PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 85(99) (2009), 63–106 DOI:10.2298/PIM0999063K

DISTRIBUTION GROUPS

Marko Kostić

Communicated by Stevan Pilipović

Abstract. We introduce distribution groups and [B_{0}, …, B_{n} , C_{0}, …, C_{n−1} ]- groups with not necessarily densely defined generators and systematically analyze relations between them.

1. Introduction

Distribution semigroups and their generators were introduced by Lions in the pioneering paper [31] and almost four decades after that, Kunstmann [28] and Wang [42] analyzed distribution semigroups with non-densely defined generators. Balabane and Emami-Rad [4]–[5] were the first who defined smooth distribution groups and applied them in the analysis of Schrödinger evolution equations in L^{p}(ℝ^{n})-type spaces. On the other hand, global integrated groups were introduced and investigated by El-Mennaoui in his doctoral dissertation [13]. We refer the reader to [3]–[6], [12]–[13], [16], [18]–[20], [26] and, especially, to the paper [33] where Miana analyzed global α-times integrated groups and smooth distribution groups in the framework of fractional calculus. It is also meaningful to accent that Keyantuo [20] briefly considered an abstract Laplacian in L^{p}(ℝ^{n})-type spaces and proved several relations between exponentially bounded integrated cosine functions and global integrated groups. For further information, see [20, Theorem 1.2, Propositions 2.1–2.2, Theorem 2.6 and Proposition 4.2]. The class of (local) convoluted C-groups extending the well known classes of integrated groups and regularized groups has been recently introduced in [26].

In a series of papers, many authors relate global integrated groups to functional calculi and proved, in such a way, different generalizations of Stone’s theorem. For various aspects in this direction, we refer to [6], [8]–[12], [14] and [16]. Further on, Galé and Miana [18] have recently introduced one-parameter groups of regular quasimultipliers within Esterle’s theory of quasimultipliers [15] and applied them in the study of regularized, distribution, integrated groups as well as holomorphic semigroups and functional calculi.

2000 Mathematics Subject Classification: Primary 47D06; Secondary 47D09, 47D60, 47D62, 47D99.

Partially supported by grant 144016 of Ministry of Science and Technological Development of Republic Serbia.

Operator-valued distribution groups considered in this article do not fall under the scope of [18, Definition 3.4] since our concept does not contain any density and growth assumptions. The assertions which link distribution groups of [18] to global integrated groups with the corresponding growth order established in [18, Propositions 3.7–3.8] with the help of the Riesz functions and the Weyl homomorphisms are no longer applicable and this is the main reason why we analyze local integrated groups. Furthermore, we focus our attention to the following system of convolution type equations (the notions and terminology are explained below):

(1.1) G ∗ (δ′⊗ I − δ ⊗ A) = 0 ⊗ I_{[D(A)]} and (δ′⊗ I − δ ⊗ A) ∗ G = 0 ⊗ I_{E},

where A is a closed linear operator acting on a Banach space E, δ′⊗ I − δ ⊗ A ∈ D′(L([D(A)], E)), G ∈ D′(L(E, [D(A)])) and I denotes the inclusion D(A) → E. Contrary to the case of distribution semigroups and distribution cosine functions (cf. [28, Theorem 3.10, pp. 844–845] and [23, Theorem 3.3]), the uniqueness of solutions of (1.1) is not satisfied. Here we stress that every operator-valued distribution G satisfying, for every φ ∈ D and x ∈ E:

(1.2) G ∈ D′(L(E)), G(φ)x ∈ D(A), AG(φ)x = G(−φ′)x, G(φ)A ⊆ AG(φ),

can be viewed as an element of the space D′(L(E, [D(A)])) which solves (1.1) (cf. also [33]). It turns out that the introduced class of [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]- groups presents a natural framework for investigation of equations involving operators satisfying (1.2). Roughly speaking, such a concept enables one to consider in a unified treatment the notions of integrated groups and regularized groups ([9]– [12]) as well as to get through to the new important relations between distribution groups and local integrated groups.

The paper is organized as follows. In Section 2, we characterize the basic structural properties of (degenerate) distribution groups, connect local integrated groups to analytic integrated semigroups, global differentiable regularized groups and establish a complex variable characterization of generators of local integrated groups. In this section, it is also proved that every generator of a local integrated group is also the generator of a distribution group. The third section is devoted to the study of (exponentially bounded) [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-groups and their subgenerators. The composition property of a [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group is proved only in the case when a subgenerator of such a group commutes with B_{1}, …, B_{n}, C_{0}, …, C_{n−1} . The loss of commutativity is disagreeable and additionally hinders our work. Section 4 is the systematic exposition of distribution groups. Our main results are Theorem 4.1 and Theorem 4.2; concerning these theorems, we would like to point out that the order of the operator-valued distribution G solving (1.1) plays a crucial role. In such a way, we notice the remarkable differences between once integrated groups and n-times integrated groups, where n ∈ ℕ and n > 1. Theorem 4.1 describes solutions of (1.1) which fulfill the condition (DG)_{4} stated below. The fundamental relationship between distribution groups and local integrated groups is established in Theorem 4.2(v) and says that the generator A of a distribution group is also the generator of a local integrated group, if ρ(A) ≠ ∅. In the present situation, the author does not know whether there exists a distribution group whose generator possesses the empty resolvent set.

The analysis of ultradistribution and (Fourier) hyperfunction groups [25] is an open problem since the argumentation presented in this paper becomes quite inoperative and cannot be employed anymore.

By E and L(E) are denoted a complex Banach space and the Banach algebra of bounded linear operators on E. For a closed linear operator A on E, D(A), N(A), R(A), ρ(A) denote its domain, kernel, range and resolvent set, respectively, while [D(A)] stands for the Banach space D(A) equipped with the graph norm. Put D_{∞} (A) := ⋂ _{n=0}^{∞} D(A^{n}) and ∥x∥_{n} := ∑ _{i=0}^{n} ∥A^{i}x∥, n ∈ ℕ, x ∈ D(A^{n}). Further, let us recall that A is stationary dense [27] if

n(A) := inf \{k ∈ ℕ_{0} : D(A^{m}) ⊆ \overline{D(A^{m+1} )} for all m ⩾ k\} < ∞.

If Y is a subspace of E, denote by A_{Y} the part of A in Y, i.e., A_{Y} = {(x, y) ∈ A : x ∈ Y, y ∈ Y }. We assume henceforth C ∈ L(E) and C is injective.

Schwartz spaces of test functions on the real line ℝ are denoted by D = C_{0}^{∞} , E = C^{∞} and S. Their strong duals are D′, E′and S′, respectively. By D_{0} we denote the subspace of D which consists of the elements supported by [0, ∞). Further on, D′(L(E)) = L(D, L(E)), E′(L(E)) = L(E, L(E)) and S′(L(E)) = L(S, L(E)) are the spaces of continuous linear functions D → L(E), E → L(E) and S → L(E), respectively, equipped with the topology of uniform convergence on bounded subsets of D, E and S, respectively; D′_{0}(L(E)), E′_{0}(L(E)) and S′_{0}(L(E)) are the subspaces of D′(L(E)), E′(L(E)) and S′(L(E)), respectively, containing the elements supported by [0, ∞). Let ρ ∈ D satisfy ∫_{−∞}^{∞} ρ(t) dt = 1 and suppρ ⊆ [0, 1]. By a regularizing sequence we mean a sequence (ρ_{n}) in D_{0} obtained by ρ_{n}(t) := nρ(nt), t ∈ ℝ, n ∈ ℕ. If K ⊆ ℝ, put D_{K} := {φ ∈ D : suppφ ⊆ K}. In this paper, the convolution of operator-valued distributions is taken in the sense of [28, Proposition 1.1]. Suppose t ∈ ℝ. A distribution δ_{t} is defined by δ_{t}(φ) := φ(t), φ ∈ D. Further, if φ ∈ D and G ∈ D′(L(E)), we define φ̌(·) := φ(−·) and Ǧ(·) := G(ˇ·). Clearly, (φ ∗ ψ)ˇ = φ̌ ∗ ψ̌ and φ̌^{(n)} = (−1)^{n}(φ^{(n)} )ˇ , φ, ψ ∈ D, n ∈ ℕ.

Let a > 0 and b > 0. The exponential region E(a, b) is defined in [1] by

E(a, b) := {λ ∈ ℂ | Reλ ⩾ b, | Imλ| ⩽ e^{a Re λ} }.

Finally, if f : ℝ → ℂ and t ∈ ℝ, put τ_{t}f(s) := f(s − t), s ∈ ℝ.

2. Structural properties of distribution groups

We need the following definition of a C-distribution semigroup.

Definition 2.1. [22] Let G ∈ D′_{0}(L(E)) and CG = GC. If

(C.D.S.1)  G(φ ∗_{0} ψ)C = G(φ) G(ψ), φ, ψ ∈ D,

where φ ∗_{0} ψ(t) = ∫_{0}^{t} φ(t − u)ψ(u) du, t ∈ ℝ, then G is called a pre-(C-DSG) and if, additionally,

(C.D.S.2)  N (G) := ⋂_{φ∈D_{0}} N(G(φ)) = {0},

then G is called a C-distribution semigroup, (C-DSG) in short.

Let G be a (C-DSG) and let T ∈ E′_{0}. Define G(T) on a subspace of E by

y = G(T)x iff G(T ∗ φ)x = G(φ)y for all φ ∈ D_{0}.

Denote its domain by D(G(T)). By (C.D.S.2), G(T) is a function. Moreover, G(T) is a closed linear operator. If φ ∈ D, put φ_{+} (t) := φ(t)H(t) and φ_{−} (t) := φ(t)H(−t), t ∈ ℝ, where H(·) is the Heaviside function. Then φ_{+} , φ_{−} ∈ E′ and the definitions of G(φ_{+} ) and G(φ_{−} ) are clear. We know that G(φ) = 0, φ ∈ D_{(−∞,0]} and that G(φ_{+} )C = G(φ), φ ∈ D [22].

The infinitesimal generator of a (C-DSG) G is defined by A := G(−δ′).

Finally, if C = I, then we also say that G is a distribution semigroup, (DSG) shortly; if this is the case, then there is no risk for confusion and we also write G for G.

Definition 2.2. An element G ∈ D′(L(E)) is called a pre-distribution group, pre-(DG) in short, if the following condition holds:

(DG)_{1}  G(φ ∗ ψ) = G(φ) G(ψ) for all φ, ψ ∈ D.

If G additionally satisfies:

(DG)_{2}  N (G) := ⋂_{φ∈D} N(G(φ)) = {0},

then G is called a distribution group, (DG) shortly. A pre-(DG) G is dense iff:

(DG)_{3}  The set R(G) := ⋃_{φ∈D} R(G(φ)) is dense in E.
Suppose G ∈ D′(L(E)) satisfies (DG)_{2} and T ∈ E′. We define G(T) by
G(T) := {(x, y) ∈ E^{2} | G(T ∗ φ)x = G(φ)y for all φ ∈ D}.

Due to (DG)_{2}, G(T) is a function and it is straightforward to see that G(T) is a closed linear operator in E.

The generator A of a (DG) G is defined by A := G(−δ′). Notice, if G is a (DG) generated by A, then (1.2) holds.

Further on, an element G ∈ D′(L(E)) is called regular (representable) if the following holds:

(DG)_{4} For every x ∈ R(G), there is a function t ↦ u(t; x), t ∈ ℝ satisfying:

u(·; x) ∈ C(ℝ : E), u(0; x) = x and G(ψ)x = ∫_{−∞}^{∞} ψ(t) u(t; x) dt, ψ ∈ D.

It is checked at once that the function u(·; x) is unique.

Example 2.1. (i) Suppose ±A are generators of C-distribution semigroups G_{±} . Put G(φ) := G_{+} (φ) + G_{−} (φ̌), φ ∈ D. Then A and G fulfill (1.2). Indeed, G ∈ D′(L(E)), G(φ)A ⊆ AG(φ), φ ∈ D, AG_{+} (φ)x = G_{+} (−φ′)x − φ(0)Cx and −AG_{−} (φ)x = G_{−} (−φ′)x − φ(0)Cx, φ ∈ D, x ∈ E (cf. [22]). So, AG(φ)x = G_{+} (−φ′)x − φ(0)Cx + G_{−} (φ̌′)x + φ̌(0)Cx = G_{+} (−φ′)x + G_{−} (−φ̌′)x = G(−φ′)x, x ∈ E, φ ∈ D. Furthermore, it can be proved the following: G(φ∗ψ)C = G(φ)G(ψ), φ, ψ ∈ D (cf. [22], [26] and [33]), ⋂_{φ∈D_{0}} N(G(φ)) = {0} and ⋂_{φ∈D_{0}} N(G(φ̌)) = {0}.

(ii) Suppose G is a (DG), P ∈ L(E), P^{2} = P and GP = P G. Set G_{P}(φ)x := G(φ)P x, φ ∈ D, x ∈ E. Then G_{P} is a pre-(DG) and N (G_{P}) = N(P).

(iii) Suppose A and G fulfill (1.2). Define G_{T} (T ∈ E′) by G_{T}(φ)x := G(T ∗φ)x, φ ∈ D, x ∈ E. Then (1.2) holds for A and G_{T}.

(iv) [9, Example 16.3] Let E := {f : ℝ → ℂ is continuous | lim_{|x|→∞} e^{x^{2}} f(x) =

0\}, ∥f∥ := sup_{x∈ℝ} |e^{x^{2}} f(x)|, f ∈ E and A := \frac{d}{dx} with maximal domain. Put (S(t)f)(x) := e^{−(x+t)^{2}} f(x+t), x ∈ ℝ, t ∈ ℝ, f ∈ E. Then S(t)f ∈ E, ∥S(t)∥ ⩽ e^{2t^{2}} , ∫_{0}^{t} S(s)f ds ∈ D(A) and A ∫_{0}^{t} S(s)f ds = S(t)f − S(0)f, t ∈ ℝ, f ∈ E. Let f ∈ E and φ ∈ D be fixed. Set G(φ)f := ∫_{−∞}^{∞} φ(t)S(t)f dt. Clearly, G ∈ D′(L(E)) and the partial integration shows G(φ)f ∈ D(A), AG(φ)f = G(−φ′)f and

(G(φ)Af − AG(φ)f)(x) = 2 ∫_{−∞}^{∞} φ(t)(x + t)e^{−(x+t)^{2}} f(x + t) dt, x ∈ ℝ.

Consequently, A does not commute with G(·) and (1.2) does not hold. Furthermore, it can be checked directly that G fulfills (DG)_{2} and that G is not regular.

(v) Let F denote the Fourier transform on the real line,

F(f)(ξ) = \frac{1}{2π} ∫_{−∞}^{∞} e^{−iξt} f(t)dt, ξ ∈ ℝ.

Suppose that E is a quasi-spectral distribution in the sense of [12, Definition 2.2] and that E can be continuously extended to S. Put F(D) := {F(φ) : φ ∈ D} and G(φ) := E(F^{−1} (φ)), φ ∈ S, where F^{−1} denotes the inverse Fourier transform. Then G ∈ S′(L(E)), G(φ∗ψ) = G(φ)G(ψ), φ, ψ ∈ S and ⋂ _{φ∈F(D)} N(G(φ)) = {0}. Suppose, additionally, that for every x ∈ E and ϕ ∈ S with ϕ(0) = 1:

(2.1)  lim_{n→∞} E(ϕ_{n})x = x, where ϕ_{n}(t) = ϕ(t/n), t ∈ ℝ, n ∈ ℕ.

Notice that (2.1) implies that E is a spectral distribution in the sense of [12, Definition 2.4] (cf. also [6, Definition 1.1]). We will show that ⋂ _{φ∈D_{0}} N(G(φ)) = {0}. Indeed, suppose ρ ∈ D, ∫_{−∞}^{∞} ρ(t) dt = 1, suppρ ⊆ [0, 1] and G(φ)x = 0, φ ∈ D_{0}, i.e., E(F^{−1} (φ))x = 0, φ ∈ D_{0}. Let ϕ(t) = F^{−1} (ρ)(t) = ∫_{−∞}^{∞} e^{iξt}ρ(ξ) dξ, t ∈ ℝ. Then ϕ ∈ S and ϕ(0) = 1. Put ρ_{n}(t) = nρ(nt) and ϕ_{n}(t) = F^{−1} (ρ_{n})(t), t ∈ ℝ, n ∈ ℕ. Clearly, ϕ_{n}(t) = ϕ(t/n), t ∈ ℝ, n ∈ ℕ and (2.1) implies x = lim_{n→∞} E(ϕ_{n})x = lim_{n→∞} E(F^{−1} (ρ_{n}))x = 0. Analogously, ⋂ _{φ∈D_{0}} N(G(φ̌)) = {0} and this implies (DG)_{2} for G.

Further on, a closed linear operator A satisfying (1.2) need not be the generator of a (DG) and this implies that relations between distribution groups and convolution type equations are, at least, quite unclear.

The proofs of the following assertions are omitted.

Lemma 2.1. Suppose G is a pre-(DG). Then Ǧ is a pre-(DG). If, in addition, G is a (DG) generated by A, then Ǧ is a (DG) generated by −A.

Proposition 2.1. [28] Let G be a pre-(DG), F := E/N (G) and q be the corresponding canonical mapping q : E → E/N (G).

(a) Let H ∈ L(D, L(F)) be defined by qG(φ) := H(φ)q for all φ ∈ D. Then H is a (DG) in F.

(b) \overline{⟨R(G)⟩}= \overline{R(G)}, where ⟨R(G)⟩ is the linear span of R(G).

(c) Assume that G is not dense. Set R := \overline{R(G)} and H := G_{|R}. Then H is a dense pre-(DG) in R. Moreover, if G is a (DG) generated by A, then H is a (DG) in R generated by A_{R}.

(d) The adjoint G^{∗} of G is a pre-(DG) in E^{∗} with N (G^{∗}) = \overline{R(G)}^{∘}. (\overline{R(G)}^{∘} is the polar of \overline{R(G)}.)

(e) If E is reflexive, then N (G) = \overline{R(G^{∗})}^{∘}.

(f) G^{∗} is a (DG) in E^{∗} iff G is a dense pre-(DG). If E is reflexive, then G^{∗} is a dense pre-(DG) in E^{∗} iff G is a (DG).

(h) N (G) ∩ ⟨R(G)⟩ = {0}.

(j) Suppose x = G(φ)y, for some φ ∈ D and y ∈ E. Set u(t; x) := G(τ_{t}φ)y, t ∈ ℝ. Then u(0; x) = x, u(·; x) ∈ C^{∞} (ℝ : E), \frac{d^{n}}{dt^{n}} u(t; x) = A^{n}u(t; x), t ∈ ℝ, n ∈ ℕ_{0} and G(ψ)x = ∫_{−∞}^{∞} ψ(t) u(t; x) dt, ψ ∈ D. Hence, G is regular.

Proposition 2.2. [28] Let G be a (DG) and let S, T ∈ E′, φ ∈ D and x ∈ E. Then:

(a) (G(φ)x, G(\underbrace{T ∗ ⋯∗ T}_{m}∗φ)x) ∈ G(T)^{m}, m ∈ ℕ.

(b) G(S)G(T) ⊆ G(S ∗ T), D(G(S)G(T)) = D(G(S ∗ T)) ∩ D(G(T)) and G(S) + G(T) ⊆ G(S + T). In general, G(S)G(T) ≠ G(S ∗ T).

(c) G(φ)G(T) ⊆ G(T)G(φ).

(d) If G is dense, its generator is densely defined.

Suppose, for the time being, that D′(L(E)) ∋ G fulfills (DG)_{3} and (DG)_{4}. Then G is a pre-(DG) iff:

(2.2) ⋃_{\begin{matrix}t∈ℝ, \\ x∈R(G) \end{matrix}} u(t; x) ⊆ R(G) and u(t + s; x) = u(t; u(s; x)), t, s ∈ ℝ, x ∈ R(G).

The necessity of (2.2) follows directly from Proposition 2.1(j). To prove the sufficiency, notice that

G(φ ∗ ψ)x =∫_{−∞}^{∞} ∫_{−∞}^{∞} [φ(t − s) ψ(s) ds]u(t; x) ds dt = ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ(t) ψ(s) u(t + s; x) ds dt

=∫_{−∞}^{∞} φ(t) ∫_{−∞}^{∞} ψ(s) u(s; u(t; x)) ds dt = ∫_{−∞}^{∞} φ(t) G(ψ) u(t; x) dt

= G(ψ) ∫_{−∞}^{∞} φ(t) u(t; x) dt = G(ψ) G(φ)x, x ∈ R(G).

The denseness of R(G) in E automatically implies (DG)_{1}.

From now on, we employ the following definition of an α-times integrated Csemigroup.

Definition 2.3. Let A be a closed operator, 0 < τ ⩽ ∞ and α > 0. If there exists a strongly continuous operator family (S(t))_{t∈[0,τ)} such that S(t)A ⊆ AS(t), t ∈ [0, τ), S(t)C = CS(t), t ∈ [0, τ), ∫_{0}^{t} S(s)x ds ∈ D(A), x ∈ E, t ∈ [0, τ) and

A ∫_{0}^{t} S(s)x ds = S(t)x − \frac{t^{α}}{Γ(α + 1)} Cx, x ∈ E, t ∈ [0, τ),

then (S(t))_{t∈[0,τ)} is called a (local, if τ < ∞) α-times integrated C-semigroup with a subgenerator A. If τ = ∞, then we say that (S(t))_{t⩾0} is an exponentially bounded, α-times integrated C-semigroup with a subgenerator A if, additionally, there exist M > 0 and ω ∈ ℝ such that ∥S(t)∥ ⩽ Me^{ωt}, t ⩾ 0.

We know (cf. [24] and [29]–[30]) that (S(t))_{t∈[0,τ)} satisfies S(t)S(s) = S(s)S(t), 0 ⩽ t, s < τ and

S(t)S(s)x = [ ∫_{0}^{t+s} − ∫_{0}^{t} − ∫_{0}^{s} ] \frac{(t + s − r)^{α−1}}{Γ(α)} S(r)Cx dr, x ∈ E, 0 ⩽ t, s, t + s < τ.

