Semigroups on Frechet spaces and equations with infinite delays

Semigroups on Frechet spaces and equations with infinite delays

T SENGADIR

Department of Mathematics, SSN College of Engineering, Old Mahabalipuram Road, Kalavakkam 603 110, India

MS received 28 February 2005

Abstract. In this paper, we show existence and uniqueness of a solution to a functional differential equation with infinite delay. We choose an appropriate Frechet space so as to cover a large class of functions to be used as initial functions to obtain existence and uniqueness of solutions.

Keywords. Functional differential equation; infinite delay; semigroup; Frechet space.

1. Introduction and preliminaries

In this paper we study linear functional differential equations with infinite delay. Consider x(t)=ax(t)+^∞

i=1

bix(t−τi), t ≥0

x(θ)=φ(θ), θ∈(−∞,0] (1.1)

wherea∈R,{b_i}^∞_i=1is an arbitrary sequence of real numbers,{τ_i}^∞_i=1is a strictly increas- ing sequence of strictly positive reals such that limi→∞τ_i = ∞andφ:(−∞,0]−→Ris continuous.

For the special case{b_i}^∞_i=1 ∈ l¹, (1.1) can be uniquely solved for any givenφ ∈BC (−∞,0], the space of all bounded real-valued continuous functions. The proof of this is indicated in Example1.2. Denote this solution byxφ. Consider the family of operators St, t≥0 on BC(−∞,0] defined as

[S_tφ](θ)=x_φ(t+θ), if t+θ >0

=φ(t+θ), if t+θ≤0. (1.2)

It is elementary to see that{S_t}is not a strongly continuous semigroup on BC(−∞,0].

We prove this as follows: IfS_t is a semigroup, we must have

tlim→0S_tφ=φ.

Given > 0, we can findδ > 0 such that Stφ−φ∞ ≤ for |t| ≤ δ. Now, let δ^∗ =min(1, δ). Considerθ1, θ2 ∈(−∞,−1] with 0 ≤θ2−θ1 ≤δ^∗. Lett =θ2−θ1. Note that 0≤t ≤δand

θ₁+t≤θ₂+δ≤θ₂+1≤0.

71

Consider

|φ(θ₂)−φ(θ₁)| ≤ |φ(t+θ₁)−φ(θ₁|

≤ |(S_tφ)(θ₁)−φ(θ₁)|

≤ S_tφ−φ_∞

≤.

Thus, we have shown thatφis uniformly continuous on(−∞,−1]. By the uniform con- tinuity ofφon [−1,0], uniform continuity ofφon(−∞,0] follows. But this is a contra- diction as there are bounded continuous functions which are not uniformly continuous.

On the space BUC(−∞,0], the space of bounded uniformly continuous functions, the family of operatorsS_tdefined by (1.2) do form a semigroup but this space is properly con- tained in BC(−∞,0]. In the literature, certain Banach spaces, which contain BC(−∞,0], which are classes of functions satisfying certain growth conditions, are used in the con- text of infinite delay equations. Refer [5, 6]. Our approach is to find a Frechet space that contains BC(−∞,0] on whichSt’s form a semigroup.

Instead of constructing a weight function which is related tobiandτi, we obtain a family of semi-norms for the initial functionφthat enable us to get estimates for the solution and also to capture a Frechet spaceFsuch that (1.2) defines a strongly continuous semigroup on F. We do not make any explicit summability assumption on the sequence{b_i}^∞_i=1, but the spaceF heavily depends on the properties of{b_i}^∞_i=1. If{b_i} ∈l¹, then BC(−∞,0]⊂F but ifb_i =¹_i, BC(−∞,0] is not contained inF.

The basic theory of finite delay differential equations is covered in [7]. [8] and [1] are some basic references for Banach phase spaces related to infinite delay equations. Consider the following examples:

Example 1.1. LetX= BC[0,∞)andA:D(A)=BC¹[0,∞)−→BC[0,∞)be defined as [Aϕ](x)=ϕ(x). DefineS_t:X−→X, t≥0, as(S_tϕ)(x)=ϕ(t+x). It is easy to see thatAsatisfies the following conditions:

(a) the resolvents(λI−A)⁻¹u=_∞

0 e^−λτS_τudτ exist forλ >0,whereIis the identity operator onX,

(b) (λI −A)⁻¹ ≤ ¹_λ and thatS_t satisfy the following conditions:

(i) S₀=I, (ii) S_t+s =S_tS_s.

