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MATRIX TRANSFORMATIONS AND GENERATORS OF
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 a nalytic semigroup and
matrix transformations mapping from a set of sequences into χ,whereχ is either of the
sets l
∞
, c
0
,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
operator Δ acting on the sequence spaces c
0
and c. Then they dealt with the fine spectrum
of the operator B(r,s) defined by a matrix band over the sequence spaces c
0
and c.In
de Malafosse [3, 5], there are results on the spectrum of the Ces
`
aro matrix C
1
and on
the matrix Δ considered as operators from s

r
to itself. Spectral proper ties of unbounded
operators are used in the theory of the sum of operators. The notion of generators of
analytic semigroup was 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]ofthesumof
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 for λ ≥ 0 (1.1)
in a reflexive Banach set of sequences E,wherey
∈ E, A and B are two closed linear operators
represented by infinite matrices with domains D(A) and D( B) included in E.
Here we are interested in some extensions of results given in [15] using similar matri-
ces A and B. Recall that the choice of these matrices was motivated by the solvability of
Hindaw i Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 67062, Pages 1–14
DOI 10.1155/JIA/2006/67062
2 Matrix and generators of analytic semigroups
a class of infinite-tr idiagonal systems. Then we study some spectral properties of A and
B considered as matrix transformations in the sets (s
1/a
,l
∞
)and(s
1/β
,l
∞
), or (s

0
1/a
,c
0
)and
(s
0
1/β
,c
0
), or (s
(c)
1/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 form χ(A)andχ(B)withχ
= l
∞
, c
0
,orc.
Here the relative boundedness with respect to A or B is not satisfied, so we are not within
the framework of the classical pertur bation theory given by Kato [14]orPazy[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. Let M
= (a

nm
)
n,m≥1
be an infinite matrix
and consider the sequence x
= (x
n
)
n≥1
. We will define the product Mx = (M
n
(x))
n≥1
with
M
n
(x) =

∞
m=1
a
nm
x
m
whenever the series are convergent for all n ≥ 1. Let s denote the set
of all complex sequences. We write ϕ, c
0
, c and l
∞
for the sets of finite, null, convergent,

and bounded sequences, respectively. For any g iven subsets X, Y of s,wewillsaythat
the operator represented by the infinite matrix M
= (a
nm
)
n,m≥1
maps X into Y, that is,
M
∈ (X,Y), if the series defined by M
n
(x) =

∞
m=1
a
nm
x
m
are convergent for all n ≥ 1and
for all x
∈ X and Mx ∈ Y for all x ∈ X. For any subset X of s,wewillwrite
MX
={y ∈ s : y = Mx for some x ∈ X}. (2.1)
If Y is a subset of s, we will denote the so-called matrix domain by
Y(M)
={x ∈ s : y = Mx ∈ Y}. (2.2)
Let X
⊂ s beaBanachspace,withnorm·
X
.ByᏮ(X), we will denote the setofall

bounded linear operators, mapping X into itself. We will say that L
∈ Ꮾ(X)ifandonlyif
L : X
→ X is a linear operator and
L
∗
Ꮾ(X)
= sup
x=0


Lx
X
/x
X

< ∞. (2.3)
It is well known that Ꮾ(X)isaBanachalgebrawiththenorm
L
∗
Ꮾ(X)
.ABanachspace
X
⊂ s is a BK space if the projection P
n
: x → x
n
from X into C is continuous for all n.
ABKspaceX
⊃ ϕ is said to have AK if for every x ∈ X, x = lim

p→∞

p
k
=1
x
k
e
k
,where
e
k
= (0, ,1, ), 1 being in the kth position. It is well known that if X has AK, then
Ꮾ(X)
= (X,X), see [9, 13, 19].
Put now U
+
={x = (x
n
)
n≥1
∈ s : x
n
> 0foralln}.Forξ = (ξ
n
)
n≥1
∈ U
+
,wewilldefine

the diagonal matrix D
ξ
= (ξ
n
δ
nm
)
n,m≥1
,(whereδ
nm
= 0foralln = m and δ
nm
= 1 other-
wise). For α
∈ U
+
,wewillwrites
α
= D
α
l
∞
,(cf.[3–10, 15]. The set s
α
is a BK space with
the norm
x
s
α
= sup

n≥1
(|x
n
|/α
n
). The set of all infinite matrices M = (a
nm
)
n,m≥1
with
M
S
α
= sup
n≥1

1
α
n
∞

m=1


a
nm


α
m


< ∞
(2.4)
B. de Malafosse and A. Medeghri 3
is a Banach algebra with identity normed by
·
S
α
.RecallthatifM ∈ (s
α
,s
α
), then
Mx
s
α
≤M
S
α
x
s
α
for all x ∈ s
α
. Thus we obtain the following result, (cf. [7]) where
we put B(s
α
) = Ꮾ(s
α
)


