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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 $(T (t))_{t \geq 0}$ be a strongly continuous semigroup on a Banach space $X$ with generator $(A, D (A))$ (Infinitesimal generator of a C0-semigroup). If $x \in D (A)$ , then $T (t) x \in D (A)$ and $A T (t) x = T (t) A x$ for every $t \geq 0$ ; moreover the orbit $t \mapsto T (t) x$ is differentiable on $(0, \infty)$ with $\frac{d}{d t} T (t) x = T (t) A x = A T (t) x$ , and right differentiable at $t = 0$ with right derivative $A x$ .

Facts & Assumptions

Given: Dependent 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) and a vector $x \in D (A)$ .

[F1]

$x \in D (A)$ means that $\frac{T (h) x - x}{h} \to A x$ as $h ↓ 0$ (Infinitesimal generator of a C0-semigroup).

[F2]

The semigroup law $T (t + h) = T (t) T (h)$ holds for all $t, h \geq 0$ , and operator norms satisfy $∥ S y ∥ \leq ∥ S ∥ ∥ y ∥$ (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 $[0, \infty)$ (A semigroup with continuity at zero is uniformly bounded on every compact time interval, Strongly continuous semigroup): for $t_{0} > 0$ there is $M$ with $∥ T (s) ∥ \leq M$ for $s \in [0, t_{0}]$ , and $T (s) z \to T (t) z$ for every $z$ as $s \to t$ .

Proof

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

1.1F2

For $t \geq 0$ and $h > 0$ , the semigroup law gives $\frac{T (t + h) x - T (t) x}{h} = T (t) \frac{T (h) x - x}{h}$ .

2.1F1F2step 1.1

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

3.1F2F3step 1.1step 2.1

Left derivative at $t > 0$ : for $0 < h < t$ one has $\frac{T (t - h) x - T (t) x}{- h} = T (t - h) \frac{T (h) x - x}{h}$ . Let $v_{h} : = \frac{T (h) x - x}{h} \to A x$ and use the local bound $M$ on $[0, t]$ from [F3]: $∥ T (t - h) v_{h} - T (t) A x ∥ \leq M ∥ v_{h} - A x ∥ + ∥ (T (t - h) - T (t)) A x ∥ \to 0$ , because $v_{h} \to A x$ and $T (t - h) \to T (t)$ strongly as $h ↓ 0$ by [F3]. Hence the left derivative at $t$ also equals $T (t) A x = A T (t) x$ .

4.1step 2.1step 3.1∎

Combining [step 2.1] and [step 3.1], for every $x \in D (A)$ and every $t \geq 0$ the orbit satisfies $T (t) x \in D (A)$ , $A T (t) x = T (t) A x$ , and $t \mapsto T (t) x$ is differentiable on $(0, \infty)$ with derivative $T (t) A x = A T (t) x$ , with right derivative $A x$ at $t = 0$ .

Depends on

Used by

Dependency tree · two levels

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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)
Roland Schnaubelt, Evolution Equations, Karlsruhe Institute of Technology (2023/24 course, complete lecture notes) (standard reference, not scraped)