Despite the above observationsS_t is not strongly semigroup onXsince the con- dition

(iii) limt→0S_tϕ=ϕ, ϕ∈X does not hold.

The proof that (iii) does not hold is similar to the proof of the analogous assertion in the case of the infinite delay equations. The Hille–Yosida theorem is not applicable precisely becauseAis not densely defined. One way to overcome this difficulty is to construct a smaller Banach space called the Hille–Yosida space for the operatorA[10] on which the restriction ofAgenerates a semigroup. In this construction, the condition (a) plays a crucial role.

But consider the operatorA:D(A)=C¹[0,∞)−→C[0,∞)defined as [Aϕ](x)= ϕ(x).Ais an ‘extension’ ofAthat does generate the semigroupSt: C[0,∞)−→C[0,∞)

defined as [St]ϕ(x) =ϕ(t+x)on the Frechet space C[0,∞). Thus,it is clear that by considering the Hille–Yosida space,a lot of useful information is lost.

Example 1.2. We now indicate the proof of assertion that for {b_i} ∈ l¹, (1.1) can be uniquely solved for any givenφ∈BC(−∞,0].

Fort∈[0, τ1], t−τ_i ∈(−∞,0] and hence forφ∈BC(−∞,0], φ(t−τ_i)is meaningful.

Definez_i ∈ C[0, τ1] asz_i = b_iφ(t−τ_i). Nowz_i∞ = |b_i|sup_t[0,τ1]|φ(t −τ_i)| ≤

|b_i|φ_∞= |b_i|φ_∞. Since{b_i}is inl¹, the series_∞

i=1z_iconverges. Hence_∞

i=1z_i converges in C[0, τ₁]. Thus, _∞

i=1φ(t −τ_i) ∈ C[0, τ₁]. Now, consider the ordinary differential equation

x(t)=ax(t)+^∞

i=1

b_iφ(t−τ_i), t ∈[0, τ₁],

x(0)=φ(0)

whose solution exists and is unique.

Now, assume the existence of a unique functionyk:(−∞, kτ1]−→Rsuch thatyk(θ)= φ(θ)forθ∈(−∞,0] and whose restriction to [0, kτ₁] is a solution to

x(t)=ax(t)+^∞

i=1

b_ix(t−τ_i), t[0, kτ₁],

x(0)=φ(0).

As before, the expression_∞

i=1biyk(t−τi)defines a continuous function on [kτ1, (k+ 1)τ1]. Now, consider

x(t)=ax(t)+^∞

i=1

b_iy_k(t−τ_i), t ∈[kτ1, (k+1)τ1], x(kτ₁)=y_k(kτ₁)

which has a unique solution z. Define y_k+1 as y_k+1(s) = y_k(s), s ∈ (−∞, kτ₁] and y_k+1(s)=z(s), s∈[kτ₁, (k+1)τ1]. Thus, we obtain a solution to (1.1) in(−∞, (k+1)τ1].

By induction and patching up of solutions, we get a unique solution to (1.1) on the whole of(−∞,∞).

LetX=BC(−∞,0] and{b_i}^∞_i=1∈l¹. Define A:D(A)=

ϕ ∈BC¹(−∞,0]:ϕ(0)=^∞

i=1

biϕ(−τi)

−→BC(−∞,0]

as

[Aϕ](x)=ϕ(x).

D(A)is not dense inXand henceAdoes not generate a semigroup. Motivated by Exam- ple 1.1, we look for a Frechet spaceF that contains BC(−∞,0] and an ‘extension’Ato Asuch thatAgenerates a semigroup onF.

Remark 1.3. The general theory of semigroups on Frechet spaces is very complicated. For example,even a bounded linear operator on a Frechet space need not generate a semigroup [12]. In [11],a generalisation of the Hille–Yosida theorem for a closed and unbounded operator in a locally convex space is proved. But the hypotheses of this theorem are not easily verified in many concrete cases. By proving various estimates for the solutionxof (1.1),we are able to capture a Frechet spaceF on which the solution of (1.1) gives rise to a semigroup.

Refer to [3], [13] and [2] for applications of semigroups on locally convex spaces to PDE’s.

