ANALYTIC SEMIGROUPSBRUNO 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
Trang 1ANALYTIC 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, where χ is 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 operatorΔ acting on the sequence spaces c0andc Then they dealt with the fine spectrum
of the operatorB(r,s) defined by a matrix band over the sequence spaces c0 andc In
de Malafosse [3,5], there are results on the spectrum of the Ces`aro matrixC1 and on the matrixΔ considered as operators from srto itself Spectral properties 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] 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
in a reflexive Banach set of sequences E, where y ∈ 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-cesA and B 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, Pages 1 14
DOI 10.1155/JIA/2006/67062
Trang 2a class of infinite-tridiagonal systems Then we study some spectral properties ofA and
B considered 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(1c) /β,c) Then we show that ( − A) and ( − B) are generators of ana-lytic semigroups, where D(A) and D(B) are of the form χ(A) and χ(B) with χ = l ∞,c0, orc Here the relative boundedness with respect to A or B is 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 =(a nm)n,m ≥1 be an infinite matrix and consider the sequencex =(x n)n ≥1 We will define the productMx =(M n(x)) n ≥1with
M n(x) =∞ m =1a nm x mwhenever the series are convergent for alln ≥1 Lets denote the set
of all complex sequences We writeϕ, c0,c and l∞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 =(a nm)n,m ≥1maps X into Y, that is,
M ∈(X,Y), if the series defined by M n(x) =∞ m =1a nm x mare convergent for alln ≥1 and for allx ∈ X and Mx ∈ Y for all x ∈ X For any subset X of s, we will write
IfY is a subset of s, we will denote the so-called matrix domain by
LetX ⊂ s be a Banach space, with norm · X ByᏮ(X), we will denote the set of all bounded linear operators, mapping X into itself We will say that L ∈ Ꮾ(X) if and only if
L : X → X is a linear operator and
L ∗
Ꮾ(X) =sup
x 0
Lx X / x X< ∞ (2.3)
It is well known thatᏮ(X) is a Banach algebra with the norm L ∗
Ꮾ(X) A Banach space
X ⊂ s is a BK space if the projection P n:x → x n fromX intoCis continuous for alln.
A BK space X ⊃ ϕ is said to have AK if for every x ∈ X, x =limp →∞p
k =1x 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 nowU+= { x =(x n)n ≥1∈ s : x n > 0 for all n } Forξ =(ξ n)n ≥1∈ U+, we will define the diagonal matrixD ξ =(ξ n δ nm)n,m ≥1, (whereδ nm =0 for alln m and δ nm =1 other-wise) Forα ∈ U+, we will writes α = D α l∞, (cf [3–10,15] The sets αis a BK space with the norm x s α =supn ≥1(| x n | /α n) The set of all infinite matricesM =(a nm)n,m ≥1with
M S α =sup
n ≥1
1
α n
∞
m =1
a nmα m< ∞ (2.4)
Trang 3is a Banach algebra with identity normed by · S α Recall that if M ∈(s α,s α), then
Mx s α ≤ M S α x s αfor allx ∈ s α Thus we obtain the following result, (cf [7]) where
we putB(s α)= Ꮾ(sα)
(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 setss0
α = D α c0ands(α c) = D α c, (cf [7]) The sets s0
α and s(α c) are BK spaces with the norm · s αands0
αhas AK It was shown in [9,10] that for any matrixM ∈(s α,s α), we get
M ∗
Ꮾ(s α)= M ∗
Ꮾ(s0
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
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] LetE be a Banach space We consider two closed linear operators A and B, whose domains are D(A) and D(B)
in-cluded inE For every x ∈ D(A)
D(B), we then define their sum Sx = Ax + Bx.
