Gen. Math. Notes, Vol. 19, No. 2, December, 2013, pp.37-58
c ICSRS Publication, 2013
ISSN 2219-7184; Copyright www.i-csrs.org
Available free online at http://www.geman.in
A Survey on Hahn Sequence Space
Murat Kiri³ci
Department of Mathematical Education
Hasan Ali Yucel Education Faculty
stanbul University, Vefa, 34470
Fatih, Istanbul, Turkey
E-mail: mkirisci@hotmail.com; murat.kirisci@istanbul.edu.tr
(Received: 5-4-13 / Accepted: 19-9-13)
Abstract
In this work, we investigate the studies related to the Hahn sequence space.
Keywords: Sequence space, beta- and gamma-duals, BK−spaces, matrix
transformations, geometric properties of sequence spaces.
1
Introduction
By a
sequence space, we understand a linear subspace of the space ω = CN of all
complex sequences which contains
We write
`∞ , c
and
c0
φ,
the set of all nitely non-zero sequences.
for the classical spaces of all bounded, convergent
bs, cs, `1 and `p , we denote the
space of all bounded, convergent, absolutely and p-absolutely convergent series,
and null sequences, respectively.
Also by
Murat Kiri³ci
38
respectively. We dene
bv, d`, σ∞ , σc , σs, dE
x = (xk ) ∈ ω :
bv =
σ∞ =
σc =
σs =
dE =
Z
E =
where
R
E
as follows:
|xk − xk−1 | < ∞ ,
k=1
∞
X
1
|xk | < ∞ ,
x = (xk ) ∈ ω :
k
k=1
n
1 X x = (xk ) ∈ ω : sup xk < ∞ ,
n n
k=1
n
X
1
xk exists ,
x = (xk ) ∈ ω : lim
n→∞ n
k=1
n X
X
k−1
x = (xk ) ∈ ω : (C − 1) −
xk = lim
1−
xk
n→∞
n
k=1
x = (xk ) ∈ ω : (k −1 xk ) ∈ E ,
x = (xk ) ∈ ω : (kxk ) ∈ E ,
d`1 =
∞
X
and
dE
and
R
E
exists
are called the dierentiated and integrated spaces of
E,
respectively.
A coordinate space (or
K−space) is a vector space of numerical sequences,
where addition and scalar multiplication are dened pointwise.
An
F K−space
is a locally convex Fréchet space which is made up of se-
quences and has the property that coordinate pro jections are continuous.
A
BK− space
is locally convex Banach space which is made up of sequences
and has the property that coordinate projections are continuous.
BK−space X is said to
kx − xk → 0 as n → ∞.
A
if
[n]
have
AK (or sectional convergence)
if and only
X has (C, 1) − AK (Cesàro-sectional convergence of order one)
x ∈ X and
 
k−1
1 − n xk , if k = 1, 2, ..., n
(Pn1 (x))k =

0
, if k = n + 1, n + 2, ...
Pn1 (x) ∈ X
and
if for all
kPn1 (x) − xkX → 0 (n → ∞).
X be an F K−space. A sequence (xk) in X is said to be weakly Cesàro
0
bounded, if [f (x1 ) + f (x2 ) + ... + f (xk )]/k is bounded for each f ∈ X , the
Let
,
A Survey on Hahn Sequence Space
dual space of
39
X.
Φ stand, for the set of all nite sequences. The space X is said to have
AD (or) be an AD space if Φ is dense in X . We note that AK ⇒ AD [3].
Let
F K−space X ⊃ Φ
x ∈ X.
An
each
Let
kx[n] k
X
for
is said to have
AB
(x[n] )
if
is bounded set in
BK−space. Then X is said to have monotone norm
m > n and kxk = sup[n] kx[n] k.
be a
D = {x ∈ Φ : kxk ≤ 1}
if
X
for
kx[m] k ≥
BK−space X , that is, D is the
intersection of the closed unit sphere(disc) with Φ. A subset E of Φ is called
a determining set for X if and only if its absolutely convex hull K is identical
with D [18].
Let
be in a
X is said to be rotund
kxk = kyk ≤ 1 in X [17].
The normed space
whenever
x 6= y and
S(λ, µ) dened
The set
if and only if
by
S(λ, µ) = {z = (zk ) ∈ ω : xz = (xk zk ) ∈ µ
is called the
k(x + y)/2k < 1,
multiplier space
for all
of the sequence spaces
υ
observe for a sequence space
with
S(λ, µ) ⊂ S(υ, µ)
λ⊃υ⊃µ
and
λ
x = (xk ) ∈ λ}
and
µ.
(1)
One can eaisly
that the inclusions
S(λ, µ) ⊂ S(λ, υ)
hold. With the notation of (1), the alpha-, beta-, gamma- and sigma-duals of
a sequence space
λ,
which are respectively denoted by
λα , λβ , λγ
and
λσ
are
dened by
λα = S(λ, `1 ),
λβ = S(λ, cs) λγ = S(λ, bs)
For each xed positive integer
k − th
place and zeros elsewhere.
X
Xf = {
conjugate is denoted by
Xf
and is given by
containing
Φ
0
bv = bs
λσ = S(λ, σs).
δ k = {0, 0, ..., 1, 0, ...}, 1 in the
F K−space X containing Φ, its
we write
Given an
f −dual or sequential dual is denoted by
(f (δ k )) : f ∈ X 0 }. An F K−space X
f
replete if X
⊂ σ(`∞ ). The space bv is semi
and its
all sequences
is said to be semi
replete, because
k,
and
[13].
λ and µ be two sequence spaces, and A = (ank ) be an innite matrix of
complex numbers ank , where k, n ∈ N. Then, we say that A denes a matrix
mapping from λ into µ, and we denote it by writing A : λ → µ if for every
sequence x = (xk ) ∈ λ. The sequence Ax = {(Ax)n }, the A-transform of x, is
Let
Murat Kiri³ci
40
in
µ;
where
(Ax)n =
X
ank xk
for each
n ∈ N.
