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✓ 26 results · all verified · 16 also independently AI-judged
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Strongly Continuous Semigroups and Hille Yosida

1 · Prerequisites

2 · Summary

This page develops the theory of strongly continuous (C0) semigroups on Banach spaces from the definition through the generation theorems, with the sign convention u′=Au and the resolvent normalised as R(λ,A)=(λI−A)−1. It begins with the semigroup axioms, local uniform boundedness, the equivalence of strong continuity with continuity at zero, and the exponential bound; the Bochner-calculus prerequisites (linearity of the Bochner integral, average convergence, the mean value inequality and the Banach-valued fundamental theorem of calculus) are proved alongside so that every integral manipulation has an exact supplier. The infinitesimal generator is defined by one-sided difference quotients and developed: domain invariance and commutation with orbits, integrated orbits, closedness of the generator and density of its domain, the resolvent of the closed operator, the resolvent identity, the Laplace-transform formula, and the resolvent power estimates.

The generation theory follows: the exponential series of a bounded operator, the Yosida approximants and the strong convergence of the resolvent and of the approximants on the domain, the convergence of the bounded Yosida semigroups to the generated semigroup, and the Hille-Yosida theorem with all resolvent powers together with its contraction shortcut. Dissipative operators, the Lumer-Phillips theorem and the maximal-dissipativity statement are proved in the norm form, and the classical, strong, mild and integral solutions of the abstract Cauchy problem are defined and related by the variation-of-constants theorem and the well-posedness theorem. A closing remark records the sign, resolvent and contraction dictionaries. The generation and automatic boundedness results assume Dependent Choice, which supplies the uniform boundedness and closed graph arguments and their Countable Choice consequences. Pure resolvent algebra instead uses bounded inverses in the resolvent definition; the equivalence with bijectivity alone is stated under DC. The norm-duality description of dissipativity and the Laplace uniqueness argument state HB separately. Classical upgrades of the inhomogeneous problem use C1 forcing or its explicit Bochner-primitive extension; Lipschitz continuity alone on an arbitrary Banach space does not supply that extension.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Strongly continuous semigroup

Definition

Let X be a Banach space over K∈{R,C} (Banach space, Real and complex scalar conventions for normed spaces) and let B(X) be the bounded linear operators on X (A bounded linear operator between normed spaces, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators). A family (T(t))t≥0⊆B(X) is a strongly continuous one-parameter semigroup (or C0-semigroup) if (i) T(0)=I; (ii) T(t+s)=T(t)T(s) for all s,t≥0; (iii) for every x∈X the orbit map t↦T(t)x is continuous from [0,∞) into X. Property (iii) says that t↦T(t) is continuous for the strong operator topology on B(X); no continuity in the operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum) is assumed or implied, and the operators need be neither isometries nor contractions. A strongly continuous group is defined analogously with R in place of [0,∞) and the functional equation holding for all s,t∈R.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

A semigroup with continuity at zero is uniformly bounded on every compact time interval

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let X be a Banach space and let (T(t))t≥0⊆B(X) satisfy T(0)=I, T(t+s)=T(t)T(s) for s,t≥0 and lim⁡t↓0T(t)x=x for every x∈X (in particular every strongly continuous semigroup satisfies these hypotheses, Strongly continuous semigroup). Then for every t0≥0, sup⁡0≤t≤t0∥T(t)∥<∞.

Facts & Assumptions

Given: A Banach space X and a family (T(t))t≥0⊆B(X) with T(0)=I, T(t+s)=T(t)T(s) for all s,t≥0, and lim⁡t↓0T(t)x=x for every x∈X. These are the hypotheses of Strongly continuous semigroup with continuity at every time weakened to continuity at 0; the item assumes Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain), carried by the cited uniform boundedness principle, and step 1.1 selects a sequence, which uses Countable Choice, a consequence of DC.

[F1]

The operator norm satisfies ∥ST∥≤∥S∥ ∥T∥ for all S,T∈B(X) (Composition satisfies |ST|\le|S|,|T|), and ∥⋅∥ is the operator norm on B(X) (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[F2]

Under DC, a pointwise bounded family of bounded linear operators between a Banach space and a normed space is norm bounded (Uniform boundedness principle).

[F3]

For every x∈X one has T(t)x→x as t↓0; this is hypothesis (iii) of Strongly continuous semigroup at the point 0, where T(0)x=x.

Proof

technique · direct, combining the uniform boundedness principle with the semigroup law: first a finite bound on a small interval $[0,\delta]$, then iteration over $\lfloor t_0/\delta\rfloor$ steps
1.1F2F3given

There are δ>0 and M0<∞ with sup⁡0≤t≤δ∥T(t)∥≤M0. Otherwise sup⁡0<t≤1/n∥T(t)∥=∞ for every n, so for each n one may select tn∈(0,1/n] with ∥T(tn)∥≥n; this selection is the only use of Countable Choice, available because DC implies ACω.

2.1F3step 1.1

For every x∈X the sequence T(tn)x converges to x, because tn≤1/n→0 and lim⁡t↓0T(t)x=x; hence the family {T(tn):n≥1} is pointwise bounded on the Banach space X.

3.1F2step 1.1step 2.1

By the uniform boundedness principle [F2] the family {T(tn)} is norm bounded, that is sup⁡n∥T(tn)∥<∞, contradicting ∥T(tn)∥≥n→∞; hence the assumed unboundedness of every right neighbourhood of 0 is impossible, proving [step 1.1].

4.1F1step 3.1algebra

Put M:=max⁡{1,M0}≥1. For t0≥0 and t∈[0,t0] write t=nδ+s with n:=⌊t/δ⌋∈N0 and s∈[0,δ). The functional equation gives T(t)=T(δ)nT(s), by induction on n from T(u+δ)=T(u)T(δ).

5.1F1step 3.1step 4.1algebra

Therefore ∥T(t)∥≤∥T(δ)∥n∥T(s)∥≤Mn+1 by [F1] and [step 1.1], where n=⌊t/δ⌋≤⌊t0/δ⌋; hence sup⁡0≤t≤t0∥T(t)∥≤M⌊t0/δ⌋+1<∞.

6.1step 5.1∎

Since t0≥0 was arbitrary, sup⁡0≤t≤t0∥T(t)∥<∞ for every t0, as required.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Continuity at time zero implies continuity of every orbit

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let X be a Banach space and (T(t))t≥0⊆B(X) satisfy T(0)=I, T(t+s)=T(t)T(s) for s,t≥0 and lim⁡t↓0T(t)x=x for every x∈X. Then (T(t)) is a strongly continuous semigroup (Strongly continuous semigroup): every orbit map t↦T(t)x is continuous on [0,∞).

Facts & Assumptions

Given: Dependent Choice; A Banach space X and a family (T(t))t≥0⊆B(X) with T(0)=I, T(t+s)=T(t)T(s) for s,t≥0 and lim⁡t↓0T(t)x=x for every x∈X.

[F1]

Local boundedness: for every t0≥0 there is M<∞ with ∥T(t)∥≤M for all t∈[0,t0]; this uses DC through the uniform boundedness principle (A semigroup with continuity at zero is uniformly bounded on every compact time interval).

[F2]

For S∈B(X) the operator norm satisfies ∥Sy∥≤∥S∥ ∥y∥ for all y, and ∥ST∥≤∥S∥ ∥T∥ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Composition satisfies |ST|\le|S|,|T|).

[F3]

The hypotheses are those of a strongly continuous semigroup with continuity required only at 0 (Strongly continuous semigroup): T(0)=I, the functional equation holds, and T(h)x→x as h↓0 for every x.

Proof

technique · direct, transporting continuity at $0$ to an arbitrary time with the functional equation and the local bound
1.1F1

Fix t0≥0 and let M be a bound for ∥T(t)∥ on [0,t0], which exists by [F1]; fix also x∈X.

1.2F2F3

Right continuity at t0: for h>0, the functional equation gives T(t0+h)x−T(t0)x=T(t0)(T(h)x−x), whose norm is at most ∥T(t0)∥ ∥T(h)x−x∥→0 as h↓0 by [F2] and [F3].

1.3F1F2F3

Left continuity at t0: for 0<h≤t0 one has T(t0)=T(t0−h+h)=T(t0−h)T(h), hence T(t0−h)x−T(t0)x=T(t0−h)(x−T(h)x) and, since t0−h∈[0,t0], ∥T(t0−h)x−T(t0)x∥≤M∥x−T(h)x∥→0 by [F1], [F2] and [F3].

2.1F3step 1.2step 1.3∎

The two one-sided limits at t0 both equal T(t0)x, so the orbit t↦T(t)x is continuous at every t0≥0; as x was arbitrary, all orbits are continuous on [0,∞), and the family is a strongly continuous semigroup as defined in [F3].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Exponential bound for a C0-semigroup

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X (Strongly continuous semigroup). Then there exist M≥1 and ω∈R with ∥T(t)∥≤Meωt for all t≥0. For X≠{0} one may take M:=sup⁡0≤s≤1∥T(s)∥ and ω:=log⁡M; for X={0} take M=1, ω=0.

Facts & Assumptions

Given: Dependent Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X (Strongly continuous semigroup).

[F1]

Local boundedness: M0:=sup⁡0≤s≤1∥T(s)∥<∞; the proof of the lemma uses DC through the uniform boundedness principle (A semigroup with continuity at zero is uniformly bounded on every compact time interval).

[F2]

The operator norm is submultiplicative: ∥ST∥≤∥S∥ ∥T∥, and ∥Sy∥≤∥S∥ ∥y∥ for every y∈X (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Composition satisfies |ST|\le|S|,|T|, A bounded linear operator between normed spaces).

[F3]

The semigroup law T(r+s)=T(r)T(s) for r,s≥0 and T(0)=I (Strongly continuous semigroup).

[F4]

The real exponential satisfies eu+v=euev, eu>0 for all real u, exp⁡:R→(0,∞) is a bijection, and eu≥1 for u≥0 (The real exponential function and the number e by a power series, The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y), The exponential is a continuous bijection from R onto (0,∞)).

Proof

technique · direct, bounding the powers of $T(1)$ by the local bound and absorbing them into an exponential
1.1F1F3F4

If X={0} take M:=1 and ω:=0: then ∥T(t)∥=0≤Meωt for every t. Otherwise X≠{0}, so ∥I∥=1 and [F1] gives M:=sup⁡0≤s≤1∥T(s)∥≥∥T(0)∥=1 with M<∞; set ω:=log⁡M, which exists by [F4] because M≥1.

2.1F2F3step 1.1

For t≥0 write t=n+s with n:=⌊t⌋∈N0 and s∈[0,1). By the semigroup law, T(t)=T(s)T(1)n, hence ∥T(t)∥≤∥T(s)∥ ∥T(1)∥n≤Mn+1 by [F2] and the choice of M.

3.1F4step 2.1

Since M≥1 and n≤t, [F4] gives Mn+1=M enlog⁡M≤M etlog⁡M=Meωt.

4.1step 1.1step 3.1∎

Combining [step 2.1] and [step 3.1], ∥T(t)∥≤Meωt for every t≥0, with M≥1 and ω∈R; in the nonzero case the displayed M=sup⁡0≤s≤1∥T(s)∥ and ω=log⁡M are the explicit choices, and in the zero-space case the bound holds trivially for M=1, ω=0.

Note. The constant ω=log⁡M need not be optimal: any larger ω also works, since eωt is nondecreasing in ω for t≥0; the growth bound ω0(T)=inf⁡{ω:∃M, ∥T(t)∥≤Meωt} is not needed here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Linearity of the Bochner integral

Statement

Let (Ω,A,μ) be a measure space, let X be a real or complex Banach space, and let f,g:Ω→X be Bochner integrable (Bochner-integrable function) with α,β scalars. Then αf+βg is Bochner integrable and ∫Ω(αf+βg) dμ=α∫Ωf dμ+β∫Ωg dμ; for a measurable set E the same identity holds with f,g replaced by f1E,g1E. The integral is therefore additive and homogeneous, and in particular well defined on differences.

Facts & Assumptions

Given: A measure space (Ω,A,μ), a real or complex Banach space X, Bochner integrable functions f,g:Ω→X, scalars α,β, and a measurable set E.

[F1]

By the definition of Bochner integrability (Bochner-integrable function) there are sequences (sn), (tn) of integrable X-valued simple functions with ∫Ω∥f−sn∥ dμ→0, ∫Ω∥g−tn∥ dμ→0, and ∫Ωf dμ=lim⁡n∫Ωsn dμ, ∫Ωg dμ=lim⁡n∫Ωtn dμ.

[F2]

The integral of an integrable Banach-valued simple function is independent of its representation, is linear, and satisfies ∥∫Es dμ∥≤∫E∥s∥ dμ for every measurable E (The Banach-valued simple integral is well defined).

[F3]

A strongly measurable h:Ω→X is Bochner integrable if and only if ∫Ω∥h∥ dμ<∞; for such h and any defining approximating sequence of integrable simple functions, the integral is the limit of the simple integrals (Bochner integrability criterion, Bochner-integrable function).

[F4]

Addition in X and scalar multiplication K×X→X are continuous (Vector addition and scalar multiplication are continuous in a normed space).

Proof

technique · direct, approximating $\alpha f+\beta g$ by the corresponding linear combinations of the defining simple functions
1.1F1F4

The functions f,g are strongly measurable by [F1]. Let un,vn be their measurable simple approximations converging pointwise outside measurable null sets N,N′; these need not be the defining L1 approximations sn,tn. The simple functions αun+βvn converge to αf+βg off N∪N′ by [F4], proving strong measurability.

1.2F1F3algebra

Norm estimate: for every n, ∥αf+βg−(αsn+βtn)∥≤∣α∣ ∥f−sn∥+∣β∣ ∥g−tn∥ pointwise, hence after integration ∫Ω∥αf+βg−(αsn+βtn)∥ dμ≤∣α∣∫Ω∥f−sn∥ dμ+∣β∣∫Ω∥g−tn∥ dμ→0; in particular ∫Ω∥αf+βg∥ dμ≤∣α∣∫Ω∥f∥ dμ+∣β∣∫Ω∥g∥ dμ<∞, since ∫∥f∥,∫∥g∥<∞ by [F3].

2.1F1F3step 1.1step 1.2

By [step 1.1], [step 1.2] and the integrability criterion [F3], the function αf+βg is Bochner integrable, and αsn+βtn is a defining sequence of integrable simple functions for it, so ∫Ω(αf+βg) dμ=lim⁡n∫Ω(αsn+βtn) dμ.

3.1F1F2F4step 2.1

Linearity of the simple integral [F2] gives ∫Ω(αsn+βtn) dμ=α∫Ωsn dμ+β∫Ωtn dμ for every n, whose right-hand side converges to α∫Ωf dμ+β∫Ωg dμ by [F1] and continuity of the vector operations [F4]; combining with [step 2.1] yields ∫Ω(αf+βg) dμ=α∫Ωf dμ+β∫Ωg dμ.

4.1F1F2F3step 3.1

Restricted form: the functions f1E and g1E are Bochner integrable, because they are strongly measurable and dominated in norm by ∥f∥ and ∥g∥ respectively, and (αf+βg)1E=α(f1E)+β(g1E) pointwise; applying [step 3.1] to the pair f1E,g1E gives ∫E(αf+βg) dμ=α∫Ef dμ+β∫Eg dμ.

5.1step 4.1∎

The integral is thus additive and homogeneous on the Bochner integrable functions: taking α=β=1 gives additivity, β=0 with α arbitrary gives homogeneity, and β=−1 shows the difference f−g is Bochner integrable with ∫Ω(f−g) dμ=∫Ωf dμ−∫Ωg dμ, so the integral is well defined on differences.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Average convergence for a continuous Banach-valued function

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue-measure interfaces. Let X be a real or complex Banach space, let a<b and let f:[a,b]→X be continuous. Then f is Bochner integrable, and for t∈[a,b) and h>0 with t+h≤b one has ∥1h∫tt+hf(s) ds−f(t)∥≤sup⁡s∈[t,t+h]∥f(s)−f(t)∥→0 as h↓0; analogously 1h∫t−htf→f(t) as h↓0 for t∈(a,b]. The same one-sided limits hold for vector-valued curves that are merely continuous at t provided they are Bochner integrable on some neighbourhood of t.

Facts & Assumptions

Given: Countable Choice; A real or complex Banach space X, real numbers a<b, and a continuous f:[a,b]→X.

[F1]

f is Bochner integrable when there are integrable X-valued simple functions sn with ∫ab∥f−sn∥ ds→0, and then ∫abf ds=lim⁡n∫absn ds; integrals over subintervals are defined through the indicators 1[t,t+h], and constant functions have the expected integrals (Bochner-integrable function, Linearity of the Bochner integral).

[F2]

The Bochner integral is linear: for Bochner integrable u,v and scalars α,β the function αu+βv is Bochner integrable with ∫(αu+βv)=α∫u+β∫v (Linearity of the Bochner integral).

[F3]

Norm inequality: ∥∫Eu ds∥≤∫E∥u∥ ds for every Bochner integrable u and measurable E (Bochner integral norm inequality).

[F4]

Continuity of f at a point t means: for every η>0 there is δ>0 with ∣s−t∣<δ, s∈[a,b], implying ∥f(s)−f(t)∥<η (Continuity of a map between metric spaces, at a point and globally, in the ε-δ form).

[F5]

The interval [a,b] is a compact metric space (Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line), and a continuous map from a compact metric space to a metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous): for every η>0 there is δ>0 such that ∥f(s)−f(u)∥<η whenever s,u∈[a,b] and ∣s−u∣<δ.

