# On the Behaviour of Singular Semigroups in Intermediate and Interpolation Spaces and Its Applications to Maximal Regularity for Degenerate Integro-Differential Evolution Equations

Authors: Alberto Favaron; Angelo Favini
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

For those semigroups, which may have power type singularities and whose generators are abstract multivalued linear operators, we characterize the behaviour with respect to a certain set of intermediate and interpolation spaces. The obtained results are then applied to provide maximal time regularity for the solutions to a wide class of degenerate integro- and non-integro-differential evolution equations in Banach spaces.


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## Body

## 1. Introduction

LetX be a complex Banach space and let {𝒯A(t)}t≥0 be a semigroup of operators on X, which is generated by a multivalued linear operator A:𝒟(A)⊆X→X and which may have a power type singularity at the origin t=0, that is,
(1)∥𝒯A(t)∥ℒ(X)≤C0tν,∀t>0,𝒯A(0)x=x,∀x∈X,
for some nonnegative constant C0 and nonpositive exponent ν, where ℒ(X) denotes the Banach algebra of all endomorphisms of X endowed with the uniform operator norm. In this context our aim here is twofold. The first is to characterize the behaviour of {𝒯A(t)}t≥0 with respect to some intermediate and interpolation spaces between X and the domain 𝒟(A) of A. The second is to investigate how this behaviour reflects on the question of maximal time regularity for the solutions to a class of degenerate integro- and non-integrodifferential initial value problems in X.The class of operators we will deal with consists precisely of those multivalued linear operatorsA whose single-valued resolvents satisfy the following estimate:
(2)∥(λI-A)-1∥ℒ(X)≤C(|λ|+1)-β,∀λ∈Σα.
Here, I is the identity operator, C is a positive constant, β∈(0,1], and Σα is the complex region {z∈C:ℜez≥-c(|ℑmz|+1)α,ℑmz∈R}, c>0, α∈[β,1]. It thus happens (cf. [1–3]) that A is the infinitesimal generator of a semigroup of linear bounded operators in X satisfying (1) with ν=να,β, where να,β=(β-1)/α.To outline the motivations of our research, let us assume for a moment thatA is a single-valued linear operator satisfying (2). It is well known that if β=1, then A is the infinitesimal generator of a bounded analytic semigroup. For this case, an extensive literature exists concerning the behaviour of {𝒯A(t)}t≥0 with respect to the real interpolation spaces (X,𝒟(A))γ,p, γ∈(0,1), p∈[1,∞], and its application to questions of maximal regularity for the solutions to nondegenerate (possibly nonautonomous) integro- and non-integrodifferential abstract Cauchy problems. See, for instance, [4–11]. Due to (1) with ν=ν1,β, the case of α=1 and β∈(0,1) is definitely worsened and the literature for it is considerably less conspicuous, although estimate of type (2), with (ℜeλ+|ℑmλ|β)-1 in place of (|λ|+1)-β, goes back even to [12, Remark p. 383] in the ambit of Abel summable semigroups admitting uniform derivatives of all orders. One of the main problems with the case β∈(0,1) is that some equivalent characterizations of (X,𝒟(A))γ,p begin to fail (cf. [13]), so that some spaces which were just real interpolation spaces between X and 𝒟(A) in the case β=1 become only intermediate spaces in the case β∈(0,1). However, avoiding questions of interpolation theory and of maximal regularity, a quite satisfactorily semigroup theory for the single-valued case with β∈(0,1) and its application to the unique solvability of some concrete partial (non-integro-) differential equations have been developed in [14–18]. Since the multivalued case embraces the single-valued one, our contribution in this field is to fill this gap, supplying a theory for the behaviour of singular semig n intermediate and interpolation spaces which, in the case β=1, reduces to that in [9, 11]. As an effect of this theory, there is the possibility of investigating questions of maximal time regularity for an entire class of nondegenerate evolution equations which does not fall within the case β=1.The case whenA is really a multivalued linear operator arises naturally when we shift our attention to degenerate evolution equations of the type considered in [1–3]. There, a semigroup theory for multivalued linear operators was introduced as a tool to handle degenerate equations by means of analogous techniques of the nondegenerate ones. Such a theory has been then successfully applied to questions of maximal regularity for the solutions to a wide class of degenerate integro- and non-integrodifferential equations. We quote [2, 19–23] where, in general and unless β=1, it is shown that the time regularity of the solutions decreases with respect to that of the data. In this respect, we mention the recent results in [20] where, under an additional condition of space regularity on the data and provided that α and β are large enough, the loss of time regularity is restored. Regrettably (cf. the appendix below), we have found some inaccuracies in [20, Section 4], and for this reason we must indicate some changes to that paper. On the other side, fortunately, the basic idea in [20] is correct and remedy can be applied to all the inappropriate items. Furthermore, unexpectedly, we will see that the more delicate approach followed in this paper not only corrects the mistakes in [20], but also gives rise to an effective improvement of the achievable results. In fact, here, we will straighten out, refine, and extend [20], enlarging the class of the admissible spaces to which the data may belong, weakening the assumption for the pair (α,β), and complicating the structure of the underlying equations. This is why we will first analyze the behaviour of the semigroup generated by A with respect to some intermediate and interpolation spaces which turn out to be equivalent only in the case β=1. Indeed, the phenomena exhibited in [13] for the single-valued case extend to the multivalued one (cf. [24]), and, until now, for the mentioned behaviour there exist no more than some partial results obtained in [2, 19, 24].We now give the detailed plan of the paper. In Section2, for a multivalued linear operator A having domain 𝒟(A) and satisfying (2), we introduce the corresponding generated semigroup {etA}t≥0. This leads us to define also the linear bounded operators [(-A)θ]∘etA, ℜeθ≥0, t>0, ([(-A)0]∘etA=etA) and to recall the fundamental estimates for their ℒ(X)-norm. For the operators [(-A)θ]∘etA a semigroup type property is proven in Proposition 1. We then introduce the spaces we will deal with in this paper, that is, the interpolation spaces (X,𝒟(A))γ,p and the spacesXAγ,p, γ∈(0,1), p∈[1,∞]. Special attention is given to the embeddings linking these two classes of spaces which, in general, are equivalent only in the case β=1. Some relations existing between the spaces XAγ,p for different values of γ and p are proven in Proposition 2 and discussed in Remarks 3–5. We conclude the section recalling the estimates proven in [19, 24] for the norms ∥[(-A)θ]∘etA∥ℒ(X;(X,𝒟(A))γ,p), ℜeθ≥0, and ∥[(-A)1]∘etA∥ℒ(Yγp;X), Yγp∈{(X,𝒟(A))γ,p,XAγ,p}. In Remarks 7 and 8 we explain why, unless we renounce to optimality, in the case β<1 these estimates can not be directly extended to the norms ∥[(-A)θ]∘etA∥ℒ(X;XAγ,p) and ∥[(-A)θ]∘etA∥ℒ(Yγp;X), ℜeθ≥1, respectively.In Section3, we investigate the behaviour of the operators [(-A)θ]∘etA with respect to both of the spaces (X,𝒟(A))γ,p and XAγ,p. First, in Proposition 9, we deal with the norms ∥[(-A)θ]∘etA∥ℒ(X;XAγ,p), ℜeθ≥0, and we show that, except for replacing (X,𝒟(A))γ,p with XAγ,∞ if p=∞ and with XAβγ,p if p∈[1,∞), the same estimates of [19] for the norms ∥[(-A)θ]∘etA∥ℒ(X;(X,𝒟(A))γ,p) continue to hold. The second significant result is Proposition 12 where, extending those in [24] to values of θ other than one, we establish estimates for the norms ∥[(-A)θ]∘etA∥ℒ(Yγp;X), ℜeθ≥1, Yγp∈{(X,𝒟(A))γ,p,XAγ,p}. As a byproduct we deduce the basic Corollary 14, which in Section 5 will be a key tool in proving the equivalence between the following problem (3) and the fixed-point equation (179). The estimates in Proposition 12 are then merged together with those in [19] to achieve estimates for the norms ∥[(-A)θ]∘etA∥ℒ((X,𝒟(A))γ,p,(X,𝒟(A))δ,p), ℜeθ≥1. In particular, two different estimates are obtained, if γ+δ<1 or not. For if γ+δ<1, then (cf. the proof of Proposition 16) we can take advantage of the reiteration theorem for interpolation spaces and obtain estimates that, unless β=1, are better than those rougher estimates derived in the general case γ,δ∈(0,1) (see Remarks 17 and 18). We stress that if β=1, θ∈N and A is single-valued, then we restore the estimates in [9]. Finally, in Proposition 20, a combination of Propositions 9 and 12 yields the estimate for the norms ∥[(-A)θ]∘etA∥ℒ(XAγ,p,XAδ,p), ℜeθ≥1. Since β<1, the spaces XAσ,q are, in general, only intermediate spaces between X and 𝒟(A) for σ∈(0,β); here the reiteration theorem does not apply and a weaker result is obtained (cf. (101)–(103)).The estimates of Section3 are applied in Section 4 to study the time regularity of those operator functions Qj, j=1,…,6, that we will need in Section 5. In particular (cf. formula (106)), we modify the definition of Q2 in [20, Section 4] in order that it is well defined, at least when acting on functions g∈Cδ([0,T];X), δ∈((2-α-β)/α,1) (cf. Corollary 26). Consequently, operators Q3 and Q4 in [20] change too, and the new Q5 and Q6 should be introduced (cf. formulae (107)–(110)). The Hölder in time regularity of the Qj’s is characterized in Lemmas 22, 24, 30, and 32 and Propositions 29 and 36. The main feature of these results is to show that the loss of regularity produced by Q2 and Q5 can be restored, in Q3 and Q6 respectively, employing the regularization property established in [20, Section 3] for a wide range of general convolution operators.In Section5 we analyze the maximal time regularity of the strict solutions v to the following class of degenerate integrodifferential equations in a complex Banach space X:
(3)ddt(Mv(t))=[λ0M+L]v(t)+∑i1=1n1‍𝒦(ki1,Li1v)(t)+∑i2=1n2‍hi2(t)yi2+f(t),t∈IT,Mv(0)=Mv0.
Here, IT=[0,T], λ0∈C, n1,n2∈N, hi2:IT→C, yi2∈X, i2=1,…,n2, whereas, Z being  another complex Banach space and 𝒫:Z×X→X being a bilinear bounded operator, ki1:IT→Z, and 𝒦(ki1,Li1v)(t)=∫0t‍𝒫(ki1(t-s),Li1v(s))ds, i1=1,…,n1. Of course, if Z=C, then 𝒫 may be the scalar multiplication in X. As M, L, and Li1, i1=1,…,n1, we take closed single-valued linear operators from X to itself, whose domains fulfill the relation 𝒟(L)⊆⋂i1=1n1‍[𝒟(M)∩𝒟(Li1)], and we require L to have a bounded inverse, allowing  M to be notinvertible.   Hence, in general, A=LM-1 is only a multivalued linear operator in X having domain 𝒟(A)=M(𝒟(L)). Assuming that A satisfies (2) and that the data ki1, hi2, yi2 and f, il=1,…,nl, l=1,2, are suitably chosen, problem (3) is then reduced to an equivalent fixed point-equation for the new unknown w=L(v-v0), v0∈𝒟(L). It is here that the results of Sections 3 and 4 play their role, leading us to Theorem 48. In that theorem,   provided that 5α+2β>6, we will prove that if ki1∈Cηi1(IT;Z), hi2∈Cσi2(IT;C), yi2∈Yγi2r, Yγi2r∈{(X,𝒟(A))γi2,r,XAγi2,r}, and f∈Cμ(IT;X) for opportunely chosen ηi1, σi2, γi2, and μ, il=1,…,nl, l=1,2, then problem (3) has a unique strict solution v∈Cτ(IT;𝒟(L)) satisfying v(0)=v0 and Lv,dMv/dt∈Cτ(IT;X), where τ=minil=1,…,nl,l=1,2{ηi1,σi2} (cf. Remark 51). Section 5 concludes with applications of Theorem 48 to integral and nonintegral subcases of (3), (cf. Theorems 52–54 and 56). We stress that Theorem 48 repairs, generalizes, and improves [20, Theorems 5.6 and 5.7], where similar results were proven only for the case (n1,n2,Yψp)=(1,1,XAψ,p) and under the stronger condition 3α+8β>10.In Section6, we give an application of Theorem 48 to a concrete case of problem (3) arising in the theory of heat conduction for materials with memory. In particular, we show how Theorem 48 characterizes the appropriate functional framework where to search for the solution of the inverse problem of recovering both v and the vector (k1,…,kr1), r1≤n1, in (3) with (i2,n2)=(i1,n1) and hi1=ki1, i1=1,…,n1.Finally, in the Appendix we explain how to amend [20, Theorems 5.6 and 5.7] in accordance to Theorem 48.