The spectral properties ofA and B are the following:
(H) there areC A,C B > 0, and ε A,ε B ∈]0,π[ such that
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 thatA and B are 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) and D(B) are densely defined in E, it is well known (cf [2]) that the bounded operator defined by
L λ = − 1
2iπ
Γ(B + zI) −1(A − λI − zI) −1dz ∀ λ > 0, (2.9)
Trang 4whereΓ 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 transformationsA and B mapping in a set of
sequences and we will show that they satisfy hypothesis (H)
3 Definition of the operatorsA and B
We will consider two infinite matrices and deal with the case whenA and B map into l∞
orc0and with the case whenA and B map into c In each case, we will study their spectral
properties
For given sequencesa =(a n)n ≥1,b =(b n)n ≥1,β =(β n)n ≥1, andγ =(γ n)n ≥1, letA and
B be the following infinite matrices:
A =
⎡
⎢
⎢
⎣
· ·
·
⎤
⎥
⎥
⎦, B =
⎡
⎢
⎢
⎣
· ·
γ n β n
⎤
⎥
⎥
3.1 The case whenA and B are operators mapping from D(A) and D(B) into E, where
E = l∞orc0 The next conditions are consequences of results given in [15] WhenE is
either of the setsl∞orc0, we assume thatA satisfies the following properties:
a ∈ U+, a nis strictly increasing, lim
n →∞ a n = ∞, (3.2a) there isM A > 0 such thatb n ≤ M A ∀ n. (3.2b) Similarly, we assume thatB satisfies the next conditions:
β ∈ U+, lim
lim
k →∞
β2k+1
(α) there is M B > 0 such thatγ2
k ≤ M B ∀ n, (β)γ2k+1 = o(1) (n −→ ∞) (3.3c)
3.2 The case whenA 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 ∈ S1,
(ii) limn →∞∞
m =1a nm = l for some l ∈ C ,
(iii) limn →∞ a nm = l m for some l m ∈ C , m =1, 2,
We will see thatD(A) = c(A) = s(1c) /a andD(B) = c(B) = s(1c) /β Here we will show that neither of the setsD(A) = s(1c) /aandD(B) = s(1c) /βis embedded in the other
Trang 5Proposition 3.2 Let a, β ∈ U+ Then
s(1c) /as(1c) /β, s(1c) /βs(1c) /a (3.4)
if and only if β/a, a/β / ∈ c.
Proof The inclusion s(1c) /a ⊂ s(1c) /β means thatI ∈(s(1c) /a,s(1c) /β) and by Lemma 2.2, we have
D β/a ∈(c,c) From Lemma 3.1, we conclude that s(1c) /a ⊂ s(1c) /β if and only ifβ/a ∈ c So
s(1c) /as(1c) /β is equivalent toβ/a / ∈ c Similarly, we have s(1c) /βs(1c) /a if and only ifa/β / ∈ c.
We assume thatA and B satisfy the following hypotheses The matrix A is defined in
(3.1) with
a ∈ U+, a nis strictly increasing, lim
n →∞ a n = ∞, (3.5a)
ForB given in (3.1), we do the following hypotheses:
β ∈ U+, lim
lim
k →∞
β2k+1
a2k+1 = ∞, lim
k →∞
β2k
This lead to the next remark
Remark 3.3 The choice of β in (3.6b) is justified byProposition 3.2and so neither of the setsD(A) = s(1c) /aandD(B) = s(1c) /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 thatA and
B are closed operators when b, γ ∈ c Finally, notice that D(B) ⊂ c means 1/β ∈ c which
is trivially satisfied in (3.6a) and it is the same forA.
4 First properties of the operatorsA and B
4.1 The case when the operatorsA and B are considered as matrix maps from D(A)
andD(B) into E, where E is equal to l∞, orc0 In this section, we will assumeA and B
satisfy (3.2) and (3.3) For the convenience of the reader, recall the following well-known results
Lemma 4.1 (i)A ∈(l∞,l∞ ) if and only if A ∈ S1.
(ii)A ∈(c0,c0) if and only if A ∈ S1and lim n →∞ a nm = 0 for each m =1, 2,
Proposition 4.2 (i)A ∈(s1/a,l∞ ) and A ∈(s0
/a,c0).
(ii)B ∈(s1/β,l∞ ) and B ∈(s01/β,c0).
Trang 6Proof (i) We have [AD1/a]nn =1 and [AD1/a]n,n+1 = b n /a n+1for alln, and [AD1/a]nm =0 otherwise Then
AD1/a
S1=sup
n
1 +b n
a n+1
So byLemma 4.1, we haveAD1/a ∈(l∞,l∞) and byLemma 2.2,A ∈(s1/a,l∞) The proof is similar forA ∈(s0
/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= γ n /β n −1for alln, and [BD1/β]nm =0 other-wise By (3.3a), (3.3c)(β), we have
γ2n+1
and by (3.2a), (3.3b) and (3.3c)(α), we get
γ2n
β2n −1 = γ2n
a2n −1
a2n −1
Then
BD1
/β
S1=sup
n
1 +γ
n
β n −1
SoBD1/β ∈(l∞,l∞) andB ∈(s1/β,l∞) Finally, we obtainB ∈(s01/β,c0) reasoning as above
We deduce that ifE = l ∞ = s1, the matrixA is defined on D(A) = s1/aandB is 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) = s0
/aandD(B) = c0(B) = s0
1/β We deduce from (3.3a), (3.3b) that in each case, neither of the setsD(A) and D(B) is embedded in
the other
4.2 The case when the operatorsA and B are considered as matrix maps from D(A)
andD(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(1c) /a,c) and B ∈(s(1c) /β,c).
