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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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