# Resolvent Positive Operators and Positive Fractional Resolvent Families

Authors: Nan-Ding Li; Ru Liu; Miao Li
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This paper is concerned with positiveα-times resolvent families on an ordered Banach space E (with normal and generating cone), where 0<α≤2. We show that a closed and densely defined operator A on E generates a positive exponentially bounded α-times resolvent family for some 0<α<1 if and only if, for some ω∈ℝ, when λ>ω,λ∈ρA, Rλ,A≥0 and supλRλ,A:λ≥ω<∞. Moreover, we obtain that when 0<α<1, a positive exponentially bounded α-times resolvent family is always analytic. While A generates a positive α-times resolvent family for some 1<α≤2 if and only if the operator λα−1λα−A−1 is completely monotonic. By using such characterizations of positivity, we investigate the positivity-preserving of positive fractional resolvent family under positive perturbations. Some examples of positive solutions to fractional differential equations are presented to illustrate our results.


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## Body

## 1. Introduction

Many linear dynamical systems can be modelled as an abstract Cauchy problem:(1)u′t=Aut,t>0,u0=x0on a Banach space and then be treated by the theory of C0-semigroups [1–3] in a unified way. Motivated by problems in probability theory and PDEs, it is necessary to investigate Cauchy problems having positive solutions to each positive initial value. For this purpose, the theory of semigroups on a Banach space with an order structure was established. Feller [4] and Phillips [5] gave the first characteristics of generators of some special positive semigroups. After that, the theory of positive operators on ordered Banach spaces has been developed systematically during the 60s and 70s [6, 7]. This led to further progress in positive semigroups during the 80s, and these developments were recorded in the first monograph on positive semigroups [8]. For more recent work, we refer to [9].PositiveC0-semigroups are closely related to resolvent positive operators. If Tt is a positive C0-semigroup on an ordered Banach space, then its generator A is resolvent positive, i.e., Rλ,A≥0 when λ>ω for some ω∈ℝ. However, a densely defined resolvent positive operator does not always generate a C0-semigroup. The first example was given in [10]. Later, Arendt [11] proved that a densely defined resolvent positive operator generates a once integrated semigroup (see Lemma 16).During the last decades, there are considerable interests on the abstract fractional Cauchy problem of orderα:
(2)Dtαut=Aut,t>0,uk0=xk,xk∈X,k=0,1,⋯,m−1,where A is a closed, densely defined linear operator on X, Dtα is the Caputo derivative, and m=α is the smallest integer greater than or equal to α. It is known that (2) is well posed if and only if A generates an α-times resolvent family [12]. The class of fractional resolvent families is a special class of resolvent families introduced by Prüss [13]. A once resolvent family is actually a C0-semigroup, and a twice resolvent family is a cosine operator function. Bajlekova has proved that the generator of an exponentially bounded α-times resolvent family is bounded if α>2 [12] (Theorem 2.6). For this reason, we restrict ourselves only to the cases that α∈0,2. For the existence of solutions to the abstract Cauchy problem of fractional order, see for examples [12, 14–16]. There are also literatures devoted to abstract semilinear fractional Cauchy problems, see, e.g., [17, 18]. We refer to the survey paper [19] and the references therein for the basic theory of abstract fractional differential equations.It is natural to consider positive solutions to fractional differential equations. For example, the positive solutions to the time fractional diffusion equation:(3)Dtαut,x=Δut,x,t>0,x∈ℝN,were discussed in [20–22], and more general equation with fractional Laplacian
(4)Dtαut,x=−−Δβut,x,t>0,x∈ℝN,was studied in [23, 24]. The positivity of the fundamental solutions was derived from detailed analysis on some special functions including Mittag-Leffler functions, Mainardi functions, Bessel functions, and Fox H-functions. The investigation of positive solutions to these concrete fractional differential equations inspires us to study positive fractional resolvent families in a unified way. To our best knowledge, the positive solutions for abstract fractional Cauchy problems were discussed only in [25] on Banach lattices, under the assumption that the operator A generates a C0-semigroup.By the theory of fractional resolvent families and subordination principles developed in [12, 26, 27], we are able to obtain the positivity of solutions to fractional Cauchy problems via an operator theoretic approach. Our main result in this paper establishes the relations between positive fractional resolvent families and resolvent positive operators on an ordered Banach space E with generating and normal cone. More precisely, we show that (Theorem 15) if A is a closed densely defined operator on E, then A generates a positive exponentially bounded α-times resolvent family for some α∈0,1 if and only if
(5)Aisresolventpositive,supλRλ,A:λ≥ω<∞,for some ω∈ℝ. Condition (5) is not enough to guarantee that A generates a C0-semigroup, see Example 7, but it is sufficient for the generator of an α-times resolvent family for 0<α<1. Our proof is based on a result given by Arendt mentioned above. As a byproduct, we derive that a positive exponentially bounded α-times resolvent family with α∈0,1 is “automatically” analytic (Theorem 17). While for α∈1,2, A generates a positive α-times resolvent family if and only if λα−1λα−A−1 is completely monotonic. Based on these characterizations, we obtain the positivity-preserving of a positive fractional resolvent family under positive perturbations of relatively bounded operators, which generalizes those for positive semigroups [28, 29]. We would like to mention that the stability of positive fractional resolvent families was studied by one of the authors recently in [30].The paper is organized as follows. In Section2, we give some introductions on fractional calculus, fractional resolvent families, and ordered Banach spaces. Several examples of fractional differential equations with positive solutions are given in Section 3. Our main results on the relations between resolvent positive operators and positive resolvent families are presented in Section 4, and the positivity of fractional resolvent families is characterized there. We consider in Section 5 the positive perturbations of positive fractional resolvent families and apply our results to two examples related to Schrödinger operators.