(s
α
,s
α
).
Lemma 2.1. For any given α
∈ U
+
, B(s
α
) = S
α
= (s
α
,s
α
).
In the same way, we will define the sets s
0
α
= D
α
c
0
and s
(c)
α
= D
α

c,(cf.[7]). The sets s
0
α
and s
(c)
α
are BK spaces with the norm ·
s
α
and s
0
α
has AK. It was shown in [9, 10]thatfor
any matrix M
∈ (s
α
,s
α
), we get
M
∗
Ꮾ(s
α
)
=M
∗
Ꮾ(s
0
α
)

=M
∗
Ꮾ(s
(c)
α
)
=M
S
α
. (2.5)
In all what follows, we will use the next lemma.
Lemma 2.2. Let α, β
∈ U
+
and let X, Y be subsets of s. Then
M
∈

D
α
X,D
β
Y

iff D
1/β
MD
α
∈ (X,Y). (2.6)
2.2. O perator generators of analytic semigroups. We recall here some results given in

Da Prato and Grisvard [2] and Labbas and Terreni [16, 17]. Let E be a Banach space. We
consider two closed linear operators A and B, whose domains are D(A)andD(B)in-
cluded in E.Foreveryx
∈ D(A)

D( B), we then define their sum Sx = Ax + Bx.
The spectral properties of A and B are the following:
(H) there are C
A
,C
B
> 0, and ε
A
,ε
B
∈ ]0,π[suchthat
ρ(A)
⊃

A
=

z ∈ C :


Arg(z)


<π− ε
A


,


(A − zI)
−1


£(E)
≤
C
A
|z|
∀
z ∈

A
−{0},
ρ(B)
⊃

B
=

z ∈ C :


Arg(z)



<π− ε
B

,


(B − zI)
−1


£(E)
≤
C
B
|z|
∀
z ∈

B
−{0},
ε
A
+ ε
B
<π.
(2.7)
It is said that A and B are generators of analytic semigroups not strongly continuous at
t
= 0andwehaveσ(A)


σ(−B) = ∅ and ρ(A)

ρ(−B) =
C
.
Thefollowingiswellknowninthecommutativecase:
(A
− ξI)
−1
(B − ηI)
−1
− (B − ηI)
−1
(A − ξI)
−1
= 0 ∀ξ ∈ ρ(A), η ∈ ρ(B), (2.8)
if D(A)andD(B) are densely defined in E,itiswellknown(cf.[2]) that the bounded
operator defined by
L
λ
=−
1
2iπ

Γ
(B + zI)
−1
(A − λI − zI)
−1
dz ∀λ>0, (2.9)

4 Matrix and generators of analytic semigroups
where Γ is an infinite-sectorial curve lying in ρ(A
− λI)

ρ(−B) coincides with (A + B −
λI)
−1
.
In the following, we will consider matrix transformations A and B mapping in a set of
sequences and we will show that they satisfy hypothesis (H).
3. Definition of the operators A and B
We will consider two infinite matrices and deal with the case when A and B map into l
∞
or c
0
and with the case when A and B map into c. In each case, we will study their spectral
properties.
For given sequences a
= (a
n
)
n≥1
, b = (b
n
)
n≥1
, β = (β
n
)
n≥1

,andγ = (γ
n
)
n≥1
,letA and
B be the follow ing infinite matr ices:
A
=
⎡
⎢
⎢
⎢
⎢
⎣
a
1
b
1
O
··
Oa
n
b
n
·
⎤
⎥
⎥
⎥
⎥

⎦
, B =
⎡
⎢
⎢
⎢
⎢
⎣
β
1
O
··
γ
n
β
n
O ·
⎤
⎥
⎥
⎥
⎥
⎦
. (3.1)
3.1. The case when A and B are operators mapping from D(A) and D(B) into E,where
E
= l
∞
or c
0

. The next conditions are consequences of results given in [15]. When E is
either of the sets l
∞
or c
0
, we assume that A satisfies the following properties:
a
∈ U
+
, a
n
is strictly increasing, lim
n→∞
a
n
=∞, (3.2a)
there is M
A
> 0suchthat


b
n


≤
M
A
∀n. (3.2b)
Similarly, we assume that B satisfies the next conditions:

β
∈ U
+
,lim
k→∞
β
2k
= L = 0, (3.3a)
lim
k→∞
β
2k+1
a
2k+1
=∞, (3.3b)
(α) there is M
B
> 0suchthat


γ
2k


≤
M
B
∀n,(β)γ
2k+1
= o(1) (n −→ ∞ ). (3.3c)

