INTEGRATED C-SEMIGROUPS OF UNBOUNDED LINEAR OPERATORS IN BANACH SPACES

Vol. 35, No. 2, 2005, 1-17

INTEGRATED C-SEMIGROUPS OF UNBOUNDED LINEAR OPERATORS IN BANACH SPACES

Ratko Kravaruˇsi´c¹, Milorad Mijatovi´c²

Abstract. A family of unbounded linear operators (S(t))t≥0 in the Banach space (E,k · k) which satisfies the composition law for an inte- gratedC−semigroup on a domainD⊂Eis introduced and investigated.

The Banach spaces (Eω,k · kω), ω >0,are used for the construction of a family of infinitesimal generatorsA^ω, ω >0 which determine an operator Acalled the infinitesimal generator of (S(t))t≥0.

AMS Mathematics Subject Classification (1991): 47D06

Key words and phrases: Integrated C−semigroup, infinitesimal genera- tors,C−pseudoresolvents

1. Introduction

Integrated semigroups of unbounded linear operator in Banach spaces have been studied in [7], [8]. This paper is a continuation of these studies. Here we use also some results of [9], [15], for n−times integrated C−semigroups and mild integratedC−existence families of bounded operators.

We proved in [7] that any integrated semigroup of unbounded linear oper- ators under additional conditions is an exponentially bounded integrated semi- group on a subspace with a possibly stronger norm. We obtain this result for the integratedC−semigroups of unbounded operators with additional condition for the operatorC.

2. Structural properties

Let (S(t))t≥0 be a family of unbounded linear operators in a Banach space (E,k · k) and let C : D(C) →E be an unbounded liner operator. Denote by D(S(t)) the domain ofS(t) and set

(1)

D=















 x∈T

s,t≥0D(S(s)S(t))

¯¯

¯¯

¯¯

¯¯

¯¯

¯¯

S(0)x= 0

S(t)xis strongly continuous fort≥0, S(t)Cx=CS(t)xfort≥0,

S(s)S(t)x=R^s

0

(S(r+t)−S(r))Cxdr

=S(t)S(s)xfort≥0.

















1Faculty of Economics, University of Banja Luka Bosnia and Hercegovina

2Department of Mathematics and Informatics, University of Novi Sad, Trg Dositeja Obradovi´ca 4, Novi Sad, Serbia and Montenegro e–mail: [email protected]

IfD6={0},then (S(t))t≥0is said to bean integratedC−semigroup of unbounded linear opetarorsinE.Note thatD⊂S(C).

The set

N ={x∈D; S(t)x= 0, t≥0}

is calleda degeneration spaceof an integratedC−semigroup of unbounded linear operators (S(t))t≥0.A semigroup (S(t))t≥0 is callednondegenerate ifN ={0}

and it is calleddegenerateotherwise.

Lemma 1. If an integratedC−semigroup of bounded linear operators(S(t))t≥0

is nondegenerate, thenC is injective (cf. [15], the proof of Lemma 2.2).

Definition 1. Forω∈R⁺= (0,∞), x∈ T

t≥0

D(S(t)),let

(2) kxkω:= sup

t≥0e^−ωtkS(t)xk and set

(3) Eω:={x∈D; kxkω<∞}. Then, k · kω is a norm onEω.

LetEωdenote the closure of the setEωunder the normk · kandS(t)|Eω is the part ofS(t) in Eωi.e.

(4) D(S(t)|Eω) ={x∈Eω; x∈D(S(t)) andS(t)x∈Eω}.

In this paper we assume that for all ω > 0, C is bounded linear operator under the normk · kandkCkω=Mω.

Proposition 1.

a) If ω1 ≤ ω2 and x ∈ D, then kxkω2 ≤ kxkω1. Hence, if ω1 ≤ ω2 then Eω1⊂Eω2.

b) Ifx∈Eω then S(t)x∈Eω and

(5) kS(t)xkω≤ 2

ωMωe^ωtkxkω. Proof.

a) Letω1≤ω2andx∈D. Then, we have kxkω2 = sup

t≥0

e^−ω2tkS(t)xk

= sup

t≥0e^−ω1t·e^(ω1−ω2)tkS(t)xk ≤sup

t≥0e^ω1tkS(t)xk=kxkω1. Thus,Eω1⊂Eω2 ifω1≤ω2.

