvious that for u and v we have the 1 following Binomial representation
0 ( 1) v v t v t q k k t t v q t u −u + = =
− for all k .Let v and define the sequence space l p( vq) by
( ) ( ) : p k v v q q p l = u= u u l .The sequence spaces are Banach spaces normed by
1 1 , p v q v p p v i q p i p u u = = +
(1.1)For Euler difference sequence spaces and sequence spaces generated by a sequence of Orlicz functions, the reader can consult Altay and Polat [6], Altay and Başar [7] and Qamaruddin and Saifi [8], respectively.
2. Main results
Theorem 2.1. The sequence space (l p vq) is a Banach space with the norm
, vq p . Proof: Let ( ( ))
( )
( ( )n ) k nu = u is a Cauchy sequence in l p( vq). Thus, for we may 0 find a positive integer N such that
, ( ) ( ) v q n r p u u −
whenever ,n r . In other words, we have N
( ) ( ) ( ) ( ) 1 1 p p v p v v q q n r n r i i p i u u u u = + − −
, for n r, N. Since ( ) ( ) ( ) ( ) , vq r n i i p n r u u u u − − for i=1, 2,3,...,v and ( ) ( ) ( ) ( ) , vq v v q q p n r n r p u u u u − − . Therefore, ( ) (uin )and ( ) )(vqun are Cauchy sequences in ℂ and lp, respectively. The
completeness of the spaces ℂ and lp show the existence of elements y i ,
1, 2,3,...,
i= v, and z=(zk)lp such that
( ) lim i i 0 n n u −y = (2.1) for i=1, 2,3,...,v and ( ) lim v n 0. p q n u −z = (2.2) Since
( ) ( ) v v q q n p n k zk z u u − − we get ( ) 0 n k k v qu z − →
as n → for all k by equation (2.2). We obtain a recursive formula for lim ( ), 1,
n n v i u+ i as n → . We have 1 1 ( ) ( ) ( ) 1 1 0 ( 1) ( 1) t v v n v n t v t n v q t v q t u u u + − − + = − = −
− and so 1 1 1 1 0 ( ) : lim ( 1) ( 1) v v t v t v v n t n v i v q t w u z y − + + − + = = = − − −
Assume that wv+1,...,wv k+ −1,1 have been established. Where k v,
( ) :lim n , 1, 2,..., 1. i n v v i w+ u + i= k−
Using these, we acquire, for 1 k v
( 0 1 1 ) ( 1) : lim ( 1) ( 1) v k t v t k t k t v v k k n v k t k t v t v k t t n v k v q t q z y w u v w − − + = + − − + − + − + = + − − = − − − =
On the other side, for k we get v 1 ( ) ( ) ( ) 0 ( 1) ( 1) t k v v n v n t v t n v k q k t v q t u u u+ − − + = − = −
− . So that ) 1 ( 0 : lim ( 1) ( 1) v v t v t v k v k t t n k k n v q t w+ u + z w+ − − = = = − − −
.Let w=
(
y1,...,y wv, v+1,wv+2,... .)
We assert that ( ) v p qwl , that is, . First, show vqw lp
1 1 1 1 0 1 1 1 1 1 0 0 1 ( ) ( 1) ( 1) ( 1) ( 1) v v t v t v q t v t v v t v t t v t t t t t v q t v v q q t t w y w y z y z − − + + = − − − − + + = = = − + − = − + − − =
Also, for k=2,3,...,v. We get
1 0 1 ( ) ( 1) ( 1) ( 1) v k v v t v t t v t v q k t k t k v k t t v k k v v q q t t w y w w z − − − − + + + = = − + = − + − + − =
Similarly, for k we acquire v 1 0 ( ) ( 1) ( 1) . v v t v t v q k t k v k t k v q t w w w z − − + + = = − + − =
Thus we have presented that vqw= z lp. It remains to prove that
( ) , v 0 q p n u w − → as n → Then, we obtain , ( ) ( ) ( ) 1 ( ) ( ) 1 lim lim lim lim 0. v q p v p n p v n v k k q q n n k v p p n v n k k q p n n n k u y u w u y u z u w = = = − + − = − − − + =
This is proof of the theorem.
Theorem 2.2. The sequence spaces
(
( ), . , v)
qp p
v q
l and
(
lp, . p)
are linearly isometric. Proof: Take in to consideration the map : ( v) pq p l l T → given by Ty u= , where ) ( k) p( vq y= y l and u=( )uk with , if 1 ; , if . k k v q k v y k v u y − k v =
The linearity of the difference operator Δ refers the linearity of T . If ( v) p q yl and Ty=u, then 1 1 1 1 , vq . v p p p p v k q k v p p k k v v p p v k q k k k p p Ty u y y y y y − = = + = = = = + = + =
This demonstrates that T is well-defined and it is also norm preserving. We presented that T is one-to-one and onto. Assume that Ty =0.