3.2. The case when A and B are operators mapping D(A) and D(B) into c. Here we need
to recall the characterization of (c,c).
Lemma 3.1. A
∈ (c,c) if and only if
(i) A
∈ S
1
,
(ii) lim
n→∞

∞
m=1
a
nm
= l for some l ∈ C,
(iii) lim
n→∞
a
nm
= l
m
for some l
m
∈ C, m = 1,2,
We will see that D(A)
= c(A) = s
(c)
1/a
and D(B) = c(B) = s

(c)
1/β
. Here we will show that
neither of the sets D(A)
= s
(c)
1/a
and D(B) = s
(c)
1/β
is embedded in the other.
B. de Malafosse and A. Medeghri 5
Proposition 3.2. Let a, β
∈ U
+
. Then
s
(c)
1/a
 s
(c)
1/β
, s
(c)
1/β
 s
(c)
1/a
(3.4)
if and only if β/a, a/β /

∈ c.
Proof. The inclusion s
(c)
1/a
⊂ s
(c)
1/β
means that I ∈ (s
(c)
1/a
,s
(c)
1/β
)andbyLemma 2.2,wehave
D
β/a
∈ (c,c). From Lemma 3.1,weconcludethats
(c)
1/a
⊂ s
(c)
1/β
if and only if β/a ∈ c.So
s
(c)
1/a
 s
(c)
1/β
is equivalent to β/a /∈ c. Similarly, we have s

(c)
1/β
 s
(c)
1/a
if and only if a/β /∈ c.
This completes the proof.

We assume that A and B satisfy the following hypotheses. The matrix A is defined in
(3.1)with
a
∈ U
+
, a
n
is strictly increasing, lim
n→∞
a
n
=∞, (3.5a)
b
∈ c. (3.5b)
For B givenin(3.1), we do the following hypotheses:
β
∈ U
+
,lim
n→∞
β
n

=∞, (3.6a)
lim
k→∞
β
2k+1
a
2k+1
=∞,lim
k→∞
β
2k
a
2k
= l = 0, (3.6b)
γ
∈ c. (3.6c)
This lead to the next remark.
Remark 3.3. The choice of β in (3.6b) is justified by Proposition 3.2 and so neither of the
sets D(A)
= s
(c)
1/a
and D(B) = s
(c)
1/β
is embedded in the other one. We will see in Proposition
5.7 and Theorem 5.8 that we need to have (3.5a)and(3.6a). Then we will see that A and
B are closed operators when b, γ
∈ c. Finally, notice that D(B) ⊂ c means 1/β ∈ c which
is trivial ly satisfied in (3.6a) and it is the same for A.

4. First properties of the operators A and B
4.1. The case when the operators A and B are considered as matrix maps from D(A)
and D(B) into E,whereE is equal to l
∞
,orc
0
. In this section, we will assume A and B
satisfy (3.2)and(3.3). For the convenience of the reader, recall the following well-know n
results.
Lemma 4.1. (i) A
∈ (l
∞
,l
∞
) if and only if A ∈ S
1
.
(ii) A
∈ (c
0
,c
0
) if and only if A ∈ S
1
and lim
n→∞
a
nm
= 0 for each m = 1,2,
Proposition 4.2. (i) A

∈ (s
1/a
,l
∞
) and A ∈ (s
0
1/a
,c
0
).
(ii) B
∈ (s
1/β
,l
∞
) and B ∈ (s
0
1/β
,c
0
).
6 Matrix and generators of analytic semigroups
Proof. (i) We have [AD
1/a
]
nn
= 1and[AD
1/a
]
n,n+1

= b
n
/a
n+1
for all n,and[AD
1/a
]
nm
= 0
otherwise. Then


AD
1/a


S
1
= sup
n

1+


b
n


a
n+1


=
O(1) (n −→ ∞ ). (4.1)
So by Lemma 4.1,wehaveAD
1/a
∈ (l
∞
,l
∞
)andbyLemma 2.2, A ∈ (s
1/a
,l
∞
). The proof is
similar for A
∈ (s
0
1/a
,c
0
) note that in this case, [AD
1/a
]
nm
→ 0(n →∞)foreachm ≥ 1.
(ii) Now [BD
1/β
]
nn
= 1and[BD

1/β
]
n,n−1
= γ
n
/β
n−1
for all n,and[BD
1/β
]
nm
= 0 other-
wise. By (3.3a), (3.3c)(β), we have
γ
2n+1
β
2n
= o(1) (n −→ ∞ ) (4.2)
and by (3.2a), (3.3b)and(3.3c)(α), we get
γ
2n
β
2n−1
=
γ
2n
a
2n−1
a
2n−1

β
2n−1
= o(1) (n −→ ∞ ). (4.3)
Then


BD
1/β


S
1
= sup
n

1+


γ
n


β
n−1

=
O(1) (n −→ ∞ ). (4.4)
So BD
1/β
∈ (l

∞
,l
∞
)andB ∈ (s
1/β
,l
∞
). Finally, we obtain B ∈ (s
0
1/β
,c
0
) reasoning as above.

