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A semigroup with unbounded generator is not norm continuous at zero

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the Lebesgue-measure interfaces. Let $(T (t))_{t \geq 0}$ be a strongly continuous semigroup on a Banach space $X$ with generator $(A, D (A))$ (Infinitesimal generator of a C0-semigroup). If $T$ is continuous at $0$ in the operator norm, i.e. $∥ T (t) - I ∥ \to 0$ as $t ↓ 0$ , then $A \in B (X)$ and $D (A) = X$ . Consequently either an unbounded generator or a proper generator domain $D (A) \neq X$ rules out operator-norm continuity at $0$ .

Facts & Assumptions

Given: Countable Choice; A strongly continuous semigroup $(T (t))_{t \geq 0}$ on a Banach space $X$ with generator $(A, D (A))$ (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup), which is continuous at $0$ in the operator norm.

[F1]

For every $ε > 0$ , operator-norm continuity at $0$ gives $τ > 0$ such that $∥ T (s) - I ∥ \leq 1 / 2$ for every $0 \leq s \leq τ$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Strongly continuous semigroup).

[F2]

The time integral of each continuous orbit is Bochner integrable; the integral is linear and satisfies $∥ \int g (s) d s ∥ \leq \int ∥ g (s) ∥ d s$ (Bochner-integrable function, Linearity of the Bochner integral, Bochner integral norm inequality). The integrated-orbits identity gives $V x \in D (A)$ and $A V x = T (t) x - x$ for $V x : = \int_{0}^{t} T (s) x d s$ (Time integrals of semigroup orbits lie in the generator domain).

[F3]

The operator space $B (X)$ is a Banach space in the operator norm and composition is submultiplicative (If (Y) is Banach then (\mathcal B(X,Y)) is Banach, Composition satisfies |ST|\le|S|,|T|, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). For $∥ K ∥ < 1$ , the Neumann series makes $I + K$ invertible in $B (X)$ (Neumann series, Unital Banach algebra); for a real Banach space the same geometric-series and telescoping argument applies.

[F4]

If $S, T \in B (X)$ , then $S T \in B (X)$ and $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ (A bounded linear operator between normed spaces, Composition satisfies |ST|\le|S|,|T|).

Proof

technique · direct: a short-time integral of the semigroup is invertible and maps onto the generator domain; the integrated-orbits identity then expresses the generator as a product of bounded operators

1.1F1F2given

If $X = {0}$ the conclusion is immediate; assume $X \neq {0}$ . By [F1] choose $t \in (0, τ]$ and define $V x : = \int_{0}^{t} T (s) x d s$ . By [F2], the operator $V$ is linear, and for every $x \in X$ , $∥ V x ∥ \leq \int_{0}^{t} ∥ T (s) x ∥ d s \leq \frac{3 t}{2} ∥ x ∥$ ; hence $V \in B (X)$ . The integrated-orbits identity gives $V X \subseteq D (A)$ .

1.2F1F2

For every $x \in X$ , linearity and the norm inequality for the Bochner integral give $∥ (V - t I) x ∥ = ∥ \int_{0}^{t} (T (s) - I) x d s ∥ \leq \int_{0}^{t} ∥ (T (s) - I) x ∥ d s \leq \frac{t}{2} ∥ x ∥$ by [F1]. Taking the supremum over $∥ x ∥ \leq 1$ yields $∥ V - t I ∥ \leq t / 2$ .

2.1F3step 1.2

Put $K : = (V - t I) / t$ . Then $V = t (I + K)$ and $∥ K ∥ \leq 1 / 2 < 1$ ; [F3] gives $(I + K)^{- 1} = \sum_{n \geq 0} (- K)^{n}$ in $B (X)$ , so $V$ is invertible and $V X = X$ .

3.1F2step 2.1

Since $V X = X$ and $V X \subseteq D (A)$ by [F2], every element of $X$ lies in $D (A)$ ; hence $D (A) = X$ .

4.1F2F4step 2.1step 3.1

For $y \in X$ , write $y = V x$ with $x = V^{- 1} y$ . The identity in [F2] gives $A y = A V x = (T (t) - I) x = (T (t) - I) V^{- 1} y .$ Thus $A = (T (t) - I) V^{- 1} \in B (X)$ by [F4].

5.1step 3.1step 4.1∎

Therefore operator-norm continuity at $0$ forces $D (A) = X$ and a bounded generator; contrapositively, an unbounded generator cannot have an operator-norm continuous semigroup at $0$ .

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Sources

Klaus-Jochen Engel and Rainer Nagel, One-Parameter Semigroups for Linear Evolution Equations, Graduate Texts in Mathematics 194 (complete author-hosted monograph) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)