## 2. Preliminaries

Let us first recall the basic definitions of fractional calculus (see [12, 31]). Let α>0, m=α, and I=0,T for some T>0. For f∈L1I, the fractional integral of order α>0 is defined by
(6)Jtαft≔gα∗ft=∫0tgαt−sfsdst>0,where
(7)gαt≔tα−1Γα,t>0,0,t≤0,and Γα is the gamma function. Set moreover g0t≔δt, the Dirac delta-function. The Caputo fractional derivative of order α>0 is defined by
(8)Dtαft≔Jm−αddtmft,if f∈L1I∩Cm−1I and gm−α∗f∈Wm,1I.The Mittag-Leffler functionEα,βz is defined by
(9)Eα,βz≔∑n=0∞znΓαn+β=12πi∫Haμα−βeμμα−zdμ,z∈ℂ,where α,β>0, Ha is the Hankel contour which starts and ends at −∞ and encircles the disc t≤z1/α counterclockwise. We use Eαt≔Eα,1t for short. The Mittag-Leffler function Eαt satisfies the fractional differential equation:
(10)DtαEαωtα=ωEαωtα.And the most important properties of these functions are associated with their Laplace integral:(11)∫0∞e−λttβ−1Eα,βstαdt=λα−βλα−s,Reλ>s1/α,and their asymptotic expansion as z⟶∞. For 0<α<2 and β>0,
(12)Eα,βz=1αz1−β/αexpz1/α+εα,βz,argz≤12απ,Eα,βz=εα,βz,arg−z<1−12απ,where
(13)εα,βz=−∑n=1N−1z−nΓβ−αn+Oz−N,as z⟶∞ and 2≤N∈ℕ.Another function is the scaled Wright-type functionψα,β with indexes 0<α<1,β≥0:
(14)ψα,βt,s≔∑n=0∞−1nsntβ−αn−1n!Γ−αn+β=12πi∫Haz−βetz−szαdz,t>0,s∈ℂ.The Wright functionψα,β satisfies ψα,βt,s≥0 for t,s>0,
(15)∫0∞e−λtψα,βt,sdt=λ−βe−λαs,s,λ>0,(16)∫0∞eλsψα,βt,sds=tα+β−1Eα,α+βλtα,t>0,λ∈ℂ.Moreover,(17)ψα,β+γt,s=gγ∗ψα,β·,st,t,s,γ>0.More details about positivity and integral representation of these two functions can be found in [32–34].Now, we turn to the abstract fractional Cauchy problem on a Banach spaceX. For a Banach space X, we denote by LX the Banach algebra of all bounded linear operators on X. Throughout this paper, we assume that A is a densely defined closed operator on X. We denote by Rλ,A≔λ−A−1 the resolvent of A at λ if λ∈ρA, the resolvent set of A. And we denote by sA≔supReλ:λ∈σA, where σA is the spectrum of A.Consider the Cauchy problem onX of order α∈0,2:
(18)Dtαut=Aut,t>0,u0=x0inadditionu′0=0ifα>1.The problem (18) is well posed if and only if the corresponding Volterra integral equation
(19)ut=x0+∫0tgαt−sAusdsis well posed in the sense of [13] (Definition 1.2). This leads to the definition of the α-times resolvent family [12].Definition 1.
Letα∈0,2. A function Sα·:0,+∞⟶LX is called an α-times resolvent family generated by A if the following conditions are satisfied:
(i)
Sαt is strongly continuous for t≥0 and Sα0=I(ii)
SαtDA⊂DA and ASαtx=SαtAx for all x∈DA, t≥0(iii)
Forx∈DA, the resolvent equation(20)Sαtx=x+gα∗SαtAx,holds for all t≥0Suppose thatA generates an α-times resolvent family Sαt. Then the Cauchy problem (18) is solvable, and its unique mild solution is given by ut=Sαtx0. And for α∈1,2, the unique mild solution to the Cauchy problem
(21)Dtαut=Aut,t>0,u0=x0,u′0=x1,is given by
(22)ut=Sαtx0+∫0tSατx1dτ.We call anα-times resolvent family Sαt exponentially bounded if there exist constants M≥1 and ω≥0 such that Sαt≤Meωt for all t≥0. If A generates an exponentially bounded α-times resolvent family, we will write for short A∈Cα.Sαt is called analytic if it admits an analytic extension to a sector Σθ0≔z∈ℂ:z≠0andargz<θ0 for some θ0∈0,π/2 and for each θ∈0,θ0 there exist constants Mθ≥1, ωθ≥0 such that Sαz≤MθeωθRezz∈Σθ. If A generates an analytic exponentially bounded α-times resolvent family, we will write for short A∈Aα.Lemma 2 ([12], Theorem 2.9).
Let0<α≤2. Then A∈Cα if and only if there are constants M≥1 and ω≥0 such that ωα,+∞⊂ρA, and there is a strongly continuous family Sαt⊂LX satisfying Sαt≤Meωt for all t≥0 and
(23)λα−1Rλα,Ax=∫0∞e−λtSαtxdt,λ>ω,x∈X.In [35] (Lemma 3), it was shown that if A generates an exponentially bounded α-times resolvent family, then A is densely defined.It is well known that the analyticity of a cosine operator function implies the boundedness of its generator. Therefore, we will restrict ourselves to the analyticity only when0<α<2. We need the following result on generating bounded analytic α-times resolvent families.Lemma 3 ([12], Corollary 2.17).