Proof

technique · direct, using explicit uniform step approximation, then linearity and the integral norm inequality
1.1F5

By [F5], f is uniformly continuous on [a,b]. Its oscillation over pairs at distance at most ℓ therefore tends to zero as ℓ↓0.

2.1F1step 1.1

For each n∈N, set ℓn=(b−a)/(n+1), xk=a+kℓn for 0≤k≤n+1, and sn=∑k<n+1f(xk)1[xk,xk+1), with the last interval including b. These are integrable simple functions and converge uniformly, hence pointwise, to f.

3.1F1step 1.1step 2.1

The norm error is at most the oscillation from step 1.1, so ∫ab∥f−sn∥≤(b−a)sup⁡∣s−u∣≤ℓn∥f(s)−f(u)∥→0. Together with the pointwise simple approximation, [F1] proves Bochner integrability of f.

4.1F1F2step 3.1

For t∈[a,b) and 0<h≤b−t, linearity gives h−1∫tt+hf−f(t)=h−1∫tt+h(f(s)−f(t)) ds, since the constant function has integral hf(t).

5.1F3step 4.1

By the norm inequality, the norm of this difference is at most h−1∫tt+h∥f(s)−f(t)∥ ds≤sup⁡s∈[t,t+h]∥f(s)−f(t)∥.

6.1F4step 5.1

Continuity at t makes the last supremum tend to zero as h↓0: given η>0, take h below a continuity radius for f at t. Hence the forward averages converge to f(t).

7.1F2F3F4step 3.1step 6.1

For t∈(a,b] and 0<h≤t−a, the same linearity and norm estimates give ∥h−1∫t−htf−f(t)∥≤sup⁡s∈[t−h,t]∥f(s)−f(t)∥→0. This proves the backward form directly.

8.1F2F3F4step 6.1step 7.1∎

If instead f is only continuous at t and Bochner integrable on a neighbourhood of t, the forward and backward estimates above still apply: local integrability supplies the integrals and continuity at t makes their norm errors vanish. Thus the stated general one-sided limits also hold.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Mean value inequality for a differentiable Banach-valued curve

Statement

Let X be a Banach space, a<b, and let φ:[a,b]→X be continuous on [a,b] and differentiable on (a,b) in the sense of Fréchet derivative between Banach spaces. If there is C≥0 with ∥φ′(s)∥≤C for all s∈(a,b), then ∥φ(b)−φ(a)∥≤C(b−a). In particular, if φ is continuous, differentiable on (a,b) and φ′=0 there, then φ is constant.

Facts & Assumptions

Given: A real or complex Banach space X (Banach space), read as a real vector space for differentiation so that the real-variable Fréchet derivative of Fréchet derivative between Banach spaces applies (the complex case uses the underlying real structure, and complex differentiability is a special case); real numbers a<b, a curve φ:[a,b]→X continuous on [a,b] and differentiable on (a,b), and a constant C≥0 with ∥φ′(s)∥≤C for all s∈(a,b).

[F1]

At each s∈(a,b) the Fréchet derivative φ′(s):R→X is bounded linear and there is a remainder with ∥φ(s+h)−φ(s)−φ′(s)h∥/∣h∣→0 as h→0 (Fréchet derivative between Banach spaces).

[F2]

The norm function is continuous on X and satisfies the reverse triangle inequality ∣∥x∥−∥y∥∣≤∥x−y∥ (The reverse triangle inequality in a normed space).

[F3]

Vector addition and scalar multiplication on X are continuous (Vector addition and scalar multiplication are continuous in a normed space), so limits of sums and scalar multiples may be taken termwise.

Proof

technique · direct, by a maximal-interval argument on the auxiliary function $d(\tau)=\|\varphi(\tau)-\varphi(r)\|-\widetilde C(\tau-r)$
1.1F2F3

Fix C~>C and r∈(a,b), and put d(τ):=∥φ(τ)−φ(r)∥−C~(τ−r) for τ∈[r,b]. By [F2] and [F3] the function d is continuous, so S:={τ∈[r,b]:d(τ)≤0} is a nonempty closed subset of [r,b], since r∈S; being nonempty and bounded above it has a supremum τ0∈S, which is therefore its maximum.

2.1F1F2step 1.1

If τ0<b then τ0>r: if τ0=r, then [F1] at r (legitimate because r>a) gives φ(r+h)=φ(r)+φ′(r)h+o(h) with ∥φ′(r)∥≤C, so by [F2] d(r+h)≤(C−C~)h+o(h)<0 for all sufficiently small h>0, contradicting the maximality of r=τ0; hence τ0∈(r,b)⊆(a,b), where φ is differentiable.

3.1F1F2step 2.1

If τ0<b, then differentiability at τ0 gives φ(τ0+h)=φ(τ0)+φ′(τ0)h+o(h) with ∥φ′(τ0)∥≤C, so for small h>0 [F2] gives d(τ0+h)≤d(τ0)+(C−C~)h+o(h)≤(C−C~)h+o(h)<0, since d(τ0)≤0; then τ0+h∈S, contradicting the maximality of τ0. Hence τ0=b.

4.1step 3.1

At τ0=b the defining inequality of S reads d(b)≤0, that is ∥φ(b)−φ(r)∥≤C~(b−r).

5.1F2F3step 4.1

Letting r↓a along a sequence: φ(r)→φ(a) by continuity and [F3], so [F2] gives ∥φ(b)−φ(r)∥→∥φ(b)−φ(a)∥, and b−r→b−a; hence ∥φ(b)−φ(a)∥≤C~(b−a).

6.1step 5.1algebra

Since C~>C was arbitrary, ∥φ(b)−φ(a)∥≤C(b−a).

7.1step 6.1algebra∎

If in addition φ′=0 on (a,b), take C=0 in [step 6.1]; then for every t∈(a,b] the same argument applied to the restriction of φ to [a,t] gives φ(t)=φ(a), and φ(a)=φ(a), so φ is constant on [a,b].

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Fundamental theorem of calculus for Banach-valued continuous curves

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue-measure interfaces. Let X be a real or complex Banach space, let a<b, and let f:[a,b]→X be continuous. Differentiation uses the underlying real structure. Then G(t):=∫atf(s) ds is differentiable on (a,b) and has the corresponding one-sided derivatives at a,b, with G′(t)=f(t), and this derivative extends continuously to [a,b]. Consequently, if φ:[a,b]→X is continuous, differentiable on (a,b) with φ′ continuous on (a,b) and extendable to a continuous X-valued function on [a,b], then ∫abφ′(s) ds=φ(b)−φ(a). The Bochner integral here is the one of Bochner-integrable function; the identities also hold for continuous curves on [0,∞) restricted to compact subintervals.

Facts & Assumptions

Given: Countable Choice; A Banach space X, real numbers a<b, a continuous f:[a,b]→X, the primitive G(t):=∫atf(s) ds for t∈[a,b], and a continuous φ:[a,b]→X differentiable on (a,b) whose derivative extends to a continuous X-valued function on [a,b].

[F1]

Average convergence (Average convergence for a continuous Banach-valued function): a continuous f:[a,b]→X is Bochner integrable, and for t∈[a,b), h>0, t+h≤b, ∥1h∫tt+hf−f(t)∥≤sup⁡[t,t+h]∥f−f(t)∥→0, with the analogous backward limit for t∈(a,b]; the same one-sided limits hold for curves continuous at t and Bochner integrable near t.

[F2]

The Bochner integral is linear, so for [c,d]⊆[a,b] the difference of primitives is ∫cdf (Linearity of the Bochner integral, Bochner-integrable function).

[F4]

Differentiability on (a,b) means Fréchet differentiability at every point of the open interval (Fréchet derivative between Banach spaces); at the endpoints only the relevant one-sided difference quotients are considered.

[F5]

Mean value inequality (Mean value inequality for a differentiable Banach-valued curve): a curve continuous on an interval, differentiable inside with derivative bounded by C, changes by at most C times the length; in particular a curve with vanishing interior derivative is constant.

Proof

technique · direct, computing the difference quotient of the primitive with the average-convergence lemma and then applying the mean value inequality to $\varphi$ minus its integral
1.1F1F2

By [F1] the continuous f is Bochner integrable on [a,b], so G(t)=∫atf is defined for every t∈[a,b]; by [F2] G(t+h)−G(t)=∫tt+hf for [t,t+h]⊆[a,b].

2.1F1F2F4step 1.1

Difference quotients of G: for t∈[a,b) and h>0 with t+h≤b, G(t+h)−G(t)h=1h∫tt+hf→f(t) by [F1]; similarly G(t)−G(t−h)h=1h∫t−htf→f(t) for t∈(a,b]. Hence G is differentiable on (a,b) with G′=f, and has the one-sided derivatives f(t) at the endpoints.

3.1F4step 2.1

G′=f is continuous on [a,b], and the existence of the one-sided derivative at a and at b makes G continuous there from the appropriate side; at interior points G is continuous by differentiability.

3.2F2F4step 2.1

Since φ′ extends to a continuous X-valued function on [a,b], denote the extension again by φ′ and put ψ(t):=φ(t)−∫atφ′(s) ds for t∈[a,b]. By [step 2.1] the primitive of φ′ is differentiable on (a,b) with derivative φ′, and ψ is differentiable on (a,b); by linearity of the derivative and of the integral, ψ′=φ′−φ′=0 on (a,b), and ψ(a)=φ(a).

4.1F5step 3.2

ψ is continuous on [a,b] and differentiable on (a,b) with ψ′=0 there; by [F5] (constant case) ψ is constant on [a,b], so ψ(b)=ψ(a)=φ(a), that is φ(b)−∫abφ′=φ(a).

5.1step 2.1step 4.1∎

Therefore ∫abφ′(s) ds=φ(b)−φ(a). The same computation, applied to each compact subinterval [a,b]⊆[0,∞) after restricting a continuous curve on [0,∞), gives the stated identity in that setting.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Infinitesimal generator of a C0-semigroup

Definition

For a Banach space X, the operator vocabulary of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores is extended as follows: an operator is a linear map A:D(A)→X on a linear subspace of X, its graph is Γ(A)={(x,Ax):x∈D(A)}⊆X×X, it is closed when this graph is closed, and densely defined when D(A) is dense in X. Use ∥(x,y)∥=∥x∥+∥y∥ on X×X and the graph norm ∥x∥A=∥x∥+∥Ax∥. These norms are equivalent to the square-sum norms in the Hilbert suppliers. The product is Banach because its two coordinate Cauchy sequences converge in X; the closed graph is therefore Banach, and x↦(x,Ax) is an isometry of the graph-norm domain onto it. Under Countable Choice, sequential closedness is equivalent to closedness: for any point in a closure, choose graph points within 1/n and pass to their limit.

Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X (Strongly continuous semigroup). Its infinitesimal generator is the linear operator A:D(A)⊆X→X with domain D(A):={x∈X: lim⁡t↓0T(t)x−xt exists in X} and Ax:=lim⁡t↓0T(t)x−xt for x∈D(A). The limit is a one-sided limit at the boundary point 0, and D(A) is a linear subspace of X (Normed subspace); the operator is recorded as the pair (A,D(A)) in the sense of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores. Neither boundedness nor closedness of A, nor density of D(A), is assumed in the definition; under its stated choice hypothesis, The generator is closed and densely defined proves that the graph of A is closed and D(A) is dense in X. The domain need not be closed in the norm of X.

The one-sided limit. For x∈D(A) the difference quotient T(t)x−xt∈X is defined for every t>0, and the defining limit is taken along t↓0 only; no two-sided limit at the boundary point 0 of [0,∞) is considered, and the vector Ax is the limit when it exists. The value Ax is unique because X is a metric space (Banach space).

The domain is a linear subspace. The zero vector lies in D(A) and A0=0. If x,y∈D(A) and α,β are scalars, then by linearity of each T(t) (A bounded linear operator between normed spaces) the difference quotient of αx+βy equals αT(t)x−xt+βT(t)y−yt for every t>0; as t↓0 this converges to αAx+βAy, because vector addition and scalar multiplication are continuous and scalar multiplication by the fixed scalars α,β is continuous. Hence αx+βy∈D(A) with A(αx+βy)=αAx+βAy, so D(A) is a linear subspace of X (Normed subspace) and A is linear on it. The operator is recorded as the pair (A,D(A)) in the vocabulary of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores.

What is not assumed. Boundedness and graph closedness of A, and density of D(A) in X, are not defining assumptions. Graph closedness and domain density are conclusions of the later theorem under its stated choice hypothesis; they do not assert that D(A) is closed in the norm of X. The semigroup axioms used here are those of Strongly continuous semigroup, in particular T(t) is everywhere defined and bounded for every t≥0.

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The generator commutes with the semigroup on its domain

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator (A,D(A)) (Infinitesimal generator of a C0-semigroup). If x∈D(A), then T(t)x∈D(A) and AT(t)x=T(t)Ax for every t≥0; moreover the orbit t↦T(t)x is differentiable on (0,∞) with ddtT(t)x=T(t)Ax=AT(t)x, and right differentiable at t=0 with right derivative Ax.

Facts & Assumptions

Given: Dependent Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator (A,D(A)) (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup) and a vector x∈D(A).

[F1]

x∈D(A) means that T(h)x−xh→Ax as h↓0 (Infinitesimal generator of a C0-semigroup).

[F2]

The semigroup law T(t+h)=T(t)T(h) holds for all t,h≥0, and operator norms satisfy ∥Sy∥≤∥S∥ ∥y∥ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Composition satisfies |ST|\le|S|,|T|).

[F3]

The family is locally bounded and the orbit of every vector is continuous on [0,∞) (A semigroup with continuity at zero is uniformly bounded on every compact time interval, Strongly continuous semigroup): for t0>0 there is M with ∥T(s)∥≤M for s∈[0,t0], and T(s)z→T(t)z for every z as s→t.

Proof

technique · direct, writing the difference quotients of the orbit as $T(\cdot)$ applied to the generator quotient
1.1F2

For t≥0 and h>0, the semigroup law gives T(t+h)x−T(t)xh=T(t)T(h)x−xh.

2.1F1F2step 1.1

Since T(h)x−xh→Ax by [F1] and T(t) is a fixed bounded operator, the right difference quotient in [step 1.1] converges to T(t)Ax as h↓0. Hence the right derivative of the orbit at t exists and equals T(t)Ax; taking t=0 shows that the right derivative at 0 is Ax and, for general t, that T(t)x∈D(A) with AT(t)x=T(t)Ax.

3.1F2F3step 1.1step 2.1

Left derivative at t>0: for 0<h<t one has T(t−h)x−T(t)x−h=T(t−h)T(h)x−xh. Let vh:=T(h)x−xh→Ax and use the local bound M on [0,t] from [F3]: ∥T(t−h)vh−T(t)Ax∥≤M∥vh−Ax∥+∥(T(t−h)−T(t))Ax∥→0, because vh→Ax and T(t−h)→T(t) strongly as h↓0 by [F3]. Hence the left derivative at t also equals T(t)Ax=AT(t)x.

4.1step 2.1step 3.1∎

Combining [step 2.1] and [step 3.1], for every x∈D(A) and every t≥0 the orbit satisfies T(t)x∈D(A), AT(t)x=T(t)Ax, and t↦T(t)x is differentiable on (0,∞) with derivative T(t)Ax=AT(t)x, with right derivative Ax at t=0.

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Time integrals of semigroup orbits lie in the generator domain

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue-measure interfaces. Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator (A,D(A)) (Infinitesimal generator of a C0-semigroup). For every x∈X and every t≥0 the Bochner integral Jtx:=∫0tT(s)x ds belongs to D(A), and AJtx=T(t)x−x. The integral is taken in the sense of Bochner-integrable function; the integrand is continuous, hence Bochner integrable on [0,t].

Facts & Assumptions

Given: Countable Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator (A,D(A)) (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup), and x∈X, t≥0.

[F1]

The generator is defined by D(A)={y:lim⁡h↓0T(h)y−yh exists} and Ay= that limit (Infinitesimal generator of a C0-semigroup).

[F2]

Every orbit s↦T(s)y is continuous on [0,∞), T(s) is linear and bounded, and T(s+r)=T(s)T(r)=T(r)T(s) for s,r≥0 (Strongly continuous semigroup, A bounded linear operator between normed spaces).

[F3]

Average convergence (Average convergence for a continuous Banach-valued function): a continuous curve on a compact interval is Bochner integrable, and its forward and backward averages converge to its value at the point.

[F4]

Linearity of the Bochner integral over measurable sets (Linearity of the Bochner integral, Bochner-integrable function), including additivity for adjacent subintervals via indicators.

[F5]

The Bochner integral is defined through integral-norm limits of integrable simple functions, and a strongly measurable function is Bochner integrable exactly when the integral of its norm is finite (Bochner-integrable function, Bochner integrability criterion); the norm inequality ∥∫Eu∥≤∫E∥u∥ holds (Bochner integral norm inequality).

[F6]

Bounded linear maps commute with Bochner integration: if S∈B(X) and u is Bochner integrable, then S∫u=∫Su (Bounded linear maps commute with Bochner integration).

[F7]

Lebesgue measure and measurability are translation invariant (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

Proof

technique · direct, differentiating the orbit integral at $0$ with the functional equation and the average-convergence lemma
1.1F2F3

The orbit s↦T(s)x is continuous on the compact interval [0,t] by [F2], hence Bochner integrable there by [F3]; therefore Jtx:=∫0tT(s)x ds is a well-defined element of X.

