ANALYTIC SEMIGROUPS
BRUNO DE MALAFOSSE AND AHMED MEDEGHRI
Received 3 May 2006; Revised 8 June 2006; Accepted 15 June 2006
We establish a relation between the notion of an operator of an analytic semigroup and matrix transformations mapping from a set of sequences intoχ, whereχis either of the sets l∞,c0, orc. We get extensions of some results given by Labbas and de Malafosse
concerning applications of the sum of operators in the nondifferential case.
Copyright © 2006 B. de Malafosse and A. Medeghri. This is an open access article dis-tributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is prop-erly cited.
1. Introduction
In this paper, we are interested in the study of operators represented by infinite matrices. Note that in [1], Altay and Bas¸ar gave some results on the fine spectrum of the difference operatorΔacting on the sequence spacesc0andc. Then they dealt with the fine spectrum
of the operatorB(r,s) defined by a matrix band over the sequence spacesc0 andc. In
de Malafosse [3,5], there are results on the spectrum of the Ces`aro matrixC1 and on
the matrixΔconsidered as operators fromsrto itself. Spectral properties of unbounded
operators are used in the theory of the sum of operators. The notion ofgenerators of analytic semigroupwas developed in this way. Recall that this theory was studied by many authors such as Da Prato and Grisvard [2,12], Fuhrman [11], Labbas and Terreni [16, 17]. Some applications can also be found in Labbas and de Malafosse [15] of the sum of operators in the theory of summability in the noncommutative case. Some results were obtained in de Malafosse [4] on the equation
Ax+Bx−λx=y forλ≥0 (1.1)
in areflexive Banach set of sequencesE, wherey∈E,AandBaretwo closed linear operators represented by infinite matrices with domainsD(A)and D(B)included inE.
Here we are interested in some extensions of results given in [15] using similar matri-cesAandB. Recall that the choice of these matrices was motivated by the solvability of
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 67062, Pages1–14
a class of infinite-tridiagonal systems. Then we study some spectral properties ofAand Bconsidered as matrix transformations in the sets (s1/a,l∞) and (s1/β,l∞), or (s01/a,c0) and
(s0
1/β,c0), or (s(1c/a),c) and (s (c)
1/β,c). Then we show that (−A) and (−B) aregenerators of
ana-lytic semigroups, whereD(A) andD(B) are of the formχ(A) andχ(B) withχ=l∞,c0, orc.
Here therelative boundedness with respect toAorBis not satisfied, so we are not within the framework of the classical perturbation theory given by Kato [14] or Pazy [18].
In this paper, we establish a relation between results in summability and basic notions used in the theory of the sum of operators. For this, we need to recall the following.
2. Preliminary results
2.1. Recall of some results in summability. LetM=(anm)n,m≥1 be an infinite matrix
and consider the sequencex=(xn)n≥1. We will define the productMx=(Mn(x))n≥1with
Mn(x)=∞m=1anmxmwhenever the series are convergent for alln≥1. Letsdenote the set
of all complex sequences. We writeϕ,c0,candl∞for the sets of finite, null, convergent, and bounded sequences, respectively. For any given subsetsX,Y of s, we will say that the operator represented by the infinite matrixM=(anm)n,m≥1maps XintoY,that is,
M∈(X,Y), if the series defined byMn(x)=∞m=1anmxmare convergent for alln≥1 and
for allx∈XandMx∈Y for allx∈X. For any subsetXofs, we will write
MX= {y∈s:y=Mx for somex∈X}. (2.1)
IfY is a subset ofs, we will denote the so-called matrix domain by
Y(M)= {x∈s:y=Mx∈Y}. (2.2)
LetX⊂sbe a Banach space, with norm · X. ByᏮ(X), we will denote theset of all
bounded linear operators, mappingX into itself.We will say thatL∈Ꮾ(X) if and only if L:X→Xis a linear operatorand
L∗
Ꮾ(X)=sup x=0
LxX/xX
<∞. (2.3)
It is well known thatᏮ(X) is a Banach algebra with the normL∗
Ꮾ(X). A Banach space
X⊂sis a BK space if the projectionPn:x→xn fromXintoCis continuous for alln.