b) Letx∈Eω.Then kS(t)xkω = sup

s≥0

e^−ωskS(s)S(t)xk=e^ωtsup

s≥0

e^−ω(t+s)kS(s)S(t)xk

=e^ωtsup

s≥0

e^−ω(s+t)

°°

° Rs 0

(S(r+t)−S(r))Cxdr

°°

°

≤e^ωtsup

s≥0

e^−ωs³Z^s

0

e^ωre^−ω(r+t)kS(r+t)Cxkdr+e^−ωt Zs

0

e^ωre^−ωrkS(r)Cxkdr´

≤e^ωtkCxkωsup

s≥0

e^−ωs³Z^s

0

e^ωrdr+e^−ωt Zs

0

e^ωrdr´

≤Mωe^ωtkxkωsup

s≥0

e^−ωs(1 +e^−ωt) Zs

0

e^ωrdr

=Mωe^ωtkxkωsup

s≥0

1

ω e^−ωs(1 +e^−ωt)(e^ωs−1)

=M_ωe^ωtkxk_ωsup

s≥0

1

ω(1 +e^−ωt)(1−e^−ωs)≤ 2

ωM_ωe^ωtkxk_ω.

2

Remark 1. By the proof of Proposition 1 b), we have e^−ω(t+s)kS(s)S(t)xk ≤ 2M_ω

ω kxkω

and

kS(s)S(t)xk ≤ 2Mωe^ω(t+s) ω kxkω.

The following additional assumption will be needed throughout the paper.

(6) For every ω > 0 and for every x ∈ D, there exists K_ω > 0 such that kxkω≥Kωkxk.

Remark 2. If for an integratedC−semigroup of unbounded linear operators (S(t))t≥0 there existt0≥0 andKt0 >0 such that

(7) kS(t0)xk ≥Kt0kxk, x∈D, then, for everyω >0

kxkω= sup

t≥0e^−ωtkS(t)xk ≥e^−ωt0kS(t0)xk ≥Kωkxk, x∈D, where Kω=e^−ωt0Kt0.

Theorem 1. Let (S(t))t≥0 be an integratedC−semigroup of unbounded linear operators inE such that:

(i) (S(t))_t≥0 is nondegenerate,

(ii) C is the bounded linear operator under the normk · kω inEω, (iii) condition (6) holds.

Then:

a) Let ω > 0 be fixed. Suppose that for every t ≥ 0, S(t)|Eω is a closed operator in Eω.

Then(Eω,k · kω)is a Banach space.

b) If S(t)is a closed operator inE, thenS(t)|Eω is a closed operator inEω

fort≥0andω >0.

Proof.

a)Recall the assumption:

If{xn} ⊂D(S(t)|Eω), kxn−xk →0 andkS(t)xn−yk →0 asn→ ∞,then x∈D(S(t)|Eω) andS(t)x=y.

Suppose {xn} ⊂Eω is a Cauchy sequence with respect to the normk · kω. For everyε >0 there exists a numberN >0 such that

(8) kxm−xnkω= sup

t≥0

e^−ωtkS(t)xm−S(t)xnk< ε, m, n > N .

By (6) we have kxm−xnk< ε

Kω, m, n > N.Hence, there existsx∈E such thatkxn−xk →0 asn→ ∞.By (8)

(9) e^−ωtkS(t)xm−S(t)xnk< ε, t≥0, m, n > N ,

that is, for t ≥ 0, {e^−ωtS(t)xn}t≥0 is a Cauchy sequence in the norm of E.

Therefore, for everyt≥0 there existsyt∈E such thatke^−ωtS(t)xn−ytk →0 asn→ ∞.Fixn > N in (9) and letm→ ∞. Then,

(10) ke^−ωtS(t)xn−ytk ≤ε . In (10)N is independent oft.

Since S(t)|Eω is closed for t ≥0, the same holds for e^−ωtS(t)|Eω, t≥ 0.

This implies x∈D(e^−ωtS(t)|Eω) =D(S(t)|Eω) andyt=e^−ωtS(t)x.Now, by (10)

(11) e^−ωtkS(t)xn−S(t)xk ≤ε, n > N, t≥0.

This implies kxn−xkω ≤ ε, for n > N and kxn −xkω → 0 as n → ∞.

Consequently,kxkω<∞.

