D O I: 1 0 .1 5 0 1 / C o m m u a 1 _ 0 0 0 0 0 0 0 7 8 7 IS S N 1 3 0 3 –5 9 9 1
VECTOR-VALUED CESÀRO SUMMABLE GENERALIZED LORENTZ SEQUENCE SPACE
O ¼GUZ O ¼GUR AND BIRSEN SA ¼GIR
Abstract. The main purpose of this paper is to introduce Cesàro summa-ble generalized Lorentz sequence space C1[d(v; p)]. We study some topologic
properties of this space and obtain some inclusion relations.
1. Introduction
Throughout this work, N; R and C denote the set of positive integers, real numbers and complex numbers, respectively. For some properties of sequences, we refer to [4; 8] :
For 1 p < 1, the Cesàro sequence space is de…ned by
Cesp= 8 < :x 2 w : 1 X j=1 1 j j X i=1 jx(i)j !p < 1 9 = ;; equipped with norm
kxk = 0 @X1 j=1 1 j j X i=1 jx(i)j !p1 A 1 p :
This space was …rst introduced by Shiue [14] : It is very useful in the theory of matrix operators and others. Later, many authors studied this space [see 1; 5; 11; 13] :
Let (E; k k) be a Banach space. The Lorentz sequence space l(p; q; E) (or lp;q(E))
for 1 p; q 1 is the collection of all sequences faig 2 c0(E) such that
kfaigkp;q= 8 < : 1 P i=1 iq=p 1 a (i) q 1=q f or 1 p < 1; 1 q < 1 supii1=p a (i) f or 1 p 1; q = 1
Received by the editors: March 18, 2016, Accepted: Aug 14, 2016. 2010 Mathematics Subject Classi…cation. 40A05, 40H05, 46A45, 46E30.
Key words and phrases. Lorentz sequence space, Cesàro summable, vector-valued space.
c 2 0 1 7 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis t ic s .
is …nite, where a (i) is non-increasing rearrangement of fkaikg (We can
inter-pret that the decreasing rearrangement a (i) is obtained by rearranging fkaikg
in decreasing order). This space was introduced by Miyazaki in [9] and examined comprehensively by Kato in [3] (see also [6; 7]):
A weight sequence v = fv(i)g is a positive decreasing sequence such that v(1) = 1; limi!1v(i) = 0 and limi!1V (i) = 1; where V (i) =
i
P
n=1v(n) for every i 2 N:
Popa [12] de…ned the generalized Lorentz sequence space d(v; p) for 0 < p < 1 as follows d(v; p) = 8 < :x = fxig 2 w : kxkv;p= sup 1 X i=1 x (i) p v(i) !1=p < 1 9 = ;; where ranges over all permutations of the positive integers and v = fv(i)g is a weight sequence. It is know that d(v; p) c0 and hence for each x 2 d(v; p) there
exists a non-increasing rearrangement fx g = fxig of x and
kxkv;p= 1 X n=1 jxij p v(i) !1 p (see [10; 12]).
Let (X; k k) be a Banach space and v = fv(k)g be a weight sequence. We introduce the vector-valued Cesáro summable generalized Lorentz sequence space C1[d(v; p)] for 0 < p < 1. The space C1[d(v; p)] is the collection of all X valued
0 sequences fxng (fxng 2 c0fXg) such that 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p
is …nite, where x (n) is non-increasing rearrangement of fkxnkg.
We shall need the following lemmas.
Lemma 1. (Hardy, Littlewood and Pólya [2]). Let faig1 i n and fbig1 i n be two
sequences of positive numbers. Then we have X i ai bi X i ai bi X i ai bi;
where faig is the non-increasing rearrangements of sequence faig1 i n and fbig
and f big are the non-increasing and non-decreasing rearrangements of sequence
fbig1 i n, respectively.
Lemma 2. (Kato [3]) Let nx( )i o be an X valued double sequence such that limi!1x( )i = 0 for each 2 N and let fxig be an X valued sequence such that
lim !1x( )i = xi (uniformly in i). Then limi!1xi= 0 and for each i 2 N
x (i) lim
!1 x
( )
(i) ;
where x (i) and
n
x( )(i) o
i are the non-increasing rearrangements of fkxikg
andn x( )i o
i, respectively.