A BK space X⊃ϕis said to have AK if for everyx∈X,x=limp→∞kp=1xkek, where
ek=(0,..., 1,...), 1 being in thekth position. It is well known that ifX has AK, then
Ꮾ(X)=(X,X), see [9,13,19]. Put nowU+= {x=(x
n)n≥1∈s:xn>0 for alln}. Forξ=(ξn)n≥1∈U+, we will define
the diagonal matrixDξ=(ξnδnm)n,m≥1, (whereδnm=0 for alln=mandδnm=1
other-wise). Forα∈U+, we will writes
α=Dαl∞, (cf. [3–10,15]. The setsαis a BK space with
the normxsα=supn≥1(|xn|/αn). The set of all infinite matricesM=(anm)n,m≥1with
MSα=sup
n≥1
1 αn
∞
m=1
is a Banach algebra with identity normed by · Sα. Recall that if M∈(sα,sα), then Mxsα≤ MSαxsαfor allx∈sα. Thus we obtain the following result, (cf. [7]) where we putB(sα)=Ꮾ(sα) (sα,sα).
Lemma2.1. For any givenα∈U+,B(s
α)=Sα=(sα,sα).
In the same way, we will define the setss0
α=Dαc0andsα(c)=Dαc, (cf. [7]).The sets s0α
ands(αc)are BK spaces with the norm · sαands
0
αhas AK. It was shown in [9,10] that for
any matrixM∈(sα,sα), we get
M∗
Ꮾ(sα)= M ∗
Ꮾ(s0
α)= M ∗
Ꮾ(s(αc))= MSα. (2.5) In all what follows, we will use the next lemma.
Lemma2.2. Letα,β∈U+and letX,Y be subsets ofs. Then
M∈DαX,DβY
iffD1/βMDα∈(X,Y). (2.6)
2.2. Operator generators of analytic semigroups. We recall here some results given in Da Prato and Grisvard [2] and Labbas and Terreni [16,17]. LetEbe a Banach space. We consider twoclosed linear operators A and B, whose domains areD(A) andD(B) in-cluded inE. For everyx∈D(A) D(B), we then define their sumSx=Ax+Bx.
The spectral properties ofAandBare the following: (H) there areCA,CB>0, andεA,εB∈]0,π[ such that
ρ(A)⊃
A
=z∈C:Arg(z)< π−εA
,
(A−zI)−1 £(E)≤
CA
|z| ∀z∈
A
−{0},
ρ(B)⊃
B
=z∈C:Arg(z)< π−εB
,
(B−zI)−1 £(E)≤
CB
|z| ∀z∈
B
−{0},
εA+εB< π.
(2.7)
It is said thatAandBare generators of analytic semigroups not strongly continuous at t=0 and we haveσ(A) σ(−B)=∅andρ(A)ρ(−B)=C.
The following is well known in the commutative case:
(A−ξI)−1(B−ηI)−1−(B−ηI)−1(A−ξI)−1=0 ∀ξ∈ρ(A), η∈ρ(B), (2.8)
ifD(A) andD(B) are densely defined inE, it is well known (cf. [2]) that the bounded operator defined by
Lλ= −21iπ
Γ(B+zI)
whereΓis an infinite-sectorial curve lying inρ(A−λI) ρ(−B) coincides with (A+B− λI)−1.
In the following, we will consider matrix transformationsAandBmapping in a set of sequences and we will show that they satisfy hypothesis (H).
3. Definition of the operatorsAandB
We will consider two infinite matrices and deal with the case whenAandBmap intol∞ orc0and with the case whenAandBmap intoc. In each case, we will study their spectral
properties.
