1. Introduction
Let
X 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,
()
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 operators
A whose single-valued resolvents satisfy the following estimate:
()
Here,
I is the identity operator,
C is a positive constant,
β ∈ (0,1], and
Σα is the complex region {
z ∈
C :
ℜe
z ≥ −
c(|
ℑm
z | + 1)
α,
ℑm
z ∈
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 that A 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 (ℜe λ+|ℑm λ|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 when A 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 Section 2, 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, ℜe θ ≥ 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 spaces, 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 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 , ℜe θ ≥ 0, and , . 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 and , ℜe θ ≥ 1, respectively.
In Section 3, we investigate the behaviour of the operators [(−A) θ] ∘etA with respect to both of the spaces (X, 𝒟(A)) γ,p and . First, in Proposition 9, we deal with the norms , ℜe θ ≥ 0, and we show that, except for replacing (X, 𝒟(A)) γ,p with if p = ∞ and with if p ∈ [1, ∞), the same estimates of [19] for the norms 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 , ℜe θ ≥ 1, . 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 , ℜe θ ≥ 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 , ℜe θ ≥ 1. Since β < 1, the spaces 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 Section 3 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 Section
5 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:
()
Here,
IT = [0,
T],
λ0 ∈
C,
n1,
n2 ∈
N,
,
,
i2 = 1, …,
n2, whereas,
Z being another complex Banach space and
𝒫 :
Z ×
X →
X being a bilinear bounded operator,
, and
,
i1 = 1, …,
n1. Of course, if
Z =
C, then
𝒫 may be the scalar multiplication in
X. As
M,
L, and
,
i1 = 1, …,
n1, we take closed single-valued linear operators from
X to itself, whose domains fulfill the relation
, and we require
L to have a bounded inverse, allowing
M to be
not invertible. 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
,
,
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 5
α + 2
β > 6, we will prove that if
,
,
,
, and
f ∈
Cμ(
IT;
X) for opportunely chosen
,
,
, 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,
d
Mv/d
t ∈
Cτ(
IT;
X), where
(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
and under the stronger condition 3
α + 8
β > 10.
In Section 6, 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 , r1 ≤ n1, in (3) with (i2, n2) = (i1, n1) and , 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
Let X 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 set ℛ(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 all x, y ∈ 𝒟(A); (iv) λAx = A(λx), for all λ ∈ C∖{0}, for all x ∈ 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}.
If A 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).
Given
𝒰 ∈
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
()
where
λA1,
A1 +
A2 and
A1A2 are m. l. operators in
X and
.
Let A 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].
If
Xj,
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
. 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]):
()
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]):
()
In particular, if 0 ∈
ρ(
A), then, since
A∘(0
I −
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,
()
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 : ℜe z ≥ −c(|ℑm z | + 1) α, ℑm z ∈ R},
-
α ∈ (0,1], c > 0, and for some exponent β ∈ (0, α] and constant C > 0 the following estimate holds:
()
Introduce the family {e
tA}
t≥0 ∈
ℒ(
X) defined by e
0A =
I and
()
where Γ⊊
Σα∖{
z ∈
C :
ℜe
z ≥ 0} is the contour parametrized by
λ = −
c(|
η | + 1)
α +
iη,
η ∈ (−
∞,
∞). Then (cf. [
1, pp. 360, 361]), {e
tA}
t≥0 is a semigroup on
X, infinitely many times strongly differentiable for
t > 0 with
()
where
. In general, no analyticity should be expected for e
tA. For if
α < 1 in (H1), then
Σα does not contain any sector Λ
ω+π/2 = {
z ∈
C∖{0} : |arg
z| <
ω +
π/2},
ω ∈ (0,
π/2), and [
15, Theorem 5.3], which extends e
tA analytically to the sector Λ
ω containing the positive real axis, is not applicable. We stress that (
9) and
𝒩((
zI −
A)
−1) =
A0,
z ∈
ρ(
A), imply
A0⊆
𝒩(e
tA) for every
t > 0, whereas
𝒩(e
0A) =
𝒩(
I) = {0}. Hence, if
A is really an m. l. operator, then {0}⊊
A0⊆⋂
t>0 𝒩(e
tA). From the semigroup property it also follows that
for
t1 ≥
t0 ≥ 0.
Now, for every
θ ∈
C such that
ℜe
θ ≥ 0 we set
()
Here, for the multivalued function (−
λ)
θ = e
θLn(−λ) we choose the principal branch holomorphic in the region
C∖{
z ∈
C :
ℜe
z ≥ 0}, where for principal branch we mean the principal determination ln |
z| +
i arg(
z) of Ln(
z). We briefly recall the main properties of operators [(−
A)
θ]
∘e
tA. Of course, [(−
A)
0]
∘e
tA = e
tA,
t > 0. As shown in [
26, p. 426], [(−
A)
k]
∘e
tA,
k ∈
N,
t > 0, is a section of (−
A)
ke
tA, so that from (
10) we get
()
Moreover (cf. [
19, formula (22)] with
θ ≥ 0 being replaced by
ℜe
θ ≥ 0), we get
()
Finally, (H1) implies the following estimates (cf. [
1,
24]):
()
where the
’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,
∞) → e
tA ∈
ℒ(
X) may be singular at the origin and the semigroup is not necessarily strongly continuous in the
X-norm on the closure
of
𝒟(
A) in
X. Notice that if
α +
β > 1, then the singularity is a
weak one, in the sense that {e
tA}
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>0 𝒩(e
tA), and if
α = 1, then
A0 =
𝒩(e
tA) for every
t > 0.
Observe that A0⊆𝒩([(−A) θ] ∘etA), ℜe θ ≥ 0, t > 0, so that A0⊆⋂t>0 𝒩([(−A) θ] ∘etA), ℜe θ ≥ 0. The operators [(−A) θ] ∘etA satisfy the following semigroup type property.
Proposition 1. Let θj ∈ C, ℜeθj ≥ 0, and let tj > 0, j = 1,2. Then
()
Proof. First, the function λ ∈ ρ(A)→(−λ) θetλ(λI − A) −1 ∈ ℒ(X) being holomorphic for every ℜe θ ≥ 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 : ℜe z ≥ 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
()
Now, after having enclosed Γ and Γ
′ on the left with an arc Δ
R 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 :
ℜe
z ≤ −
c′}, the arc Δ
R may be parametrized in polar coordinates by
ℜe
z = −
c′ +
Rcos
φ,
ℑm
z =
Rsin
φ,
φ ∈ (
π/2,3
π/2). Then, for every
z ∈ Δ
R we have
()
Since
t > 0 and
φ ∈ (
π/2,3
π/2), the right-hand side of the latter inequality goes to zero as
R goes to infinity, so that
for every
ℜe
θ ≥ 0 and
t > 0. The residue theorem together with the fact that Γ
′ lies to the right of Γ thus yields
and
. Replacing these identities in (
16) and using the equality
which is satisfied for the principal branch of the function (−
λ)
θ = e
θLn(−λ), we finally find
()
The right-hand side being precisely
, the proof is complete.