IfρA⊂λ:Reλ>0 and there exists a constant C such that
(24)Rλ,A≤CReλ,Reλ>0,then A∈Aα for any α∈0,1.The next lemma is concerned with the perturbation of fractional resolvent families [12] (Theorems 2.25 and 2.26).Lemma 4.
(i)
IfA∈Cα for some 1≤α≤2 and B∈LX, then A+B∈Cα(ii)
IfA∈Aα for some 0<α<2 and B is a closed linear operator satisfying DB⊃DA and(25)Bx≤aAx+bx,x∈DA
Then there existsδ>0 such that if 0≤a≤δ then A+B∈Aα.The following subordination principles are very important forC0-semigroups and fractional resolvent families. The first one comes from [36]; the second one comes from [12] (Theorems 3.1 and 3.3).Lemma 5 (subordination principle).
(i)
IfA generates a bounded C0-semigroup Tt, then for 0<α<1, −−Aα generates the analytic C0-semigroup Tαt given by(26)Tαt=∫0∞ψα,0s,tTsds,t>0(ii)
Let0<β<α≤2 and A generate an exponentially bounded α-times resolvent family Sαt. Then A generates the analytic exponentially bounded β-times resolvent family Sβt, which is given by(27)Sβt=∫0∞ψβα,1−βαt,sSαsds,t>0Finally, we give a brief introduction on ordered Banach spaces, as well as the positive operators, completely monotonic functions, and Bernstein functions on ordered Banach spaces. For details, we refer to [1, 8, 11, 37].Definition 6.
Let E be a real Banach space. By a positive cone in E, we understand a closed subset E+ of E satisfying E++E+⊂E+ and λE+⊂E+ for all λ>0. The order relations on E are defined by x≥y iff x−y∈E+. And moreover, the cone E+ is said to be generating if E=E+−E+; the cone E+ is said to be normal if there is some constant c such that z≤y≤x always implies y≤cmaxz,x.
Associated to each ordered Banach spaceE, there is an ordering on its dual space E′ with the dual cone E′+ defined by
(28)E′+=ω∈E′:ωx≥0forallx∈E+.Definition 7.
An operatorT∈LE is called positive if TE+⊂E+ and will be denoted by T≥0. A function f:Λ⟶E is called positive if fλ∈E+ for every λ∈Λ and will be denoted by f≥0.
A functionf∈C∞0,+∞,E is called completely monotonic if
(29)−1nfnλ≥0,∀λ>0.
A functionφ∈C∞0,+∞,E is called a Bernstein function if φλ≥0 for any λ>0 and φ′ is completely monotonic.
The exponential functione−axa>0 and the Mittag-Leffler function Eα,β−axa>0 with 0<α<1,α≤β, are scalar completely monotonic functions, and xα0<α<1 is a Bernstein function.Lemma 8.
Letf,g be completely monotonic functions on E and φ be a scalar Bernstein function. Then
(i)
f+g,f·g are completely monotonic(ii)
f∘φ is completely monotonic


## 3. Positive Solutions to Fractional Differential Equations

In this section, we give several examples of fractional differential equations with positive solutions. These are motivations for us to consider positive fractional resolvent families on an ordered Banach space and then investigate the positive solutions to fractional differential equations in a unified way.Example 1.
If0<α<1, the solution to the following scalar fractional differential equation
(30)Dtαut=λut,t>0,u0=u0
is given byut=u0Eαλtα. It is obvious that ut≥0 for t>0 if λ>0 and u0≥0. In case that λ<0, since the function Eα−t is completely monotonic for t≥0 when 0<α<1, we also have the solution ut positive if the initial data u0≥0.
Now, let1<α≤2. The solution to
(31)Dtαut=λut,t>0,u0=u0,u′0=u1,is given by
(32)ut=u0Eαλtα+u1Eα,2λtα.
As in the case of0<α<1, it is obvious that ut≥0 if λ>0 and u0,u1≥0. However, if λ<0, the Mittag-Leffler function Eαλt1<α<2 has a finite number of zeros on the positive real axis and thus is not positive [33, 38].Example 2.
LetE be an ordered Banach space with normal and generating cone and let A be a linear bounded operator on E. For 0<α≤2, consider the fractional differential equation
(33)Dtαut=Aut,t>0,u0=u0inadditionu′0=u1ifα>1.
The solution is given by(34)ut=EαtαAu0=∑n=0∞tαnΓαn+1Anu0,if 0<α≤1 or
(35)ut=EαtαAu0+Eα,2tαAu1=∑n=0∞tαnΓαn+1Anu0+∑n=0∞tαnΓαn+2Anu1,if 1<α≤2. If A≥0, the solution ut remains positive if both u0 and u1 are positive. If A=λI, a constant operator on E, the solution ut is positive if u0,u1≥0 when either 0<α≤1 and λ∈ℝ or 1<α≤2 and λ>0.Example 3.
Let0<α<1. Consider the following fractional transport equation
(36)Dtαut,x=−∂∂xut,x,t>0,u0,x=fx.