For given sequencesa=(an)n≥1,b=(bn)n≥1,β=(βn)n≥1, andγ=(γn)n≥1, letAand
Bbe the following infinite matrices:
A=
⎡ ⎢ ⎢ ⎢ ⎢ ⎣
a1 b1 O
· ·
O an bn
·
⎤ ⎥ ⎥ ⎥ ⎥
⎦, B=
⎡ ⎢ ⎢ ⎢ ⎢ ⎣
β1 O
· · γn βn
O ·
⎤ ⎥ ⎥ ⎥ ⎥
⎦. (3.1)
3.1. The case whenAandBare operators mapping fromD(A)andD(B)intoE, where
E=l∞orc0. The next conditions are consequences of results given in [15]. WhenEis either of the setsl∞orc0, we assume thatAsatisfies the following properties:
a∈U+, anis strictly increasing, nlim→∞an= ∞, (3.2a)
there isMA>0 such thatbn≤MA∀n. (3.2b)
Similarly, we assume thatBsatisfies the next conditions:
β∈U+, lim
k→∞β2k=L=0, (3.3a)
lim
k→∞ β2k+1
a2k+1= ∞,
(3.3b)
(α) there isMB>0 such thatγ2k≤MB∀n, (β)γ2k+1=o(1) (n−→ ∞). (3.3c)
3.2. The case whenAandBare operators mappingD(A)andD(B)intoc. Here we need to recall the characterization of (c,c).
Lemma3.1. A∈(c,c)if and only if (i)A∈S1,
(ii) limn→∞∞m=1anm=lfor somel∈C,
(iii) limn→∞anm=lmfor somelm∈C,m=1, 2,....
We will see thatD(A)=c(A)=s(1c/a) andD(B)=c(B)=s (c)
1/β. Here we will show that
Proposition3.2. Leta,β∈U+. Then
s(1c/a) s (c) 1/β, s
(c) 1/βs
(c)
1/a (3.4)
if and only ifβ/a,a/β /∈c. Proof. The inclusion s(1c/a) ⊂s
(c)
1/β means thatI∈(s (c) 1/a,s
(c)
1/β) and by Lemma 2.2, we have
Dβ/a∈(c,c). From Lemma 3.1, we conclude that s(1c/a) ⊂s (c)
1/β if and only ifβ/a∈c. So
s(1c/a) s (c)
1/β is equivalent toβ/a /∈c. Similarly, we have s (c) 1/βs
(c)
1/a if and only ifa/β /∈c.
This completes the proof.
We assume thatAandBsatisfy the following hypotheses. The matrixAis defined in (3.1) with
a∈U+, anis strictly increasing, nlim→∞an= ∞, (3.5a)
b∈c. (3.5b)
ForBgiven in (3.1), we do the following hypotheses:
β∈U+, lim
n→∞βn= ∞, (3.6a)
lim
k→∞ β2k+1
a2k+1 = ∞,
lim
k→∞ β2k
a2k =l=0, (3.6b)
γ∈c. (3.6c)
This lead to the next remark.
Remark 3.3. The choice ofβin (3.6b) is justified byProposition 3.2and so neither of the setsD(A)=s(1c/a) andD(B)=s
(c)
1/βis embedded in the other one. We will see inProposition
5.7andTheorem 5.8that we need to have (3.5a) and (3.6a). Then we will see thatAand Bare closed operators whenb,γ∈c. Finally, notice thatD(B)⊂cmeans 1/β∈cwhich is trivially satisfied in (3.6a) and it is the same forA.
4. First properties of the operatorsAandB
4.1. The case when the operatorsAandBare considered as matrix maps fromD(A)
andD(B)intoE, whereEis equal tol∞, orc0. In this section, we will assumeAandB satisfy (3.2) and (3.3). For the convenience of the reader, recall the following well-known results.
Lemma4.1. (i)A∈(l∞,l∞)if and only ifA∈S1.