For an m. l. operator
A satisfying (H1) we introduce now the spaces (
X,
𝒟(
A))
γ,p and
. We first specify a topology on
𝒟(
A) equipping it with the norm ∥
x∥
𝒟(A) = inf
y∈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
for every
x ∈
X1. If
X1↪
X2↪
X1, that is, if
X1 =
X2 and the norms
and
are 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
. 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
,
q ∈ [1,
∞), and
. Let
p0,
p1 ∈ [1,
∞) or let
p0 =
p1 =
∞, and for
γ ∈ (0,1) define
if
p0,
p1 ∈ [1,
∞) and
p =
∞ if
p0 =
p1 =
∞. Let us set
()
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
the completion of
𝒟(
A) relative to
X and endowing it with the norm
in [
28, p. 14], we get (
X,
𝒟(
A))
0,∞≅
X and
. 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:
()
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
()
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
,
,
j,
k = 1,2, then
,
γ0 ∈ (0,1),
p0 ∈ [1,
∞], and
()
As a consequence of this general result and the identity
()
from the third in (
21) we find that if
T ∈
ℒ(
X) is such that
and
, then
,
γj ∈ (0,1),
pj ∈ [1,
∞],
j = 0,1, 2, and the following estimate holds:
()
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 fixed
x ∈
𝒟(
A) the map
T(
λ) =
λx satisfies ∥
T∥
ℒ(C;X) = ∥
x∥
X, ∥
T∥
ℒ(C;𝒟(A)) = ∥
x∥
𝒟(A) and
. Then (
22) with
,
and
yields the interpolation inequality:
()
with
c0 being the positive constant depending on
γ and
p such that
.
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
()
From (
26), using (
22) with
,
j = 1,2, and
, it then follows for every
γ ∈ (0,1) and
p ∈ [1,
∞]
()
where
c1 is the positive constant depending on
γ and
p such that
.
For
γ ∈ (0,1) and
p ∈ [1,
∞] we now define the Banach spaces
by
()
It is a well-known fact that if
A is single-valued and
β = 1 in (H1), then
(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:
()
()
Then, as in the single-valued case,
if
β = 1 in (H1). More precisely (see the proof of [
24, Proposition 4.3]), if
,
γ ∈ (0,1),
p ∈ [1,
∞], then
()
whereas if
x ∈ (
X,
𝒟(
A))
γ,p,
γ ∈ (1 −
β, 1),
p ∈ [1,
∞], then
()
with
c2 being a positive constant depending on
β,
γ and
p.
By setting
δ =
γ +
β − 1,
γ ∈ (1 −
β, 1), from (
30) it follows
()
Then, if
β ∈ (0,1), the spaces
,
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
,
δ ∈ (0,
β),
p ∈ [1,
∞], are only intermediate or just interpolation spaces between
X and
𝒟(
A).
Notice that , γ ∈ (0,1), p ∈ [1, ∞]. Indeed, assume that there exists x ≠ 0 such that 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 , contradicting . 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 . This is true, instead, if β = 1. In this case the topological direct sum 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 every
γ ∈ (0,1) and
p ∈ [1,
∞] from (
27), (
29), and (
31) it follows
()
Hence, for
γ ∈ (0,1) and
p ∈ [1,
∞] we may rewrite (
27) and (
34) more compactly as
()
where
and
c3 is equal to
c1(
C + 1)
1−γCγ or 2
c1(
C + 1)
1−γCγ according that
or
.
With the exception of the case
β = 1, in general it is not clear if embeddings analogous to (
20) hold even for the spaces
. In fact, using (
20), (
29), and (
30) we can only prove that if
γ ∈ (1 −
β, 1) and 1 ≤
q ≤
p ≤
∞, then
()
whereas if 1 −
β <
γ2 <
γ1 < 1 and
p1,
p2 ∈ [1,
∞], then
()
What can be proved without invoking (
20), (
29), and (
30) and using only the definition of the norm
is instead the following result, which extends to the spaces
the embeddings
, and
, 0 <
γ2 <
γ1 < 1,
p ∈ [1,
∞] (cf. (
20) with (
p1,
p2) = (
p,
p) and (
p1,
p2) = (
∞,
p)).
Proposition 2. Let A be an m. l. operator satisfying the resolvent condition (H1). Then the following embeddings hold for every 0 < γ2 < γ1 < 1 and p ∈ [1, ∞]:
()
()
Proof. If β = 1 in (H1), then there is nothing to prove since 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 , 0 < γ2 < γ1 < 1, we write
()
where
()
(
a1,
b1,
a2,
b2) = (0,1, 1,
∞). Using the first inequality in (
26) we find
()
where
c4 = 2
1−β(
C + 1)(
γ2p)
−1/p. Concerning
I2, instead, using
γ2 −
γ1 < 0, we get
()
Summing up (
40)–(
43) and setting
c5 = [(
c4)
p + 1]
1/p, it thus follows
, completing the proof of (
38) in the case
p ∈ [1,
∞). Let
p =
∞. For every
, 0 <
γ2 <
γ1 < 1, we write
()
where
,
j = 3,4,
U3 = (0,1),
U4 = [1,
∞). Again, the first inequality in (
26) yields
()
Instead, using
γ2 −
γ1 < 0, we have
()
Summing up (
44)–(
46) and setting
c6 = 2
1−β(
C + 1), we thus find
. 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
, 0 <
γ2 <
γ1 < 1, we write
, where
I1 and
I2 are defined by (
41). Hence, the same computations as in (
42) yield
()
As far as
I2 is concerned, instead, we have
()
where
c7 = [(
γ1 −
γ2)
p]
−1/p. Summing up (
47) and (
48) and setting
c8 = [(
c4)
p + (
c7)
p]
1/p, we deduce
. The proof is complete.
Remark 3. Notice that (37) with p1 = p2 = p yields , 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 in the case β < 1 is that it is not clear if it holds , 1 ≤ q < p ≤ ∞. In fact, the embedding
()
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
, 0 <
γ2 <
γ1 < 1 (see the proof of [
27, Theorem 1.3.3(e)] taking there
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
, we will be faced with two problems. The first is that we do not know if the function
,
ξ ∈ (0,
∞),
x ∈
X, is monotone decreasing, which would allow us to prove
,
γ ∈ (0,1),
p ∈ [1,
∞). For if
g(
ξ) was monotone decreasing, then for every
ξ ∈ (0,
∞) and
,
γ ∈ (0,1),
p ∈ [1,
∞), we would find
()
where
c9 = (
γp)
−1/p. Taking the supremum with respect to
ξ ∈ (0,
∞) in the latter inequality, we would get
, proving
,
γ ∈ (0,1),
p ∈ [1,
∞). The second problem is that the function
ξγg(
ξ) is not necessarily bounded for
,
γ ∈ (0,1),
p ∈ [1,
∞), precluding us to prove
,
γ ∈ (0,1),
q ∈ [1,
p). Indeed, from (
35) we can only find
, 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
()
If now in addition
g(
ξ) were also monotone decreasing, in order that
, from the latter inequality we would get
, completing the proof of
,
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
.
Remark 5. Let 0 < γ2 < γ1 < 1 be fixed and for every p ∈ [1, ∞] and let us set and . 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 , from (20) we deduce that the sets Ap and Bp are related by the following inclusions in which 1 < q1 < q2 < ∞:
()
Now let
β < 1. As observed in Remark
4, in this case the embedding
, 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 . 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 . Later, we will need this fact.
Proposition 6. Let A be as in Proposition 2. If γ ∈ (1 − β, 1); then {etA} t≥0 is strongly continuous in the X-norm on for every p ∈ [1, ∞].