1.2F2F6

By [F6] applied to T(h) and [F2], T(h)Jtx=∫0tT(h)T(s)x ds=∫0tT(s+h)x ds.

1.3F3F5F7

Shift identity for continuous integrands: if g is continuous on [0,t+h], then ∫0tg(s+h) ds=∫ht+hg(u) du. For an integrable simple function s=∑jcj1Ej the identity holds termwise, since translating Ej∩[h,t+h] back by h gives Ej−h intersected with [0,t], a set of the same measure by [F7]; for a nonnegative measurable function it follows by taking the supremum of the pairings of dominated simple functions, and for a Bochner integrable g it follows by applying the scalar case to the nonnegative integrable ∥g−sn∥ and the simple case to sn along a defining approximation with ∫ht+h∥g−sn∥→0 [F5]. A continuous g on the compact interval is Bochner integrable by [F3].

2.1F2F4step 1.2step 1.3

Adding the identity ∫0t+h=∫0h+∫ht+h=∫0t+∫tt+h of [F4], [steps 1.2 and 1.3] give (T(h)−I)Jtx=∫ht+hT(u)x du−∫0tT(u)x du=∫tt+hT(u)x du−∫0hT(u)x du.

3.1F2F3step 2.1

Dividing by h>0 and applying the average-convergence limits of [F3] to the continuous orbit at the points t and 0 (where T(0)x=x) yields T(h)Jtx−Jtxh=1h∫tt+hT(u)x du−1h∫0hT(u)x du⟶T(t)x−x.

4.1F1step 3.1

By the definition of the generator [F1], the convergence of these right difference quotients means exactly that Jtx∈D(A) and AJtx=T(t)x−x.

5.1step 1.1step 4.1∎

Since x∈X and t≥0 were arbitrary, for every x and every t the integral Jtx lies in D(A) and A∫0tT(s)x ds=T(t)x−x; at t=0 both sides are 0.

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The generator is closed and densely defined

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator (A,D(A)) (Infinitesimal generator of a C0-semigroup). Then A is a closed linear operator and D(A) is dense in X (Densely defined, closed and closable operators, and cores).

Facts & Assumptions

Given: Dependent Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator (A,D(A)) (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup).

[F1]

Time integrals of orbits lie in the generator domain: for every x∈X and t>0 the Bochner integral Jtx=∫0tT(s)x ds satisfies Jtx∈D(A) and AJtx=T(t)x−x (Time integrals of semigroup orbits lie in the generator domain, Bochner-integrable function).

[F2]

Average convergence (Average convergence for a continuous Banach-valued function): a continuous curve on a compact interval is Bochner integrable, and 1h∫0hT(s)y ds→y as h↓0 for every y∈X; hence 1tJtx→x for every x.

[F3]

The semigroup is locally bounded and all orbits are continuous, and T(h)z→z as h↓0 for every z (A semigroup with continuity at zero is uniformly bounded on every compact time interval, Continuity at time zero implies continuity of every orbit); the norm inequality for Bochner integrals bounds ∥∫0hT(s)z ds∥≤hsup⁡0≤s≤h∥T(s)∥ ∥z∥. For x∈D(A) the orbit is differentiable with derivative T(s)Ax (The generator commutes with the semigroup on its domain), so the fundamental theorem of calculus for Banach-valued continuous curves gives ∫0hT(s)Ax ds=T(h)x−x (Fundamental theorem of calculus for Banach-valued continuous curves).

[F4]

Closedness and density of a linear operator are the graph and domain conditions of Densely defined, closed and closable operators, and cores with the operator vocabulary of Unbounded linear operators: domain, graph and extension.

Proof

technique · direct: density from the orbit averages, closedness by passing a convergent sequence through the integral identity
1.1F1F2

Density: for x∈X and t>0, [F1] gives Jtx∈D(A), and by [F2] 1tJtx→x as t↓0; hence x lies in the closure of D(A). Since x was arbitrary, D(A) is dense in X.

1.2F3

Closedness: suppose xn∈D(A) with xn→x and Axn→y in X. For fixed h>0 and every n, the orbit of xn is differentiable with derivative T(s)Axn, so [F3] gives T(h)xn−xn=∫0hT(s)Axn ds.

2.1F3step 1.2

As n→∞, the left-hand side tends to T(h)x−x, because T(h) is bounded and the orbit of x is continuous; the right-hand side tends to ∫0hT(s)y ds, because ∥∫0hT(s)(Axn−y) ds∥≤hsup⁡0≤s≤h∥T(s)∥ ∥Axn−y∥→0 by the local bound [F3]. Hence T(h)x−x=∫0hT(s)y ds for every h>0.

3.1F2step 2.1

Dividing by h and using the average-convergence limit of [F2] for the continuous curve s↦T(s)y gives T(h)x−xh=1h∫0hT(s)y ds→y as h↓0. By the definition of the generator, x∈D(A) and Ax=y; hence the graph of A contains the limits of all convergent graph sequences, and under DC (hence Countable Choice) the closure-sequence criterion in Infinitesimal generator of a C0-semigroup makes the graph closed.

4.1F4step 1.1step 3.1∎

Together with [step 1.1], the generator of a strongly continuous semigroup is a closed and densely defined linear operator, in the sense of [F4].

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Resolvent and spectrum of a closed operator on a Banach space

Definition

Let X be a Banach space over K∈{R,C} (Banach space, Real and complex scalar conventions for normed spaces) and let A:D(A)⊆X→X be a closed linear operator: D(A) is a linear subspace and Γ(A)={(x,Ax):x∈D(A)} is closed in X×X with norm ∥(x,y)∥=∥x∥+∥y∥. This extends the Hilbert-space vocabulary of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores to Banach spaces. A scalar λ∈K belongs to the resolvent set ρ(A) if λI−A:D(A)→X is bijective and its inverse is bounded on X. Under Dependent Choice, boundedness of the inverse follows from bijectivity by Closed graph theorem; thus under DC this is equivalent to the bijectivity-only convention. For λ∈ρ(A) the bounded operator R(λ,A):=(λI−A)−1∈B(X) is the resolvent, and σ(A):=K∖ρ(A) is the spectrum. One has R(λ,A)X=D(A), R(λ,A)(λI−A)y=y for y∈D(A), and λR(λ,A)x−x=AR(λ,A)x for x∈X; in particular AR(λ,A)=λR(λ,A)−I∈B(X). This is the Banach-space form of the Hilbert-space vocabulary Resolvent and spectrum of an unbounded operator; the shift convention λI−A is the same, with the closed graph theorem replacing the Hilbert-space bounded-inverse convention.

Boundedness under DC. Assume Dependent Choice. If λ∈K and λI−A is bijective, then λI−A is a closed operator: its graph is the image of the graph Γ(A)⊆X⊕X (Unbounded linear operators: domain, graph and extension) under the homeomorphism (y,z)↦(y,λy−z) of X⊕X with inverse (u,v)↦(u,λu−v). Hence (λI−A)−1:X→X has closed graph and is everywhere defined, so the closed graph theorem (Closed graph theorem, which assumes DC) makes it bounded; this is the one place where an axiom beyond ZF enters, and it is the DC carried by the cited theorem. The Banach-space vocabulary is Banach space with the scalar convention of Real and complex scalar conventions for normed spaces; boundedness and the space B(X) are those of A bounded linear operator between normed spaces.

Elementary identities. Let λ∈ρ(A) and R:=R(λ,A). Since R is the inverse of the bijection λI−A:D(A)→X, its range is D(A): RX=D(A). It satisfies R(λI−A)y=y(y∈D(A)),(λI−A)Rx=x(x∈X). The first identity says R(λ,A)(λI−A)=ID(A) and the second says (λI−A)R(λ,A)=IX. Reading the second identity as λRx−ARx=x and rearranging gives λR(λ,A)x−x=AR(λ,A)x(x∈X), that is AR(λ,A)=λR(λ,A)−I as everywhere-defined operators X→X; in particular AR(λ,A)∈B(X) even though A itself need not be bounded. The same identities hold for every λ∈ρ(A), and σ(A):=K∖ρ(A) collects the scalars for which λI−A fails to be bijective or has an unbounded inverse. Under DC, the latter possibility is excluded by the closed graph argument above. When X is a Hilbert space this is the Banach-space form of Resolvent and spectrum of an unbounded operator, with the same shift convention λI−A; the closed graph theorem replaces the Hilbert-space convention that the resolvent is bounded by definition.

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Resolvent identity for closed operators

Statement

Let A:D(A)⊆X→X be a closed linear operator on a Banach space X, with resolvent R(λ,A)=(λI−A)−1 on ρ(A) (Resolvent and spectrum of a closed operator on a Banach space). For all λ,μ∈ρ(A), R(λ,A)−R(μ,A)=(μ−λ)R(λ,A)R(μ,A), and consequently R(λ,A)R(μ,A)=R(μ,A)R(λ,A).

Facts & Assumptions

Given: A closed linear operator A:D(A)⊆X→X on a Banach space X, scalars λ,μ∈ρ(A), and the resolvents R(λ,A),R(μ,A)∈B(X) (Resolvent and spectrum of a closed operator on a Banach space).

[F1]

R(λ,A)=(λI−A)−1 satisfies R(λ,A)X=D(A), R(λ,A)(λI−A)y=y for y∈D(A) and (λI−A)R(λ,A)x=x for x∈X; the same holds with μ in place of λ (Resolvent and spectrum of a closed operator on a Banach space). Ranges of resolvents lie in D(A), and composition with the bounded maps R(λ,A),R(μ,A) is associative and bilinear wherever defined (A bounded linear operator between normed spaces, Unbounded linear operators: domain, graph and extension).

Proof

technique · direct, factoring the difference of the two inverse shifts through the common domain $D(A)$
1.1F1algebra

For x∈X the vector R(μ,A)x lies in D(A) by [F1], so (λI−A)R(μ,A)x and (μI−A)R(μ,A)x are defined and differ by (λ−μ)R(μ,A)x; applying the bounded linear map R(λ,A) and using linearity gives (λ−μ)R(λ,A)R(μ,A)x=R(λ,A)(λI−A)R(μ,A)x−R(λ,A)(μI−A)R(μ,A)x.

2.1F1step 1.1

The first term on the right equals R(μ,A)x, because R(λ,A)(λI−A)y=y for y∈D(A) and R(μ,A)x∈D(A); the second equals R(λ,A)x, because (μI−A)R(μ,A)=I on X. Hence (λ−μ)R(λ,A)R(μ,A)x=R(μ,A)x−R(λ,A)x, that is R(λ,A)x−R(μ,A)x=(μ−λ)R(λ,A)R(μ,A)x.

3.1step 2.1

Since x∈X was arbitrary, R(λ,A)−R(μ,A)=(μ−λ)R(λ,A)R(μ,A).

4.1step 3.1algebra∎

Interchanging λ and μ gives R(μ,A)−R(λ,A)=(λ−μ)R(μ,A)R(λ,A); adding the two identities yields 0=(μ−λ)[R(λ,A)R(μ,A)−R(μ,A)R(λ,A)]. For λ≠μ this gives R(λ,A)R(μ,A)=R(μ,A)R(λ,A), and for λ=μ the equality is trivial.

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Laplace transform formula for the resolvent

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator A, and let M≥1, ω∈R satisfy ∥T(t)∥≤Meωt for all t≥0 (Exponential bound for a C0-semigroup). Then for every real λ>ω: λ∈ρ(A); for every x∈X the improper Bochner integral ∫0∞e−λtT(t)x dt:=lim⁡R→∞∫0Re−λtT(t)x dt converges in X; and R(λ,A)x=∫0∞e−λtT(t)x dt,∥R(λ,A)∥≤Mλ−ω. In particular (ω,∞)⊆ρ(A).

Facts & Assumptions

Given: Dependent Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator A, constants M≥1, ω∈R with ∥T(t)∥≤Meωt, and a real λ>ω; for R>0, JRx:=∫0Re−λtT(t)x dt.

[F1]

The exponential bound and the Bochner framework: continuous curves on compact intervals are Bochner integrable, ∥∫Eh∥≤∫E∥h∥, and X is complete (Exponential bound for a C0-semigroup, Bochner-integrable function, Bochner integrability criterion, Bochner integral norm inequality).

[F2]

The generator is closed and densely defined (The generator is closed and densely defined), for x∈D(A) the orbit is differentiable with T(t)x∈D(A) and T(t)Ax=AT(t)x (The generator commutes with the semigroup on its domain), and the fundamental theorem of calculus holds for continuous curves with continuous derivative (Fundamental theorem of calculus for Banach-valued continuous curves, Infinitesimal generator of a C0-semigroup).

[F3]

Linearity of the Bochner integral (Linearity of the Bochner integral), and average convergence for continuous curves (Average convergence for a continuous Banach-valued function).

[F4]

Resolvent vocabulary: ρ(A) consists of the scalars with λI−A bijective and bounded inverse, and then R(λ,A)=(λI−A)−1∈B(X) (Resolvent and spectrum of a closed operator on a Banach space).

Proof

technique · direct: define the Laplace integral as a norm limit of finite integrals, verify the two inverse identities on $D(A)$, and transfer them to $X$ by closedness and density
1.1F1F3

Convergence and bound: for 0<R<R′ the curves e−λtT(t)x are continuous, hence Bochner integrable on compacts, and [F1] gives ∥JR′x−JRx∥≤∫RR′Me(ω−λ)t∥x∥ dt≤M∥x∥e(ω−λ)Rλ−ω→0 as R→∞. Thus (JRx)R is Cauchy for every x, so Jx:=lim⁡R→∞JRx exists, and the same estimate at R=0 gives ∥Jx∥≤Mλ−ω∥x∥; the map x↦Jx is linear.

1.2F2

For x∈D(A) the curve g(t):=e−λtT(t)x is differentiable with g′(t)=−λe−λtT(t)x+e−λtT(t)Ax=−e−λt(λI−A)T(t)x by [F2]. The pair curve t↦(e−λtT(t)x,e−λtT(t)Ax) is continuous with values in the closed graph Γ(A)⊆X⊕X, so its Bochner integral lies in Γ(A): approximate the pair uniformly on [0,R] by step functions sampled at partition points. Each simple integral is a finite linear combination of graph vectors, hence lies in the graph; the coordinate integrals converge by the norm inequality, and closedness retains their limit. Thus consequently JRx∈D(A) and AJRx=∫0Re−λtT(t)Ax dt.

2.1F2step 1.2

Hence, for x∈D(A), (λI−A)JRx=λJRx−AJRx=∫0Rg′(t) dt⋅(−1); more explicitly λJRx−AJRx=∫0Re−λt(λT(t)x−T(t)Ax)dt=−∫0Rg′(t) dt=x−e−λRT(R)x by the fundamental theorem of calculus [F2], and ∥e−λRT(R)x∥≤Me(ω−λ)R∥x∥→0 as R→∞.

3.1F2step 1.1step 2.1

Passing to the limit R→∞ in the identity of [step 2.1]: JRx→Jx and (λI−A)JRx→x; since A is closed (hence λI−A is closed), the pair limit gives Jx∈D(A) and (λI−A)Jx=x for every x∈D(A).

3.2F3step 2.1

Likewise J(λI−A)x=lim⁡R→∞JR(λI−A)x=lim⁡R→∞(λJRx−AJRx)=x for every x∈D(A), by the same computation as [step 2.1]; note that (λI−A)x∈X is a fixed vector to which the definition of J applies.

4.1F2step 3.1step 3.2

λI−A is injective: if (λI−A)x=0 for x∈D(A), then x=J(λI−A)x=J0=0 by [step 3.2]. It is also surjective: for y∈X choose yn∈D(A) with yn→y (density, [F2]); then (Jyn) converges to Jy and (λI−A)Jyn=yn→y, so closedness of λI−A gives Jy∈D(A) and (λI−A)Jy=y.

5.1F4step 1.1step 4.1∎

Therefore λI−A:D(A)→X is bijective with inverse J, which is bounded with ∥J∥≤M/(λ−ω); hence λ∈ρ(A) and R(λ,A)=J=∫0∞e−λtT(t)x dt as an improper Bochner integral, with ∥R(λ,A)∥≤M/(λ−ω). As λ>ω was arbitrary, (ω,∞)⊆ρ(A).

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Resolvent power estimates for semigroup generators

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator A and let M≥1, ω∈R satisfy ∥T(t)∥≤Meωt (Exponential bound for a C0-semigroup). Then for every real λ>ω and every integer m≥1, R(λ,A)mx=1(m−1)!∫0∞sm−1e−λsT(s)x ds(x∈X), the integral converging absolutely, and ∥R(λ,A)m∥≤M(λ−ω)m.

Facts & Assumptions

Given: Dependent Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator A and ∥T(t)∥≤Meωt (Exponential bound for a C0-semigroup); real λ>ω and an integer m≥1.

[F1]

Laplace formula: for real μ>ω, μ∈ρ(A) and R(μ,A)x=∫0∞e−μtT(t)x dt for every x, with ∥R(μ,A)∥≤M/(μ−ω) (Laplace transform formula for the resolvent).

[F2]

Resolvent identity: R(μ,A)−R(λ,A)=(λ−μ)R(μ,A)R(λ,A) (Resolvent identity for closed operators).

[F3]

Bochner-integral toolkit: linearity, the norm inequality, and the fact that continuous curves on compact intervals are Bochner integrable; improper integrals of ∥T(t)x∥-dominated curves converge by the usual Cauchy estimate (Linearity of the Bochner integral, Bochner integral norm inequality, Bochner-integrable function).