In general, a subgenerator of (S(t))_{t∈[0,τ)} is not unique but, in the case C = I, every subgenerator is unique and coincides with the (integral) generator of (S(t))_{t∈[0,τ)} , defined by

\{ (x, y) ∈ E^{2} : S(t)x − \frac{t^{α}}{Γ(α + 1)} Cx = ∫_{0}^{t} S(s)y ds for all t ∈ [0, τ)\} .

We refer the reader to [43]–[44] for the definition of a local regularized semigroup and its generator. Suppose n ∈ ℕ and τ ∈ (0, ∞); then it is well known [44] that a closed linear operator A generates a local n-times integrated semigroup on [0, τ) if and only if ρ(A) ≠ ∅ and A generates a local R(λ : A)^{n}-semigroup on [0, τ), where λ ∈ ρ(A).

Definition 2.4. [26] Let A and B be closed linear operators, τ ∈ (0, ∞] and α > 0. A strongly continuous operator family (S(t))_{t∈(−τ,τ)} is called a (local, if τ < ∞) α-times integrated group generated by A if:

(i) (S_{+} (t) := S(t))_{t∈[0,τ)} and (S_{−} (t) := S(−t))_{t∈[0,τ)} are (local) α-times integrated semigroups generated by A and B, respectively, and

(ii) for every x ∈ E and t, s ∈ (−τ, τ) with t < 0 < s:

S(t)S(s)x=S(s)S(t)x=\{ \begin{matrix}∫_{t+s}^{s} \frac{(r−t−s)^{α−1}}{Γ(α)} S(r)x dr+∫_{t}^{0} \frac{(t+s−r)^{α−1}}{Γ(α)} S(r)x dr, t+s ⩾ 0, \\ ∫_{t}^{t+s} \frac{(t+s−r)^{α−1}}{Γ(α)} S(r)x dr+∫_{0}^{s} \frac{(r−t−s)^{α−1}}{Γ(α)} S(r)x dr, t+s < 0. \end{matrix}

Lemma 2.2. [26] (i) Let (S(t))_{t∈(−τ,τ)} be an α-times integrated group generated by A, for some α > 0 and τ ∈ (0, ∞]. Put Š(t) := S(−t), t ∈ (−τ, τ). Then (Š(t))_{t∈(−τ,τ)} is an α-times integrated group generated by B.

(ii) Suppose τ ∈ (0, ∞], α > 0 and A is the generator of an α-times integrated group (S(t))_{t∈(−τ,τ)} . Then there exist a > 0 and b > 0 so that:

(ii.1) E(a, b) ⊆ ρ(A) ∩ ρ(B), R(λ : A)S(t) = S(t)R(λ : A), λ ∈ E(a, b), t ∈

(−τ, 0] and R(λ : B)S(s) = S(s)R(λ : B), λ ∈ E(a, b), s ∈ [0, τ). (ii.2) S(t)A ⊆ AS(t), t ∈ (−τ, 0] and S(s)B ⊆ BS(s), s ∈ [0, τ).

(iii) Suppose β > α > 0 and A is the generator of an α-times integrated group (S_{α} (t))_{t∈(−τ,τ)} . Put

S_{β}(t) := \{ \begin{matrix}∫_{0}^{t} \frac{(t−s)^{β−α−1}}{Γ(β−α)} S_{α} (s)x ds, & t ∈ [0, τ), x ∈ E \\ ∫_{0}^{−t} \frac{(−t−s)^{β−α−1}}{Γ(β−α)} S_{α} (−s)x ds, & t ∈ (−τ, 0), x ∈ E. \end{matrix}

Then (S_{β}(t))_{t∈(−τ,τ)} is a β-times integrated group generated by A.

(iv) Let α > 0, τ ∈ (0, ∞] and let (S(t))_{t∈(−τ,τ)} be an α-times integrated group generated by A. Then B = −A.

(v) Suppose 0 < τ ⩽ ∞, α > 0 and ±A generate α-times integrated semigroups (S_{±} (t))_{t∈[0,τ)} . Then A generates an α-times integrated group (S(t))_{t∈(−τ,τ)} given by: S(t) =: S_{+} (t), t ∈ [0, τ), S(t) := S_{−} (−t), t ∈ (−τ, 0).

(vi) Suppose 0 < τ ⩽ ∞ and α > 0. A strongly continuous operator family (S(t))_{t∈(−τ,τ)} is an α-times integrated group generated by A if and only if (S_{±} (t))_{t∈[0,τ)} are α-times integrated semigroups generated by ±A.

Proposition 2.3. Suppose ±A generate distribution semigroups G_{±} and put G(φ) := G_{+} (φ) + G_{−} (φ̌), φ ∈ D. Then G is a (DG) generated by A.

Proof. By the standard arguments, we have that there exists n ∈ ℕ such that, for every k ∈ ℕ, ±A generate (2^{k}n)-times integrated semigroups (S_{±}^{k} (t))_{t∈[0,2^{k} τ)} .

Then one obtains

G_{+} (φ)x = (−1)^{2^{k} n} ∫_{0}^{∞} φ^{(2^{k} n)} (t)S_{+}^{k} (t)x dt and

G_{−} (φ)x = (−1)^{2^{k} n} ∫_{0}^{∞} φ^{(2^{k} n)} (t)S_{−}^{k} (t)x dt, φ ∈ D_{(−∞,2^{k} τ)} .

In order to prove that G is a (DG) generated by A, suppose x ∈ N (G). Then, for every φ ∈ D_{0}, G(φ)x = 0 and this implies G_{+} (φ)x = 0, φ ∈ D_{0}. Since G_{+} is a (DSG) generated by A, we have x = 0 and (DG)_{2} holds for G. Note that Lemma 2.2(vi) implies that, for every k ∈ ℕ, A generates a local (2^{k}n)-times integrated group (S^{k}(t))_{t∈(−2^{k} τ,2^{k} τ)} . Now one can repeat literally the arguments given in the proof of [33, Theorem 6] in order to conclude that (DG)_{1} holds for all φ, ψ ∈ D_{(−2^{k−1} τ,2^{k−1} τ)} . Hence, G satisfies (DG)_{1}. It remains to prove that B = A, where B is the generator of G. Suppose (x, y) ∈ B. Then G(−φ′)x = G(φ)y, φ ∈ D, i.e., G_{+} (−φ′)x+G_{−} ( ˇ −φ′ )x = G_{+} (φ)x+G_{−} (φ̌)x, φ ∈ D. This, in particular, holds for every φ ∈ D_{0} and one obtains G_{+} (−φ′)x = G_{+} (φ)x, φ ∈ D_{0}. In other words, B ⊆ A. Assume now (x, y) ∈ A. Then the definition of G and [28, Lemma 3.6] imply:

G(φ)y = G(φ)Ax = G_{+} (φ)Ax + G_{−} (φ̌)Ax
= G_{+} (−φ′)x − φ(0)x − G_{−} (−φ̌′)x + φ(0)x
= G_{+} (−φ′)x + G_{−} ( ˇ −φ′ )x = G(−φ′)x, φ ∈ D.

This gives A ⊆ B and ends the proof of proposition. □

The previous theorem implies that a wide class of multiplication operators acting on L^{p}(ℝ^{n})-type spaces can be used for the construction of distribution groups. In particular, several examples presented in [1] offers one to construct local once integrated groups which can be explicitly calculated.

The following corollary is an immediate consequence of Lemma 2.2, [30, Theorems 2.1–2.2] and [43, Corollary 2.7].

Corollary 2.1. (a) Suppose α > 0, τ ∈ (0, ∞] and A generates an α-times integrated group (S_{α} (t))_{t∈(−τ,τ)} . Then, for every a ∈ (0, \frac{τ}{α} ), there exist b > 0 and M > 0 so that:

(2.3)  E(a, b) ⊆ ρ(±A) and ∥R(λ : ±A)∥ ⩽ M|λ|^{α} , λ ∈ E(a, b).

(b) Suppose there exist a > 0, b > 0, M > 0 and α > −1 so that (2.3) holds. Then, for every β > α + 1 and τ = a(β − α − 1), A generates a local β-times integrated group (S_{β}(t))_{t∈(−τ,τ)} .

(c) Suppose n ∈ ℕ, k ∈ ℕ, τ ∈ (0, ∞) and A generates a local n-times integrated group (S_{n}(t))_{t∈(−τ,τ)} . Then A generates a local (kn)-times integrated group (S_{kn} (t))_{t∈(−kτ,kτ)} .

Suppose α ∈ (0, ∞), α ∉ ℕ and f ∈ S. Put n = ⌈α⌉ := inf{k ∈ ℤ : k ⩾ α}.

Recall [35], the Weyl fractional derivatives W_{+}^{α} and W_{−}^{α} of order α are defined by:

W_{+}^{α} f(t) := \frac{(−1)^{n}}{Γ(n − α)} \frac{d^{n}}{dt^{n}} ∫_{t}^{∞} (s − t)^{n−α−1} f(s) ds, t ∈ ℝ

W_{−}^{α} f(t) := \frac{1}{Γ(n − α)} \frac{d^{n}}{dt^{n}} ∫_{−∞}^{t} (t − s)^{n−α−1} f(s) ds, t ∈ ℝ.

If α = n ∈ ℕ, put W_{+}^{n} := (−1)^{n} \frac{d^{n}}{dt^{n}} and W_{−}^{n} := \frac{d^{n}}{dt^{n}} . Then we know [33] that W_{±}^{α+β} = W_{±}^{α} W_{±}^{β}, α > 0, β > 0.

The following result can be attributed to Miana.

Theorem 2.1. [33] Suppose α > 0 and (S(t))_{t∈ℝ} is an α-times integrated group generated by A. Put G(φ)x := ∫_{0}^{∞} W_{+}^{α} φ(t)S(t)x dt+∫_{0}^{∞} W_{+}^{α} φ̌(t)S(−t)x dt, φ ∈ D, x ∈ E. Then G is a (DG) generated by A.

Notice that, in the case α = n ∈ ℕ, we have the following equality:

G(φ)x = (−1)^{n} ∫_{0}^{∞} φ^{(n)} (t)S(t)x dt + ∫_{−∞}^{0} φ^{(n)} (t)S(t)x dt.

We refer the reader to [9, Section XXI] and [24]–[25] for the basic material concerning analytic integrated semigroups.

Remark 2.1. Let α > 0, ω ⩾ 0 and let A be the generator of an α-times integrated group (S_{α} (t))_{t∈ℝ} with ∥S_{α} (t)∥ = O(e^{ω|t|} ), t ∈ ℝ. Due to [33, Theorem 8], A^{2} generates an exponentially bounded, (\frac{α}{2} )-times integrated semigroup (V_{α} (t))_{t⩾0} given by V_{α} (t) := \frac{1}{\sqrt{4tπ}} ∫_{−∞}^{∞} e^{−s^{2}/4t} S_{α} (s) ds. If Rez > 0, then one can define V_{α} (z)

by V_{α} (z) := \frac{1}{\sqrt{4zπ}} ∫_{−∞}^{∞} e^{−s^{2}/4z} S_{α} (s) ds (\sqrt{1} = 1). Arguing as in the proof of [25, Theorem 11] (cf. also [3, p. 220], [25, Proposition 8] and [26, Proposition 2.4]), we have that (V_{α} (t))_{t⩾0} is an exponentially bounded, analytic (\frac{α}{2} )-times integrated semigroup of angle \frac{π}{2} .

The next theorem clarifies an interesting relation between integrated groups and global differentiable regularized groups.

Theorem 2.2. Suppose α > 0, τ ∈ (0, ∞], b ∈ (0, 1) and A generates an α-times integrated group (S_{α} (t))_{t∈(−τ,τ)} . Then, for every γ ∈ (0, arctan(cos(b\frac{π}{2} ))), there exist two analytic operator families (T_{b,+} (t))_{t∈Σ_{γ}} ⊆ L(E) and (T_{b,−} (t))_{t∈Σ_{γ}} ⊆ L(E) so that:

(a) For every t ∈ Σ_{γ}, T_{b,+} (t) and T_{b,−} (t) are injective operators.

(b) For every t_{1} ∈ Σ_{γ} and t_{2} ∈ Σ_{γ}, A generates a global (T_{b,+} (t_{1})T_{b,−} (t_{2}))- group (V_{b,t_{1} ,t_{2}} (s))_{s∈ℝ} .

(c) For every x ∈ E, t_{1} ∈ Σ_{γ} and t_{2} ∈ Σ_{γ}, the mapping s ↦ V_{b,t_{1} ,t_{2}} (s)x,

s ∈ ℝ is infinitely differentiable in (−∞, 0) ∪ (0, ∞).

Proof. Due to Corollary 2.1, there exist c > 0, d > 0 and M > 0 so that E(c, d) ⊆ ρ(±A) and that ∥R(λ : ±A)∥ ⩽ M|λ|^{α}, λ ∈ E(c, d). Choose a number a ∈ (0, \frac{π}{2} ) such that b ∈ (0, \frac{π}{2(π−a)} ) and that γ ∈ (0, arctan(cos(b(π − a)))). It is clear that there are numbers d ∈ (0, 1] and ω ∈ (d + 1, ∞) so that

Ω_{a,d} := {z ∈ ℂ : |z| ⩽ d} ∪ {re^{iθ} : r > 0, θ ∈ [−a, a]} ⊆ ρ(A − ω) ∩ ρ(−A − ω).

Let the curve Γ_{a,d} = ∂Ω_{a,d} be oriented upwards. Define T_{b,±} (t), t ∈ Σ_{γ} by:

T_{b,±} (t)x := \frac{1}{2πi} ∫_{Γ_{a,d}} e^{−t(−λ)^{b}} R(λ : ±A − ω)x dλ, x ∈ E.

Applying the arguments given in Section 2 of [39], one can deduce that (T_{b,±} (t))_{t∈Σ_{γ}} are analytic operator families and that, for every t ∈ Σ_{γ}, T_{b,+} (t) and T_{b,−} (t) are injective operators. Clearly,

T_{b,+} (t_{1})(−A − ω) ⊆ (−A − ω)T_{b,+} (t_{1}),
T_{b,−} (t_{2})(A − ω) ⊆ (A − ω)T_{b,−} (t_{2}), t_{1}, t_{2} ∈ Σ_{γ}.

It is straightforward to prove that T_{b,+} (t_{1})T_{b,−} (t_{2}) = T_{b,−} (t_{2})T_{b,+} (t_{1}), t_{1}, t_{2} ∈ Σ_{γ} and the arguments given in [21] shows that ±A−ω are generators of global T_{b,±} (t)- semigroups (U_{b,t,±} (s))_{s⩾0} . Suppose t_{1}, t_{2} ∈ Σ_{γ} and x ∈ E. Then one obtains

T_{b,−} (t_{2})(U_{b,t_{1} ,+} (s)x − T_{b,+} (t_{1})x)

= T_{b,−} (t_{2})(A − ω) ∫_{0}^{s} U_{b,t_{1} ,+} (v)x dv = (A − ω)T_{b,−} (t_{2}) ∫_{0}^{s} U_{b,t_{1} ,+} (v)x dv.

Hence,

(A−ω) ∫_{0}^{s} (T_{b,−} (t_{2})U_{b,t_{1} ,+} (v))x dv = T_{b,−} (t_{2})U_{b,t_{1} ,+} (s)x−T_{b,+} (t_{1})T_{b,−} (t_{2})x, s ⩾ 0.

Furthermore, [T_{b,−} (t_{2})U_{b,t_{1} ,+} (s)]T_{b,+} (t_{1}) = T_{b,+} (t_{1})[T_{b,−} (t_{2})U_{b,t_{1} ,+} (s)], s ⩾ 0, and [T_{b,−} (t_{2})U_{b,t_{1} ,+} (s)](A−ω) ⊆ (A−ω)[T_{b,−} (t_{2})U_{b,t_{1} ,+} (s)], s ⩾ 0. This simply implies that (T_{b,−} (t_{2})U_{b,t_{1} ,+} (s))_{s⩾0} is a global (T_{b,+} (t_{1})T_{b,−} (t_{2}))-semigroup generated by (T_{b,+} (t_{1})T_{b,−} (t_{2}))^{−1} (A−ω)(T_{b,+} (t_{1})T_{b,−} (t_{2})) = A−ω. So, (e^{ωs} T_{b,−} (t_{2})U_{b,t_{1} ,+} (s))_{s⩾0} is a global (T_{b,+} (t_{1})T_{b,−} (t_{2}))-semigroup generated by A. Analogically, for every t_{1}, t_{2} ∈ Σ_{γ}, (e^{ωs} T_{b,+} (t_{1})U_{b,t_{2} ,−} (s))_{s⩾0} is a global (T_{b,+} (t_{1})T_{b,−} (t_{2}))-semigroup generated by −A. Hence, for every t_{1} ∈ Σ_{γ} and t_{2} ∈ Σ_{γ}, A generates a global (T_{b,+} (t_{1})T_{b,−} (t_{2}))-group (V_{b,t_{1} ,t_{2}} (s))_{s∈ℝ} given by:

V_{b,t_{1} ,t_{2}} (s) := \{ \begin{matrix}e^{ωs} T_{b,−} (t_{2}) U_{b,t_{1} ,+} (s), & s ⩾ 0, \\ e^{−ωs} T_{b,+} (t_{1}) U_{b,t_{2} ,−} (−s), & s < 0. \end{matrix}

The mapping s ↦ V_{b,t_{1} ,t_{2}} (s)x, s ∈ ℝ is infinitely differentiable in (−∞, 0) ∪ (0, ∞) since the mappings s ↦ T_{b,−} (t_{2})U_{b,t_{1} ,+} (s)x and s ↦ T_{b,+} (t_{1})U_{b,t_{2} ,−} (s)x are infinitely differentiable in s > 0 [21]. The proof is completed. □

3. [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-groups

We introduce [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-groups as follows.

Definition 3.1. Let A be a closed linear operator. Suppose, further, 0 < τ ⩽ ∞, n ∈ ℕ and B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ∈ L(E). A strongly continuous operator family (S(t))_{t∈(−τ,τ)} is a [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group with a subgenerator A if and only if:

(a) A ∫_{0}^{t} S(s)x ds = S(t)x + ∑_{j=0}^{n} t^{j} B_{j} x, t ∈ (−τ, τ), x ∈ E and

(b) AS(t)x − S(t)Ax = ∑_{j=0}^{n−1} t^{j} C_{j} x, t ∈ (−τ, τ), x ∈ D(A).

It is said that (S(t))_{t∈(−τ,τ)} is non-degenerate if the assumption S(t)x = 0, for all t ∈ (−τ, τ), implies x = 0. Define the integral generator of a non-degenerate [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group (S(t))_{t∈(−τ,τ)} by:

Â = \{ (x, y) ∈ E^{2} : S(t)x + ∑_{j=0}^{n} t^{j} B_{j} x − ∑_{j=0}^{n−1} \frac{t^{j+1}}{j + 1} C_{j} x = ∫_{0}^{t} S(s)yds, t ∈ (−τ, τ)\} .

The integral generator Â of a non-degenerate [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group (S(t))_{t∈(−τ,τ)} is a function and it is straightforward to see that Â is a closed linear operator which is an extension of any subgenerator of (S(t))_{t∈(−τ,τ)} . Further on, the injectiveness of B_{i} for some i ∈ {0, …, n} implies that (S(t))_{t∈(−τ,τ)} is nondegenerate. In general, a subgenerator A of (S(t))_{t∈(−τ,τ)} does not commute with S(·) and the set of all subgenerators of (S(t))_{t∈(−τ,τ)} need not be monomial. Let us show this by the following illustrative example.

Example 3.1. (i) Let

E := ℝ^{2}, A(x_{1}, x_{2}) := (x_{1} − x_{2}, 0),

B_{0}(x_{1}, x_{2}) := (0, −x_{2}), B_{1}(x_{1}, x_{2}) := (−x_{1} − x_{2}, −x_{1}), B_{2}(x_{1}, x_{2}) := (0, 0),

C_{0}(x_{1}, x_{2}) := (−x_{2}, 0), C_{1}(x_{1}, x_{2}) := (−x_{1} + x_{2}, −x_{1} + x_{2}),
S(t)(x_{1}, x_{2}) := (tx_{1}, tx_{1} + x_{2}), t ∈ ℝ, (x_{1}, x_{2}) ∈ E.

It is straightforward to verify that (S(t))_{t∈ℝ} is a [B_{0}, B_{1}, B_{2}, C_{0}, C_{1}]-group with a subgenerator A and that: S(t)S(s) = S(s)S(t) iff t = s, S(t)D ≠ DS(t), t ∈ ℝ, D ∈ {B_{1}, C_{0}, C_{1}} and D_{0}D_{1} ≠ D_{1}D_{0}, D_{i} ∈ {B_{i}, C_{i}}, i = 1, 2.

(ii) Suppose C_{j} = 0, j = 0, …, n − 1 and the bounded linear operators B_{j}, j = 0, …, n fulfill E ≠ \overline{∑ _{i=0}^{n} R(B_{i})}. Put S(t)x := − ∑ _{j=0}^{n} t^{j} B_{j} x, t ∈ (−τ, τ), x ∈ E and denote by Λ the family of all closed subspaces of E containing \overline{∑ _{i=0}^{n} R(B_{i})}. If F ∈ Λ, define a closed linear operator A_{F} by D(A_{F}) := F and A_{F} x := 0, x ∈ D(A_{F}). Then A_{F} is a subgenerator of a [B_{0}, …, B_{n}, \underbrace{0, …, 0}_{n}]-group (S(t))_{t∈(−τ,τ)} .

Remark 3.1. (i) Assume n ∈ ℕ, τ ∈ (0, ∞] and A generates an n-times integrated group (S(t))_{t∈(−τ,τ)} . Put \overline{S}(t) := S(t), t ∈ [0, τ) and \overline{S}(t) := (−1)^{n}S(t),

t ∈ (−τ, 0). Then (\overline{S}(t))_{t∈(−τ,τ)} is a [ \underbrace{0, …, 0}_{n}, \frac{(−1)}{n!} I, \underbrace{0, …, 0}_{n}]-group having A as a

subgenerator.