Proof It is enough to notice that by (3.5) we have (1 +b n /a n+1)n ≥1∈ c Then fromLemma 3.1, we conclude thatA ∈(s(1c) /a,c) We also have by (3.6),
γ n
We will see inTheorem 5.8(i), (ii) thatD(A) = c(A) = s(1c) /aandD(B) = c(B) = s(1c) /β
Trang 75 The matricesA and B as operator generators of an analytic semigroup
In this section, we will show thatA and B are generators of analytic semigroup in each
caseE = l∞,E = c0, orE = c.
5.1 The case whenA and B are considered as matrix maps from D(A) and D(B) into E,
whereE = l∞orc0 In this section,A and B satisfy (3.2) and (3.3) The next result was shown in [15] in the case whenA ∈(s1/a,l∞) andB ∈(s1/β,l∞) witha n = a n,a > 1, and β
was defined byβ2n =1 andβ2n+1 =(2n + 1)! for all n, so we omit the proof.
Proposition 5.1 In the space l∞ , the two linear operators A and B are closed and satisfy the following:
(i)D(A) = s1/a ,
(ii)D(B) = s1/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)ande(− Bt)not strongly con-tinuous att =0 We have similar results whenA and B are matrix maps into c0 We require some elementary lemmas whose proofs are left to the reader
Lemma 5.2 Let ε ∈]0,π/2[ and let x0> 0 be a real Then
x0− λ ≥ x0sinε ∀ λ ∈ C withArg(λ) ≥ ε. (5.2)
Lemma 5.3 Let x0> 0 be a real Then
x0− λ ≥ | λ |sinθ ∀ λ = | λ | e iθ ∈ 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(s0
α)=(s0
α,s0
α) As we have seen in (2.5), for any matrix
C ∈(s0
α,s0
α), we have C ∗
Ꮾ(s0
α)= C ∗
(s0
α,s0
α)= C S α
Proposition 5.4 (i) Let ε A ∈]0,π/2[ For every λ ∈ C with |Arg(λ) | ≥ ε A , the infinite matrix A − λI considered as an operator in s01/a is invertible and
(A − λI) −1∈c0,s0
/a
Trang 8
(ii) Let ε B ∈]0,π/2[ For every μ ∈ C with |Arg(μ) | ≤ π − ε B , the infinite matrix B + μI considered as an operator in s01/β is invertible and
(B + μI) −1∈c0,s01/β
Proof (i) Fix ε A ∈]0,π/2[ and consider the infinite-sectorial set
ε A
= λ ∈ C:Arg(λ)< ε A
For anyλ / ∈ε A, put
χ n = b n
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 ByLemma 5.2, we have
χ n ≤ M A
a nsinε A ∀ n and all λ / ∈
ε A
Sincea ntends to infinity asn tends to infinity, there is n0such that
χ n ≤1
2 ∀ n ≥ n0 and allλ / ∈
ε A
Consider now the infinite matrix
T λ =
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
·
1
·
·
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
whereT λis the matrix of ordern0defined by [T λ]nn =1 for 1≤ n ≤ n0; [T λ]n,n+1 = χ nfor
1≤ n ≤ n0−1, and [T λ]nm =0 otherwise Elementary calculations give
T λ −1=
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
1 − χ1 χ1χ2 − χ1χ2χ3 · · ·
1 − χ n χ n χ n+1 ·
·
0
1 χ n0−1
1
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Trang 9PuttingᏭλ= D λ(A − λI) Tλ we easily get [Ꮽλ]nn =1 for alln, [Ꮽ λ]n,n+1 = χ nforn ≥ n0, and [Ꮽλ]nm =0 otherwise Then byLemma 2.2and since the sequencea is increasing, we
get
I −Ꮽλ ∗
(s0
/a,s0
/a)= I −Ꮽλ
S1/a =sup
n ≥ n0
χ n a n
a n+1
≤1
2 ∀ λ / ∈
ε A
SinceᏭλ=Ꮽλ− I + I ∈(s0
/a,s0
/a) and (5.12) holds,Ꮽλ is invertible in the Banach al-gebra of all bounded operators Ꮾ(s0
1/a)=(s01/a,s01/a) mapping s01/a to itself and Ꮽ−1
λ ∈
(s0/a,s0/a) Then for any given y ∈ c0, we successively get y = D λ y =(y n /(a n − λ)) n ≥1∈
s0/a,Ꮽ−1
λ y ∈ s0/a,Tλ(Ꮽ−1
λ y )∈ s0/a, and (A − λI) −1= T λᏭ −1
λ D λ ∈(c0,s0/a) So we have shown (i)
(ii) Forε B ∈]0,π/2[, let Σ B = { μ ∈ C:|Arg(μ) | ≤ π − ε B }and put
χ n = γ n
To deal with the inverse ofB + μI, we need to study the sequences | γ2k+1 | /β2kand| χ2 k | β2k /
β2k −1 By (3.3a), (3.3b), we have
γ2k+1
On the other hand, for everyμ ∈ΣB, we get
χ
2k β2k
β2k −1 ≤ M B
β2ksinε B
β2k
β2k −1= M B
sinε B
1
β2k −1
1
β2k −1 = a2k −1
β2k −1
1
From (5.14) and (5.16), we deduce that there isn1such that
γ2k+1 1
β2k ≤1
2sinε B for 2k + 1 ≥ n1,
χ
2k β2k
β2k −1 ≤1
2 for 2k ≥ n1∀ μ ∈ΣB.
