PII. S0161171203204324 http://ijmms.hindawi.com © Hindawi Publishing Corp.
ON SOME BK SPACES
BRUNO DE MALAFOSSE
Received 25 April 2002
We characterize the spacessα(∆),sα◦(∆), andsα(c)(∆)and we deal with some sets generalizing the well-known setsw0(λ),w∞(λ),w(λ),c0(λ),c∞(λ), andc(λ). 2000 Mathematics Subject Classification: 46A45, 40C05.
1. Notations and preliminary results. For a given infinite matrix A = (anm)n,m≥1, the operatorsAn, for any integern≥1, are defined by
An(X)=
∞
m=1
anmxm, (1.1)
whereX=(xn)n≥1is the series intervening in the second member being con-vergent. So, we are led to the study of the infinite linear system
An(X)=bn, n=1,2,..., (1.2)
whereB=(bn)n≥1is a one-column matrix andXthe unknown, see [2,3,5,6,7,9]. Equation (1.2) can be written in the formAX=B, whereAX=(An(X))n≥1. In
this paper, we will also considerAas an operator from a sequence space into another sequence space.
A Banach spaceEof complex sequences with the normEis a BK space if each projectionPn:X→PnXis continuous. A BK spaceEis said to have AK (see [8]) if for everyB=(bn)n≥1,B=∞n=1bmem, that is,
∞
m=N+1 bmem
E
→0 (n→ ∞). (1.3)
We shall writes,c,c0, andl∞for the sets of all complex, convergent sequences,
sequences convergent to zero, and bounded sequences, respectively. We shall writecs andl1 for the sets of convergent and absolutely convergent series, respectively. We will use the set
U+∗=unn≥1∈s|un>0∀n
Using Wilansky’s notations [12], we define, for any sequence α=(αn)n≥1∈ U+∗and for any set of sequencesE, the set
α∗E=
(xn)n≥1∈s| xn αn
n∈E
. (1.5)
Writing
α∗E=
sα◦ ifE=c0, sα(c) ifE=c, sα ifE=l∞,
(1.6)
we have for instance
α∗c0=sα◦=
xnn≥1∈s|xn=o
αnn→ ∞. (1.7)
Each of the spacesα∗E, whereE∈ {c0,c,l∞}, is a BK space normed by
Xsα=sup
n≥1 xn αn
, (1.8)
andsα◦ has AK.
Now, letα=(αn)n≥1andβ=(βn)n≥1∈U+∗. We shall writeSα,βfor the set of infinite matricesA=(anm)n,m≥1such that
anmαmm≥1∈l1 ∀n≥1,
∞
m=1
anmαm=O
βn (n → ∞). (1.9)
The setSα,βis a Banach space with the norm
ASα,β=sup
n≥1 ∞
m=1
anmαm βn
. (1.10)
LetEandF be any subsets ofs. WhenAmapsE intoF, we writeA∈(E,F), see [10]. So, for everyX∈E,AX∈F (AX∈F will mean that for eachn≥1, the series defined byyn=∞m=1anmxmis convergent and(yn)n≥1∈F). It has been proved in [8] thatA∈(sα,sβ)if and only ifA∈Sα,β. So, we can write (sα,sβ)=Sα,β.
Whensα=sβ, we obtain the unital Banach algebraSα,β=Sα, (see [2,3,9]) normed byASα= ASα,α.
We also haveA∈(sα,sα)if and only ifA∈Sα. IfI−ASα<1, we say that
Ifα=(rn)n
≥1, thenΓα,Sα,sα,sα◦, andsα(c)are replaced byΓr,Sr,sr,sr◦, and sr(c), respectively, (see [2,3,5,6,7,8,9]). Whenr=1, we obtains1=l∞,s◦1=c0, ands1(c)=c, and puttinge=(1,1,...), we haveS1=Se. It is well known, see [10], that
s1,s1=c0,s1=c,s1=S1. (1.11)
We writeen=(0,...,1,...)(where 1 is in thenth position). For any subsetEofs, we put
AE= {Y∈s| ∃X∈E Y=AX}. (1.12)
IfF is a subset ofs, we denote
F(A)=FA= {X∈s|Y=AX∈F}. (1.13)
We can see thatF(A)=A−1F.
2. Setssα(∆),sα(◦ ∆), andsα(c)(∆). In this section, we will give necessary and sufficient conditions permitting us to write the setssα(∆),sα◦(∆), andsα(c)(∆) by means of the spacessξ,s◦ξ, ors
(c)
ξ . For this, we need to study the sequence C(α)α.