(ii) Let A be a subgenerator of a [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group (S(t))_{t∈(−τ,τ)} . Put Š(t) := S(−t), t ∈ (−τ, τ), B̌_{j} := (−1)^{j} B_{j} and Č_{j} := (−1)^{j+1} C_{j}. Then (Š(t))_{t∈(−τ,τ)} is a [B̌_{0}, …, B̌_{n} , Č_{0}, …, Č_{n−1} ]-group with a subgenerator −A.

(iii) Let k ∈ ℕ and let D_{1}, …, D_{k} ∈ L(E). For a given i ∈ {1, …, k}, put \overline{D_{i}} := ∏ _{j=1}^{i} D_{j}. Define S_{i}(·), i ∈ \{0, …, k\} recursively by:

S_{0}(t)x := S(t)x, …, S_{i}(t)x := ∫_{0}^{t} S_{i−1} (s)D_{i}x ds, x ∈ E, t ∈ (−τ, τ),

and suppose, additionally, that D_{i}A ⊆ AD_{i}, i ∈ {1, …, k}. By a simple induction argument, one can deduce that, for every i ∈ {1, …, k}, (S_{i}(t))_{t∈(−τ,τ)} is a

[ \underbrace{0, …, 0}_{i}, \frac{0!B_{0} \overline{D_{i}}}{i!} , …, \frac{n!B_{n} \overline{D_{i}}}{(i+n)!} , \underbrace{0, …, 0}_{i}, \frac{0!C_{0} \overline{D_{i}}}{i!} , \frac{(n−1)!C_{n−1} \overline{D_{i}}}{(i+n−1)!} ]-group with a subgener-

ator A.

(iv) Suppose A generates a C-regularized group (T (t))_{t∈ℝ} in the sense of [9, Definition 7.2]. Put T_{k}(t)x := ∫_{0}^{t} \frac{(t−s)^{k−1}}{(k−1)!} T (s)x ds, t ∈ ℝ, x ∈ E, k ∈ ℕ. Then (T_{k}(t))_{t∈ℝ} is a [\underbrace{0, …, 0}_{k}, −\frac{1}{k!} C, \underbrace{0, …, 0}_{k}]-group having A as a subgenerator.

Suppose A is closed, B_{0}, ⋯, B_{n} ∈ L(E) and define

ρ_{B_{0} ,...,B_{n}} (A) := \{ λ ∈ ℂ : R( ∑_{j=0}^{n} \frac{j!}{λ^{j}} B_{j} ) ⊆ R(λ − A)\} .

The following profiling of exponentially bounded [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]- groups can be simply proved ([24]).

Proposition 3.1. (i) Let A be a subgenerator of a [B_{0}^{+} , …, B_{n}^{+} , C_{0}^{+} , …, C_{n−1}^{+} ]- group (S(t))_{t∈ℝ} satisfying ∥S(t)∥ ⩽ Me^{ω|t|} , t ∈ ℝ, for some M > 0 and ω ⩾ 0. Set B_{j}^{−} := (−1)^{j} B_{j}^{+} and C_{j}^{−} := (−1)^{j+1} C_{j}^{+} . Then:

(i1) ρ_{B_{0}^{+} ,...,B_{n}^{+}} (A) ∩ ρ_{B_{0}^{−} ,...,B_{n}^{−}} (−A) ⊇ {λ ∈ ℂ : Reλ > ω},

(i2) ∫_{0}^{∞} e^{−λt} S(±t)x dt = −(λ ∓ A)^{−1} ∑ _{j=0}^{n} \frac{j!}{λ^{j}} B_{j}^{±} x, Reλ > ω, x ∈ E and

(i3) ±A ∫_{0}^{∞} e^{−λt} S(±t)x dt − ∫_{0}^{∞} e^{−λt} S(±t) ± Axdt = ∑ _{j=0}^{n−1} \frac{j!}{λ^{j+1}} C_{j}^{±} x, Reλ > ω, x ∈ D(A).

(ii) Suppose A is a closed operator and (S(t))_{t∈ℝ} is a strongly continuous operator family satisfying ∥S(t)∥ ⩽ Me^{ω|t|} , t ∈ ℝ, for some M > 0 and ω ⩾ 0. If (i1), (i2) and (i3) hold, then (S(t))_{t∈ℝ} is a [B_{0}^{+} , …, B_{n}^{+} , C_{0}^{+} , …, C_{n−1}^{+} ]-group with a subgenerator A.

Suppose n ∈ ℕ. If A is a closed operator and B_{0}, ⋯, B_{n} ∈ L(E), then we define linear operators Y_{i}, i ∈ {0, …, n} recursively by:

Y_{0} := B_{0}, Y_{i+1} := (i + 1)!B_{i+1} + AY_{i}, i ∈ {0, …, n − 1}.

Note that Y_{1} is closed and that the assumption 0 ∈ ρ(A) simply implies the closedness of Y_{i}, i ∈ {0, …, n}.

Proposition 3.2. Suppose τ ∈ (0, ∞], n ∈ ℕ ∖ {1} and A is a subgenerator of a [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group (S(t))_{t∈(−τ,τ)} . Then:

(i) iB_{i}x − C_{i−1} x ∈ D(A), x ∈ D(A), i ∈ {1, …, n}, A(iB_{i}x − C_{i−1} x) = iB_{i}Ax − iC_{i}x, i ∈ {1, …, n − 1} and A(nB_{n}x − C_{n−1} x) = nB_{n}Ax, x ∈ D(A).

(ii) D(A^{k}) ⊆ ⋂ _{j=0}^{k} D(Y_{j}), k ∈ \{0, …, n\} and Y_{k}x = −( \frac{d^{k}}{dt^{k}} S(t)x)_{t=0} , x ∈ D(A^{k}), k ∈ {0, …, n}.

(iii) For every k ∈ {0, …, n − 1} and x ∈ D(A^{k+1} ):

(3.1)  C_{k}x + \frac{1}{k!}AY_{k}(x) = \frac{1}{k!}Y_{k}(Ax).

(iv) If R(B_{0})⊆D(A), then Y_{2} is closed, D(A^{k}) ⊆ ⋂ _{j=0}^{k+1} D(Y_{j}), k ∈{0, ..., n−1}, (3.1) holds for every k ∈ {1, …, n − 1} and x ∈ D(A^{k}) and there exists an appropriate constant M > 0 so that ∥Y_{k+1} x∥ ⩽ M∥x∥_{k}, k ∈ {0, …, n − 1}, x ∈ D(A^{k}).

(v) A(−Y_{n}x + Y_{n−1} Ax) = −n!B_{n}Ax, x ∈ D(A^{n}); if R(B_{0}) ⊆ D(A), then AY_{n}x = Y_{n}Ax, x ∈ D(A^{n}).

Proof. Suppose x ∈ D(A). Clearly, \frac{d}{dt}S(t)x = AS(t)x − ∑ _{i=1}^{n} it^{i−1} B_{i}x, t ∈ (−τ, τ) and

∑_{i=0}^{n−1} t^{i}C_{i}x = AS(t)x−S(t)Ax = AS(t)x−[A∫_{0}^{t}S(s)Ax ds−∑_{i=0}^{n} t^{i}B_{i}Ax], t ∈ (−τ, τ).

Hence,

∑_{i=0}^{n−1} t^{i}C_{i}x − ∑_{i=0}^{n} t^{i}B_{i}Ax = A[S(t)x − ∫_{0}^{t} S(s)Ax ds], t ∈ (−τ, τ).

Since

\frac{d}{dt}[S(t)x − ∫_{0}^{t} S(s)Ax ds] = AS(t)x − ∑_{i=1}^{n} it^{i−1} B_{i}x − S(t)Ax

= ∑_{i=0}^{n−1} t^{i}C_{i}x − ∑_{i=1}^{n} it^{i−1} B_{i}x, t ∈ (−τ, τ),

the closedness of A implies

∑_{i=0}^{n−1} t^{i}C_{i}x − ∑_{i=1}^{n} it^{i−1} B_{i}x = \frac{d}{dt}[S(t)x − ∫_{0}^{t} S(s)Ax ds] ∈ D(A), t ∈ (−τ, τ)

and:

(3.2) A[ ∑_{i=0}^{n−1} t^{i}C_{i}x − ∑_{i=1}^{n} it^{i−1} B_{i}x] = ∑_{i=1}^{n−1} it^{i−1} C_{i}x − ∑_{i=1}^{n} it^{i−1} B_{i}Ax, t ∈ (−τ, τ).

Differentiate (3.2) sufficiently many times in order to see that iB_{i}x−C_{i−1} x ∈ D(A), i ∈ {1, …, n} and that (i) holds. To prove (ii), notice that the closedness of A and argumentation used in the proof of (i) enable one to conclude that the mapping t ↦ \frac{d^{k}}{dt^{k}} S(t)x, t ∈ (−τ, τ) is k-times continuously differentiable for every k ∈ {0, …, n} and x ∈ D(A^{k}). Fix a k ∈ {0, …, n}; then we obtain:

(3.3)  \frac{d^{l+1}}{dt^{l+1}} S(t)x = A\frac{d^{l}}{dt^{l}}S(t)x − ∑_{j=l+1}^{n} j ⋯ (j − l)t^{j−l−1} B_{j} x, t ∈ (−τ, τ),

for every l ∈ {0, …, k − 1}.

Since Y_{0} = B_{0}, the proof of (ii) follows by induction.

Suppose now x ∈ D(A^{k+1} ). Then the mapping t ↦ S(t)Ax, t ∈ (−τ, τ) is k-times continuously differentiable. Since C_{k}x = \frac{1}{k!} \frac{d^{k}}{dt^{k}} [AS(t)x − S(t)Ax], t ∈ (−τ, τ), we have that the mapping t ↦ AS(t)x, t ∈ (−τ, τ) is k-times continuously differentiable and the closedness of A implies that \frac{d^{k}}{dt^{k}} AS(t)x = A\frac{d^{k}}{dt^{k}} S(t)x, t ∈ (−τ, τ) and that C_{k}x = \frac{1}{k!} [A\frac{d^{k}}{dt^{k}} S(t)x − \frac{d^{k}}{dt^{k}} S(t)Ax], ∈ (−τ, τ). Put t = 0 in the last equality to finish the proof of (iii).

To prove (iv), notice that R(B_{0}) ⊆ D(A) and that the Closed Graph Theorem implies Y_{1} = AB_{0} + B_{1} ∈ L(E); the closedness of Y_{2} simply follows from this fact. Suppose now x ∈ D(A^{k}). Since \frac{d}{dt}S(t)x − S(t)Ax = ∑ _{i=0}^{n−1} t^{i}C_{i}x − ∑ _{i=1}^{n} it^{i−1} B_{i}x, t ∈ (−τ, τ), one concludes that

\frac{d^{k}}{dt^{k}}S(t)x − \frac{d^{k−1}}{dt^{k−1}} S(t)Ax = \frac{d^{k−1}}{dt^{k−1}} [ ∑_{i=0}^{n−1} t^{i}C_{i}x − ∑_{i=1}^{n} it^{i−1} B_{i}x], t ∈ (−τ, τ).

This implies −Y_{k}x + Y_{k−1} Ax = (k − 1)!C_{k−1} x − k!B_{k}x and an employment of (i) gives −Y_{k}x + Y_{k−1} Ax ∈ D(A) and:

(3.4) A[−Y_{k}x + Y_{k−1} Ax] = k!C_{k}x − k!B_{k}Ax, k ∈ {1, …, n − 1}, x ∈ D(A^{k}).

Because R(B_{0}) ⊆ D(A), one concludes inductively from (3.4) that Y_{k}x ∈ D(A), x ∈ D(A^{k}), k ∈ {0, …, n − 1}, i.e., D(A^{k}) ⊆ ⋂ _{i=0}^{k+1} D(Y_{i}), k ∈ {0, …, n − 1} and (3.4) implies k!C_{k}x + AY_{k}x = k!B_{k}Ax + AY_{k−1} Ax = Y_{k}Ax, k ∈ {1, …, n − 1}, x ∈ D(A^{k}). The existence of a constant M > 0 satisfying ∥Y_{k+1} x∥ ⩽ M∥x∥_{k}, k ∈ {0, …, n − 1}, x ∈ D(A^{k}) essentially follows from an application of (3.1) and an induction argument. This ends the proof of (iv) while the proof of (v) follows simply from that of (iv). □

Remark 3.2. (a) Suppose τ ∈ (0, ∞] and A is a subgenerator of a [B_{0}, B_{1}, C_{0}]- group (S(t))_{t∈[0,τ)} . Arguing as in the proof of Proposition 3.2, one obtains A(B_{1}x− C_{0}x) = B_{1}Ax, x ∈ D(A) and AY_{1}x = Y_{1}Ax, x ∈ D(A^{2}). Furthermore, if R(B_{0}) ⊆ D(A), then AY_{1}x = Y_{1}Ax, x ∈ D(A).

(b) The next question is motivated by the analysis of Arendt, El-Mennaoui and Keyantuo [1]: If A is a subgenerator of a [\underbrace{0, …, 0}_{n}, \frac{−1}{n!} I, C_{0}, …, C_{n−1} ]-group

(S(t))_{t∈[0,τ)} , n ∈ ℕ, 0 < τ ⩽ ∞, does S(t)A ⊆ AS(t), t ∈ (−τ, τ)? The answer is affirmative and we will show this only in the non-trivial case n > 1. Indeed, S(0) = 0 and this implies C_{0}x = 0, x ∈ D(A). By Proposition 3.2(i), we have AC_{i−1} x = iC_{i}x, i ∈ {1, …, n − 1}, x ∈ D(A). Inductively, C_{i}x = 0, i ∈ {1, …, n − 1}, x ∈ D(A) and an immediate consequence is S(t)A ⊆ AS(t), t ∈ (−τ, τ).

(c) Let A be a subgenerator of a [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group (S(t))_{t∈(−τ,τ)} , n ⩾ 2, (S(t))_{t∈(−τ,τ)} non-degenerate and Â the integral generator of (S(t))_{t∈(−τ,τ)} . Then iB_{i}x − C_{i−1} x ∈ D(A), x ∈ D(Â), i ∈ {1, …, n}, A(iB_{i}x − C_{i−1} x) = iB_{i} Âx − iC_{i}x, i ∈ {1, …, n − 1} and A(nB_{n}x − C_{n−1} x) = nB_{n} Âx, x ∈ D(Â). To prove this, suppose (x, y) ∈ Â. Clearly,

A ∫_{0}^{t} S(s)x ds = ∫_{0}^{t}( S(s)y + ∑_{j=0}^{n−1} s^{j} C_{j} x) ds, t ∈ (−τ, τ).

Differentiate this equality to obtain that S(t)x ∈ D(A) and that AS(t)x = S(t)y + ∑ _{j=0}^{n−1} t^{j} C_{j} x, t ∈ (−τ, τ). Hence,

A[ ∫_{0}^{t} S(s)y ds + ∑_{j=0}^{n−1} \frac{t^{j+1}}{j + 1} C_{j} x − ∑_{j=0}^{n} t^{j} B_{j} x] = S(t)y + ∑_{j=0}^{n−1} t^{j} C_{j} x ,

A[ ∑_{j=0}^{n−1} \frac{t^{j+1}}{j + 1} C_{j} x − ∑_{j=0}^{n} t^{j} B_{j} x] = ∑_{j=0}^{n−1} t^{j} C_{j} x − ∑_{j=0}^{n} t^{j} B_{j} y , t ∈ (−τ, τ).

Differentiation of the previous equality leads us to the desired assertion. Notice that (c) extends Proposition 3.2(i) to non-degenerate groups and that, in the case n = 1, A(B_{1}x − C_{0}x) = B_{1}Ax, x ∈ D(Â).

Proposition 3.3. Suppose 0 < τ ⩽ ∞ and A is a subgenerator of a [B_{0}, …, B_{n},

C_{0}, …, C_{n−1} ]-group (S(t))_{t∈(−τ,τ)} . If

(3.5)  C_{j} A ⊆ AC_{j} , j = 0, …, n − 1 and B_{j} A ⊆ AB_{j} , j = 1, …, n,

then, for every x ∈ E,

S(t)S(s)x = ∑_{j=1}^{n} [ ∫_{0}^{s} j(t + s − r)^{j−1} B_{j} S(r)x dr − ∫_{t}^{t+s} j(t + s − r)^{j−1} S(r)B_{j} x dr]

− ∑_{j=0}^{n−1} [ ∫_{0}^{s}(t + s − r)^{j} C_{j} S(r)x dr + t^{j} C_{j} ∫_{0}^{s} S(r)x dr] (3.6)

−S(t + s)B_{0}x − ∑_{j=0}^{n−1} ∑_{i=0}^{n} ∫_{0}^{s}(t + s − r)^{j} r^{i}drC_{j} B_{i}x, t, s ∈ (−τ, τ), |t + s| < τ.

Proof. Suppose, for a moment, y ∈ D(A) and t ∈ (−τ, τ). Then ∫_{0}^{t} AS(s)y ds = S(t)y + ∑ _{j=0}^{n} t^{j} B_{j} y, i.e., ∫_{0}^{t} [S(s)Ay + ∑ _{j=0}^{n−1} s^{j} C_{j} y]ds = S(t)y + ∑ _{j=0}^{n} t^{j} B_{j} y. Hence,

\frac{d}{dt}S(t)y = S(t)Ay + ∑_{j=0}^{n−1} t^{j} C_{j} y − ∑_{j=1}^{n} jt^{j−1} B_{j} y, t ∈ (−τ, τ).

Fix an x ∈ E and t, s ∈ (−τ, τ) with |t + s| < τ. Define afterwards the function f : (t + s − τ, t + s + τ) ∩ (−τ, τ) → E by f(r) := S(t + s − r) ∫_{0}^{r} S(s)x ds. Then we obtain:

\frac{d}{dr}f(r) = \frac{d}{dr}[S(t+s−r)∫_{0}^{r}S(v)x dv] = S(t+s−r)S(r)x − [S(t+s−r)A∫_{0}^{r}S(v)x dv

+ ∑_{j=0}^{n−1} (t+s−r)^{j} C_{j} ∫_{0}^{r} S(v)x dv − ∑_{j=1}^{n} j(t+s−r)^{j−1} B_{j} ∫_{0}^{r} S(v)x dv]

= S(t+s−r)S(r)x − S(t+s−r)[S(r)x + ∑_{j=0}^{n} r^{j} B_{j} x]

− ∑_{j=0}^{n−1} (t+s−r)^{j} C_{j} ∫_{0}^{r} S(v)x dv + ∑_{j=1}^{n} j(t+s−r)^{j−1} B_{j} ∫_{0}^{r} S(v)x dv

= − ∑_{j=0}^{n−1} (t+s−r)^{j} C_{j} ∫_{0}^{r} S(v)x dv + ∑_{j=1}^{n} j(t+s−r)^{j−1} B_{j} ∫_{0}^{r} S(v)x dv

− ∑_{j=0}^{n} r^{j} S(t+s−r)B_{j} x,

for all r ∈ (t + s − τ, t + s + τ) ∩ (−τ, τ). Integrate the last equality with respect to r from 0 to s to conclude that:

(3.7) S(t) ∫_{0}^{s} S(v)x dv = − ∑_{j=0}^{n−1} C_{j} ∫_{0}^{s}(t + s − r)^{j} ∫_{0}^{r} S(v)x dv dr

+ ∑_{j=1}^{n} jB_{j} ∫_{0}^{s}(t + s − r)^{j−1} ∫_{0}^{r} S(v)x dv dr − ∑_{j=0}^{n} ∫_{0}^{s} r^{j} S(t + s − r)B_{j} x dr.

Thereby,

S(t)S(s)x = S(t)[A ∫_{0}^{s} S(v)x dv − ∑_{j=0}^{n} s^{j} B_{j} x]

= [AS(t) ∫_{0}^{s} S(v)x dv − ∑_{j=0}^{n−1} t^{j} C_{j} ∫_{0}^{s} S(v)x dv] − ∑_{j=0}^{n} s^{j} S(t)B_{j} x

80 KOSTIĆ

= A[ − ∑_{j=0}^{n−1} C_{j} ∫_{0}^{s}(t + s − r)^{j} ∫_{0}^{r} S(v)x dv dr

+ ∑_{j=1}^{n} jB_{j} ∫_{0}^{s}(t + s − r)^{j−1} ∫_{0}^{r} S(v)x dv dr − ∑_{j=0}^{n} ∫_{0}^{s} r^{j} S(t + s − r)B_{j} x dr]

− ∑_{j=0}^{n−1} t^{j} C_{j} ∫_{0}^{s} S(v)x dv − ∑_{j=0}^{n} s^{j} S(t)B_{j} x.

