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Analytic Semigroups and Linear Evolution Equations — Examples

1 · Prerequisites

2 · Summary

These companions compute the analytic-semigroup theory of the main page on explicit operators and mark its sharp boundaries. The exponential series of a bounded operator is verified to be an entire semigroup generated by that operator, with the sectorial-extension criterion and unboundedness witnesses on sectors; the Dirichlet heat semigroup is realised through its eigenbasis, with the spectral series, the contraction bound and the smoothing constants $(m / (e t))^{m}$ computed coefficientwise. Multiplication operators supply bounded and unbounded sectorial examples, including a nonselfadjoint generator whose spectrum is the essential range of its multiplier, showing that self-adjointness is sufficient but not necessary for bounded analytic generation.

The counterexamples separate the notions: the translation semigroup on $L^{2}$ fails to be analytic, the sector angle depends on the sign convention, an analytic semigroup need not be norm continuous at the vertex, and a time-discontinuous forcing blocks classical regularity at its jump even under the stationary compatibility condition. A final abstract sequence-space example shows that positive-time smoothing into $D (A^{m})$ is an operator-theoretic statement about a diagonal generator and names Sobolev derivatives only after an elliptic-regularity identification.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passOpen item page →

The analytic semigroup generated by a bounded operator

Statement

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.

Let $X$ be a complex Banach space and let $A \in B (X)$ . The exponential series $E (z) : = \sum_{n \geq 0} \frac{z^{n}}{n!} A^{n}$ (The exponential series of a bounded operator) converges in the operator norm for every $z \in C$ and defines an entire function $E : C \to B (X)$ with $E (z + w) = E (z) E (w)$ ; its restriction $T : = E ∣_{[0, \infty)}$ is a uniformly continuous strongly continuous semigroup with generator the bounded operator $A$ . Moreover:

(1) $T$ admits a bounded analytic extension to $Σ_{δ}$ for some $δ \in (0, π / 2]$ if and only if $A$ satisfies the sectorial resolvent condition with exponent $δ$ in the $e^{t A}$ convention (Sectorial operator with the semigroup sign convention), equivalently $σ (A) \cap Σ_{π / 2 + δ} = \emptyset$ and the sectorial resolvent bound holds on $Σ_{π / 2 + δ - ε}$ ;

(2) $E$ can be unbounded on every sector of positive angle even though it is entire: for the nilpotent Jordan block $A = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ on $C^{2}$ one has $E (z) = I + z A$ , which is unbounded along every ray, and for $A = 1$ on $C$ one has $E (z) = e^{z}$ , which is unbounded on every sector that meets the open right half-plane;

(3) if $X$ is a complex Hilbert space and the numerical range (Numerical range and numerical radius) of $A$ lies in the closed sector ${μ : ∣ \arg (- μ) ∣ \leq π / 2 - θ} \cup {0}$ for some $θ \in (0, π / 2)$ , then $T$ is bounded analytic on $Σ_{θ}$ , so its maximal analytic angle is at least $θ$ . The example shows that sector boundedness of the exponential is a property of $A$ , not a consequence of boundedness of $A$ .

For a real Banach space the complex-time assertions are applied to $A_{C}$ on the canonical complexification; its real-time restriction is the original exponential semigroup (Canonical Banach complexification of a real Banach space).

Facts & Assumptions

Given: A Banach space $X$ , an operator $A \in B (X)$ , the exponential series $E (z) = \sum_{n \geq 0} z^{n} A^{n} / n!$ , and the restriction $T = E ∣_{[0, \infty)}$ .

[L1]

$E (t) = \sum_{n \geq 0} t^{n} A^{n} / n!$ converges absolutely in operator norm, uniformly on compact subsets of $R$ , satisfies $E (0) = I$ , $E (t + s) = E (t) E (s)$ , $∥ E (t) ∥ \leq e^{∣ t ∣ ∥ A ∥}$ , $E^{'} (t) = A E (t) = E (t) A$ , and $(E (t) - I) / t \to A$ in operator norm; $E$ is a uniformly continuous strongly continuous group whose generator is $A$ (The exponential series of a bounded operator).

[L2]

The bounded operator $A$ satisfies the sectorial resolvent condition with exponent $δ$ exactly when $Σ_{π / 2 + δ} \subseteq ρ (A)$ and for every $ε \in (0, δ)$ there is $M_{ε}$ with $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ on $Σ_{π / 2 + δ - ε}$ ; the characterization theorem makes this equivalent to generation of a bounded analytic semigroup of angle $δ$ by $A$ (Sectorial operator with the semigroup sign convention, Sectorial resolvent characterisation of bounded analytic semigroups).

[L3]

For a bounded operator the resolvent set contains ${∣ λ ∣ > ∥ A ∥}$ with $∥ R (λ, A) ∥ \leq (∣ λ ∣ - ∥ A ∥)^{- 1}$ by the Neumann series, and $R (\cdot, A)$ is holomorphic on the open set $ρ (A)$ (Neumann series and small perturbations of bounded inverses, Resolvent identity and holomorphy for a closed operator, Spectrum and resolvent of a bounded operator, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[L4]

The numerical range $W (T)$ is the set of $⟨ T x, x ⟩$ on the unit sphere (Numerical range and numerical radius).

[L5]

A bounded analytic semigroup of angle $δ$ has generator the infinitesimal generator of its restriction to $[0, \infty)$ (Complex sector and bounded analytic semigroup, Infinitesimal generator of a C0-semigroup, Strongly continuous semigroup).

[L6]

Two strongly continuous semigroups with the same generator coincide when the generator is bounded: if $U$ is such a semigroup and $x \in X = D (A)$ , then $U (t) x - x = \int_{0}^{t} A U (s) x d s$ , so $t \mapsto U (t) x$ solves $u^{'} = A u$ , $u (0) = x$ , and $E (- t) u (t)$ has vanishing derivative by the fundamental theorem, whence $u (t) = E (t) x$ (The exponential series of a bounded operator, Fundamental theorem of calculus for Banach-valued continuous curves).

Verification

technique · direct

1.1L1L5L6givenalgebra

The exponential and its generator. By [L1] the series $E (z)$ converges for every complex $z$ (the scalar series $\sum ∣ z ∣^{n} ∥ A ∥^{n} / n!$ bounds it), its derivative series converges uniformly on $∣ z ∣ \leq R$ since it is bounded by $∥ A ∥ e^{R ∥ A ∥}$ . For the difference quotients the remainder after the linear term is bounded by $∣ h ∣ ∥ A ∥^{2} e^{(R + ∣ h ∣) ∥ A ∥} / 2$ , so $E^{'} (z) = A E (z)$ and $E$ is entire. Absolute convergence permits regrouping the double product series; the binomial identity then gives $E (z + w) = E (z) E (w)$ , and its restriction $T$ is a uniformly continuous strongly continuous semigroup with $(T (t) - I) / t \to A$ , so its generator is the bounded operator $A$ with $D (A) = X$ ; [L6] identifies $T$ with any other strongly continuous semigroup having generator $A$ .

1.2L1L3givenalgebra

Unboundedness examples. For $A = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ on $C^{2}$ the series terminates: $A^{0} = I$ , $A^{1} = A$ and $A^{n} = 0$ for $n \geq 2$ , so $E (z) = I + z A$ ; on the unit vector $e_{2}$ one has $E (z) e_{2} = (z, 1)$ with Euclidean norm $\sqrt{1 + ∣ z ∣^{2}}$ , so $∥ E (z) ∥ \geq \sqrt{1 + ∣ z ∣^{2}}$ by [L3] and the operator norm is unbounded along every ray; for $A = 1$ on $C$ the series gives $E (z) = e^{z}$ with $∣ E (z) ∣ = e^{Re z}$ , which is unbounded on every sector that meets the open right half-plane.

1.3L2L3given

The resolvent estimate remains a separate condition. For bounded $A$ , the Neumann-series estimate of [L3] controls $R (λ, A)$ when $∣ λ ∣ > 2 ∥ A ∥$ , but it does not control behavior as $λ \to 0$ . The sectorial resolvent condition in [L2] therefore retains the separate $O (∣ λ ∣^{- 1})$ bound on every smaller sector; spectrum avoidance alone is not used to infer it.

2.1step 1.1step 1.3L2L5L6givenalgebra

The sectorial extension criterion. By [L2], for bounded $A$ the sectorial resolvent condition with exponent $δ$ is exactly the conjunction of $Σ_{π / 2 + δ} \subseteq ρ (A)$ and the resolvent bounds on the smaller sectors; step 1.3 explains why the bound at the vertex must remain explicit. The same characterization makes this condition equivalent to generation of a bounded analytic semigroup of angle $δ$ . By [step 1.1], $T$ is the semigroup generated by $A$ , so it has such an extension exactly under that sectorial condition; the spectrum formulation in (1) is just the equivalent resolvent-set clause together with the bound.

3.1step 1.1step 2.1L3L4givenalgebra∎

The numerical range case. Assume $X$ is a complex Hilbert space and $W (A)$ lies in the closed sector $C_{θ} : = {μ : ∣ \arg (- μ) ∣ \leq π / 2 - θ} \cup {0}$ with $θ \in (0, π / 2)$ ; for $λ \in Σ_{π / 2 + θ - ε}$ and $u \neq 0$ the normalised value $z_{u} : = ⟨ A u, u ⟩ / ∥ u ∥^{2}$ lies in $C_{θ}$ by [L4] and the angular distance from $λ$ to $C_{θ}$ is at least $ε$ , so $∣ λ - z_{u} ∣ \geq ∣ λ ∣ \sin ε$ and Cauchy-Schwarz gives $∥ (λ I - A) u ∥ \geq ∣ λ ∣ \sin ε ∥ u ∥$ ; hence $λ I - A$ is injective with closed range and, since for $∣ λ ∣ > ∥ A ∥$ it is invertible by [L3] and the set of surjectivity points is open (Neumann series) and closed in the connected complement of $C_{θ}$ : if $λ_{n} \to λ$ there, the positive distance of $λ$ to $C_{θ}$ bounds $∥ R (λ_{n}, A) ∥$ uniformly for large $n$ , and the resolvent identity makes these inverses Cauchy in operator norm. Their limit $R$ satisfies $(λ I - A) R = R (λ I - A) = I$ by boundedness of $A$ . Thus surjectivity is closed, every such $λ$ is in $ρ (A)$ with $∥ R (λ, A) ∥ \leq 1 / (∣ λ ∣ \sin ε)$ ; by [step 2.1] the sectorial condition with exponent $θ$ holds and $T$ extends to a bounded analytic semigroup on $Σ_{θ}$ , so its maximal analytic angle is at least $θ$ .

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The analytic Dirichlet heat semigroup

Statement

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.