It remains to prove thatx∈D.Sincexn∈D by (1), we have

(12) S(s)S(t)x_n= Zs

0

(S(r+t)−S(r))Cx_ndr ,

By Remark 1 we have

kS(s)S(t)xn−S(s)S(t)xk ≤ 2M_ωe^ω(t+s)

ω kxn−xkω. Now, fix s, t≥0.Then by (5)

kS(t)Cxn−S(t)Cxk ≤Mωe^ωtkxn−xkω. It implies

°°

° Zs

0

(S(r+t)−S(r))Cxndr− Zs

0

(S(r+t)−S(r))Cxdr

°°

°

≤ Zs

0

k(S(r+t)−S(r))C(xn−x)kdr

≤Mω

ω (e^ω(s+t)+e^ωs)kxn−xkω→0 asn→ ∞. Further

(13) kS(s)S(t)x− Zs

0

(S(r+t)−S(r))Cxdrk →0 asn→ ∞.

SincekS(t)xn−S(t)xk ≤e^ωtkxn−xkω andkxn−xkω→0 asn→ ∞,we have kS(t)xn−S(t)xk →0 asn→ ∞. On the other handS(s)|Eω is closed inEω, by (13), we obtain

S(t)x∈D(S(s)) andS(s)S(t)x= Zs

0

(S(r+t)−S(r))Cxdr .

Lett1≥0.Then,

(14)

kS(t)x−S(t1)xk ≤ kS(t)x−S(t)xnk+kS(t)xn−S(t1)xnk +kS(t1)xn−S(t1)xk ≤e^ωtkxn−xkω

+kS(t)xn−S(t1)xnk+e^ωt1kxn−xkω. Forε >0 andnsufficiently large chooseδ >0 such thate^ωt< e^ωt1+εand

kS(t)xn−S(t1)xnk< ε, for 0<|t−t1| < δ . Then (14) follows thatS(t)xis strongly continuous fort≥0.

Clearly, it holds

kS(t)Cx−CS(t)xk ≤ kS(t)Cx−S(t)Cxnk+kCS(t)xn−CS(t)xk

≤Mωe^ωtkxn−xkω+2M_ω²e^ωt

ωKω kxn−xkω=Mωe^ωt

³

1 + 2Mω

ωKω

´

kxn−xkω< ε fornsufficiently large.

It is easy to see that

kS(0)xk=kS(0)x−S(0)xnk ≤ kx−xnkω< ε fornsufficiently large. HenceS(0)x= 0 and x∈D.

b) We have S(t)|Eω ⊂ S(t), t ≥ 0, so if S(t) is closed, then S(t)|Eω is closable, with the closureS(t)|E_ω.Ifx∈D(S(t)|E_ω),then there is a sequence {xn} ⊂D(S(t)|Eω),andy∈Eωsuch thatkxn−xk →0 andkS(t)xn−yk →0 as n→ ∞.Thenx∈Eωand sinceS(t) is closed,x∈D(S(t)) andS(t)x=y∈Eω. Thusx∈D(S(t)|Eω),andS(t)|Eω is a closed operator inEω. 2

3. Family of C−pseudoresolvents

In this section we suppose that for a nondegenerate integratedC−semigroups (S(t))t≥0of unbounded linear operators for every ω >0 hold:

(i)The operatorC is bounded under the normk · kω inEω. (ii) There existsKω>0 such that

kxk ≤ 1

Kωkxkω.

(iii)The operatorS(t)|Eω is closed inEωfort≥0 andω >0.

Then, we haveE^k·k_ω ^ω=Eω.

Definition 2. For fixedω >0 andλ∈C, Reλ > ω define

R^ω(λ)x=λ Z∞

0

e^−λtS(t)xdt, x∈Eω.

Observe that

°°

°λ Z∞

0

e^−λtS(t)xdt

°°

°≤ |λ|

Z∞

0

e^−tReλkS(t)xkdt≤ |λ|

Kω

Z∞

0

e^−tReλkS(t)xkωdt

≤ 2Mω|λ|

ωK_ω kxkω

Z∞

0

e^(ω−Reλ)tdt= 2Mω|λ|

ωK_ω(Reλ−ω)kxkω.

Thus, the integral is an improper Riemann integral converging absolutely in the norm of E. Observe thatR^ω(λ) is in general unbounded in (E,k · k) and that its domain isEω.