We need the following definitions and results in the next section.

DEFINITION 1.4

(i) A topological vector spaceXis said to be a Frechet space if its topology is generated by a family of countable semi-norms{qi}^∞_i=1andX is complete with respect to the family{qi}^∞_i=1.

(ii) A linear mapS:X →X is said to be bounded if for everyi ∈ N, there are finitely many indicesj1, j2, . . . , jmand a constantCsuch that for allφ∈X,

qi(Sφ)≤Cmax(qj₁(φ), qj₂(φ), . . . , qjm(φ)).

The basic theory of Frechet spaces and the proof of the following proposition can be found in [14].

PROPOSITION 1.5

A linear mapS:X→Xis continuous if and only if it is bounded.

DEFINITION 1.6

A family of bounded linear operators{S_t:t ≥0}onXis said to be a strongly continuous semigroup if the properties (i), (ii) and (iii) of Example 1.1 hold.

PROPOSITION 1.7

Let l¹(X) denote the Banach space of all sequences {x_i}^∞_i=1 of elements of a Banach X such that_∞

i=1x_i < ∞. Let C([0, T];l¹)be the Banach space of all continuous functionsh: [0, T]−→l¹. The Banach spacel¹(C([0, T]))is isometrically embedded in C([0, T];l¹).

Proof. LetG∈l¹(C([0, T])). DefineG^∗: [0, T]−→l¹(R)as [G^∗(s)]⁽ⁱ⁾=G⁽ⁱ⁾(s). It is given that

∞ i=1

sup{|G⁽ⁱ⁾(s)|:s∈[0, T]}<∞.

It is easy to check thatG^∗is a bounded function and that sup

_∞

i=1

|[G^∗(s)]⁽ⁱ⁾|:s∈[0, T]

=^∞

i=1

sup{|G⁽ⁱ⁾(s)|:s∈[0, T]}.

It remains to be shown that

tlim→t₀

∞ i=1

|[G^∗(t)]⁽ⁱ⁾−[G^∗(t0)]⁽ⁱ⁾| =0.

By the hypothesis for eachi, lim_t→t0|[G^∗(t)]⁽ⁱ⁾−G^∗(t₀)]⁽ⁱ⁾| =0 and for a given >0, there existsK∈Nsuch that

∞ i=K+1

sup{|G⁽ⁱ⁾(s)|:s∈[0, T]}< /3.

Further, there existsδ >0 such that|t−t0|< δimplies that K

i=1

|[G^∗(t)]⁽ⁱ⁾−[G^∗(t0)]⁽ⁱ⁾|< /3.

Hence for allt, t₀in [0, T] with|t−t₀|< δ, ∞

i=1

|[G^∗(t)]^[(i)−[G^∗(t0)]^[(i)| = K i=1

|[G^∗(t)]^[(i)−[G^∗(t0)]^[(i)|

+ ^∞

i=K+1

|[G^∗(t)]^[(i)−[G^∗(t0)]^[(i)| ≤/3

+ ^∞

i=K+1

|[G^∗(t)]^[(i)| + ^∞

i=K+1

|[G^∗(t0)]^[(i)|

≤/3+/3+/3=. The result is proved.

The following definitions and statements on Frechet space valued Riemann integral are found in [9].

Theorem 1.8. LetX be a Frechet space and letu: [a, b] −→ X be continuous. The integralb

a u(t)dt ∈Xcan be defined uniquely which has the following properties:

(i) for every continuous linear functionalx^∗:X−→R, x^∗

b

a u(t)dt = b

a x^∗(u(t))dt, (ii) for every seminormqk,

q_{k b}

a u(t)dt ≤ _b

aq_k(u(t))dt, (iii) _b

a u(t)dt+_c

b u(t)dt =_c

au(t)dt,

(iv) _b

a[u(t)+v(t)]dt =_b

a u(t)dt+_b

a v(t)dt, (v) c_b

a u(t)dt =_b

a cu(t)dt.

DEFINITION 1.9

LetJbe a sub interval ofR. A functionu:J −→Xis said to be differentiable att0∈J if there existsy ∈Xwith the following property: for every >0 andk∈N, there exists δ >0 such that for allh∈Rwith|t0+h|< δandt0+h∈J, we have

q_k

u(t0+h)−u(t0)

h −y < .

