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Direct and Inverse Fractional Abstract Cauchy Problems

Mathematics 2019, 7(12), 1165; https://doi.org/10.3390/math7121165

Article
Fractional Cauchy Problems for Infinite Interval Case-II
by Mohammed Al Horani 1, Mauro Fabrizio 2 , Angelo Favini 2,* and Hiroki Tanabe 3
1
Department of Mathematics, The University of Jordan, Amman 11942, Jordan
2
Dipartimento di Matematica, Università di Bologna, Piazza di Porta S. Donato 5, 40126 Bologna, Italy
3
Takarazuka, Hirai Sanso 12-13, Osaka 665-0817, Japan
*
Author to whom correspondence should be addressed.
Received: 29 October 2019 / Accepted: 19 November 2019 / Published: 2 December 2019

## Abstract

:
We consider fractional abstract Cauchy problems on infinite intervals. A fractional abstract Cauchy problem for possibly degenerate equations in Banach spaces is considered. This form of degeneration may be strong and some convenient assumptions about the involved operators are required to handle the direct problem. Required conditions on spaces are also given, guaranteeing the existence and uniqueness of solutions. The fractional powers of the involved operator $B X$ have been investigated in the space which consists of continuous functions u on $[ 0 , ∞ )$ without assuming $u ( 0 ) = 0$. This enables us to refine some previous results and obtain the required abstract results when the operator $B X$ is not necessarily densely defined.
Keywords:
fractional derivative; abstract Cauchy problem; evolution equations; degenerate equations
MSC:
26A33

## 1. Introduction

In recent years, many studies were devoted to the problem of recovering the solution u to
$B M u − L u = f$
where B, M, and L are closed linear operators on the complex Banach space E with $D ( L ) ⊆ D ( M )$, $0 ∈ ρ ( L )$, $f ∈ E$ and u is unknown. The first approach to handle existence and uniqueness of the solution u to Equation (1) was given by Favini-Yagi  (see in particular the monograph ). By using the real interpolation space $( E , D ( B ) ) θ , ∞$, $0 < θ < 1$ (see [3,4]), suitable assumptions on the operators B, M, L guarantee that Equation (1) has a unique solution. This result was improved by Favini, Lorenzi, and Tanabe  (see also [6,7,8]).
In all cases, the basic assumptions read as follows:
(H$1$)
Operator B has a resolvent $( z − B ) − 1$ for any $z ∈ C$, Re$z < a$, $a > 0$ satisfying
$∥ ( z − B ) − 1 ∥ L ( E ) ≤ c | R e z | + 1 , R e z < a .$
(H$2$)
Operators L, M satisfy
$∥ M ( z M − L ) − 1 ∥ L ( E ) ≤ c ( | z | + 1 ) β$
for any $z ∈ Σ α : = z ∈ C : R e z ≥ − c ( 1 + | I m z | ) α , c > 0 , 0 < β ≤ α ≤ 1$.
(H$3$)
Let A be the possibly multivalued linear operator $A = L M − 1$, $D ( A ) = M ( D ( L ) )$. Then, A and B commute in the resolvent sense:
$B − 1 A − 1 = A − 1 B − 1 .$
Very recently, Al Horani et al. , see also , generalized the previous results to the interpolation space $( E , D ( B ) ) θ , p$, $1 < p ≤ ∞$, i.e.,
Lemma 1.
Let B, M, L be three closed linear operators on the complex Banach space E satisfying (H$1$)–(H$3$), $0 < β ≤ α ≤ 1$, $α + β > 1$. Then, for all $f ∈ ( E , D ( B ) ) θ , p$, $2 − α − β < θ < 1$, $1 < p ≤ ∞$, Equation (1) admits a unique solution u such that $L u$, $B M u ∈ ( E , D ( B ) ) θ , p$.
There are many choices of the operator B verifying Assumption (H$1$). In , the authors handled the abstract equation of the form
$D t α ˜ ( M u ( t ) ) − L u ( t ) = f ( t ) , 0 ≤ t ≤ T < ∞$
in the Banach space X with initial condition $g 1 − α ˜ ∗ M y ( 0 ) = 0$, where
$g β ( t ) = 1 Γ ( β ) t β − 1 t > 0 , 0 t ≤ 0 , β ≥ 0 ,$
and $Γ ( β )$ is the Gamma function.
For Riemann–Liouville derivative $D t α ˜$ of order $α ˜$, we address the monograph  (see also [12,13]). Very recent applications concerning Caputo fractional derivative operator are also discussed in  by the same authors using a completely different method than Sviridyuk’s group (see [15,16]). Some related topics can be found in [17,18,19].
In , the authors extended the results of direct and inverse problems, given in , to degenerate differential equations on the half line $[ 0 , ∞ )$. Precisely, let X be a complex Banach space and
endowed with the sup norm. If $B X$ is the operator defined by
and $M , L$ are two closed linear operators in the complex Banach space E satisfying
$∥ M ( λ M − L ) − 1 ∥ ≤ c ˜ ( | λ | + 1 ) β ∀ λ ∈ Σ α = { λ ; Re λ ≥ − c ( 1 + | Im λ | ) α } , c > 0 , c ˜ > 0 ,$
$0 < β ≤ α ≤ 1$, $0 < α ˜ < 1$, then for all $f ∈ ( E , D ( B X α ˜ ) ) θ , p , 2 − α − β < θ < 1 , 1 < p ≤ ∞$, equation
$B X α ˜ M u − L u = f$
admits a unique solution u. Moreover, $L u , B X α ˜ M u ∈ ( E , D ( B X α ˜ ) ) ω , p , ω = θ + α + β − 2$.
In this paper, we refine our results in  by investigating the fractional power of the operator $B X$ in the space of continuous functions u defined on $[ 0 , ∞ )$ without assuming $u ( 0 ) = 0$, i.e.,
where X is a complex Banach space. In this case, $B X$ is not densely defined. In such a case, it is not known whether $B X α B X β = B X α + β$ is true or not, since in the proof of Lemma A2 of T. Kato  it seems it is essentially used that A is densely defined. To obtain our results on such a new space E, we should investigate the previous fractional power problem in case $A = B X$.
The interpolation space $( E , D ( B X α ˜ ) ) θ , p$, $0 < θ < 1$, $p ∈ ( 1 , ∞ ]$ could be characterized by using the famous results of P. Grisvard. Since the operator $B X$ is of type $( π / 2 , 1 )$ and $α ˜ ∈ ( 0 , 1 )$, $− B X α ˜$ is the infinitesimal generator of an analytic semigroup $e − t B X α ˜ t > 0$ (see , Proposition 0.9, p. 19), the interpolation space $( E , D ( B X α ˜ ) ) θ , p$ could be characterized by
$( E , D ( B X α ˜ ) ) θ , p = u ∈ E ; ∥ t 1 − θ B X α ˜ e − t B X α ˜ u ∥ L p ∗ ( E ) + ∥ u ∥ E < ∞ = u ∈ E ; ∥ ξ θ B X α ˜ ( ξ + B X α ˜ ) − 1 u ∥ L p ∗ ( E ) < ∞ ,$
where $L p ∗ ( E )$ denotes the space of all strongly measurable $E −$valued functions f on $( 0 , ∞ )$ such that
$∫ 0 ∞ ∥ f ( t ) ∥ E p d t t < ∞ , 1 ≤ p < ∞ , ∥ f ( t ) ∥ L ∞ ∗ ( E ) = sup 0 < t < ∞ ∥ f ( t ) ∥ E , p = ∞ .$
The following lemma is also needed:
Lemma 2.
$∥ f ∗ g ∥ L ∗ p ≤ ∥ f ∥ L ∗ p ∥ g ∥ L ∗ 1$$∀ f ∈ L ∗ p ( R + )$, $∀ g ∈ L ∗ 1 ( R + )$
Section 2 is devoted to our main results. In Section 3, we present our conclusions and remarks.