## 2. Multivalued Linear Operators, Singular Semigroups, and the Spaces(X,𝒟(A))γ,p and XAγ,p

LetX be a complex Banach space endowed with norm ∥·∥X and let 𝒫(X) be the collection of all the subsets of X. For a number λ∈C and elements 𝒰,𝒱,𝒲∈𝒫(X)∖∅, λ𝒰, and 𝒱+𝒲 denote the subsets of X defined by {λu:u∈𝒰} and {v+w:v∈𝒱,w∈𝒲}, respectively. Then, a mapping A from X into 𝒫(X) is called a multivalued linear operator in X if its domain 𝒟(A)={x∈X:Ax≠∅} is a linear subspace of X and A satisfies the following: (i) Ax+Ay⊂A(x+y), for all x,y∈𝒟(A);  (ii) λAx⊂A(λx), for all λ∈C, for all x∈𝒟(A). From now on, the shortening m. l. will be always used for multivalued linear.The setℛ(A)=⋃x∈𝒟(A)‍Ax is called the range of A. If ℛ(A)=X, then A is said to be surjective. The following properties of a m. l. operator A are immediate consequences of its definition (cf. [1, Theorems 2.1 and 2.2]):  (iii) Ax+Ay=A(x+y), for allx,y∈𝒟(A); (iv) λAx=A(λx), for allλ∈C∖{0}, for allx∈𝒟(A);  (v) A0 is a linear subspace of X and Ax=y+A0 for any y∈Ax, x∈𝒟(A). In particular, A is single-valued if and only if A0={0}.IfA is an m. l. operator in X, then its inverse A-1 is defined to be the operator having domain 𝒟(A-1)=ℛ(A) such that A-1y={x∈𝒟(A):y∈Ax}, y∈𝒟(A-1). A-1 is an m. l. operator in X too, and (A-1)-1=A. The set A-10={x∈𝒟(A):0∈Ax} is called the kernel of A and denoted by 𝒩(A). If 𝒩(A)={0}; that is, if A-1 is single-valued, then A is said to be injective. Observe that  (v)  yields Ax=A0 if and only if x∈𝒩(A).Given𝒰∈𝒫(X)∖∅, we write A(𝒰)=⋃u∈𝒰∩𝒟(A)‍Au, so that, in particular, A(X)=A(𝒟(A))=ℛ(A). If Aj, j=1,2 are m. l. operators in X and λ∈C, then the scalar multiplication λA1, the sum A1+A2, and the product A1A2 are defined by
(4)𝒟(λA1)=𝒟(A1),(λA1)x=λA1x,x∈𝒟(λA1),𝒟(A1+A2)=𝒟(A1)∩𝒟(A2),(A1+A2)x=A1x+A2x,x∈𝒟(A1+A2),𝒟(A1A2)={x∈𝒟(A2):A1(A2x)≠∅},(A1A2)x=A1(A2x),x∈𝒟(A1A2),
where   λA1, A1+A2 and A1A2 are m. l. operators in X and (A1A2)-1=A2-1A1-1.LetA and B be m. l. operators in X. We write A⊂B if 𝒟(A)⊆𝒟(B) and Ax⊆Bx for every x∈𝒟(A). Clearly, A⊂B⊂A if and only if A=B. If A⊂B and Ax=Bx for every x∈𝒟(A), then B is called an extension of A. If a linear single-valued operator S has domain 𝒟(S)=𝒟(A) and S⊂A, that is, Sx∈Ax for every x∈𝒟(A), then S is called a section of A. With an arbitrary section S, it holds Ax=Sx+A0, x∈𝒟(A), and ℛ(A)=ℛ(S)+A0, but this latter sum may or may not be direct (cf. [25, p. 14]). A method for constructing sections is provided in [25, Proposition I.5.2].IfXj, j=1,2, are two complex Banach spaces, then the linear space of all bounded single-valued linear operators L from X1=𝒟(L) to X2 is denoted by ℒ(X1;X2) (ℒ(X1) if X1=X2) and it is equipped with the uniform operator norm ∥L∥ℒ(X1;X2)=sup∥x∥X1≤1∥Lx∥X2=infK≥0{∥Lx∥X2≤K∥x∥X1:x∈X1}. Then the resolvent set ρ(A) of a m. l. operator A is defined to be the set {z∈C:(zI-A)-1∈ℒ(X)}, with I being the identity operator in X. The basic properties of the resolvent set of single-valued linear operators hold the same for m. l. operators. First, if ρ(A)≠∅, then A is closed; that is, its graph {(x,y)∈X×X:x∈𝒟(A),y∈Ax} is closed (cf. [25, p. 43]). Further (cf. [1, Theorem 2.6]), ρ(A) is an open set and the operator function z∈ρ(A)→(zI-A)-1∈ℒ(X) is holomorphic. Finally (cf. [1, formula (2.1)]), the resolvent equation (λ2-λ1)(λ1I-A)-1(λ2I-A)-1=(λ1I-A)-1-(λ2I-A)-1, λ1,λ2∈ρ(A), is satisfied, too. Unlike the single-valued case, instead, for z∈ρ(A)  the following inclusions hold (cf. [1, Theorem 2.7]):
(5)(zI-A)-1A⊂z(zI-A)-1-I⊂A(zI-A)-1.
Then, in general, z(zI-A)-1-I, z∈ρ(A), is only a bounded section of the m. l. operator A(zI-A)-1. Throughout this paper, we denote this bounded section by A∘(zI-A)-1, but we warn the reader that here A∘ does not necessarily denote a section of A itself. Of course, if A is single-valued, then A∘(zI-A)-1 reduces to A(zI-A)-1. Notice that (5) implies that (zI-A)-1A, z∈ρ(A), is single-valued on 𝒟(A) and (zI-A)-1Ax=(zI-A)-1y with any y∈Ax, x∈𝒟(A). Another difference with the single-valued case is that for every z∈ρ(A)  it  holds 𝒩((zI-A)-1)=A0. Indeed, ((zI-A)-1)-10=(zI-A)0=A0. Therefore, in the m. l. case, {0}⊊𝒩((zI-A)-1), z∈ρ(A). However (cf. [24, Lemma 2.1]), if 0∈ρ(A), then 𝒩(A∘(zI-A)-1)={0}, and, in addition, x∉A0 if and only if A∘(zI-A)-1x∉A0, z∈ρ(A). We also recall that for every λ1,λ2∈ρ(A) the following slight variants of the resolvent equation hold (cf. [24, Lemma 2.2]):
(6)(λ2-λ1)(λ1I-A)-1A∘(λ2I-A)-1=A∘(λ1I-A)-1-A∘(λ2I-A)-1,(λ2-λ1)A∘(λ1I-A)-1(λ2I-A)-1=A∘(λ1I-A)-1-A∘(λ2I-A)-1.
In particular, if 0∈ρ(A), then, since A∘(0I-A)-1=-I, the first in (6) with (λ1,λ2)=(0,λ) yields λ(-A)-1A∘(λI-A)-1=-I-A∘(λI-A)-1=-λ(λI-A)-1; that is,
(7)A-1A∘(λI-A)-1=(λI-A)-1,λ∈ρ(A).
Let (A,𝒟(A)) be a m. l. operator in X satisfying the following resolvent condition:  (H1)
ρ(A) contains a region Σα={z∈C:ℜez≥-c(|ℑmz|+1)α,ℑmz∈R},α∈(0,1], c>0, and for some exponent β∈(0,α] and constant C>0  the following estimate holds:
(8)∥(λI-A)-1∥ℒ(X)≤C(|λ|+1)-β,∀λ∈Σα.  Introduce the family {etA}t≥0∈ℒ(X) defined by e0A=I and
(9)etA=12πi∫Γ‍etλ(λI-A)-1dλ,t>0,
where Γ⊊Σα∖{z∈C:ℜez≥0} is the contour parametrized by λ=-c(|η|+1)α+iη, η∈(-∞,∞). Then (cf. [1, pp. 360, 361]), {etA}t≥0 is a semigroup on X, infinitely many times strongly differentiable for t>0 with
(10)DtketA=12πi∫Γ‍λketλ(λI-A)-1dλ,DtketA=t>0,k∈N={1,2,…},
where Dtk=dk/dtk. In general, no analyticity should be expected for etA. For if α<1 in (H1), then Σα does not contain any sector Λω+π/2={z∈C∖{0}:|argz|<ω+π/2}, ω∈(0,π/2), and [15, Theorem 5.3], which extends etA analytically to the sector Λω containing the positive real axis, is not applicable. We stress that (9) and 𝒩((zI-A)-1)=A0, z∈ρ(A), imply A0⊆𝒩(etA) for every t>0, whereas 𝒩(e0A)=𝒩(I)={0}. Hence, if A is really an m. l. operator, then {0}⊊A0⊆⋂t>0‍𝒩(etA). From the semigroup property it also follows that 𝒩(et0A)⊆𝒩(et1A) for t1≥t0≥0.Now, for everyθ∈C such that ℜeθ≥0 we set
(11)[(-A)θ]∘etA=12πi∫Γ‍(-λ)θetλ(λI-A)-1dλ,t>0.
Here, for the multivalued function (-λ)θ=eθLn(-λ) we choose the principal branch holomorphic in the region C∖{z∈C:ℜez≥0}, where for principal branch we mean the principal determination ln|z|+iarg(z) of Ln(z). We briefly recall the main properties of operators [(-A)θ]∘etA. Of course, [(-A)0]∘etA=etA, t>0. As shown in [26, p. 426], [(-A)k]∘etA, k∈N, t>0, is a section of (-A)ketA, so that from (10) we get
(12)(-1)kDtketA=[(-A)k]∘etA⊂(-A)ketA,t>0,k∈N.
Moreover (cf. [19, formula (22)] with θ≥0 being replaced by ℜeθ≥0), we get
(13)[(-A)θ]∘etA-[(-A)θ]∘esA=-∫st‍[(-A)θ+1]∘eξAdξ,ℜeθ≥0,0<s<t.
Finally, (H1) implies the following estimates (cf. [1, 24]):
(14)∥[(-A)θ]∘etA∥ℒ(X)≤c~α,β,θt(β-ℜeθ-1)/α,ℜeθ≥0,t>0,
where the c~α,β,θ’s are positive constants depending on α, β, and θ. Thus, letting θ=0 in (14), we see that if β∈(0,1), then the operator function t∈(0,∞)→etA∈ℒ(X) may be singular at the origin and the semigroup is not necessarily strongly continuous in the X-norm on the closure 𝒟(A)¯ of 𝒟(A) in X. Notice that if α+β>1, then the singularity is a weak one, in the sense that {etA}t≥0 is integrable in norm in any interval [0,τ], τ>0. Further (cf. [24, Lemma 3.9]), if α+β>1, then A0=⋂t>0‍𝒩(etA), and if α=1, then A0=𝒩(etA) for every t>0.Observe thatA0⊆𝒩([(-A)θ]∘etA), ℜeθ≥0, t>0, so that A0⊆⋂t>0‍𝒩([(-A)θ]∘etA), ℜeθ≥0. The operators [(-A)θ]∘etA satisfy the following semigroup type property.Proposition 1.
Letθj∈C, ℜeθj≥0, and let tj>0, j=1,2. Then
(15)[(-A)θ1]∘et1A[(-A)θ2]∘et2A=[(-A)θ1+θ2]∘e(t1+t2)A.Proof.
First, the functionλ∈ρ(A)→(-λ)θetλ(λI-A)-1∈ℒ(X) being holomorphic for every ℜeθ≥0 and t>0, and the contour Γ in (11) with (θ,t)=(θ2,t2) can be replaced with the contour Γ′⊊Σα∖{z∈C:ℜez≥0} parametrized by μ=-c′(|η|+1)α+iη, η∈(-∞,∞), c′∈(0,c), and lies to the right of Γ. Then, for every x∈X, from the resolvent equation we obtain
(16)[(-A)θ1]∘et1A[(-A)θ2]∘et2Ax=(12πi)2∫Γ‍(-λ)θ1et1λ×[∫Γ′‍(-μ)θ2et2μ(λI-A)-1(μI-A)-1xdμ]dλ=(12πi)2∫Γ‍(-λ)θ1et1λ(λI-A)-1×[(∫Γ′‍(-μ)θ2et2μ(μ-λ)-1dμ)x]dλ-(12πi)2∫Γ′‍(-μ)θ2et2μ(μI-A)-1×[(∫Γ‍(-λ)θ1et1λ(λ-μ)-1dλ)x]dμ.
Now, after having enclosed Γ and Γ′ on the left with an arc ΔR of the circle {z∈C:|z+c′|=R}, R>c-c′, we apply the residue theorem and let R go to infinity. To this purpose, we observe that since the contours Γ and Γ′ both lie in the half-plane {z∈C:ℜez≤-c′}, the arc ΔR may be parametrized in polar coordinates by ℜez=-c′+Rcosφ, ℑmz=Rsinφ, φ∈(π/2,3π/2). Then, for every z∈ΔR we have
(17)|(-z)θetz|=|z|ℜeθe-ℑmθarg(-z)etℜez≤(R+c′)ℜeθe(π/2)|ℑmθ|e-tc′etRcosφ.
Since t>0 and φ∈(π/2,3π/2), the right-hand side of the latter inequality goes to zero as R goes to infinity, so that limR→∞,z∈ΔR(-z)θetz=0 for every ℜeθ≥0 and t>0. The residue theorem together with the fact that Γ′ lies to the right of Γ thus yields ∫Γ′(-μ)θ2et2μ(μ-λ)-1dμ=2πi(-λ)θ2et2λ and ∫Γ‍(-λ)θ1et1λ(λ-μ)-1dλ=0. Replacing these identities in (16) and using the equality (-λ)θ1(-λ)θ2=(-λ)θ1+θ2 which is satisfied for the principal branch of the function (-λ)θ=eθLn(-λ), we finally find
(18)[(-A)θ1]∘et1A[(-A)θ2]∘et2Ax=12πi∫Γ‍(-λ)θ1+θ2e(t1+t2)λ(λI-A)-1xdλ.
The right-hand side being precisely [(-A)θ1+θ2]∘e(t1+t2)Ax, the proof is complete.For an m. l. operatorA satisfying (H1) we introduce now the spaces (X,𝒟(A))γ,p and XAγ,p. We first specify a topology on 𝒟(A) equipping it with the norm ∥x∥𝒟(A)=infy∈Ax∥y∥X, x∈𝒟(A). Since A-1∈ℒ(X), this norm is equivalent to the graph norm and makes 𝒟(A) a complex Banach space (cf. [2, Proposition 1.11]). As X1 and X2 being given normed complex linear spaces, we will write X1↪X2 if X1⊆X2 and there exists a positive constant C0 such that ∥x∥X2≤C0∥x∥X1 for every x∈X1. If X1↪X2↪X1, that is,  if X1=X2 and the norms ∥·∥X1 and ∥·∥X2are equivalent, then we will write X1≅X2. Of course, 𝒟(A) with the norm ∥·∥𝒟(A) satisfies 𝒟(A)↪X. In fact, if x∈𝒟(A), then for every y∈Ax we have x=A-1y, so that ∥x∥X≤∥A-1∥ℒ(X)∥y∥X≤C∥y∥X. Taking the infimum with respect to y∈Ax, we thus find ∥x∥X≤C∥x∥𝒟(A) for every x∈𝒟(A). If Y is a Banach space, we denote by C((0,∞);Y) the set of all continuos functions from (0,∞) to Y, and for a Y-valued strongly measurable function g(ξ), ξ∈(0,∞), we set ∥g(ξ)∥Lq*(Y)=(∫0∞‍∥g(ξ)∥Yq(dξ/ξ))1/q, q∈[1,∞), and ∥g(ξ)∥L∞*(Y)=supξ∈(0,∞)∥g(ξ)∥Y. Let p0,p1∈[1,∞) or let p0=p1=∞, and for γ∈(0,1) define p-1=(1-γ)p0-1+γp1-1 if p0,p1∈[1,∞) and p=∞   if p0=p1=∞.  Let us set
(19)(X,𝒟(A))γ,p={∥ξγ-1v1(ξ)∥Lp1*(𝒟(A))x∈X:x=v0(ξ)+v1(ξ),ξ∈(0,∞),v0∈C((0,∞);X),v1∈C((0,∞);𝒟(A)),∥ξγv0(ξ)∥Lp0*(X)+∥ξγ-1v1(ξ)∥Lp1*(𝒟(A))<∞},∥x∥(X,𝒟(A))γ,p=infv0,v1{∥ξγv0(ξ)∥Lp0*(X)+∥ξγ-1v1(ξ)∥Lp1*(𝒟(A))}.
This characterization of the spaces (X,𝒟(A))γ,p is that obtained by the so-called “mean-methods”, and it is equivalent to that performed by the “K-method" (cf. [27, Theorem 1.5.2 and Remark 1.5.2/2]) and the “trace-method” (cf. [27, Theorem 1.8.2]). Then, due to [27, Theorem 1.3.3], for every γ∈(0,1) and p∈[1,∞] the space (X,𝒟(A))γ,p is an exact real interpolation space of exponent γ between X and 𝒟(A). Observe that by exchanging the role of X and 𝒟(A) and performing the transformation ξ=τ-1, we get (X,𝒟(A))γ,p=(𝒟(A),X)1-γ,p. Also, if 𝒟(A)=X, then (X,𝒟(A))γ,p≅X (cf. [27, Theorem 1.3.3(f)]). The definition of the spaces (X,𝒟(A))γ,p is meaningful even for the limiting cases (γ,p)=(i,∞), i=0,1, whereas (X,𝒟(A))i,p, i=0,1, p∈[1,∞), reduces to the zero element of X. In particular (cf. [28, pp. 10–15]), denoting by Y~X the completion of 𝒟(A) relative to X and endowing it with the norm ∥·∥Y~X in [28, p. 14], we get (X,𝒟(A))0,∞≅X and (X,𝒟(A))1,∞≅Y~X. Let γ1∈(0,1) and let pj∈[1,∞], j=1,2. Then, for γ2∈(0,γ1) and qj∈[1,pj], j=1,2,   the following chain of embeddings holds:
(20)𝒟(A)↪(X,𝒟(A))1,∞↪(X,𝒟(A))γ1,1↪(X,𝒟(A))γ1,q1↪(X,𝒟(A))γ1,p1↪(X,𝒟(A))γ2,1↪(X,𝒟(A))γ2,q2↪(X,𝒟(A))γ2,p2↪𝒟(A)¯.
Let γ∈[0,1]. Recall that a Banach space E is said to be of class J(γ,X,𝒟(A))∩K(γ,X,𝒟(A)) and shortened to E∈J(γ)∩K(γ), if E is an intermediate space between (X,𝒟(A))γ,∞ and (X,𝒟(A))γ,1, that is, if (X,𝒟(A))γ,1↪E↪(X,𝒟(A))γ,∞. From (20) it thus follows that (X,𝒟(A))γ,p∈J(γ)∩K(γ), for every γ∈(0,1) and p∈[1,∞]. Moreover, since (X,𝒟(A))i,1={0}, i=0,1, and (X,𝒟(A))0,∞≅X, we have 𝒟(A)∈J(1)∩K(1) and X∈J(0)∩K(0). Then   (cf. [28, p. 12], [27, Theorem 1.10.2], and [9, Section 1.2.3]), for γj∈(0,1) and pj∈[1,∞], j=0,1,2, the reiteration theorem yields
(21)((X,𝒟(A))γ1,p1,(X,𝒟(A))γ2,p2)γ0,p0≅(X,𝒟(A))(1-γ0)γ1+γ0γ2,p0,((X,𝒟(A))γ1,p1,𝒟(A))γ0,p0≅(X,𝒟(A))(1-γ0)γ1+γ0,p0,(X,(X,𝒟(A))γ2,p2)γ0,p0≅(X,𝒟(A))γ0γ2,p0.
Finally (cf. [29, Theorem 1.II and Remark 1.III]), we recall that if X1 and X2 are two complex Banach spaces and T∈ℒ(X1;X2) is such that T∈ℒ(Y1k;Y2k), Yjk⊆Xj, j,k=1,2, then T∈ℒ((Y11,Y12)γ0,p0;(Y21,Y22)γ0,p0), γ0∈(0,1), p0∈[1,∞], and
(22)∥T∥ℒ((Y11,Y12)γ0,p0;(Y21,Y22)γ0,p0)≤∥T∥ℒ(Y11;Y21)1-γ0∥T∥ℒ(Y12;Y22)γ0.
As a consequence of this general result and the identity
(23)((X,𝒟(A))γ1,p1,X)γ0,p0=(X,(X,𝒟(A))γ1,p1)1-γ0,p0,
from the third in (21) we find that if T∈ℒ(X) is such that T∈ℒ(X;(X,𝒟(A))γ1,p1) and T∈ℒ((X,𝒟(A))γ2,p2;X), then T∈ℒ((X,𝒟(A))γ0γ2,p0;(X,𝒟(A))(1-γ0)γ1,p0), γj∈(0,1), pj∈[1,∞], j=0,1,2, and the following estimate holds:
(24)∥T∥ℒ((X,𝒟(A))γ0γ2,p0;(X,𝒟(A))(1-γ0)γ1,p0)≤∥T∥ℒ(X;(X,𝒟(A))γ1,p1)1-γ0∥T∥ℒ((X,𝒟(A))γ2,p2;X)γ0.
Notice that here γ0γ2+(1-γ0)γ1∈(min{γ1,γ2},max{γ1,γ2})⊊(0,1) for every γ0∈(0,1). Therefore, if we let γ=γ0γ2 and let δ=(1-γ0)γ1, then γ+δ<1, γ1=δ/(1-γ0)>δ, and γ2=γ/γ0>γ. Hence, in order that the additional inequalities γj<1, j=1,2, are satisfied, we have to choose γ0∈(γ,1-δ). As we will see this simple observation will be the key for the proof of the second estimates (90) in the following Proposition 16.We recall that for every fixedx∈𝒟(A) the map T(λ)=λx satisfies ∥T∥ℒ(C;X)=∥x∥X, ∥T∥ℒ(C;𝒟(A))=∥x∥𝒟(A) and ∥T∥ℒ(C,(X,𝒟(A))γ,p)=∥x∥(X,𝒟(A))γ,p. Then (22) with X1=Y11=Y12=C, X2=Y21=X and Y22=𝒟(A) yields the interpolation inequality:
(25)x∥x∥(X,𝒟(A))γ,p≤c0∥x∥X1-γ∥x∥𝒟(A)γ,x∈𝒟(A),γ∈(0,1),p∈[1,∞],
with c0 being the positive constant depending on γ and p such that ∥λ∥(C,C)γ,p≤c0|λ|.As another application of (22) and for further needs, we also recall that if A satisfies (H1), then A∘(zI-A)-1 satisfies the estimate (cf. [24, formulae (4.16) and (4.17)]).Consider(26)∥A∘(zI-A)-1∥ℒ(X)≤(C+1)(|z|+1)1-β,∀z∈Σα,∥A∘(zI-A)-1∥ℒ(𝒟(A);X)≤C(|z|+1)-β,∀z∈Σα.
From (26), using (22) with Xj=Yj1=Y22=X, j=1,2, and Y12=𝒟(A), it then follows for every γ∈(0,1) and p∈[1,∞](27)∥A∘(zI-A)-1∥ℒ((X,𝒟(A))γ,p;X)≤c1(C+1)1-γCγ(|z|+1)1-β-γ,∀z∈Σα,
where c1 is the positive constant depending on γ and p such that ∥x∥X≤c1∥x∥(X,X)γ,p.Forγ∈(0,1) and p∈[1,∞] we now define the Banach spaces XAγ,p by
(28)XAγ,p={x∈X:[x]XAγ,p:=∥ξγA∘(ξI-A)-1x∥Lp*(X)<∞},∥x∥XAγ,p=∥x∥X+[x]XAγ,p.
It is a well-known fact that if A is single-valued and β=1 in (H1), then (X,𝒟(A))γ,p≅XAγ,p (cf. [30, Theorem 3.1] and [27, Theorem 1.14.2]). On the contrary, if β∈(0,1), then such an equivalence is no longer true, as first observed in [13, Theorem 2] for single-valued operators and, in the case p=∞, in [2, Theorem 1.12] for the m. l. ones. Recently, extending [13] to m. l. operators and [2] to p∈[1,∞], in [24, Proposition 4.3] it has been shown that the following embedding relations hold:
(29)XAγ,p↪(X,𝒟(A))γ,p,γ∈(0,1),p∈[1,∞],(30)(X,𝒟(A))γ,p↪XAγ+β-1,p,γ∈(1-β,1),p∈[1,∞].
Then, as in the single-valued case, (X,𝒟(A))γ,p≅XAγ,p if β=1 in (H1). More precisely (see the proof of [24, Proposition 4.3]), if x∈XAγ,p, γ∈(0,1), p∈[1,∞], then
(31)∥x∥(X,𝒟(A))γ,p≤2∥x∥XAγ,p,
whereas if x∈(X,𝒟(A))γ,p, γ∈(1-β,1), p∈[1,∞], then
(32)∥x∥XAγ+β-1,p≤c2∥x∥(X,𝒟(A))γ,p,
with c2 being a positive constant depending on β, γ and p.By settingδ=γ+β-1, γ∈(1-β,1), from (30) it follows
(33)𝒟(A)↪(X,𝒟(A))1+δ-β,p↪XAδ,p↪X,δ∈(0,β),p∈[1,∞].
Then, if β∈(0,1), the spaces XAδ,p, δ∈(0,1), p∈[1,∞], are intermediate spaces between X and 𝒟(A) only for δ∈(0,β), whereas, when δ∈[β,1), they may be smaller than 𝒟(A). In any case, when β∈(0,1), it is not known if the spaces XAδ,p, δ∈(0,β), p∈[1,∞], are only intermediate or just interpolation spaces between X and 𝒟(A).Notice that[XAγ,p∩A0]={0}, γ∈(0,1), p∈[1,∞]. Indeed, assume that there exists x≠0 such that x∈[XAγ,p∩A0] for some γ∈(0,1) and p∈[1,∞]. Then, since x∈A0=𝒩((zI-A)-1), z∈ρ(A), we have A∘(ξI-A)-1x=ξ(ξ-A)-1x-x=-x for every ξ>0 and [x]XAγ,p=∥ξγ∥Lp*(X)∥x∥X=∞, contradicting x∈XAγ,p. This property plays a key role in the proof of many of the results in [24]. Further, due to (30), it implies that [𝒟(A)∩A0]=[(X,𝒟(A))γ,p∩A0]={0}, γ∈(1-β,1), p∈[1,∞]. On the contrary, since {0} may be a proper subset of [(X,𝒟(A))γ,p∩A0] for γ∈(0,1-β], β<1, in general it is not true that [𝒟(A)¯∩A0]={0}. This is true, instead, if β=1. In this case the topological direct sum X0=𝒟(A)¯⊕A0 is a closed subspace of X, and if X is reflexive, it coincides with the whole X (cf. [3, Theorems 2.4 and 2.6]).For everyγ∈(0,1) and p∈[1,∞] from (27), (29), and (31) it follows
(34)∥A∘(zI-A)-1∥ℒ(XAγ,p;X)≤2c1(C+1)1-γCγ(|z|+1)1-β-γ,∀z∈Σα.
Hence, for γ∈(0,1) and p∈[1,∞] we may rewrite (27) and (34) more compactly as
(35)∥A∘(zI-A)-1∥ℒ(Yγp;X)≤c3(|z|+1)1-β-γ,∀z∈Σα,
where Yγp∈{(X,𝒟(A))γ,p,XAγ,p} and c3 is equal to c1(C+1)1-γCγ or 2c1(C+1)1-γCγ  according that   Yγp=(X,𝒟(A))γ,p or Yγp=XAγ,p.With the exception of the caseβ=1, in general it is not clear if embeddings analogous to (20) hold even for the spaces XAγ,p. In fact, using (20), (29), and (30) we can only prove that if γ∈(1-β,1) and 1≤q≤p≤∞, then
(36)XAγ,q↪(X,𝒟(A))γ,q↪(X,𝒟(A))γ,q↪XAγ+β-1,p,
whereas if 1-β<γ2<γ1<1 and p1,p2∈[1,∞], then
(37)XAγ1,p1↪(X,𝒟(A))γ1,p1↪(X,𝒟(A))γ2,p2↪XAγ2+β-1,p2.
What can be proved without invoking (20), (29), and (30) and using only the definition of the norm ∥·∥XAγ,p is instead the following result, which extends to the spaces XAγ,p the embeddings (X,𝒟(A))γ1,p↪(X,𝒟(A))γ2,p, and (X,𝒟(A))γ1,∞↪(X,𝒟(A))γ2,p, 0<γ2<γ1<1, p∈[1,∞] (cf. (20) with (p1,p2)=(p,p) and (p1,p2)=(∞,p)).Proposition 2.
LetA be an m. l. operator satisfying the resolvent condition (H1). Then the following embeddings hold for every 0<γ2<γ1<1 and p∈[1,∞]:
(38)XAγ1,p↪XAγ2,p,(39)XAγ1,∞↪XAγ2,p.Proof.
Ifβ=1 in (H1), then there is nothing to prove since (X,𝒟(A))γ,p≅XAγ,p and both (38) and (39) follow from (20). Therefore, without loss of generality, we assume that β∈(0,α] is such that β<α if α=1. We begin by proving (38). Let first p∈[1,∞). For every x∈XAγ1,p, 0<γ2<γ1<1, we write
(40)[x]XAγ2,pp=I1+I2,
where
(41)Ij=∫ajbj‍∥ξγ2A∘(ξI-A)-1x∥Xpdξξ,j=1,2,(a1,b1,a2,b2)=(0,1,1,∞). Using the first inequality in (26) we find
(42)I1≤(C+1)p∥x∥Xp∫01‍ξγ2p-1(ξ+1)(1-β)pdξ≤2(1-β)p(C+1)p∥x∥Xp∫01‍ξγ2p-1dξ≤[c4∥x∥XAγ1,p]p,
where c4=21-β(C+1)(γ2p)-1/p. Concerning I2, instead, using γ2-γ1<0, we get
(43)I2=∫1∞‍ξ(γ2-γ1)p∥ξγ1A∘(ξI-A)-1x∥Xpdξξ≤∫1∞‍∥ξγ1A∘(ξI-A)-1x∥Xpdξξ≤[x]XAγ1,pp≤∥x∥XAγ1,pp.
Summing up (40)–(43) and setting c5=[(c4)p+1]1/p, it thus follows ∥x∥XAγ2,p=∥x∥X+[x]XAγ2,p≤(1+c5)∥x∥XAγ1,p, completing the proof of (38) in the case p∈[1,∞). Let p=∞. For every x∈XAγ1,∞, 0<γ2<γ1<1, we write
(44)[x]XAγ2,∞=max{I3,I4},
where Ij=supξ∈Uj∥ξγ2A∘(ξI-A)-1x∥X, j=3,4, U3=(0,1), U4=[1,∞). Again, the first inequality in (26) yields
(45)I3≤(C+1)∥x∥Xsupξ∈(0,1)[ξγ2(ξ+1)1-β]≤21-β(C+1)∥x∥XAγ1,∞.
Instead, using γ2-γ1<0, we have
(46)I4=supξ∈[1,∞)ξγ2-γ1∥ξγ1A∘(ξI-A)-1x∥X≤[x]XAγ1,∞≤∥x∥XAγ1,∞.
Summing up (44)–(46) and setting c6=21-β(C+1), we thus find ∥x∥XAγ2,∞=∥x∥X+[x]XAγ2,∞≤(1+c6)∥x∥XAγ1,∞. This completes the proof of (38) for the case p=∞. We now prove (39). Due to (38) with p=∞, it suffices to assume that p∈[1,∞). As above, for every x∈XAγ1,∞, 0<γ2<γ1<1, we write [x]XAγ2,pp=I1+I2, where I1 and I2 are defined by (41). Hence, the same computations as in (42) yield
(47)I1≤[c4∥x∥XAγ1,∞]p.
As far as I2 is concerned, instead, we have
(48)I2≤[x]XAγ1,∞p∫1∞‍ξ(γ2-γ1)p-1dξ≤[c7∥x∥XAγ1,∞]p,
where c7=[(γ1-γ2)p]-1/p. Summing up (47) and (48) and setting c8=[(c4)p+(c7)p]1/p, we deduce ∥x∥XAγ2,p≤(1+c8)∥x∥XAγ1,∞. The proof is complete.Remark 3.
Notice that (37) with p1=p2=p yields XAγ1,p↪XAγ2+β-1,p, 1-β<γ2<γ1<1, and this latter embedding is less accurate than (38).Remark 4.
The main problem for extending (20) to the spaces XAγ,p in the case β<1 is that it is not clear if it holds XAγ,q↪XAγ,p, 1≤q<p≤∞. In fact, the embedding
(49)(X,𝒟(A))γ,q↪(X,𝒟(A))γ,q,γ∈(0,1),1≤q<p≤∞,
is a consequence of the property of the functional K entering the definition of the interpolation spaces (X,𝒟(A))γ,p through the “K-method”, and in particular of its monotonicity (see the proof of [27, Theorem 1.3.3(c),(d)]). With embedding (49) at hands, to derive (20) it thus suffices to prove that (X,𝒟(A))γ1,∞↪(X,𝒟(A))γ2,1, 0<γ2<γ1<1 (see the proof of [27, Theorem 1.3.3(e)] taking there (A0,A1,θ,θ~)=(𝒟(A),X,1-γ1,1-γ2) and using (𝒟(A),X)1-γ,p=(X,𝒟(A))γ,p).  If we try to repeat the proof of (49) for the spaces XAγ,p, we will be faced with two problems. The first is that we do not know if the function g(ξ)=∥A∘(ξI-A)-1x∥X, ξ∈(0,∞), x∈X, is monotone decreasing, which would allow us to prove XAγ,p↪XAγ,∞, γ∈(0,1), p∈[1,∞). For if g(ξ) was monotone decreasing, then for every ξ∈(0,∞) and x∈XAγ,p, γ∈(0,1), p∈[1,∞), we would find
(50)ξγg(ξ)=c9(∫0ξ‍μγpdμμ)1/pg(ξ)≤c9(∫0ξ‍[μγg(μ)]pdμμ)1/p≤c9[x]XAγ,p,
where c9=(γp)-1/p. Taking the supremum with respect to ξ∈(0,∞) in the latter inequality, we would get [x]XAγ,∞≤c9[x]XAγ,p, proving XAγ,p↪XAγ,∞, γ∈(0,1), p∈[1,∞). The second problem is that the function ξγg(ξ) is not necessarily bounded for x∈XAγ,p, γ∈(0,1), p∈[1,∞), precluding us to prove XAγ,q↪XAγ,p, γ∈(0,1), q∈[1,p). Indeed, from (35) we can only find ξγg(ξ)≤c3ξγ(ξ+1)1-β-γ∥x∥XAγ,p, and when β<1, the right-hand side of this inequality goes to infinity as ξ goes to infinity. On the contrary, if ξγg(ξ) were bounded, then for every 1≤q<p<∞ we would obtain
(51)[x]XAγ,pp=∫0∞‍[ξγg(ξ)]pdξξ≤(supξ∈(0,∞)ξγg(ξ))p-q∫0∞‍[ξγg(ξ)]qdξξ=[x]XAγ,∞p-q[x]XAγ,pp.
If now in addition g(ξ) were also monotone decreasing, in order that [x]XAγ,∞≤c9[x]XAγ,q, from the latter inequality we would get [x]XAγ,p≤(c9)(p-q)/p[x]XAγ,q, completing the proof of XAγ,q↪XAγ,p, γ∈(0,1), 1≤q<p<∞. Due to the former computations, we can thus conclude that in the case β<1 the quoted problems are the main obstacles which prevent us to extend (49) and, as its consequence, (20) to the spaces XAγ,p.Remark 5.
Let0<γ2<γ1<1 be fixed and for every p∈[1,∞] and let us set Ap=XAγ2,p and Bp=XAγ1,p. We thus have the two families of sets 𝒜={Ap}p∈[1,∞] and ℬ={Bp}p∈[1,∞]. Let first β=1. In this case, since (X,𝒟(A))γ,p≅XAγ,p, from (20) we deduce that the sets Ap and Bp are related by the following inclusions in which 1<q1<q2<∞:
(52)B1⊆Bq1⊆Bq2⊆B∞⊆A1⊆Aq1⊆Aq2⊆A∞.
Now let β<1. As observed in Remark 4, in this case the embedding XAγ,q↪XAγ,p, 1≤q<p≤∞, may be not satisfied and the chain of inclusions (52) could not take place. However, (38) and (39) hold true and for every p∈[1,∞], and we have Bp⊆Ap and B∞⊆Ap.We have already pointed out that{etA}t≥0 may be not strongly continuous in the X-norm on 𝒟(A)¯. On the contrary, the following result (cf. [24, Proposition 5.2] for the proof) shows thatthe things are finer on (X,𝒟(A))γ,p and XAγ,p. Later, we will need this fact.Proposition 6.
LetA be as in  Proposition 2. If γ∈(1-β,1); then {etA}t≥0 is strongly continuous in the X-norm on Yγp∈{(X,𝒟(A))γ,p,XAγ,p} for every p∈[1,∞].We conclude the section listing some estimates for the operators[(-A)θ]∘etA defined by (11) with respect to the spaces (X,𝒟(A))γ,p and XAγ,p. First, in [19, Lemma 3.1] it is shown that [(-A)θ]∘etAx∈𝒟(A) for every x∈X and that the estimate ∥[(-A)θ]∘etAx∥𝒟(A)≤∥[(-A)θ+1]∘etAx∥X is satisfied. Hence, using (14), we get
(53)∥[(-A)θ]∘etA∥ℒ(X;𝒟(A))≤c~α,β,θ+1t(β-ℜeθ-2)/α,∥[(-A)θ]∘etA∥ℒ(X;𝒟(A))ℜeθ≥0,t>0.
Combining (14) and (53) with (25) and letting c10=c0(cα,β,θ)1-γ(cα,β,θ+1)γ, it thus follows (cf. [19, Proposition 3.1]) that for every γ∈(0,1) and p∈[1,∞]  the following estimate   holds:
(54)∥[(-A)θ]∘etA∥ℒ(X;(X,𝒟(A))γ,p)≤c10t(β-γ-ℜeθ-1)/α,ℜeθ≥0,t>0.Remark 7.
We stress that ifβ<1, then we can not derive an estimate for the ℒ(X;XAγ,p)-norm of [(-A)θ]∘etA simply by replacing (X,𝒟(A))γ,p with XAγ,p in (54). This is for two reasons. First, when γ∈[β,1), we are not assured that [(-A)θ]∘etAx∈XAγ,p for every x∈X. For if γ∈[β,1), then the space XAγ,p may be smaller than the domain 𝒟(A) to which [(-A)θ]∘etAx belongs by virtue of [19, Lemma 3.1]. The second reason is that, even limiting to γ∈(0,β) in order that 𝒟(A)↪XAγ,p, from (31) we only get ∥[(-A)θ]∘etAx∥(X,𝒟(A))γ,p≤2∥[(-A)θ]∘etAx∥XAγ,p, x∈X, and we do not know if the right-hand side can be bounded from above by some constant times t(β-γ-ℜeθ-1)/α∥x∥X. Of course, we can employ (32), but in this way all that we can reach is the estimate
(55)∥[(-A)θ]∘etA∥ℒ(X;XAγ+β-1,p)≤c11t(β-γ-ℜeθ-1)/α,ℜeθ≥0,t>0,
where c11=c2c10, γ∈(1-β,1) and p∈[1,∞]. Letting δ=γ+β-1, (55) can be rewritten equivalently as
(56)∥[(-A)θ]∘etA∥ℒ(X;XAδ,p)≤c11t(2β-δ-ℜeθ-2)/α,ℜeθ≥0,t>0,
where δ∈(0,β) and p∈[1,∞]. When β<1, there are good motivations to believe that estimate (56) is not the best one. In fact, for instance, when (θ,p)=(0,∞), (56) leads us to an estimate which is rougher than the estimate
(57)∥etA∥ℒ(X;XAδ,∞)≤c12t(β-δ-1)/α,δ∈(0,1),t>0,
as shown in [2, Proposition 3.2], with c12 being a positive constant depending on α, β, and δ. Also, (57) ensures that etAx, x∈X, belongs to XAδ,∞ for every δ∈(0,1) and not only for δ∈(0,β) as (56) suggests. Furthermore, due to (31), estimate (57) yields (54) with (θ,γ,p)=(0,δ,∞). This leads us to believe that (57) can be improved and that estimate (54) holds the same if XAγ,∞ is taken in place of (X,𝒟(A))γ,∞.Now letYγp∈{(X,𝒟(A))γ,p,XAγ,p}, γ∈(0,1), p∈[1,∞]. As far as the estimates for the ℒ(Yγp;X)-norm of operators [(-A)θ]∘etA are concerned, instead, at the moment only the following estimates for the case θ=1 are available (cf. [24, Lemma 5.1]):
(58)∥[(-A)1]∘etA∥ℒ(Yγp;X)≤c13t(β+γ-2)/α,t>0,γ∈(0,1),p∈[1,∞],
with c13 being a positive constant depending on α, β, γ, and p. Estimates (58) are successfully applied in [24, Corollary 5.4] to prove that if α+β>1, then the map t→etA is Hölder continuous from [0,∞) to ℒ(Yγp;X), γ∈(2-α-β,1), p∈[1,∞], with Hölder exponent σ=(α+β+γ-2)/α. In Section 3 we will extend (58), proving some estimates for the ℒ(Yγp;X)-norm of [(-A)θ]∘etA, ℜeθ≥1, which reduce to (58) in the case θ=1.Remark 8.
Observe that an estimate for the norm∥[(-A)θ]∘etA∥ℒ(Yγp;X), ℜeθ≥1, t>0, Yγp∈{(X,𝒟(A))γ,p,XAγ,p}, γ∈(0,1), p∈[1,∞], can be obtained combining (14), (15), and (58). Indeed, using (15), for every ℜeθ≥1, t>0 and x∈Yγp, we have
(59)∥[(-A)θ]∘etAx∥X=∥[(-A)θ-1]∘e(t/2)A[(-A)1]∘e(t/2)Ax∥X≤∥[(-A)θ-1]∘e(t/2)A∥ℒ(X)∥[(-A)1]∘e(t/2)Ax∥X.
Therefore, due to (14) and (58), from (59) we deduce that
(60)∥[(-A)θ]∘etA∥ℒ(Yγp;X)≤c14t(2β+γ-ℜeθ-2)/α,ℜeθ≥1,t>0,
where γ∈(0,1), p∈[1,∞] and c14=2(2+ℜeθ-γ-2β)/αc~α,β,θ-1c13. As we will see in the next section estimate (60) is not optimal, in the sense that the negative exponent (2β+γ-ℜeθ-2)/α can be refined; of course, unless β=1. The main reason to believe that (60) can be improved is that its derivation consists of two steps: the first in which [(-A)θ]∘etA is decomposed with the help of (15), and the second in which (60) is obtained combining estimates of very different nature, such as (14) and (58). It is thus to be expected that in this double step derivation some regularity goes missing and that a better result can be reached analyzing more detailedly [(-A)θ]∘etAx for x∈Yγp.


