# Space ofω-Periodic Limit Functions and Its Applications to an Abstract Cauchy Problem

Authors: Rui Xie; Chuanyi Zhang
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We introduce a new space consisting of what we callω-periodic limit functions. We investigate some properties of the new function space. In particular, we study inclusion relations among asymptotically periodic type function spaces. Finally, we apply the ω-periodic limit functions to investigate the existence and uniqueness of asymptotically ω-periodic mild solutions of an abstract Cauchy problem.


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## Body

## 1. Introduction

Let(X,·) be a Banach space. In this paper, we denote by R+ the interval [0,∞) and by Cb(R+,X) the space consisting of bounded and continuous functions from R+ into X, endowed with the uniform convergence norm ·∞. Set C0(R+,X)= {f∈Cb(R+,X): limt→∞f(t)=0} and Pω(R+,X)= {f∈Cb(R+,X): fisωperiodic}. A function f∈Cb(R+,X) is said to be asymptotically ω-periodic if it can be expressed as f=g+h, where g∈Pω(R+,X) and h∈C0(R+,X). The subspace of Cb(R+,X) consisting of the asymptotically ω-periodic functions will be denoted by APω(R+,X).Letf=g+h∈APω(R+,X). Since g∈Pω(R+,X) and h∈C0(R+,X), one gets(1)limt→∞ft+ω-ft=limt→∞ht+ω-ht=0.The converse is not true. The authors in [1] provided two examples to show that there exists a bounded and continuous function which satisfies (1) but is not asymptotically ω-periodic. At the same time, (1) leads authors in [2] to propose the following definition.Definition 1 (see [2, Definition 3.1]).
A functionf∈Cb(R+,X) is said to be S-asymptotically ω-periodic if there exists ω>0 such that limt→∞(f(t+ω)-f(t))=0. The subspace of Cb(R+,X) consisting of the S-asymptotically ω-periodic functions will be denoted by SAPω(R+,X).The following concept ofS-asymptotic ω-periodicity in the Stepanov sense is introduced in [3, Definition 1.3].Definition 2.
A functionf∈BSp(R+,X) is called S-asymptotically ω-periodic in the Stepanov sense if ∫tt+1‍f(s+ω)-f(s)pds→0 as t→∞. The subspace of BSp(R+,X) consisting of the S-asymptotically ω-periodic functions in the Stepanov sense will be denoted by SAPωp(R+,X).LetSAPωu(R+,X)=SAPω(R+,X)∩Cu(R+,X), where Cu(R+,X) consist of those functions from Cb(R+,X) which are uniformly continuous on R+. We have the following proper inclusions:(2)APωR+,X⊂SAPωu(R+,X)⊂SAPω(R+,X)⊂SAPωp(R+,X).References [1, Examples 2.1 and 2.2] and [2, Examples 3.1 and 3.2] show that there exists a function f∈SAPωu(R+,X) but not in APω(R+,X). We will give an S-asymptotically ω-periodic function which is not uniformly continuous in Section 3. Reference [3, Example 2.3] shows that there exists a function f∈SAPωp(R+,X) but not in SAPω(R+,X). Moreover, the function f in this example is bounded continuous. That is, f∈SAPωp(R+,X)∩Cb(R+,X) but not in SAPω(R+,X). Henríquez et al. in [2] and Pierri in [4] gave some conditions under which an S-asymptotically ω-periodic function is asymptotically ω-periodic. Later, Henríquez in [3] showed that, if f∈SAPωp(R+,X) and f is uniformly continuous on R+, then f∈SAPωu(R+,X). For some qualitative properties of SAPω(R+,X)(SAPωp(R+,X)), we refer the reader to [2, 3]. For their applications we refer the reader to [2–10].In this paper, we will introduce a new space consisting of what we callω-periodic limit functions. Our aim to propose the new function space is twofold. On the one hand, ω-periodic limit functions generalize asymptotically ω-periodic functions in a different way from S-asymptotically ω-periodic functions and have some relationship with asymptotically ω-periodic functions (in the Stepanov sense). On the other hand, the ω-periodic limit functions contribute to studying the existence and uniqueness of asymptotically ω-periodic mild solutions of some abstract Cauchy problems.The paper is organized as follows. In Section2, we define the space of ω-periodic limit functions and investigate its properties. In Section 3, we discuss inclusion relations among asymptotically periodic type function spaces. Finally, in Section 4 we study the existence and uniqueness of asymptotically ω-periodic mild solutions of an abstract Cauchy problem with ω-periodic limit coefficients.