By the subordination principle (Lemma5(ii)), on X=L1ℝ, the solution is given by
(37)ut,x=∫0∞ψα,1−αt,sfx−sds;while on X=L1ℝ+, the solution is given by
(38)ut,x=ψα,1−αt,·∗fx=∫0xψα,1−αt,sfx−sds.
See also [32] (Example 12). Since ψα,βt,s≥0 when α∈0,1 and β>0, the solution ut,x is positive if fx≥0.Example 4.
Let0<α<1; consider the following fractional diffusion equation:
(39)Dtαut,x=Δut,x,t>0,x∈ℝN,u0,x=fxon LpℝN1≤p<∞. When α=1, this is the classical heat equation. It is well known that the Laplacian operator Δ generates a bounded C0-semigroup Tt on LpℝN given by
(40)Ttfx=14πtN/2∫ℝNe−x−y2/4tfydy.
Thus again by the subordination principle, the solution to the above fractional diffusion equation (38) is given by
(41)ut,x=∫0∞ψα,1−αt,s4πsN/2∫ℝNe−x−y2/4sfydyds,which is positive if f≥0.Example 5.
Let1<α≤2. Consider the following fractional wave equation:
(42)Dtαut,x=Δut,x,t>0,x∈ℝ,u0,x=fx,ut0,x=gxon Lpℝ1≤p<∞. This is exactly the wave equation on the real line; its solution is given by d’Alembert’s formula
(43)u2t,x=fx+t+fx−t2+12∫x−tx+tgsds,which is positive if both f and g are positive. Thus, by the subordination principle and (21), the solution to (41) is given by
(44)ut,x=12∫0∞ψα2,1−α2t,sfx+s+fx−sds+12∫0t∫0∞ψα2,1−α2τ,sgx+s+gx−sdsdτ=12∫0∞ψα2,1−α2t,sfx+s+fx−sds+12∫0∞ψα2,2−α2t,sgx+s+gx−sds,
where we have used (16). So, the solution to (41) is positive if both f and g are positive.Remark 9.
The fundamental solution for fractional diffusion equations (38) and (41) on ℝN was first given by Schneider and Wyss in [22], where it was also shown that the fundamental solution changes sign in dimension N≥2ifα>1. And in the case α>1 and N=1, the positivity of the fundamental solution was proved in [21].Example 6.
Consider the following fractional differential equation for fractional Laplacian:(45)Dtαut,x=−−Δβut,x,t>0,x∈ℝN,u0,x=fxon LpℝN1≤p<∞, where α,β∈0,1. Let Tt be the C0-semigroup generated by the Laplacian Δ as in (39), by Lemma 5(i), −−Δβ generates an analytic semigroup Tβt for every β∈0,1, which is given by
(46)Tβtfx=∫0∞ψβ,0τ,tTτfxdτ.
Therefore, the solution for (44) can be represented by
(47)ut,x=∫0∞ψα,1−αt,sTβsfxds=∫0∞∫0∞ψα,1−αt,sψβ,0τ,sTτfxdτds,which is positive if f≥0.Remark 10.
Kemppainen gave the fundamental solution for (44) in terms of the Fox H-functions in [23] and its positivity. It turns out that the fundamental solution is positive only if α∈0,1andβ∈0,1, orα∈1,2, α≤2β≤2andN=1.


## 4. Resolvent Positive Operators and Positive Fractional Resolvent Families

Throughout the rest of this paper, we assume thatE is an ordered Banach space with normal and generating cone E+. We will investigate in this section the relations between resolvent positive operators and the generators of positive fractional resolvent families.Definition 11.
An operatorA on E is called resolvent positive if there exists ω∈ℂ such that ω,∞⊂ρAandRλ,A≥0 (i.e., Rλ,AE+⊆E+) for all λ>ω.
Anα-times resolvent family Sαt is called positive if Sαt≥0 for t≥0.
We denote byA∈C+α to mean that A generates a positive exponentially bounded α-times resolvent family. And by A∈A+α, we mean that A generates an analytic exponentially bounded α-times resolvent family which is positive on 0,∞.Remark 12.
(i)
LetA be a resolvent positive operator. Then(48)sA=infw∈ℝ:w,∞⊂ρA,Rλ,A≥0forallλ>w,and the function R·,A is decreasing on sA,∞ [1, 11]
(ii)
IfA generates a positive α-times resolvent family Sαt, then (17) has a positive mild solution Sαtx0 for every initial value x0∈E+, and by (21), the mild solutions to (20) are also positive if x0,x1∈E+(iii)
By the subordination principle (Lemma5), if A∈C+α, then A∈A+β for every 0<β<α. More general results for the positivity-preserving concerning the fractional powers will be given in Proposition 18The positivity of anα-times resolvent family is characterized in the following result.Theorem 13.
Letα∈0,2 and A be the generator of an exponentially bounded α-times resolvent family Sαt on E.
(i)
Ifα∈1,2, then Sαt is positive for t≥0 if and only if(49)−1kdkdλkλα−1λα−A−1≥0,∀k∈ℕ∪0,for λ large enough
(ii)
Ifα∈0,1, then Sαt is positive for t≥0 if and only if Rλ,A≥0 for λ large enoughProof.
(i)
Suppose that∥Sαt∥≤Meωt for some constants M and ω, then by Lemma 2λα∈ρA for λ>ω and (22) holds; it then follows that(50)−1kdkdλkλα−1λα−A−1=∫0∞e−λttkSαtxdt.