(2)
k
Throughout the text, for short we suppose that the summation without limits
1 to ∞. By (λ : µ), we denote the class of all matrices A such that
A : λ → µ. Thus, A ∈ (λ : µ) if and only if the series on the right side of (2)
converges for each n ∈ N and each x ∈ λ and we have Ax = {(Ax)n }n∈N ∈ µ
for all x ∈ λ. A sequence x is said to be A-summable to l if Ax converges to l
which is called the A-limit of x.
runs from
Lemma 1.1. ([18], Theorem 7.2.7) Let X be an F K−space with X ⊃ Φ.
Then,
(i) X β ⊂ X γ ⊂ X f .
(ii) If X has AK , X β = X f .
(iii) If X has AD, X β = X γ .
2
Hahn Sequence Space
Hahn [7] introduced the
(
h=
x:
BK−space h
∞
X
of all sequences
∆xk = xk − xk+1 ,
k|∆xk | < ∞
for all
kxkh =
lim xk = 0 ,
and
k ∈ N.
k→∞
X
The following norm
k|∆xk | + sup |xk |
k
k
was dened on the space
h
2.1) dened a new norm on
by Hahn [7] (and also [6]). Rao ([12], Proposition
h
as
kxk =
Hahn proved following properties of
Lemma 2.1.
R
(iii) hβ = σ∞ .
P
k|∆xk |.
the space h:
k
(i) h is a Banach space.
(ii) h ⊂ `1 ∩ c0 .
such that
)
k=1
where
x = (xk )
A Survey on Hahn Sequence Space
Clearly,
h
is a
BK−space
41
[6].
In [6], Goes and Goes studied functional analytic properties of the
bv0 ∩ d`1 .
BK−space
Additionally, Goes and Goes considered the arithmetic means of se-
bv0 ∩ P
d`1 , and used an important fact which the sequence
n
−1
of arithmetic means (n
k=1 xk ) of an x ∈ bv0 is a quasiconvex null sequence.
bv0
quences in
and
Rao [12] studied some geometric properties of Hahn sequence space and
gave the characterizations of some classes of matrix transformations.
Now we give some additional properties of
h
which proved by Goes and
Goes [6].
Theorem 2.2. ([6], Theorem 3.2) h = `1 ∩ bv = `1 ∩ bv0 .
R
Proof.
R
k = 1, 2, ...
For
k∆xk = xk+1 + ∆(kxk ).
Hence
x∈h
(3)
implies
∞>
∞
X
k|∆xk | ≥
k=1
∞
X
|∆(kxk )| −
k=1
∞
X
|xk+1 |.
k=1
h ⊂ `1 by PartR of (ii)
therefore h ⊂ `1 ∩
bv .
The last series is convergent since
also
P∞
k=1
|∆(kxk )| < ∞
and
Conversely, (3) implies for
∞>
∞
X
|xk+1 | +
k=1
Thus,
`1 ∩
R
bv ⊂ h.
∞
X
x ∈ `1 ∩
|∆(kxk )| ≥
k=1
R
bv
∞
X
of Lemma 2.1. Hence
that
k|∆xk |
k=1
Hence, we have shown that
and
lim xk = 0.
k→∞
h = `1 ∩
R
bv .
The second
equality in the theorem follows now from Lemma 2.1(ii).
Lemma 2.3. ([6], Lemma 3.3) If X and Y are β−dual (σ−dual) Köthe
spaces, then X ∩ Y is also a β−dual (σ−dual) Köthe space.
Proof.
Köthe space if and only if
X
and
ζ = β or ζ = σ , then E
E = (E ) ≡ E ζζ ([5], p.139, Theorem
We use the known fact that if
Y
are
ζ−dual
ζ ζ
is a
ζ−dual
3). Hence if
Köthe spaces, then
(X ∩ Y )ζζ = (X ζ + Y ζ )ζζζ = (X ζ + Y ζ )ζ = X ζζ ∩ Y ζζ = X ∩ Y.
Murat Kiri³ci
42
Theorem 2.4. ([6], Theorem 3.4) h = (σ∞ )β .
Proof.
By Part (iii) of Lemma 2.1,
hβ = σ∞ .
Hence by the remark in the
beginning of the last proof it is enough to show that
h = `1 ∩
bv = (cs)β .
space. In fact: By Theorem 2.2,
and
R
β
bv = (d(cs))
since
R
bv
h
β−dual Köthe
β
known `1 = (c0 )
is a
and as is well
Theorem 2.5. ([6], Theorem 3.5) h is a BK−space with AK .
Proof. `1
R
P
P
bv0 with norms kxk =
k |xk | and kxk =
k |∆(kxk )| respectively are BK−spaces with AK . Hence by Theorem 2.2 and since the
intersection of two BK−spaces with AK is again a BK−space with AK ([15],
and
p.500), the theorem follows.
This result is found in [12] with a dierent norm.
Remark 2.6. ([6], 3.6) Let E be a BK−space with AK , and let E 0 be
the conjugate space of E , i.e. the space of linear continuous functionals on
0
E . It is known that E can be identied with E β through the isomorphism
0
β
ϕ ∈ E ⇔ (ϕ(ei ))∞
i=1 ∈ E , where
1 , if k = i
0 , if k =
6 i
i
(e )k =
This is true because E has AK if and only if every ϕ ∈ E can be written in
the form
0
ϕ(x) =
∞
X
(x ∈ E);
xk y k ,
k=1
where y ∈ E β [21]. As usual we write E β = E if E is a BK−space with AK .
0
Analogously we have E σ = E if E is a BK−space with (C, 1) − AK .
0
Theorem 2.7. ([6], Theorem 3.7)
(i) h = σ∞ .
0
(ii) (σ0 ) = h.
0
Proof.