Proof

technique · direct: differentiate the Laplace representation in $\lambda$, identify the derivative with $-R(\lambda,A)^2$ by the resolvent identity, and iterate
1.1F1F3

For λ>ω the integrand t↦e−λtT(t)x is dominated in norm by Me(ω−λ)t∥x∥; for a fixed λ0∈(ω,λ) the tails satisfy ∫R∞te(ω−λ0)t dt→0 as R→∞, which is the uniform-in-h domination used below.

1.2F1F2

The function μ↦R(μ,A)x is differentiable on (ω,∞) with derivative −R(λ,A)2x: by [F2], R(λ+h,A)−R(λ,A)hx=−R(λ+h,A)R(λ,A)x, and R(λ+h,A)→R(λ,A) in operator norm as h→0 because ∥R(λ+h,A)−R(λ,A)∥≤∣h∣ ∥R(λ+h,A)∥ ∥R(λ,A)∥ and ∥R(λ+h,A)∥ is bounded near λ by [F1].

2.1F1F3step 1.1

The same derivative computed from the integral is −∫0∞te−λtT(t)x dt: the difference quotient is ∫0∞e−ht−1he−λtT(t)x dt, whose integrands converge pointwise to −te−λtT(t)x and are dominated by te(ω−λ0)tM∥x∥ for ∣h∣ small and λ0∈(ω,λ]; splitting the integral at R and using uniform convergence on [0,R] for the mean-value estimate ∣(e−ht−1)/h∣≤te∣h∣t and the tail estimate of [step 1.1] passes the limit through the improper integral.

3.1step 1.2step 2.1

Comparing [step 1.2] and [step 2.1]: R(λ,A)2x=∫0∞te−λtT(t)x dt for every x, and the integral converges absolutely.

4.1F1F2step 1.2step 2.1step 3.1

Induction on m gives R(λ,A)mx=1(m−1)!∫0∞sm−1e−λsT(s)x ds: the case m=1 is [F1] and the case m=2 is [step 3.1]. Assume the formula for m; since the resolvents commute, ddλR(λ,A)m=−mR(λ,A)m+1 by [step 1.2], while differentiating the integral representation in λ (the same tail-splitting argument as [step 2.1], with domination sme(ω−λ0)sM∥x∥ with ω<λ0<λ and ∣h∣<λ−λ0) gives −1(m−1)!∫0∞sme−λsT(s)x ds. Equating the two expressions yields the formula for m+1.

5.1F1F3step 4.1∎

Norm bound: by [F3] and M≥1, ∥R(λ,A)mx∥≤1(m−1)!∫0∞sm−1e(ω−λ)sM∥x∥ ds=M∥x∥(λ−ω)m, the scalar integral is (m−1)!/(λ−ω)m: integration by parts on [0,R] gives Im=(m−1)Im−1/(λ−ω) after R→∞, with I1=1/(λ−ω) and vanishing polynomial-exponential boundary terms. For all nearby difference quotients use a strictly smaller parameter λ0∈(ω,λ); polynomial times e−(λ0−ω)s is integrable by the same recurrence. Thus taking the supremum over ∥x∥≤1 gives ∥R(λ,A)m∥≤M/(λ−ω)m.

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The exponential series of a bounded operator

Statement

Let X be a real or complex Banach space and let A∈B(X) (A bounded linear operator between normed spaces). For t∈R define E(t):=∑n=0∞tnn!An. Then the series converges absolutely in the operator norm ∥⋅∥ of B(X), uniformly for t in compact subsets of R; E(t)∈B(X) with ∥E(t)∥≤e∣t∣∥A∥; E(0)=I and E(t+s)=E(t)E(s) for all s,t∈R; t↦E(t) is of class C∞ in the operator norm with E′(t)=AE(t)=E(t)A, and ∥E(t)−I−tA∥≤t22∥A∥2e∣t∣∥A∥, so that E(t)−It→A in operator norm as t→0. In particular E is a uniformly continuous (hence strongly continuous) group of bounded operators on X whose generator is the bounded operator A.

Facts & Assumptions

Given: A real or complex Banach space X, an operator A∈B(X), and for t∈R and N∈N the partial sums sN(t):=∑n=0Ntnn!An and the series E(t):=∑n=0∞tnn!An.

[F2]

Composition in B(X) is associative and bilinear, I is its identity, and ∥ST∥≤∥S∥ ∥T∥ for all S,T∈B(X) (Composition satisfies |ST|\le|S|,|T|); consequently ∥An∥≤∥A∥n for every n≥0. Completeness of B(X) for the operator norm [F1] together with these facts is all the structure used below; no separate Banach-algebra packaging is needed, and the estimates are identical over R and C.

[F3]

In a Banach space a series converges whenever it converges absolutely, i.e. whenever the series of norms converges; its partial sums are then Cauchy (Series criterion for Banach spaces, An absolutely convergent series has Cauchy partial sums, Series and absolute convergence in a normed space).

[F4]

For real u≥0 the exponential series satisfies ∑n≥0un/n!=eu<∞, and eu>0 (The real exponential function and the number e by a power series, The exponential is a continuous bijection from R onto (0,∞)); by n!≥2 (n−2)! for n≥2, its tail obeys ∑n≥2un/n!≤u22eu.

Proof

technique · direct, by norm estimates on the exponential series and an elementary Cauchy-product lemma
1.1F2F4

The series converges absolutely for every real t: by [F2] the general term obeys ∥tnAn/n!∥≤∣t∣n∥A∥n/n!, so with u:=∣t∣ ∥A∥ the comparison series ∑nun/n! is the scalar exponential of [F4] and converges.

1.2F1F2F3

Cauchy-product step. If ∑kxk and ∑jyj converge absolutely in B(X) and zn:=∑k+j=nxkyj, then ∑nzn converges absolutely with sum (∑kxk)(∑jyj). Indeed ∑n∥zn∥≤∑n∑k+j=n∥xk∥ ∥yj∥=(∑k∥xk∥)(∑j∥yj∥)<∞ by [F2], so ∑nzn converges by [F3]. Writing PN:=∑k≤Nxk, QN:=∑j≤Nyj, TN:=∑n≤Nzn, the product PNQN=∑n≤2N∑k+j=n, k,j≤Nxkyj differs from T2N only by the terms with k>N or j>N, so ∥PNQN−T2N∥≤(∑k>N∥xk∥)(∑j∥yj∥)+(∑k∥xk∥)(∑j>N∥yj∥)→0 by [F2] and absolute convergence; since PNQN→(∑xk)(∑yj) by continuity of the product, the subsequence T2N converges to that product, and a subsequence of a convergent sequence has the same limit, so ∑nzn=(∑kxk)(∑jyj).

2.1F3F4step 1.1

Hence E(t):=∑n≥0tnAn/n!∈B(X) is defined for every t by [F3], the family of series is dominated by the convergent scalar series ∑nRn∥A∥n/n! on every compact interval ∣t∣≤R, so the convergence is uniform there and in particular t↦E(t) is continuous in operator norm; and ∥E(t)∥≤∑n∣t∣n∥A∥n/n!=e∣t∣ ∥A∥<∞.

2.2step 1.2step 1.1algebra

Applying [step 1.2] to xk=tkk!Ak and yj=sjj!Aj, whose series converge absolutely by [step 1.1], gives E(t)E(s)=∑n(∑k+j=ntksjk!j!)An=∑n(t+s)nn!An=E(t+s), where AkAj=Ak+j and the binomial theorem in the commutative subalgebra generated by A were used.

2.3F2F4step 1.1algebra

For h≠0 and n≥1, the binomial expansion gives (t+h)n−tnh−ntn−1=∑j=2n(nj)hj−1tn−j; subtracting the two absolutely convergent series and using ∑n≥1ntn−1n!An=AE(t), the difference quotient obeys ∥E(t+h)−E(t)h−AE(t)∥≤∑n≥2∑j=2n(nj)∣h∣j−1∣t∣n−j∥A∥nn!=(∑j≥2∣h∣j−1∥A∥jj!)(∑m≥0∣t∣m∥A∥mm!)≤∣h∣ ∥A∥22e∣h∣ ∥A∥e∣t∣ ∥A∥, which tends to 0 as h→0; the rearrangement of the nonnegative double series is legitimate and the tail estimate is [F4].

3.1F1F2step 2.1

At t=0 all terms with n≥1 vanish, so E(0)=I; and since multiplication is continuous in the operator norm by [F1] and [F2], the product of the partial sums converges, which is what the next steps quantify.

3.2F2F4step 2.1

The quadratic remainder is ∥E(t)−I−tA∥=∥∑n≥2tnn!An∥≤∑n≥2∣t∣n∥A∥nn!≤t2∥A∥22e∣t∣ ∥A∥ by [F2] and the tail bound of [F4], since n!≥2(n−2)! for n≥2.

3.3step 2.3F2algebra

Therefore E is differentiable on R with E′(t)=AE(t); since A commutes with every power An, continuity of multiplication and [step 2.1] give AE(t)=lim⁡NA sN(t)=lim⁡NsN(t)A=E(t)A as well. Iterating, if E is m times differentiable with E(m)=AmE, then t↦AmE(t) is differentiable with derivative AmE′(t)=Am+1E(t) because Am is bounded; hence E(m)=AmE for all m, that is E is C∞ with E(m)(t)=AmE(t).

4.1step 2.1step 2.2step 3.2step 3.3∎

Taking t→0 in [step 3.2] gives ∥E(t)−It−A∥≤∣t∣ ∥A∥22e∣t∣ ∥A∥→0, so the difference quotients of E at 0 converge to A in operator norm; consequently E is a uniformly continuous group: by [step 2.2] E(0)=I and E(t)E(−t)=E(0)=I=E(−t)E(t), so each E(t) is invertible with inverse E(−t), and t↦E(t) is norm continuous on R by [step 2.1], hence strongly continuous, and the difference-quotient limit at 0 identifies its generator with the bounded operator A.

Notes. The same estimates give ∥E(t)∥≤e∣t∣ ∥A∥ and show the series converges in operator norm uniformly on compact t-intervals; nothing here uses a choice principle, and the zero space X={0} is included through the estimates ∥A∥=0, E(t)=I=0.

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The Yosida resolvent converges strongly to the identity

Statement

Let A:D(A)⊆X→X be closed and densely defined on a Banach space X, and let M≥1, ω∈R be such that (ω,∞)⊆ρ(A) and ∥R(λ,A)n∥≤M(λ−ω)−n for all real λ>ω and every n≥1 (Resolvent and spectrum of a closed operator on a Banach space). Then, as λ→∞ along the reals, λR(λ,A)x→x for every x∈X, and λR(λ,A)Ax→Ax for every x∈D(A).

Facts & Assumptions

Given: A closed densely defined operator A on a Banach space X with (ω,∞)⊆ρ(A) and ∥R(λ,A)n∥≤M(λ−ω)−n for all real λ>ω and n≥1 (Resolvent and spectrum of a closed operator on a Banach space).

[F1]

Resolvent identities: R(λ,A)X=D(A), R(λ,A)(λI−A)y=y for y∈D(A), and λR(λ,A)x−x=AR(λ,A)x for x∈X; in particular λR(λ,A)y=y+R(λ,A)Ay for y∈D(A), since λR(λ,A)y=R(λ,A)((λI−A)y+Ay)=y+R(λ,A)Ay (Resolvent and spectrum of a closed operator on a Banach space).

[F2]

First power estimate: ∥R(λ,A)∥≤M/(λ−ω) for λ>ω (Resolvent power estimates for semigroup generators is the source of this estimate in the semigroup case; here it is assumed directly).

[F3]

D(A) is dense in X, and operators of the form λR(λ,A) are bounded (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

technique · direct: the identity on the dense domain and a uniform bound, then a three-epsilon argument
1.1F1F2

For y∈D(A) and λ>ω, [F1] gives λR(λ,A)y=y+R(λ,A)Ay, hence by [F2] ∥λR(λ,A)y−y∥≤Mλ−ω∥Ay∥→0 as λ→∞.

1.2F2

The operators Bλ:=λR(λ,A) are uniformly norm bounded for large λ: ∥Bλ∥≤λMλ−ω≤2M for λ≥2ω when ω>0, and ∥Bλ∥≤M for ω≤0 and λ>0.

2.1F3step 1.1step 1.2

For arbitrary x∈X: given ε>0, choose y∈D(A) with ∥x−y∥<ε/(4M+2) by density [F3]; by [step 1.1] choose λ0 with ∥λR(λ,A)y−y∥<ε/2 for λ>λ0; then for such λ, ∥λR(λ,A)x−x∥≤∥Bλ(x−y)∥+∥λR(λ,A)y−y∥+∥y−x∥<2Mε4M+2+ε2+ε4M+2<ε. Hence λR(λ,A)x→x for every x∈X.

3.1F1step 2.1∎

The second statement is [step 2.1] applied to the vector Ax∈X: λR(λ,A)Ax→Ax; by [F1] and [F2] this is the same as λAR(λ,A)x=λ2R(λ,A)x−λx→Ax for x∈D(A).

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Yosida approximants

Definition

Let A:D(A)⊆X→X be closed and densely defined on a Banach space X with (ω,∞)⊆ρ(A), and let R(λ,A)=(λI−A)−1 (Resolvent and spectrum of a closed operator on a Banach space). For real λ>ω the Yosida approximant of A at λ is Aλ:=λAR(λ,A)=λ2R(λ,A)−λI∈B(X). The equality with λ2R(λ,A)−λI uses AR(λ,A)=λR(λ,A)−I; in particular Aλ is a bounded everywhere defined operator. Distinct approximants commute, AλAμ=AμAλ, and each commutes with R(μ,A).

The two formulae agree. Since λ∈ρ(A), the resolvent identity AR(λ,A)=λR(λ,A)−I of Resolvent and spectrum of a closed operator on a Banach space gives Aλ=λAR(λ,A)=λ(λR(λ,A)−I)=λ2R(λ,A)−λI, and the right-hand side is a sum of bounded everywhere defined operators (A bounded linear operator between normed spaces). Hence each Aλ is bounded with D(Aλ)=X, and it is meant as a bounded approximation of the possibly unbounded A; the sense in which Aλx→Ax for x∈D(A) is a theorem proved on this page, not part of the definition.

Commutativity. For real λ,μ>ω the resolvents commute, R(λ,A)R(μ,A)=R(μ,A)R(λ,A), by Resolvent identity for closed operators. Therefore AλAμ=(λ2R(λ,A)−λI)(μ2R(μ,A)−μI) is symmetric in λ and μ: the only mixed term that is not obviously symmetric is λ2μ2R(λ,A)R(μ,A), and that product is symmetric by the resolvent identity. Hence distinct approximants commute, and for the same reason AλR(μ,A)=λ(λR(λ,A)−I)R(μ,A)=λ2R(λ,A)R(μ,A)−λR(μ,A) equals R(μ,A)Aλ, since Aλ is the product of R(λ,A) with bounded operators and resolvents at λ and μ commute. The closedness and density of A are hypotheses, not conclusions drawn from a generator theorem. The resolvent vocabulary is Resolvent and spectrum of a closed operator on a Banach space, and the definition itself asserts no approximation property.

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Yosida approximants are bounded and converge on the domain

Statement

Under the hypotheses of The Yosida resolvent converges strongly to the identity, let Aλ=λAR(λ,A) be the Yosida approximants (Yosida approximants). Then ∥Aλ∥≤λ2Mλ−ω+∣λ∣ for λ>ω, and Aλx→Ax as λ→∞ for every x∈D(A).

Facts & Assumptions

Given: The hypotheses of The Yosida resolvent converges strongly to the identity on the closed densely defined operator A, and the Yosida approximants Aλ=λAR(λ,A)=λ2R(λ,A)−λI (Yosida approximants).

[F1]

Aλ=λ2R(λ,A)−λI is bounded with ∥Aλ∥≤λ2∥R(λ,A)∥+∣λ∣ (Yosida approximants, The operator norm is a norm on the space of bounded linear operators, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F2]

∥R(λ,A)∥≤M/(λ−ω) and λR(λ,A)x→x for all x∈X, λR(λ,A)Ax→Ax for x∈D(A) (the bound is a hypothesis of The Yosida resolvent converges strongly to the identity, which proves both convergence conclusions).

[F3]

For x∈D(A) one has AR(λ,A)x=R(λ,A)Ax: both equal λR(λ,A)x−x by the resolvent identity (Resolvent and spectrum of a closed operator on a Banach space).

Proof

technique · direct: the norm bound from the identity form, and pointwise convergence from the strong resolvent convergence
1.1F1F2

By [F1] and [F2], ∥Aλ∥≤λ2Mλ−ω+∣λ∣ for λ>ω, which is the asserted bound.

1.2F2F3

For x∈D(A), Aλx=λAR(λ,A)x=λR(λ,A)Ax by [F3]; by [F2] the right-hand side converges to Ax as λ→∞. Hence Aλx→Ax for every x∈D(A).

2.1step 1.1step 1.2∎

Both conclusions hold for every λ>ω, and no uniform convergence on all of D(A) or on bounded sets is claimed.

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Bounded Yosida semigroups converge to the generated semigroup

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let A:D(A)⊆X→X be closed and densely defined on a Banach space X and let M≥1, ω∈R satisfy (ω,∞)⊆ρ(A) and ∥R(λ,A)n∥≤M(λ−ω)−n for all real λ>ω, n≥1. Let Aλ=λAR(λ,A) be the Yosida approximants (Yosida approximants) and let Eλ(t):=etAλ be the bounded-operator exponentials of The exponential series of a bounded operator. Then for every x∈X the limit T(t)x:=lim⁡λ→∞Eλ(t)x exists, uniformly for t in compact subsets of [0,∞), and T is a strongly continuous semigroup on X with ∥T(t)∥≤Meωt and generator A.