## 2. Main Results

Let X be a complex Banach space and
$∥ u ∥ E = sup 0 ≤ t < ∞ ∥ u ( t ) ∥ X .$
Let $B X$ be an operator defined by
$B X u = u ′ for u ∈ D ( B X ) .$
Let $f ∈ E$, $Re λ > 0$. Consider the problem
$d d t u ( t ) + λ u ( t ) = f ( t ) , 0 < t < ∞ , u ( 0 ) = 0 .$
The solution is
$u ( t ) = ∫ 0 t e − λ ( t − s ) f ( s ) d s ,$
and
$∥ u ( t ) ∥ = ∫ 0 t e − λ ( t − s ) f ( s ) d s ≤ ∫ 0 t e − Re λ ( t − s ) d s ∥ f ∥ E = 1 − e − Re λ t Re λ ∥ f ∥ E ≤ 1 Re λ ∥ f ∥ E .$
Hence, u is bounded in $[ 0 , ∞ )$, and so is $u ′ = f − λ u$. This implies that u is uniformly continuous in $[ 0 , ∞ )$. Furthermore, $u ′ = f − λ u$ is uniformly continuous. Therefore, $u ∈ D ( B X )$ and $( B X + λ ) u = f$. Since $f ∈ E$ is arbitrary, one concludes that $R ( B X + λ ) = E$ and
$( B X + λ ) − 1 f ( t ) = ∫ 0 t e − λ ( t − s ) f ( s ) d s ,$
$∥ ( B X + λ ) − 1 ∥ L ( E ) ≤ 1 Re λ ∀ λ : Re λ > 0 .$
Here, we make some preparations. Suppose that A is a not necessarily densely defined closed linear operator in a Banach space X satisfying
(i)
$ρ ( − A ) ⊃ { λ ; | arg λ | < π − ω }$, $0 < ω < π ;$
(ii)
$λ ( λ + A ) − 1$ is uniformly bounded in each smaller sector ${ λ ; | arg λ | < π − ω − ϵ } , 0 < ϵ < π − ω$; and
(iii)
$∥ λ ( λ + A ) − 1 ∥ L ( X ) ≤ M$ for $λ > 0$ with some $M > 0$.
The first Assumption (i) is equivalent to $ρ ( A ) ⊃ { λ ; ω < | arg λ | ≤ π }$.
Case$0 ∈ ρ ( A )$. Set for $α > 0$
$R α ( λ ) = − 1 2 π i ∫ C ( λ + z α ) − 1 ( z − A ) − 1 d z , λ ≥ 0 ,$
where C runs in the resolvent set of A from $+ ∞ e − i θ$ to $+ ∞ e i θ$, $ω < θ ≤ π$, avoiding the negative real axis and 0, where $+ ∞ e ± i ∞ = lim r → ∞ r e ± i ∞$. Let $λ , μ ≥ 0$. Let $C ′$ be another contour which has the same property as C and is located to the right of C without intersecting C. Then,
$R α ( λ ) R α ( μ ) = 1 2 π i ∫ C ( λ + z α ) − 1 ( z − A ) − 1 d z 1 2 π i ∫ C ′ ( μ + ζ α ) − 1 ( ζ − A ) − 1 d ζ = 1 2 π i 2 ∫ C ′ ∫ C ( λ + z α ) − 1 ( μ + ζ α ) − 1 ( z − A ) − 1 ( ζ − A ) − 1 d z d ζ .$
$R α ( λ ) R α ( μ ) = 1 2 π i 2 ∫ C ′ ∫ C ( λ + z α ) − 1 ( μ + ζ α ) − 1 ( z − A ) − 1 − ( ζ − A ) − 1 ζ − z d z d ζ = 1 2 π i 2 ∫ C ′ ∫ C ( λ + z α ) − 1 ( μ + ζ α ) − 1 ( z − A ) − 1 ζ − z d z d ζ − 1 2 π i 2 ∫ C ′ ∫ C ( λ + z α ) − 1 ( μ + ζ α ) − 1 ( ζ − A ) − 1 ζ − z d z d ζ = 1 2 π i ∫ C ( λ + z α ) − 1 1 2 π i ∫ C ′ ( μ + ζ α ) − 1 ζ − z d ζ ( z − A ) − 1 d z − 1 2 π i ∫ C ′ 1 2 π i ∫ C ( λ + z α ) − 1 ζ − z d z ( μ + ζ α ) − 1 ( ζ − A ) − 1 d ζ = − 1 2 π i ∫ C ′ ( λ + ζ α ) − 1 ( μ + ζ α ) − 1 ( ζ − A ) − 1 d ζ = − 1 2 π i ∫ C ′ ( λ + ζ α ) − 1 − ( μ + ζ α ) − 1 μ − λ ( ζ − A ) − 1 d ζ .$
This yields
$( μ − λ ) R α ( λ ) R α ( μ ) = − 1 2 π i ∫ C ′ ( λ + ζ α ) − 1 ( ζ − A ) − 1 d ζ + 1 2 π i ∫ C ′ ( μ + ζ α ) − 1 ( ζ − A ) − 1 d ζ = R α ( λ ) − R α ( μ ) .$
Hence, ${ R α ( λ ) , λ ≥ 0 }$ is a pseudo resolvent:
$R α ( λ ) − R α ( μ ) = ( μ − λ ) R α ( λ ) R α ( μ ) ,$
and
$R α ( 0 ) = − 1 2 π i ∫ C z − α ( z − A ) − 1 d z .$
For $α > 0 , β > 0$,
$R α ( 0 ) R β ( 0 ) = 1 2 π i 2 ∫ C ′ ∫ C z − α ζ − β ( z − A ) − 1 ( ζ − A ) − 1 d z d ζ = 1 2 π i 2 ∫ C ′ ∫ C z − α ζ − β ( z − A ) − 1 − ( ζ − A ) − 1 ζ − z d z d ζ = 1 2 π i 2 ∫ C ′ ∫ C z − α ζ − β ( z − A ) − 1 ζ − z d z d ζ − 1 2 π i 2 ∫ C ′ ∫ C z − α ζ − β ( ζ − A ) − 1 ζ − z d z d ζ .$
The first term of the last side vanishes, and the second term is equal to
$− 1 2 π i ∫ C ′ 1 2 π i ∫ C z − α ζ − z d z ζ − β ( ζ − A ) − 1 d ζ = − 1 2 π i ∫ C ′ ζ − α − β ( ζ − A ) − 1 d ζ = R α + β ( 0 ) .$
Therefore, the following formula is obtained:
$R α ( 0 ) R β ( 0 ) = R α + β ( 0 ) , α > 0 , β > 0 .$
By virtue of Cauchy’s representation formula of holomorphic functions, one has
$R 1 ( 0 ) = − 1 2 π i ∫ C z − 1 ( z − A ) − 1 d z = A − 1 .$
Let $0 < α < 1$. Then,
$R 1 − α ( 0 ) R α ( 0 ) = R α ( 0 ) R 1 − α ( 0 ) = R 1 ( 0 ) = A − 1 .$
Therefore, if $R α ( 0 ) u = 0$, then $A − 1 u = R 1 − α ( 0 ) R α ( 0 ) u = 0$. This implies $u = 0$. Hence, $R α ( 0 )$ has an inverse. Since
$R α ( 0 ) u − R α ( λ ) u = λ R α ( 0 ) R α ( λ ) u ,$
$R α ( λ ) u = 0$ implies $R α ( 0 ) u = 0$, and hence $u = 0$, $R α ( λ )$ has an inverse $∀ λ ≥ 0$. Since $∀ u ∈ X$
$R α ( λ ) u − R α ( μ ) u = ( μ − λ ) R α ( λ ) R α ( μ ) u ,$
one observes $R α ( μ ) u ∈ R ( R α ( λ ) ) = D ( R α ( λ ) − 1 )$, and
$u − R α ( λ ) − 1 R α ( μ ) u = ( μ − λ ) R α ( μ ) u .$
Let $v ∈ D ( R α ( μ ) − 1 ) = R ( R α ( μ ) )$ and $v = R α ( μ ) u$. Then, $u = R α ( μ ) − 1 v$, $v ∈ D ( R α ( λ ) − 1 )$ and
$R α ( μ ) − 1 v − R α ( λ ) − 1 v = ( μ − λ ) v .$
This yields
$R α ( μ ) − 1 v − μ v = R α ( λ ) − 1 v − λ v , ∀ v ∈ D ( R α ( λ ) − 1 ) = D ( R α ( μ ) − 1 ) .$
Set $A α = R α ( λ ) − 1 − λ$ for some $λ ≥ 0$. Then, $A α = R α ( λ ) − 1 − λ ∀ λ ≥ 0$, and $R α ( λ ) − 1 = A α + λ$. This implies
$R α ( λ ) = ( λ + A α ) − 1 .$
Furthermore, $A α = R α ( 0 ) − 1$ and
$( A α ) − 1 = R α ( 0 ) = − 1 2 π i ∫ C z − α ( z − A ) − 1 d z .$
Letting $α = 1$, one gets $( A 1 ) − 1 = R 1 ( 0 ) = A − 1$. Hence, $A 1 = A$. Thus, writing $( A α ) − 1 = A − α$
$A − α = R α ( 0 ) = − 1 2 π i ∫ C z − α ( z − A ) − 1 d z , α > 0 , A − α A − β = R α ( 0 ) R β ( 0 ) = R α + β ( 0 ) = A − α − β , α > 0 , β > 0 .$
It is not difficult to show that the following relation holds if $0 < α < 1$:
$( λ + A α ) − 1 = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + A ) − 1 d μ , λ ≥ 0 .$
Proposition 1.
Let A be a not necessarily densely defined closed linear operator in a Banach space X satisfying (i)–(iii) and $0 ∈ ρ ( A )$. Then, the fractional power $A α$ of A is defined for $α > 0$, and the followings hold:
$A α A β = A α + β , α > 0 , β > 0 , A 1 = A , A − α = ( A α ) − 1 i s b o u n d e d$
and Equation (10) holds.
General case In what follows, we assume (i)–(iii). Let $0 < α < 1$. Set for $λ > 0$
$R α ( λ ) = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + A ) − 1 d μ .$