BK−space with AK and by Lemma 2.1(ii)
0
h = σ∞ . Hence by Remark 2.6, h = σ∞ .
(ii) It is known that σ0 is a BK−space with AK ([21], p.61) with the norm
n
X
1
kxk = sup xk .
n n (i) By Theorem 2.5,
h
is a
β
k=1
A Survey on Hahn Sequence Space
Hence, again by Remark 2.6,
43
0
(σ0 )β = (σ0 ) .
It remains to be shown that
β
(σ0 ) = h.
R
σ0 =
β
β
β
this implies (σ0 ) = (c0 ) ∩ [d(cs)] = h
R
(d(cs))β = bv .
By Zeller([21], Theorem 1.1),
R
c0 + cs.
Hence
σ0 = c0 + d(cs)
(c0 )β = `
by Theorem 2.2, since
and
and
Remark 2.8. ([6], p.97) σ0 ⊂ σc ⊂ σ∞ and (σ0 )β = (σ∞ )β = h (by
Theorem 2.4 and 2.7) imply σcβ = h (see also [19], p.268).
Proposition 2.9. ([12], Proposition 2) The space h is not rotund.
Proof.
x = (1, 0, 0, ...) and y = (1/2, 1/2, 0, 0, ...). Then x and y are
in h. Also kxk = kyk = 1. Obviously x 6= y . But x + y = (3/2, 1/2, 0, ...) and
k(x + y)/2k = 1. Therefore, h is not rotund.
We take
Proposition 2.10. ([12], Proposition 3) The unit disc in the space h has
extreme points.
Proof.
For each xed
k = 1, 2, ...,
we write
k
1X i
s =
δ.
k i=1
k
Let
y = (yk )
h
X
be any point of
ksk + yk =
ksk + yk ≤ 1 and ksk − yk ≤ 1.
υ|∆yv | + k 1/k + yk − yk+1 such that
(4)
But
(5)
υ≥1,υ6=k
and
ksk − yk =
X
υ|∆yv | + k 1/k − yk + yk+1 .
(6)
υ≥1,υ6=k
ksk + yk ≤ 1 implies k|1/k + yk − yk+1 | ≤ 1 and ksk − yk ≤ 1 implies
k|1/k − yk + yk+1 | ≤ 1. Consequently yk = yk+1 . But when this holds we have
Hence
k|1/k + yk − yk+1 | = 1 = k|1/k − yk + yk+1 |.
ksk − yk ≤ 1 hold, all other terms in the
sums of (5) and (6) must be zero. That is, ∆yv = 0, (υ 6= k). This together
with the equality yk = yk+1 gives yk = constant. For y = (yk ) ∈ h, yn =
P∞
P∞
k=n (yk − yk+1 ) so that n|yn | ≤
k=n k|yk − yk+1 | which converges to zero.
Thus, |yn | = O(1/n) and so yυ → 0 as υ → ∞. Hence yυ = 0, (υ = 1, 2, ...).
k
k
k
Thus ks + yk ≤ 1 and ks − yk ≤ 1 imply that y = (0, 0, 0, ...). Hence, s is
an extreme point of the unit disc in h (for each xed k = 1, 2, ...).
Thus in order that
ksk + yk ≤ 1
and
Murat Kiri³ci
44
Proposition 2.11. ([12], Proposition 4) Let sk be dened as in (4) for all
k ∈ N. Consider the set E = {sk : k = 1, 2, ...}. The E is a determining set
for the space h.
Pm
k
Proof.
Let x ∈ D . Then x ∈ Φ and kxk ≤ 1. Consequently, x =
k=1 xk δ =
Pm
P
m
k
k=1 tk s where tk = k(xk − xk+1 ), (k = 1, 2, ...). Also,
k=1 |tk | ≤ kxk ≤ 1.
Therefore, x ∈ K , the absolutely convex hull of E . Thus
D ⊂ K.
On the other hand, let
Writing
x = (x1 , x2 , ...),
x ∈ K.
Then
(7)
x=
Pm
k=1 tk s
k
with
Pm
k=1
|tk | ≤ 1.
we observe that
tm
t2
+ ··· + ,
2
m
t2 t3
tm
=
+ + ··· + ,
2
3
m
x1 = t1 +
x2
.
.
.
.
.
.
.
.
.
tm
,
m
= xm+2 = · · · = 0.
xm =
xm+1
Hence
kxk =
P∞
k=1
k|xk − xk+1 | =
Pm
k=1
|tk | ≤ 1.
Thus we have
K ⊂ D.
Combining (7) and (8) it follows that
set for the space
K = D.
(8)
Therefore,
E
is a determining
h.
Proposition 2.12. ([12], Proposition 5) Let X be any F K−space which
contains Φ. Then X includes h if and only if E = {sk : k = 1, 2, ...} is
bounded in X .
Proof.
This is an immediate consequences of our Proposition 2.11 and Theorem
8.2.4 of [18].
Lemma 2.13. ([13], Lemma 1) Every semi replete space contains the Hahn
space h.
Proof.
X be any
f
[12]. Thus X
= hf .
hf ⇒ h ⊂ X .
Let
semi replete space. Then
Since
h
has
AD,
X f ⊂ σ(`∞ ).
we have by Theorem
σ(`∞ ) = hf
f
8.6.1 [18], X
⊂
But
A Survey on Hahn Sequence Space
45
Lemma 2.14. ([13], Lemma 2) A sequence z = (zk ) belongs to σ(`∞ ) if
and only if z β is semi replete.
Proof.
Suppose that
σ(`∞ ).
But
zβ
has
z ∈ σ(`∞ ).
AK under the
Then
h[σ(`∞ )]β ⊂ z β
so that
z ββ ⊂ hβ =
sequence of norms
P0 (x) = kzxkcs , (x ∈ z β ), I = (Ik ), x = (xk ).
Pn (x) = |xn |,
(x ∈ z β ), n ≥ 1.
Hence
z βf ⊂ σ(`∞ ).
