PII. S0161171204209401 http://ijmms.hindawi.com © Hindawi Publishing Corp.
CALCULATIONS ON SOME SEQUENCE SPACES
BRUNO DE MALAFOSSE
Received 16 September 2002
We deal with space of sequences generalizing the well-known spacesw∞p(λ),c∞(λ,µ), re-placing the operatorsC(λ)and∆(µ)by their transposes. We get generalizations of results concerning the strong matrix domain of an infinite matrixA.
2000 Mathematics Subject Classification: 46A45, 40C05.
1. Notations and preliminary results. For a given infinite matrixA=(anm)n,m≥1, the operatorsAnare defined, for any integern≥1, by
An(X)= ∞
m=1
anmxm, (1.1)
whereX=(xn)n≥1, the series intervening in the second member being convergent. 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 [1,2,3,4,5,6,7,
8,10]. Equation (1.2) can be written in the formAX=B, whereAX=(An(X))n≥1. In this paper, we will also consider Aan operator from a sequence space into another sequence space.
A Banach space E of complex sequences with the norm E is a BK space if each projectionPnX=xnis continuous for allX∈E. A BK spaceEis said to have AK, (see [12,13]), ifB=∞m=1bmem, for everyB=(bn)n≥1∈E, (withen=(0,...,1,...), 1 being in thenth position), that is,
∞
m=N+1 bmem
E
→0 (n→ ∞). (1.3)
We will writesfor the set of all complex sequences,l∞,c,c0for the sets of bounded, convergent, and null sequences, respectively. We will denote bycsandl1 the sets of convergent and absolutely convergent series, respectively.
In all that follows we will use the set
U+∗=unn≥1∈s|un>0∀n
From Wilansky’s notations [15], we define for any sequence
α=αnn≥1∈U+∗, (1.5)
and for any set of sequencesE, the set
1
α −1
∗E=xnn≥1∈s xnαn
n∈E
. (1.6)
We will writeα∗Einstead of(1/α)−1∗Efor short. So we get
α∗E=
sα◦ ifE=c0, sα(c) ifE=c, sα ifE=l∞.
(1.7)
We have for instance
α∗c0=sα◦=xnn≥1∈s|xn=o
αnn → ∞. (1.8)
Each of the spacesα∗E, whereE∈ {c0,c,l∞}, is a BK space normed by
Xsα=sup
n≥1 xn αn
, (1.9)
andsα◦ has AK.
Now letα= (αn)n≥1 and β=(βn)n≥1∈U+∗. Sα,β is the set of infinite matrices A=(anm)n,m≥1such that
anmαmm≥1∈l1 ∀n≥1, ∞
m=1
anmαm
=Oβn (n → ∞). (1.10)
Sα,βis a Banach space with the norm
ASα,β=sup
n≥1 ∞
m=1
anmαm βn
. (1.11)
LetEandFbe any subsets ofs. WhenAmapsEintoF, we will writeA∈(E,F), see [11]. So for everyX∈E,AX∈F, (AX∈F will mean that for eachn≥1 the series defined byyn=∞m=1anmxm is convergent and(yn)n≥1∈F). It has been proved in [9] that A∈(sα,sβ)if and only ifA∈Sα,β. So we can write that(sα,sβ)=Sα,β.
Whensα=sβ, we obtain the unital Banach algebraSα,β=Sα, (see [1,2, 3,5,6,10]) normed byASα= ASα,α.
Ifα=(rn)n≥1,Γ
α,Sα,sα,sα◦, andsα(c)are replaced byΓr,Sr,sr,s◦r, andsr(c), respec-tively, (see [1,2,3,5,6,10]). Whenr=1, we obtains1=l∞,s1◦=c0, ands
(c)
1 =c, and puttinge=(1,1,...), we haveS1=Se. It is well known, see [11], that
s1,s1=c0,s1=c,s1=S1. (1.12)
For any subsetEofs, we put
AE= {Y∈s| ∃X∈E, Y=AX}. (1.13)
IfF is a subset ofs, we will denote
F(A)=FA= {X∈s|Y =AX∈F}. (1.14)
We can see thatF(A)=A−1F.