## 2. Space ofω-Periodic Limit Functions

Definition 3.
Letf∈Cb(R+,X) and ω>0. We call fω-periodic limit if g(t)=limn→∞f(t+nω) is well defined for each t∈R+, where n∈N. The collection of such functions will be denoted by PωL(R+,X).Remark 4.
The functiong in Definition 3 is measurable but not necessarily continuous.We list some basic properties ofω-periodic limit functions in the following proposition, its proof is obvious, and so we omit it.Proposition 5.
Iff, f1, and f2 are ω-periodic limit and g(t)=limn→∞f(t+nω) is well defined for each t∈R+, then the following statements are true: (1)
f1+f2 is ω-periodic limit;(2)
cf is ω-periodic limit for every scalar c;(3)
g(t)=g(t+ω) for each t∈R+;(4)
g is bounded on R+; moreover g∞≤f∞;(5)
fa(t)=f(t+a) is ω-periodic limit for each fixed a∈R+.Remark 6.
It follows from Proposition5(1)(2) that PωL(R+,X) is a linear subspace of Cb(R+,X).Remark 7.
Because of Proposition5(3), we give name of ω-periodic limit for functions in Definition 3.Theorem 8.
P
ω
L
(
R
+
,
X
) is a Banach space.Proof.
Let{fk} be a sequence of ω-periodic limit functions such that (3)limk→∞fk(t)=f(t)uniformly in t∈R+.
By the definition of theω-periodic limit function, we have (4)limn→∞fk(t+nω)=gk(t)for each k∈N and each t∈R+. So for each t∈R+ the sequence of {gk(t)} is a Cauchy sequence in X because of the following inequality: (5)gk(t)-ght≤gkt-fkt+nω+fkt+nω-fht+nω+fh(t+nω)-gh(t).Thus the sequence {gk(t)} converges to a function g(t) pointwise.
To show thatf∈PωL(R+,X) we only need to show that (6)limn→∞f(t+nω)=g(t)pointwise on R+. But this comes from the following inequality: (7)f(t+nω)-g(t)≤f(t+nω)-fk(t+nω)+fk(t+nω)-gk(t)+gk(t)-g(t).Letf=g+h∈APω(R+,X), where g∈Pω(R+,X) and h∈C0(R+,X). For each t∈R+, one has (8)limn→∞ft+nω=limn→∞gt+ht+nω=gt,which shows that APω(R+,X)⊂PωL(R+,X).Next we give an example to show that there exists anω-periodic limit function which is not asymptotically ω-periodic. For concision, we only consider the case ω=2. Similarly one can exhibit examples for the general case ω>0.Example 9 (example of a functionf∈PωL(R+,X) but not in APω(R+,X)).
Consider the setK={{kn}⊂(0,1):kn>kn+1,kn→0(n→∞),n∈N}. For {kn}∈K, define the function f{kn}:R+→R by (9)f{kn}(t)=1,t=2n-1,n∈N;0,t∈0,2∪2n+1-kn∪2n+1+kn,n∈N;linear,inbetween.If we define the function g:R+→R by(10)g(t)=1,t=2n-1,n∈N;0,otherwise,then we have g(t)=limm→∞f{kn}(t+2m) for each t∈R+, m∈N. So f{kn} is a 2-periodic limit function.