We conclude the section listing some estimates for the operators [(−
A)
θ]
∘e
tA defined by (
11) with respect to the spaces (
X,
𝒟(
A))
γ,p and
. First, in [
19, Lemma 3.1] it is shown that [(−
A)
θ]
∘e
tAx ∈
𝒟(
A) for every
x ∈
X and that the estimate
is satisfied. Hence, using (
14), we get
()
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:
()
Remark 7. We stress that if β < 1, then we can not derive an estimate for the -norm of [(−A) θ] ∘etA simply by replacing (X, 𝒟(A)) γ,p with in (54). This is for two reasons. First, when γ ∈ [β, 1), we are not assured that for every x ∈ X. For if γ ∈ [β, 1), then the space 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 , from (31) we only get , 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
()
where
c11 =
c2c10,
γ ∈ (1 −
β, 1) and
p ∈ [1,
∞]. Letting
δ =
γ +
β − 1, (
55) can be rewritten equivalently as
()
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
()
as shown in [
2, Proposition 3.2], with
c12 being a positive constant depending on
α,
β, and
δ. Also, (
57) ensures that e
tAx,
x ∈
X, belongs to
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
is taken in place of (
X,
𝒟(
A))
γ,∞.
Now let
,
γ ∈ (0,1),
p ∈ [1,
∞]. As far as the estimates for the
-norm of operators [(−
A)
θ]
∘e
tA are concerned, instead, at the moment only the following estimates for the case
θ = 1 are available (cf. [
24, Lemma 5.1]):
()
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 → e
tA is Hölder continuous from [0,
∞) to
,
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
-norm of [(−
A)
θ]
∘e
tA,
ℜe
θ ≥ 1, which reduce to (
58) in the case
θ = 1.
Remark 8. Observe that an estimate for the norm , ℜeθ ≥ 1, t > 0, , γ ∈ (0,1), p ∈ [1, ∞], can be obtained combining (14), (15), and (58). Indeed, using (15), for every ℜe θ ≥ 1, t > 0 and , we have
()
Therefore, due to (
14) and (
58), from (
59) we deduce that
()
where
γ ∈ (0,1),
p ∈ [1,
∞] and
. As we will see in the next section estimate (
60) is not optimal, in the sense that the negative exponent (2
β +
γ −
ℜe
θ − 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]
∘e
tA 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]
∘e
tAx for
.
3. Behaviour of [(−A) θ] ∘etA in (X, 𝒟(A)) γ,p and
According to Remark 7 we begin by improving (54), showing that the same estimate holds with (X, 𝒟(A)) γ,p being replaced by if p = ∞ and by 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. Let ℜeθ ≥ 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
()
()
Proof. If β = 1, then 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, ℜe θ ≥ 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
()
()
Of course, from estimate (
54) we find
()
with
cγ,p being such that
,
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
()
Here we have used twice the equality ∫
Γ (−
λ)
θe
tλ(
λ −
ξ)
−1d
λ = 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
,
λ ∈
Σα, it follows that
()
Now, since
ℜe
λ ≤ −
c < 0 for every
λ ∈ Γ and since
ξ ∈ (0,
∞), we have
()
Therefore, for every
λ ∈ Γ and
ξ ∈ (0,
∞) the following inequality holds:
()
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(2
π)
−1e
(π/2)|ℑm θ|cγ, we thus find (here we use also that on Γ it holds |
λ | ≥
c, so that
ℜe
λ = −
c(|
ℑm
λ | + 1)
α ≥ −
c(1 +
c−1)
α |
λ|
α):
()
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
()
where
,
E(
χ),
χ > 0, being the Euler gamma function
. Then, summing up (
65) and (
71), from (
63) it follows that
()
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
()
where
,
j = 1,2, (
a1,
b1,
a2,
b2) = (0,1, 1,
∞). First, (
35) with
yields
()
Therefore, since (
ξ + 1)
1−β−γ ≤
cβ,γ for every
ξ ∈ (0,1], where
cβ,γ = 2
1−β−γ or
cβ,γ = 1 according that
γ ∈ (0,1 −
β) or
γ ∈ [1 −
β, 1), from (
54), we deduce that
()
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
()
where
c20 =
c18[(1 −
β)
γp]
−1/p. Summing up (
73)–(
76), it thus follows that
()
where
c21 = [(
c19)
p + (
c20)
p]
1/p. Finally, (
65) and (
77) lead us to
()
Since
x ∈
X was arbitrary, this completes the proof of (
62) with
c16 =
cγ,pc10 +
c21.
Remark 10. If θ = 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 if p = ∞ and 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 , p ∈ [1, ∞].
Remark 11. We recall that when β < 1 the spaces , 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 eventually smaller than 𝒟(A), for (62) to hold we have to consider only spaces , p ∈ [1, ∞), bigger than 𝒟(A). In fact, letting σ = βγ, we have σ ∈ (0, β) for every γ ∈ (0,1).
In accordance with Remark 8 we now improve estimate (58).
Proposition 12. Let ℜeθ ≥ 1, γ ∈ (0,1), p ∈ [1, ∞] and let . Then, there exists a positive constant c22 depending on α, β, γ, θ, and p such that
()
Proof. First, using the identity A∘(zI − A) −1 = z(zI − A) −1 − I, z ∈ Σα, for every x ∈ X, we rewrite [(−A) θ] ∘etAx, ℜe θ ≥ 0, in the following way:
()
Here we have used ∫
Γ (−
λ)
θ−1e
tλd
λ = 0, which follows from the Cauchy formula applied to (−
λ)
θe
tλ 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,
ℜe
θ ≥ 1,
γ ∈ (0,1), and
p ∈ [1,
∞] be fixed and let
x be an arbitrary element of
. From (
35) it then follows that
()
where
c23 = (2
π)
−1e
(π/2)|ℑm θ|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:
()
where
or
according that
γ ∈ (0,1 −
β] or
γ ∈ (1 −
β, 1) ((0,1 −
β] =
∅ if
β = 1). Therefore, setting
, (
81) and (
82) yield
()
with
cα being as in (
70). Finally, the transformation
cαtμα =
s in the last integral leads us to the following estimate:
()
where
,
E(
χ),
χ > 0, is the Euler’s gamma function. Notice that here
ℜe
θ ≥ 1 implies
ℜe
θ + 1 −
β −
γ ≥ 2 −
β −
γ > 0 for every
β ∈ (0,1] and
γ ∈ (0,1), so that
E((
ℜe
θ + 1 −
β −
γ)/
α) makes sense. Since (
84) is satisfied for every arbitrary element
, 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 ℜe θ ≥ 1, the following inequality holds:
()
Then,
,
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. Let α + β > 1 in (H1). Then, for every x ∈ X the following equalities hold:
()
Proof. The assertion is obvious for t = 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
()
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
, due to
I ⊂
AA−1. Then,
):
()
where
cα,β,γ =
α(
α +
β +
γ − 2)
−1. We now recall that (cf. [
24, formula (3.21)])
()
with (−
A)
−ζ being the negative fractional powers of −
A defined by (cf. [
24, Section 3]) (2
πi)
−1∫
Γ (−
λ)
−ζ(
λI −
A)
−1d
λ,
ℜe
ζ > 1 −
β. To complete the proof it thus suffices to apply (
89) with
ζ = 1 to (
87) and to recall that [(−
A)
0]
∘e
tA = e
tA,
t > 0. Notice that the integral on the right-hand side of (
86) is convergent, too. In fact, from (
14), it follows that
.