Facts & Assumptions

Given: Dependent Choice; A closed densely defined operator A on a Banach space X with (ω,∞)⊆ρ(A) and ∥R(λ,A)n∥≤M(λ−ω)−n for all real λ>ω and n≥1; the Yosida approximants Aλ=λAR(λ,A) and the exponentials Eλ(t)=etAλ of The exponential series of a bounded operator (Yosida approximants, Resolvent and spectrum of a closed operator on a Banach space).

[F1]

Exponential series: Eλ(t)=∑j≥0tjj!Aλj converges in operator norm, ∥Eλ(t)∥≤e∣t∣ ∥Aλ∥, Eλ(0)=I, Eλ(t+s)=Eλ(t)Eλ(s), Eλ is C1 with Eλ′(t)=AλEλ(t), so Eλ(t)x−x=∫0tEλ(s)Aλx ds for every x by the fundamental theorem of calculus (The exponential series of a bounded operator, Fundamental theorem of calculus for Banach-valued continuous curves).

[F2]

The approximants satisfy the norm bound of Yosida approximants are bounded and converge on the domain and Aλx→Ax for every x∈D(A); distinct approximants commute and so do the exponentials (Yosida approximants, Composition satisfies |ST|\le|S|,|T|).

[F3]

A is closed and D(A) is dense by hypothesis; since Aλ=λ2R(λ,A)−λI and R(λ,A) commutes with I, the binomial Cauchy-product argument in the exponential-series proof, applied to the commuting bounded operators −λI and λ2R(λ,A), gives Eλ(t)=e−λtetλ2R(λ,A), so the resolvent power estimates yield ∥Eλ(t)∥≤e−λt∑j≥0(tλ2)jj!M(λ−ω)j=Meωtλ/(λ−ω) for t≥0. [F1, F2]

[F4]

Average convergence and strong continuity: a continuous curve is Bochner integrable and its forward averages converge to its value (Average convergence for a continuous Banach-valued function); continuity at 0 plus the semigroup law gives continuity of every orbit (Continuity at time zero implies continuity of every orbit, Strongly continuous semigroup).

[F5]

Laplace formula: a strongly continuous semigroup with ∥T(t)∥≤Meωt has (ω,∞)⊆ρ(its generator) (Laplace transform formula for the resolvent).

Proof

technique · direct: uniform bounds and a Cauchy estimate on the dense domain, then passage to the limit, and identification of the generator by uniqueness of the resolvent
1.1F3

Uniform bound. For λ>max⁡{ω,0} and t≥0, [F3] gives ∥Eλ(t)∥≤Meωtλ/(λ−ω). For every T0<∞, ωtλ/(λ−ω)→ωt uniformly on 0≤t≤T0. Thus the displayed majorants converge uniformly to Meωt there; they are uniformly bounded for large λ, and lim sup⁡λ→∞∥Eλ(t)∥≤Meωt for each fixed t, regardless of the sign of ω.

2.1F1F2step 1.1

Cauchy estimate on D(A). For x∈D(A) and λ,μ>ω, the exponentials commute and the FTC gives Eλ(t)x−Eμ(t)x=−∫0tdds[Eλ(t−s)Eμ(s)x]ds=∫0tEλ(t−s)Eμ(s)(Aλx−Aμx) ds; hence ∥Eλ(t)x−Eμ(t)x∥≤t C(t)2∥Aλx−Aμx∥ with C(t):=sup⁡λ large,0≤s≤t∥Eλ(s)∥ finite by [step 1.1]. Since Aλx→Ax by [F2], the family (Eλ(t)x)λ is Cauchy, uniformly for t in compact intervals.

3.1F2F3step 1.1step 2.1

The limit and its bound. For x∈X and y∈D(A) close to x, ∥Eλ(t)x−Eμ(t)x∥≤∥Eλ(t)(x−y)∥+∥Eμ(t)(x−y)∥+∥Eλ(t)y−Eμ(t)y∥, and the first two terms are small uniformly in λ,μ and t in compacts by [step 1.1] while the last is small by [step 2.1]; density [F3] gives convergence uniformly on compact t-intervals for every x. The limit orbit is continuous on each compact interval: for any point, bound its increment by the two uniform approximation errors and the increment of one continuous approximating orbit. Define T(t)x:=lim⁡λEλ(t)x; then T(t) is linear and bounded with ∥T(t)x∥=lim⁡λ∥Eλ(t)x∥≤Meωt∥x∥ by [step 1.1].

4.1F1step 3.1

Semigroup law. For t,s≥0 and x∈X, Eλ(t+s)x=Eλ(t)Eλ(s)x→T(t)T(s)x by [step 3.1] and the uniform bound on compacts, while Eλ(t+s)x→T(t+s)x; hence T(t+s)=T(t)T(s), and T(0)=I.

4.2F1F4step 3.1

Strong continuity. For x∈D(A) and t in a compact interval, Eλ(t)x−x=∫0tEλ(s)Aλx ds=∫0tEλ(s)Ax ds+∫0tEλ(s)(Aλx−Ax) ds, and the two integrals tend to ∫0tT(s)Ax ds and 0 uniformly, because Eλ(s)→T(s) strongly uniformly on the interval by [step 3.1] and Aλx→Ax; hence T(t)x−x=∫0tT(s)Ax ds→0 as t↓0 by average convergence [F4]. With the local bound of [step 3.1] this extends from the dense domain to all x, so T(t)x→x as t↓0; by the semigroup law and [F4] every orbit is continuous, so T is a strongly continuous semigroup with bound ∥T(t)∥≤Meωt.

5.1F4step 4.2

The generator contains A. The identity of [step 4.2] shows T(t)x−x=∫0tT(s)Ax ds for x∈D(A); dividing by t and using average convergence for the continuous curve s↦T(s)Ax gives T(t)x−xt→Ax. Hence D(A)⊆D(B) and Bx=Ax for x∈D(A), where B is the generator of T.

6.1F2F5step 5.1∎

A=B. By [F5] applied to T and its bound, every real λ>ω lies in ρ(B); it also lies in ρ(A) by hypothesis, and λI−A=λI−B on D(A). Thus λI−A:D(A)→X and λI−B:D(B)→X are both bijections agreeing on D(A): given x∈D(B), (λI−B)x=(λI−A)y for some y∈D(A), and since (λI−B)y=(λI−A)y=(λI−B)x while λI−B is injective, x=y∈D(A); hence D(A)=D(B) and A=B.

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Hille-Yosida generation theorem

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let A:D(A)⊆X→X be a closed and densely defined linear operator on a Banach space X and let M≥1, ω∈R. Then A generates a strongly continuous semigroup (T(t))t≥0 with ∥T(t)∥≤Meωt for all t≥0 if and only if both: (i) (ω,∞)⊆ρ(A), and (ii) ∥R(λ,A)n∥≤M(λ−ω)−n for every real λ>ω and every n≥1 (Resolvent and spectrum of a closed operator on a Banach space). All resolvent powers are required in general; the first power alone guarantees all power estimates when M=1, the exponentially rescaled contraction case, where all higher power estimates follow from the single one by submultiplicativity (treated later on this page).

Facts & Assumptions

Given: Dependent Choice; A closed densely defined linear operator A on a Banach space X and constants M≥1, ω∈R (Resolvent and spectrum of a closed operator on a Banach space, The generator is closed and densely defined).

[F1]

Sufficiency: under (i) (ω,∞)⊆ρ(A) and (ii) ∥R(λ,A)n∥≤M(λ−ω)−n for all real λ>ω, n≥1, the operator A generates a strongly continuous semigroup with ∥T(t)∥≤Meωt (Bounded Yosida semigroups converge to the generated semigroup).

[F2]

Necessity of the location of the spectrum and of the first estimate: if T is a strongly continuous semigroup with generator A and ∥T(t)∥≤Meωt, then A is closed and densely defined, (ω,∞)⊆ρ(A), and ∥R(λ,A)∥≤M/(λ−ω) (The generator is closed and densely defined, Laplace transform formula for the resolvent, Strongly continuous semigroup).

[F3]

Necessity of all powers: under the hypotheses of [F2], R(λ,A)mx=1(m−1)!∫0∞sm−1e−λsT(s)x ds and ∥R(λ,A)m∥≤M(λ−ω)−m for every m≥1 (Resolvent power estimates for semigroup generators).

Proof

technique · direct: sufficiency is the Yosida construction and necessity is read off from the Laplace representation of the resolvent
1.1F1

(Sufficiency.) Assume (i) and (ii). Then all hypotheses of [F1] hold, so A generates a strongly continuous semigroup (T(t))t≥0 with ∥T(t)∥≤Meωt for all t≥0.

1.2F2

(Necessity, domain and spectrum.) Assume conversely that A generates T with ∥T(t)∥≤Meωt. Then A is closed with dense domain by [F2]; the Laplace-transform formula for the resolvent gives λ∈ρ(A) and ∥R(λ,A)∥≤M/(λ−ω) for every real λ>ω; in particular (ω,∞)⊆ρ(A), which is (i).

1.3F3

(Necessity, all powers.) Under the same hypothesis, [F3] gives the integral representation of every power R(λ,A)m and the estimate ∥R(λ,A)m∥≤M(λ−ω)−m for all m≥1 and real λ>ω, which is (ii).

2.1step 1.1step 1.2step 1.3∎

Combining [step 1.1] with [steps 1.2-1.3]: A generates a strongly continuous semigroup with ∥T(t)∥≤Meωt if and only if (i) and (ii) hold. The general theorem retains all power estimates; the case where the first estimate alone suffices (M=1, ω=0) is isolated as the next corollary.

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Contraction Hille-Yosida theorem

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let A:D(A)⊆X→X be closed and densely defined on a Banach space X. Then A generates a strongly continuous semigroup of contractions (∥T(t)∥≤1 for all t≥0) if and only if (0,∞)⊆ρ(A) and ∥λR(λ,A)∥≤1 for all λ>0, equivalently ∥R(λ,A)∥≤1/λ for all λ>0. In this case the first-power estimate implies all the power estimates ∥R(λ,A)n∥≤λ−n by submultiplicativity, so no separate power condition is needed; when X is complex, every z with Re⁡z>0 belongs to ρ(A) and ∥R(z,A)n∥≤(Re⁡z)−n for every n≥1.

Facts & Assumptions

Given: Dependent Choice; A closed densely defined operator A on a Banach space X (Hille-Yosida generation theorem).

[F1]

Hille-Yosida with M=1, ω=0: A generates a strongly continuous semigroup with ∥T(t)∥≤1 for all t≥0 if and only if (0,∞)⊆ρ(A) and ∥R(λ,A)n∥≤λ−n for every real λ>0 and every n≥1 (Hille-Yosida generation theorem).

[F2]

The operator norm is submultiplicative, ∥ST∥≤∥S∥ ∥T∥ (Composition satisfies |ST|\le|S|,|T|), and is a norm on B(X) (The operator norm is a norm on the space of bounded linear operators).

[F3]

The complex exponential satisfies ∣e−zt∣=e−tRe⁡z and ddte−zt=−ze−zt (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0, The complex exponential is entire and its complex derivative is itself). The Laplace inverse argument and resolvent differentiation for real parameters are proved in Laplace transform formula for the resolvent and Resolvent power estimates for semigroup generators; their complex extension is derived in step 2.1.

Proof

technique · direct: specialise the general theorem and observe that the first-power estimate implies all power estimates by submultiplicativity
1.1F1

Suppose A generates a contraction semigroup, ∥T(t)∥≤1. Then [F1] with M=1, ω=0 gives (0,∞)⊆ρ(A) and ∥R(λ,A)n∥≤λ−n for all n≥1; in particular the first-power estimate ∥λR(λ,A)∥≤1, equivalently ∥R(λ,A)∥≤1/λ, holds for all λ>0.

1.2F1F2

Conversely, suppose (0,∞)⊆ρ(A) and ∥λR(λ,A)∥≤1, i.e. ∥R(λ,A)∥≤1/λ, for all λ>0. By submultiplicativity [F2], ∥R(λ,A)n∥≤∥R(λ,A)∥n≤λ−n for every n≥1; hence the power conditions of [F1] hold with M=1, ω=0, and A generates a strongly continuous semigroup of contractions.

2.1F1F3step 1.2

Let X be complex, a:=Re⁡z>0, and Jzx:=∫0∞e−zsT(s)x ds. By [F3] its tail norm is at most e−aR∥x∥/a. For x∈D(A), integration of the derivative of e−zsT(s)x, with the closed-graph integration argument of the Laplace theorem, gives (zI−A)Jzx=x=Jz(zI−A)x; density and closedness extend the first identity to every x∈X, exactly as in that theorem, so z∈ρ(A) and R(z,A)=Jz. For small real h, the resolvent identity gives dR(z+h,A)/dh∣h=0=−R(z,A)2 in operator norm. Differentiating the integral along this real increment is justified by splitting off its tail and dominating by ske−as/2∥x∥ for each needed order; induction gives R(z,A)nx=(n−1)!−1∫0∞sn−1e−zsT(s)x ds. The scalar integral of the norm majorant is (n−1)!/an, by the integration-by-parts recurrence of the power-estimate proof. Hence ∥R(z,A)n∥≤a−n, the precise complex half-plane estimate.

3.1step 1.1step 1.2step 2.1∎

Therefore contractivity of the generated semigroup is equivalent to (0,∞)⊆ρ(A) and ∥λR(λ,A)∥≤1 for all λ>0, and in this case the single estimate forces all the resolvent power bounds.

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Dissipative operator

Definition

Let X be a Banach space over K∈{R,C} and let A:D(A)⊆X→X be a linear operator with domain D(A) (Unbounded linear operators: domain, graph and extension). A is dissipative if ∥(λI−A)x∥≥λ∥x∥for all λ>0 and all x∈D(A); equivalently ∥x−αAx∥≥∥x∥ for all α>0 and x∈D(A). A dissipative operator has injective λI−A for every λ>0 and ∥(λI−A)−1z∥≤λ−1∥z∥ on the range of λI−A; no surjectivity, closedness or density is implied. If X is a Hilbert space (Hilbert space), then A is dissipative if and only if Re ⟨Ax,x⟩≤0 for every x∈D(A): from the norm dissipativity inequality, squaring gives Re ⟨Ax,x⟩≤∥Ax∥2/(2λ) for every λ>0, so letting λ→∞ yields Re ⟨Ax,x⟩≤0. Conversely, if this real-part inequality holds, expanding ∥(λI−A)x∥2=λ2∥x∥2−2λRe ⟨Ax,x⟩+∥Ax∥2 gives the norm dissipativity inequality. Finally, under the Hahn-Banach extension principle HB (The real dominated-extension principle as an additional hypothesis over ZF) dissipativity is equivalent to the norm-duality form: for every x∈D(A) there exists x∗∈X∗ with ∥x∗∥=∥x∥, x∗(x)=∥x∥2 and Re x∗(Ax)≤0; the equivalence is the two-dimensional argument of [T] Lemma 11.19, where HB produces the norming functionals and supplies the extension from span{x,Ax} (Relative dual norming, point separation, and recovery of the norm), and the extraction of the limit uses compactness of the finite-dimensional dual unit ball (For n≥1 every bounded sequence in Rn has a convergent subsequence).

Two normalisations of the defining inequality. Putting α=1/λ shows that the displayed inequality is equivalent to ∥x−αAx∥≥∥x∥ for all α>0 and x∈D(A). Since λx=(λI−A)x+Ax, the inequality ∥(λI−A)x∥≥λ∥x∥ is in turn equivalent to the one-sided estimate λ∥x∥≤∥(λI−A)x∥ used below.

Injectivity and the inverse bound. If (λI−A)x=0 for some λ>0 and x∈D(A), then λ∥x∥≤∥(λI−A)x∥=0, so x=0: each λI−A is injective. If z=(λI−A)x lies in the range, then ∥x∥≤λ−1∥z∥, so the inverse defined on the range satisfies ∥(λI−A)−1z∥≤λ−1∥z∥. No surjectivity onto X, no closedness of A and no density of D(A) is asserted, and none is implied.

The real-part form on a Hilbert space. Let X be a Hilbert space. If A is dissipative and x∈D(A), then for every λ>0 the expansion ∥(λI−A)x∥2=λ2∥x∥2−2λRe⁡⟨Ax,x⟩+∥Ax∥2 gives 2λRe⁡⟨Ax,x⟩≤∥Ax∥2, that is Re⁡⟨Ax,x⟩≤∥Ax∥2/(2λ); letting λ→∞ yields Re⁡⟨Ax,x⟩≤0. Conversely, if Re⁡⟨Ax,x⟩≤0 for all x∈D(A), the same expansion gives ∥(λI−A)x∥2≥λ2∥x∥2 for every λ>0, so A is dissipative. This real-part form is choice-free; the Hilbert-space vocabulary comes from Hilbert space.

The norm-duality form under HB. Assume the Hahn-Banach extension principle (The real dominated-extension principle as an additional hypothesis over ZF) and let x∈D(A). We claim that A is dissipative if and only if for every x∈D(A) there is x∗∈X∗ with ∥x∗∥=∥x∥, x∗(x)=∥x∥2 and Re⁡x∗(Ax)≤0.