Theorem 2. Fix ω >0 andλ∈Cwith Reλ > ω.

a) (i)R^ω(λ)(Eω)⊂Eω. Moreover, ω(Reλ−ω)

2Mω|λ| kR^ω(λ)xkω≤ kxkω, x∈Eω. (ii) R^ω(λ)x∈D(S(t)C)and

S(t)CR^ω(λ)x=R^ω(λ)S(t)Cx=R^ω(λ)CS(t)x, t≥0, x∈Eω. b) (i) For everyx∈Eω, kxkR^ω <∞,where

(15) kxkR^ω:= sup

n∈N0

sup

λ>0

(λ−ω)ⁿ⁺¹ n!

°°

°

³R^ω(λ) λ

´_(n) Cx

°°

°, λ > ω .

The norm k · k_R^ω is equivalent to the norm k · k_ω.

(ii) If ω1 ≤ω2 and Reλ > ω2, thenR^ω1(λ)x=R^ω2(λ)x, x∈Eω. Thus, as operators in E, R^ω1(λ)⊂R^ω2(λ)if Reλ > ω2.

Proof.

a) (i)Lett≥0 andx∈Eω.Then, kS(s)xkω≤ 2Mωe^ωs

ω kxkω<∞ which implies

°°

°λ Z∞

0

e^−λtS(t)xdt

°°

°ω≤ |λ|

Z∞

0

e^−tReλkS(t)xk_ωdt

≤ |λ|

Z∞

0

e^(ω−Reλ)t2Mω

ω kxkωdt≤ 2Mω|λ|

ω(Reλ−ω)kxkω<∞.

(ii) We obtain R^ω(λ)x ∈ Eω ⊂ D(S(t)C). Since S(t) is closed under the normk · kandS(t)C=CS(t),then holds

S(t)CR^ω(λ)x=R^ω(λ)S(t)Cx=R^ω(λ)CS(t)x, x∈Eω.

b) (i)We will show that, for everyx∈Eω, R^ω(λ)x∈D.Theorem 2a) implies that R^ω(λ)x∈ T

t≥0

D(S(t)) andS(t)R^ω(λ)x=R^ω(λ)S(t)x, t≥0.

It follows R^ω(λ)S(t)x ∈ T

s≥0

D(S(s)) and also S(t)R^ω(λ)x ∈ T

s≥0

D(S(s)).

Thus,R^ω(λ)x∈ T

s,t≥0

D(S(s)S(t)).

Therefore

S(s)S(t)R^ω(λ)x=R^ω(λ)S(s)S(t)x

=λ Z∞

0

e^−λpS(p)S(s)S(t)xdp=λ Z∞

0

e^−λpS(p) Zs

0

(S(r+t)−S(r))Cxdr dp

= Zs

0

(S(r+t)−S(r))CR^ω(λ)x dr .

Moreover,S(t)R^ω(λ)x=R^ω(λ)S(t)ximplies

S(0)R^ω(λ)x= Z∞

0

e^−λsS(0)S(s)x ds= 0.

We will prove lim

t→t1

S(t)R^ω(λ)x = S(t1)R^ω(λ)x, x ∈ Eω. For x∈ Eω and s≥0, using strong continuity, we have

kS(t)S(s)x−S(t1)S(s)xk →0 as t→t1. Remark 1 implies

kS(t)S(s)xk ≤ 2Mωe^ωs

ω e^ωtkxkω≤ 2Mωe^ωs

ω (e^ωt1+ε)kxkω,

for sufficiently small |t−t1|. The dominated convergence theorem for vector valued integrals implies

t→tlim1

S(t)R^ω(λ)x= lim

t→t1

λ Z∞

0

e^−λsS(t)S(s)xds=λ Z∞

0

e^−λs lim

t→t1

S(t)S(s)xds

=λ Z∞

0

e^−λsS(t1)S(s)xds=λS(t1) Z∞

0

e^−λsS(s)xds=S(t1)R^ω(λ)x.

By (i),kR^ω(λ)xkω<∞and ω(Reλ−ω)

2|λ|Mω kR^ω(λ)xkω≤ kxkω. ThusR^ω(λ) is a bounded linear operator with respect to the normk · kω.

(ii) Letx∈Eω andλ > ω.Then, forn∈N0,

(16) ³R^ω(λ)

λ

´_(n)

Cx= (−1)ⁿ Z∞

0

tⁿe^−λtS(t)Cxdt ,

and

°°

°

³R^ω(λ) λ

´_(n) Cx

°°

°≤ Z∞

0

tⁿe^−λtkS(t)Cxkdt≤ Z∞

0

tⁿe^−λt 1 Kω

kS(t)Cxk_ωdt

≤ 2Mω

ωKω

Z∞

0

tⁿe^{−(λ−ω)t}kxkωdt= 2Mω

ωKω

n!