2. MAIN RESULTS
Theorem 1. The space C1[d(v; p)] for 0 < p < 1 is a linear space over the …eld
K = R or C:
Proof. Let x; y 2 C1[d(v; p)]. Since v is non-increasing, the non-increasing
re-arrangements of v is itself. Thus, using the inequality P
i ai bi P i ai bi from Lemma 1, we have 1 X k=1 " 1 k k X n=1 x (n)+ y (n) #p v(k) 1 X k=1 " 1 k k X n=1 x (n) + y (n) #p v(k) D 1 X k=1 " 1 k k X n=1 x (n) #p v(k) +D 1 X k=1 " 1 k k X n=1 y (n) #p v(k) D 1 X k=1 " 1 k k X n=1 x (n) #p v(k) +D 1 X k=1 " 1 k k X n=1 y (n) #p v(k) < 1;
where D = max 1; 2p 1 . Here x (n) ; y (n) and x (n)+ y (n)
denote the non-increasing rearrangements of the sequences fkxnkg ; fkynkg and
fkxn+ ynkg, respectively. Let 2 K: Hence we get 1 X k=1 " 1 k k X n=1 x (n) #p v(k) = 1 X k=1 " j j k k X n=1 x (n) #p v(k) = j jp 1 X k=1 " 1 k k X n=1 x (n) #p v(k) < 1:
This shows that x + y 2 C1[d(v; p)] ; x 2 C1[d(v; p)] and so C1[d(v; p)] is a
linear space.
Theorem 2. The space C1[d(v; p)] for 1 p < 1 is normed space with the norm
kxkC;v;p= 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p ;
where x (n) denotes the non-increasing rearrangements of fkxnkg.
Proof. It is clear that k0kC;v;p= 0. Let kxkC;v;p = 0. Then we have1k k
P
n=1
x (n) =
0 for all k 2 N: Hence we get x (n) = 0 for all n 2 N and so x = 0:
Let x; y 2 C1[d(v; p)] : Since weight sequence v is decreasing, the non-increasing
rearrangements of v is itself. Thus, using the inequality P
i ai bi P i ai bi from Lemma 1, we have kx + ykC;v;p = 1 X k=1 " 1 k k X n=1 x (n)+ y (n) #p v(k) !1 p 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p + 1 X k=1 " 1 k k X n=1 y (n) #p v(k) !1 p 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p + 1 X k=1 " 1 k k X n=1 y (n) #p v(k) !1 p = kxkC;v;p+ kykC;v;p ;
where x (n) ; y (n) and x (n)+ y (n) denote the non-increasing
re-arrangements of fkxnkg ; fkynkg and fkxn+ ynkg, respectively.
Let be an element in K and let x be a vector in C1[d(v; p)] : Hence we have
k xkC;v;p = 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p = j j 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p = j j kxkC;v;p:
Theorem 3. The space C1[d(v; p)] for 1 p < 1 is complete with respect to its
Proof. Let x(s) be an arbitrary Cauchy sequence in C
1[d(v; p)] with x(s) =
n x(s)n
o1
n=1for all s 2 N: Then we have
lim s;t!1 x (s) x(t) C;v;p= lims;t!1 1 X k=1 " 1 k k X n=1 x(s) s;t(n) x (t) s;t(n) #p v(k) !1 p = 0; (1) where n x(s) s;t(n) x (t) s;t(n) o
denotes the non-increasing rearrangement of n
x(s)n x(t)n
o
. Hence we obtain lims;t!1 x(s)s;t(n) x (t) s;t(n) = 0 for each n 2 N and so n x(s)n o
; for a …xed n 2 N; is a Cauchy sequence in X:
Then, there exists xn2 X such that x(s)n ! xn as s ! 1. Let x = fxng :
Since limn!1x(s)n = 0 for each s 2 N, by Lemma 2 we have limn!1xn = 0:
Therefore we can choose the non-increasing rearrangement n x t(n) x (t) t(n) o n of n xn x(t)n o
n: Also, for an arbitrary " > 0 there exists N 2 N such that
1 X k=1 " 1 k k X n=1 x(s) s;t(n) x (t) s;t(n) #p v(k) !1 p < " (2)
for s; t > N . Let t be an arbitrary positive integer with t > N and …xed. If we put yn(s)= x(s)n x(t)n and yn= xn x(t)n ,
then we have lim
n!1y
(s)
n = 0 for each s 2 N and slim
!1y
(s)
n = yn (uniformly in n).