Let $Ω \subseteq R^{n}$ be nonempty and bounded open and let $A = Δ_{D}$ be the Dirichlet Laplacian with its eigenbasis ${e_{j}}_{j \geq 1}$ and eigenvalues $- λ_{j}$ of Discrete spectrum of a symmetric elliptic Dirichlet operator (The Dirichlet Laplacian generates an analytic heat semigroup), so that $a (e_{j}, v) = λ_{j} (e_{j}, v)$ for every $v \in H_{0}^{1} (Ω)$ and $λ_{j} \to + \infty$ . Then the heat semigroup is given by the spectral series $T (t) f = \sum_{j \geq 1} e^{- λ_{j} t} (f, e_{j})_{L^{2}} e_{j}, f \in L^{2} (Ω),$ converging in $L^{2} (Ω)$ for every $t > 0$ , with $∥ T (t) ∥ \leq e^{- λ_{1} t}$ when $λ_{1} > 0$ , and for every $m \geq 1$ $A^{m} T (t) f = \sum_{j \geq 1} (- λ_{j})^{m} e^{- λ_{j} t} (f, e_{j})_{L^{2}} e_{j}, ∥ A^{m} T (t) ∥ \leq (\frac{m}{e t})^{m} .$ The bound is the spectral-coefficient form of Smoothing estimates for the semigroup generated by a sectorial operator and exhibits the $t^{- m}$ singularity as the supremum of $s^{m} e^{- s t}$ , $s \geq 0$ . For every nonempty bounded open $Ω$ and every $t > 0$ , the abstract conclusion is $T (t) f \in D (A^{m})$ . Its spatial consequence $T (t) f \in H^{2 m} (Ω)$ also holds when $n \geq 2$ and $Ω$ is a bounded $C^{2 m}$ domain, by Global $H^{2}$ Dirichlet regularity and Higher-order boundary regularity for Dirichlet problems; when $n = 1$ , it holds for every bounded open $Ω$ by the distributional derivative argument in step 4.1. The statement carries the Axiom of Choice and Countable Choice inherited from Discrete spectrum of a symmetric elliptic Dirichlet operator.

Facts & Assumptions

Given: A nonempty bounded open set $Ω \subseteq R^{n}$ ; the symmetric Dirichlet form $a (u, v) = \int_{Ω} \nabla u \cdot \overline{\nabla v} d x$ on $V = H_{0}^{1} (Ω)$ with associated operator $A$ defined by $a (u, v) = - (A u, v)_{L^{2}}$ (The $L^{2}$ operator associated with a symmetric elliptic form), identified with $Δ_{D}$ ; the eigenbasis ${e_{j}}_{j \geq 1} \subseteq H_{0}^{1} (Ω)$ of $L^{2} (Ω)$ and eigenvalues $λ_{1} \leq λ_{2} \leq \dots$ with $λ_{j} \to + \infty$ , orthonormal in $L^{2} (Ω)$ and satisfying $a (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ ; the coefficients $c_{j} : = (f, e_{j})_{L^{2}}$ of $f \in L^{2} (Ω)$ ; and the semigroup $T$ generated by $A$ .

[L1]

The form-norm expansion: for every $u \in H_{0}^{1} (Ω)$ the series $\sum_{j} (u, e_{j})_{L^{2}} e_{j}$ converges to $u$ in the $H_{0}^{1}$ norm and $a (u, u) = \sum_{j} λ_{j} ∣ (u, e_{j})_{L^{2}} ∣^{2}$ , while for every $f \in L^{2} (Ω)$ the series $\sum_{j} (f, e_{j})_{L^{2}} e_{j}$ converges to $f$ in $L^{2} (Ω)$ with $∥ f ∥_{L^{2}}^{2} = \sum_{j} ∣ (f, e_{j})_{L^{2}} ∣^{2}$ ; this assumes the Axiom of Choice and Countable Choice (Eigenbasis expansion in the form norm).

[L2]

The eigenbasis of the symmetric elliptic Dirichlet operator: there is an orthonormal basis ${e_{j}}$ of $L^{2} (Ω)$ with $e_{j} \in H_{0}^{1} (Ω)$ and $a (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}}$ for every $v \in H_{0}^{1} (Ω)$ , where $λ_{1} \leq λ_{2} \leq \dots$ are real with $λ_{j} \to + \infty$ , each repeated according to finite multiplicity; the Axiom of Choice and Countable Choice are assumed (Discrete spectrum of a symmetric elliptic Dirichlet operator, The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L3]

For the principal Dirichlet form $a_{0}$ the operator $A$ of the weak identity $a_{0} (u, v) = - (A u, v)_{L^{2}}$ is densely defined and self-adjoint with $⟨ A u, u ⟩ = - a_{0} (u, u) \leq 0$ , hence $A = Δ_{D}$ and generates a contraction analytic semigroup of maximal allowed angle $π / 2$ (The Dirichlet Laplacian generates an analytic heat semigroup).

[L4]

For a strongly continuous semigroup $S$ with generator $G$ and $∥ S (t) ∥ \leq M e^{ω t}$ , every real $λ > ω$ lies in $ρ (G)$ and $R (λ, G) x = \int_{0}^{\infty} e^{- λ t} S (t) x d t$ as an improper Bochner integral (Laplace transform formula for the resolvent).

[L5]

A generator of a strongly continuous semigroup is closed and densely defined (The generator is closed and densely defined).

[L6]

For a sectorial $A$ of angle $δ$ with vertex $0$ and generated semigroup $T$ , one has $T (t) X \subseteq D (A^{m})$ for every $t > 0$ , $m \geq 1$ , and the contour semigroup is the unique exponentially bounded strongly continuous semigroup with generator $A$ (Smoothing estimates for the semigroup generated by a sectorial operator, The generator of the contour semigroup is the sectorial operator).

[L7]

For the Dirichlet Laplacian on a bounded domain the first eigenvalue satisfies $λ_{1} > 0$ (The Poincare constant is the reciprocal square root of the first Dirichlet eigenvalue).

[L8]

Global $H^{2}$ Dirichlet regularity, under Countable Choice: for $n \geq 2$ , a bounded $C^{2}$ domain, $a^{i j} \in W^{1, \infty} (Ω)$ , bounded lower-order coefficients, and $f \in L^{2}$ , a weak zero-Dirichlet solution lies in $H^{2}$ ; the constant coefficients of $Δ_{D}$ meet these coefficient hypotheses (Global $H^{2}$ Dirichlet regularity).

[L9]

Higher-order boundary regularity, under Countable Choice: for $n \geq 2$ , a bounded $C^{k + 2}$ domain, $a^{i j} \in W^{k + 1, \infty}$ , lower-order coefficients in $W^{k, \infty}$ , and $f \in H^{k}$ , a weak zero-Dirichlet solution lies in $H^{k + 2}$ ; the constant coefficients of $Δ_{D}$ meet these coefficient hypotheses (Higher-order boundary regularity for Dirichlet problems).

[L10]

If strongly measurable $X$ -valued functions converge pointwise almost everywhere in norm and are dominated in norm by one integrable scalar function, then their Bochner integrals converge in norm (Bochner dominated convergence theorem).

Proof

technique · direct

1.1L1L2givenalgebra

Diagonal action and domain characterisation. Since $a (e_{j}, v) = λ_{j} (e_{j}, v)_{L^{2}}$ for all $v \in H_{0}^{1} (Ω)$ , the defining identity gives $e_{j} \in D (A)$ with $A e_{j} = - λ_{j} e_{j}$ ; for $u \in L^{2} (Ω)$ with coefficients $c_{j} = (u, e_{j})_{L^{2}}$ one has $u \in D (A)$ if and only if $\sum_{j} λ_{j}^{2} ∣ c_{j} ∣^{2} < \infty$ , and then $A u = - \sum_{j} λ_{j} c_{j} e_{j}$ : if $u \in D (A)$ then $(A u, e_{j})_{L^{2}} = - a (u, e_{j}) = - λ_{j} (u, e_{j})_{L^{2}}$ by symmetry and [L1] gives $\sum_{j} λ_{j}^{2} ∣ c_{j} ∣^{2} = ∥ A u ∥_{L^{2}}^{2} < \infty$ together with $A u = - \sum_{j} λ_{j} c_{j} e_{j}$ , while conversely $\sum_{j} λ_{j}^{2} ∣ c_{j} ∣^{2} < \infty$ makes $u_{N} : = \sum_{j \leq N} c_{j} e_{j}$ Cauchy in the form norm by [L1], hence convergent in $H_{0}^{1} (Ω)$ to a class equal to $u$ in $L^{2} (Ω)$ , and for $v \in H_{0}^{1} (Ω)$ continuity of $a$ in the $H^{1}$ norm gives $a (u, v) = \lim_{N} a (u_{N}, v) = \sum_{j} λ_{j} c_{j} \overline{(v, e_{j})_{L^{2}}}$ , the scalar series converging absolutely by Cauchy-Schwarz and [L1], so $u \in D (A)$ with $A u = - \sum_{j} λ_{j} c_{j} e_{j}$ .

2.1step 1.1L1L7givenalgebra

The spectral series defines the semigroup. For $t > 0$ put $S (t) f : = \sum_{j} e^{- λ_{j} t} c_{j} e_{j}$ and $S (0) f : = f$ : the series converges in $L^{2} (Ω)$ because the weights $e^{- λ_{j} t}$ are bounded in $j$ and $\sum_{j} ∣ c_{j} ∣^{2} = ∥ f ∥_{L^{2}}^{2}$ by [L1], the family is linear with $∥ S (t) ∥ \leq \sup_{j} e^{- λ_{j} t} \leq e^{- λ_{1} t}$ when $λ_{1} > 0$ by [L7], $S (t) S (s) = S (t + s)$ is coefficientwise, and $∥ S (t) f - f ∥_{L^{2}}^{2} = \sum_{j} (e^{- λ_{j} t} - 1)^{2} ∣ c_{j} ∣^{2} \to 0$ as $t ↓ 0$ by dominated convergence; each $S (t) f$ lies in $D (A)$ with $A S (t) f = - \sum_{j} λ_{j} e^{- λ_{j} t} c_{j} e_{j}$ by the criterion of [step 1.1], since $\sum_{j} λ_{j}^{2} e^{- 2 λ_{j} t} ∣ c_{j} ∣^{2} \leq \sup_{j} (λ_{j}^{2} e^{- 2 λ_{j} t}) ∥ f ∥_{L^{2}}^{2} < \infty$ .

3.1step 2.1L7givenalgebra

Powers and smoothing constants. For $m \geq 1$ the criterion of [step 1.1] applied inductively along the diagonal action gives $A^{m} S (t) f = \sum_{j} (- λ_{j})^{m} e^{- λ_{j} t} c_{j} e_{j}$ with $S (t) f \in D (A^{m})$ , and $λ_{j} \geq λ_{1} > 0$ by [L7] gives $∥ A^{m} S (t) ∥ \leq \sup_{s \geq 0} s^{m} e^{- s t} = (m / (e t))^{m}$ , the supremum being at $s = m / t$ by one-variable calculus; the same identities hold for $T$ once $T = S$ is established.