We deduce that if E = l
∞
= s
1
, the matrix A is defined on D(A) = s
1/a
and B is defined
on D(B)
= s
1/β
. It can be shown that l
∞
(A) = s
1/a
and l
∞

(B) = s
1/β
.WhenE = c
0
,wewill
see in Theorem 5.6(i), (ii) that D(A)
= c
0
(A) = s
0
1/a
and D( B) = c
0
(B) = s
0
1/β
.Wededuce
from (3.3a), (3.3b) that in each case, neither of the sets D(A)andD(B)isembeddedin
the other.
4.2. The case when the operators A and B are considered as matrix maps from D(A)
and D(B) into c. We assume that A and B satisfy (3.5)and(3.6). From the preceding, we
immediately get the following.
Proposition 4.3. A
∈ (s
(c)
1/a
,c) and B ∈ (s
(c)
1/β
,c).

Proof. It is enough to notice that by (3.5)wehave(1+b
n
/a
n+1
)
n≥1
∈ c.ThenfromLemma
3.1,weconcludethatA
∈ (s
(c)
1/a
,c). We also have by (3.6),
γ
n
β
n−1
−→ 0(n −→ ∞ ) (4.5)
so BD
1/β
∈ (c,c)andB ∈ (s
(c)
1/β
,c). 
We will see in Theorem 5.8(i), (ii) that D(A) = c(A) = s
(c)
1/a
and D(B) = c(B) = s
(c)
1/β
.

B. de Malafosse and A. Medeghri 7
5. The matrices A and B as operator generators of an analytic semigroup
In this section, we will show that A and B are generators of analytic semigroup in each
case E
= l
∞
, E = c
0
,orE = c.
5.1. The case when A and B are considered as matrix maps from D(A) and D(B) into E,
where E
= l
∞
or c
0
. In this section, A and B satisfy (3.2)and(3.3).Thenextresultwas
shown in [15] in the case when A
∈ (s
1/a
,l
∞
)andB ∈ (s
1/β
,l
∞
)witha
n
= a
n
, a>1, and β

was defined by β
2n
= 1andβ
2n+1
= (2n +1)!foralln,soweomittheproof.
Proposition 5.1. In the space l
∞
, the two linear operators A and B are closed and satisfy
the following:
(i) D(A)
= s
1/a
,
(ii) D(B)
= s
1/β
,
(iii)
D( A) = l
∞
, D(B) = l
∞
,
(iv) there are ε
A
, ε
B
> 0 (with ε
A
+ ε

B
<π) such that


(A − λI)
−1


∗
Ꮾ(l
∞
)
≤
M
|λ|
∀
λ = 0,


Arg(λ)


≥
ε
A
,


(B + μI)
−1



∗
Ꮾ(l
∞
)
≤
M
|μ|
∀
μ = 0,


Arg(μ)


≤
π − ε
B
.
(5.1)
This result shows that
−A and −B satisfy hypothesis (H) and σ(−A)

σ(B) = ∅.So
−A and −B are generators of the analytic semigroups e
(−At)
and e
(−Bt)
not strongly con-

tinuous at t
= 0. We have similar results when A and B are matrix maps into c
0
.Werequire
some elementary lemmas whose proofs are left to the reader.
Lemma 5.2. Let ε
∈ ]0,π/2[ and let x
0
> 0 beareal.Then


x
0
− λ


≥
x
0
sinε ∀λ ∈ C with


Arg(λ)


≥
ε. (5.2)
Lemma 5.3. Let x
0
> 0 beareal.Then



x
0
− λ


≥|
λ|sinθ ∀λ =|λ|e
iθ
/∈ R
−
,


x
0
− λ


≥|
λ|∀λ ∈ R
−
.
(5.3)
We can state the following result where we will use the fact that for any α
∈ U
+
, since
s

0
α
is a BK space with AK, we have Ꮾ(s
0
α
) = (s
0
α
,s
0
α
). As we have seen in (2.5), for any matrix
C
∈ (s
0
α
,s
0
α
), we have C
∗
Ꮾ(s
0
α
)
=C
∗
(s
0
α