2. Some properties of the operators∆+andΣ+. Here we will deal with the operators represented byC+(λ)and∆+(λ).
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)
So, we putC+(λ)=C(λ)t. It can be proved that the matrix∆(λ)=(cnm)n,m≥ 1with
cnm =
λn ifm=n,
−λn−1 ifm=n−1, n≥2, 0 otherwise,
(2.3)
is the inverse ofC(λ), see [12,14]. Similarly, we put∆+(λ)=∆(λ)t. Ifλ=e, we get the well-known operator of first difference represented by∆(e)=∆and it is usually written asΣ=C(e). Note that∆=Σ−1andΣbelong to any given spaceSRwithR >1. WritingDλ=(λnδnm)n,m≥1, (whereδnm=0 forn≠mandδnn=1 otherwise), we have
∆+(λ)=Dλ∆+. So for any givenα∈U+∗, we see that if(αn−1/αn)|λn/λn−1| =O(1), then∆+(λ)∈(s(α/|λ|),sα). Since Ker∆+(λ)≠0, we are lead to define the set
sα∗
∆+(λ)=sα∆+(λ)s(α/|λ|)=X=xn
n≥1∈s(α/|λ|)|∆+(λ)X∈sα
It can easily be seen that
s(α/|λ|)∗
∆+(e)=s∗ (α/|λ|)
∆+=s∗ α
∆+(λ). (2.5)
2.1. Properties of the sequenceC(α)α. We will use the following sets:
C1=
α∈U+∗ αn1
n
k=1 αk
=O(1) (n→ ∞) ,
C=
α∈U+∗ αn1
n
k=1 αk
∈c
,
C1+=
α∈U+∗
cs 1 αn
∞
k=n αk
=O(1) (n → ∞) ,
Γ=α∈U+∗ lim n→∞
αn−1
αn <1
,
Γ+=α∈U+∗ lim n→∞
αn+1
αn <1
.
(2.6)
Note thatα∈Γ+if and only if 1/α∈Γ. We will see inProposition 2.1that ifα∈C1,α tends to infinity. On the other hand, we see that∆∈Γαimpliesα∈Γ andα∈Γ if and only if there is an integerq≥1 such that
γq(α)= sup n≥q+1
αn−1
αn <1. (2.7)
We obtain the following results in which we put[C(α)α]n=(nk=1αk)/αn.
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, (2.8)
(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[, (2.9)
Proof. Assume thatαn−1/αn→0. Then there is an integerNsuch that
n≥N+1⇒αn−1 αn ≤
1
2. (2.10)
So there exists a realK >0 such thatαn≥K2nfor allnand
αk αn =
αk αk+1···
αn−1 αn ≤
1
2 n−k
forN≤k≤n−1. (2.11)
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
, (2.12)
and sincen−1k=N(1/2)n−k=1−(1/2)n−N→1,(n→ ∞), we deduce that
1 αn
n−1
k=1 αk
=O(1) (2.13)
and([C(α)α]n)∈l∞. Using the identity
C(α)αn=
α1+···+αn−1 αn−1
αn−1 αn +1=
C(α)αn−1
αn−1
αn +1, (2.14)
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. (2.15)
(ii) is a direct consequence of the identity (2.14). (iii) We putΣn=
n
k=1αk. Then for a realM >1,
C(α)αn=
Σn
Σn−Σn−1≤M ∀n.
(2.16)
SoΣn≥(M/(M−1))Σn−1andΣn≥
M/(M−1)n−1α1∀n. Therefore, from
α1 αn
M
M−1 n−1
≤C(α)αn=
Σn
αn≤M, (2.17)
(iv) Ifα∈Γ, there is an integerq≥1 for which
k≥q+1 implies αk−1
αk ≤χ <1, withχ=γq(α). (2.18)
So there is a realM>0 for which
αn≥M
χn ∀n≥q+1. (2.19)
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. (2.20)
Using (2.19), we getσnq≤(1/M)χn(q
k=1αk). So
C(α)αn≤a+bχn (2.21)
witha=1/(1−χ)andb=(1/M)(qk=1αk).