Further, the elementyis denoted byu(t0).

DEFINITION 1.10

LetJbe a sub interval ofR. A functionu:J −→Xis said to be continuously differentiable onJ, ifuis differentiable at every pointt0 ∈ J and the function mappingt tou(t)is continuous onJ. The class of all such functions is denoted by C¹(J;X).

Fundamental theorem of integral calculus

Ifu: [a, b]−→Xis continuously differentiable, then u(a)−u(b)=

_b

a u(t)dt.

A Frechet phase space

For a givenϕ∈C((−∞,0])define the family of seminorms{.k:k∈N}as ϕk =sup{|ϕ(θ)|:θ∈[−k,0]}.

Let{b_i}^∞_i=1and{τ_i}^∞_i=1be as in the beginning of this section. Givenk∈N, definen(k)∈N as the smallest positive integer such thatτ_i ≥kτ₁for alli≥n(k).

DefineF as

F = {ϕ ∈C(−∞,0]:pk(ϕ) <∞ for all k∈N}, where the seminormsp_kare defined as follows:

p_k(ϕ)= ^∞

i=n(k)

sup{|b_iϕ(s−τ_i)|:s∈[0, kτ₁]}.

PROPOSITION 1.11

The spaceF equipped with the topology generated by the family of seminorms{ · k:k∈ N} ∪ {pk:k∈N}is a Frechet space.

Proof. Letφ_j be a Cauchy sequence inF. Clearly, there existsφ∈C(−∞,0] such that φ_jconverges toφuniformly on every compact set of(−∞,0]. Consider the Banach space l¹(C([0, kτ₁])). For everyj ∈N, defineG_j ∈l¹(C([0,kτ₁]))as

G⁽ⁱ⁾_j (s)=bn(k)+(i−1)φj(s−τn(k)+(i−1)), i∈N.

By the hypothesis,Gjis a Cauchy sequence inl¹(C([0, kτ1]))and hence converges to some G∈ l¹(C([0, kτ₁])). For fixediands, lim_j→∞G⁽ⁱ⁾_j (s) =G⁽ⁱ⁾(s)and henceG⁽ⁱ⁾(s)= b_n(k)+(i−1)φ(s−τ_n(k)+(i−1)). This implies thatp_k(φ) <∞andφ_j converges toφinF. The proof is complete.

Now we state the definition of a mild solution of an abstract Cauchy problem in Frechet spaces from [4].

LetXbe a Frechet space whose topology is given by the family of seminorms{q_k:k∈N}

andD(A)be a subset ofX. For a closed operatorA:D(A)−→Xandφ∈X, consider the abstract Cauchy problem

du dt =Au,

u(0)=φ. (1.3)

DEFINITION 1.12

A functionu(·) ∈ C([0,∞),D(A))∩C¹([0,∞), X)that satisfies (1.3) is said to be a solution of Problem (1.3). A functionu(·)∈C([0,∞), X)is said to be a mild solution of (1.3), ifv(t)=_t

0u(s)ds∈D(A), for allt ≥0, and d(v(·))

dt (t)=A(v(t))+φ, t ≥0. (1.4)

2. Main results

LetF be as in the previous section. DefineD(A)= {ϕ∈F:ϕ∈F andϕ(0)=Lϕ}and A:D(A)−→F asAϕ=ϕwhere

Lϕ=aϕ(0)+^∞

i=1

biϕ(−τi).

DEFINITION 2.1

We say that a functionx:R−→Ris a solution to (1.1) if the following hold:

(i) xis continuous andx(θ)=φ(θ)for allθ∈(−∞,0].

(ii) The restriction ofxto [0,∞)is continuously differentiable.

(iii) x(t)=ax(t)+_∞

i=1b_ix(t−τ_i)for allt≥0.

Remark 2.2. Note that our definition of the solution does not imply thatxis differentiable from the left att =0.

Theorem 2.3. LetF, LandAbe as in the beginning of this section. ThenAgenerates a strongly continuous semigroup{S_t:t ≥0}of bounded linear operators onF. Further, for a givenφ∈F,the mapx:R→Rdefined as

x(t)=φ(t), t ∈(−∞,0], x(t)=[Stφ](0), t∈(0,∞) is a unique solution to(1.1).