## 3. Behaviour of[(-A)θ]∘etA in (X,𝒟(A))γ,p and XAγ,p

According to Remark7 we begin by improving (54), showing that the same estimate holds with (X,𝒟(A))γ,p being replaced by XAγ,∞ if p=∞ and by XAβγ,p if p∈[1,∞). Throughout this and the next section, A will be an m. l. operator in X having nonempty domain 𝒟(A) and satisfying the resolvent condition (H1) of Section 2.Proposition 9.
Letℜeθ≥0, γ∈(0,1) and let p∈[1,∞]. Then, there exist positive constants cj, j=15,16, depending on α, β, γ, θ, and p such that
(61)∥[(-A)θ]∘etA∥ℒ(X;XAγ,∞)≤c15t(β-γ-ℜeθ-1)/α,t>0,p=∞,(62)∥[(-A)θ]∘etA∥ℒ(X;XAβγ,p)≤c16t(β-γ-ℜeθ-1)/α,t>0,p∈[1,∞).Proof.
Ifβ=1, then (X,𝒟(A))γ,p≅XAγ,p and (61) and (62) with cj=c2c10, j=15,16, follow by taking β=1 in (32) and (54). Therefore, without the loss of generality, we assume that β∈(0,α] is such that β<α if α=1. Let θ∈C, ℜeθ≥0, γ∈(0,1), and p∈[1,∞) be fixed and let x be an arbitrary element of X. Then, for every t>0 we have
(63)∥[(-A)θ]∘etAx∥XAγ,∞=∥[(-A)θ]∘etAx∥X+∥ξγA∘(ξI-A)-1[(-A)θ]∘etAx∥L∞*(X),(64)∥[(-A)θ]∘etAx∥XAβγ,p=∥[(-A)θ]∘etAx∥X+∥ξβγA∘(ξI-A)-1[(-A)θ]∘etAx∥Lp*(X).
Of course, from estimate (54) we find
(65)∥[(-A)θ]∘etAx∥X≤cγ,pc10∥x∥Xt(β-γ-ℜeθ-1)/α,t>0,
with cγ,p being such that ∥y∥X≤cγ,p∥y∥(X,𝒟(A))γ,p, y∈(X,𝒟(A))γ,p, p∈[1,∞]. It thus suffices to investigate only the second terms on the right-hand side of (63) and (64). We begin by proving (61). First, using the second identity in (6), for every ξ∈(0,∞) we get
(66)ξγA∘(ξI-A)-1[(-A)θ]∘etAx=12πi∫Γ‍ξγ(-λ)θetλA∘(ξI-A)-1(λI-A)-1xdλ=ξγ[12πi∫Γ‍(-λ)θetλ(λ-ξ)-1dλ]A∘(ξI-A)-1x-12πi∫Γ‍ξγ(-λ)θetλ(λ-ξ)-1A∘(λI-A)-1xdλ=-12πi∫Γ‍ξγ(-λ)θetλ(λ-ξ)-1[λ(λI-A)-1-I]xdλ=12πi∫Γ‍ξγ(-λ)θ+1etλ(λ-ξ)-1(λI-A)-1xdλ.
Here we have used twice the equality ∫Γ‍(-λ)θetλ(λ-ξ)-1dλ=0, ξ∈(0,∞), which follows from Cauchy’s formula after having enclosed Γ on the left with an arc of the circle {z∈C:|z+c|=R}, R>0, and letting R to infinity. From (66), using ∥(λI-A)-1∥ℒ(X)≤C(|λ|+1)-β≤C|λ|-β, λ∈Σα, it follows that
(67)∥ξγA∘(ξI-A)-1[(-A)θ]∘etAx∥X≤C(2π)-1∥x∥X×∫Γ‍ξγ|λ|1+ℜeθ-βe-ℑmθarg(-λ)etℜeλ|λ-ξ|-1|dλ|≤C(2π)-1e(π/2)|ℑmθ|∥x∥X×∫Γ‍(ξ|λ|)γ|λ|γ+ℜeθ-βetℜeλ|1-(ξλ)|-1|dλ|.
Now, since ℜeλ≤-c<0 for every λ∈Γ and since ξ∈(0,∞), we have
(68)|1-(ξλ)|=|1-(ξλ¯|λ|2)|=[1+(ξ|λ|)2-2ξℜeλ|λ|2]1/2≥[1+(ξ|λ|)2]1/2.
Therefore, for every λ∈Γ and ξ∈(0,∞)  the following inequality  holds:
(69)(ξ|λ|)γ|1-(ξλ)|-1≤(ξ|λ|)γ[1+(ξ|λ|)2]-1/2≤γγ/2(1-γ)(1-γ)/2=:cγ,
where we have used the fact that the function f(s)=sγ(1+s2)-1/2, s≥0, γ∈(0,1), attains its maximum value cγ at the point sγ=γ1/2(1-γ)-1/2. Coming back to (67) and setting c17=C(2π)-1e(π/2)|ℑmθ|cγ, we thus find (here we use also that on Γ it holds |λ|≥c, so that ℜeλ=-c(|ℑmλ|+1)α≥-c(1+c-1)α|λ|α):
(70)∥ξγA∘(ξI-A)-1[(-A)θ]∘etAx∥X≤c17∥x∥X∫Γ‍|λ|γ+ℜeθ-βetℜeλ|dλ|≤c17∥x∥X∫Γ‍|λ|γ+ℜeθ-βe-c(1+c-1)αt|λ|α|dλ|≤2c17∥x∥X∫0∞‍μγ+ℜeθ-βe-cαtμαdμ,
where cα=c(1+c-1)α. Finally, taking the supremum with respect to ξ∈(0,∞) in (70) and performing the transformation cαtμα=s in the integral on the right, we obtain
(71)∥ξγA∘(ξI-A)-1[(-A)θ]∘etAx∥L∞*(X)≤c18∥x∥Xt(β-γ-ℜeθ-1)/α,
where c18=2c17α-1cα(β-γ-ℜeθ-1)/αE((γ+ℜeθ+1-β)/α), E(χ), χ>0, being the Euler gamma function ∫0∞‍sχ-1e-sds. Then, summing up (65) and (71), from (63) it follows that
(72)∥[(-A)θ]∘etAx∥XAγ,∞≤(cγ,∞c10+c18)∥x∥Xt(β-γ-ℜeθ-1)/α,∥[(-A)θ]∘etAx∥XAγ,∞≤x(cγ,∞c10+c18)ℜeθ≥0,t>0.
Since x∈X was arbitrary, this completes the proof of (61) with c15=cγ,∞c10+c18. Let us now prove (62). For every p∈[1,∞) we write
(73)∥ξβγA∘(ξI-A)-1[(-A)θ]∘etAx∥Lp*(X)p=I1+I2,
where Ij=∫ajbj‍∥ξβγA∘(ξI-A)-1[(-A)θ]∘etAx∥Xp(dξ/ξ), j=1,2, (a1,b1,a2,b2)=(0,1,1,∞). First, (35) with Yγp=(X,𝒟(A))γ,p yields
(74)I1≤∥[(-A)θ]∘etAx∥(X,𝒟(A))γ,pp∫01‍ξβγp-1[c3(ξ+1)1-β-γ]pdξ.
Therefore, since (ξ+1)1-β-γ≤cβ,γ for every ξ∈(0,1], where cβ,γ=21-β-γ or cβ,γ=1  according that   γ∈(0,1-β) or γ∈[1-β,1), from (54), we deduce that
(75)I1≤[cβ,γc3]p∥[(-A)θ]∘etAx∥(X,𝒟(A))γ,pp∫01‍ξβγp-1dξ=[c19∥x∥Xt(β-γ-ℜeθ-1)/α]p,
with c19=cβ,γc3c10(βγp)-1/p. As far as I2 is concerned, exploiting (71) and recalling that we have assumed β<1, we obtain
(76)I2=∫1∞‍ξ(β-1)γp∥ξγA∘(ξI-A)-1[(-A)θ]∘etAx∥Xpdξξ≤[c18∥x∥Xt(β-γ-ℜeθ-1)/α]p∫1∞‍ξ(β-1)γp-1dξ≤[c20∥x∥Xt(β-γ-ℜeθ-1)/α]p,
where c20=c18[(1-β)γp]-1/p. Summing up (73)–(76), it thus follows that
(77)∥ξβγA∘(ξI-A)-1[(-A)θ]∘etAx∥Lp*(X)≤c21∥x∥Xt(β-γ-ℜeθ-1)/α,
where c21=[(c19)p+(c20)p]1/p. Finally, (65) and (77) lead us to
(78)∥[(-A)θ]∘etAx∥XAβγ,p≤(cγ,pc10+c21)∥x∥Xt(β-γ-ℜeθ-1)/α,∥[(-A)θ]∘etAx∥XAβγ,p≤x(cγ,pc10+c21)ℜeθ≥0,t>0.
Since x∈X was arbitrary, this completes the proof of (62) with c16=cγ,pc10+c21.Remark 10.
Ifθ=0, then (61) is precisely the estimate (57). In this sense our result improves [2] and shows that (54) holds the same with (X,𝒟(A))γ,p being replaced with XAγ,∞ if p=∞ and XAβγ,p and if p∈[1,∞). Also, when β<1, (61) and (62) are in two aspects better than the estimate (55) deduced from (54) with the help of (32). First, here we do not need to restrict γ to (1-β,1). Further, despite limiting γ to (1-β,1), (61) and (62) show that [(-A)θ]∘etAx, ℜeθ≥0, t>0, x∈X, enjoys more regularity than that predicted by (55). For, since when β<1 it   holds 0<γ+β-1<βγ<γ, from (38) and (39) it follows XAγ,∞↪XAβγ,p↪XAγ+β-1,p, p∈[1,∞].Remark 11.
We recall that whenβ<1 the spaces XAσ,p, σ∈(0,1), p∈[1,∞], are intermediate spaces between X and 𝒟(A) for σ∈(0,β), but they may be contained in 𝒟(A) for σ∈[β,1). Therefore, whereas (61) is satisfied for spaces XAσ,∞ eventually smaller than 𝒟(A), for (62) to hold we have to consider only spaces XAσ,p, p∈[1,∞), bigger than 𝒟(A). In fact, letting σ=βγ, we have σ∈(0,β) for every γ∈(0,1).In accordance with Remark8 we now improve estimate (58).Proposition 12.
Letℜeθ≥1, γ∈(0,1), p∈[1,∞] and let Yγp∈{(X,𝒟(A))γ,p,XAγ,p}. Then, there exists a positive constant c22 depending on α, β, γ, θ, and p such that
(79)∥[(-A)θ]∘etA∥ℒ(Yγp;X)≤c22t(β+γ-ℜeθ-1)/α,t>0.Proof.
First, using the identityA∘(zI-A)-1=z(zI-A)-1-I, z∈Σα, for every x∈X, we rewrite [(-A)θ]∘etAx, ℜeθ≥0, in the following way:
(80)[(-A)θ]∘etAx=-12πi∫Γ‍(-λ)θ-1etλλ(λI-A)-1xdλ=-12πi∫Γ‍(-λ)θ-1etλ[A∘(λI-A)-1x+I]xdλ=-12πi∫Γ‍(-λ)θ-1etλA∘(λI-A)-1xdλ,t>0.
Here we have used ∫Γ‍(-λ)θ-1etλdλ=0, which follows from the Cauchy formula applied to (-λ)θetλ after having enclosed Γ on the left with an arc of the circle {z∈C:|z+c|=R}, R>0, and letting R to infinity. Let now θ∈C, ℜeθ≥1, γ∈(0,1), and p∈[1,∞] be fixed and let x be an arbitrary element of Yγp. From (35) it then follows that
(81)∥[(-A)θ]∘etAx∥X≤c23∥x∥Yγp∫Γ‍|λ|ℜeθ-1etℜeλ(|λ|+1)1-β-γ|dλ|,t>0,
where c23=(2π)-1e(π/2)|ℑmθ|c3. Now, recalling that |λ|≥c>0 for every λ∈Γ, we have |λ|≤|λ|+1≤(1+c-1)|λ|, λ∈Γ. As a consequence, the following inequality holds:
(82)(|λ|+1)1-β-γ≤c~β,γ|λ|1-β-γ,∀λ∈Γ,
where c~β,γ=(1+c-1)1-β-γ or c~β,γ=1  according   that γ∈(0,1-β] or γ∈(1-β,1) ((0,1-β]=∅ if β=1). Therefore, setting c24=2c~β,γc23, (81) and (82) yield
(83)∥[(-A)θ]∘etAx∥X≤c24∥x∥Yγp∫0∞‍μℜeθ-β-γe-cαtμαdμ,t>0,
with cα being as in (70). Finally, the transformation cαtμα=s in the last integral leads us to the following estimate:
(84)∥[(-A)θ]∘etAx∥X≤c25∥x∥Yγpt(β+γ-ℜeθ-1)/α,t>0,
where c25=c24α-1cα(β+γ-ℜeθ-1)/αE((ℜeθ+1-β-γ)/α), E(χ), χ>0, is the Euler’s gamma function. Notice that here ℜeθ≥1 implies ℜeθ+1-β-γ≥2-β-γ>0 for every β∈(0,1] and γ∈(0,1), so that E((ℜeθ+1-β-γ)/α) makes sense. Since (84) is satisfied for every arbitrary element x∈Yγp, the proof is complete with c22=c25.Remark 13.
Estimate (79) is better than (60) obtained in Remark 8 using (14), (15), and (58). In fact, for every β∈(0,α], α∈(0,1], γ∈(0,1) and ℜeθ≥1, the following inequality holds:
(85)ρ1:=(2β+γ-ℜeθ-2)α≤(β+γ-ℜeθ-1)α:=ρ2<0.
Then, tρ2≤tρ1, t∈(0,1], and (79) is more accurate than (60) for small values of t.Estimate (79) with θ=1 yields the following result which we will need in Section 5 to prove the equivalence between problem (170) and the fixed-point equation (179).Corollary 14.
Letα+β>1 in (H1). Then, for every x∈X  the following equalities hold:
(86)A-1(etA-I)x=(etA-I)A-1x=∫0t‍e(t-s)Axds,t≥0.Proof.
The assertion is obvious fort=0. Let t>0 and let x∈X. Commuting A-1∈ℒ(X) with the integral sign, from (9) and the resolvent equation, we have A-1etAx=etAA-1x, which proves the first equality in (86). To prove the second equality, we first write
(87)(etA-I)A-1x=∫0t‍[DrerA]r=t-sA-1xds=-∫0t‍[(-A)1]∘e(t-s)AA-1xds,
and we show that the latter integral is convergent. Indeed, since α+β>1, we may consider A-1x∈𝒟(A) as an element of (X,𝒟(A))γ,p, where γ∈(2-α-β,1) and p∈[1,∞]. With this choice for γ, from (79) with θ=1 and (25) we obtain (here we use also ∥A-1x∥𝒟(A)=infy∈A(A-1x)∥y∥X=infy∈(AA-1)x∥y∥X=∥x∥𝒟(AA-1)≤∥x∥X, due to I⊂AA-1. Then, ∥A-1x∥(X,𝒟(A))γ,p≤c0∥A-1x∥X1-γ∥A-1x∥𝒟(A)γ≤c0∥A-1∥ℒ(X)1-γ∥x∥X):
(88)∥∫0t‍[(-A)1]∘e(t-s)AA-1xds∥X≤c22∥A-1x∥(X,𝒟(A))γ,p∫0t‍(t-s)(β+γ-2)/αds≤c22cα,β,γc0∥A-1∥ℒ(X)1-γ∥x∥Xt(α+β+γ-2)/α,
where cα,β,γ=α(α+β+γ-2)-1. We now recall that (cf. [24, formula (3.21)])
(89)[(-A)1]∘etA(-A)-ζ=[(-A)1-ζ]∘etA,ℜeζ∈(1-β,1],t>0,
with (-A)-ζ being the negative fractional powers of -A defined by (cf. [24, Section 3]) (2πi)-1∫Γ‍(-λ)-ζ(λI-A)-1dλ, ℜeζ>1-β. To complete the proof it thus suffices to apply (89) with ζ=1 to (87) and to recall that [(-A)0]∘etA=etA, t>0. Notice that the integral on the right-hand side of (86) is convergent, too. In fact, from (14), it follows that ∥∫0t‍e(t-s)Axds∥X≤c~α,β,0∥x∥X∫0t‍(t-s)(β-1)/αds=α(α+β-1)-1c~α,β,0∥x∥Xt(α+β-1)/α.Remark 15.
In particular, from (86) it follows that if α+β>1, then ∫0t‍e(t-s)Axds∈𝒟(A) for every x∈X and (etA-I)x⊆A∫0t‍e(t-s)Axds. This extends to m. l. operators satisfying (H1) the well-known result for sectorial single-valued linear operators (see, for instance, [9, Proposition 2.1.4(ii)] and [11, Proposition 1.2(ii)]).With the help of (54) and Proposition 12, we can now derive the following interpolation estimates (90) for the operators [(-A)θ]∘etA, ℜeθ≥1, which are considered as operators from (X,𝒟(A))γ,p to (X,𝒟(A))δ,p. As we will see in the proof of Proposition 16, here the fact that the spaces (X,𝒟(A))σ,p are real interpolation spaces between X and 𝒟(A) plays a key role. For it allows us to exploit the interpolation inequality (24) in the derivation of our estimates in the case γ+δ<1.Proposition 16.
Letℜeθ≥1, γ,δ∈(0,1), and p∈[1,∞]. Then, there exist positive constants cj, j=26,27, depending on α, β, γ, δ, θ, and p such that for every t>0(90)∥[(-A)θ]∘etA∥ℒ((X,𝒟(A))γ,p;(X,𝒟(A))δ,p)≤{c26t(2β+γ-δ-ℜeθ-2)/α,γ,δ∈(0,1),c27t(β+γ-δ-ℜeθ-1)/α,ifγ+δ<1.Proof.
For brevity, we will use the shorteningsYσp=(X,𝒟(A))σ,p, σ∈(0,1), p∈[1,∞]. We begin by proving the first estimate in (90). Let θ∈C, ℜeθ≥1, γ,δ∈(0,1) and p∈[1,∞] be fixed and let x be an arbitrary element of Yγp. Moreover, let ζ and ζ′ be two arbitrary complex numbers such that θ=ζ+ζ′ and whose real parts satisfy ℜeζ≥0 and ℜeζ′≥1. From the decomposition formula (15) it then follows for every t>0:
(91)∥[(-A)θ]∘etAx∥Yδp=∥[(-A)ζ]∘e(t/2)A[(-A)ζ′]∘e(t/2)Ax∥Yδp≤∥[(-A)ζ]∘e(t/2)A∥ℒ(X;Yδp)∥[(-A)ζ′]∘e(t/2)Ax∥X≤∥[(-A)ζ]∘e(t/2)A∥ℒ(X;Yδp)∥[(-A)ζ′]∘e(t/2)A∥ℒ(Yγp;X)∥x∥Yγp.
Therefore, using (54) and (79) with the triplet (θ,γ,t) being equal to (ζ,δ,t/2) and (ζ′,γ,t/2), respectively, from (91) and ℜeθ=ℜeζ+ℜeζ′, we deduce that
(92)∥[(-A)θ]∘etAx∥Yδp≤c10c22(t2)(β-δ-ℜeζ-1)/α(t2)(β+γ-ℜeζ′-1)/α∥x∥Yγp≤c26t(2β+γ-δ-ℜeθ-2)/α∥x∥Yγp,t>0,
where c26=2(2+ℜeθ+δ-γ-2β)/αc10c22. This completes the proof of the first estimate in (90), due to the arbitrariness of x∈Yγp. Let us now prove the second estimate in (90). Let θ∈C, ℜeθ≥1, γ,δ∈(0,1), γ+δ<1, and p∈[1,∞] be fixed. Using γ+δ<1, we fix γ2∈(γ/(1-δ),1)⊊(γ,1), and we let γ1=(γ2δ)/(γ2-γ). Clearly, since γ2∈(γ/(1-δ),1), we have γ1∈(δ,1). In addition, it holds:
(93)1-δ>γ1-δγ1=(γ2δγ2-γ-δ)(γ2-γγ2δ)=γγ2>γ.
Due to (93), we now set γ0=γ/γ2=(γ1-δ)/γ1∈(γ,1-δ), so that γ=γ0γ2 and δ=(1-γ0)γ1. From (24) with p0=p it thus follows that
(94)∥[(-A)θ]∘etA∥ℒ(Yγp;Yδp)≤∥[(-A)θ]∘etA∥ℒ(X;Yγ1p1)1-γ0∥[(-A)θ]∘etA∥ℒ(Yγ2p2;X)γ0,t>0,
where pj∈[1,∞], j=1,2. Applying (54) and (79) with the pair (γ,p) being replaced with (γ1,p1) and (γ2,p2), respectively, from (94) we finally obtain
(95)∥[(-A)θ]∘etA∥ℒ(Yγp;Yδp)≤[c10t(β-γ1-ℜeθ-1)/α]1-γ0[c22t(β+γ2-ℜeθ-1)/α]γ0≤(c10)1-γ0(c22)γ0t[β+γ0γ2-(1-γ0)γ1-ℜeθ-1]/α=(c10)δ/γ1(c22)γ/γ2t(β+γ-δ-ℜeθ-1)/α,t>0.
This completes the proof of the second estimate in (90) with c27=(c10)δ/γ1(c22)γ/γ2.Remark 17.
We stress that ifβ<1 and γ+δ<1, then the first estimate in (90) is rougher than the second one for small values of t, which justify our special attention to the case γ+δ<1. Indeed, if β<1, then for every ℜeθ≥1 the following inequality  holds:
(96)ρ3:=(2β+γ-δ-ℜeθ-2)α<(β+γ-δ-ℜeθ-1)α=:ρ4<0,
so that tρ4≤tρ3 for t∈(0,1]. In other words, if β and γ+δ are both less than one, then the second estimate in (90) establishes that the norm ∥[(-A)θ]∘etA∥ℒ((X,𝒟(A))γ,p;(X,𝒟(A))δ,p), ℜeθ≥1, may blow up as t goes to 0, but with an order of singularity lower than that predicted by the first estimate. In this sense, though less general, the second estimate in (90) is better than the first one.Remark 18.
The reason why the second estimate in (90) yields a better exponent than the first one is the same mentioned in Remark 8. That is, while the first estimate is obtained in two steps: decomposing [(-A)θ]∘etA through (15) and then applying (54) and (79), the second estimate is essentially derived in a single step, using (24).The following Remark19 points out why, with the exception of the case when β=1 and A is single-valued, to prove (90) we can not proceed as in [9, Proposition 2.2.9].Remark 19.
In the optimal caseβ=1, the exponents in both estimates (90) coincide equals to ν=γ-δ-ℜeθ. Hence, in this special case, the assumption γ+δ<1 does not give any enhancement. Also, if we further assume that θ∈N, then we restore the same estimates as in [9, Proposition 2.2.9(i)]. In this respect, our result extends [9] to the m. l. case, even though our proof really differs from that in [9]. For, there, the norms in the spaces (X,𝒟(A))σ,p are replaced with the norms in the spaces 𝒟A(σ,p), with the latter being the spaces of all x∈X such that ∥x∥𝒟A(σ,p)=∥x∥X+[x]𝒟A(σ,p)<∞, where [x]𝒟A(σ,p)=∥ξ(2-β-σ)/α[(-A)1]∘eξA∥Lp*(X). It is well known that if β=1 and A is single-valued, then (X,𝒟(A))σ,p≅𝒟A(σ,p) (cf. [31, Theorem 3], [9, Proposition 2.2.2] and [27, Theorem 1.14.5]). On the contrary, if (α,β)≠(1,1) and/or A is really an m. l. operator, such equivalence is no longer true and we have
(97)XAσ,p↪(X,𝒟(A))σ,p↪𝒟A(ασ,p),p∈[1,∞),XAσ,∞↪(X,𝒟(A))σ,∞↪𝒟A(σ,∞),p=∞.
Differently from the spaces XAσ,p and as a consequence of A0⊆⋂t>0‍𝒩([(-A)1]∘etA), the spaces 𝒟A(σ,p) contain A0. It can thus be shown that if α+β>1, then for every σ∈(2-α-β,1) and     φ∈(0,(α+β+σ-2)/α) (here (α+β+σ-2)/α<1, since σ<1≤2-β)   the following embeddings hold:
(98){0}∪[𝒟A(σ,p)∖A0]↪XAφ,p↪(X,𝒟(A))φ,p,p∈[1,∞),{0}∪[𝒟A(σ,∞)∖A0]↪XA(α+β+σ-2)/α,∞↪(X,𝒟(A))(α+β+σ-2)/α,∞,
with {0}∪[𝒟A(σ,p)∖A0] being endowed with the norm of 𝒟A(σ,p). Obviously, due to (29), it suffices to prove the embeddings on the right of (97) and on the left of (98). It is out of the aims of this paper to go into the details of these proofs, and for them we refer the readers to [24, Proposition 6.3]. Here we want only to make clear that, with the exception of the case when β=1 and A is single-valued, embeddings (97) and (98) prevent us from carrying out the proof of estimates (90) simply by repeating the computations in [9]. Notice that, due to the property [XAσ,p∩A0]={0}, from the second embeddings in (97) and (98) it follows that if α+β>1 and σ∈(2-α-β,1), then
(99)XAσ,∞↪{0}∪[𝒟A(σ,∞)∖A0]↪XA(α+β+σ-2)/α,∞.
Since  (α+β+σ-2)/α≤σ (indeed, α≤1≤(2-β-σ)/(1-σ) implies α+β+σ-2≤ασ), (99) agrees with (38) for p=∞. In addition, if 2α+β>2 and σ∈((2-α-β)/α,1), then the first embeddings in (97) and (98) yield for every φ∈(0,(α+β+ασ-2)/α) the following:
(100)XAσ,p↪{0}∪[𝒟A(ασ,p)∖A0]↪XAφ,p,p∈[1,∞).
Since φ<(α+β+ασ-2)/α≤σ, (100) agrees with (38) for p∈[1,∞). Furthermore, if β=1, then from (29), (30), and (99) it follows that (X,𝒟(A))σ,∞≅XAσ,∞≅{0}∪[𝒟A(σ,∞)∖A0], σ∈(0,1). This confirms that in the real m. l. case the equivalence between XAσ,p, (X,𝒟(A))σ,p and 𝒟A(σ,p) does not hold even when β=1.Using Propositions9 and 12, we now obtain estimates for the operators [(-A)θ]∘etA, ℜeθ≥1, considered as operators from XAγ,p to  XAδ,p. Clearly, since β<1 the spaces XAσ,p may be not real interpolation spaces between X and 𝒟(A), we can not proceed as in the proof of the second estimate in (90) and a weaker result has to be expected.Proposition 20.
Letℜeθ≥1, γ,δ∈(0,1), and p∈[1,∞]. Then, there exist positive constants cj, j=28,29,30, depending on α, β, γ, δ, θ, and p such that
(101)∥[(-A)θ]∘etA∥ℒ(XAγ,∞;XAδ,∞)≤c28t(2β+γ-δ-ℜeθ-2)/α,p=∞,t>0,(102)∥[(-A)θ]∘etA∥ℒ(XAγ,p;XAβδ,p)≤c29t(2β+γ-δ-ℜeθ-2)/α,p∈[1,∞),t>0.
Moreover, if γ∈(0,1) and δ∈(1-β,1) are such that γ+δ<1, then
(103)∥[(-A)θ]∘etA∥ℒ(XAγ,p;XAδ+β-1,p)≤c30t(β+γ-δ-ℜe-1)/α,p∈[1,∞],t>0.Proof.
Due to (61) and (79), in order to prove (101) and (102) it suffices to repeat the same computations as in (91) and (92), with the pair ((X,𝒟(A))γ,p,(X,𝒟(A))δ,p) being replaced with (XAγ,∞,XAδ,∞) or with (XAγ,p,XAβδ,p) provided that p=∞ or p∈[1,∞). In this way we derive (101) and (102) with cj+13=2(2+ℜeθ+δ-γ-2β)/αcjc22, j=15,16. As far as (103) is concerned, we recall that if Xj, j=1,…,4, are four Banach spaces such that Xj↪Xj+2, j=1,2, and L∈ℒ(X3;X2), then L∈ℒ(X1;X4) with ∥L∥ℒ(X1;X4)≤C1C2∥L∥ℒ(X3;X2), C1 and C2 being the positive constants such that ∥x∥Xj+2≤Cj∥x∥Xj, x∈Xj, j=1,2. Applying this result to L=[(-A)θ]∘etA with (X1,X2,X3,X4)=(XAγ,p,(X,𝒟(A))δ,p,(X,𝒟(A))γ,p,XAδ+β-1,p), from (29)–(32) and the second estimate in (90) we deduce (103) with c30=2c2c27. This completes the proof.Remark 21.
The assumptionγ+δ<1 with γ∈(0,1) and δ∈(1-β,1) implies that γ∈(0,1-δ)⊊(0,β). Therefore (cf. Remark 11), we conclude that for (103) to hold we have to consider [(-A)θ]∘etA, ℜeθ≥1, as an operator between the intermediate spaces XAγ,p and XAɛ,p, where γ,ε∈(0,β), ɛ=δ+β-1, δ∈(1-β,1), γ+δ<1.