(v) Ifα∈Γ+, there areχ∈]0,1[and an integerq≥1 such that
αk αk−1≤χ
fork≥q. (2.22)
Then for everyn≥q, we have
1 αn
∞
k=n αk
=∞
k=n
αk αn ≤1+
∞
k=n+1 k−n−1
i=0
αk−i αk−i−1 ≤
∞
k=n
χk−n=O(1). (2.23)
This gives the conclusion.
Remark2.2. Note that as a direct consequence ofProposition 2.1, we haveC1!C1+=
Γ!Γ+=φ.
Remark2.3. The conditionα∈C1 does not imply thatα∈Γ, see [8].
2.2. Some new properties of the operators∆and∆+. In the following we will use some lemmas, the next one is well known, see [15].
Lemma2.4. The conditionA∈(c0,c0)is equivalent to
A∈S1, lim
n anm=0 for eachm≥1.
(2.24)
Lemma2.5. If∆+is bijective fromsαinto itself, thenα∈cs.
Proof. Assume thatα∉cs, that is,nαn= ∞. Two cases are possible. (1)e∈Ker∆+!sα. Then∆+cannot be bijective fromsαinto itself.
X=(xn,0)n≥1insα. Then there is a unique scalarx1such that
xn,0=x1− n−1
k=1
αk. (2.25)
So
xn
i,0
αni = αn1i
x1−ni−1 k=1
αk
→ ∞ asi→ ∞, (2.26)
andX∉sα, which is contradictory.
We conclude that each of the propertiese∈Ker∆+!sαande∉Ker∆+!sαis impos-sible and∆+is not bijective fromsαinto itself. This proves the lemma.
Lemma2.6. For everyX∈c0,Σ+(∆+X)=Xand for everyX∈cs,∆+(Σ+X)=X. Proof. It can easily be seen that
Σ+∆+X n=
∞
m=n
xm−xm+1=xn ∀X∈c0,
∆+Σ+X n=
∞
m=n xm−
∞
m=n+1
xm=xn ∀X∈cs.
(2.27)
We can assert the following result, in which we putα+ =(αn+1)n≥1 and sα◦∗(∆+)= sα(◦ ∆+)!s◦
α. Note that from (2.5) we have
s∗α
∆+(e)=s∗ α
∆+=sα∆+sα. (2.28)
Theorem2.7. (i) (a)sα(∆)=sαif and only ifα∈C1, (b) sα(◦ ∆)=sα◦ if and only ifα∈C1,
(c) sα(c)(∆)=sα(c)if and only ifα∈C.
(ii) (a)α∈C1 if and only ifsα+(∆+)=sαand∆+is surjective fromsαintosα+, (b) α∈C1+if and only ifsα∗(∆+)=sαand∆+is bijective fromsαintosα,
(c) α∈C1+implies thats◦∗α (∆+)=s◦αand∆+is bijective froms◦αintosα◦. (iii) α∈C1+if and only ifsα(Σ+)=sαandsα(Σ+)=sαimpliessα(◦ Σ+)=s◦α.
Proof. (i) has been proved in [8].