Remark 15. In particular, from (86) it follows that if α + β > 1, then for every x ∈ X and . 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, ℜe θ ≥ 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. Let ℜeθ ≥ 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
()
Proof. For brevity, we will use the shortenings , σ ∈ (0,1), p ∈ [1, ∞]. We begin by proving the first estimate in (90). Let θ ∈ C, ℜe θ ≥ 1, γ, δ ∈ (0,1) and p ∈ [1, ∞] be fixed and let x be an arbitrary element of . Moreover, let ζ and ζ′ be two arbitrary complex numbers such that θ = ζ + ζ′ and whose real parts satisfy ℜe ζ ≥ 0 and ℜe ζ′ ≥ 1. From the decomposition formula (15) it then follows for every t > 0:
()
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
ℜe
θ =
ℜe
ζ +
ℜe
ζ′, we deduce that
()
where
c26 = 2
(2+ℜe θ+δ−γ−2β)/αc10c22. This completes the proof of the first estimate in (
90), due to the arbitrariness of
. Let us now prove the second estimate in (
90). Let
θ ∈
C,
ℜe
θ ≥ 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:
()
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
()
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
()
This completes the proof of the second estimate in (
90) with
.
Remark 17. We stress that if β < 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 ℜe θ ≥ 1 the following inequality holds:
()
so that
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
,
ℜe
θ ≥ 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 Remark 19 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 case β = 1, the exponents in both estimates (90) coincide equals to ν = γ − δ − ℜe θ. 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 , where . 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
()
Differently from the spaces
and as a consequence of
A0⊆⋂
t>0 𝒩([(−
A)
1]
∘e
tA), 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:
()
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
, from the second embeddings in (
97) and (
98) it follows that if
α +
β > 1 and
σ ∈ (2 −
α −
β, 1), then
()
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 2
α +
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:
()
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
,
σ ∈ (0,1). This confirms that in the real m. l. case the equivalence between
, (
X,
𝒟(
A))
σ,p and
𝒟A(
σ,
p) does not hold even when
β = 1.
Using Propositions 9 and 12, we now obtain estimates for the operators [(−A) θ] ∘etA, ℜe θ ≥ 1, considered as operators from to . Clearly, since β < 1 the spaces 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. Let ℜeθ ≥ 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
()
()
Moreover, if
γ ∈ (0,1) and
δ ∈ (1 −
β, 1) are such that
γ +
δ < 1, then
()
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 or with provided that p = ∞ or p ∈ [1, ∞). In this way we derive (101) and (102) with cj+13 = 2(2+ℜe θ+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 , C1 and C2 being the positive constants such that , x ∈ Xj, j = 1,2. Applying this result to L = [(−A) θ] ∘etA with , from (29)–(32) and the second estimate in (90) we deduce (103) with c30 = 2c2c27. This completes the proof.
Remark 21. The assumption γ + 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, ℜe θ ≥ 1, as an operator between the intermediate spaces and , 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 Section
5. 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 = sup
t∈[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
. We endow the subspace
,
δ ∈ [0,1) with the norm ∥·∥
δ,a,b;Z. Further, for
k ∈
N and
δ ∈ (0,1) we set
,
(
), and
,
. Recall that if 0 ≤
δ2 ≤
δ1 ≤ 1, then
and
,
. Finally, given three complex Banach spaces
,
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
, we denote by
𝒦 the convolution operator
()
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
. As usual, for every
q ∈ [1,
∞], we will denote by
q′ the conjugate exponent of
q.
Now let
X3 =
X and introduce the following linear operators
Qj,
j = 1, …, 6, where
,
j = 1,2, 5,
,
l = 3,6,
k = 1,2,
,
,
,
p ∈ [1,
∞], and
t ∈ [0,
T],
T > 0 as follows:
()
()
()
()
()
()
with
g4y being the function from [0,
T] to
defined by (
g4y)(
t) =
g4(
t)
y. We will find conditions on
,
j = 1,2, 5,
l = 3,6,
k = 1,2, in order that
,
and
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. Let α + β > 1 in (H1). Then, for every δ1 ∈ (0, (α + β − 1)/α), the operator Q1 defined by (105) maps into , and for every t ∈ [0, T] satisfies the following estimate, where p ∈ (α/(α + β − 1 − αδ1), ∞) as follows:
()
Here
C1(
t) is a nondecreasing function of
t depending also on
α,
β,
δ1, and
p′.
Proof. Let , 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
()
where
. 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
()
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
, where
and
. Reasoning as in (
112) and using the inequality
,
μ ∈ (0,1], we get
()
Similarly, but taking advantage from
, we obtain
()
Thus, letting
from (
114) and (
115) it follows that
()
Finally, summing up (
113) and (
116) and using
, we derive (
111) with
. This completes the proof.
Remark 23. We stress that if we renounce to its Hölder regularity, then for Q1g1 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)), , t ∈ [0, T].
Lemma 24. Let 3α + β > 3 in (H1). Then, for every δ2 ∈ ((3 − 2α − β)/α, 1), the operator Q2 defined by (106) maps into , ν2 = (αδ2 + 2α + β − 3)/α ∈ (0, δ2], and for every t ∈ [0, T] it satisfies the following estimate:
()
Here
C2(
t) is a nondecreasing function of
t depending also on
α,
β, and
δ2.
Proof. Denote by the number (1 − α)/α. In particular, since 3α + β > 3 implies α ∈ (2/3,1], we have . Let t ∈ [0, T], , δ2 ∈ ((3 − 2α − β)/α, 1), and ν2 = (αδ2 + 2α + β − 3)/α ∈ (0, δ2]. We notice that and (αδ2 + β − 3)/α = ν2 − 2. Then, using (14) with θ = 1, for every τ ∈ [0, t] we obtain
()
where
. Hence
()
Now let (since [
Q2g2](0) = 0, the case
t1 = 0 follows from (
118) with
τ =
t2) 0 <
t1 <
t2 ≤
t. We have
, where for a function
g : [0,
T] →
X we set
()
First, using (
13) with (
s,
t,
θ) = (
t1 −
s,
t2 −
s, 1),
s ∈ (0,
t1), and (
14) with
θ = 2, and letting
, we get
()
Let us turn to
. We first observe that the integral
is convergent. For,
, where
is equal to
if
β = 1 and to
if
β ∈ (0,1). Thus, we may rewrite it as
. Consequently,
()
where we have used [
t2(
t2 −
t1)
−1]
(β−1)/α ≤ 1 and
. As far as
is concerned, instead, reasoning as in the derivation of (
118) we find
()
Then, summing up (
121)–(
123) and letting
, we obtain
()
Finally, (
119) and (
124) yield (
117) with
.
Remark 25.
In particular, Lemma 24 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 Corollary 14, 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 , t ∈ [0, T].
Corollary 26.