If $ϵ > 0$, $A ϵ = A + ϵ$ satisfies (i)–(iii), and $0 ∈ ρ ( A ϵ )$. Hence, $A ϵ α$ is defined, and
$( λ + A ϵ α ) − 1 = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + A ϵ ) − 1 d μ ,$
$( λ + A ϵ α ) − 1 − ( μ + A ϵ α ) − 1 = ( μ − λ ) ( λ + A ϵ α ) − 1 ( μ + A ϵ α ) − 1 , λ > 0 , μ > 0 .$
In view of (iii)
$∥ ( μ + A ϵ ) − 1 − ( μ + A ) − 1 ∥ X = ∥ − ϵ ( μ + ϵ + A ) − 1 ( μ + A ) − 1 ∥ X ≤ ϵ M μ + ϵ M μ .$
Therefore, with the aid of the dominated convergence theorem one obtains from Equations (17) and (18)
$( λ + A ϵ α ) − 1 → R α ( λ ) , λ > 0 a s ϵ → 0 .$
Thus, we have obtained:
Proposition 2.
Let A be a not necessarily densely defined closed linear operator in a Banach space X satisfying(i)–(iii). Then, the bounded operator valued function ${ R α ( λ ) , λ > 0 }$ defined by Equation (19) is a pseudo resolvent:
$R α ( λ ) − R α ( μ ) = ( μ − λ ) R α ( λ ) R α ( μ ) , λ > 0 , μ > 0 .$
Case$A = B X$. Let $ϵ > 0$. Then, $B X + ϵ > 0$ has a bounded inverse and
$∥ ( B X + ϵ + λ ) − 1 ∥ L ( E ) ≤ 1 Re λ + ϵ < 1 Re λ ∀ λ : Re λ ≥ 0 .$
By virtue of Proposition 1, the fractional power $( B X + ϵ ) α$ of $B X + ϵ$ is defined for $α > 0$, and the followings hold:
and for $0 < α < 1$
$( λ + ( B X + ϵ ) α ) − 1 = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + B X + ϵ ) − 1 d μ , λ ≥ 0 .$
Especially,
$( B X + ϵ ) − α = sin π α π ∫ 0 ∞ μ − α ( μ + B X + ϵ ) − 1 d μ .$
Therefore,
$( λ + ( B X + ϵ ) α ) − 1 f ( t ) = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + B X + ϵ ) − 1 f ( t ) d μ = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ∫ 0 t e − ( μ + ϵ ) ( t − s ) f ( s ) d s d μ = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − ( μ + ϵ ) ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s$
and
$( B X + ϵ ) − α f ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ − α e − ( μ + ϵ ) ( t − s ) d μ f ( s ) d s .$
By the change of the independent variable $μ ( t − s ) = τ$,
$∫ 0 ∞ μ − α e − ( μ + ϵ ) ( t − s ) d μ = e − ϵ ( t − s ) ∫ 0 ∞ μ − α e − μ ( t − s ) d μ = e − ϵ ( t − s ) ∫ 0 ∞ ( t − s ) α τ − α e − τ ( t − s ) − 1 d τ = ( t − s ) α − 1 e − ϵ ( t − s ) ∫ 0 ∞ τ − α e − τ d τ = ( t − s ) α − 1 e − ϵ ( t − s ) Γ ( 1 − α ) .$
Hence, using $Γ ( α ) Γ ( 1 − α ) = π / sin π α$, one observes
$( B X + ϵ ) − α f ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 e − ϵ ( t − s ) f ( s ) d s .$
Set for $λ > 0$,
$R ( λ ) = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + B X ) − 1 d μ .$
Then, in view of Equation (23),
$R ( λ ) − R ( μ ) = ( μ − λ ) R ( λ ) R ( μ ) , λ > 0 , μ > 0 .$
From Equations (12) and (31), it follows that
$∥ R ( λ ) ∥ L ( E ) ≤ sin π α π ∫ 0 ∞ μ α − 1 ( λ + μ α cos π α ) 2 + ( μ sin π α ) 2 d μ .$
For $f ∈ E$, in view of Equations (11) and (31),
$R ( λ ) f ( t ) = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + B X ) − 1 f ( t ) d μ = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ∫ 0 t e − μ ( t − s ) f ( s ) d s d μ = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s .$
Using
$Γ ( α ) Γ ( 1 − α ) = π sin π α , ∫ 0 ∞ μ − α e − ( t − s ) μ d μ = Γ ( 1 − α ) ( t − s ) α − 1 ,$
one deduces from Equation (34)
$R ( λ ) f ( t ) − 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s − sin π α π ∫ 0 t ∫ 0 ∞ μ − α e − μ ( t − s ) d μ f ( s ) d s = sin π α π ∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α − μ − α e − μ ( t − s ) d μ f ( s ) d s .$
One has
$∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α − μ − α e − μ ( t − s ) d μ f ( s ) d s X ≤ ∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α − μ − α e − μ ( t − s ) d μ ∥ f ( s ) ∥ X d s ≤ ∫ 0 t ∫ 0 ∞ μ α ( λ + μ α cos π α ) 2 + ( μ α sin π α ) 2 − μ − α e − μ ( t − s ) d μ d s ∥ f ∥ E .$
Since
$μ α ( λ + μ α cos π α ) 2 + ( μ α sin π α ) 2 − μ − α e − μ ( t − s ) ≤ μ α ( λ + μ α cos π α ) 2 + ( μ α sin π α ) 2 + μ − α e − μ ( t − s ) ≤ 1 ( sin π α ) 2 + 1 μ − α e − μ ( t − s ) , ∫ 0 t ∫ 0 ∞ μ − α e − μ ( t − s ) d μ d s = Γ ( 1 − α ) ∫ 0 t ( t − s ) α − 1 d s = Γ ( 1 − α ) t α α < ∞$
and
the last right hand-side of Inequality (36) tends to 0 as $λ → 0$. Therefore,
Suppose $R ( λ ) f = 0 ∃ λ > 0$. Then, $R ( λ ) f = 0 ∀ λ > 0$, i.e., $R ( λ ) f ( t ) ≡ 0 ∀ λ > 0$. Hence, by virtue of Equation (37),
$1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s ≡ 0 .$
This yields
$0 ≡ ∫ 0 t ( t − τ ) − α ∫ 0 τ ( τ − s ) α − 1 f ( s ) d s d τ = ∫ 0 t ∫ s t ( t − τ ) − α ( τ − s ) α − 1 d τ f ( s ) d s = B ( 1 − α , α ) ∫ 0 t f ( s ) d s ⟹ f ( t ) ≡ 0 .$
Therefore, $R ( λ )$ has an inverse $∀ λ > 0$. Set $B X α = R ( λ ) − 1 − λ$ for some $λ > 0$. Then, $B X α = R ( λ ) − 1 − λ$, $∀ λ > 0$, and $R ( λ ) = ( λ + B X α ) − 1 , ∀ λ > 0$. Hence, in view of Equation (31),
$( λ + B X α ) − 1 = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + B X ) − 1 d μ , λ > 0 .$
By virtue of Equations (34) and (38), one observes that, for $λ > 0$ and $f ∈ E$,
$( λ + B X α ) − 1 f ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s , 0 < t < ∞ .$
Since
noting Equations (28) and (39), one observes that, if $λ > 0$,
$( λ + ( B X + ϵ ) α ) − 1 f ( t ) → ( λ + B X α ) − 1 f ( t )$
uniformly in $[ 0 , T ]$, $0 < T < ∞$ as $ϵ → 0$. By virtue of Equations (12) and (38), one deduces
$∥ ( λ + B X α ) − 1 ∥ L ( E ) ≤ sin π α π ∫ 0 ∞ μ α − 1 λ 2 + 2 λ μ α cos π α + μ 2 α d μ = sin π α π ∫ 0 ∞ r α − 1 1 + 2 r α cos π α + r 2 α d r 1 λ = 1 λ , λ > 0 .$
For an arbitrary $λ$ with Re$λ > 0$, let $μ$ be so large that $μ > | λ | 2 / ( 2 R e λ )$. Then, $μ > | μ − λ |$. One has
$( λ + B X α ) − 1 = ( μ + B X α + λ − μ ) − 1 = I + ( λ − μ ) ( μ + B X α ) − 1 ( μ + B X α ) − 1 = ( μ + B X α ) − 1 I + ( λ − μ ) ( μ + B X α ) − 1 − 1 = ( μ + B X α ) − 1 ∑ n = 0 ∞ ( − 1 ) n ( λ − μ ) n ( μ + B X α ) − n .$
Hence,