2.1. Properties of the sequenceC(α)α. Here, we will deal with the opera-tors represented byC(λ)and∆(λ), see [2,5,7,8,9].
Let
U=unn≥1∈s|un≠0∀n
. (2.1)
We defineC(λ)=(cnm)n,m≥1, forλ=(λn)n≥1∈U, by
cnm=
1
λn ifm≤n, 0 otherwise.
(2.2)
It can be proved that the matrix∆(λ)=(cnm)n,m≥1, with
cnm =
λn ifm=n,
−λn−1 ifm=n−1, n≥2, 0 otherwise,
(2.3)
We use the following sets:
C=
α∈U+∗|C(α)α= 1
αn n
k=1 αk
n≥1 ∈c
,
C1=α∈U+∗|C(α)α∈s1=l∞,
Γ=
α∈U+∗|nlim→∞
αn−1 αn
<1
.
(2.4)
Note that∆∈Γαimpliesα∈Γ. It can be easily seen thatα∈Γ if and only if there is an integerq≥1 such that
γq(α)= sup n≥q+1
αn−1 αn
<1. (2.5)
See [7].
In order to express the following results, we will denote by[C(α)α]n (in-stead of[C(α)]n(α)) thenth coordinate ofC(α)α. We get the following propo-sition.
Proposition2.1. Letα∈U+∗. Then (i) αn−1/αn→0if and only if[C(α)α]n→1;
(ii) (a) α∈Cimplies that(αn−1/αn)n≥1∈c,
(b) [C(α)α]n→limplies thatαn−1/αn→1−1/l;
(iii) ifα∈C1, there areK >0andγ >1such that
αn≥Kγn ∀n; (2.6)
(iv) the conditionα∈Γ implies thatα∈C1and there exists a realb >0such that
C(α)αn≤ 1
1−χ+bχn forn≥q+1, χ=γq(α)∈]0,1[. (2.7)
Proof. (i) Assume thatαn−1/αn→0. Then there is an integerNsuch that
n≥N+1⇒αn−1 αn ≤
1
2. (2.8)
So, there exists a realK >0 such thatαn≥K2nfor allnand
αk αn =
αk αk+1···
αn−1 αn ≤
1 2
n−k
Then
1 αn
n−1
k=1 αk
= 1
αn N−1
k=1 αk
+n−1
k=N αk αn
≤ 1 K2n
N−1
k=1 αk
+n−1
k=N 1 2
n−k ;
(2.10)
and since
n−1
k=N 1 2
n−k
=1− 12 n−N
→1 (n→ ∞), (2.11)
we deduce that
1 αn
n−1
k=1 αk
=O(1), C(α)αn
∈l∞. (2.12)
Using the identity
C(α)αn=
α1+···+αn−1 αn−1
αn−1 αn +1
=C(α)αn−1 αn−1
αn
+1,
(2.13)
we get[C(α)α]n→1. This proves the necessity. Conversely, if[C(α)α]n→1, then
αn−1 αn =
C(α)αn−1
C(α)αn−1
→0. (2.14)
Assertion (ii) is a direct consequence of identity (2.14). (iii) We putΣn=
n
k=1αk. Then for a realM >1,
C(α)αn= Σn
Σn−Σn−1≤M ∀n.
(2.15)
So, Σn ≥(M/(M−1))Σn−1 and Σn≥(M/(M−1))n−1α1 for all n. Therefore, from
α1 αn
M M−1
n−1
≤C(α)αn=Σαnn ≤M, (2.16)
(iv) Ifα∈Γ, then there is an integerq≥1 for which
k≥q+1 ⇒αk−1
αk ≤χ <1 withχ=γq(α). (2.17)
So, there is a realM>0 for which
αn≥M
χn ∀n≥q+1. (2.18)
Writingσnq=1/αn(qk=1αk)anddn=[C(α)α]n−σnq, we get
dn= 1 αn
n
k=q+1 αk
=1+
n−1
j=q+1 n−j
k=1 αn−k αn−k+1
≤ n
j=q+1
χn−j≤ 1 1−χ.
(2.19)
And using (2.18), we get
σnq≤ 1 Mχn
q
k=1 αk
. (2.20)
So
C(α)αn≤a+bχn (2.21)
witha=1/(1−χ)andb=(1/M)(qk=1αk).
Remark2.2. Note thatα∈C1 does not imply thatα∈Γ.