Besides, fixingφ ∈ F and definingu: [0,∞) → F as u(t) = S_tφ, u(·)is a mild solution to the abstract Cauchy problem (1.3).

We need the following lemmas to prove Theorem 2.3 and we actually define the semi- group via the solution to (1.1).

Lemma 2.4. Letφ∈F. The problem(1.1)has a unique solutionx:R→R.

Further,for eachk∈ Nthere exists a finite subset of{.k:k∈ N} ∪ {p_k:k∈ N}

and a constantC_k ≥0 such that

sup{|x(s)|:s∈[0, kτ1]} ≤Ckmax{q(φ):q∈ }. (2.1) Proof. Considert ∈[0, τ1]. Clearly,t−τ_i ≤t−τ1for alli∈N. Thus fort∈[0, τ1], t− t_i ≤0. Definey1: [0, τ1]→Ras the unique solution of the initial value problem.

x(t)=ax(t)+^∞

i=1

b_iφ(t−τ_i),

x(0)=φ(0). (2.2)

Note that asφ ∈F, t →_∞

i=1biφ(t−τi)defines a continuous function on [0, τ1]. We have

y1(t)=φ(0)e^at+e^at t

0

e^−as _∞

i=1

biφ(s−τi)

ds.

Definex₁:(−∞, τ₁]→Ras

x₁(s)=φ(s), s∈(−∞,0]

=y₁(s), s∈[0, τ₁].

In the remaining part of the proof we shall assume thata=0. The estimates fora =0 are easier to obtain. Clearly,

sup{|x₁(t)|:t ∈[0, τ₁]} ≤

s∈sup[0,τ₁]

e^as

φ1+

s∈sup[0,τ₁]

e^as−1 a

p₁(φ).

(2.3) Here, note that forr > 0,^ear_a⁻¹ >0 for alla =0. Now we claim that for eachk ∈ N, there exists a functionxk:(−∞, kτ1]→Rwith the following properties:

(i) For eacht ∈[0, kτ₁], ∞

i=1

b_ix_k(t−τ_i)

converges and this summation defines a continuous function on [0, kτ₁].

(ii) x_kis the unique solution to x(t)=ax(t)+^∞

i=1

b_ix(t−τ_i), for t ∈[0, kτ1]

x(θ)=φ(θ), for t ∈(−∞,0]. (2.4)

(iii) There exists a finite subset of{ · k:k∈N} ∪ {pk:k∈N}and a constantCk≥0 such that

sup{|x_k(s)|:s∈[0, kτ₁]} ≤C_kmax{q(φ):q ∈ }. We prove this by induction onk. The casek=1 is already proved.

Assuming that our claim is true for arbitraryk∈ N, we show that the claim is true for k+1. DefineI_k =[(k−1)τ₁, kτ₁] andx_kIkas

xkIk =sup{|xk(s)|:s∈[(k−1)τ1, kτ1]}.

Fors∈I_k+1andi∈N, s−τ_i ∈(−∞, kτ1] and so we can consider the summation ∞

i=1

bixk(s−τi)=

n(k+1)−1 i=1

bixk(s−τi)+ ^∞

i=n(k+1)

bixk(s−τi).

The first summation involves only finitely many terms. For s ∈ [kτ1, (k +1)τ1] and i≥n(k+1), s−τi < s−(k+1)τ1≤0 and hence

∞ i=n(k+1)

b_ix_k(s−τ_i)= ^∞

i=n(k+1)

b_iφ(s−τ_i).

So, fors∈I_k+1, the summation_∞

i=1b_ix_k(s−τ_i)defines a continuous function onI_k+1. For a givenk∈N− {1}, definem(k)as the smallest positive integer such that

−m(k) <min{(k−1)τ1−τi:i=1,2, . . . , n(k)−1}.

The following estimates follow from the definition of the seminorms and the integersm(k):

sup

∞ i=n(k+1)

bixk(s−τi)

:s∈[kτ1, (k+1)τ1]

≤pk+1(φ). (2.5)

Moreover, sup

n(k+1)−1 i=n

bixk(s−τi)

:s∈[kτ1, (k+1)τ1]

≤

n(k+1)−1 i=1

|b_i| ×max{φm(k+1),x_kIk}. (2.6)

Definey_k+1: [kτ₁, (k+1)τ₁]→Ras y_k+1(t)=x_k(kτ1)e^a(t−kτ1)+e^at

t kτ1

e^−as _∞

i=1

b_ix_k(s−τ_i)

ds.