(ii)(a) Sufficiency. If∆+ is surjective from sα intosα+, then for every B∈sα+ the solutions of∆+X=Binsαare given by
xn+1=x1− n
k=1
wherex1is arbitrary. If we takeB=α+, we getxn=x1−n
k=2αk. So
xn αn=
x1 αn−
1 αn
n
k=2 αk
=O(1). (2.30)
Takingx1= −α1, we conclude that(n−k=11αk)/αn=O(1)andα∈C1. Conversely, assume thatα∈C1. From the inequality
αn−1 αn ≤
1 αn
n
k=1 αk
=O(1), (2.31)
we deduce thatαn−1/αn=O(1)and ∆+∈(sα,sα+). Then for any givenB∈sα+, the solutions of the equation∆+X=Bare given byx1= −uand
−xn=u+ n−1
k=1
bk forn≥2, (2.32)
whereuis an arbitrary scalar. So there exists a realK >0 such that
xn αn =
u+n−1 k=1bk
αn ≤
|u|+Knk=2αk
αn =O(1) (2.33)
andX∈sα. We conclude that∆+is surjective fromsαintosα+. (ii)(b) Necessity. Assume thatα∈C1+. Then∆+∈(sα,sα), since
αn+1 αn ≤
1 αn
∞
k=n αk
=O(1) (n→ ∞). (2.34)
Further, fromsα⊂cs, we deduce, usingLemma 2.4, that for any givenB∈sα,
∆+Σ+B=B. (2.35)
On the other hand,Σ+B=(∞
k=nbk)n≥1∈sα, sinceα∈C1+. So∆+is surjective fromsα intosα. Finally,∆+is injective because the equation
∆+X=O (2.36)
admits the unique solutionX=Oinsα, since
Ker∆+=uet|u∈C (2.37)
Sufficiency. For every B ∈sα, the equation ∆+X =B admits a unique solution in sα. Then fromLemma 2.5,α∈csand sincesα⊂cs, we deduce fromLemma 2.6that X=Σ+B∈sαis the unique solution of∆+X=B. TakingB=α, we getΣ+α∈sα, that is, α∈C1+.
(ii)(c) Ifα∈C1+,∆+is bijective fromsα◦into itself. Indeed, we haveD1/α∆+Dα∈(c0,c0) from (2.34) andLemma 2.4. Furthermore, sinceα∈C1+we havesα◦⊂csand for every B∈sα◦,
∆+Σ+B=B. (2.38)
FromLemma 2.4, we haveΣ+∈(sα,s◦ α)◦ , so the equation∆+X=B admits the solution X0=Σ+B ins◦
α and we have proved that∆+ is surjective fromsα◦ into itself. Finally, α∈C1+ implies thatet∉sα◦, so Ker∆+
!
s◦α= {0}and we conclude that∆+is bijective fromsα◦ into itself.
(iii) comes from (ii), sinceα∈C1+if and only if∆+is bijective fromsαinto itself and
Σ+∆+X=∆+Σ+X=X ∀X∈sα. (2.39)
As a direct consequence ofTheorem 2.7we obtain the following results.
Corollary2.8. LetRbe any real>0. Then
R >1⇐⇒sR(∆)=sR⇐⇒s◦R(∆)=sR◦⇐⇒sR
∆+=sR. (2.40)
Proof. From (i) and (ii) inTheorem 2.7, we see that it is enough to prove thatα= (Rn)n≥1∈C1if and only ifR >1. We have(Rn)n≥1∈C1 if and only ifR≠1 and
R−n n
k=1 Rk
= 1
1−RR
−n+1− R
1−R=O(1) asn → ∞. (2.41)
This means thatR >1 and the corollary is proved.
Using the notationα−=(1,α1,α2,...,αn−1,...)we get the next result.
Corollary2.9. Letα∈U+∗andµ∈U. Then (i) α/|µ| ∈C1 if and only if
sα∆+(µ)=s(α/|µ|)−, (2.42)
(ii) α/|µ| ∈C1+if and only if
sα∗
∆+(µ)=s(α/|µ|). (2.43)
Proof. First we have
Indeed,
X∈sα∆+(µ)⇐⇒Dµ∆+X∈sα⇐⇒∆+X∈s(α/|µ|)⇐⇒X∈s(α/|µ|)∆+. (2.45)
Now, if α/|µ| ∈ C1, from (i) in Theorem 2.7, we have s(α/|µ|)(∆+) = s(α/|µ|)− and sα(∆+(µ))=s(α/|µ|)−. Conversely, assume sα(∆+(µ))=s(α/|µ|)−. Reasoning as above, we gets(α/|µ|)(∆+)=s(α/|µ|)−, and using (i) inTheorem 2.7we conclude thatα/|µ| ∈C1 and (i) holds.