2.2. New properties of the operator represented by∆. Throughout this paper, we will denote byDξthe infinite diagonal matrix(ξnδnm)n,m≥1for any given sequenceξ=(ξn)n≥1. Now, we require some lemmas.
Lemma 2.3. The condition ∆∈(sα,s◦ α)◦ is equivalent to∆α =D1/α∆Dα ∈ (c0,c0)and∆∈(sα(c),s(c)α )implies∆α=D1/α∆Dα∈(c,c).
Proof. First,Dαis bijective fromc0intosα◦. In fact, the equationDαX=B, for everyB=(bn)n∈sα◦, admits a unique solutionX=D1/αB=(bn/αn)n∈c0. Suppose now that∆∈(sα,s◦ α)◦ . Then for everyX∈c0, we get successivelyX= DαX∈sα◦, ∆X∈sα◦, and ∆α=D1/α∆Dα∈(c0,c0). Conversely, assume that ∆α∈(c0,c0)and letX∈sα◦. ThenX=DαXwithX∈c0. So,∆X∈Dαc0=sα◦ and∆∈(s◦α,s◦α). By a similar reasoning, we get∆∈(sα(c),sα(c))⇒∆α∈(c,c).
Lemma2.4. The conditionA∈(c,c)is equivalent to the following conditions: (i) A∈S1;
(ii) (anm)n≥1∈cfor eachm≥1; (iii) (∞m=1anm)n≥1∈c.
If for any given sequenceX=(xn)n∈c, with limnxn=l,An(X)is conver-gent for allnand limnAn(X)=l, it is written that
limX=A−limX, (2.22)
andAis called a Toeplitz matrix. We also have the next result.
Lemma2.5. The operatorA∈(c,c)is a Toeplitz matrix if and only if (i) A∈S1;
(ii) limnanm=0for eachm≥1; (iii) limn(∞m=1anm)=1.
Now, we can assert the following theorem.
Theorem2.6. We have successively (i) sα(∆)=sαif and only ifα∈C1; (ii) sα(◦ ∆)=s◦αif and only ifα∈C1; (iii) sα(c)(∆)=sα(c)if and only ifα∈C;
(iv) ∆α=D1/α∆Dαis bijective fromcinto itself withlimX=∆α−limXif and only if
αn−1
αn →0. (2.23)
Proof. (i) We havesα(∆)=sαif and only if∆,Σ∈(sα,sα). This means that ∆,Σ∈Sα, that is,
∆Sα=sup
n≥1
1+αn−1 αn
<∞, ΣSα=sup
n≥1
C(α)αn<∞. (2.24)
Since 0< αn−1/αn≤[C(α)α]n, we deduce that∆,Σ∈Sαif and only ifΣSα<
∞, that is,α∈C1.
(ii) FromLemma 2.3, ifsα(◦ ∆)=sα◦, then∆α=D1/α∆Dα∈(c0,c0). So,∆α∈ (c0,l∞)=S1and since∆α=(dnm)n,m≥1with
dnm=
1 ifm=n,
−αn−1
αn ifm=n−1, 0 otherwise,
we deduce thatαn−1/αn=O(1), n→ ∞. Further, sα(◦ ∆)=s◦α implies Σα = D1/αΣDα∈(c0,c0)andΣα∈(c0,l∞)=S1. SinceΣα=(σnm)n,m≥1with
σnm=
αm
αn ifm≤n, 0 ifm > n,
(2.26)
we deduce that
sup n≥1 1
αn n
k=1 αk
<∞, (2.27)
that is,α∈C1. Conversely, assume that α∈C1. First, ∆∈(sα,s◦ α)◦ . Indeed, from the inequality
αn−1 αn ≤supn≥1
C(α)αn
<∞, (2.28)
we deduce that for everyX∈sα◦,xn/αn=o(1),
xn−xn−1 αn =
xn αn−
xn−1 αn−1
αn−1
αn =o(1) (2.29)
and∆X∈sα◦. Further, takeB=(bn)n≥1∈sα◦. Then there existsν=(νn)n≥1∈c0 such thatbn=αnνn. We must prove that the equation∆X=Badmits a unique solution in the spaces◦α. First, we obtain
X=ΣB= n
k=1 αkνk
n≥1
. (2.30)
In order to show thatX=(xn)n≥1∈sα◦, we will consider any givenε >0. From
Proposition 2.1(iii), the conditionα∈C1 implies thatαn→ ∞. So, there exists an integerNsuch that
Sn= 1 αn
N
k=1 αkνk
≤2ε forn≥N,
sup n≥N+1
νk≤ ε
2 supn≥1
C(α)αn .