,s
0
α
)
=C
S
α
.
Proposition 5.4. (i) Let ε
A
∈ ]0,π/2[.Foreveryλ ∈ C with |Arg(λ)|≥ε
A
, the infinite
matrix A
− λI considered as an operator in s
0
1/a
is invertible and
(A
− λI)
−1
∈

c
0
,s
0
1/a

. (5.4)

8 Matrix and generators of analytic semigroups
(ii) Let ε
B
∈ ]0,π/2[.Foreveryμ ∈ C with |Arg(μ)|≤π − ε
B
, the infinite matrix B + μI
considered as an operator in s
0
1/β
is invertible and
(B + μI)
−1
∈

c
0
,s
0
1/β

. (5.5)
Proof. (i) Fix ε
A
∈ ]0,π/2[ and consider the infinite-sectorial set

ε
A
=

λ ∈ C :



Arg(λ)


<ε
A

. (5.6)
For any λ/
∈

ε
A
,put
χ
n
=
b
n
a
n
− λ
(5.7)
and D

λ
= D
(1/(a
n

−λ))
n
.Then[D

λ
(A − λI)]
nn
= 1, [D

λ
(A − λI)]
n,n+1
= χ
n
for all n and
[D

λ
(A − λI)]
nm
= 0 otherwise. By Lemma 5.2,wehave


χ
n


≤
M
A

a
n
sinε
A
∀n and all λ/∈

ε
A
. (5.8)
Since a
n
tends to infinity as n tends to infinit y, there is n
0
such that


χ
n


≤
1
2
∀n ≥ n
0
and all λ/∈

ε
A
. (5.9)

Consider now the infinite matr ix

T
λ
=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
·
T
−1
λ
· 0
···
1
01
·

·
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
, (5.10)
where T
λ
is the matrix of order n
0
defined by [T
λ
]
nn
= 1for1≤ n ≤ n
0
;[T
λ

]
n,n+1
= χ
n
for
1
≤ n ≤ n
0
− 1, and [T
λ
]
nm
= 0 otherwise. Elementary calculations give
T
−1
λ
=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎝
1 −χ
1
χ
1
χ
2
−χ
1
χ
2
χ
3
·· ·
1 −χ
2
χ
2
χ
3
·· ·
1 −χ
n
χ
n
χ
n+1
·

·
0
1 χ
n
0
−1
1
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
. (5.11)
B. de Malafosse and A. Medeghri 9
Putting Ꮽ
λ
= D

λ

(A − λI)

T
λ
we easily get [Ꮽ
λ
]
nn
= 1foralln,[Ꮽ
λ
]
n,n+1
= χ
n
for n ≥ n
0
,
and [Ꮽ
λ
]
nm
= 0 otherwise. Then by Lemma 2.2 and since the sequence a is increasing , we
get


I − Ꮽ
λ


∗

(s
0
1/a
,s
0
1/a
)
=


I − Ꮽ
λ


S
1/a
= sup
n≥n
0



χ
n


a
n
a
n+1


≤
1
2
∀λ/∈

ε
A
. (5.12)
Since Ꮽ
λ
= Ꮽ
λ
− I + I ∈ (s
0
1/a
,s
0
1/a
)and(5.12)holds,Ꮽ
λ
is invertible in the Banach al-
gebra of all bounded operators Ꮾ(s
0
1/a
) = (s
0
1/a
,s
0

1/a
)mappings
0
1/a
to itself and Ꮽ
−1
λ
∈
(s
0
1/a
,s
0
1/a
). Then for any given y ∈ c
0
, we successively get y

= D

λ
y = (y
n
/(a
n
− λ))
n≥1
∈
s
0

1/a
, Ꮽ
−1
λ
y

∈ s
0
1/a
,

T
λ
(Ꮽ
−1
λ
y

) ∈ s
0
1/a
,and(A − λI)
−1
=

T
λ
Ꮽ
−1
λ

D

λ
∈ (c
0
,s
0
1/a
). So we have
shown (i).
(ii) For ε
B
∈ ]0,π/2[, let Σ
B
={μ ∈ C : |Arg(μ)|≤π − ε
B
} and put
χ

n
=
γ
n
β
n
+ μ
. (5.13)
To deal with the inverse of B + μI, we need to study the sequences
|γ
2k+1

|/β
2k
and |χ

2k
|β
2k
/
β
2k−1
.By(3.3a), (3.3b), we have
γ
2k+1
β
2k
−→ 0(k −→ ∞ ). (5.14)
On the other hand, for every μ
∈ Σ
B
,weget


χ

2k


β
2k
β

2k−1
≤
M
B
β
2k
sinε
B
β
2k
β
2k−1
=
M
B
sinε
B
1
β
2k−1
, (5.15)
1
β
2k−1
=
a
2k−1
β
2k−1
1

a
2k−1
= o(1) (k −→ ∞ ). (5.16)
From (5.14)and(5.16), we deduce that there is n
1
such that