(ii)α/|µ| ∈C1+implies that∆+is bijective froms(α/|µ|)into itself. Thus
sα∗
∆+(µ)=s∗ (α/|µ|)
∆+=s(α/|µ|). (2.46)
This proves the necessity. Conversely, assume thatsα∗(∆+(µ))=s(α/|µ|). Thens(α/|µ|)(∗ ∆+) =s(α/|µ|)and fromTheorem 2.7(ii)(b),α/|µ| ∈C1+and (ii) holds.
2.3. Spaceswαp(λ)andwα+p(λ)for given realp >0. Here we will define sets gener-alizing the well-known sets
w∞(λ)p =X∈s|C(λ)|X|p∈l∞,
w0p(λ)=
X∈s|C(λ)|X|p∈c0, (2.47)
see [9,12, 13, 14, 15]. It is proved that each of the setsw0p=w0p((n)n)and w∞p = w∞((n)n)p is ap-normed FK space for 0< p <1 (i.e., a complete linear metric space for which each projectionPnis continuous) and a BK space for 1≤p <∞with respect to the norm
X =
sup ν≥1
1 2ν
2ν+1−1
n=2ν
xnp
if 0< p <1,
sup ν≥1
1 2ν
2ν+1−1
n=2ν
xnp 1/p
if 1≤p <∞.
(2.48)
The setw0phas the property AK, (i.e., everyX=(xn)n≥1∈w0phas a unique representa-tionX=∞n=1xnent) and every sequenceX=(xn)n≥1∈wphas a unique representation
X=let+ ∞
n=1
xn−letn, (2.49)
wherel∈Cis such thatX−let∈wp
0, (see [4]). Now, letα∈U+∗andλ∈U+∗. We have
wα(λ)p =X∈s|C(λ)|X|p∈sα,
wα+p(λ)=X∈s|C+(λ)|X|p∈sα,
wα◦p(λ)=X∈s|C(λ)|X|p∈sα◦
,
wα◦+p(λ)=
X∈s|C+(λ)|X|p∈s◦ α
.
(2.50)
Theorem2.10. (i) (a)The conditionα∈C1+is equivalent to
wα+p(λ)=s(αλ)1/p. (2.51)
(b) Ifα∈C1+, then
wα◦p(λ)=s(αλ)◦ 1/p. (2.52)
(ii) (a)The conditionαλ∈C1 is equivalent to
wα(λ)p =s(αλ)1/p. (2.53)
(b) Ifαλ∈C1, then
wα◦+p(λ)=s(αλ)◦ 1/p. (2.54)
Proof. Assume thatα∈C1+. SinceC+(λ)=Σ+D1/λ, we have
wα+p(λ)=X|Σ+D1/λ|X|p∈sα=X|D1/λ|X|p∈sαΣ+, (2.55)
and sinceα∈C1+impliessα(Σ+)=sα, we conclude that
wα+p(λ)=X| |X|p∈Dλsα=sαλ=s(αλ)1/p. (2.56)
Conversely, we have(αλ)1/p∈s(αλ)
1/p=wα+p(λ). So
C+(λ)(αλ)1/pp= ∞
k=n αkλk
λk
n≥1
∈sα, (2.57)
that is,α∈C1+and we have proved (i). We obtain (i)(b) by reasoning as above. (ii) Assume thatαλ∈C1. Then
Since∆(λ)=∆Dλ, we get ∆(λ)sα =∆sαλ. Now, from αλ∈C1 we deduce that ∆ is bijective fromsαλ into itself andwα(λ)p =s(αλ)1/p. Conversely, assume thatwα(λ)p =
s(αλ)1/p. Then(αλ)1/p∈s(αλ)1/pimplies that
C(λ)(αλ)∈sα, (2.59)
and sinceD1/αC(λ)(αλ)∈s1=l∞, we conclude thatC(αλ)(αλ)∈l∞. The proof of (ii)(b) follows the same lines as in the proof of the necessity in (ii) replacingsαλbysαλ◦ .