On the other hand, if we choosebn=2n+1, cn=2n+1-kn, then we have bn-cn=kn→0 as n→∞, while f{kn}(bn)-f{kn}(cn)=1. This proves that f{kn} is not uniformly continuous on R+. So f{kn} is not asymptotically 2-periodic.Fixp∈[1,∞) and q denotes the conjugate exponent of p.The Bochner transformfb(t,s), t∈R+, s∈[0,1] of a function f(t) on R+, with values in X is defined by fb(t,s)=f(t+s).Letf:R+→X be a measurable function. We say that f is a Stepanov bounded function, with the exponent p, if sup⁡t∈R+∫tt+1‍f(s)pds<∞. The collection of such functions will be denoted by BSp(R+,X).The spaceBSp(R+,X) endowed with the norm (11)fSp=fbL∞(R+,Lp(0,1;X))=sup⁡t∈R+∫tt+1‍fspds1/pis a Banach space.Define the subspaces ofBSp(R+,X) by SpPω(R+,X)= {f∈BSp(R+,X): ∫tt+1‍f(s+ω)-f(s)pds=0,t∈R+} and BS0p(R+,X)= {f∈BSp(R+,X): limt→∞∫tt+1‍f(s)pds=0}. A function f∈BSp(R+,X) is called asymptotically ω-periodic in the Stepanov sense if it can be expressed as f=g+h, where g∈SpPω(R+,X) and h∈BS0p(R+,X). The subspace of BSp(R+,X) consisting of the asymptotically ω-periodic functions in the Stepanov sense will be denoted by SpAPω(R+,X).Letf∈PωL(R+,X). Then there exists a measurable function g such that g(t)=limn→∞f(t+nω) for each t∈R+. By Proposition 5(3)(4), we know that g∈SpPω(R+,X). If we denote h(s)=f(s)-g(s), s∈R+, it is not hard to show that h∈BS0p(R+,X). Thus, f∈SpAPω(R+,X).Therefore,PωL(R+,X)⊂SpAPω(R+,X). It is easy to know that the inclusion is proper.Now we have the following relationship between them:(12)APω(R+,X)⊂PωLR+,X⊂SpAPωR+,X.The following proposition is a part of [11, Theorem 2.1]. For the sake of completeness, we include the proof in Appendix.Proposition 10.
Letf∈PωL(R+,X) and g(t)=limn→∞f(t+nω) be well defined for each t∈R+. If g(t)=limn→∞f(t+nω) uniformly on [0,ω], then f∈APω(R+,X).Corollary 11.
Letf∈PωL(R+,X) and g(t)=limn→∞f(t+nω)be well defined for each t∈R+. If g(t)=limn→∞f(t+nω) uniformly on R+, then f∈APω(R+,X).Proposition 12.
Iff∈SpAPω(R+,X) and f is uniformly continuous on R+, then f∈APω(R+,X).Proof.
We have thatf is bounded [3, Proposition 2.2]. Let f=g+h, where g∈SpPω(R+,X) and h∈BS0p(R+,X). Note that (13)∫tt+1‍fs+nω-gspds1/p≤∫tt+1‍fs+nω-gs+nωpds1/p+∫tt+1‍gs+nω-gspds1/p≤∫tt+1‍fs+nω-gs+nωpds1/p+∑k=1n‍∫tt+1‍gs+kω-gs+k-1ωpds1/p=∫tt+1‍hs+nωpds1/p+∑k=1n‍∫t+k-1ωt+k-1ω+1‍gs+ω-gspds1/p=∫t+nωt+nω+1‍hspds1/p.Let ɛ>0. Since h∈BS0p(R+,X), we can choose N1∈N such that (14)∫t+nωt+nω+1‍hspds1/p<ɛwhen n≥N1 for all t∈R+. Therefore,(15)sup⁡t∈R+∫tt+1‍fs+nω-gspds1/p⟶0as n→∞. We denote fn(s)=f(s+nω), s∈R+. For ɛ>0 given, we select N2∈N such that (16)∫tt+1‍fns-gspds1/p<ɛ2when n≥N2 for all t∈R+. Then we have(17)∫tt+1‍fns-fn+pspds1/p≤∫tt+1‍fns-gspds1/p+∫tt+1‍fn+ps-gspds1/p<ɛwhen n≥N2 for all t∈R+ and any p∈N. We will prove that {fn} is a