$( λ + B X α ) − 1 L ( E ) ≤ ( μ + B X α ) − 1 L ( E ) ∑ n = 0 ∞ λ − μ n ( μ + B X α ) − 1 L ( E ) n = 1 μ ∑ n = 0 ∞ | λ − μ | μ n = 1 μ 1 1 − | λ − μ | / μ = 1 μ − | λ − μ | .$
Since
$1 μ − | λ − μ | = μ + | λ − μ | μ 2 − | λ − μ | 2 = μ + | λ − μ | μ 2 − μ 2 − 2 μ R e λ + | λ | 2 = μ + | λ − μ | 2 μ R e λ − | λ | 2 = 1 + | λ / μ − 1 | 2 R e λ − | λ | 2 / μ ⟶ 1 R e λ a s μ ⟶ ∞ ,$
one concludes
$ρ ( B X α ) ⊃ { λ ; Re λ < 0 } and ∥ ( λ + B X α ) − 1 ∥ L ( E ) ≤ 1 Re λ , Re λ > 0 .$
One has $∀ λ > 0$, $∀ f ∈ D ( B X α )$
$R ( λ ) − 1 f = λ f + B X α f ⟹ λ R ( λ ) f + R ( λ ) B X α f = f .$
Hence,
$R ( λ ) B X α f ( t ) = f ( t ) − λ R ( λ ) f ( t ) , t > 0 .$
Since Equation (37) holds for any $f ∈ E$, one has
$R ( λ ) B X α f ( t ) ⟶ 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 B X α f ( s ) d s , ∀ t ∈ ( 0 , ∞ ) , λ R ( λ ) f ( t ) ⟶ 0 , ∀ t ∈ ( 0 , ∞ ) .$
Therefore, one obtains
$1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 B X α f ( s ) d s = f ( t ) .$
This implies
$1 Γ ( α ) ∫ 0 t ( t − τ ) − α ∫ 0 τ ( τ − s ) α − 1 B X α f ( s ) d s d τ = ∫ 0 t ( t − τ ) − α f ( τ ) d τ .$
The left hand side is equal to
$1 Γ ( α ) ∫ 0 t ∫ s t ( t − τ ) − α ( τ − s ) α − 1 d τ B X α f ( s ) d s = Γ ( 1 − α ) ∫ 0 t B X α f ( s ) d s .$
Hence,
$∫ 0 t B X α f ( s ) d s = 1 Γ ( 1 − α ) ∫ 0 t ( t − s ) − α f ( s ) d s .$
From this, it follows that
$∫ 0 t ( t − τ ) α − 1 ∫ 0 τ B X α f ( s ) d s d τ = 1 Γ ( 1 − α ) ∫ 0 t ( t − τ ) α − 1 ∫ 0 τ ( τ − s ) − α f ( s ) d s d τ .$
By the change of the order of the integration
$1 α ∫ 0 t ( t − s ) α B X α f ( s ) d s = Γ ( α ) ∫ 0 t f ( s ) d s .$
By the differentiation of both sides
$∫ 0 t ( t − s ) α − 1 B X α f ( s ) d s = Γ ( α ) f ( t ) .$
Therefore, $B X α$ has an inverse $B X − α$, and for $f ∈ D ( B X − α )$
$B X − α f ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s , 0 < t < ∞ .$
Consequently, the following proposition is established:
Proposition 3.
Let $B X$ be the operator defined by Equations (7) and (8). Then, $B X$ satisfies $ρ ( B X ) ⊃ { λ ; Re λ < 0 }$ and Equations (11) and (12) hold. The fractional power $B X α , 0 < α < 1 ,$ of $B X$ is defined implicitly by Equation (38) or Equation (39). $B X α$ has an inverse $B X − α$ and for $f ∈ D ( B X − α )$ Equation (41) holds.
Especially if $f ∈ D ( B X − α )$, then the function $∫ 0 t ( t − s ) α − 1 f ( s ) d s$ belongs to E. The converse is given in the next proposition.
Proposition 4.
Suppose that both functions f and $1 Γ ( α ) ∫ 0 · ( · − s ) α − 1 f ( s ) d s , 0 < α < 1 ,$ belong to E. Then, $f ∈ D ( B X − α )$ and Equation (41) holds.
Proof.
As a preparation, we first consider the case of a finite interval. Let $0 < T < ∞$. Let
$D ( B X ) = { u ∈ C 1 ( [ 0 , T ] ; X ) ; u ( 0 ) = 0 } , B X u = u ′ .$
Then, $∀ λ ∈ C$
$( λ + B X ) − 1 f ( t ) = ∫ 0 t e − λ ( t − s ) f ( s ) d s , especially B X − 1 f ( t ) = ∫ 0 t f ( s ) d s , 0 ≤ t ≤ T .$
Therefore, $B X$ satisfies the assumptions of Proposition 1 with $ω = π / 2$, and hence its fractional power $B X α$ is defined for $0 < α < 1$, and we have:
$( λ + B X α ) − 1 = sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ( μ + B X ) − 1 d μ , λ ≥ 0 .$
Analogously to Equation (40), the following statement is established:
$ρ ( B X α ) ⊃ { λ ; Re λ < 0 } and ∥ ( λ + B X α ) − 1 ∥ L ( C ( [ 0 , T ] ; X ) ) ≤ 1 Re λ , Re λ > 0 .$
It follows from Equations (42) and (43) that for $f ∈ C ( [ 0 , T ] ; X )$, $λ ≥ 0$
$( λ + B X α ) − 1 f ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s , 0 ≤ t ≤ T .$
Especially if $λ = 0$,
$B X − α f ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ − α e − μ ( t − s ) d μ f ( s ) d s = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s , 0 ≤ t ≤ T .$
For $f ∈ C ( [ 0 , T ] ; X )$ and $λ > 0$,
$( λ + B X α ) − 1 ( λ B X − α + 1 ) f = ( λ + B X α ) − 1 ( λ + B X α ) B X − α f = B X − α f .$
From Equation (45) with f replaced by $( λ B X − α + 1 ) f$ and Equation (46), it follows that
$( λ + B X α ) − 1 ( λ B X − α + 1 ) f ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ ( λ B X − α + 1 ) f ( s ) d s = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ λ Γ ( α ) ∫ 0 s ( s − σ ) α − 1 f ( σ ) d σ + f ( s ) d s = sin π α π λ Γ ( α ) ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ ∫ 0 s ( s − σ ) α − 1 f ( σ ) d σ d s + sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s .$
By the changes of the order of integration,
$∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ ∫ 0 s ( s − σ ) α − 1 f ( σ ) d σ d s = ∫ 0 t ∫ σ t ∫ 0 ∞ μ α e − μ ( t − s ) ( s − σ ) α − 1 λ 2 + 2 λ μ α cos π α + μ 2 α d μ d s f ( σ ) d σ = ∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ∫ σ t e − μ ( t − s ) ( s − σ ) α − 1 d s d μ f ( σ ) d σ .$
Substituting Equation (49) (with s and $σ$ interchanged) into Equation (48), one deduces
$( λ + B X α ) − 1 ( λ B X − α + 1 ) f ( t ) = sin π α π λ Γ ( α ) ∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α ∫ s t e − μ ( t − σ ) ( σ − s ) α − 1 d σ d μ f ( s ) d s ∗ + sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) λ 2 + 2 λ μ α cos π α + μ 2 α d μ f ( s ) d s = sin π α π ∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α × λ Γ ( α ) ∫ s t e − μ ( t − σ ) ( σ − s ) α − 1 d σ + e − μ ( t − s ) d μ f ( s ) d s .$
From Equations (46), (47), and (50), it follows that the following equality holds $∀ f ∈ C ( [ 0 , T ] ; X )$:
$sin π α π ∫ 0 t ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α λ Γ ( α ) ∫ s t e − μ ( t − σ ) ( σ − s ) α − 1 d σ + e − μ ( t − s ) d μ f ( s ) d s = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s , 0 ≤ t ≤ T .$
This yields that, for $< t ≤ T$:
$sin π α π ∫ 0 ∞ μ α λ 2 + 2 λ μ α cos π α + μ 2 α λ Γ ( α ) ∫ 0 t e − μ ( t − σ ) σ α − 1 d σ + e − μ t d μ = t α − 1 Γ ( α ) .$
Since $T > 0$ is arbitrary, one concludes that Equation (51) holds for $0 < t < ∞$.