3. New sets of sequences of the form[A1,A2]. In this section, we will deal with the sets
A1(λ),A2(µ)=X∈s|A1(λ)A2(µ)X∈sα, (3.1)
whereA1andA2are of the formC(ξ), C+(ξ),∆(ξ), or∆+(ξ)and we give necessary conditions to get[A1(λ),A2(µ)]in the formsγ.
Letλandµ∈U+∗. For simplification, we will write throughout this section
A1,A2=A1(λ),A2(µ)=X∈s|A1(λ)A2(µ)X∈sα (3.2)
for any matrices
A1(λ)∈∆(λ),∆+(λ),C(λ),C+(λ),
A2(µ)∈∆(µ),∆+(µ),C(µ),C+(µ). (3.3)
So we have for instance
[C,∆]=X∈s|C(λ)∆(µ)X∈sα=wα(λ)∆(µ),.... (3.4)
In all that follows, the conditionsξ∈Γ, or 1/η∈Γ for any given sequencesξandηcan be replaced by the conditionsξ∈C1 andη∈C1+.
3.1. Spaces[C,C],[C,∆],[∆,C], and[∆,∆]. For the convenience of the reader we will write the following identities, whereA1(λ)andA2(µ)are lower triangles and we will use the conventionµ0=0:
[C,C]= X∈s
λn1
n
m=1 µm1
m
k=1 xk
=αnO(1) ,
[C,∆]=
X∈s λn1 n
k=1
µkxk−µk−1xk−1=αnO(1) ,
[∆,C]= X∈s
−λn−1
µn−11
n−1
k=1 xi
+λn µn1
n
k=1 xi
=αnO(1) ,
Note that forα=eandλ=µ,[C,∆]is the well-known set of sequences that are strongly bounded, denoted byc∞(λ), see [9,12,13,14,15]. We get the following result.
Theorem3.1. (i)Ifαλandαλµ∈Γ, then
[C,C]=s(αλµ), (3.6)
(ii)ifαλ∈Γ, then
[C,∆]=s(α(λ/µ)), (3.7)
(iii)ifαandαµ/λ∈Γ, then
[∆,C]=s(α(µ/λ)), (3.8)
(iv)ifαandα/λ∈Γ, then
[∆,∆]=s(α(µ/λ)). (3.9)
Proof. We have for any givenX
C(λ)C(µ)X∈sα (3.10)
if and only ifC(µ)X∈sαC(λ)=s(αλ), sinceαλ∈Γ. So we get
X∈∆(µ)sαλ (3.11)
and the conditionαλµ∈Γ implies∆(µ)sαλ=s(αλµ), which permits us to conclude (i). (ii) Now, for any givenX, the conditionC(λ)(|∆(µ)X|)∈sαis equivalent to
∆(µ)X∈∆(λ)sα=∆sαλ=sαλ, (3.12)
sinceαλ∈Γ. Thus
X∈C(µ)sαλ=D1/µΣsαλ=s(α(λ/µ)). (3.13)
(iii) Similarly,∆(λ)(|C(µ)X|)∈sαif and only if C(µ)X∈sα
∆(λ)=C(λ)sα=D1/λΣsα=s(α/λ), (3.14)
sinceα∈Γ. So
X∈∆(µ)s(α/λ)=∆s(αµ/λ). (3.15)
We conclude sinceαµ/λ∈Γ implies that∆s(αµ/λ)=s(αµ/λ). (iv) Here,
∆(λ)∆(µ)X∈sα if and only if∆(µ)X∈C(λ)sα=s(α/λ), (3.16)
ifα∈Γ. Thus we have
X∈C(µ)s(α/λ)=s(α/λµ) (3.17)
Remark3.2. If we define
A1,A20=
X∈s|A1(λ)A2(µ)X∈s◦α
, (3.18)
we get the same results as inTheorem 3.1, replacing in each case (i), (ii), (iii), and (iv)sξ bysξ◦.