## 4. Hölder Regularity of Some Operator Functions

Here, we study the Hölder regularity of those operator functions that we will need in Section5. From now on, with (Z,∥·∥Z) being a complex Banach space, C([a,b];Z)=C0([a,b];Z) and Cδ([a,b];Z), δ∈(0,1), a<b, denote, respectively, the spaces of all continuous and δ-Hölder continuous functions from [a,b] into Z endowed with the norms ∥g∥0,a,b;Z=supt∈[a,b]∥g(t)∥Z and ∥g∥δ,a,b;Z=∥g∥0,a,b;Z+|g|δ,a,b;Z, where |g|δ,a,b;Z is the seminorm supa≤t1<t2≤b(t2-t1)-δ∥g(t2)-g(t1)∥Z. We endow the subspace C0δ([a,b];Z)={g∈Cδ([a,b];Z):g(a)=0}, δ∈[0,1) with the norm ∥·∥δ,a,b;Z. Further, for k∈N and δ∈(0,1) we set Ck([a,b];Z)={g∈C([a,b];Z):Dtkg∈C([a,b];Z)}, ∥g∥k,a,b;Z=∑0k‍∥Dtjg∥0,a,b;Z (Dt0=I), and Ck+δ([a,b];Z)={g∈Ck([a,b];Z):Dtkg∈Cδ([a,b];Z)}, ∥g∥k+δ,a,b;Z=∥g∥k,a,b;Z+|Dtkg|δ,a,b;Z. Recall that if 0≤δ2≤δ1≤1, then Cδ1([a,b];Z)↪Cδ2([a,b];Z) and ∥g∥δ2,a,b;Z≤max{1,(b-a)δ1-δ2}∥g∥δ1,a,b;Z, g∈Cδ1([a,b];X). Finally, given three complex Banach spaces (Xk,∥·∥Xk), k=1,2,3, and a bilinear bounded operator 𝒫 from X1×X2 to X3 with norm C0, that is, 𝒫∈ℬ(X1×X2;X3) and ∥𝒫∥ℬ(X1×X2;X3)=sup∥xk∥Xk=1,k=1,2∥𝒫(x1,x2)∥X3=C0, we denote by 𝒦 the convolution operator
(104)𝒦(v1,v2)(t)=∫0t‍𝒫(v1(t-r),v2(r))dr,t∈[0,b],b>0,
where vk:[0,b]→Xk, k=1,2. Of course, if (X1,X2)=(C,X3) and if 𝒫 is the scalar multiplication in X3, that is, 𝒫(z,x)=zx, z∈C, x∈X3, then C0=1 and 𝒦 reduces to the usual convolution operator 𝒦(v1,v2)(t)=∫0t‍v1(t-r)v2(r)dr. As usual, for every q∈[1,∞], we will denote by q′ the conjugate exponent of q.Now letX3=X and introduce the following linear operators Qj, j=1,…,6, where gj∈Cδj([0,T];X), j=1,2,5, glk∈Cδlk([0,T],Xk), l=3,6, k=1,2, g4∈Cδ4([0,T];C), y∈Yγp, Yγp∈{(X,𝒟(A))γ,p,XAγ,p}, p∈[1,∞], and t∈[0,T], T>0 as follows:
(105)[Q1g1](t):=∫0t‍e(t-s)Ag1(s)ds,(106)[Q2g2](t):=∫0t‍[(-A)1]∘e(t-s)A[g2(s)-g2(t)]ds,(107)[Q3(g31,g32)](t):=[Q2𝒦(g31,g32)](t),(108)[Q4(g4,y)](t):=[Q2(g4y)](t),(109)[Q5g5](t):=[etA-I]g5(t),(110)[Q6(g61,g62)](t):=[Q5𝒦(g61,g62)](t),
with g4y being the function from [0,T] to Yγp defined by (g4y)(t)=g4(t)y. We will find conditions on δj,δlk,δ4,γ∈(0,1), j=1,2,5, l=3,6, k=1,2, in order that Qjgj∈Cτj([0,T];X), Ql(gl1,gl2)∈Cτl([0,T];X) and Q4(g4,y)∈Cτ4([0,T];X) for opportunely chosen τj,τl,τ4∈(0,1). We emphasize of the presence of the increment g2(s)-g2(t) inside the integral defining Q2g2. As we will see, and differently from Q1, it is just this presence which makes Q2g2 well-defined for smooth enough functions g2. This is the reason why the operator Q2 as it was defined in [20, formula (4.12)] can make no sense and has to be replaced with that defined by the present (106) (cf. the appendix below). We begin our analysis on the Qj’s with the following result proven in [20, Lemma 4.1]. Since we will need it later, here, removing some misprints in [20], we report its short proof for the reader’s convenience.Lemma 22.
Letα+β>1 in (H1). Then, for every δ1∈(0,(α+β-1)/α), the operator Q1 defined by (105) maps Cδ1([0,T];X) into C0δ1([0,T];X), and for every t∈[0,T] satisfies the following estimate, where p∈(α/(α+β-1-αδ1),∞) as follows:
(111)∥Q1g1∥δ1,0,t;X≤C1(t)(∫0t‍∥g1∥δ1,0,s;Xpds)1/p.
Here C1(t) is a nondecreasing function of t depending also on α, β, δ1, and p′.Proof.
Letg1∈Cδ1([0,T];X), δ1∈(0,(α+β-1)/α), and t∈[0,T]. From (14) and the Hölder inequality with p∈(α/(α+β-1-αδ1),∞)⊊(1,∞), for any τ∈[0,t], we deduce that
(112)∥[Q1g1](τ)∥X≤c~α,β,0∫0τ‍(τ-s)(β-1)/α∥g1∥0,0,s;Xds≤c31τ[α-(1-β)p′]/(αp′)(∫0τ∥g1∥δ1,0,s;Xpds‍)1/p≤c31τ[α-(1+αδ1-β)p′]/(αp′)τδ1(∫0τ∥g1∥δ1,0,s;Xpds‍)1/p,
where c31=c~α,β,0α1/p′[α-(1-β)p′]-1/p′. Here α-(1+αδ1-β)p′>0, since p′∈(1,α/(1+αδ1-β)). For 1-1/p>1-(α+β-1-αδ1)/α=(1+αδ1-β)/α. passing to the supremum with respect to τ∈[0,t] in (112) we thus find
(113)∥Q1g1∥0,0,t;X≤c31t[α-(1+αδ1-β)p′]/(αp′)tδ1(∫0t∥g1∥δ1,0,s;Xpds‍)1/p.
Now let  (since [Q1g1](0)=0, the case t1=0 follows from (112) with τ=t2)    0<t1<t2≤t. The change of variable t-s=r in (105) leads us to [Q1g1](t2)-[Q1g1](t1)=∑k=12‍Ik;t1,t2,g1, where I1;t1,t2,g1:=∫t1t2‍erAg1(t2-r)dr and I2;t1,t2,g1:=∫0t1‍erA[g1(t2-r)-g1(t1-r)]dr. Reasoning as in (112) and using the inequality t2μ-t1μ≤(t2-t1)μ, μ∈(0,1], we get
(114)∥I1;t1,t2,g1∥X≤c31(t2-t1)[α-(1-β)p′]/(αp′)(∫t1t2‍∥g1∥δ1,0,t2-r;Xpdr)1/p≤c31t2[α-(1+αδ1-β)p′]/(αp′)(t2-t1)δ1(∫0t‍∥g1∥δ1,0,t-r;Xpdr)1/p.
Similarly, but taking advantage from g1∈Cδ1([0,T];X), we obtain
(115)∥I2;t1,t2,g1∥X≤c~α,β,0(t2-t1)δ1∫0t1‍r(β-1)/α|g1|δ1,0,t2-r;Xdr≤c31t1[α-(1+αδ1-β)p′]/(αp′)t1δ1(t2-t1)δ1(∫0t‍∥g1∥δ1,0,t-r;Xpdr)1/p.
Thus, letting c~1(t)=c31t[α-(1+αδ1-β)p′]/(αp′) from (114) and (115) it follows that
(116)∥[Q1g1](t2)-[Q1g1](t1)∥X≤c~1(t)(tδ1+1)(t2-t1)δ1(∫0t∥g1∥δ1,0,t-r;Xpdr‍)1/p.
Finally, summing up (113) and (116) and using ∫0t‍∥g1∥δ1,0,t-r;Xpdr=∫0t‍∥g1∥δ1,0,s;Xpds, we derive (111) with C1(t)=c~1(t)(2tδ1+1). This completes the proof.Remark 23.
We stress that if we renounce to its Hölder regularity, then forQ1g1   to be well-defined it suffices that α and β are as in Lemma 22 and that g1 is merely in C([0,T];X). In fact (see the last part of the proof of Corollary 14, replacing there x with g1(s)), ∥[Q1g1](t)∥X≤α(α+β-1)-1c~α,β,0∥g1∥0,0,t;Xt(α+β-1)/α, t∈[0,T].Lemma 24.
Let3α+β>3 in (H1). Then, for every δ2∈((3-2α-β)/α,1), the operator Q2 defined by (106) maps Cδ2([0,T];X) into C0ν2([0,T];X), ν2=(αδ2+2α+β-3)/α∈(0,δ2], and for every t∈[0,T] it satisfies the following estimate:
(117)∥Q2g2∥ν2,0,t;X≤C2(t)|g2|δ2,0,t;X.
Here C2(t) is a nondecreasing function of t depending also on α, β, and δ2.Proof.
Denote byα¯ the number (1-α)/α. In particular, since 3α+β>3 implies α∈(2/3,1], we have α¯∈[0,1/2). Let t∈[0,T], g2∈Cδ2([0,T];X), δ2∈((3-2α-β)/α,1), and ν2=(αδ2+2α+β-3)/α∈(0,δ2]. We notice that (αδ2+β-2)/α=ν2+α¯-1 and (αδ2+β-3)/α=ν2-2. Then, using (14) with θ=1, for every τ∈[0,t] we obtain
(118)∥[Q2g2](τ)∥X≤c~α,β,1|g2|δ2,0,τ;X∫0τ‍(τ-s)(αδ2+β-2)/αds=c32|g2|δ2,0,τ;Xτν2+α¯,
where c32=c~α,β,1(ν2+α¯)-1. Hence
(119)∥Q2g2∥0,0,t;X≤c32|g2|δ2,0,t;Xtν2+α¯.
Now let (since [Q2g2](0)=0, the case t1=0 follows from (118) with τ=t2)  0<t1<t2≤t. We have [Q2g2](t2)-[Q2g2](t1)=∑‍k=13Jk;t1,t2,g2, where for a function g:[0,T]→X we set
(120)J1;t1,t2,g:=∫0t1‍{[(-A)1]∘e(t2-s)A-[(-A)1]∘e(t1-s)A}×[g(s)-g(t1)]ds,J2;t1,t2,g:=∫0t1‍[(-A)1]∘e(t2-s)A[g(t1)-g(t2)]ds,J3;t1,t2,g:=∫t1t2‍[(-A)1]∘e(t2-s)A[g(s)-g(t2)]ds.
First, using (13) with (s,t,θ)=(t1-s,t2-s,1), s∈(0,t1), and (14) with θ=2, and letting (c33,c34)=(c~α,β,2(1-ν2)-1,c33ν2-1), we get
(121)∥J1;t1,t2,g2∥X≤c~α,β,2|g2|δ2,0,t1;X∫0t1‍[∫t1-st2-s‍ξ(β-3)/αdξ](t1-s)δ2ds≤c~α,β,2|g2|δ2,0,t1;X∫0t1‍[∫t1-st2-s‍ξ(αδ2+β-3)/αdξ]ds=c33|g2|δ2,0,t1;X∫0t1‍[(t1-s)ν2-1-(t2-s)ν2-1]ds=c34|g2|δ2,0,t1;X[t1ν2+(t2-t1)ν2-t2ν2]≤c34|g2|δ2,0,t2;X(t2-t1)ν2.
Let us turn to J2;t1,t2,g2. We first observe that the integral ∫0t1‍[(-A)1]∘e(t2-s)Ads is convergent.   For, ∥∫0t1‍[(-A)1]∘e(t2-s)Ads∥X≤c~α,β,1∫0t1‍(t2-s)(β-2)/αds≤Cα,β,t1,t2, where Cα,β,t1,t2 is equal to c~α,β,1ln[t2(t2-t1)-1] if β=1 and to α(2-α-β)-1c~α,β,1[(t2-t1)(α+β-2)/α-t2(α+β-2)/α] if β∈(0,1). Thus, we may rewrite it as -∫t2t2-t1‍[(-A)1]∘erAdr=∫t2t2-t1‍DrerAdr=e(t2-t1)A-et2A. Consequently,
(122)∥J2;t1,t2,g2∥X≤c~α,β,0[(t2-t1)(β-1)/α+t2(β-1)/α]|g2|δ2,0,t2;X(t2-t1)δ2≤c~α,β,0{1+[t2(t2-t1)-1](β-1)/α}×|g2|δ2,0,t2;X(t2-t1)(αδ2+β-1)/α≤2c~α,β,0|g2|δ2,0,t2;X(t2-t1)ν2+2α¯,
where we have used [t2(t2-t1)-1](β-1)/α≤1 and (αδ2+β-1)/α=ν2+2α¯. As far as J3;t1,t2,g2 is concerned, instead, reasoning as in the derivation of (118) we find
(123)∥J3;t1,t2,g2∥X≤c~α,β,1|g2|δ2,0,t2;X∫t1t2‍(t2-s)ν2+α¯-1ds=c32|g2|δ2,0,t2;X(t2-t1)ν2+α¯.
Then, summing up (121)–(123) and letting c~2(t)=c34+2c~α,β,0t2α¯+c32tα¯, we obtain
(124)∥[Q2g2](t2)-[Q2g2](t1)∥X≤∑k=13∥Jk;t1,t2,g2∥X≤c~2(t)|g2|δ2,0,t;X(t2-t1)ν2.
Finally, (119) and (124) yield (117) with C2(t)=c32tν2+α¯+c~2(t).Remark 25.
In particular, Lemma24 establishes that, with the exception of the case β=1 in which ν2=δ2, Q2 produces a loss of regularity equal to δ2-ν2=(3-2α-β)/α.As Corollary14, the next result will be needed to prove the equivalence between problem (170) and the fixed-point equation (179). From now on, if A-1∈ℒ(X) and g∈Cδ([0,T];X), δ∈[0,1), with A-1g we will always mean the function in Cδ([0,T];𝒟(A)) defined by (A-1g)(t)=A-1(g(t)). Notice that ∥A-1g∥δ,0,t;𝒟(A)≤∥g∥δ,0,t;X, t∈[0,T].Corollary 26.
(i)
Let2α+β>2 in (H1). Then, for every g∈Cδ([0,T];X), δ∈((2-α-β)/α,1),
(125)A-1[Q2g](t)=-∫0t‍e(t-s)A[g(s)-g(t)]ds,t∈[0,T].
(ii)
Letα+β>1 in (H1). Then, for every g∈C([0,T];X)(126)[Q2(A-1g)](t)=-∫0t‍e(t-s)A[g(s)-g(t)]ds,t∈[0,T].Proof.
Of course, it suffices to assume thatt∈(0,T]. Let us first prove (i). So, let 2α+β>2, g∈Cδ([0,T];X), δ∈((2-α-β)/α,1), and t∈(0,T], and we observe that both sides of (125) are well defined. Indeed, replacing the pair (g2,δ2) with (g,δ), from (118) we get
(127)∥[Q2g](t)∥X≤c~α,β,1α(αδ+α+β-2)-1|g|δ,0,t;Xt(αδ+α+β-2)/α.
On the other side, It,g=∫0t‍e(t-s)A[g(s)-g(t)]ds satisfies
(128)∥It,g∥X≤c~α,β,0|g|δ,0,t;X∫0t‍(t-s)(αδ+β-1)/αds≤c35|g|δ,0,t;Xt(α+αδ+β-1)/α,
where c35=α(αδ+α+β-1)-1c~α,β,0. Then, commuting A-1∈ℒ(X) with the integral signs, using (80) with θ=1, and taking into account (7), we find
(129)A-1[Q2g2](t)=A-1∫0t‍[-12πi∫Γ‍e(t-s)λA∘(λI-A)-1dλ][g2(s)-g2(t)]ds=-∫0t‍[12πi∫Γ‍e(t-s)λA-1A∘(λI-A)-1dλ][g2(s)-g2(t)]ds=-∫0t‍[12πi∫Γ‍e(t-s)λ(λI-A)-1dλ][g2(s)-g2(t)]ds.
Since (2πi)-1∫Γ‍e(t-s)λ(λI-A)-1dλ=e(t-s)A, the proof of (125) is complete. We now prove (ii). Let α+β>1, g∈C([0,T];X) and t∈(0,T]. Then, for every γ∈(2-α-β,1), the same reasonings made to derive (88), except for replacing x with g(s)-g(t), yield
(130)∥[Q2(A-1g)](t)∥X≤2c22cα,β,γc0∥A-1∥ℒ(X)1-γ∥g∥0,0,t;Xt(α+β+γ-2)/α.
Hence, [Q2(A-1g)](t) being meaningful, we obtain (126) simply applying to it formula (89) with ζ=1 and then using [(-A)0]∘e(t-s)A=e(t-s)A, s∈(0,t). In particular, a better estimate than (130) holds. For, [Q2(A-1g)](t)=-∫0t‍e(t-s)A[g(s)-g(t)]ds satisfies
(131)∥[Q2(A-1g)](t)∥X≤2c~α,β,0∥g∥0,0,t;X∫0t‍(t-s)(β-1)/αds≤2c36∥g∥0,0,t;Xt(α+β-1)/α,
where c36=α(α+β-1)-1c~α,β,0. The proof is complete.Let us now examine the operatorQ3 defined by (107). To this purpose we need the following result which is proved in [20, Corollary 3.2].Lemma 27.
Letδ3k∈(0,1), k=1,2, be such that σ3=δ31+δ32∈(0,1/p′), p∈(1/(1-δ31),∞). Then the convolution operator 𝒦 defined by (104) maps Cδ31([0,T];X1)×Cδ32([0,T];X2) into C0σ3([0,T];X), and for every t∈[0,T] satisfies the following estimate:
(132)∥𝒦(g31,g32)∥σ3,0,t;X≤t-σ3+1/p′c~3(t)∥g31∥δ31,0,t;X1(∫0t∥g32∥δ32,0,s;X2pds‍)1/p.
Here c~3(t) is a nondecreasing function of t depending also on δ31 and δ32. Further, in the cases δ31∈(0,1), δ32=0, and δ31=δ32=0, the following estimates hold, respectively, as follows:
(133)∥𝒦(g31,g32)∥δ31,0,t;X≤C0t1-δ31(1+tδ31)∥g31∥δ31,0,t;X1∥g32∥0,0,t;X2,∥𝒦(g31,g32)∥0,0,t;X≤C0t∥g31∥0,0,t;X1∥g32∥0,0,t;X2.From Lemmas24 and 27 we obtain the following Lemma 28.Lemma 28.
Letα and β be as in  Lemma 24. Then, for every δ31∈((3-2α-β)/α,1) and δ32∈(0,1) such that σ3=δ31+δ32∈((3-2α-β)/α,1/p′), p∈(1/(1-δ31),∞), the operator Q3 defined by (107) maps Cδ31([0,T];X1)×Cδ32([0,T];X2) into C0ν3([0,T];X), ν3=(ασ3+2α+β-3)/α, and for every t∈[0,T] satisfies the following estimate:
(134)∥Q3(g31,g32)∥ν3,0,t;X≤t-σ3+1/p′C2(t)c~3(t)∥g31∥δ31,0,t;X1(∫0t‍∥g32∥δ32,0,s;X2pds)1/p.Proof.
First, ifδ31∈((3-2α-β)/α,1) and p∈(1/(1-δ31),∞), then 1/p′∈(δ31,1)⊊((3-2α-β)/α,1). Consequently, the assumption σ3=δ31+δ32∈((3-2α-β)/α,1/p′), δ32∈(0,1), makes sense. Now, Lemma 27 yields 𝒦(g31,g32)∈C0σ3([0,T];X) for any pair (g31,g32)∈Cδ31([0,T];X1)×Cδ32([0,T];X2). Then, recalling that Q3(g31,g32)=Q2𝒦(g31,g32), the assertion follows from Lemma 24, with δ2 and g2 being replaced by σ3 and 𝒦(g31,g32), respectively. Finally, (134) follows from (117) and (132).We can now restore the loss of regularity produced byQ2.Proposition 29.
Let5α+2β>6 in (H1). Then, for every δ3∈((3-2α-β)/α,1/2), the operator Q3 defined by (107) maps Cδ3([0,T];X1)×Cδ3([0,T];X2) into C0δ3([0,T];X), and for every t∈[0,T] satisfies the following estimate, where p∈(1/(1-2δ3),∞) and C3(t)=C2(t)c~3(t)max{1,t(αδ3+2α+β-3)/α}:
(135)∥Q3(g31,g32)∥δ3,0,t;X≤t1-2δ3-1/pC3(t)∥g31∥δ3,0,t;X1(∫0t‍∥g32∥δ3,0,s;X2pds)1/p.Proof.
Letδ3∈((3-2α-β)/α,1/2) and let p∈(1/(1-2δ3),∞)⊊(1/(1-δ3),∞). Then, 2δ3∈((6-4α-2β)/α,1/p′)⊆((3-2α-β)/α,1/p′). We are thus in position to apply Lemma 28 with δ31=δ32=δ3 from which we deduce that Q3 maps Cδ3([0,T];X1)×Cδ3([0,T];X2) into C0ν3([0,T];X), ν3=(2αδ3+2α+β-3)/α. But, since our choice for δ3 implies ν3>δ3, we a fortiori have the fact that Q3 maps Cδ3([0,T];X1)×Cδ3([0,T];X2) into C0δ3([0,T];X). Finally, (135) follows from (134) and the estimate ∥g∥γ,0,t;X≤max{1,tδ-γ}∥g∥δ,0,t;X, g∈Cδ([0,T];X), δ≥γ.The next Lemma30 concerns the operator Q4. Its proof is similar to that of Lemma 24, but with the essential difference that the presence of y∈Yγp allows us to use estimate (79) in place of (14). As a consequence and provided to choose γ large enough, we will achieve a better result in which any loss of regularity is observed.Lemma 30.
Let2α+β>2 in (H1) and r∈[1,∞]. Then, for every δ4∈(0,1) and γ∈(3-2α-β,1) the operator Q4 defined by (108) maps Cδ4([0,T];C)×Yγr, Yγr∈{(X,𝒟(A))γ,r,XAγ,r}, into C0δ4([0,T];X), and for every t∈[0,T] satisfies the following estimate:
(136)∥Q4(g4,y)∥δ4,0,t;X≤C4(t)t(2α+β+γ-3)/α|g4|δ4,0,t;C∥y∥Yγr.
Here C4(t) is a nondecreasing function of t depending on α, β, δ4, γ and r.Proof.
Lett∈[0,T], g4∈Cδ4([0,T];C), δ4∈(0,1), and y∈Yγr, γ∈(3-2α-β,1), r∈[1,∞]. As in the proof of Lemma 24 we set α¯=(1-α)/α and we observe that, since 2α+β>2 implies α∈(1/2,1], here α¯∈[0,1). Furthermore, we denote by σα,β,γ the number (2α+β+γ-3)/α∈(0,1), so that the exponents (β+γ-2)/α and (β+γ-3)/α appearing in (79) with θ=1 and θ=2 may be rewritten, as σα,β,γ+α¯-1 and σα,β,γ-2, respectively. Then, using (79) with θ=1, we obtain
(137)∥[Q4(g4,y)](τ)∥X≤c22|g4|δ4,0,τ;C∥y∥Yγr∫0τ‍(τ-s)δ4+σα,β,γ+α¯-1ds=c37|g4|δ4,0,τ;C∥y∥Yγrτδ4+σα,β,γ+α¯,∀τ∈[0,t],
where c37=c22(δ4+σα,β,γ+α¯)-1. Hence, taking the supremum with respect to τ∈[0,t], one has
(138)∥Q4(g4,y)∥0,0,t;X≤c37|g4|δ4,0,t;C∥y∥Yγrtδ4+σα,β,γ+α¯.
Now, let (since [Q4(g4,y)](0)=0, the case t1=0 follows from (137) with τ=t2)  0<t1<t2≤t. We have [Q4(g4,y)](t2)-[Q4(g4,y)](t1)=∑‍k=13Jk;t1,t2,g4y, the Jk;t1,t2,g’s, g:[0,T]→X, being as in (120). Using (13) with (s,t,θ)=(t1-s,t2-s,1), s∈(0,t1), and (79) with θ=2, and letting (c38,c39)=(c22(1-δ4)-1,c38δ4-1), we get
(139)∥J1;t1,t2,g4y∥X≤c22|g4|δ4,0,t1;C∥y∥Yγr×∫0t1[∫t1-st2-s‍ξσα,β,γ-2dξ](t1-s)δ4ds‍≤c22|g4|δ4,0,t1;C∥y∥Yγr∫0t1‍[∫t1-st2-s‍ξδ4+σα,β,γ-2dξ]ds≤c22|g4|δ4,0,t1;C∥y∥Yγr×∫0t1‍(t2-s)σα,β,γ[∫t1-st2-s‍ξδ4-2dξ]ds≤c38|g4|δ4,0,t2;C∥y∥Yγrt2σα,β,γ×∫0t1‍[(t1-s)δ4-1-(t2-s)δ4-1]ds=c39|g4|δ4,0,t2;C∥y∥Yγrt2σα,β,γ[t1δ4+(t2-t1)δ4-t2δ4]≤c39|g4|δ4,0,t2;C∥y∥Yγrt2σα,β,γ(t2-t1)δ4.
Now, let us examine Jk;t1,t2,g4y, k=2,3. First, using (79) with θ=1, we find
(140)∥J2;t1,t2,g4y∥X≤c22|g4|δ4,0,t2;C∥y∥Yγr[∫0t1‍(t2-s)σα,β,γ+α¯-1ds](t2-t1)δ4=c40|g4|δ4,0,t2;C∥y∥Yγr[t2σα,β,γ+α¯-(t2-t1)σα,β,γ+α¯](t2-t1)δ4≤c40|g4|δ4,0,t2;C∥y∥Yγrt2σα,β,γ+α¯(t2-t1)δ4.
Instead, the same computations made to derive (137) yield
(141)∥J3;t1,t2,g4y∥X≤c22|g4|δ4,0,t2;C∥y∥Yγr∫t1t2‍(t2-s)δ4+σα,β,γ+α¯-1ds=c37|g4|δ4,0,t2;C∥y∥Yγr(t2-t1)δ4+σα,β,γ+α¯.
From (139)–(141) and ∥[Q4(g4,y)](t2)-[Q4(g4,y)](t1)∥X≤∑k=13‍∥Jk;t1,t2,g4y∥X, it follows that
(142)∥[Q4(g4,y)](t2)-[Q4(g4,y)](t1)∥X≤c~4(t)tσα,β,γ|g4|δ4,0,t;C∥y∥Yγr(t2-t1)δ4,
where c~4(t)=c39+(c37+c40)tα¯. Finally, summing up (138) and (142) we get (136) with C4(t)=c37tδ4+α¯+c~4(t). The proof is complete.Remark 31.
Notice that ifYγr=XAγ,r, then in order to be sure that the conclusions of Lemma 30 hold with y which really belongs to some intermediate space between X and 𝒟(A) we have to choose γ∈(3-2α-β,β). This is possible, provided that the stronger assumption 2α+β>3-β≥2 is satisfied. Otherwise, if 2α+β∈(2,3-β], β<1, then γ∈(3-2α-β,1)⊊[β,1) and y may be contained in 𝒟(A).Finally, for the operatorQ5 we have the following result. Again a loss of regularity is exhibited, even though of an amount smaller than that in Lemma 24 (cf. Remark 33).Lemma 32.
Let2α+β>2 in (H1). Then, for every δ5∈((2-α-β)/α,1), the operator Q5 defined by (109) maps C0δ5([0,T];X) into C0ν5([0,T];X), ν5=(αδ5+α+β-2)/α∈(0,δ5], and for every t∈[0,T] satisfies the following estimate:
(143)∥Q5g5∥ν5,0,t;X≤C5(t)|g5|δ5,0,t;X.
Here C5(t) is a nondecreasing function of t depending also on α, β, and δ5.Proof.
Letg5∈C0δ5([0,T];X), δ5∈((2-α-β)/α,1), and ν5=(αδ5+α+β-2)/α∈(0,δ5]. We still let α¯=(1-α)/α and as in Lemma 30 we have α¯∈[0,1). Further, observe that δ5+(β-1)/α=ν5+α¯∈(0,δ5]. Let t∈[0,T]. Then, using (14) and g5(0)=0, we get
(144)∥Q5g5∥0,0,t;X≤supτ∈[0,t][c~α,β,0τ(β-1)/α+1]|g5|δ5,0,τ;Xτδ5≤[c~α,β,0+t(1-β)/α]|g5|δ5,0,t;Xtν5+α¯.
Now, let  (since [Q5g5](0)=0, the case t1=0 follows from (144) and ∥[Q5g5](t2)∥X≤∥Q5g5∥0,0,t2;X)   0<t1<t2≤t. We have [Q5g5](t2)-[Q5g5](t1)=∑k=13‍Uk;t1,t2,g5, where for a function g:[0,T]→X we let
(145)U1;t1,t2,g:=et2A[g(t2)-g(t1)],U2;t1,t2,g:=(et2A-et1A)g(t1),U3;t1,t2,g:=g(t1)-g(t2).
First, since t2(β-1)/α≤(t2-t1)(β-1)/α for every β∈(0,1], we deduce that
(146)∥U1;t1,t2,g5∥X≤c~α,β,0t2(β-1)/α|g5|δ5,0,t2;X(t2-t1)δ5≤c~α,β,0|g5|δ5,0,t2;X(t2-t1)ν5+α¯.
As far as U2;t1,t2,g5 is concerned, instead, rewriting et2A-et1A as -∫t1t2‍[(-A)1]∘erAdr and using both g5(0)=0 and (αδ5+β-2)/α=ν5-1, it follows that
(147)∥U2;t1,t2,g5∥X≤c~α,β,1|g5|δ5,0,t1;Xt1δ5∫t1t2‍r(β-2)/αdr≤c~α,β,1|g5|δ5,0,t1;X∫t1t2‍rν5-1dr≤c~α,β,1ν5-1|g5|δ5,0,t1;X(t2ν5-t1ν5)≤c~α,β,1ν5-1|g5|δ5,0,t1;X(t2-t1)ν5.
Then, since ∥U3;t1,t2,g5∥X≤|g5|δ5,0,t2;X(t2-t1)δ5, from (146) and (147) we find
(148)∥[Q5g5](t2)-[Q5g5](t1)∥X≤∑k=13∥Uk;t1,t2,g5∥X≤c~5(t)|g5|δ5,0,t;X(t2-t1)ν5,
where c~5(t)=c~α,β,0tα¯+c~α,β,1ν5-1+tδ5-ν5. Summing up (144) and (148) we obtain (143) with C5(t)=[c~α,β,0+t(1-β)/α]tν5+α¯+c~5(t). This completes the proof.Remark 33.
Thus, with the exception ofβ=1, Q5 produces a loss of regularity equal to δ5-ν5=(2-α-β)/α≤(3-2α-β)/α. In this sense  Q5  behaves better than Q2.Remark 34.
Notice that, under the weaker assumptionsα+β>1 and g5∈C([0,T];X), (86) with x=g5(t), t∈[0,T], yields A-1[Q5g5](t)=[Q5(A-1g5)](t)=∫0t‍e(t-s)Ag5(t)ds.Similarly as we have done in Proposition29 for restoring the loss of regularity produced by Q2, we now show how Lemma 27 allows to restore that produced by Q5. We begin with the following version of Lemma 28 relative to Q6, and which is obtained combining Lemma 27 with Lemma 32 instead of Lemma 24.Lemma 35.
Letα and β be as in  Lemma 32. Then, for every δ61∈((2-α-β)/α,1) and δ62∈(0,1) such that σ6=δ61+δ62∈((2-α-β)/α,1/p′), p∈(1/(1-δ61),∞), the operator Q6 defined by (110) maps Cδ61([0,T];X1)×Cδ62([0,T];X2) into C0ν6([0,T];X), ν6=(ασ6+α+β-2)/α, and for every t∈[0,T] satisfies the following estimate:
(149)∥Q6(g61,g62)∥ν6,0,t;X≤t-σ6+1/p′C5(t)c~3(t)∥g61∥δ61,0,t;X1(∫0t∥g62∥δ62,0,s;X2pds‍)1/p.Proof.
First, ifδ61∈((2-α-β)/α,1) and p∈(1/(1-δ61),∞), then 1/p′∈(δ61,1)⊊((2-α-β)/α,1). Consequently, the assumption σ6=δ61+δ62∈((2-α-β)/α,1/p′)  makes sense, provided to choose δ62∈(0,1) small enough. Lemma 27 then yields 𝒦(g61,g62)∈C0σ6([0,T];X) for any pair (g61,g62)∈Cδ61([0,T];X1)×Cδ62([0,T];X2). Then, since Q6(g61,g62)=Q5𝒦(g61,g62), the assertion follows from Lemma 32, with the pair (δ5,g5) being replaced by (σ6,𝒦(g61,g62)). Finally, (149) follows from (143) and (132).From Lemma35 we obtain the analogous of Proposition 29 for Q6.Proposition 36.
Let3α+2β>4 in (H1). Then, for every δ6∈((2-α-β)/α,1/2), the operator Q6 defined by (110) maps Cδ6([0,T];X1)×Cδ6([0,T];X2) into C0δ6([0,T];X), and for every t∈[0,T] satisfies the following estimate, where p∈(1/(1-2δ6),∞) and C6(t)=C5(t)c~3(t)max{1,t(αδ6+α+β-2)/α}:
(150)∥Q6(g61,g62)∥δ6,0,t;X≤t1-2δ6-1/pC6(t)∥g61∥δ6,0,t;X1(∫0t‍∥g62∥δ6,0,s;X2pds)1/p.Proof.
Letδ6∈((2-α-β)/α,1/2) and p∈(1/(1-2δ6),∞)⊊(1/(1-δ6),∞). Then, 2δ6∈((4-2α-2β)/α,1/p′)⊆((2-α-β)/α,1/p′) and we can apply Lemma 35 with δ6k=δ6, k=1,2. We thus deduce that Q6 maps Cδ6([0,T];X1)×Cδ6([0,T];X2) into C0ν6([0,T];X), ν6=(2αδ6+α+β-2)/α. But, since δ6>(2-α-β)/α implies ν6>δ6, we a fortiori have the fact that Q6 maps Cδ6([0,T];X1)×Cδ6([0,T];X2) into C0δ6([0,T];X). Finally, (150) follows from (149) and ∥Q6(g61,g62)∥δ6,0,t;X≤max{1,tν6-δ6}∥Q6(g61,g62)∥ν6,0,t;X.In Section6 we will also encounter Q5 acting on functions which enjoy some space regularity, that is, functions g5 which are Hölder continuous in time with values on Yγr∈{(X,𝒟(A))γ,r,XAγ,r}. In this case Lemma 32 can be refined, and the loss of regularity produced by Q5 is naturally restored by the additional condition of space regularity on g5. In some sense, the forthcoming Corollary 38 is the analogous of Lemma 30, where the function g4y involved in the definition of Q4(g4,y) (cf. (108)) was of class Cδ4([0,T];Yγr).Lemma 37.
Letα+β>1 in (H1) and Yγr∈{(X,𝒟(A))γ,r,XAγ,r}, γ∈(2-α-β,1), r∈[1,∞]. Then, for every δ5∈(0,(α+β+γ-2)/α], the operator Q5 defined by (109) maps Cδ5([0,T];Yγr) into C0δ5([0,T],X), and for every t∈[0,T] satisfies the following estimate:
(151)∥Q5g5∥δ5,0,t;X≤c41t(α+β+γ-2-αδ5)/α(2tδ5+1)∥g5∥δ5,0,t;Yγr.
Here c41 is a positive constant depending on α, β, γ, and r.Proof.
Letγ∈(2-α-β,1)⊆(1-β,1) and let χα,β,γ be the number (α+β+γ-2)/α∈(0,1), so that the exponent (β+γ-2)/α in (79) with θ=1 is equal to χα,β,γ-1. Let g5∈Cδ5([0,T];Yγr), δ5∈(0,χα,β,γ], r∈[1,∞]. Since [Q5g5](0)=0, we assume that t∈(0,T] and we observe that, due to Propositions 6 and 12, [Q5g5](t) is rewritten as follows:
(152)[Q5g5](t)=[etA-I]g5(t)=limɛ→0+[etA-eɛA]g5(t)=limɛ→0+∫ɛt‍DsesAg5(t)ds=-limɛ→0+∫ɛt‍[(-A)1]∘esAg5(t)ds=-∫0t‍[(-A)1]∘esAg5(t)ds.
Indeed, for every ɛ∈[0,t) and x∈Yγr, (79) with θ=1 yields
(153)∥∫ɛt‍[(-A)1]∘esAxds∥X≤c22∥x∥Yγr∫ɛt‍sχα,β,γ-1ds≤c41∥x∥Yγr(t-ɛ)χα,β,γ,
where c41=c22χα,β,γ-1. From (152) and (153) with (ɛ,t,x)=(0,τ,g5(τ)) we thus get
(154)∥Q5g5∥0,0,t;X=supτ∈[0,t]∥[Q5g5](τ)∥X≤c41∥g5∥0,0,t;Yγrtχα,β,γ.
Now, let 0≤t1<t2≤t. From (152) it follows that [Q5g5](t2)-[Q5g5](t1)=-∑k=12‍Vk;t1,t2,g5, where for every function g:[0,T]→Yγp we have set
(155)V1;t1,t2,g≔∫0t1‍[(-A)1]∘esA[g(t2)-g(t1)]ds,V2;t1,t2,g:=∫t1t2‍[(-A)1]∘esAg(t2)ds.
Hence, using (153) with the triplet (ɛ,t,x) being replaced by (0,t1,g5(t2)-g5(t1)) and (t1,t2,g5(t2)), respectively, we deduce that
(156)∥V1;t1,t2,g5∥X≤c41|g5|δ5,0,t2;Yγrt1χα,β,γ(t2-t1)δ5,∥V2;t1,t2,g5∥X≤c41∥g5∥0,0,t2;Yγr(t2-t1)χα,β,γ.
As a consequence, since δ5∈(0,χα,β,γ],
(157)∥[Q5g5](t2)-[Q5g5](t1)∥X≤c41tχα,β,γ-δ5(tδ5+1)∥g5∥δ5,0,t;Yγr(t2-t1)δ5.
Summing up (154) and (157), we obtain (151). The proof is complete.Since in Lemma37 it is not required that g5(0)=0, the special case of the constant function g5(t)=x∈Yγp, t∈[0,T], is admissible, and we obtain the following result.Corollary 38.
Letα, β, and Yγr be as in  Lemma 37, and let x∈Yγr, γ∈(2-α-β,1), and r∈[1,∞]. Then, for every δ7∈(0,(α+β+γ-2)/α], the function [Q7x](·):=(e·A-I)x belongs to C0δ7([0,T];X), and for every t∈[0,T] satisfies the estimate
(158)∥Q7x∥δ7,0,t;X≤c41t(α+β+γ-2-αδ7)/α(tδ7+1)c41∥x∥Yγr.Proof.
Letg5(t)=x in the proof of Lemma 37, and observe that V1,t1,t2,g5 reduces to the zero element of X. Estimate (158) then follows from (154) and the second estimate  in (156).For later purposes, we conclude the section with the following remark.Remark 39.
The condition5α+2β>6 in (H1) required in Proposition 29 is the strongest among the conditions for the pair (α,β) required in Corollary 14 and the other results of this section. Indeed,
(159)5α+2β>6⟹3α+2β>6-2α≥4⟹3α+β>4-β≥3⟹2α+β>3-α≥2⟹α+β>2-α≥1.
Hence, if 5α+2β>6, then Corollary 14 and all the results from Lemma 22 to Corollary 38 are applicable. Next we will make large usage of this fact, but we warn the reader that, for brevity and regarding it as acquired, we will not mention it anymore.


