# On the Differentiability

Authors: Marat V. Markin
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

For the evolution equationy′(t)=Ay(t) with a scalar type spectral operator A in a Banach space, conditions on A are found that are necessary and sufficient for all weak solutions of the equation on [0,∞) to be strongly infinite differentiable on [0,∞) or [0,∞). Certain effects of smoothness improvement of the weak solutions are analyzed.


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## Body

## 1. Introduction

Consider the evolution equation(1.1)y′(t)=Ay(t)
with a scalar type  spectral operatorA in a complex Banach space X.Following [1], we understand by a weak solution of equation (1.1) on an interval [0,T) (0<T≤∞) such a vector function  y:[0,T)→X that (1)
y(·) is strongly continuous on [0,T);(2)
for anyg*∈D(A*),
(1.2)ddt〈y(t),g*〉=〈y(t),A*g*〉,t∈[0,T).(D(·) is the domain of an operator, A* is the operator adjoint to A, and 〈·,·〉 is the pairing between the space X and its dual X*).As was shown [2], Theorem 4.2, the general weak  solution of (1.1) on [0,T) (0<T≤∞) is as  follows: (1.3)y(t)=etAf,t∈[0,T),f∈⋂0≤t<TD(etA),
the operator exponentials etA, 0≤t<T, defined in the sense of the operational calculus for scalar type spectral operators (see Section 2).Here, we are to find conditions onA that are necessary and sufficient for all weak solutions of (1.1) on [0,∞) to be strongly infinite differentiable on [0,∞) or (0,∞).This goal attained; all the principal results of paper [3] and the corresponding ones of [4] obtain their natural generalizations.


## 2. Preliminaries

Henceforth, unless specified otherwise, letA be a scalar type spectral operator in a complex Banach space X with a norm ∥·∥, EA(·) its spectral measure (the resolution of the identity), and σ(A) the operator's spectrum, with the latter being the support for the former.Observe that, in a Hilbert space, thescalar type spectral operators are those similar to the normal ones [5].For scalar type spectral operators, there has been developed anoperational calculus for Borel measurable functions defined on σ(A) [6, 7]. With F(·) being such a function, a new scalar type spectral operator(2.1)F(A)=def∫σ(A)F(λ)dEA(λ)
is defined as follows: (2.2)F(A)f=deflim⁡n→∞Fn(A)f,f∈D(F(A)),D(F(A))=def{f∈X∣lim⁡n→∞Fn(A)fexists},
where (2.3)Fn(⋅)=defF(⋅)χ{λ∈σ(A)∣|F(λ)|≤n}(⋅),n=1,2,…,
(χα(·) is the characteristic function of a set α) and (2.4)Fn(A)=def∫σ(A)Fn(λ)dEA(λ),n=1,2,…,
are bounded scalar type spectral operators on X defined in the same manner as for normal operators (see, e.g., [8, 9]).In particular,(2.5)A=∫σ(A)λdEA(λ).
The properties of the spectral measure, EA(·), and the operational calculus underlying our discourse are exhaustively delineated in [6, 7]. We will outline here a few noteworthy facts.Due to its strong countable additivity, the spectral measureEA(·) is bounded [10], that is, there is an M>0 such that(2.6)‖EA(α)‖≤M,α∈B
(ℬ is the σ-algebra of Borel sets in the complex plane ℂ).Observe that∥·∥ is used in (2.6) to designate the norm in the space of bounded linear operators on X. We will adhere to this rather common economy of symbols adopting the same notation for the norm in the dual space X* as well.For anyf∈X and g*∈X*, let v(f,g*,·) be the total variation of the complex-valued measure 〈EA(·)f,g*〉 on ℬ.As we discussed in [4, 11], with F(·) being an arbitrary Borel measurable function on ℂ, for any f∈D(F(A)), g*∈X*, and δ∈ℬ,(2.7)∫δ|F(λ)|dv(f,g*,λ)≤4M‖EA(δ)F(A)f‖‖g*‖,
where M>0 is the constant from (2.6).In particular, forF(λ)=1, λ∈ℂ, and δ=ℂ (or any Borel set containing σ(A)), we have(2.8)v(f,g*,δ)≤4M‖f‖‖g*‖,f∈X,g*∈X*.
Further, for a Borel measurable nonnegative function F(·) on ℂ, a δ∈ℬ, and a sequence {Δn}n=1∞ of pairwise disjoint Borel sets in ℂ,(2.9)∫δF(λ)dv(EA(∪n=1∞Δn)f,g*,λ)=∑n=1∞∫δ∩ΔnF(λ)dv(EA(Δn)f,g*,λ),f∈X,g*∈X*.
Indeed, since for the spectral measure [6, 7] (2.10)EA(α)EA(β)=EA(α∩β),α,β∈B,
for the total variation, we have(2.11)v(EA(β)f,g*,α)=v(f,g*,α∩β),α,β∈B.
Whence, due to nonnegativity of F(·) [12], (2.12)∫δF(λ)dv(EA(∪n=1∞Δn)f,g*,λ)=∫δ∩∪n=1∞ΔnF(λ)dv(f,g*,λ)=∑n=1∞∫δ∩ΔnF(λ)dv(f,g*,λ)=∑n=1∞∫δ∩ΔnF(λ)dv(EA(Δn)f,g*,λ).
We shall need the following regions in the complex plane: (2.13)Lb+=def{λ∈C∣Reλ≥max⁡(0,b+ln⁡|Im⁡λ|)},Lb-,b+=def{λ∈C∣Reλ≤min⁡(0,-b-ln⁡|Im⁡λ|)orReλ≥max⁡(0,b+ln⁡|Im⁡λ|)},
where b+ and b- are positive constants.The termsspectral measure and operational calculus frequently referred to will be abbreviated to s.m. and o.c., respectively.