Cauchy sequence in norm ·∞. In fact, if we assume the contrary, there exist ε0>0, s′∈R+, and sequences {nk}⊂N such that nk→∞ as k→∞ and {pk}⊂N such that (18)fnks′-fnk+pk(s′)≥ε0.Since f∈Cu(R+,X), fn∈Cu(R+,X). In particular, fnk,fnk+pk∈Cu(R+,X). So there exists a δ>0 such that f(s1)-f(s2)<ε0/3 when s1-s2≤δ, s1,s2∈R+. We assume that δ<1. For s-s′≤δ, s∈R+, we have (19)fnks′-fnks<ε03,fnk+pks′-fnk+pks<ε03.Thus,(20)fnk(s)-fnk+pk(s)≥fnks′-fnk+pks′-fnks′-fnks-fnk+pk(s)-fnk+pks′>ε03when s∈[s′,s′+δ]. This implies that (21)∫s′s′+1‍fnk(s)-fnk+pkspds≥∫s′s′+δ‍fnk(s)-fnk+pkspds≥ε03pδ.Thus,(22)∫s′s′+1‍fnks-fnk+pkspds1/p≥ε03δ1/p,which is a contradiction. Thus {fn} is a Cauchy sequence in norm ·∞. Hence, there exists a function g~∈Cu(R+,X) such that fn→g~ as n→∞. Then we have (23)sup⁡t∈R+∫tt+1‍fs+nω-g~spds1/p⟶0as n→∞. Combining with (15), we get(24)∫tt+1‍gs-g~spds1/p=0for every t∈R+. From (25)∫tt+1‍g~s+ω-g~spds1/p≤∫tt+1‍g~s+ω-gs+ωpds1/p+∫tt+1‍gs+ω-gspds1/p+∫tt+1‍gs-g~spds1/p,one gets(26)∫tt+1‍g~s+ω-g~spds1/p=0for every t∈R+. Since g~∈Cu(R+,X), then g~(·+ω)∈Cu(R+,X) and g~(·+ω)-g~(·)∈Cu(R+,X). Combining with (26), we get g~(s+ω)=g~(s) for every s∈R+. Thus, g~∈Pω(R+,X). Denote h~=f-g~∈Cu(R+,X). By (24), we have (27)∫tt+1‍h~spds1/p=∫tt+1‍fs-g~spds1/p=∫tt+1‍fs-gspds1/p=∫tt+1‍hspds1/p.From h∈BS0p(R+,X), we have h~∈BS0p(R+,X). Note that h~∈Cu(R+,X). We can show that h~∈C0(R+,X). If we assume the contrary, there are a constant ε0>0 and a nondecreasing sequence {tn} such that tn→∞ as n→∞ and h~(tn)≥ε0. Since h~∈Cu(R+,X), there exists a δ>0 such that h~(s1)-h~(s2)<ε0/2 when s1-s2≤δ, s1,s2∈R+. We assume that δ<1. Therefore, (28)h~t≥h~tn-h~tn-h~t>ε02for t∈[tn,tn+δ]. This implies that (29)∫tntn+1‍h~(s)pds≥∫tntn+δ‍h~(s)pds≥ε02pδ,which is a contradiction. Therefore, f=g~+h~∈APω(R+,X).Remark 13.
In view of (12) and Proposition 12, it is interesting to know if there exists a function f∈SpAPω(R+,X)∩Cb(R+,X) but not in PωL(R+,X).Corollary 14.
Iff∈PωL(R+,X) and f is uniformly continuous on R+, then f∈APω(R+,X).Remark 15.
If a functionf∈PωL(R+,X) and is also uniformly continuous, then f∈APω(R+,X). So the range of f is relatively compact. It is interesting to know if there exists a function f∈PωL(R+,X) but the range of f is not relatively compact.


## 3. Inclusion Relations among Asymptotically Periodic Type Function Spaces

In this section, we mainly discuss inclusion relations for these asymptotically periodic type function spaces in (2) and (12).Letf=g+h∈SpAPω(R+,X), where g∈SpPω(R+,X) and h∈BS0p(R+,X). From (30)∫tt+1‍ft+ω-fspds1/p≤∫tt+1‍gs+ω-gspds1/p+∫tt+1‍hs+ω-hspds1/p≤∫tt+1‍hs+ωpds1/p+∫tt+1‍hspds1/p,one gets(31)limt→∞∫tt+1‍ft+ω-fspds1/p=0,which shows SpAPω(R+,X)⊂SAPωp(R+,X).Example 16 (example of a functionf∈PωL(R+,X) but not in SAPω(R+,X)).