The necessity of our assertion follows immediately from the above identity. The sufficiency follows from the Post-Widder inversion formula:(51)Sαtx=limk⟶∞−1k1k!ktk+1Sα∧kkt=limk⟶∞1k!ktk+1−1kdkdλkλα−1λα−A−1λ=k/tx(ii)
The necessity follows from (22) and the positivity of Sαt. It remains to show the sufficiency in the case α∈0,1. This follows from the asymptotic formula below (see [13] (Proposition 2.11)):(52)Sαt=limn⟶∞1n!∑k=1n+1bk,n+1αI−tnαA−kx,where bj,kα are given by the recurrence relations
(53)b1,1α=1,bj,kα=k−1−jαbj,k−1α+αj−1bj−1,k−1α,1≤j≤k,k=2,3,⋯,bj,kα=0,j>k,k=1,2,⋯,and the fact that the coefficients bj,kα≥0 when α∈0,1 and A is resolvent positive.Remark 14.
When1<α≤2, Rλ,A≥0 for λ large enough does not imply that Sαt≥0 for all t≥0. Since the scalar function cosax oscillates for every a∈ℝ, it is known that for 1<α<2, the function Eα−ax has finite zeros on the positive real axis if a>0 (see also Example 1) while λ+a−1 remains positive if λ is large enough.Some examples are given in [11] to illustrate that a densely defined resolvent positive operator A may not generate a C0-semigroup. To guarantee such operator generates a C0-semigroup, some additional conditions are needed. Among them are the interior of E+ which is nonempty, DA+ is cofinal in E+, the resolvent of A is bounded below, and so on. However, we will show in our main theorem that a densely defined resolvent positive operator always generates a fractional resolvent family if a uniform boundedness condition of its resolvents is satisfied, and such condition is also necessary for the operator to generate a fractional resolvent family.Theorem 15.
LetA be a closed densely defined operator on E. Then A∈C+α for some α∈0,1 if and only if A is resolvent positive and there exists some ω∈ℝ such that ω,∞⊂ρA, and
(54)supλRλ,A:λ≥ω<∞.
To prove the theorem, we need the following result from [11], which says that a densely defined resolvent positive operator can always generate a once integrated semigroup.Lemma 16.
LetA be a resolvent positive operator on an ordered Banach space X. IfDA is dense, then A generates a once integrated semigroup St such that
(55)Rλ,A=∫0∞e−λtdSt,λ>ω,when ω>sA. Moreover, St satisfies S0=0, 0≤Ss≤St when 0≤s≤t and limt⟶∞St=−A−1 if sA<0.Proof of Theorem15.
First, we show the sufficiency. Suppose thatA is a densely defined resolvent positive operator satisfying (53). Set B=A−ωI, then B is resolvent positive with sB<0. For λ>0, we have
(56)λRλ,B=λRλ+ω,A=λ+ωRλ+ω,A−ωRλ+ω,A;it then follows
(57)λRλ,B≤supμRμ,A:μ≥ω+Cω·Rω,Afor some constant C since Rλ+ω,A≤Rω,A (Remark 12(i)) and E+ is normal. Therefore, we obtain
(58)supλRλ,B:λ>0<∞.
By Lemma16, B generates integrated semigroup St such that
(59)Rλ,B=∫0∞e−λtdSt,Reλ>0,and St satisfies S0=0, 0≤Ss≤St when 0≤s≤t and limt⟶∞St=R0,B. It follows from (58) that for λ satisfying Reλ>0, x∈E+ and x′∈E+′(60)Rλ,Bx,x′≤∫0∞e−λtdStx,x′=∫0∞e−λtdStx,x′≤∫0∞e−ReλtdStx,x′=RReλ,Bx,x′.
SinceE+ is generating and normal, so is E+′ [39]. For every x′∈E′, there is a decomposition x′=x1′−x2′ with x1′,x2′∈E+′ satisfying maxx1′,x2′≤M′x′ for some constant M′≥1. Thus, for x∈E+,
(61)Rλ,Bx=supx′∈E′,∥x′∥≤1Rλ,Bx,x′≤2supy′∈E+′,∥y′∥≤M′Rλ,Bx,y′≤2supy′∈E+′,∥y′∥≤M′RReλ,Bx,y′≤2M′RReλ,Bx.
Again, sinceE+ is generating, there exists another constant M>1 such that for every x∈E, there is a decomposition x=x1−x2 with x1,x2∈E+ satisfying maxx1,x2≤Mx. Thus, for x∈E,
(62)Rλ,Bx≤2M′RReλ,Bx1+RReλ,Bx2≤4MM′RReλ,B·x.
Therefore, we obtain(63)sup∥ReλRλ,B∥:Reλ>0≤4MM′sup∥μRμ,B∥:μ>0<∞.
Thanks to this relation, we conclude by Lemma3 that B∈Aα for any α∈0,1. Since A=B+ωI, by Lemma 4, A also generates analytic α-times resolvent family Sαt. The positivity of Sαt follows from Theorem 13 since A is resolvent positive.
Conversely, ifA generates a positive exponentially bounded α-times resolvent family Sαt satisfying Sαt≤Meω0t for all t≥0, by Theorem 13, A is resolvent positive and by Lemma 2,
(64)λα−1Rλα,A=∫0∞e−λtSαtdt,λ>ω0.
It then follows(65)λα−1Rλα,A≤Mλ−ω0,λ>ω0.