(2.31)
WritingRn=1/αn|n
k=N+1αkνk|, we conclude that
Rn≤
sup N+1≤k≤n
νk
C(α)αn≤ ε
Finally, we obtain
xn αn =
αn1
N
k=1 αkνk
+ 1
αn n
k=N+1 αkνk
≤Sn+Rn≤ε forn≥N,
(2.33)
andX∈s0 α.
(iii) As above,s(c)α (∆)=sα(c)if and only if∆α,Σα∈(c,c); and fromLemma 2.4, we have∆α∈(c,c)if and only if(αn−1/αn)n∈c. In fact, we have∆α∈S1and n
m=1dnm=1+αn−1/αn tends to a limit asn→ ∞. Afterwards,Σα∈(c,c)is equivalent to
(a) Σα∈S1, that is,α∈C1;
(b) limn(αm/αn)=0 for allm≥1; (c) α∈C.
FromProposition 2.1(iii), (c) implies thatαntends to infinity, so (c) implies (a) and (b). Finally, from Proposition 2.1(ii), we conclude that α∈Cimplies (αn−1/αn)n∈c. This completes the proof of (iii).
(iv) FromLemma 2.5, it can be easily verified that∆α∈(c,c)and limX= ∆α−limXif and only ifαn−1/αn→0. We conclude, using (iii), sinceαn−1/αn=
o(1)implies thatα∈C.
Remark2.7. InTheorem 2.6(iv), we see thatΣα∈(c,c)and limX=Σα− limXif and only ifαn−1/αn→0. In fact, we must have for eachm≥1,σnm= αm/αn=o(1) (n→ ∞)and
lim n
n
m=1 σnm
=lim
n 1+n−1
m=1 αm αn
=1, (2.34)
and fromProposition 2.1(i), the previous property is satisfied if and only if αn−1/αn→0.
Remark 2.8. It can be seen that the condition(αn−1/αn)n ∈c does not imply thatα∈C1. It is enough to consider C(e)e=(n)n∉c0.
The next corollary is a direct consequence of the previous results.
Corollary2.9. Consider the following properties: (i) α∈C;
(ii) sα(c)(∆)=sα(c); (iii) α∈Γ; (iv) α∈C1;
(v) sα(∆)=sα; (vi) sα◦(∆)=s◦α.
Then (i)(ii)⇒(iii)⇒(iv)(v)(vi).
Corollary2.10. (i)(sα−sα)(◦ ∆)=sα−s◦
αif and only ifα∈C1, (ii)(sα(c)−sα)(◦ ∆)=sα(c)−s◦αif and only ifα∈C,
(iii)α∈Cimplies(sα−s(c)
α )(∆)=sα−s(c)α .
Proof. (i) If∆is bijective fromsα−sα◦into itself, then for everyB∈sα−sα◦, we haveX=ΣB∈sα−s◦
α. Sinceα∈sα−sα◦, we conclude thatΣα∈sα, that is, C(α)α∈l∞. Conversely, fromTheorem 2.6(i) and (ii), it can be easily seen that ∆is bijective fromsα tosαand fromsα◦ tos◦α, sinceα∈C1. So, ∆is bijective fromsα−sα◦ tosα−sα◦.
(ii) Suppose that∆is bijective fromsα(c)−sα◦into itself. Reasoning as above, we haveα∈s(c)α −sα◦andΣα∈sα(c), soD1/αΣα=C(α)α∈c. Conversely, using
Theorem 2.6(i) and (iii), we see that∆is bijective fromsα(c)tosα(c)and fromsα◦ tosα◦ sinceα∈CandC⊂C1. So, (sα(c)−sα)(◦ ∆)=sα(c)−sα◦.
Similarly, (iii) comes from the fact that∆is bijective froms(c)α into itself and fromsαinto itself, sinceα∈C.
Remark2.11. Assume that limn→∞[C(α)α]n=l. Then
xn
αn →Limplies
xn−xn−1 αn →
L
l. (2.35)
Indeed, fromProposition 2.1(ii) (b),αn−1/αn→1−(1/l)and
xn−xn−1 αn =
xn αn−
xn−1 αn−1
αn−1
αn →L−L 1− 1 l
=L
l. (2.36)
3. Generalization to the setssr(∆h)andsα(∆h)forhreal. In this section, we consider the operator∆h, wherehis a real, and give among other things a necessary and sufficient condition to havesα(∆h)=sα.