(ii)A∈(c0,c0)if and only ifA∈S1andlimn→∞anm=0for eachm=1, 2,....
Proposition4.2. (i)A∈(s1/a,l∞)andA∈(s01/a,c0).
Proof. (i) We have [AD1/a]nn=1 and [AD1/a]n,n+1=bn/an+1for alln, and [AD1/a]nm=0
otherwise. Then
AD1/a S1=sup
n
1 +bn an+1
=O(1) (n−→ ∞). (4.1)
So byLemma 4.1, we haveAD1/a∈(l∞,l∞) and byLemma 2.2,A∈(s1/a,l∞). The proof is
similar forA∈(s0
1/a,c0) note that in this case, [AD1/a]nm→0 (n→ ∞) for eachm≥1.
(ii) Now [BD1/β]nn=1 and [BD1/β]n,n−1=γn/βn−1for alln, and [BD1/β]nm=0
other-wise. By (3.3a), (3.3c)(β), we have γ2n+1
β2n =o(1) (n−→ ∞) (4.2)
and by (3.2a), (3.3b) and (3.3c)(α), we get γ2n
β2n−1 =
γ2n
a2n−1
a2n−1
β2n−1 =o(1) (n−→ ∞). (4.3)
Then
BD1
/βS1=sup n
1 +γn βn−1
=O(1) (n−→ ∞). (4.4)
SoBD1/β∈(l∞,l∞) andB∈(s1/β,l∞). Finally, we obtainB∈(s01/β,c0) reasoning as above.
We deduce that ifE=l∞=s1, the matrixAis defined onD(A)=s1/aandBis defined
onD(B)=s1/β. It can be shown thatl∞(A)=s1/aandl∞(B)=s1/β. WhenE=c0, we will
see inTheorem 5.6(i), (ii) thatD(A)=c0(A)=s01/aandD(B)=c0(B)=s01/β. We deduce
from (3.3a), (3.3b) that in each case, neither of the setsD(A) andD(B) is embedded in the other.
4.2. The case when the operatorsAandBare considered as matrix maps fromD(A)
andD(B)intoc. We assume thatAandBsatisfy (3.5) and (3.6). From the preceding, we immediately get the following.
Proposition4.3. A∈(s1(c/a),c)andB∈(s (c) 1/β,c).
Proof. It is enough to notice that by (3.5) we have (1 +bn/an+1)n≥1∈c. Then fromLemma
3.1, we conclude thatA∈(s(1c/a),c). We also have by (3.6),
γn
βn−1 −→
0 (n−→ ∞) (4.5)
soBD1/β∈(c,c) andB∈(s(1c/β),c).
5. The matricesAandBas operator generators of an analytic semigroup
In this section, we will show thatAandBare generators of analytic semigroup in each caseE=l∞,E=c0, orE=c.
5.1. The case whenAandBare considered as matrix maps fromD(A)andD(B)intoE, whereE=l∞orc0. In this section,AandBsatisfy (3.2) and (3.3). The next result was shown in [15] in the case whenA∈(s1/a,l∞) andB∈(s1/β,l∞) withan=an,a >1, andβ
was defined byβ2n=1 andβ2n+1=(2n+ 1)! for alln, so we omit the proof.
Proposition5.1. In the spacel∞, the two linear operatorsAandBare closed and satisfy the following:
(i)D(A)=s1/a,
(ii)D(B)=s1/β,
(iii)D(A)=l∞,D(B)=l∞,
(iv)there areεA,εB>0(withεA+εB< π) such that
(A−λI)−1∗
Ꮾ(l∞)≤
M
|λ| ∀λ=0, Arg(λ)≥εA, (B+μI)−1∗
Ꮾ(l∞)≤
M
|μ| ∀μ=0, Arg(μ)≤π−εB.