Taking into consideration (3.5), we get:

= − ∑_{j=0}^{n−1} C_{j} ∫_{0}^{s}(t + s − r)^{j} [S(r)x + ∑_{i=0}^{n} r^{i}B_{i}x]dr

+ ∑_{j=1}^{n} jB_{j} ∫_{0}^{s}(t + s − r)^{j−1} [S(r)x + ∑_{i=0}^{n} r^{i}B_{i}x]dr

− A ∑_{j=0}^{n} ∫_{0}^{s} r^{j} S(t+s−r)B_{j} x dr − ∑_{j=0}^{n−1} t^{j} C_{j} ∫_{0}^{s} S(v)x dv − ∑_{j=0}^{n} s^{j} S(t)B_{j} x. (3.8)

Observing that:

A ∫_{0}^{s}S(t+s−r)B_{0}x dr = A ∫_{t}^{t+s} S(v)B_{0}x dv = A[ ∫_{0}^{t+s}S(v)B_{0}x dv −∫_{0}^{t}S(v)B_{0}x dv]

= S(t + s)B_{0}x + ∑_{i=0}^{n} (t + s)^{i}B_{i}B_{0}x − S(t)B_{0}x − ∑_{i=0}^{n} t^{i}B_{i}B_{0}x

and that

A ∫_{0}^{s} r^{j} S(t + s − r)B_{j} x dr = A ∫_{t}^{t+s}(t + s − v)^{j} S(v)B_{j} x dv
= A[ − s^{j} ∫_{0}^{t} S(σ)B_{j} x dσ + ∫_{t}^{t+s} j(t + s − v)^{j−1} ∫_{0}^{v} S(σ)B_{j} x dσ dv]
= −s^{j} [S(t)B_{j} x + ∑_{i=0}^{n} t^{i}B_{i}B_{j} x] + ∫_{t}^{t+s}j(t+s−v)^{j−1} [S(v)B_{j} x + ∑_{i=0}^{n} v^{i}B_{i}B_{j} x]dv,

for all j = 1, …, n, (3.8) implies:

S(t)S(s)x = − ∑_{j=0}^{n−1} ∫_{0}^{s}(t + s − r)^{j} C_{j} S(r)x dr − ∑_{j=0}^{n−1} ∫_{0}^{s}(t + s − r)^{j} ∑_{i=0}^{n} r^{i}C_{j} B_{i}x dr

+ ∑_{j=1}^{n} j ∫_{0}^{s}(t + s − r)^{j−1} B_{j} S(r)x dr + ∑_{j=1}^{n} j ∫_{0}^{s}(t + s − r)^{j−1} ∑_{i=0}^{n} r^{i}B_{j} B_{i}x dr

− S(t + s)B_{0}x − ∑_{i=0}^{n} (t + s)^{i}B_{i}B_{0}x + S(t)B_{0}x + ∑_{i=0}^{n} t^{i}B_{i}B_{0}x

+∑_{j=1}^{n} s^{j} [S(t)B_{j} x+∑_{i=0}^{n} t^{i}B_{i}B_{j} x]−∑_{j=1}^{n} j∫_{t}^{t+s}(t+s−v)^{j−1} [S(v)B_{j} x+∑_{i=0}^{n} v^{i}B_{i}B_{j} x]dv

(3.9)  − ∑_{j=0}^{n−1} t^{j} C_{j} ∫_{0}^{s} S(v)x dv − ∑_{j=0}^{n} s^{j} S(t)B_{j} x.

Clearly, S(t)B_{0}x + ∑ _{j=1}^{n} s^{j} S(t)B_{j} x − ∑ _{j=0}^{n} s^{j} S(t)B_{j} x = 0 and:

− ∑_{j=1}^{n} ∑_{i=0}^{n} ∫_{t}^{t+s} j(t + s − v)^{j−1} v^{i}dvB_{i}B_{j} x = − ∑_{j=1}^{n} ∑_{i=0}^{n} ∫_{0}^{s}(t + s − r)^{i}jr^{j−1} drB_{i}B_{j} x

= − ∑_{j=1}^{n} ∑_{i=0}^{n} t^{i}s^{j} B_{i}B_{j} x − ∑_{j=1}^{n} ∑_{i=1}^{n} ∫_{0}^{s} i(t + s − r)^{i−1} r^{j} drB_{i}B_{j} x

= − ∑_{j=1}^{n} ∑_{i=0}^{n} t^{i}s^{j} B_{i}B_{j} x − ∑_{j=1}^{n} ∑_{i=1}^{n} ∫_{0}^{s} j(t + s − r)^{j−1} r^{i}drB_{j} B_{i}x.

Therefore,

(3.10) − ∑_{j=1}^{n} ∑_{i=0}^{n} ∫_{t}^{t+s} j(t + s − v)^{j−1} v^{i}dvB_{i}B_{j} x + ∑_{j=1}^{n} ∑_{i=0}^{n} s^{j} t^{i}B_{i}B_{j} x

+ ∑_{j=1}^{n} ∑_{i=0}^{n} ∫_{0}^{s} j(t + s − r)^{j−1} r^{i}drB_{j} B_{i}x

= ∑_{j=1}^{n} ∫_{0}^{s} j(t + s − r)^{j−1} drB_{j} B_{0}x = ∑_{j=1}^{n} [(t + s)^{j} − t^{j}]B_{j} B_{0}x.

Finally, (3.6) follows from an application of (3.9) and (3.10). □

Remark 3.3. The composition property does not remain true if the condition (3.5) is neglected. Namely, let A, B_{0}, B_{1}, B_{2}, C_{0}, C_{1} and (S(t))_{t∈ℝ} possess the same meaning as in Example 3.1(i). Then (S(t))_{t∈ℝ} is a [B_{0}, B_{1}, B_{2}, C_{0}, C_{1}]-group with a subgenerator A and a tedious matrix computation shows that (3.5) and (3.6) are not valid. Moreover, ρ_{B_{0} ,B_{1} ,B_{2}} (A) ⊇ {λ ∈ ℂ : Reλ > 0} and R(B_{0})+R(B_{1})+R(B_{2}) ⊈ R(1 − A) (see Proposition 3.1).

4. Connections between distribution groups and local integrated groups

In order to establish a satisfactory relationship of distribution groups with local integrated groups, we need the following definition introduced by Tanaka and Okazawa in [40] (cf. [40, Definition 4]):

(△) Suppose n ∈ ℕ and τ ∈ (0, ∞]. A strongly continuous operator family (S(t))_{t∈[0,τ)} is called a (local) n-times integrated semigroup if:

(i) S(t)S(s)x = [∫_{0}^{t+s} − ∫_{0}^{t} − ∫_{0}^{s}] \frac{(t+s−r)^{n−1}}{(n−1)!} S(r)x dr, x ∈ E, 0 ⩽ t, s, t+s < τ, and S(0) = 0,

(ii) S(t)x = 0 for every t ∈ [0, τ) implies x = 0.

Suppose (S(t))_{t∈[0,τ)} is an n-times integrated semigroup in the sense of (△). The infinitesimal generator A_{0} of (S(t))_{t∈[0,τ)} is defined in [40] by

D(A_{0}) := \{ x ∈ ⋃_{σ∈(0,τ]} C^{n}(σ) : lim_{h→0+} \frac{S^{(n)} (h)x − x}{h} exists\} ,

A_{0}x := lim_{h→0+} \frac{S^{(n)} (h)x − x}{h} , x ∈ D(A_{0}),

where C^{n}(σ) := {x ∈ E | S(·)x : [0, σ) → E is n-times continuously differentiable}.

The infinitesimal generator A_{0} of (S(t))_{t∈[0,τ)} is a closable linear operator and the closure of A_{0}, \overline{A_{0}}, is said to be the complete infinitesimal generator, c.i.g in short, of (S(t))_{t∈[0,τ)} .

Suppose (S(t))_{t∈[0,τ)} is a (local) n-times integrated semigroup in the sense of Definition 2.3. Then (S(t))_{t∈[0,τ)} is an n-times integrated semigroup in the sense of (△); in general, the converse statement does not hold (see [1], [24, Proposition 2.1],

[28] and [40, Proposition 4.5]).

Theorem 4.1. (a) Suppose G ∈ D′(L(E)) and A is a closed linear operator so that (1.2) holds. Then, for every τ ∈ (0, ∞), there exist n_{0} = n_{0}(τ) ∈ ℕ and B_{0}, …, B_{n_{0}} , C_{0}, …, C_{n_{0} −1} ∈ L(E) such that A is a subgenerator of a [B_{0}, …, B_{n_{0}} , C_{0}, …, C_{n_{0} −1} ]-group (S_{τ}(t))_{t∈(−τ,τ)} satisfying S_{τ}(t)x ∈ D(A) for all x ∈ E and t ∈ (−τ, τ).

(b) Let G and A be as in the formulation of (a) and let A_{1} = A\overline{R(G)} . Suppose, in addition, that G is regular and put S(t) := S_{τ}(t), t ∈ (−τ, τ), where (S_{τ}(t))_{t∈(−τ,τ)} is a [B_{0}, …, B_{n_{0}} , C_{0}, …, C_{n_{0} −1} ]-group constructed in (a). Then:

(b1) R(G) ⊆ ⋂ _{i=0}^{n_{0}} D(Y_{i}), Y_{n_{0}} x = −x, x ∈ R(G), the function t ↦ u(t; x), t ∈ ℝ is infinitely differentiable, u(t; x) ∈ D_{∞} (A) and \frac{d^{n}}{dt^{n}} u(t; x) = u(t; A^{n}x),

t ∈ ℝ, x ∈ R(G), n ∈ ℕ.

(b2) If n_{0} = 1, then ±A_{1} generate local once integrated semigroups (S_{±}^{1} (t))_{t∈[0,τ)} in R(G) given by S_{±}^{1} (t)x := S(±t)(±x) ± B_{0}x, x ∈ \overline{R(G)}, t ∈ [0, τ). Fur-

thermore, A_{1} generates a C_{0}-group in \overline{R(G)}.

(b3) Suppose n_{0} = 2 and put S_{±}^{2} (t)x := S(±t)x + B_{0}x + t(±AB_{0} ± B_{1})x, t ∈ [0, τ), x ∈ \overline{R(G)}. Then S_{±}^{2} (t) ∈ L(\overline{R(G)}),

( ∫_{0}^{t} S_{±}^{2} (s)x ds, S_{±}^{2} (t)x − \frac{t^{2}}{2}x) ∈ ±A_{1}, x ∈ \overline{R(G)}, t ∈ [0, τ),

S_{±}^{2} (t)A_{1}x = A_{1}S_{±}^{2} (t)x, t ∈ [0, τ), x ∈ R(G), S_{±}^{2} (t)x ∈ D(A), x ∈ \overline{R(G)}, the mapping t ↦ \frac{d}{dt}S_{±}^{2} (t)x, t ∈ [0, τ) is continuously differentiable for every x ∈ \overline{R(G)}, \overline{R(G)} ⊆ ⋂ _{i=0}^{2} D(Y_{i}) and Y_{2}x = −x, x ∈ \overline{R(G)}. Furthermore, ±A_{1} are generators of local once integrated semigroups (\frac{d}{dt}S_{±}^{2} (t))_{t∈[0,τ)} .

(b4) Suppose n_{0} ⩾ 3,

(4.1)  \overline{R(G)} ⊆ ⋂_{i=2}^{n_{0} −1} D(Y_{i})

and there exists M > 0 with

(4.2)  ∥Y_{i}x∥ ⩽ M∥x∥, x ∈ \overline{R(G)}, i = 2, …, n_{0} − 1.

The following holds: \overline{R(G)} ⊆ D(Y_{n_{0}} ) and Y_{n_{0}} x = −x, x ∈ \overline{R(G)}. Set S_{+}^{n_{0}} (t)x := S(t)x + ∑ _{i=0}^{n_{0} −1} \frac{t^{i}}{i!} Y_{i}x and S_{−}^{n_{0}} (t)x := (−1)^{n_{0}} S(−t)x +

∑ _{i=0}^{n_{0} −1} \frac{(−1)^{n_{0} +i} t^{i}}{i!} Y_{i}x, x ∈ \overline{R(G)}, t ∈ [0, τ). Then: S_{±}^{n_{0}} (t) ∈ L(\overline{R(G)}), ( ∫_{0}^{t} S_{±}^{n_{0}} (s)x ds, S_{±}^{n_{0}} (t)x−\frac{t^{n_{0}}}{n_{0} !} x) ∈ ±A_{1}, x ∈ \overline{R(G)}, t ∈ [0, τ), S_{±}^{n_{0}} (t)A_{1}x = A_{1}S_{±}^{n_{0}} (t)x, t ∈ [0, τ), x ∈ R(G) and S_{±}^{n_{0}} (t)x ∈ D(A), x ∈ \overline{R(G)}. Set

A′_{n_{0} −1,±} := \{(x, y) ∈ ±A_{1} : C_{i}x + \frac{1}{i!}AY_{i}x = ±\frac{1}{i!}Y_{i}y, i = 2, …, n_{0} − 1\}.

Then A′_{n_{0} −1,±} are generators of local (n_{0} −1)-times integrated semigroups (\frac{d}{dt}S_{±}^{n_{0}} (t))_{t∈[0,τ)} .

(b5) Suppose n_{0} ⩾ 3 and ρ(A) ≠ ∅. Then, for every τ_{0} ∈ (0, ∞), there exists n(τ_{0}) ∈ ℕ such that A_{1} generates a local n(τ_{0})-times integrated group on (−τ_{0}, τ_{0}).

Proof. (a) Let τ ∈ (0, ∞) be chosen arbitrarily. Since AG(φ)x = G(−φ′)x, φ ∈ D, x ∈ E we have G ∈ D′(L(E, [D(A)])). An employment of [32, Theorem 2.1.1] implies that there exist an integer n_{0} = n_{0}(τ) and a continuous function S_{τ} : [−τ, τ] → L(E, [D(A)]) such that G(φ)x = (−1)^{n_{0}} ∫_{−τ}^{τ} φ^{(n_{0} )} (t)S_{τ}(t)x dt, φ ∈ D_{(−τ,τ)} , x ∈ E. We obtain:

AG(φ)x = (−1)^{n_{0}} ∫_{−τ}^{τ} φ^{(n_{0} )} (t)AS_{τ}(t)x dt = (−1)^{n_{0} +1} ∫_{−τ}^{τ} φ^{(n_{0} +1)} (t) ∫_{0}^{t} AS_{τ}(s)x ds dt

= G(−φ′)x = (−1)^{n_{0} +1} ∫_{−τ}^{τ} φ^{(n_{0} +1)} (t)S_{τ}(t)x dt, φ ∈ D_{(−τ,τ)} , x ∈ E.

An immediate consequence is:

∫_{−τ}^{τ} φ^{(n_{0} +1)} (t)[ ∫_{0}^{t} AS_{τ}(s)x ds − S_{τ}(t)x]dt = 0, φ ∈ D_{(−τ,τ)} , x ∈ E.

The well-known arguments of distribution theory (cf. for instance [17, Lemma 8.1.1] or apply the Hanh–Banach theorem and [36, Theorem 5.10, p. 80]) imply that there exist B_{0}, …, B_{n_{0}} ∈ L(E) which satisfy (a) of Definition 3.1. Similarly, if x ∈ D(A), then G(φ)Ax = AG(φ)x, φ ∈ D and we get:

∫_{−τ}^{τ} φ^{(n_{0} )} (t)[AS_{τ}(t)x − S_{τ}(t)Ax] dt = 0, φ ∈ D_{(−τ,τ)} , x ∈ E.

Thus, there exist C_{0}, …, C_{n_{0} −1} ∈ L(E) which satisfy

AS_{τ}(t)x − S_{τ}(t)Ax = ∑_{j=0}^{n−1} t^{j} C_{j} x,

for all t ∈ (−τ, τ) and x ∈ D(A). To prove (b1), we need the following notion from [23]. Suppose ζ ∈ D and ∫_{−∞}^{∞} ζ(t) dt = 1. Given φ ∈ D, we define I_{ζ}(φ) by:

I_{ζ}(φ)(t) := ∫_{−∞}^{t} [φ(u) − ζ(u) ∫_{−∞}^{∞} φ(v) dv]du, t ∈ ℝ.

Then we have: I_{ζ}(φ) ∈ D, I_{ζ}(φ′) = φ and \frac{d}{dt}I_{ζ}(φ)(t) = φ(t) − ζ(t) ∫_{−∞}^{∞} φ(v) dv, t ∈ ℝ [23]. Suppose x ∈ R(G). Since AG(φ)x = G(−φ′)x, φ ∈ D one concludes − ∫_{−∞}^{∞} φ′(t) u(t; x) dt = A ∫_{−∞}^{∞} φ(t) u(t; x) dt, φ ∈ D and the partial integration gives:

(4.3)  A ∫_{−∞}^{∞} φ′(t) ∫_{0}^{t} u(s; x) ds dt = ∫_{−∞}^{∞} φ′(t) u(t; x) dt, φ ∈ D.

Suppose (ρ_{n}) is a regularizing sequence and put φ_{n} = I_{ζ}(ρ_{n}) in (4.3) in order to see that:

A ∫_{−∞}^{∞} [ρ_{n}(t) − ζ(t)] ∫_{0}^{t} u(s; x) ds dt = ∫_{−∞}^{∞} [ρ_{n}(t) − ζ(t)]u(t; x) dt.

The closedness of A and u(0; x) = x imply, for every ζ ∈ D with ∫_{−∞}^{∞} ζ(t) dt = 1:

(4.4)  A ∫_{−∞}^{∞} ζ(t) ∫_{0}^{t} u(s; x) ds dt = ∫_{−∞}^{∞} ζ(t) u(t; x) dt − x.

It is evident that, for every t ∈ ℝ, there exists a sequence (ζ_{n}) in D so that ∫_{−∞}^{∞} ζ_{n}(t) dt = 1, n ∈ ℕ and that lim_{n→∞} ζ_{n} = δ_{t}, in the sense of distributions.

Put ζ_{n} in (4.4). As above, the closedness of A implies ∫_{0}^{t} u(s; x)ds ∈ D(A) and A ∫_{0}^{t} u(s; x) ds = u(t; x) − x, t ∈ ℝ. Inductively,

(4.5) A ∫_{0}^{t} \frac{(t − s)^{k}}{k!} u(s; x) ds = ∫_{0}^{t} \frac{(t − s)^{k−1}}{(k − 1)!} u(s; x) ds − \frac{t^{k}}{k!}x, t ∈ ℝ, k ∈ ℕ.

Clearly, Ax ∈ R(G) and A commutes with G(·). Hence,

(4.6)  A ∫_{−∞}^{∞} φ(t)u(t; x) dt = ∫_{−∞}^{∞} φ(t) u(t; Ax) dt, φ ∈ D.

An application of (4.6) gives u(t; x) ∈ D(A), Au(t; x) = u(t; Ax), t ∈ ℝ and this implies u(t; x) ∈ D_{∞} (A), t ∈ ℝ. Since A ∫_{0}^{t} u(s; x) ds = u(t; x) − x, t ∈ ℝ one obtains by induction that the function t ↦ u(t; x), t ∈ ℝ is infinitely differentiable and that \frac{d^{n}}{dt^{n}} u(t; x) = u(t; A^{n}x), t ∈ ℝ, x ∈ R(G), n ∈ ℕ. Furthermore,

(4.7)  A ∫_{0}^{t} \frac{(t − s)^{n_{0} −1}}{(n_{0} − 1)!} u(s; x) ds = ∫_{0}^{t} \frac{(t − s)^{n_{0} −1}}{(n_{0} − 1)!} u(s; Ax) ds.

Since

G(φ)x = (−1)^{n_{0}} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)S(t)x dt = ∫_{−∞}^{∞} φ(t) u(t; x) dt

= (−1)^{n_{0}} ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{0}^{t} \frac{(t − s)^{n_{0} −1}}{(n_{0} − 1)!} u(s; x) ds dt, φ ∈ D_{(−τ,τ)} ,

there is a subset {y_{0}(x), …, y_{n_{0} −1} (x)} of E such that:

(4.8)  S(t)x − ∫_{0}^{t} \frac{(t − s)^{n_{0} −1}}{(n_{0} − 1)!} u(s; x) ds = − ∑_{i=0}^{n_{0} −1} t^{i}y_{i}(x), t ∈ (−τ, τ).

Put t = 0 to obtain y_{0}(x) = B_{0}x. A consequence of (4.8) is:

(4.9) ∫_{0}^{t} S(s)x ds − ∫_{0}^{t} \frac{(t − s)^{n_{0}}}{n_{0}!} u(s; x) ds = − ∑_{i=0}^{n_{0} −1} \frac{t^{i+1}}{i + 1} y_{i}(x), t ∈ (−τ, τ).

Due to (4.5), one can apply A on both sides of (4.9) in order to see that, for every t ∈ (−τ, τ):

[S(t)x + ∑_{i=0}^{n_{0}} t^{i}B_{i}x] − ∫_{0}^{t} \frac{(t − s)^{n_{0} −1}}{(n_{0} − 1)!} u(s; x) ds + \frac{t^{n_{0}}}{n_{0}!} x = −A ∑_{i=0}^{n_{0} −1} \frac{t^{i+1}}{i + 1} y_{i}(x).

Returning to (4.8) implies:

(4.10) − ∑_{i=0}^{n_{0} −1} t^{i}y_{i}(x) + ∑_{i=0}^{n_{0}} t^{i}B_{i}x + \frac{t^{n_{0}}}{n_{0}!} x = −A ∑_{i=0}^{n_{0} −1} \frac{t^{i+1}}{i + 1} y_{i}(x), t ∈ (−τ, τ).

Since A is closed, one can differentiate (4.10) sufficiently many times to obtain that: x ∈ ⋂ _{i=0}^{n_{0}} D(Y_{i}), Y_{n_{0}} x = −x and y_{i}(x) = \frac{1}{i!} Y_{i}x, i ∈ \{0, …, n_{0} − 1\}. This completes the proof of (b1).