3.2. Sets[∆,∆+],[∆,C+],[C,∆+],[∆+∆],[∆+,C],[∆+∆+],[C+,C],[C+,∆],[C+,∆+],
and[C+,C+]. We get immediately from the definitions of the operators∆(ξ),∆+(η), C(ξ), andC+(η), the following:
∆,∆+=X|λnµnxn−µn+1xn+
1−λn−1µn−1xn−1−µnxn=αnO(1)
,
∆,C+= X|λn
∞ i=n xi µi −λn−1 ∞ i=n−1 xi µi
=αnO(1) ,
C,∆+= X λn1 n
k=1
µkxk−µk+1xk+1=αnO(1) ,
∆+,∆=X|λnµnxn−µn−1xn−
1−λn+1µn+1xn+1−µnxn=αnO(1)
,
∆+,C= X
λnµnn
i=1 xi
−λn+µn+11
n+1 i=1 xi
=αnO(1) ,
∆+,∆+=X|λnµnxn−µn+1xn+1−λn+1µn+1xn+1−µn+2xn+2=αnO(1),
C+,C= X ∞ k=n 1 λk µk1 k i=1 xi
=αnO(1) ,
C+,∆= X ∞ k=n 1
λkµkxk−µk−1xk−1 =αnO(1) ,
C+,∆+= X ∞ k=n 1
λkµkxk−µk+1xk+1 =αnO(1) ,
C+,C+= X ∞ k=n 1 λk ∞ i=k xi µi
=αnO(1) .
(3.19)
We can assert the following result, in which we do the conventionαn=1 forn≤0.
Theorem3.3. (i)Assume thatα∈Γ. Then
∆,∆+=s(α/λµ)− if α λµ∈Γ,
∆,C+=s(α(µ/λ)) if λ α∈Γ.
(3.20)
(ii)The conditionsαλ∈Γ andαλ/µ∈Γ together imply
(iii)The conditionα/λ∈Γ implies
∆+,∆=s(α
n−1/µnλn−1)n=s(1/µ(α/λ)−). (3.22)
(iv)Ifα/λandµ(α/λ)−=(µn(αn−1/λn−1))n∈Γ, then
∆+,C=sµ(α/λ)−. (3.23)
(v)Ifα/λand1/µ(α/λ)−=(αn−1/µnλn−1)n∈Γ, then
∆+,∆+=s((α/λ)−
/µ)−=s(αn−2/λn−2µn−1)n. (3.24)
(vi)If1/αandαλµ∈Γ, then
C+,C=s(αλµ). (3.25)
(vii)If1/αandαλ∈Γ, then
C+,∆=s(α(λ/µ)). (3.26)
(viii)If1/αandα(λ/µ)∈Γ, then
C+,∆+=s(α(λ/µ))−. (3.27)
(ix)If1/αand1/αλ∈Γ, then
C+,C+=s(αλµ). (3.28)
Proof. (i) First, for any givenX, the condition∆(λ)(|∆+(µ)X|)∈sαis equivalent to ∆+(µ)X∈sα∆(λ)=s(α/λ), (3.29)
sinceα∈Γ. SoX∈sαλ(∆+(µ))and applyingCorollary 2.9, we conclude the first part of the proof of (i).