Sufficiency. If such x∗ is given and λ>0, then λ∥x∥2=λRe⁡x∗(x)=Re⁡x∗((λI−A)x)+Re⁡x∗(Ax)≤Re⁡x∗((λI−A)x)≤∥x∗∥ ∥(λI−A)x∥=∥x∥ ∥(λI−A)x∥. For x≠0 this is λ∥x∥≤∥(λI−A)x∥; for x=0 it is trivial. Hence A is dissipative.

Necessity. Fix x∈D(A); for x=0 take x∗=0, so assume x≠0 and put M:=span⁡{x,Ax}, a subspace of finite dimension at most two over the scalar field K. For each positive integer λ=n the vector (λI−A)x is nonzero, because (λI−A) is injective. Construct a sequence yn∈M∗ without simultaneously choosing functionals on X. Fix a basis of the finite-dimensional space M. The coordinate vectors of functionals of norm at most one form a closed bounded subset of a finite real coordinate space: ∣y(u)∣≤∥u∥ for every u∈M is an intersection of closed conditions, and each basis evaluation is bounded. For each n, intersect this set with y((nI−A)x)=∥(nI−A)x∥. The intersection is nonempty by Relative dual norming, point separation, and recovery of the norm applied to M, and compact by Heine-Borel in Rn: with the Euclidean metric a subset of Rn is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line. Select its lexicographically least coordinate vector by minimizing its real coordinates successively; each minimum exists because the corresponding projected compact set is nonempty. This finite deterministic procedure defines yn for all n, with norm one and the required norming identity, without Countable Choice. Write yλ=yn, λ=n, below. Then Re⁡yλ(Ax)=λRe⁡yλ(x)−∥(λI−A)x∥≤λ∣yλ(x)∣−λ∥x∥≤0, and likewise yλ(x)=∥(λI−A)x∥/λ+yλ(Ax)/λ, where ∣∥(λI−A)x∥/λ−∥x∥∣≤∥Ax∥/λ→0 and ∣yλ(Ax)∣≤∥Ax∥; hence yλ(x)→∥x∥.

Passing to a subsequence. The restrictions yn lie in the unit ball of the dual of the finite-dimensional space M, which is sequentially compact: after choosing coordinates for M∗, the coordinates of a functional amount to a bounded sequence in a Euclidean space, and For n≥1 every bounded sequence in Rn has a convergent subsequence extracts a convergent subsequence. Take λn→∞ along a subsequence on which yλn converges to some y∈M∗. Then ∥y∥≤1, Re⁡y(Ax)≤0 (a closed condition), and y(x)=lim⁡nyλn(x)=∥x∥; consequently ∥y∥=1 because ∣y(x)∣=∥x∥ with x≠0.

Extension to X. The real part g:=Re⁡y is a real-linear functional on the real vector space M with ∣g(u)∣≤∥u∥ for u∈M; here the real structure of X is the one underlying the complex case as well. Apply HB, with the sublinear functional p(u)=∥u∥, to extend g to a real-linear G:X→R satisfying G(u)≤∥u∥ for all u∈X; then ∣G(u)∣≤∥u∥ by applying the inequality to −u. In the real case set x~:=G. In the complex case set x~(u):=G(u)−iG(iu); then x~(iu)=G(iu)+iG(u)=ix~(u); together with real linearity this proves complex linearity, and its real part is G. For ∥u∥≤1 put z=x~(u). If z≠0, take c=z‾/∣z∣, so ∣c∣=1 and x~(cu)=cz=∣z∣ is real. Thus ∣z∣=G(cu)≤∥cu∥=∥u∥≤1; if z=0 the same bound is immediate. Thus ∥x~∥≤1 in either case, and x∗:=∥x∥ x~ satisfies ∥x∗∥=∥x∥ because x~(x), having real part G(x)=g(x)=∥x∥ and modulus at most ∥x∥, equals the positive real number ∥x∥. Finally Re⁡x∗(Ax)=∥x∥ G(Ax)=∥x∥Re⁡y(Ax)≤0, as required. The definition and both elementary forms are choice-free; only the norm-duality form uses HB and the finite-dimensional compactness above.

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Lumer-Phillips generation theorem

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let A:D(A)⊆X→X be a densely defined dissipative operator on a Banach space X (Dissipative operator). Then the following are equivalent: (a) A generates a strongly continuous semigroup of contractions; (b) Ran(λ0I−A)=X for some λ0>0; (c) Ran(λI−A)=X for every λ>0. In that case A is closed, (0,∞)⊆ρ(A), ∥R(λ,A)∥≤1/λ for all λ>0, and A is maximal dissipative (it has no proper dissipative extension).

Facts & Assumptions

Given: Dependent Choice; A densely defined dissipative operator A:D(A)⊆X→X on a Banach space X (Dissipative operator, Strongly continuous semigroup).

[F1]

Dissipativity means ∥(λI−A)x∥≥λ∥x∥ for all λ>0 and x∈D(A); hence each λI−A is injective and ∥(λI−A)−1z∥≤λ−1∥z∥ on the range of λI−A (Dissipative operator).

[F2]

Contraction Hille-Yosida: a closed densely defined operator with (0,∞)⊆ρ(A) and ∥R(λ,A)∥≤1/λ for all λ>0 generates a strongly continuous semigroup of contractions; conversely the generator of a contraction semigroup has (0,∞)⊆ρ(A) and ∥R(λ,A)∥≤1/λ (Contraction Hille-Yosida theorem).

[F3]

Resolvent identity: R(λ,A)−R(λ0,A)=(λ0−λ)R(λ,A)R(λ0,A) (Resolvent identity for closed operators). Consequently, for ∣λ0−λ∣<1/∥R(λ0,A)∥ the series ∑k≥0(λ0−λ)kR(λ0,A)k+1 converges in B(X) and its sum is the inverse of λI−A: writing R0=R(λ0,A) and Q=I+(λ−λ0)R0, one has (λI−A)R0=Q and, on D(A), λI−A=Q(λ0I−A). Thus R0Q−1 is a two-sided inverse and has range in D(A). In the Neumann series Q−1=∑k≥0(λ0−λ)kR0k, the operators commute with R0 by taking limits of polynomials, giving the displayed series. Indeed, for X≠{0}, B(X) is complete for the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach) with submultiplicative composition (Composition satisfies |ST|\le|S|,|T|, The operator norm as the least bound and as the unit-sphere or unit-ball supremum), so the Neumann-series computation applies — for complex X through Neumann series and the unital Banach-algebra structure (Unital Banach algebra), and for real X by the identical telescoping computation. An operator with a bounded everywhere-defined inverse is closed: the inverse graph is the zero set of the continuous map (z,y)↦y−Rz, and swapping graph coordinates gives the graph of the original operator. A scalar shift of its graph is a homeomorphism. Hence A is closed as soon as λI−A has such an inverse; a closed bijective operator with bounded inverse lies in the resolvent set (Resolvent and spectrum of a closed operator on a Banach space).

[F4]

Operators of the form λI−A with λ>0 are closed when A is closed, and A is closed as soon as some λI−A has a bounded everywhere-defined inverse; generators are closed and densely defined (The generator is closed and densely defined, Resolvent and spectrum of a closed operator on a Banach space).

Proof

technique · direct: dissipativity gives injectivity and the inverse bound, surjectivity at one $\lambda_0$ propagates by the resolvent series, and the contraction generation theorem finishes
1.1F1F4

If X={0} then D(A)=X and all claims hold for the unique zero operator and semigroup; hence assume X≠{0}. (b)⇒ closedness, λ0∈ρ(A). Assume Ran⁡(λ0I−A)=X for some λ0>0. By [F1] λ0I−A is injective with ∥(λ0I−A)−1z∥≤λ0−1∥z∥ for all z∈X; thus its inverse is a bounded everywhere-defined operator and λ0I−A is bijective, so λ0∈ρ(A) and A is closed by [F4].

1.2F2

(a)⇒(b),(c). If A generates a contraction semigroup, [F2] gives (0,∞)⊆ρ(A) with ∥R(λ,A)∥≤1/λ; therefore every λI−A, λ>0, is bijective onto X, which is (c) and, taking e.g. λ=1, also (b). Trivially (c)⇒(b).

2.1F1F3step 1.1

Propagation to (0,∞). With λ0∈ρ(A) and ∥R(λ0,A)∥≤1/λ0, the series of [F3] converges for ∣λ−λ0∣<λ0 and represents R(λ,A); hence (0,2λ0)⊆ρ(A). Dissipativity now gives ∥R(λ,A)∥≤1/λ for every λ∈ρ(A)∩(0,∞) by [F1], in particular on (0,2λ0). Replacing λ0 by any λ1∈(λ0,2λ0), the same argument gives (0,2λ1)⊆ρ(A) with λ1>λ0; iterating with the explicit points λk=(3/2)kλ0, each inside the previous interval (0,2λk−1), gives (0,2λk)⊆ρ(A) for all k. Since λk→∞, this yields (0,∞)⊆ρ(A) and ∥R(λ,A)∥≤1/λ for all λ>0. In particular λI−A is surjective for every λ>0, so (c) holds.

3.1F2step 1.1step 2.1

(b)⇒(a). A is closed and densely defined by hypothesis and [step 1.1]; [step 2.1] supplies (0,∞)⊆ρ(A) and ∥R(λ,A)∥≤1/λ; hence [F2] makes A the generator of a strongly continuous semigroup of contractions.

3.2F1step 2.1

Maximal dissipativity. Let A′⊇A be a dissipative extension and fix λ>0. By [step 2.1], λI−A maps D(A) onto X; given x∈D(A′), choose y∈D(A) with (λI−A)y=(λI−A′)x. Since A′ agrees with A on D(A), (λI−A′)y=(λI−A)y=(λI−A′)x, and injectivity of λI−A′ by [F1] gives x=y∈D(A). Hence D(A′)=D(A) and A′=A: A has no proper dissipative extension.

4.1step 1.1step 2.1step 3.1step 1.2step 3.2∎

Combining the implications: (a), (b) and (c) are equivalent for a densely defined dissipative operator, and in that case A is closed, (0,∞)⊆ρ(A), ∥R(λ,A)∥≤1/λ, and A is maximal dissipative.

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The variation-of-constants integral is continuous for integrable forcing

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for Lebesgue time integration. Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with constants M≥1, ω∈R and ∥T(t)∥≤Meωt (Exponential bound for a C0-semigroup). Let T0>0 and let f:(0,T0)→X be Bochner integrable with ∫0T0∥f(s)∥ ds<∞ (Bochner-integrable function). Then uf(t):=∫0tT(t−s)f(s) ds(0≤t≤T0) is a well-defined element of X, the map t↦uf(t) is continuous on [0,T0], and ∥uf(t)∥≤MT0eωt∫0t∥f(s)∥ ds for a constant MT0 depending only on M,ω,T0 and the local bound of ∥T∥ on [0,T0].

Facts & Assumptions

Given: Countable Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with ∥T(t)∥≤Meωt for some M≥1, ω∈R (Exponential bound for a C0-semigroup); T0>0; a Bochner integrable f:(0,T0)→X with ∫0T0∥f(s)∥ ds<∞; and uf(t):=∫0tT(t−s)f(s) ds.

[F1]

The exponential bound makes K:=sup⁡0≤r≤T0∥T(r)∥ finite, since ∥T(r)∥≤Meωr≤Memax⁡(ω,0)T0. [thm-exponential-bound-for-a-c-zero-semigroup]

[F2]

Bochner integrability of f supplies integrable simple functions g with ∫(0,T0)∥f−g∥ arbitrarily small and a strong-measurability approximation (Bochner-integrable function, Strongly measurable Banach-valued function); a strongly measurable h with ∫∥h∥<∞ is Bochner integrable (Bochner integrability criterion).

[F3]

Linearity of the Bochner integral and the norm inequality ∥∫Eh∥≤∫E∥h∥ (Linearity of the Bochner integral, Bochner integral norm inequality, Bochner-integrable function).

[F4]

Absolute continuity of the scalar integral: for every ε>0 there is δ>0 with ∫E∥f∥<ε whenever λ(E)<δ (Absolute continuity of the integral).

[F5]

The orbit map of every vector is continuous on [0,∞) and T(h)→I strongly: T(h)x→x for every x (Strongly continuous semigroup).

Proof

technique · direct: well-definedness from the Bochner criterion, then a uniform estimate $\int\|(T(h)-I)f\|\to0$ obtained from a single simple approximation, and a splitting of the increment
1.1F1F2F3F5

Fix t∈[0,T0]. For each measurable simple approximation gn=∑jxnj1Enj to f, the map s↦T(t−s)gn(s) is strongly measurable: for each of its finitely many values xnj, approximate the continuous curve s↦T(t−s)xnj uniformly by step functions on equal partitions of [0,t], then multiply by 1Enj. Choose a partition size by its least integer giving error below 1/n on all finitely many curves. The resulting measurable simple function approximates T(t−s)gn(s) uniformly within 1/n. As gn(s)→f(s) off a null set and ∥T(t−s)∥≤K, these approximants converge pointwise there to T(t−s)f(s). Its norm is bounded by K∥f(s)∥, so [F2] gives Bochner integrability and [F3] gives ∥uf(t)∥≤Memax⁡(−ω,0)T0eωt∫0t∥f(s)∥ ds.

1.2F1F2F5

Claim: ρ(h):=∫0T0∥(T(h)−I)f(s)∥ ds→0 as h↓0. Given ε>0, choose an integrable simple function g=∑j=1mxj1Ej with ∫0T0∥f−g∥<ε/(2(K+1)) by [F2]; then ρ(h)≤(K+1)∫∥f−g∥+∑jλ(Ej)∥(T(h)−I)xj∥, and the finite sum tends to 0 as h↓0 because T(h)xj→xj for each j by [F5]. Hence lim sup⁡hρ(h)≤ε/2<ε, and ε was arbitrary.

2.1F3step 1.1

Increment splitting: for 0≤t<t+h≤T0, linearity [F3] and the semigroup law give uf(t+h)−uf(t)=∫0t+hT(t+h−s)f(s) ds−∫0tT(t−s)f(s) ds=∫tt+hT(t+h−s)f(s) ds+∫0tT(t−s)(T(h)−I)f(s) ds, where in the last term T(t−s+h)=T(t−s)T(h).

3.1F1F3F4step 1.2step 2.1

Taking norms in [step 2.1] and using K from [F1] and the norm inequality [F3]: ∥uf(t+h)−uf(t)∥≤K∫tt+h∥f(s)∥ ds+Kρ(h)→0 as h↓0, the first term by absolute continuity [F4] and the second by [step 1.2]. The backward increment is bounded by K∫t−ht∥f∥+Kρ(h) by the same splitting with t−h in place of t, hence also tends to 0.

4.1F1F3F4step 3.1

Continuity at the endpoints: ∥uf(h)−uf(0)∥≤K∫0h∥f∥→0 by [F3], [F4], and at t=T0 the backward bound of [step 3.1] applies; hence t↦uf(t) is continuous on the closed interval [0,T0], and the estimate of [step 1.1] is the stated bound with MT0:=Memax⁡(−ω,0)T0.

5.1step 1.1step 1.2step 3.1step 4.1∎

The claims of the statement follow: uf is well defined, continuous on [0,T0], and bounded by MT0eωt∫0t∥f∥; no compactness of the range of f and no choice beyond the declared Bochner framework was used.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Classical, strong and mild abstract Cauchy solutions

Definition

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue time integrals. Let A:D(A)⊆X→X be a linear operator on a real or complex Banach space X, with the Banach graph conventions of Infinitesimal generator of a C0-semigroup. Let T0>0, x∈X and let f:(0,T0)→X be Bochner integrable (Bochner-integrable function). Consider the abstract Cauchy problem u′(t)=Au(t)+f(t) for 0<t<T0, u(0)=x; endpoint equations are imposed only when f is continuously extended to [0,T0]. Derivatives use the underlying real structure, with one-sided derivatives at the endpoints. (1) A classical solution is a function u∈C1([0,T0];X) with u(t)∈D(A) for every t∈[0,T0], Au∈C([0,T0];X), u′(t)=Au(t)+f(t) for 0<t<T0 and u(0)=x; here membership in D(A) is with the graph norm and the derivative is the Fréchet derivative of a curve (Fréchet derivative between Banach spaces). (2) A strong solution is a continuous u∈C([0,T0];X) with u(t)∈D(A) for all t, u(0)=x, and Au∈C([0,T0];X), such that u(t)=x+∫0t(Au(s)+f(s)) ds for every t. (3) When A generates a strongly continuous semigroup T satisfying an exponential norm bound, a mild solution (variation-of-constants solution) is the continuous function given by the convergent Bochner integral u(t)=T(t)x+∫0tT(t−s)f(s) ds, which is well defined by The variation-of-constants integral is continuous for integrable forcing. (4) An integral (integrated) solution is a continuous u with ∫0tu(s) ds∈D(A) and u(t)=x+A∫0tu(s) ds+∫0tf(s) ds for all t; for f=0 this is the integrated form of the homogeneous problem. These are four formulations, with overlaps and equivalences under additional hypotheses: neither differentiability nor membership of u(t) in D(A) is asserted by the mild definition, and the definitions do not assign a derivative to a mild solution.

Here “strong solution” means the graph-continuous integral notion explicitly stated above; some sources use that term for an almost-everywhere differential notion instead. Engel–Nagel II.6.3 calls the homogeneous integral formulation “mild”, whereas the variation-of-constants terminology here follows Schnaubelt Definition 2.11. The mild and integral formulations coincide for generators by the variation-of-constants theorem; they are not asserted to be different classes. Under DC the exponential bound required for the mild formulation is automatic, by Exponential bound for a C0-semigroup.