From (2.5) and (2.6), we get the estimate sup

∞ i=1

b_ix_k(s−τ_i)

:s∈[kτ₁, (k+1)τ₁]

≤

_n(k+1)−1

i=1

|bi| ×max{φm(k+1),xkI_k} +pk+1(φ)

. (2.7)

Definingx_k+1:(−∞, (k+1)τ₁]→Ras x_k+1(s)=x_k(s), s ∈(−∞, kτ₁]

=y_k+1(s), s∈(kτ₁, (k+1)τ₁],

our claims (i) and (ii) are proved. Now, we proceed to prove (iii) fork+1. From the definition ofy_k+1and the estimate (2.7), we get, fork∈N,

sup{|x_k+1(s)|:s∈[kτ1, (k+1)τ1]} = x_k+1Ik+1

=sup{|yk+1(s)|:s∈[kτ1, (k+1)τ1]}

≤

s∈sup[0,kτ₁]

e^as

x_kIk +

s∈sup[0,τ₁]

e^a(k+1)s−1 a

×

_n(k+1)−1

i=1

|b_i| ×max{φm(k+1),x_kIk} +p_k+1(φ)

. (2.8)

Since

x_kIk ≤sup{|x_k(s)|:s∈[0, kτ1]}

the assertion (iii) fork+1 follows from the estimate (2.8) and hence the assertion (iii) fork.

The solution to (1.1) is obtained by patching the functionsxk. Uniqueness ofx now follows.

Lemma 2.5. Letϕ ∈ F andx:R→Rbe a continuous function such thatx(θ)=ϕ(θ) for allθ ∈ (−∞,0]. Defineu: [0,∞) → C((−∞,0])as [u(t)](θ) = x(t+θ). Then u∈C([0,∞);F ).

Proof. Fixt ∈ [0, jτ₁]. It is trivial to check thatu(t)∈ C((−∞,0]). We also have the estimate

u(t)k =sup{θ∈[−k,0]:|[u(t)](θ)|}

≤max(ϕk,sup{|x(t)|:t ∈[0, jτ1}). (2.9)

Now we show thatp_k(u(t)) <∞for eachk ∈N. Lets ∈[0, kτ₁] andt ∈[0, jτ₁]. It is clear thatn(k+j)≥n(k)and we have the following assertions:

i≥n(k)⇒(s−τi)≤0 and

n(k)≤i < n(k+j)⇒ −(k+j)τ₁≤t+s−τ_i ≤jτ₁. Thus, for everyK≥n(k+j), we have

K i=n(k)

b_i[u(t)](s−τ_i)=

n(k+j)−1 i=n(k)

b_i[u(t)](s−τ_i)

+ K i=n(k+j)

b_i[u(t)](s−τ_i)

=

n(k+j)−1 i=n(k)

b_ix(t+s−τ_i)

+ ^K

i=n(k+j)

b_iφ(t+s−τ_i).

Now choosing a positive integerm≥(k+j)τ1, we have the estimate p_k(u(t))≤

n(k+j) i=1

|b_i|max(sup{|x(t)|:t ∈[0, jτ_i},φm)+p_(k+j)(φ).

(2.10) Equations (2.9) and (2.10) show thatu(t)∈F.

Next, we show the continuity ofuat an arbitraryt₀ ∈ [0,∞). Let > 0 be given.

By the uniform continuity ofxon [−k−1+t₀, t₀+1], there existsδ >0 such that for p, q∈[−k−1+t₀, t₀+1] with|p−q|< δ,|x(p)−x(q)|< . Takeδ^∗ =min(1, δ) and considerswith|t₀−s| < δ^∗. Clearly, for anyθ ∈ [−k,0], t₀+θands+θ both belong to [−k−1+t₀, t₀+1] and hence

sup{|x(t0+θ)−x(s+θ)|:θ ∈[−k,0]}< . That is, given >0, there is aδ^∗>0 such that

u(s)−u(t0)k<

whenever|s−t0|< δ^∗.

Next, we claim that for a given >0 andt0 ∈ [0, jτ1], there existsδ > 0 such that pk(u(t0)−u(t)) < whenevert0∈[0, jτ1] satisfies|t−t0|< δ.