Let us come back to Example9. For t∈[2(n-1),2n], we have(32)max⁡2(n-1)≤t≤2nfknt+2-fknt=kn-1-knkn=kn-1kn-1,where n≥2. If we choose k~n=1/mn, m>1, then {k~n}∈K. So f{k~n} is a 2-periodic limit function.
On the other hand, we have(33)max⁡2(n-1)≤t≤2nfk~nt+2-fk~nt=m-1for n≥2, which shows that f{k~n} is not S-asymptotically 2-periodic.Example 17 (example of a functionf∈SAPω(R+,X) but not in SAPωu(R+,X)).
Consider Example9 again. If {kn}∈K satisfies the condition (H): kn-1/kn→1n→∞, then by (32), one has (34)max⁡2(n-1)≤t≤2nfknt+2-fknt⟶0n⟶∞,which shows that f{kn} is S-asymptotically 2-periodic. For example, let k¯n=1/(n+1)α, α>0. Clearly {k¯n}∈K and {k¯n} also satisfies the condition (H). So f{k¯n} is S-asymptotically 2-periodic. On the other hand, f{k¯n} is not uniformly continuous. Thus f{k¯n} is an S-asymptotically 2-periodic function which is not uniformly continuous.The following proposition plays a role in the example below.Proposition 18.
Iff∈SpAPω(R+,X), then limT→∞∫TT+ω‍(∫tt+1‍f(s)pds)1/pdt=M, where M is a nonnegative constant.Proof.
Letf=g+h, where g∈SpPω(R+,X) and h∈BS0p(R+,X). We first prove that there exists a nonnegative constant M such that limT→∞∫TT+ω‍(∫tt+1‍g(s)pds)1/pdt=M. Note that (35)∫tt+1‍gspds1/p≤∫tt+1‍gs+ω-gspds1/p+∫tt+1‍gs+ωpds1/p≤∫tt+1‍gs+ωpds1/p.In the same way, one gets(36)∫tt+1‍gs+ωpds1/p≤∫tt+1‍gspds1/p.Thus, we obtain(37)∫tt+1‍gspds1/p=∫tt+1‍gs+ωpds1/p.Let F(t)=(∫tt+1‍g(s)pds)1/p. By (37), we have (38)Ft=∫tt+1‍gs+ωpds1/p=∫t+ωt+ω+1‍gspds1/p=F(t+ω).In view of g∈SpPω(R+,X), F(t)≤gSp for each t∈R+. Therefore, there exists a nonnegative constant M such that ∫TT+ω‍F(t)dt=M for each T∈R+. Thus,(39)limT→∞∫TT+ω‍F(t)dt=M.Since h∈BS0p(R+,X), we have(40)limt→∞∫tt+1‍hspds1/p=0.Moreover, one gets(41)limT→∞∫TT+ω‍∫tt+1‍hspds1/pdt=0.Note that(42)∫TT+ω‍∫tt+1‍fspds1/pdt≤∫TT+ω‍∫tt+1‍gspds1/pdt+∫TT+ω‍∫tt+1‍hspds1/pdt,∫TT+ω‍∫tt+1‍fspds1/pdt≥∫TT+ω‍∫tt+1‍gspds1/pdt-∫TT+ω‍∫tt+1‍hspds1/pdt.By (39), (41), and (42), we have (43)limT→∞∫TT+ω‍∫tt+1‍fspds1/pdt=M.Example 19 (example of a functionf∈SAPωu(R+,X) but not in SpAPω(R+,X)).
Letf:R+→R be the function defined by (44)f(t)=0,t=2n2,n=0,1,2,…;2,t=2n+12,n=0,1,2,…;linear,inbetween.Fix ω>0. Let an=(2n)2, bn=(2n+1)2. Then we have bn-an=4n+1→∞(n→∞) and an+1-bn= 4n+3→∞(n→∞). For any ɛ>0, we can choose N∈N such that 4N+1>ω and 2ω/4N+1<ɛ. Then(45)sup⁡t≥4N2ft+ω-ft≤2ω4N+1<ɛ,which shows that f is S-asymptotically ω-periodic. From (45) we know that f is uniformly continuous. Thus, f∈SAPωu(R+,X).
