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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.

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