Therefore
zβ
is semi replete.
z β is semi replete. Then z βf ⊂ σ(`∞ ). Since
= z βf . So z ∈ z ββ ⊂ σ(`∞ ). This completes the
Conversely, suppose that
zβ
has
AK ,
we have
z ββ
proof.
h.
Theorem 2.15. ([13], p.45) The intersection of all semi replete spaces in
Proof.
I
Let
denote the intersection of all semi replete spaces.
I contains h. On the
{z β : z ∈ σ(`∞ )} and so I
2.13,
T
other hand, by Lemma 2.14,
is contained in
[σ(`∞ )]β = h.
I
By Lemma
is contained in
Thus
I = h.
This
proves the theorem.
Let
X
F K−space
be an
with
X ⊃ Φ.
Then
B + = X f γ = B + (X) = {z ∈ ω : (z [n] ) is bounded in X}
= {z ∈ ω : (zn f (δ [n] ) ∈ bs, ∀f ∈ X 0 }.
Also we write
B = B+ ∩ X .
Let
space with monotone norm has
X is an AB−space if and only if B = X .
AB (see Theorem 10.3.12 of [18]).
Any
Lemma 2.16. ([14], Lemma 2) hf = σ(`∞ ).
Proof. hβ = σ(`∞ )
We have
β
f
h =h
by Lemma 2.1 ([7] and [6]). Also
. Therefore
f
h = σ(`∞ ).
h
has
AK
([12] and [6]).
This completes the proof.
Theorem 2.17. ([14], Theorem 1) Let Y be any F K−space ⊃ Φ. Then
Y ⊃ h if and only if the sequence (δ (k) ) is weakly Cesàro bounded.
Proof.
We know that
h has AK .
Since every
AK−space is AD
ing two sided implications establish the result.
[3], the follow-
Murat Kiri³ci
46
Y ⊃ h ⇔ Y f ⊂ hf since h has AD and hence by using Theorem 8.6.1.
⇔ Y f ⊂ σ(`∞ ) by Lemma 2.16
0
⇔ for each f ∈ Y , the topological dual of Y, f (δ (k) ) ∈ σ(`∞ )
f (δ (1) ) + f (δ (2) ) + ... + f (δ (k) )
⇔
∈ `∞
k
f (δ (1) ) + f (δ (2) ) + ... + f (δ (k) )
⇔
is bounded.
k
⇔
The sequence
(δ (k) )
in [18]
is weakly Cesàro bounded.
This completes the proof.
Theorem 2.18. ([14], Theorem 2) Suppose that h is a closed subspace of
an F K−space X . Then B + (X) ⊂ h.
Proof.
AD.
Note that
c0
has
AK .
σ(c0 ) has AK . Consequently σ(c0 ) has
[σ(c0 )] = [σ(c0 )]γ . By, Theorem 10.3.5 of [18]
Hence
Therefore by Lemma 1.1,
β
and Lemma 2.16, we have
B + (X) = B + (h) = hf γ = (hf )γ = (σ(`∞ ))γ .
(σ(`∞ ))γ ⊂ (σ(c0 ))γ = (σ(c0 ))β and (σ(c0 ))β = h
B + (X) ⊂ h. This completes the proof.
But
(See p.97 [6]).
Hence
Theorem 2.19. ([14], Theorem 3) Let X be an AK−space including Φ.
Then X ⊃ h if and only if B + (X) ⊃ h.
Proof.
(Necessity): Suppose that
X ⊃ h.
Then the inclusion,
B + (X) ⊃ B + (h)
(9)
holds by monotonicity Theorem 10.2.9 of [18]. By Theorem 10.3.4 of [18], we
have
B + (h) = hf γ = h
(10)
B + (X) ⊃ B + (h) = h.
B + (X) ⊃ h. We have
From (9) and (10), we obtain
(Suciency): Suppose that
hγ ⊃ [B + (X)]γ .
But
h
has
AK
and so
h
has
AD.
(11)
Therefore
hβ = hγ = hf .
(12)
A Survey on Hahn Sequence Space
47
But always
B + (X) = X f γ .
(13)
hγ ⊃ (X f γ )γ = (X f )γγ ⊃ X f . Thus from (12), hf ⊃ X f .
8.6.1 of [18]; since h has AD we conclude that X ⊃ h. This
From (11) and (13),
Now by Theorem
completes the proof.
Theorem 2.20. ([14], Theorem 4) The space h has monotone norm.
Proof.
Let
m > n.
It follows from
|xn | ≤ |xn − xn+1 | + |xn+1 − xn+2 | + ... + |xm−1 − xm | + |xm |
that
kx
(n)
k≤
X
n−1
k|xk − xk+1 | + n|xn − xn+1 | + ... + (m − 1)|xm−1 − xm | + m|xm | = kx(m) k
k=1
The sequence
limn→∞ x
(n)
(kx(n) k)
being monotone increasing, it thus follows from
x =
that
kxk = lim kx(n) k = sup kx(n) k.
n→∞
(n)
Rao and Subramanian [14] dened semi-Hahn space as below:
An
F K−space X
is called semi-Hahn if
X f ⊂ σ(`∞ ).
In other words
0
f (δ (k) ) ∈ σ(`∞ ), ∀f ∈ X
f (δ 1 ) + f (δ 2 ) + ... + f (δ k )
⇔ {
} ∈ `∞
k
f (δ 1 ) + f (δ 2 ) + ... + f (δ k )
} is bounded
⇔ {
k
0
for each
f ∈X.
Example 2.21. ([14], p.169) The Hahn space is semi Hahn. Indeed, if h
is a Hahn space, then, hf = σ(`∞ ) by Lemma 2.16.
Lemma 2.22. ([18], 4.3.7) Let z be a sequence. Then (z β , p) is an AK−space
with p = (pk ) : k ∈ N, where
X
m
p0 (x) = sup zk xk ,
m
pn (x) = |xn |.
k=1
For any k such that zk 6= 0, pk may be omitted. If z ∈ Φ, p0 may be omitted.
Murat Kiri³ci
48
Theorem 2.23. ([14], Theorem 5) z β is semi-Hahn if and only if z ∈ σ(`∞ ).
Proof. Step 1. Suppose that z β is semi-Hahn. z β has AK by Lemma 2.22.
z ββ = (z β )f by Theorem 7.2.7 of [18]. So z β is
ββ
semi-Hahn if and only if z
⊂ σ(`∞ ). But then z ∈ z ββ ⊂ σ(`∞ ).
Step 2. Conversely, let z ∈ σ(`∞ ). Then z β ⊃ {σ(`∞ )}β and z ββ ⊂ σ(`∞ )ββ =
hβ = σ(`∞ ). But (z β )f = z ββ . Hence (z β )f ⊂ σ(`∞ ) which gives that z β is
Hence
zβ = zf .
Therefore
semi-Hahn. This completes the proof.
Theorem 2.24. ([14], Theorem 6) Every semi-Hahn space contains h.
Proof.
Let
X
be any semi-Hahn space. Then, one can see that
⇒
⇒
⇒
⇒
X f ⊂ σ(`∞ ).
0
f (δ (k) ) ∈ σ(`∞ ), ∀f ∈ X
δ (k) is weakly Cesaro bounded
X ⊃ h by Theorem 2.17.
w.r. to X
Theorem 2.25. ([14], Theorem 7) The intersection of all semi-Hahn spaces
is h.
Proof.
Let
I
be the intersection of all semi-Hahn spaces. Then the intersection
I ⊂ ∩{z β : z ∈ σ(`∞ )} = {σ(`∞ )}β = h
(14)
By Theorem 2.24,
h⊂I
From 14 and 14, we get
(15)
I = h.
Corollary 2.26. ([14], p.170) The smallest semi-Hahn space is h.
3
Matrix Transformations
Now we give some matrix transformations:
Theorem 3.1. ([12] Proposition 6) A ∈ (h : c0 ) if and only if
lim ank = 0,
n→∞
(k = 1, 2, ...)
k
1 X
sup anυ < ∞.
n,k k υ=1
(16)
(17)
A Survey on Hahn Sequence Space
Proof.
49
c0 is a
BK−space. So, we invoke Theorem 8.3-4 of [18] and conclude that A ∈ (h : c0 )
if and only if the columns of A are in c0 , and A(E) is a bounded subset of c0 .
nP
o
k
k
k
Here, we recall that E = {s : k = 1, 2, ...}. But As =
a
/k
:
n
=
1,
2,
...
.
υ=1 nυ
So, A ∈ (h : c0 ) if and only if (16) and (17) hold.
We have by Theorem 2.5 that
h
is a
BK−space
with
AK .
Also
Omitting the proofs, we formulate the following results.
Theorem 3.2. ([12], Proposition 7) A ∈ (h : c) if and only if (17) holds
and
lim ank exists, (k = 1, 2, ...)
n→∞
(18)
Theorem 3.3. ([12], Proposition 8) A ∈ (h : `∞ ) if and only if (17) holds.
Theorem 3.4. ([12], Proposition 9) A ∈ (h : `1 ) if and only if
∞
X
|ank | converges, (k = 1, 2, ...)
(19)
n=1
∞ k
1 X X
anυ < ∞.
sup
k k n=1 υ=1
(20)
Theorem 3.5. ([12], Proposition 10) A ∈ (h : h) if and only if (16) holds
and
∞
X
n|ank − an+1,k | converges, (k = 1, 2, ...)
(21)
k
∞
1 X X
sup
n
(anυ − an+1,υ ) < ∞.
k k n=1
υ=1
(22)
n=1
4
The Hahn Sequence Space of Fuzzy Numbers
In this section, we introduce the sequence space
h(F ) called the Hahn sequence
space of fuzzy numbers [2].
The concept of fuzzy sets and fuzzy set operations was rst introduced
by Zadeh [20].
Sequences of fuzzy numbers have been discussed by Aytar
and Pehlivan [1], Mursaleen and Ba³arr [9], Nanda [10] and many others.
Murat Kiri³ci
50
The study of Hahn sequence space was initiated by Rao [12] with certain
specic purpose in Banach space theory. Talo and Ba³ar [16] gave the idea of
determining the dual of sequence space of fuzzy numbers by using the concept
of convergence of a series of fuzzy numbers.
Denition 4.1. A fuzzy number is a fuzzy set on the real axis, i.e., a map-
ping u : R → [0, 1] which satises the following conditions:
(i) u is normal, i.e., there exists an x0 ∈ R such that u(x0 ) = 1.
(ii) u is fuzzy convex, i.e., u[λx+(1−λ)y] ≥ min{u(x), u(y)} for all x, y ∈ R
and for all λ ∈ [0, 1].
(iii) u is upper semi continuous.
(iv) The set [u]0 = {x ∈ R : u(x) > 0} is compact [20], where {x ∈ R : u(x) > 0}
denotes the closure of the set {x ∈ R : u(x) > 0} in the usual topology of
R.
We denote the set of all fuzzy numbers on R by E and called it as the space
0
of fuzzy numbers. The λ−level set [u]λ of u ∈ E is dened by
0
[u]λ =
{t ∈ R : u(t) ≥ λ} , (0 < λ ≤ 1),
{t ∈ R : u(t) > λ} , (λ = 0).
The set [u]λ is closed bounded and non-empty interval for each λ ∈ [0, 1] which
is dened by [u]λ = [u− (λ), u+ (λ)]. Since each r ∈ R can be regarded as a fuzzy
number r dened by
r=
R
can be embedded in
E
0
.
1 , (x = r),
0 , (x 6= r),
Let
u, w ∈ E
0
and
k ∈ R.
0
E by
The operations
addition, scalar multiplication and product dened on
u + v = w ⇔ [w]λ = [u]λ + [v]λ for all λ ∈ [0, 1]
⇔ [w]− (λ) = [u− (λ), v − (λ)] and [w]+ (λ) = [u+ (λ), v + (λ)]
[ku]λ = k[u]λ
for all
λ ∈ [0, 1]
and
uv = w ⇔ [w]λ = [u]λ [v]λ
for all
for all
λ ∈ [0, 1],
where it is immediate that
[w]− (λ) = min{u− (λ)v − (λ), u− (λ)v + (λ), u+ (λ)v − (λ), u+ (λ)v + (λ)}
λ ∈ [0, 1]
A Survey on Hahn Sequence Space
51
and
[w]+ (λ) = max{u− (λ)v − (λ), u− (λ)v + (λ), u+ (λ)v − (λ), u+ (λ)v + (λ)}
for all
λ ∈ [0, 1].
W
Let
be the set of all closed and bounded intervals
with endpoints
A
and
A
i.e.,
A = [A, A].
A
Dene the relation
of real numbers
d
on
W
by
d(A, B) = max{|A − B|, |A − B|}.
W [10] and (W, d) is a complete
0
Now we can dene the metric D on E by means of a Hausdor
Then it can be observed that
metric space [4].
metric
d
d
is a metric on
as
−
−
+
+
D(u, v) = sup d([u]λ , [v]λ ) = sup |u (λ) − v (λ)|, |u (λ) − v (λ)| .
λ∈[0,1]
0
(E , D)
λ∈[0,1]
is a complete metric space ([11] Theorem 2.1).
One can extend the
natural order relation on the real line to intervals as follows:
A≤B
if and only if
The partial order relation on
E
0
|u|
and
of a fuzzy number
|u|(t) =
and
A ≤ B.
is dened as follows:
u ≤ v ⇔ [u]λ ≤ [v]λ ⇔ u− (λ) ≤ v − (λ)
An absolute value
A≤B
u
u+ (λ) ≤ v + (λ)
for all
λ ∈ [0, 1].
is dened by
max{u(t), u(−t)} , (t ≥ 0),
0
, (t < 0).
0
λ−level set [|u|]λ of the absolute value of u ∈ E is in the form [|u|]λ , where
|u|− (λ) = max{0, u− (λ), u+ (λ)} and |u|+ (λ) = max{|u− (λ)|, |u+ (λ)|}. The
0
absolute value |uv| of u, v ∈ E satises the following inequalities [16]
|uv|− (λ) ≤ |uv|+ (λ) ≤ max{|u|− (λ)|v|− (λ), |u|− (λ)|v|+ (λ), |u|+ (λ)|v|− (λ), |u|+ (λ)|v|+ (λ)}
In the sequel, we require the following denitions and lemmas.
Denition 4.2. A sequence
u = (uk ) of fuzzy numbers is a function u from
0
the set N into the set E . The fuzzy number uk denotes the value of the function
at k ∈ N and is called the k th term of the sequence. Let w(F ) denote the set
of all sequences of fuzzy numbers.
Murat Kiri³ci
52
Lemma 4.3. The following statements hold:
(i) D(uv, 0) ≤ D(u, 0)D(v, 0) for all u, v ∈ E .
0
(ii) If uk → u as k → ∞ then D(uk , 0) → D(u, 0) as k → ∞; where
(uk ) ∈ w(F ).
Denition
4.4. A sequence u = (uk ) ∈ w(F ) is called convergent with limit
0
u ∈ E if and only if for every ε > 0 there exists an n0 = n0 (ε) ∈ N such that
D(uk , u) < ε for all k ≥ n0 .
If the sequence (uk ) ∈ w(F ) converges to a fuzzy number u then by the
+
denition of D the sequences of functions {u−
k (λ)} and {uk (λ)} are uniformly
convergent to u− (λ) and u+ (λ) in [0, 1], respectively.
Denition 4.5. A sequence u(uk ) ∈ w(F ) is called bounded if and only if
the set of all fuzzy numbers consisting of the terms of the sequence (uk ) is a
bounded set. That is to say that a sequence (uk ) ∈ w(F ) is said to be bounded
if and only if there exist two fuzzy numbers m and M such that m ≤ uk ≤ M
for all k ∈ N.
Denition 4.6. Let (uk ) ∈ w(F ). PThen the expression
n
P
uk is called a
the sequences
series of fuzzy numbers. Denote Sn = k=0 uk for all n ∈ N. If P
(Sn ) converges to a fuzzy number uPthen we say that the series
uk of fuzzy
n
u
=
u
which
implies
as
n
→ ∞ that
numbers
converges
to
u
and
write
k=0 k
Pn
Pn
−
−
+
+
k=0 uk (λ) → uk (λ) and
k=0 uk (λ) → uk (λ) uniformly in λ ∈
P[0,−1]. Con−
+
versely,
if
the
fuzzy
numbers
u
=
{[u
(λ),
u
(λ)]
:
λ
∈
[0,
1]},
uk (λ) and
k
k
k
P +
−
+
uk (λ) converge uniformly inPλ then u = {[u (λ), u (λ)] : λ ∈ [0, 1]} denes
a fuzzy number such that u = uk .
We say otherwise the series of fuzzy numbers diverges. Additionally if the
sequence
(Sn )
is bounded.
is bounded then we say that the series
By
cs(F )
and
bs(F ),
P
uk
of fuzzy numbers
we denote the sets of all convergent and
bounded series of fuzzy numbers, respectively.
Lemma 4.7. Let for the series of functions
P
uk (x) and k vk (x) there
exists an n0 ∈ N such that |uk (x)| ≤ vk (x) for all k ≥ n0 and for all x ∈ [a, b]
with uk : [a, b] → R and
R. If the series converges uniformly
P vk : [a, b] →P
in [a, b] then the series k |uk (x)| and k |vk (x)| are uniformly convergent in
[a, b].