- (i)
Let 2α + β > 2 in (H1). Then, for every g ∈ Cδ([0, T]; X), δ ∈ ((2 − α − β)/α, 1),
()
- (ii)
Let α + β > 1 in (H1). Then, for every g ∈ C([0, T]; X)
()
Proof. Of course, it suffices to assume that t ∈ (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
()
On the other side,
satisfies
()
where
. Then, commuting
A−1 ∈
ℒ(
X) with the integral signs, using (
80) with
θ = 1, and taking into account (
7), we find
()
Since (2
πi)
−1∫
Γ e
(t−s)λ(
λI −
A)
−1d
λ = 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
()
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,
satisfies
()
where
. The proof is complete.
Let us now examine the operator Q3 defined by (107). To this purpose we need the following result which is proved in [20, Corollary 3.2].
Lemma 27. Let , k = 1,2, be such that , . Then the convolution operator 𝒦 defined by (104) maps into , and for every t ∈ [0, T] satisfies the following estimate:
()
Here
is a nondecreasing function of
t depending also on
and
. Further, in the cases
,
, and
, the following estimates hold, respectively, as follows:
()
From Lemmas 24 and 27 we obtain the following Lemma 28.
Lemma 28. Let α and β be as in Lemma 24. Then, for every and such that , , the operator Q3 defined by (107) maps into , ν3 = (ασ3 + 2α + β − 3)/α, and for every t ∈ [0, T] satisfies the following estimate:
()
Proof. First, if and , then . Consequently, the assumption , , makes sense. Now, Lemma 27 yields for any pair . Then, recalling that , the assertion follows from Lemma 24, with δ2 and g2 being replaced by σ3 and , respectively. Finally, (134) follows from (117) and (132).
We can now restore the loss of regularity produced by Q2.
Proposition 29. Let 5α + 2β > 6 in (H1). Then, for every δ3 ∈ ((3 − 2α − β)/α, 1/2), the operator Q3 defined by (107) maps into , and for every t ∈ [0, T] satisfies the following estimate, where p ∈ (1/(1 − 2δ3), ∞) and :
()
Proof. Let δ3 ∈ ((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 from which we deduce that Q3 maps into , ν3 = (2αδ3 + 2α + β − 3)/α. But, since our choice for δ3 implies ν3 > δ3, we a fortiori have the fact that Q3 maps into . 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 Lemma 30 concerns the operator Q4. Its proof is similar to that of Lemma 24, but with the essential difference that the presence of 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. Let 2α + 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 , , into , and for every t ∈ [0, T] satisfies the following estimate:
()
Here
C4(
t) is a nondecreasing function of
t depending on
α,
β,
δ4,
γ and
r.
Proof. Let t ∈ [0, T], , δ4 ∈ (0,1), and , γ ∈ (3 − 2α − β, 1), r ∈ [1, ∞]. As in the proof of Lemma 24 we set and we observe that, since 2α + β > 2 implies α ∈ (1/2,1], here . 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 and σα,β,γ − 2, respectively. Then, using (79) with θ = 1, we obtain
()
where
. Hence, taking the supremum with respect to
τ ∈ [0,
t], one has
()
Now, let (since [
Q4(
g4,
y)](0) = 0, the case
t1 = 0 follows from (
137) with
τ =
t2) 0 <
t1 <
t2 ≤
t. We have
, the
’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
, we get
()
Now, let us examine
,
k = 2,3. First, using (
79) with
θ = 1, we find
()
Instead, the same computations made to derive (
137) yield
()
From (
139)–(
141) and
, it follows that
()
where
. Finally, summing up (
138) and (
142) we get (
136) with
. The proof is complete.
Remark 31. Notice that if , 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 operator Q5 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. Let 2α + β > 2 in (H1). Then, for every δ5 ∈ ((2 − α − β)/α, 1), the operator Q5 defined by (109) maps into , ν5 = (αδ5 + α + β − 2)/α ∈ (0, δ5], and for every t ∈ [0, T] satisfies the following estimate:
()
Here
C5(
t) is a nondecreasing function of
t depending also on
α,
β, and
δ5.
Proof. Let , δ5 ∈ ((2 − α − β)/α, 1), and ν5 = (αδ5 + α + β − 2)/α ∈ (0, δ5]. We still let and as in Lemma 30 we have . Further, observe that . Let t ∈ [0, T]. Then, using (14) and g5(0) = 0, we get
()
Now, let (since [
Q5g5](0) = 0, the case
t1 = 0 follows from (
144) and
) 0 <
t1 <
t2 ≤
t. We have
, where for a function
g : [0,
T] →
X we let
()
First, since
for every
β ∈ (0,1], we deduce that
()
As far as
is concerned, instead, rewriting
as
and using both
g5(0) = 0 and (
αδ5 +
β − 2)/
α =
ν5 − 1, it follows that
()
Then, since
, from (
146) and (
147) we find
()
where
. Summing up (
144) and (
148) we obtain (
143) with
. This completes the proof.
Remark 33. Thus, with the exception of β = 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 assumptions α + β > 1 and g5 ∈ C([0, T]; X), (86) with x = g5(t), t ∈ [0, T], yields .
Similarly as we have done in Proposition 29 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. Let α and β be as in Lemma 32. Then, for every and such that , , the operator Q6 defined by (110) maps into , ν6 = (ασ6 + α + β − 2)/α, and for every t ∈ [0, T] satisfies the following estimate:
()
Proof. First, if and , then . Consequently, the assumption makes sense, provided to choose small enough. Lemma 27 then yields for any pair . Then, since , the assertion follows from Lemma 32, with the pair (δ5, g5) being replaced by . Finally, (149) follows from (143) and (132).
From Lemma 35 we obtain the analogous of Proposition 29 for Q6.
Proposition 36. Let 3α + 2β > 4 in (H1). Then, for every δ6 ∈ ((2 − α − β)/α, 1/2), the operator Q6 defined by (110) maps into , and for every t ∈ [0, T] satisfies the following estimate, where p ∈ (1/(1 − 2δ6), ∞) and :
()
Proof. Let δ6 ∈ ((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 , k = 1,2. We thus deduce that Q6 maps into , 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 into . Finally, (150) follows from (149) and .
In Section 6 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 . 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 .
Lemma 37. Let α + β > 1 in (H1) and , γ ∈ (2 − α − β, 1), r ∈ [1, ∞]. Then, for every δ5 ∈ (0, (α + β + γ − 2)/α], the operator Q5 defined by (109) maps into , and for every t ∈ [0, T] satisfies the following estimate:
()
Here
c41 is a positive constant depending on
α,
β,
γ, and
r.
Proof. Let γ ∈ (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 , 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:
()
Indeed, for every
ɛ ∈ [0,
t) and
, (
79) with
θ = 1 yields
()
where
. From (
152) and (
153) with (
ɛ,
t,
x) = (0,
τ,
g5(
τ)) we thus get
()
Now, let 0 ≤
t1 <
t2 ≤
t. From (
152) it follows that
, where for every function
we have set
()
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
()
As a consequence, since
δ5 ∈ (0,
χα,β,γ],
()
Summing up (
154) and (
157), we obtain (
151). The proof is complete.
Since in Lemma 37 it is not required that g5(0) = 0, the special case of the constant function , t ∈ [0, T], is admissible, and we obtain the following result.
