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An analytic semigroup need not be norm continuous at zero

Statement refuted

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain) for the cited integral and semigroup suppliers.

Assume the Axiom of Choice (The Axiom of Choice) for the Dirichlet eigenbasis witness.

The claim that a bounded analytic semigroup is norm continuous at the vertex, i.e. that $∥ T (t) - I ∥ \to 0$ as $t ↓ 0$ whenever $T$ is a bounded analytic semigroup, is false. Let $Ω$ be a nonempty bounded open set and let $A = Δ_{D}$ be the Dirichlet Laplacian with heat semigroup $T$ (The Dirichlet Laplacian generates an analytic heat semigroup, The analytic Dirichlet heat semigroup). Then $T$ is a bounded analytic semigroup of angle $π / 2$ , but $T (t) \to I$ fails in the operator norm as $t ↓ 0$ : for every $t > 0$ and every eigenfunction $e_{j}$ with eigenvalue $- λ_{j}$ , $∥ (T (t) - I) e_{j} ∥_{2} = ∣ e^{- λ_{j} t} - 1 ∣ ∥ e_{j} ∥_{2},$ and the right-hand side tends to $1$ as $j \to \infty$ for fixed $t > 0$ because $λ_{j} \to + \infty$ ; hence $∥ T (t) - I ∥ \geq 1$ for every $t > 0$ . In particular analyticity improves regularity in the time variable at positive times (Analytic semigroups are operator-norm differentiable away from zero) but does not upgrade strong continuity at the vertex to norm continuity; for a bounded generator the reverse conclusion holds (The analytic semigroup generated by a bounded operator).

Facts & Assumptions

Given: The Axiom of Choice; a nonempty bounded open set $Ω \subseteq R^{n}$ ; the Dirichlet Laplacian $A = Δ_{D}$ with its heat semigroup $T$ and eigenbasis ${e_{j}}$ with eigenvalues $- λ_{j}$ , $λ_{j} \to + \infty$ , normalised by $∥ e_{j} ∥_{L^{2}} = 1$ ; and a fixed $t > 0$ .

[L1]

$A = Δ_{D}$ generates a contraction analytic semigroup $T$ of maximal allowed angle $π / 2$ , hence a bounded analytic semigroup of angle $π / 2$ (The Dirichlet Laplacian generates an analytic heat semigroup).

[L2]

The heat semigroup is given by the spectral series $T (s) f = \sum_{j} e^{- λ_{j} s} (f, e_{j})_{L^{2}} e_{j}$ with $A e_{j} = - λ_{j} e_{j}$ , so $T (s) e_{j} = e^{- λ_{j} s} e_{j}$ for every $s > 0$ , and the eigenvalues satisfy $λ_{j} \to + \infty$ (The analytic Dirichlet heat semigroup, Discrete spectrum of a symmetric elliptic Dirichlet operator).

[L3]

In the setting of the smoothing theorem, $t \mapsto T (t)$ is of class $C^{\infty}$ on $(0, \infty)$ in the operator norm, with $\frac{d}{d t} T (t) = A T (t)$ ; no norm continuity or differentiability at $0$ is asserted, and for an unbounded generator it fails (Analytic semigroups are operator-norm differentiable away from zero).

[L4]

For a bounded operator $A \in B (X)$ the exponential series defines a uniformly continuous strongly continuous semigroup with generator $A$ ; in particular norm continuity at the vertex holds for bounded generators (The analytic semigroup generated by a bounded operator, A bounded linear operator between normed spaces).

[L5]

The operator norm is $∥ S ∥ = \sup {∥ S x ∥ : ∥ x ∥ \leq 1}$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Counterexample

technique · direct

1.1L2L5givenalgebra

The eigenfunction computation. For $t > 0$ and each basis eigenfunction $e_{j}$ of [L2], $T (t) e_{j} = e^{- λ_{j} t} e_{j}$ , so $(T (t) - I) e_{j} = (e^{- λ_{j} t} - 1) e_{j}$ and therefore $∥ (T (t) - I) e_{j} ∥_{L^{2}} = ∣ e^{- λ_{j} t} - 1 ∣ ∥ e_{j} ∥_{L^{2}} = ∣ e^{- λ_{j} t} - 1 ∣$ ; since $λ_{j} \to + \infty$ and $t > 0$ fixed, $λ_{j} t \to + \infty$ and $e^{- λ_{j} t} \to 0$ , so $∣ e^{- λ_{j} t} - 1 ∣ \to 1$ along $j$ , and $∥ T (t) - I ∥ \geq \sup_{j} ∣ e^{- λ_{j} t} - 1 ∣ = 1$ by [L5].

2.1step 1.1L1L3given

The semigroup is analytic but not norm continuous at zero. By [L1] $T$ is a bounded analytic semigroup of angle $π / 2$ generated by $A$ , so [L3] makes $t \mapsto T (t)$ operator-norm differentiable at every $t > 0$ , while [step 1.1] shows $∥ T (t) - I ∥ \geq 1$ for every $t > 0$ ; hence $T (t) \to I$ fails in operator norm as $t ↓ 0$ , and the failure is attached to the vertex, not to the analyticity on the open sector.

3.1step 1.1L2L4givenalgebra∎

Contrast with bounded generators. If the generator $A$ were bounded, [L4] would make $T$ uniformly continuous, in particular $∥ T (t) - I ∥ \to 0$ ; the computation of [step 1.1] together with $A e_{j} = - λ_{j} e_{j}$ from [L2] shows $∥ A e_{j} ∥_{L^{2}} = λ_{j} \to + \infty$ on the unit vectors $e_{j}$ , so the Dirichlet Laplacian is unbounded and the two conclusions are consistent; thus norm continuity at the vertex is not a consequence of analyticity but fails exactly for the unbounded-generator case, and the displayed estimate $∥ T (t) - I ∥ \geq 1$ is the explicit witness.

The same witness shows that $T (t) \to I$ in the norm topology fails maximally: the distance from $T (t)$ to the identity is at least $1$ along the eigenbasis. Strong continuity at the vertex is nevertheless asserted, since $T (t) f \to f$ in norm for each fixed $f$ ; only the uniform-in- $f$ statement fails.

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