We return to the case of the infinite interval $( 0 , ∞ )$. Suppose that both functions f and $u ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s$ belong to E. One has by virtue of Equation (39) with $λ = 1$
$( 1 + B X α ) − 1 u ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) 1 + 2 μ α cos π α + μ 2 α d μ u ( s ) d s = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) 1 + 2 μ α cos π α + μ 2 α d μ 1 Γ ( α ) ∫ 0 s ( s − σ ) α − 1 f ( σ ) d σ d s = sin π α π Γ ( α ) ∫ 0 t ∫ σ t ∫ 0 ∞ μ α e − μ ( t − s ) 1 + 2 μ α cos π α + μ 2 α d μ ( s − σ ) α − 1 d s f ( σ ) d σ = sin π α π ∫ 0 t ∫ 0 ∞ μ α 1 + 2 μ α cos π α + μ 2 α 1 Γ ( α ) ∫ σ t e − μ ( t − s ) ( s − σ ) α − 1 d s d μ f ( σ ) d σ ,$
and
$( 1 + B X α ) − 1 f ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ α e − μ ( t − s ) 1 + 2 μ α cos π α + μ 2 α d μ f ( s ) d s .$
Adding Equations (52) and (53), and using Equation (51) with $λ = 1$, one observes
$( 1 + B X α ) − 1 ( u + f ) ( t ) = sin π α π ∫ 0 t ∫ 0 ∞ μ α 1 + 2 μ α cos π α + μ 2 α 1 Γ ( α ) ∫ s t e − μ ( t − σ ) ( σ − s ) α − 1 d σ + e − μ ( t − s ) d μ f ( s ) d s = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 f ( s ) d s = u ( t ) .$
This yields that $u ∈ D ( B X α )$ and $u + f = ( I + B X α ) u = u + B X α u$. Consequently, $f ∈ D ( B X − α )$ and $B X − α f = u$. □
In view of Propositions 3 and 4, the following statement is obtained:
Corollary 1.
Let $f ∈ E$. Then, $f ∈ D ( B X − α )$ if and only if $∫ 0 · ( · − s ) α − 1 f ( s ) d s ∈ E$. For $f ∈ D ( B X − α )$, Equation (41) holds.
For $f ∈ E , α > 0 , β > 0$,
$1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 1 Γ ( β ) ∫ 0 s ( s − σ ) β − 1 f ( σ ) d σ d s = 1 Γ ( α ) Γ ( β ) ∫ 0 t ∫ σ t ( t − s ) α − 1 ( s − σ ) β − 1 d s f ( σ ) d σ = 1 Γ ( α + β ) ∫ 0 t ( t − σ ) α + β − 1 f ( σ ) d σ .$
Suppose $f ∈ D ( B X − β ) , 0 < β < 1$. Then, in view of Corollary 1 $, ∫ 0 · ( · − s ) β − 1 f ( s ) d s ∈ E$. Hence,
$1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 ( B X − β f ) ( s ) d σ d s = 1 Γ ( α + β ) ∫ 0 t ( t − σ ) α + β − 1 f ( σ ) d σ .$
Therefore, under the assumption $f ∈ D ( B X − β )$ $B X − β f ∈ D ( B X − α )$ if and only if $f ∈ D ( B X − α − β )$, and in this case $B X − α B X − β f = B X − α − β f$ holds. In particular, it is obtained that
$B X − α B X − β ⊂ B X − α − β .$
Problem $β > α ⟹ D ( B X − β ) ⊂ D ( B X − α ) ?$
Let $0 < α ˜ < 1$, $0 < β ≤ α ≤ 1$. Let L and M be densely defined closed linear operators in X such that $0 ∈ ρ ( L )$, $D ( L ) ⊂ D ( M )$ and
$( H ) ∥ M ( λ M − L ) − 1 ∥ L ( X ) ≤ c ˜ ( | λ | + 1 ) β , c ˜ > 0 , ∀ λ ∈ Σ α = { λ ; Re λ ≥ − c ( 1 + | Im λ | ) α } , c > 0 .$
Consider the equation
$B X α ˜ M u − L u = f .$
Let $a 0$ and a be such that
$0 < a 0 < min { c , 1 } , c < a < c + a 0 .$
Equation (56) is equivalent to
$( B X α ˜ + a 0 ) M u − ( L + a 0 M ) u = f .$
Since $− a 0 > − c$, in view of (H) $M ( L + a 0 M ) − 1 ∈ L ( X )$ and if $Re ( λ − a 0 ) ≥ − c ( 1 + | Im ( λ − a 0 ) | ) α$,
$∥ M ( ( λ − a 0 ) M − L ) − 1 ∥ L ( X ) ≤ c ˜ ( | λ − a 0 | + 1 ) β ,$
i.e., if $Re λ ≥ a 0 − c ( 1 + | Im λ | ) α$,
$∥ M ( λ M − ( L + a 0 M ) ) − 1 ∥ L ( X ) ≤ c ˜ ( | λ − a 0 | + 1 ) β .$
Since $| λ − a 0 | + 1 ≥ | λ | − a 0 + 1 = | λ | + ( 1 − a 0 ) ≥ ( 1 − a 0 ) ( | λ | + 1 )$,
$c ˜ ( | λ − a 0 | + 1 ) β ≤ c ˜ ( 1 − a 0 ) β ( | λ | + 1 ) β = c 1 ( | λ | + 1 ) β , c 1 = c ˜ ( 1 − a 0 ) − β .$
Therefore,
$∥ M ( λ M − ( L + a 0 M ) ) − 1 ∥ L ( X ) ≤ c 1 ( | λ | + 1 ) β ∀ λ : Re λ ≥ a 0 − c ( 1 + | Im λ | ) α .$
The inequality in Equation (40) implies
$∥ ( B X α ˜ + a 0 − λ ) − 1 ∥ L ( E ) ≤ 1 a 0 − Re λ ∀ λ : Re λ < a 0 .$
By virtue of Equation (38),
$( B X α ˜ + a 0 ) − 1 = sin π α ˜ π ∫ 0 ∞ μ α ˜ a 0 2 + 2 a 0 μ α ˜ cos π α ˜ + μ 2 α ˜ ( B X + μ ) − 1 d μ .$
Let $f ∈ L p ( 0 , ∞ ; X )$ and $T ∈ L ( X )$. Since
$( T ( B X + μ ) − 1 f ) ( t ) = T · ( ( B X + μ ) − 1 f ) ( t ) = T ∫ 0 t e − μ ( t − s ) f ( s ) d s = ∫ 0 t e − μ ( t − s ) T f ( s ) d s = ∫ 0 t e − μ ( t − s ) ( T f ) ( s ) d s = ( ( B X + μ ) − 1 T f ) ( t ) ,$
one observes $T ( B X + μ ) − 1 f = ( B X + μ ) − 1 T f$, i.e., $T ( B X + μ ) − 1 = ( B X + μ ) − 1 T$. Therefore, with the aid of Equation (62), one obtains
$T ( B X α ˜ + a 0 ) − 1 = ( B X α ˜ + a 0 ) − 1 T .$
Applying this to $T = M ( L + a 0 M ) − 1$, one obtains
$( B X α ˜ + a 0 ) − 1 M ( L + a 0 M ) − 1 = M ( L + a 0 M ) − 1 ( B X α ˜ + a 0 ) − 1 .$
Let $Γ$ be the curve
$Γ = { z = a − c ( 1 + | y | ) α + i y , y ∈ R } .$
In view of Equation (57), one has $a 0 < c < a , a − c < a 0$. Hence, if $λ = a − c ( 1 + | y | ) α + i y ∈ Γ$ with $y ∈ R$, one has
$a 0 − c ( 1 + | Im λ | ) α = a 0 − c ( 1 + | y | ) α < a − c ( 1 + | y | ) α = Re λ ≤ a − c < a 0 .$
Therefore, Equations (60) and (61) hold on $Γ$.
We verify
$∥ ( B X α ˜ + a 0 − λ ) − 1 ∥ L ( E ) ≤ c 2 1 + | Re λ | ∀ λ ∈ Γ , c 2 = a − c + 1 a 0 − a + c .$
We show that, if $λ ∈ Γ$, the following inequality holds:
$a 0 − Re λ ≥ a 0 − a + c a − c + 1 ( 1 + | Re λ | ) .$
Note here $a 0 − a + c > 0$ and $a − c + 1 > a − c > 0$ in view of Equation (57). Hence,
$a 0 − a 0 − a + c a − c + 1 = a 0 a − a 0 c + a − c a − c + 1 = ( a 0 + 1 ) ( a − c ) a − c + 1 > 0 .$
This yields that
$a 0 − Re λ − a 0 − a + c a − c + 1 ( 1 + Re λ ) = a 0 − Re λ − a 0 − a + c a − c + 1 − a 0 − a + c a − c + 1 Re λ = ( a 0 + 1 ) ( a − c ) a − c + 1 − a 0 + 1 a − c + 1 Re λ = a 0 + 1 a − c + 1 ( a − c − Re λ ) ≥ 0$
if $λ ∈ Γ$. Therefore, Equation (66) holds if $λ ∈ Γ$ and $Re λ ≥ 0$. Recalling that $a 0 < 1$ we see if $Re λ ≤ 0$,
$a 0 − Re λ = a 0 + | Re λ | ≥ a 0 ( 1 + | Re λ | ) .$
Therefore, recalling Equation (67) one observes that Equation (66) holds also in the case $Re λ ≤ 0$.
Since $Re λ ≤ a − c < a 0$ if $λ ∈ Γ$, it follows from Equations (61) and (66) that Equation (65) holds.
Set $B = B X α ˜ + a 0$, $T = M ( L + a 0 M ) − 1$. Then, Equations (61) and (64) are expressed as