(λ−ω)ⁿ⁺¹kxkω. This implies

ωKω

2Mω

sup

n∈N0

sup

λ>ω

(λ−ω)ⁿ⁺¹ n!

°°

°

³R^ω(λ) λ

´_(n) Cx

°°

°≤ kxkω.

We will use the following assertion (cf. [3]):

Letf(t) be continuous and bounded. Ifλ→ ∞,;n→ ∞so that n λ−ω →t, then,

(λ−ω)ⁿ⁺¹ n!

Z∞

0

e^{−(λ−ω)s}sⁿf(s)ds→f(t). By (16)

(λ−ω)ⁿ⁺¹ n!

³R^ω(λ) λ

´_(n)

Cx= (−1)ⁿ(λ−ω)ⁿ⁺¹ n!

Z∞

0

e^{−(λ−ω)s}sⁿe^−ωsS(s)Cxds

and by using the preceding statement, we obtain e^−ωtS(t)Cx= lim

n→∞λ→∞

(−1)ⁿ(λ−ω)ⁿ⁺¹ n!

³R^ω(λ) λ

´_(n) Cx .

Fort≥0

e^−ωtkS(t)Cxk ≤ lim

n→∞sup

λ>ω

°°

°(λ−ω)ⁿ⁺¹ n!

³R^ω(λ) λ

´_(n) Cx

°°

°

≤ sup

n∈N0

sup

λ>ω

(λ−ω)ⁿ⁺¹ n!

°°

°

³R^ω(λ) λ

´_(n) Cx

°°

° and

kxkω≤ sup

n∈N0

sup

λ>ω

(λ−ω)ⁿ⁺¹ n!

°°

°

³R^ω(λ) λ

´_(n) Cx

°°

°.

(ii) Obviously,Eω₁ ⊆Eω₂ ifω1≤ω2.Forx∈Eω₁ andReλ > ω2 the oper- atorsR^ω1(λ) and R^ω2(λ) are defined andR^ω1(λ)x=R^ω2(λ)x.ThusR^ω1(λ)⊂

R^ω2(λ). 2

4. Family of infinitesimal generators

Definition 3. A function R(·)defined on a subsetD(R) of the complex plane with values in L(E)is called C−pseudoresolvent if it comutes withC and sat- isfies the equation

(17) (µ−λ)R(λ)R(µ) =R(λ)C−R(µ)C, (λ, µ∈D(R)).

R(·)is said to be nondegenerate if R(λ)x= 0 for allλ∈D(R)impliesx= 0.

Theorem 3. The family of operators (R^ω(λ))Reλ>ω on Eω, ω > 0 is the C−pseudoresolvent i.e.

(µ−λ)R^ω(λ)R^ω(µ) =R^ω(λ)C−R^ω(µ)C, Reλ > ω, Reµ > ω . Proof. Note that the operatorC is bounded under the normk · kω and

CR^ω(λ) =R^ω(λ)C .

Fixω >0.We will show that the family of operators (R^ω(λ))Reλ>ω satisfies equation (17). Let λ, µ ∈ C, λ 6= µ, Reλ, Reµ > ω, and x ∈ Eω. Then R^ω(λ)R^ω(µ) is well defined because ((R^ω(µ))(Eω)⊂Eω.We have

(18) R^ω(λ)R^ω(µ)x=λ Z∞

0

e^−λsS(s)R^ω(µ)xds

=λµ Z∞

0

e^−λs Z∞

0

e^−µtS(s)S(t)x dt ds

=λµ Z∞

0

e^−λs Z∞

0

e^−µt Zs

0

(S(r+t)−S(r))Cxdrdtds

= 1

λ−µ h

λµ(λ−µ)

³Z^∞

0

e^−λs Z∞

0

e^−µt Zs

0

S(r+t)Cxdrdtds

− Z∞

0

e^−λs Z∞

0

e^−µt Zs

0

S(r)Cxdrdtds

´i

= 1

λ−µ[λµ(λ−µ)(I1−I2)]. By using Theorem 2a) and the change of variables, we obtain

(19) I1=

Z∞

0

e^−λs Z∞

0

e^µt Zs

0

S(r+t)Cxdrdtds= Z∞

0

e^−λs Z∞

0

e^−µt Zs+t

t

S(v)Cxdvdtds

= Z∞

0

Zv

0

Z∞

v−t

e^−λse^−µtS(v)Cxdsdtdv= 1 λ

Z∞

0

Zv

0

e^−λ(v−t)e^−µtS(v)Cxdtdv

= 1

λ(λ−µ) Z∞

0

e^−λv(e^(λ−µ)v−1)S(v)Cxdv= 1 λ(λ−µ)