Chooseω>sA such that ω1/α>ω0, then for λ≥ω1/α, we have λα∈ρA with
(66)λαRλα,A=λα−1λ−ω0Rλα,A+ω0λα−1Rλα,A≤M+ω0Mω1/α−ω0,and thus, (53) follows.It is remarkable that condition (53) is independent of α∈0,1. So, if A∈C+α for some 0<α<1, then A∈C+β for all 0<β<1. On the other hand, by the subordination principle (Lemma 5(ii)), A∈Cβ implies that A∈Aα for all 0<α<β≤2. Therefore, from the “positivity” of an exponentially bounded α-times resolvent family with α∈0,1, one can derive its “analyticity.”Theorem 17.
Letα∈0,1. If A∈C+α, then A∈A+β for all β∈0,1. In particular, A∈C+α implies A∈A+α.The next example provides an example of a resolvent positive operator satisfying (53). By Theorem 15, such operator can generate α-times resolvent family for every α∈0,1; however, it is not the generator of a C0-semigroup.Example 7.
LetE=C00,1=f∈C0,1:f0=0, 0<θ<1 and A:DA⟶E defined by
(67)Afx=−f′x+θxfx,x∈0,1,0,x=0,with DA=f∈C10,1:f′0=f0=0. Then the following assertions hold:
(i)
A is resolvent positive and satisfies(68)supμRμ,A:μ≥0≤11−θ(ii)
A generates an α-times resolvent family for every α∈0,1 but not a C0-semigroup(iii)
A generates a β-times integrated semigroup Tβt for every β>0 which is given by(69)Tβtfx=∫0x∧tt−sβ−1Γβxθx−s1−θfx−sds.
The proof of (i) was given in [11]. By Theorem 15, we obtain that A generates an α-times resolvent family for any α∈0,1. But as showed in [11], A is not a generator of a C0-semigroup. This is (ii).
Finally, if we defineTβt by (68), it is not hard to show that Tβt is bounded on E. By a routine calculation, one can verify that Tβt is the β-times integrated semigroup generated by A.By the theory of fractional powers developed by Martínez and Sanz [40], we are able to show that the fractional power of a resolvent positive operator is also resolvent positive.Proposition 18.
Suppose thatA is a resolvent positive operator satisfying 0,∞⊂ρA and supλRλ,A:λ>0<∞. Then for 0<β≤1, the operator −−Aβis resolvent positive and
(70)supλRλ,−−Aβ:λ>0≤supλRλ,A:λ>0.
If in additionDA is dense, then so is D−−Aβ and −−Aβ∈A+α for all α∈0,1.Proof.
By [40] (Proposition 5.3.2), we have the uniform boundedness of the resolvent of the operator −x−Aβ and the representation of its resolvents:
(71)Rλ,−−Aβ=sinβππ∫0∞μβλ2+2λμβcosβπ+μ2βRμ,Adμ,λ>0;from this formula, one can derive the positivity of Rλ,−−Aβ from the fact that A is resolvent positive. The last assertion follows immediately from Theorems 15 and 17.Remark 19.
(i)
From Proposition18, one immediately has the positivity of the solution to (44) with positive initial data(ii)
The subordination relations between the fractional resolvent families generated byA and −−Aβ were established in [27], while the positivity of the subordination function was discussed in [41]
At the end of this section, we consider the following fractional inhomogeneous equation onE with α∈0,1:
(72)Dtαut=Aut+ft,t∈0,τ,u0=x0,where x0∈E and f∈C0,τ,E. Recall that u is a mild solution of (71) if u∈C0,τ,E, gα∗ut∈DA for t∈0,τ and
(73)ut=x0+gα∗ft+Agα∗ut,t∈0,τ.Proposition 20.
LetA be a densely defined resolvent positive operator satisfying (53), x0∈Eandf∈C0,τ,E.
(i)
Ifut is a mild solution of (71), then(74)ut=Sαtx0+ddtgα∗Sα∗ft,t∈0,τ,where Sαt is the α-times resolvent family generated by A(ii)
Iff=f0+∫0tf′sds with f′∈L10,τ,E, then (71) has a unique mild solution(75)ut=Sαtx0+gα∗Sαtf0+gα∗Sα∗f′t,t∈0,τ.
If in additionx0,f0∈E+ and f′∈L10,τ,E+, then ut≥0 for all t∈0,τ.
(iii)
Iff=f0+g1−α∗ht with h∈L10,τ,E, then (71) has a unique mild solution(76)ut=Sαtx0+gα∗Sαtf0+Sα∗ht,t∈0,τ.
If in additionx0,f0∈E+ and h∈L10,τ,E+, then ut≥0 for all t∈0,τ.Proof.
Our assumptions imply thatA generates a positive α-times resolvent family Sαt for every α∈0,1 by Theorem 15. The assertion (i) was proved in [27]. The first half of (ii) follows from [13] (Proposition 1.2); the second half follows from the positivity of Sαt, x0, f0, and f′.
Now, we prove (iii). Iff=f0+g1−α∗ht and ut is a mild solution to (71), we have by (i)
(77)ut=Sαtx0+ddtgα∗Sα∗ft=Sαtx0+ddtg1+α∗Sαtf0+gα∗Sα∗g1−α∗ht=Sαtx0+gα∗Sαtf0+Sα∗ht;this is (75). Conversely, if ut is given by the above formula, then ut is continuous by [1] (Proposition 1.3.4), and it is obvious that gα∗ut∈DA with
(78)Agα∗ut=Agα∗Sαtx0+Agα∗gα∗Sαtf0+Agα∗Sα∗ht=Sαtx0−x0+gα∗Sαtf0−g1+αtf0+Sα∗ht−g1∗ht=ut−x0−g1+αtf0−g1−α∗gα∗ht=ut−x0−gα∗ft.