$B − 1 T = T B − 1 ,$
$∥ ( B − λ ) − 1 ∥ L ( E ) ≤ 1 a 0 − Re λ ∀ λ : Re λ < a 0 ,$
respectively. Let $v = ( L + a 0 M ) u$ be the new unknown variable. Then, $M u = M ( L + a 0 M ) − 1 v = T v$ and Equation (58) is expressed as
$B T v − v = f .$
Our candidate of the solution to Equation (70) is
$v = ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z .$
We have
$z T − 1 = z M ( L + a 0 M ) − 1 − 1 = [ z M − ( L + a 0 M ) ] ( L + a 0 M ) − 1 .$
If $z = a − c ( 1 + | y | ) α + i y ∈ Γ$, then $Re z = a − c ( 1 + | Im z | ) α > a 0 − c ( 1 + | Im z | ) α$. Hence, in view of Equation (60)
$∥ M ( z M − ( L + a 0 M ) ) − 1 ∥ L ( X ) ≤ c 1 ( | z | + 1 ) β .$
Therefore, if $z ∈ Γ$,
$∥ ( z T − 1 ) − 1 ∥ L ( E ) = ∥ ( z T − 1 ) − 1 ∥ L ( X ) = ∥ ( L + a 0 M ) ( z M − ( L + a 0 M ) ) − 1 ∥ L ( X ) = ∥ { z M − ( z M − ( L + a 0 M ) ) } ( z M − ( L + a 0 M ) ) − 1 ∥ L ( X ) = ∥ z M ( z M − ( L + a 0 M ) ) − 1 − I ∥ L ( X ) ≤ ∥ z M ( z M − ( L + a 0 M ) ) − 1 ∥ L ( X ) + 1 ≤ c 1 | z | ( | z | + 1 ) β + 1 ≤ c 1 ( | z | + 1 ) 1 − β + 1 ≤ c 2 ( | z | + 1 ) 1 − β , c 2 = c 1 + 1 .$
This yields
$∥ v ∥ E ≤ c 2 2 π ∫ Γ | z | − ( 1 + θ ) ( 1 + | z | ) 1 − β | z | θ ∥ B ( B − z ) − 1 f ∥ E | d z | .$
Let $z = a − c ( 1 + | y | ) α + i y ∈ Γ$. From
$B ( B − z ) − 1 = B ( B + 1 + | y | ) − 1 + ( 1 + | y | + z ) ( B − z ) − 1 B ( B + 1 + | y | ) − 1 = 1 + ( 1 + | y | + z ) ( B − z ) − 1 B ( B + 1 + | y | ) − 1 ,$
it follows that
$∥ B ( B − z ) − 1 f ∥ E = ∥ 1 + ( 1 + | y | + z ) ( B − z ) − 1 B ( B + 1 + | y | ) − 1 f ∥ E ≤ ∥ 1 + ( 1 + | y | + z ) ( B − z ) − 1 ∥ L ( E ) ∥ B ( B + 1 + | y | ) − 1 f ∥ E ≤ 1 + 1 + | y | + | z | a 0 − Re z ∥ B ( B + 1 + | y | ) − 1 f ∥ E .$
Since
$1 + | y | + | z | = 1 + | y | + | a − c ( 1 + | y | ) α + i y | ≤ 1 + | y | + a + c ( 1 + | y | ) α + | y | = 1 + 2 | y | + a + c ( 1 + | y | ) α ,$
one has
$1 + 1 + | y | + | z | a 0 − Re z ≤ 1 + 1 + 2 | y | + a + c ( 1 + | y | ) α a 0 − a + c ( 1 + | y | ) α = a 0 + 1 + 2 | y | + 2 c ( 1 + | y | ) α a 0 − a + c ( 1 + | y | ) α ,$
and
$a 0 + 1 + 2 | y | + 2 c ( 1 + | y | ) α ≤ ( a 0 + 1 ) ( 1 + | y | ) + 2 ( 1 + | y | ) + 2 c ( 1 + | y | ) = ( a 0 + 3 + 2 c ) ( 1 + | y | ) .$
Since $a 0 < a$,
$a 0 − a + c ( 1 + | y | ) α ≥ c ( 1 + | y | ) α − ( a − a 0 ) ( 1 + | y | ) α = ( c − a + a 0 ) ( 1 + | y | ) α .$
Note here that $c − a + a 0 > 0$ (cf. Equation (57)). Hence,
$1 + 1 + | y | + | z | a 0 − Re z ≤ ( a 0 + 3 + 2 c ) ( 1 + | y | ) ( c − a + a 0 ) ( 1 + | y | ) α = c 3 ( 1 + | y | ) 1 − α , c 3 = a 0 + 3 + 2 c c − a + a 0 .$
From Equations (74) and (75), it follows that
$∥ B ( B − z ) − 1 f ∥ E ≤ c 3 ( 1 + | y | ) 1 − α ∥ B ( B + 1 + | y | ) − 1 f ∥ E .$
For $z = a − c ( 1 + | y | ) α + i y ∈ Γ , y ∈ R$, one has
$| z | = | a − c ( 1 + | y | ) α + i y | ≤ a + c ( 1 + | y | ) α + | y | ≤ a + c ( 1 + | y | ) + | y | = a + c + ( c + 1 ) | y | ≤ c 4 ( 1 + | y | ) , c 4 = max { a , 1 } + c ,$
$( 1 + | z | ) 1 − β ≤ ( 1 + c 4 ( 1 + | y | ) ) 1 − β ≤ ( 1 + | y | + c 4 ( 1 + | y | ) ) 1 − β = c 5 ( 1 + | y | ) 1 − β , c 5 = ( 1 + c 4 ) 1 − β ,$
$| 1 + | y | + z | ≤ 1 + | y | + | z | ≤ 1 + | y | + c 4 ( 1 + | y | ) = ( 1 + c 4 ) ( 1 + | y | ) = c 6 ( 1 + | y | ) , c 6 = 1 + c 4 = 1 + max { a , 1 } + c .$
Here, we show that $∃ c 7$ such that
$| z | ≥ c 7 ( 1 + | y | ) ∀ z ∈ Γ : z = a − c ( 1 + | y | ) α + i y , y ∈ R .$
Proof.
Let $0 < b < a − c ( ⟺ c − b < c < a − b )$.
(i)
Case $| a − c ( 1 + | y | ) α | ≤ b$. In this case,
$| a − c ( 1 + | y | ) α | ≤ b ⟺ − b ≤ a − c ( 1 + | y | ) α ≤ b ⟹ a − b ≤ c ( 1 + | y | ) α ⟺ a − b c ≤ ( 1 + | y | ) α ⟺ a − b c 1 / α ≤ 1 + | y | ⟹ | y | ≥ a − b c 1 / α − 1 ≡ δ > 0 .$
Hence,
$δ 1 + δ ( 1 + | y | ) = δ 1 + δ + δ | y | 1 + δ ≤ | y | 1 + δ + δ | y | 1 + δ = | y | .$
Therefore,
$| z | = | a − c ( 1 + | y | ) α + i y | ≥ | y | ≥ δ 1 + δ ( 1 + | y | ) .$
(ii)
Case $a − c ( 1 + | y | ) α > b$. In this case
$| z | ≥ | a − c ( 1 + | y | ) α | = a − c ( 1 + | y | ) α > b ,$
and
$a − c ( 1 + | y | ) α > b ⟺ ( 1 + | y | ) α < a − b c ⟺ 1 + | y | < a − b c 1 / α ⟺ c a − b 1 / α ( 1 + | y | ) < 1 .$
Therefore,
$| z | > b > b c a − b 1 / α ( 1 + | y | ) .$
(iii)
Case $c ( 1 + | y | ) α − a > b$. In this case
$c ( 1 + | y | ) α > a + b ⟺ ( 1 + | y | ) α > a + b c ⟺ | y | > a + b c 1 / α − 1 ≡ γ > 0 .$
Hence,
$γ γ + 1 ( 1 + | y | ) = γ γ + 1 + γ γ + 1 | y | < | y | γ + 1 + γ γ + 1 | y | = | y | ≤ | z | .$
Thus Equation (80) holds with $c 7 = min δ 1 + δ , b c a − b 1 / α , γ γ + 1$.
□
Hence, from Equations (73) and (76)–(80), it follows that
$∥ v ∥ E ≤ c 8 ∫ Γ ( 1 + | y | ) 1 − α − β − θ ( 1 + | y | ) θ ∥ B ( B + 1 + | y | ) − 1 f ∥ E | d z | ,$
where $c 8 = c 2 c 7 − ( 1 + θ ) c 5 c 4 θ c 3 / 2 π$. For $z = a − c ( 1 + | y | ) α + i y , y ≥ 0$
$| d z | = | − c α ( 1 + y ) α − 1 d y + i d y | = { ( c α ) 2 ( 1 + y ) 2 ( α − 1 ) + 1 } 1 / 2 d y ≤ ( ( c α ) 2 + 1 ) 1 / 2 d y .$
Therefore,
$∥ v ∥ E ≤ 2 ( ( c α ) 2 + 1 ) 1 / 2 c 8 ∫ 0 ∞ ( 1 + y ) 1 − α − β − θ ( 1 + y ) θ ∥ B ( B + 1 + y ) − 1 f ∥ E d y = 2 ( ( c α ) 2 + 1 ) 1 / 2 c 8 ∫ 1 ∞ y 2 − α − β − θ y θ ∥ B ( B + y ) − 1 f ∥ E d y y ≤ 2 ( ( c α ) 2 + 1 ) 1 / 2 c 8 p − 1 ( θ + α + β − 2 ) p ( p − 1 ) / p ∫ 1 ∞ y θ p ∥ B ( B + y ) − 1 f ∥ E p d y y 1 / p ≤ 2 ( ( c α ) 2 + 1 ) 1 / 2 c 8 p − 1 ( θ + α + β − 2 ) p ( p − 1 ) / p ∥ f ∥ ( E , D ( B ) ) θ , p .$
Thus, it has been shown that v is well defined by Equation (71) if $f ∈ ( E , D ( B ) ) θ , p$.
Next, we show that v satisfies Equation (70). We show that the following inequality holds with some constant $c 9$:
$1 + | a − c ( 1 + | y | ) α | ≥ c 9 ( 1 + | y | ) α ∀ y ∈ R .$
Proof.
(i) Case $| a − c ( 1 + | y | ) α | ≤ 1 / 2$. Since
$| a − c ( 1 + | y | ) α | ≤ 1 2 ⟺ − 1 2 ≤ a − c ( 1 + | y | ) α ≤ 1 2 ⟹ − 1 ≤ 2 a − 2 c ( 1 + | y | ) α ⟺ 2 c ( 1 + | y | ) α ≤ 2 a + 1 ⟺ 2 c 2 a + 1 ( 1 + | y | ) α ≤ 1 ,$
one observes