3.2step 1.1step 2.1L3L4L5L6L10givenalgebra

The generator is $A$ , so $S = T$ . For $λ > 0$ and $g \in L^{2} (Ω)$ with coefficients $g_{j}$ , the series $u : = \sum_{j} g_{j} (λ + λ_{j})^{- 1} e_{j}$ satisfies $\sum_{j} λ_{j}^{2} ∣ u_{j} ∣^{2} \leq \sum_{j} ∣ g_{j} ∣^{2} < \infty$ , so $u \in D (A)$ with $(λ I - A) u = \sum_{j} (λ + λ_{j}) u_{j} e_{j} = g$ by [step 1.1], while $(λ I - A) u = 0$ forces $(λ + λ_{j}) (u, e_{j})_{L^{2}} = 0$ for all $j$ and hence $u = 0$ ; therefore $R (λ, A) g = \sum_{j} (λ + λ_{j})^{- 1} g_{j} e_{j}$ with $∥ R (λ, A) ∥ \leq 1 / (λ + λ_{1}) \leq 1 / λ$ . To compare with the Laplace transform of $S$ , for $N \geq 1$ put $S_{N} (t) g : = \sum_{j \leq N} e^{- λ_{j} t} g_{j} e_{j}$ . Parseval and $λ_{j} \geq 0$ give $∥ S_{N} (t) g ∥_{L^{2}}^{2} = \sum_{j \leq N} e^{- 2 λ_{j} t} ∣ g_{j} ∣^{2} \leq ∥ g ∥_{L^{2}}^{2}$ ; by [L1], $S_{N} (t) g \to S (t) g$ in $L^{2}$ for every $t \geq 0$ . Thus the functions $t \mapsto e^{- λ t} S_{N} (t) g$ converge pointwise in norm and are dominated by the integrable scalar function $t \mapsto e^{- λ t} ∥ g ∥_{L^{2}}$ , so [L10] gives $\int_{0}^{\infty} e^{- λ t} S (t) g d t = \lim_{N \to \infty} \int_{0}^{\infty} e^{- λ t} S_{N} (t) g d t = \lim_{N \to \infty} \sum_{j \leq N} (λ + λ_{j})^{- 1} g_{j} e_{j} = R (λ, A) g,$ where the last limit holds in $L^{2}$ since $\sum_{j} ∣ g_{j} ∣^{2} / (λ + λ_{j})^{2} < \infty$ . By [L4] the left side is $R (λ, G_{S}) g$ , where $G_{S}$ is the generator of $S$ . Since $G_{S}$ is closed as a generator by [L5] and $A$ is closed as the generator of $T$ by [L3, L5], their common everywhere-defined inverse $R (λ, A) = R (λ, G_{S})$ gives $D (G_{S}) = R (λ, A) L^{2} = D (A)$ and $G_{S} = A$ ; finally [L6] supplies the unique exponentially bounded semigroup with generator $A$ , and both $S$ and the heat semigroup $T$ of [L3] are exponentially bounded strongly continuous semigroups with generator $A$ , so $S = T$ and the spectral series represents the heat semigroup.

4.1step 3.1L3L6L8L9givenalgebra∎

Abstract smoothing and the spatial reading. By [L6] the semigroup $T$ generated by the sectorial operator $A$ of [L3] satisfies $T (t) L^{2} \subseteq D (A^{m})$ for every $t > 0$ and $m \geq 1$ , which is the abstract content of the identities of [step 3.1]. For $n \geq 2$ and a bounded $C^{2 m}$ domain, put $v : = T (t) f$ and $w_{r} : = A^{r} v$ for $0 \leq r \leq m$ . Since $v \in D (A^{m})$ , each $w_{r} \in D (A) \subseteq H_{0}^{1} (Ω)$ for $r < m$ , $w_{m} \in L^{2}$ , and $A w_{r} = w_{r + 1}$ . Thus $- Δ w_{m - 1} = - w_{m} \in L^{2}$ , so [L8] gives $w_{m - 1} \in H^{2}$ . Descending from $r = m - 2$ to $r = 0$ , if $w_{r + 1} \in H^{2 (m - r - 1)}$ , then [L9] with $k = 2 (m - r - 1)$ applies to $- Δ w_{r} = - w_{r + 1}$ ; its domain requirement is $C^{2 (m - r)}$ , supplied by $C^{2 m}$ , and its coefficient requirements hold for the Laplacian. It follows that $w_{r} \in H^{2 (m - r)}$ , in particular $v \in H^{2 m} (Ω)$ . For $n = 1$ and any bounded open $Ω$ , $A = \partial_{x}^{2}$ distributionally. Each $w_{r} \in D (A) \subseteq H_{0}^{1} (Ω)$ for $r < m$ satisfies $w_{r}^{''} = w_{r + 1} \in L^{2}$ ; induction gives $w_{r} = v^{(2 r)}$ distributionally, while $w_{r}^{'} \in L^{2}$ for $r < m$ and $w_{m} \in L^{2}$ . Hence all weak derivatives of $v$ through order $2 m$ lie in $L^{2}$ , so $v \in H^{2 m} (Ω)$ without boundary regularity. The abstract $D (A^{m})$ conclusion remains valid in every dimension for arbitrary bounded open $Ω$ .

Remarks

The series for $T (t) f$ is the coefficientwise functional calculus of the self-adjoint generator along its eigenbasis; the universal scalar bound $(m / (e t))^{m}$ gives an explicit instance of the generic estimate $C_{m} t^{- m}$ of [L6]. The actual norm is $\sup_{j} λ_{j}^{m} e^{- λ_{j} t}$ , which can be strictly smaller because the eigenvalues are discrete. The statement inherits the Axiom of Choice and Countable Choice from the spectral suppliers [L2] and [L7] and no further choice principle beyond Dependent Choice is used in the verification.

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The sectorial multiplication operator

Statement

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.

Let $(Ω, μ)$ be a $σ$ -finite measure space with $μ (Ω) > 0$ , let $q : Ω \to R$ be measurable with $q \leq 0$ $μ$ -a.e., and let $A = M_{q}$ on $H = L^{2} (μ)$ have maximal domain $D (A) = {f \in L^{2} (μ) : q f \in L^{2} (μ)}$ . Then:

(1) $A$ is self-adjoint and densely defined;

(2) for every $λ \in C ∖ (- \infty, 0]$ , the multiplication operator $M_{(λ - q)^{- 1}}$ is bounded and is the inverse of $λ I - A$ , so $R (λ, A) f = (λ - q)^{- 1} f$ ;

(3) $∥ R (λ, A) ∥ = {ess sup}_{x} ∣ λ - q (x) ∣^{- 1} \leq \frac{1}{dist (λ, (- \infty, 0])} .$ Thus $∥ R (λ, A) ∥ \leq 1 / ∣ λ ∣$ when $Re λ \geq 0$ and $∥ R (λ, A) ∥ \leq 1 / ∣ Im λ ∣$ when $Re λ < 0$ ; in particular on $Σ_{π - ε}$ it is at most $1 / (∣ λ ∣ \sin ε)$ . These are sharp uniform bounds over all $q \leq 0$ , but equality for a fixed $q$ is not asserted. Consequently $A$ is sectorial with maximal exponent $π / 2$ in the $e^{t A}$ convention;

(4) $T (t) f = e^{t q} f$ is a strongly continuous contraction semigroup: $∥ e^{t q} ∥_{\infty} \leq 1$ , and $T (t) f \to f$ in $L^{2}$ by dominated convergence as $t ↓ 0$ . For each real $λ > 0$ , its Bochner Laplace integral acts pointwise as $\int_{0}^{\infty} e^{- λ t} e^{t q} f d t = (λ - q)^{- 1} f$ . Thus the Laplace-transform formula Laplace transform formula for the resolvent gives $R (λ, G) = M_{(λ - q)^{- 1}} = R (λ, A)$ , where $G$ is the generator of $T$ ; equality of resolvents at one point implies $G = A$ , including equality with the stated maximal domain. The computation is pointwise; a nonreal bounded multiplier requires its own argument. Dependent Choice is assumed for the semigroup suppliers; Countable Choice is inherited from the $L^{2}$ Hilbert-space and adjoint vocabulary.

Facts & Assumptions

Given: A $σ$ -finite measure space $(Ω, μ)$ with $μ (Ω) > 0$ , a measurable $q \leq 0$ a.e., the Hilbert space $H = L^{2} (μ)$ with its integral inner product, and $A = M_{q}$ with $D (A) = {f \in H : q f \in H}$ .

[L1]

$L^{2} (μ)$ is the quotient of $L^{2} (μ)$ by the a.e. zero functions, and its integral pairing makes it a complex Hilbert space under Countable Choice (The space $L^{p} (μ)$ as the quotient by null functions, $L^{2}$ with the integral pairing is a Hilbert space, Hilbert space).

[L2]

The adjoint domain consists of the vectors $y$ for which $x \mapsto ⟨ T x, y ⟩$ is bounded on $D (T)$ (Adjoint of a densely defined operator).

[L3]

$z \in ρ (T)$ when $z I - T : D (T) \to H$ is bijective with bounded inverse $R (z, T)$ (Resolvent and spectrum of a closed operator on a Banach space).

[L4]

The operator norm of a bounded operator is the supremum of $∥ T x ∥$ over the unit ball, so $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[L5]

$A$ is sectorial of angle $δ \in (0, π / 2]$ at vertex $0$ when $Σ_{π / 2 + δ} \subseteq ρ (A)$ with $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ ∣$ on $Σ_{π / 2 + δ - ε}$ for every $ε \in (0, δ)$ ; the equivalent conditions of the characterisation theorem then give a bounded analytic semigroup of angle $δ$ generated by $A$ (Sectorial operator with the semigroup sign convention, Sectorial resolvent characterisation of bounded analytic semigroups).

[L6]

A strongly continuous semigroup is a family $T (t)$ of bounded operators with $T (0) = I$ , $T (t + s) = T (t) T (s)$ and continuous orbits, and its generator is defined by the right difference quotients (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup).

[L7]

For a strongly continuous semigroup with generator $G$ and bound $∥ T (t) ∥ \leq M e^{ω t}$ one has $R (λ, G) x = \int_{0}^{\infty} e^{- λ t} T (t) x d t$ for every real $λ > ω$ (Laplace transform formula for the resolvent).

[L8]

Dominated convergence gives $L^{2}$ convergence from pointwise convergence with a fixed $L^{2}$ majorant (Dominated convergence).

Verification

technique · direct

1.1L1L2L8givenalgebra

The operator is densely defined and self-adjoint. For $f \in H$ the truncations $f_{n} : = f 1_{{∣ q ∣ \leq n}}$ lie in $D (A)$ and $f_{n} \to f$ in $H$ by [L8], so $D (A)$ is dense; for $f, g \in D (A)$ the identity $⟨ M_{q} f, g ⟩ = \int q f \overline{g} d μ = ⟨ f, M_{q} g ⟩$ shows symmetry. If $g \in D (A^{*})$ , then for every $h \in D (A)$ the adjoint relation gives $\int h (q \overline{g} - \overline{A^{*} g}) d μ = 0$ . Put $r : = A^{*} g - q g$ and $E_{n} : = {∣ q ∣ \leq n, ∣ r ∣ \leq n}$ , and define $h_{n} : = 1_{E_{n}} r$ . Then $h_{n}$ is in $L^{2}$ , since $∥ h_{n} ∥_{2} \leq ∥ A^{*} g ∥_{2} + n ∥ g ∥_{2}$ , while $q h_{n} \in L^{2}$ because $∣ q ∣ \leq n$ on its support, so $h_{n} \in D (A)$ . The adjoint identity with $h = h_{n}$ yields $0 = \int h_{n} (q \overline{g} - \overline{A^{*} g}) d μ = - \int_{E_{n}} ∣ r ∣^{2} d μ$ . The sets $E_{n}$ increase to full measure because $q$ is finite-valued and $g, A^{*} g \in L^{2}$ , so $r = 0$ almost everywhere. Thus $q g = A^{*} g \in H$ , and $g \in D (A)$ with $A^{*} g = A g$ ; therefore $D (A^{*}) = D (A)$ and $A = A^{*}$ by [L2].