## 3. Differentiability of a Particular Weak Solution

Proposition 3.1.
Letn=1,2,… and I be a subinterval of an interval [0,T)(0<T≤∞). A weak solution y(·) of (1.1) on [0,T) is n times strongly differentiable on I if and only if
(3.1)y(t)∈D(An),t∈I,
in which case,
(3.2)y(k)(t)=Aky(t),k=1,2,…,n,t∈I.Proof.
“Only if” part.
Letn=1,2,… and suppose that a weak solution y(·) of (1.1) on [0,T) is n times strongly differentiable on a subinterval I⊆[0,T).
Then for anyg*∈D(A*),
(3.3)〈y′(t),g*〉=ddt〈y(t),g*〉=〈y(t),A*g*〉,t∈I.
Hence, by the closedness of the operator A (cf. [1]),
(3.4)y(t)∈D(A),y′(t)=Ay(t),t∈I.
This concludes proving the “only if” part for n=1, in which case, as is to be noted, the subinterval I can shrink to a single pointt∈[0,T).
Letn≥2 and let the interval I be not a singleton. Then differentiating (3.4) at an arbitrary t∈I, we obtain
(3.5)y′′(t)=lim⁡Δt→0y′(t+Δt)-y′(t)Δt=lim⁡Δt→0Ay(t+Δt)-y(t)Δt,
with the increments Δt being such that t+Δ∈I.
Since(3.6)lim⁡Δt→0y(t+Δt)-y(t)Δt=y′(t),t∈I,
by the closedness of A, we infer that
(3.7)y′(t)∈D(A),y′′(t)=Ay′(t),t∈I.
Considering (3.4) and (3.7),
(3.8)y(t)∈D(A2),y′′(t)=A2y(t),t∈I.
Continuing inductively in this manner, we arrive at
(3.9)y(t)∈D(Ak),y(k)(t)=Aky(t),k=1,2,…,n,t∈I.
“If” part.
Lety(·) be a weak solution of (1.1) on an interval [0,T) (0<T≤∞) such that for an n=1,2,… and a subinterval I⊆[0,T),
(3.10)y(t)∈D(An),t∈I.
As was discussed (see Section 1),
(3.11)y(t)=etAf,t∈[0,T),
with some f∈⋂0≤t<TD(etA).
The fact thatetAf∈D(An), t∈I, implies by the properties of the o.c. and [2], Proposition 3.1, that, for any g*∈X*,
(3.12)∫σ(A)|λ|ketReλdv(f,g*,λ)<∞,k=1,…,n,t∈I.
Given a k=1,…,n and an arbitrary t∈I, let us choose a segment [a,b]⊂I (a<b) so that t is its midpoint if 0<t<T or a=0 if t=0. For increments Δt such that a≤t+Δt≤b and any g*∈X*, we have
(3.13)|〈Ak-1y(t+Δt)-Ak-1y(t)Δt-Aky(t),g*〉|bythepropertiesoftheo.c.;=|〈∫σ(A)[λk-1e(t+Δt)λ-λk-1etλΔt-λketλ]dEA(λ)f,g*〉|bythepropertiesoftheo.c.;=|∫σ(A)[λk-1e(t+Δt)λ-λk-1etλΔt-λketλ]d〈EA(λ)f,g*〉|≤∫σ(A)|λk-1e(t+Δt)λ-λk-1etλΔt-λketλ|dv(f,g*,λ)bytheLebesgueDominatedConvergenceTheorem;⟶0asΔt⟶0.
Indeed, for any k=1,…,n and an arbitrary λ∈σ(A),
(3.14)|λk-1e(t+Δt)λ-λk-1etλΔt-λketλ|≤|λk-1e(t+Δt)λ-λk-1etλΔt|+|λketλ|bytheTotalChangeTheorem;≤2|λ|kmax⁡a≤s≤besReλ≤2{|λ|keaReλifReλ<0,|λ|kebReλifReλ≥0,by(3.12),consideringthata,b∈I;∈L1(σ(A),v(f,g*,⋅)),|λk-1e(t+Δt)λ-λk-1etλΔt-λketλ|⟶0asΔt→0.
Therefore,  for anyg*∈X*,
(3.15)dkdtk〈y(t),g*〉=〈Aky(t),g*〉,k=1,…,n,t∈I.
For Δm:={λ∈ℂ∣|λ|≤m}, m=1,2,…, let us fix an arbitrary k=1,…,n and consider the sequence of vector functions
(3.16)ym(t):=EA(Δm)Aky(t)=EA(Δm)AketAf,m=1,2,…,t∈I.
Since f∈D(AketA), t∈I, by the properties of the o.c.,
(3.17)EA(Δm)AketAf=[AEA(Δm)]ketAEA(Δm)f,m=1,2,…,t∈I.
Due to the boundedness of Δm, m=1,2,…, by the properties of the o.c., AEA(Δm), m=1,2,…, is a bounded linear operator on X (∥AEA(Δm)∥≤4Mm, m=1,2,…, [7]). Hence, the vector functions
(3.18)ym(t)=EA(Δm)AketAf,m=1,2,…,t∈I,
are strongly continuous on I.
For an arbitrary segment[a,b]⊆I, we have
(3.19)sup⁡a≤t≤b‖AketAf-EA(Δm)AketAf‖bythepropertiesoftheo.c.;=sup⁡a≤t≤b‖∫{λ∈σ(A)∣|λ|>m}λketλdEA(λ)f‖asfollowsfromtheHahn-BanachTheorem;=sup⁡a≤t≤bsup⁡{g*∈X*∣∥g*∥=1}|〈∫{λ∈σ(A)∣|λ|>m}λketλdEA(λ)f,g*〉|bythepropertiesoftheo.c.;=sup⁡a≤t≤bsup⁡{g*∈X*∣∥g*∥=1}|∫{λ∈σ(A)∣|λ|>m}λketλd〈EA(λ)f,g*〉|≤sup⁡a≤t≤bsup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|>m}|λ|ketReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}sup⁡a≤t≤b[∫{λ∈σ(A)∣|λ|>m,Reλ≤0}|λ|ketReλdv(f,g*,λ)+∫{λ∈σ(A)∣|λ|>m,Reλ>0}|λ|ketReλdv(f,g*,λ)]≤sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|>m,Reλ≤0}|λ|keaReλdv(f,g*,λ)+sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|>m,Reλ>0}|λ|mebReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|>m}|λ|keaReλdv(f,g*,λ)+sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|>m}|λ|kebReλdv(f,g*,λ)by(2.7);≤4M[sup⁡{g*∈X*∣∥g*∥=1}‖EA({λ∈σ(A)∣|λ|>m})AkeaAf‖‖g*‖+sup⁡{g*∈X*∣∥g*∥=1}‖EA({λ∈σ(A)∣|λ|>m})AkebAf‖‖g*‖]=4M[‖EA({λ∈σ(A)∣|λ|>m})AkeaAf‖+‖EA({λ∈σ(A)∣|λ|>m})AkebAf‖]bythestrongcontinuityofthes.m.;⟶0asm⟶∞.
Hence, for any k=1,…,n, the vector function AketAf, t∈I, is strongly continuous on I, being the uniform limit of the sequence of strongly continuous on I vector functions {ym(·)}m=1∞ on any segment [a,b]⊆I.
Let us fix an arbitrarya∈I and integrate (3.15) for k=1 between a and an arbitrary t∈I. Considering the strong continuity of AetAf, t∈I, we have
(3.20)〈etAf-eaAf,g*〉=〈∫atAesAfds,g*〉,g*∈X*.
Whence, as follows from the Hahn-Banach Theorem,
(3.21)etAf-eaAf=∫atAesAfds,t∈I.
By the strong continuity of AetAf, t∈I,
(3.22)ddtetAf=AetAf,t∈I.
Consequently, by (3.15), for n=2,
(3.23)ddt〈ddtetAf,g*〉=〈A2etAf,g*〉,t∈I.
Whence, analogously,
(3.24)d2dt2etAf=A2etAf,t∈I.
Continuing inductively in this manner, we infer that, for any k=1,…,n,
(3.25)dkdtketAf=AketAf,t∈I.Corollary 3.2.
LetI be a subinterval of an interval [0,T)(0<T≤∞). A weak solution y(·) of (1.1) on [0,T) is strongly infinite differentiable on I if and only if
(3.26)y(t)∈C∞(A)=def⋂n=1∞D(An),t∈I,
in which case,
(3.27)y(n)(t)=Any(t),n=1,2,…,t∈I.Thus, we have obtained generalizations of Proposition 4.1 and Corollary 4.1 of [3], respectively.