$1 + | a − c ( 1 + | y | ) α | ≥ 1 ≥ 2 c 2 a + 1 ( 1 + | y | ) α .$
(ii) Case $a − c ( 1 + | y | ) α > 1 / 2$ (this can occur only in case $a − c > 1 / 2$). Since
one has
$1 + a − c ( 1 + | y | ) α > c ( 1 + a ) ( 1 + | y | ) α a − 1 / 2 − c ( 1 + | y | ) α = 1 + a a − 1 / 2 − 1 c ( 1 + | y | ) α = 3 / 2 a − 1 / 2 c ( 1 + | y | ) α = 3 c 2 a − 1 ( 1 + | y | ) α .$
Therefore,
$1 + | a − c ( 1 + | y | ) α | = 1 + a − c ( 1 + | y | ) α ≥ 3 c 2 a − 1 ( 1 + | y | ) α .$
(iii) Case $c ( 1 + | y | ) α − a > 1 / 2$. In this case,
$1 + | a − c ( 1 + | y | ) α | = 1 + c ( 1 + | y | ) α − a .$
If $a ≤ 1$,
$1 + c ( 1 + | y | ) α − a ≥ c ( 1 + | y | ) α .$
If $a > 1$,
From Equations (84) and (85), it follows that
$1 + | a − c ( 1 + | y | ) α | = 1 + c ( 1 + | y | ) α − a ≥ min { c , c − a + 1 } ( 1 + | y | ) α .$
Note that $c − a + 1 > 0$ since $a < c + a 0 < c + 1$.
Consequently, it has been proved that Equation (81) holds with
$c 9 = min 2 c 2 a + 1 , 3 c 2 a − 1 , min { c , c − a + 1 } if a − c > 1 / 2 , min 2 c 2 a + 1 , min { c , c − a + 1 } if a − c ≤ 1 / 2 .$
□
Next, we show that v satisfies Equation (70). Since
$z − 1 T ( z T − 1 ) − 1 B ( B − z ) − 1 = z − 2 ( z T − 1 + 1 ) ( z T − 1 ) − 1 ( B − z + z ) ( B − z ) − 1 = z − 2 { 1 + ( z T − 1 ) − 1 } { 1 + z ( B − z ) − 1 } = z − 2 + z − 1 ( B − z ) − 1 + z − 2 ( z T − 1 ) − 1 + z − 1 ( z T − 1 ) − 1 ( B − z ) − 1 ,$
we have
$T v = ( 2 π i ) − 1 ∫ Γ z − 1 T ( z T − 1 ) − 1 B ( B − z ) − 1 f d z = ( 2 π i ) − 1 ∫ Γ z − 2 f d z + ( 2 π i ) − 1 ∫ Γ z − 1 ( B − z ) − 1 f d z + ( 2 π i ) − 1 ∫ Γ z − 2 ( z T − 1 ) − 1 f d z + ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 ( B − z ) − 1 f d z .$
Clearly, $( 2 π i ) − 1 ∫ Γ z − 2 f d z = 0$. By assumption $M { ( z − a 0 ) M − L } − 1$ is holomorphc in $Re z ≥ a 0 − c ( 1 + | Im z | ) α$. Since
$z T − 1 = z M ( L + a 0 M ) − 1 − 1 = { z M − ( L + a 0 M ) } ( L + a 0 M ) − 1 = ( z M − L − a 0 M ) ( L + a 0 M ) − 1 , ( z T − 1 ) − 1 = ( L + a 0 M ) ( z M − L − a 0 M ) − 1 = ( L − ( z − a 0 ) M + z M ) { ( z − a 0 ) M − L } − 1 = { z M − ( ( z − a 0 ) M − L ) } { ( z − a 0 ) M − L } − 1 = z M { ( z − a 0 ) M − L } − 1 − 1 ,$
$( z T − 1 ) − 1$ is also holomorphic in $Re z ≥ a 0 − c ( 1 + | Im z | ) α$. If $z ∈ Γ$, then
$Re z = a − c ( 1 + | Im z | ) α > a 0 − c ( 1 + | Im z | ) α .$
Hence, $Γ$ lies in the region where $( z T − 1 ) − 1$ is holomorphc. If
$Re z ≥ a 0 − c ( 1 + | Im z | ) α ⟺ Re ( z − a 0 ) ≥ − c ( 1 + | Im ( z − a 0 ) | ) α ,$
then
$∥ M { ( z − a 0 ) M − L } − 1 ∥ L ( X ) ≤ c ˜ ( | z − a 0 | + 1 ) β .$
Hence,
Therefore,
Hence, one observes
$∫ Γ z − 2 ( z T − 1 ) − 1 d z = 0 .$
Let R be a large positive number. The set $Γ ∩ { | z | = R }$ consists of two points $z 1 , z 2$. Let $Γ R$ be the closed curve which consists of the part of $Γ$ in the disk $| z | ≤ R$ and the part of the circle $| z | = R$ in $Re z ≤ Re z 1 = Re z 2$. Since $( B − z ) − 1$ is holomorphic in the region $Re z < a 0$, which contains the closed set surrounded by $Γ R$,
$( 2 π i ) − 1 ∫ Γ R z − 1 ( B − z ) − 1 d z = B − 1 .$
Since $z k = a − c ( 1 + | Im z k | ) α + i Im z k$ and $| z k | = R$, $k = 1 , 2$, one has
$a − c ( 1 + | Im z k | ) α 2 + ( Im z k ) 2 = R 2 , k = 1 , 2 .$
This implies $| Im z k | = O ( R )$ as $R → ∞$, and hence $| Re z k | = O ( R α )$ as $R → ∞$, $k = 1 , 2$. Therefore, by virtue of Equation (69) for $z ∈ Γ R ∩ { | z | = R }$
$∥ ( B − z ) − 1 ∥ L ( E ) ≤ 1 a 0 − Re z = 1 a 0 + | Re z | ≤ 1 a 0 + | Re z k | = O ( R − α ) ( k = 1 , 2 ) ,$
as $R → ∞$. Letting $R → ∞$ in Equation (90), one observes
$( 2 π i ) − 1 ∫ Γ z − 1 ( B − z ) − 1 d z = B − 1 .$
From Equations (87), (89), and (91), one obtains
$T v = B − 1 f + ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 ( B − z ) − 1 f d z ,$
and hence
$B T v = f + ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z = f + v .$
Thus, we have established Equation (70).
Our next step is to establish the maximal regularity of solutions to Equation (56) or, equivalently, to Equation (58). By observing the resolvent identity
$( B + t ) − 1 ( B − z ) − 1 = − ( t + z ) − 1 { ( B + t ) − 1 − ( B − z ) − 1 } = − ( t + z ) − 1 ( B + t ) − 1 + ( t + z ) − 1 ( B − z ) − 1 ,$
we get, for $t > 0$,
$B ( B + t ) − 1 ( B − z ) − 1 = [ I − t ( B + t ) − 1 ] ( B − z ) − 1 = ( B − z ) − 1 − t ( B + t ) − 1 ( B − z ) − 1 = ( B − z ) − 1 − t [ − ( t + z ) − 1 ( B + t ) − 1 + ( t + z ) − 1 ( B − z ) − 1 ] = ( B − z ) − 1 + t ( t + z ) − 1 ( B + t ) − 1 − t ( t + z ) − 1 ( B − z ) − 1 .$
Hence,
$( B + t ) − 1 v = ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 B ( B + t ) − 1 ( B − z ) − 1 f d z = ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 ( B − z ) − 1 + t ( t + z ) − 1 ( B + t ) − 1 − t ( t + z ) − 1 ( B − z ) − 1 f d z = ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 ( B − z ) − 1 f d z + ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 t ( t + z ) − 1 ( B + t ) − 1 f d z − ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 t ( t + z ) − 1 ( B − z ) − 1 f d z .$
Therefore, we deduce
$B ( B + t ) − 1 v = v + ( 2 π i ) − 1 ∫ Γ z − 1 ( t + z ) − 1 ( z T − 1 ) − 1 d z t B ( B + t ) − 1 f − ( 2 π i ) − 1 ∫ Γ z − 1 t ( t + z ) − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z = v + J 1 ( f , t ) + J 2 ( f , t ) , t ∈ R + .$
One observes that $( z T − 1 ) − 1 = z M ( ( z − a 0 ) M − L ) − 1 − I$ is holomorphic in ${ z ; Re z ≥ a 0 − c ( 1 + | Im z | α ) }$. Hence, the integrand of $J 1 ( f , t )$ is holomorphic in ${ z ; Re z ≥ a − c ( 1 + | Im z | α ) }$ and its norm is $O ( | z | − 1 − β )$ as $| z | → ∞$ in view of Equation (88). Therefore, $J 1 ( f , t ) = 0$ for any $( t , f ) ∈ ( R + , E )$. Moreover, $J 2 ( f , t )$ satisfies