First, recall that we can associate to any power seriesf (z)=∞k=0akzk, de-fined in the open disk|z|< R, the upper triangular infinite matrixA=ϕ(f )∈
0<r <RSr defined by
ϕ(f )=
a0 a1 a2 · a0 a1 ·
0 a0 ·
·
(3.1)
(see [3,4,5]). Practically, we will writeϕ[f (z)]instead ofϕ(f ). We have the following lemma.
Lemma3.1. (i)The mapϕ:f→Ais an isomorphism from the algebra of the power series defined in|z|< Rinto the algebra of the corresponding matrices
(ii)Letf (z)=∞k=0akzk, witha0≠0, and assume that1/f (z)=
∞ k=0akzk admitsR>0as radius of convergence. Then
ϕ 1 f
=ϕ(f )−1∈ " 0<r <R
Sr. (3.2)
Now, forh∈R−N, we define (see [13])
−h+k−1 k
=−h(−h+1)···(−h+k−1)
k! ifk >0,
−h+k−1 k
=1 ifk=0,
(3.3)
and putting∆+=∆t, we get for anyh∈R,
∆+h=ϕ(1−z)h=ϕ ∞
k=0
−h+k−1 k
zk
for|z|<1. (3.4)
Then if∆h=(τnm)n,m,
τnm=
−h+n−m−1 n−m
ifm≤n,
0 ifm > n.
(3.5)
Using the isomorphismϕ, we get the following proposition.
Proposition3.2(see [5]). (i)The operator represented by∆is bijective from sr into itself for every r >1, and∆+ is bijective fromsr into itself for allr,
0< r <1.
(ii)The operator∆+ is surjective and not injective fromsr into itself for all
r >1.
(iii)For allr≠1and for every integerµ≥1,(∆+)hsr=sr. (iv)We have successively
(α) ifhis a real greater than0andh∉N, then∆hmapssr into itself whenr ≥1, but not for0< r <1; if−1< h <0, then∆hmapssr into itself whenr >1, but not forr=1;
(β) ifh >0andh∉N, then(∆+)hmapssr into itself when0< r≤1, but not ifr >1; if −1< h <0, then (∆+)h mapssr into itself for 0< r <1, but not forr=1.
(v)Lethbe any given integer≥1, Then
A∈sr∆h,sr⇐⇒sup n≥1 ∞
m=1
anmrm−n
<∞ ∀r >1,
A∈'sr∆+h,sr(⇐⇒sup
n≥1 ∞
m=1
anmrm−n
<∞ ∀r∈]0,1[.
(vi)For every integerh≥1,
s1⊂s1∆h⊂s(n
h)n≥1⊂ )
r >1
sr. (3.7)
(vii)Ifh >0andh∉N, thenqis the greatest integer strictly less than(h+1). For allr >1,
Ker'∆+h()
sr=spanV1,V2,...,Vq, (3.8)
where
V1=et, V2=A11,A12,...t,
V3=0,A22,A3,...2 t,..., Vq='0,0,...,Aqq−−11,Aqq−1,...,Aqn−1,... (t
; (3.9)
Aji =i!/(i−j)!, with 0≤j≤i, being the number of permutations ofithings takenj at a time.
We give here an extension of the previous results, wheresris replaced bysα.
Proposition3.3. Lethbe a real greater than 0. The conditionsα(∆h)=sα is equivalent to
γn(h)= 1 αn
n−1
k=1
h+n−k−1 n−k
αk
=O(1) (n→ ∞). (3.10)
Proof. The operator∆h is bijective fromsα into itself if and only if∆h, Σh∈(sα,sα). We have∆h∈(sα,sα)if and only if
D1/α∆hDα∈S1, (3.11)
and using (3.5), we deduce that∆h∈(sα,sα)if and only if
1 αn
n
k=1
−h+n−k−1 n−k
αk=O(1). (3.12)
Further,(Σt)h=ϕ[(1−z)−h], where
ϕ(z)=1+
∞
n=1
h+n−1 n
So,D1/αΣhDα∈S1if and only if (3.10) holds. Finally, sinceh >0, we have
−h+n−k−1 n−k
≤
h+n−k−1 n−k
fork=1,2,...,n−1, (3.14)
and we conclude since (3.10) implies (3.12).
We deduce immediately the next result.