Denition 4.8.
P
k
(Weierstrass M Test) Let uk : [a, b] → R are given.
If therePexists an Mk ≥ 0 such that |uP
k (x)| ≤ Mk for all k ∈ N and the
series k Mk converges then the series k uk (x) is uniformly and absolutely
convergent in [a, b]
A Survey on Hahn Sequence Space
53
Denition 4.9. A mapping T from X1 and X2 is said to be fuzzy isometric
if d2 (T x, T y) = d1 (x, y) for all x, y ∈ X1 . The space X1 is said to be fuzzy
isometric with the space X2 if there exists a bijective fuzzy isometry from X1
onto X2 and write X1 ∼
= X2 . The spaces X1 and X2 are then called fuzzy
isometric spaces.
The following spaces are needed.
(uk ) ∈ w(F ) : sup D(uk , 0) < ∞ ,
`∞ (F ) =
k∈N
o
0
(uk ) ∈ w(F ) : ∃` ∈ E lim D(uk , `) = 0 ,
k→∞
n
o
c0 (F ) =
(uk ) ∈ w(F ) : lim D(uk , 0) = 0 ,
k→∞
)
(
X
`p (F ) =
(uk ) ∈ w(F ) :
D(uk , 0) < ∞ .
c(F ) =
n
k
Let
A
denote the matrix
ank =
A = (ank )
dened by
n(−1)n−k , n − 1 ≤ k ≤ n
0
, 1≤k ≤n−1
or
k>n
y = (yk ) which will be frequently used as the A−transform
x = (xk ), i.e.,
Dene the sequence
of a sequence
yk = (Ax)k = k(xk − xk−1 ) k ≥ 1
(23)
h(F ) and h∞ (F ) as the sets of all sequences such
A−transforms are in `(F ) and `∞ (F ) that is,
X
h(F ) = u = (uk ) ∈ ω(F ) :
D[(Au)k , 0] < ∞ and lim D[uk , 0] = 0
We introduce the sets
that their
k→∞
k
h∞ (F ) =
u = (uk ) ∈ ω(F ) : sup D[(Au)k , 0] < ∞ .
k
by
Example 4.10. ([2], Example 3.1) Consider the sequence u = (uk ) dened
uk =
X
D[(Au)k , 0] =
1 , 1≤k≤n
0 , k>n
X
D[k(uk − uk−1 ), 0] = 0
which is convergent. Also limk→∞ D(uk , 0) = 0. Hence, u ∈ h(F ).
Murat Kiri³ci
54
Theorem 4.11. ([2], Theorem 3.2) h(F ) and h∞ (F ) are complete metric
spaces with the metrics dh and dh∞ dened by
X
dh(u, v) =
D[(Au)k , (Av)k ]
k
dh∞ (u, v) = sup D[(Au)k , (Av)k ]
k∈N
respectively, where u = (uk ) and v = (vk ) are the elements of the spaces h(F )
or h∞ (F ).
Proof.
Let
{ui }
be any Cauchy sequence in the space
(i)
(i)
(i)
{u0 , u1 , u2 , · · · }.
n0 (ε)
ε > 0,
Then for a given
h(F ),
where
ui =
there exists a positive integer
such that
dh(ui , uj ) =
X
D[(Au)in , (Au)jn ] < ε
(24)
n
for
i, j ≥ n0 (ε).
We obtain for each xed
n∈N
from (24) that
D[(Au)in , (Au)jn ] < ε
for every
i, j ≥ n0 (ε).
m
X
We obtain for each xed
n∈N
from (24) that
D[(Au)in , (Au)jn ] ≤ dh(ui , uj ) < ε.
(25)
k=0
i ≥ n0 (ε)
Take any
and takin limit as
j→∞
rst and next
m→∞
in (24),
we obtain
dh(ui , u) < ε.
Finally, we proceed to prove
h(F ),
u ∈ h(F ).
Since
{ui }
is a Cauchy sequence in
we have
X
D[(Au)ik , 0] ≤ ε
and
k
lim [(Au)ik , 0] = 0.
k→∞
Now
D[(Au)k , 0] ≤ D[(Au)k , (Au)ik ] + D[(Au)ik , (Au)jk ] + D[(Au)jk , 0].
Hence,
D[(Au)k , 0] ≤
X
k
X
D[(Au)ik , (Au)jk ] +
k
limk→∞ [(Au)ik , 0] = 0. Hence, u ∈ h(F ).
Cauchy sequence, the space h(F ) is complete.
Also from (25),
arbitrary
D[(Au)k , (Au)ik ] +
X
D[(Au)jk , 0] < ε.
k
Since,
{ui }
is an
A Survey on Hahn Sequence Space
55
Theorem 4.12. ([2], Denition 3.3) The space h(F ) is isomorphic to the
space `1 (F ).
Proof.
Consider the transformation
T (x).
To prove the fact
T dened from h(F ) to `1 (F ) by x 7→ y =
h(f ) ∼
= `1 (F ), we show the existence of a bijection
between the spaces h(F ) and `1 (F ). We can nd that only one x ∈ h(F ) with
T x = y . This means that T is injective.
Let y ∈ `1 (F ). Dene the sequence x = (xk ) such that (Ax)k = yk for all k ∈ N.
P
P
dh(x, 0) = k D[(Ax)k , 0] = k D[yk , 0] < ∞. Thus, x ∈ h(F ).
Consequently, T is bijective and is isometric. Therefore, h(F ) and `1 (F )
Then
are
isomorphic.
Theorem 4.13. ([2], Theorem 3.4) Let d denote the set of all sequences of
fuzzy numbers dened as follows
X
k|xk − xk−1 | < ∞ and x ∈ c0 (F ) .
d = x = (xk ) ∈ w(F ) :
k
Then, the set d is identical to the set h(F ).
Proof.