## 4. Differentiability of Weak Solutions

Theorem 4.1.
Every weak solution of (1.1) on [0,∞) is strongly infinite differentiable on [0,∞) if and only if there is a b+>0 such that the set σ(A)∖ℒb+ is bounded.Proof.
“If” part.
Letb+>0 be such that the set σ(A)∖ℒb+ is bounded and y(·) a weak solution of (1.1) on [0,∞). Then (see Section 1)
(4.1)y(t)=etAf,0≤t<∞,
with some f∈⋂0≤t<∞D(etA).
For anyn=1,2,…, t≥0 and an arbitrary g*∈X*,
(4.2)∫σ(A)|λ|netReλdv(f,g*,λ)=∫σ(A)∖Lb+|λ|netReλdv(f,g*,λ)+∫σ(A)∩Lb+|λ|netReλdv(f,g*,λ)<∞.
Indeed, the former integral is finite due to the boundedness of the set σ(A)∖ℒb+, the finiteness of the measure v(f,g*,·) and the continuity of the integrated function on ℂ.
For the latter one, we have(4.3)∫σ(A)∩Lb+|λ|netReλdv(f,g*,λ)≤∫σ(A)∩Lb+[|Reλ|+|Im⁡⁡λ|]netReλdv(f,g*,λ)forλ∈σ(A)∩Lb+,Reλ≥0,|Im⁡⁡λ|≤eb+-1Reλ;≤∫σ(A)∩Lb+[Reλ+eb+-1Reλ]netReλdv(f,g*,λ)sincex≤ex,x≥0;≤∫σ(A)∩Lb+[b+eb+-1Reλ+eb+-1Reλ]netReλdv(f,g*,λ)=[b++1]n∫σ(A)e[nb+-1+t]Reλdv(f,g*,λ)sincef∈⋂0≤t<∞D(etA),by[2],Proposition3.1;<∞.
Further, for any n=1,2,… and t≥0,
(4.4)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∖Lb+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)+sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∩Lb+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)⟶0asm⟶∞.
Indeed, since, due to the boundedness of the set σ(A)∖ℒb+, the set {λ∈σ(A)∖ℒb+∣|λ|netReλ>m} is void for all sufficiently large m's,
(4.5)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∖Lb+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)⟶0asm→∞.
In addition to this
(4.6)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∩Lb+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)analogouslyto(4.3);≤sup⁡{g*∈X*∣∥g*∥=1}[b++1]n∫{λ∈σ(A)∩Lb+∣|λ|netReλ>m}e[nb+-1+t]Reλdv(f,g*,λ)sincef∈⋂0≤t<∞D(etA),by(2.7);≤[b++1]nsup⁡{g*∈X*∣∥g*∥=1}4M‖EA({λ∈σ(A)∩Lb+∣|λ|netReλ>m})e(nb+-1+t)Af‖‖g*‖=4M[b++1]n‖EA({λ∈σ(A)∩Lb+∣|λ|netReλ>m})e(nb+-1+t)Af‖bythestrongcontinuityofthes.m.;⟶0asm⟶∞.
By the properties of theo.c. and [2], Proposition 3.1, (4.2) and (4.4) imply that
(4.7)y(t)=etAf∈C∞(A),0≤t<∞.
Then, by Corollary3.2, y(·) is strongly infinite differentiable on [0,∞).
“Only if” part.
We will prove this part by contrapositive.
Assume that, for anyb+>0, the set σ(A)∖ℒb+ is unbounded.
In particular, for anyn=1,2,…, unbounded is the set σ(A)∖ℒ(2n)-1.
Hence, we can choose a sequence of points in the complex plane{λn}n=1∞ as follows:
(4.8)λn∈σ(A),n=1,2,…,Reλn<max⁡(0,(2n)-1ln⁡|Im⁡λ|),n=1,2,…,λ0:=0,|λn|>max⁡[n4,|λn-1|],n=1,2,….
The latter implies in particular that the pointsλn are distinct (λi≠λj, i≠j).
Since, for anyn=1,2,…, the set
(4.9){λ∈C∣Reλ<max⁡(0,(2n)-1ln⁡|Im⁡λ|),|λ|>max⁡[n4,|λn-1|]}
is open in ℂ, there exists an ɛn>0 such that this set along with the point λn contains the open disk of radius ɛn centered at λn(4.10)Δn:={λ∈C∣|λ-λn|<ɛn}.
Then, for any λ∈Δn, n=1,2,…,
(4.11)Reλ<max⁡(0,(2n)-1ln⁡|Im⁡λ|),|λ|>max⁡[n4,|λn-1|].
Further, since the points λn, n=1,2,…, are distinct, we can regard the radii of the disks, ɛn, n=1,2,…, to be small enough so that
(4.12)0<ɛn<1n,n=1,2,…,Δi∩Δj=∅,i≠j(i.e.,thedisksarepairwisedisjoint).
Whence, by the properties of the s.m.,
(4.13)EA(Δi)EA(Δj)=0,i≠j.
Observe also, that the subspaces EA(Δn)X, n=1,2,…, are nontrivial since Δn∩σ(A)≠∅, with Δn being an open set.
By choosing a unit vectoren∈EA(Δn)X (∥en∥=1), n=1,2,…, we obtain a vector sequence such that
(4.14)EA(Δi)ej=δijei,i,j=1,2,…
(δij is the Kronecker delta symbol).
As is readily verified, (4.14) implies that the vectors {e1,e2,…} are linearly independent.
Moreover, there is anɛ>0 such that
(4.15)dn:=dist⁡(en,span⁡({ek∣k=1,2,…,k≠n}))≥ɛ,n=1,2,….
Indeed, the opposite implies the existence of a subsequence {dn(k)}k=1∞ such that
(4.16)dn(k)⟶0ask⟶∞.