Corollary 38. Let α, β, and be as in Lemma 37, and let , 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 , and for every t ∈ [0, T] satisfies the estimate
()
Proof. Let g5(t) = x in the proof of Lemma 37, and observe that 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 condition 5α + 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,
()
Hence, if 5
α + 2
β > 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:
()
Here
𝒦 is the convolution operator (
104) in which (
X1,
X2,
X3) = (
Z,
X,
X), whereas
M,
L, and
,
i1 = 1, …,
n1, are closed single-valued linear operators from
X to itself, whose domains fulfill the relation
. Further, we assume that
()
whereas we allow
M to have
no bounded inverse. Hence, in general,
A : =
LM−1 is only the m. l. operator defined by
()
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,
,
i1 = 1, …,
n1. As far as the data vector
is concerned, at the moment, we only assume
λ0 ∈
C,
v0 ∈
𝒟(
M),
,
,
,
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
,
,
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,
,
ψ ∈ (0,1),
q ∈ [1,
∞], will always denote one between the spaces (
X,
𝒟(
A))
ψ,q and
,
A being defined by (
162). That is,
. To avoid confusion, if more than a single
is involved in some statement, that is, if we write
,
j = 1, …,
n,
n ∈
N, then it is understood that the same choice has been made for all the
in the sense that
or
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 : ℜe z ≥ −c(|ℑm z | + 1) α, ℑm z ∈ R}, α ∈ (0,1], c > 0, and for some exponent β ∈ (0, α] and constant C > 0 the estimate 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
,
i1 = 1, …,
n1, yields to consider also various nonintegrodifferential degenerate problems. For instance, those corresponding to
and
,
il = 1, …,
nl,
l = 1,2, respectively:
()
()
Although (
164) differs from (
163) only in the fact that
f is replaced with
; nevertheless a very different result is achieved when the
’s are assumed to belong to
, at least for opportunely chosen
,
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
’s. The same phenomenon is carried over into the integrodifferential case for the following problems, corresponding to
,
i2 = 1, …,
n2, and
λ0 =
f = 0:
()
()
t ∈
IT. When
β < 1, the loss of time regularity for the pair (
Lv,
DtMv) with respect to that of the vector
in problem (
165) (cf. [
22, Theorem 7.1] and [
23, Theorem 2.1] for
n1 = 1) can be restored in problem (
166) assuming that
,
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
’s and
’s.
We stress that, if β = 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
()
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
,
Dtv(0) = 0,
v(0) =
v0, whereas when
M =
I and
,
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 {e
tA}
t≥0 defined by e
0A =
I and (
9) and satisfying (
14). Assuming that
v0 ∈
𝒟(
L), we let
()
Then, by setting
()
where
A−1 =
ML−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:
()
Now let 2
α +
β > 2, and assume that
,
, and
f ∈
Cμ(
IT;
X), where
,
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
()
Indeed,
δ being the smallest Hölder exponent, for every
il = 1, …,
nl,
l = 1,2, we have
and
,
(cf. Lemma
27 for the case
with the pair
being replaced by (in fact, since
,
i1 = 1, …,
n1, if
w ∈
C(
IT;
X), then
, whereas the constant functions
,
t ∈
IT,
i1 = 1, …,
n1, obviously belong to
C(
IT;
X))
and
, 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
()
with
Q1 being the operator defined by (
105). Further (cf. [
2, Remark p. 55] with
u0 = 0, and where
A∘e
tA stands for
Dte
tA = −[(−
A)
1]
∘e
tA) the derivative of
A−1w is given by
()
with
Q2 being the operator in (
106). Notice that
Q2Fw is well defined by virtue of (
127) with
g =
Fw. Now let
and
where
,
i2 = 1, …,
n2, and
r ∈ [1,
∞]. Since
,
i1 = 1, …,
n1, from (
169) it thus follows that
Fw(0): =
x0 is independent on
w and
()
Indeed (cf. (
20) or (
38)), we have
,
i2 = 1, …,
n2, and
, the embeddings being equalities for those between the numbers
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
and the differential equation in (
170) is satisfied even at
t = 0. To see this, we observe that
()
where
,
, and
. First, from Proposition
6 we get
. On the other side, using
Fw ∈
Cδ(
IT;
X),
δ ∈ ((2 −
α −
β)/
α, 1)⊆((1 −
β)/
α, 1), we obtain
()
so that
. Finally, (
127) with
g =
Fw yields
, too. Formula (
173) thus holds at
t = 0 with
.
Remark 40. In [2, Remark p. 55], formula (173) was extended until t = 0 only under the more restrictive assumption , 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 , 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
()
where, according to the notation in Corollary
38, we have set [
Q7x0](
t) = (e
tA −
I)
x0. In particular,
w(0) = 0. We conclude that, under the previous assumptions on the pair (
α,
β) and on the data vector
, 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 functions
and
,
i2 = 1, …,
n2, defined by
()
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
()
the functions
wl,
l = 0,1, and the operator
Rw being defined by
()
()
()
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
satisfy the assumptions below (
170) and (
173). Then, as before,
and
,
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
()
Instead, due to the definition of
Q3 and
Q4, using (
125) we obtain
()
Therefore, from (
183), (
184), and the definition (
105) of
Q1 it follows that
()
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
and
, respectively. Consequently
()
Hence, commuting
A−1 ∈
ℒ(
X) with both the integral sign and the semigroup, one has
()
Similarly, since Remark
34 and formula (
125) yield
()
we find that
()
i1 = 1, …,
n1. In conclusion, from (
187) and (
189) it follows that
()
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, for 5
α + 2
β > 6,
β ∈ (0,
α],
α ∈ (0,1], and
ν ∈ ((3 − 2
α −
β)/
α, 1),
Iα,β,ν⊆((3 − 2
α −
β)/
α, 1/2)⊆(0,1/2) will denote the interval defined by
()
Clearly, if
ν,
ρ ∈ ((3 − 2
α −
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 , , i1 = 1, …, n1, and let . Then, for every fixed δ ∈ Iα,β,η, the operator R defined by (182) maps continuously Cδ(IT; X) into , and for every t ∈ IT satisfies the following estimate, where p ∈ (1/(1 − 2δ), ∞):
()
Here
c42(
T) is a positive constant depending only on
T,
λ0,
α,
β,
,
δ,
p,
and
,
i1 = 1, …,
n1.
Proof. Let , , i1 = 1, …, n1, and let us fix an arbitrary number δ ∈ Iα,β,η, where . In particular, since δ ≤ η ≤ ηi, we have with , i1 = 1, …, n1. Now let w ∈ Cδ(IT; X) and t ∈ IT. First, formula (186) being applicable, we rewrite (182) as
()
Now, we notice that 5
α + 2
β > 6 implies that
()
Since (1 −
β)/
α ≤ (2 −
α −
β)/
α ≤ (3 − 2
α −
β)/
α, from (
194) it follows that
δ ∈
Iα,β,η⊆((3 − 2
α −
β)/
α, 1/2)⊆((2 −
α −
β)/
α, 1/2)⊊((1 −
β)/
α, (
α +
β − 1)/
α), and, consequently,
()
We conclude (cf. Remark
39) that Lemma
22 and Propositions
29 and
36 are applicable with
δ ∈
Iα,β,η and
p ∈ (1/(1 − 2
δ),
∞). Then, using estimates (
111), (
135), and (
150) with the pair (
g1,
δ1) and the quintuplets
,
l = 3,6, being replaced, respectively, by (
w,
δ) and (indeed, since
, if
w ∈
Cδ(
IT;
X), then
with
,
i1 = 1, …,
n1)
,
i1 = 1, …,
n1, from (
193) we finally obtain
()
Here we have set
, 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 Lemma 42 the Hölder exponents are such that 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 .