Thus by Lemma 2 we get
y (n) lim
s!1 y
(s) s(n)
for each n 2 N; that is, x t(n) x (t) t(n) slim!1 x (s) s;t(n) x (t) s;t(n) (3)
for each n 2 N. Hence, by (2); (3) we get x x(t) C;v;p = 1 X k=1 " 1 k k X n=1 x t(n) x (t) t(n) #p v(k) !1 p 1 X k=1 " 1 k k X n=1 lim s!1 x (s) s;t(n) x (t) s;t(n) #p v(k) !1 p = lim s!1 1 X k=1 " 1 k k X n=1 x(s) s;t(n) x (t) s;t(n) #p v(k) !1 p < ":
Also, since C1[d(v; p)] is a linear space we have fxng =
n xn x(N )n o +nx(N )n o 2 C1[d(v; p)] : Hence the space C1[d(v; p)] is complete with respect to its norm.
Theorem 4. Let 1 < p < 1: Then, the inclusion d(v; p) C1[d(v; p)] holds.
Proof. Let x 2 d(v; p). Then there exists T > 0 such that lim m!1 m X n=1 x (n) p v(n) !1 p = 1 X n=1 x (n) p v(n) !1 p T < 1;
where x (n) denotes the non-increasing rearrangements of fkxnkg. Since 1
P
k=1 1
kp < 1 for 1 < p < 1 and v is decreasing; we get
1 X k=1 " 1 k k X n=1 x (n) #p v(k) = 1 X k=1 1 kp " k X n=1 x (n) #p v(k) max 1; 2p 1 1 X k=1 1 kp " k X n=1 x (n) p v(n) # T max 1; 2p 1 1 X k=1 1 kp < 1: This completes the proof.
Proof. Let x 2 C1[d(v; p)] and let x (n) denotes the non-increasing
rearrange-ment of fkxnkg. Since v(k) is decreasing we have 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p m X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p x (m) m X k=1 v(k) !1 p x (m) (v(m)) 1 pmp1
for every m 2 N: Hence we get
x (m) (v(m)) 1 pm 1p 1 X k=1 " 1 k k X n=1 x (n) #p v(k) !1 p (v(m)) p1kxk C;v;p
for every m 2 N: Thus
1 X k=1 " 1 k k X n=1 x (n) #q v(k) = 1 X k=1 " 1 k k X n=1 x (n) #q p" 1 k k X n=1 x (n) #p v(k) 1 X k=1 " 1 k k X n=1 (v(n)) 1pkxk C;v;p #q p" 1 k k X n=1 x (n) #p v(k) (v(n)) p1kxk C;v;p q p 1X k=1 " 1 k k X n=1 x (n) #p v(k) < 1:
This implies that x 2 C1[d(v; q)] :
Comment. If we put 4mx instead of x, where m 2 N and 40x
k = fxkg ; 4xk=
xk xk+1; 4mxk = 4m 1xk 4m 1xk+1= m
P
v=1
( 1)v mv xk+v for all k 2 N in the
de…nition of C1[d(v; p)], we obtain Cesàro summable generalized Lorentz di¤erence
sequence space C1[d(v; 4m; p)] of order m: It can be shown that the sequence space
C1[d(v; 4m; p)] is a Banach space with norm
kxkC;v;4m;p= m X k=1 x (k) + 1 X k=1 " 1 k k X n=1 4mx (n) #p v(k) !1 p ; where 4mx
(n) denotes the non-increasing rearrangements of fk4mxnkg, and
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Current address : O. O¼gur: Giresun University, Art and Science Faculty, Department of Math-ematics, Güre, Giresun, TURKEY
E-mail address : [email protected]
Current address : B. Sa¼g¬r: Ondokuz May¬s University, Art and Science Faculty, Department of Mathematics, Kurupelit campus, Samsun, TURKEY