Then, for any k=1,2,…, by selecting a vector fn(k)∈span({ej∣j=1,2,…,j≠n(k)}) such that ∥en(k)-fn(k)∥<dn(k)+1/k, considering (4.14) and (2.6), we arrive at the contradiction:
(4.17)1=‖en(k)‖=‖EA(Δn(k))(en(k)-fn(k))‖≤M‖en(k)-fn(k)‖⟶0ask⟶∞.
As follows from the Hahn-Banach Theorem, for any n=1,2,…, there is an en*∈X* such that
(4.18)‖en*‖=1,〈ei,ej*〉=δijdi,i,j=1,2,….
Concerning the sequence of the real parts, {Reλn}n=1∞, there are two possibilities: it is either bounded or unbounded above.
Suppose that the sequence{Reλn}n=1∞ is bounded above, that is, there is such an ω>0 that
(4.19)Reλn≤ω,n=1,2,….
Let
(4.20)f:=∑n=1∞1n2en.
By (4.14),
(4.21)EA(∪n=1∞Δn)f=f,EA(Δn)f=1n2en,n=1,2,….
For any t≥0 and an arbitrary g*∈X*,
(4.22)∫σ(A)etReλdv(f,g*,λ)by(4.21);=∫σ(A)etReλdv(EA(∪n=1∞Δn)f,g*,λ)by(2.9);=∑n=1∞∫ΔnetReλdv(EA(Δn)f,g*,λ)by(4.21);=∑n=1∞1n2∫ΔnetReλdv(en,g*,λ)forλ∈Δn,by(4.12),and(4.19),Reλ=Reλn+(Reλ-Reλn)≤Reλn+|λ-λn|≤ω+ɛn≤ω+1;≤et(ω+1)∑n=1∞1n2v(en,g*,Δn)by(2.8);≤et(ω+1)∑n=1∞1n24M‖en‖‖g*‖=4Met(ω+1)‖g*‖∑n=1∞1n2<∞.
Similarly, for any t≥0,
(4.23)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣etReλ>n}etReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}et(ω+1)∑n=1∞1n2∫Δn∩{λ∈σ(A)∣etReλ>n}1dv(en,g*,λ)by(2.9)and(4.21);=et(ω+1)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣etReλ>n}1v(f,g*,λ)by(2.7);≤et(ω+1)sup⁡{g*∈X*∣∥g*∥=1}4M‖EA({λ∈σ(A)∣etReλ>n})f‖‖g*‖=4Met(ω+1)‖EA({λ∈σ(A)∣etReλ>n})f‖bythestrongcontinuityofthes.m.;⟶0asn⟶∞.
By [2], Proposition 3.1, (4.22) and (4.23) imply that f∈⋂0≤t<∞D(etA).
Therefore (see Section1), y(t):=etAf, 0≤t<∞, is a weak solution of (1.1) on [0,∞).
Let(4.24)h*:=∑n=1∞1n2en*∈X*,
(cf. (4.18)).
Considering (4.18) and (4.15), we have
(4.25)〈en,h*〉=dnn2≥ɛn2,n=1,2,….
Similarly to (4.22),
(4.26)∫σ(A)|λ|dv(f,h*,λ)=∑n=1∞1n2∫Δn|λ|dv(en,h*,λ)forλ∈Δn,by(4.11),|λ|≥n4;≥∑n=1∞n2v(en,h*,Δn)≥∑n=1∞n2|〈EA(Δn)en,h*〉|by(4.14)and(4.25);≥∑n=1∞n2ɛn2=∞.
Whence, by [2], Proposition 3.1, y(0)=f∉D(A).
Consequently, by Proposition3.1, the weak solution y(t)=etAf, 0≤t<∞, of (1.1) on [0,∞) is not strongly differentiable at 0.
Now, assume that the sequence{Reλn}n=1∞ is unbounded above.
Therefore, there is a subsequence{Reλn(k)}k=1∞ such that
(4.27)Reλn(k)≥k,k=1,2,….
Then the vector
(4.28)f:=∑k=1∞e-n(k)Reλn(k)en(k)
is well defined.
By (4.14),
(4.29)EA(∪k=1∞Δn(k))f=f,EA(Δn(k))f=e-n(k)Reλn(k)en(k),k=1,2,….
For any t≥0 and an arbitrary g*∈X*, similar to (4.22),
(4.30)∫σ(A)etReλdv(f,g*,λ)=∑k=1∞e-n(k)Reλn(k)∫Δn(k)etReλdv(en(k),g*,λ)forλ∈Δn(k),by(4.12),Reλ=Reλn(k)+(Reλ-Reλn(k))≤Reλn(k)+|λ-λn(k)|≤Reλn(k)+1;≤et∑k=1∞e-[n(k)-t]Reλn(k)v(en(k),g*,Δn(k))by(2.8);≤et∑k=1∞e-[n(k)-t]Reλn(k)4M‖en(k)‖‖g*‖=4Met‖g*‖∑k=1∞e-[n(k)-t]Reλn(k)bytheComparisonTest;<∞.
Indeed, for all sufficiently large k's, n(k)-t≥1 and by (4.27),
(4.31)e-[n(k)-t]Reλn(k)≤e-k.
Analogously, for an arbitrary t≥0,
(4.32)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣etReλ>n}etReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}et∑k=1∞e-[n(k)-t]Reλn(k)∫Δn(k)∩{λ∈σ(A)∣etReλ>n}1dv(en(k),g*,λ)=etsup⁡{g*∈X*∣∥g*∥=1}∑k=1∞e-[(n(k)/2)-t]Reλn(k)e-(n(k)/2)Reλ(k)∫Δn(k)∩{λ∈σ(A)∣etReλ>n}1dv(en(k),g*,λ)dueto(4.27),thereisanL>0suchthate-[(n(k)/2)-t]Reλn(k)≤L,k=1,2,…;≤Letsup⁡{g*∈X*∣∥g*∥=1}∑k=1∞e-(n(k)/2)Reλn(k)∫Δn(k)∩{λ∈σ(A)∣etReλ>n}1dv(en(k),g*,λ)forh:=∑k=1∞e-(n(k)/2)Reλn(k)en(k),by(4.14)and(2.9);=Let∫{λ∈σ(A)∣etReλ>n}1dv(h,g*,λ)by(2.7);≤Letsup⁡{g*∈∣∥g*∥=1}4M‖EA({λ∈σ(A)∣etReλ>n})h‖‖g*‖=4LMet‖EA({λ∈σ(A)∣etReλ>n})h‖bythestrongcontinuityofthes.m.;⟶0asn⟶∞.
From (4.30) and (4.32), by [2], Proposition 3.1, we infer that f∈⋂0≤t<∞D(etA).
Therefore (see Section1), y(t):=etAf, 0≤t<∞, is a weak solution of (1.1) on [0,∞).
For anyλ∈Δn(k), k=1,2,…, by (4.11), (4.12), and (4.27),
(4.33)Reλ=Reλn(k)-(Reλn(k)-Reλ)≥Reλn(k)-|Reλn(k)-Reλ|≥Reλn(k)-ɛn(k)≥Reλn(k)-1n(k)≥k-1≥0,Reλ<max⁡(0,(2n(k))-1ln⁡|Im⁡λ|).