## 5. Application to Maximal Time Regularity

The results of the previous sections are here applied to correct, refine, and extend the results in [20] concerning the maximal time regularity of the solutions to a class of degenerate abstract evolution equations. Let (X,∥·∥X) and (Z,∥·∥Z) be two complex Banach spaces, and consider the following degenerate first-order integrodifferential Cauchy problem for v:IT→X, where IT=[0,T], T>0, and n1,n2∈N:
(160)Dt(Mv(t))=[λ0M+L]v(t)+∑i1=1n1‍𝒦(ki1,Li1v)(t)+∑i2=1n2‍hi2(t)yi2+f(t),t∈IT,Mv(0)=Mv0.
Here 𝒦 is the convolution operator (104) in which (X1,X2,X3)=(Z,X,X), whereas M, L, and Li1, i1=1,…,n1, are closed single-valued linear operators from X to itself, whose domains fulfill the relation 𝒟(L)⊆⋂i1=1n1‍[𝒟(M)∩𝒟(Li1)]. Further, we assume that
(161)LadmitsacontinuousinverseoperatorL-1∈ℒ(X),i.e.,0∈ρ(L),
whereas we allow M to have no bounded inverse. Hence, in general, A:=LM-1 is only the m. l. operator defined by
(162)𝒟(A)={x∈𝒟(M-1):L(M-1x)≠∅}={x∈ℛ(M):M-1x∩𝒟(L)≠∅}={x∈ℛ(M):thereexistsy∈𝒟(L)suchthaty∈M-1x}={x∈ℛ(M):x=Myforsomey∈𝒟(L)}=M(𝒟(L)),Ax=⋃y∈M-1x∩𝒟(L)‍Ly={Ly:y∈𝒟(L)suchthatx=My},x∈𝒟(A).
Therefore, problem (160) can not be reduced, via the change of unknown u=Mv, to an integrodifferential problem related to single-valued linear operators. On the contrary, due to (161) and the closed graph theorem, ML-1,Li1L-1∈ℒ(X), i1=1,…,n1. As far as the data vector (λ0,v0,k1,…,kn1,h1,…,hn2,y1,…,yn2,f) is concerned, at the moment, we only assume λ0∈C, v0∈𝒟(M), ki1:IT→Z, hi2:IT→C, yi2∈X, il=1,…,nl, l=1,2, and f:IT→X, in order that (160) makes sense in X. This minimal assumptions will be refined later. In general, only strict solutions v to (160) shall be investigated, where (cf. [22, 23]) by a strict solution v to (160) we mean that, 𝒟(L) being endowed with the graph norm ∥·∥𝒟(L)=∥·∥X+∥L·∥X, v∈C(IT;𝒟(L)), Mv∈C1(IT;X), and (160) holds. Clearly, if M-1 is really a m. l. operator, then Mv(0)=Mv0 does not necessarily mean v(0)=v0, but only v(0)-v0∈M-10. As we will see below, if v0∈𝒟(L) and the data ki1, hi2, yi2 and f, il=1,…,nl, l=1,2, satisfy suitable assumptions, then for a strict solution v to (160) it just holds v(0)=v0. Throughout the section, Yψq, ψ∈(0,1), q∈[1,∞], will always denote one between the spaces (X,𝒟(A))ψ,q and XAψ,q, A being defined by (162). That is, Yψq∈{(X,𝒟(A))ψ,q,XAψ,q}. To avoid confusion, if more than a single Yψq is involved in some statement, that is, if we write xj∈Yψjq, j=1,…,n, n∈N, then it is understood that the same choice has been made for all the Yψjq in the sense that Yψjq=(X,𝒟(A))ψj,q or Yψjq=XAψj,q for every j=1,…,n.According to [2, Section 1.6], we recall that the M-modified resolvent set ρM(L) of L is defined to be the set {z∈C:(zM-L)-1∈ℒ(X)}. The bounded operator (zM-L)-1 is called the M modified resolvent of L. It is easy to prove that ρM(L)⊆ρ(A) and that M(zM-L)-1=(zI-A)-1, z∈ρM(L) (cf. [2, Theorem 1.14]). With the notion of M-modified resolvent of L at hand, we assume that(H2)
ρM(L) contains a region Σα={z∈C:ℜez≥-c(|ℑmz|+1)α,ℑmz∈R}, α∈(0,1], c>0, and for some exponent β∈(0,α] and constant C>0 the estimate ∥M(λM-L)-1∥ℒ(X)≤C(|λ|+1)-β holds for every λ∈Σα.Before we proceed with our analysis we remark that, due to the wide range of choices for the data vector, problem (160) contains many subcases at its interior. So, in spite of the case when at least one between the ki’s is different from zero and problem (160) is really an integrodifferential one, the choice ki1=0, i1=1,…,n1, yields to consider also various nonintegrodifferential degenerate problems. For instance, those corresponding to λ0=ki1=hi2=0 and λ0=ki1=f=0, il=1,…,nl, l=1,2, respectively:
(163)Dt(Mv(t))=Lv(t)+f(t),t∈IT,Mv(0)=Mv0,(164)Dt(Mv(t))=Lv(t)+∑i2=1n2hi2(t)yi2,t∈IT,Mv(0)=Mv0.
Although (164) differs from (163) only in the fact that f is replaced with ∑i2=1n2‍hi2(t)yi2; nevertheless a very different result is achieved when the yi2’s are assumed to belong to Yγi2r, at least for opportunely chosen γi2∈(0,1), i2=1,…,n2. As we will see (cf. Remark 51 and Theorem 56), in this situation the loss of time regularity for the pair (Lv,DtMv) with respect to that of f, typical of the case β<1 in (H2) (see [21, Theorem 9], [2, Theorem 3.26], and [22, Theorem 7.2]), can be restored in order that (Lv,DtMv) possesses the maximal time regularity which is the minimal between the time regularities of the hi2’s. The same phenomenon is carried over into the integrodifferential case for the following problems, corresponding to λ0=hi2=0, i2=1,…,n2, and λ0=f=0:
(165)Dt(Mv(t))=Lv(t)+∑i1=1n1𝒦(ki1,Li1v)(t)+f(t),Mv(0)=Mv0,(166)Dt(Mv(t))=Lv(t)+∑i1=1n1𝒦(ki1,Li1v)(t)+∑i2=1n2hi2(t)yi2,Mv(0)=Mv0,t∈IT. When β<1, the loss of time regularity for the pair (Lv,DtMv) with respect to that of the vector (k1,…,kn1,f) in problem (165) (cf. [22, Theorem 7.1] and [23, Theorem 2.1] for n1=1) can be restored in problem (166) assuming that yi2∈Yγi2r, i2=1,…,n2. In this context (cf. Remark 51 and Theorem 53) the pair (Lv,DtMv) has the maximal time regularity which is the minimal between the time regularities of the ki1’s and hi2’s.We stress that, ifβ=1, then no loss of time regularity is observed and all the quoted results agree with the well-known theory of maximal regularity in spaces of continuous functions for the nondegenerate version of (160), corresponding to the case when M=I and L generates an analytic semigroup. Hence, roughly speaking, one can verify the consistency of any result on problem (160) with condition (H2) simply by letting β=1 on its statement, and then checking if it is compatible with those for the nondegenerate case. To this purpose, we recall that the question of maximal regularity for the nondegenerate (possibly nonautonomous) version of (160) has been deeply investigated by several authors. See, for instance, [4, 6–8, 10, 32] for problem (165) with (M,β,n1)=(I,1,1) and [9, 11] for problem (163) with (M,β)=(I,1).Finally, assumption (161) excludes the case of L=0 in (160), so that our results cannot be compared with those in [5, 33, 34]. There, assuming that the bilinear bounded operator 𝒫 underlying the definition of 𝒦 is the scalar multiplication in X, the problem
(167)Dtv(t)=𝒦(k1,L1v)(t)+f(t),t∈IT,v(0)=v0
is treated under the following assumptions: (i) L1 is a closed densely defined linear operator generating an analytic semigroup; (ii) k1:[0,∞)→R is absolutely Laplace transformable. Observe that, if (k1,f)=(1,0), then problem (167) reduces to the abstact wave equation Dt2v(t)=L1v(t), Dtv(0)=0, v(0)=v0, whereas when M=I and λ0=ki1=hi2=f=0, il=1,…,nl, l=1,2, problem (160) reduces to the abstract heat equation Dtv(t)=Lv(t), v(0)=v0. In other words, whereas [5, 33, 34] are concerned with the hyperbolic case, here we are concerned with the parabolic one.Let us now come back to problem (160). Of course, assumption (H2) implies that the operator A defined by (162) satisfies (H1), so that it generates a semigroup {etA}t≥0 defined by e0A=I and (9) and satisfying (14). Assuming that v0∈𝒟(L), we let
(168)w=L(v-v0)⟺v=L-1w+v0.
Then, by setting
(169)Fw(t)=λ0A-1w(t)+∑i1=1n1[𝒦(ki1,Si1w)(t)+𝒦(ki1,Li1v0)(t)]+∑i2=1n2hi2(t)yi2+v1+f(t),t∈IT,
where A-1=ML-1∈ℒ(X), Si1=Li1L-1∈ℒ(X), i1=1,…,n1, and v1=(λ0M+L)v0, we see that v is a strict solution to (160) if and only if w satisfies  (indeed, if v∈C(IT;𝒟(L)), then ∥w(t)-w(s)∥X=∥L[v(t)-v(s)]∥X≤∥v(t)-v(s)∥𝒟(L)→0 as s→t, t,s∈IT, that is,  w∈C(IT;X). Conversely, if w∈C(IT;X), then v=L-1w+v0∈𝒟(L) and ∥v(t)-v(s)∥𝒟(L)≤(∥L-1∥ℒ(X)+1)∥w(t)-w(s)∥X→0 as s→t, t,s∈IT, that is, v∈C(IT;𝒟(L)). Finally, since Mv=A-1w+Mv0, we have Mv∈C1(IT;X) if and only if A-1w∈C1(IT;X))  w∈C(IT;X), A-1w∈C1(IT;X), and solves to the following problem:
(170)Dt(A-1w(t))=w(t)+Fw(t)∈A(A-1w(t))+Fw(t),t∈IT,A-1w(0)=0(i.e.,w(0)∈A0).
Now let 2α+β>2, and assume that ki1∈Cηi1(IT;Z), hi2∈Cσi2(IT;C), and f∈Cμ(IT;X), where ηi1,σi2,μ∈(2-α-β/α,1), il=1,…,nl, l=1,2. Then, if w∈C(IT;X) is a solution to (170) such that A-1w∈C1(IT;X), the function Fw satisfies
(171)Fw∈Cδ(IT;X),δ=minik=1,…,nk,k=1,2{ηi1,σi2,μ}∈(2-α-βα,1).
Indeed, δ being the smallest Hölder exponent, for every il=1,…,nl, l=1,2, we have A-1w,hi2yi2,f∈Cδ(IT;X) and 𝒦(ki1,Si1w), 𝒦(ki1,Li1v0)∈C0ηi1(IT;X)↪C0δ(IT;X) (cf. Lemma 27 for the case (δ31,δ32,X1,X2)=(ηi1,0,Z,X) with the pair (g31,g32) being replaced by  (in fact, since Si1=Li1L-1∈ℒ(X), i1=1,…,n1, if w∈C(IT;X), then Si1w∈C(IT;X), whereas the constant functions κi1(t)=Li1v0, t∈IT, i1=1,…,n1, obviously belong to C(IT;X))   (ki1,Si1w) and (ki1,Li1v0), resp.). Consequently (cf. [2, Theorem 3.7 and Remark p. 54] with u0=0), the solution A-1w to the multivalued evolution problem Dt(A-1w)∈A(A-1w)+Fw, A-1w(0)=0 is necessarily of the form
(172)A-1w(t)=[Q1Fw](t),t∈IT,
with Q1 being the operator defined by (105). Further (cf. [2, Remark p. 55] with u0=0, and where A∘etA stands for DtetA=-[(-A)1]∘etA) the derivative of A-1w is given by
(173)Dt(A-1w(t))=etAFw(t)-[Q2Fw](t),t∈IT∖{0},
with Q2 being the operator in (106). Notice that Q2Fw is well defined by virtue of (127) with g=Fw. Now let yi2∈Yγi2r and v1+f(0)∈Yφr where γi2,φ∈(1-β,1), i2=1,…,n2, and r∈[1,∞]. Since A-1w(0)=𝒦(ki1,Si1w)(0)=𝒦(ki1,Li1v0)(0)=0, i1=1,…,n1, from (169) it thus follows that Fw(0):=x0 is independent on w and
(174)x0=∑i2=1n2hi2(0)yi2+v1+f(0)∈Yγr,γ=mini2=1,…,n2{γi2,φ}∈(1-β,1).
Indeed (cf. (20) or (38)), we have Yγi2r↪Yγr, i2=1,…,n2, and Yφr↪Yγr, the embeddings being equalities for those between the numbers γ1,…,γn2 and φ which are equal to γ. Then, under these assumptions on the data, formula (173) for Dt(A-1w(t)) can be extended until t=0. For, we have limt→0+Dt(A-1w(t))=x0∈A0+x0 and the differential equation in (170) is satisfied even at t=0. To see this, we observe that
(175)∥Dt(A-1w(t))-x0∥X≤I1(t)+I2,w(t)+I3,w(t),t∈IT∖{0},
where I1(t)=∥(etA-I)x0∥X, I2,w(t)=∥etA[Fw(t)-x0]∥X, and I3,w(t)=∥[Q2Fw](t)∥X. First, from Proposition 6 we get limt→0+I1(t)=0. On the other side, using Fw∈Cδ(IT;X), δ∈((2-α-β)/α,1)⊆((1-β)/α,1), we obtain
(176)I2,w(t)≤c~α,β,0∥Fw∥δ,0,t;Xt(αδ+β-1)/α,t∈IT∖{0},
so that limt→0+I2,w(t)=0. Finally, (127) with g=Fw yields limt→0+I3,w(t)=0, too. Formula (173) thus holds at t=0 with Dt(A-1w(0))=limt→0+Dt(A-1w(t))=x0.Remark 40.
In [2, Remark p. 55], formula (173) was extended until t=0 only under the more restrictive assumption x0∈XAγ,∞, γ∈(1-β,1). Indeed [24, Proposition 5.2] was not available at the time of [2] and only the strong continuity of {etA}t≥0 in the X-norm on the spaces XAγ,∞, γ∈(1-β,1), was known (cf. [2, Theorem 3.3]). Notice that in the case of problem (163) the element x0 reduces to Lv0+f(0), so that in the nondegenerate case (M,β)=(I,1) we get back the classical assumption Lv0+f(0)∈(X,𝒟(L))γ,r, γ∈(0,1), r∈[1,∞] (see, for instance, [9, Theorem 4.3.1(iii)] and [11, Theorem 4.5]).Since (170) implies that w(t)=Dt(A-1w(t))-Fw(t), from (173) we thus find that
(177)w(t)=[Q7x0](t)+(etA-I)[Fw(t)-x0]-[Q2Fw](t),t∈IT,
where, according to the notation in Corollary 38, we have set [Q7x0](t)=(etA-I)x0. In particular, w(0)=0. We conclude that, under the previous assumptions on the pair (α,β) and on the data vector (k1,…,kn1,h1,…,hn2,y1,…,yn2,f,v1), if w∈C(IT;X) solves (170), then necessarily w∈C0(IT;X). As a consequence (cf. (168)), the strict solution v to (160) satisfies the initial condition just in the sense v(0)=v0.Introduce the functionsf~:IT→X and h~i2:IT→Yγi2r, i2=1,…,n2, defined by
(178)f~(t)=f(t)-f(0),h~i2(t)=[hi2(t)-hi2(0)]yi2,t∈IT.
Then, replacing Fw with the right-hand side of (169), using (174), and recalling the definitions of the operators Qj, j=2,…,6, in (106)–(110), from (177) we deduce that w∈C0(IT;X) solves the fixed-point equation
(179)w=w0+w1+Rw,
the functions wl, l=0,1, and the operator Rw being defined by
(180)w0:=Q7x0+∑i1=1n1Q6(ki1,Li1v0)+∑i2=1n2Q5h~i2+Q5f~,(181)w1:=-∑i1=1n1Q3(ki1,Li1v0)-∑i2=1n2Q4(hi2,yi2)-Q2f,(182)Rw:=λ0[Q5(A-1w)-Q2(A-1w)]+∑i1=1n1[Q6(ki1,Si1w)-Q3(ki1,Si1w)].
Conversely, let w∈C0(IT;X) be a solution to the fixed-point equation (179), and assume that the pair (α,β) and the data vector (k1,…,kn1,h1,…,hn2,y1,…,yn2,f,v1) satisfy the assumptions below (170) and (173). Then, as before, 𝒦(ki1,Si1w),𝒦(ki1,Li1v0)∈C0δ(IT;X) and hi2yi2,f∈Cδ(IT;X), il=1,…,nl, l=1,2, δ∈((2-α-β)/α,1) being as in (171). We apply A-1∈ℒ(X) to both sides of (179), and we show that A-1w satisfies (172) with Fw∈C(IT;X) as in (169), so that A-1w is a solution to problem (170). To this purpose, we take into account Corollaries 14 and 26. Let t∈IT. First (cf. Remark 34 and recall that Q6(·,·)=Q5𝒦(·,·)), using (86), (174), and (178), we get
(183)A-1w0(t)=∫0t‍e(t-s)A[x0+∑i1=1n1𝒦(ki1,Li1v0)(t)+∑i2=1n2h~i2(t)+f~(t)]ds=∫0t‍e(t-s)A[v1+∑i1=1n1𝒦(ki1,Li1v0)(t)+∑i2=1n2hi2(t)yi2+f(t)]ds.Instead, due to the definition ofQ3 and Q4, using (125) we obtain
(184)A-1w1(t)=∫0t‍e(t-s)A{∑i1=1n1[𝒦(ki1,Li1v0)(s)-𝒦(ki1,Li1v0)(t)]+∑i2=1n2[hi2(s)-hi2(t)]yi2+f(s)-f(t)}ds.
Therefore, from (183), (184), and the definition (105) of Q1 it follows that
(185)A-1[w0+w1](t)=[Q1(v1+∑i1=1n1𝒦(ki1,Li1v0)+∑i2=1n2hi2yi2+f)](t),
the left-hand side being well-defined due to Remark 23. As far as A-1[Rw](t) is concerned, we first observe that, w being in C(IT;X), from formula (126) and Remark 34 it follows that [Q2(A-1w)](t) and [Q5(A-1w)](t) are both well defined and equal to -∫0t‍e(t-s)A[w(s)-w(t)]ds and ∫0t‍e(t-s)Aw(t)ds, respectively. Consequently
(186)[Q5(A-1w)-Q2(A-1w)](t)=∫0t‍e(t-s)Aw(s)ds=[Q1w](t).
Hence, commuting A-1∈ℒ(X) with both the integral sign and the semigroup, one has
(187)A-1[Q5(A-1w)-Q2(A-1w)](t)=[Q1(A-1w)](t).
Similarly, since Remark 34 and formula (125) yield
(188)A-1[Q6(ki1,Si1w)](t)=∫0t‍e(t-s)A𝒦(ki1,Si1w)(t)ds,A-1[Q3(ki1,Si1w)](t)=-∫0t‍e(t-s)A[𝒦(ki1,Si1w)(s)-𝒦(ki1,Si1w)(t)]ds,
we find that
(189)A-1[Q6(ki1,Si1w)-Q3(ki1,Si1w)](t)=[Q1𝒦(ki1,Si1w)](t),i1=1,…,n1. In conclusion, from (187) and (189) it follows that
(190)A-1[Rw](t)=[Q1(λ0A-1w+∑i1=1n1𝒦(ki1,Si1w))](t).
Summing up (185) and (190), we finally obtain A-1w(t)=[Q1Fw](t), Fw being as in (169). This completes the proof of the equivalence between problem (170) and the fixed point equation (179), provided that the data satisfy the mentioned assumptions.Remark 41.
We can summarize the previous reasonings as follows: problem (160) has been reduced to the fixed-point equation (179) for the new unknow w=L(v-v0), v0∈𝒟(L). This fixed-point argument is similar to that first successfully applied in [4, 7, 8, 32] to problem (165) with (M,β,n1)=(I,1,1) and then generalized in [23] to the degenerate case. A different approach has been followed in [6, 10] for the nondegenerate case and in [22] for the degenerate one. There, assuming that k1 is absolutely Laplace transformable (cf. [6, 22]) or of bounded variation (cf. [10]), problem (165) with n1=1 is solved by constructing its relative resolvent operator. We quote also [35] where the method of constructing the fundamental solution for the equation without the integral term is applied to a class of  concrete degenerate integrodifferential equations.From now on, for5α+2β>6, β∈(0,α], α∈(0,1], and ν∈((3-2α-β)/α,1), Iα,β,ν⊆((3-2α-β)/α,1/2)⊆(0,1/2) will denote the interval defined by
(191)Iα,β,ν={(3-2α-βα,ν],ifν∈(3-2α-βα,12),(3-2α-βα,12),ifν∈[12,1).
Clearly, if ν,ρ∈((3-2α-β)/α,1), ν≤ρ, then Iα,β,ν⊆Iα,β,ρ.Lemma 42.
Assume (161), and let 5α+2β>6 in (H2). Assume that ki1∈Cηi1(IT;Z), ηi1∈((3-2α-β)/α,1), i1=1,…,n1, and let η=mini1=1,…,n1ηi1. Then, for every fixed δ∈Iα,β,η, the operator R defined by (182) maps continuously Cδ(IT;X) into C0δ(IT;X), and for every t∈IT satisfies the following estimate, where p∈(1/(1-2δ),∞):
(192)∥Rw∥δ,0,t;X≤c42(T)(∫0t∥w∥δ,0,s;Xpds‍)1/p,w∈Cδ(IT;X).
Here c42(T) is a positive constant depending only on T, λ0, α, β, ηi1, δ, p, ∥ki1∥ηi1,0,T;Z and ∥Si1∥ℒ(X), i1=1,…,n1.Proof.
Letki1∈Cηi(IT;Z), ηi1∈((3-2α-β)/α,1), i1=1,…,n1, and let us fix an arbitrary number δ∈Iα,β,η, where η=mini1=1,…,n1ηi. In particular, since δ≤η≤ηi, we have ki1∈Cδ(IT;Z) with ∥ki1∥δ,0,t;Z≤max{1,tηi1-δ}∥ki1∥ηi1,0,t;Z, i1=1,…,n1. Now let w∈Cδ(IT;X) and t∈IT. First, formula (186) being applicable, we rewrite (182) as
(193)Rw=λ0Q1w+∑i1=1n1[Q6(ki1,Si1w)-Q3(ki,Siw)].
Now, we notice that 5α+2β>6 implies that
(194)α+β-1α=5α+2β-3α-22α>4-3α2α≥12.
Since (1-β)/α≤(2-α-β)/α≤(3-2α-β)/α, from (194) it follows that δ∈Iα,β,η⊆((3-2α-β)/α,1/2)⊆((2-α-β)/α,1/2)⊊((1-β)/α,(α+β-1)/α), and, consequently,
(195)αα+β-1-αδ<11-2δ.
We conclude (cf. Remark 39) that Lemma 22 and Propositions 29 and 36 are applicable with δ∈Iα,β,η and p∈(1/(1-2δ),∞). Then, using estimates (111), (135), and (150) with the pair (g1,δ1) and the quintuplets (gl1,gl2,δl,X1,X2), l=3,6, being replaced, respectively, by (w,δ) and  (indeed, since Si1=Li1L-1∈ℒ(X), if w∈Cδ(IT;X), then Si1w∈Cδ(IT;X) with ∥Si1w∥δ,0,t;X≤∥Si1∥ℒ(X)∥w∥δ,0,t;X, i1=1,…,n1) (ki1,Si1w,δ,Z,X),   i1=1,…,n1, from (193) we finally obtain