³R^ω(µ)Cx

µ −R^ω(λ)Cx λ

´ ,

(20) I₂= Z∞

0

e^−λs Z∞

0

e^−µt Zs

0

S(r)Cxdrdtds= 1 µ

Z∞

0

e^−λs Zs

0

S(r)Cxdrds

= 1 µ

Z∞

0

S(r)Cx Z∞

r

e^−λsdsdr= 1 µ

Z∞

0

e^−λrS(r)Cx Z∞

r

e^−λ(s−r)dsdr

= 1 λµ

Z∞

0

e^−λrS(r)Cxdr=R^ω(λ)Cx λ²µ . Thus (18), (19) and (20) imply

(21) R^ω(λ)R^ω(µ)x

= 1

λ−µ h

λµ(λ−µ)³ 1 λ(λ−µ)

³R^ω(µ)Cx

µ −R^ω(λ)Cx λ

´

−R^ω(λ)Cx λ²µ

´i

= 1

λ−µ h

µ³R^ω(µ)Cx

µ −R^ω(λ)Cx λ

´

−λ−µ

λ R^ω(λ)Cxi

= 1

µ−λ

³

R^ω(λ)Cx−R^ω(µ)Cx

´

and the family of operators (R^ω(λ))Reλ>ω satisfies equation (17). 2

Lemma 2.

(i) The null space

N(R^ω(λ)) ={x∈Eω; R^ω(λ)x= 0}

is independent of the choice ofλwith Reλ > ω.

(ii) The inverseC⁻¹(Range(R^ω(λ)), Reλ > ω,is independent of the choice of λ.

Proof.

(i)Letx∈ N(R^ω(λ)).Then (17) implies

CR^ω(µ)x=CR^ω(λ)x+ (λ−µ)R^ω(µ)R^ω(λ)x= 0, x∈Eω, Reλ, Reµ > ω . The operator C is injective and we have R^ω(µ)x = 0 for Reµ > ω. Then N(R^ω(λ)) =N(R^ω(µ)).

(ii)Letx∈C⁻¹(Range(R^ω(λ))).Then there existsy∈Eωsuch thatCx= R^ω(λ)y forReλ > ω. ForReµ > ω(λ6=µ) we have

C²x=CR^ω(λ)y=CR^ω(µ)y−(λ−µ)R^ω(µ)R^ω(λ)y

=R^ω(µ)(Cy−(λ−µ)R^ω(λ)y) =R^ω(µ)(Cy−(λ−µ)Cx) =CR^ω(µ)(y−(λ−µ)x). SinceC is injective, we obtain

Cx=R^ω(µ)z for z=y−(λ−µ)x . Therefore,

x∈C⁻¹(Range(R^ω(µ)), Reµ > ω .

2

Lemma 3.

(i) The null space N(C−λR^ω(λ))is independent ofλwith Reλ > ω.

(ii) The inverseC⁻¹(Range(C−λR^ω(λ)))is independent ofλwithReλ > ω.

Proof.

(i)ForN(C−λR^ω(λ)) we haveCx−λR^ω(λ)x= 0. Hence, R^ω(µ)Cx−λR^ω(µ)R^ω(λ)x= 0

and

CR^ω(µ)x− λ

λ−µ(CR^ω(µ)x−CR^ω(λ)x) = 0. SinceC is injective, we haveR^ω(µ)x− λ

λ−µ(R^ω(µ)x−R^ω(λ)x= 0, λ6=µ.By multiplying both sides of the equality byλ−µit follows

λR^ω(µ)x−µR^ω(µ)x−λR^ω(µ)x+λR^ω(λ)x= 0 and

λR^ω(λ)x=µR^ω(µ)x . Therefore,

Cx−µR^ω(µ)x= 0, Reµ > ω . (ii) Letx∈C⁻¹(Range(C−λR^ω(λ))).Then

(22) Cx=Cy−λR^ω(λ)y

for some y ∈ Eω and Reλ > ω. We will show that for z =x+µ

λ(y−x) the following holds

Cx=Cz−µR^ω(µ)z, Reµ > ω . By using (22) we obtain

C²x=C²y−λR^ω(λ)Cy

=C²y−λ[R^ω(µ)Cy−(λ−µ)R^ω(µ)R^ω(λ)y]