Hence, for λ∈Δn(k), k=1,2,…,
(4.34)|λ|≥|Im⁡λ|≥e2n(k)Reλ≥e2n(k)(Reλn(k)-1/n(k)).
Using this  estimate, we obtain
(4.35)∫σ(A)|λ|dv(f,h*,λ)=∑k=1∞e-n(k)Reλn(k)∫Δn(k)|λ|dv(en(k),h*,λ)≥∑k=1∞e-n(k)Reλn(k)e2n(k)(Reλn(k)-1/n(k))v(en(k),h*,Δn(k))=e-2∑k=1∞en(k)Reλn(k)|〈EA(Δn(k))en(k),h*〉|by(4.14),(4.25),and(4.27);≥e-2ɛ∑k=1∞en(k)n(k)2=∞.
Whence, by [2], Proposition 3.1, y(0)=f∉D(A).
Therefore, by Proposition3.1, the weak solution y(t)=etAf, 0≤t<∞, of (1.1) on [0,∞) is not strongly differentiable at 0.
With every possibility concerning{Reλn}n=1∞ considered, we infer  that the opposite to the assumption that, for a certain b+>0, the set σ(A)∖ℒb+ is bounded, allows to single out a weak solution of (1.1) on [0,∞) that is not strongly differentiable at 0, much less strongly infinite differentiable on [0,∞).
Thus, the “only if” part has been proved by contrapositive.Theorem 5.1 of [3] has been generalized.Theorem 4.2.
Every weak solution of (1.1) on [0,∞) is strongly infinite differentiable on (0,∞) if and only if there is a b+>0 such that, for any b->0, the set σ(A)∖ℒb-,b+ is bounded.Proof.
“If” part.
Letb+>0 be such that, for an arbitrary b->0, the set σ(A)∖ℒb-,b+ is bounded.
Lety(·) be a weak solution of (1.1) on [0,∞). Then (see Section 1)
(4.36)y(t)=etAf,0≤t<∞,
with some f∈⋂0≤t<∞D(etA).
For anyn=1,2,…, t>0, and an arbitrary g*∈X*,
(4.37)∫σ(A)|λ|netReλdv(f,g*,λ)=∫σ(A)∖Lb-,b+|λ|netReλdv(f,g*,λ)+∫{λ∈σ(A)∩Lb-,b+∣Reλ≥0}|λ|netReλdv(f,g*,λ)+∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}|λ|netReλdv(f,g*,λ)<∞.
The first integral in this sum is finite due the boundedness of the setσ(A)∖ℒb-,b+, the finiteness of the measure v(f,g*,·), and the continuity of the integrated function on ℂ.
The finiteness of the second integral is proved in absolutely the same manner as for the corresponding integral in the proof of the “if” part of Theorem4.1.
Finally for the third one, considering thatb->0 is arbitrary and setting b-:=nt-1, we have
(4.38)∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}|λ|netReλdv(f,g*,λ)≤∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}[|Reλ|+|Im⁡λ|]netReλdv(f,g*,λ)forλ∈σ(A)∩Lb-,b+withReλ<0,|Im⁡λ|≤eb--1[-Reλ];≤∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}[-Reλ+eb--1[-Reλ]]netReλdv(f,g*,λ)sincex≤ex,x≥0;≤∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}[b-eb--1[-Reλ]+eb--1[-Reλ]]netReλdv(f,g*,λ)=[b-+1]n∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}e[nb--1-t][-Reλ]dv(f,g*,λ)recallthatb-:=nt-1;=[b-+1]n∫{λ∈σ(A)∩Lb-,b+∣Reλ<0}1dv(f,g*,λ)≤[b-+1]nv(f,g*,σ(A))by(2.8);≤4M[b-+1]n‖f‖‖g*‖<∞.
Further, for any n=1,2,… and t>0,
(4.39)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∖Lb-,b+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)+sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∩Lb-,b+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)⟶0asm⟶∞.
Indeed, since, due to the boundedness of the set σ(A)∖ℒb-,b+, the set {λ∈σ(A)∖ℒb-,b+∣|λ|netReλ>m} is void for all sufficiently large m's,
(4.40)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∖Lb-,b+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)⟶0asm⟶∞.
Similarly to the “if” part of Theorem 4.1 and (4.38), we have
(4.41)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∩Lb-,b+∣|λ|netReλ>m}|λ|netReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}[b++1]n∫{λ∈σ(A)∩Lb-,b+∣Reλ≥0,|λ|netReλ>m}e[nb+-1+t]Reλdv(f,g*,λ)+sup⁡{g*∈X*∣∥g*∥=1}[b-+1]n∫{λ∈σ(A)∩Lb-,b+∣Reλ<0,|λ|netReλ>m}e[nb--1-t][-Reλ]dv(f,g*,λ)withb-:=nt-1;=sup⁡{g*∈X*∣∥g*∥=1}[b++1]n∫{λ∈σ(A)∩Lb-,b+∣Reλ≥0,|λ|netReλ>m}e[nb+-1+t]Reλdv(f,g*,λ)+sup⁡{g*∈X*∣∥g*∥=1}[b-+1]n∫{λ∈σ(A)∩Lb-,b+∣Reλ<0,|λ|netReλ>m}1dv(f,g*,λ)sincef∈⋂0≤t<∞D(etA),by(2.7);≤[b++1]nsup⁡{g*∈X*∣∥g*∥=1}4M‖EA({λ∈σ(A)∩Lb-,b+∣Reλ≥0,|λ|netReλ>m})e[nb+-1+t]Af‖‖g*‖+[b-+1]nsup⁡{g*∈X*∣∥g*∥=1}4M‖EA({λ∈σ(A)∩Lb-,b+∣Reλ<0,|λ|netReλ>m})f‖‖g*‖=4M[b++1]n‖EA({λ∈σ(A)∩Lb-,b+∣Reλ≥0,|λ|netReλ>m})e[nb+-1+t]Af‖+4M[b-+1]n‖EA({λ∈σ(A)∩Lb-,b+∣Reλ<0,|λ|netReλ>m})f‖bythestrongcontinuityofthes.m.;⟶0asm⟶∞.