(196)∥Rw∥δ,0,t;X≤∥λ0Q1w∥δ,0,t;X+∑l=3,6,i1=1,…,n1∥Ql(ki1,Si1w)∥δ,0,t;X≤c42(T)(∫0t∥w∥δ,0,s;Xpds‍)1/p.
Here we have set c42(T)=|λ0|C1(T)+T1-2δ-1/p∑l=3,6,i1=1,…,n1Cl(T)∥ki1∥δ,0,T;Z∥Si1∥ℒ(X), where Cl(T), l=1,3,6, are the values at t=T of the functions Cl(t) in Lemma 22 and Propositions 29 and 36. This completes the proof.Remark 43.
Assume that in Lemma42 the Hölder exponents ηi1∈((3-2α-β)/α,1) are such that η=mini1=1,…,n1ηi belongs to ((3-2α-β)/α,1/2). In this case (cf. (191)), the choice δ=η is admissible, and the meaning of Lemma 42 is that the operator R defined by (182) preserves the minimal of the time regularities of k1,…,kn1.Corollary 44.
Let the assumptions of  Lemma42 be satisfied, and let η and R be as there. Then, for every fixed δ∈Iα,β,η, the sequence {Rn}n=0∞ (R0=I, Rn=RRn-1, n∈N) satisfies the following estimates, where w∈Cδ(IT;X) and p∈(1/(1-2δ),∞):
(197)∥Rnw∥δ,0,t;X≤[c42(T)]n(tnn!)1/p∥w∥δ,0,T;X,∥Rnw∥δ,0,t;X≤[c42O]t∈IT,n∈N∪{0}.Proof.
Reasoning as in [23, p. 468], we prove (197) by induction. Since for every fixed δ∈Iα,β,η the operator R maps Cδ(IT;X) in C0δ(IT;X), replacing w with Rnw in (192) and introducing the sequence of scalar nonnegative nondecreasing functions {φn}n=0∞ defined by φn(t)=∥Rnw∥δ,0,t;X, t∈IT, from (192) we obtain
(198)φn+1(t)≤c42(T)(∫0t|φn(s)|pds‍)1/p,φn+1(t)≤c42(0)t∈IT,n∈N∪{0}.
Then, applying to (198) an induction argument in which the first step of the induction follows from (192), we immediately deduce the following estimates:
(199)φn(t)≤[c42(T)]n(tnn!)1/p∥w∥δ,0,T;X,φ(t)≤[c4(T)]t∈IT,n∈N∪{0}.
The proof is complete.Lemma 45.
Let5α+2β>6 in (H2) and v0∈⋂i1=1n1‍𝒟(Li1). Assume that ki1∈Cηi1(IT;Z), hi2∈Cσi2(IT;C), and yi2∈Yγi2r, where ηi1,σi2∈((3-2α-β)/α,1), γi2∈(3-2α-β,1), il=1,…,nl, l=1,2, and r∈[1,∞]. Let τ1=minil=1,…,nl,l=1,2{ηi1,σi2}. Then, for every fixed δ∈Iα,β,τ1, the function w1 defined by (181) belongs to C0δ(IT;X), provided that f∈Cμ(IT;X), μ∈[δ+μα,β,1), μα,β=(3-2α-β)/α.Proof.
Let us fixδ∈Iα,β,τ1, τ1=minil=1,…,nl,l=1,2{ηi1,σi2}. Of course, ki1∈Cδ(IT;Z) and hi2∈Cδ(IT;C), il=1,…,nl, l=1,2. Then, Proposition 29 and Lemma 30 applied with the quintuplets (g31,g32,δ3,X1,X2) and the quadruplet (g4,y,δ4,γ) being replaced, respectively, by (the constant functions κi1(t)=Li1v0, t∈IT, i=1,…,n1, being obviously of class Cδ(IT;X)) (ki1,Li1v0,δ,Z,X) and (hi2,yi2,δ,γi2), imply that Q3(ki1,Li1v0),Q4(hi2,yi2)∈C0δ(IT;X), il=1,…,nl, l=1,2. Now, since δ∈Iα,β,τ1⊆((3-2α-β)/α,1/2)⊆(0,1/2), the number δ+μα,β satisfies
(200)3-2α-βα<δ≤δ+μα,β<6-3α-2β2α<1,
and assumption f∈Cμ(IT;X), μ∈[δ+μα,β,1), is meaningful. Lemma 24 with (g2,δ2)=(f,μ) then yields Q2f∈C0να,β,μ(IT;X), να,β,μ=(αμ+2α+β-3)/α. Since να,β,μ≥να,β,δ+μα,β=δ, we get Q2f∈C0δ(IT;X), too. Summing up, we get the assertion.Before considering the functionw0 in (180), we introduce the following notation. In the sequel, for 3α+2β>4, β∈(0,α], α∈(0,1], and ν∈((2-α-β)/α,1), Jα,β,ν⊆((2-α-β)/α,1/2)⊆(0,1/2) will denote the interval
(201)Jα,β,ν={(2-α-βα,ν],ifν∈(2-α-βα,12),(2-α-βα,12),ifν∈[12,1).
Notice that, since (2-α-β)/α≤(3-2α-β)/α, if the stronger condition 5α+2β>6 is satisfied, then (191) and (201) yield Iα,β,ν⊆Jα,β,ν for every fixed ν∈((3-2α-β)/α,1). The introduction of the intervals Jα,β,ν is justified by Lemma 46, which requires a weaker condition on the pair (α,β) than the one in Lemmas 42 and 45.Lemma 46.
Let3α+2β>4 in (H2), and let v0∈𝒟(L). Assume that ki1∈Cηi1(IT;X), hi2∈Cσi2(IT;C), yi2∈Yγi2r, and v1+f(0)∈Yφr, where ηi1,σi2∈((2-α-β)/α,1), γi2,φ∈(4-2α-2β,1), il=1,…,nl, l=1,2, r∈[1,∞], and v1=(λ0M+L)v0. Let γ=mini2=1,…,n2{γi2,φ} and τ0=minil=1,…,nl,l=1,2{ηi1,σi2,χα,β,γ}, where χα,β,γ=(α+β+γ-2)/α. Then, for every fixed δ∈Jα,β,τ0, the function w0 defined by (180) belongs to C0δ(IT;X), provided that f∈Cμ(IT;X), μ∈[δ+ϱα,β,1),ϱα,β=(2-α-β)/α.Proof.
Observe that (cf. (159)) all the results from Lemma 32 to Corollary 38 will be applicable. First, since 2α+2β>4-α≥3, the choice γi2,φ∈(4-2α-2β,1), i2=1,…,n2, is meaningful. Moreover, since γ=mini2=1,…,n2{γi2,φ}∈(4-2α-2β,1), the number χα,β,γ=(α+β+γ-2)/α satisfies χα,β,γ∈((2-α-β)/α,1). Hence, τ0=minil=1,…,nl,l=1,2{ηi1,σi2,χα,β,γ}∈((2-α-β)/α,1), too, and Jα,β,τ0 is well defined. Now, let δ∈Jα,β,τ0 be fixed. Due to (20) or (38), the element x0 defined by (174) belongs to Yγr, whereas the functions h~i2 defined by (178) are of class C0δ(IT;Yγi2r)↪C0δ(IT;Yγr). Then, since γ∈(4-2α-2β,1)⊆(2-α-β,1), from Lemma 37 and Corollary 38 applied with the pairs (g5,δ5) and (x,δ7) being replaced by (h~i2,δ) and (x0,δ), respectively, we deduce that Q5h~i2,Q7x0∈C0δ(IT;X), i2=1,…,n2. In addition, since the ki1’s and the constant functions κi1(t)=Li1v0 belong to Cδ(IT;X), from Proposition 36 applied with (g61,g62,X1,X2)=(ki1,Li1v0,Z,X), it follows that Q6(ki1,Li1v0)∈C0δ(IT;X), i1=1,…,n1. Finally, since δ∈Jα,β,τ0⊆((2-α-β)/α,1/2), the number δ+ϱα,β satisfies
(202)2-α-βα<δ≤δ+ϱα,β<4-α-2β2α<1,
and the assumption f∈Cμ(IT;X), μ∈[δ+ϱα,β,1), makes sense. Then, the function f~=f-f(0) being of class C0μ(IT;X), Lemma 32 applied with (g5,δ5)=(f~,μ) yields Q5f~∈C0ν~α,β,μ(IT;X), ν~α,β,μ=(αμ+α+β-2)/α. Since ν~α,β,μ≥ν~α,β,δ+ϱα,β=δ, we conclude that Q5f~∈C0δ(IT;X), too. Summing up, we get the assertion.Remark 47.
We stress that, ifβ∈(0,1) in (H2), then 0<ρα,β≤μα,β, so that in both Lemmas 45 and 46 we have to assume that f∈Cμ(IT;X) with μ>δ. This is necessary in order to restore the loss of regularity produced by the operators Q2 and Q5.We can now prove the main results of the section.Theorem 48.
Assume (161) and v0∈𝒟(L), and let 5α+2β>6 in (H2). Assume that ki1∈Cηi1(IT;Z), hi2∈Cσi2(IT;C), yi2∈Yγi2r, and v1+f(0)∈Yφr, where ηi1,σi2∈((3-2α-β)/α,1), γi2,φ∈(5-3α-2β,1), il=1,…,nl, l=1,2, r∈[1,∞], and v1=(λ0M+L)v0. Let γ=mini2=1,…,n2{γi2,φ} and τ=minil=1,…,nl,l=1,2{ηi1,σi2,χα,β,γ}, where χα,β,γ=(α+β+γ-2)/α. Then, for every fixed δ∈Iα,β,τ problem (160) admits a unique strict solution v∈Cδ(IT;𝒟(L)) satisfying v(0)=v0 and such that Lv,DtMv∈Cδ(IT;X), provided that f∈Cμ(IT;X), μ∈[δ+μα,β,1), μα,β=(3-2α-β)/α.Proof.
Of course, due to (159), the assumption γi2,φ∈(5-3α-2β,1), i2=1,…,n2, makes sense. In addition, since γ=mini2=1,…,n2{γi2,φ}∈(5-3α-2β,1), we have χα,β,γ=(α+β+γ-2)/α∈((3-2α-β)/α,1). Therefore, by virtue of the choice of the Hölder exponents ηi1 and σi2, the number τ=minil=1,…,nl,l=1,2{ηi1,σi2,χα,β,γ} belongs to ((3-2α-β)/α,1) too, and the interval Iα,β,τ is well defined. Further, the numbers η, τ1, and τ0 being as in the statements of Lemmas 42, 45, and 46, respectively, we have τ=τ0≤τ1≤η. As a consequence, since Iα,β,τ⊆Iα,β,τ1⊆Iα,β,η and Iα,β,τ⊆Jα,β,τ, all the mentioned lemmas are applicable with δ∈Iα,β,τ. To this purpose, we stress that since ((3-2α-β)/α,1)⊆((2-α-β)/α,1) and (5-3α-2β,1)⊆(4-2α-2β,1)⊆(3-2α-β,1), the conditions for the applicability of both Lemmas 45 and 46 are fulfilled. Hence, now let δ∈Iα,β,τ being fixed. First, due to Lemma 42, the operator R~=R|C0δ(IT;X), R~g=Rg, g∈C0δ(IT;X), a fortiori maps C0δ(IT;X) into itself. Then, C0δ(IT;X) being endowed with the same norm ∥·∥δ,0,T;X of Cδ(IT;X), from (197) we obtain the estimates
(203)∥R~n∥ℒ(C0δ(IT;X))≤[c42(T)]n(Tnn!)1/p,n∈N∪{0},∥R~n∥ℒ(C0δ(IT;X))≤[c42(T)(TTT)]np∈(11-2δ,∞).
In particular, (203) yields that ∑n=0∞‍R~n converges in ℒ(C0δ(IT;X)). From generalized Neumann’s Theorem it thus follows that 1∈ρ(R~), the inverse (I-R~)-1∈ℒ(C0δ(IT;X)) being precisely ∑n=0∞‍R~n. Since Lemmas 45 and 46 (both applied with (observe  here that if μ∈[δ+μα,β,1), then the exponent ν~α,β,μ in the last part of the proof of Lemma 46 satisfies ν~α,β,μ≥ν~α,β,δ+μα,β≥ν~α,β,δ+ϱα,β=δ. For, να,β,δ+μα,β=(αδ+1-α)/α=δ+(1-α)/α)f∈Cμ(IT;X), μ∈[δ+μα,β,1)⊆[δ+ρα,β,1)) imply that w0,w1∈C0δ(IT;X), we conclude that the fixed-point equation (179) admits the unique solution
(204)w=∑n=0∞R~n(w0+w1)∈C0δ(IT;X).
Observe now that the data vector (k1,…,kn1,h1,…,hn2,f,y1,…,yn2,v1+f(0)) satisfies all the assumptions which were needed to show the equivalence between the fixed-point equation (179) and problem (170). Indeed, δ≤τ and δ≤δ+μα,β≤μ imply, respectively, that ki1∈Cδ(IT;Z),hi2∈Cδ(IT;C) and f∈Cδ(IT;X),il=1,…,nl,l=1,2, whereas, as in Lemma 46, γ=mini2=1,…,n2{γi2,φ} implies that yi2,v1+f(0)∈Yγr. Therefore, since A-1∈ℒ(X), if w∈C0δ(IT;X) is the solution to the fixed-point equation (179), then A-1w∈C0δ(IT;X), too, and the function Fw defined by (169) satisfies
(205)Fw∈Cδ(IT;X),x0=Fw(0)=∑i2=1n2hi2(0)yi2+v1+f(0)∈Yγr,
where δ∈Iα,β,τ⊊(2-α-β)/α,1), γ∈(5-3α-2β,1)⊊(1-β,1), and r∈[1,∞]. Consequently, recalling (168), we have proved that problem (160) has a unique strict global solution v=L-1w+v0∈Cδ(IT;𝒟(L)) satisfying v(0)=L-1w(0)+v0=v0 and such that Lv=w+Lv0∈Cδ(IT;X). As far as the regularity of DtMv is concerned, instead, it suffices to observe that (168), (170), w∈C0δ(IT;X), and Fw∈Cδ(IT;X) yield
(206)DtMv=DtA-1w=w+Fw∈Cδ(IT;X).
The proof is complete.Remark 49.
Theorem48 improves the faulty Thereoms 5.6 and 5.7 in [20] in two aspects. First, the assumption 3α+8β>10 is weakened to    5α+2β>6. In fact, 3α+8β>10 implies that 5α+2β=3α+8β+2α-6β>10-4α≥6. Hence, in the special case α=1, the constraint β>7/8 in [20] reduces to the definitely weaker β>1/2. Second, in [20], only for n1=n2=1 and opportunely chosen γ<β, the data y1 and v1+f(0) were assumed to belong to the intermediate spaces XAγ,r, whereas here, removing the assumption γ<β and considering the general case n1,n2∈N, we allow y1,…,yn2 and v1+f(0) to belong also to the interpolation spaces (X,𝒟(A))γ,r. To emphasize how much these aspects are decisive, let α=1 in Theorem 48. Then, if β∈(1/2,2/3] and the choice XAψ,r is understood for Yψr, we have γi2,φ∈(2-2β,1)⊊[β,1), and the spaces XAγi2,r and XAφ,r, i2=1,…,n2, may be smaller than 𝒟(A). However, the choice Yψr=(X,𝒟(A))ψ,r being admissible, in this situation too we can solve problem (160) with the data in spaces larger than 𝒟(A). Further, since 2/3<7/8, in this case the results in [20] would not be applicable. These observations lead us to conclude that the more delicate approach followed in this paper with respect to that in [20, Sections 4 and 5], and especially the sharper results of the present Sections 3 and 4, yield a valuable refinement in the treatment of questions of maximal time regularity for the strict solutions to (160); of course, unless that the not too much significant case β=1 is assumed in (H2).Remark 50.
The assumption5α+2β>6 in (H2) implies that β∈((6-5α)/2,α]⊆(1/2,1] and α∈(6/7,1]. In particular, if α=1, then Theorem 48 holds with β∈(1/2,1], ηi1,σi2∈(1-β,1), γi2,φ∈(2-2β,1), il=1,…,nl, l=1,2, and μ1,β=1-β. Hence, γ∈(2-2β,1), χ1,β,γ=β+γ-1∈(1-β,β), and δ∈I1,β,τ with τ∈(1-β,β), where
(207)I1,β,τ=(1-β,τ],ifτ∈(1-β,12),I1,β,τ=(1-β,12),ifτ∈[12,β).
Clearly, if β=1, then 5α+2β>6 is redundant, and Theorem 48 holds with ηi1,σi2,γi2,φ∈(0,1),il=1,…,nl,l=1,2,μ1,1=0,γ=χ1,1,γ∈(0,1), and δ∈I1,1,τ, τ∈(0,1), where I1,1,τ=(0,τ] if τ∈(0,1/2) and I1,1,τ=(0,1/2) if  τ∈[1/2,1).Remark 51.
Observe that, if theηi1’s and σi2’s are assumed to vary in the smaller interval Uα,β:=((3-2α-β)/α,(α+β-1)/α), then φ and the γi2’s can be chosen such that τ=minil=1,…,nl,l=1,2{ηi1,σi2}. To this purpose, letting ρ=maxil=1,…,nl,l=1,2{ηi1,σi2}∈Uα,β, it suffices to take γi2,φ∈Vα,β,ρ, i2=1,…,n2, where Vα,β,ρ:=[2+αρ-α-β,1)⊊(5-3α-2β,1). Then γ=mini2=1,…,n2{γi2,φ}∈Vα,β,ρ and χα,β,γ=(α+β+γ-2)/α≥ρ. In other words, provided that the data vector (y1,…,yn2,v1+f(0)) is smooth enough, the pair (Lv,DtMv) has the maximal time regularities which is the minimal between the time regularities of the ki1’s and hi2’s.We conclude with the results which follow from Theorem48 for problems (163)–(166).Theorem 52.
Assume (161) and v0∈𝒟(L), and let 5α+2β>6 in (H2). Assume that ki1∈Cηi1(IT;Z) and Lv0+f(0)∈Yγr, where ηi1∈((3-2α-β)/α,1), i1=1,…,n1, γ∈(5-3α-2β,1), and r∈[1,∞]. Let τ=mini1=1,…,n1{ηi1,χα,β,γ}, where χα,β,γ=(α+β+γ-2)/α. Then, for every fixed δ∈Iα,β,τ problem (165) admits a unique strict solution v∈Cδ(IT;𝒟(L)) satisfying v(0)=v0 and such that Lv,DtMv∈Cδ(IT;X), provided that f∈Cμ(IT;X), μ∈[δ+μα,β,1), μα,β=(3-2α-β)/α.Proof.
Repeat the proofs of Lemmas42, 45, and 46, Corollary 44, and Theorem 48, letting there λ0=hi2=0, i2=1,…,n2. To this purpose, observe that (169) and (174) reduce to Fw(t)=∑‍i1=1n1[𝒦(ki1,Si1w)(t)+𝒦(ki1,Li1v0)(t)]+Lv0+f(t) and x0=Lv0+f(0). Consequently, (180)–(182) change to w0=Q7x0+∑‍i1=1n1Q6(ki1,Li1v0)+Q5f~, w1=-∑‍i1=1n1Q3(ki1,Li1v0)-Q2f, and Rw=∑‍i1=1n1[Q6(ki1,Si1w)-Q3(ki1,Si1w)].Theorem 53.
Assume (161) and v0∈𝒟(L), and let 5α+2β>6 in (H2). Assume that ki1∈Cηi1(IT;Z), hi2∈Cσi2(IT;C), yi2∈Yγi2r, and Lv0∈Yφr, where ηi1,σi2∈((3-2α-β)/α,1), γi2,φ∈(5-3α-2β,1), il=1,…,nl, l=1,2, and r∈[1,∞]. Let γ=mini2=1,…,n2{γi2,φ} and τ=minil=1,…,nl,l=1,2{ηi1,σi2,χα,β,γ}, where χα,β,γ=(α+β+γ-2)/α. Then, for every fixed δ∈Iα,β,τ problem (166) admits a unique strict solution v∈Cδ(IT;𝒟(L)) satisfying v(0)=v0 and such that Lv,DtMv∈Cδ(IT;X).Proof.
Letλ0=f=0 in the proofs of Lemmas 42, 45, and 46, Corollary 44, and Theorem 48. In this case, (169) and (174) reduce to Fw(t)=∑‍i1=1n1[𝒦(ki1,Si1w)(t)+𝒦(ki1,Li1v0)(t)]+∑‍i2=1n2hi2(t)yi2+Lv0 and x0=∑‍i2=1n2hi2(0)yi2+Lv0. Hence, (180)–(182) change to w0=Q7x0+∑‍i1=1n1Q6(ki1,Li1v0)+∑‍i2=1n2Q5h~i2,w1=-∑‍i1=1n1Q3(ki1,Li1v0)-∑‍i2=1n2Q4(hi2,yi2), and Rw=∑‍i1=1n1[Q6(ki1,Si1w)-Q3(ki1,Si1w)].Let us now turn to the degenerate differential problems (163) and (164).Theorem 54.
Assume (161) and v0∈𝒟(L), and let 5α+2β>6 in (H2). Assume that Lv0+f(0)∈Yγr, γ∈(5-3α-2β,1), r∈[1,∞], and let χα,β,γ=(α+β+γ-2)/α. Then, for every fixed δ∈Iα,β,χα,β,γ problem (163) admits a unique strict global solution v∈Cδ(IT;𝒟(L)) satisfying v(0)=v0 and such that Lv,DtMv∈Cδ(IT;X), provided that f∈Cμ(IT;X), μ∈[δ+μα,β,1), μα,β=(3-2α-β)/α.Proof.
Letλ0=ki1=hi2=0,il=1,…,nl,l=1,2, in problem (160) and formulae (169), (174) and, (179)–(182). Then, Fw(t)=Lv0+f(t),x0=Lv0+f(0) and w=w0+w1=Q7x0+Q5f~-Q2f. Consequently, Lemma 42 and Corollary 44 are unneeded, and the proof of Theorem 48 simplifies as follows. First, due to γ∈(5-3α-2β,1) we have χα,β,γ∈((3-2α-β)/α,1), and the interval Iα,β,χα,β,γ is well defined. Hence, let δ∈Iα,β,χα,β,γ being fixed. Since (cf. (200)) f∈Cμ(IT;X), μ∈[δ+μα,β,1)⊊((3-2α-β)/α,1), reasoning as in the last part of the proof of Lemma 45 we get Q2f∈C0δ(IT;X). Moreover (see the proof of Lemma 46), since x0∈Yγr, γ∈(5-3α-2β,1)⊆(2-α-β,1) and f~∈C0μ(IT;X), μ∈[δ+μα,β,1)⊆[δ+ϱα,β,1), ϱα,β=(2-α-β)/α, Corollary 38 and Lemma 32 applied with (x,δ7)=(x0,δ) and (g5,δ5)=(f~,δ+ϱα,β) yield Q7x0,Q5f~∈C0δ(IT;X). Summing up, we find that w∈C0δ(IT;X). The assertion then follows from v=L-1w+v0 and (cf. (206)) DtMv=w+Lv0+f.Remark 55.
We refer to [19, Theorem 5.3] for a result of both time  and space regularity for problem (163). There, provided that ψ and δ are opportunely chosen and the data satisfy assumptions similar to those in Theorem 54, it is shown that DtMv∈Cδ(IT;(X,𝒟(A))ψ,r), and that the higher is the order ψ of the interpolation space where we look for space regularity, the lower is the Hölder exponent δ of regularity in time. Notice that Lv=DtMv-f has no space regularity, unless f has too.Theorem 56.
Assume (161) and v0∈𝒟(L), and let 5α+2β>6 in (H2). Assume that hi2∈Cσi2(IT;C), yi2∈Yγi2r, and Lv0∈Yφr, where σi2∈((3-2α-β)/α,1), γi2,φ∈(5-3α-2β,1), i2=1,…,n2, and r∈[1,∞]. Let γ=mini2=1,…,n2{γi2,φ} and τ=mini2=1,…,n2{σi2,χα,β,γ}, where χα,β,γ=(α+β+γ-2)/α. Then, for every fixed δ∈Iα,β,τ, problem (164) admits a unique strict global solution v∈Cδ(IT;𝒟(L)) satisfying v(0)=v0 and such that Lv,DtMv∈Cδ(IT;X).Proof.
Letλ0=ki1=f=0, i1=1,…,n1, in problem (160) and formulae (169), (174), and (179)–(182). Then, Fw(t)=∑i2=1n2‍hi2(t)yi2+Lv0,x0=∑i2=1n2‍hi2(0)yi2+Lv0 and w=w0+w1=Q7x0+∑i2=1n2‍Q5h~i2-∑i2=1n2‍Q4(hi2,yi2). Therefore, as in Theorem 54, we do not need Lemma 42 and Corollary 44, and the proof of Theorem 48 simplifies as follows. Again, γ=mini2=1,…,n2{γi2,φ}∈(5-3α-2β,1) implies that χα,β,γ∈((3-2α-β)/α,1), so that τ=mini2=1,…,n2{σi2,χα,β,γ}∈((3-2α-β)/α,1), and the interval Iα,β,τ is well defined. Let δ∈Iα,β,τ be fixed. First (see the proof of Lemma 45), since γi2∈(5-3α-2β,1)⊆(3-2α-β,1), Lemma 30 applied with (g4,y,δ4,γ)=(hi2,yi2,δ,γi2) yields Q4(hi2,yi2)∈C0δ(IT;X), i2=1,…,n2. On the other side (see the proof of Lemma 46), since x0∈Yγr and h~i2∈C0δ(IT;Yγi2r)↪C0δ(IT;Yγr), γ∈(5-3α-2β,1)⊆(2-α-β,1), from Lemma 37 and Corollary 38 applied with (g5,δ5)=(h~i2,δ) and (x,δ7)=(x0,δ) we deduce that Q5h~i2,Q7x0∈C0δ(IT;X),i2=1,…,n2. Summing up, we find that w∈C0δ(IT;X), and the assertion again follows from v=L-1w+v0 and (cf. (206)) DtMv=w+Lv0+∑i2=1n2‍hi2yi2.


