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The translation semigroup is not strongly continuous on L-infinity

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the $L^{\infty}$ Banach-space interface. Let $X = L^{\infty} (R)$ with the essential-supremum norm (The space $L^{p} (μ)$ as the quotient by null functions, The essential supremum of a measurable function with respect to a measure) and let $(T (t) f) (s) : = f (s + t)$ (Translation of a function on $R^{n}$ ). Then $T (0) = I$ , $T (t + s) = T (t) T (s)$ and every $T (t)$ is a linear isometry of $X$ , but $(T (t))_{t \geq 0}$ is not strongly continuous: for $f : = 1_{(- \infty, 0]}$ and every $t > 0$ one has $∥ T (t) f - f ∥_{\infty} = 1$ , so $T (t) f \to̸ f$ as $t ↓ 0$ . This is the endpoint excluded by the finite- $p$ translation theorem.

Refuted claim. Every one-parameter family $(T (t))_{t \geq 0}$ of linear isometries of $L^{\infty} (R)$ with $T (0) = I$ and $T (t + s) = T (t) T (s)$ is strongly continuous. The right translation semigroup below satisfies all the algebraic hypotheses and all the isometry properties but fails strong continuity at $0$ ; this is the $p = \infty$ endpoint excluded from the finite- $p$ translation theorem.

Facts & Assumptions

Given: Countable Choice; $X = L^{\infty} (R)$ with the essential-supremum norm, and $(T (t) f) (s) : = f (s + t)$ for $t \geq 0$ , $f \in X$ , $s \in R$ ; the function $f : = 1_{(- \infty, 0]}$ .

[F1]

$L^{\infty} (R)$ consists of almost-everywhere classes of measurable functions with $∥ f ∥_{\infty} = \inf {M : ∣ f ∣ \leq M a.e.}$ , and $T (t)$ is well defined on classes because a translation preserves null sets (The space $L^{p} (μ)$ as the quotient by null functions, The essential supremum of a measurable function with respect to a measure).

[F2]

Translation of functions is defined by $(τ_{h} f) (s) = f (s - h)$ (Translation of a function on $R^{n}$ ), so with $T (t) f = τ_{- t} f$ the shift acts by $s \mapsto s + t$ as displayed.

[F3]

A strongly continuous semigroup must satisfy $T (0) = I$ , the semigroup law, and continuity of every orbit on $[0, \infty)$ , in particular at $0$ (Strongly continuous semigroup).

Counterexample

technique · direct computation with the step function $f=\mathbf 1_{(-\infty,0]}$

1.1F2algebra

$T (0) = I$ and $T (t + s) = T (t) T (s)$ for $s, t \geq 0$ : indeed $(T (0) f) (s) = f (s) = f (s)$ and $(T (t) T (s) f) (u) = (T (s) f) (u + t) = f (u + t + s) = (T (t + s) f) (u)$ for every $u$ .

1.2F1F2

Each $T (t)$ is linear and an isometry of $X$ : $∣ (T (t) f) (s) ∣ = ∣ f (s + t) ∣$ , and $s \mapsto s + t$ is a bijection of $R$ carrying null sets to null sets, so $∥ T (t) f ∥_{\infty} = ∥ f ∥_{\infty}$ .

2.1F1F2step 1.1

For $f = 1_{(- \infty, 0]}$ and $t > 0$ one has $T (t) f = 1_{(- \infty, - t]}$ , hence $(T (t) f - f) (s) = - 1$ for $s \in (- t, 0]$ and $= 0$ for $s \notin (- t, 0]$ ; the difference is the indicator of the interval $(- t, 0]$ up to a sign, and this interval has positive measure, so $∥ T (t) f - f ∥_{\infty} = 1$ .

3.1F3step 1.1step 1.2step 2.1∎

Consequently $T (t) f$ does not converge to $f$ in the norm of $X$ as $t ↓ 0$ : the distance stays equal to $1$ for every $t > 0$ , whereas any limit must have distance tending to $0$ . Since $f = T (0) f$ , the orbit of $f$ is not continuous at $0$ , so the family fails hypothesis (iii) of [F3] and is not a strongly continuous semigroup, despite satisfying all the algebraic axioms and consisting of linear isometries.

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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)