Corollary3.4. Lethbe an integer greater than or equal to 1. The following properties are equivalent:
(i) α∈C1; (ii) sα(∆)=sα; (iii) sα(∆h)=sα; (iv) C(α)(Σh−1α)∈l
∞.
Proof. From the proof of Proposition 3.3, sα(∆h)= sα is equivalent to D1/αΣhDα=C(α)Σh−1Dα∈S1, that is,C(α)(Σh−1α)∈l
∞. So, (iii) and (iv) are
equivalent. It remains to prove that (ii)(iii). Ifsα(∆)=sα,∆and consequently ∆hare bijective fromsαinto itself and condition (iii) holds. Conversely, assume thatsα(∆h)=sαholds. Then (3.10) holds, and since
h+n−k−1 n−k
≥1 fork=1,2,...,n−1, (3.15)
we deduce that
C(α)αn≤γn(h)=O(1), n → ∞. (3.16)
So, (i) holds and (ii) is satisfied.
4. Generalization of well-known sets. In this section, we see that under some conditions, the spaces wα(λ)* , w*α(λ)◦ , w*α∗(λ), cα(λ,µ)+ , c+α(λ,µ)◦ , and
+
cα∗(λ,µ)can be written by means of the setssξors◦ξ.
4.1. Setswα(λ)* ,w*α(λ)◦ , andw*α∗(λ). We recall some definitions and prop-erties of some spaces. For every sequenceX=(xn)n, we define|X| =(|xn|)n and
*
wα(λ)=X∈s|C(λ)|X|∈sα, *
wα(λ)◦ =X∈s|C(λ)|X|∈sα◦
, *
wα∗(λ)=X∈s|X−let∈w*α(λ)◦ for somel∈C.
For instance, we see that
* wα(λ)=
X=xnn∈s|sup n≥1 1
λnαn n
k=1
xk<∞
. (4.2)
If there existA, B >0 such that A < αn < B for all n, we get the well-known spaceswα(λ)* =w∞(λ),w*α(λ)◦ =w0(λ), andw*α∗(λ)=w(λ)(see [12]). It has been proved that ifλis a strictly increasing sequence of reals tending to infinity,w0(λ)andw∞(λ)are BK spaces andw0(λ)has AK, with respect to
the norm
X =C(λ)|X|l∞=sup n
1
λn n
k=1
xk (4.3)
(see [1]).
We have the next result.
Theorem4.1. Letαandλbe any sequences ofU+∗. (i)Consider the following properties:
(a) αn−1λn−1/αnλn→0;
(b) sα(c)(C(λ))=s(c)αλ; (c) αλ∈C1;
(d) wα(λ)* =sαλ; (e) w*α(λ)◦ =sαλ◦ ; (f) w*α∗(λ)=sαλ◦ .
Then (a)⇒(b), (c)(d), and (c)⇒(e) and (f).
(ii)Ifαλ∈C1,*wα(λ),*wα(λ)◦ , andw*α∗(λ)are BK spaces with respect to the norm
Xsαλ=sup n≥1
xn αnλn
, (4.4)
andw*α(λ)◦ =w*α∗(λ)has AK.
Proof. (i) First, we prove that (a)⇒(b). We have
sα(c)
C(λ)=∆(λ)s(c)α =∆Dλsα(c)=∆sαλ(c), (4.5)
and fromProposition 2.1(i) andTheorem 2.6(iii), we get successivelyαλ∈C, ∆sαλ(c)=s
(c)
αλ, and (b) holds.
(c)(d). Assume that (c) holds. Then
*
Since ∆(λ)=∆Dλ, we get ∆(λ)sα =∆sαλ. Now, using (c), we see that ∆ is bijective fromsαλinto itself andwα(λ)=sαλ. Conversely, assume thatwα(λ)= sαλ. Thenαλ∈sαλimplies thatC(λ)(αλ)∈sα, and sinceD1/αC(λ)(αλ)∈s1= l∞, we conclude thatC(αλ)(αλ)∈l∞. The proof of (c)⇒(e) follows on the same
lines of the proof of (c)⇒(d) replacingsαλbysαλ◦ .
We prove that (c) implies (f). TakeX∈w*α∗(λ). There is a complex numberl such that
C(λ)X−let∈s◦α. (4.7)
So
X−let∈∆(λ)s◦
α=∆sαλ,◦ (4.8)
and fromTheorem 2.6(ii),∆s◦αλ=sαλ◦ . Now, since (c) holds, we deduce from
Proposition 2.1(iii) thatαnλn→ ∞andlet∈s◦
αλ. We conclude thatX∈w*α∗(λ) if and only ifX∈let+s◦
αλ=sαλ◦ .