We have∆(λ)(|C+(µ)X|)∈sαif and only if
C+(µ)X∈C(λ)sα=D1/λΣsα. (3.30)
Sinceα∈Γ, we haveΣsα=sαandD1/λΣsα=s(α/λ). Then, forα/λ∈Γ+,X∈[∆,C+]if and only if
X∈w(α/λ)+1 (µ)=s(α(µ/λ)). (3.31)
(ii) For any givenX,C(λ)(|∆+(µ)X|)∈sαis equivalent to
∆+(µ)X∈w1
α(λ), (3.32)
and sinceαλ∈Γ we havewα(λ)1 =sαλ. So
X∈sαλ∆+(µ)=s(α(λ/µ))− (3.33)
(iii) Here,∆+(λ)(|∆(µ)X|)∈sαif and only if
∆(µ)X∈sα
∆+(λ)=s(α/λ)−, (3.34)
sinceα/λ∈Γ. Thus
X∈C(µ)s(α/λ)−=D1/µΣs(α/λ)−=s(αn−1/λn−1µn) (3.35)
if(α/λ)−∈Γ, that is,α/λ∈Γ. (iv) Ifα/λ∈Γ, we get
∆+(λ)C(µ)X∈sα⇐⇒C(µ)X∈sα∆+(λ)
=s(α/λ)−⇐⇒X∈∆(µ)s(α/λ)−. (3.36)
Sinceµ(α/λ)−∈Γ, we conclude that[∆+,C]=s(µ(α/λ)− ). (v) One has
∆+,∆+=X|∆+(µ)X∈sα∆+(λ), (3.37)
and sinceα/λ∈Γ, we get
sα∆+(λ)=s(α/λ)−. (3.38)
We deduce that ifα/λ∈Γ,
∆+,∆+=s(α/λ)−∆+(µ). (3.39)
Then, fromCorollary 2.9, ifα/λ∈Γ and(α/λ)−/µ=(αn−1/λn−1µn)n∈Γ,
s(α/λ)−∆+(µ)=s((α/λ)−/µ)−=s(αn−2/λn−2µn−1)n. (3.40)
(vi) We have
C+(λ)C(µ)X∈sα⇐⇒C(µ)X∈wα+1(λ), (3.41)
and sinceα∈Γ+, we havew+1
α (λ)=sαλ. Then forαλµ∈Γ,X∈[C+,C]if and only if
X∈∆(µ)sαλ=s(αλµ). (3.42)
(vii) The conditionC+(λ)(|∆(µ)X|)∈sαis equivalent to
∆(µ)X∈wα+1(λ), (3.43)
and sinceα∈Γ+, we havew+1
α (λ)=sαλ. Thus
X∈sαλ∆(µ)=D1/µΣsαλ=s(α(λ/µ)), (3.44)
(viii) First, we have
C+,∆+=X|∆+(µ)X∈w+1 α (λ)
, (3.45)
and the conditionα∈Γ+implies thatw+1
α (λ)=sαλ. Thus
C+,∆+=X|∆+(µ)X∈sαλ=sαλ∆+(µ), (3.46)
and we conclude since
sαλ∆+(µ)=s(αλ/µ)− for αλ
µ ∈Γ. (3.47)
(ix) Ifα∈Γ+,
C+,C+=X|C+(µ)X∈wα+1(λ)=sαλ
=wαλ+1(µ). (3.48)
We conclude thatwαλ+1(µ)=s(αλµ), sinceαλ∈Γ+.
Remark3.4. Note that inTheorem 3.3, we have[A1,A2]=sα(A1A2)=(sα(A1))A2 forA1∈ {∆(λ),∆+(λ),C(λ),C+(λ)}andA2∈ {∆(µ),∆+(µ),C(µ),C+(µ)}. For instance, we have
[∆,C]= X
λn
µn− λn−1 µn−1
n−1
i=1 xi+λn
µnxn=αnO(1)
forαµλ ∈Γ. (3.49)
Similarly, under the corresponding conditions given in Theorems3.1and3.3, we get
[∆,∆]=X| −λn−1µn−2xn−2+µn−1λn+λn−1xn−1−λnµnxn=αnO(1),
∆,C+= X
λn
µnxn+
λn−λn−1 ∞
m=n−1 xm
µm =αnO(1) ,
∆,∆+=X| −λn−1µn−1xn−1+µnλn+λn−1xn−λnµn+1xn+1=αnO(1),
∆+,∆=X| −λnµn−1xn−1+λn+λn+1µnxn−λn+1µn+1xn+1=αnO(1).
(3.50)
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Bruno de Malafosse: Laboratoire de Mathématiques Appliquées du Havre (LMAH), Université du Havre, Institut Universitaire de Technologie du Havre, 76610 Le Havre, France