=C²y−λR^ω(µ)(Cy−(λ−µ)R^ω(λ)y)

=C²y−λR^ω(µ)

³

Cy−λ−µ

λ C(y−x)

´

=C

³

Cy−λR^ω(µ)

³ x+µ

λ(y−x)

´´

. SinceC is injective, we have

Cx=Cy−λR^ω(µ)³ x+µ

λ(y−x)´ and after multiplication with µ

λ we obtain 0 = µ

λC(y−x)−µR^ω(µ)

³ x+µ

λ(y−x)

´ .

Finally

Cx=C³ λ+µ

λ(y−x)−µR^ω(µ)³ x+µ

λ(y−x)´´

and therefore

Cx=Cz−µR^ωµ(z), for z=x+µ

λ(y−x), µ6=λ, Reλ > ω . Note that

kR^ω(λ)Cxkω≤ 2|λ|Mω

ω(Reλ−ω)kxkω, Reλ > ω, ω∈Eω.

and the operatorsR^ω(λ)Care bounded under the norm k · kω. 2

Theorem 4. For the family of operators (R^ω(λ))Reλ>ω it holds:

(i) There exists some linear operatorB^ω such thatλI−B^ω is injective and

(23)





Range(R^ω(λ))⊂D(B^ω),

R^ω(λ)(λI−B^ω)⊂(λI−B^ω)R^ω(λ) =C, for allλwithReλ > ω,

if and only if

N(R^ω(λ)) ={0}.

(ii) The largest operator which satisfies (23) is the closed linear operatorA^ω defined by

(24)









D(A^ω) :=C⁻¹[Range(R^ω(λ)] ={x∈Eω; Cx∈Range(R^ω(λ))},

A^ωx:= (λ−((R^ω(λ))⁻¹)Cx, x∈D(A^ω), which is independent ofλ, Reλ > ω . (iii) IfB^ω satisfies (23) thenC⁻¹B^ωC=A^ω, where

D(C⁻¹B^ωC) ={x∈Eω; Cx∈D(B^ω)andB^ωCx∈Range(C)}.

In particularyC⁻¹A^ωC=A^ω.

Proof. (see [11] Theorem 3.4) 2

Let D(A) = S

ω>0

D(A^ω), where A^ω is given in Theorem 4. For x∈ D(A) let ω > 0 such that x ∈ D(A^ω). There exists y ∈ Eω such that Cx = R^ω(λ)y, Reλ > ω.We define

(25) Ax:=λx−y .

We call A theinfinitesimal generator of the once integrated C−semigroup of unbounded linear operators (S(t))t≥0.

It is clear that x∈D(A) impliesx∈D(A^ω) for someω >0 and Ax=λx−y=λx−(R^ω(λ))⁻¹Cx=A^ωx .

Fory∈Eωwe haveCx=R^ω(λ)y.Thus, the operatorAis well defined and D(A|Eω) =A^ω.

It is easy to show thatD(A) is a subspace ofE andAis a linear operator.

Theorem 5.

(i) Ifω1≤ω2 thenA^ω1 ⊂A^ω2.

(ii) For allx∈Eω the resolvent equation (λI−A)y=x, Reλ > ω has a unique solution belonging toEω andy=C⁻¹R^ω(λ)x.

(iii) Let ω >0.Then for t≥0, S(t)(D(A^ω))⊂D(A^ω)and S(t)A^ωx=A^ωS(t)x, x∈D(A^ω).

(iv) The operator A is closed under the topology induced by the normk · kω

and

CA^ω⊂A^ωC . (v) For allt≥0 andx∈D(A), R^t

0

S(r)xdr∈D(A). The functiont→S(t)x is differentiable of t for t >0. It holdsS⁰(t)x−Cx =S(t)Ax, or equivalenty, S(t)x−tCx=R^t

0

S(r)Axdr, t >0.

Proof.

(i) Letω1≤ω2 andx∈D(A^ω1). Then we havex∈Range(C⁻¹R^ω(λ)) = C⁻¹Range(R^ω(λ)) andx=C⁻¹R^ω(λ)y,for somey∈Eω1 andReλ > ω1.It is clear thatω1≤ω2 impliesR^ω1(λ)< R^ω2(λ) and

C⁻¹R^ω1(λ)y=C⁻¹R^ω2(λ)y.