Assertion (ii) is a direct consequence of (i).
4.2. Setscα(λ,µ)+ ,c+α(λ,µ)◦ , andc+α∗(λ,µ). Let α=(αn)n∈U+∗ be a given sequence, we consider now forλ∈U,µ∈sthe space
+
cα(λ,µ)=wα(λ)∆(µ)=
X∈s|∆(µ)X∈wα(λ). (4.9)
It is easy to see that
+
cα(λ,µ)=X∈s|C(λ)∆(µ)X∈sα, (4.10)
that is,
+ cα(λ,µ)=
X=xnn∈s|sup n≥2 1
λnαn n
k=2
µkxk−µk−1xk−1<∞ , (4.11)
see [1]. Similarly, we define the following sets:
+
c◦α(λ,µ)=X∈s|C(λ)∆(µ)X∈s◦α
, +
c∗α(λ,µ)=X∈s|X−let∈c+◦α(λ,µ)for somel∈C. (4.12)
Recall that ifλ=µ, it is written thatc0(λ)=(w0(λ))∆(λ),
andc∞(λ)=(w∞(λ))∆(λ), see [11]. It can be easily seen that
c0(λ)=+ce◦(λ,λ), c∞(λ)=+ce(λ,λ), c(λ)=c+∗e(λ,λ). (4.14)
These sets of sequences are called strongly convergent to 0, strongly conver-gent, and strongly bounded. Ifλ∈U+∗ is a sequence strictly increasing to infinity,c(λ)is a Banach space with respect to
Xc∞(λ)=sup n≥1 1
λn n
k=1
λkxk−λk−1xk−1 (4.15)
with the conventionx0=0. Each of the spacesc0(λ),c(λ), andc∞(λ)is a BK
space, relatively to the previous norm (see [1]). The setc0(λ)has AK and every X∈c(λ)has a unique representation given by
X=let+
∞
k=1
xk−letk, (4.16)
whereX−let∈c0. The scalarlis called the strongc(λ)-limit of the sequence X.
We obtain the next result.
Theorem4.2. Letα,λ, andµbe sequences ofU+∗. (i)Consider the following properties:
(a) αλ∈C1;
(b) cα(λ,µ)+ =sα(λ/µ); (c) c+◦α(λ,µ)=sα(λ/µ)◦ ;
(d) c+∗α(λ,µ)= {X∈s|X−let∈s◦
α(λ/µ)for somel∈C}. Then (a)(b) and (a)⇒(c) and (d).
(ii)Ifαλ∈C1, thencα(λ,µ)+ ,c+α(λ,µ)◦ , andc+α∗(λ,µ)are BK spaces with respect to the norm
Xsα(λ/µ)=sup n≥1
µnxn αnλn
. (4.17)
The setc+◦α(λ,µ) has AK and everyX∈c+α∗(λ,µ)has a unique representation given by (4.16), whereX−let∈s◦
α(λ/µ).
Proof. We show that (a)⇒(b). TakeX∈cα(λ,µ). We have∆(µ)X∈wα(λ), which is equivalent to
X∈C(µ)sαλ=D1/µΣsαλ, (4.18)
and using Theorem 2.6(i),∆and consequently Σare bijective from sαλ into itself. So,Σsαλ=sαλ andX∈D1/µΣsαλ=sα(λ/µ). We conclude that (b) holds. We prove that (b) implies (a). First, putαλ,µ, =((−1)n(λn/µn)αn)n
,
αλ,µ ∈ sα(λ/µ) = cα(λ,µ)+ = sα(λ/µ), and since ∆(µ) = ∆Dµ and Dµαλ,µ, = ((−1)nλnαn)n
≥1, we get|∆(µ),αλ,µ| =(ξn)n≥1, with
ξn= λ1α1
ifn=1, λn−1αn−1+λnαn ifn≥2.
(4.19)
From (b), we deduce thatΣ|∆(µ)αλ,µ| ∈, sαλ. This means that
Cn = 1 αnλn
λ1α1+n k=2
λk−1αk−1+λkαk
=O(1), n → ∞. (4.20)
From the inequality
C(αλ)(αλ)n≤Cn, (4.21)
we obtain (a). The proof of (a)⇒(c) follows on the same lines of the proof of (a)⇒(b) withsαreplaced bysα◦.