By Proposition 1.6, and the definition ofF, for a givenϕ ∈ F, G_k: [0, kτ₁]→ l¹(R) defined as [G_k(s)]⁽ⁱ⁾=b_n(k)+(i−1)ϕ(s−τ_n(k)+(i−1))is a continuous function. Consider

K i=n(k)

b_i[u(t₀)−u(t)](s−τ_i)=

n(k+j)−1 i=n(k)

b_i[u(t₀)−u(t)](s−τ_i)

+ K i=n(k+j)

bi[u(t0)−u(t)](s−τi)

=

n(k+j)−1 i=n(k)

bi[x(t0+s−τi)−x(t+s−τi)]

= K i=n(k+j)

b_i[φ(t0+s−τ_i)−φ(t+s−τ_i)]

and hence ∞ i=n(k)

b_i[u(t₀)−u(t)](s−τ_i)=

n(k+j)−1 i=n(k)

b_i[x(t₀+s−τ_i)−x(t+s−τ_i)]

+ ^∞

i=n(k+j)

b_i[φ(t₀+s−τ_i)−φ(t+s−τ_i)].

By the uniform continuity ofG_n(k+j)on [0, (k+j)τ₁], given >0, there exists aδ₁>0 such that|p−q|< δ₁implies that

∞ i=n(k+j)

|b_i||[φ(p−τ_i)−φ(q−τ_i)]|< /2.

So, fort₀andt ∈[0, jτ₁] with|t₀−t|< δ₁, sup

_∞

i=n(k+j)

|bi||[φ(p−τi)−φ(q−τi)]|:s∈[0, kτ1]

< /2.

Since the expressionn(k+j)−1

i=n(k) b_i[x(t₀+s−τ_i)−x(t+s−τ_i)] involves evaluation of xover a compact set, given >0, there existsδ2>0 such that

n(k+j)−1 i=n(k)

sup{|b_i||[x(t0+s−τ_i)−x(t+s−τ_i)]|}< /2

whenever t₀, t ∈ [0, jτ₁] and |t₀−t| < δ₂. Our claim is now proved by taking δ = min(δ1, δ2).

Proof of Theorem 2.3. By Lemma 2.4, for a givenφ∈F, we have a solutionx_φ:R→R to (1.1).

DefineS_t:F →F as

[S_tφ](θ)=x_φ(t+θ), t+θ >0

=φ(t+θ), t+θ≤0.

Lemma 2.5 shows thatS_tφ∈F for allt ≥0 and the mapu(t)=S_tφis a continuous function from [0,∞)intoF. From the definition ofS_t one can verify linearity of eachS_t and the propertySt+s =StSs. We need only to check that eachSt is a bounded linear map onF. This follows from the estimates (2.9), 2.10) and (2.1)

Next, we have

[u(t)](θ)=φ(t+θ), t+θ≤0 [u(t)](θ)=φ(0)+

_t+θ

0

L(u(s))ds, t +θ >0 (2.11)

and using (2.11), it is easy to see thatuis indeed a mild solution to (1.3).

The Banach space Cg defined below was the phase space used in [4] for the study of infinite delay equations of which (1.1) is a special case. An interesting observation is that F contains Cg.

Example 2.6. Letg:(−∞,0] → [1,∞)be a continuous non-increasing function such that

∞ i=1

b_ig(−τ_i) <∞. (2.12)

Consider the space Cgof allϕ∈C(−∞,0]→Rsuch that ϕgdef

=sup |ϕ(θ)|

g(θ) :θ∈(−∞,0]

<∞. Then Cg⊂F.

Proof. Consider|b_iϕ(s−τ_i)| = |b_i|^|ϕ(s_g(s⁻₋^τ_τiⁱ⁾₎^|g(s−τ_i). Fors∈[0, kτ₁] andi≥n(k),0≥ s−τ_i ≥ −τ_i and hence

sup{|b_iϕ(s−τ_i)|:s∈[0, kτ₁]} ≤(φg)|b_i|g(−τ_i).

Now, the assertion follows from (2.12).

Acknowledgement

Major part of this work was done while the author was with the Department of Mathemat- ics, Indian Institute of Science, Bangalore. The author acknowledges the referee whose comments greatly improved the presentation of the paper and who pointed out many mis- prints.

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