To prove (b2), fix an x ∈ R(G). Put S_{+}^{n_{0}} (t)x = S(t)x + ∑ _{i=0}^{n_{0} −1} \frac{t^{i}}{i!} Y_{i}(x), t ∈

[0, τ). By (4.8), S_{+}^{n_{0}} (t)x = ∫_{0}^{t} \frac{(t−s)^{n_{0} −1}}{(n_{0} −1)!} u(s; x) ds, t ∈ [0, τ) and an employment of (4.7) implies AS_{+}^{n_{0}} (t)x = S_{+}^{n_{0}} (t)Ax, t ∈ [0, τ). To prove (b2), suppose n_{0} = 1. Then S_{+}^{1} (t)x = S(t)x + B_{0}x, t ∈ [0, τ), x ∈ \overline{R(G)}. By the proof of (b1), one yields S_{+}^{1} (t)x = ∫_{0}^{t} u(s; x) ds, t ∈ [0, τ), x ∈ R(G). Accordingly, S_{+}^{1} (t)(\overline{R(G)}) ⊆ \overline{R(G)}, t ∈ [0, τ). By (4.5), ( ∫_{0}^{t} S_{+}^{1} (s)x ds, S_{+}^{1} (t)x − tx) ∈ A_{1}, t ∈ [0, τ), x ∈ R(G) and the closedness of A implies ( ∫_{0}^{t} S_{+}^{1} (s)x ds, S_{+}^{1} (t)x − tx) ∈ A_{1}, t ∈ [0, τ), x ∈ \overline{R(G)}. Clearly, S_{+}^{1} (t)A_{1} ⊆ A_{1}S_{+}^{1} (t), t ∈ [0, τ) and this proves that (S_{+}^{1} (t))_{t∈[0,τ)} is a once integrated semigroup generated by A_{1}. The similar arguments (see also the proof of (b3)) work for −A_{1} and (S_{−}^{1} (t))_{t∈[0,τ)} . To prove that A_{1} generates a C_{0}-group in \overline{R(G)}, we argue as follows. Since ( ∫_{0}^{t} S_{+}^{1} (s)x ds, S_{+}^{1} (t)x − tx) ∈ A_{1}, t ∈ [0, τ), x ∈ \overline{R(G)} and R(S_{+}^{1} (t)) ⊆ D(A), t ∈ [0, τ) one gets that the mapping t ↦ \frac{d}{dt}S_{+}^{1} (t)x, t ∈ [0, τ) is continuously differentiable for every x ∈ \overline{R(G)} and that \frac{d}{dt}S_{+}^{1} (t)x = AS_{1}(t)x + x, t ∈ [0, τ), x ∈ \overline{R(G)}. Moreover, it can be easily checked that, for every fixed x ∈ \overline{R(G)}, the function u(t) = S_{+}^{1} (t)x, t ∈ [0, τ) is a unique solution of the problem:

C_{1}(τ) : \{ \begin{matrix}u ∈ C([0, τ) : [D(A_{1})]) ∩ C^{1}([0, τ) : \overline{R(G)}), \\ u′(t) = A_{1}u(t) + x, t ∈ [0, τ), \\ u(0) = 0. \end{matrix}

An application of [1, Theorem 1.2] gives that A_{1} generates a C_{0}-semigroup in \overline{R(G)}. Similarly, −A_{1} generates a C_{0}-semigroup in \overline{R(G)} and this clearly implies that A_{1} generates a C_{0}-group in \overline{R(G)}.

To prove (b3), note that the proof of (b1) implies that S_{+}^{2} (t)x = ∫_{0}^{t}(t − s)u(s; x) ds, t ∈ [0, τ), x ∈ R(G). So, S_{+}^{2} (t)(\overline{R(G)}) ⊆ \overline{R(G)}, t ∈ [0, τ). Note also that S(0) = −B_{0} and that the Closed Graph Theorem gives S_{+}^{2} (t) ∈ L(\overline{R(G)}), t ∈ [0, τ). Next, the closedness of A and (4.5) imply ( ∫_{0}^{t} S_{+}^{2} (s)x ds, S_{+}^{2} (t) − \frac{t^{2}}{2} x) ∈ A_{1}, x ∈ \overline{R(G)}, t ∈ [0, τ). Since ∫_{0}^{t} S_{+}^{2} (s)x ds ∈ D(A), x ∈ \overline{R(G)}, t ∈ [0, τ) and R(B_{0}) ⊆ D(A), we immediately obtain AB_{0}x + B_{1}x ∈ D(A), x ∈ \overline{R(G)}. Further on,

A ∫_{0}^{t} S_{+}^{2} (s)x ds = S(t)x + B_{0}x + tB_{1}x + t^{2}B_{2}x + tAB_{0}x + \frac{t^{2}}{2}A(AB_{0}x + B_{1}x)

= S_{+}^{2} (t)x − \frac{t^{2}}{2}x, x ∈ \overline{R(G)}, t ∈ [0, τ).

Therefore, A(AB_{0}x + B_{1}x) = −x − 2B_{2}x, x ∈ \overline{R(G)}, and in conclusion, one

yields: S_{+}^{2} (t)x ∈ D(A), AS_{+}^{2} (t)x = A(S(t)x + B_{0}x) + t(−x − 2B_{2}x), x ∈ \overline{R(G)}, t ∈ [0, τ), \overline{R(G)} ⊆ ⋂ _{i=0}^{2} D(Y_{i}) and Y_{2}x = −x, x ∈ \overline{R(G)}. Suppose x ∈ D(A_{1}). Since R(B_{0}) ⊆ D(A) and AS_{+}^{2} (t)x−S_{+}^{2} (t)Ax = [AS(t)x+AB_{0}x+tA(AB_{0}+B_{1})x]− [S(t)Ax + B_{0}Ax + t(AB_{0} + B_{1})Ax] = tC_{1}x + t[A(AB_{0} + B_{1})x − (AB_{0} + B_{1})Ax], t ∈ [0, τ), Proposition 3.2(iv) immediately implies (S_{+}^{2} (t)x, S_{+}^{2} (t)A_{1}x) ∈ A_{1}, t ∈ [0, τ). So, (S_{+}^{2} (t))_{t∈[0,τ)} is a twice integrated semigroup generated by A_{1}. Because R(S_{+}^{2} (t)) ⊆ D(A), t ∈ [0, τ), the mapping t ↦ S_{+}^{2} (t)x is continuously differentiable for every fixed x ∈ \overline{R(G)} and the following holds: \frac{d}{dt}S_{+}^{2} (t)x = AS_{+}^{2} (t)x + tx = A(S(t)x + B_{0}x) − 2tB_{2}x, t ∈ [0, τ), x ∈ \overline{R(G)}. Then it is straightforward to see that \frac{d}{dt}S_{+}^{2} (t) ∈ L(\overline{R(G)}), t ∈ [0, τ) and that ( ∫_{0}^{t}(\frac{d}{ds} S_{+}^{2} (s)x) ds, (\frac{d}{dt}S_{+}^{2} (t)x) − tx) ∈ A_{1}, t ∈ [0, τ). Suppose now x ∈ D(A_{1}). Then \frac{d}{dt}S_{+}^{2} (t)x = AS_{+}^{2} (t)x + tx = S_{+}^{2} (t)Ax + tx ∈ D(A) and A\frac{d}{dt}S_{+}^{2} (t)x = AS_{+}^{2} (t)Ax + tAx = \frac{d}{dt}S_{+}^{2} (t)Ax, t ∈ [0, τ), ( \frac{d}{dt}S_{+}^{2} (t)x, \frac{d}{dt}S_{+}^{2} (t)A_{1}x) ∈ A_{1}, t ∈ [0, τ), and consequently, ( \frac{d}{dt}S_{+}^{2} (t))_{t∈[0,τ)} is a once integrated semigroup generated by A_{1}. In order to obtain the corresponding statement for the operator −A_{1} and (S_{−}^{2} (t))_{t∈[0,τ)} , notice the following facts: (1.2) holds for −A and Ǧ, Ǧ fulfills (DG)_{4} with u(ˇ·; x), Ǧ(φ)x = (−1)^{n_{0}} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)[(−1)^{n_{0}} Š(t)]x dt, x ∈ E, φ ∈ D and ((−1)^{n_{0}} Š(t))_{t∈(−τ,τ)} is a [(−1)^{n_{0}} B_{0}, …, (−1)^{n_{0} +n_{0}} B_{n_{0}} , (−1)^{n_{0} +1} C_{0}, …, (−1)^{n_{0} +n_{0}} C_{n_{0} −1} ]-group with a subgenerator −A.

In order to prove (b4), assume x ∈ \overline{R(G)}. Let (x_{n}) be a sequence in R(G) with lim_{n→∞} x_{n} = x. Due to (4.2) and (b1), lim_{n→∞} Y_{n_{0} −1} (x_{n}) = Y_{n_{0} −1} x and lim_{n→∞} AY_{n_{0} −1} (x_{n}) = −x − n_{0}!B_{n_{0}} x. Hence, Y_{n_{0} −1} x ∈ D(A), x ∈ D(Y_{n_{0}} ) and Y_{n_{0}} x = −x as claimed. This yields S_{±}^{n_{0}} (t)x ∈ D(A), x ∈ \overline{R(G)}. As in the proofs of (b1), (b2) and (b3), one obtains S_{±}^{n_{0}} (t) ∈ L(\overline{R(G)}), ( ∫_{0}^{t} S_{±}^{n_{0}} (s)x ds, S_{±}^{n_{0}} (t)x − \frac{t^{n_{0}}}{n_{0} !} x) ∈ ±A_{1}, x ∈ \overline{R(G)}, t ∈ [0, τ) and S_{±}^{n_{0}} (t)A_{1}x = A_{1}S_{±}^{n_{0}} (t)x, t ∈ [0, τ), x ∈ R(G). We will sketch the rest of the proof of (b4) only for A and S(·). Suppose t, s ∈ [0, τ) and t + s < τ. Since AS_{+}^{n_{0}} (·)x = S_{+}^{n_{0}} (·)Ax, x ∈ R(G), one can repeat literally the arguments given in the proof of [29, Propostion 2.4] in order to conclude that:

(4.11) S_{+}^{n_{0}} (t)S_{+}^{n_{0}} (s)x = [ ∫_{0}^{t+s} − ∫_{0}^{t} − ∫_{0}^{s} ] \frac{(t + s − r)^{n_{0} −1}}{(n_{0} − 1)!} S_{+}^{2} (r)xdr, x ∈ R(G).

The standard limit procedure implies that (4.11) remains true for every x ∈ \overline{R(G)} and t, s ∈ [0, τ) with t + s < τ. It is straightforward to verify that (S_{+}^{n_{0}} (t))_{t∈[0,τ)} ⊆ L(\overline{R(G)}) is a local n_{0}-times integrated semigroup in the sense of (△). To prove that A′_{n_{0} −1,+} is the generator of a local n_{0}-times integrated semigroup (S_{+}^{n_{0}} (t))_{t∈[0,τ)} in the sense of Definition 3.1, we argue as follows. First of all, let us observe that A′_{n_{0} −1,+} is a closed operator and that the arguments employed in the proof of Proposition 3.2 also show that D(A^{n_{0}} )∩D(A_{1}) ⊆ C^{n_{0}} (τ). Suppose now x ∈ D(A_{0}), where A_{0} is the infinitesimal generator of (S_{+}^{n_{0}} (t))_{t∈[0,τ)} . This implies the existence of a number σ ∈ (0, τ) so that the mapping t ↦ S_{+}^{n_{0}} (t)x, t ∈ [0, σ) is n_{0}-times continuously differentiable and that A_{0}x = lim_{t→0+} ( \frac{d^{n_{0}}}{dt^{n_{0}}} (S_{+}^{n_{0}} (t)x) − x)/t. On the other hand, the closedness of A offers one to show that \frac{d^{k}}{dt^{k}} S(t)x ∈ D(A) and that \frac{d^{k+1}}{dt^{k+1}} S(t)x = A\frac{d^{k}}{dt^{k}} S(t)x − ∑ _{j=k+1}^{n} j ⋯ (j − k)t^{j−k−1} B_{j} x, for every t ∈ [0, σ)

and k ∈ \{0, …, n_{0} − 1\}. Therefore, x ∈ ⋂ _{i=0}^{n_{0}} D(Y_{i}) and Y_{k}x = −( \frac{d^{k}}{dt^{k}} S(t)x)_{t=0} , k ∈ {0, …, n_{0}}. Moreover,

\frac{d^{n_{0}}}{dt^{n_{0}}} (S_{+}^{n_{0}} (t)x) − x = A[ \frac{d^{n_{0} −1}}{dt^{n_{0} −1}} S_{+}^{n_{0}} (t)x] − n_{0}!B_{n_{0}} x − x

= A[ \frac{d^{n_{0} −1}}{dt^{n_{0} −1}} S_{+}^{n_{0}} (t)x] + AY_{n_{0} −1} x = A[ \frac{d^{n_{0} −1}}{dt^{n_{0} −1}} S_{+}^{n_{0}} (t)x − ( \frac{d^{n_{0} −1}}{dt^{n_{0} −1}} S(t)x)_{t=0} x].

It is also evident that

x = ( \frac{d^{n_{0}}}{dt^{n_{0}}} S(t)x)_{t=0} = lim_{t→0+} \frac{1}{t}( \frac{d^{n_{0} −1}}{dt^{n_{0} −1}} S_{+}^{n_{0}} (t)x − ( \frac{d^{n_{0} −1}}{dt^{n_{0} −1}} S(t)x)_{t=0} x) .

The closedness of A implies x ∈ D(A_{1}), A_{0}x = A_{1}x and, because of that, \overline{A_{0}} ⊆ A_{1}. Further on, R(G) ⊆ D(A^{n_{0}} ) ∩ D(A_{1}) ⊆ C^{n_{0}} (τ) and an application of [40, Proposition 4.5] gives ( ∫_{0}^{t} S_{+}^{n_{0}} (s)x ds, S_{+}^{n_{0}} (t)x − \frac{t^{n_{0}}}{n_{0} !} x) ∈ \overline{A_{0}}, x ∈ \overline{R(G)}, t ∈ [0, τ) and S_{+}^{n_{0}} (t)\overline{A_{0}}x = \overline{A_{0}}S_{+}^{n_{0}} (t)x, t ∈ [0, τ), x ∈ D(\overline{A_{0}}). Suppose (x, y) ∈ \overline{A_{0}}. Then

0 = AS_{+}^{n_{0}} (t)x − S_{+}^{n_{0}} (t)y = ∑_{i=0}^{n_{0} −1} t^{i}C_{i}x + ∑_{i=0}^{n_{0} −1} \frac{t^{i}}{i!} AY_{i}x − ∑_{i=0}^{n_{0} −1} \frac{t^{i}}{i!} Y_{i}y, t ∈ [0, τ)

and this implies \overline{A_{0}} ⊆ A′_{n_{0} −1,+} . Further, fix an x ∈ D(A′_{n_{0} −1,+} ) and notice that AS_{+}^{n_{0}} (t)x = S_{+}^{n_{0}} (t)Ax, t ∈ [0, τ) and that

A′_{n_{0} −1,+} ∋ ( ∫_{0}^{t} S_{+}^{n_{0}} (s)x ds, S_{+}^{n_{0}} (t)x − \frac{t^{n_{0}}}{n_{0}!}x) = ( ∫_{0}^{t} S_{+}^{n_{0}} (s)xd s, A_{1} ∫_{0}^{t} S_{+}^{n_{0}} (s)x ds)

= ( ∫_{0}^{t} S_{+}^{n_{0}} (s)x ds, ∫_{0}^{t} S_{+}^{n_{0}} (s)Ax ds) , t ∈ [0, τ).

This implies

C_{i} ∫_{0}^{t} S_{+}^{n_{0}} (s)x ds + \frac{1}{i!}AY_{i} ∫_{0}^{t} S_{+}^{n_{0}} (s)x ds = \frac{1}{i!}Y_{i} ∫_{0}^{t} S_{+}^{n_{0}} (s)Ax ds,

t ∈ [0, τ), i ∈ {2, …, n_{0} − 1}.

Differentiate this equality to obtain that C_{i}S_{+}^{n_{0}} (t)x+ \frac{1}{i!} AY_{i}S_{+}^{n_{0}} (t)x = \frac{1}{i!} Y_{i}S_{+}^{n_{0}} (t)Ax, t ∈ [0, τ), i ∈ {2, …, n_{0} − 1}. Thus, S_{+}^{n_{0}} (t)A′_{n_{0} −1,+} ⊆ A′_{n_{0} −1,+} S_{+}^{n_{0}} (t), t ∈ [0, τ) and A′_{n_{0} −1,+} is the generator of a local n_{0}-times integrated semigroup (S_{+}^{n_{0}} (t))_{t∈[0,τ)} in the sense of Definition 2.3. An application of the arguments given in the proof of [24, Proposition 2.1] gives \overline{A_{0}} = A′_{n_{0} −1,+} . Since R(S_{+}^{n_{0}} (t)) ⊆ D(A), the mapping t ↦ S_{+}^{n_{0}} (t)x, t ∈ [0, τ) is continuously differentiable for every fixed x ∈ \overline{R(G)} and

\frac{d}{dt}S_{+}^{n_{0}} (t)x = AS_{+}^{n_{0}} (t)x + \frac{t^{n_{0} −1}}{(n_{0} − 1)!} x = AS(t)x + ∑_{i=0}^{n_{0} −1} \frac{t^{i}}{i!} AY_{i}x + \frac{t^{n_{0} −1}}{(n_{0} − 1)!} x,

t ∈ [0, τ), x ∈ \overline{R(G)}. Then it can be easily verified that (\frac{d}{dt}S_{+}^{n_{0}} (t))_{t∈[0,τ)} ⊆ L(\overline{R(G)}) is a local (n_{0} − 1)-times integrated semigroup in the sense of (△). The c.i.g of (\frac{d}{dt}S_{+}^{n_{0}} (t))_{t∈[0,τ)} is \overline{A_{0}} = A′_{n_{0} −1,+} and an application of [40, Proposition 4.5] enables one to conclude that A′_{n_{0} −1,+} is the generator of a local (n_{0} − 1)-times integrated semigroup (\frac{d}{dt}S_{+}^{n_{0}} (t))_{t∈[0,τ)} in the sense of Definition 2.3.

To prove (b5), suppose λ ∈ ρ(A) and set A_{λ} = A − λ, G_{λ} = e^{−λ·} G and u_{λ}(·; x) = e^{−λ·} u(·; x), x ∈ R(G) = R(G_{λ}). It is straightforward to check that A_{λ} and G_{λ} fulfill (1.2) and that G_{λ} is regular with G_{λ}(φ)x = ∫_{−∞}^{∞} φ(t)u_{λ}(t; x) dt, φ ∈ D, x ∈ R(G_{λ}). Clearly,

G_{λ}(φ)x = G(e^{−λ·} φ)x = (−1)^{n_{0}} ∫_{−τ}^{τ}(e^{−λ·} φ)^{(n_{0} )} (t)S(t)x dt

= (−1)^{n_{0}} ∫_{−τ}^{τ} ∑_{i=0}^{n_{0}} (−1)^{n_{0} −i} \binom{n_{0}}{i} λ^{n_{0} −i} e^{−λt} φ^{(i)} (t)S(t)x dt

= ∑_{i=0}^{n_{0}} (−1)^{i} \binom{n_{0}}{i} λ^{n_{0} −i} ∫_{−τ}^{τ} φ^{(i)} (t)(e^{−λt} S(t)x) dt = (−1)^{n_{0}} ∫_{−τ}^{τ} φ^{(n_{0} )} (t)e^{−λt} S(t)x dt

+ ∑_{i=1}^{n_{0}} (−1)^{i} \binom{n_{0}}{i} λ^{n_{0} −i} (−1)^{n_{0} −i} ∫_{−τ}^{τ} φ^{(n_{0} )} (t) ∫_{0}^{t} \frac{(t − s)^{n_{0} −i−1}}{(n_{0} − i − 1)!} e^{−λs} S(s)x ds dt

= (−1)^{n_{0}} ∫_{−τ}^{τ}φ^{(n_{0} )} (t)[e^{−λt} S(t)x + ∑_{i=1}^{n_{0}} \binom{n_{0}}{i} λ^{n_{0} −i} ∫_{0}^{t} \frac{(t − s)^{n_{0} −i−1}}{(n_{0} − i − 1)!} e^{−λs} S(s)x ds]dt,

for every φ ∈ D_{(−τ,τ)} and x ∈ E. Put, for every t ∈ (−τ, τ) and x ∈ E :

S_{λ}(t)x := e^{−λt} S(t)x + ∑_{i=1}^{n_{0}} \binom{n_{0}}{i} λ^{n_{0} −i} ∫_{0}^{t} \frac{(t − s)^{n_{0} −i−1}}{(n_{0} − i − 1)!} e^{−λs} S(s)x ds.