Corollary 44. Let the assumptions of Lemma 42 be satisfied, and let η and R be as there. Then, for every fixed δ ∈ Iα,β,η, the sequence (R0 = I, Rn = RRn−1, n ∈ N) satisfies the following estimates, where w ∈ Cδ(IT; X) and p ∈ (1/(1 − 2δ), ∞):
()
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 , replacing w with Rnw in (192) and introducing the sequence of scalar nonnegative nondecreasing functions defined by , t ∈ IT, from (192) we obtain
()
Then, applying to (
198) an induction argument in which the first step of the induction follows from (
192), we immediately deduce the following estimates:
()
The proof is complete.
Lemma 45. Let 5α + 2β > 6 in (H2) and . Assume that , , and , where , , il = 1, …, nl, l = 1,2, and r ∈ [1, ∞]. Let . Then, for every fixed , the function w1 defined by (181) belongs to , provided that f ∈ Cμ(IT; X), μ ∈ [δ + μα,β, 1), μα,β = (3 − 2α − β)/α.
Proof. Let us fix , . Of course, and , il = 1, …, nl, l = 1,2. Then, Proposition 29 and Lemma 30 applied with the quintuplets and the quadruplet (g4, y, δ4, γ) being replaced, respectively, by (the constant functions , t ∈ IT, i = 1, …, n1, being obviously of class Cδ(IT; X)) and , imply that , il = 1, …, nl, l = 1,2. Now, since , the number δ + μα,β satisfies
()
and assumption
f ∈
Cμ(
IT;
X),
μ ∈ [
δ +
μα,β, 1), is meaningful. Lemma
24 with (
g2,
δ2) = (
f,
μ) then yields
,
να,β,μ = (
αμ + 2
α +
β − 3)/
α. Since
, we get
, too. Summing up, we get the assertion.
Before considering the function
w0 in (
180), we introduce the following notation. In the sequel, for 3
α + 2
β > 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
()
Notice that, since (2 −
α −
β)/
α ≤ (3 − 2
α −
β)/
α, if the stronger condition 5
α + 2
β > 6 is satisfied, then (
191) and (
201) yield
Iα,β,ν⊆
Jα,β,ν for every fixed
ν ∈ ((3 − 2
α −
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. Let 3α + 2β > 4 in (H2), and let v0 ∈ 𝒟(L). Assume that , , , and , where , , il = 1, …, nl, l = 1,2, r ∈ [1, ∞], and v1 = (λ0M + L)v0. Let and , where χα,β,γ = (α + β + γ − 2)/α. Then, for every fixed , the function w0 defined by (180) belongs to , 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 , i2 = 1, …, n2, is meaningful. Moreover, since , the number χα,β,γ = (α + β + γ − 2)/α satisfies χα,β,γ ∈ ((2 − α − β)/α, 1). Hence, , too, and is well defined. Now, let be fixed. Due to (20) or (38), the element x0 defined by (174) belongs to , whereas the functions defined by (178) are of class . 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 and (x0, δ), respectively, we deduce that , i2 = 1, …, n2. In addition, since the ’s and the constant functions belong to Cδ(IT; X), from Proposition 36 applied with , it follows that , i1 = 1, …, n1. Finally, since , the number δ + ϱα,β satisfies
()
and the assumption
f ∈
Cμ(
IT;
X),
μ ∈ [
δ +
ϱα,β, 1), makes sense. Then, the function
being of class
, Lemma
32 applied with
yields
,
. Since
, we conclude that
, too. Summing up, we get the assertion.
Remark 47. We stress that, if β ∈ (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 , , , and , where , , il = 1, …, nl, l = 1,2, r ∈ [1, ∞], and v1 = (λ0M + L)v0. Let and , 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 , i2 = 1, …, n2, makes sense. In addition, since , 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 and , the number 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 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 , , , a fortiori maps into itself. Then, being endowed with the same norm ∥·∥δ,0,T;X of Cδ(IT; X), from (197) we obtain the estimates
()
In particular, (
203) yields that
converges in
. From generalized Neumann’s Theorem it thus follows that
, the inverse
being precisely
. Since Lemmas
45 and
46 (both applied with (observe here that if
μ ∈ [
δ +
μα,β, 1), then the exponent
in the last part of the proof of Lemma
46 satisfies
. For,
)
f ∈
Cμ(
IT;
X),
μ ∈ [
δ +
μα,β, 1)⊆[
δ +
ρα,β, 1)) imply that
, we conclude that the fixed-point equation (
179) admits the unique solution
()
Observe now that the data vector
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
,
and
f ∈
Cδ(
IT;
X),
il = 1, …,
nl,
l = 1,2, whereas, as in Lemma
46,
implies that
. Therefore, since
A−1 ∈
ℒ(
X), if
is the solution to the fixed-point equation (
179), then
, too, and the function
Fw defined by (
169) satisfies
()
where
δ ∈
Iα,β,τ⊊(2 −
α −
β)/
α, 1),
γ ∈ (5 − 3
α − 2
β, 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),
, and
Fw ∈
Cδ(
IT;
X) yield
()
The proof is complete.
Remark 49. Theorem 48 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 , whereas here, removing the assumption γ < β and considering the general case n1, n2 ∈ N, we allow 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 is understood for , we have , and the spaces and , i2 = 1, …, n2, may be smaller than 𝒟(A). However, the choice 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 assumption 5α + 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], , , 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
()
Clearly, if
β = 1, then 5
α + 2
β > 6 is redundant, and Theorem
48 holds with
,
,
,
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 the ’s and ’s are assumed to vary in the smaller interval Uα,β : = ((3 − 2α − β)/α, (α + β − 1)/α), then φ and the ’s can be chosen such that . To this purpose, letting , it suffices to take , i2 = 1, …, n2, where Vα,β,ρ : = [2 + αρ − α − β, 1)⊊(5 − 3α − 2β, 1). Then and χα,β,γ = (α + β + γ − 2)/α ≥ ρ. In other words, provided that the data vector is smooth enough, the pair (Lv, DtMv) has the maximal time regularities which is the minimal between the time regularities of the ’s and ’s.
We conclude with the results which follow from Theorem 48 for problems (163)–(166).
Theorem 52. Assume (161) and v0 ∈ 𝒟(L), and let 5α + 2β > 6 in (H2). Assume that and , where , i1 = 1, …, n1, γ ∈ (5 − 3α − 2β, 1), and r ∈ [1, ∞]. Let , 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 Lemmas 42, 45, and 46, Corollary 44, and Theorem 48, letting there , i2 = 1, …, n2. To this purpose, observe that (169) and (174) reduce to and x0 = Lv0 + f(0). Consequently, (180)–(182) change to , , and .
Theorem 53. Assume (161) and v0 ∈ 𝒟(L), and let 5α + 2β > 6 in (H2). Assume that , , , and , where , , il = 1, …, nl, l = 1,2, and r ∈ [1, ∞]. Let and , 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. Let λ0 = f = 0 in the proofs of Lemmas 42, 45, and 46, Corollary 44, and Theorem 48. In this case, (169) and (174) reduce to and . Hence, (180)–(182) change to ,, and .