Then, we obtain 0 v qyk = for all k 1, (2.3) 1 2 ... v 0. y = y = = y = (2.4)
We show that the difference equation (2.3) with initial conditions (2.4) refers that 0
k
y = for all k 1,, that is, y =
(
0, 0,...)
. Therefore, T is one-to-one.Assume that u=(uk)lp. Describe the sequence y=(yk) as follows. Let yk = for uk
, 1, 2,..., .
v
k v q k
u + = u k = v
The succeeding terms of the sequence y is then showed recursively by
1 1 1 1 0 ( 1) ( 1) v v t v t v v t t v q t y u u − − + + + = = − − −
0 1 1 ( 1) ( 1) , 1 ( 1) v k t v t v k t k t v v k k v k t k t t v t t v q t v q v k t u u y k v y − − + + = + − − + − + = − + − − = − − −
and 1 0 ( 1) ( 1) , . v v t v t v k v k t k t v q t y u y k v − − + + + = = − − −
Utilizing a similar argument as in the proof of the previous theorem, we prove that
v
qyk uk v+
=
Thus, we obtain 1 1 . p p v v q p q k k p k v k p p y y u u = + = = = =
So that ylp( . Since T is onto, ( )qv) l p vq and lp are linearly isometric.
Definition 2.3. An Orlicz function is a continuous, convex function and nondecreasing
)
)
: 0, 0,
M → such that M z =( ) 0, if and only if z = , 0 M u ( ) 0, and
( ) as
M u → u→ . M is said to fulfil − condition if there exists a positive 2 constant K such that M(2 )z KM z( ) for all z . Let 0 =(Mk) be a sequence of Orlicz functions meeting the − condition [9]. An Orlicz function M has been defined 2 in [10] also see [11] for a more general representation in thise direction in the following from:
0
( ) ( )
u
M u =
p t dtwhere p , know as the kernel of M , is right-differentable for t0, ( )p t 0, (0)p =0
for t0, p is nondecreasing, and t→ , ( )p t → .
Lindenstrauss and Tzafriri [12] have utilized the view of Orlicz function to find the sequence space,
(
)
1 ) ( ) : ,for some 0 , ( k p p k k k u u M u l = = =
which is a Banach Spaces with respect to the norm
(
)
1 ( )k inf 0 : k k 1 . k u M u = =
The space l( )is closely related to space lp, which is an Orlicz sequence space with ( ) p, for 1 .
M u = u p
Describe the sequence spaces as:
(
)
1 ) ( ) : ,for some 0 , ( k p p k k k u u M u l = = =
) ( ) : )
( , vq vq p(
p u uk
l = = ul .
Theorem 2.4. Let =(Mk)be a sequence of Orlicz functions fulfil the − condition. 2 If
(
)
1 p k k k M u =
(2.5)for all t,0 then lp( vq) lp( ,vq).
Proof: Assume that condition (2.5) exists and let u=(uk) . Then, we get lp( vq)
1 . p v q k k u =
(2.6) The convergence of 1 p v q k k u =
implies that lim vq k 0. k u =Thus, we can find n such that vquk for all k N1 . Let
1 1
max v ,... v ,1 .
q q N
K = u u −
Then vquk for all k . For K , utilizing the monotonicity of 0 M , we get k
(
v)
(
)
k q k k
M u M K for all k . This inequality shows that
(
)
(
)
1 1 . p p v k q k k k k M u M K = =
This estimate proves that vqu lp( ) that is, ulp( , By equation (2.5) vq). Therefore, the inclusion (lp qv) lp( , holds. vq)
3. Results and discussion
Peralta [5] studied l p( vq) and checked the topological properties of this space. Later Karakaş et al. [4] defined difference operator v
q
. We used Peralta' s [5] studies and extented it by used the generalized difference operator . We generated the difference vq sequence space l p( vq) and
,
.
q v
p , and investigated some of their properties. We showed
that, if l p( vq) is equipped with an appropriate norm
,
.
q v
p is a Banach space. We
further more showed that, the sequence spaces
(
( ), . , v)
q p p v q l and(
p, .)
p l are linearlyisometric. It is shown that lp( qv) lp( , . Where vq) a family of Orlicz functions, is coincides the − condition2 .
References
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