Classical solution. A classical solution is C1 on the closed interval and takes values in the domain of A at every time, including the endpoints; the equation u′(t)=Au(t)+f(t) is required pointwise on (0,T0), and the derivative is the Fréchet derivative of the curve (Fréchet derivative between Banach spaces). Since A is generally unbounded, membership u(t)∈D(A) and continuity of Au are genuine restrictions. Together with continuity of u, they mean graph-norm continuity in D(A) (Unbounded linear operators: domain, graph and extension).

Strong solution. A strong solution need not be differentiable; instead Au is continuous, and the equation is imposed in integrated form u(t)=x+∫0t(Au(s)+f(s))ds, which makes sense because Au is continuous and f is Bochner integrable (Bochner-integrable function). If Au+f on (0,T0) extends continuously to [0,T0], this identity gives u∈C1([0,T0];X), with u′=Au+f in the interior and one-sided endpoint derivatives equal to the extension values, by Fundamental theorem of calculus for Banach-valued continuous curves. Thus strong and classical solutions coincide when f extends continuously to [0,T0]; continuity only on (0,T0) does not ensure endpoint derivatives.

Mild solution. The mild solution is the explicit variation-of-constants function u(t)=T(t)x+∫0tT(t−s)f(s) ds, which is a well-defined continuous X-valued function on [0,T0] by The variation-of-constants integral is continuous for integrable forcing; here f need only be Bochner integrable, and no differentiability, no membership of u(t) in D(A) and no pointwise equation are asserted.

Integral (integrated) solution. An integral solution replaces differentiability by a weaker regularity: the primitive ∫0tu(s) ds lies in D(A) for every t, and u(t)=x+A∫0tu(s) ds+∫0tf(s) ds holds. For f=0 this is the integrated form of the homogeneous problem u′=Au, u(0)=x; its advantage is that it only evaluates A on the primitive, which always lies in D(A) when u is a mild solution of the homogeneous problem.

The definitions do not by themselves assert implications between the four notions beyond the elementary ones visible above, and they do not assign a derivative to a mild solution. The precise equivalence results under additional hypotheses, and the separation of the notions when those hypotheses fail, are proved as theorems and exhibited by counterexamples on the companion pages.

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Well-posedness of the abstract Cauchy problem is equivalent to generation

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let A:D(A)⊆X→X be a closed linear operator on a Banach space X and consider the homogeneous problem u′(t)=Au(t), u(0)=x. Let (EU) be the statement that for every x∈D(A) there exists exactly one classical solution u(⋅,x) on [0,∞) (Classical, strong and mild abstract Cauchy solutions). Then the following conditions are equivalent: (a) A generates a strongly continuous semigroup; (b) (EU) holds and ρ(A)≠∅; (c) (EU) holds and there is a sequence λn↑∞ with (λnI−A)D(A)=X for every n; (d) (EU) holds, D(A) is dense, and for every sequence xn∈D(A) with xn→0 one has u(t,xn)→0 uniformly for t in compact subsets of [0,∞). Condition (d) is the definition of well-posedness of the abstract Cauchy problem; it is existence plus uniqueness plus continuous dependence on the initial datum in the uniform topology on compact time intervals. If any (hence all) holds, then u(t,x)=T(t)x for the generated semigroup T.

Facts & Assumptions

Given: A closed linear operator A:D(A)⊆X→X on a Banach space X (Unbounded linear operators: domain, graph and extension, Densely defined, closed and closable operators, and cores), and the condition (EU) that for every x∈D(A) there is exactly one classical solution u(⋅,x) of u′=Au, u(0)=x on [0,∞) (Classical, strong and mild abstract Cauchy solutions). Write X1:=(D(A),∥⋅∥A) for the graph-norm space, A1x:=Ax with D(A1):=D(A2)={x∈D(A):Ax∈D(A)}, and B for a generator when it exists. The proof assumes Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain), carried by the closed graph theorem Closed graph theorem used in [F4]; the sequence selections in steps 1.3 and 1.4 are instances of Countable Choice, a consequence of DC.

[F1]

Since A is closed, its graph and the graph-norm space X1 are Banach by the explicit Banach graph convention in Infinitesimal generator of a C0-semigroup. The operator A1 is closed on X1: if xn→x and Axn→y in X1, convergence in X and closedness of A give Ax=y; then x∈D(A2) and A1x=y. This uses the assumed closedness, not a generator theorem.

[F2]

For a generator the following hold: local boundedness on compact time intervals, orbit continuity, the Laplace formula R(λ,A)x=∫0∞e−λtT(t)x dt for real λ>ω whenever ∥T(t)∥≤Meωt, hence (ω,∞)⊆ρ(A), and D(A) is dense with A closed (Exponential bound for a C0-semigroup, Laplace transform formula for the resolvent, The generator is closed and densely defined, A semigroup with continuity at zero is uniformly bounded on every compact time interval, Continuity at time zero implies continuity of every orbit).

[F3]

For a strongly continuous semigroup generated by A: T(t)D(A)⊆D(A), AT(t)x=T(t)Ax on D(A), and A∫0tT(s)x ds=T(t)x−x (The generator commutes with the semigroup on its domain, Time integrals of semigroup orbits lie in the generator domain).

[F4]

Closed graph theorem: an everywhere defined linear map between Banach spaces with closed graph is bounded (Closed graph theorem, under DC); consequently a closed operator whose graph-norm domain is complete has the properties used below.

Proof

technique · direct, following the standard reduction to the graph-norm space $X_1$: first an internal lemma identifying (EU) with generation on $X_1$, then the four implications
1.1given

Internal lemma (reduction to X1), part 1: (EU) gives a semigroup on X1. Put T1(t)x:=u(t,x) for x∈X1. Uniqueness makes T1(t) linear and gives T1(0)=I, T1(t+s)=T1(t)T1(s); the classical equation u′=Au shows that t↦T1(t)x is continuous into X1 for each x∈X1.

1.2F2F3

(a)⇒(b),(c),(d). If A generates a semigroup T with ∥T(t)∥≤Meωt: for x∈D(A) the orbit T(⋅)x is a classical solution and any classical solution v satisfies v≡T(⋅)x by the rigidity computation dds[T(t−s)v(s)]=0, so (EU) holds; [F2] gives (ω,∞)⊆ρ(A) and (λn−A)D(A)=X for any λn↑∞ in that half-line, so (c) holds and ρ(A)≠∅; density and the local bound sup⁡0≤s≤t0∥T(s)∥ give, for xn→0 in D(A), sup⁡0≤t≤t0∥u(t,xn)∥≤sup⁡0≤s≤t0∥T(s)∥ ∥xn∥→0, which is (d).

1.3F1F3F4

(d)⇒(a). Define T(t)x:=u(t,x) for x∈D(A); uniqueness makes each T(t) linear, and the semigroup law holds on D(A) by uniqueness of solutions. The continuous-dependence hypothesis (d) transfers to a local bound: if no δ>0 had ∥T(t)x∥≤1 for all x∈D(A) with ∥x∥≤δ and all t∈[0,1], then choosing xn with ∥xn∥≤1/n and ∥T(tn)xn∥>1 would give a sequence xn→0 with u(tn,xn)↛0 uniformly, contradicting (d); hence sup⁡0≤t≤1∥T(t)x∥≤∥x∥/δ for all x∈D(A). Since D(A) is dense, each T(t) extends uniquely to a bounded operator on X with the same bound, and the semigroup law and strong continuity extend by density, using ∥T(t)∥≤(M1)n+1 on [0,n] with M1:=max⁡{1,1/δ} from the semigroup law. The generator B of the extension satisfies A⊆B, because on D(A) the difference quotients are those of the classical solutions and converge to Ax. The extension leaves D(A) invariant, so D(A) is a core of B: for x∈D(B) choose xj∈D(A) with xj→x; then 1t∫0tT(s)xj ds lies in the graph-norm closure of D(A) (the integrand is D(A)-valued and graph-norm continuous) and the integrated-orbits identity 1t∫0tT(s)x ds→x in graph norm as t↓0 together with 1t∫0tT(s)xj ds→1t∫0tT(s)x ds in graph norm shows x∈D(A)‾∥⋅∥B. Since A is closed, A⊆B and D(A) is a core of B, every graph limit from D(A) remains in Γ(A), so A=B.

2.1F1F4step 1.1

Boundedness of T1(t). For any Banach space Z, C([0,t],Z) is Banach in the supremum norm: a uniformly Cauchy sequence converges pointwise by completeness, uniformly by its common Cauchy estimates, and its uniform limit is continuous by the three-term increment estimate. Fix t>0 and consider Φ:X1→C([0,t],X1), Φ(x):=T1(⋅)x. Its graph is closed: if xn→x in X1 and Φ(xn)→f uniformly in X1, then the integral identity T1(s)xn=xn+∫0sAT1(r)xn dr passes to the limit in X and gives f(s)=x+∫0sAf(r) dr for s≤t; the extension f~(s):=T1(s−t)f(t) for s>t, =f(s) for s≤t, then solves (ACP) with initial value x, so f~=T1(⋅)x by uniqueness and f=Φ(x). By [F4] Φ is bounded on the Banach space X1, hence T1(t)∈B(X1) and T1 is a strongly continuous semigroup on X1.

3.1F1F4step 2.1

The generator of T1 is A1. First AT1(t)x=T1(t)Ax for x∈D(A1): the curve f(t):=x+∫0tT1(s)Ax ds is differentiable with f′=T1(t)Ax and satisfies Af=f′ (move A inside the integral by the closed-graph argument in X), so f=T1(⋅)x by uniqueness and AT1(t)x=T1(t)Ax. Hence for x∈D(A1) the quotient 1t(T1(t)x−x) converges to Ax in X and its A-image converges to A2x in X; that is, the convergence holds in X1, so A1⊆B. Conversely, if x∈D(B), then A1t(T1(t)x−x) converges in X and 1t(T1(t)x−x)→Ax in X; closedness of A gives Ax∈D(A), that is x∈D(A1). Thus B=A1.

4.1givenstep 3.1F4

(b)⇒(a). Let λ∈ρ(A). For x∈X one has x∈D(A) iff (λ−A)−1x∈D(A1), and Ax=(λ−A)A1(λ−A)−1x for x∈D(A): indeed (λ−A)−1x=R and AR=λR−x, and AR∈D(A) exactly when x∈D(A). Thus S:=(λ−A)−1:X→X1 is a bounded isomorphism with bounded inverse λ−A:X1→X, and A=S−1A1S with D(A)=S−1D(A1). By [step 3.1] and the internal lemma, A1 generates T1 on X1; then T(t):=S−1T1(t)S is a strongly continuous semigroup on X whose generator is S−1A1S=A, because the difference quotients of T are those of T1 conjugated by the bounded isomorphism S.

4.2F1F2F4step 1.1step 2.1step 3.1

(c)⇒(b). By [step 3.1] the operator A1 generates T1 on X1 (via [step 1.1] and [step 2.1]), so by [F2] its resolvent set contains a half-line (ω1,∞); choose λ=λn>ω1 with (λ−A)D(A)=X. If Ax=λx for some x∈D(A), then Ax∈D(A), so x∈D(A1) and A1x=λx; since λ∈ρ(A1) this forces x=0. Hence λ−A is injective and, by hypothesis, surjective, so it is bijective; being closed it has bounded inverse by [F4], and λ∈ρ(A).

5.1step 1.2step 4.1step 4.2step 1.3∎

All implications are established, so (a)-(d) are equivalent; and in each direction the solution is u(t,x)=T(t)x for the generated semigroup, as asserted.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Variation of constants for the inhomogeneous abstract Cauchy problem

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain). Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator A (Infinitesimal generator of a C0-semigroup), let T0>0, x∈X, and let f:(0,T0)→X be Bochner integrable with ∫0T0∥f(s)∥ ds<∞. Then: (1) (Duhamel rigidity) every classical solution u of u′=Au+f, u(0)=x on [0,T0] satisfies u(t)=T(t)x+∫0tT(t−s)f(s) ds(0≤t≤T0). (2) The formula defines a continuous u, which is the unique mild solution and the unique integral solution of the problem in the sense of Classical, strong and mild abstract Cauchy solutions: if v is an integral solution, then v=u. (3) (classical upgrade) If in addition x∈D(A) and f extends to [0,T0] either as a C1 curve or in the form f(t)=f(0)+∫0tg(s) ds for some Bochner integrable g:(0,T0)→X, then u is a classical solution: u∈C1([0,T0];X), u(t)∈D(A) for all t, u(0)=x and u′(t)=Au(t)+f(t) pointwise. Mere continuity, or mere Lipschitz continuity on an arbitrary Banach space, is not asserted to give a classical solution. A Lipschitz curve is covered by (3) when it additionally has the displayed Bochner derivative representation.

Facts & Assumptions

Given: Dependent Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator A (Infinitesimal generator of a C0-semigroup); T0>0, x∈X, and a Bochner integrable f:(0,T0)→X with ∫0T0∥f∥<∞; the continuous function u(t):=T(t)x+∫0tT(t−s)f(s) ds (Classical, strong and mild abstract Cauchy solutions, The variation-of-constants integral is continuous for integrable forcing).

[F1]

The formula defines a continuous X-valued function on [0,T0], and the norm inequality, linearity and Bochner framework of the integral apply; the exponential bound gives a local bound K=sup⁡0≤r≤T0∥T(r)∥ (The variation-of-constants integral is continuous for integrable forcing, Linearity of the Bochner integral, Bochner integral norm inequality, Bochner-integrable function).

[F2]

For x∈D(A) the orbit is differentiable with T(s)x∈D(A), AT(s)x=T(s)Ax, and the primitive of an orbit satisfies A∫0tT(s)y ds=T(t)y−y for every y∈X (The generator commutes with the semigroup on its domain, Time integrals of semigroup orbits lie in the generator domain); the fundamental theorem of calculus applies to continuous curves with continuous derivative (Fundamental theorem of calculus for Banach-valued continuous curves).

[F3]

DC supplies the local operator bound by Exponential bound for a C0-semigroup. A continuous graph-valued curve has a graph-valued integral by sampled step approximation and closedness, as proved in Laplace transform formula for the resolvent. Uniform continuity of a continuous curve on a compact interval is Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous.

Proof

technique · direct: the product rule for $s\mapsto T(t-s)u(s)$ gives Duhamel's formula; the integral solution identity is verified for simple forcing and transferred by closedness; uniqueness follows from a vanishing integral
1.1F2

Duhamel rigidity. Let u be a classical solution and fix t∈[0,T0]. The curve g(s):=T(t−s)u(s) on [0,t] is differentiable: g′(s)=T(t−s)u′(s)−AT(t−s)u(s)=T(t−s)(u′(s)−Au(s))=T(t−s)f(s), where u(s)∈D(A) and [F2] was used. The classical conditions make u′−Au continuous on [0,T0], so f agrees on the interior with this continuous extension. Its product with T(t−s) is continuous: an increment is bounded by K∥f(s)−f(s0)∥+∥(T(t−s)−T(t−s0))f(s0)∥, which tends to zero. Thus g′ extends continuously to the endpoints, and by the fundamental theorem of calculus [F2], u(t)=g(t)=g(0)+∫0tg′(s) ds=T(t)x+∫0tT(t−s)f(s) ds.

1.2F1

The formula is continuous and well defined. By [F1] u is a well-defined continuous function on [0,T0]; this is the mild solution of the problem in the sense of Classical, strong and mild abstract Cauchy solutions.

1.3F1F2F3given

First take f(s)=x01E(s). Put Jry=∫0rT(q)y dq. The needed exchange of vector integrals is justified directly: on the compact triangle 0≤s≤r≤t, the curve T(r−s)x0 is uniformly continuous. On a fine square grid approximate it uniformly by finitely valued functions sampled at points of the intersecting triangle cells, and multiply by 1E(s)1s≤r. Scalar Fubini (Fubini's theorem for L^1 functions on a sigma-finite product) applies to each indicator coefficient. Both iterated integral errors are at most t2 times the uniform approximation error by [F1], so exchange remains valid in the limit. Consequently ∫0tu(r) dr=Jtx+∫E∩(0,t)Jt−sx0 ds. The latter integral lies in D(A) and its A-image is ∫E∩(0,t)(T(t−s)x0−x0) ds: the pair (Jt−sx0,AJt−sx0) is continuous by [F2], and sampled step approximations, multiplied by 1E, have graph-valued integrals; closedness of A retains the limiting pair. Thus A∫0tu=u(t)−x−∫0tf. Finite linearity proves this for every integrable simple fn. For general f, take defining simple fn with ∫∥f−fn∥→0. The local bound K gives sup⁡t∥un(t)−u(t)∥≤K∫∥f−fn∥→0, so both coordinates of the graph pair converge: ∫0tun→∫0tu and A∫0tun→u(t)−x−∫0tf. Closedness proves the integral-solution identity for u.

1.4F2

Uniqueness among integral solutions. Let v be an integral solution of u′=Au+f, u(0)=x, and put d:=v−u, which is continuous with ∫0td∈D(A) and satisfies d(t)=A∫0td(r) dr for all t. For fixed t define h(s):=T(t−s)∫0sd(r) dr; then h is differentiable with h′(s)=T(t−s)(d(s)−A∫0sd)=0 by [F2] and the equation for d, so h is constant and ∫0td(r) dr=h(t)−h(0)=0 (using T(0)=I and the primitive's value 0 at s=0). Hence ∫0td=0 for every t; differentiating in t with the fundamental theorem of calculus gives d(t)=0 for all t, so v=u.