## 6. An Application to a Concrete Case

Theorem48 is here applied to determine the right functional framework where to search for the solution of an inverse problem arising in the theory of heat conduction for materials with memory. To this purpose, let Ω⊊RN,N∈N, be a bounded domain with boundary ∂Ω of class C1,1 (cf. [36, p. 94]). If Ω represents a rigid thermal body with memory, then the linearized theory of heat flow yields the following equations linking the internal energy e, the heat flux q=(q1,…,qN), and the temperature Θ (cf. [32, 37–40]):
(208)e(t,x)=e0+a(0,x)Θ(t,x)+∫0t‍Dta(t-s,x)Θ(s,x)ds,qj(t,x)=-∑i=1r1bi(0)Ci,j(x;Dx)Θ(t,x)-∑i=1r1∫0t‍Dtbi(t-s)Ci,j(x;Dx)Θ(s,x)ds,j=1,…,N,Dte(t,x)=-divxq(t,x)+g(t,x)=-∑j=1NDxjqj(t,x)+g(t,x).
Here t∈IT, IT=[0,T], T>0, x=(x1,…,xN)∈Ω, r1∈N, e0∈R, and Dt=∂/∂t, whereas the Ci,j(x;Dx)’s represent the first-order linear differential operators
(209)Ci,j(x;Dx)=∑k=1Nci,j,k(x)Dxk,x∈Ω¯,i=1,…,r1,j=1,…,N,
where ci,j,k∈C1(Ω¯;R) and Dxk=∂/∂xk, i=1,…,r1, j,k=1,…,N. According to the terminology of [39, 40], the functions a, bi, i=1,…,r1, and g are called, respectively, the energy-temperature relaxation function, the heat conduction relaxation functions, and the heat supply function and we assume that they satisfy the following conditions:
(210)Dtka(·,x)∈C(IT;R),k=0,1,2,a(0,x)≥0,x∈Ω,(211)Dtkbi∈C(IT;R),k=0,1,i=1,…,r1,g∈C1(IT×Ω;R).
Notice that, different from [32, 37–40], here the energy-temperature relaxation function a is assumed to depend also on the spatial variable x∈Ω. In physical terms, this is equivalent to say that Ω represents a rigid inhomogeneous material with memory. Furthermore, in contrast with the quoted papers where only the cases r1=1 and C1,j(x;Dx)=Dxj are treated, here we have assumed that the history record of Ω is kept by an arbitrary number r1∈N of heat conduction relaxation functions and that the Ci,j’s are the more general first-order differential operators defined in (209).By setting(212)a~j,k=∑i=1r1bi(0)ci,j,k∈C1(Ω¯;R),j,k=1,…,N,
from (208) and (209), it thus follows that the temperature Θ must satisfy the following equation:
(213)a(0,x)DtΘ(t,x)+Dta(0,x)Θ(t,x)+∫0t‍Dt2a(t-s,x)Θ(s,x)ds-g(t,x)=∑j,k=1N‍Dxj[a~j,k(x)DxkΘ(t,x)]+∑i=1r1‍∫0t‍Dtbi(t-s)∑j=1N‍DxjCi,j(x;Dx)Θ(s,x)ds.
Let us now assume that a is of the following special form:
(214)a(t,x)=∑n=12mn(x)un(t),(t,x)∈IT×Ω,
where the functions mn and un, n=1,2, satisfy the following conditions (cf. (210)):
(215)mn∈L∞(Ω),n=1,2,m1≥0,m2>0,(216)un∈C2(IT;R),n=1,2,u2(0)=0,u1(0)>0,Dtu2(0)>0.
Here, Lq(Ω)=Lq(Ω;R),q∈[1,∞], is the usual Lq space with norm ∥·∥q;Ω (cf. [36, Chapter 7]). Using m2,u1(0),Dtu2(0)>0, for t∈IT and x∈Ω we now set
(217)a0(x)=-[u1(0)]-1m2(x)Dtu2(0)<0,(218)aj,k(x)=[u1(0)]-1a~j,k(x),j,k=1,…,N,(219)L(x;Dx)=∑j,k=1NDxj[aj,k(x)Dxk]+a0(x),(220)Li(x;Dx)=[u1(0)]-1∑j=1NDxjCi,j(x;Dx),i=1,…,r1,(221)Lr1+n(x;Dx)=Lr1+n(x)=[u1(0)]-1mn(x),n=1,2,(222)ki(t)=Dtbi(t),i=1,…,r1,kr1+n(t)=-Dt2un(t),n=1,2,(223)g~(t,x)=[u1(0)]-1g(t,x),λ0=-[u1(0)]-1Dtu1(0)∈R.
Then, since (214)–(216) yield a(0,x)=m1(x)u1(0) and Dtka(t,x)=∑n=12‍mn(x)Dtkun(t), k=1,2, if we multiply both sides of (213) by [u1(0)]-1 and use (218)–(223), we are led to the following basic differential equation for the temperature Θ, where n1=r1+2:
(224)Dt[m1(x)Θ(t,x)]=λ0m1(x)Θ(t,x)+L(x;Dx)Θ(t,x)+g~(t,x)+∑i=1n1∫0t‍ki(t-s)Li(x;Dx)Θ(s,x)ds,t∈IT,x∈Ω.
We endow this differential equation with the initial condition Θ(0,x)=Θ0(x),x∈Ω, and the Dirichlet boundary condition Θ(t,x)=0, t∈IT, x∈∂Ω.We now suppress the dependence onx∈Ω, and we transform (224) in a degenerate integrodifferential Cauchy problem in a Banach space X. To this purpose, for every fixed q∈(1,∞)  and observing that  mn∈L∞(Ω) implies that ∥mnu∥q;Ω≤∥mn∥∞;Ω∥u∥q;Ω for every u∈Lq(Ω), n=1,2, we set
(225)X=𝒟(M)=𝒟(Lr1+n)=Lq(Ω),n=1,2,(226)𝒟(L)=Wq2(Ω)∩W°q1(Ω),𝒟(Li)=Wq2(Ω),i=1,…,r1,(227)M,Lr1+n∈ℒ(X),Mu=m1u,Lr1+nu=Lr1+n(x)u,u∈X,n=1,2,(228)L:𝒟(L)⊆X⟶X,Lu=L(x;Dx)u,u∈𝒟(L),(229)Li:𝒟(Li)⊆X⟶X,Liu=Li(x;Dx)u,u∈𝒟(Li),i=1,…,r1.
Here (cf. [36, Chapter 7]), Wqk(Ω)=Wqk(Ω;R), k∈N∪{0}, q∈(1,∞), denotes the usual Sobolev space endowed with the norm ∥·∥k,q;Ω ((Wq0(Ω),∥·∥0,q;Ω)=(Lq(Ω),∥·∥q;Ω)), whereas W°qk(Ω) denotes the completion of C0∞(Ω;R) in Wqk(Ω), C0∞(Ω;R) being the set of all real-valued infinitely differentiable functions having compact support in Ω. We further assume that there exists positive constant Λi, i=0,…,r1, such that for every (x,ξ)∈Ω¯×RN the following inequalities hold:
(230)∑j,k=1Nci,j,k(x)ξjξk≥Λi|ξ|2,i=1,…,r1,∑j,k=1r1bi(0)Λi≥Λ0,
where |ξ|2=∑l=1N‍ξl2. Therefore, from (212), (218), and (230) we get
(231)∑j,k=1Naj,k(x)ξjξk=[u1(0)]-1∑i=1r1bi(0)∑j,k=1Nci,j,kξjξk≥[u1(0)]-1Λ0|ξ|2.
From (225)–(231) it follows that M,L, and Li, i=1,…,n1, are closed linear operators from X to itself, and the relation 𝒟(L)⊊⋂i=1n1‍[𝒟(M)∩𝒟(Li)]=Wq2(Ω) holds. In addition, due to (212), (217), (218), and (231), from [36, Theorem 9.15 and Lemma 9.17], it follows that for every fixed q∈(1,∞) the operator L admits an inverse operator L-1∈ℒ(X;Wq2(Ω)). Hence, a fortiori, L-1∈ℒ(X) and so condition (161) is satisfied (observe also that L-1∈ℒ(X;Wq2(Ω)) implies that the norms ∥·∥2,q;Ω and ∥·∥𝒟(L)=∥·∥q;Ω+∥L·∥q;Ω are equivalent on 𝒟(L). In fact, if v∈𝒟(L), then ∥v∥2,q;Ω=∥L-1Lv∥2,q;Ω≤∥L-1∥ℒ(X;Wq2(Ω))∥v∥𝒟(L)≤ C~∥L-1∥ℒ(X;Wq2(Ω))∥v∥2,q;Ω, C~ being a positive constant depending on maxj,k=1,…,N∥aj,k∥C1(Ω¯;R)). The closed graph theorem then yield ML-1,LiL-1∈ℒ(X),i=1,…,n1. Moreover (cf. [19, formula (77)], and [41, formula (2.16)]), the following estimate holds (of course, here X=Lq(Ω;R) is replaced with the more general X=Lq(Ω;C)):
(232)∥M(λM-L)-1∥ℒ(X)≤C(|λ|+1)-β,∀λ∈Σ1,β=1q,
where Σ1={z∈C:ℜez≥-c(|ℑmz|+1),ℑmz∈R}, c being a suitable positive constant depending on q and ∥m1∥∞;Ω. Hence, condition (H2) is satisfied with X=Lq(Ω) and (α,β)=(1,1/q). Notice that, since m1 may have zeros in Ω, M-1 is in general a m. l. operator, so that A=LM-1 is determined by (cf. (162)):
(233)𝒟(A)=M(𝒟(L))={m1v:v∈𝒟(L)},Au={Lv:v∈𝒟(L)suchthatu=m1v},u∈𝒟(A).
Using the convolution operator 𝒦 in (104) in which for the bilinear operator 𝒫 we take the scalar multiplication in X, from (224)–(229) we finally obtain that the temperature Θ(t)=Θ(t,·) solves the following degenerate integrodifferential Cauchy problem in X:
(234)Dt(MΘ(t))=[λ0M+L]Θ(t)+∑i=1n1𝒦(ki,LiΘ)(t)+g~(t),t∈IT,Θ(0)=Θ0.
Now, assume for a moment that we are interested in solving the inverse problem of recovering both the temperature Θ and the memory kernels k1,…,kr1 in (234). Clearly, due to (222), if we recover k1,…,kr1, then the heat conduction relaxation functions b1,…,br1 will be known too, unless of the r1 arbitrary constants bi(0), i=1,…,r1. Indeed, b1(t)=bi(0)+∫0t‍ki(s)ds, t∈IT. To solve such an inverse problem, we need r1 additional informations other than the initial condition Θ(0)=Θ0, which, in general, suffices only to guarantee the well-posedness of the direct problem of recovering Θ in (234). Suppose then that the following additional pieces of information are given:
(235)Ψj[MΘ(t)]=gj(t),t∈IT,j=1,…,r1,
where Ψj∈X*=ℒ(X;R) and gj∈C2+νj(IT;R), νj∈(0,1),j=1,…,r1. We will search for a solution vector (Θ,k1,…,kr1) of the inverse problem (234) and (235) such that Θ∈C1+δ(IT;𝒟(L)) and kj∈Cηj(IT;R), j=1,…,r1, with the Hölder exponents δ and ηj,j=1,…,r1, to be made precise in the sequel. We stress that here we will not solve completely the mentioned inverse problem. For, its detailed treatment would lead us out of the aims of this paper. Our intention here is only to highlight how the main results of Section 5 allow to determine the correct functional framework in which the solution of the inverse problem has to be searched. However, a complete treatment of the inverse problem will be the object of a future paper.Assuming thatΘ∈C1+δ(IT;𝒟(L)) solves (234), we introduce the new unknown
(236)v(t,x)=DtΘ(t,x)⟺Θ(t,x)=Θ0(x)+∫0t‍v(s,x)ds.
Then, differentiating (234) with respect to time and using
(237)Dt𝒦(ki,LiΘ)(t)=Dt∫0t‍ki(t-s)LiΘ(s)ds=Dt∫0t‍ki(s)LiΘ(t-s)ds=ki(t)LiΘ(0)+∫0t‍ki(s)LiDtΘ(t-s)ds=ki(t)LiΘ0+∫0t‍ki(t-s)Liv(s)ds,
we find that v∈Cδ(IT;𝒟(L)) solves the following degenerate integrodifferential problem:
(238)Dt(Mv(t))=[λ0M+L]v(t)+∑i=1n1[𝒦(ki,Liv)(t)+ki(t)yi]+f(t),t∈IT,Mv(0)=Mv0,
where yi=LiΘ0,i=1,…,n1,f=Dtg~ and Mv0=[λ0M+L]Θ0+g~(0,·)  (indeed, since M is the multiplication operator by the function m1 independent of t, from the differential equation in (234) with t=0 we get Mv(0)=MDtΘ(0)=[λ0M+L]Θ(0)+g~(0)).  Of course, (238) is the special case (i1,i2,n2)=(i,i,n1), hi=ki, i=1…,n1, of problem (160).Conversely, assume thatv∈Cδ(IT;𝒟(L)) solves (238). Then, the function Θ defined by (236) belongs to C1+δ(IT;𝒟(L)) and solves (234). Indeed, using the fact that m1 does not depend on time and that M, L, and Li, i=1,…,n1, are closed, we obtain
(239)Dt(MΘ(t))-[λ0M+L]Θ(t)-∑i=1n1𝒦(ki,LiΘ)(t)-g~(t)=Dt[M(Θ0+∫0t‍v(s)ds)]-[λ0M+L][Θ0+∫0t‍v(s)ds]-∑i=1n1∫0t‍ki(t-s)Li[Θ0+∫0s‍v(ξ)dξ]ds-g~(0)-∫0t‍Dsg~(s)ds=Mv(t)-[λ0M+L]Θ0-∫0t‍[λ0M+L]v(s)ds-∑i=1n1∫0t‍ki(t-s)LiΘ0ds-∑i=1n1∫0t‍ki(t-s)[∫0s‍Liv(ξ)dξ]ds-g~(0)-∫0t‍f(s)ds.
Now, observe that
(240)Mv(t)=Mv0+∫0t‍Ds(Mv(s))ds=[λ0M+L]Θ0+g~(0)+∫0t‍Ds(Mv(s))ds,(241)∫0t‍ki(t-s)LiΘ0ds=∫0t‍ki(s)LiΘ0ds=∫0t‍ki(s)yids,i=1,…,n1,
whereas an application of Fubini’s theorem combined with the changes of variables ξ=s-r, r-s=τ and t-s=ζ easily yields for every i=1,…,n1 the following:
(242)∫0t‍ki(t-s)[∫0s‍Liv(ξ)dξ]ds=∫0t‍ki(t-s)[∫0s‍Liv(s-r)dr]ds=∫0t‍[∫st‍ki(t-r)Liv(r-s)dr]ds=∫0t‍[∫0t-s‍ki(t-s-τ)Liv(τ)dτ]ds=∫0t‍𝒦(ki,Liv)(t-s)ds=∫0t‍𝒦(ki,Liv)(ζ)dζ=∫0t‍𝒦(ki,Liv)(s)ds.
Therefore, replacing (240)–(242) in (239), it follows for every t∈IT that
(243)Dt(MΘ(t))-[λ0M+L]Θ(t)-∑i=1n1𝒦(ki,LiΘ)(t)-g~(t)=∫0t‍{∑i=1n1𝒦Ds(Mv(s))-[λ0M+L]v(s)-∑i=1n1[𝒦(ki,Liv)(s)+ki(s)yi]-f(s)∑i=1n1𝒦}ds,
and the latter integral is equal to zero by virtue of (238). Since from (236) it follows that Θ(0)=Θ0, we have thus shown that (234) and (238) are equivalent. Such an equivalence is the first step in solving the mentioned inverse problem of recovering the vector (Θ,k1,…,kr1) with the help of the additional information (235).Let us now apply the linear functionalΨj, j=1…,r1, to (238). Using
(244)Ψj[Dtk(Mv(t))]=Ψj[MDtk+1Θ(t)]=Dtk+1Ψj[MΘ(t)]=Dtk+1gj(t),k=0,1,
we thus find the following system of r1 equations for the r1 unknown k1,…,kr1:
(245)∑i=1r1Ψj[yi]ki(t)=Nj(t)-Ψj[Lv]-∑i=1n1Ψj[𝒦(ki,Liv)(t)],j=1,…,r1,
where we have set (recall that kr1+n=-Dtun, n=1,2, are known)
(246)Nj(t)=(Dt-λ0)Dtgj(t)-Ψj[f(t)]-∑n=12Ψj[yr1+n]kr1+n(t),j=1,…,r1.
Therefore, if the matrix 𝒰:=𝒰Ψ1,…,Ψr1y1,…,yr1=(Ψi[yj])i,j=1,…,r1 has determinant det𝒰≠0, then from Cramer’s formula it follows that the solution (k1,…,kr1) of (245) is given by
(247)kj(t)=[det𝒰]-1∑k=1r1{∑i=1n1ΨkNk(t)-Ψk[Lv]-∑i=1n1Ψk[𝒦(ki,Liv)(t)]}𝒰k,j=:R~j(v,k1,…,kr1)(t),j=1,…,r1,
with 𝒰k,j, k,j=1,…,r1, being the cofactor of the element Ψk[yj] of 𝒰 (with the convention that 𝒰1,1=1 in the case of r1=1). We have thus found a system of r1 fixed-point equations for the r1 unknown k1,…,kr1.Now, letYψr∈{(X,𝒟(A))ψ,r,XAψ,r}, ψ∈(0,1), r∈[1,∞], where A is as in (233). Assume that v0 in the initial condition Mv(0)=Mv0 belongs to 𝒟(L) and that
(248)ki∈Cηi(IT,R),f∈Cμ(IT;X),ηi,μ∈(1q′,1),i=1,…,n1,yi∈Yγip,v1+f(0)∈Yφp,γi,φ∈(1q′,1),p∈[1,∞],i=1,…,n1,
where v1=(λ0M+L)v0 and q′ is the conjugate exponent of q∈(1,∞). Then (cf. (179) with (i1,i2,n2)=(i,i,n1), (α,β,Z)=(1,1/q,R), and hi=ki, i=1,…,n1), problem (238) is equivalent to the fixed-point equation
(249)w(t)=R(w,k1,…,kr1)(t)+∑l=01wl(k1,…,kr1)(t)=:T(w,k1,…,kr1)(t),
where w=L(v-v0) and
(250)w0(k1,…,kr1)=Q7x0+∑i=1n1[Q6(ki,Liv0)+Q5k~i]+Q5f~,w1(k1,…,kr1)=-∑i=1n1[Q3(ki,Liv0)+Q4(ki,yi)]-Q2f,R(w,k1,…,kr1)=λ0[Q5(A-1w)-Q2(A-1w)]+∑i=1n1[Q6(ki,Siw)-Q3(ki,Siw)].
Here, the Qj’s, j=2,…,6, are defined by (106)–(110), Si=LiL-1, and the functions f~, k~i and Q7x0 are defined by f~(t)=f(t)-f(0), k~i(t)=[ki(t)-ki(0)]yi, and [Q7x0](t)=(etA-I)x0, respectively, where (cf. (174)) x0=∑i=1n1‍ki(0)yi+v1+f(0).Then, sincev=L-1w+v0, if we set Rj(w,k1,…,kr1)=R~j(L-1w+v0,k1,…,kr1), j=1,…,r1, and
(251)Ξ(w,k1,…,kr1)=(T(w,k1,…,kr1),R1(w,k1,…,kr1),…,Rri(w,k1,…,kr1)),
from (247) and (249) we deduce that to solve the inverse problems (234) and (235) for the unknown vector (Θ,k1,…,kr1), it suffices to show that the fixed-point equation
(252)(w,k1,…,kr1)=Ξ(w,k1,…,kr1)
has a unique solution. In general, this will be done by proving that Ξ is a contraction map in the Banach space
(253)Zδ,η1,…,ηr1=Cδ(IT;X)×Cη1(IT;R)×⋯×Cηr1(IT;R),∥(f0,f1,…,fr1)∥Zδ,η1,…,ηr1=∥f0∥δ,0,T;X+∥f1∥η1,0,T;R+⋯+∥fr1∥ηr1,0,T;R,
at least for opportunely chosen Hölder exponents δ∈(0,1) and ηi∈(1/q′,1), i=1,…,r1, and, eventually, sufficiently small values of T>0. It is just in the choice of δ and the ηi’s that the main result of Section 5 plays a key role. The Hölder exponents have to be chosen so that the direct problem (234) in which the ki’s are assumed to be known is well posed. Due to the shown equivalence between problems (234) and (238), the well-posedness of the direct problem (234) is then a consequence of Theorem 48 and formula (236). More precisely, recalling Remark 50 for the case α=1, an application of that theorem yields the following maximal time regularity result for the solution Θ of (234).Theorem 57.
LetX, 𝒟(M), 𝒟(L), and 𝒟(Li), i=1,…,n1, n1=r1+2, r1∈N, be defined by (225) and (226) with q∈(1,2). Let M,L, and Li, i=1,…,n1, be defined by (227)–(229) through (209), (212), and (215)–(221), and let (230) and (231) be satisfied. Further, let (A,𝒟(A)) be defined by (233), and let Yψr∈{(X,𝒟(A))ψ,r,XAψ,r}, ψ∈(0,1), r∈[1,∞]. Let ηi∈(1/q′,1) and γi,φ∈(2/q′,1), i=1,…,n1, and assume that
(254)ki∈Cηi(IT;R),i=1,…,n1,Θ0∈𝒟(L),(λ0M+L)Θ0+g~(0,·)=Mv0forsomev0∈𝒟(L),LiΘ0∈Yγir,v1+Dtg~(0,·)∈Yφr,i=1,…,n1,r∈[1,∞],
where ki, i=1,…,n1,g~ and λ0 are defined by (222) and (223) through (211) and (216), whereas v1=(λ0M+L)v0. Let γ=mini=1,…,n1{γi,φ} and τ=mini=1,…,n1{ηi,γ-1/q′}, and let I1,1/q,τ⊆(1/q′,1/2) be the interval defined by (cf. (207) with β=1/q)
(255)I1,1/q,τ=(1q′,τ],ifτ∈(1q′,12),I1,1/q,τ=(1q′,12),ifτ∈[12,1).
Then, for every fixed δ∈I1,1/q,τ problem (234), or, equivalently, problem (224), admits a unique strict solution Θ∈C1+δ(IT;𝒟(L)) satisfying DtΘ(0)=v0 and such that DtMΘ,LΘ∈C1+δ(IT;X), provided that Dtg~∈Cμ(IT;X), μ∈[δ+1/q′,1).Proof.
Apply Theorem48 with (i1,i2,n2)=(i,i,n1), (α,β,Z)=(1,1/q,R), and hi=ki, i=1,…,n1, to the equivalent problem (238). Since M is the multiplication operator by the function m1 independent of t, the assertion then follows from DtΘ=v∈Cδ(IT;𝒟(L)), DtΘ(0)=v(0), DtLΘ=Lv∈Cδ(IT;X) and Dt2MΘ=DtMv∈Cδ(IT;X).Larger values ofq in Theorem 57 can be obtained assuming more smoothness and some order of vanishing for the function m1. In fact, let m1∈C1(Ω¯) be such that the following estimate holds for some positive constant K:
(256)|∇m1(x)|:={∑j=1N[Dxjm1(x)]2}1/2≤K[m1(x)]ϑ,|∇m1(x)|{∑j=1N[Dxjm1(x)]2}x∈Ω¯,ϑ∈(0,1).
Then (232) holds with β=1/q being replaced by (cf. [41, formulae (3.23) and (4.41)]):
(257)β=12-ϑ,ifq∈(2-ϑ,2),β=2q(2-ϑ),ifq∈[2,∞).
(precisely, in [41, formula (3.23)] it is shown that (|λ|+1)∥Mu∥q;Ωq(2-ϑ)/2≤Cq[∥f∥q;Ω∥Mu∥q;Ω-1+q(2-ϑ)/2+∥f∥q;Ωq(2-ϑ)/2], where u=(λM-L)-1f and q∈[2,∞). Using (cf. [41, formula (2.15)]) ∥Mu∥q;Ω≤∥m1∥∞;Ω∥u∥q;Ω≤C∥m1∥∞;Ω∥f∥q;Ω, we thus find that (|λ|+1)∥Mu∥q;Ωq(2-ϑ)/2≤Cq[(C∥m1∥∞;Ω)-1+q(2-ϑ)/2+1]∥f∥q;Ωq(2-ϑ)/2; that is, ∥M(λM-L)-1∥ℒ(X)≤{Cq[(C∥m1∥∞;Ω)-1+q(2-ϑ)/2+1]}2/[q(2-ϑ)](|λ|+1)-2/[q(2-ϑ)]).  Under (256) we thus find the following better result, where q may be greater than two.Theorem 58.
Let (256) holds, and let X, (M,𝒟(M)), (L,𝒟(L)), (Li,𝒟(Li)), i=1,…,n1, be as in  Theorem 57, but with q∈(2-ϑ,2)∪[2,4/(2-ϑ)). Let (254) be fulfilled, but with ηi∈(1-β,1) and γi,φ∈(2-2β,1), i=1,…,n1, where β is as in (257). Let γ=mini=1,…,n1{γi,φ} and τ=mini=1,…,n1{ηi,β+γ-1}, and let I1,β,τ be as in (207). Then, for every fixed δ∈I1,β,τ problem (234), or, equivalently, problem (224), admits a unique strict solution Θ∈C1+δ(IT;𝒟(L)) satisfying DtΘ(0)=v0 and such that DtMΘ,LΘ∈C1+δ(IT;X), provided that Dtg~∈Cμ(IT;X), μ∈[δ+1-β,1).Proof.
It suffices to observe that for everyϑ∈(0,1) and q∈(2-ϑ,2)∪[2,4/(2-ϑ)), the number β in (257) satisfies β>1/2. Hence, proceeding as in the proofs of Theorem 57, except for replacing there β=1/q with β as in (257), we get the assertion.



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*Source: 275494-2013-12-10.xml*