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 , γ ∈ (5 − 3α − 2β, 1), r ∈ [1, ∞], and let χα,β,γ = (α + β + γ − 2)/α. Then, for every fixed 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. Let , 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 . 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 is well defined. Hence, let 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 . Moreover (see the proof of Lemma 46), since , γ ∈ (5 − 3α − 2β, 1)⊆(2 − α − β, 1) and , 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 yield . Summing up, we find that . 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 , , and , where , , i2 = 1, …, n2, and r ∈ [1, ∞]. Let and , 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. Let , i1 = 1, …, n1, in problem (160) and formulae (169), (174), and (179)–(182). Then, , and . Therefore, as in Theorem 54, we do not need Lemma 42 and Corollary 44, and the proof of Theorem 48 simplifies as follows. Again, implies that χα,β,γ ∈ ((3 − 2α − β)/α, 1), so that , and the interval Iα,β,τ is well defined. Let δ ∈ Iα,β,τ be fixed. First (see the proof of Lemma 45), since , Lemma 30 applied with yields , i2 = 1, …, n2. On the other side (see the proof of Lemma 46), since and , γ ∈ (5 − 3α − 2β, 1)⊆(2 − α − β, 1), from Lemma 37 and Corollary 38 applied with and (x, δ7) = (x0, δ) we deduce that ,, i2 = 1, …, n2. Summing up, we find that , and the assertion again follows from v = L−1w + v0 and (cf. (206)) .
6. An Application to a Concrete Case
Theorem
48 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]):
()
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
()
where
and
,
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:
()
()
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
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
()
from (
208) and (
209), it thus follows that the temperature Θ must satisfy the following equation:
()
Let us now assume that
a is of the following special form:
()
where the functions
mn and
un,
n = 1,2, satisfy the following conditions (cf. (
210)):
()
()
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
()
()
()
()
()
()
()
Then, since (
214)–(
216) yield
a(0,
x) =
m1(
x)
u1(0) and
,
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:
()
We endow this differential equation with the initial condition Θ(0,
x) = Θ
0(
x),
x ∈
Ω, and the Dirichlet boundary condition Θ(
t,
x) = 0,
t ∈
IT,
x ∈
∂Ω.
We now suppress the dependence on
x ∈
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
for every
u ∈
Lq(
Ω),
n = 1,2, we set
()
()
()
()
()
Here (cf. [
36, Chapter 7]),
,
k ∈
N ∪ {0},
q ∈ (1,
∞), denotes the usual Sobolev space endowed with the norm ∥·∥
k,q;Ω (
), whereas
denotes the completion of
in
,
being the set of all real-valued infinitely differentiable functions having compact support in
Ω. We further assume that there exists positive constant Λ
i,
i = 0, …,
r1, such that for every
the following inequalities hold:
()
where
. Therefore, from (
212), (
218), and (
230) we get
()
From (
225)–(
231) it follows that
M,
L, and
Li,
i = 1, …,
n1, are closed linear operators from
X to itself, and the relation
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
. Hence, a fortiori,
L−1 ∈
ℒ(
X) and so condition (
161) is satisfied (observe also that
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
,
being a positive constant depending on
). 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)):
()
where
Σ1 = {
z ∈
C :
ℜe
z ≥ −
c(|
ℑm
z | + 1),
ℑm
z ∈
R},
c being a suitable positive constant depending on
q and
. 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)):
()
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:
()
Now, assume for a moment that we are interested in solving the
inverse problem of recovering both the temperature Θ and the memory kernels
in (
234). Clearly, due to (
222), if we recover
, then the heat conduction relaxation functions
will be known too, unless of the
r1 arbitrary constants
bi(0),
i = 1, …,
r1. Indeed,
,
t ∈
IT. To solve such an inverse problem, we need
r1 additional informations other than the initial condition Θ(0) = Θ
0, 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:
()
where Ψ
j ∈
X* =
ℒ(
X;
R) and
,
νj ∈ (0,1),
j = 1, …,
r1. We will search for a solution vector
of the inverse problem (
234) and (
235) such that Θ ∈
C1+δ(
IT;
𝒟(
L)) and
,
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 that Θ ∈
C1+δ(
IT;
𝒟(
L)) solves (
234), we introduce the new unknown
()
Then, differentiating (
234) with respect to time and using
()
we find that
v ∈
Cδ(
IT;
𝒟(
L)) solves the following degenerate integrodifferential problem:
()
where
yi =
LiΘ
0,
i = 1, …,
n1,
and
(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
). Of course, (
238) is the special case (
i1,
i2,
n2) = (
i,
i,
n1),
hi =
ki,
i = 1 … ,
n1, of problem (
160).
Conversely, assume that
v ∈
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
()
Now, observe that
()
()
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:
()
Therefore, replacing (
240)–(
242) in (
239), it follows for every
t ∈
IT that
()
and the latter integral is equal to zero by virtue of (
238). Since from (
236) it follows that Θ(0) = Θ
0, 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
with the help of the additional information (
235).
Let us now apply the linear functional Ψ
j,
j = 1 … ,
r1, to (
238). Using
()
we thus find the following system of
r1 equations for the
r1 unknown
:
()
where we have set (recall that
,
n = 1,2, are known)
()
Therefore, if the matrix
has determinant det
𝒰 ≠ 0, then from Cramer’s formula it follows that the solution
of (
245) is given by
()
with
𝒰k,j,
k,
j = 1, …,
r1, being the cofactor of the element Ψ
k[
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
.
Now, let
,
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
()
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
()
where
w =
L(
v −
v0) and
()
Here, the
Qj’s,
j = 2, …, 6, are defined by (
106)–(
110),
Si =
LiL−1, and the functions
,
and
Q7x0 are defined by
,
, and [
Q7x0](
t) = (e
tA −
I)
x0, respectively, where (cf. (
174))
.
Then, since
v =
L−1w +
v0, if we set
,
j = 1, …,
r1, and
()
from (
247) and (
249) we deduce that to solve the inverse problems (
234) and (
235) for the unknown vector
, it suffices to show that the fixed-point equation
()
has a unique solution. In general, this will be done by proving that Ξ is a contraction map in the Banach space
()
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. Let X, 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 , ψ ∈ (0,1), r ∈ [1, ∞]. Let ηi ∈ (1/q′, 1) and γi, φ ∈ (2/q′, 1), i = 1, …, n1, and assume that
()
where
ki,
and
λ0 are defined by (
222) and (
223) through (
211) and (
216), whereas
v1 = (
λ0M +
L)
v0. Let
and
, and let
I1,1/q,τ⊆(1/
q′, 1/2) be the interval defined by (cf. (
207) with
β = 1/
q)
()
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
,
μ ∈ [
δ + 1/
q′, 1).
Proof. Apply Theorem 48 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 .
Larger values of
q in Theorem
57 can be obtained assuming more smoothness and some order of vanishing for the function
m1. In fact, let
be such that the following estimate holds for some positive constant
K:
()
Then (
232) holds with
β = 1/
q being replaced by (cf. [
41, formulae (3.23) and (4.41)]):
()
(precisely, in [
41, formula (3.23)] it is shown that
, where
u = (
λM −
L)
−1f and
q ∈ [2,
∞). Using (cf. [
41, formula (2.15)])
, we thus find that
; that is,
). 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 and , 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 , μ ∈ [δ + 1 − β, 1).
Proof. It suffices to observe that for every ϑ ∈ (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.