By the properties of the o.c. and [2], Proposition 3.1, (4.37) and (4.39) imply that
(4.42)y(t)=etAf∈C∞(A),0<t<∞.
Whence, by Corollary 3.2, y(·) is strongly infinite differentiable on (0,∞).
“Only if” part.
As well as in Theorem4.1, we will prove this part by contrapositive.
Thus, we assume that, for anyb+>0, there is such a b->0 that the set σ(A)∖ℒb-,b+ is unbounded.
Let us show that this assumption can even be strengthened. To wit: there is such ab->0 that, for any b+>0, the set σ(A)∖ℒb-,b+ is unbounded.
Indeed, there are two possibilities:(1)
for a certainb->0, the set {λ∈σ(A)∣-b-ln|Imλ|<Reλ≤0} is unbounded;(2)
for anyb->0, the set {λ∈σ(A)∣-b-ln|Imλ|<Reλ≤0} is bounded.
In the first case, the setσ(A)∖ℒb-,b+ is unbounded with the very b->0, for which the set {λ∈σ(A)∣-b-ln|Imλ|<Reλ≤0} is unbounded, and an arbitrary b+>0.
In the second case, based on the premise we infer that, for anyb+>0, the set {λ∈σ(A)∣0<Reλ<b+ln|Imλ|} is unbounded. Then so is the set σ(A)∖ℒb-,b+ for any b->0 and b+>0.
Thus, let us fix ab->0 such that the set σ(A)∖ℒb-,b+ is unbounded for an arbitrary b+>0.
In particular, for anyn=1,2,…, the set σ(A)∖ℒb-,(2n)-1 is unbounded.
Hence, we can select a sequence of points in the complex plane{λn}n=1∞ in the following manner:
(4.43)λn∈σ(A),n=1,2,…,min⁡(0,-b-ln⁡|Im⁡λ|)<Reλn<max⁡(0,(2n)-1ln⁡|Im⁡λ|)n=1,2,…,λ0:=0,|λn|>max⁡[n4,|λn-1|],n=1,2,….
The latter, in particular, implies that the points λn are distinct.
Since, for anyn=1,2,…, the set
(4.44){λ∈C∣min⁡(0,-b-ln⁡|Im⁡λ|)<Reλ<max⁡(0,(2n)-1ln⁡|Im⁡λ|),|λ|>max⁡[n4,|λn-1|]}
is open in ℂ, there exists such an ɛn>0 that this set, along with the point λn, contains the open disk of radius ɛn centered at λn(4.45)Δn:={λ∈C∣|λ-λn|<ɛn}.
Hence, for any λ∈Δn, n=1,2,…,
(4.46)min⁡(0,-b-ln⁡|Im⁡λ|)<Reλ<max⁡(0,(2n)-1ln⁡|Im⁡λ|),|λ|>max⁡[n4,|λn-1|].
Since the points λn are distinct, we can regard the radii of the disks, ɛn, to be small enough so that
(4.47)0<ɛn<1n,n=1,2,…,Δi∩Δj=∅,i≠j.
By the properties of thes.m., the latter implies
(4.48)EA(Δi)EA(Δj)=0,i≠j.
Observe that the subspaces EA(Δn)X, n=1,2,…, are nontrivial since Δn∩σ(A)≠∅, Δn being an open set.
By choosing a unit vectoren∈EA(Δn)X (∥en∥=1), n=1,2,…, we obtain a vector sequence such that
(4.49)EA(Δi)ej=δijei.
In the same manner as in the proof of Theorem 4.1, one can show that there is an ɛ>0 such that
(4.50)dn:=dist⁡(en,span⁡({ek∣k=1,2,…,k≠n}))≥ɛ,n=1,2,….
As follows from the Hahn-Banach Theorem, for any n=1,2,…, there is an en*∈X* such that
(4.51)‖en*‖=1,〈ei,ej*〉=δijdi.
Concerning the sequence of the real parts, {Reλn}n=1∞, there are two possibilities: it is either bounded or unbounded.
First, assume that the sequence{Reλn}n=1∞ is bounded, that is, there is an ω>0 such that
(4.52)|Reλn|≤ω,n=1,2,….
Let
(4.53)f:=∑n=1∞1n2en.
By (4.49),
(4.54)EA(∪n=1∞Δn)f=f,EA(Δn)f=1n2en,n=1,2,….
In absolutely the same fashion as it was done in the case of bounded above sequence {Reλn}n=1∞ in the proof of the “only if” part of Theorem 4.1, it is shown that f∈⋂0≤t<∞D(etA).
Therefore (see Section1), y(t):=etAf, 0≤t<∞, is a weak solution of (1.1) on [0,∞).
Let(4.55)h*:=∑n=1∞1n2en*∈X*
(cf. (4.51)).
Taking (4.50) and (4.51) into account, we have
(4.56)〈en,h*〉=dnn2≥ɛn2,n=1,2,….
Thus,
(4.57)∫σ(A)|λ|eReλdv(f,h*,λ)considering(4.54),by(2.9);=∑n=1∞1n2∫Δn|λ|eReλdv(en,h*,λ)forλ∈Δn,by(4.46),|λ|≥n4;forλ∈Δn,by(4.47)and(4.52),Reλ=Reλn-(Reλn-Reλ)≥Reλn-|Reλn-Reλ|≥-ω-ɛn≥-ω-1;≥e-(ω+1)∑n=1∞n2v(en,h*,Δn)≥e-(ω+1)∑n=1∞n2|〈EA(Δn)en,h*〉|by(4.49)and(4.56);≥e-(ω+1)∑n=1∞n2ɛn2=∞.