On the other hand, there existN1,N2∈N such that 0≤f(t)<1, t∈[(2n)2-ω/2, (2n)2+ω/2+1] when n≥N1 and 1<f(t)≤2, t∈[(2n+1)2-ω/2, (2n+1)2+ω/2+1] when n≥N2.
So, whenn≥max⁡{N1,N2}, one gets(46)∫(2n)2-ω/2(2n)2+ω/2‍∫tt+1‍fspds1/pdt<ω,∫(2n+1)2-ω/2(2n+1)2+ω/2‍∫tt+1‍fspds1/pdt>ω.By (46) the limit of ∫TT+ω‍(∫tt+1‍f(s)pds)1/pdt does not exist. By Proposition 18, f is not asymptotically ω-periodic in the Stepanov sense.Now we can write down the following diagram which summaries asymptotically periodic type function spaces and their inclusion relations:(47)APωR+,X⟹SAPωuR+,X⟹SAPωR+,X⇓⇓PωLR+,X⟹SpAPωR+,X⟹SAPωpR+,X.In view of Examples16 and 19, PωL(R+,X)(SpAPω(R+,X)) and SAPωu(R+,X)(SAPω(R+,X)) do not contain each other.


## 4. Existence of Asymptoticallyω-Periodic Solutions of an Abstract Cauchy Problem

In this section we apply the results of Section2 on PωL(R+,X) to investigate the existence and uniqueness of asymptotically ω-periodic mild solutions for the following abstract Cauchy problem:(48)x′t=Ax(t)+ft,xt,t∈R+;x(0)=x0∈X,where A is the infinitesimal generator of an exponentially stable C0-semigroup (T(t))t≥0; that is, there exist M>0 and r>0 such that T(t)≤Me-rt for all t∈R+.The following definition generalizes Definition3.Definition 20.
A joint continuous functionf:R+×X→X is called ω-periodic limit in t∈R+ uniformly for x in bounded subsets of X if for every bounded subset K of X, {f(t,x):t∈R+,x∈K} is bounded and limn→∞f(t+nω,x)=g(t,x) exists for each t∈R+ and each x∈K. Denote by PωL(R+×X,X) the set of all such functions.The following is a composition theorem ofω-periodic limit functions.Theorem 21.
Letf:R+×X→X be ω-periodic limit in t∈R+ uniformly for x in bounded subsets of X and assume that f satisfies a Lipschitz condition in x uniformly in t∈R+:(49)f(t,x)-f(t,y)≤Lx-y,for all x,y∈X and t∈R+, where L is a positive constant. Let φ:R+→X be ω-periodic limit. Then the function F:R+→X defined by F(t)=f(t,φ(t)) is ω-periodic limit.Proof.
Sinceφ is a ω-periodic limit function, we have(50)limn→∞φt+nω=ϕ(t)for each t∈R+.
By Proposition5(4), we can choose a bounded subset K of X such that φ(t), ϕ(t)∈K for all t∈R+. Thus F(t) is bounded.
On the other hand, we have(51)limn→∞f(t+nω,x)=g(t,x)for each t∈R+ and each x∈K.
Let us consider the functionG:R+→X defined by G(t)=g(t,ϕ(t)). Note that (52)F(t+nω)-G(t)≤f(t+nω,φ(t+nω))-f(t+nω,ϕ(t))+f(t+nω,ϕ(t))-g(t,ϕ(t))≤Lφ(t+nω)-ϕ(t)+f(t+nω,ϕ(t))-g(t,ϕ(t)).
We deduce from (50) and (51) that (53)limn→∞F(t+nω)=G(t)for each t∈R+, which finishes the proof.Definition 22.
A functionx∈Cb(R+,X) is called a mild solution of problem (48) if (54)x(t)=T(t)x0+∫0t‍T(t-s)f(s,x(s))ds,t∈R+.Now we can establish the main result of this section.Theorem 23.
Letf:R+×X→X be ω-periodic limit in t∈R+ uniformly for x in bounded subsets of X and assume that f satisfies a Lipschitz condition in x uniformly in t∈R+:(55)f(t,x)-f(t,y)≤Lx-y,for all x,y∈X and t∈R+, where L is a positive constant. If ML<r, then there exists a unique asymptotically ω-periodic mild solution of problem (48).Proof.