We show that (a) implies (d). TakeX∈c+∗α(λ,µ). There existsl∈Csuch that
∆(µ)X−let∈w*α(λ),◦ (4.22)
and from (c)⇒(e) inTheorem 4.1, we have*wα(λ)◦ =s◦αλ. So
X−let∈C(µ)s◦
αλ=D1/µΣs◦αλ, (4.23)
and fromTheorem 2.6(ii),Σsαλ◦ =sαλ◦ , andD1/µΣsαλ◦ =sα(λ/µ)◦ , we conclude that X∈c+∗α(λ,µ)if and only ifX∈let+s◦
α(λ/µ)for somel∈C.
Assertion (ii) is a direct consequence of (i) and of the fact that for every X∈c+∗α(λ), we have
X−let− N
k=1
xk−letk
sα(λ/µ)
= sup n≥N+1
µnxn−l αnλn
=o(1), N → ∞.
(4.24)
We deduce immediately the following corollary.
Corollary4.3. Assume thatα,λ,µ∈U+∗. (i)Ifαλ∈C1andµ∈l∞, then
+
(ii)Then
λ∈Γ ⇒λ∈C1 ⇒c0(λ)=sλ◦, c∞(λ)=sλ. (4.26)
Proof. (i) Sinceµ∈l∞, we deduce, usingProposition 2.1(iii), that there are
K >0 andγ >1 such that
αnλn µn ≥Kγ
n ∀n. (4.27)
So,let∈s◦
α(λ/µ)and (4.25) holds. (ii) comes fromTheorem 4.2 sinceΓ ⊂C1.
Example4.4. We denote bye,the base of the natural system of logarithms. From the well-known Stirling formula, we have
nn+1/2 n! ∼e,
n√1
2π, (4.28)
sos(nn+(1/2)/n!)n=s,e. Further,λ=(n
n/n!)n∈Γ since
λn−1 λn =e,
−(n−1)ln(1+1/(n−1)) →1 ,
e <1. (4.29)
We conclude that
+ ce◦
nn
n!
n, 1 √
n
n
=se◦,, +ce
nn n!
n, 1 √
n
n
=s,e. (4.30)
References
[1] A. M. Al-Jarrah and E. Malkowsky,BK spaces, bases and linear operators, Rend. Circ. Mat. Palermo (2) Suppl. (1998), no. 52, 177–191.
[2] B. de Malafosse,Some properties of the Cesàro operator in the spacesr, Comm. Fac. Sci. Univ. Ankara Ser. A1Math. Statist.48(1999), no. 1-2, 53–71. [3] ,Bases in sequence spaces and expansion of a function in a series of power
series, Mat. Vesnik52(2000), no. 3-4, 99–112.
[4] ,Application of the sum of operators in the commutative case to the infinite matrix theory, Soochow J. Math.27(2001), no. 4, 405–421.
[5] ,Properties of some sets of sequences and application to the spaces of bounded difference sequences of orderµ, Hokkaido Math. J.31(2002), no. 2, 283–299.
[6] ,Recent results in the infinite matrix theory, and application to Hill equa-tion, Demonstratio Math.35(2002), no. 1, 11–26.
[7] ,Variation of an element in the matrix of the first difference operator and matrix transformations, Novi Sad J. Math.32(2002), no. 1, 141–158. [8] B. de Malafosse and E. Malkowsky,Sequence spaces and inverse of an infinite
matrix, Rend. Circ. Mat. Palermo (2)51(2002), no. 2, 277–294.
[10] I. J. Maddox,Infinite Matrices of Operators, Lecture Notes in Mathematics, vol. 786, Springer-Verlag, Berlin, 1980.
[11] F. Móricz,OnΛ-strong convergence of numerical sequences and Fourier series, Acta Math. Hungar.54(1989), no. 3-4, 319–327.
[12] A. Wilansky,Summability Through Functional Analysis, North-Holland Mathe-matics Studies, vol. 85, North-Holland Publishing, Amsterdam, 1984. [13] K. Zeller and W. Beekmann,Theorie der Limitierungsverfahren, Zweite, erweiterte
und verbesserte Auflage. Ergebnisse der Mathematik und ihrer Grenzgebi-ete, vol. 15, Springer-Verlag, Berlin, 1970 (German).
Bruno de Malafosse: Laboratoire de Mathematiques Appliquées Havrais, Institut Uni-versitaire de Technologie Le Havre, Université du Havre, BP 4006, 76610 Le Havre, France