Whence, by the properties of the o.c. and [2], Proposition 3.1, y(1)=eAf∉D(A).
Consequently, by Proposition3.1, the weak solution y(t)=etAf,0≤t<∞, of (1.1) on [0,∞) is not once strongly differentiable on (0,∞).
Now, assume that the sequence{Reλn}n=1∞ is unbounded. Therefore, there is a subsequence {Reλn(k)}k=1∞ such that
(4.58)Reλn(k)⟶∞orReλn(k)⟶-∞ask⟶∞.
Suppose that Reλn(k)→∞ as k→∞.
Without restricting generality, we can regard that(4.59)Reλn(k)≥k,k=1,2,….
Considering this case just like the analogous one in the proof of the “only if” part of Theorem 4.1, one can show that the vector
(4.60)f:=∑k=1∞e-n(k)Reλn(k)en(k)
belongs to ⋂0≤t<∞D(etA) and, hence, (see Section 1) y(t):=etAf, 0≤t<∞, is a weak solution of (1.1) on [0,∞).
For anyλ∈Δn(k), k=1,2,…, by (4.46), (4.47), and (4.59),
(4.61)Reλ=Reλn(k)-(Reλn(k)-Reλ)≥Reλn(k)-|Reλn(k)-Reλ|≥Reλn(k)-ɛn(k)≥Reλn(k)-1n(k)≥k-1≥0,Reλ<max⁡(0,(2n(k))-1ln⁡|Im⁡λ|),
Hence, for λ∈Δn(k), k=1,2,…,
(4.62)|λ|≥|Im⁡λ|≥e2n(k)Reλ≥e2n(k)(Reλn(k)-1/n(k)).
Using this estimate, we have
(4.63)∫σ(A)|λ|eReλdv(f,h*,λ)considering(4.49),by(2.9);=∑k=1∞e-n(k)Reλn(k)∫Δn(k)|λ|eReλdv(en(k),h*,λ)≥∑k=1∞e-n(k)Reλn(k)e2n(k)(Reλn(k)-1/n(k))v(en(k),h*,Δn(k))=e-2∑k=1∞en(k)Reλn(k)|〈EA(Δn(k))en(k),h*〉|by(4.49),(4.56),and(4.59);≥e-2ɛ∑k=1∞en(k)n(k)2=∞.
Whence, by the properties of the o.c. and [2], Proposition 3.1, y(1)=eAf∉D(A).
Therefore, by Proposition3.1, the weak solution y(t)=etAf, 0≤t<∞, of (1.1) on [0,∞) is not once strongly differentiable on (0,∞).
Now, suppose thatReλn(k)→-∞ as k→∞.
Without restricting generality, we can regard that(4.64)Reλn(k)≤-k,k=1,2,….
Let
(4.65)f:=∑k=1∞1k2en(k).
For any t≥0 and an arbitrary g*∈X*,
(4.66)∫σ(A)etReλdv(f,g*,λ)considering(4.49),by(2.9);=∑k=1∞1k2∫Δn(k)etReλdv(en(k),g*,λ)forλ∈Δn(k),by(4.42)and(4.64),Reλ=Reλn(k)+(Reλ-Reλn(k))≤Reλn(k)+|λ-λn(k)|≤-k+1≤0;≤∑k=1∞1k2v(en(k),g*,Δn(k))by(2.8);≤∑k=1∞1k24M‖en(k)‖‖g*‖=4M‖g*‖∑k=1∞1k2<∞.
Analogously, for an arbitrary t≥0,
(4.67)sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣etReλ>n}etReλdv(f,g*,λ)≤sup⁡{g*∈X*∣∥g*∥=1}∑k=1∞1k2∫Δn(k)∩{λ∈σ(A)∣etReλ>n}1dv(en(k),g*,λ)=sup⁡{g*∈X*∣∥g*∥=1}∫{λ∈σ(A)∣etReλ>n}1dv(f,g*,λ)by(2.7);≤sup⁡{g*∈X*∣∥g*∥=1}4M‖EA({λ∈σ(A)∣etReλ>n})f‖‖g*‖=4M‖EA({λ∈σ(A)∣etReλ>n})f‖bythestrongcontinuityofthes.m.;⟶0asn⟶∞.
From (4.66) and (4.67), by [2], Proposition 3.1, we infer that f∈⋂0≤t<∞D(etA).
Therefore (see Section1), y(t):=etAf, 0≤t<∞, is a weak solution of (1.1) on [0,∞).
Let(4.68)h*:=∑k=1∞1k2en(k)*∈X*
(cf. (4.51)).
Taking into account (4.50) and (4.51), we have
(4.69)〈en(k),h*〉≥ɛk2,k=1,2,….
For any λ∈Δn(k), k=1,2,…, by (4.47) and (4.64),
(4.70)Reλ=Reλn(k)+(Reλ-Reλn(k))≤Reλn(k)+|Reλ-Reλn(k)|≤Reλn(k)+ɛn(k)≤Reλn(k)+1≤-k+1≤0,Reλ>min⁡(0,-b-ln⁡|Im⁡λ|),
Hence, for λ∈Δn(k), k=1,2,…,
(4.71)|λ|≥|Im⁡λ|≥eb--1[-Reλ].
Using these estimates, we have
(4.72)∫σ(A)|λ|e(b--1/2)Reλdv(f,h*,λ)considering(4.49),by(2.9);=∑k=1∞1k2∫Δn(k)|λ|e(b--1/2)Reλdv(en(k),h*,λ)≥∑k=1∞1k2∫Δn(k)e(b--1/2)[-Reλ]dv(en(k),h*,λ)≥∑k=1∞e(b--1/2)[k-1]k2v(en(k),h*,Δn(k))≥e-b--1/2∑k=1∞e(b--1/2)kk2|〈EA(Δn(k))en(k),h*〉|by(4.49),(4.64),and(4.69);≥e-b--1/2ɛ∑k=1∞e(b--1/2)kk4=∞.
Whence, by the properties of the o.c. and [2], Proposition 3.1, y(b-/2)=e(b--1/2)Af∉D(A).
Therefore, by Proposition3.1, the weak solution y(t)=etAf, 0≤t<∞, of (1.1) on [0,∞) is not once strongly differentiable on (0,∞).
With every possibility concerning{Reλn}n=1∞ considered, we infer that the opposite to the assumption that there is a b+>0 such that, for any b->0, the set σ(A)∖ℒb-,b+ is bounded, allows to single out a weak solution of (1.1) on [0,∞) that is not once strongly differentiable on (0,∞), much less strongly infinite differentiable on (0,∞).
Thus, the “only if” part has been proved by contrapositive.Theorem 5.2 of [3] and Theorem 4.1 of [4] have been generalized.