We define the operatorΓ on the space APω(R+,X) by (56)Γxt=Ttx0+∫0t‍Tt-sf(s,x(s))ds=T(t)x0+v(t).We will show that Γx∈APω(R+,X). Since T(t)x0∈C0(R+,X)⊂APω(R+,X), it remains to show that v∈APω(R+,X). We denote F(s)=f(s,x(s)) for short. In view of Theorem 21, if x∈APω(R+,X)⊂PωL(R+,X), then F∈PωL(R+,X). By the definition of the ω-periodic limit function, (57)G(t)=limn→∞F(t+nω)is well defined for each t∈R+. Moreover, by Proposition 5(3)(4), there exists a positive constant K such that G∞≤F∞≤K and G(t)=G(t+ω). Note that (58)v(t+nω)=∫0t+nω‍Tt+nω-sFsds=∫-nωt‍T(t-s)F(s+nω)ds=∫-nω0‍T(t-s)F(s+nω)ds+∫0t‍T(t-s)F(s+nω)ds=∫0nω‍T(t+s)F(nω-s)ds+∫0t‍T(t-s)F(s+nω)ds=I1(t,n)+I2(t,n).Next we will prove that I1(t,n) is a Cauchy sequence in X for each t∈R+. Let ɛ>0. For any p∈N, n∈N, one has(59)I1(t,n+p)-I1(t,n)=∫0n+pω‍Tt+sFn+pω-sds-∫0nω‍Tt+sFnω-sds=∫nωn+pω‍Tt+sFn+pω-sds+∫0nω‍Tt+sFn+pω-s-Fnω-sds=I3(t,n,p)+I4(t,n,p).Now we estimate the term I3(t,n,p): (60)I3(t,n,p)≤∫nω(n+p)ω‍T(t+s)F((n+p)ω-s)ds≤KM∫nω(n+p)ω‍e-r(t+s)ds≤KM∫nω∞‍e-r(t+s)ds≤KMre-rnω.We can choose N1∈N such that KM/re-rnω<ɛ when n≥N1. That is, I3(t,n,p)<ɛ when n≥N1 uniformly for t∈R+.
Forn≥N1, we consider (61)I4(t,n,p)=∫0N1ω‍Tt+sFn+pω-s-Fnω-sds+∫N1ωnω‍Tt+sFn+pω-s-Fnω-sds=I5(t,n,p)+I6(t,n,p).Then we have (62)I5(t,n,p)≤∫0N1ω‍T(t+s)hhhhh·F((n+p)ω-s)-F(nω-s)ds=∫0N1ω‍T(t+N1ω-s)hhhhh·F((n-N1+p)ω+s)-F((n-N1)ω+s)ds≤M∫0N1ω‍e-r(N1ω-s)F((n-N1)ω+s)-G(s)ds+M∫0N1ω‍e-r(N1ω-s)F((n-N1+p)ω+s)-G(s)ds.For each s∈[0,N1ω], we have (63)e-r(N1ω-s)F((n-N1)ω+s)-G(s)≤2Ke-rN1ω-s,∫0N1ω‍2Ke-r(N1ω-s)ds=2Kr1-e-rN1ω.Since F∈PωL(R+,X), for each s∈[0,N1ω], one has (64)e-r(N1ω-s)F((n-N1)ω+s)-G(s)⟶0as n→∞. By Lebesgue’s Dominated Convergence Theorem, we obtain (65)limn→∞∫0N1ω‍e-r(N1ω-s)F((n-N1)ω+s)-G(s)ds=0.Moreover, we have (66)limn→∞∫0N1ω‍e-r(N1ω-s)hhhhhhh·F((n-N1+p)ω+s)-G(s)ds=0.Thus, we can select N2∈N(N2>N1) such that I5(t,n,p)<ɛ when n≥N2 uniformly for t∈R+.
Next we estimate the termI6(t,n,p): (67)I6(t,n,p)≤∫N1ωnω‍T(t+s)hhhhhhhhh·F((n+p)ω-s)-F(nω-s)ds≤2KM∫N1ωnω‍e-r(t+s)ds≤2KMre-rN1ω<2ɛuniformly for t∈R+.