2.1F1F2step 1.3

Assume x∈D(A) and f(t)=f(0)+∫0tg(s) ds with g Bochner integrable; the C1 case is g=f′. Write v(t)=∫0tT(r)f(t−r) dr (reflecting equal partitions under s=t−r gives the same sampled sums, hence the same Bochner integral, for this continuous integrand). Substituting the primitive representation and exchanging the triangle integrals gives v(t)=∫0t[T(r)f(0)+∫0rT(r−s)g(s) ds] dr. This exchange follows by the same grid argument as step 1.3 for simple g, and by L1 approximation for general g: both errors are at most KT0∫∥g−gn∥. The integrand is continuous by [F1] applied to g, so the FTC gives v′(t)=T(t)f(0)+∫0tT(t−s)g(s) ds, continuously on [0,T0]. Since T(t)x is C1 by [F2], u=T(⋅)x+v is C1.

3.1F2step 1.3step 2.1given

Since u is an integral solution by [step 1.3], subtraction gives A∫tt+hu(r) dr=u(t+h)−u(t)−∫tt+hf(r) dr for all 0≤t<t+h≤T0; at t=T0 use the corresponding backward difference. Divide by h: on the right, u(t+h)−u(t)h→u′(t) and 1h∫tt+hf→f(t) by average convergence for the continuous f, so the right side tends to u′(t)−f(t); on the left, 1h∫tt+hu→u(t) by average convergence for the continuous u. Since A is closed, the limit pair (u(t),u′(t)−f(t)) lies in the graph of A; hence u(t)∈D(A) and Au(t)=u′(t)−f(t), that is u′(t)=Au(t)+f(t) pointwise, and u is a classical solution.

4.1step 1.1step 1.4step 3.1∎

Claims (1), (2) and (3) are [step 1.1], [steps 1.2-1.4] and [steps 2.1, 3.1]; the classical upgrade holds for the stated C1 or Bochner-primitive forcing.

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Uniqueness of the scalar Laplace transform in the exponential-growth class

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue-measure interfaces. Let K∈{R,C} and let f:[0,∞)→K be continuous with ∣f(t)∣≤Ceσt for some C≥0, σ∈R and all t≥0. If the Laplace transform vanishes on a right half-line, ∫0∞e−λtf(t) dt=0for every real λ>σ, then f(t)=0 for every t≥0.

Facts & Assumptions

Given: Countable Choice; A real or complex-valued continuous f:[0,∞)→K with ∣f(t)∣≤Ceσt for some C≥0, σ∈R and all t≥0, and ∫0∞e−λtf(t) dt=0 for every real λ>σ; for λ>σ the integrand is dominated by Ce−(λ−σ)t and the integral exists as a Lebesgue integral over [0,∞).

[F1]

If a<b, φ is C1 and injective with φ′≠0 on a neighbourhood of [a,b], and the continuous function h is defined on an interval containing φ([a,b]), then ∫min⁡φmax⁡φh=∫abh(φ(t))∣φ′(t)∣ dt (In one dimension the compact-Jordan formula is substitution over the unoriented image interval with the absolute derivative). This substitution is stated for Riemann integrals; on the compact intervals used below all its integrands are continuous, hence bounded and Riemann integrable, and A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral identifies those integrals with their Lebesgue integrals under Countable Choice.

[F2]

Polynomials are uniformly dense in C([0,1],R): for every continuous real φ on [0,1] and η>0 there is a polynomial p with sup⁡[0,1]∣p−φ∣<η (Polynomials are uniformly dense in C([0,1],R)).

[F3]

The Lebesgue integral is linear on L1 and satisfies ∣∫h∣≤∫∣h∣; the integral over a measurable set is defined by restricting each real positive/negative and imaginary component, giving ∫Eh=∫h1E (The Lebesgue integral is linear on L1(μ), The modulus of an integral is bounded by the integral of the modulus, The class L1(μ) of integrable functions, Integrable real and complex functions, and their integrals); Integral over a measurable subset alone supplies only the nonnegative convention.

Proof

technique · direct, reducing to real $f$, converting the Laplace moments into moments of a continuous function on $[0,1]$, and applying polynomial density
1.1F3algebra

It suffices to prove the theorem for real-valued f: if f is complex-valued, then Re⁡f and Im⁡f are continuous, satisfy the same bound ∣Re⁡f∣,∣Im⁡f∣≤Ceσt, and by [F3] have ∫0∞e−λtRe⁡f(t) dt=Re⁡0=0 and likewise for Im⁡f for every real λ>σ.

1.2F3algebra

Assume f real. Fix λ0>max⁡{σ,0} and put δ:=λ0−σ>0 and F(t):=e−λ0tf(t). Then F is continuous with ∣F(t)∣≤Ce−δt for t≥0, so F∈L1(0,∞); moreover for every integer k≥0 the number λ:=λ0+k+1 exceeds σ and ∫0∞e−(k+1)tF(t) dt=∫0∞e−λtf(t) dt=0.

1.3F3algebra

Put g(x):=F(−ln⁡x) for x∈(0,1] and g(0):=0. Then g is continuous on [0,1]: it is continuous on (0,1] as a composition, and ∣g(x)∣=∣F(−ln⁡x)∣≤Cxδ→0 as x↓0 because δ>0, matching g(0)=0; also ∣g(x)∣≤Cxδ≤C on [0,1], so g∈L1(0,1) and ∫01∣g∣ dx≤C.

2.1F1step 1.3

For T>0 and k≥0, [F1] applied on [0,T] to φ(t)=e−t and the continuous h(x)=xkg(x) on [0,1] gives ∫e−T1xkg(x) dx=∫0Te−(k+1)tF(t) dt.

3.1F3step 1.2step 2.1algebra

Letting T→∞ in [step 2.1]: the right-hand side tends to ∫0∞e−(k+1)tF(t) dt because its tail is bounded by ∫T∞Ce−(δ+k+1)t dt≤Ce−(δ+k+1)T/(δ+k+1)→0; the left-hand side tends to ∫01xkg(x) dx because the missing part satisfies ∣∫0e−Txkg(x) dx∣≤C∫0e−Txδ+k dx≤Ce−(k+δ+1)T→0; by [step 1.2] the limits are 0, so ∫01xkg(x) dx=0 for every integer k≥0.

4.1F2F3step 1.3step 3.1

Every continuous real φ on [0,1] satisfies ∫01φ(x)g(x) dx=0: fix η>0 and, by [F2], choose a polynomial p with sup⁡[0,1]∣φ−p∣≤η/(1+∫01∣g∣ dx); then ∣∫01(φ−p)g dx∣≤sup⁡∣φ−p∣∫01∣g∣ dx<η by [F3], while ∫01pg dx=0 by [step 3.1]; hence ∣∫01φg dx∣<η for every η>0, so the integral vanishes.

5.1step 4.1algebra

The function g vanishes identically on [0,1]: otherwise g(x0)≠0 for some x0∈(0,1] with g(x0)>0 (or <0), and by continuity there is an interval J⊆[0,1] of positive length with g>0 on J (respectively g<0 on J); choosing a continuous nonnegative bump φ supported in J with φ(x0)>0 gives φg≥0, positive at x0 and continuous, so ∫01φg dx>0 (respectively <0), contradicting [step 4.1].

6.1step 1.1step 1.3step 5.1∎

Consequently F(t)=g(e−t)=0 for every t≥0, whence f(t)=eλ0tF(t)=0 for every t≥0 in the real case; the complex case follows by applying the real case to Re⁡f and Im⁡f as in [step 1.1].

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Laplace uniqueness identifies two exponentially bounded semigroups

Statement

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain) and the Hahn-Banach extension principle HB (The real dominated-extension principle as an additional hypothesis over ZF). Let (S(t))t≥0 and (T(t))t≥0 be strongly continuous semigroups on a Banach space X with generators A and B and resolvents RA,RB (Resolvent and spectrum of a closed operator on a Banach space), and suppose there are M≥1, ω∈R with ∥S(t)∥,∥T(t)∥≤Meωt for all t≥0. If RA(λ)=RB(λ) for every real λ>ω, then S(t)=T(t) for every t≥0. In particular two strongly continuous semigroups with the same generator coincide.

Facts & Assumptions

Given: Dependent Choice; The Hahn-Banach extension principle HB (The real dominated-extension principle as an additional hypothesis over ZF); strongly continuous semigroups (S(t))t≥0, (T(t))t≥0 on a Banach space X with generators A,B and resolvents RA,RB (Strongly continuous semigroup, Resolvent and spectrum of a closed operator on a Banach space); M≥1, ω∈R with ∥S(t)∥,∥T(t)∥≤Meωt (Exponential bound for a C0-semigroup); and RA(λ)=RB(λ) for every real λ>ω.

[F1]

Laplace formula: for real λ>ω, λ lies in the resolvent sets of both generators and RA(λ)x=∫0∞e−λtS(t)x dt, RB(λ)x=∫0∞e−λtT(t)x dt (Laplace transform formula for the resolvent).

[F2]

Bounded linear functionals and, more generally, bounded linear maps commute with Bochner integrals: x∗(∫h)=∫x∗∘h (Bounded linear maps commute with Bochner integration, Bochner-integrable function).

[F3]

Scalar Laplace uniqueness: a continuous scalar function φ with ∣φ(t)∣≤Ceσt whose Laplace transform vanishes for every real λ>σ is identically zero (Uniqueness of the scalar Laplace transform in the exponential-growth class).

[F4]

Point separation and norming under HB: for every x≠0 there is x∗∈X∗ with ∥x∗∥=1 and x∗(x)=∥x∥, so the dual separates points (Relative dual norming, point separation, and recovery of the norm).

Proof

technique · direct: scalarise the difference of the two semigroups by a functional and apply scalar Laplace uniqueness
1.1given

Put D(t):=S(t)−T(t) for t≥0. For fixed x∈X and x∗∈X∗ the scalar function φ(t):=x∗(D(t)x) is continuous and satisfies ∣φ(t)∣≤2Meωt∥x∗∥ ∥x∥, because S,T are strongly continuous and exponentially bounded.

2.1F1F2step 1.1

For real λ>ω, [F1] and [F2] give ∫0∞e−λtφ(t) dt=x∗(∫0∞e−λtD(t)x dt)=x∗[(RA(λ)−RB(λ))x]=0.

3.1F3step 1.1step 2.1

By scalar Laplace uniqueness [F3] applied with σ:=ω and C:=2M∥x∗∥∥x∥, the continuous function φ vanishes identically: x∗(S(t)x)=x∗(T(t)x) for every t≥0.

4.1F4step 3.1

Since x∗∈X∗ was arbitrary, the dual separates points of X (using HB, [F4]), so S(t)x=T(t)x for every x and every t≥0; that is, S(t)=T(t) for all t.

5.1F1step 4.1∎

If moreover A=B, then both semigroups have exponential bounds and, taking a common pair M,ω for the two bounds (for instance the maxima of the respective constants), their resolvents agree on (ω,∞) because both are given by the Laplace formula for the same operator; [step 4.1] then gives S=T.

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Restriction to a closed invariant subspace is a C0-semigroup and its generator is the part

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue-measure interfaces. Let (T(t))t≥0 be a strongly continuous semigroup on a Banach space X with generator (A,D(A)), and let Y⊆X be a closed linear subspace (Normed subspace, A closed subspace of a Banach space is Banach) such that T(t)Y⊆Y for every t≥0. Then the restrictions TY(t):=T(t)∣Y form a strongly continuous semigroup on the Banach space Y, and its generator is the part AY of A in Y: D(AY)={y∈D(A)∩Y: Ay∈Y} and AYy=Ay. In particular the generator of the restricted semigroup is the restriction of A to that domain.

Facts & Assumptions

Given: Countable Choice; A strongly continuous semigroup (T(t))t≥0 on a Banach space X with generator (A,D(A)) (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup), and a closed linear subspace Y⊆X with T(t)Y⊆Y for every t≥0.

[F1]

A closed linear subspace of a Banach space is a Banach space for the restricted norm (A closed subspace of a Banach space is Banach, Normed subspace), and convergence in the norm of Y is the same as convergence in X for vectors of Y.

[F2]

The generator is defined by right difference quotients: y∈D(A) exactly when T(h)y−yh converges as h↓0, and then the limit is Ay (Infinitesimal generator of a C0-semigroup).

[F3]

Time integrals of orbits lie in the generator domain: for y∈X and t>0 the Bochner integral Jty=∫0tT(s)y ds satisfies Jty∈D(A) and AJty=T(t)y−y (Time integrals of semigroup orbits lie in the generator domain, Bochner-integrable function).

Proof

technique · direct: inheritance of the semigroup properties, then comparison of the two difference-quotient limits through the closed subspace
1.1F1

The restrictions TY(t):=T(t)∣Y are bounded linear maps of Y into itself by hypothesis, with TY(0)=IY and TY(t+s)=TY(t)TY(s) inherited from T. For each y∈Y the orbit t↦TY(t)y is continuous into Y, because it is continuous into X and the norm of Y is the restriction of the norm of X by [F1]; hence TY is a strongly continuous semigroup on the Banach space Y.

1.2F1F2

Let B denote the generator of TY. If y∈D(A)∩Y and Ay∈Y, then T(t)y∈Y for all t, so the difference quotients TY(h)y−yh=T(h)y−yh lie in Y and converge in X to Ay∈Y; by [F1] they converge in Y to Ay. Therefore y∈D(B) and By=Ay.

1.3F1F2

Conversely, if y∈D(B), then by definition TY(h)y−yh→By in Y, hence also in X by [F1]; the same vectors are the difference quotients of T, so y∈D(A) and Ay=By. In particular Ay=By∈Y, so D(B)⊆{y∈D(A)∩Y:Ay∈Y}.

2.1F1F3step 1.2step 1.3∎

The two inclusions give D(B)={y∈D(A)∩Y:Ay∈Y} with By=Ay, that is, the generator of TY is the part AY of A in Y; for reference, this domain is dense in Y, since for y∈Y and t>0 the integral Jty lies in D(A)∩Y with AJty=T(t)y−y∈Y by [F3] and the Y-valued Bochner integral stays in the closed subspace Y, while 1tJty→y as t↓0.

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Semigroup sign and generator conventions

Statement

Generation and automatic exponential-bound results below are understood under the DC hypotheses of their supplier theorems.

This track consistently writes the abstract evolution equation as u′=Au+f and defines the generator by Ax=lim⁡t↓0(T(t)x−x)/t; the heat flow on a Dirichlet domain is therefore generated by A=ΔD, the Dirichlet Laplacian (the negative of the L2 operator associated with the Dirichlet energy form), not by −ΔD. Sources writing u′+Bu=0 or ut+Au=0 use B=−A or generate e−tA: Brezis's maximal-monotone chapter is stated for u′+Au=0 with A m-accretive, so its A corresponds to −A here, and Pazy-type statements u˙+Au=0 translate the same way. Resolvents are normalised as R(λ,A)=(λI−A)−1; a source using (A−λ)−1 has the opposite sign, and its resolvent equals −R(λ,A) at the same parameter, so the norms of all powers are unchanged. The contraction case ∥T(t)∥≤1 corresponds to M=1, ω=0; boundedness (M>1) is not the same as contractivity, and the general generation theorem keeps all resolvent powers.

The convention of this track. The abstract evolution equation is written u′(t)=Au(t)+f(t),u(0)=x, and the generator is defined by Ax=lim⁡t↓0(T(t)x−x)/t on its domain (Infinitesimal generator of a C0-semigroup). The resolvent is normalised as R(λ,A)=(λI−A)−1. If ∥T(t)∥≤Meωt for all t≥0, then for every real λ>ω the Laplace representation is R(λ,A)x=∫0∞e−λtT(t)x dt (Resolvent and spectrum of a closed operator on a Banach space, Laplace transform formula for the resolvent); membership in ρ(A) alone does not guarantee convergence of this integral.

Translation dictionary. A source that writes the homogeneous equation as u′+Bu=0 or ut+Au=0 is using the opposite sign: its B equals −A in this track, and its solutions are e−tB in its own notation, that is etA here. In the same way a source whose resolvent is (A−λ)−1 instead of (λI−A)−1 has the opposite shift convention; (A−λI)−1=−R(λ,A) at the same λ; its nth power is (−1)nR(λ,A)n, so its norm is unchanged. Negating the operator itself is a separate change of generator sign.

Heat flow. With this convention the Dirichlet heat flow ∂tu=Δu on a Dirichlet domain is generated by A=ΔD, the Dirichlet Laplacian (the negative of the L2 operator associated with the Dirichlet energy form), and not by −ΔD; the sign of the generator is the sign of the spatial operator in the equation, not its negative. The dissipativity used by Lumer-Phillips is therefore the inequality Re⁡⟨ΔDu,u⟩≤0 on the domain of the Dirichlet Laplacian (Infinitesimal generator of a C0-semigroup names the generator whose behaviour is being discussed), and this is the identification used downstream when the heat semigroup is realised from the Laplacian.

Contraction versus boundedness. The contraction case is the pair M=1, ω=0 of the general generation theorem; a bound with M>1 alone does not imply that the semigroup is contractive, and the general theorem keeps all resolvent power estimates, the first estimate alone guaranteeing all powers when M=1 (with exponential rescaling if ω≠0). This dictionary is the one applied when Hille-Yosida generation theorem is specialised to A=ΔD and when variation of constants is written in the form u(t)=T(t)x+∫0tT(t−s)f(s) ds.

5 · Examples, counterexamples and false statements

None yet.

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