(5.17)
As in (i), define the matricesD μ = D(1/(β n+μ) n)and
R μ =
⎛
⎜
⎜
⎜
⎜
⎜
⎜
·
R −1
1
·
·
⎞
⎟
⎟
⎟
⎟
⎟
⎟
Trang 10whereR μis the matrix of ordern1−1 defined by [R μ]nn =1 for alln, by [R μ]n,n −1= χ n for
2≤ n ≤ n1−1, and by [R μ]nm =0 otherwise Then elementary calculations show that the matrixᏮμ= R μ D μ(B + μI) is defined by [Ꮾ μ]nn =1 for alln, by [Ꮾ μ]n,n −1= χ nforn ≥ n1, and by [Ꮾμ]nm =0 otherwise Then for anyμ ∈ΣB,
I −Ꮾμ ∗
(s0
/β,s0
/β)=sup
n ≥ n1
χ n β n
β n −1
=max
τ1,τ2
where
τ1= sup
k ≥ n1/2
χ
2k β2k
β2k −1
, τ2= sup
k ≥(n1−1)/2
χ
2k+1β2k+1
β2k (5.20)
ByLemma 5.2, we get| β2k+μ | ≥ β2ksinε Bfor allμ ∈ΣBandτ1≤1/2 Then
τ2≤1
2sinε B 1
β2k+1sinε B β2k+1 =1
2 for 2k ≥ n1. (5.21) This implies I −Ꮾμ ∗
(s0
/β,s0
/β)≤1/2 Reasoning as in (i) with (B + μI) −1=Ꮾ−1
μ Rμ D
μ, we conclude thatB + μI considered as an operator from s01/β intoc0 is invertible and (B + μI) −1∈(c0,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) = s0
/a ∀ λ ∈ C, Arg(λ) ≥ ε A,
c0(B + μI) = s01/β ∀ μ ∈ C, Arg(μ) ≤ π − ε B (5.22)
We immediately obtain the next result
Theorem 5.6 In the space c0, the two linear operators A and B are closed and satisfy the following:
(i)D(A) = c0(A) = s0
/a ,
(ii)D(B) = c0(B) = s0
1/β ,
(iii)D(A) c0, D(B) c0,
(iv) there are ε A , ε B > 0 (with ε A+ε B < π ) such that
(A − λI) −1 ∗
Ꮾ(c0 )≤ M
| λ | ∀ λ 0, Arg(λ) ≥ ε A,
(B + μI) −1 ∗
Ꮾ(c0 )≤ 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 tox =(x n)n ≥1inc0, asp tends to infinity, where x p ∈ s01/afor allp Then Ax p → y (p →
∞) inc0, that is for anyn, we have A n(x p)→ A n(x) = y n(p → ∞) It remains to show that
x ∈ s01/a For this, note that sinceb ∈ l∞andx ∈ c0, we conclude thata n x n = y n − b n x n+1
tends to a zero asn tends to infinity The proof for B is similar.
... data-page="7">5 The matricesA and B as operator generators of an analytic semigroup
In this section, we will show thatA and B are generators of analytic semigroup in each... consider matrix transformations< i>A and B mapping in a set of< /i>
sequences and we will show that they satisfy hypothesis (H)
3 Definition of the operatorsA and B... π.
(2.7)
It is said thatA and B are generators of analytic semigroups not strongly continuous at
t =0 and we haveσ(A)
σ(