(5.1)
This result shows that−Aand−Bsatisfy hypothesis (H) andσ(−A) σ(B)=∅. So −Aand−Bare generators of the analytic semigroupse(−At)ande(−Bt)not strongly
con-tinuous att=0. We have similar results whenAandBare matrix maps intoc0. We require
some elementary lemmas whose proofs are left to the reader. Lemma5.2. Letε∈]0,π/2[and letx0>0be a real. Then
x0−λ≥x0sinε ∀λ∈C withArg(λ)≥ε. (5.2)
Lemma5.3. Letx0>0be a real. Then
x0−λ≥ |λ|sinθ ∀λ= |λ|eiθ∈/ R−,
x0−λ≥ |λ| ∀λ∈R−. (5.3)
We can state the following result where we will use the fact that for anyα∈U+, since
s0
αis a BK space with AK, we haveᏮ(s0α)=(s0α,s0α). As we have seen in (2.5), for any matrix
C∈(s0
α,s0α), we haveC∗Ꮾ(s0 α)= C
∗
(s0
α,s0α)= CSα.
Proposition5.4. (i)LetεA∈]0,π/2[. For everyλ∈Cwith|Arg(λ)| ≥εA, the infinite
matrixA−λIconsidered as an operator ins01/ais invertible and
(A−λI)−1∈c 0,s01/a
(ii)LetεB∈]0,π/2[. For everyμ∈Cwith|Arg(μ)| ≤π−εB, the infinite matrixB+μI
considered as an operator ins01/βis invertible and
(B+μI)−1∈c 0,s01/β
. (5.5)
Proof. (i) FixεA∈]0,π/2[ and consider the infinite-sectorial set
εA
=λ∈C:Arg(λ)< εA
. (5.6)
For anyλ /∈εA, put
χn=abn
n−λ (5.7)
and Dλ=D(1/(an−λ))n. Then [Dλ(A−λI)]nn =1, [Dλ(A−λI)]n,n+1=χn for all n and
[Dλ(A−λI)]nm=0 otherwise. ByLemma 5.2, we have
χn≤ MA
ansinεA ∀n and all λ /∈
εA
. (5.8)
Sinceantends to infinity asntends to infinity, there isn0such that
χn≤1
2 ∀n≥n0 and allλ /∈
εA
. (5.9)
Consider now the infinite matrix
Tλ= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ·
Tλ−1 · 0
· · · 1 0 1 · · ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ , (5.10)
whereTλis the matrix of ordern0defined by [Tλ]nn=1 for 1≤n≤n0; [Tλ]n,n+1=χnfor
1≤n≤n0−1, and [Tλ]nm=0 otherwise. Elementary calculations give
Tλ−1=
⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
1 −χ1 χ1χ2 −χ1χ2χ3 · · ·
1 −χ2 χ2χ3 · · ·
1 −χn χnχn+1 ·
·
0
PuttingᏭλ=Dλ(A−λI)Tλ we easily get [Ꮽλ]nn=1 for alln, [Ꮽλ]n,n+1=χnforn≥n0,
and [Ꮽλ]nm=0 otherwise. Then byLemma 2.2and since the sequenceais increasing, we
get
I−Ꮽλ∗
(s0 1/a,s01/a)=
I−Ꮽλ
S1/a=sup
n≥n0
χn an
an+1
≤1
2 ∀λ /∈
εA
. (5.12)
SinceᏭλ=Ꮽλ−I+I∈(s01/a,s01/a) and (5.12) holds,Ꮽλ is invertible in the Banach
al-gebra of all bounded operators Ꮾ(s01/a)=(s01/a,s01/a) mapping s01/a to itself and Ꮽ−λ1∈
(s01/a,s01/a). Then for any given y∈c0, we successively get y=Dλy=(yn/(an−λ))n≥1∈
s01/a,Ꮽ−λ1y∈s01/a,Tλ(Ꮽ−λ1y)∈s01/a, and (A−λI)−1=TλᏭλ−1Dλ∈(c0,s01/a). So we have
shown (i).
(ii) ForεB∈]0,π/2[, letΣB= {μ∈C:|Arg(μ)| ≤π−εB}and put
χn= γn
βn+μ. (5.13)
To deal with the inverse ofB+μI, we need to study the sequences|γ2k+1|/β2kand|χ2k|β2k/
β2k−1. By (3.3a), (3.3b), we have
γ2k+1
β2k −→0 (k−→ ∞). (5.14)
On the other hand, for everyμ∈ΣB, we get
χ
2k
β2k
β2k−1 ≤
MB
β2ksinεB
β2k
β2k−1=
MB
sinεB
1 β2k−1
, (5.15)
1 β2k−1 =
a2k−1
β2k−1
1
a2k−1=o(1) (k−→ ∞). (5.16)
From (5.14) and (5.16), we deduce that there isn1such that
γ2k+1 1 β2k≤
1
2sinεB for 2k+ 1≥n1,
χ
2k
β2k
β2k−1 ≤
1
2 for 2k≥n1∀μ∈ΣB.
(5.17)
As in (i), define the matricesDμ=D(1/(βn+μ)n)and
Rμ= ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝
· R−1
μ · 0
· · · 1
0 1
· ·
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠
whereRμis the matrix of ordern1−1 defined by [Rμ]nn=1 for alln, by [Rμ]n,n−1=χnfor
2≤n≤n1−1, and by [Rμ]nm=0 otherwise. Then elementary calculations show that the
matrixᏮμ=RμDμ(B+μI) is defined by [Ꮾμ]nn=1 for alln, by [Ꮾμ]n,n−1=χnforn≥n1,
and by [Ꮾμ]nm=0 otherwise. Then for anyμ∈ΣB,
I−Ꮾμ∗
(s0
1/β,s01/β)=nsup≥n 1
χn βn βn−1
=maxτ1,τ2
, (5.19)
where
τ1= sup k≥n1/2
χ
2k
β2k
β2k−1
, τ2= sup
k≥(n1−1)/2
χ
2k+1
β2k+1
β2k .
(5.20)
ByLemma 5.2, we get|β2k+μ| ≥β2ksinεBfor allμ∈ΣBandτ1≤1/2. Then
τ2≤1
2sinεB 1
β2k+1sinεBβ2k+1=
1
2 for 2k≥n1. (5.21)
This impliesI−Ꮾμ∗(s0
1/β,s01/β)≤1/2. Reasoning as in (i) with (B+μI)
−1=Ꮾ−1
μ RμDμ, we
conclude thatB+μI considered as an operator froms01/β intoc0 is invertible and (B+
μI)−1∈(c
0,s01/β) for allμ∈ΣB. This concludes the proof.
Remark 5.5. As a direct consequence of the preceding, it is trivial that
c0(A−λI)=s01/a ∀λ∈C, Arg(λ)≥εA,
c0(B+μI)=s01/β ∀μ∈C, Arg(μ)≤π−εB.
(5.22)
We immediately obtain the next result.
Theorem5.6. In the spacec0, the two linear operatorsAandBare closed and satisfy the
following:
(i)D(A)=c0(A)=s01/a,
(ii)D(B)=c0(B)=s01/β,
(iii)D(A)=c0,D(B)=c0,
(iv)there areεA,εB>0(withεA+εB< π) such that (A−λI)−1∗
Ꮾ(c0)≤
M
|λ| ∀λ=0, Arg(λ)≥εA,
(B+μI)−1∗
Ꮾ(c0)≤
M
|μ| ∀μ=0, Arg(μ)≤π−εB.
(5.23)
Proof. Show thatAis a closed operator. For this, consider a sequencexp=(xnp)n≥1
tend-ing tox=(xn)n≥1inc0, asptends to infinity, wherexp∈s01/afor allp. ThenAxp→y(p→
∞) inc0, that is for anyn, we haveAn(xp)→An(x)=yn(p→ ∞). It remains to show that
x∈s01/a. For this, note that sinceb∈l∞andx∈c0, we conclude thatanxn=yn−bnxn+1
(i)⇒(ii). First by Proposition 4.2, we have A∈(s01/a,c0). It remains to show that
c0(A)=s01/a. Letx∈s10/a. Since (bn/an+1)n≥1∈c0, we deduce that
An(x)=anxn+ bn
an+1an+1xn+1=o
(1) (n−→ ∞) (5.24)
and we have shown thats01/a⊂c0(A).
Now let x∈c0(A). Then y=Ax∈c0. In the proof of Proposition 5.4, we can take
λ=0. Indeed, there isn0such thatχn= |bn|/an≤1/2 for alln≥n0. Theny∈c0implies
D1/ay=(yn/an)n≥1∈s01/a, Ꮽ−1D1/ay∈s01/a, and x=A−1y=T0Ꮽ−01D1/ay∈s01/a. This
shows thatc0(A)⊂s01/aand sinces01/a⊂c0(A), we conclude thatc0(A)=s01/a. The proof is
similar forB.
(iii) Letx=(1/an)n≥1∈c0and assumexp=(xnp)n≥1tend toxins01/a, that is,
x
p−xs1/a=sup
n
an
xnp− 1
an
−→0 (p−→ ∞). (5.25)
Sinceantends to infinity, we should havexnp→1/anandanxnp→1 (p→ ∞) for alln.
This contradicts the fact thatxp∈s0
1/afor all p. The reasoning is the same forB. So (iii)
holds.
(iv) ByProposition 5.4(i), we have seen that for anyλ /∈εA (A−λI) −1∈(c
0,s01/a),
Tλ∈S1and using the notation of the proof ofProposition 5.4, we get
(A−λI)−1y
l∞=TλᏭ
−1
λ yl∞≤TλS1Ꮽ −1 λ S1D
λS1yl∞. (5.26) Then byLemma 5.3, we successively get
D
λS1=sup n≥1
1
an−λ
≤
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
1
|λ|sinθ forλ= |λ|eiθ∈/R−, 1
|λ| forλ∈R−.
(5.27)
Now as we have seen in (5.9), we haveI−ᏭλS1=supn≥n0(|χn|)≤1/2 and then in the Banach algebraS1, we easily get
Ꮽ−1 λ S1≤
∞
m=0
I−Ꮽ
λmS1≤ ∞
m=0
2−m=2. (5.28)
Finally, by (5.9), (5.10), and (5.11), we have supλ /∈
εATλS1<∞and we conclude that (A−λI)−1∗
Ꮾ(c0)≤M/|λ| for allλ /∈
εA. We get similar results forB+μI. This
con-cludes the proof of (iv).
Proposition5.7. (i)LetεA∈]0,π/2[. For everyλwith|Arg(λ)| ≥ε, the infinite matrix
A−λIconsidered as an operator ins(1c/a) is invertible and
(A−λI)−1∈$c,s(1c/a) %
. (5.29)
(ii) Let εB∈]0,π/2[. For everyμ with|Arg(μ)| ≤π−εB, the infinite matrix B+μI
considered as an operator ins(1c/β) is invertible and
(B+μI)−1∈$c,s(c) 1/β %
. (5.30)
Proof. (i) First, it can easily be seen that since
1 + bn an−λ
an
an+1 −→1 (n−→ ∞)∀λ /∈
εA
(5.31)
andᏭλ∈S1/a, we haveᏭλ∈(s(1c/a),s (c)
1/a). Then using the notation ofProposition 5.4, there
isn0such that
I−Ꮽλ∗
Ꮾ(s(1c/a))=
I−Ꮽλ∗
S1/a=sup
n≥n0
χn an
an+1
≤1
2 ∀λ /∈
εA
. (5.32)
Then Ꮽ−λ1∈Ꮾ(s(1c/a)) S1/a, and sinceᏭ−λ1∈S1/a is an infinite matrix, we haveᏭ−λ1∈
(s(1c/a),s (c)
1/a). Then for any y∈c, we successively get Dλy=(yn/(an−λ))n≥1∈s(1c/a),
Ꮽ−1
λ (Dλy)∈s (c)
1/a, and Tλ[Ꮽλ−1(Dλy)]∈s (c)
1/a. We conclude that (A−λI)−1∈(c,s (c) 1/a) for
everyλwith|Arg(λ)| ≥ε.
(ii) Reasoning as inProposition 5.4, there areM,M>0 such that for everyμ∈ΣB,
γ2k
β2k+μ β2k
β2k−1 ≤
M β2ksinεB
β2k
β2k−1 =o(1) (k−→ ∞),
γ2k+1
β2k+1+μ
β2k+1
β2k ≤
M β2k+1sinεB
β2k+1
β2k =o(1) (k−→ ∞).
(5.33)
We deduce that there isn1with
τ1= sup k≥n1/2
γ2k
β2k+μ β2k
β2k−1≤
1
2, τ2=k≥(supn1−1)/2
γ2k+1
β2k+1+μ
β2k+1
β2k ≤
1
Now by (5.33), we have
γn
βn+μ
βn
βn−1 −→0 (n−→ ∞) (5.35)
andᏮμ∈(s(1c/β),s (c)
1/β) for allμ∈ΣB. Thus I−Ꮾμ∗
(s(1c/β),s (c) 1/β)=
I−Ꮾμ∗
S1/β=sup
n≥n1
γn
βn+μ
βn
βn−1
=maxτ1,τ2
≤1 2
(5.36)
andᏮ−1 μ ∈(s
(c) 1/β,s
(c)
1/β) for allμ∈ΣB. Reasoning as in (i) with (B+μI)−1=Ꮾ−μ1RμDμ, we
conclude thatB+μIconsidered as operator froms1(c/β) tocis invertible and (B+μI)−1∈
(c,s(1c/β)) for allμ∈ΣB. This completes the proof.
We can state the following.
Theorem5.8. In the space c, the two linear operatorsAandBare closed and satisfy the following:
(i)D(A)=c(A)=s(1c/a),
(ii)D(B)=c(B)=s(1c/β),
(iii)D(A)=c,D(B)=c,
(iv)there areεA,εB>0(withεA+εB< π) such that (A−λI)−1∗
Ꮾ(c)≤
M
|λ| ∀λ=0, Arg(λ)≥εA, (B+μI)−1∗
Ꮾ(c)≤
M
|μ| ∀μ=0, Arg(μ)≤π−εB.
(5.37)
Proof. Show thatAis a closed operator. For this, consider a sequencexp=(xnp)n≥1
tend-ing tox=(xn)n≥1inc, asptends to infinity, wherexp∈s(1c/a)for allp. ThenAxp→y(p→∞)
inc, that is,An(xp)→An(x)=yn(p→ ∞) for alln. It remains to show thatx∈s(1c/a). For
this, note that sinceb∈candx∈c, we conclude thatanxn=yn−bnxn+1tends to a limit
asntends to infinity. The proof forBis similar. The proof of statements (i) and (ii) comes fromProposition 4.3and follows the same lines as that forTheorem 5.6(i), (ii).
(iii) Follows exactly the same lines as that in the proof given in the case whenE=l∞ in [15, Proposition 3, page 196].
The proof of (iv) is a consequence ofProposition 5.7(i) and follows exactly the same
lines as that in the proof ofTheorem 5.6.
Acknowledgments
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Bruno de Malafosse: Laboratoire de Math´ematiques Appliqu´ees du Havre (LMAH), Universit´e du Havre, BP 4006 IUT Le Havre, 76610 Le Havre, France
E-mail address:bdemalaf@wanadoo.fr
Ahmed Medeghri: Laboratoire de Math´ematiques Pures et Appliqu´ees, Universit´e de Mostaganem, BP 227, Mostaganem 27000, Algeria