1.2L1L6L8givenalgebra

The contraction semigroup. For $f \in H$ define $T (t) f : = e^{t q} f$ for $t \geq 0$ ; since $q \leq 0$ one has $∣ e^{t q} ∣ \leq 1$ and $∥ T (t) f ∥_{2} \leq ∥ f ∥_{2}$ with $T (t)$ linear and $T (0) = I$ , the functional equation is immediate from $e^{(t + s) q} = e^{t q} e^{s q}$ , and $∣ e^{t q} - 1 ∣^{2} ∣ f ∣^{2} \leq 4 ∣ f ∣^{2}$ together with pointwise convergence $e^{t q} \to 1$ gives $∥ T (t) f - f ∥_{2} \to 0$ as $t ↓ 0$ by [L8]; hence $T$ is a strongly continuous semigroup of contractions by [L6].

2.1step 1.1L1L3L4givenalgebra

The resolvent formula and its norm. Fix $λ \notin (- \infty, 0]$ and put $m_{λ} : = (λ - q)^{- 1}$ , bounded with $∣ m_{λ} (x) ∣ \leq 1 / dist (λ, (- \infty, 0])$ because $q (x) \in (- \infty, 0]$ a.e.; the multiplication operator $M_{m_{λ}}$ is bounded with $∥ M_{m_{λ}} ∥ = {ess sup}_{x} ∣ m_{λ} (x) ∣$ (the upper bound from [L4] and the lower bound by testing on the indicator of a finite-measure subset of ${∣ m_{λ} ∣ > s - ε}$ ); also $q m_{λ} = λ m_{λ} - 1$ is bounded, so $M_{m_{λ}} H \subset D (A)$ . The pointwise identities $(λ - q) m_{λ} = 1 = m_{λ} (λ - q)$ give $(λ I - A) M_{m_{λ}} f = f$ for $f \in H$ and $M_{m_{λ}} (λ I - A) g = g$ for $g \in D (A)$ ; hence $M_{m_{λ}}$ is the two-sided inverse of $λ I - A$ , so $λ \in ρ (A)$ with $R (λ, A) = M_{m_{λ}}$ and $∥ R (λ, A) ∥ = {ess sup}_{x} ∣ λ - q (x) ∣^{- 1} \leq 1 / dist (λ, (- \infty, 0])$ .

3.1step 2.1L5givenalgebra

Sectoriality with exponent $π / 2$ . By [step 2.1] every $λ \notin (- \infty, 0]$ , in particular every $λ \in Σ_{π} = C ∖ (- \infty, 0]$ , lies in $ρ (A)$ with $∥ R (λ, A) ∥ \leq 1 / dist (λ, (- \infty, 0])$ ; for $Re λ \geq 0$ the nearest point of $(- \infty, 0]$ is the origin and the distance is $∣ λ ∣$ , for $Re λ < 0$ it is $∣ Im λ ∣$ , and on $Σ_{π - ε}$ the distance is at least $∣ λ ∣ \sin ε$ (for $∣ \arg λ ∣ \leq π / 2$ the distance is $∣ λ ∣$ , and for $π / 2 \leq ∣ \arg λ ∣ \leq π - ε$ it is $∣ λ ∣ \sin (π - ∣ \arg λ ∣) \geq ∣ λ ∣ \sin ε$ ); hence $A$ is sectorial of angle $π / 2$ by [L5], and as the sectorial exponent is capped at $π / 2$ by the definition this exponent is maximal.

4.1step 1.2step 2.1L5L7givenalgebra∎

The generator is $A$ . For real $λ > 0$ the Bochner integral $v = \int_{0}^{\infty} e^{- λ t} T (t) f d t$ exists by the contraction bound. For $h \in H$ , Cauchy–Schwarz and $q \leq 0$ give $\int_{0}^{\infty} \int_{Ω} e^{- (λ - q (x)) t} ∣ f (x) h (x) ∣ d μ d t \leq λ^{- 1} ∥ f ∥_{2} ∥ h ∥_{2}$ . Scalar Fubini (Fubini's theorem for L^1 functions on a sigma-finite product) therefore gives $⟨ v, h ⟩ = \int_{Ω} (λ - q)^{- 1} f \overline{h} d μ$ . A bounded linear functional commutes with the Bochner integral by its simple-function definition, so this equality identifies $v = M_{(λ - q)^{- 1}} f$ . By [L7], $R (λ, G) = M_{(λ - q)^{- 1}} = R (λ, A)$ . The common inverse has range $D (G) = D (A)$ and determines both operators, so $G = A$ . Thus $T$ is the real-time restriction of the bounded analytic semigroup generated by $A$ , by [L5] and uniqueness of real-time semigroups.

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The translation semigroup is not analytic

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.

On $X = L^{p} (R)$ with $1 \leq p < \infty$ , let $T (t) f : = f (\cdot + t)$ be the right-translation semigroup (Continuous compactly supported functions are translation-continuous in $L^{p}$ , $C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ). Then $(T (t))_{t \geq 0}$ is a strongly continuous semigroup of isometries, its generator is $A f = f^{'} with D (A) = W^{1, p} (R) = {f \in L^{p} (R) : f^{'} \in L^{p} (R)},$ the derivative being the weak derivative (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms), and $(T (t))_{t \geq 0}$ is not an analytic semigroup in the sense of Complex sector and bounded analytic semigroup. Two independent obstructions are recorded: (1) (range) for $f \in L^{p} ∖ W^{1, p}$ and every $t > 0$ one has $T (t) f \notin D (A)$ because a weak derivative of a translate would translate back to a weak derivative of $f$ ; but analytic semigroups satisfy $T (t) X \subseteq D (A)$ for $t > 0$ (Cauchy estimates for an analytic semigroup give generator power bounds); (2) (spectrum) $σ (A) = i R$ , which meets every sector $Σ_{π / 2 + δ}$ , so $A$ fails the sectorial resolvent condition (Sectorial operator with the semigroup sign convention); the bounded analytic semigroup characterization Sectorial resolvent characterisation of bounded analytic semigroups therefore rules out bounded analytic generation. The range obstruction in (1) rules out even an analytic semigroup extension. Countable Choice is inherited from the $L^{p}$ translation-continuity, density and Sobolev vocabulary (Continuous compactly supported functions are translation-continuous in $L^{p}$ , $C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ , Integer-order Sobolev spaces and their norms); no further choice principle beyond Dependent Choice is used.

Refuted claim. A strongly continuous semigroup of isometries on $L^{p} (R)$ generated by a first-order differential operator is analytic, at least after shrinking the sector. The right-translation semigroup is a semigroup of isometries whose generator is differentiation, yet no positive time maps all of $L^{p}$ into the domain $W^{1, p}$ , which analytic semigroups are required to do; independently, the imaginary-axis spectrum blocks every sectorial resolvent estimate.

Facts & Assumptions

Given: $1 \leq p < \infty$ , $X = L^{p} (R)$ with the quotient $L^{p}$ norm, the right-translation family $T (t) f = f (\cdot + t)$ in the convention of Translation of a function on $R^{n}$ , and the generator $(A, D (A))$ of Infinitesimal generator of a C0-semigroup.

[L1]

For $f \in C_{c} (R)$ one has $∥ τ_{h} f - f ∥_{p} \to 0$ as $h \to 0$ , and $C_{c}^{\infty} (R)$ is dense in $L^{p} (R)$ (both assume Countable Choice) (Continuous compactly supported functions are translation-continuous in $L^{p}$ , $C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (R^{n})$ for $1 \leq p < \infty$ ).

[L2]

$C_{c}^{\infty} (R)$ is dense in $W^{1, p} (R)$ in the Sobolev norm (assuming Countable Choice) (Compactly supported smooth functions are dense in W^{k,p}(R^n), Integer-order Sobolev spaces and their norms).

[L3]

$v = D^{1} u$ weakly means $\int_{R} u φ^{'} = - \int_{R} v φ$ for every $φ \in C_{c}^{\infty} (R)$ ; $W^{1, p} (R)$ consists of the $L^{p}$ classes with weak derivative in $L^{p}$ (Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms).

[L4]

The $L^{p}$ norm is translation invariant, Lebesgue measure and measurability are translation invariant, and Fubini applies to absolutely integrable integrands on products of $σ$ -finite spaces (The space $L^{p} (μ)$ as the quotient by null functions, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Fubini's theorem for L^1 functions on a sigma-finite product).

[L5]

A strongly continuous semigroup is a family with $T (0) = I$ , $T (t + s) = T (t) T (s)$ and continuous orbits; its generator has domain consisting of the vectors with convergent right difference quotients (Strongly continuous semigroup, Infinitesimal generator of a C0-semigroup).

[L6]

A bounded analytic semigroup of angle $δ > 0$ with generator $A$ satisfies $T (t) X \subseteq D (A^{m})$ and $T^{(m)} (t) = A^{m} T (t)$ for every $t > 0$ and $m \geq 1$ (Cauchy estimates for an analytic semigroup give generator power bounds, Complex sector and bounded analytic semigroup).

[L7]

$A$ is sectorial of angle $δ > 0$ at vertex $0$ only if $Σ_{π / 2 + δ} \subseteq ρ (A)$ ; the conditions (a)-(e) of the characterisation theorem are equivalent, so a sectorial operator generates a bounded analytic semigroup on each smaller sector (Sectorial operator with the semigroup sign convention, Sectorial resolvent characterisation of bounded analytic semigroups).

[L8]

A strongly measurable curve is Bochner integrable exactly when the integral of its norm is finite, and the Bochner integral obeys the norm inequality (Bochner integrability criterion, Bochner integral norm inequality, Bochner-integrable function).

Counterexample

technique · direct

1.1L1L4L5givenalgebra

Strong continuity and isometries. Each $T (t)$ is linear and, by translation invariance of the $L^{p}$ norm [L4], an isometry with $∥ T (t) f ∥_{p} = ∥ f ∥_{p}$ ; the functional equation $T (t + s) = T (t) T (s)$ and $T (0) = I$ are immediate, and the orbit of $f \in C_{c} (R)$ is continuous at $0$ by [L1]; for general $f \in L^{p}$ and $ε > 0$ choose $φ \in C_{c}^{\infty} (R)$ with $∥ f - φ ∥_{p} < ε$ by [L1] and estimate $∥ T (t) f - f ∥_{p} \leq ∥ T (t) (f - φ) ∥_{p} + ∥ T (t) φ - φ ∥_{p} + ∥ φ - f ∥_{p} \leq 2 ε + ∥ T (t) φ - φ ∥_{p}$ , so strong continuity extends to all of $X$ ; hence $(T (t))_{t \geq 0}$ is a strongly continuous semigroup of isometries.