Let
x ∈ h(F ). Then
X
D((Ax)k , 0) < ∞
and
k
lim D[xk , 0] = 0.
k→∞
(26)
Using (23),
X
k
D(yk , 0) < ∞
and
lim D[xk , 0] = 0.
k→∞
D(yk , 0) = supλ∈[0,1] max{|yk− (λ)|, |yk+ (λ)|}. Now, max{|yk− (λ)|, |yk+ (λ)|} ≤
P
P
is
k D(yk , 0) < ∞. This implies that
k |yk | < ∞. That
k k|xk − xk−1 | <
P
∞. AlsoP
from (26), x ∈ c0 (F ). Thus, x ∈ d. Then,
k k|xk − xk−1 | < ∞.
P
+
−
(λ)|,
|y
That is
|y
|
<
∞
.
Therefore
k
max{|y
k
k (λ)|} converges for
k k
P k
λ ∈ [0, 1]. This gives for λ = 0,
k D(yk , 0) < ∞. Also (xk ) ∈ c0 (F ) implies
limk→∞ D(xk , 0) = 0. This completes the proof.
We have,
P
k
P
Denition 4.14. ([2], Denition 3.5) The α−dual, β−dual and γ−dual
S(F )α , S(F )β and S(F )γ of a set S(F ) ⊂ w(F ) are dened by
{S(F )}α = (uk) ∈ w(F ) : (uk vk ) ∈ `1 (F ) for all (vk ) ∈ S(F ) ,
{S(F )}β = (uk) ∈ w(F ) : (uk vk ) ∈ cs(F ) for all (vk ) ∈ S(F ) ,
{S(F )}γ = (uk) ∈ w(F ) : (uk vk ) ∈ bs(F ) for all (vk ) ∈ S(F ) ,
Murat Kiri³ci
56
Denition 4.15. ([2], Denition 3.6) Let B denote the matrix B = (bnk )
dened by
bnk =
1/n , 1 ≤ k ≤ n
0 ,
otherwise
Dene the sequence
Pk y = (yk ) by the B−transform of a sequence x = (xk ), i.e.,
i=1 xi /k for all k ∈ N.
yk = (Bx)k =
The Cesàro space of `∞ (F ) is the set of all sequences such that their
B−transforms are in `∞ (F ). That is,
σ(`∞ (F )) = x = (xk ) : sup D[(Bx)k , 0] < ∞ .
k
Theorem 4.16. ([2], Theorem 3.7) σ(`∞ (F )) is a complete metric space
with the metric
dσ (u, v) = sup D[(Bu)k , (Bv)k ],
k
where u = (uk ) and v = (vk ) are the elements of the space σ(`∞ (F )).
Theorem 4.17. ([2], Theorem 3.8) The β− and γ−dual of the set h(F ) is
the set σ(`∞ (F )).
Proof. Let (uk ) ∈ h(F ) and (vk ) ∈ σ(`∞ (F )). If (uk ) ∈ h(F ), then limk→∞ D[uk , 0] =
0. Therefore for given ε > 0 there exists n0 such that D(uk , 0) < ε. If
(vk ) ∈ σ(`∞ (F )), then supk D[(Bv)k , 0] < ∞. Thus, D(vk , 0) < ∞ for all k
and n. Hence, there exists a M > 0 such that D(vk , 0) < M for all k and n.
Now,
|(uk )− (λ)| ≤ D(uk , 0) ≤ D(uk , 0)D(vk , 0) < εM.
P
P
+
−
Weierstrass Test yields that
k (uk ) (λ) converge
k (uk ) (λ) and
P
β
and hence
k uk converges. Thus σ(`∞ (F )) ⊂ h (F ).
(vk ) ∈ hβ (F ).
uniformly
P
k uk vk converges
(uk ) ∈ h(F ). This also holds for the sequence (uk ) of fuzzy numbers
+
−
dened by uk = χ[−1, 1] for all k ∈ N. Since u (λ) = −1 and u (λ) = 1 for
k
k
all λ ∈ [0, 1], the series
X
X
+
+
−
+
−
−
+
(uk vk )+ (λ) =
max{u−
k (λ)vk (λ), uk (λ)vk (λ), uk (λ)vk (λ), uk (λ)vk (λ)}
Conversely, suppose that
Then the series
for all
k
k
=
X
=
X
max{−vk− (λ), −vk+ (λ), vk− (λ), vk+ (λ)}
k
max{|vk− (λ)|, |vk+ (λ)|}
k
converges uniformly.
and
Thus
hβ (F ) = σ(`∞ (F )).
supk D[(Bv)k , 0] < ∞.
This completes the proof.
Hence,
(vk ) ∈ σ(`∞ (F ))
A Survey on Hahn Sequence Space
5
57
Conclusion
Hahn dened the space
h
and gave its some general properties.
Goes and
Goes [6] studied the functional analytic properties of this space. The study of
Hahn sequence space was initiated by Rao [12] with certain specic purpose in
Banach space theory.
Also Rao [12] computed some matrix transformations.
Rao and Srinivasalu [13] introduce a new class of sequence spaces called semi
replete spaces.
Rao and Subramanian [14] dened the semi Hahn space and
proved that the intersection of all semi Hahn spaces is the Hahn space. Bal-
h(F ) called
β− and γ−duals
asubramanian and Pandiarani[2] dened the new sequence space
the Hahn sequence space of fuzzy numbers and proved that
of
h(F )
is the Cesàro space of the set of all fuzzy bounded sequences.
Determine the matrix domain
α−, β−
and
γ−duals
hA of arbitrary triangles A and compute their
and characterize matrix transformations on them into
the classical sequence spaces and almost convergent sequence space may happen new results. Also it may study the some geometric properties of this space.
Acknowledgements:
I would like to express thanks to Professor Feyzi
Ba³ar, Department of Mathematics, Fatih University, Buyukcekmece, Istanbul34500, Turkey, for his valuable help on some results and the useful comments
which improved the presentation paper.
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