$J 2 ( f , t ) = − ( 2 π i ) − 1 ∫ Γ z − 1 ( t + z − z ) ( t + z ) − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z = − ( 2 π i ) − 1 ∫ Γ z − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z + ( 2 π i ) − 1 ∫ Γ ( t + z ) − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z = − v + ( 2 π i ) − 1 ∫ Γ ( t + z ) − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z .$
Thus, we have obtained
$B ( B + t ) − 1 v = ( 2 π i ) − 1 ∫ Γ ( t + z ) − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z .$
Setting $ω = θ + α + β − 2$, we can estimate $t ω ∥ B ( B + t ) − 1 v ∥$ taking into account the identity
$B ( B − z ) − 1 = B ( B + 1 + | y | ) − 1 + ( 1 + | y | + z ) ( B − z ) − 1 B ( B + 1 + | y | ) − 1 = 1 + ( 1 + | y | + z ) ( B − z ) − 1 B ( B + 1 + | y | ) − 1 , z ∈ Γ : z = a − c ( 1 + | y | ) α + i y , y ∈ R .$
Here, we show that the following inequality holds for $t > 0$, $y ∈ R$:
$| t + a − c ( 1 + | y | ) α + i y | ≥ c 10 ( t + 1 + | y | ) , c 10 = min { 1 , a − c } max { 2 2 , 2 c + 1 } .$
Proof.
(i) Case $| t + a − c ( 1 + | y | ) α | < ( t + a − c ) / 2$. Recalling $a > c$, $0 < α ≤ 1$, we deduce
$| t + a − c ( 1 + | y | ) α | < ( t + a − c ) / 2 ⟹ t + a − c ( 1 + | y | ) α < ( t + a − c ) / 2 ⟹ ( t + a + c ) / 2 < c ( 1 + | y | ) α ≤ c ( 1 + | y | ) ⟹ ( t + a − c ) / 2 < c | y | ⟹ t + a − c < 2 c | y | .$
Hence,
$min { 1 , a − c } ( t + 1 ) ≤ t + a − c < 2 c | y | .$
This implies
$min { 1 , a − c } ( t + 1 + | y | ) < 2 c | y | + | y | = ( 2 c + 1 ) | y | .$
Therefore,
$| t + a − c ( 1 + | y | ) α + i y | ≥ | y | > min { 1 , a − c } 2 c + 1 ( t + 1 + | y | ) .$
(ii) Case $| t + a − c ( 1 + | y | ) α | ≥ ( t + a − c ) / 2$. From
$| t + a − c ( 1 + | y | ) α + i y | 2 = ( t + a − c ( 1 + | y | ) α ) 2 + y 2 ≥ ( t + a − c ) 2 / 4 + y 2 > 1 8 ( 2 ( t + a − c ) 2 + 2 y 2 ) ≥ 1 8 ( t + a − c + | y | ) 2 ≥ 1 8 min { 1 , a − c } ( t + 1 + | y | ) 2$
it follows that
$| t + a − c ( 1 + | y | ) α + i y | ≥ min { 1 , a − c } 2 2 ( t + 1 + | y | ) .$
□
From Equations (72), (76), (78), (93), and (94), it follows that
$t ω ∥ B ( B + t ) − 1 v ∥ E = t ω ( 2 π i ) − 1 ∫ Γ ( t + z ) − 1 ( z T − 1 ) − 1 B ( B − z ) − 1 f d z E ≤ t ω 2 π ∫ Γ c 2 c 5 ( 1 + | y | ) 1 − β c 3 ( 1 + | y | ) 1 − α c 10 ( t + 1 + | y | ) ∥ B ( B + 1 + | y | ) − 1 f ∥ E | d z | = t ω 2 π c 2 c 5 c 3 c 10 ∫ Γ ( 1 + | y | ) 2 − α − β t + 1 + | y | ∥ B ( B + 1 + | y | ) − 1 f ∥ E | d z | ≤ c 11 t ω ∫ 0 ∞ ( 1 + y ) 2 − α − β t + 1 + y ∥ B ( B + 1 + y ) − 1 f ∥ E d y ,$
where
$c 11 = 1 π c 2 c 5 c 3 c 10 ( ( c α ) 2 + 1 ) 1 / 2 .$
One has
$t ω ∫ 0 ∞ ( 1 + y ) 2 − α − β t + 1 + y ∥ B ( B + 1 + y ) − 1 f ∥ E d y = t ω ∫ 1 ∞ y 3 − α − β − θ t + y y θ ∥ B ( B + y ) − 1 f ∥ E d y y = ∫ 1 ∞ ( t y − 1 ) α + β + θ − 2 t y − 1 + 1 y θ ∥ B ( B + y ) − 1 f ∥ E d y y = ∫ 1 ∞ g ( t y − 1 ) f 1 ( y ) d y y ,$
where $f 1 ( y ) = y θ ∥ B ( B + y ) − 1 f ∥ E$, $g ( y ) = y θ + α + β − 2 1 + y$. Applying Lemma 2 and using Equation (96) one obtains
$∫ 0 ∞ t ω ∫ 0 ∞ ( 1 + y ) 2 − α − β t + 1 + y ∥ B ( B + 1 + y ) − 1 f ∥ E d y p d t t 1 / p ≤ ∫ 0 ∞ ∫ 0 ∞ g ( t y − 1 ) f 1 ( y ) d y y p d t t 1 / p ≤ ∫ 0 ∞ g ( t ) d t t ∫ 0 ∞ f 1 ( y ) p d y y 1 / p .$
With the aid of the change of the independent variable $s = ( 1 + t ) − 1$, one observes
$∫ 0 ∞ g ( t ) d t t = ∫ 0 ∞ t θ + α + β − 2 1 + t d t t = ∫ 0 1 ( 1 − s ) θ + α + β − 3 s − θ − α − β + 2 d s = Γ ( θ + α + β − 2 ) Γ ( 3 − θ − α − β ) = Γ ( ω ) Γ ( 1 − ω ) ,$
and
$∫ 0 ∞ f 1 ( y ) p d y y = ∫ 0 ∞ y θ p ∥ B ( B + y ) − 1 f ∥ E p d y y .$
Hence, one obtains from Equation (97)
$∫ 0 ∞ t ω ∫ 0 ∞ ( 1 + y ) 2 − α − β t + 1 + y ∥ B ( B + 1 + y ) − 1 f ∥ E d y p d t t 1 / p ≤ Γ ( ω ) Γ ( 1 − ω ) ∫ 0 ∞ y θ p ∥ B ( B + y ) − 1 f ∥ E p d y y 1 / p .$
It follows from Equations (95) and (98) that
$∫ 0 ∞ t ω ∥ B ( B + t ) − 1 v ∥ E p d t t 1 / p ≤ ∫ 0 ∞ c 11 t ω ∫ 0 ∞ ( 1 + y ) 2 − α − β t + 1 + y ∥ B ( B + 1 + y ) − 1 f ∥ E d y p d t t 1 / p = c 11 ∫ 0 ∞ t ω ∫ 0 ∞ ( 1 + y ) 2 − α − β t + 1 + y ∥ B ( B + 1 + y ) − 1 f ∥ E d y p d t t 1 / p ≤ c 11 Γ ( ω ) Γ ( 1 − ω ) ∫ 0 ∞ y θ p ∥ B ( B + y ) − 1 f ∥ E p d y y 1 / p .$
Hence, $v = ( L + a 0 M ) u ∈ ( E , D ( B ) ) ω , p$. This implies $B T v = f + v ∈ ( E , D ( B ) ) ω , p$. Since $B T v = B M ( L + a 0 M ) − 1 v = B M u$, one has $B M u ∈ ( E , D ( B ) ) ω , p$:
$∫ 0 ∞ t ω p ∥ B ( B + t ) − 1 B M u ∥ p d t t < ∞ .$
Hence,
$∫ 0 ∞ t ω p ∥ B ( B + t ) − 1 M u ∥ p d t t = ∫ 0 ∞ t ω p ∥ B − 1 B ( B + t ) − 1 B M u ∥ p d t t ≤ ∥ B − 1 ∥ p ∫ 0 ∞ t ω p ∥ B ( B + t ) − 1 B M u ∥ p d t t < ∞ ,$
i.e., $M u ∈ ( E , D ( B ) ) ω , p$. In view of $v = ( L + a 0 M ) u ∈ ( E , D ( B ) ) ω , p$ it follows that $L u ∈ ( E , D ( B ) ) ω , p$. Therefore, $B X α ˜ M u = L u + f ∈ ( E , D ( B ) ) ω , p$. Thus, the following result is established:
Theorem 1.
Let $M , L$ be two closed linear operators in the complex Banach space E satisfying $( H )$, and let $0 < β ≤ α ≤ 1$. Let $B X$ be the operator defined by Equations (7) and (8) and $0 < α ˜ < 1$. Then, for all $f ∈ ( E , D ( B X α ˜ ) ) θ , p , 2 − α − β < θ < 1 , 1 < p ≤ ∞$ equation $B X α ˜ M u + L u = f$ admits a unique solution u. Moreover, $L u , B X α ˜ M u ∈ ( E , D ( B X α ˜ ) ) ω , p , ω = θ + α + β − 2$.

## 3. Conclusions

The fractional powers of the involved operator $B X$ are investigated in the space of continuous functions which do not necessarily vanish at the origin. This enables us to prove some previous results in the case where the involved operator $B X$ is not necessarily densely defined. Precisely, a fractional abstract Cauchy problem for possibly degenerate equations in Banach spaces is considered and refined.

## Author Contributions

All authors have equally contributed to this work. All authors wrote, read, and approved the final manuscript.

## Funding

This research received no external funding.

## Conflicts of Interest

The authors declare no conflict of interest.

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