1.2L2L3L4L5givenalgebra

The generator is differentiation on $W^{1, p}$ . If the difference quotient of $f$ converges in $L^{p}$ to $g$ , then for every test function $φ$ the substitution $y = x + h$ and translation invariance give $\int g φ = \lim_{h} \int f (y) \frac{φ (y - h) - φ (y)}{h} d y = - \int f φ^{'}$ (uniform convergence of the difference quotients of $φ$ with compactly supported domination), so $g = D^{1} f$ weakly and $f \in W^{1, p} (R)$ by [L3]; conversely for $f \in W^{1, p}$ and $φ \in C_{c}^{\infty}$ one has $∥ \frac{T (h) φ - φ}{h} - φ^{'} ∥_{p} \to 0$ while the shift inequality $∥ (T (h) - I) u ∥_{p} \leq ∣ h ∣ ∥ u^{'} ∥_{p}$ for $u \in W^{1, p}$ follows from $u (x + h) - u (x) = \int_{0}^{h} u^{'} (x + s) d s$ for smooth $u$ and extends by the density [L2]; given $δ > 0$ choose $φ \in C_{c}^{\infty}$ with $∥ f - φ ∥_{W^{1, p}} < δ$ , write $u = f - φ$ , and estimate $∥ \frac{T (h) f - f}{h} - f^{'} ∥_{p} \leq ∥ \frac{T (h) φ - φ}{h} - φ^{'} ∥_{p} + 2 ∥ u^{'} ∥_{p} \leq ∥ \frac{T (h) φ - φ}{h} - φ^{'} ∥_{p} + 2 δ$ , so the limsup is $0$ and $W^{1, p} (R) \subseteq D (A)$ with $A f = f^{'}$ ; hence $D (A) = W^{1, p} (R)$ .

1.3L3L4givenalgebra

The range obstruction. Translation commutes with weak differentiation: if $v = D^{1} u$ and $h \in R$ , then for every test function $φ$ the identities $\int u (x + h) φ^{'} (x) d x = \int u (y) φ^{'} (y - h) d y = - \int v (y) φ (y - h) d y = - \int v (x + h) φ (x) d x$ show $v (\cdot + h) = D^{1} (u (\cdot + h))$ ; hence if $T (t) f = f (\cdot + t) \in W^{1, p} (R)$ for some $t > 0$ , then $f = T (- t) (T (t) f)$ is a translate of a $W^{1, p}$ function and lies in $W^{1, p} (R)$ , a contradiction whenever $f \in L^{p} ∖ W^{1, p}$ ; such $f$ exist, for instance $f = 1_{[0, 1]}$ , whose would-be weak derivative must vanish a.e. off the two jump points by testing away from them, hence a.e. everywhere, which is incompatible with $\int_{R} f φ^{'} = φ (1) - φ (0)$ for a test function with $φ (1) \neq φ (0)$ ; therefore $T (t) f \notin D (A)$ for every $t > 0$ .

2.1step 1.2step 1.3L6givenalgebra

The translation semigroup is not analytic. If $T$ had any analytic extension, norm differentiability at $t > 0$ would give $h^{- 1} (T (h) - I) T (t) y = h^{- 1} (T (t + h) - T (t)) y \to T^{'} (t) y$ for every $y$ . The generator definition [L5] would therefore put $T (t) y$ in $D (A)$ , with no global sector bound required; [step 1.3] exhibits a vector $f \in X$ with $T (t) f \notin D (A)$ for every $t > 0$ , while [step 1.2] identifies $D (A) = W^{1, p}$ as the domain of the generator, so $(T (t))_{t \geq 0}$ admits no analytic extension with generator $A$ , bounded or not.

2.2step 1.1step 1.2L3L4L8givenalgebra

The half-planes lie in the resolvent set. For $Re λ > 0$ and $f \in X$ put $R_{λ} f : = \int_{0}^{\infty} e^{- λ t} T (t) f d t$ and for $Re λ < 0$ put $R_{λ} f : = - \int_{0}^{\infty} e^{λ t} T (- t) f d t$ ; both are absolutely convergent Bochner integrals by [L8] because the isometries of [step 1.1] give $∥ e^{- λ t} T (t) f ∥ \leq e^{- Re λ t} ∥ f ∥$ , and the substitution $y = x + t$ (respectively $y = x - t$ ) with [L4] gives for every test function $φ$ the identity $\int (R_{λ} f) φ^{'} = \int f φ - λ \int (R_{λ} f) φ$ in both cases, so $R_{λ} f \in W^{1, p} (R)$ with weak derivative $λ R_{λ} f - f$ , that is $(λ I - A) R_{λ} f = f$ with $A$ as in [step 1.2]; conversely, for $g \in W^{1, p} (R)$ the same Fubini computation with the weak-derivative identity of [L3] gives $\int (R_{λ} (λ g - g^{'})) φ = \int g φ$ for every test function $φ$ , so $R_{λ} (λ I - A) g = g$ ; hence $λ \in ρ (A)$ with $R (λ, A) = R_{λ}$ for every $λ \notin i R$ .

2.3step 1.2L3L4givenalgebra

The imaginary axis lies in the spectrum. Fix $ξ \in R$ , choose $φ \in C_{c}^{\infty} (R)$ with $φ \equiv̸ 0$ and set $u_{N} (x) : = φ (x / N) e^{i ξ x} \in W^{1, p} (R)$ for $N \geq 1$ ; then $(A - i ξ) u_{N} = u_{N}^{'} - i ξ u_{N} = N^{- 1} φ^{'} (x / N) e^{i ξ x}$ by [step 1.2], and by the substitution $x = N y$ in the Lebesgue integrals $∥ (A - i ξ) u_{N} ∥_{p} = N^{1 / p - 1} ∥ φ^{'} ∥_{p}$ while $∥ u_{N} ∥_{p} = N^{1 / p} ∥ φ ∥_{p}$ , so the ratios $∥ (A - i ξ) u_{N} ∥_{p} / ∥ u_{N} ∥_{p} = N^{- 1} ∥ φ^{'} ∥_{p} / ∥ φ ∥_{p}$ tend to $0$ ; if $i ξ$ lay in $ρ (A)$ the bounded inverse would give the positive lower bound $∥ (A - i ξ) u ∥_{p} \geq ∥ u ∥_{p} / ∥ R (i ξ, A) ∥$ for all $u \in D (A)$ , a contradiction for large $N$ ; hence $i ξ \in σ (A)$ for every $ξ$ , that is $i R \subseteq σ (A)$ .

3.1step 2.1step 2.2step 2.3L7givenalgebra∎

The spectral obstruction and the conclusion. By [step 2.2] and [step 2.3] one has $σ (A) = i R$ ; since the point $i$ has argument $π / 2$ , it lies in $Σ_{π / 2 + δ}$ for every $δ > 0$ , so no sector $Σ_{π / 2 + δ}$ is contained in $ρ (A)$ and [L7] rules out sectoriality of every positive exponent; the characterisation theorem [L7] therefore rules out generation of a bounded analytic semigroup, independently of the range obstruction of [step 2.1], which already excludes every analytic extension; the restriction $p < \infty$ is essential, since right translation is not strongly continuous on $L^{\infty}$ , and the argument uses no choice principle beyond Dependent Choice, which implies the Countable Choice required by the $L^{p}$ translation-continuity, density and Sobolev vocabulary.

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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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The sector changes under the sign convention

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.

The statement "replacing $A$ by $- A$ preserves sectoriality at vertex $0$ and the sector angle, merely inverting the generated semigroup" is false. Witness: $X = C$ and $A = 1$ . Then $σ (A) = {1}$ , and $1 \in Σ_{π / 2 + δ}$ for every $δ > 0$ , so $Σ_{π / 2 + δ} \subseteq̸ ρ (A)$ and $A$ is not sectorial in the $e^{t A}$ convention (Sectorial operator with the semigroup sign convention); the generated semigroup is $e^{t}$ , which is not bounded. However $- A = - 1$ satisfies $Σ_{π / 2 + δ} \subseteq ρ (- A) = C ∖ {- 1}$ for every $δ < π / 2$ , with $∥ R (λ, - A) ∥ = 1 / ∣ λ + 1 ∣ \leq M_{ε} / ∣ λ ∣$ on $Σ_{π / 2 + δ - ε}$ where $M_{ε} = 1 / \sin ε$ ; hence $- A$ is sectorial of angle $π / 2$ and generates the bounded analytic semigroup $e^{- t}$ (Sectorial resolvent characterisation of bounded analytic semigroups). Thus sectoriality is an oriented condition located on the spectral side; the dictionary of Sectorial operator with the semigroup sign convention must be applied to the operator that actually appears, and a signless citation of " $A$ is sectorial" changes the sector by reflection through the origin.

Refuted claim. If $A$ is sectorial at vertex $0$ in the $e^{t A}$ convention then so is $- A$ , with the same sector angle, and the generated semigroup is merely replaced by its inverse. The one-dimensional operator $A = 1$ has $σ (A) = {1}$ inside every sector $Σ_{π / 2 + δ}$ , so it is not sectorial at vertex $0$ , while $- A = - 1$ is sectorial of the maximal angle $π / 2$ and generates the contractive semigroup $e^{- t}$ .

Facts & Assumptions

Given: The one-dimensional complex Banach space $X = C$ , the bounded operators $A = 1$ and $- A = - 1$ acting as multiplication on $X$ , and the open sectors $Σ_{γ} = {λ \neq 0 : ∣ \arg λ ∣ < γ}$ .

[L1]

$A$ is sectorial of angle $δ \in (0, π / 2]$ at vertex $ω$ in the $e^{t A}$ convention when $ω + Σ_{π / 2 + δ} \subseteq ρ (A)$ with $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ - ω ∣$ on $ω + Σ_{π / 2 + δ - ε}$ for every $ε \in (0, δ)$ (Sectorial operator with the semigroup sign convention).

[L2]

For a bounded operator $A \in B (X)$ the series $E (t) = \sum_{n \geq 0} t^{n} A^{n} / n!$ is a strongly continuous group of bounded operators whose generator is $A$ , with $∥ E (t) ∥ \leq e^{∣ t ∣ ∥ A ∥}$ (The exponential series of a bounded operator, A bounded linear operator between normed spaces).

[L3]

The conditions (a)-(e) of the sectorial resolvent characterisation are equivalent, so a densely defined closed operator sectorial at vertex $0$ with a positive exponent generates a bounded analytic semigroup (Sectorial resolvent characterisation of bounded analytic semigroups, Complex sector and bounded analytic semigroup).

[L4]

The operator norm is submultiplicative and $∣ R (λ, - 1) ∣ = 1 / ∣ λ + 1 ∣$ in the one-dimensional space (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Counterexample

technique · direct

1.1L1L2L4givenalgebra

The witness $A = 1$ is not sectorial. In $X = C$ the operator $λ I - A$ is multiplication by $λ - 1$ , which is invertible exactly for $λ \neq 1$ , so $ρ (A) = C ∖ {1}$ and $σ (A) = {1}$ ; since $1 \in Σ_{π / 2 + δ}$ for every $δ > 0$ , no sector with vertex $0$ , $Σ_{π / 2 + δ}$ , is contained in $ρ (A)$ and [L1] rules out sectoriality at vertex $0$ of every positive exponent; moreover $A$ is bounded with $∥ A ∥ = 1$ , so [L2] gives the generated semigroup $E (t) = e^{t}$ with $∥ E (t) ∥ = e^{t}$ , which is unbounded on $[0, \infty)$ .

1.2L1L4givenalgebra

The reflected operator $- A = - 1$ is sectorial of angle $π / 2$ . Here $λ I + A$ is multiplication by $λ + 1$ , invertible for $λ \neq - 1$ , so $ρ (- A) = C ∖ {- 1}$ and $R (λ, - A) = (λ + 1)^{- 1}$ ; the point $- 1$ has argument $π$ while every $λ \in Σ_{π / 2 + δ}$ has $∣ \arg λ ∣ < π / 2 + δ \leq π$ for $δ \leq π / 2$ , so $- 1 \notin Σ_{π / 2 + δ}$ and $Σ_{π / 2 + δ} \subseteq ρ (- A)$ ; for $λ = r e^{i α} \in Σ_{π / 2 + δ - ε}$ with $ε < δ \leq π / 2$ one has $∣ α ∣ \leq π / 2 + δ - ε < π - ε$ , hence $\cos ∣ α ∣ \geq \cos (π - ε) = - \cos ε$ and $∣ λ + 1 ∣^{2} = r^{2} + 2 r \cos ∣ α ∣ + 1 \geq r^{2} - 2 r \cos ε + 1 = (r \cos ε - 1)^{2} + r^{2} \sin^{2} ε \geq r^{2} \sin^{2} ε$ ; hence $∥ R (λ, - A) ∥ = 1 / ∣ λ + 1 ∣ \leq M_{ε} / ∣ λ ∣$ with $M_{ε} = 1 / \sin ε$ , and [L1] makes $- A$ sectorial of angle $π / 2$ .

2.1step 1.1step 1.2L1L2L3givenalgebra∎

The refutation. The two computations show that $A = 1$ is not sectorial in the $e^{t A}$ convention while $- A = - 1$ is sectorial of the endpoint angle $π / 2$ , so replacing $A$ by $- A$ does not preserve sectoriality or the sector angle; by [L3] the sectorial operator $- A$ generates a bounded analytic semigroup, which by [L2] is $t \mapsto e^{- t}$ with norm $e^{- t} \leq 1$ , the inverse of the unbounded semigroup $e^{t}$ generated by $A$ ; the dictionary of [L1] therefore has to be applied to the operator that actually occurs, and the functions here are explicit, so no choice principle beyond Dependent Choice is used.

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A sectorial nonselfadjoint multiplication generator

Statement

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). Let $(Ω, μ)$ be a $σ$ -finite measure space with $μ (Ω) > 0$ , let $θ \in (0, π / 2)$ , and let $q : Ω \to C$ be measurable and essentially bounded, with essential range contained in the closed left sector ${ζ : ∣ \arg (- ζ) ∣ \leq π / 2 - θ} \cup {0}$ . Then $A = M_{q}$ on $L^{2} (μ)$ satisfies the sectorial resolvent condition with exponent $θ$ in the $e^{t A}$ convention and generates the bounded analytic semigroup $T (z) f = e^{z q} f$ on $Σ_{θ}$ ; its maximal analytic angle is at least $θ$ . Its spectrum is $essran (q)$ (as proved directly in step 1.2). Whenever $q$ is nonreal on a set of positive measure, the generator $A$ is nonselfadjoint: $A^{*} = M_{\overset{ˉ}{q}} \neq M_{q} = A$ (both are bounded operators on all of $L^{2} (μ)$ ), so $A$ is not a self-adjoint semigroup generator. Self-adjoint nonpositive generation is a sufficient route to bounded analytic semigroups (Self-adjoint nonpositive operators generate bounded analytic semigroups), but this example shows that self-adjointness is not necessary: the multiplier is nonselfadjoint and still generates a bounded analytic semigroup. It is the bounded-operator companion to the form-generated theorem Form-generated sectorial elliptic semigroups.

Facts & Assumptions

Given: The Axiom of Choice; a $σ$ -finite measure space $(Ω, μ)$ with $μ (Ω) > 0$ ; a number $θ \in (0, π / 2)$ ; a measurable essentially bounded $q : Ω \to C$ whose essential range lies in the closed left sector ${ζ : ∣ \arg (- ζ) ∣ \leq π / 2 - θ} \cup {0}$ ; the bounded multiplication operator $A = M_{q}$ on the complex Hilbert space $H = L^{2} (μ)$ ; and the family $T (z) f = e^{z q} f$ .

[L1]

For $δ \in (0, π / 2]$ and $ω \in R$ an operator is sectorial of angle $δ$ with vertex $ω$ in the $e^{t A}$ convention when $ω + Σ_{π / 2 + δ} \subseteq ρ (A)$ and $∥ R (λ, A) ∥ \leq M_{ε} / ∣ λ - ω ∣$ on each $ω + Σ_{π / 2 + δ - ε}$ , where $Σ_{γ}$ is the open sector of half-angle $γ$ around the positive real axis (Sectorial operator with the semigroup sign convention, Complex sector and bounded analytic semigroup).

[L2]

A bounded linear operator $T \in B (H)$ is one with a finite bound $∥ T x ∥ \leq C ∥ x ∥$ , and the operator norm is $∥ T ∥ = \sup {∥ T x ∥ : ∥ x ∥ \leq 1}$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[L3]

A densely defined $A$ is sectorial of angle $θ$ with vertex $0$ if and only if it generates a bounded analytic semigroup $(T (z))_{z \in Σ_{θ} \cup {0}}$ of angle $θ$ with generator $A$ (Sectorial resolvent characterisation of bounded analytic semigroups).

[L4]

For a bounded operator $A$ the exponential series $E (z) = \sum_{n \geq 0} z^{n} A^{n} / n!$ converges in operator norm for every $z \in C$ , defines an entire function with $E (z + w) = E (z) E (w)$ whose restriction is a strongly continuous semigroup with generator $A$ , and $E$ extends boundedly analytically to $Σ_{δ}$ exactly when the sectorial resolvent condition with exponent $δ$ holds (The analytic semigroup generated by a bounded operator).

[L5]

The Hilbert adjoint of a bounded operator is the unique $T^{*}$ with $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ ; for multiplication operators $⟨ M_{q} f, g ⟩ = \int q f \overline{g} d μ$ and $⟨ f, M_{\overset{ˉ}{q}} g ⟩ = \int f \overline{\overset{ˉ}{q} g} d μ$ show $(M_{q})^{*} = M_{\overset{ˉ}{q}}$ (The Hilbert-space adjoint of a bounded operator).

[L6]

On a $σ$ -finite measure space every positive-measure measurable set contains a finite-measure measurable subset of positive measure, and $L^{2}$ functions are almost-everywhere classes (Finite, sigma-finite, and semifinite measures, The space $L^{p} (μ)$ as the quotient by null functions).

[L7]

The special case of real $q \leq 0$ is the companion multiplication example: its resolvents are the bounded multiplications by $(λ - q)^{- 1}$ and it is sectorial with maximal exponent $π / 2$ (The sectorial multiplication operator).

[L8]

A densely defined self-adjoint nonpositive operator generates a contractive bounded analytic semigroup (Self-adjoint nonpositive operators generate bounded analytic semigroups).

[L9]

The form-generated theorem gives a complementary generation route for operators associated with closed sectorial forms and assumes no symmetry (Form-generated sectorial elliptic semigroups).

[L10]

The Axiom of Choice supplies a choice function for the countable family of nonempty sets of finite-measure positive-measure subsets used in step 1.2 (The Axiom of Choice).

Verification

technique · direct

1.1L1L2givenalgebra

The resolvent bound. Since $A = M_{q}$ is bounded with $D (A) = H$ , for $λ \neq 0$ the operator $λ I - A = M_{λ - q}$ has the two-sided inverse $M_{(λ - q)^{- 1}}$ as soon as $(λ - q)^{- 1}$ is essentially bounded; fix $ε \in (0, θ)$ and $λ \in Σ_{π / 2 + θ - ε}$ . The essential-range definition implies $q (x) \in essran (q)$ almost everywhere: every value outside the essential range has a neighbourhood with null preimage; a countable rational-ball base covers that complement by countably many such neighbourhoods, so its preimage is null. Hence for almost every $x$ the value $q (x)$ lies in the closed sector ${ζ : ∣ \arg (- ζ) ∣ \leq π / 2 - θ} \cup {0}$ whose boundary rays have arguments $\pm (π / 2 + θ)$ , while $∣ \arg λ ∣ \leq π / 2 + θ - ε$ . For nonzero $ζ = q (x)$ let $γ \in [0, π]$ be the principal angle between $λ$ and $ζ$ ; the sector geometry gives $γ \geq ε$ . If $γ \leq π / 2$ , then $∣ λ - ζ ∣^{2} = ∣ λ ∣^{2} \sin^{2} γ + (∣ λ ∣ \cos γ - ∣ ζ ∣)^{2} \geq ∣ λ ∣^{2} \sin^{2} ε$ . If $γ \geq π / 2$ , then $\cos γ \leq 0$ and $∣ λ - ζ ∣^{2} = ∣ λ ∣^{2} + ∣ ζ ∣^{2} - 2 ∣ λ ∣ ∣ ζ ∣ \cos γ \geq ∣ λ ∣^{2} \geq ∣ λ ∣^{2} \sin^{2} ε$ ; for $ζ = 0$ the same lower bound follows from $∣ λ - ζ ∣ = ∣ λ ∣$ . Thus $∣ λ - q (x) ∣ \geq ∣ λ ∣ \sin ε$ almost everywhere; for each $f \in H$ , $∥ M_{(λ - q)^{- 1}} f ∥_{2}^{2} = \int_{Ω} ∣ λ - q ∣^{- 2} ∣ f ∣^{2} d μ \leq (∣ λ ∣ \sin ε)^{- 2} ∥ f ∥_{2}^{2},$ so $∥ M_{(λ - q)^{- 1}} ∥ \leq 1 / (∣ λ ∣ \sin ε)$ and $R (λ, A) = M_{(λ - q)^{- 1}}$ with $λ \in ρ (A)$ .

1.2L6L10givenalgebra

The spectrum is the essential range. If $λ \notin essran (q)$ then by definition of the essential range there is $δ > 0$ with $μ ({∣ q - λ ∣ < δ}) = 0$ , so $∣ λ - q ∣ \geq δ$ almost everywhere and $M_{(λ - q)^{- 1}}$ is a bounded inverse of $λ I - A$ , hence $λ \in ρ (A)$ ; conversely, if $λ \in essran (q)$ then for every integer $n \geq 1$ the set $E_{n} : = {∣ q - λ ∣ < 1 / n}$ has positive measure, so by $σ$ -finiteness it contains a measurable $F_{n}$ with $0 < μ (F_{n}) < \infty$ , and $f_{n} : = μ (F_{n})^{- 1 / 2} 1_{F_{n}}$ is a unit vector with $∥ (λ I - A) f_{n} ∥_{2}^{2} = \int_{F_{n}} ∣ λ - q ∣^{2} d μ / μ (F_{n}) \leq 1 / n^{2}$ ; a bounded inverse $R$ of $λ I - A$ would give $1 = ∥ f_{n} ∥_{2} \leq ∥ R ∥ / n$ for every $n$ , impossible, so $λ \notin ρ (A)$ and $σ (A) = essran (q)$ .

1.3L5L6givenalgebra

Nonselfadjointness. If $q$ is nonreal on a set of positive measure, [L6] supplies a finite-measure subset $F$ of that set with $μ (F) > 0$ . Then $1_{F} \in L^{2} (μ)$ and $(M_{q} - M_{\overset{ˉ}{q}}) 1_{F} = (q - \overset{ˉ}{q}) 1_{F}$ is a nonzero $L^{2}$ class, so $M_{q} \neq M_{\overset{ˉ}{q}}$ . By [L5] $(M_{q})^{*} = M_{\overset{ˉ}{q}}$ ; hence $A^{*} \neq A$ and $A$ is not self-adjoint.

2.1step 1.1L1L3L4givenalgebra

Generation and the explicit semigroup. By [step 1.1] the sectorial resolvent condition of [L1] holds with exponent $θ$ : for every $ε \in (0, θ)$ the bound $∥ R (λ, A) ∥ \leq 1 / (∣ λ ∣ \sin ε)$ holds on $Σ_{π / 2 + θ - ε}$ ; hence by [L4] the exponential series $E (z) = \sum_{n} z^{n} A^{n} / n!$ extends boundedly analytically to $Σ_{θ}$ and is generated by $A$ ; moreover $\sum_{n} z^{n} q^{n} / n!$ converges in essential supremum norm to $e^{z q}$ because $q$ is essentially bounded, so $E (z) = M_{e^{z q}}$ , that is $E (z) f = e^{z q} f$ ; on $Σ_{θ}$ one has $Re (z q) \leq 0$ almost everywhere, since the angle between $z$ and $q (x)$ is at least $π / 2$ , so $∣ e^{z q} ∣ \leq 1$ and the family is bounded on every $Σ_{δ^{'}}$ with $δ^{'} < θ$ ; therefore the maximal analytic angle of $A$ is at least $θ$ , by the definition of the angle as the supremum of the admissible exponents in [L1] and [L3].

3.1step 1.2step 1.3step 2.1L7L8L9givenalgebra∎

Assembly. By [L8], self-adjoint nonpositive generation is a sufficient route to bounded analytic semigroups. Here [step 2.1] shows that $A = M_{q}$ generates the bounded analytic semigroup $T (z) f = e^{z q} f$ with maximal angle at least $θ$ , while [step 1.3] shows that $A$ is nonselfadjoint when $q$ is nonreal on a set of positive measure; thus self-adjointness is not necessary. By [step 1.2] its spectrum is $essran (q)$ , [L7] is the real multiplier special case, and [L9] supplies the complementary form-based generation context without a symmetry restriction.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passOpen item page →

A time-discontinuous forcing blocks classical regularity at its jump

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.

Let $X = C$ , $A = 0$ (a bounded generator), $0 < t_{0} < b$ , and $f : = 1_{[t_{0}, b]}$ , which is bounded and measurable but neither continuous nor H"older at $t_{0}$ . The mild solution of $u^{'} = A u + f = 0 \cdot u + f$ , $u (0) = 0$ , is $u (t) = \int_{0}^{t} 1_{[t_{0}, b]} (s) d s = {\begin{cases} 0, & 0 \leq t \leq t_{0}, \\ t - t_{0}, & t_{0} < t \leq b . \end{cases}$ It is continuous and Lipschitz and satisfies $u^{'} (t) = f (t)$ for $t \neq t_{0}$ , but it is not differentiable at $t_{0}$ (left derivative $0$ , right derivative $1$ ) and hence is not a classical solution on $[0, b]$ in the sense of Classical, strong and mild abstract Cauchy solutions. Thus $u \in C ([0, b], X) \cap C^{1} ((0, b] ∖ {t_{0}}, X)$ but $u \notin C^{1} ([0, b], X)$ , and the H"older-continuity hypothesis in Classical regularity for Holder-continuous forcing under initial compatibility cannot be lowered to mere boundedness. For continuous forcing with modulus $ω_{f} (σ) : = \sup_{∣ s - r ∣ \leq σ} ∥ f (s) - f (r) ∥$ , the analytic smoothing bound $∥ A T (σ) ∥ \leq C σ^{- 1}$ (Smoothing estimates for the semigroup generated by a sectorial operator) makes $\int_{0}^{b} ω_{f} (σ) σ^{- 1} d σ < \infty$ sufficient for the singular generator integral in the Duhamel cancellation estimate; H"older continuity is one way to meet this condition. This is a sufficient condition for that estimate, not a necessary condition for classicality, and the jump is outside its continuous-Dini hypothesis. The present $A = 0$ example shows directly that bounded measurable forcing alone does not suffice: the jump makes the mild solution nondifferentiable. The example isolates this failure of time regularity with a trivial initial datum.

Refuted claim. With $A = 0$ on $C$ , a bounded measurable forcing produces a classical solution of $u^{'} = A u + f$ . The jump forcing $f = 1_{[t_{0}, b]}$ satisfies $f \in L^{\infty} (0, b)$ and the mild solution exists, but its left and right derivatives at the jump $t_{0}$ disagree, so the mild solution is not even differentiable there and the classical notion fails without any additional time regularity of $f$ .

Facts & Assumptions

Given: $X = C$ , the zero operator $A = 0$ with $D (A) = X$ , the numbers $0 < t_{0} < b$ , the indicator $f = 1_{[t_{0}, b]}$ and the datum $x = 0$ .

[L1]

The infinitesimal generator of a strongly continuous semigroup $(T (t))_{t \geq 0}$ is $A x : = \lim_{t ↓ 0} (T (t) x - x) / t$ on its domain (Infinitesimal generator of a C0-semigroup).

[L2]

For a strongly continuous semigroup with generator $A$ and a Bochner integrable $f$ with $\int_{0}^{T_{0}} ∥ f ∥ < \infty$ , the formula $u (t) = T (t) x + \int_{0}^{t} T (t - s) f (s) d s$ defines the unique mild solution and the unique integral solution of $u^{'} = A u + f$ , $u (0) = x$ (Variation of constants for the inhomogeneous abstract Cauchy problem, Bochner-integrable function).

[L3]

A classical solution is a $u \in C^{1} ([0, T_{0}]; X)$ with $u (t) \in D (A)$ for every $t$ , $A u \in C ([0, T_{0}]; X)$ , $u^{'} (t) = A u (t) + f (t)$ for $0 < t < T_{0}$ and $u (0) = x$ ; endpoint equations are imposed only when $f$ extends continuously to $[0, T_{0}]$ (Classical, strong and mild abstract Cauchy solutions).

[L4]

For $A$ sectorial with semigroup bounds $c_{0}, c_{1}$ and $f \in C^{α} ([0, b], X)$ one has $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ for the Duhamel term $v_{1} (t) = \int_{0}^{t} T (t - s) (f (s) - f (t)) d s$ (Analytic Duhamel cancellation removes the generator singularity).

[L5]

The classical regularity theorem assumes $x \in D (A)$ and $f \in C^{α} ([0, b], X)$ for some $α \in (0, 1)$ and concludes classicality of the mild solution (Classical regularity for Holder-continuous forcing under initial compatibility).

[L6]

For the analytic contour semigroup generated by a sectorial operator with vertex $0$ , $T (σ) X \subseteq D (A)$ and $∥ A T (σ) ∥ \leq C σ^{- 1}$ for $σ > 0$ , with $C$ depending on the sectoriality bounds (Smoothing estimates for the semigroup generated by a sectorial operator).

Counterexample

technique · direct

1.1L1L2givenalgebra

The semigroup and the mild solution. On $X$ the operator $A = 0$ has domain $X$ and generates the identity semigroup $T (t) = I$ : for every $x$ the difference quotient $(T (t) x - x) / t = 0$ converges to $0 = A x$ , so $D (A) = X$ and the generator is $0$ ; the forcing $f = 1_{[t_{0}, b]}$ is bounded and measurable, hence Bochner integrable on $(0, b)$ with $\int_{0}^{b} ∥ f ∥ = b - t_{0} < \infty$ , and [L2] with $x = 0$ gives the unique mild solution $u (t) = \int_{0}^{t} 1_{[t_{0}, b]} (s) d s$ , that is $u (t) = 0$ for $0 \leq t \leq t_{0}$ and $u (t) = t - t_{0}$ for $t_{0} < t \leq b$ ; this $u$ is continuous, equals $0$ at the origin and is Lipschitz with constant $1$ on $[0, b]$ .

2.1step 1.1givenalgebra

The differentiability failure at the jump. For $0 < h < t_{0}$ the left difference quotient of $u$ at $t_{0}$ is $(u (t_{0}) - u (t_{0} - h)) / h = 0$ , while for $0 < h < b - t_{0}$ the right quotient is $(u (t_{0} + h) - u (t_{0})) / h = h / h = 1$ ; hence the one-sided limits differ and $u$ is not differentiable at $t_{0}$ , although on each open piece $u^{'} (t) = f (t)$ (the derivative is $0$ on $(0, t_{0})$ and $1$ on $(t_{0}, b)$ ), so $u \in C^{1} ((0, b] ∖ {t_{0}}, X)$ and $u \notin C^{1} ([0, b], X)$ .

3.1step 2.1L3L4L5L6givenalgebra∎

Why boundedness is not enough. A classical solution on $[0, b]$ must be $C^{1}$ on the closed interval with $A u$ continuous and $u^{'} (t) = A u (t) + f (t) = f (t)$ for $0 < t < b$ by [L3], so [step 2.1] shows that this mild solution is not classical even though the generator $A = 0$ is bounded, the datum is trivial and $f$ is bounded; therefore the H"older hypothesis of [L5] cannot be weakened to mere boundedness. For a continuous forcing with modulus $ω_{f} (σ) : = \sup_{∣ s - r ∣ \leq σ} ∥ f (s) - f (r) ∥$ , [L6] bounds the generator integrand in the cancellation term by $C ω_{f} (σ) σ^{- 1}$ . Thus $\int_{0}^{b} ω_{f} (σ) σ^{- 1} d σ < \infty$ is sufficient for that cancellation estimate; in the H"older case it yields $C [f]_{α} t^{α} / α$ . This sufficient estimate is not a necessary characterization of classicality. In the present example $A = 0$ , hence $A T (σ) = 0$ ; the failure follows directly from the unequal one-sided derivatives in [step 2.1], not from a singular generator kernel. All functions here are explicit, so no choice principle beyond Dependent Choice is used.

Remarks

The same witness works in any nonzero Banach space after multiplying both $f$ and $u$ by a fixed nonzero vector.

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Abstract smoothing does not imply a spatial derivative without a PDE realisation

Statement

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.

Put $N_{\geq 1} : = {n \in N : n \geq 1}$ . Let $X = ℓ^{2} (N_{\geq 1})$ and let $A$ be the diagonal operator with $D (A) = {x \in ℓ^{2} : \sum_{n} n^{4} ∣ x_{n} ∣^{2} < \infty}$ , $(A x)_{n} = - n^{2} x_{n}$ , which is self-adjoint and sectorial of angle $π / 2$ (Self-adjoint nonpositive operators generate bounded analytic semigroups). Its semigroup is $T (t) x = (e^{- n^{2} t} x_{n})_{n \geq 1}$ , and for every $t > 0$ and every $m \geq 1$ one has $T (t) x \in D (A^{m})$ with $A^{m} T (t) x = ((- n^{2})^{m} e^{- n^{2} t} x_{n})_{n}$ for every $x \in ℓ^{2}$ , since $\sup_{n \geq 1} n^{2 m} e^{- n^{2} t} \leq \sup_{s \geq 0} s^{m} e^{- s t} = (m / (e t))^{m}$ is finite. In particular $x = (1 / n)_{n} \notin D (A)$ satisfies $T (t) x \in D (A^{m})$ for every $m$ and every $t > 0$ . Nevertheless $ℓ^{2} (N_{\geq 1})$ carries no spatial variables: $A^{m}$ is an abstract sequence operator, and the inclusion $T (t) ℓ^{2} \subseteq ⋂_{m} D (A^{m})$ is a purely operator-theoretic smoothing statement. Only after identifying the abstract sequence operator with a differential operator through an elliptic-regularity theorem does $D (A^{m})$ name Sobolev derivatives (Abstract generator-domain smoothing becomes spatial regularity only after domain identification).

Facts & Assumptions

Given: The complex Hilbert space $X = ℓ^{2} (N_{\geq 1})$ with inner product $⟨ x, y ⟩ = \sum_{n} x_{n} \overline{y_{n}}$ , norm $∥ x ∥ = (\sum_{n} ∣ x_{n} ∣^{2})^{1 / 2}$ and standard orthonormal basis $e_{n}$ ; the diagonal operator $A$ with $D (A) = {x \in X : \sum_{n} n^{4} ∣ x_{n} ∣^{2} < \infty}$ and $(A x)_{n} = - n^{2} x_{n}$ ; the diagonal family $T (t) x = (e^{- n^{2} t} x_{n})_{n \geq 1}$ for $t \geq 0$ ; and the iterated domains $D (A^{m}) = {x \in D (A^{m - 1}) : A x \in D (A^{m - 1})}$ .

[L1]

$ℓ^{2} (N_{\geq 1})$ is a complex Hilbert space with the standard orthonormal basis: the trigonometric system is an orthonormal basis of $L^{2} (T; C)$ (The trigonometric system is complete in $L^{2}$ of the torus, $L^{2}$ with the integral pairing is a Hilbert space), and the Fourier coefficient map of an orthonormal basis is a linear isometry onto the corresponding $ℓ^{2}$ space, which is therefore complete (A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set, Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ , The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L2]

A self-adjoint densely defined operator with $⟨ A x, x ⟩ \leq 0$ is sectorial of angle $π / 2$ with vertex $0$ and generates a bounded analytic semigroup of angle $π / 2$ (Self-adjoint nonpositive operators generate bounded analytic semigroups).

[L3]

For a sectorial operator the generated semigroup satisfies $T (t) X \subseteq D (A^{m})$ for every $t > 0$ , $m \geq 1$ , and the contour semigroup is the unique exponentially bounded strongly continuous semigroup with that generator (Smoothing estimates for the semigroup generated by a sectorial operator, The generator of the contour semigroup is the sectorial operator, Abstract parabolic smoothing for mild solutions).

[L4]

The graph domains $D (A^{m})$ carry the graph norm and are recursively defined; the remark on domain identification records that they acquire a spatial meaning only through an elliptic-regularity theorem (Abstract generator-domain smoothing becomes spatial regularity only after domain identification, A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Verification

technique · direct

1.1L1givenalgebra

The diagonal operator is self-adjoint and nonpositive. $D (A)$ contains the finitely supported vectors, hence is dense in $X$ by [L1]; for $x, y \in D (A)$ the series $⟨ A x, y ⟩ = \sum_{n} (- n^{2}) x_{n} \overline{y_{n}}$ converges absolutely and equals $\overline{⟨ A y, x ⟩}$ because the diagonal entries are real, so $A$ is symmetric. If $y \in D (A^{*})$ , the adjoint identity tested against each $e_{n} \in D (A)$ gives $(A^{*} y)_{n} = - n^{2} y_{n}$ for every $n$ ; since $A^{*} y \in X = ℓ^{2} (N_{\geq 1})$ and $(e_{n})$ is an orthonormal basis by [L1], Parseval gives $\sum_{n} n^{4} ∣ y_{n} ∣^{2} = ∥ A^{*} y ∥^{2} < \infty$ , so $y \in D (A)$ . Symmetry gives the reverse inclusion $D (A) \subseteq D (A^{*})$ , hence $D (A^{*}) = D (A)$ and $A$ is self-adjoint. Finally $⟨ A x, x ⟩ = - \sum_{n} n^{2} ∣ x_{n} ∣^{2} \leq 0$ for every $x \in D (A)$ .

1.2L1givenalgebra

The diagonal family is the semigroup generated by $A$ . For $t \geq 0$ one has $∥ T (t) x ∥^{2} = \sum_{n} e^{- 2 n^{2} t} ∣ x_{n} ∣^{2} \leq e^{- 2 t} ∥ x ∥^{2}$ , so $T (t)$ is a contraction for $t \geq 0$ ; the functional equation is coefficientwise and strong continuity at $0$ follows from $∥ T (t) x - x ∥^{2} = \sum_{n} (1 - e^{- n^{2} t})^{2} ∣ x_{n} ∣^{2} \to 0$ by dominated convergence; for $x \in D (A)$ the difference quotients satisfy $∥ (T (t) x - x) / t - A x ∥^{2} = \sum_{n} (\frac{1 - e^{- n^{2} t}}{n^{2} t} - 1)^{2} n^{4} ∣ x_{n} ∣^{2} \to 0$ by dominated convergence, since $∣ \frac{1 - e^{- s}}{s} - 1 ∣ \leq 1$ for $s \geq 0$ and $\sum_{n} n^{4} ∣ x_{n} ∣^{2} < \infty$ , so $A$ is contained in the generator; conversely, if $x$ lies in the domain $G$ of the generator then for each $n$ continuity of the $n$ -th coordinate functional gives $G x_{n} = \lim_{t ↓ 0} (e^{- n^{2} t} - 1) x_{n} / t = - n^{2} x_{n}$ , so $\sum_{n} n^{4} ∣ x_{n} ∣^{2} = ∥ G x ∥^{2} < \infty$ and $x \in D (A)$ with $A x = G x$ ; hence the generator of $T$ is exactly $A$ .

2.1step 1.1step 1.2L2L3givenalgebra

The abstract semigroup is this diagonal semigroup. By [step 1.1] $A$ is self-adjoint and nonpositive, so [L2] makes $A$ sectorial of angle $π / 2$ and the generator of a bounded analytic semigroup, while [step 1.2] exhibits $T$ as an exponentially bounded strongly continuous semigroup with generator $A$ ; by the uniqueness in [L3] these semigroups coincide, so the diagonal family $T (t) x = (e^{- n^{2} t} x_{n})$ is the semigroup generated by $A$ , which is the assertion of the statement.

3.1step 2.1L3givenalgebra

The iterated domains and the smoothing identities. By induction from [step 1.2] the graph domain is $D (A^{m}) = {x \in X : \sum_{n} n^{4 m} ∣ x_{n} ∣^{2} < \infty}$ with $(A^{m} x)_{n} = (- n^{2})^{m} x_{n}$ : the case $m = 1$ is the definition of $D (A)$ , and if the description holds for $m$ then $A^{m} x \in D (A)$ exactly when $\sum_{n} n^{4} ∣ (A^{m} x)_{n} ∣^{2} = \sum_{n} n^{4 m + 4} ∣ x_{n} ∣^{2} < \infty$ ; consequently for $t > 0$ the vector $T (t) x$ has $A^{m} T (t) x = ((- n^{2})^{m} e^{- n^{2} t} x_{n})_{n}$ and $∥ A^{m} T (t) x ∥^{2} = \sum_{n} n^{4 m} e^{- 2 n^{2} t} ∣ x_{n} ∣^{2} \leq \sup_{n} (n^{2 m} e^{- n^{2} t})^{2} ∥ x ∥^{2} < \infty$ , so $T (t) x \in D (A^{m})$ and $∥ A^{m} T (t) ∥ \leq \sup_{n \geq 1} n^{2 m} e^{- n^{2} t} \leq \sup_{s \geq 0} s^{m} e^{- s t} = (m / (e t))^{m}$ ; this reproduces the abstract membership $T (t) X \subseteq D (A^{m})$ of [L3] with an explicit constant.

4.1step 3.1givenalgebra

The witness is not in $D (A)$ but is smoothed. For $x = (1 / n)_{n \geq 1}$ one has $\sum_{n} ∣ x_{n} ∣^{2} = \sum_{n} n^{- 2} < \infty$ , so $x \in X$ , while $\sum_{n} n^{4} ∣ x_{n} ∣^{2} = \sum_{n} n^{2} = + \infty$ , so $x \notin D (A)$ by [step 3.1]; for every $t > 0$ and every $m \geq 1$ , however, $\sum_{n} n^{4 m} e^{- 2 n^{2} t} n^{- 2} < \infty$ because the exponential decay dominates every polynomial, so $T (t) x \in D (A^{m})$ with the series of [step 3.1], and the smoothing thus raises the abstract regularity of a vector that is not even in the domain of $A$ .

5.1step 4.1L4givenalgebra∎

No spatial derivative is produced. The statements of steps 3.1 and 4.1 are identities between sequences: $A^{m}$ acts by the multiplier $(- n^{2})^{m}$ and the index $n$ carries no spatial or differential meaning, so the inclusion $T (t) ℓ^{2} \subseteq ⋂_{m} D (A^{m})$ is purely operator-theoretic; by [L4] the graph domain $D (A^{m})$ acquires the interpretation of Sobolev derivatives only after an elliptic-regularity theorem identifies $A$ with a differential operator, and no such identification is present for this diagonal sequence operator, which is why the example is the companion witness to that remark.

Analytic Semigroups and Linear Evolution Equations — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The analytic semigroup generated by a bounded operator

Statement

Facts & Assumptions

Verification

The analytic Dirichlet heat semigroup

Statement

Facts & Assumptions

Proof

Remarks

The sectorial multiplication operator

Statement

Facts & Assumptions

Verification

The translation semigroup is not analytic

Statement refuted

Facts & Assumptions

Counterexample

An analytic semigroup need not be norm continuous at zero

Statement refuted

Facts & Assumptions

Counterexample

The sector changes under the sign convention

Statement refuted

Facts & Assumptions

Counterexample

A sectorial nonselfadjoint multiplication generator

Statement

Facts & Assumptions

Verification

A time-discontinuous forcing blocks classical regularity at its jump

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Abstract smoothing does not imply a spatial derivative without a PDE realisation

Statement

Facts & Assumptions

Verification

Sources