Then the mapping S_{λ} : (−τ, τ) → L(E, [D(A_{λ})]) is continuous and G_{λ}(φ)x = (−1)^{n_{0}} ∫_{−τ}^{τ} φ^{(n_{0} )} (t)S_{λ}(t)x dt, φ ∈ D_{(−τ,τ)} , x ∈ E. The proof of (a) implies that there exist bounded linear operators B_{0}^{λ}, …, B_{n_{0}}^{λ} , C_{0}^{λ}, …, C_{n_{0} −1}^{λ} such that A_{λ} is a subgenerator of a [B_{0}^{λ}, …, B_{n_{0}}^{λ} , C_{0}^{λ}, …, C_{n_{0} −1}^{λ} ]-group (S_{λ}(t))_{t∈(−τ,τ)} . Define Y_{i}^{λ} recursively by: Y_{0}^{λ} = B_{0}^{λ} and Y_{i+1}^{λ} := (i+1)!B_{i+1}^{λ} +A_{λ}Y_{i}^{λ}, i ∈ {0, …, n_{0}−1}. Since 0 ∈ ρ(A_{λ}), we have that Y_{i}^{λ} is closed, i = 1, …, n_{0}. Suppose, for the time being, x ∈ \overline{R(G)} and (x_{n}) is a sequence in R(G) such that lim_{n→∞} x_{n} = x. A consequence of Y_{n_{0}}^{λ} x_{n} = −x_{n}, n ∈ ℕ is lim_{n→∞} A_{λ}Y_{n_{0} −1}^{λ} x_{n} = −x − n_{0}!B_{n_{0}}^{λ} x and the boundedness of A_{λ}^{−1} implies lim_{n→∞} Y_{n_{0} −1}^{λ} x_{n} = A_{λ}^{−1} (−x − n_{0}!B_{n_{0}}^{λ} x). Continuing this procedure enables one to establish that, for every i = 1, …, n_{0} − 1, lim_{n→∞} Y_{i}^{λ}x_{n} exists. The closedness of Y_{i}^{λ} yields x ∈ ⋂ _{i=0}^{n_{0}} D(Y_{i}^{λ}) and Y_{n_{0}}^{λ} x = −x. Put A_{1,λ} = (A_{λ})\overline{R(G_{λ} )} and

Cx = A_{λ}^{−(n_{0} −1)} x, x ∈ \overline{R(G)}. Because G_{λ}A_{λ} ⊆ A_{λ}G_{λ}, we have A_{λ}^{−k} G_{λ} = G_{λ}A_{λ}^{−k} , k ∈ ℕ, A_{λ}^{−k} (R(G)) ⊆ R(G), k ∈ ℕ and A_{λ}^{−k} (\overline{R(G)}) ⊆ \overline{R(G)}, k ∈ ℕ. This offers one to see that 0 ∈ ρ(A_{1,λ} ) and that C ∈ L(\overline{R(G)}) is injective. Assume now x ∈ D(A_{1,λ}^{n_{0} −1} ). Then A_{1,λ}^{n_{0} −1} x ∈ \overline{R(G)} and this gives x = A_{λ}^{−(n_{0} −1)} (A_{λ}^{n_{0} −1} x) = C(A_{λ}^{n_{0} −1} x) ∈ R(C). Hence, D(A_{1,λ}^{n_{0} −1} ) ⊆ R(C). Proceeding as in the proof of Proposition 3.2, one obtains that the mapping t ↦ S_{λ}(t)x, t ∈ (−τ, τ) is n_{0}-times continuously differentiable and that there exists a function M : (−τ, τ) → (0, ∞), independent of x, so that ∥\frac{d^{n_{0}}}{dt^{n_{0}}} S_{λ}(t)x∥ ⩽ M(|t|) ∑ _{i=0}^{n_{0}} ∥A_{λ}^{i}x∥, t ∈ (−τ, τ). Put u^{λ}(t; x) = \frac{d^{n_{0}}}{dt^{n_{0}}} S_{λ}(t)x, t ∈ [0, τ), x ∈ D(A_{λ}^{n_{0} −1} ) and T (t)x = u^{λ}(t; Cx), t ∈ [0, τ), x ∈ \overline{R(G)}. Due to Proposition 3.2, D(A_{λ}^{n_{0} −1} ) ⊆ D(Y_{λ}^{n_{0}} ) and u^{λ}(0; x) = −Y_{λ}^{n_{0}} x, x ∈ D(A_{λ}^{n_{0} −1} ). Moreover, R(C) ⊆ \overline{R(G)} ∩ D(A_{λ}^{n_{0} −1} ) and this implies u^{λ}(0; Cx) = −Y_{λ}^{n_{0}} Cx = Cx. The mapping t ↦ T (t)x, t ∈ [0, τ) is continuous for every fixed x ∈ \overline{R(G)} and ∥T (t)x∥ = ∥u^{λ}(t; A_{λ}^{−(n_{0} −1)} x)∥ ⩽ M(t) ∑ _{i=0}^{n_{0} −1} ∥A_{λ}^{−1} ∥^{i}∥x∥, t ∈ [0, τ), x ∈ \overline{R(G)}. The partial integration shows G_{λ}(φ)x = ∫_{−τ}^{τ} φ(t)u^{λ}(t; x) dt, φ ∈ D_{[0,τ)}, x ∈ D(A_{λ}^{n_{0} −1} ) and this implies u^{λ}(t; x) ∈ \overline{R(G)}, t ∈ [0, τ), x ∈ D(A_{λ}^{n_{0} −1} ). Therefore, T (t)x ∈ \overline{R(G)}, t ∈ [0, τ), x ∈ \overline{R(G)} and T (t) ∈ L(\overline{R(G)}), t ∈ [0, τ). As in the proof of (b1), one concludes A_{λ} ∫_{0}^{t} u^{λ}(s; Cx) ds = A_{λ} ∫_{0}^{t} T (s)xds = u^{λ}(t; Cx) − u^{λ}(0; Cx) = T (t)x − Cx, t ∈ [0, τ), x ∈ \overline{R(G)} and that u^{λ}(t; A_{λ}x) = A_{λ}u^{λ}(t; x), t ∈ [0, τ), x ∈ D(A_{λ}^{n_{0}} ). Due to the previous equality, we have T (t)A_{1,λ} ⊆ A_{1,λ} T (t) and T (t)C = CT (t), t ∈ [0, τ). Now it is straightforward to prove that the abstract Cauchy problem:

\{ \begin{matrix}v ∈ C([0, τ) : [D(A_{1,λ} )]) ∩ C^{1}([0, τ) : \overline{R(G)}), \\ v′(t) = A_{1,λ} v(t) + Cx, t ∈ [0, τ), \\ v(0) = 0, \end{matrix}

possesses a unique solution for every x ∈ \overline{R(G)}, given by v(t) = ∫_{0}^{t} T (s)x ds, t ∈ [0, τ), x ∈ \overline{R(G)}. This simply implies that the abstract Cauchy problem:

(ACP, τ) : \{ \begin{matrix}f ∈ C([0, τ) : [D(A_{1,λ} )]) ∩ C^{1}([0, τ) : \overline{R(G)}), \\ f′(t) = A_{1,λ} f(t), t ∈ [0, τ), \\ f(0) = x, \end{matrix}

has a unique solution for every x ∈ C(D(A_{1,λ} )) and that A_{1,λ} is the integral generator of a local C-semigroup (T (t))_{t∈[0,τ)} . As before, D(A_{λ}^{n_{0}} ) ⊆ C(D(A_{λ})) and an application of [44, Theorem 4.4] shows that A_{1,λ} generates a local (n_{0}−1)-times integrated semigroup on [0, τ). A rescaling result for local integrated semigroups (cf. for instance [1]) implies that A_{1} generates a local (n_{0} − 1)-times integrated semigroup on [0, τ). Analogously, −A_{1} generates a local (n_{0} − 1)-times integrated semigroup on [0, τ) and the proof ends an employment of Corollary 2.1. □

Theorem 4.2. Let G be a (DG) generated by A. Then the group (S(t))_{t∈(−τ,τ)} , constructed in Theorem 4.1(a), is non-degenerate. If n_{0} = 1, then A generates a C_{0}-group. If n_{0} = 2, then:

(a) (S_{±}^{1} (t) := ±A(S(±t)x + B_{0}x) − 2tB_{2}x)_{t∈[0,τ)} are local once integrated semigroups in the sense of (△).

(b) The c.i.g of (S_{+}^{1} (t))_{t∈[0,τ)} , resp., (S_{−}^{1} (t))_{t∈[0,τ)} is A\overline{R(G)} , resp., (−A)\overline{R(G)} .

(c) Suppose A is densely defined or λ − A is surjective for some λ ∈ ℂ. Then ±A are generators of local once integrated semigroups (S_{±}^{1} (t))_{t∈[0,τ)} .

Furthermore:

(i) For every x ∈ E and φ, ψ ∈ D_{(−τ,τ)} with suppφ + suppψ ⊆ (−τ, τ):

(4.12) G(φ) G(ψ)x = ∑_{i=0}^{n_{0}} (−1)^{i+1} i! ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} −i)} (s)S(t + s)B_{i}x ds dt.

(ii) Y_{n_{0}} x = −x, x ∈ ⋂ _{i=0}^{n_{0}} D(Y_{i}).

(iii) Suppose x ∈ D(A^{n_{0} −1} ). Then {x, Ax} ⊆ ⋂ _{i=2}^{n_{0}} D(Y_{i}), Y_{n_{0}} x = −x,

Y_{n_{0}} Ax = −Ax and D(A^{n_{0} −1} ) ⊆ \overline{R(G)}. (iv) A is stationary dense with n(A) ⩽ n_{0} − 1.

(v) If ρ(A) ≠ ∅, then for every τ_{0} ∈ (0, ∞), there is an n(τ_{0}) ∈ ℕ so that A generates a local n(τ_{0})-times integrated group on (−τ_{0}, τ_{0}).

(vi) G is dense iff D_{∞} (A) is dense in E. In the case ρ(A) ≠ ∅, G is dense iff

A is densely defined. (vii) ⋂ _{φ∈D_{0}} N(G(φ)) = {0} and ⋂ _{φ∈D_{0}} N(G(φ̌)) = {0}.

Proof. Assume S(t)x = 0, t ∈ (−τ, τ). This implies G(ψ)x = 0, ψ ∈ D_{(−τ,τ)}

and G(ρ)x = lim_{n→∞} G(ρ ∗ ρ_{n})x = lim_{n→∞} G(ρ)G(ρ_{n})x = 0, ρ ∈ D, where (ρ_{n}) is a regularizing sequence. Owing to (DG)_{2}, one can deduce that x ∈ N (G) and that (S(t))_{t∈(−τ,τ)} is non-degenerate. Put now S_{1}(t)x = S(t)x + B_{0}x, t ∈ (−τ, τ), x ∈ E. We will prove that (S_{1}(t))_{t∈[0,τ)} is a once integrated semigroup generated by A. First of all, note that S_{1}(t)A ⊆ AS_{1}(t), t ∈ (−τ, τ) and that S_{1} : (−τ, τ) → L(E, [D(A)]) is continuous. This clearly implies \frac{d}{dt}S_{1}(t)x = AS_{1}(t)x + Bx, t ∈ (−τ, τ), x ∈ E, where B = −B_{1} − AB_{0} ∈ L(E). Further, ∫_{0}^{t} S_{1}(s)xds ∈ D(A), t ∈ (−τ, τ), x ∈ E and one gets that A ∫_{0}^{t} S_{1}(s)x ds = A ∫_{0}^{t}(S(s)x + B_{0}x) ds = S(t)x + B_{0}x+tB_{1}x+tAB_{0}x = S_{1}(t)x−tBx, t ∈ (−τ, τ), x ∈ E and that (S_{1}(t))_{t∈(−τ,τ)} is a [0, −B, 0]-group with a subgenerator A. We will prove that B = I. Suppose ζ, η ∈ D_{(−τ/4,τ/4)} and (ρ_{n}) is a regularizing sequence. We know [23] that suppI_{ζ}(φ) ⊆ [min(−τ/4, inf(suppφ)), max(τ/4, sup(suppφ))] and that there exists k ∈ ℕ such that suppI_{ζ}(ρ_{n}) ∪ suppI_{η}(ρ_{n}) ⊆ [−τ/4, τ/4], n ⩾ k. Fix an x ∈ E. By (DG)_{1} (see also the equation (4.21) given below), one gets that, for every φ, ψ ∈ D_{(−τ/4,τ/4)} :

(4.13) ∫_{−∞}^{∞} φ′(t) ∫_{−∞}^{∞} ψ′(s)S_{1}(t)S_{1}(s)x ds dt = − ∫_{−∞}^{∞} φ′(t) ∫_{−∞}^{∞} ψ(s)S_{1}(t + s)x ds dt.

Put φ = I_{ζ}(ρ_{n}), n ⩾ k in (4.13). Then one obtains, for every φ, ψ ∈ D_{(−τ/4,τ/4)} :

∫_{−∞}^{∞} [ρ_{n}(t) − ζ(t)] ∫_{−∞}^{∞} ψ′(s)S_{1}(t)S_{1}(s)x ds dt

= − ∫_{−∞}^{∞} [ρ_{n}(t) − ζ(t)] ∫_{−∞}^{∞} ψ(s)S_{1}(t + s)x ds dt.

Letting n → ∞ and applying the partial integration, one concludes that, for every ψ ∈ D_{(−τ/4,τ/4)} :

(4.14) − ∫_{−∞}^{∞} ζ(t) ∫_{−∞}^{∞} ψ′(s)S_{1}(t)S_{1}(s)x ds dt

= ∫_{−∞}^{∞} ψ′(s) ∫_{0}^{s} S_{1}(v)x dv ds − ∫_{−∞}^{∞} ζ(t) ∫_{−∞}^{∞} ψ′(s) ∫_{0}^{s} S_{1}(t + v)x dv ds dt.

Plug ψ = I_{η}(ρ_{n}), n ⩾ k into (4.14). We get, for every ψ ∈ D_{(−τ/4,τ/4)} :

− ∫_{−∞}^{∞} ζ(t) ∫_{−∞}^{∞} [ρ_{n}(s) − η(s)]S_{1}(t)S_{1}(s)x ds dt

= ∫_{−∞}^{∞} [ρ_{n}(s)−η(s)] ∫_{0}^{s}S_{1}(v)x dv ds − ∫_{−∞}^{∞} ζ(t) ∫_{−∞}^{∞} [ρ_{n}(s)−η(s)] ∫_{0}^{s} S_{1}(t+v)x dv dt ds.

The standard limit procedure leads us to the next equality:

(4.15) ∫_{−∞}^{∞} ζ(t) η(s)S_{1}(t)S_{1}(s)x ds dt

= − ∫_{−∞}^{∞} η(s) ∫_{0}^{s} S_{1}(v)x dv ds + ∫_{−∞}^{∞} ζ(t) η(s) ∫_{0}^{s} S_{1}(t + v)x dv dt ds.

Let t, s ∈ (−τ/4, τ/4) be fixed and let (ζ_{n})_{n∈ℕ} and (η_{n})_{n∈ℕ} be sequences in D_{(−τ/4,τ/4)} satisfying ∫_{−∞}^{∞} ζ_{n}(t) dt = 1, ∫_{−∞}^{∞} η_{n}(t) dt = 1, n ∈ ℕ, lim_{n→∞} ζ_{n} = δ_{t} and lim_{n→∞} η_{n} = δ_{s}, in the sense of distributions. By virtue of (4.15), we have:

(4.16)  S_{1}(t)S_{1}(s)x = [ ∫_{0}^{t+s} − ∫_{0}^{t} − ∫_{0}^{s} ]S_{1}(r)x dr.

Notice that (4.16) implies

S_{1}(t)( \frac{d}{dr}S_{1}(r)x)_{|r=s} = S_{1}(t + s)x − S_{1}(s)x,
S_{1}(t)[AS_{1}(s)x + Bx] = S_{1}(t + s)x − S_{1}(s)x.

Since S_{1}(t)A ⊆ AS_{1}(t), t ∈ (−τ, τ), one yields:

A[ ∫_{0}^{t+s} − ∫_{0}^{t} − ∫_{0}^{s} ]S_{1}(r)x dr + S_{1}(t)Bx = S_{1}(t + s)x − S_{1}(s)x, i.e.,

S_{1}(t + s)x − (t + s)Bx − S_{1}(t)x + tBx − S_{1}(s)x + sBx + S_{1}(t)Bx

= S_{1}(t + s)x − S_{1}(s)x.

So, S_{1}(v)[Bx − x] = 0, v ∈ (−τ/4, τ/4). Since G(φ) = − ∫_{−∞}^{∞} φ′(v)S_{1}(v)x dv, φ ∈ D_{(−τ/4,τ/4)} , one can easily conclude that (S_{1}(t))_{t∈(−τ/4,τ/4)} is a non-degenerate operator family. Hence, B = I and (S_{1}(t))_{t∈[0,τ)} is a once integrated semigroup generated by A. Analogously, (−S(−t) − B_{0})_{t∈[0,τ)} is a once integrated semigroup generated by −A and one can repeat literally the arguments given in the proof of Theorem 4.1(b2) in order to see that A generates a C_{0}-group. Suppose now n_{0} = 2 and denote A_{1} = A\overline{R(G)} . We will only prove that A_{1} is the c.i.g of (S_{+}^{1} (t))_{t∈[0,τ)} . Evidently, AB_{0} + B_{1} ∈ L(E), G(φ)x = ∫_{−∞}^{∞} φ′′(t)[S(t)x + B_{0}x + t(AB_{0} + B_{1})x]dt, φ ∈ D_{(−τ,τ)} , x ∈ E and the mapping t ↦ S(t)x + B_{0}x + t(AB_{0} + B_{1})x, t ∈ [0, τ) is continuously differentiable with

\frac{d}{dt}[S(t)x + B_{0}x + t(AB_{0} + B_{1})x] = AS(t)x − B_{1}x − 2tB_{2}x + (AB_{0} + B_{1})x,

t ∈ [0, τ), x ∈ E. Therefore,

(4.17)  G(φ)x = − ∫_{−∞}^{∞} φ′(t)S_{+}^{1} (t)x dt, φ ∈ D_{[0,τ)}, x ∈ E.

Suppose x ∈ E, φ, ψ ∈ D_{[0,τ)} and suppφ + suppψ ⊆ [0, τ). Since G satisfies (DG)_{1} (see also (4.21)), we obtain

∫_{−∞}^{∞} φ′(t) ∫_{−∞}^{∞} ψ′(s)S_{+}^{1} (t)S_{+}^{1} (s)x dt ds = − ∫_{−∞}^{∞} φ′(t) ∫_{−∞}^{∞} ψ(s)S_{+}^{1} (t + s)x dt ds.

Arguing as in the case n_{0} = 1, one gets, for every t, s ∈ [0, τ) with t + s < τ:

S_{+}^{1} (t)S_{+}^{1} (s)x = [ ∫_{0}^{t+s} − ∫_{0}^{t} − ∫_{0}^{s} ]S_{+}^{1} (r)x dr.

Further on, S_{+}^{1} (0) = 0 and the mapping t ↦ S_{+}^{1} (t)x, t ∈ [0, τ) is continuous. It can be simply verified that (S_{+}^{1} (t))_{t∈[0,τ)} is a non-degenerate operator family, and consequently, (S_{+}^{1} (t))_{t∈[0,τ)} is a local once integrated semigroup in the sense of (△). Suppose x ∈ D(A_{0}). Then there exists σ ∈ (0, τ] such that the mapping t ↦ S_{+}^{1} (t)x, t ∈ [0, σ) is continuously differentiable and that A_{0}x = lim_{t→0+} ( \frac{d}{dt}(S_{+}^{1} (t)x)−x)/t. The partial integration and (4.17) yield:

(4.18)  G(φ)x = ∫_{−∞}^{∞} φ(t)( \frac{d}{dt}S_{+}^{1} (t)x) t, φ ∈ D_{[0,σ)} .

Owing to (4.18) and Theorem 4.1(b3), we get lim_{n→∞} G(ρ_{n})x = x ∈ \overline{R(G)}, S_{+}^{2} (t)x = S(t)x + B_{0}x + t(AB_{0} + B_{1})x ∈ \overline{R(G)}, t ∈ [0, τ) and \overline{R(G)} ∋ \frac{d}{dt}S_{+}^{2} (t)x = AS(t)x−B_{1}x−2tB_{2}x+(AB_{0}+B_{1})x = S_{+}^{1} (t)x, t ∈ [0, τ). Consequently, \frac{d}{dt}S_{+}^{1} (t)x ∈ \overline{R(G)}, t ∈ [0, σ), A_{0}x = lim_{t→0+} ( \frac{d}{dt}(S_{+}^{1} (t)x) − x)/t ∈ \overline{R(G)} and:

(4.19)  \{x, A_{0}x\} ⊆ \overline{R(G)}.

Further, \frac{d}{dt}S(t)x = AS(t)x − B_{1}x − 2tB_{2}x, t ∈ [0, τ),

\frac{d}{dt}S_{+}^{1} (t)x + 2B_{2}x = lim_{h→0} \frac{A[S(t + h)x − S(0)x] − A[S(t)x − S(0)x]}{h}

= lim_{h→0} A\frac{S(t + h)x − S(t)x}{h} , t ∈ [0, σ)

and

lim_{h→0} \frac{S(t + h)x − S(t)x}{h} = AS(t)x − B_{1}x − 2tB_{2}x, t ∈ [0, τ).

The closedness of A gives AS(t)x−B_{1}x−2tB_{2}x ∈ D(A), t ∈ [0, σ) and A[AS(t)x− B_{1}x − 2tB_{2}x] = \frac{d}{dt}S_{+}^{1} (t)x + 2B_{2}x, t ∈ [0, σ). Put t = 0 in the previous equality to obtain A(AB_{0} + B_{1})x = −x − 2B_{2}x. Hence,

A_{0}x = lim_{t→0+} \frac{\frac{d}{dt}(S_{+}^{1} (t)x) − x}{t} = lim_{t→0+} \frac{A[AS(t)x − B_{1}x − 2tB_{2}x] − 2B_{2}x − x}{t}

= lim_{t→0+} \frac{A[AS(t)x − B_{1}x − 2tB_{2}x] + A(AB_{0} + B_{1})x}{t}

= lim_{t→0+} A\frac{A[S(t)x − S(0)x] − 2tB_{2}x}{t} .

On the other hand,

lim_{t→0+} \frac{A[S(t)x − S(0)x] − 2tB_{2}x}{t} = lim_{t→0+} \frac{S_{+}^{1} (t)x − S_{+}^{1} (0)x}{t} = ( \frac{d}{dt}S_{+}^{1} (t)x)_{t=0} = x.

Therefore, x ∈ D(A), A_{0}x = Ax, A_{0} ⊆ A and (4.19) enables one to see that A_{0} ⊆ A_{1} and that \overline{A_{0}} ⊆ A_{1}. Furthermore, Theorem 4.1(b3) shows that A_{1} is the generator of a once integrated semigroup ( \frac{d}{dt}S_{+}^{2} (t))_{t∈[0,τ)} ⊆ L(\overline{R(G)}) in the sense of Definition 2.3. Accordingly, ( \frac{d}{dt}S_{+}^{2} (t))_{t∈[0,τ)} ⊆ L(\overline{R(G)}) is a local once integrated semigroup in the sense of (△) and it can be easily proved that the c.i.g of ( \frac{d}{dt}S_{+}^{2} (t))_{t∈[0,τ)} is A_{1}. But, the c.i.g of (S_{+}^{1} (t))_{t∈[0,τ)} is an extension of the c.i.g of ( \frac{d}{dt}S_{+}^{2} (t))_{t∈[0,τ)} . Hence, A_{1} ⊆ \overline{A_{0}} and A_{1} = \overline{A_{0}}. Further on, it is straightforward to see that \frac{d}{dt}S_{+}^{1} (t)x = AS(t)Ax−B_{1}Ax−2tB_{2}Ax−(2B_{2}−C_{1})x, t ∈ [0, τ), x ∈ D(A). Due to [40, Lemma 4.3(b)], we obtain that x = ( \frac{d}{dt}S_{+}^{1} (t)x)_{t=0} , x ∈ D(A) and an immediate consequence of this equality and (4.18) is lim_{n→∞} G(ρ_{n})x = x, x ∈ D(A). By Theorem 4.1(b3), we have D(A) ⊆ \overline{R(G)} ⊆ ⋂ _{i=0}^{2} D(Y_{i}) and Y_{2}x = −x, x ∈ D(A). Suppose x ∈ D(A). By Proposition 3.2(iv), Ax ∈ D(Y_{1}), C_{1}x + AY_{1}x = Y_{1}Ax and, because of that, 2B_{2}x + Y_{1}Ax = C_{1}x + Y_{2}x = C_{1}x − x. Now an application of Proposition 3.2(i) shows that Y_{1}Ax = −(2B_{2}x−C_{1}x)−x ∈ D(A) and that AY_{1}Ax = −2B_{2}Ax−Ax. In other words, Ax ∈ ⋂ _{i=0}^{2} D(Y_{i}) and Y_{2}Ax = −Ax.