γ
2k+1


1
β
2k
≤
1
2
sinε
B
for 2k +1≥ n
1
,


χ

2k



β
2k
β
2k−1
≤
1
2
for 2k
≥ n
1
∀μ ∈ Σ
B
.
(5.17)
As in (i), define the matrices D

μ
= D
(1/(β
n
+μ)
n
)
and

R
μ
=
⎛
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
·
R
−1
μ
· 0
···
1
01
·
·
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎠
, (5.18)
10 Matrix and generators of analytic semigroups
where R
μ
is the matrix of order n
1
− 1definedby[R
μ
]
nn
= 1foralln,by[R
μ
]
n,n−1
= χ

n
for
2
≤ n ≤ n
1
− 1, and by [R
μ
]
nm
= 0 otherwise. Then elementary calculations show that the

matrix Ꮾ
μ
=

R
μ
D

μ
(B + μI)isdefinedby[Ꮾ
μ
]
nn
= 1foralln,by[Ꮾ
μ
]
n,n−1
= χ

n
for n ≥ n
1
,
and by [Ꮾ
μ
]
nm
= 0 otherwise. Then for any μ ∈ Σ
B
,



I − Ꮾ
μ


∗
(s
0
1/β
,s
0
1/β
)
= sup
n≥n
1



χ

n


β
n
β
n−1


=
max

τ
1
,τ
2

, (5.19)
where
τ
1
= sup
k≥n
1
/2


χ

2k


β
2k
β
2k−1
, τ
2
= sup

k≥(n
1
−1)/2


χ

2k+1


β
2k+1
β
2k
. (5.20)
By Lemma 5.2,weget
|β
2k
+ μ|≥β
2k
sinε
B
for all μ ∈ Σ
B
and τ
1
≤ 1/2. Then
τ
2
≤

1
2
sinε
B
1
β
2k+1
sinε
B
β
2k+1
=
1
2
for 2k
≥ n
1
. (5.21)
This implies
I − Ꮾ
μ

∗
(s
0
1/β
,s
0
1/β
)

≤ 1/2. Reasoning as in (i) with (B + μI)
−1
= Ꮾ
−1
μ

R
μ
D

μ
,we
conclude that B + μI considered as an operator from s
0
1/β
into c
0
is invertible and (B +
μI)
−1
∈ (c
0
,s
0
1/β
)forallμ ∈ Σ
B
. This concludes the proof. 
Remark 5.5. As a direct consequence of the preceding, it is trivial that
c

0
(A − λI) = s
0
1/a
∀λ ∈ C,


Arg(λ)


≥
ε
A
,
c
0
(B + μI) = s
0
1/β
∀μ ∈ C,


Arg(μ)


≤
π − ε
B
.
(5.22)

We immediately obtain the next result.
Theorem 5.6. In the space c
0
, the two linear operators A and B areclosedandsatisfythe
following:
(i) D(A)
= c
0
(A) = s
0
1/a
,
(ii) D(B)
= c
0
(B) = s
0
1/β
,
(iii)
D( A) = c
0
, D(B) = c
0
,
(iv) there are ε
A
, ε
B
> 0 (with ε

A
+ ε
B
<π) such that


(A − λI)
−1


∗
Ꮾ(c
0
)
≤
M
|λ|
∀
λ = 0,


Arg(λ)


≥
ε
A
,



(B + μI)
−1


∗
Ꮾ(c
0
)
≤
M
|μ|
∀
μ = 0,


Arg(μ)


≤
π − ε
B
.
(5.23)
Proof. Show that A is a closed operator. For this, consider a sequence x

p
= (x
np
)
n≥1

tend-
ing to x
= (x
n
)
n≥1
in c
0
,asp tends to infinity, w here x

p
∈ s
0
1/a
for all p.ThenAx

p
→ y (p →
∞
)inc
0
,thatisforanyn,wehaveA
n
(x

p
) → A
n
(x) = y
n

(p →∞). It remains to show that
x
∈ s
0
1/a
. For this, note that since b ∈ l
∞
and x ∈ c
0
,weconcludethata
n
x
n
= y
n
− b
n
x
n+1
tends to a zero as n tends to infinity. The proof for B is similar.
B. de Malafosse and A. Medeghri 11
(i)
⇒(ii). First by Proposition 4.2,wehaveA ∈ (s
0
1/a
,c
0
). It remains to show that
c
0

(A) = s
0
1/a
.Letx ∈ s
0
1/a
.Since(b
n
/a
n+1
)
n≥1
∈ c
0
,wededucethat
A
n
(x) = a
n
x
n
+
b
n
a
n+1
a
n+1
x
n+1