Thus,ut is a mild solution of (71).


## 5. Positive Perturbations of Positive Fractional Resolvent Families

In this section, we consider the positive perturbations of positive fractional resolvent families. Let us begin with bounded perturbations. IfA∈Cα for some α∈1,2 on a Banach space and B is a bounded operator, we know by Lemma 4 that A+B∈Cα. We show in the following result that the positivity can be preserved if the perturbation is positive.Theorem 21.
Letα∈1,2 and A∈C+α. If B is a bounded positive operator on E, then A+B∈C+α.Proof.
LetSαt be the α-times resolvent family generated by A. By Lemma 4, A+B generates an α-times resolvent family Sαt;A+B, and it is given by a modified Dyson-Phillips series:
(79)Sαt;A+B=∑n=0∞Sα,nt,with Sα,0t=Sαt, and
(80)Sα,nt=∫0tgα−1∗Sαt−sBSα,n−1sds,n=1,2,⋯.
From the above representation formula, one can conclude thatSαt;A+B≥0 by induction.For unbounded perturbations, we first consider a perturbation of Miyadera-Voigt-type [42].Theorem 22.
Letα∈1,2. Suppose that A generates a positive α-times resolvent family Sαt satisfying Sαt≤Meωt, andB:DA⟶E is a positive operator. If there are constants μ>ω and γ∈0,1 such that
(81)∫0∞e−μrB∫0rgα−1r−sSαsxdsdr≤γx,x∈DA,then A+B∈C+α.Proof.
It follows from [42] (Theorem 3.1) that A+B generates an α-times resolvent family Sαt;A+B. It remains to show the positivity of Sαt;A+B. By Theorem 13, it is enough to show that
(82)−1kdkdλkλα−1Rλα,A+B≥0,k=0,1,2,⋯,for λ large enough.
We first show that our assumptions implyBRλα,A≤γ<1 for λ≥μ. Indeed, since ωα,∞⊂ρA by Lemma 2, the operator BRλα,A:E⟶E is bounded and positive for λ≥μ. For x∈DA, using (80), we get
(83)BRλα,Ax=B∫0∞e−λtgα−1∗Sαtxdt≤∫0∞e−λtBgα−1∗Sαtxdt≤∫0∞e−μtBgα−1∗Sαtxdt≤γx.
The positivity ofSαt implies that λα−1Rλα,A is completely monotonic by Theorem 13. Since Rλα,A=λ1−α·λα−1Rλα,A, by Lemma 8, Rλα,A is completely monotonic and so is BRλα,A since B:DA⟶E is positive. Thus, (81) follows from Lemma 8, the identity
(84)λα−1Rλα,A+B=λα−1Rλα,AI−BRλα,A−1=λα−1Rλα,A∑n=0∞BRλα,An,the complete monotonicity of λα−1Rλα,A and Rλα,A, and the absolute convergence of the series ∑n=0∞BRλα,An since BRλα,A<1 for λ≥μ.Next, we consider the relative bounded perturbations. The following lemma is needed.Lemma 23 (see [11]).
LetA be a resolvent positive operator and B:DA⟶E a positive operator. If rBRλ,A<1 for some λ>sA, then A+B with domain DA is a resolvent positive operator and sA+B<λ. Moreover, if supμRμ,A:μ≥λ<∞, then also supμRμ,A+B:μ≥λ<∞.Theorem 24.
Letα∈0,2. Suppose that A∈C+α, B:DB⟶E is a positive and closed linear operator such that DB⊃DA and there are constants a,b≥0 such that
(85)Bx≤aAx+bx,x∈DA.
If any of the following conditions is satisfied:(i)
Ifα∈0,1(ii)
Ifα∈1,2 and Sαt is analytic
then there existsδ>0, such that if 0≤a≤δ, then A+B∈A+α.Proof.
SinceA∈C+α, it follows from Theorem 15 and its proof that there exists ω>0 such that
(86)M≔supλRλ,A:λ≥ω<∞.
By (84), for x∈X and λ>ω,
(87)BRλ,Ax≤aARλ,Ax+bRλ,Ax=aλRλ,Ax−x+bRλ,Ax≤aM+1x+bMλx.
If we choose0<δ1<1/1+M, then BRλ,A<1 when λ≥ω1 for some ω1≥ω.
Now, suppose thatα∈0,1. Using Lemma 23, we also have
(88)supλRλ,A+B:λ≥ω1<∞.
Applying Theorem15 to the operator A+B, we obtain that A+B∈C+α, which implies that A+B∈A+α by Theorem 17.
If (ii) holds, thenSαt is analytic. By Lemma 4, there is some δ2>0 such that if 0≤a≤δ2, then A+B∈Aα. Choose δ=minδ1,δ2, then when 0≤a≤δ2, A+B∈Aα and BRλ,A<1 when λ≥ω1. The positivity can be proved as in the proof of Theorem 22.Remark 25.