Let us prove (c). First of all, suppose λ ∈ ℂ and λ − A is surjective. Assume x = (λ − A)y, for some y ∈ D(A). We obtain E = ⋂ _{i=0}^{2} D(Y_{i}) ∋ x and Y_{2}x = Y_{2}(λy − Ay) = −λy + Ay = −x. Proceeding as in the proof of (b3) of Theorem 4.1, one gets that

A ∫_{0}^{t} S_{+}^{2} (s)x ds = S(t)x + B_{0}x + tB_{1}x + t^{2}B_{2}x + tAB_{0}x + \frac{t^{2}}{2}A(AB_{0}x + B_{1}x)

= S_{+}^{2} (t)x − \frac{t^{2}}{2}x, x ∈ E, t ∈ [0, τ).

This implies ∫_{0}^{t} S_{+}^{1} (s)x ds ∈ D(A), t ∈ [0, τ) and A ∫_{0}^{t} S_{+}^{1} (s)x ds = S_{+}^{1} (t)x − tx, x ∈ E, t ∈ [0, τ). Assume x ∈ D(A). Due to Proposition 3.2(i), we get S_{+}^{1} (t)x = (S(t) − S(0))Ax − t(2B_{2}x − C_{1}x) ∈ D(A), t ∈ [0, τ) and AS_{+}^{1} (t)x = A(S(t) − S(0))Ax − 2tB_{2}Ax = S_{+}^{1} (t)Ax, t ∈ [0, τ). Suppose now that A is densely defined. Since D(A) ⊆ \overline{R(G)}, we automatically obtain that \overline{R(G)} = E and that A is the c.i.g of (S_{+}^{1} (t))_{t∈[0,τ)} . Due to [40, Proposition 4.5], (S_{+}^{1} (t))_{t∈[0,τ)} is a local once integrated semigroup in the sense of Definition 2.3. To prove (i), suppose x ∈ E, φ, ψ ∈ D_{(−τ,τ)} and suppφ + suppψ ⊆ (−τ, τ). Note that:

G(φ)G(ψ)x = ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} )} (s)S(t)S(s)x ds dt

= − ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s)S(t) ∫_{0}^{s} S(v)x dv ds dt.

Repeating literally the arguments given in the proof of Proposition 3.3, one obtains (3.7) and the last equality implies:

G(φ)G(ψ)x = − ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s)[ − ∑_{j=0}^{n_{0} −1} ∫_{0}^{s}(t + s − r)^{j} ∫_{0}^{r} C_{j} S(v)x dv dr

+ ∑_{j=1}^{n_{0}} j ∫_{0}^{s}(t+s−r)^{j−1} ∫_{0}^{r} B_{j} S(v)x dv dr − ∑_{j=0}^{n_{0}} ∫_{0}^{s} r^{j} S(t+s−r)B_{j} x dr]ds dt.

Noticing that ∫_{−∞}^{∞} φ^{(n)} (t)t^{j} dt = 0, n ∈ ℕ, j ∈ ℕ_{0}, n > j, one can deduce that:

I_{1} := ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s) ∑_{j=0}^{n_{0} −1} ∫_{0}^{s}(t + s − r)^{j} ∫_{0}^{r} C_{j} S(v)x dv dr ds dt = 0.

As a matter of fact,

I_{1} = ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s)×

96 KOSTIĆ

× ∑_{j=0}^{n_{0} −1} ∫_{0}^{s} ∑_{\begin{matrix}(k_{1} ,k_{2} ,k_{3} )∈ℕ_{0}^{3} \\ k_{1} +k_{2} +k_{3} =j \end{matrix}} \frac{j!}{k_{1}!k_{2}!k_{3}!}t^{k_{1}} s^{k_{2}} (−r)^{k_{3}} ∫_{0}^{r} C_{j} S(v)x dv dr ds dt.

Suppose j ∈ {0, …, n_{0} − 1}, (k_{1}, k_{2}, k_{3}) ∈ ℕ_{0}^{3} and k_{1} + k_{2} + k_{3} = j. One gets:

∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s) ∫_{0}^{s} \frac{j!}{k_{1}!k_{2}!k_{3}!}t^{k_{1}} s^{k_{2}} (−r)^{k_{3}} ∫_{0}^{r} C_{j} S(v)x dv dr ds dt

= ∫_{−∞}^{∞} φ^{(n_{0} )} (t)t^{k_{1}} dt ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s) ∫_{0}^{s} \frac{j!}{k_{1}!k_{2}!k_{3}!}s^{k_{2}} (−r)^{k_{3}} ∫_{0}^{r} C_{j} S(v)x dv dr ds = 0.

Hence, I_{1} = 0. Analogically,

∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s) ∑_{j=1}^{n_{0}} j ∫_{0}^{s}(t + s − r)^{j−1} ∫_{0}^{r} B_{j} S(v)x dv dr ds dt = 0

and we obtain:

G(φ)G(ψ)x = ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s) ∑_{j=0}^{n_{0}} ∫_{0}^{s} r^{j} S(t + s − r)B_{j} x dr ds dt

= ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ^{(n_{0} +1)} (s) ∑_{j=0}^{n_{0}} ∫_{t}^{t+s}(t + s − r)^{j} S(r)B_{j} x dr ds dt. (4.20)

Put, for every t ∈ (−τ, τ) and j ∈ {1, …, n_{0} + 1}:

g_{j,t} (s) := ∫_{t}^{t+s}(t + s − r)^{j−1} S(r)B_{j} x dr, s ∈ (−τ − t, τ − t).

It is straightforward to check that \frac{d}{ds} g_{j,t} (s) = (j − 1) ∫_{t}^{t+s} (t + s − r)^{j−2} S(r)B_{j} x dr, j > 1, s ∈ (−τ − t, τ − t) and that \frac{d}{ds} g_{1,t}(s) = S(t + s)B_{1}x, s ∈ (−τ − t, τ − t). The partial integration and (4.20) imply:

G(φ)G(ψ)x = − ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)ψ^{(n_{0} )} (s) ∑_{j=1}^{n_{0}} ∫_{t}^{t+s} j(t + s − r)^{j−1} S(r)B_{j} x dr ds dt

− ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)ψ^{(n_{0} )} (s)S(t + s)B_{0}x ds dt.

Apply again the partial integration in order to see that:

G(φ)G(ψ)x = − ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)ψ^{(n_{0} )} (s)S(t + s)B_{0}x ds dt

+ ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)ψ^{(n_{0} −1)} (s)S(t + s)B_{1}x ds dt

+ ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ^{(n_{0} )} (t)ψ^{(n_{0} −1)} (s) ∑_{j=2}^{n_{0}} j(j − 1) ∫_{t}^{t+s}(t + s − r)^{j−2} S(r)B_{j} x dr ds dt.

Continuing this procedure, we finally obtain (4.12).

To prove (ii), suppose φ, ψ ∈ D_{(−τ,τ)} and suppφ+suppψ ⊆ (−τ, τ). Evidently,

G(φ ∗ ψ)x = (−1)^{n_{0}}  ∫_{−∞}^{∞} (φ ∗ ψ)^{(n_{0} )} (t)S(t)x dt = (−1)^{n_{0}} ∫_{−∞}^{∞} (φ^{(n_{0} )} ∗ ψ)(t)S(t)x dt

= (−1)^{n_{0}} ∫_{−∞}^{∞} [ ∫_{−∞}^{∞} φ^{(n_{0} )} (t − s)ψ(s) ds]S(t)x dt (4.21)

= (−1)^{n_{0}} ∫_{−∞}^{∞} φ^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ(s)S(t + s)x ds dt, x ∈ E.

Due to (4.21), we have:

(4.22) G(ψ)x = lim_{n→∞} G(ψ ∗ρ_{n})x = lim_{n→∞} (−1)^{n_{0}} ∫_{−∞}^{∞} ρ_{n}^{(n_{0} )} (t) ∫_{−∞}^{∞} ψ(s)S(t+s)x ds dt,

x ∈ E. Combining (DG)_{1}, (4.12) and (4.22) gives:

(4.23)  (−1)^{n_{0}} G(ψ)x = ∑_{i=0}^{n_{0}} (−1)^{i+1} i!G(ψ^{(n_{0} −i)} )B_{i}x, x ∈ E.

Suppose now x ∈ ⋂ _{i=0}^{n_{0}} D(Y_{i}). A consequence of the definition of Y_{n_{0}} and (1.2) is n_{0}!G(ψ)B_{n_{0}} x+AG(ψ)Y_{n_{0} −1} x = n_{0}!G(ψ)B_{n_{0}} x−G(ψ′)Y_{n_{0} −1} x = G(ψ)Y_{n_{0}} x. If n_{0} ⩾ 2, then we obtain n_{0}!G(ψ)B_{n_{0}} x − G(ψ′)(AY_{n_{0} −2} x + (n_{0} − 1)!B_{n_{0} −1} x) = G(ψ)Y_{n_{0}} x and n_{0}!G(ψ)B_{n_{0}} x − (n_{0} − 1)!G(ψ′)B_{n_{0} −1} x + G(ψ′′)Y_{n_{0} −2} x = G(ψ)Y_{n_{0}} x. By the definition of Y_{i} and (1.2), one concludes inductively:

(4.24)  ∑_{i=0}^{n_{0}} (−1)^{n_{0} +i} i!G(ψ^{(n_{0} −i)} )B_{i}x = G(ψ)Y_{n_{0}} x.

This equality and (4.23) imply G(ψ)(Y_{n_{0}} x + x) = 0; a simple consequence is G(η)(Y_{n_{0}} x + x) = 0, η ∈ D and the proof of (ii) finishes an application of (DG)_{2}.

To prove (iii), one can argue as in the proof of (b5) of Theorem 4.1. We sketch the proof for the sake of completeness. Fix an x ∈ D(A^{n_{0} −1} ). Since S : (−τ, τ) → L(E, [D(A)]) is continuous, one can argue as in the proof of Proposition 3.2 in order to see that the mapping t ↦ S(t)x, t ∈ (−τ, τ) is n_{0}-times continuously differentiable and that there exists a function M : (−τ, τ) → (0, ∞) satisfying ∥ \frac{d^{n_{0}}}{dt^{n_{0}}} S(t)x∥ ⩽ M(t)∥x∥_{n_{0} −1} , t ∈ (−τ, τ). Furthermore, (3.3) holds for every l ∈ \{0, …, n_{0} − 1\} and one obtains inductively Y_{k}x = −( \frac{d^{k}}{dt^{k}} S(t)x)_{t=0} , k ∈ \{0, …, n\}. Denote u(t; x) = \frac{d^{n_{0}}}{dt^{n_{0}}} S(t)x, t ∈ (−τ, τ); then the partial integration shows G(φ)x = ∫_{−∞}^{∞} φ(t) u(t; x)x dt, φ ∈ D_{(−τ,τ)} . The previous equality and (ii)

imply lim_{n→∞} G(ρ_{n})x = u(0; x) = −Y_{n_{0}} x = x ∈ \overline{R(G)}. Therefore, D(A^{n_{0} −1} ) ⊆ \overline{R(G)}.

Further on, Proposition 3.2(iv) implies

C_{n_{0} −1} x + \frac{1}{(n_{0} − 1)!} AY_{n_{0} −1} x = \frac{1}{(n_{0} − 1)!} Y_{n_{0} −1} (Ax), i.e.,
C_{n_{0} −1} x + \frac{1}{(n_{0} − 1)!} [−x − n_{0}!B_{n_{0}} x] = \frac{1}{(n_{0} − 1)!} Y_{n_{0} −1} (Ax).

Due to Proposition 3.2(i), Y_{n_{0} −1} (Ax) ∈ D(A) and a simple computation gives Y_{n_{0}} Ax = −Ax which finishes the proof of (iii).

Further on, let us observe that (iii) implies D(A^{n}) ⊆ D(A^{n_{0} −1} ) ⊆ \overline{R(G)} ⊆ \overline{D_{∞} (A)} ⊆ \overline{D(A^{n+1} )}, for every n ∈ ℕ such that n ⩾ n_{0} − 1. Hence, A is stationary dense and n(A) ⩽ n_{0} − 1.

To prove (v), suppose λ ∈ ρ(A). We will prove that A generates a local (n_{0} −1)-times integrated group on (−τ, τ). Repeating literally the arguments given in the proof of Theorem 4.1, one gets A ∫_{0}^{t} u(s; x) ds = u(t; x) − x, t ∈ (−τ, τ), x ∈ D(A^{n_{0} −1} ) and Au(t; x) = u(t; Ax), t ∈ (−τ, τ), x ∈ D(A^{n_{0}} ). Put S^{n_{0} −1} (t)x = u(t; R(λ : A)^{n_{0} −1} x), t ∈ [0, τ), x ∈ E. Clearly, the mapping t ↦ S^{n_{0} −1} (t)x, t ∈ [0, τ) is continuous for every x ∈ E and an induction argument shows that, for every k ∈ ℕ_{0}, there exists an appropriate constant M(k, λ) ∈ (0, ∞) which fulfills ∥A^{k}R(λ : A)^{k}x∥ ⩽ M(k, λ)∥x∥, x ∈ E. This implies

∥S^{n_{0} −1} (t)x∥ = ∥u(t; R(λ : A)^{n_{0} −1} x)∥ ⩽ M(t)∥R(λ : A)^{n_{0} −1} x∥_{n_{0} −1}

⩽ M(t) ∑_{i=0}^{n_{0} −1} M(i, λ)∥R(λ : A)∥^{n_{0} −1−i} ∥x∥,

x ∈ E and S^{n_{0} −1} (t) ∈ L(E), t ∈ [0, τ). In order to simplify the notation, denote C = R(λ : A)^{n_{0} −1} . We have A ∫_{0}^{t} S^{n_{0} −1} (s)x ds = A ∫_{0}^{t} u(s; Cx) ds = u(t; Cx) − Cx = S^{n_{0} −1} (t)x − Cx, t ∈ [0, τ), x ∈ E. Since Au(t; x) = u(t; Ax), t ∈ (−τ, τ), x ∈ D(A^{n_{0}} ), one easily obtains S^{n_{0} −1} (t)A ⊆ AS^{n_{0} −1} (t), S^{n_{0} −1} (t)R(λ : A) = R(λ : A)S^{n_{0} −1} (t) and, by induction, S^{n_{0} −1} (t)C = CS^{n_{0} −1} (t), t ∈ [0, τ). Now it is straightforward to prove that the abstract Cauchy problem:

\{ \begin{matrix}v ∈ C([0, τ) : [D(A)]) ∩ C^{1}([0, τ) : E), \\ v′(t) = Av(t) + Cx, t ∈ [0, τ), \\ v(0) = 0, \end{matrix}

has a unique solution for every x ∈ E, given by v(t) = ∫_{0}^{t} S^{n_{0} −1} (s)x ds, t ∈ [0, τ), x ∈ E. By [44, Theorem 4.4], A generates a local (n_{0} − 1)-times integrated semigroup on [0, τ). Since −A generates a (DG) Ǧ, we also obtain that −A generates a local (n_{0} − 1)-times integrated semigroup on [0, τ) and Lemma 2.2(v) implies that A generates a local (n_{0} − 1)-times integrated group on (−τ, τ). Thus, (v) is a consequence of Corollary 2.1(c).

To prove (vi), notice that the assumption \overline{R(G)} = E and R(G) ⊆ D_{∞} (A) imply that D_{∞} (A) is dense in E. The converse statement is obvious since \overline{D_{∞} (A)} ⊆ \overline{D(A^{n_{0} −1} )} ⊆ \overline{R(G)} (cf. the proofs of (iii) and (iv)). In the case ρ(A) ≠ ∅, the denseness of \overline{D_{∞} (A)} in E is equivalent to the denseness of D(A) in E (see, for example, [27]) and the proof of (vi) completes a routine argument.

It remains to be proved (vii). Suppose G(φ)x = 0, φ ∈ D_{0}. This implies (−1)^{n_{0}} ∫_{−∞}^{∞} φ^{n_{0}} (t)S(t)x dt = 0, φ ∈ D_{[0,τ)} and the existence of bounded linear operators D_{0}, …, D_{n_{0} −1} ∈ L(E) satisfying S(t)x = ∑ _{j=0}^{n_{0} −1} t^{j} D_{j} x, t ∈ [0, τ). Hence,

(4.25)  A ∑_{j=0}^{n_{0} −1} \frac{t^{j+1}}{j + 1} D_{j} x = ∑_{j=0}^{n_{0} −1} t^{j} D_{j} x + ∑_{j=0}^{n_{0}} t^{j} B_{j} x, t ∈ [0, τ).

Substitute t = 0 in (4.25) to obtain D_{0} = −B_{0}. Differentiating (4.25), it is straightforward to see that: x ∈ ⋂ _{i=0}^{n_{0}} D(Y_{i}), ⋃ _{i=0}^{n_{0} −1} \{D_{i}x\} ⊆ D(A), D_{i}x = \frac{(−1)}{i!} Y_{i}x, i = 1, …, n_{0} − 1 and A(D_{n_{0} −1} x) = n_{0}B_{n_{0}} x. This implies (n_{0}B_{n_{0}} + \frac{1}{(n_{0} −1)!} AY_{n_{0} −1} )x = 0, i.e., x ∈ N(Y_{n_{0}} ). Due to (ii), x = 0 and ⋂ _{φ∈D_{0}} N(G(φ)) = {0}. The second equality in (vii) follows by passing to −A and Ǧ. □

Example 4.1. Let E := L^{∞} (ℝ) and let A := \frac{d}{dt} with maximal domain. Then A is not densely defined and generates a once integrated group (S_{1}(t))_{t∈ℝ} given by (S_{1}(t)f)(s) := ∫_{0}^{t} f(r + s) dr, s ∈ ℝ, t ∈ ℝ (cf. also [12, Example 4.1]). Put S_{2}(t)f := ∫_{0}^{t} S_{1}(s)f ds, t ⩾ 0, f ∈ E, S_{2}(t)f := ∫_{0}^{−t} S_{1}(−s)f ds, t < 0, f ∈ E and G(φ)f := ∫_{−∞}^{∞} φ′′(t)S_{2}(t)f dt, φ ∈ D, f ∈ E. Then (S_{2}(t))_{t∈ℝ} is a twice integrated group generated by A, the mapping S_{2} : ℝ → L(E, [D(A)]) is continuous and G is a non-dense (DG)vgenerated by A (cf. Theorem 4.2 with n_{0} = 2). We would like to point out that there exists f ∈ D(A) such that Af ∉ \overline{R(G)}. Suppose contrarily that R(A) ⊆ \overline{R(G)}. By Theorem 6.2, D(A) ⊆ \overline{R(G)} and we obtain (λ − A)f ∈ \overline{R(G)}, λ ∈ ℂ, f ∈ D(A). Since ℂ ∖ iℝ ⊆ ρ(A), one yields E = \overline{R(G)} and the contradiction is obvious. Hence, Theorem 4.2 implies that (S_{1}(t))_{t⩾0} is a once integrated semigroup generated by A in the sense of Definition 2.3 and that the c.i.g of (S_{1}(t))_{t∈ℝ} is A\overline{R(G)} (≠ A). Furthermore, \overline{R(G)} ⊊ ⋂ _{i=0}^{2} D(Y_{i}) = E.

Proposition 4.1. Suppose G_{1} and G_{2} are distribution groups generated by A and ρ(A) ≠ ∅. Then G_{1} = G_{2}.

Proof. Suppose x ∈ E, λ ∈ ρ(A) and φ ∈ D_{(−τ,τ)} , for some τ ∈ (0, ∞). We will prove that G_{1}(φ)x = G_{2}(φ)x. Clearly, G_{i} ∈ D′(L(E, [D(A)])), i = 1, 2 and an application of [32, Theorem 2.1.1] gives that there exist n_{1} ∈ ℕ, n_{2} ∈ ℕ and continuous mappings S_{i} : (−τ, τ) → L(E, [D(A)]), i = 1, 2 so that G_{i}(ψ)x = (−1)^{n_{i}} ∫_{−∞}^{∞} ψ^{(n_{i} )} (t)S_{i}(t)x dt, ψ ∈ D_{(−τ,τ)} , x ∈ E, i = 1, 2. The proof of Theorem 4.1 shows that there are bounded linear operators

B_{0}, …, B_{n_{1}} , \overline{B_{0}}, …, \overline{B_{n_{2}}} , C_{0}, …, C_{n_{1} −1} , \overline{C_{0}}, …, \overline{C_{n_{2} −1}}

such that (S_{1}(t))_{t∈(−τ,τ)} , resp., (S_{2}(t))_{t∈(−τ,τ)} is a [B_{0}, …, B_{n_{1}} , C_{0}, …, C_{n_{1} −1} ]- group, resp., [ \overline{B_{0}}, …, \overline{B_{n_{2}}} , \overline{C_{0}}, …, \overline{C_{n_{2} −1}} ]-group with a subgenerator A. Without loss of generality, we may assume n_{1} = n_{2}. Further on, the proof of Theorem 4.2 implies that ( \frac{d^{n_{1}}}{dt^{n_{1}}} S_{i}(t)R(λ : A)^{n_{1} −1} )_{t∈[0,τ)} , i = 1, 2 are local R(λ : A)^{n_{1} −1} - semigroups generated by A. Hence, there exist x_{0}, …, x_{n_{1} −1} ∈ E which satisfies S_{1}(t)R(λ : A)^{n_{1} −1} x − S_{2}(t)R(λ : A)^{n_{1} −1} x = ∑ _{i=0}^{n_{1} −1} t^{i}x_{i}, t ∈ [0, τ). An immediate consequence is:

R(λ : A)^{n_{1} −1} G_{1}(φ)x − R(λ : A)^{n_{1} −1} G_{2}(φ)x

= G_{1}(φ)R(λ : A)^{n_{1} −1} x − G_{2}(φ)R(λ : A)^{n_{1} −1} x

= (−1)^{n_{1}} ∫_{−∞}^{∞} φ^{(n_{1} )} (t) ∑_{i=0}^{n_{1} −1} t^{i}x_{i} dt = 0,

which clearly implies G_{1}(φ)x = G_{2}(φ)x. □

Remark 4.1. (i) Suppose A generates a (DG) G and ρ(A) ≠ ∅. Then there exist a > 0 and b > 0 such that E(a, b) ⊆ ρ(±A) and that the next representation formula holds for G:

G(φ)x = \frac{1}{2πi} ∫_{Γ} ∫_{−∞}^{∞} φ(t)[e^{λt} R(λ : A)x + e^{−λt} R(λ : −A)x]dtdλ,

x ∈ E, φ ∈ D, where we assume that the curve Γ = ∂E(a, b) is oriented upwards.