## 5. Certain Effects of Smoothness Improvement

As we observed in the proofs of the “only if” parts of Theorems4.1 and 4.2, the opposits to the “if” parts’ premises imply that there is a weak solution of (1.1) on [0,∞), which is not strongly differentiable at 0 or, respectively, once strongly differentiable on (0,∞).Therefore, the case of finite strong differentiability of the weak solutions is superfluous and we obtain the following effects of smoothness improvement.Proposition 5.1.
If every weak solution of (1.1) on [0,∞) is strongly differentiable at 0, then all of them are strongly infinite differentiable on [0,∞).Proposition 5.2.
If every weak solution of (1.1) on [0,∞) is once strongly differentiable on (0,∞), then all of them are strongly infinite differentiable on (0,∞).These statements generalize Propositions 6.1 and 6.2 of [3], respectively, the latter agreeing with the case when A is a linear operator (not necessarily spectral of scalar type) generating a C0-semigroup (cf. [1, 13]).


## 6. Final Remarks

Due to thescalar type spectrality of the operator A, all the above criteria are formulated exclusively in terms of its spectrum, no restrictions on the operator's resolvent behavior necessary, which makes them inherently qualitative and more transparent than similar results for C0-semigroups (cf. [14]).If a scalar type spectralA generates a C0-semigroup (cf. [15]), we immediately obtain the results of paper [4] regarding infinite differentiability.



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