(i)
(84) can be replaced by the following condition: the operator B is closed with DB⊃DA and(89)Bx≤aAx+bx,x∈K,where K is a core of the generator A, i.e., x,Ax;x∈K¯=x,Ax;x∈DA. In fact, for any x∈DA, there exists a sequence of elements xn∈K, such that xn,Axn⟶x,Ax, n⟶∞. Thus, xn⟶x and Axn⟶Ax. Since both xn and Axn are Cauchy sequences, by (88), Bxn is also a Cauchy sequence. By the closedness of B, Bxn⟶Bx. For any xn∈K, we have by (88) that Bxn≤aAxn+bxn. By letting n⟶∞, one gets (84) for every x∈DA(ii)
It is known that ifA is closed and B satisfies (84) with a<1, then A+B,DA is a closed operator [2] (Ch. III, Lemma 2.4)(iii)
Assume thatA∈C+α for some α∈0,1 and B:DA⟶E is a positive operator. If for any ε>0, there exists bε>0 such that(90)Bx≤εAx+bεx,∀x∈DA
Then, it follows from Theorems15 and 17 that A+B∈A+α.
We end this paper with two examples related to Schrödinger operators [43, 44] to illustrate our perturbation theory.Example 8.
LetE=L1ℝN. Consider the following fractional Schrödinger equation:
(91)Dtαut,x=Δut,x+Vxut,x,t>0,x∈ℝN,u0,x=u0x,where Δ is the Laplacian and V is a potential function such that Vx≥0. Let A be the closure of the Laplacian Δ defined on C0∞ℝN. As a perturbation, we choose
(92)Bfx≔Vxfxforf∈DB≔f∈E:V·f·∈E.
It is clear thatB is positive and closed on E. Moreover, we assume that the potential function V∈KN, the Kato class, that is,
(93)limδ⟶0supx∈ℝN∫x−y≤δx−y2−NVydy=0,N≥3,limδ⟶0supx∈ℝ2∫x−y≤δlnx−y−1Vydy=0,N=2,supx∈ℝ∫x−y≤1Vydy<∞,N=1.
It was proved in [43] (Theorem 4.14) that for every ε>0, there exists Cε>0 such that
(94)Vf1≤εΔf1+Cεf1,∀f∈DA,if and only if V∈KN. By applying our Theorem 24, we conclude that if V∈KN, then A+B=Δ+V∈A+α for α∈0,1 on E. Therefore, for u0x≥0, there is a positive mild solution in E=L1ℝN to the fractional Schrödinger equation (90) if α∈0,1.
This is also true ifE=LpℝN for 1<p<∞, since the Schrödinger operator Δ+V also generates C0-semigroup on LpℝN.
Next, we consider the positive solution for (90) in the case of 1<α<2. By the analysis in Example 5, this is possible only if N=1. Since the Laplacian Δ generates a cosine operator function,
(95)Ctux=ux+t+ux−t2,on E=L1ℝ, Δ∈C+α for 0<α<2 by the subordination principle. Provided the potential function V≥0 and V∈K1, then it follows from (93) and Theorem 24 that Δ+V∈A+α.Remark 26.
By using the higher-order Kato classKβ introduced by Davies and Hinz [45], similar results can be obtained for the fractional Laplacian operator with positive potentials in K2β0<β<1:
(96)Dtαut,x=−−Δβut,x+Vxut,x,t>0,x∈ℝN,u0,x=u0x.
While the higher-order Kato classK2β is defined by
(97)Kβ≔V:limδ⟶0supx∈ℝN∫x−y≤δx−y2β−NVydy=0,when 0<2β<N. Proposition 18 provides one way to show −−Δβ∈A+α for α∈0,1. To apply our perturbation theory, we can use the following fact from [46]: V∈K2β if and only if for every ε>0 there exists Cε>0 such that
(98)Vf1≤ε−Δβf1+Cεf1,∀f∈D−Δβ.It is known [43] that for N≥2,
(99)V:supx∈ℝN∫x−y≤1Vypdy<∞⊂KN⊂V:supx∈ℝN∫x−y≤1Vydy<∞,if p>N/2. And if Vx=fx is a spherically symmetric function when N≥3, then V∈KN if and only if
(100)supx≥2∫x−y≤1fydy<∞,∫01rfrdr<∞.For examples, the solution for (90) is positivity-preserving if the potential V is of the following form:
(i)
x−a with 0≤a<2 if N≥3, in particular the Coulomb potential x−1 when N=3(ii)
x−2logx−b with b>1 if N≥3(iii)
∑i<j1/xi−xj on ℝ3N with xi∈ℝ3 for i=1,⋯,N, the N-body HamiltonianAt the borderlineLN/2⊄KN, however, by using the Sobolev estimates, we can obtain the positivity-preserving of (90) with potentials in LN/2.Example 9.
We assumeE=LpℝN, 3≤N∈ℤ+, 1≤p<N/2andα∈0,1. Let A and B be defined as in Example 8 with V∈LN/2ℝN and Vx≥0 for all x∈ℝN. We will show that B satisfies (89). Let Vn be the truncation of V:
(101)Vnx=Vx,ifVx≤n,0,otherwise.
By Hölder’s inequality, we have for everyf∈C0∞ℝN,
(102)Vfp≤V−Vnfp+Vnfp≤V−VnN/2fr+Vn∞fp,where 2/N+1/r=1/p and p<r; furthermore, by Sobolev’s inequality,
(103)fr≤CΔfp,∀f∈C0∞ℝN.
Thus, we obtain(104)Vfp≤CV−VnN/2Δfp+Vn∞fp,∀f∈C0∞ℝN.
Takingn large enough, one gets (89). Thus, A+B∈A+α.



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