Thus,I1(t,n+p)-I1(t,n)≤I3(t,n,p)+I5(t,n,p)+I6(t,n,p)<4ɛ when n≥N2. This shows that I1(t,n) is a Cauchy sequence. So we can denote h(t)=limn→∞I1(t,n) for each t∈R+. We also have that h(t)=limn→∞I1(t,n) uniformly for t∈R+.
Now we consider the termI2(t,n). Clearly ∫0t‍T(t-s)G(s)ds is well defined for each t∈R+. For mω≤t<(m+1)ω, m∈N, one has (68)I2(t,n)-∫0t‍T(t-s)G(s)ds≤∫0t‍T(t-s)F(s+nω)-G(s)ds≤M∫0t‍e-r(t-s)F(s+nω)-G(s)ds≤M∫0mω‍e-r(t-s)F(s+nω)-G(s)ds+M∫mωt‍F(s+nω)-G(s)ds≤M∑k=0m-1‍∫kω(k+1)ω‍e-r(t-s)F(s+nω)-G(s)ds+M∫mω(m+1)ω‍F(s+nω)-G(s)ds≤M∑k=0m-1‍e-rt-k+1ω·∫kω(k+1)ω‍F(s+nω)-G(s)ds+M∫mω(m+1)ω‍F(s+nω)-G(s)ds.For each s∈[0,ω], we have F(s+nω)-G(s) →0 as n→∞ and F(s+nω)-G(s) ≤2K. By Lebesgue’s Dominated Convergence Theorem, we obtain (69)limn→∞∫0ω‍F(s+nω)-G(s)ds=0.For ɛ>0 given, we select N3∈N such that (70)∫0ω‍F(s+nω)-G(s)ds<ɛwhen n≥N3. For any i∈N, one has (71)∫iω(i+1)ω‍F(s+nω)-G(s)ds=∫0ω‍F(s+iω+nω)-G(s+iω)ds=∫0ω‍F(s+iω+nω)-G(s)ds<ɛwhen n≥N3. Thus, (72)I2(t,n)-∫0t‍T(t-s)G(s)ds≤M∑k=0m-1‍e-r(t-(k+1)ω)ɛ+Mɛ≤11-e-rω+1Mɛwhen n≥N3 uniformly for t∈R+. That is, (73)limn→∞I2(t,n)=∫0t‍T(t-s)G(s)dsuniformly for t∈R+. Now we have limn→∞v(t+nω)=limn→∞I1(t,n) +limn→∞I2(t,n)= h(t)+∫0t‍T(t-s)G(s)ds uniformly for t∈R+. By Corollary 11, we get v∈APω(R+,X).
The spaceAPω(R+,X) is a Banach space. For the sake of completeness, we include the proof of the completeness of the space APω(R+,X) in Appendix.
Finally, forx1,x2∈APω(R+,X), one has (74)Γx1-Γx2∞≤MLrx1-x2∞,which shows that Γ is a contraction. To complete the proof of the theorem we only need to invoke the contraction mapping principle.Finally, we provide an example to illustrate our results.Example 24.
Consider the following partial differential equations:(75)∂∂tut,x=∂2∂x2ut,x+atfut,x,t∈R+,x∈0,π;u(t,0)=u(t,π)=0,t∈R+;u(0,x)=g(x),x∈[0,π],where a∈PωL(R+,R), g:[0,π]→R, and f:R→R are appropriate functions. In addition, f satisfies(76)|f(x)-f(y)|≤Lx-y,x,y∈R,where L>0. In what follows we consider X=L2([0,π]) and let A be the operator given by Au=u′′ with domain D(A)={u∈X:u′′∈X,u(0)=u(π)=0}. It is well known that A is the infinitesimal generator of an analytic semigroup (T(t))t≥0 on X. Moreover, A has discrete spectrum with eigenvalues -n2, n∈N, and corresponding normalized eigenfunctions given by zn(ξ)=(2/π)1/2sin(nξ). Furthermore, {zn:n∈N} is an orthonormal basis of X and T(t)x=∑n=1∞‍e-n2t〈x,zn〉zn for x∈X. Thus, we have T(t)≤e-t for every t∈R+. Therefore, if a∞L<1, (75) has a unique asymptotically ω-periodic mild solution by Theorem 23.



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