(ii) Suppose G ∈ D′(L(E)) is regular, A is a closed linear operator so that (1.2) holds and there are no non-trivial solutions of the abstract Cauchy problem:

(ACP_{1}) : \{ \begin{matrix}u ∈ C(ℝ : [D(A)]) ∩ C^{1}(ℝ : E), \\ u′(t) = Au(t), t ∈ ℝ, \\ u(0) = x, \end{matrix}

when x = 0 (cf. Theorem 4.1). Then G(φ∗ψ)x = G(φ) G(ψ)x, x ∈ \overline{R(G)}, φ, ψ ∈ D. To show this, let us point out that G(φ ∗ ψ)x = ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ(t) ψ(s) u(t + s; x) ds dt and G(ψ)G(φ)x=∫_{−∞}^{∞} φ(t)G(ψ)u(t; x) dt, x ∈ R(G), φ, ψ ∈ D. Since G(·) ∈ L(E), the consideration is over if we prove that G(ψ)u(t; x) = ∫_{−∞}^{∞} ψ(s)u(t + s; x) ds, ψ ∈ D, x ∈ R(G), t ∈ ℝ. Put, for fixed ψ ∈ D and x ∈ R(G), f(t) := G(ψ)u(t; x)− ∫_{−∞}^{∞} ψ(s)u(t + s; x) ds, t ∈ ℝ. Then

A ∫_{0}^{t} f(s) ds = G(ψ)[u(t; x) − x] − ∫_{−∞}^{∞} ψ(s)A ∫_{s}^{t+s} u(r; x) dr ds

= G(ψ)[u(t; x) − x] − ∫_{−∞}^{∞} ψ(s)[u(t + s; x) − u(s; x)] ds = f(t), t ∈ ℝ.

So, the function u(t) = ∫_{0}^{t} f(s)ds, t ∈ ℝ solves (ACP_{1}) and u(0) = 0. This proves f ≡ 0.

(iii) Suppose G ∈ D′(L(E)) is regular, (1.2) holds for A and G, τ ∈ (0, ∞) and ρ(A) ≠ ∅. Set G_{1} := G_{|\overline{R(G)}} . Then G_{1} is a dense (DG) in \overline{R(G)} generated by A_{1}. To this end, we employ the same terminology as in the proof of Theorem 4.1(b5); without loss of generality, we can assume that 0 ∈ ρ(A) so that A_{λ} = A, u_{λ} = u and G_{λ} = G. Suppose (ρ_{n}) is a regularizing sequence. Choose an arbitrary τ ∈ (0, ∞) and notice that

C^{2}G(φ ∗ ψ)x = CG(φ ∗ ψ)Cx = C ∫_{−∞}^{∞} (φ ∗ ψ)(t) u(t; Cx) dt

= C ∫_{−∞}^{∞} (φ ∗ ψ)(t) T (t)x dt = C ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ(t) ψ(s) T (t + s)x ds dt

= ∫_{−∞}^{∞} ∫_{−∞}^{∞} φ(t) ψ(s) T (t) T (s)x ds dt = G(φ)CG(ψ)Cx = C^{2}G(φ) G(ψ)x,

for every x ∈ \overline{R(G)} and φ, ψ ∈ D_{[0,τ)} with suppφ + suppψ ⊆ [0, τ). The injectiveness of C combining with the argumentation used in the proof of Theorem 4.1(b1) enables one to deduce that G(φ ∗ ψ)x = G(φ)G(ψ)x, φ, ψ ∈ D, x ∈ \overline{R(G)} and that G_{1} ∈ D′(L(\overline{R(G)})) satisfies (DG)_{1}. The assumption G_{1}(φ)x = 0, φ ∈ D implies G_{1}(φ)Cx = ∫_{−∞}^{∞} φ(t)u(t; Cx) dt = ∫_{−∞}^{∞} φ(t)T (t)x dt = 0, for every φ ∈ D_{[0,τ)} and Cx = T (0)x = lim_{n→∞} G_{1}(ρ_{n})Cx = 0. So, x = 0 and G_{1} is a (DG) in \overline{R(G)}. It can be easily seen that G_{1} is generated by A_{1}.

(iv) Suppose G and A are as in (ii) and let λ ∈ ρ(A). Then the [B_{0}, …, B_{n_{0}} , C_{0}, …, C_{n_{0} −1} ]-group (S(t))_{t∈(−τ,τ)} , constructed in Theorem 4.1(a), satisfies (4.1), (4.2) as well as Y_{n_{0}} x = −x, x ∈ \overline{R(G)}. Namely, (ii) shows that G_{1} = G_{|\overline{R(G)}} is a

(DG) in \overline{R(G)} generated by A_{1}. The proof of Theorem 4.1(b5) implies that A_{1} generates a local n_{0}-times integrated group (S_{n_{0}} (t))_{t∈(−τ,τ)} in L(\overline{R(G)}) and it is not difficult to show that G_{1}(φ)x = (−1)^{n_{0}} ∫_{0}^{∞} φ^{(n_{0} )} (t)S_{n_{0}} (t)x dt+∫_{−∞}^{0} φ^{(n_{0} )} (t)S_{n_{0}} (t)x dt, φ ∈ D_{(−τ,τ)} , x ∈ \overline{R(G)}. Hence, ∫_{0}^{∞} φ^{(n_{0} )} (t)S_{n_{0}} (t)x dt = ∫_{0}^{∞} φ^{(n_{0} )} (t)S(t)x dt, φ ∈ D_{[0,τ)}, x ∈ \overline{R(G)} and an application of [17, Theorem 8.1.1] gives the existence of operators D_{i} ∈ L(\overline{R(G)}, E), i = 0, …, n_{0} − 1 satisfying S(t)x = S_{n_{0}} (t)x + ∑ _{i=0}^{n_{0} −1} t^{i}D_{i}x, t ∈ [0, τ), x ∈ \overline{R(G)}. Since A ∫_{0}^{t} S(s)x ds = S(t)x + ∑ _{i=0}^{n_{0}} t^{i}B_{i}x, t ∈ [0, τ), x ∈ E one obtains

A ∑_{i=0}^{n_{0} −1} \frac{t^{i+1}}{i + 1} D_{i}x = ∑_{i=0}^{n_{0} −1} t^{i}(D_{i}x + B_{i}x) + t^{n_{0}} B_{n_{0}} x + \frac{t^{n_{0}}}{n_{0}!}x, t ∈ [0, τ), x ∈ \overline{R(G)}.

This implies \overline{R(G)} ⊆ D(Y_{n_{0}} ), D_{i}x = \frac{(−1)}{i!} Y_{i}x, i = 1, …, n_{0} − 1, (4.1), (4.2) and Y_{n_{0}} x = −x, x ∈ \overline{R(G)}.

(iv) Suppose G is a (DG) and φ ∈ D. Then G(φ) = G(φ_{+} )+G(φ_{−} ) if and only if {G(φ_{+} ), G(φ_{−} )} ⊆ L(E) if and only if G(φ_{+} ) ∈ L(E). Namely, the assumption G(φ) = G(φ_{+} ) + G(φ_{−} ) immediately implies D(G(φ_{+} )) = D(G(φ_{−} )) = E and the Closed Graph Theorem gives G(φ_{+} ) ∈ L(E) and G(φ_{−} ) ∈ L(E). Clearly, {G(φ_{+} ), G(φ_{−} )} ⊆ L(E) implies G(φ_{+} ) ∈ L(E). Suppose now G(φ_{+} ) ∈ L(E). We will show that G(φ_{−} ) ∈ L(E) and that G(φ_{−} ) = G(φ)−G(φ_{+} ). Fix an x ∈ E and notice that G(φ ∗ ψ)x = G(ψ)G(φ)x, ψ ∈ D implies G(φ_{+} ∗ ψ)x + G(φ_{−} ∗ ψ)x = G(ψ)G(φ)x, ψ ∈ D. Since x ∈ D(G(φ_{+} )), we obtain G(φ_{−} ∗ ψ)x = G(ψ)[G(φ)x − G(φ_{+} )x]. So, x ∈ D(G(φ_{−} )) and G(φ_{−} )x = G(φ)x − G(φ_{+} )x.

Proposition 4.2. Suppose G is a (DG) and G(φ_{+} ) ∈ L(E), φ ∈ D. Put G_{+} (φ) := G(φ_{+} ) and G_{−} (φ) := G((φ̌)_{−} ), φ ∈ D. Then ±A are generators of distribution semigroups G_{±} .

Proof. Owing to the previous remark, we have G_{±} (φ) ∈ L(E), φ ∈ D and G(φ) = G_{+} (φ) + G_{−} (φ̌), φ ∈ D. Evidently, suppG_{+} ∪ suppG_{−} ⊆ [0, ∞), G_{±} ∈ D′_{0}(L(E)) and Theorem 4.2(vii) implies ⋂ _{φ∈D_{0}} N(G_{±} (φ)) = {0}. Since (φ∗_{0}ψ)_{+} = φ_{+} ∗ ψ_{+} , φ, ψ ∈ D, Proposition 2.2 yields that G_{+} is a pre-(DSG). Analogously, G_{−} is a pre-(DSG) and one obtains that G_{+} , resp., G_{−} is a (DSG). Designate by A_{+} , resp., A_{−} , the generator of G_{+} , resp., G_{−} . Then it is straightforward to verify that A_{±} are extensions of ±A. The proof is completed if one shows:

A_{+} ⊆ A and A_{−} ⊆ −A.

We will first prove that A_{+} = −A_{−} . To see this, suppose x ∈ E and φ, ψ ∈ D. Then one obtains: G(φ_{+} ∗ ψ)x = G(ψ)G(φ_{+} )x, G_{+} (φ_{+} ∗ ψ)x + G_{−} ((φ_{+} ∗ ψ)ˇ )x = (G_{+} (ψ) + G_{−} (ψ̌)) G_{+} (φ)x and:

(4.26)  G_{+} (φ_{+} ∗ ψ)x + G_{−} ((φ_{+} ∗ ψ)ˇ )x = G_{+} (φ ∗_{0} ψ)x + G_{−} (ψ̌) G_{+} (φ)x.

Further, notice that (ψ ∗ φ_{+} − φ ∗ ψ_{−} − φ ∗_{0} ψ)(t) = 0, t ⩾ 0. The last equality and (4.26) give G_{+} (ψ ∗ φ_{+} )x = G_{+} (φ ∗ ψ_{−} )x + G_{+} (φ ∗_{0} ψ)x and:

(4.27)  G_{+} (φ ∗ (ψ̌)_{−} )x + G_{−} ((φ_{+} ∗ ψ̌)ˇ )x = G_{−} (ψ) G_{+} (φ)x.

Suppose now (x, y) ∈ D(A_{+} ), a > 0, ψ ∈ D_{(a,∞)} and (ρ_{n}) is a regularizing sequence satisfying suppρ_{n} ⊆ [0, \frac{1}{n}], n ∈ ℕ. Since G_{+} (−φ′)x = G_{+} (φ)y, φ ∈ D_{0}, (4.27) enables one to establish the following equalities:

(4.28) G_{+} (ρ_{n} ∗ (ψ̌)_{−} )y + G_{−} ((ρ_{n} ∗ ψ̌)ˇ )y = G_{−} (ψ) G_{+} (ρ_{n})y

= G_{−} (ψ) G_{+} (−ρ′_{n})x = G_{+} (−ρ′_{n} ∗ (ψ̌)_{−} )x + G_{−} (( − ρ′_{n} ∗ ψ̌)ˇ )x.

Clearly, supp(ρ_{n} ∗ (ψ̌)_{−} ) ∪ supp(−ρ′_{n} ∗ (ψ̌)_{−} ) ⊆ [0, \frac{1}{n}] + (−∞, −a) ⊆ (−∞, 0], n ⩾ \frac{1}{a} and an application of (4.28) gives:

(4.29)  G_{−} ((ρ_{n} ∗ ψ̌)ˇ )y = G_{−} (( − ρ_{n} ∗ (ψ̌)′)ˇ )x.

Letting n → ∞ in (4.29), one concludes that G_{−} (ψ)y = −G_{−} (((ψ̌)′)ˇ )x = G_{−} (ψ′)x and, as a matter of routine, one can see that the previous equalities remain true for every ψ ∈ D_{0}. In conclusion, one gets (x, −y) ∈ A_{−} and A_{+} ⊆ −A_{−} ; analogously, A_{+} ⊃ −A_{−} and this finally gives A_{+} = −A_{−} . Taking into account [28, Lemma 3.6], one can deduce the following:

G(φ)A_{+} x = G_{+} (φ)A_{+} x + G_{−} (φ̌)A_{+} x = G_{+} (−φ′)x − φ(0)x − A_{−} G_{−} (φ̌)x
= G_{+} (−φ′)x − φ(0)x − (G_{−} (−φ̌′)x − φ̌(0)x)
= G_{+} (−φ′)x + G_{−} ( ˇ −φ′ )x = G(−φ′)x, φ ∈ D.

Hence, (x, A_{+} x) ∈ A, A_{+} ⊆ A and A_{+} = A. □

Remark 4.2. Suppose G is a (DG) generated by A and ρ(A) ≠ ∅. Due to Lemma 2.2 and Theorem 4.2, we have that A, resp., −A, is the generator of a (DSG) G_{1}, resp., G_{2}. Obviously, G(φ) = G_{1}(φ)+G_{2}(φ̌), φ ∈ D and G_{1}(φ) G_{2}(ψ) = G_{2}(ψ) G_{1}(φ), φ, ψ ∈ D. Let x ∈ E and φ ∈ D be fixed. We will prove that G_{+} (φ) = G(φ_{+} ) = G_{1}(φ). To this end, it is enough to show G(ψ ∗ φ_{+} )x = G(ψ) G_{1}(φ)x, ψ ∈ D, i.e.,

(4.30) G_{1}(ψ ∗ φ_{+} )x + G_{2}((ψ ∗ φ_{+} )ˇ )x = G_{1}(φ) G_{2}(ψ̌)x + G_{1}(φ) G_{1}(ψ), ψ ∈ D.

Notice that the proof of [33, Theorem 6] (see [33, (9), p. 61]) enables one to see that G_{1}(φ ∗ ψ_{−} )x + G_{2}((φ_{+} ∗ ψ)ˇ )x = G_{1}(φ) G_{2}(ψ̌)x, ψ ∈ D. As in the proof of Proposition 4.2, one has (ψ ∗ φ_{+} − φ ∗ ψ_{−} − φ ∗_{0} ψ)(t) = 0, t ⩾ 0, ψ ∈ D, which gives G_{1}(ψ ∗ φ_{+} )x = G_{1}(φ ∗ ψ_{−} )x + G_{1}(φ ∗_{0} ψ)x = G_{1}(φ ∗ ψ_{−} )x + G_{1}(φ) G_{1}(ψ)x, ψ ∈ D. Hence,

G_{1}(ψ ∗ φ_{+} )x + G_{2}((ψ ∗ φ_{+} )ˇ )x = G_{1}(φ ∗ ψ_{−} )x + G_{1}(φ) G_{1}(ψ)x + G_{2}((ψ ∗ φ_{+} )ˇ )x
= G_{1}(φ ∗ ψ_{−} )x + G_{1}(φ) G_{1}(ψ)x + G_{1}(φ) G_{2}(ψ̌)x − G_{1}(φ ∗ ψ_{−} )x
= G_{1}(φ) G_{2}(ψ̌)x + G_{1}(φ) G_{1}(ψ)x, ψ ∈ D

and this proves (4.30). Accordingly, A is the generator of G_{+} = G_{1} and the previous remark implies that G(φ_{−} ) = G_{2}(φ̌) ∈ L(E), φ ∈ D and that G((ˇ·)_{−} ) is a (DSG) generated by −A.

Theorem 4.3. Suppose B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ∈ L(E) and A is a subgenerator of a [B_{0}, …, B_{n}, C_{0}, …, C_{n−1} ]-group (S(t))_{t∈ℝ} . Set

G(φ)x := (−1)^{n} ∫_{−∞}^{∞} φ^{(n)} (t)S(t)x dt, φ ∈ D, x ∈ E.

Then:

(a) (1.2) holds and (4.12) holds for every φ ∈ D and ψ ∈ D.

(b) N (G) ⊆ N(Y_{n}) and, in particular, the injectiveness of Y_{n} implies (DG)_{2} for G.

(c) For every φ ∈ D and ψ ∈ D, N(Y_{n}) ⊆ N(G(φ)G(ψ)); especially, if G is regular, then N(Y_{n}) = N (G).

(d) Assume B_{0} = ⋯= B_{n−1} = 0 and B_{n} = −\frac{1}{n!} I. Then G is a (DG) generated by A.

Proof. To prove (a), notice that G ∈ D′(L(E)). Suppose x ∈ E and φ ∈ D. Then AG(φ)x = G(−φ′)x; indeed,

G(φ)x = (−1)^{n} ∫_{−∞}^{∞} φ^{(n)} (t)S(t)x dt = (−1)^{n+1} ∫_{−∞}^{∞} φ^{(n+1)} (t) ∫_{0}^{t} S(s)x ds dt ∈ D(A)

and

AG(φ)x = (−1)^{n+1} ∫_{−∞}^{∞} φ^{(n+1)} (t)[S(t)x + ∑_{j=0}^{n} t^{j} B_{j} x]dt

= (−1)^{n+1} ∫_{−∞}^{∞} φ^{(n+1)} (t)S(t)x dt = G(−φ′)x.

Further,

G(φ)Ax = (−1)^{n+1} ∫_{−∞}^{∞} φ^{(n+1)} (t) ∫_{0}^{t} S(s)Ax ds dt

= (−1)^{n+1} ∫_{−∞}^{∞} φ^{(n+1)} (t) ∫_{0}^{t} [AS(s)x − ∑_{j=0}^{n−1} s^{j} C_{j} x]ds dt

= (−1)^{n+1} ∫_{−∞}^{∞} φ^{(n+1)} (t)A ∫_{0}^{t} S(s)x ds dt = AG(φ)x, x ∈ D(A).

Hence, G(φ)A ⊆ AG(φ) and (1.2) holds. The proofs of (4.12) and (b) are contained in those of Theorem 4.2. Let φ ∈ D, ψ ∈ D and let x ∈ N(Y_{n}) be fixed. Arguing as in the proof of Theorem 4.2, one gets the validity of (4.24). Hence,

0 = ∑_{i=0}^{n} (−1)^{i+1} i!G((φ^{(n)} ∗ ψ)^{(n−i)} )B_{i}x = ∑_{i=0}^{n} (−1)^{i+1} i!G(φ^{(n)} ∗ ψ^{(n−i)} )B_{i}x

= ∑_{i=0}^{n} (−1)^{i+1} i!(−1)^{n} ∫_{−∞}^{∞} φ^{(n)} (t) ∫_{−∞}^{∞} ψ^{(n−i)} (s)S(t + s)B_{i}x ds dt.

Owing to (4.12), G(φ)G(ψ)x = 0 and N(Y_{n}) ⊆ N(G(φ)G(ψ)). Let (ρ_{k}) be a regularizing sequence and G be regular. Then G(ψ)x = lim_{k→∞} G(ρ_{k})G(ψ)x = 0 and x ∈ N (G). This proves N(Y_{n}) ⊆ N (G), and due to (b), we have N (G) ⊆ N(Y_{n}). The proof of (c) is completed.

To prove (d), notice that the proof of Theorem 4.2 implies (4.21) for G. Since B_{0} = ⋯= B_{n−1} = 0 and B_{n} = −\frac{1}{n!} I, we immediately obtain (DG)_{1} from (4.12). Clearly, Y_{n} = n!B_{n} = −I and (DG)_{2} follows from an application of (b). Hence, G is a (DG). Put now \overline{S}(t) := S(t), t ⩾ 0 and \overline{S}(t) := (−1)^{n}S(t), t < 0. It is straightforward to verify that (\overline{S}(t))_{t∈ℝ} is an n-times integrated group generated by A. Furthermore, it is evident that

G(φ)x = (−1)^{n} ∫_{0}^{∞} φ^{(n)} (t)\overline{S}(t)x dt + ∫_{−∞}^{0} φ^{(n)} (t)\overline{S}(t)x dt, φ ∈ D, x ∈ E.

By the proof of Theorem 2.1, we have that G is generated by A. □

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Faculty of Technical Sciences (Received 27 12 2008) University of Novi Sad 21000 Novi Sad Serbia

    marco.s@verat.net