= o(1) (n −→ ∞ ) (5.24)
and we have shown that s
0
1/a
⊂ c
0
(A).
Now let x
∈ c
0
(A). Then y = Ax ∈ c
0
.IntheproofofProposition 5.4,wecantake
λ
= 0. Indeed, there is n
0
such that χ
n
=|b
n
|/a
n
≤ 1/2foralln ≥ n
0
.Theny ∈ c
0
implies
D
1/a
y = (y

n
/a
n
)
n≥1
∈ s
0
1/a
, Ꮽ
−1
D
1/a
y ∈ s
0
1/a
,andx = A
−1
y =

T
0
Ꮽ
−1
0
D
1/a
y ∈ s
0
1/a
. This

shows that c
0
(A) ⊂ s
0
1/a
and since s
0
1/a
⊂ c
0
(A), we conclude that c
0
(A) = s
0
1/a
.Theproofis
similar for B.
(iii) Let x
= (1/a
n
)
n≥1
∈ c
0
and assume x

p
= (x
np
)

n≥1
tend to x in s
0
1/a
, that is,


x

p
− x


s
1/a
= sup
n

a
n




x
np
−
1
a
n






−→
0(p −→ ∞ ). (5.25)
Since a
n
tends to infinity, we should have x
np
→ 1/a
n
and a
n
x
np
→ 1(p →∞)foralln.
This contradicts the fact that x

p
∈ s
0
1/a
for all p. The reasoning is the same for B. So (iii)
holds.
(iv) By Proposition 5.4(i), we have seen that for any λ/
∈

ε

A
(A − λI)
−1
∈ (c
0
,s
0
1/a
),

T
λ
∈ S
1
and using the notation of the proof of Proposition 5.4,weget


(A − λI)
−1
y


l
∞
=



T
λ

Ꮽ
−1
λ
y



l
∞
≤



T
λ


S
1


Ꮽ
−1
λ


S
1



D

λ


S
1
y
l
∞
. (5.26)
Then by Lemma 5.3, we successively get


D

λ


S
1
= sup
n≥1




1

a

n
− λ





≤
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
1
|λ|
sinθ for λ =|λ|e
iθ
/∈ R
−
,
1
|λ|
for λ ∈ R
−

.
(5.27)
Nowaswehaveseenin(5.9), we have
I − Ꮽ
λ

S
1
= sup
n≥n
0
(|χ
n
|) ≤ 1/2 and then in the
Banach algebra S
1
, we easily get


Ꮽ
−1
λ


S
1
≤
∞

m=0




I − Ꮽ
λ



m
S
1
≤
∞

m=0
2
−m
= 2. (5.28)
Finally, by (5.9), (5.10), and (5.11), we have sup
λ/∈

ε
A


T
λ

S
1

< ∞ and we conclude that
(A − λI)
−1

∗
Ꮾ(c
0
)
≤ M/|λ| for all λ/∈

ε
A
. We get similar results for B + μI. This con-
cludes the proof of (iv).

5.2. The case when A and B are matrix maps from D(A) and D(B) into c. In this section,
A and B satisfy conditions given in (3.5)and(3.6). We will state results similar to those
given in Section 5.1,whenc
0
is replaced by c.
12 Matrix and generators of analytic semigroups
Proposition 5.7. (i) Let ε
A
∈]0,π/2[.Foreveryλ with |Arg(λ)|≥ε, the infinite matrix
A
− λI considered as an operator in s
(c)
1/a
is invertible and
(A

− λI)
−1
∈

c,s
(c)
1/a

. (5.29)
(ii) Let ε
B
∈ ]0,π/2[.Foreveryμ with |Arg(μ)|≤π − ε
B
, the infinite matrix B + μI
considered as an operator in s
(c)
1/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+
b

n
a
n
− λ
a
n
a
n+1
−→ 1(n −→ ∞ ) ∀λ/∈

ε
A
(5.31)
and Ꮽ
λ
∈ S
1/a
,wehaveᏭ
λ
∈ (s
(c)
1/a
,s
(c)
1/a
). Then using the notation of Proposition 5.4,there
is n

0
such that



I − Ꮽ
λ


∗
Ꮾ(s
(c)
1/a
)
=


I − Ꮽ
λ


∗
S
1/a
= sup
n≥n

0



χ
n



a
n
a
n+1

≤
1
2
∀λ/∈

ε
A
. (5.32)
Then Ꮽ
−1
λ
∈ Ꮾ(s
(c)
1/a
)

S
1/a
, and since Ꮽ
−1
λ
∈ S
1/a

is an infinite matrix, we have Ꮽ
−1
λ
∈
(s
(c)
1/a
,s
(c)
1/a
). Then for any y ∈ c, we successively get D