Hence

Cx=R^ω2(λ)y, x∈D(A^ω1).

ThenA^ω1x=λx−y=A^ω2(x), Reλ > ω2, x∈D(A^ω1).It impliesA^ω1⊂A^ω2. (ii) We will show thaty=C⁻¹R^ω(λ)Cx∈Eω is the unique solution of the resolvent equation. Forx∈Eω andReλ > ω we have

(λI−A^ω)C⁻¹R^ω(λ)x= [λI−(λI−(R^ω(λ)C)⁻¹C]C⁻¹R^ω(λ)x=x.

ThenA|E_ω=A^ω implies (ii).

(iii)Forx∈D(A),letω >0 such thatx∈D(A^ω).Therefore we have S(t)Ax=S(t)A^ωx=S(t)(λx−y)

=λS(t)x−S(t)y=A^ωS(t)x=AS(t)x, where Cx=R^ω(λ)y forReλ > ω.

(iv)The operatorA^ωis closed under the norm k · kω (Theorem 4) and CA^ωx=C(λx−y) =λCx−Cy=A^ωCx.

(v)Letω >0 andt≥0 be fixed. Then

e^−ωs

°°

°S(s)C Zt

0

S(r)xdr

°°

°≤e^−ωs Zt

0

°°

°S(s)S(r)Cx

°°

°dr

≤e^−ωs2Mωe^ωs ω kxkω

Zt

0

dr≤ 2Mωe^ωt

ω² kxkω<∞.

Hence,R^t

0

S(r)xdr∈Eω.There exists

R^ω(λ) Zt

0

S(r)xdr=λ Z∞

0

e^−λsS(s) Zt

0

S(r)xdrds .

Lety ∈Eω such thatCx=R^ω(λ)y.The operator A^ω is closed and forA^ωx= λx−y we have

Z∞

0

S(r)A^ωxdr=A^ω Zt

0

S(r)xdr=λ Zt

0

S(r)xdr− Zt

0

S(r)ydr .

ThereforeR^t

0

S(r)xdr∈D(A^ω) andR^t

0

S(r)xdr∈D(A) becauseA|Eω=A^ω. We obtain, forx∈D(A^ω), Cx=R^ω(λ)y for some y ∈Eω with Reλ > ω andA^ωx=λx−y.By Fubini’s theorem it holds (cf. [7])

S(t+h)−S(t)

h Cx= λ

µ

³

S(t+h) Z∞

0

e^−λrS(r)ydr−S(t) Z∞

0

e^−λrS(r)ydr

´

= e^λh−1 h e^λt

Z∞

0

e^−λvS(v)Cydv−e^λ(t+h) h

t+hZ

0

e^−λvS(v)Cydv+e^λt h

Zt

0

e^−λvS(v)Cydv.

Leth→0. We have

(26) S⁰(t)x=e^λtC²x−f⁰(t)

where

f(t) =e^λt Zt

0

e^−λvS(v)Cydv.

Differentiating, it follows

f⁰(t) =λe^λt Zt

0

e^−λvS(v)Cydv+e^λt·e^−λtS(t)Cy

and (26) implies

(27) S⁰(t)Cx=e^λtC²x−λe^λt Zt

0

e^−λvS(v)Cydv−S(t)Cy .

Therefore,

e^λtC²x−λe^λt Zt

0

e^−λvS(v)Cydv=λe^λt

³Z^∞

0

e^−λvS(v)Cydv− Zt

0

e^−λvS(v)Cydv

´

λ Z∞

0

e^−λpS(p+t)Cydp=λ Z∞

0

e^−λp(S⁰(p)S(t)y+S(p)Cy)dv

=λS(t) Z∞

0

e^−λpS⁰(p)ydp+λ Z∞

0

e^−λpS(p)Cydp=λS(t)Cx+C²x.

SinceS(t)C=CS(t) and by using (27) we obtain

CS⁰(t)x=C²x+λCS(t)x−CS(t)y . The operatorC is injective and we have

S⁰(t)x=Cx+λS(t)x−S(t)y.

Therefore

S⁰(t)x=Cx+S(t)A^ωx, ω >0, and

S(t)A^ωx=S⁰(t)x−Cx . SinceA=A^ω onEω, it implies

Zt

0

S(r)Axdr=S(t)x−tCx, t >0.

2

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Received by the editors March 1, 2004