λ
y = (y
n
/(a
n
− λ))
n≥1
∈ s
(c)
1/a
,
Ꮽ
−1
λ
(D

λ
y) ∈ s

(c)
1/a
,and

T
λ
[Ꮽ
−1
λ
(D

λ
y)] ∈ s
(c)
1/a
.Weconcludethat(A − λI)
−1
∈ (c,s
(c)
1/a
)for
ever y λ with
|Arg(λ)|≥ε.
(ii) Reasoning as in Proposition 5.4,thereareM, M

> 0suchthatforeveryμ ∈ Σ
B
,





γ
2k
β
2k
+ μ




β
2k
β
2k−1
≤
M
β
2k
sinε
B
β
2k
β
2k−1
= o(1) (k −→ ∞ ),





γ
2k+1
β
2k+1
+ μ




β
2k+1
β
2k
≤
M

β
2k+1
sinε
B
β
2k+1
β
2k
= o(1) (k −→ ∞ ).
(5.33)
We deduce that there is n

1
with

τ
1
= sup
k≥n

1
/2




γ
2k
β
2k
+ μ




β
2k
β
2k−1
≤
1
2
, τ
2
= sup

k≥(n

1
−1)/2




γ
2k+1
β
2k+1
+ μ




β
2k+1
β
2k
≤
1
2
. (5.34)
B. de Malafosse and A. Medeghri 13
Now by (5.33), we have





γ
n
β
n
+ μ




β
n
β
n−1
−→ 0(n −→ ∞ ) (5.35)
and Ꮾ
μ
∈ (s
(c)
1/β
,s
(c)
1/β
)forallμ ∈ Σ
B
.Thus


I − Ꮾ
μ



∗
(s
(c)
1/β
,s
(c)
1/β
)
=


I − Ꮾ
μ


∗
S
1/β
= sup
n≥n

1





γ

n
β
n
+ μ




β
n
β
n−1

=
max

τ
1
,τ
2

≤
1
2
(5.36)
and Ꮾ
−1
μ
∈ (s
(c)

1/β
,s
(c)
1/β
)forallμ ∈ Σ
B
. Reasoning as in (i) w ith (B + μI)
−1
= Ꮾ
−1
μ

R
μ
D

μ
,we
conclude that B + μI considered as operator from s
(c)
1/β
to c is invertible and (B + μI)
−1
∈
(c,s
(c)
1/β
)forallμ ∈ Σ
B
. This completes the proof. 

We can state the follow ing.
Theorem 5.8. In the space c, the two linear operators A and B are closed and satisfy the
following:
(i) D(A)
= c(A) = s
(c)
1/a
,
(ii) D(B)
= c(B) = s
(c)
1/β
,
(iii)
D( A) = c, D(B) = c,
(iv) there are ε
A
, ε
B
> 0 (with ε
A
+ ε
B
<π) such that


(A − λI)
−1



∗
Ꮾ(c)
≤
M
|λ|
∀
λ = 0,


Arg(λ)


≥
ε
A
,


(B + μI)
−1


∗
Ꮾ(c)
≤
M
|μ|
∀
μ = 0,



Arg(μ)


≤
π − ε
B
.
(5.37)
Proof. Show that A is a closed operator. For this, consider a sequence x

p
= (x
np
)
n≥1
tend-
ing to x
= (x
n
)
n≥1
in c,asp tends to infinity, where x

p
∈ s
(c)
1/a
for all p.ThenAx


p
→y(p→∞)
in c, that is, A
n
(x

p
) → A
n
(x) = y
n
(p →∞)foralln. It remains to show that x ∈ s
(c)
1/a
.For
this, note that since b
∈ c and x ∈ c,weconcludethata
n
x
n
= y
n
− b
n
x
n+1
tends to a limit
as n tends to infinity. The proof for B is similar. The proof of statements (i) and (ii) comes
from Proposition 4.3 and follows the same lines as that for Theorem 5.6(i), (ii).
(iii) Follows exactly the same lines as that in the proof given in the case when E

= l
∞
in [15, Proposition 3, page 196].
The proof of (iv) is a consequence of Proposition 5.7(i) and follows exactly the same
lines as that in the proof of Theorem 5.6.

Acknowledgments
The authors thank the referee for his contribution to improve the presentation of the
paper. The work of the second author is supported by the Agence Universitaire de la
Francophonie (AUF).
14 Matrix and generators of analytic semigroups
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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:
Ahmed Medeghri: Laboratoire de Math
´
ematiques Pures et Appliqu
´
ees, Universit
´
e de Mostaganem,
BP 227, Mostaganem 27000, Algeria
E-mail address: